Skip navigation.

4th Quarter, Math Level 2 (10th grade math)

00.00 Start Here (Math Level 2)

 

Course Description

The fundamental purpose of Mathematics II is to formalize and extend the mathematics that students learned in 9th grade.  Students will focus on quadratic expressions, equations, and functions but also be introduced to radical, piecewise, absolute value, inverse and exponential functions. They will extend the set of rational numbers to the set of complex numbers and link probability and data through conditional probability and counting methods. Finally, they will study similarity and right triangle trigonometry and circles with their quadratic algebraic representations. The course ties together the algebraic and geometric ideas studied. Students should experience mathematics as a coherent, useful, and logical subject that makes use of their ability to make sense of problem situations.

Class Overview

This integrated Secondary Mathematics II course is based on the New Utah State Standards Initiative.

Credit

This course is worth .25 credits, or nine weeks of Mathematics I. There are four Mathematics II quarter classes available. Taking all four will add up to one credit or one year of Mathematics I. In order to earn credit for each quarter, you must commit to following the EHS Honor Code: "As a student of the Electronic High School, I agree to turn in my assignments in a timely manner, do my own work, not share my work with others, and treat all students, teachers, and staff with respect." This course is for tenth grade students. After completing the work for the class, students must pass a proctored final exam to earn credit. There is not a paper-based textbook assigned for this course. If you find that having a textbook is useful, you can check out a textbook from most local libraries. You can also search for topics on the Internet to find many useful resources.

Prerequisites

You should have successfully completed 9th grade math.

Supplies needed

  • Graph paper (this can also be downloaded)
  • Scientific or Graphing Calculator (You can download scientific and graphing calculator simulators or find online versions, but as you will need a graphing calculator for the rest of your high school career, you may consider buying one now.)
  • Access to a printer to print the daily assignments is vital. Most assignments are NOT interactive and must be printed out to complete.
  • Ability to scan or photograph a completed assignment to submit electronically.

Organization of Secondary Math Level 2

Units: There are 11 units for the full credit of Secondary Math 2.

Quarter 1 has four units.
Quarter 2 has one unit.
Quarter 3 has three units.
​Quarter 4 has three units.

Schedule: When you enroll in a quarter class, you are given 10 weeks to finish all the requirements needed to earn the credit for a quarter. There are no "required" due dates for the assignments in this course. However, there is a pacing guide provided for you that will help you stay on track to being successful and finishing the course within the 10 week time frame. The pacing guide is located in the Syllabus in Module 1. Before you begin, go over the pacing guide to help you set up your own due dates for the assignments. Give your parents permission to nag you about it. You don't want to be one of those students who does a whole lot of work, but never finishes the course.

This Quarter Class

The units in this class have lessons, assignments, quizzes and a unit test.

Lessons: Each lesson provides instruction on a given topic. Many include instructional videos (hosted on YouTube) and one or more assignments for independent practice.

Assignments: Print, then complete the practice worksheet, showing how you arrived at your answer. In order to earn full points on each problem, you are required to show step by step the process needed to arrive at your answer. Then circle your answer so it is easily seen. I do give partial credit for showing your work--so ALWAYS show your work even if you think it is obvious. Assignments are submitted by uploading the digital assignment through the course website by following the instructions within each assignment. Under some circumstances, you may snail-mail a hard copy of the assignment to the instructor. If you choose this option, be sure to make a copy for yourself, as the instructor will NOT return your assignment. Also, please send the instructor an e-mail if you must mail an assignment.

Quizzes: Some quizzes are taken online. Others are more like the assignments. Print, then complete the quiz, showing all of your work as to how you arrived at your answer. In order to earn full points on each problem, you are required to show step by step the process needed to arrive at your answer. Then circle your answer so it is easily seen. I do give partial credit for showing your work--so ALWAYS show your work even if you think it is obvious. Quizzes are submitted by uploading the digital assignment through the course website by following the instructions within each quiz.

Proctored Final

Each quarter class has a proctored final exam and is worth 25% of the final grade.

Information about the Final Exam

  1. You must get approval from your teacher before you are allowed to take the final exam.
  2. You must complete every assignment and have an overall grade of C in the course to be approved to take the final exam.
  3. The final exam is a comprehensive exam that must be taken with an approved proctor.
  4. You are allowed to have a page of notes. You also will need a calculator and scratch paper.
  5. You must pass the final exam with a 60% in order to pass the class.
  6. The final exam is worth 25% of your final grade.
  7. The exam is timed. You will have 2.0 hours to complete the exam. You must finish it in one attempt.

Final Grade

Assignments and quizzes are worth 75% of the final grade. The proctored final test is worth 25% of the final grade.

Grading Scale

You earn a grade based on a modified total points percentage method. This means that the total number of points you earned is divided by the total number of points possible, times 100%. That will make up 75% of your final grade. The final exam is the remaining 25%. These scores are combined for a total percentage of the class. This percentage is translated into a grade based on this standard scale:

94-100% A
90-93% A-
87-89% B+
83-86% B
80-82% B-
77-79% C+
73-76% C
70-72% C-
67-69% D+
63-66% D
60-62% D-
0-59% no credit

00.00 *Student supplies for Secondary Math 2 (Math2)

There is no textbook assigned for this course. If you find that having a textbook is useful, you will want to select a textbook that aligns with the Common Core State Standards Initiative. You can purchase textbooks online, at some bookstores (selection is limited), or you can check out a textbook from most local libraries. Supplies needed:

  • Graph paper (this can also be downloaded)
  • Tracing paper or transparency film (if you use transparencies you will need transparency markers).
  • Straight edge (A ruler is an example of a straight edge. I have used a scrap of mat board. A piece of paper will not work! My favorite is a drafting ruler with the cork on the bottom; this gives a very clean line, and doesn't slide around.)
  • Compass (There is a very wide range of compass prices. You don't need a drafting quality compass, but you will be happier if you don't buy the cheapest compass you can find. You want something that doesn't change size as you draw with it.)
  • Graphing Calculator (You can download graphing calculator simulators, but as you will need a graphing calculator for the rest of your high school career, you may as well buy one now.)
  • Geometric software.

 

00.01 Curriculum Standards (Math Level 2)

Overview information on the Utah Mathematics Level II Core is here.

00.01.01 Student Software Needs

 

Students need access to a robust internet connection and a modern web browser.

This class may also require the Apple QuickTime plug-in to view media.

For students using a school-issued Chromebook, ask your technical support folks to download the QuickTime plug-in and enable the plug-in for your Chromebook.

$0.00

00.02 About Me (Math Level 2)

teacher-scored 10 points possible 10 minutes

About Me Assignment: This assignment gives me, as your teacher, a chance to get to know you better! To complete and submit this assignment copy the material between the asterisks into a blank word-processing document. Answer the questions using complete sentences, appropriate punctuation and sentence structure. Please write your answers in either BOLD or a {\color{Blue}DIFFERENT } {\color{Blue}COLOR }. Save the document. Finally, select all, copy, then paste the entire document into the box that opens when you click to submit this assignment.

**********************************************************************************

1. What is your full name, what name do you prefer to go by, your parent's/guardian's names, and contact information for both you and your parents? (email addresses and phone numbers.)

2. What high school do you attend and what grade are you in? What is the name of the last math class you completed?

3. Why have you chosen to take this math class with EHS?

4. What is your counselor's full name and contact information?

5. Have you read the EHS Honor Code and do you commit to following it? EHS Honor Code "As a student of the Electronic High School, I agree to turn in my assignments in a timely manner, do my own work, not share my work with others, and treat all students, teachers, and staff with respect."

6. Are you committed to finishing the class within the 10 week time frame, completing your final exam in week 9?

7. Now tell me about you! What are your likes/dislikes etc. Please be sure to include anything you think I need to know as your teacher.

*************************************************************************

I am excited to learn more about you!

Grading criteria:

1. All requested information is included.

2. Complete sentences, correct punctuation and correct grammar are used.

Pacing: complete this by the end of Week 1 of your enrollment date for this class.


09.00 Circles (Math Level 2)

The goal of all math classes is for students to become mathematically proficient. In particular, students need to be able to:

  • Make sense of problems and persevere in solving them.
  • Reason abstractly and quantitatively.
  • Construct viable arguments and critique the reasoning of others.
  • Model with mathematics.
  • Use appropriate tools strategically.
  • Attend to precision.
  • Look for and make use of structure.
  • Look for and express regularity in repeated reasoning.

As you work through this unit, keep these proficiencies in mind!

This unit focuses on the parts of the circle and the equation of a circle.

At the conclusion of this unit you must be able to state without hesitation that:

  • I can write the equation of a circle given various information.
  • I can find the center and radius of a circle given by an equation by completing the square.
  • I can prove all circles are similar.
  • I can solve problems involving the inscribe and circumscribed angles.
  • I can construct the line tangent to circle at a given point.
  • I can inscribe a regular hexagon in a circle.
  • I can inscribe an equilateral triangle in a circle.
  • I know that a line is tangent to a circle at a given point if it is perpendicular to the radius drawn to that point.
  • I can construct tangents to a circle through a given point
  • I can prove that tangent segments from the same point are equal in length.
  • I can solve problems involving tangents of circles.
  • I understand radian measure.
  • I can convert from degree to radian measure.
  • I can convert from radian to degree measure.
  • I can find the perimeter and area of regular polygons.

09.01 Derive the Equation of a Circle (Math Level 2)

Derive the equation of a circle using the Pythagorean Theorem.

A circle is the set of points in a plane that lie a fixed distance, called the radius, from any point, called the center. The diameter is the length of a line segment passing through the center whose endpoints are on the circle. In addition, a circle can be formed by the intersection of a cone and a plane that is perpendicular to the axis of the cone:

Using this definition, and the distance formula, we should be able to determine the equation for a circle on the Cartesian grid. It will just take a little bit of algebra.

How would you do this? Start by selecting an arbitrary center point, located at (h, k). These are the standard variables used to describe center and the location of the center of a circle. Recall, these are the same variables used to denote the vertex of a parabola.

Next, select an arbitrary distance from the center. Say, length r. Again, this is the standard variable used for the length of the radius of a circle.

Now, think about this for a second. It may be useful to draw this image.

Before we begin, let’s make certain that you can graph circles in the coordinate plane.

Please view the following video:

This circle is defined by the set of points (x, y) such that these points are a distance r from the center of the circle. Okay?

The first thing we need to do is find the length of a and b. First, we need to find the coordinates for the intersection point of sides a and b. It would be (x, k). Make certain you understand how that was determined!

Now, we can use the distance formula to find the length of a the light blue segment:

= = = = \dpi{100} \fn_phv \left | x-k\right |

Now, for b the length of the turquoise segment:

= = = =

We know from the Pythagorean theorem that r2 is equal to a2 + b2.

If we plug the values we calculated for a and b into the Pythagorean theorem we find:

This is the equation of a circle!

Question: Why did I drop the absolute values when I squared a and b?

In this form, the center and radius are apparent. For example, given the equation (x−2)2+(y+5)2=16 we have,

In this case, the center is (2,−5) and r=4.

All right, so you now can find the equation of any circle you wish. Let's find the equation of the circle with radius 15 centered at point (2, -7). Literally, all we need to do is plug in these values: r = 15, h = 2 and k = -7.

Pretty simple!!

Now what if you are given the graph of a circle. Could you find the equation of that circle?

Given this graph of a circle, write the equation of the circle in standard form.

To write the equation of this circle in standard form, we need to know the center point and the length of the radius. Point A is the center of this circle. What are the coordinates of point A?

Point A = (-5,3).

Now we need the length of the radius. The radius goes from the center to the edge of the circle. Count how far it is from the center to the edge. It is 3. So the radius is 3. Now we have all we need to write the equation of this circle.

(x-h)^{2}+(y-k)^{2}= r^{2}

(x-(-5))^{2}+(y-3)^{2}=3^{2}

(x+5)^{2}+(y-3)^{2}=9

That's it!!

Now what if you are given the center point and a point on the circle? Can you still find the equation of that circle? YES!!

Consider the following problem. A given circle is centered at (5, 6) and point (-2, -18), lies on this circle. What is the equation of this circle?

We know that h = 5 and k = 6. These are given. We just need to find r. r is the distance between the center and all the points on the circle. In particular, r is the distance between point (5, 6) and point (-2, -18). Use the distance formula to find this distance.

Now that we have found the radius, we can write the equation for this circle.

How would you graph this circle?

Start by plotting the point (5, 6) on a Cartesian grid. To do this right, you could use a compass. But for this class, it is probably okay to plot the points 25 units above, below, to the right, and to the left of the center, then estimate the circle as best as you can. In the image, each grid line is equal to 5 units.

Also, notice that with a circle, as with a parabola, the standard forms are the forms that are most useful, and not necessarily the form that you usually see in all algebra classes. For example, most of these are not actually functions; each allowed value of x usually produces two values of y.

Your turn! (Seriously, try this without looking at the answer!!)

  1. Find the equation of the circle centered at (6, -9) with point (-9, -1) on the edge of the circle. Draw the graph of this circle.

Answer

2.

Please view the following videos for another example of finding the equation of a circle when given a graph and when given the center and a point on the circle:

09.01.01 Derive the Equation of a Circle (Math Level 2)

Next, we will learn to find the equation of a circle given the endpoints of a diameter.

Again we will need the center point of the circle and the length of the radius to be able to write the equation of the circle.The process to write this equation is shown in the video below. Here is an outline of the steps needed to find the center and the radius.

  1. First find the center of the circle. We know that the diameter is a chord that runs through the center of the circle. The center of the circle will be the midpoint of the diameter. We will use the midpoint formula to find the midpoint of the diameter.

\fn_phv m=\left ( \frac{x_{1}+x_{2}}{2} \right ),\left ( \frac{y_{1}+y_{2}}{2} \right )

    2. Second we need to find the length of the radius. We are given the endpoints of the diameter and we know that the radius is half the length of the diameter. We will need to use the distance formula to find the length of the diameter and then take half of that and that will be the length of the radius.

Ok we are now ready to see how this is done!

Please view the following videos:

09.01.02 Derive the Equation of a Circle (Math Level 2)

If after completing this topic you can state without hesitation that...

  • I can write the equation of a circle given various information.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

09.01.03 Derive the Equation of a Circle - Worksheet (Math Level 2)

teacher-scored 60 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 1 of your enrollment date for this class.


09.02 Find the Center and Radius of a Circle given by an Equation (Math Level 2)

Find the center and radius of a circle given by an equation by completing the square.

If you are given the equation of a circle in standard form, you should be able to readily read the center and the radius from the equation. Given the following equation for a circle, find the center and the radius.

(x-1)^{2}+(y-10)^{2} = 25

Reading this from the equation, we find that the center is located at (1, 10). Also, the radius squared is 25. Therefore, the radius is 5.

If you are given this equation, instead, (x+1)^{2}+(y-10)^{2} = 25, the radius is still 5, but the center is now (-1, 10).

Remember, If you are adding the terms in the given equation, you must take the negative value to find the coordinates of the center!

What if you are given an equation in general form, and asked to find the radius and center? How do you solve that problem?

Given the following equation in general form for a circle, find the center and the radius.

x^{2}+y^{2}+18x-2y+18=0

We can’t read these values readily from this equation. We need the equation to be in standard form. We can convert this general form to standard form by completing the squares. We learned how to complete the square back in quarter 2, but let's review a bit.

To complete the squares for both the x and y terms we take the following steps.

Step 1: Start by grouping the x terms and the y terms, leaving a blank space to put the number that completes the square. Also move the constant to the right side of the equation by subtracting 18:

(x^{2}+18x+____)+(y^{2}-2y+____)=-18

Step 2: Complete the square for each grouping.

a) Be sure the coefficients of x2 and y2 are both 1. They are. Therefore, we do not need to divide by anything. We can move onto the next step.

b) For both the x and y groupings, take the value of "b" , divide it by 2 and square it. \left ( \frac{b}{2} \right )^{2}Then add that value to both sides of the equation.

For the x grouping the value of "b" is 18: \left ( \frac{18}{2} \right )^{2}=\left ( 9 \right )^{2}={\color{Blue} 81}

For the y grouping the value of "b" is -2: \left ( \frac{-2}{2} \right )^{2}=\left ( -1 \right )^{2}={\color{Green} 1}

(x^{2}+18x+{\color{Blue} 81})+ (y^{2}-2y+{\color{Green} 1})=-18+{\color{Blue} 81}+{\color{Green} 1}

Step 3: Factor both the x grouping and the y grouping and combine the terms on the right.

(x+9)(x+9)+(y-1)(y-1)=64

Because the factors are exactly the same for both the x and y groupings, we can write them as one factor squared.

(x+9)^{2}+(y-1)^{2}=64

Look! We now have the equation of the circle written in standard form!!!

Now we can find the location of the center and the value of the radius from the equation. The center is located at (–9, 1) and the radius squared is 64, so the radius is 8. Not too bad! Think you can do this yourself?

Find the center and radius of the circle given by the equation:

x^{2}+y^{2} - 20x+14y+113 = 0

Seriously, try these without looking at the answer!!

Answer

The circle is centered at (10, –7) and has a radius of 6.

If after completing this topic you can state without hesitation that...

  • I can find the center and radius of a circle given by an equation by completing the square.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

I highly recommend you view the following videos before you complete the assignment.

09.02 Find the Center and Radius of a Circle given by an Equation - videos (Math Level 2)

09.02 Find the Center and Radius of a Circle given by an Equation - Worksheet (Math Level 2)

teacher-scored 72 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 1 of your enrollment date for this class.


09.03 Prove All Circles are Similar (Math Level 2)

Prove all circles are similar.

The material for this lesson is from Illustriative Mathematics under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License (illustrativemathematics.org/illustrations/1368).

In this lesson we need to prove that all circles are similar. This means that given any two circles there is a sequence of transformations of the plane reflections, rotations, translations, and dilations transforming one to the other. If we give the plane a coordinate system, then the circles have equations and the transformations can be described explicitly with equations. The goal of this task is to work on showing that all circles are similar using these two different methods, the first visual and the second algebraic.

Solution 1: Transformations

This first approach to the problem uses transformations of the plane to move the given circle to the circle of radius 1 centered at (0,0).

a. To move the center of the circle to (0,0) we need to translate by −a in the x-direction and −b in the y-direction. This is pictured below:i

In order to transform our circle of radius r centered at (0,0) into a circle of radius 1 centered at (0,0) we can apply a dilation, with center (0,0), with scale factor \dpi{100} \fn_phv \frac{1}{r}. This is pictured below in this case r >1 so this is a contraction:

The green arrows show where selected points on the blue circle of radius r map to on the red circle of radius 1.

b. Part (a) applies to any circle in the plane. If C1 and C2 are two circles then by part (a) both C1 and C2 are similar to the circle of radius 1 with center (0,0). Therefore C1 is similar to C2.

Solution 2: Algebraic manipulations

This solution works with the equations of circles and models the geometric transformations with algebraic manipulations.

a. The circle of radius r centered at (a, b) is given by the solutions of the equation

(x-a)^{2} + (y-b)^{2} = r^{2}

To transform this into the equation for the circle of radius 1 we may first remove the a from x−a and the b from y−b by substituting x′ = x – a and y′ = y − b. With these new coordinates our equation becomes

({x}')^{2}+({y}')^{2}=r^2}

This is the equation of a circle of radius r centered at (0,0) in x′-y′ coordinates. We can now scale x′ and y′ taking x′′= \frac{{x}'}{r} and \frac{{x}''}{r} and, with these coordinates, our equation becomes

{x}''+y''=1

the equation of a circle of radius 1 in x′′-y′′ coordinates. Note, comparing this method to the first solution, that moving from x-y coordinates to x′-y′ coordinates corresponds to translating the circle by −a in the x-direction and −b in the y-direction. The second substitution x′′ = \frac{x''}{r} and y'' = \frac{y''}{r}corresponds to the dilation by a factor of \frac{1}{r}.

 b. Part (a) applies to any circle in the plane. If C1 and C2 are two circles then by part (a) both C1 and C2 are similar to the circle of radius 1 with center (0, 0). Therefore, C1 is similar to C2.

Done!

Let's try 2 examples.

To prove any two circles are similar, only a translation (slide) and dilation (enlargement or reduction) are necessary. This can always be done by using the differences in the center coordinates to determine the translation and determining the quotient of the radii for the dilation ratio.

We will show that two given circles are similar by describing the transformations (translations and dilations) needed to get the first circle directly onto the second circle.

1. Circle C has its center at (-1,2) and has a radius of 3. Circle D has its center at (3,4) and has a radius of 5. Describe the needed translation and ratio of dilation to get Circle C to be directly on Circle D to prove they are similar.

(It is helpful to graph the circles to visualize what is happening.)

To translate Circle C to Circle D, we will need to move the center 4 units to the right and 2 units up. Then to dilate circle C to the size of circle D, we need to use the ratio of the radius of Circle D to Circle C. 5/3.

SOLUTION:

 

Translate Circle C 4 units right and 2 units up. Dilate Circle C using ratio 5/3.

 

2. Circle C has its center at (0, 2) and has a radius of 6. Circle D has its center at (0, –6) and has a radius of 2.

To translate Circle C to Circle D, we will need to move the center down 8 units. Then to dilate circle C to the size of circle D, we need to use the ratio of the radius of Circle D to Circle C: 2/6 = 1/3.

SOLUTION:

Translate Circle C down 8 units. Dilate Circle C using ratio 1/3.

If after completing this topic you can state without hesitation that...

  • I can prove all circles are similar.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

09.03 Prove All Circles are Similar - Worksheet (Math Level 2)

teacher-scored 58 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 2 of your enrollment date for this class.


09.04 Circle Geometry Angles (Math Level 2)

Identify, central angles, inscribed angles, circumscribed angles, chords, secants, tangents, radii, diameters, major and minor arcs, and semicircles and solve problems involving circles and their component parts.

In this lesson we will look at the relationships between angles and circles.

Central Angle and Arcs

A central angle is an angle whose vertex lies on the center of the circle. Not too complicated? There are two arcs formed by this angle: a smaller arc called the minor arc, and a larger arc called the major arc. Arcs are denoted with the symbol . Minor arcs are denoted by the endpoints. Major arcs are denoted by the endpoints and an additional point on the arc. In the image below, the minor arc is \widehat{JS} and the major arc is \widehat{JZS}.

There are also two arcs formed by a secant, chord, or any angle that intersects a circle. The fact I am mentioning it here should tell you something.

The measure of a minor arc is, by definition, the measure of the central angle that forms the arc. The measure is denoted by a letter m. In the image above m\widehat{JS}=\angle JOS. The full circle is 360°. The minor arc and the major arc with the same endpoints make up a complete circle. Therefore, the measure of a major arc is 360° minus the measure of the minor arc. In the image above m\widehat{JZS}=360^{\circ} -m\widehat{JZ} = 360^{\circ}-m\angle JOS

A semicircle is an arc formed by a diameter; therefore, its measure is 180°. A semicircle needs to be labeled with three points, in the same way as a major arc is labeled. A major arc is any arc that is greater than a semicircle, and a minor arc is any arc that is less than a semicircle. In addition to the measure of the arc, an arc also has a length. This is the distance traveled from one end point of an arc to another. We will learn how to find the value of this later. In the meantime, it is important to note that for two arcs to be congruent, not only must the arcs have the same measure, they must also have the same length. How can two arcs have the same measure and different lengths? Or even the same length and different measures? We will start with the first, as this is easier to visualize. We will get back to the second after we learn how to find arc lengths.

Consider the following. In the image below, the minor arc \widehat{JS} has the same measure as the minor arc \widehat{FM} since they have the same central angle \angleJOS. However, it should be fairly clear that the distance along the circle from point J to point S is longer than the distance along the circle from point F to point M.

On the other hand, you may remember that all circles are similar, and that the similarity ratio is the ratio of the radii, which may lead you to make the very insightful supposition that while the arcs have different lengths, these lengths have the same ratio as the radii of the circles. Or specifically,

length \widehat{FM} : length \widehat{JS} = OF : OJ.

Just something to think about.

Now, back to congruent arcs. If two arcs in the same circle or congruent circles have equal measures, these arcs are congruent (further evidence that the length of the arc is somehow related to the radius of the circle, since two circles are congruent if the radii are congruent).

In the image below, \overline{HX} and \overline{NV} are diameters of ⊙C. Since these diameters intersect at point C, two pairs of vertical angles are formed: \angleVCX and \angleHCN are vertical angles, and \angleHCV and \angleNCX are vertical angles.

 

It has been a while, but hopefully you still remember that vertical angles are congruent. Therefore, \angleVCX \cong \angleHCN and \angleHCV \cong \angleNCX. Since the measures of the arcs are equal to the measures of the central angles, the measures of the arcs formed by these angles are equal, in particular m\widehat{HN} = m\widehat{VX} and m\widehat{HV} = m\widehat{NX}.

Since these arcs all lie on the same circle, the arcs with equal measures are congruent. Or, \widehat{HN} = \widehat{VX} and \widehat{HV} = \widehat{NX}.

Good?
 

Click on the link below for additional instruction on central and inscribed angles.

Inscribed Angles

An inscribed angle is an angle whose vertex lies on the circumference of the circle. In the image \angleJKS, \angleJDS, and \angleJZS are all inscribed angles that form the same arc, \widehat{JS}.

I claim that not only are these angles all congruent, but that the measure of an inscribed angle is equal to one half the measure of the central angle that forms the same arc, namely \angleJOS.

Looking at these angles, it is plausible that these angles are equal, and they do look to be about half the measure of the central angle, but you do not have to take my word for it! You can prove it!

We do have to prove each case separately. The three cases are shown in the image above. The center of the circle can lie inside of the inscribed angle, as in \angleJDS. The center of the circle can lie on one of the rays forming the inscribed angle, as in \angleJZS. Finally, the center of the circle can lie outside of the inscribed angle, as in \angleJKS.

We will start with the center lying along one of the rays, as this case is simpler than the other two.

Given ⊙O, with inscribed angle \angleJZS such that point O is collinear with points J and Z, as shown. Prove that m\angle JZS = \frac{1}{2}\cdot m\angle JOS.

Whenever proving properties of circles, always look for things you can do with the radii. Since all radii in a circle are congruent, we can find congruent triangles, we can find isosceles triangles, and we can find congruent isosceles triangles. These are the shapes you want to look for. Once you have found a triangle you can use, the rest of the proof should follow. Okay.

Remember, I mentioned this was the easier of the three conditions, mostly because you already had the triangle you needed.

We will do one more as an example. This is mostly algebra, so I am not going to write up a formal proof. Instead, I am going to explain how to derive a proof by starting with what I know, moving to what I want to prove, then working backwards from there.

Given ⊙O, with inscribed angle \angleJKS such that point O is lies outside of \angleJKS, as shown. Prove that m\angleJKS = m\angleJOS.

Okay, start with what we have. We have two radii and two segments that we know nothing about. We need to connect the radii, which are congruent, with the unknown segments and make some useful triangles.

If we construct \overline{KO} we will form two triangles that connect our unknown segments to the already existing radii, so construct this.

The isosceles triangles formed are \bigtriangleupKOS and \bigtriangleupKOJ. This means that m\angleOKS = m\angleOSK, and m\angleOKJ = m\angleOJK. Also, the sum of the three angles are equal to 180°, so with a little bit of algebra, we can find equations for the vertex angles.

Stop for a second. Do we want to find equations for the vertex angles? We don't really care about either \angleKOS or \angleOKJ. What we do care about are \angleJKS and \angleJOS, so let's jump to the end, and work backwards.

We can see that \overline{KS} cuts \angleOKJ into \angleOKS and \angleJKS. Therefore, we need equations for \angleOKJ and \angleOKS. We also can see that \overline{OJ} cuts \angleKOS into \angleKOJ and \angleJOS, so we need equations for \angleKOS and \angleKOJ.

All right, we probably have enough information, so let's write it down.

m\angleKOJ + m\angleOKJ + m\angleOJK = 180° (Equation 1)

m\angleKOS + m\angleOKS + m\angleOSK =180° (Equation 2)

These are the sums of the internal angles of the two triangles. Plus we know that

m\angleOKJ = m\angleOJK (Equation 3)

m\angleOKS = m\angleOSK (Equation 4)

We want to keep \angleOKJ and \angleOKS, so substitute Equations 3 and 4 into Equations 1 and 2, eliminating \angleOJK and \angleOSK.

m\angleKOJ + m\angleOKJ + m\angleOKJ = 180°

m\angleKOJ + 2\cdotm\angleOKJ = 180° (Equation 5)

m\angleKOS + m\angleOKS + m\angleOKS =180°

m\angleKOS + 2\cdotm\angleOKS =180° (Equation 6)

We can set these equations equal.

m\angleKOJ + 2\cdotm\angleOKJ = m\angleKOS + 2\cdotm\angleOKS (Equation 7)

Okay, let's stop again and take stock. We need \angleOKJ and \angleOKS to find \angleJKS; we need \angleKOS and \angleKOJ to find \angleJOS. Cool! All four of these angles are in the equation above. Let's write down the two equations relating the angles we have to the angles we need.

m\angleOKJ = m\angleOKS + m\angleJKS (Equation 8)

m\angleKOS = m\angleKOJ + m\angleJOS (Equation 9)

From here it is just algebra. We actually have three equations we need to solve for two terms. I am going to start by substituting Equation 9 into Equation 7.

m\angleKOJ + 2\cdotm\angleOKJ = m\angleKOJ + m\angleJOS + 2\cdotm\angleOKS

2\cdotm\angleOKJ = m\angleJOS + 2\cdotm\angleOKS (Equation 10)

That was nice: the m\angleKOJ term subtracted off of both sides leaving behind a m\angleJOS term. We know that eventually we want a term like 1\cdotm∠JOS, so divide equation 10 by 2.

m\angleOKJ = \frac{1}{2}m\angleJOS + m\angleOKS (Equation 11)

Now, substitute Equation 8 into Equation 11.

m\angleOKS + m\angleJKS = \frac{1}{2}m\angleJOS + m\angleOKS

m\angleJKS = \frac{1}{2}m\angleJOS (eq. 12)

Again, the unwanted term nicely subtracted off of both sides, and we proved that the measure of the inscribed angle was half the measure of the central angle that intersects the same arc.

Now that we know where we want to go, and how to get there, it is a good idea to write this up formally, so here is the formal proof.

Given:O, with inscribed angle \angleJKS such that point O lies outside of \angleJKS, as shown.

Prove: m\angleJKS = m\angleJOS.

It is worth pointing out that the ancient Greeks did not have algebra. This was developed by the Arabs. Therefore, every algebraic step must be justified using the measurement axioms. This only took a couple more steps than the previous proof, but it was a good thing I knew where I was going before I started!

Circumscribed Angles

A circumscribed angle is an angle whose rays are tangent to the circle. In the image below, \angleJRS is circumscribed about ⊙O at points J and S.

As you may suspect, there is a relationship between the measure of the circumscribed angle and the central angle that forms the same arc. So, let's calculate it.

Consider the example above.

The circumscribed angle \angleJRS and the central angle \angleJOS form a quadrilateral. We already know that the sum of the angles of a quadrilateral equals 360°. The rays of the circumscribed angle are perpendicular to the rays of the central angle. Therefore the angles formed are right angles. In particular, m\angleOJR = m\angleOSR = 90°, so we can solve this problem.

m\angleJRS + m\angleJOS + m\angleOJR + m\angleOSR = 360°

m\angleJRS + m\angleJOS + 90° + 90° = 360°

m\angleJRS + m\angleJOS + 180° = 360°

m\angleJRS + m\angleJOS = 180°

m\angleJRS = 180° - m\angleJOS

Therefore, the measure of a tangential angle is equal to 180° minus the measure of the central angle that forms the same arc.

Now we have found the relationship between the inscribed and circumscribed angles and the central angle. The measure of the central angle is equal to the measure of the minor arc the central angle forms. Therefore, it is also possible to determine the measures of the inscribed and circumscribed angles if we know the measure of the arc formed by these angles.

Some Practice Problems

These concepts shouldn't be too difficult. Let's see if you can solve some problems.

Consider the figure shown here.

1. Name the objects shown in the image.

a. What are the names of three chords?
b. What are the names of three secants?
c. What are the names of three tangents?
d. What are the names of three radii?
e. What are the names of two diameters?
f. What are the names of three minor arcs?
g. What are the names of three major arcs?
h. What are the names of three semicircles?
i. What are the names of three central angles?
j .What are the names of three inscribed angles?
k. What are the names of two circumscribed angles?

2. Let m\angleVCX = 57°

a. Find m\angleHCN.
b. Find m\angleNCX.
c. Find m\angleHCV.
d. Find mVX.
e. Find mVHX.
f. Find mHV.
g. Find mHNV.
h. Find mHAX.
i. Find m\angleHAN.
j. Find m\angleNAX.
k. Find m\angleHAX.
l. Find m\angleHTN.
m. Find m\angleNPX.

Answers and Explanations

1. Name the objects shown in the image:

a. What are the names of three chords?

A chord is any segment whose endpoints are on the circumference of the circle. Chords are named in the same way as segments, by the endpoints with a line over the top. The order of the endpoints does not matter. There are five chords in the image; you were asked to name three. Since there are two ways to name each of these segments, I listed both. You would only need one member from three of the pairs listed below.

  • \overline{AH} or \overline{HA}
  • \overline{AN} or \overline{NA}
  • \overline{XA} or \overline{AX}
  • \overline{XH} or \overline{HX} and
  • \overline{VN} or \overline{NV}

b. What are the names of three secants?

This was a trickier problem, because the secants are not as clear in the image. However, a secant is a line that intersects a circle at two points. This means that we are looking for the lines containing the chords shown. Lines are named in the same way as chords, but with a double arrow on top, instead of a line. Just as there are five chords in the image, there are five secants. However, there are more than two ways to name some of these secants, so I listed all possible names. Again, you would need one member from three of the five groups below.

  • or ;
  • or ;
  • or ;
  • , , , , or ; and
  • , , , , or .

c. What are the names of three tangents?

A tangent is a line that intersects the circle in one point. Since a tangent is perpendicular to the radii, we can identify the tangents by the right angle. There are only three tangents shown, so you would need all three of the tangents listed. Again, I listed all possible names for each line. You need one member from each set.

  • or ;
  • , , , , or ; and
  • or .

d. What are the names of three radii?

  • or ;
  • or ;
  • or ; and
  • or .

e. What are the names of two diameters?

A diameter is a chord that goes through the center of the circle. There are two diameters in the image. Again, since these can be named in two ways, I listed both. You only needed one member of each pair.

  • or ; and
  • or .

f. What are the names of three minor arcs?

A minor arc is an arc whose measure is less than a semicircle. There are eight minor arcs in the image. Minor arcs are named by the endpoints with an arc over the top. The order doesn't matter, so there are sixteen names listed here; you only need one member from three of the pairs.

  • \widehat{HA} or \widehat{AH}
  • \widehat{AV} or \widehat{VA}
  • \widehat{VX} or \widehat{XV}
  • \widehat{XN} or \widehat{NX}
  • \widehat{HV} or \widehat{VH}
  • \widehat{AX} or \widehat{XA} and
  • \widehat{NA} or \widehat{AN}

g. What are the names of three major arcs?

A major arc is an arc whose measure is greater than a semicircle. There are eight major arcs in the image. Major arcs are named by three points (the end points, and a point in the middle) with an arc over the top. This means that there are several names for the eight major arcs. I have listed all of them. You only need to have one member from three of the eight groups.

  • \widehat{HNA}, \widehat{HXA}, \widehat{HVA}, \widehat{AXH}, \widehat{ANH} or \widehat{AVH}
  • \widehat{AHV}, \widehat{ANV}, \widehat{AXV}, \widehat{VXA}, \widehat{VNA} or \widehat{VHA}
  • \widehat{VAX}, \widehat{VHX}, \widehat{VNX}, \widehat{XNV}, \widehat{XHV} or \widehat{XAV}
  • \widehat{XVN}, \widehat{XAN}, \widehat{XHN}, \widehat{NHX}, \widehat{NAX} or \widehat{NVX}
  • \widehat{NXH}, \widehat{NVH}, \widehat{NAH}, \widehat{HAN}, \widehat{HVN} or \widehat{HXN}
  • \widehat{HNV}, \widehat{HNV}, \widehat{VXH} or \widehat{VNH}
  • \widehat{ANX}, \widehat{NVA}, \widehat{XNA} or \widehat{XHA} and
  • \widehat{NXA}, \widehat{NVA}, \widehat{AVN} or \widehat{AXN}

h.  What are the names of three semicircles?

Semicircles are arcs formed by diameters. Since there are two diameters in the images, there are four semicircles. The semicircles are named in the same way as a major arc, and, just like a major arc, there are many ways to name each of the semicircles. You will need one member from three of the groups listed here.

  • \widehat{HAX}, \widehat{HVX}, \widehat{XVA} or \widehat{XAH}
  • \widehat{HNX} or \widehat{XNH}
  • \widehat{NHV}, \widehat{NAV}, \widehat{VAN} or \widehat{VHN} and
  • \widehat{NXV} or \widehat{VXN}

i.  What are the names of three central angles?

A central angle is an angle whose vertex is the center of the circle. In the image there are actually six central angles. Each angle can be named in a couple ways. You will need one member from three of the pairs listed here.

  • \angleVCX or \angleXVC
  • \angleXCN or \angleNCX
  • \angleNCH or \angleHCN
  • \angleHCV or \angleVCH

Then there are two central straight angles. We don't normally consider straight angles, but they are used in the same way to solve problems in a circle as any other central angle, so the two central straight angles are

  • \angleHCX or \angleXCH and
  • \angleNCV or \angleVCN.

 j.  What are the names of three inscribed angles?

An inscribed angle is an angle whose vertex lies on the circumference. There are six of these in the image, and each one can be named in more than one way. You need to have one member from three of the groups shown.

  • \angleHAN or \angleNAH;
  • \angleHAX or \angleXAH;
  • \angleNAX or \angleXAN;
  • \angleAHX, \angleAHC, \angleCHA or \angleXHA;
  • \angleANV, ∠ANC, \angleCNA or \angleVNA; and
  • \angleHXA, \angleCXA, \angleAXC or \angleAXH.

k. What are the names of two circumscribed angles?

The circumscribed angles are angles whose rays are tangent to the circle. In the image there are two such angles, so you will need both of these. However, as before, there are several ways to name these angles. Therefore, you will only need one member of each group listed below.

  • \angleHTN, \angleHTP, \angleNTH or \anglePTH; and
  • \angleTPX, \angleNPX, \angleXPN or \angleXPT.

2. Let m\angleVCX = 57°

a. Find m\angleHCN.

\angleHCN and \angleVCX are vertical angles, and vertical angles are congruent. Therefore m\angleHCN = 57°.

b. Find m\angleNCX.

\angleNCX and \angleVCX are supplementary. Since the sum of two supplemental angles is 180°, m\angleNCX = 180° - m\angleVCX = 180° - 57° = 123°.

c. Find m\angleHCV.

\angleHCV and \angleVCX are also supplementary, so m\angleHCV = 123°. Alternately, \angleHCV and \angleNCX are vertical angles, so they are congruent and their measures are equal.

d. Find m\widehat{VX}.

The measure of a minor arc is equal to the measure of the central angle that intersects it. Since \angleVCX is the central angle that forms \widehat{VX}, m\widehat{VX} = m\angleVCX = 57°.

e. Find m\widehat{VHX}.

The measure of a major arc is 360° minus the measure of the minor arc with the same endpoints. Therefore m\widehat{VHX} = 360^{\circ}57^{\circ} = 303^{\circ}

f. Find m\widehat{HV}.

This is another minor arc, so the m\widehat{HV} is equal to the central angle that forms it. The central angle forming \widehat{HV} is \angleHCV, which we already determined to be 123°. Therefore m\widehat{HV} = m\angleHCV = 123°.

g. Find m\widehat{HNV}.

This is another major arc, so its measure is 360° -m\widehat{HV}= 360° – 123° = 237°.

h. Find m\widehat{HAX}.

\widehat{HAX} is a semicircle. The measure of a semicircle is 180°. Therefore, m\widehat{HAX} = 180°.

i. Find m\angleHAN.

\angleHAN is an inscribed angle that forms \widehat{HN}. The measure of an inscribed angle is equal to one half the measure of the central angle that forms the same arc. Therefore, m\angleHAN = \frac{1}{2} x m\angleHCN = \frac{1}{2} x 57° = 28.5°.

j. Find m\angleNAX.

\angleNAX is an inscribed angle that forms \widehat{NX}. The central angle that also forms this arc is \angleNCX. Therefore, m\angleNAX = \frac{1}{2} x m∠NCX = \frac{1}{2} x 123° = 61.5°.

k. Find m\angleHAX.

\angleHAX is an inscribed angle that forms a semicircle. The "central angle" that also forms this semicircle is the straight angle \angleHCX, or the diameter \widehat{HX}. A straight angle has a measure of 180°, so this angle has a measure of \frac{1}{2} x 180° or 90°. This is actually an important property. An inscribed angle on a diameter has a measure of 90°.

l. Find m\angleHTN.

\angleHTN is a circumscribed angle that forms \widehat{HN}. The measure of a circumscribed angle is 180° minus the measure of the central angle that forms the same arc, or 180° - m\angleHCN = 180° - 57° = 123°.

m. Find m\angleNPX.

\angleNPX is a circumscribed angle that forms \widehat{NX}. Therefore, m\angleNPX = 180° - m\angleNCX = 180° - 123° = 57°.

Wow! That was long, but not too hard. Are you ready to do this on your own?

Inscribed Quadrilaterals

For a figure to be inscribed in a circle, each vertex of the figure must lie on the circumference of the circle. You already know how to construct rectangles and squares using circumscribed circles, so it should not surprise you that these shapes can be easily inscribed within a circle. What other quadrilaterals can be inscribed in a circle?

Squares? Rectangles? Parallelograms? Trapezoids? Kites? Random figures?

Yes for squares. Yes for rectangles. Looks like a no for parallelograms, but you shouldn't take my word for it. We should prove this instead.

Yes for some trapezoids, but not in general, only isosceles trapezoids; again, we can prove this. Yes, for kites, although there are some limits to the shapes allowed. In fact, there appears to be a wide variety of shapes that we can inscribe into a circle.

Okay, why can we inscribe some shapes and not others? What do the shapes we can inscribe have in common?

Consider the irregular quadrilateral inscribed in  W. Label this QRST, and construct the radii from the center to each vertex.

Consider the center angle  QWS. We know that mQTS = 1/2 x mQWS. In the same way, mQRS =1/2 x (360° - mQWS). Therefore, the sum of opposite angles of an inscribed quadrilateral is mQTS + mQRS = 1/2 x mQWS + 1/2 x (360° - mQWS).

Combining terms we get: mQTS + mQRS = 1/2 x (mQWS + 360° - mQWS).

The m∠QWS terms cancel, leaving mQTS + mQRS = 1/2 x (360°) = 180°.

As this is not a special figure, the sum of opposite angles of a quadrilateral inscribed in a circle is equal to 180°. That is, opposite angles of a quadrilateral inscribed in a circle are supplementary.

Since opposite angles in a parallelogram are equal, the only parallelograms which meet the above criteria are rectangles.

For trapezoids, opposite angles are supplementary if the trapezoid is an isosceles trapezoid, so only isosceles trapezoids can be inscribed into a circle.

What about the kite? Can you prove that the vertex angles lie on a diameter, and that the non-vertex angles are right angles?

If after completing this topic you can state without hesitation that...

  • I can solve problems involving the inscribe and circumscribed angles and quadrilaterals.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

I highly recommend you view the following videos showing examples of solving problems with inscribed and circumscribed angles.

09.04 Circle Geometry Angles - Worksheet (Math Level 2)

teacher-scored 96 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 2 of your enrollment date for this class.


09.05 Tangent Lines (Math Level 2)

Understand that a line is tangent to a circle at a given point if it is perpendicular to the radius drawn to that point, construct tangents to a circle through a given point, prove that tangent segments from the same point are equal in length and solve problems involving tangents of circles.

The material for this lesson is from EngageNY Module 5 Lesson 11 under a Creative Commons Attribution Non-Commercial Share-Alike license.

In this lesson we will learn about the relationship between tangent lines and circles.
 
First, let’s look at some pictures of a line and a circle:

As you can see, a line can intersect a circle twice, only once, or not at all.

Definitions

Secant Line: A secant line to a circle is a line that intersects a circle in exactly two points.

Tangent Line: A tangent line to a circle is a line in the same plane that intersects the circle in one and only one point.

Tangent Segment: A segment is said to be a tangent segment to a circle if the line it is contained in is tangent to the circle, and one of its endpoints is the point where the line intersects the circle.

In the figure below:

  • The point where the tangent line intersects the circle is called the point of tangency (P).
  • A tangent line to a circle is perpendicular to the radius of the circle drawn to the point of tangency.
  • A line through a point on a circle is tangent at the point if, and only if, it is perpendicular to the radius drawn to the point of tangency.

Now that we have the definitions we can get to work!

To begin with, please watch the following videos. Take notes and make certain you understand the material!

If after completing this topic you can state without hesitation that...

  • I know that a line is tangent to a circle at a given point if it is perpendicular to the radius drawn to that point.
  • I can construct tangents to a circle through a given point
  • I can prove that tangent segments from the same point are equal in length.
  • I can solve problems involving tangents of circles.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

09.05 Tangent Lines - Worksheet (Math Level 2)

teacher-scored 56 points possible 30 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 3 of your enrollment date for this class.


09.06 Arc Length, Radians and Degrees (Math Level 2)

Understand radian measure and convert from degree to radian measure and from radian to degree measure.

You know that there can be different units of measurement to measure the same thing. For example, length can be measured in feet and meters and temperature can be measured in degrees Celsius and degrees Fahrenheit. We often use formulas to convert between different units of measurement.

There are also two ways to measure angles. You know how to measure them in degrees. Now you will learn how to measure them in radians and how to convert between these two measurements. While degree measure is used in everyday activities such as building construction and surveying land, radian measure is used for many calculations, such as the speed and distance traveled by satellites above the Earth’s surface—including the International Space Station! It is important to be able to measure angles in radians as well as in degrees and to be able to convert between the two systems.

Remember that a central angle is an angle whose vertex is at the center of a circle. In the circle below, the center is point O, the length of the radius is r, and  is a central angle.

Notice that  cuts off or determines an arc  that has length s. The radian measure of a central angle, often denoted by the Greek letter theta (), is defined to be the ratio of the arc length to the length of the radius. So the radian measure of  is given by:

Arc length, s, and radius, r, must be in the same units.

EXAMPLE 1:

Problem:  What is the radian measure of central angle AOB?

In this circle, r = 3 inches. The arc determined by  has length s = 6 inches. Substitute these into the formula.

 

Answer:     The measure of central angle AOB is 2 radians.

EXAMPLE 2:

Problem:  Find the measure of central angle AOB in radians.

In this circle, r =1 cm. The arc determined by  has length  cm. Substitute these into the formula.

Answer:   The measure of central angle AOB is  radians.

 

en.wikipedia.org/wiki/File:Radian_cropped_color.svg

en.wikipedia.org/wiki/File:Radian_cropped_color.svg

As we just learned, radian measure of a central angle is defined as the ratio of the arc length to the length of the radius.

We can solve this equation for s, to find an equation that will help us find the measure of an arc length. Multiply both sides by r, the radius.

This definition leads to the arc length formula: s=r\theta. The measure of the central angle, , must be in radians to use this formula.

Radians and degrees are connected by the relationship \fn_phv 360^{\circ}=2\pi radians. If you wish to convert from degrees to radians, multiply the degree measure by \frac{\pi }{180^{\circ}}. If you wish to convert from radians to degrees, multiply the radian measure by \frac{180^{\circ} }{\pi }.

If after completing this topic you can state without hesitation that...

  • I understand radian measure.
  • I can convert from degree to radian measure.
  • I can convert from radian to degree measure.

…you are ready for the next topic  Otherwise, go back and review the material before moving on.

Each topic is divided into sections that include the following:

  • Warm Up - questions to answer to see if you are ready for the lesson.
  • Presentation - high quality video with excellent illustrations that teaches the topic.
  • Worked Examples - examples that are worked out step-by-step with narration.
  • Practice - quiz problems on the topic covered.
  • Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material 

09.06 Arc Length, Radians and Degrees - Worksheet (Math Level 2)

teacher-scored 66 points possible 30 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 3 of your enrollment date for this class.


09.06 Unit 09 Review Quiz - (Math Level 2)

computer-scored 81 points possible 35 minutes
Pacing: complete this by the end of Week 3 of your enrollment date for this class.


10.00 Volume (Math Level 2)

The goal of all math classes is for students to become mathematically proficient. In particular, students need to be able to:

  • Make sense of problems and persevere in solving them.
  • Reason abstractly and quantitatively.
  • Construct viable arguments and critique the reasoning of others.
  • Model with mathematics.
  • Use appropriate tools strategically.
  • Attend to precision.
  • Look for and make use of structure.
  • Look for and express regularity in repeated reasoning.

As you work through this unit, keep these proficiencies in mind!

This unit focuses on the deriving the equation for the circumference of a circle and the volume of geometric solids.

At the conclusion of this unit you must be able to state without hesitation that:

  • I can illustrate that the circumference of a circle is \fn_phv 2\pi r.
  • I can derive the Formula for the Circumference of a Circle.
  • I can identify geometric solids.
  • I can find the volume of geometric solids.
  • I can find the volume of a composite geometric solid.

10.01 Derive the Formula for the Circumference of a Circle (Math Level 2)

Illustrate that the circumference of a circle is 2πr and derive the formula for the circumference of a circle.

Illustration that the circumference of a circle is \fn_phv 2\pi r:

The circumference of a circle is the distance around it. The term is used when measuring physical objects, as well as when considering abstract geometric forms. When a circle's radius is 1, its circumference is 2\pi.

Click on the link below to understand this relationship better:

Click on the link below:

Derive the Formula for the Circumference of a Circle

Example 1

Below is a picture of a circle of diameter 1, labeled C1, and diameter d=2r, labeled C2:

In the case pictured, d is larger than 1. All circles are similar and in this case the scale factor going from the circle of diameter 1 to the circle of diameter 2r is 2r. The circumference of a circle is a one-dimensional measurement and so it scales in the same way as the diameters:

Since the circumference of C1 is \pi by definition, it follows from the above equation that the circumference of C2 is 2\pi r.

Example 2

"CircleArea" by Jim.belk - Own work. Licensed under Public domain via Wikimedia Commons - commons.wikimedia.org/wiki/File:CircleArea.svg#mediaviewer/File:CircleArea.svg

The chart below summarizes the derivation of the area of a circle using this method.

If after completing this topic you can state without hesitation that...

  • I can illustrate that the circumference of a circle is \fn_phv 2\pi r.
  • I can derive the formula for the circumference of a circle.
  • I can derive the formula for the area of a circle as \pi r^{2}.

…you are ready for the next topic! Otherwise, go back and review the material before moving on.

10.01 Derive the Formula for the Circumference of a Circle - Worksheet (Math Level 2)

teacher-scored 66 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 4 of your enrollment date for this class.


10.02 Use Volume Formulas for Cylinders, Pyramids, Cones, and Spheres (Math Level 2)

Identify geometric solids and find the volume of geometric solids and a composite geometric solids.

Living in a two-dimensional world would be pretty boring. Thankfully, all of the physical objects that you see and use every day—computers, phones, cars, shoes—exist in three dimensions. They all have length, width, and height. (Even very thin objects like a piece of paper are three-dimensional. The thickness of a piece of paper may be a fraction of a millimeter, but it does exist.)

In the world of geometry, it is common to see three-dimensional figures. In mathematics, a flat side of a three-dimensional figure is called a face. Polyhedrons are shapes that have four or more faces, each one being a polygon. These include cubes, prisms, and pyramids. Sometimes you may even see single figures that are composites of two of these figures.

Three-dimensional solids have length, width, and height. You use a measurement called volume to figure out the amount of space that these solids take up. To find the volume of a specific geometric solid, you can use a volume formula that is specific to that solid. Sometimes, you will encounter composite geometric solids. These are solids that combine two or more basic solids. To find the volume of these, identify the simpler solids that make up the composite figure, find the volumes of those solids, and combine them as needed.

If after completing this topic you can state without hesitation that...

  • I can identify geometric solids.
  • I can find the volume of geometric solids.
  • I can find the volume of a composite geometric solid.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

Each topic is divided into sections that include the following:

  • Warm Up - questions to answer to see if you are ready for the lesson.
  • Presentation - high quality video with excellent illustrations that teaches the topic.
  • Worked Examples - examples that are worked out step-by-step with narration.
  • Practice - quiz problems on the topic covered.
  • Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material!!

teacher-scored 60 points possible 30 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 4 of your enrollment date for this class.


10.02.01 Unit 10 Review Quiz - (Math Level 2)

computer-scored 57 points possible 35 minutes
Pacing: complete this by the end of Week 4 of your enrollment date for this class.


11.00 Probability (Math Level 2)

The goal of all math classes is for students to become mathematically proficient. In particular, students need to be able to:

  • Make sense of problems and persevere in solving them.
  • Reason abstractly and quantitatively.
  • Construct viable arguments and critique the reasoning of others.
  • Model with mathematics.
  • Use appropriate tools strategically.
  • Attend to precision.
  • Look for and make use of structure.
  • Look for and express regularity in repeated reasoning.

As you work through this unit, keep these proficiencies in mind!

This unit focuses probability. You will learn the usefullness of Venn diagrams, how to compute probability of independent and dependent events, how to calculate permutations and combinations, about conditional probability, the multiplication and addition rules, how two-way frequency tables to calculate conditional probability and finally, how to use samples to estimate probabilities.

At the conclusion of this unit you must be able to state without hesitation that:

  • I can use Venn diagrams to demonstrate the intersection and union of sets.
  • I can use the Fundamental Counting Principle to determine the size of the sample space for simple and compound events.
  • I can determine the probability of simple and compound events.
  • I can calculate the probability of independent events.
  • I can use the Fundamental Counting Principle to compute permutations and combinations.
  • I can calculate the probability of dependent events.
  • I know the difference between independent and dependent events.
  • I can use the Multiplication Rule to calculate probability.
  • I can calculate conditional probability.
  • I can use the addition rules to calculate the probability of disjoint events.
  • I can use two-way frequency tables of data to compute conditional probabilities.
  • I can use sample to estimate probabilities.

11.01 Probability Venn Diagrams. Unions and Intersections - (Math Level 2)

Use Venn diagrams to demonstrate the intersection and union of sets.

The material for this lesson is from The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICE‑EM), the education division of the Australian Mathematical Sciences Institute (AMSI), 2010 (except where otherwise indicated). This work is licensed under the Creative Commons Attribution‑NonCommercial-NoDerivs 3.0 Unported License. 2011

Diagrams make mathematics easier because they help us to see the whole situation at a glance. The English mathematician John Venn (1834–1923) began using diagrams to represent sets. His diagrams are now called Venn diagrams.

In most problems involving sets, it is convenient to choose a larger set that contains all of the elements in all of the sets being considered. This larger set is called the universal set, and is usually given the symbol E. In a Venn diagram, the universal set is generally drawn as a large rectangle, and then other sets are represented by circles within this rectangle.

For example, if V = { vowels }, we could choose the universal set as E = { letters of the alphabet } and all the letters of the alphabet would then need to be placed somewhere within the rectangle, as shown below.

In the Venn diagram below, the universal set is E = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }, and each of these numbers has been placed somewhere within the rectangle.

The region inside the circle represents the set A of odd whole numbers between 0 and 10. Thus we place the numbers 1, 3, 5, 7 and 9 inside the circle, because A = {1, 3, 5, 7, 9}. Outside the circle we place the other numbers 0, 2, 4, 6, 8 and 10 that are in E but not in A.

Representing subsets on a Venn diagram

When we know that S is a subset of T, we place the circle representing S inside the circle representing T. For example, let S = { 0, 1, 2 }, and T = { 0, 1, 2, 3, 4 }. Then S is a subset of T, as illustrated in the Venn diagram below.

The intersection of two sets

The intersection of two sets A and B consists of all elements belonging to A and to B. This is written as A \cap B. For example, some musicians are singers and some play an instrument.

If A = { singers } and B = { instrumentalists }, then A \cap B = { singers who play an instrument }.

Here is an example using letters.

If V = { vowels } and F = { letters in ‘dingo’ }, then V \cap F = { i, o }.

Intersection and the word ‘and’

The word ‘and’ tells us that there is an intersection of two sets. For example:

{ singers } \cap { instrumentalists } = { people who sing and play an instrument }

{ vowels } \cap { letters of ‘dingo’ } = { letters that are vowels and are in ‘dingo’ }

The union of two sets

The union of two sets A and B consists of all elements belonging to A or to B. This is written as A \cup B. Elements belonging to both set belong to the union. Continuing with the example of singers and instrumentalists:

If A = { singers } and B = { instrumentalists }, then A \cup B = { musical performers }.

In the case of the sets of letters:

If V = { vowels } and F = { letters in ‘dingo’ }, then V \cup F = { a, e, i, o, u, d, n, g }.

Union and the word ‘or’ 

The word ‘or’ tells us that there is a union of two sets. For example:

{ singers } \cup { instrumentalists } = { people who sing or play an instrument }

{ vowels } \cup { letters in ‘dingo’ } = { letters that are vowels or are in ‘dingo’ }

The word ‘or’ in mathematics always means ‘and/or’, so there is no need to add ‘or both’ to these descriptions of the unions. For example,

If A = { 0, 2, 4, 6, 8, 10, 12, 14 } and B = { 0, 3, 6, 9, 12 }, then A \cup B = { 0, 2, 3, 4, 6, 8, 9, 10, 12, 14 }.

Representing the intersection and union on a Venn diagram

The Venn diagram below shows the two sets:

A = { 1, 3, 5, 7, 9 } and B = { 1, 2, 3, 4, 5 }.

  • The numbers 1, 3 and 5 lie in both sets, so we place them in the overlapping region of the two circles.
  • The remaining numbers in A are 7 and 9. These are placed inside A, but outside B.
  • The remaining numbers in B are 2 and 4. These are placed inside B, but outside A.

Thus the overlapping region represents the intersection A \cap B = { 1, 3, 5 }, and the two circles together represent the union A \cup B = { 1, 2, 3, 4, 5, 7, 9 }.

If after completing this topic you can state without hesitation that...

  • I can use Venn diagrams to demonstrate the intersection and union of sets.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

I highly recommend you view the following websites before moving on to the assignment.

11.01 Probability Venn Diagrams. Unions and Intersections - Worksheet (Math Level 2)

teacher-scored 30 points possible 30 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 5 of your enrollment date for this class.


11.02 Introduction to Probability (Math Level 2)

Use the Fundamental Counting Principle to determine the size of the sample space for simple and compound events and determine the probability of simple and compound events.

                                      

 

 

 

 

 

 

 

 

Probability provides a measure of how likely it is that something will occur. It is a number between and including the numbers 0 and 1. It can be written as a fraction, a decimal, or a percent.

Picking numbers randomly means that there is no specific order in which they are chosen. Many games use dice or spinners to generate numbers randomly. If you understand how to calculate probabilities, you can make thoughtful decisions about how to play these games by knowing the likelihood of various outcomes.

First you need to know some terms related to probability.

Random situations are those which cannot be predicted with certainty. However, by using probability, a measure of how likely it is that a random situation will turn out a particular way, we may still be able to make some predictions about those situations. For example, many games use dice or spinners to generate numbers randomly. If we understand how to calculate probabilities, we can make thoughtful decisions about how to play these games by knowing the likelihood of various outcomes.

Defining Events

First we need to introduce some terms. When working with probability, a random action or series of actions is called a trial. An outcome is the result of a trial, and an event is a particular collection of outcomes. Events are usually described using a common characteristic of the outcomes.

Probability helps us understand random, unpredictable situations where multiple outcomes are possible. It is a measure of the likelihood of an event, and it depends on the ratio of event and possible outcomes, if all those outcomes are equally likely:

The Fundamental Counting Principle is a shortcut to finding the size of the sample space when there are many trials and outcomes:  

When a trial consists of more than one random element, the number of outcomes in the sample space is equal to the product of the number of outcomes for each random element.

If after completing this topic you can state without hesitation that...

  • I can use the Fundamental Counting Principle to determine the size of the sample space for simple and compound events.
  • I can determine the probability of simple and compound events.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

Each topic is divided into sections that include the following:

  • Warm Up - questions to answer to see if you are ready for the lesson.
  • Presentation - high quality video with excellent illustrations that teaches the topic.
  • Worked Examples - examples that are worked out step-by-step with narration.
  • Practice - quiz problems on the topic covered.
  • Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material!!

I highly recommend that you click on the links below and watch the videos before continuing:

teacher-scored 75 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 5 of your enrollment date for this class.


11.03 Probability of Independent Events (Math Level 2)

Calculate the probability of independent events.

Some probability situations involve more than one event. When the events do not affect one another, they are known as independent events. Independent events can include repeating an action like rolling a die more than once, or using two different random elements, such as flipping a coin and spinning a spinner. Many other situations can involve independent events as well. In order to calculate probabilities accurately, we need to know if one event influences the outcome of other events.

Independent Events

The main characteristic of a situation with independent events is that the original state of the situation doesn't change when one event occurs. There are two ways in which this happens:

Two (or more) events are independent if the occurrence of one event doesn’t change the probability that the other event occurs. There are two kinds of situations when this happens:

  • When the random action doesn’t remove an outcome (such as rolling a die or flipping a coin multiple times, or performing random actions that have no connection to each other like drawing a card and then rolling a die); and
  • When the random action does remove a possible outcome, but the outcome is replaced before the action happens again (such as pulling a card and then putting it back in the deck). This is referred to as drawing with replacement.

Here are examples of each case:

Situation Events Why the events are independent
You roll a die, and if it's not a 6, you roll again. What's the probability of getting a 6 on the second roll?

First roll is not a 6.

Second roll is a 6.                                    

The fact that the first roll is not a 6 doesn't change the probability that the second roll will be a 6. (Some people like to say, "the die doesn't remember what you rolled before.")
You pull a marble from a bag with 2 red, 2 white, and 1 green marble. You note the color, put it back, and then pull another marble. What's the probability of pulling red marbles both times?

First pull is red.

Second pull is red.                                           

The events are independent because you replaced the first marble you pull and returned the bag to its original state.

 

You draw a card from a deck of 52 cards, and then you roll a die. What’s the probability of drawing a 2 and then rolling a 2?

The card draw is a 2.

The die roll is a 2.

Although the card is not replaced after drawing, the die roll does not depend on the cards, so no possible outcomes have been removed. Regardless of the outcome of the card draw, the probability for the die roll will not be affected.

 

Let's look more closely at the second example. On the first pull, the probability of getting red is \small \frac{2}{5}, because there are 5 marbles and 2 of them are red. If that red marble is replaced (put back in the bag),  the probability for the second pull is still \small \frac{2}{5}, and that means the events are independent. The result of one trial does not affect the result of another.

But what would happen if the first marble wasn't put back in the bag? Then the probability of getting red will be different for the second pull. If one red marble is removed, on the second pull the probability is now \small \frac{1}{4} because there are only 4 marbles and 1 is red.

Now let's go back to the first example. Suppose a die were rolled 15 times without getting a 6. On the next roll, is the probability of rolling a 6 still \small \frac{1}{6}, or is it greater? Some people may believe the next roll is more likely to be a 6, because "I'm due for a 6!" However, this is incorrect—the die cannot remember what was rolled before. While it's somewhat unusual to roll 16 times without getting a 6, the probability of getting a 6 after 15 of those rolls have gone by is the same as getting a 6 on any other roll.

Probability of Independent Events


Let's look at the sample and event spaces for the examples given in the previous section.

  • You roll a die twice. What's the probability of getting a 6 on the second roll but not on the first?

In this example, a die is rolled twice. The chart below shows all the outcomes. The white die is the first roll and the gray die is the second roll.

There are 6 possible outcomes for the first roll, and for each of those, there are 6 possible outcomes for the second roll. There are 6 • 6, or 36, possible outcomes:

Sample space: {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

The event space is all the outcomes for which the first roll was not 6, and the second roll was 6. For the first roll, there were 5 possible outcomes that are not 6. For each of those, there is 1 possible outcome that is a 6. So there are 5 • 1 or 5 outcomes in the event space:

Event space: {(1,6), (2,6), (3,6), (4,6), (5,6)}

Notice that the size of the sample space for both rolls is the product of the size of the sample space for each roll. Similarly, the size of the event space for two rolls is the product of the size of the event space for each roll.

So, the probablilty of the first event, not rolling a 6, is 5/6. The probablity of the second event, rolling a 6, is 1/6. The probability of not rolling a 6, then rolling a 6 is 5/6 • 1/6 = 5/36.

We can derive a formula from these data. Since the event space for a situation can be found by multiplying the event spaces of each independent event, and the sample space for the situation can be found by multiplying the sample spaces of each independent event, we have:

This is true for all situations with independent events. It can be extended to more than two events, as well.

If A and B are independent events, P(A and B) = P(A) • P(B).

In general, for any number of independent events, the probability that all the events happen is the product of the probabilities that the individual events happen.

Summary:

Two (or more) events are independent if the occurrence of one event doesn’t change the probability that the other event occurs. There are two kinds of situations when this happens:

1) When the random action doesn’t remove an outcome (such as rolling a die or flipping a coin multiple times, or performing random actions that have no connection to each other like drawing a card and then rolling a die); and

2) when the random action does remove a possible outcome, but the outcome is replaced before the action happens again (such as pulling a card and then putting it back in the deck).

When events are independent, the probability that they all happen is equal to the product of the probabilities that each individual event happens.

If after completing this topic you can state without hesitation that...

  • I can calculate the probability of independent events.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

Each topic is divided into sections that include the following:

  • Warm Up - questions to answer to see if you are ready for the lesson.
  • Presentation - high quality video with excellent illustrations that teaches the topic.
  • Worked Examples - examples that are worked out step-by-step with narration.
  • Practice - quiz problems on the topic covered.
  • Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material!!

teacher-scored 60 points possible 30 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 5 of your enrollment date for this class.


11.04 Permutations and Combinations (Math Level 2)

Use the Fundamental Counting Principle to compute permutations and combinations.

          

http://upload.wikimedia.org/wikibooks/en/7/79/Permutation.png

Some probability situations involve multiple events. When one of the events affects the others, they are called dependent events. For example, when items are chosen from a list or group and are not replaced, the first choice reduces the options available for later choices.

DEPENDENT EVENTS:

Two events are dependent if the original state of the situation changes from one event to the other event, and this alters the probability of the second event.

Dependent events occur when an action removes a possible outcome, and the outcome is not replaced before a second action takes place.

This is referred to as choosing without replacement.

So one way to tell whether events are dependent or independent is to find out whether a removed outcome is replaced (making them independent) or not replaced (making them dependent). Here are some examples:

Situation Events Why the events are dependent
At a party, you draw four names of the guests to form a team of four people. What’s the probability that John, Perna, Tosho, and Lee will be on a team together?

Draw John                          

Draw Perna

Draw Tosho

Draw Lee

Once you draw a name, you don’t put that name back into the pool of names you draw from. Each time, there is one fewer name in the sample space, and (if the event continues to occur) one fewer name in the event space. The probability that the event happens changes with each new draw.
You pull a marble from a bag with 2 red, 2 white, and 1 green marble. You hold onto it and then pull another marble. What's the probability of pulling a red marble and then pulling the green marble?

First pull is red.

Second pull is green.

The events are dependent because you don’t replace the first marble you pull. There is one fewer marble in the sample space, so the probability of picking a green one is different for the second pull than it was for the first pull.

 

You draw two cards from a standard deck of 52 cards. (In a standard deck, each card has a suit—hearts, clubs, diamond, or spades—and a rank—Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, or King. What’s the probability both cards are 2s?

One card is a 2.

The other card is a 2.                                       

Since two cards are drawn, the probability that the first is a 2 is different from the probability that the second is a 2. (Fewer cards to choose from gives a smaller sample space.)

 

PERMUTATIONS AND COMBINATIONS

One thing we know about situations involving dependent events is that one action removes possible outcomes from future actions. There's another important issue to consider about the outcomes of dependent events: how are they organized? Must we make a list, noting the order in which they occurred, or do we just lump them together and ignore the order?

Consider the three examples given before, and think about whether order matters:

Situation Events Does the order matter?
At a party, you draw four names of the guests to form a team of four people. What’s the probability that John, Perna, Tosho, and Lee will be on a team together?

Draw John

Draw Perna

Draw Tosho                

Draw Lee

Order doesn’t matter. Those four people will be on the same team whether you draw John, Perna, Tosho, and then Lee, or Perna, Tosho, Lee, and finally John.
You pull a marble from a bag with 2 red, 2 white, and 1 green marble. You hold onto it and then pull another marble. What's the probability of pulling a red marble and then pulling the green marble?

First pull is red.                                          

Second pull is green.                                                                 

Order matters. Pulling a green and then a red is not a successful outcome for this situation.
You draw two cards from a standard deck of 52 cards. (In a standard deck, each card has a suit—hearts, clubs, diamond, or spades—and a rank—Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, or King. What’s the probability both cards are 2s?

One card is a 2.

The other card is a 2.

Order does not matter. The outcome is satisfied whether you draw the 2 of hearts and then the 2 of clubs, or the 2 of clubs and then the 2 of hearts.

 

In situations that create groups of objects (such as people, marbles, or cards), we need to know whether their order matters or not. Otherwise we can't find the sample and event spaces accurately.

Consider the team example. In the sample space, the outcome John, Perna, Tosho, Lee is the same as the outcome Perna, Lee, John, Tosho—there is no difference between the teams created, even though the members are named in a different order. On the other hand, suppose the first person drawn had to be the one to toss a water balloon, the second had to be the one who caught it, the third had to be the one who pops it (if it’s still intact), and the fourth is the one who tries to catch the water in a cup! If John is a terrible balloon thrower but Perna is great at it, it would be better for the team to be Perna, Lee, John, Tosho (so Perna throws to Lee) than John, Perna, Tosho, Lee (John throws to Perna). Order matters.

EXAMPLE:

Problem: A bag contains 5 marbles, colored white, red, blue, purple and green. Find the size of the sample space if you pull two marbles, without replacement, in two ways: 1) Order matters. 2) Order does not matter.

1. List the possibilities for the first pull. To save space, just use the initials of the colors (since they’re all different).

      First pull

W   R   B   P   G

2. Now add the second pull. Since we first want to look at when order matters, just add the second pull after the first one. Remember that this is without replacement, so you can’t repeat a color.          

Second Pull                          First Pull

  W R B P G
W   RW BW PW GW
R WR   BR PR GR
B WB RB   PB GB
P WP RP BP   GP
G WG RG BG PG  

Sample space (order matters): {RW, BW, PW, GW, WR, BR, PR, GR, WB, RB, PB, GB, WP, RP, BP, GP, WG, RG, BG, PG}

3. Now, how is this different when order doesn’t matter? WR and RW become the same outcome, as do WB and BW, and so on. In the diagram here, each outcome is paired with the equivalent outcome. Since you only need one of each pair, there are half as many outcomes.

Sample space (order doesn’t matter): {RW, BW, PW, GW, BR, PR, GR, PB, GB, GP}

ANSWER: When order matters, the sample space has 20 outcomes. When order doesn’t matter, the sample space has 10 outcomes.

There are two possible ways to arrange or combine outcomes of dependent events. Permutations are groupings in which the order of items matters. Combinations are groupings in which content matters but order does not.

So, basically to compute a permutation (where order matters), you will multipy the number of choices together. When computing a combination (where order does not matter), you will mulitply the number of choices together, but then divide that by the number of ways the choices can be made.

Distinguish between Permutations and Combinations

We know that for permutations, order matters and that for combinations, order doesn’t matter. Permutations are typically lists (order matters) and combinations are typically groups (order doesn’t matter). Permutation means arrangement or rearrangement of things. Combination means selection of some or all of a number of different items.  

In the chart below, read the material in the first column and then place a √ in the column it describes:

  Permutation Combination

1. You were given a new locker with a three-digit combination. If you forget the combination, how many possible tries would you have?

   

2. Five friends went to a movie. They were able to find a row with five empty seats. How many different ways can they be seated in the row?

   

3. How many ways can the letters in the word MATH be arranged?

   

4. A pizza place offers six different toppings for their two-topping pizzas. How many different pizzas can be made?

   

5. In how many ways can 20 CD’s be arranged on a shelf? 

   

6. Only five members of the nine-member tennis team can travel to the away meet. How many choices does the coach have?

   

7. You were given a new locker with a three-digit combination. If you forget the combination, how many possible tries would you have?

   

8. A college basketball team has 12 players. How many different choices does the coach have to pick the five starters?

   

9. Four of the fastest students ran in a race to see who was fastest. In how many ways can they finish 1st, 2nd, 3rd, or 4th?

   

 

The answers are at the end of the lesson material.

There’s an easier way to write permutation and combination formulas, using an idea called factorials. A factorial is a product of all the whole numbers from 1 to a given number.  The symbol ! after a number is used to represent this product. For example, 3! = 3 • 2 • 1, and 7! = 7 • 6 • 5 • 4 • 3 • 2 • 1. So, in general, n! = n • (n − 1) • … • 2 • 1. Special note: 0! is defined to be 1.

The number of permutations formula starts out like n!, but it ends with (n − k + 1) instead of 1. We need to remove the factors from (n − k) to 1 from the product. We can do that by dividing by (n − k)!

Then, for combinations, we divide that result by k • (k − 1) • … • 2 • 1, or k!

Using Factorials

When choosing k of n objects, knowing if the order matters or not, the following formulas can be used:

Number of permutations =

Number of combinations =

Many calculators have a factorial (!) key or command. To find the number of permutations of choosing 20 of 24 objects, entering 24! ÷ 4! is a lot faster and easier than 24 • 23 • 23 • 21 • 20 • 19 • 18 • 17 • 16 • 15 • 14 • 13 • 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5. (Then again, if only 4 choices are made, it may still be easier to enter 24 • 23 • 22 • 21.)

Let's try these formulas on a problem. First, we'll use the Fundamental Counting Principle.

EXAMPLE:

Problem:     A school organization has 30 members. Four members will be chosen at random for an interview with the school newspaper about the group. How many groups of four are possible?

1. First, decide if this is a permutation or a combination situation.

It's a combination.

There’s no reason for any person to be considered different from any other, based on the order chosen, so this is a combination.

2. There are 30 possibilities for the first pick. Then 29 possibilities for the second person, 28 for the third, and 27 for the fourth. The Fundamental Counting Principle says we multiply these outcomes to get the total number of possibilities.

However, that product gives us the number of permutations, when order matters. We need to take all the arrangements of four particular people and use just one representative of each.

For four people, there are 4 choices for the first to be listed, 3 choices for the second, 2 for the third, and then only 1 choice for the last in the list. The Fundamental Counting Principle again tells us how many times a group of 4 people will show up in the permutations list.

Divide by the product provided by the Fundamental Counting Principle.

ANSWER: There are 27,405 different groups of 4 people possible from the 30 members!

Now we'll solve the same problem with the factorial formula:

You are asked to choose 4 out of the 30 people to make the committee, the order does not matter. So we need to use the combination formula:

Where "n" is the number of total items and "k" is how many of the items you are choosing.

Both methods produced the same answer!

Summary

Two (or more) events are dependent if the probability for one event changes when the success of the other event is determined. This typically happens when the random action for one event removes a possible outcome and the outcome is not replaced before the action for the other event happens.

To find the sample and event spaces for these situations, consider whether the events involve permutations (order matters) or combinations (order does not matter.) There are two ways to calculate sample and event space without listing them all and counting, with the Fundamental Counting Principal and with factorial formulas.

The Fundamental Counting Principle allows us to find the number of permutations and combinations as follows:

When choosing k of n objects, the number of permutations is

When choosing k of n objects, the number of combinations is

Factorial formulas calculate permutations and combinations this way:

When choosing k of n objects, the number of permutations =

When choosing k of n objects, the number of combinations =

Each topic is divided into sections that include the following:

  • Warm Up - questions to answer to see if you are ready for the lesson.
  • Presentation - high quality video with excellent illustrations that teaches the topic.
  • Worked Examples - examples that are worked out step-by-step with narration.
  • Practice - quiz problems on the topic covered.
  • Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material!!

It is very important that you learn to distinguish between permutations and combinations. To that end, click on the links below for some additional videos to view after completing the NROC material.

11.04 Permutations and Combinations - Extra Videos (Math Level 2)

Answers:

  1. combination
  2. permutation
  3. permutation
  4. combination
  5. permutation
  6. combination
  7. combination
  8. combination
  9. permuation

If after completing this topic you can state without hesitation that...

  • I can use the Fundamental Counting Principle to compute permutations and combinations.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

11.04 Permutations and Combinations - Worksheet (Math Level 2)

teacher-scored 56 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 6 of your enrollment date for this class.


11.05 Probability of Dependent Events (Math Level 2)

Calculate the probability of dependent events.

The probability for any kind of event—simple, compound, independent, dependent—always follows this basic formula:

The probability is the ratio of the sizes of the event and sample spaces. For some situations, like independent and dependent events, there are ways to calculate these numbers without going through the process of finding and counting possible outcomes one by one, which is sometimes tedious and prone to mistakes.

There are several ways to find the probabilities of dependent events. The sequence of events can be treated as a whole, in which case the Fundamental Counting Principle or the permutations and combinations formulas are used to find the sizes of the event and sample spaces.

Or each event can be treated separately, in which case the probabilities of each event are multiplied to find the probability of the entire chain of events:

If A and B are dependent events, P(A and B) = P(A) • P(B after A) where P(B after A) is the probability that B occurs after A has occurred.

Regardless of method, it's important to note that when events are dependent, the size of both the event and sample spaces can change over time, because the occurrence of one event affects the outcomes of later events.

If after completing this topic you can state without hesitation that...

  • I can calculate the probability of dependent events.

…you are ready for the next topic! Otherwise, go back and review the material before moving on.

Each topic is divided into sections that include the following:

  • Warm Up - questions to answer to see if you are ready for the lesson.
  • Presentation - high quality video with excellent illustrations that teaches the topic.
  • Worked Examples - examples that are worked out step-by-step with narration.
  • Practice - quiz problems on the topic covered.
  • Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material!!

11.05 Probability of Dependent Events - Worksheet (Math Level 2)

teacher-scored 60 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 6 of your enrollment date for this class.


11.06 Multiplication Rule (Math Level 2)

Know the difference between independent and dependent events and use the Multiplication Rule to calculate probability.

The material for this lesson is from:

Licensed under CK-12 Foundation is licensed under Creative Commons AttributionNonCommercial 3.0 Unported (CC BY-NC 3.0)Terms of UseAttribution
 

The intersection of two events is the event that includes all outcomes that are in both of the original events. The symbol for intersection is \cap.

The Multiplication Rule states that for two events A and B:

\fn_phv P(A\cap B)= P(B|A)P(A)=P(A|B)P(B)

Example: 10% of the emails that Michelle receives are spam emails. Her spam filter catches spam 95% of the time. Her spam filter misidentifies non-spam as spam 2% of the time. What is the probability of an email chosen at random being spam and being correctly identified as spam by her spam filter?

Click on the link below and watch the video.

11.06 Multiplication Rule (Math Level 2)

Guidance

Consider two events A and B. The shaded section of the Venn diagram below is the outcomes shared by events A and B. It is called the intersection of events A and B or A \cap B. Note that B \cap A is equivalent to A \cap B.

Sometimes you can calculate P(A \cap B) directly, especially if you know all of the outcomes in the sample space. Other times, you will only have partial information about the sample space and the events. 

If two events A and B are dependent, then you can find the conditional probability of A given B (or the conditional probability of B given A):

\fn_phv P(A|B) = \frac{P(A\cap B)}{P(B)}                               P(B|A) = \frac{P(A\cap B)}{P(A)}

You can rewrite the above equations to solve for P(A\cap B):

P(A\cap B) = P(A|B)P(B)                  P(A\cap B) = P(B|A)P(A)

These formulas are known as the Multiplication Rule.

What if the events are independent? The Multiplication Rule will still work, but it can be simplified. Recall that if two events are independent , then the result of one event has no impact on the result of the other event. For independent events A and B, P(A|B)=P(A) and P(B|A)=P(B). By substitution, the above formulas become:

P(A\cap B) = P(A)P(B)                       P(A\cap B) = P(B)P(A)

Note that now, these two formulas are identical.

If you don't know whether or not two events are independent or dependent, you can always use the Multiplication Rule for calculating the probability of the intersection of the two events.  P(A \cap B)= P(A)P(B) is just a special case of the Multiplication Rule.

If the events are... You can use the formula...
Independent P(A \cap B)= P(A)P(B)
Independent or Dependent

P(A \cap B)= P(A|B)P(B)

or

P(A \cap B)= P(B|A)P(A)

 

Remember that often you will be able to calculate  P(A \cap B) directly. However, sometimes you will only be given information about the conditional probabilities of the events or the probabilities of the individual events. In those cases, the Multiplication Rule is helpful.

Example A

If Mark goes to the store, the probability that he buys ice cream is 30%. The probability that he goes to the store is 10%. What is the probability of him going to the store and buying ice cream?

Solution: The first sentence of the problem is a statement of conditional probability. You could restate it as “the probability of Mark buying ice cream given that Mark has gone to the store is 30%”. Let S be the event that Mark goes to the store. Let I be the event that Mark buys ice cream. Rewrite the statements and question in the problem in terms of S and I:

P(I|S) = 30% = 0.30

P(S) = 10% or 0.10

P(S \cap I) = ????

In order to figure out the probability of the intersection of the events, use the Multiplication Rule.

P(S \cap I) = P(I \cap S) = P(I|S)(P(S) = 0.30 x 0.10 = 0.03 or 3%.

There is a 3% chance that Mark will go to the store and buy ice cream.

Example B

Consider the experiment of choosing a card from a deck, keeping it, and then choosing a second card from the deck. Let A be the event that the first card is a diamond. Let B be the event that the second card is a red card. Find P(B \cap A).

Solution: By the Multiplication Rule, P(B \cap A) = P(B|A)P(A). Consider each of these probabilities separately.

  • P(A) is the probability that the first card is a diamond. There are 13 diamonds in the deck of 52 cards, so P(A)  = \frac{13}{52}.
  • P(B|A) the probability that the second card is a red card given that the first card was a diamond. After the first card was chosen, there are 51 cards left in the deck. 25 of them are red since the first card was a diamond. Therefore, P(B|A) = \frac{25}{51}.

P(B \cap A) = P(B|A)P(A) = \frac{25}{51}\cdot \frac{13}{52}=\frac{325}{2652}=\frac{25}{204}\sim 0.12

The probability is approximately equal to 12%.

Vocabulary

The union of two events is the event that includes all outcomes that are in either or both of the original events. The symbol for union is \cup.

The intersection of two events is the event that includes all outcomes that are in both of the original events. The symbol for intersection is \cap.

The probability of an event is the chance of the event occurring.

The Multiplication Rule states that for two events A and B, P(A \cap B) = P(B|A)P(A) = P(A|B)P(A).

A Venn diagram is a way to visualize sample spaces, events, and outcomes.

Guided Practice

  1. 0.1% of the population is said to have a new disease. A test is developed to test for the disease. 98% of people without the disease will receive a negative test result. 99.5% of people with the disease will receive a positive test result. A random person who was tested for the disease is chosen. What is the probability that the chosen person does not have the disease and got a negative test result?
  2. P(C|D) = 0.8 and P(C|D) = 0.5. What is P(C \cap D)
  3. Using the information from the previous problem, if P(D|C) = 0.6 what is P(C) what is?

Answers:

  1. 97.902%. This means that the probability that a random person who had this test done doesn't have the disease and got a negative test result is 97.902%. Most of the people who took the test for the disease will not have it and will get a negative test result. 
  2. 0.4
  3. Approximately 0.67

If after completing this topic you can state without hesitation that...

  • I know the difference between independent and dependent events.
  • I can use the Multiplication Rule to calculate probability.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

teacher-scored 56 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 7 of your enrollment date for this class.


11.07 Conditional Probability (Math Level 2)

Calculate conditional probability.

Conditional Probability Conditional probability is the term for the probability of event A given, or conditional upon, event B.

For example, you toss two coins. The first coin shows TAILS and the other coin fell off your desk out of sight. Now, what is the probability that they are both TAILS? Since you already know that the coin that is on your desk is TAILS, the probability of getting TAILS on the second coin is \small \fn_phv \small \frac{1}{2}.

If this were a “normal” coin-tossing problem, the probability that you would get two tails is (HH, HT, TH, TT) = \small \frac{1}{4}. With conditional probability, the probability is affected by what happened previously. Conditional probability is written P(B|A) and is read the probability of B given A. All of this is true ONLY if the events are dependent! Independent events, by definition, can’t be conditional on another event.

  • This is the formula for conditional probability: \small P(B|A)=\frac{P(A \cap B)}{P(A)}

Remember that the probability of a given event is:

In this case, we are treating the “condition or given” as a new sample space (denominator). So, what is the size of the event space? It is the probability of A and B.

Yeah, I know, ugly!

Let’s look at an example and see how it all works.

The probability that it is Friday and that a student is absent is 5%, or \small \frac{5}{100}, or 0.05. Since there are five school days in a week, the probability that it is Friday is \small \frac{1}{5} or 0.2. What is the probability that a student is absent given that today is Friday?

  • P(Absent | Friday) = \small \frac{P(Friday and Absent)}{P(Friday)} = \small \frac{.05}{0.2} = 0.25 = 25%

The trick is keeping track of each of the parts! How about another one?

At North High School, the probability that a student takes US History and French is 0.053. The probability that a student takes US History is 0.42. What is the probability that a student takes French given that the student is taking US History?

  • P(French | US History) = \small \frac{P(US History and French)}{P(US History)}\small \frac{0.053}{0.42} = 0.126 = 12.6%

Not so bad! Here is a Venn diagram of this situation:

The blue circle is the probability of A. The red circle is the probability of B. What about the probability of B given A? Since we already know that A has occurred, we will only focus on the blue circle. That becomes the new sample space. So the probability of B given A is: 

In the examples above the “numerator” was provided. Things aren’t always so accommodating. A survey was taken of 100 students’ sandwich preference. The table below summarizes the gathered data:

Find the probability that a person will chose turkey, given they want it on wheat bread. This is an example of conditional probability. We want to find the probability of event B given, or conditional upon, event A. Or we want to find the probability of a student choosing turkey, given they want the sandwich made with wheat bread.

  • The formula for conditional probability is: P(B|A) = \small \frac{P(A|B)}{P(A)}

The “given or condition” is the new sample space or the denominator. What is the probability of having a sandwich made of wheat? Looking at the table, we can see that there were 30 wheat sandwiches out of the 100 sandwiches made. P(wheat) = \small \frac{30}{100}.

Now, what about the probability of a turkey sandwich on wheat bread? There were 18 out of the 100 students who made that choice. P(turkey and wheat) = \small \frac{18}{30}.

Therefore, P(Turkey | Wheat) = \small \frac{\frac{18}{30}}{\frac{30}{100}} = \frac{18}{100} = 0.18.

You could also draw a tree diagram to represent the data after computing the various independent probabilities:

Chose white: \small \frac{70}{100} = 0.70

Chose wheat: \small \frac{30}{100} = 0.30

Chose ham: \small \frac{26}{100} = 0.26

Chose turkey: \small \frac{74}{100} = 0.74

Chose ham on white: \small \frac{14}{70} = 0.20

Chose ham on wheat: \dpi{100} \fn_phv \small \frac{12}{30} = 0.40

Chose turkey on white: \small \frac{56}{70} = 0.80

Chose turkey on wheat: \small \frac{18}{30} = 0.60

Make certain you can see where each of the numbers came from and why! You can compute conditional probabilities directly from tree diagrams. The probability of a turkey sandwich on wheat bread is circled. P(Turkey | Wheat) = 0.60 x 0.30 = 0.18!

Try one on your own!

Two cards are chosen at random without replacement from a well-shuffled deck. What is the probability that the second card drawn is also a king given that the first one drawn was a king? Don’t peak!

The new sample space is the probability of drawing a king on the first draw.

  • P(king) = \small \frac{4}{52}. Simple enough.

The probability of drawing two kings in a row:

  • P(king and king) = \small \frac{4}{52}\cdot \frac{3}{51}

Conditional probability = P(king|king) = .

Sweet!

I highly recommend that you click on the links below and watch the videos before continuing.

If after completing this topic you can state without hesitation that...

  • I can calculate conditional probability.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

teacher-scored 56 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 7 of your enrollment date for this class.


11.08 Addition Rule (Math Level 2)

Use the addition rules to calculate the probability of disjoint events.

The material for this lesson is from Elgin Community College under a Creative Commons License (faculty.elgin.edu/dkernler/statistics/ch05/Chapter5.pdf)

The first situation we want to look at is when two events have no outcomes in common. We call events like this disjoint events - also commonly known as mutually exclusive events.

Here's an example of a Venn diagram showing two disjoint outcomes, E and F.

Let's continue this a little further and put points on the chart to indicate outcomes.

Looking at the picture, we can clearly see that P(E) = \small \dpi{100} \fn_phv \small \frac{5}{15} = \small \frac{1}{3}, since there are 5 outcomes in E, and 15 total outcomes. Similarly, P(F) also is \small \frac{1}{3}.

Next, we want to consider all of the events that are in either E or F. In probability, we call that event E or F. So in our example, P(E or F) = \small \frac{10}{15} = \small \frac{2}{3}.

But, we could just see that from the picture! Just count the dots that are E and add to it the number of dots in F.

In general, we can create a rule. We'll call it...

The Addition Rule for Disjoint Events: If E and F are disjoint (mutually exclusive) events, then P(E or F) = P(E) + P(F)

OK – time for an example.

Suppose we have a family with three children, and we consider the gender of those three children. Let B represent a boy and G represent a girl. Here is the sample space:

S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, BBB}

There are thus eight outcomes in this experiment.

Now, let's define the following events:

E = the family has exactly two boys

F = the family has exactly one boy

Describe the event "E or F" and find its probability.

ANSWER: In this case, the event is that a family has either exactly two boys OR exactly one boy. From the sample space, we can see that there are two events with exactly two boys. So, the P(E) = \small \frac{2}{8} = \small \frac{1}{4}. There are also two events with exactly one boy. So, the P(F) = \small \frac{2}{8} = \small \frac{1}{4}. We can now calculate the desired probability:

P(E or F) = P(E) + P(F) = \small \frac{1}{4}+\frac{1}{4} = \frac{2}{4} = \frac{1}{2}

Of course, there are often cases when two events do have outcomes in common, so we'll need a more robust rule for that case.

General Addition Rule

What happens when two events do have outcomes in common?

Well, let's consider the example below. In this case, P(E) = \small \frac{4}{10} = \small \frac{2}{5}, and P(F) = \small \frac{5}{10} = \small \frac{1}{2}, but P(E or F) isn't \small \frac{9}{10}. Can you see why?

The key here is the two outcomes in the middle where E and F overlap. Officially, we call this region the event E and F. It's all the outcomes that are in both E and F. In our visual example:

In this case, to find P(E or F), we'll need to add up the outcomes in E with the outcomes in F, and then subtract the duplicates we counted that are in E and F. We call this the General Addition Rule.

The General Addition Rule: P(E or F) = P(E) + P(F) - P(E and F)

Let's try a couple quick examples.

Let's consider a deck of standard playing cards.

Example 1: Suppose we draw one card at random from the deck and define the following events:

G = the card drawn is an ace

H = the card drawn is a king

Use these definitions to find P(G or H).

Is there any overlap? Would there be any in the P(G or H)?

No!

P(G) = \frac{4}{52}=\frac{1}{13} and P(H) = \frac{4}{52}=\frac{1}{13}

P(G or H) = \frac{1}{13} + \frac{1}{13} = \frac{2}{13}

So the key idea and the difference between these two examples – when you're finding P(E or F), be sure to look for outcomes that E and F have in common.

If after completing this topic you can state without hesitation that...

  • I can use the addition rules to calculate the probability of disjoint events.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

11.08 Addition Rule - videos (Math Level 2)

11.08 Addition Rule - Worksheet (Math Level 2)

teacher-scored 72 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 7 of your enrollment date for this class.


11.09 Two-Way Frequency Tables of Data and Conditional Probabilities

Use two-way frequency tables of data to compute conditional probabilities.

The material for this lesson is from DIGI 203 Algebra 1 (d203algebra.wikispaces.com/Guided+Learning+1B) under a Creative Commons Attribution Share-Alike 3.0 License.

A two-way frequency table is used to compare different categories of data. The table below was created from a survey where 50 students were asked what foreign language they were taking. In the table below, the different categories being compared are gender (boys and girls), and subject taken (Spanish, French, and German). The table can provide different pieces of information. For example, if you wanted to know how many girls are currently taking French, you would look at the "girls" row and move to the "French" column. You can see that it is 12. If you wanted to know the number of total boys who are taking a foreign language, you would look at the "boys" row, and go to the "total" column to find out that 20 boys are taking a foreign language.

Looking at the number of girls taking French is called a joint frequency, because you are combining two different categories (gender and foreign language). Looking at the total number of boys taking a foreign language in general is called a marginal frequency because you are only looking at one category (gender).

The marginal frequency (the number of times a certain response is given) is shown in red. The joint frequency (the number of times a certain response is given by a certain group) is shown in blue.

We can interpret the data even further by creating what is called a two-way relative frequency table. In this type of table, the percent of a joint frequency or marginal frequency is compared to a total. A two-way relative frequency table can be created with respect to the table total, the table rows, or the table columns.

When a two-way relative frequency table is created with respect to the table total, all of the entries in the table are divided by the overall table total. The entries are then replaced with the decimal that represents the percentage.

When a two-way relative frequency table is created with respect to the rows, the data entries in each row are divided by the total for that row. For example, in order to create the relative frequency for the Boys row, take each entry in that row (10, 2, 8, and 20), and divide by the total for the Boys row (20). This is where the 0.5, 0.1, 0.4, and 1.00 comes from in the table on the right.

When a two-way relative frequency table is created with respect to the columns, the data entries in each column are divided by the total for that column. For example, in order to create the relative frequency for the Spanish column, take each entry in that column (10, 15, and 25), and divide by the total for the Spanish column (25). This is where the 0.4, 0.6, and 1.00 comes from in the table on the right.

Example: Create a two-way frequency table and answer the questions that follow. Twenty five kids were asked what they like to do on a hot day. Ten of the 13 girls said they like to go to the pool. Eight of the boys said they like to play video games indoors. Create the two-way frequency table. Then create a two-way relative frequency table with respect to the table total.

a) How many boys prefer to go to the pool?

b) What percentage of girls out of all the kids prefer to watch video games?

c) What percentage of the kids are boys?

a) 4 boys
b) 12%
c) 48%

Click on the link below to watch the video.

Practice Problems

This piqued your interest in biking to school and you took a survey in your class of 30 students and the results are as follows:

a. How many boys are in the set?

b. How many girls are in the set?

c. How many students can bike?

d. How many girls can't bike?

e. Create a two-way frequency table (with respect to the table total).

f. With respect to the total what percent of girls could bike?

g. Create a two-way frequency table (with respect to the rows).

h. What percent of boys could bike?

i. Create a two-way frequency table (with respect to the columns).

j. What percent of people that could bike are girls?
 

There are 150 children at a local pool are signed up in the aquatic sports of which 61 signed up for swim team. There were a total of 52 children that signed for water polo and 28 of them also signed up for swim team. Use the two-way frequency table below to summarizing the data.

a. How many children did not sign up for water polo? Does this represent a marginal or joint frequenc

b. How many students did not sign up for either water polo or swim team? Does this represent a marginal or joint frequency?

You are in charge of boosting profits for the concession stand at the basketball games. You decide to do some research. On Friday night as each person entered the basketball game, you counted how many of the 400 people had a hotdog, a slice of pizza, soda and bottled water. You found that out of 182 people that had Pizza, 120 chose soda and only four did not have a drink. 50 people bought a hotdog and a soda. 158 people bought hotdogs. Ten people walked in without any food or a drink. The stand sold 200 sodas and 140 waters. Using this data, fill in the two way frequency table and determine the marginal frequencies of the data.

a. What was the joint frequency of people who ordered pizza AND a drink?

b. Based on the data, is it more likely that people will order pizza without a drink or a hot dog without a drink? Justify your answer using information from the frequency table.

(Answers are at the end of the lesson).
 

Quick Quiz

Click on the link below:

Answers to Practice Problems

1. Biking to School

a. 11

b. 19

c. 16

d. 10

e.

f. 30%

g.

h. 64%

i.

j. 56%

2. Aquatic Sports

a.  98

b. marginal

c.  65

d. joint

3. Concession Stand Sales

a. 98%

b.  Hot dog/no drink = 41%

If after completing this topic you can state without hesitation that...

  • I can use two-way frequency tables of data to compute conditional probabilities.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

teacher-scored 60 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 8 of your enrollment date for this class.


11.10 Using Sample to Estimate Probabilities (Math Level 2)

Use samples to estimate probabilities.

We have been using samples to calculate probabilities. A sample is useful for many situations that arise. However, most of the time you will be interested in an entire group – not just members of your class or school.

In the example from the previous lesson we discussed the issue of biking to school using the following table:

You could replicate this in your math class. You could, with some work, replicate it for your entire school. With additional effort and cooperation, you could replicate it throughout your district or all the schools in your state. Can you imagine the effort and cooperation needed to replicate it for all high school students? All high school students in this case are called the population. A population includes ALL the participants or items that are involved. A sample is a subset taken from the population.

Using sampling helps us to estimate the probability of an event if it is sufficiently large and the sample is selected randomly.

View the following video from the Khan Academy Collection:

Click on the link below and CAREFULLY study the sample problem:

\large {\color{Red} Assignment} \large {\color{Red} 11.10}

Estimating the Mean State Area

The table below gives the areas (in thousands of square miles) for each of the “lower 48” states. This serves as the population for this study. Your task involves taking small samples from this population and using the sample mean to estimate the mean area for the population of states by following the steps indicated below.

Your challenge is to discover some important properties of random samples, properties that illustrate why random sampling is the key to getting good statistical information about a population. In this task, unlike “real life” situations, you will have all of the population data at hand, and will use it to see how random sampling works. Your classmates will have the same task, and you will be combining your data with theirs. In the first part of the task, you will select a sample of states by a method of your choice. For the second part you will follow a specified procedure.

Procedure #1: Choose your own sample

1. By any quick method you like, select 5 states that you think represent the 48 (perhaps by tossing 5 grains of rice on a map of the states and selecting the states on which they fall; shutting your eyes and pointing your finger at a spot on the map, repeating the process until 5 states are selected; systematically selecting a state from the northeast, the south, the mid-west, all east of the Mississippi, and two states from west of the Mississippi).

2. Find the areas of these 5 states and calculate the mean for your sample.

3. Repeat this process 4 more times. Each time randomly select 5 more states and calculate the mean of the areas for those 5 states. You will have 5 means when you are finished.

4. Construct a dot plot of the 5 sample means.

Procedure #2: Use random sampling

5. Number the 48 states from 1 to 48. Then use a random number table or a random number generator to obtain 5 random numbers between 1 and 48, and then find the states corresponding to these numbers.

6. Find the areas of these 5 states and calculate the mean for your random sample.

7. Repeat this process 4 more times and calculate the mean for each set of 5 states you randomly select until you have 5 sample means.

8. Construct a dot plot of the sample means from the random samples.

9. Compare the plots produced in steps c and f. Where are the centers? Which has greater spread?

10. Repeat steps 1-6 for random samples of size 10 and compare the plots. What differences, if any, do you see in the plots? What feature or features appear to stay the same?

11. Find the actual mean state area using the data from all the states. Summarize at least two important points concerning the value of random sampling.

State Area
Texas 269
California 164
Montana 147
New Mexico 122
Arizona 114
Nevada 111
Colorado 104
Oregon 98
Wyoming 97
Michigan 97
Minnesota 87
Utah 85
Idaho 84
Kansas 82
Nebraska 77
South Dakota 77
Washington 71
North Dakota 71
Oklahoma 70
Missouri 70
Florida 66
Wisconsin 65
Georgia 59
Illinois 58
Iowa 56
New York 55
North Carolina 54
Arkansas 53
Alabama 52
Louisiana 52
Mississippi 48
Pennsylvania 46
Ohio 45
Virginia 43
Tennessee 42
Kentucky 40
Indiana 36
Maine 35
South Carolina 32
West Virginia 24
Maryland 12
Massachusetts 11
Vermont 10
New Hampshire 9
New Jersey 9
Connecticut 6
Delaware 2
Rhode Island 2

 

Conclusion

Random samples can produce distributions of sample means that center at the population mean, and that the variation in the sample means will decrease noticeably as the sample size increases. Random sampling is a “fair” way to select the sample because each item in the population gets an equal chance of being selected.  

If after completing this topic you can state without hesitation that...

  • I can use sample to estimate probabilities,

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

teacher-scored 45 points possible 30 minutes

Complete the task at the end of the lesson material, "Estimating the Mean State Area" on your own piece of paper. Make sure you answer all questions in the task.

When you have completed the task, digitize your paper by either scanning it and saving it to your computer, or by taking a digital picture, up close and clear. Then submit the file under the assignment link on your math home page.

Pacing: complete this by the end of Week 8 of your enrollment date for this class.


11.10.01 Unit 11 Review Quiz - (Math Level 2)

computer-scored 50 points possible 35 minutes
Pacing: complete this by the end of Week 8 of your enrollment date for this class.