## Engage NY Eureka Math Algebra 2 Module 3 Lesson 23 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 23 Mathematical Modeling Exercise Answer Key

Mathematical Modeling Exercises:

Exercise 1.

Working with a partner, you are going to gather some data, analyze the data, and find a function to use to model the data. Be prepared to justify your choice of function to the class.

a. Gather your data: For each trial, roll the beans from the cup to the paper plate. Count the number of beans that land marked-side up, and add that many beans to the cup. Record the data in the table below. Continue muntil you have either completed 10 trials or the number of beans at the start of the trial exceeds the number that you have.

Answer:

b. Based on the context in which you gathered this data, which type of function would best model your data points?

Ans:

Since the number of beans we add at each toss is roughly half of the beans we had at the start of that turn, the data should be modeled by an exponential function.

c. Plot the data: Plot the trial number on the horizontal axis and the number of beans in the cup at the start of the trial on the vertical axis. Be sure to label the axes appropriately and to choose a reasonable scale for the axes.

Answer:

d. Analyze the data: Which type of function best fits your data? Explain your reasoning.

Answer:

Students should see that the data follows a clear pattern of exponential growth, and they should decide to model it with an exponential function.

e. Model the data: Enter the data into the calculator and use the appropriate type of regression to find an equation that fits this data. Round the constants to two decimal places.

Answer:

Answers will vary but should be of the form f(t) = a(b^{t}), where a is near 1 and b is near 1.5.

Exercise 2.

This time, we are going to start with 50 beans in your cup. Roll the beans onto the plate and remove any beans that land marked-side up. Repeat until you have no beans remaining.

a. Gather your data: For each trial, roll the beans from the cup to the paper plate. Count the number of beans that land marked-side up, and remove that many beans from the plate. Record the data in the table below. Repeat until you have no beans remaining.

Answer:

b. Plot the data: Plot the trial number on the horizontal axis and the number of beans in the cup at the start of the trial on the vertical axis. Be sure to label the axes appropriately and choose a reasonable scale for the axes.

Answer:

c. Analyze the data: Which type of function would best fit your data? Explain your reasoning.

Answer:

Students should see that the data follows a clear pattern of exponential decay, and they should decide to model it with an exponential function.

d. Make a prediction: What do you expect the values of a and b to be for your function? Explain your reasoning.

Answer:

Using an exponential function f(t) = a(b^{t}), the value of a is the initial number of beans, so we should expect a to be near 50. The number of beans decreases by half each time, so we would expect b = 0. 5.

e. Model the data: Enter the data into the calculator. Do not enter your final data point of O beans. Use the appropriate type of regression to find an equation that fits this data. Round the constants to two decimal places.

Answer:

Answers will vary but should be of the form f(t) = a(b^{t}), where a is near 50 and b is near 0.5.

### Eureka Math Algebra 2 Module 3 Lesson 23 Problem Set Answer Key

Question 1.

For this problem, we consider three scenarios for which data have been collected and functions have been found to model the data, where a, b, c, d, p, q, r, s, t, and u are positive real number constants.

i. The function f(t) = a(b^{t}) models the original bean activity (Mathematical Modeling Exercise 1). Each bean is painted or marked on one side, and we start with one bean in the cup. A trial consists of throwing the beans in the cup and adding one more bean for each bean that lands marked-side up.

ii. The function g(t) = c . d^{t} models a modified bean activity. Each bean is painted or marked on one side, and we start with one bean in the cup. A trial consists of throwing the beans in the cup and adding two more beans for each bean that lands marked-side up.

iii. The function h(t) = p . q^{t} models the dice activity from the Exit Ticket. Start with one six-sided die In the cup. A trial consists of rolling the dice in the cup and adding one more die to the cup for each die that lands with a 6 showing.

iv. The function j(t) = r . s^{t} models a modified dice activity. Start with one six-sided die in the cup. A trial consists of rolling the dice in the cup and adding one more die to the cup for each die that lands with a 5 or a 6 showing.

v. The function k(t) = u . v^{t} models a modified dice activity. Start with one six-sided die in the cup. A trial consists of rolling the dice in the cup and adding one more die to the cup for each die that lands with an even number showing.

a. What values do you expect for a, c, p, r, and u?

Answer:

The values of these four constants should each be around 1 because the first data point in all four cases is (1, 0).

b. What value do you expect for the base b in the function f(t) = a . b^{t} in scenario (i)?

Answer:

We know from the class activity that b ≈ 1.5 because the number of beans grows by roughly half of the current amount at each trial.

c. What value do you expect for the base d in the function g(t) = c . d^{t} in scenario (ii)?

Answer:

Suppose we have 4 beans in the cup. We should expect half of them to land marked-side up. Then, we would add 2 . 2 = 4 beans to the cup, doubling the amount that we had. This is true for any number of beans; if we had n beans, then \(\frac{n}{2}\) should land marked-side up, so we would add n beans, doubling the amount. Thus, values of the function g should double, so d ≈ 2.

d. What value do you expect for the base q in the function h(t) = p . q^{t} in scenario (iii)?

Answer:

For this function h, we expect that the quantity increases by \(\frac{1}{6}\) of the current quantity at each trial. Then,

q ≈ 1 + \(\frac{1}{6}\), so q ≈ \(\frac{7}{6}\)

e. What value do you expec:t for the base s in the function i(t) = r st in scenario (iv)?

Answer:

The probability of rolling a 5 or 6 is \(\frac{1}{3}\),so we would expect that the number of dice increases by \(\frac{1}{3}\) of the current quantity at each trial. Thus, we expect s ≈ 1 + \(\frac{1}{3}\), which means that s ≈ \(\frac{4}{3}\)

f. What value do you expect for the base y in the function k(t) = u . v^{t} in scenario (v)?

Answer:

The probability of rolling an even number on a six-sided die is the same as the probability of getting the marked-side up on a bean, so we would expect that f and k are the same function. Thus, v ≈ 1.5.

g. The following graphs represent the four functions f, g, h, and j. Identify which graph represents which function.

Answer:

Question 2.

Teams 1, 2, and 3 gathered data as shown in the tables below, and each team modeled the data using an exponential function of the form f(t) = a . b^{t}.

a. Which team should have the highest value of b? Which team should have the lowest value of b? Explain how you know.

Answer:

The larger the value of the base b, the larger the function will be at t = 10. Since team 3 had the most beans at the end (they used their 50 beans first), their equation should have the highest value of the base b. Team 2 has the smallest number of beans after 10 trials, so team 2 should have the smallest value of the base b.

b. Use a graphing calculator to find the equation that best fits each set of data. Do the equations of the functions provide evidence that your answer In part (a) is correct?

Answer:

Team 1’s equation: f^{1}(t) = 0.7378(1.5334)^{t}

Team 2’s equation: f^{2}(t) = 0.6495(1.4245)^{t}

Team 3’s equation: f^{3}(t) = 1.0328(1.7068)^{t}

As predicted, team 2 has the smallest value of b, and team 3 has the largest value of b.

Question 3.

Omar has devised an activity in which he starts with 15 dice in his cup. A trial consists of rolling the dice in the cup and adding one more die to the cup for each die that lands with a 1, 2, or 3 showing.

a. Find a function f(t) = a(b^{t}) that Omar would expect to model his data.

Answer:

f(t) = 15(\(\frac{3}{2}\))^{t}

b. Solve the equation f(t) = 30. What does the solution mean?

Answer:

15(\(\frac{3}{2}\))^{t} = 30

(\(\frac{3}{2}\))^{t} = 2

t = \(\frac{\log (2)}{\log \left(\frac{3}{2}\right)}\)

t ≈ 1.71

So, Omar should have more than 30 dice by the second trial, after rolling and adding dice twice.

c. Omar wants to know in advance how many trials it should take for his initial quantity of 15 dice to double. He uses properties of exponents and logarithms to rewrite the function from part (a) as the exponential function f(t) = 15 (2^{t . log2(\(\frac{3}{2}\))(})). Has Omar correctly applied the properties o exponents and logarithms to obtain an equivalent expression for his original equation in part (a)? Explain how you know.

Answer:

Yes. The expressions are equal by the property of exponents.

(2^{t . log2(\(\frac{3}{2}\))(})) = (2^{log2(\(\frac{3}{2}\))(}))^{t} = (\(\frac{3}{2}\))^{t}

Thus,

15(2^{t . log2(\(\frac{3}{2}\))(})) = 15(\(\frac{3}{2}\))^{t}

d. Explain how the modified formula from part (c) allows Omar to easily find the expected amount of time, t, for the initial quantity of dice to double.

Answer:

The quantity is doubled at a time tfor which t . log_{2}(\(\frac{3}{2}\)) = 1. Thus, we solve the equation t . log_{2}(\(\frac{3}{2}\)) = 1 to find t = \(\frac{1}{\log _{2}\left(\frac{3}{2}\right)}\) ≈ 1.71. This agrees with our answer to part (b).

Question 4.

Brenna has devised an activity in which she starts with 10 dice in her cup. A trial consists of rolling the dice In the cup and adding one more die to the cup for each die that lands with a 6 showing.

a. Find a function f(t) = a(b^{t}) that you would expect to model her data.

Answer:

f(t) = 10(\(\frac{7}{6}\))^{t}

b. Solve the equation f(t) = 30. What does your solution mean?

Answer:

10(\(\frac{7}{6}\))^{t} = 30

(\(\frac{7}{6}\))^{t} = 3

t = \(\frac{\log (3)}{\log \left(\frac{7}{6}\right)}\)

t ≈ 7.13

Brenna’s quantity of dice should reach 30 by the eighth trial.

c. Brenna wants to know in advance how many trials it should take for her initial quantity of 10 dice to triple. Use properties of exponents and logarithms to rewrite your function from part (a) as an exponential function of the form ((t) = a(3^{ct}).

Answer:

d. Explain how your formula from part (c) allows you to easily find the expected amount of time, t, for the initial quantity of dice to triple.

Answer:

The quantity triples when 3^{t.log3(\(\frac{7}{6}\))} = 3, so that t . log_{3}(\(\frac{7}{6}\)) = 1. Then, we solve that equation to find the value of t: t = \(\frac{1}{\log _{3}\left(\frac{7}{6}\right)}\) ≈ 7.13

e. Rewrite the formula for the function f using a base-10 exponential function.

Answer:

Since \(\frac{7}{6}\) = 10^{log(\(\frac{7}{6}\))}, we have (\(\frac{7}{6}\))^{t} = (10^{log(\(\frac{7}{6}\))})^{t} = 10^{t . log(\(\frac{7}{6}\))} Then, f(t) = 10 (10^{t . log(\(\frac{7}{6}\))}).

f. Use your formula from part (e) to find out how many trials it should take for the quantity of dice to grow to 100 dice.

Answer:

The quantity will be 100 when 10^{t . log(\(\frac{7}{6}\)) = 10, so that t. log(\(\frac{7}{6}\)) = 1. Then, t = \(\frac{1}{\log \left(\frac{7}{6}\right)}\) ≈ 14.94 so that the quantity should exceed 100 by the 15th trial.}

5.

Suppose that one bacteria population can be modeled by the function P_{1} (t) = 500(2^{t}) and a second bacteria population can be modeled by the function P_{2}(t) = 500(2.83^{t}), where t measures time in hours. Keep four digits of accuracy for decimal approximations of logarithmic values.

a. What does the 500 mean in each function?

Answer:

In each function, the 500 means that each population has 500 bacteria at the onset of the experiment.

b. Which population should double first? Explain how you know.

Answer:

Since 2.83 > 2, the second population is growing at a faster rate than the first, so it should double more quickly.

c. How many hours and minutes should it take for the first population to double?

Answer:

The first population doubles every hour, since the base of the exponential function is 2. Thus, the first population doubles in one hour.

d. Rewrite the formula for P_{2}(t) in the form P_{2}(t) = a(2^{ct}), for some real numbers a and c.

Answer:

P_{2}(t) = 500(283)^{t}

= 500(2^{log2(283)})^{t}

= 500(2^{t.2(2.83)})

e. Use your formula in part (d) to find the time, t, in hours and minutes until the second population doubles.

Answer:

The second population doubles when t . log_{2}(2.83) = 1, which happens when t = \(\frac{1}{\log _{2}(2.83)}\) Thus, t ≈ 0.6663 hours, so the population doubled after approxImately 40 minutes.

Question 6.

Copper has antibacterial properties, and it has been shown that direct contact with copper alloy C11000 at 20°C kills 99.9% of all methicillin-resistant Staphylococcus aureus (MRSA) bacteria in about 75 minutes. Keep four digits of accuracy for decimal approximations of logarithmic values.

a. A function that models a population of 1,000 MRSA bacteria t minutes after coming in contact with copper alloy C11000 is P(t) = 1000(0.912)^{t} What does the base 0. 912 mean in this scenario?

Answer:

The base 0.912 means that 91.2% of the MRSA bacteria remain at the end of each minute.

b. Rewrite the formula for P as an exponential function with base \(\frac{1}{2}\).

Answer:

Since 0.912 = (\(\frac{1}{2}\))^{log\(\frac{1}{2}\)(0.912)}

= (\(\frac{1}{2}\))^{-log2(0.912)}, we have

P(t) = 1000(0.912^{t}) = 1000(\(\frac{1}{2}\))^{-t. log2(0.912)}

c. Explain how your formula from part (b) allows you to easily find the time it takes for the population of MRSA to be reduced by half.

Answer:

The population of MRSA is reduced by half when the exponent is 1. This happens when -t . log_{2}(0.912) = 1, so t = –\(-\frac{1}{\log _{2}(0.912)}\) ≈ 7.52. Thus, half of the MRSA bacteria die every 7\(\frac{1}{2}\) minutes.

### Eureka Math Algebra 2 Module 3 Lesson 23 Exit Ticket Answer Key

Question 1.

Suppose that you were to repeat the bean activity, but in place of beans, you were to use six-sided dice. Starting with one die, each time a die is rolled with a 6 showing, you add a new die to your cup.

a. Would the number of dice in your cup grow more quickly or more slowly than the number of beans did? Explain how you know.

Answer:

The number of dice in the cup should grow much more slowly because the probability of rolling a 6 is \(\frac{1}{6}\), while the probability of flipping a bean marked-side up is roughly \(\frac{1}{2}\).

Thus, the beans should land marked-side up, and thus, increase the number of beans in our cup about half of the time, while the dice would show a 6 only \(\frac{1}{6}\) of the time. As an example, it could be expected to take one or two flips of the first bean to get a bean to show the marked side, causing us to add one, but it could be expected to take six rolls of the first die to get a 6, causing us to add another die.

b. A sketch of one sample of data from the bean activity is shown below. On the same axes, draw a rough sketch of how you would expect the curve through the data points from the dice activity to look.

Answer: