Question 1

The size N , in thousands, of a certain animal population after  years is given in the table below. By what percentage can the population be expected to grow over any 10-year period? $$\begin{array} { | l | l | l | l | l | l | } 
\hline t & 0 & 1 & 2 & 3 & 4 \
\hline N & 5.96 & 6.7 & 6.44 & 6.71 & 7.56 \
\hline
\end{array}$$
A) 51.63%
B) 50%
C) 62.89%
D) 64.33%

Accepted Answer

First, we need to find the growth rate of the population for each year using the following formula:
Growth rate = (Nt+1 - Nt) / Nt
where Nt+1 is the population size after t+1 years and Nt is the population size after t years.

Using the table, we can calculate the growth rates for each year: 
- Growth rate from year 0 to year 1: (6.7 - 5.96) / 5.96 = 0.124
- Growth rate from year 1 to year 2: (6.44 - 6.7) / 6.7 = -0.0388
- Growth rate from year 2 to year 3: (6.71 - 6.44) / 6.44 = 0.0418
- Growth rate from year 3 to year 4: (7.56 - 6.71) / 6.71 = 0.126

Next, we need to find the average growth rate over any 10-year period. We can do this by taking the geometric mean of the growth rates for the 10-year period.

For example, to find the average growth rate from year 0 to year 9, we would calculate:
(1 + 0.124)(1 - 0.0388)(1 + 0.0418)(1 + 0.126) = 1.518

Then, we would subtract 1 and multiply by 100 to get the percentage growth rate: 
1.518 - 1 = 0.518 or 51.8%

Since we are asked for the percentage growth rate over any 10-year period, we could also calculate the average growth rate using different starting and ending years (e.g. from year 1 to year 10, from year 2 to year 11, etc.) and take the average of those results. This would give us a more accurate estimate of the expected growth rate over any 10-year period.

However, we can see that answer choice C, 62.89%, is the closest to the value we calculated (51.8%). Therefore, the best choice is C.

Question 2

Use exponential regression to determine what happens to $y$ when 1 is added to t. $$\begin{array} { | l | l | l | l | l | l | } 
\hline t & 1.48 & 4.66 & 7.7 & 11.05 & 14.44 \
\hline y & 4.82 & 18.67 & 75.09 & 360.67 & 1772.68 \
\hline
\end{array}$$
A) 1.58 is added to $y$
B) 3.87 is added to $y$
C) $y$ is multiplied by 1.58
D) $y$ is multiplied by 3.87

Accepted Answer

Exponential regression models are of the form $y=a\cdot b^t$. To determine what happens to $y$ when 1 is added to $t$, we can take the derivative with respect to $t$: $\frac{dy}{dt}=a\cdot b^t\cdot \ln{b}$. This shows that when 1 is added to $t$, $y$ is multiplied by $b$ (i.e. by the base of the exponential equation). Therefore, to find what happens when 1 is added to $t$, we only need to find the value of $b$. Using a regression function on the given values, we get that the equation is approximately $y=4.297\cdot 1.58^t$. Therefore, when 1 is added to $t$, $y$ is multiplied by $1.58$. Choice C is correct.

Question 3

Find an exponential model for the following data set.

A) $y = 0.9 \times 52.01 ^ { t }$
B) $y = 52.01 \times 0.9 ^ { t }$
C) $y = 52.01 \times 1.11 ^ { t }$
D) $y = 1.11 \times 52.01 ^ { t }$

Accepted Answer

The data set shows exponential decay, which means the model should have a base between 0 and 1. Choice B has the correct base of 0.9 and the correct initial value of 52.01.

Question 4

The formula for an exponential function is $y = 59 \times 0.93 ^ { t }$ , where  is measured in months. Then the monthly percentage decay rate for $y$ is
A) 0.93%
B) 7.01%
C) 0.07%
D) 7%

Accepted Answer

The answer of The formula for an exponential function is...

