## Engage NY Eureka Math Algebra 2 Module 3 Lesson 27 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 27 Opening Exercise Answer Key

Opening Exercise:

The following table contains U.S. population data for the two most recent census years, 2000 and 2010.

Census Year | U.S. Population(in millions) |

2000 | 281.4 |

2010 | 308.7 |

a. Steve thinks the data should be modeled by a linear function.

i. What is the average rate of change in population per year according to this data?

Answer:

The average rate of change is \(\frac{308.7-281.4}{2010-2000}\) = 2.73 million people per year.

ii. Write a formula for a linear function, L, to estimate the population t years since the year 2000.

Answer:

L(t) = 2.73t + 281.4

b. Phillip thinks the data should be modeled by an exponential function.

i. What is the growth rate of the population per year according to this data?

Answer:

Since \(\frac{308.7}{281.4}\) = 1.097, the population will increase by the factor 1.097 every 10 years. To determine the yearly rate, we would need to express 1.097 as the product of 10 equal numbers (e.g., 1.097^{\(\frac{1}{10}\)} 1.097^{\(\frac{1}{10}\)} … 1. 097^{\(\frac{1}{10}\)} ten times). The annual rate would be 1.097^{\(\frac{1}{10}\)}, which is approximately 1.0093.

ii. Write a formula for an exponential function, E, to estimate the population t years since the year 2000.

Answer:

Start with E(t) = a . b^{t}. Substitute (0, 281.4) into the formula to solve for a.

281.4 = a . b^{0}

Thus, a = 281.4.

Next, substitute the value of a and the ordered pair (10,308.7) into the formula to solve for b.

308.7 = 281.4b^{10}

b^{10} = 1.097

b = \(\sqrt[10]{1.097}\)

Thus, b = 1.0093 when you round to the ten-thousandths place and

E(t) = 281.4(1.0093)^{t}

c. Who has the correct model? How do you know?

Answer:

You cannot determine who has the correct model without additional information. However, populations over longer intervals of time tend to grow exponentially if environmental factors do not limit the growth, so Phillip’s model is likely to be more appropriate.

### Eureka Math Algebra 2 Module 3 Lesson 27 Mathematical Modeling Exercises Answer Key

Mathematical Modeling Exercises 1 – 14:

This challenge continues to examine U.S. census data to select and refine a model for the population of the United States over time.

Exercise 1.

The following table contains additional U.S. census population data. Would it be more appropriate to model this data with a linear or an exponential function? Explain your reasoning.

Answer:

It is not clear by looking at a graph of this data whether it lies on an exponential curve or a line. However, from the context, we know that populations tend to grow as a constant factor of the previous population, so we should use an exponentialfunction to model it. The graph below uses t = 0 to represent the year 1900. OR

The differences between consecutive population values do not remain constant and in fact get larger as time goes on, but the quotients of consecutive population values are nearly constant around 1. 1. This indicates that a linear model is not appropriate but an exponential model is.

Exercise 2.

Use a calculator’s regression capability to find a function, f, that models the U.S. Census Bureau data from 1900 to 2010.

Answer:

Using a graphing calculator and letting the year 1900 corresponding to t = 0 gives the following exponential regression equation.

P(t) = 81.1(1.0126)^{t}

Exercise 3.

Find the growth factor for each 10-year period and record it in the table below. What do you observe about these growth factors?

Answer:

The growth factors are fairly constant around 1.1.

Exercise 4.

For which decade is the 10-year growth factor the lowest? What factors do you think caused that decrease?

Answer:

The 10-year growth factor is lowest in the 1930’s, which is the decade of the Great Depression.

Exercise 5.

Find an average 10-year growth factor for the population data in the table. What does that number represent? Use the average growth factor to find an exponential function, g, that can model this data.

Answer:

Averaging the 10-year growth factors gives 1.136; using our previous form of an exponentialfunction; this means that the growth rate r satisfies 1 + r = 1.136, so r = 0.136. This represents a 13.6% population increase every ten years. The function g has an initial value g(0) = 76.2, so g is then given by g(t) = 76.2(1.136)^{\(\frac{t}{10}\)}, where t represents year since 1900.

Exercise 6.

You have now computed three potential models for the population of the United States over time: functions E, f, and g. Which one do you expect would be the most accurate model based on how they were created? Explain your reasoning.

Answer:

Student responses will vary. Potential responses:

→ I expect that function f that we found through exponential regression on the calculator is the most accurate because it uses all of the data points to compute the coefficients of the function.

→ I expect that the function E is most accurate because it uses only the most recent population values.

Exercise 7.

Summarize the three formulas for exponential models that you have found so far: Write the formula, the initial populations, and the growth rates indicated by each function. What is different between the structures of these three functions?

Answer:

We have the three models:

→ E(t) = 281.4(1.0093)^{t}: Population is 281.4 million in the year 2000; annual growth rate is 0.93%.

→ f(t) = 81. 1(1.0126)^{t}: Population is 81.1 million in the year 1900; annual growth rate is 1.26%.

→ g(t) = 76.2(13.6)^{\(\frac{t}{10}\)} Population is 76.2 million in the year 1900; 10-year growth rate is 13.6%.

In function E, t = 0 corresponds to the year 2000, while in functions f and g, t = 0 represents the year 1900. Function g is expressed in terms of a 10-year growth factor instead of an annual growth factor as in functions E and f. Function E has the year 2000 corresponding to t = 0, while in functions f and g the year t = 0 represents the year 1900.

Exercise 8.

Rewrite the functions E, f, and g as needed in terms of an annual growth rate.

Answer:

We need to use properties of exponents to rewrite g.

g(t) = 76.2(1.136))^{\(\frac{t}{10}\)}

= 76.2 ((1.136))^{\(\frac{1}{10}\)})^{t})

≈ 76.2(1.0128)^{t}

Exercise 9.

Transform the functions as needed so that the time t = 0 represents the same year in functions E, f, and g. Then compare the values of the initial populations and annual growth rates indicated by each function.

Answer:

In function E, t = 0 represents the year 2000, and in functions f and g, t = 0 represents the year 1900.

Thus, we need to translate function E horizontally to the right by 100 years, giving a new function:

E(t) = 281.4(1.0093)^{t – 100}

= 281.4(1.0093)^{-100}(1.0093)t

≈ 111.5(1.0093)^{t}

Then we have the three functions:

E(t) = 111.5(1.0093))^{t}

f(t) = 81.1(1.0126))^{t}

g(t) = 76.2(1.0128))^{t}

→ Function E has the largest initial population and the smallest growth rate at 0.93% increase per year.

→ Function g has the smallest initial population and the largest growth rate at 1.28% increase per year.

Exercise 10.

Which of the three functions is the best model to use for the U.S. census data from 1900 to 2010? Explain your reasoning.

Answer:

Student responses will vary.

Possible response:

Graphing all three functions together with the data, we see that function f appears to be the closest to all of the data points.

Exercise 11.

The U.S. Census Bureau website http://www.census.gov/popclock displays the current estimate of both the United States and world populations.

a. What is today’s current estimated population of the U.S.?

Answer:

This will vary by the date. The solution shown here uses the population 318.7 million and the date August 16, 2014.

b. If time t = 0 represents the year 1900, what is the value of t for today’s date? Give your answer to two decimal places.

Answer:

August 16 is the 228th day of the year, so the time is t = 114 + \(\frac{228}{365}\). We use t = 114.62

c. Which of the functions E, f, and g gives the best estimate of today’s population? Does that match what you expected? Justify your reasoning.

Answer:

E(114.62) = 322.2

f(114.62) = 340.7

g(114.62) = 327.4

The function E gives the closest value to today’s estimated population, but all three functions produce estimates that are too high. Possible response: I had expected that function f, which was obtained through regression, to produce the closest population estimate, so this is a surprise.

d. With your group, discuss some possible reasons for the discrepancy between what you expected in Exercise 8 and the results of part (c) above.

Answer:

Student responses will vary.

Exercise 12.

Use the model that most accurately predicted today’s population In Exercise 9, part (c) to predict when the U.S. population will reach half a billion.

Answer:

Haifa billion is 500 million. Set the formula for E equal to 500 and solve for t.

111.5(1.0093)^{t} = 500

1.0093^{t} = \(\frac{500}{111.5}\)

1.0093^{t} = 4.4843

log(1.0093)^{t} = log(4.4843)

t log(1.0093) = log(4.4843)

t = \(\frac{\log (4.4843)}{\log (1.0093)}\)

t ≈ 162

Assuming the same rate of growth, the population will reach half a billion people 162 years from the year 19%, in the year 2062.

Exercise 13.

Based on your work so far, do you think this is an accurate prediction? Justify your reasoning.

Answer:

Student responses will vary. Possible response: From what we know of population growth, the data should most likely be fit with an exponential function, however, the growth rate appears to be decreasing because the models that use all of the census data produce estimates for the current population that are too high.

I think the population will reach half a billion sometime after the year 2062 because the U.S. Census Bureau expects the growth rate to slow down. Perhaps the United States is reaching its capacity and cannot sustain the same exponential rate of growth into the future.

Exercise 14.

Here is a graph of the U.S. population since the census began in 1790. Which type of function would best model this data? Explain your reasoning.

Answer:

The shape of the curie indicates that an exponential model would be the best choice. You could model the data for short periods of time using a series of piecewise linear functions, but the average rate of change in the early years is clearly less than that in later years. A linear model would also not make sense because at some point in the past you would have had a negative number of people living in the U.S.

Exercises 15 – 16:

Exercise 15.

The graph below shows the population of New York City during a time of rapid population growth.

Answer:

Finn averaged the 10-year growth rates and wrote the function f(t) = 33131\((1.44)^{\frac{t}{10}}\), where t is the time in years since 1790.

Gwen used the regression features on a graphing calculator and got the function g(t) = 48661(1.036)^{t}, where t is the time in years since 1790.

a. Rewrite each function to determine the annual growth rate for Finn’s model and Gwen’s model.

Answer:

Finn’s function: f(t) = 33131(\(1.44^{\frac{1}{10}}\))^{t} The annual growth rate is 3.7%. Gwen’s function has a growth rate of 3.6%.

b. What is the predicted population in the year 1790 for each model?

Answer:

It will be the value of the function when t = 0. Finn: f(0) = 33131. Gwen: g(0) = 48661.

c. Lenny calculated an exponential regression using his graphing calculator and got the same growth rate as Gwen, but his initial population was very close to 0. Explain what data Lenny may have used to find his function.

Answer:

He may have used the actual year for his time values; where Gwen represented year 1790 by t = 0, Lenny may have represented year 1790 by t = 1790. If you translate Gwen’s function 1790 units to the right write the resulting function in the form f(t) = a ht, the value of a would-be very small.

48661(1.036)^{t – 1790} = \(\frac{48661(1.036)^{t}}{1.036^{1790}}\) and \(\frac{48661}{1.036^{1790}}\) ≈ 1.56 × 10^{-23}

d. When does Gwen’s function predict the population will reach 1, 000, 000? How does this compare to the graph?

Answer:

Solve the equation: 48661(1.036)^{t} = 1000000.

1.036^{t} = \(\frac{1000000}{48661}\)

log(1.036)^{t} = log(\(\frac{1000000}{48661}\))

t log(1.036) = log(\(\frac{1000000}{48661}\))

t = \(\frac{\log \left(\frac{1000000}{48661}\right)}{\log (1.036)}\)

t ≈ 85.5

Gwen’s model predicts that the population will exceed one million after 86 years, which would be during the year 1867. It appears that the population was close to one million around 1870 so the model does a fairly good job of estimating the population.

e. Based on the graph, do you think an exponential growth function would be useful for predicting the population of New York in the years after 1950?

Answer:

The graph appears to be increasing but curving downwards, and an exponential model with a base greater than 1 would always be increasing at an increasing rate, so its graph would curve upwards. The difference between the function and the data would be increasing, so this is probably not an appropriate model.

Exercise16.

Suppose each function below represents the population of a different U.S. city since the year 1900.

a. Complete the table below. Use the properties of exponents to rewrite expressions as needed to help support your answers.

Answer:

b. Could the function E(t) = 6520(1.219)^{\(\frac{t}{10}\)}, where t is years since 2000 also represent the population of one of these cities? Use the properties of exponents to support your answer.

Answer:

Yes, it could represent the population in the city with function D. The expression 1.219^{\(\frac{t}{10}\)} for any real number t. Also, E(0) ≈ D(100), which would make sense if the point of reference in time is 100 years apart.

c. Which cities are growing in size, and which are decreasing according to these models?

Answer:

The cities represented by functions A, B, and D are growing because their base value is greater than 1. The city represented by function C is shrinking because 1 – 0.01 is less than 1.

d. Which of these functions might realistically represent city population growth over an extended period of time?

Answer:

Based on the United States and New York City data, it is unlikely that a city in the United States could sustain a 50% growth rate every two years for an extended period of time as indicated by function B and its predicted population in the year 2000. The other functions seem more realistic, with annual growth or decay rates similar to other city populations we examined.

### Eureka Math Algebra 2 Module 3 Lesson 27 Problem Set Answer Key

Question 1.

Does each pair of formulas described below represent the same sequence? Justify your reasoning.

a. a_{n + 1} = \(\frac{2}{3}\)a_{n}, a_{0}= -1 and b_{n} = -(\(\frac{2}{3}\))^{n} for n ≥ 0

Answer:

Yes, checking the first few terms in each sequence gives the same values. Both sequences start with -1 and are repeatedly multiplied by \(\frac{2}{3}\).

b. a_{n} = 2a_{n – 1} + 3, a_{0} = 3 and b_{n} = 2(n – 1)^{3} + 4(n – 1) + 3 for n ≥ 1.

Answer:

No, the first two terms are the same, but the third term is different.

c. a = \(\frac{1}{3}\)(3)^{n} for n ≥ 0 and b_{n} = 3^{n – 2} for n ≥ 0.

AAnswer:

Yes, the first terms are equal a_{0} = \(\frac{1}{3}\) and b_{0} = 3^{-1} = \(\frac{1}{3}\), and the next term is found by multiplying the previous term by 3 in both sequences.

Question 2.

Tina is saving her babysitting money. She has $500 in the bank, and each month she deposits another $100. Her account earns 2% interest compounded monthly.

a. Complete the table showing how much money she has in the bank for the first four months.

Month | Amount(in dollars) |

1 | |

2 | |

3 | |

4 |

Answer:

b. Write a recursive sequence for the amount of money she has in her account after n months.

Answer:

a_{1} = 500, a_{n + 1} = a_{n}(1 + \(\frac{0.02}{12}\)) + 100

Question 3.

Assume each table represents values of an exponential function of the form f(t) = a(b)^{ct} where b is a positive real number and a and c are real numbers. Use the information in each table to write a formula for f in terms of t for parts (a) – (d).

a.

Answer:

f(t) = 10(5)^{\(\frac{t}{4}\)}

b.

Answer:

f(t) = 1000(0.75)^{\(\frac{t}{4}\)}

c.

Answer:

f(t) = 4.287\(\left(\frac{9}{5}\right)^{\frac{t}{2}}\)

d.

Answer:

f(t) = 62.5\(\left(\frac{4}{5}\right)^{\frac{t}{3}}\)

e. Rewrite the expressions for each function in parts (a) – (d) to determine the annual growth or decay rate.

Answer:

For part (a), 5^{\(\frac{t}{4}\)} = (5^{\(\frac{1}{4}\)})^{t} s0 the annual growth factor is 5^{\(\frac{1}{4}\)} ≈ 1.495, and the annual growth rate is 49.5%.

For part (b), 0.75^{\(\frac{t}{5}\)} = (o.75^{\(\frac{1}{5}\)})^{t} ≈ 0.596 so, the annual growth factor is 0.75^{\(\frac{1}{5}\)} ≈ 0.596, so the annual growth rate is

-40.4%, meaning that the quantity is decaying at a rate of 40.4%.

For part (c), (\(\frac{9}{5}\))^{\(\frac{t}{2}\)} = ((\(\frac{9}{5}\))^{\(\frac{1}{2}\)})^{t} so the annual growth factor is (^{\(\frac{t}{2}\)})^{\(\frac{1}{2}\)} ≈ 1.312 and the annual growth rate is 31.2%.

For part (a), (\(\frac{4}{5}\))^{\(\frac{t}{3}\)} = ((\(\frac{4}{5}\))^{\(\frac{1}{3}\)})^{t} so the annual growth factor is ()3 ≈ 0.928 and the annual growth rate is -0.072, which is a decay rate of 7.2%.

f. For parts (a) and (t), determine when the value of the function is double its initial amount.

Answer:

For part (a), solve the equation 2 = 5^{\(\frac{t}{4}\)} for t.

log(2) = log(5^{\(\frac{t}{4}\)})

\(\frac{t}{4}=\frac{\log (2)}{\log (5)}\)

t = 4\(\left(\frac{\log (2)}{\log (5)}\right)\)

t ≈ 1.723

For part(c), solve the equation 2 = \(\left(\frac{9}{5}\right)^{\frac{t}{2}}\) for t. The solution is 2.358.

g. For parts (b) and (d), determine when the value of the function is half of its initial function.

Answer:

For part (b), solve the equation \(\frac{1}{2}\) = (o.75)^{\(\frac{t}{5}\)} for t. The solution is 12.047.

For part (d), solve the equation \(\frac{1}{2}\) = \(\left(\frac{4}{5}\right)^{\frac{t}{3}}\) for t. The solution is 9.319.

Question 4.

When examining the data in Example 1, Juan noticed the population doubled every five years and wrote the formula P(t) = 100(2)^{\(\frac{t}{5}\)}. Use the properties of exponents to show that both functions grow at the same rate per year.

Answer:

Using properties of exponents, 100(2)^{\(\frac{t}{5}\)} = 100 \(\left(2^{\frac{1}{5}}\right)^{t}\). The annual growth is 2^{\(\frac{t}{5}\)}. In the other function, the annual growth is 4)^{\(\frac{1}{10}\)} = \(\left(4^{\frac{1}{2}}\right)^{\frac{1}{5}}\) = 2)^{\(\frac{1}{5}\)}.

Question 5.

The growth of a tree seedling over a short period of time can be modeled by an exponential function. Suppose the tree starts out 3 feet tall and its height increases by 15% per year. When will the tree be 25 feet tall?

Answer:

We model the growth of the seedling by h(t) = 3(1.15)^{t}, where t is measured in years, and we find that

3(1.15)^{t} = 25 when t = \(\frac{\log \left(\frac{25}{3}\right)}{\log (1.15)}\) so t ≈ 15.17 years. The tree will be 25 feet tall when itis 15 years and 2 months old.

Question 6.

Loggerhead turtles reproduce every 2 – 4 years, laying approximately 120 eggs in a clutch. Studying the local population, a biologist records the following data in the second and fourth years of her study:

a. Find an exponential model that describes the loggerhead turtle population in year t.

Answer:

From the table, we see that P(2) = 50 and P(4) = 1250. So, the growth rate over two years is \(\frac{1250}{50}\) = 25. Since P(2) = 50, and P(t) = P_{0}(25)^{\(\frac{t}{2}\)}, we know that 50 = p_{0}(25), so P_{0} = 2. Then 50r^{2} = p_{0}r^{4}, so 50r^{2} = 1250. Thus, r^{2} = 25 and then r = 5. Since 50 = p_{0}r^{2}, we see that P_{0} = 2. Therefore, P(t) = 2(5^{t})

b. According to your model, when will the population of loggerhead turtles be over 5,000? Give your answer in years and months.

Answer:

2(5^{t}) = 5000

5^{t} = 2500

t log(5) = log(2500)

t = \(\frac{\log (2500)}{\log (5)}\)

t ≈ 4.86

The population of loggerhead turtles will be over 5,000 after year 4.86, which is roughly 4 years and 11 months.

Question 7.

The radioactive isotope seaborgium-266 has a half-life of 30 seconds, which means that if you have a sample of A grams of seaborgium-266, then after 30 seconds half of the sample has decayed (meaning it has turned into another element), and \(\frac{A}{2}\) only grams of seaborgium-266 remain. This decay happens continuously.

a. Define a sequence a_{0}, a_{1}, a_{2}, …….. so that a_{n} represents the amount of a 100-gram sample that remains after n minutes.

Answer:

In one minute, the sample has been reduced by half two times, leaving only \(\frac{1}{4}\)

of the sample. We can represent this by the sequence a_{n} = 100(\(\frac{1}{2}\))^{2n} = 100(\(\frac{1}{2}\))^{n}. (Either form is acceptable.)

b. Define a function a(t) that describes the amount of a 100-gram sample of seaborgium-266 that remains after t minutes.

Answer:

a(t)=100(\(\frac{1}{4}\))^{t} = 1oo(\(\frac{1}{2}\))^{2t}

c. Do your sequence from part (a) and your function from part (b) model the same thing? Explain how you know.

Answer:

The function models the amount of seaborgium-266 as it constantly decreases every fraction of a second, and the sequence models the amount of seaborgium-266 that remains only in 30-second intervals. They model nearly the same thing, but not quite. The function is continuous and the sequence is discrete.

d. How many minutes does it take for less than 1 g of seaborgium-266 to remain from the original 100 g sample? Give your answer to the nearest minute.

Answer:

The sequence is a_{0} = 100, a_{1} = 25, a_{2} = 6.25, a_{3} = 1.5625, a_{4} = 0.390625, so after 4 minutes there is less than 1 g of the original sample remaining.

Question 8.

Strontium-90, magnesium-28, and bismuth all decay radioactively at different rates. Use the data provided in the graphs and tables below to answer the questions that follow.

a. Which element decays most rapidly? How do you know?

Answer:

Magnesium-28 decays most rapidly. It loses half its amount every 21 hours.

b. Write an exponential function for each element that shows how much of a 100 g sample will remain after t days. Rewrite these expressions to show precisely how their exponential decay rates compare to confirm your answer to part (a).

Answer:

→ Strontium-90: We model the remaining quantity by f(t) = 100(\(\frac{1}{2}\))^{\(\frac{t}{\frac{25}{24}}\)} where t is in days.

Rewriting the expression gives a gro wth factor of \(\left(\frac{1}{2}\right)^{\frac{24}{25}}\) ≈ 0.514, so f(t) = 100(0.514)^{t}.

→ Magnesium-28: We model the remaining quantity by f(t) = 100 (\(\frac{1}{2}\))^{\(\frac{t}{\frac{21}{24}}\)} where t is in days.

Rewriting the expression give a gro wth factor of (\(\frac{1}{2}\)) 0.453, so f(t) = 100(0.453)1

→ Bismuth: We model the remaining quantity by f(t) = 100(\(\frac{1}{2}\))^{\(\frac{t}{5}\)} where t is in days. Rewriting the

expression gives a growth factor (\(\frac{1}{2}\))^{\(\frac{1}{5}\)} ≈ 0.871, so f(t) = 100(0.871)^{t}.

The function with the smallest daily growth factor is decaying the fastest, so magnesium-24 decays the fastest.

Question 9.

The growth of two different species of fish in a lake can be modeled by the functions shown below where t is time in months since January 2000. Assume these models will be valid for at least 5 years.

Fish A: f(t) = 5,000(1.3)^{t}

Fish B: g(t) = 10000(1.1)^{t}

According to these models, explain why the fish population modeled by function f will eventually catch up to the fish population modeled by function g. Determine precisely when this will occur.

Answer:

The fish population with the larger growth rate will eventually exceed the population with a smaller growth rate, so eventually there will be a larger population of Fish A.

Solve the equation f(t) = g(t) for t to determine when the populations will be equal. After that point in time, the population of Fish A will exceed the population of Fish B.

The solution is

5000(1.3)^{t} = 10000(1.1)^{t}

\(\frac{(1.3)^{t}}{(1.1)^{t}}\) = 2

\(\left(\frac{1.3}{1.1}\right)^{t}\) = 2

t = \(\frac{\log (2)}{\log \left(\frac{1.3}{1.1}\right)}\)

t ≈ 4.15

During the fourth year, the population of Fish A will catch up to and then exceed the population of Fish B.

Question 10.

When looking at U.S. minimum wage data, you can consider the nominal minimum wage, which is the amount paid in dollars for an hour of work in the given year. You can also consider the minimum wage adjusted for inflation. Below are a table showing the nominal minimum wage and a graph of the data when the minimum wage is adjusted for inflation. Do you think an exponential function would be an appropriate model for either situation? Explain your reasoning.

Answer:

Student solutions will vary. The inflation-adjusted minimum wage is dearly not exponential because it does not strictly increase or decrease. The other data when graphed does appear roughly exponential, and a good model would be f(t) = 0.40(1.044)^{t}

Question 11.

A dangerous bacterial compound forms in a closed environment but is immediately detected. An initial detection reading suggests the concentration of bacteria in the closed environment is one percent of the fatal exposure level. Two hours later, the concentration has increased to four percent of the fatal exposure level.

a. Develop an exponential model that gives the percentage of fatal exposure level in terms of the number of hours passed.

Answer:

P(t) = 1 . \(\left(\frac{4}{1}\right)^{\frac{t}{2}}\)

= 4^{\(\frac{t}{2}\)}

= 2^{t}

b. Doctors and toxicology professionals estimate that exposure to two-thirds of the bacteria’s fatal concentration level will begin to cause sickness. Offer a time limit (to the nearest minute) for the inhabitants of the infected environment to evacuate in order to avoid sickness.

Answer:

66.66 = 2^{t}

log(66.66) = t. log(2)

t = \(\frac{\log (66.66)}{\log (2)}\) ≈ 6.0587

Inhabitants should evacuate before 6 hours and 3 minutes.

c. A more conservative approach is to evacuate the infected environment before bacteria concentration levels reach 45% of the fatal level. Offer a time limit (to the nearest minute) for evacuation in this circumstance.

Answer:

2^{t} = 45

t log(2) = log(45)

t = \(\frac{\log (45)}{\log (2)}\) ≈ 5.492

Inhabitants should evacuate within 5 hours and 30 minutes.

d. To the nearest minute, when will the infected environment reach 100% of the fatal level of bacteria concentration

Answer:

t . log(2) = log(100)

t = \(\frac{2}{\log (2)}\)

t ≈ 6.644

The infected environment will reach 100% of the fatal level of bacteria in 6 hours and 39 minutes..

Question 12.

Data for the number of users at two different social media companies is given below. Assuming an exponential growth rate, which company is adding users at a faster annual rate? Explain how you know.

Answer:

Company A: The number of users (in millions) can be modeled by A(t) = a \(\left(\frac{185}{54}\right)^{\frac{t}{2}}\) where a is the initial amount and t is time in years since 2010.

Company B: The number of users (in millions) can be modeled by B(t) = b\(\left(\frac{1056}{360}\right)^{\frac{t}{3}}\) where b is the initial amount and t is time in years since 2009.

Rewriting the expressions, you can see that Company A’s annual growth factor is \(\left(\frac{185}{54}\right)^{\frac{1}{2}}\) ≈ 1.851, and Company B’s annual growth factor is \(\left(\frac{1056}{360}\right)^{\frac{1}{3}}\) ≈ 1.432. Thus, Company A is growing at the faster rate of 85.1% compared to Company B’s 43.2%.

### Eureka Math Algebra 2 Module 3 Lesson 27 Exit Ticket Answer Key

Question 1.

The table below gives the average annual cost (e.g., tuition, room, and board) for four-year public colleges and universities. Explain why a linear model might not be appropriate for this situation.

Answer:

A linear function would not be appropriate because the average rate of change is not constant.

Question 2.

Write an exponential function to model this situation.

Answer:

If you calculate the growth factor every 10 years, you get the following values.

1981 – 1991: \(\frac{5243}{2550}\) = 2.056

1991 – 2001: \(\frac{8653}{5243}\) = 1.650

2001 – 2011: \(\frac{15918}{8653}\) = 1.840

The average of these growth factors is 1.85.

Then the average annual cost in dollars t years after 1981 is C(t) = 2 550(1.85)^{\(\frac{t}{10}\)}.

Question 3.

Use the properties of exponents to rewrite the function from Problem 2 to determine an annual growth rate.

Answer:

We know that 2550(1.85)^{\(\frac{t}{10}\)} = 2250 (1.85^{\(\frac{1}{10}\)})^{t} and 1.85^{\(\frac{1}{10}\)} ≈ 1.063. Thus the annual growth rate is 6.3%.

Question 4.

If this trend continues, when will the average annual cost exceed $35, 000?

Answer:

We need to solve the equation C(t) = 35000 for t.

2550(1.85)^{\(\frac{t}{10}\)} = 35000

(1.85)^{\(\frac{t}{10}\)} = 13.725

log((1.85)^{\(\frac{t}{10}\)}) = log(13.725)

\(\frac{t}{10}=\frac{\log (13.725)}{\log (1.85)}\)

t = 10\(\left(\frac{\log (13.725)}{\log (1.85)}\right)\)

t ≈ 42.6

The cost will exceed $35,000 after 43 years, in the year 2024.