TEKS Support |
This unit contains activities that support the following knowledge and skills elements of the TEKS.
(1) (A) |
X |
(4) (A) |
X |
(1) (B) |
X |
(4) (B) |
X |
(1) (C) |
X |
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(2) (A) |
X |
(8) (A) |
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(2) (B) |
X |
(8) (B) |
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(2) (C) |
(8) (C) |
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(2) (D) |
X |
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(3) (A) |
X |
(9) (A) |
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(3) (B) |
X |
(9) (B) |
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(3) (C) |
X |
The mathematical prerequisites for this unit are
The mathematical topics included or taught in this unit are
The equipment list for this unit is
Teacher Notes |
"Simple Model"Context Overview |
The unit examines testing strategies for sampling pairs of individuals when cost is a consideration. Pooling of samples may require only one test and therefore save money, or may require three tests, losing money in the process. Whether pairs of samples should be initially pooled for testing depends on the probability that a single result is positive.
Students gather data by performing Monte Carlo experiments assuming different probabilities, and try to determine the mathematical model for this situation. They verify empirical results, deriving a theoretical relationship between the incidence of occurrence and the expected number of tests. Finally, they determine the "break-even" point, where its more cost-effective to test the samples individually than to pool the samples and test them in pairs.
Mathematical Development |
Deciding when it is appropriate to pool samples is explored through discussion of the modeling process, concrete activity, and calculator simulation. The concept of expected value plays a major role in establishing the condition for the problem. Data analysis, using least squares regression, determines the constants that produce the best fit for linear, quadratic, and exponential functions. Examination of the residual patterns provides a basis for choosing which of these functions best describes the data. Probability area models are introduced to verify that the relationship between the probability of testing positive and the expected number of tests required is quadratic, and to determine the exact quadratic that models the situation. Finally, methods for solving the resulting quadratic equation are explored, and the problem of determining when it is cost-effective is solved.
Preparation Reading"Anabolic Steroids: Use and Effect" |
The reading provides some background into the problem of detecting steroid use and the physiological effects of taking anabolic steroids. The intent of the reading assignment is to generate a discussion of the use of steroids in the general population, and to build some understanding for the context of testing for steroid use. Prior to class, students should read the contents of this article, and be prepared to discuss some of the following questions:
In addition, you might want to facilitate a discussion about current events that involve steroid testing or steroid use. Possibilities include: the Tour de France bicycling competition from the summer of 1998, Mark McGwire admitting that he takes Androä , Ben Johnson having his life ban from track and field upheld, or steroid testing of athletes at the local high school or college. Specific topics might include how the tests are administered, the stigma of a positive result, fairness issues when one sport bans the use of a substance and another allows it, "professional" wrestling and steroid use, or even talk-show spectacles about people who have to have a leg amputated from excessive steroid use.
Another option for you might be to have students research the topic of steroid use, using the sources cited below, the Internet, or other materials they find on public health. Have them bring in an interesting article on the subject to share with the class, or present a report on their research at various points in the unit.
The following resources were used to generate the Preparation Reading, and contain information you may find useful.
Wright, J.E., and V. Cowart, Anabolic Steroids: Altered States, Carmel, Indiana: Benchmark Press, 1990.
Donohue, T., and N. Johnson, Foul Play: Drug Abuse in Sports, New York: Basil Blackwell Ltd, 1986.
Meer, J., Drugs and Sports, New York: Chelsea House Publishers, 1987.
DiscussionModeling Process |
Having brought the issue of steroid testing to the students awareness, you need to shift their thinking to the mathematical nature of testing samples in pairs. The discussion of the modeling process has probably taken place in previous units, and begins with the students developing a well-defined question. While students may be considering many questions, cost is a critical factor in trying to determine when to pool samples.
The second step of the modeling process is to identify key features or assumptions that will be part of the development of the model, and to introduce variables into the problem.
The third step of the modeling process is to begin to examine the mathematical nature of the problem and establish a relationship between the two variables.
At this point, you can do one of two things:
1) announce to students that they will begin their exploration by trying to find out if the relationship between p and E is linear, and begin Activity 1, or
2) you can ask them how they might verify their conjecture to the number of tests needed when the probability of testing positive in the population is 0.5, and then begin Activity 1.
Activity 1"Its Only a Test" |
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= 1, 2, 3, 4, 5, 7, 8
Materials needed (per group): |
Handout 1: Data Table for Part 2 (one per student) Opaque bag containing tiles, chips or beadstwo colors, representing a positive or negative test, and containing the same number of each color object (to model the situation in which p = 0.50) Graphing calculators with DRACULA program (Handout 3) loaded into memory or Handout 2: Data Table for Part 4 (one per student) |
Part 1 of the activity is a review of the previous discussion. There are several ways in which you can proceed. You might let the students struggle with the questions from Part 1, and then have the discussion of the modeling process, with students recording answers for future reference. You might have the discussion first, and give students several minutes to answer these questions to make sure they understand what was discussed. Depending on the time in class, you could have the discussion in class and send the students home to formally answer the questions as part of a homework assignment.
Part 2 is where the students actually do the experiment. Model for the students how to perform the simulation; have two students playing the roles of the athletes, and you be the tester. One student should draw an object and put it back; then the other student should draw an object and put it back. Upon consulting with each other, they should announce whether the pooled test would be positive or negative. If positive, you select one of the students for an individual test, and he or she tells you the result. Finally, you record the results of the test on one row of the data table on Handout 1the actual status determined by the color of the object drawn, the results of the test determined by the pooled result and which individual is tested first. Check to see if they understand what they are doing before allowing the students to begin the experiment.
In Part 3, students have the opportunity to answer the question, "How many tests should we expect when the probability of testing positive is 0.50?" Groups must communicate their results so the class data can be averaged together. The easiest way to do this is to have each group post its average number of tests on an overhead transparency or the board. Discuss the answers to Part 3 after the students have had a chance to think about them and formulate their own responses. Make sure that they understand that the results of the simulation indicate that it is possible to have the average number of tests be 2, but its not likely that we should expect the number of tests to average out to be 2. We need more data to minimize the fluctuations caused by individual trial results.
Part 4 has two versions to provide a little flexibility for individual teachers situations. The DRACULA simulation uses a calculator program to simulate the testing game with a larger number of trials; the commands that make up the program are included in the unit material as Handout 3. The other version, which uses coin tosses to simulate the testing game, is provided in case the calculators arent available, or if time is too limited to fully explore the activity. Handout 2 data table is for use with this version. In either case, students should discover two things about this situation: 1) increasing the number of trials to the simulation gives less individual fluctuation to the results, and 2) whatever the expected number of tests actually is, its slightly higher than 2 and not equal to 2.
Supplemental Activity 1"Middle of a Model" (Manipulative Version) |
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In case the graphing calculators are not available for Activity 2, an optional development is provided in which students repeat the testing game from Activity 1. Each group models one of the probabilities mentioned in Activity 1 by adjusting the relative numbers of objects from which they are drawing. They need to repeat the experiment 100 times, so that the data will be fairly patterned. Then they share their data as a class, and the rest of the activity is the same as Activity 1. Refer to those teacher notes for more details.
Homework 1"Lets Rumble" |
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This is a collection of questions related to the testing game introduced in Activity 1. In discussing it with students afterward, ask students to connect Question 5 to the simulation done in class. Had they done 1000 trials, instead of 10 or 100, there would be even less fluctuation in the results and they would be even closer to the actual expected number of tests.
Activity 2"Middle of a Model" |
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This activity should begin with a continuation of the discussion of the modeling process. Weve determined that a probability of 0.5 for testing positive doesnt yield two tests as the expected number. But does that mean that the behavior is linear, and that particular point isnt exactly on the line? Or is the behavior nonlinear? And, if it isnt linear, the natural question that begs to be asked is: What is it?
At this point, you can distribute Activity 2 and the graphing calculators. Each group should be assigned a probability; 8 groups of students can run simulations with p = 0.1, 0.2, 0.3, 0.4, 0.6, 0.7, 0.8, and 0.9. Remember that we already have data for p = 0.5, and the expected number of tests is determined when p = 0 and p = 1. The directions instruct the students to enter 1000 as the number of trials; this will take several minutes. You could build in some short "sponge" activity while the students are waiting, or you could tell them to enter in a smaller number, like 100. The smaller number of trials will get an answer back much faster, but the data set wont be as smoothly patterned. For the purposes of this activity and the rest of the unit, it doesnt matter which way you go.
After the groups have run the simulation, a member from each group should record the groups result on an overhead transparency or the board, and all students should record the class data on their activity pages in the data table. In analyzing the pattern, students are asked to "fit" the data with a line that starts at (0,1) and ends at (1,3). You may need to review how to find the equation of a line from two points. Make sure students understand the difference between observed value (numbers gotten from the calculator simulation) and expected value (numbers gotten by evaluating the line equation for the various values of p).
Homework 2"Expect to Be Worth Something" |
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The first part of the homework activity will require students to have a pair of dice. The activity is meant to provide them an opportunity to explore the problems various outcomes and payoffs before tackling the problem by mathematical analysis. An option for you is to do that part of the homework activity in class and let students do the analysis when they get home. Be sure that students understand the definition of probabilistic worth and expected value, or at least call to their attention the fact that the boldface words are actually definitions, and tell you how to proceed with the calculations.
Activity 3Finally, A Model |
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Materials needed |
Graphing calculators Handouts 4, 5, and 6 |
The purpose of this activity is to use regression analysis to determine if a linear, exponential, or quadratic function is the "best" type to use as a model. Were continuing to use the data that have been collected on the expected number of tests and the probability of steroid use. Remind students that the process used in the previous activity (find an equation, calculate predicted values, calculate residuals, and examine the residual plot) is going to stay the same, but this time, the equations will be determined from regression analysis.
If students have never done regression, or were unsure of what they were doing, take the opportunity to show them how to do the linear regression (Part 1). Even if they are pretty clear of the idea, they may have forgotten how to use the graphing calculators to do that task, and reviewing with them the steps on the calculator isnt a bad thing. Having done that, the students should proceed to do parts 2 and 3 on their own, so that they can discover the nature of the quadratic relationship themselves. Encourage them to explore other types of mathematical functions as well, even though thats labeled an optional part.
Students should continue to Part 4 and answer the questions based upon the work that theyve done. Have a brief discussion at the end, asking students to:
Homework 3Its a Good Fit |
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There are several options for the teacher in proceeding with this assignment. If students have access to graphing calculators, this assignment is a nice opportunity for them to practice developing linear models and determining if they truly are linear. If the students dont have access to graphing calculators outside class, several options are still available.
Activity 4Which Model Fits the Best? |
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Students work in groups to analyze several data sets out of the context of steroid testing, and must determine the nature of the mathematical relationship among the data sets. There are a couple of ways to proceed with this activity:
Activity 5Verifying the Model as Quadratic |
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= 16, 17, 18
Materials needed |
Transparency 1 |
This activity illustrates the idea of probability as an area. The specific example, which is the focus for Part 1, is the original testing problem, in which the assumption is that 50% of the population is taking steroids. Depending on your students, you can either lead the class through a discussion of the questions that form Activity 1 (with them recording the answers as they go) or let them work on that part for about 10-15 minutes and have a discussion of the material at that time. Its important that the students identify each area and its meaning with respect to the probability conditions. Be sure that they verbalize the condition and understand why the values are being multiplied. Then proceed to Part 2, in which the problem is generalized to a condition that the probability of steroid use in the population is p, instead of 0.5. Students may struggle with the algebraic simplifications needed for Questions 5 and 7, but thats the "punch line" to verifying the model as being quadratic, and well worth the effort. If needed, use Transparency 1 to review the questions from Part 2 in the follow-up discussion.
Remind your class that the probability that each of two events happens is calculated as follows:
P(A and B) = P(A) · P(B½A).
That is, "the probability of event A and event B both happening is equal to the product of the probability of event A times the probability of event B knowing that event A has already happened. If the probabilities of event A and event B are not related (independent events), the probability of "B given that A has occurred" is simply the probability of event B. It is only because we have assumed that the two individual tests are independent that we can state:
P(A and B) = P(A) · P(B).
Note: As you saw in the unit Imperfect Testing, the area model of probability can also be represented using probability trees. Though they are not discussed on the student pages of this unit, you may want to encourage students to use both methods on some problems.
Supplemental Activity 5Introduction to the Parabola |
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Materials needed |
Handout 7 Transparency 2 |
This activity is designated as a supplemental activity only because it doesnt specifically address the problem of steroid testing and the cost-effectiveness of pooling blood samples. However, it is strongly recommended that a day be spent working on the various parts of this activity if the students have never seen a quadratic function, or if they havent mastered the concepts of translations and stretches of quadratics and the distinction between standard-form and vertex-form equations for parabolas.
In part 1, students concentrate on the graphs of the parent function y = x2 and translations on that function. Introduce them to the parent function by having them fill in the table of values and plot the points on the graph grid on Handout 7. Recall the concepts of domain and range, and reinforce the range limitation by asking the class to find the x-value that yields y = 1. Encourage discussion about "pairs" of points (x, y) and (x, y), introducing the concept and terminology of line symmetry. Include several fractions and decimals in your table. Graph the function and label the vertex. (Use Transparency 2 to help you with the instruction, if needed). Use the graphing calculator to verify the drawing, and use the TRACE feature to verify the values recorded in the table.
Allow students to work in their groups to explore the transformations. There are four different transformations, so each student in the group can explore one of the shifts. Students should come together as a group to discuss their findings, and then answer Question 3. End the work for Part 1 by reviewing:
1) the shape of the parent parabola graph (y = x2),
2) the rules governing lateral translations of functions and
3) the domains and ranges of parabolas in the form y k = (x h)2.
In Part 2, students concentrate on how the graph of the parent function y = x2 is affected by stretch transformation and combinations of stretches and shifts. Once again, the work is divided into four parts, and each student in the group should explore two examples on his or her own. Then, the groups should collaborate to compare their results and draft responses to Questions 2 and 3. In the class discussion that follows the group explorations, ask if such stretch transformations change the domain or range of the parabola. (A horizontal stretch transformation would never change the domain or range, but a vertical stretch would change the range for parabolas whose vertex points are not on the x-axis.) Ask students if it matters whether one does the shift first and then the stretch, or the stretch first and then the shift. Close the discussion by reviewing how stretches in in both vertical and horizontal directions can be done.
In Part 3, students are asked to examine the two forms for the equation of a parabola: standard form and vertex form. They explore the parallel manipulation of the equation and its graph to discover how to change the equation from one form to another. Allow students to work through Questions 1-3, then discuss their findings before proceeding to Question 4, which gives students an additional opportunity to practice.
Homework 5Quadratic Nature of Pairing Samples |
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This assignment asks students to examine the expected value calculation again, first using a specific value for the probability of testing positive (0.3) and then examining the model developed. Students are led to examine how changes in the probability affect the model geometrically (how do the various shapes adjust?) and analytically (how do the various pieces of the equation behave?).
A really effective way of exploring Question 1 would be to construct an area model on the computer using a dynamic drawing program like Geometers Sketchpad. Display the areas of the three regions representing the three different tests, and the probabilities associated with those regions. Then drag the intersection point from the middle of the square (Point E) toward the upper left or lower right corner and watch what happens to the areas and the calculations!
Activity 6Solving the Problem |
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Closure on the problem of the cost-effectiveness of pooling blood samples is reached by examining the solution to the questions: "What is the break-even point? When is it cheaper to pool blood samples? When is it cheaper to test samples individually?" Various methods are explored for solving the quadratic equation describing the condition that our model should have a value equal to 2. Part 1 is pretty straightforward, using the calculators capability to determine intersection of two graphs. You might want to show your students the CALC menu feature that locates the intersection of two curves at this same time.
Part 2 examines the method of solving a quadratic equation called "completing the square." Students are led through this process and then asked to solve the modeling problem by applying their method. Let students work through Part 2; discuss individual aspects of the process as needed. After students solve the problem in Question 10, have them take a minute to compare their answers to the ones obtained in Part 1; they should verify the solution.
Finally, in Part 3, students are asked to generalize the work they did in Part 2 and to derive the quadratic formula. Let them work through Question 1, and have them tell you the steps involved. Make sure that there is no ambiguity or confusion in their process or thinking. Then let them write their answers to Question 2 before proceeding to the general equation. Discuss their work on Question 3 as needed, before letting the students go on to Question 4.
Homework 6Practice With Quadratics |
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This assignment should be an opportunity for students to practice what they have discovered in the previous work, and a chance for you to see which students really "got it" and which are still struggling with the algebra developed in this last activity.
HANDOUT 1Activity 1 Part 2 |
Data Table
Trial No. |
1st Persons Actual Status |
2nd Persons |
Result of Pooled Test |
Result of 1st Individual Test |
Result of 2nd Individual Test |
Number of Tests Needed |
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1 |
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2 |
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3 |
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4 |
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5 |
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6 |
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7 |
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8 |
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9 |
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10 |
Average Number |
HANDOUT 2Activity 1 Part 4 |
Data Table
Trial |
Number |
Trial |
Number |
Trial |
Number |
Trial |
Number of Tests |
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1 |
26 |
51 |
76 |
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2 |
27 |
53 |
77 |
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3 |
28 |
53 |
78 |
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4 |
29 |
54 |
79 |
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5 |
30 |
55 |
80 |
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6 |
31 |
56 |
81 |
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7 |
32 |
57 |
82 |
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8 |
33 |
58 |
83 |
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9 |
34 |
59 |
84 |
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10 |
35 |
60 |
85 |
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11 |
36 |
61 |
86 |
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12 |
37 |
62 |
87 |
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13 |
38 |
63 |
88 |
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14 |
39 |
64 |
89 |
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15 |
40 |
65 |
90 |
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16 |
41 |
66 |
91 |
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17 |
42 |
67 |
92 |
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18 |
43 |
68 |
93 |
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19 |
44 |
69 |
94 |
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20 |
45 |
70 |
95 |
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21 |
46 |
71 |
96 |
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22 |
47 |
72 |
97 |
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23 |
48 |
73 |
98 |
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24 |
49 |
74 |
99 |
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25 |
50 |
75 |
100 |
Average Number of Tests: |
HANDOUT 3Dracula Program |
(written for a TI-82 calculator)
PROGRAM:DRACULA
:0® C
:Disp "WHAT IS P?"
:Prompt P
:Disp "HOW MANY TRIALS?"
:Prompt N
:For(I,1,N,1)
:rand® A
:rand® B
:If A> P and B> P
:C+1® C
:If A< P
:C+3® C
:If A> P and B< P
:C+2® C
:End
:Disp "AVERAGE NUMBER"
:Disp "WAS"
:Disp C/N
:Stop
HANDOUT 4Linear Regression for Activity 3 |
Equation: E =
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Observed |
Expected |
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0.0 |
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0.1 |
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0.2 |
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0.3 |
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0.4 |
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0.5 |
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0.6 |
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0.7 |
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0.8 |
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0.9 |
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1.0 |


Exponential Regression
HANDOUT 5Exponential Regression for Activity 3 |
Equation: E =
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Observed |
Expected |
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0.0 |
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0.1 |
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0.2 |
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0.3 |
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0.4 |
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0.5 |
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0.6 |
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0.7 |
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0.8 |
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0.9 |
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1.0 |


Exponential Regression
HANDOUT 6Quadratic Regression for Activity 3 |
Equation: E =
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Observed |
Expected |
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0.0 |
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0.1 |
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0.2 |
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0.3 |
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0.4 |
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0.5 |
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0.6 |
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0.7 |
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0.8 |
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0.9 |
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1.0 |


Exponential Regression
HANDOUT 7Graph Grid for Supplemental Activity 5 |

Annotated Student Materials |
Preparation ReadingAnabolic Steroids: Use and Effect |
Anabolic steroids, synthetic compounds created to act like the male hormone testosterone, remain in the forefront of discussions regarding athletes and sports. Many athletes such as bodybuilders, weight lifters, runners, swimmers, and football players believe that steroids will give them strength advantages.
Actually, the development of anabolic steroids was an important scientific breakthrough. There are, in fact, four legal uses for the drugs. They are effective in treating certain forms of anemia, some kinds of cancer, pituitary dwarfism, and serious hormone disturbances.
By virtue of their specific chemical structure (which is totally different from any vitamin or amino acid), anabolic steroids can stimulate the genetic apparatus (DNA) within the cell of the nucleus. The DNA reacts to this stimulus by directing the production of specific new proteins. It is very important to understand that the biological response that occurs is dependent on the location and number of receptors. If the receptor is muscle, there will be a tissue-building effect. If the receptor is in the brain, there may be noticeable psychological and even behavioral effects. There are, incidentally, more receptors in "sexual" tissue than in muscle, which raises concern about the prospect of hypertrophy or cancer of the prostate gland.
Although the development of bigger muscles is a result of the anabolic component of the steroid, there is also an androgenic associated with the steroid. Androgenics cause growth of facial hair, lengthening of the vocal cords that results in the voice "breaking", acne, and the premature closure of the space between parts of the bodys long bones. Anabolic steroids can have harmful effects on the liver and the cardiovascular system (causing a higher risk for developing high blood pressure and blood-clotting problems). There is growing concern that flooding a young adolescent with synthetic sex hormones may disrupt not only physical changes but also normal psychological and emotional maturation that occur during adolescence.
Several studies have placed the incidence of anabolic-androgenic steroid use among high-school students at 6 to 7%. Figures as high as 11 percent of 11th-grade boys using steroids have been reported, meaning that between 250,000 and 500,000 high-school students have used steroids. Even more disturbing are reports that approximately 40 percent who have used steroids have used five or more cycles, and approximately 40 percent began using steroids before the age of 16. As expected, the majority of teenage users are participants in school athletics, but as many as a third of the male users are not involved in sports and take these drugs in an attempt to enhance their appearance.
There have been only a few surveys among the male college population, all of which show an estimated 2 percent usage. However, among male college athletes, the estimated usage rises to between 5 and 17%. The problem is not restricted to males. As many as 1 to 2 percent of senior high school girls are users. Between 3 and 5 percent of Olympic and professional female athletes have admitted to using them at some time during their careers.
Source: Write, J. E., and V. Coward, Anabolic Steroids: Altered States, Carmel, Indiana: Benchmark Press, 1990.
Activity 1Its Only a Test |
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= 1, 2, 3, 4, 5, 6, 7, 8
Were trying to determine when its cheaper to "pool" blood samples in testing for steroid use.
The likelihood, p, that someone is actually using steroids in the population, and the expected number of tests, E, that would be needed.
The number of tests will always be one, two, or three tests.
One test every time, because the pooled sample would come up negative (clean).
Most of the time, only one test would be needed. Once in a while, a second (or third) test would be required. The average number would be close to one.
Most of the time, three tests would be neededthe pooled sample test would come up positive, and the first individual test would come up positive. Some of the time, only two tests would be needed, since the first individual test might be negative. The average number would be close to three.
Three tests every time, because the pooled sample would always come up positive, and the first individual test would also come up positive.
I guess that two tests would be needed. Since p = 0.5 is halfway between none of the people and all of the people, I expect that halfway between one test and three tests should be my answer.
The nature of the relationship between the two variables is linear; the expected number of tests increases proportionally to the population occurrence.
In groups, you are going to explore the problem of what happens when half the population is taking steroids. Two people will play the role of Olympic athletes, who have to be tested for steroid use. The third person will be the lab worker performing the test. Go through the simulation 10 times, recording the actual status of each athlete, the result of the pooled test, and the result of any individual tests, in the data table provided for this activity. Use marks of + for a positive result or for a negative result. Then, determine the number of tests needed for each trial, and finally the average number of tests needed for the 10 trials.
In the table below, record each groups results for the average number of tests needed for 10 trials. Answers will vary; sample results provided in the table.
Group |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Avg. No. of Tests |
2.3 |
1.9 |
1.8 |
2.4 |
2.4 |
1.9 |
2.5 |
2.4 |
2.2 |
2.5 |
When the average number of tests was under 2 tests, since testing the two athletes individually would always require 2 tests.
Yes. There were specific trials that required only 1 test, and, in certain groups, the average number < 2 tests.
Probably not. There were more group results that were > 2 tests, and the average of those results was » 2.2 tests. However, given the range of results, its hard to be sure. Some of the groups may have been very unlucky in choosing which of the individual samples to test first when the pooled sample came up positive.
Possible answers include: not replacing the first persons chip before drawing the second, having the chips stick to each other, nonuniform weights or shapes for the objects drawn, not mixing or shaking up the objects well enough, or looking at the objects as they are being drawn.
Do more trials. Specifically, if each group did the test 100 times, the range of results might narrow down a bit, and the average of those results would get close to the "true" results.
Each group will run the DRACULA program. When asked the first question, "What is P? P = ?", enter in the value 0.5 (remember, were studying what happens when half the population is taking steroids). When asked the second question, "How many trials? N = ?", enter in the value 100. After a few seconds, the calculator will provide the average number of tests needed.
In the table below, record each groups results for the average number of tests needed for 100 trials.
Answers will vary; sample results provided in the table.
Group |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Avg. No. of Tests |
2.16 |
2.21 |
2.31 |
2.04 |
2.24 |
2.37 |
2.33 |
2.16 |
2.26 |
2.18 |
No. The range of possible results when only 10 trials were taken was [1.8, 2.5]. When the simulation was repeated with 100 trials, the range became [2.04, 2.37]. In every situation, the average number of tests was greater than 2, and, when you consider the total number of tests (n = 1000), the average » 2.23. Incidentally, the average of the total number of tests before was » 2.2 as well. I thought that the results were supposed to be an average of 2 tests. I wasnt sure if the way the simulation was being run, or a streak of bad luck, might be causing me to get a different answer. But now, Im convinced that its not supposed to average to 2 tests, and, if I needed even more convincing, I would simply repeat the simulation again, doing an even larger number of trials.
You will repeat the simulation from todays class, only this time using pennies. Well agree that a penny that comes up tails will be a negative test result, and a penny that comes up heads will be a positive test result. You will need someone else to play the role of the tester, but you might have figured out there is a shortcut to doing this:
Record the results of the coin-toss simulation in the data table provided. Calculate the average number of tests needed for your simulation, and come to class tomorrow ready to share your results.
In the table below, record 10 students results for the average number of tests needed for 100 trials.
Answers will vary; sample results provided in the table.
Group |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Avg. No. of Tests |
2.16 |
2.21 |
2.31 |
2.04 |
2.24 |
2.37 |
2.33 |
2.16 |
2.26 |
2.18 |
No, the range of possible results when only 10 trials were taken was [1.8, 2.5]. When the simulation was repeated with 100 trials, the range became [2.04, 2.37]. In every situation, the average number of tests was greater than 2, and, when you consider the total number of tests (n = 1000), the average » 2.23. Incidentally, the average of the total number of tests before was » 2.2 as well. I thought that the results were supposed to be an average of 2 tests. I wasnt sure if the way the simulation was being run, or a streak of bad luck, might be causing me to get a different answer. But now, Im convinced that its not supposed to average to 2 tests, and, if I needed even more convincing, I would simply repeat the simulation again, doing an even larger number of trials.
Supplemental Activity 1Middle of a Model (Manipulative Version) |
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|
In Activity 1, we assumed that the probability was 0.50 for a person selected at random to be taking steroids. This activity will explore how different probability rates affect the number of tests required.
Your group will be assigned a particular probability to use, and then you will do the Testing Game again. Modify the number of each kind of marker used, so that it models the probability your group is working with. Then do the simulation 100 times and determine the average number of tests. When done, record your results in a table and share them with other groups according to directions given.
Record the class results for the average number of tests needed for 100 trials in the space below the appropriate probability of occurrence. Use a table similar to the data table in Handout 2, Activity 1.
Answers will vary; sample data provided in the table.
Probability (p) |
0.0 |
0.1 |
0.2 |
0.3 |
0.4 |
0.5 |
Avg. No. of Tests (E) |
1.00 |
1.25 |
1.56 |
1.85 |
2.06 |
2.24 |
Probability (p) |
0.6 |
0.7 |
0.8 |
0.9 |
1.0 |
|
Avg. No. of Tests (E) |
2.41 |
2.69 |
2.79 |
2.86 |
3.00 |
|

It appears to be fairly linear, starting at (0,1) and going to (1,3).
Slope = (3 1) / (1 0) = 2
E = 2p + 1 or y = 2x + 1
Error = Observed Expected
Probability |
Observed |
Expected |
Error |
0.0 |
1.00 |
1.00 |
0.00 |
0.1 |
1.25 |
1.20 |
+ 0.05 |
0.2 |
1.56 |
1.40 |
+ 0.16 |
0.3 |
1.85 |
1.60 |
+ 0.25 |
0.4 |
2.06 |
1.80 |
+ 0.26 |
0.5 |
2.24 |
2.00 |
+ 0.24 |
0.6 |
2.41 |
2.20 |
+ 0.21 |
0.7 |
2.69 |
2.40 |
+ 0.29 |
0.8 |
2.79 |
2.60 |
+ 0.19 |
|
0.9 |
2.86 |
2.80 |
+ 0.06 |
1.0 |
3.00 |
3.00 |
0.00 |

No. The residual plot has no points below the zero line and is not showing a random pattern.
Homework 1Lets Rumble |
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|
(2)($160) = $320
Based on an average of 2.23 tests per pair, it would cost (2.23)($160) = $356.80
Trial Number |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Number of Tests |
1 |
3 |
3 |
1 |
1 |
1 |
3 |
1 |
1 |
3 |
Probably not. The situation in which two tests were required never came up.
3 green, 1 blue
3 green, 17 blue
23 green, 77 blue
1 green and 3 blue
2/3 or 0.67; 1/3 or 0.33
0%; 100%
No, but that would be the most likely outcome.
No, but that (very unlikely) event would still be the most likely outcome.
The second option is more likely.
Approximately 50. This would naturally draw the overall percentage closer to 50% (i.e. 58 out of 110, if exactly 50 of the next 100 tosses are heads). Be careful not to assume that the data will automatically compensate for the initial results.
Not replacing the drawn chip changes the probability of drawing a green chip on the second draw. We want to model the situation in which both athletes have the same probability of testing positive.
Many states have lotteries in which a collection of six numbers is chosen from 44 to 51 different possible numbers. In that situation, the numbers shouldnt be replaced.
Activity 2Middle of a Model |
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|
= 6, 9, 10, 11
In Activity 1, we assumed that the probability was 0.50 for a person selected at random to be taking steroids. This activity will explore how different probability rates affect the expected number of tests required.
The teacher will assign a particular probability to each group, which will then run the DRACULA program. When prompted by the calculator program, give it the probability (decimal, please!), and set the number of trials to be 1000. Settle in for a couple of minutes; after the calculator has done all those trials, record your results in the table below and share them with other groups according to directions given.
In the tables below, record the class results for the average number of tests needed for 1000 trials in the space below the appropriate probability of occurrence:
Answers will vary; sample data provided in the table.
Probability (p) |
0.0 |
0.1 |
0.2 |
0.3 |
0.4 |
0.5 |
Avg. No. of Tests (E) |
1.000 |
1.300 |
1.554 |
1.879 |
2.042 |
2.237 |
Probability (p) |
0.6 |
0.7 |
0.8 |
0.9 |
1.0 |
|
Avg. No. of Tests (E) |
2.418 |
2.632 |
2.727 |
2.873 |
3.000 |
|

It appears to be fairly linear, starting at (0,1) and going to (1,3).
Slope = (3 1) / (1 0) = 2
E = 2p + 1 or y = 2x + 1
Errors = Observed Expected
|
Probability |
Observed |
Expected Value |
Error |
|
0.0 |
1.000 |
1.000 |
0.000 |
|
0.1 |
1.300 |
1.200 |
+ 0.100 |
|
0.2 |
1.554 |
1.400 |
+ 0.154 |
|
0.3 |
1.879 |
1.600 |
+ 0.279 |
|
0.4 |
2.042 |
1.800 |
+ 0.242 |
|
0.5 |
2.237 |
2.000 |
+ 0.237 |
|
0.6 |
2.418 |
2.200 |
+ 0.218 |
|
0.7 |
2.632 |
2.400 |
+ 0.232 |
|
0.8 |
2.727 |
2.600 |
+ 0.127 |
|
0.9 |
2.873 |
2.800 |
+ 0.073 |
|
1.0 |
3.000 |
3.000 |
0.000 |

No. The residual plot has no points below the zero line and is not showing a random pattern.
Homework 2Expect to Be Worth Something |
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|
Find the expected value.
Answers will vary. Possibilities include "12," since it comes up in a lot of different ways; "36," since it is the biggest outcome.
|
Outcome for Dice 2 |
||||||
1 |
2 |
3 |
4 |
5 |
6 |
|
1 |
1 |
2 |
3 |
4 |
5 |
6 |
2 |
2 |
4 |
6 |
8 |
10 |
12 |
3 |
3 |
6 |
9 |
12 |
15 |
18 |
4 |
4 |
8 |
12 |
16 |
20 |
24 |
5 |
5 |
10 |
15 |
20 |
25 |
30 |
6 |
6 |
12 |
18 |
24 |
30 |
36 |
|
Value |
Prob. |
Worth |
|
1 |
1/36 |
1/36 » 0.028 |
|
2 |
2/36 |
4/36 » 0.111 |
|
3 |
2/36 |
6/36 » 0.167 |
|
4 |
3/36 |
12/36 » 0.333 |
|
5 |
2/36 |
10/36 » 0.278 |
|
6 |
4/36 |
24/36 » 0.667 |
|
8 |
2/36 |
16/36 » 0.444 |
|
9 |
1/36 |
9/36 » 0.250 |
|
10 |
2/36 |
20/36 » 0.556 |
|
12 |
4/36 |
48/36 » 1.33 |
|
15 |
2/36 |
30/36 » 0.833 |
|
16 |
1/36 |
16/36 » 0.444 |
|
18 |
2/36 |
36/36 » 1.00 |
|
20 |
2/36 |
40/36 » 1.11 |
|
24 |
2/36 |
48/36 » 1.33 |
|
25 |
1/36 |
25/36 » 0.833 |
|
30 |
2/36 |
60/36 » 1.67 |
|
36 |
1/36 |
36/36 » 1.00 |
I would predict "30" every time, since it provides the greatest worth, and there is no penalty for incorrect guesses.
$1000 (0.0001) + $300 (0.0002) + $20 (0.005) $1 (1.000) = $0.74
Each time you "play" this raffle, you can expect to lose about 74 cents.
(1/36) + (4/36) + (6/36) + (12/36) + (10/36) + (24/36) + (16/36) + (9/36) + (20/36) + (48/36) + (30/36) + (16/36) + (36/36) + (40/36) + (48/36) + (25/36) + (60/36) + (36/36) = 49/4
As a long-term behavior, the average payoff for the dice game is around 12.25 points.
|
Number of Tests |
1 |
2 |
3 |
|
Proportion of samples |
0.28 |
0.31 |
0.41 |
The average number of blood tests needed would be calculated in this way:
(1) (0.28) + (2) (0.31) + (3) (0.41) = 2.13.
So, the school would have been better off testing the athletes individually.
The number of tests in each logical category is the "value," and the fraction of times those categories come up is the "probability." The expected value for this situation is the average number of tests required, which indicates whether it is cost-effective (below 2.00) or not.
Activity 3Finally, A Model |
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|
= 12, 14, 15
In modeling whether to pool samples when testing for drug use, we need to determine how the variables are related. We speculated that increases in the probability of steroid use (p) produce proportional increases in the expected number of tests required (E). Even though the data look linear, we found that the line connecting the first and last data points was a poor model.
In this activity, you are going to use the calculators statistical regression ability to fit the data in the "best" way possible. The menu that you will use allows you to "fit" data with functions that are linear, exponential, quadratic, or any other kind you might want to try. Regression equations will balance the number of data points on either side of the line or curve, and will minimize the sum of the squares of the residuals. Modelers must also pay attention to which kind of equation to use. To determine that, you must look for a random pattern in the residual plot (errors vs. probability).
Repeat the steps from the previous part of the activity, using exponential regression.
HANDOUT 4Answers |
Equation: E = 1.972p + 1.165

Prob. |
Observed |
Expected |
Errors |
0.0 |
1.000 |
1.165 |
0.165 |
|
0.1 |
1.300 |
1.362 |
0.062 |
0.2 |
1.554 |
1.5601 |
0.006 |
0.3 |
1.879 |
1.757 |
0.122 |
0.4 |
2.042 |
1.954 |
0.088 |
0.5 |
2.237 |
2.151 |
0.086 |
0.6 |
2.418 |
2.348 |
0.070 |
0.7 |
2.632 |
2.546 |
0.086 |
0.8 |
2.727 |
2.743 |
0.016 |
0.9 |
2.873 |
2.940 |
0.067 |
1.0 |
3.000 |
3.137 |
0.137 |

HANDOUT 5Answers |
Equation: E = 1.228 (2.767)p

Prob. |
Observed |
Expected |
Errors |
0.0 |
1.000 |
1.228 |
0.228 |
0.1 |
1.300 |
1.360 |
0.060 |
0.2 |
1.554 |
1.506 |
0.048 |
0.3 |
1.879 |
1.667 |
0.212 |
0.4 |
2.042 |
1.846 |
0.196 |
0.5 |
2.237 |
2.043 |
0.194 |
0.6 |
2.418 |
2.262 |
0.156 |
0.7 |
2.632 |
2.505 |
0.127 |
0.8 |
2.727 |
2.773 |
0.046 |
0.9 |
2.873 |
3.070 |
0.197 |
1.0 |
3.000 |
3.399 |
0.399 |

Repeat the steps from the previous parts of the activity, using quadratic regression.
HANDOUT 6Answers |
Equation: E = 1.028p2 + 3.0p + 1.01

Prob. |
Observed |
Expected |
Errors |
0.0 |
1.000 |
1.011 |
0.011 |
0.1 |
1.300 |
1.301 |
0.001 |
0.2 |
1.554 |
1.570 |
0.016 |
0.3 |
1.879 |
1.818 |
0.061 |
0.4 |
2.042 |
2.046 |
0.004 |
0.5 |
2.237 |
2.254 |
0.017 |
0.6 |
2.418 |
2.441 |
0.023 |
0.7 |
2.632 |
2.607 |
0.025 |
0.8 |
2.727 |
2.753 |
0.026 |
0.9 |
2.873 |
2.878 |
0.005 |
1.0 |
3.000 |
2.983 |
0.017 |

Optional Part: There are more types of regression built into the calculator you are using. Try some of the other methods, but be sure to follow the steps as you did earlier. And remember to record everything as part of this activity work.
Its quadratic. The model for these data is described by the equation:
E = 1.028p2 + 3p + 1.01
When the data were "fit" with a quadratic equation, the residual pattern was random. The actual equation was found by doing quadratic regression on the calculator, so it is the type of regression that minimizes the sum of the squares of the errors.
No, the nature of pooling samples is quadratic; the number of trials wont change that.
Yes, the regression equation was based on the data that were generated by the calculator program having done 1000 trials. Unless we got the exact same numbers when doing only 10 trials, we would get a slightly different equation.
It is likely that the experimental results will more closely approach the expected value as the number of trials increases.
Homework 3Its a Good Fit |
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|
Below are two data sets. Graph each separately; be sure to indicate the scales used. For each graph, find the equation for the line that you think models the data most closely. Plot the line, then plot the residuals. Does there seem to be a pattern to the residuals? Do you think a line is the best kind of equation to fit these data?
x |
4 |
10 |
22 |
32 |
36 |
51 |
62 |
68 |
78 |
89 |
96 |
y |
39 |
66 |
102 |
145 |
192 |
233 |
285 |
355 |
414 |
481 |
533 |


The line of best fit is roughly y = 5.1x + 8.2 The residuals seem to "curve upward" (line is not the best kind of equation).
x |
4 |
6 |
20 |
30 |
40 |
47 |
60 |
70 |
79 |
92 |
99 |
y |
65 |
58 |
52 |
46 |
44 |
32 |
31 |
22 |
18 |
14 |
13 |


The line of best fit is approximately y = 0.54x + 63. The residuals seem to be scattered around the zero line, so a line is the best kind of equation to use.
Activity 4Which Model Fits Best? |
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|
Weve seen how the kinds of models that you use to describe the patterns in data are variedlinear, exponential, quadratic, and so on. Each can be used to fit the data, but one will work best by producing a random residual plot.
Work on the data set(s) assigned to your group. Prepare an analysis of your data, make the graphs of your data and the residuals, and state the model that best fits your data. Write a summary of your results and a logical argument defending your choice of model for your groups presentation to the class. If the data are nonlinear, explain what might cause the rate of change to not be constant.
Volume of a Drop of Water From a BeralTM Pipet
(A Beral pipet is similar to a medicine dropper and is used in chemistry experiments.)
Number of Drops |
20 |
30 |
40 |
20 |
7 |
14 |
21 |
5 |
35 |
55 |
Volume (ml) |
0.92 |
1.38 |
1.85 |
0.88 |
0.30 |
0.60 |
0.90 |
0.20 |
1.52 |
2.48 |


This is strongly linear: y = 0.029 + 0.046x
Oil Leaking from a Container
Time (hr) |
0.5 |
1.5 |
2.5 |
3.5 |
Rate (liters/hr) |
32 |
27 |
24 |
22 |


The line of best fit is y = 3.3x + 32.85. From the graph, you can see a definite pattern. A quadratic function will probably fit it (since a quadratic can be found to fit most data sets with only 4 points). It is not linear because less pressure pushes the oil out of the hole as the amount of oil decreases.


The best quadratic fit is y = 0.75x2 6.3x + 35
U.S. Electricity Consumption
E is billions of kilowatt-hr/year.
(Years are set up so that 1900 is year 0)
Years |
12 |
17 |
20 |
29 |
36 |
45 |
55 |
60 |
65 |
70 |
80 |
87 |
E |
12 |
25 |
39 |
92 |
109 |
222 |
547 |
755 |
1055 |
1531 |
2286 |
2455 |


This graph is obviously not linear. The increase in the use of electricity has not been constant since 1990. The best fit may be exponential, although the curve seems to follow a logistic pattern toward the end. (If you drop the last two data points, the exponential fit is much better.)
A Falling Stone
Time (sec) |
0 |
1 |
2 |
3 |
4 |
5 |
Distance (meters) |
0 |
4.7 |
19.8 |
45.1 |
80.3 |
124.9 |


This is quadratic: y = 4.99x2 + 0.05x 0.15
One Swing of a Pendulum
Length (cm) |
10 |
20 |
30 |
40 |
50 |
60 |
70 |
80 |
Time (sec) |
0.6 |
0.9 |
1.1 |
1.4 |
1.4 |
1.5 |
1.7 |
1.8 |


If students use the power function, they will find that the function y = 0.19x0.52 approximates the data nicely. (The actual relationship is very nearly a square-root function.)
Activity 5Verifying the Model as Quadratic |
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|
= 16, 17, 18
Weve seen how the data collected on the expected number of tests are best described by a quadratic function. To be completely satisfied with this result, it would be nice to understand why it is quadratic in nature. Think back to the original simulation for this problem (Testing Game). We assumed that the probability of a positive test occurring would be 0.50. Lets see how the problem relates to the square below, which is divided into regions. The sides of the square all have a length of 1. The total area represents a total probability of 1. The rows represent the first persons probabilities on the test, and the columns represent the second persons probabilities.

Both people testing negative
(0.5)(0.5) = 0.25
When the combined sample was positive and first of the two athletes tested negative, the second athlete had to be positive. Then exactly two tests were needed. Thus, since the first person gets the first individual test, two tests are needed when only the second person tests positive.
DEHG
(0.5)(0.5) = 0.25
When the combined sample was positive and the first athlete was positive, then three tests are needed. Whenever the first person tests positive, three tests will be needed.
ABED and BCFE, or ACFD
(0.5)(0.5) + (0.5)(0.5) = 0.50 or (1.00)(0.5) = 0.50
One. Since you would always need 1, 2, or 3 tests, it would happen 100% of the time. Therefore, the probability should be 1.
1(0.25) + 2(0.25) + 3(0.50) = 2.25.
In Activity 1, the expected value for a probability p = 0.5 was around 2.23. That was an experimental result based on the performance of the simulation; the answer of 2.25 is a theoretical result based on probability.
The area of the region actually corresponds to the value of the probability. Each region corresponds to two conditions (test result for first and second persons), and, in finding the overall probability of both events taking place, one must multiply the individual probabilities together. The dimensions of the regions correspond to the actual probabilities for those events, and, in calculating area, one must multiply the dimensions together.
To generalize the problem, introduce variables to represent the probability that one of the people tests positive for steroid use.
Use the expression 1 p
Lets go through the same analysis as before, but this time we will use the variable p to represent the incidence of steroid use in the population rather than a specific number like 0.5. The diagram provided at right will help you answer the following questions.

(1 p)2
DEHG, because this area represents a positive pooled sample and a negative first individual test. The probability for that case is represented by p (1 p).
ACFD; students may also say there are twoBCFE and ABED. The probability for that situation is represented by p.
(1 p)2 + (1 p)p + p(1 p) + p2
(1 p)2 + (1 p)p + p(1 p) + p2 = 1 2p + p2 + p p2 + p p2 + p2 = 1
1(1 p)2 + 2(1 p)p + 3(p(1 p) + p2)
1(1 p)2 + 2(1 p)p + 3(p(1 p) + p2) = 1(1 2p + p2) + 2(p p2) + 3(p p2 + p2) = 1 2p + p2 + 2p 2p2 + 3p = 1 + 3p p2
The model developed by using quadratic regression was given by the equation E = 1.028p2 + 3p + 1.01. You can see that they are both quadratic, they have exactly the same linear coefficient, and the other two coefficients are almost exactly the same.
Supplemental Activity 5Introduction to the Parabola |
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|
x |
0 |
1 |
3 |
1 |
2 |
2 |
cant |
1.414 |
2/3 |
1/2 |
2.24 |
2.24 |
y |
0 |
1 |
9 |
1 |
4 |
4 |
4 |
2 |
4/9 |
1/4 |
5 |
5 |
Person 1: y = (x + 3)2
x |
3 |
2 |
0 |
4 |
1 |
5 |
cant |
»1.59 |
2 1/3 |
2 1/2 |
»0.76 |
»5.3 |
y |
0 |
1 |
9 |
1 |
4 |
4 |
4 |
2 |
4/9 |
1/4 |
5 |
5 |
Person 2: y = x2 3
x |
0 |
1 |
1 |
2 |
2 |
3 |
3 |
|
|
1/2 |
3/2 |
3/2 |
y |
3 |
2 |
2 |
1 |
1 |
6 |
6 |
2 |
2 |
1 1/4 |
3/4 |
3/4 |
Person 3: y = (x 2)2
x |
2 |
1 |
1 |
3 |
0 |
4 |
cant |
2+ |
8/3 |
5/2 |
2+ |
6 |
y |
0 |
1 |
9 |
1 |
4 |
4 |
4 |
2 |
4/9 |
1/4 |
5 |
16 |
Person 4: y = x2 + 2
x |
0 |
1 |
1 |
2 |
2 |
3 |
3 |
|
|
1/2 |
3/2 |
3/2 |
y |
2 |
3 |
3 |
6 |
6 |
11 |
11 |
7 |
7 |
9/4 |
17/4 |
17/4 |
Person 1: Shifts 3 units to the left |
Person 3: Shifts 2 units to the right |
Person 2: Shifts down 3 units |
Person 4: Shifts up 2 units |
Person 1: (3, 0) |
Person 3: (2, 0) |
Person 2: (0, 3) |
Person 4: (0, 2) |
Adding/subtracting before the squaring operation shifts to the left/right.
Adding/subtracting after the squaring operation shifts up/down.

You can stretch functions by multiplying or dividing by a number in particular places in equations. Recall what the basic parabola, y = x2, looks like; a table of values has been listed again to help you get started.
y = x2
x |
0 |
1 |
1 |
2 |
2 |
3 |
3 |
y |
0 |
1 |
1 |
4 |
4 |
9 |
9 |
|
Person 1: y/2 = x2 |
Person 2: y = (x 2)2 1 |
|||||||||||||||
x |
0 |
1 |
1 |
2 |
2 |
3 |
3 |
x |
2 |
1 |
3 |
0 |
4 |
5 |
1 |
|
y |
0 |
2 |
2 |
8 |
8 |
18 |
18 |
y |
1 |
0 |
0 |
3 |
3 |
8 |
8 |
|
| y = (x/2)2 | y = (x/2 2)2 1 | |||||||||||||||
|
x |
0 |
2 |
2 |
4 |
4 |
6 |
6 |
|
x |
4 |
2 |
6 |
0 |
8 |
10 |
2 |
y |
0 |
1 |
1 |
4 |
4 |
9 |
9 |
y |
1 |
0 |
0 |
3 |
3 |
8 |
8 |
|
|
Person 3: y = 3x2 |
Person 4: y = (x/4)2 |
||||||||||||||||
x |
0 |
1 |
1 |
2 |
2 |
3 |
3 |
x |
0 |
4 |
4 |
8 |
8 |
12 |
12 |
||
y |
0 |
3 |
3 |
12 |
12 |
27 |
27 |
y |
0 |
1 |
1 |
4 |
4 |
9 |
9 |
||
| y = (x/2 1)2 2 | y = (x 1)2 2 | ||||||||||||||||
x |
4 |
2 |
6 |
0 |
8 |
10 |
2 |
x |
2 |
1 |
4 |
0 |
1 |
3 |
2 |
||
y |
1 |
2 |
2 |
1 |
7 |
14 |
2 |
y |
1 |
2 |
7 |
1 |
2 |
2 |
7 |
||
Person 1: |
First one is thinner; |
Person 2: |
First one is shifted to the right by 2 and down 1; second one is shifted the same, but opens wider. |
|
Person 2: |
First one is narrower; second one is shifted to the right 1 and down 2, and also wider. |
Person 4: |
First one opens wider; second one is shifted to right 1 and down 2. |
Multiplying the x-variable by a number greater than 1 (or dividing the y-variable) will "stretch" the graph, making it steeper and the U-shape narrower. Dividing the x-variable by a number greater than 1 (or multiplying the y-variable) will "shrink" the graph, making the U-shape wider.


The form y = a(x h)2 + k is called the vertex form for a parabola because the graph has its vertex at (h, k). The form y = ax2 + bx + c, where a, b, and c are numbers, is called the standard form for a parabola.
The standard graph was first shifted to the right by 3 units, then stretched by a factor of 2. Finally, it was shifted down 1 unit.
y = 2(x 3)2 1 |
Given problem |
= 2(x2 6x + 9) 1 |
Exponents first; square the binomial using FOIL |
= 2x2 12x + 18 1 |
Distribute multiplication over the parentheses |
= 2x2 12x + 17 |
Combine like terms (two constant terms) |
y = 2x2 12x + 17
"a" is the same in both so a = 2
Teacher Note: Proof of the above statement follows.
y = a(x hk)2 + k y = ax2 2ahx + (ah2 + k) |
The coordinates of the vertex are (3, 1), do h = 3, and k = 1
y = 2(x 3)2 1
y = 3(x + 2)2 14
y = 1.5x2 9x + 17.5
Homework 5Quadratic Nature of Pairing Samples |
![]() |
|

3(0.3)(1) + 2(0.7)(0.3) + 1(0.7)(0.7) = 1.81
The expected value went down, which agrees with the intuition of the problem (less people taking steroids will mean more single tests, because the pooled sample will test negative) and the nature of the model developed (function is strictly increasing on the interval [0..1], so, if p decreases, E should go down as well).
The rectangle ACFD, which represents the region in which three tests are needed, is thinner. The rectangle DEGH, which represents the region in which two tests are needed, is longer and thinner. The square EFIH, which represents the region in which only one test is needed, is larger. Shifting a number of "three tests" trials to "needing one test" means that the average number of tests should go down, which is what weve been calling the expected value. While the area of DEGH didnt change much (0.25 to 0.21), eventually it would be much thinner and would also decrease appreciably.
If p = 0, square EFIH would cover the entire region ACIG and the expected number of tests would be 1(1)(1) = 1. If p = 1, square ABED would cover the entire region ACIG and the expected number of tests would be 3(1)(1) = 3. Both of those results agree with the actual testing problem.

(0.5)2 + 3(0.5) + 1 = 2.25
(0.3)2 + 3(0.3) + 1 = 1.81
They are the same.
(0.0)2 + 3(0.0) + 1 = 1.00
(1.0)2 + 3(1.0) + 1 = 3.00
As p gets closer to 0, the term that gives the expected value its greatest influence (+ 3p) goes toward 0. The term that adjusts the expected value downward (p2) also goes to 0. Therefore, the expected value heads toward the constant +1.
As p gets closer to 1, the first and third terms will cancel each other out and the expected value will head toward the value of the middle term.
p |
p2 |
+3p |
+1 |
0.0 |
0.00 |
0.00 |
+1.00 |
0.1 |
0.01 |
+0.30 |
+1.00 |
0.2 |
0.04 |
0.60 |
+1.00 |
0.3 |
0.09 |
+0.90 |
+1.00 |
0.4 |
0.16 |
+1.20 |
+1.00 |
0.5 |
0.25 |
+1.50 |
+1.00 |
0.6 |
0.36 |
+1.80 |
+1.00 |
0.7 |
0.49 |
+2.10 |
+1.00 |
0.8 |
0.64 |
+2.40 |
+1.00 |
0.9 |
0.81 |
+2.70 |
+1.00 |
1.0 |
1.00 |
3.00 |
+1.00 |
1 EFIH + 1 DEHG + 1 ABDE + 1 BCFE = 1 (sum of probabilities)
1 DEHG + 1 ABDE = 1 p
2 ABDE + 1 BCFE = 2 p
1 ABDE = p2
Therefore, 1 EFIH + 2 DEHG + 3 ABDE + 3 BCFE = p2 + 3p + 1
Activity 6Solving the Problem |
![]() |
|
= 13, 19, 20
The final task in modeling the pooling of blood samples is to answer the question, "When is it cost-effective to pool samples? When is it going to save us money?"
E = p2 + 3p + 1
2 tests would be required per pair. To save money, we should keep the average number of tests below 2.
Answers will vary. One possibility is: xmin = 0.1, xmax = 1.1, xscl = 0.1, ymin = 0.1, ymax = 3.1, yscl = 0.5
The expected value graph curved slightly downward, starting at 1 and going up to 3. The condition graph is a horizontal line at 2.
p » 0.38; if the populations proportion of positive tests is at least 38%, then pooled testing costs more.
In previous work in this unit, you learned how to use area models to obtain the quadratic model p2 + 3p + 1. We used the formula to predict the expected number of tests when p = 0.5. Setting up the equation p2 + 3p + 1 = 2 allows us to work backward, and determine what the probability would have to be if we want the expected number of tests to be two (break-even point). This is an important enough area of mathematics that we will actually develop two methods for solving this quadratic equation. The first method is called completing the square. It is based on the same idea as the area models.
2(x + 3)2 + 2
List the order in which you evaluate this expression, and then evaluate the expression using x = 4.
Inside parentheses, do the addition |
4 + 3 = 7 |
Exponents |
72 = 49 |
Multiplication |
49 * 2 = 98 |
Final addition |
98 + 2 = 100 |
Cancel final addition operation by subtracting 2 |
2(x + 3)2 = 18 |
Cancel the multiplication by dividing by 2 |
(x + 3)2 = 9 |
Cancel the squaring by finding thesquare root |
x + 3 = 3 or 3 |
Cancel the addition operation by subtracting 3 |
x = 0 or 6 |
The second method for solving quadratic equations uses the vertex form.

The importance of the vertex form for a quadratic equation is this: if you can express a quadratic equation in the form y = a(x h)2 + k, you can solve it by undoing the order of operations. Now, lets concentrate on writing quadratics in vertex form.


All graphs are tangent to the x-axis.

+4; y = (x 2)2
x2 + 8x + 16 = (x + 4)2
x2 10x + 25 = (x 5)2
Half the linear coefficient is the constant in the square form.

2.25 or 9/4
x2 3x + 9/4 = 10 + 2.25
(x 3/2)2 = 12.25
+ 1.5 = 5 or
+ 1.5 = 2
First, it will be easier to work out the steps if the leading term is made positive and the constants are combined on one side of the equation.
Multiply through by 1 to get: |
p2 3p 1 = 2 |
Add one to each side to make it: |
p2 3p = 1 |
You need to increase the expression by (half the value of the linear term)2 to complete the square.
Add 2.25 to both sides to get: |
p2 3p + 2.25 = 1.25 |
Rewrite it as: |
(p 1.5) = 1.25 |
Find the square root on both sides: |
p 1.5 = + |
Add 1.5 to both sides |
p = + |
Note: Only one of the answers makes sense in the context of the probability of testing positive for steroid use.
The process of completing the square applies as well to the general form of the quadratic equation, ax2 + bx + c = 0. By solving the quadratic equation in the general case, the quadratic formula is obtained.
x2 + 10x + 9 = 0 x2 + 10x = 0 9 x2 + 10x + 25 = 9 + 25 = 16 (x + 5)2 = 16 x + 5 = + 4 x = 5 + 4 = 9 or 1 |
Steps should include (i) completing the square by isolating the constant term, making the leading coefficient 1 (by dividing by a), and adding to both sides the square of half the new linear coefficient; and (ii) solving by taking the square roots and adding or subtracting, as indicated by arrow diagrams.
The quadratic formula: x = Note to Teacher: Quadratic Formula Derivation
|
x =
=
=
or 
AssessmentWorking With Quadratics |
![]() |
|
1 or 5
4/3 or 2/3
x2 6x + 9
9x2 6x + 1
64; (x 8)2
49/4; (x + 7/2)2
Sample answers: Squaring (x 5) will give a middle term of 10x. The graph of (x 5)2 is tangent to the x-axis, while the graph of x2 + 25 is not. Evaluate each expression for a particular number, and you will not get them to equal each other (except at x = 0).
For each of the following, solve by completing the square.
x2 + 4x + (2)2 = 12 + 4
(x + 2)2 = 16
= + 
(x + 2) = +4
x = 2 + 4 = 2 or x = 2 4 = 6
x2 + 8x = 11
x2 + 8x + (4)2 = 11 + 16
(x + 4)2 = 27
= 
x + 4 = +
x = 4 + 
x2 2x 6 = 0
x2 2x = 6
x2 2x + 1 = 6 + 1
(x 1)2 = 7
= 
x 1 = + 
x = 1 + 

x2 + 8x
16
(x + 4)2
x2 + 8x + 16
2x2 + 5x 9 = 0.
x =
= 1.21 or 3.71
Unit Summary |
In this unit, the problem of combining blood samples in testing for steroid use was examined. The central question was to determine when it would be cheaper to test the pooled sample and when it would be cheaper to test samples individually. Assumptions were made in developing the model; they included assuming the cost of testing to be fixed and assuming that positive tests always correctly detected the presence of steroids. The logic of the problem dictated that the outcome would always be one, two, or three tests, which is a discrete range, but average results of long-term behaviors yielded data that seemed to be continuous over the same range. Simulations were performed and data collected. The data were then analyzed to determine the mathematical relationship that connected the probability of testing positive to the expected number of tests (on average). The model was verified by examining the situation in probabilistic terms, and then a solution was obtained for the "break-even" point.
Mathematical Summary |
In simulating the testing for steroid use, first with manipulatives and then using a calculator program, the result of any one trial was always 1, 2, or 3 tests, but the average was any number between 1 and 3, inclusive. This distinction points to the fact that the event is discrete, but the long-term behavior isnt. The law of large numbers came into play in the simulation, as a group of students did 10 trials, with lots of fluctuation on the average number of tests. The calculator programs were run with many more trials, and still provided fluctuations on the average number of tests. The interval containing all of the answers was much smaller, provided a better estimate of the expected value.
The ideas of probabilistic worth and expected value came into consideration because the trials had varying "payoffs." Sometimes a single trial needed a single test, while, at other times, a single trial needed three tests. The idea to "weigh" the trials with the number of tests needed is where probabilistic worth was introduced. Averaging all of those weighted results gave us an idea of what to expect as long-term behavior, or a gauge of what to expect on an individual trial, and that was the expected value for this situation.
Regression analysis was applied to the data collected to disprove the conjecture that the behavior in the data was linear. Further analysis showed that the relationship between the variables was actually quadratic. In all cases, it was the pattern in the residual plot that either said "Hey, we found the right model!" (random dots above and below the zero line) or "Nice try, but no cigar!" (pattern to the plot).
To verify the model, probability area models were used to explore the problem analytically. Since two people were being tested and they could be either positive or negative in their individual test results, the four regions of the square represented the four possible situations. The area of a region represented the probability that the situation represented by that region would happen. Multiplying the area by the number of tests would give the probabilistic worth for that outcome. Adding all those answers together gave the expected value for the testing situation. The difference is that this was the "theoretical" or expected value, not the results obtained from "experimental" averages. An expression was derived describing E(p), the way in which the expected value depends on the probability of steroid use in the population, and it turned out that the expression was quadratic.
Having discovered a new mathematical relationship, sometimes called a quadratic and other times called a parabola, some exploration into their properties was provided. The "parent graph"
y = x2
formed a "U" shape with the vertex located at the origin. Translation transformations were investigated. Replacing the x with an expression x h moved the graph over h units to the right. If h was a negative number, it moved the graph to the left instead. Similarly, replacing the y with an expression y k moved the graph up k units. If k was a negative number, it moved the graph down instead. The quadratic
(y k) = (x h)2
has a vertex at point (h, k).
Stretching transformations were also investigated, in which the x2 and y occurrences were multiplied or divided by numbers. The effect of that act was to cause the graphs to open wider or close narrower. The vertex form of the quadratic
y = a(x h)2 + k
is translated by h and k and stretched by a.
The algebra of quadratics was also explored. Vertex form for a quadratic, which looks like
y k = a(x k)2
and standard form, which looks like
y = ax2 + bx + c
were introduced, and methods to go from one form to the other were practiced. Solving quadratic equations, which was necessary to actually get the answer for which we had built the entire model, is such an important skill that three different methods were introduced. The first method was to graph each side of the equation as a separate function and TRACE to the intersection point. The second method was a technique called "completing the square." The third method was to develop a general formula, called the quadratic formula

which allows you to find the solutions by simply plugging in the three numbers from the standard equation.
Key Concepts |
Average Number of Tests: Long-term behavior of the testing game found by taking the total number of tests and dividing it by the number of trials that were done.
Domain, Range for the Testing Problem: The domain would be the allowed values for the probability of testing positive, [0...1]. The range would be the expected number of tests needed, [1...3]
Probabilistic Worth: Mathematical quantity that is defined as the product of the outcome times the probability that the outcome takes place.
Expected Value: Sum of all the probabilistic worths; describes the average payoff for an event that might happen a lot of times.
Law of Large Numbers: A concept in mathematics that, says that, as the number of trials gets large, the average (experimental) behavior will tend to be closer to the expected (theoretical) behavior.
Regression: An analysis that "fits" data to mathematical functions by minimizing the sum of the squares of errors
Residuals: How far off a prediction made by a mathematical function is from the observed data its trying to predict.
Stretch Transformations: When an expression or answer is multiplied or divided by a number, it has the effect of distorting a graph in the vertical or horizontal direction by stretching or shrinking it.
Shift Transformations: When a number is added to (or subtracted from) a variable in a function, it has the effect of moving the graph either to the left/right (x-variable) or up/down (y-variable).
Vertex: The place where a parabola reaches the bottom or top of its curve.
Line Symmetry: A line that goes through the vertex and divides the parabola into two equal (mirror image) sides
Domain, Range for a Quadratic: The domain is the allowed values for x; since all values are allowed, the domain is all real numbers. The range is the possible answers for y, which is determined by whether the quadratic is upright or inverted and where the vertex point is located.
Vertex-Form Equation of a Quadratic: y k = a(x k)2
Standard-Form Equation of a Quadratic: y = ax2 + bx + c
Completing the Square: Technique for solving a quadratic equation by putting the quadratic expression into vertex form. Then, steps for working backward to solve the equation become step-by-step arithmetic processes.
Quadratic Formula: Formula for solving a quadratic equation ax2 + bx + c = 0. The solution is found by evaluating the formula:
x = 
Solution to Short Modeling Practice |
Mini-Modeling Problem: Linear and Quadratic Equations |
|
t (sec) |
x (ft) |
|
0.0 |
0.0 |
|
0.1 |
2.9 |
|
0.2 |
5.8 |
|
0.3 |
8.7 |
|
0.4 |
11.6 |
|
0.5 |
14.5 |
|
0.6 |
17.4 |
|
0.7 |
20.2 |
|
0.8 |
23.1 |
|
0.9 |
26.0 |

Students can deduce the two parameters for the linear graph: the slope and the intercept. The intercept is equal to 0, the distance at time t = 0 sec. The slope can be found by taking any pair of points, such as (0, 0.0) and (0.5, 14.5):

Help students to avoid confusion with the typical notation used above, where y denotes the vertical axis and x denotes the horizontal axis. They should not be confused in this problem where the horizontal distance is the variable x and the independent variable is t. Thus, the linear equation for the horizontal distance is
x = (29.0 ft/sec) t + 0
Some students may perform a linear regression on the data to obtain:
| Slope = 28.9 Intercept = 0.03 r2 = 0.99999 |
Approximately the same results. |
|
t (sec) |
y |
|
0.0 |
7.0 |
|
0.1 |
10.3 |
|
0.2 |
13.3 |
|
0.3 |
15.9 |
|
0.4 |
18.2 |
|
0.5 |
20.2 |
|
0.6 |
21.9 |
|
0.7 |
23.3 |
|
0.8 |
24.3 |
|
0.9 |
25.1 |

Students can suppose it is quadratic and determine the coefficients. In this case, it is probably most efficient to perform a quadratic regression to fit the data to an equation of the form y = ax2 + bx + c, yielding:
a = 15.8 b = 34.3 c = 7.0 r2 = 0.99998 |
The strong correlation coefficient with the large number of data values indicates that the relationship probably is indeed quadratic. So the equation describing the vertical motion can be written:
y = (15.8) t2 + (34.3) t + (7.0)
55 ft = (29.0 ft/sec) t
t =
= 1.90 sec
Substitute this time value into the equation for the vertical height and see whether the ball would have had the correct height when it reached the plane of the goal.
y = (15.8)(1.90)2 + (34.3)(1.90) + (7.0) = 15.1 ft
Since the basket is at y = 10 ft, Kareems shot would have missed the goal, being over 5 feet high when reaching the basket. (As noted in the student text, we have considerably simplified the test for accuracy of Kareems shot here: ignoring the lateral aim, as well as the fine points about how the ball actually strikes the rim of the basket, and so forth. If the calculations had shown the ball to be within about 0.2 feet of the target, we would probably be inclined to say that the shot was "on target." You can ask your students for their opinions on this point: "How close must one be to make the basket?")
The ball struck the TV equipment somewhere within its trajectory. So, we can at least say that the TV equipment must have been within the range of y-values predicted by the equation for the vertical distance. Students can graph the equation for y from above and find that the ball could have gone as high as 25.6 feet (or 7.8 meters) on its way to the basket. Thus, it is possible that the TV equipment was safely above the 7 meter limit and still interfere with Kareems shot. We cannot say with certainty that the equipment was below the 7 meter clearance.
Solutions to Practice and Review Problems |
Exercise 1
50
Exercise 2
6.1
Exercise 3
60%
Exercise 4
30
Exercise 5
Exercise 6
y =
x + 
Exercise 7
Probability = 
Probability = 
Probability = 0.0001 (rounded)
Probability = 
Probability = 
Probability = 0.001 (rounded)
Exercise 8
Probability = 
Probability = 
Probability = 0.257, or 25.7 (rounded)
The possible outcomes of this procedure are that you receive 1) Brand A, 2) Brand B, 3) Brand C, or 4) Brand D.
Yes, these four outcomes are mutually exclusive. This means that you cannot be given one calculator and have it be both Brand A AND Brand C, for example. Any outcome can fall only into one and only one of the four possible outcomes listed above.
Prob {Brand A OR Brand B} = Prob {Brand A} + Prob {Brand B}
First, the student must determine the probability of getting Brand B. As was done in Part a...
Probability = 
Probability = 0.429, or 42.9% (rounded)
Then,
Prob {Brand A OR Brand B} = 0.257 + 0.429
Prob {Brand A OR Brand B} = 0.686
Or,
Prob {Brand A OR Brand B} = 
Prob {Brand A OR Brand B} = 0.686
Exercise 9
No, they are not equally likelyat least you hope that there is more of a tendency to be not infested, rather than simply a 50-50 chance of infestation.
Probability = 
Probability = 
Probability = 0.4, or 40%
Exercise 10
A |
A |
|
A |
AA |
AA |
B |
AB |
AB |
So for AA,
Probability = 
Probability = 
Probability = 0.5
And, similarly for AB.
Outcome |
Probability |
AA |
0.5 |
AB |
0.5 |
Similarly, you would expect two children to have gene pairs of type AB, or blood type AB.
Since none of the outcomes yielded a gene pair of BB, none of the children would be expected to have blood type B.
EXERCISE 11
» 0.000025, or about 0.0025% Exercise 12
y = 80,000(0.92)x ; about 52,727 people
Exercise 13

Exercise 14
New area = New length × New width
New area = (120 + x)(80 + x)
New area = 9600 + 200x + x2
New area = Present area + 20% of Present area
New area = (120 × 80) + 0.20 (120 × 80)V
New area = 11,520
Equate this to the result of Part a, and simplify to get it into the form of a quadratic equation.
9600 + 200x + x2 = 11,520
x2 + 200x 1920 = 0
x = 
x = 
x = 100 ± 1/2 
x = 9.18 or 209.18
Thus the farmer would need to plow an additional 9 meters of width and length to obtain a 20% increase in the plowed area. The negative root is not meaningful in this case.
Exercise 15
16t2 + 1000 = 0
t2 = 
t = ±
» 7.9 or 7.9
It will take about 7.9 seconds.
T = ±
» 9.68 or 9.68
It will take about 9.68 seconds.
Exercise 16
Total cost = Setup fee + Cost per ft2 (Height ´ Width)
T = 150 + 10 (H )(2H + 5)
T = 150 + 50H + 20H2
4000 = 150 + 50H + 200H2
20H2 + 50H 3850 = 0
One way to solve this equation is to use the quadratic formula, where a = 20, b = 50, and c = - 3850.
H = 
H = 
H = 1.25 ± 1/40 
H = 12.68 ft or 15.18 ft
So, since the negative result is meaningless here, the result tells us that a sign 12.68 ft high is the maximum that can be afforded from this billboard company.
Exercise 17
For d = 0, c = 1000.
For d = 5, c = 600.
For d = 10, c = 300.
For d = 15, c = 100.
Note that the domain of this application requires that only values for d between 0 meters and 20 (thousand) feet are valid.
10 = 2d2 90d + 1000
2d2 90d + 990 = 0
The values for d that satisfy the above equation can be found using the quadratic formula, where a = 2, b = 90, and c = 990.
d = 
d = 
d = 22.5 ± 1/4 
d = 19.1 thousand feet or 25.9 thousand feet
Since the second value is not in the domain of our function, the first value is the correct answer: At d = 19.1 thousand feet the concentration is 10 ppm.
Exercise 18
Since this is basically a task in substitution and solving a quadratic, little guidance is offered the students. They should substitute P = 60 watts, V = 110 volts, and R = 30 ohms into the equation and solve the quadratic equation for I.
P = VI I 2R
60 = 110 I (30) I 2
30 I 2 110 I + 60 = 0
Well solve this quadratic using the quadratic formula, where a = 30, b = 110, and c = 60.
I = 
I = 
I = 1.833 ± 1/60 
I = 3.00 amperes or 0.67 amperes
So, a current of either 0.67 ampere or 3.00 amperes will produce 60 watts of power at the load.
Exercise 19
Area = Length × Width
245 = (20 2x)(x)
Multiply out the terms and put the result in standard form.
38 = 20x 2x2
2x2 20x + 38 = 0

x = 
x = 
x = 5 ± 1/4 
x = 5 ± 2.45
x = 7.45 in. and 2.55 in.
Each root "fits" the problem here!
Exercise 20
(40 ft)2 = 8(25 ft)h 4h2
4h2 200h + 1600 = 0

h = 
h = 
h = 25 ± 1/8 
h = 40 ft or 10 ft
The value of 40 ft probably wont be accepted since that would describe an arch that was nearly a complete circle above the ground. You can draw a picture for your students that looks like the one shown above. Thus the second solution is the one we want: h = 10 ft. This depicts an arch like the one shown in the text, whose center height is 10 ft above the ground.