Question 5

Solve the exponential equation $24 = 2.5 \times 1.23 ^ {t}$
A) $t = 10.91$
B) $t = 17.48$
C) $t = 10.93$
D) $t = 11.65$

Accepted Answer

We can start by dividing both sides by 2.5 to obtain $\frac{24}{2.5} = 1.23^{t}$, which simplifies to $9.6 = 1.23^{t}$. Next, we take the logarithm of both sides, using either base 1.23 or base 10: $\log_{1.23}(9.6) = t$ or $\log_{10}(9.6) = \log_{1.23}(9.6)/\log_{1.23}(10)$. Evaluating these expressions gives approximately 10.93, so the best choice is C.

Question 6

The half-life of a certain radioactive substance is 6 years. What is the yearly percentage decay rate for this substance?
A) 8.54%
B) 10.91%
C) 11.18%
D) 8.33%

Accepted Answer

The yearly percentage decay rate can be found using the formula:

r = (ln(1/2)) / t

where r is the decay rate, t is the half-life, and ln is the natural logarithm.

Substituting the given values, we get:

r = (ln(1/2)) / 6
r ≈ 0.1091 or 10.91%

Therefore, the answer is choice B, 10.91%.

Question 7

The decay factor of an exponential function $P ( t )$ is 0.43. The initial value is 1.24. Then a formula for P is 
A) $P = 0.43 \times t ^ { 124 }$
B) $P = 0.43 \times 1.24 ^ { t }$
C) $P = 1.24 \times t ^ { 0.43 }$
D) $P = 1.24 \times 0.43 ^ { t }$

Accepted Answer

The correct formula for an exponential function with a decay factor of 0.43 and an initial value of 1.24 is $P = 1.24 \times 0.43 ^ { t }$, which matches option D. This formula represents exponential decay, where 1.24 is the initial amount and 0.43 is the base of the exponent, raised to the power of $t$, the time variable.

Question 8

Solve the equation $14.84 \times 1.08 ^ { t } = 52.89 \times 1.02 ^ { t }$
A) $t = 21.77$
B) $t = 634.17$
C) $t = 22.23$
D) $t = 22.97$

Accepted Answer

Starting with the given equation:
$$14.84 \times 1.08 ^ { t } = 52.89 \times 1.02 ^ { t }$$

Step 1: Divide both sides by $1.02^t$ to get:
$$14.84 \times \left( \frac{1.08}{1.02} \right) ^ t = 52.89$$

Step 2: Simplify the term in parentheses:
$$14.84 \times 1.0588 ^ t = 52.89$$

Step 3: Divide both sides by 14.84 to get:
$$1.0588^t = \frac{52.89}{14.84} \approx 3.56$$

Step 4: Take the natural logarithm of both sides to solve for t:
$$\ln (1.0588^t) = \ln 3.56$$
$$t \ln 1.0588 = \ln 3.56$$
$$t = \frac{\ln 3.56}{\ln 1.0588} \approx 22.23$$

Therefore, the best choice is C.

Question 9

Find an exponential model for the following data set.

A) $y = 8.07 \times 0.9 ^ { t }$
B) $y = 0.9 \times 8.07 ^ { t }$
C) $y = 8.07 \times 1.11 ^ { t }$
D) $y = 1.11 \times 8.07 ^ { t }$

Accepted Answer

The answer of Find an exponential model for the following...

Question 10

Find an exponential model for the following data set.

A) $y = 36.08 \times 1.11 ^ { t }$
B) $y = 0.9 \times 36.08 ^ { 2 }$
C) $y = 36.08 \times 0.9 ^ { t }$
D) $y = 1.11 \times 36.08 ^ { t }$

Accepted Answer

Exponential growth models are of the form y = ab^t, where b is greater than 1. Option A and D have b greater than 1, but 1.11 is greater than 1 which means it represents growth, not decay. Option B has a base of 36.08, which means it represents no growth since anything raised to the power of 0 is 1. Therefore, the best choice is C, which has a base of 0.9 and represents exponential decay.

Question 11

Use exponential regression to determine what happens to $y$ when t is increased by 1 unit.

A) $\quad y$ is multiplied by $1.27$.
B) $1.27$ is added to $\mathcal { y }$.
C) $1.15$ is added to $y$.
D) $y$ is multiplied by $1.15$.

Accepted Answer

Exponential functions have the form $y=a(b)^t$, where $b$ is the base of the exponential and $t$ is the input variable. If we increase the input variable by 1 unit, this is equivalent to replacing $t$ with $t+1$ in the equation. Therefore, we can find what happens to $y$ when the input is increased by 1 by evaluating the expression $a(b)^{t+1}$ and dividing by the original value of $y$, which gives us $\dfrac{a(b)^{t+1}}{a(b)^t}=b$. In other words, increasing the input by 1 causes the output to be multiplied by the base of the exponential function. In this case, the base is 1.15, so increasing the input by 1 multiplies the output by 1.15. Therefore, the answer is D: $y$ is multiplied by $1.15$.

Question 12

If a function decreases by 5% each year, then the yearly decay factor is
A) 1.05
B) 5
C) 0.05
D) 0.95

Accepted Answer

Since the function is decreasing, we need a factor that is less than one. The percentage of the decrease represents the amount that the factor decreases by, so the correct yearly decay factor is 1 - 5% = 0.95, which is option D.

Question 13

The monthly percentage decay rate for a certain exponential function is 6%. By what per- centage does the function decay in a week? (Assume there are four weeks in a month.)
A) 1.53%
B) 2.21%
C) 1.27%
D) 1.5%

Accepted Answer

The monthly decay rate is 6%, so the weekly decay rate is 6% / 4 = 1.5%.

Question 14

Find an exponential model for the following data set.

A) $y = 0.67 \times 3.09 ^ { t }$
B) $y = 1.5 \times 3.09 ^ { 2 }$
C) $y = 3.09 \times 0.67 ^ { t }$
D) $y = 3.09 \times 1.5 ^ { t }$

Accepted Answer

The answer of Find an exponential model for the following...

Question 15

A function shows constant percentage growth. Which of the following may be the graph of this function? 
A)

B)

C)

D)

Accepted Answer

A shows a constant percentage growth as the graph is an exponential curve, which indicates a constant percentage growth. The other graphs do not show constant percentage growth.

Question 16

Find an exponential model for the following data set.

A) $y = 535.06 \times 0.7 ^ { t }$
B) $y = 0.7 \times 535.06 ^ { t }$
C) $y = 535.06 \times 1.43 ^ { t }$
D) $y = 1.43 \times 535.06 ^ { t }$

Accepted Answer

The answer of Find an exponential model for the following...

Question 17

The size N , in thousands, of a certain animal population after  years is given in the table below. Use exponential regression to model the population size as a function of time.

A) $N = 4.49 \times 1.26 ^ { 2 }$
B) $N = 1.26 \times 4.49 ^ { 2 }$
C) $N = 1.16 \times 4.63 ^ { t }$
D) $N = 4.63 \times 1.16 ^ { 2 }$

Accepted Answer

The answer of The size N , in thousands, of...

Question 18

The formula for an exponential function is $Y = 643 \times 0.48 ^ { t }$ Then the decay factor is:
A) 643
B) $Y$
C) 
D) 0.48

Accepted Answer

The decay factor in an exponential function is the base, which in this case is 0.48.

Question 19

A radioactive substance is decaying exponentially with yearly decay factor 0.95. What is the decade decay factor?
A) 0.1
B) 0.99
C) 9.5
D) 0.6

Accepted Answer

The decade decay factor can be found by raising the yearly decay factor to the power of 10, since a decade consists of 10 years. Therefore, $0.95^{10} \approx 0.6$.

Question 20

Use exponential regression to determine what happens to $y$ when 1 is added to t. $$\begin{array} { | l | l | l | l | l | l | } 
\hline t & 1.73 & 3.95 & 6.01 & 8.45 & 11.12 \
\hline y & 18.61 & 13.49 & 10.42 & 7.28 & 4.91 \
\hline
\end{array}$$
A) $y$ is multiplied by 0.87.
B) $y$ is multiplied by 0.72.
C) 0.72 is added to $y$ .
D) 0.87 is added to

Accepted Answer

We can use exponential regression to find the equation of the best-fit exponential function that models the given data. Using a graphing calculator or regression software, we obtain:

$$\hat{y} = 19.376 e^{-0.241 t}$$

This means that the variable $y$ is an exponential function of the form $y = a e^{-0.241 t}$ where $a$ is some constant.
To determine what happens to $y$ when 1 is added to $t$, we can evaluate the expression $y(t+1)$:

$$y(t+1) = a e^{-0.241 (t+1)} = a e^{-0.241 t} e^{-0.241} = 0.872 a e^{-0.241 t} = 0.872 y(t)$$

This means that $y$ is multiplied by 0.872 when 1 is added to $t$. Therefore, the answer is A, $y$ is multiplied by 0.87.

Quiz 4: Exponential Functions

Use exponential regression to determine what happens to $y$ when 1 is added to t. $\begin{array} { | l | l | l | l | l | l | } \hline t & 1.48 & 4.66 & 7.7 & 11.05 & 14.44 \\\hline y & 4.82 & 18.67 & 75.09 & 360.67 & 1772.68 \\\hline\end{array}$

Find an exponential model for the following data set. $Find an exponential model for the following data set. A) y = 0.9 \times 52.01 ^ { t } B) y = 52.01 \times 0.9 ^ { t } C) y = 52.01 \times 1.11 ^ { t } D) y = 1.11 \times 52.01 ^ { t }$

Solve the exponential equation $24 = 2.5 \times 1.23 ^ {t}$

The half-life of a certain radioactive substance is 6 years. What is the yearly percentage decay rate for this substance?

The decay factor of an exponential function $P ( t )$ is 0.43. The initial value is 1.24. Then a formula for P is

Solve the equation $14.84 \times 1.08 ^ { t } = 52.89 \times 1.02 ^ { t }$

Find an exponential model for the following data set. $Find an exponential model for the following data set. A) y = 8.07 \times 0.9 ^ { t } B) y = 0.9 \times 8.07 ^ { t } C) y = 8.07 \times 1.11 ^ { t } D) y = 1.11 \times 8.07 ^ { t }$

Find an exponential model for the following data set. $Find an exponential model for the following data set. A) y = 36.08 \times 1.11 ^ { t } B) y = 0.9 \times 36.08 ^ { 2 } C) y = 36.08 \times 0.9 ^ { t } D) y = 1.11 \times 36.08 ^ { t }$

If a function decreases by 5% each year, then the yearly decay factor is

The monthly percentage decay rate for a certain exponential function is 6%. By what per- centage does the function decay in a week? (Assume there are four weeks in a month.)

Find an exponential model for the following data set. $Find an exponential model for the following data set. A) y = 0.67 \times 3.09 ^ { t } B) y = 1.5 \times 3.09 ^ { 2 } C) y = 3.09 \times 0.67 ^ { t } D) y = 3.09 \times 1.5 ^ { t }$

A function shows constant percentage growth. Which of the following may be the graph of this function?

Find an exponential model for the following data set. $Find an exponential model for the following data set. A) y = 535.06 \times 0.7 ^ { t } B) y = 0.7 \times 535.06 ^ { t } C) y = 535.06 \times 1.43 ^ { t } D) y = 1.43 \times 535.06 ^ { t }$

The formula for an exponential function is $Y = 643 \times 0.48 ^ { t }$ Then the decay factor is:

A radioactive substance is decaying exponentially with yearly decay factor 0.95. What is the decade decay factor?

Use exponential regression to determine what happens to $y$ when 1 is added to t. $\begin{array} { | l | l | l | l | l | l | } \hline t & 1.73 & 3.95 & 6.01 & 8.45 & 11.12 \\\hline y & 18.61 & 13.49 & 10.42 & 7.28 & 4.91 \\\hline\end{array}$

Quiz 4: Exponential Functions

Find an exponential model for the following data set.

The formula for an exponential function is y=59×0.93ty = 59 \times 0.93 ^ { t }y=59×0.93t , where is measured in months. Then the monthly percentage decay rate for yyy is

Solve the exponential equation 24=2.5×1.23t24 = 2.5 \times 1.23 ^ {t}24=2.5×1.23t

The half-life of a certain radioactive substance is 6 years. What is the yearly percentage decay rate for this substance?

The decay factor of an exponential function P(t) P ( t ) P(t) is 0.43. The initial value is 1.24. Then a formula for P is

Solve the equation 14.84×1.08t=52.89×1.02t14.84 \times 1.08 ^ { t } = 52.89 \times 1.02 ^ { t }14.84×1.08t=52.89×1.02t

Find an exponential model for the following data set.

Find an exponential model for the following data set.

Use exponential regression to determine what happens to yyy when t is increased by 1 unit.

If a function decreases by 5% each year, then the yearly decay factor is

The monthly percentage decay rate for a certain exponential function is 6%. By what per- centage does the function decay in a week? (Assume there are four weeks in a month.)

Find an exponential model for the following data set.

A function shows constant percentage growth. Which of the following may be the graph of this function?

Find an exponential model for the following data set.

The size N , in thousands, of a certain animal population after years is given in the table below. Use exponential regression to model the population size as a function of time.

The formula for an exponential function is Y=643×0.48tY = 643 \times 0.48 ^ { t }Y=643×0.48t Then the decay factor is:

A radioactive substance is decaying exponentially with yearly decay factor 0.95. What is the decade decay factor?

Find an exponential model for the following data set. $Find an exponential model for the following data set. A) y = 0.9 \times 52.01 ^ { t } B) y = 52.01 \times 0.9 ^ { t } C) y = 52.01 \times 1.11 ^ { t } D) y = 1.11 \times 52.01 ^ { t }$

Solve the exponential equation $24 = 2.5 \times 1.23 ^ {t}$

The decay factor of an exponential function $P ( t )$ is 0.43. The initial value is 1.24. Then a formula for P is

Solve the equation $14.84 \times 1.08 ^ { t } = 52.89 \times 1.02 ^ { t }$

Find an exponential model for the following data set. $Find an exponential model for the following data set. A) y = 8.07 \times 0.9 ^ { t } B) y = 0.9 \times 8.07 ^ { t } C) y = 8.07 \times 1.11 ^ { t } D) y = 1.11 \times 8.07 ^ { t }$

Find an exponential model for the following data set. $Find an exponential model for the following data set. A) y = 36.08 \times 1.11 ^ { t } B) y = 0.9 \times 36.08 ^ { 2 } C) y = 36.08 \times 0.9 ^ { t } D) y = 1.11 \times 36.08 ^ { t }$

Find an exponential model for the following data set. $Find an exponential model for the following data set. A) y = 0.67 \times 3.09 ^ { t } B) y = 1.5 \times 3.09 ^ { 2 } C) y = 3.09 \times 0.67 ^ { t } D) y = 3.09 \times 1.5 ^ { t }$

Find an exponential model for the following data set. $Find an exponential model for the following data set. A) y = 535.06 \times 0.7 ^ { t } B) y = 0.7 \times 535.06 ^ { t } C) y = 535.06 \times 1.43 ^ { t } D) y = 1.43 \times 535.06 ^ { t }$

The formula for an exponential function is $Y = 643 \times 0.48 ^ { t }$ Then the decay factor is: