Question 1

A certain radioactive isotope decays at a rate of 0.175% annually. Determine the half-life of this isotope, to the nearest year.
A) 396 yr
B) 172 yr
C) 4 yr
D) 286 yr

Accepted Answer

The half-life of a radioactive isotope can be calculated using the formula $T_{1/2} = \frac{\ln(2)}{r}$, where $r$ is the decay rate per period. Given a decay rate of 0.175% annually, $r = 0.00175$ per year. Thus, $T_{1/2} = \frac{\ln(2)}{0.00175} \approx 396$ years.

Question 2

An artifact is discovered at a certain site. If it has 53% of the carbon-14 it originally contained, what is the approximate age of the artifact to the nearest year? (carbon-14 decays at the rate of 0.0125% annually.)
A) 2206 yr
B) 4240 yr
C) 5079 yr
D) 3760 yr

Accepted Answer

The decay of carbon-14 can be modeled using the formula N(t) = N0 * (1 - 0.000125)^t, where N(t) is the remaining amount of carbon-14, N0 is the initial amount, and t is the time in years. Solving for t when N(t)/N0 = 0.53 gives approximately 5079 years.

Question 3

Assume the cost of a gallon of milk is $3.40. With continuous compounding, find the time it would take the cost to be 4 times as much (to the nearest tenth of a year), at an annual inflation rate of 6%.
A) 0.1 yr
B) 8.5 yr
C) 0.0 yr
D) 23.1 yr

Accepted Answer

The formula for continuous compounding is $A = Pe^{rt}$, where $A$ is the amount after time $t$, $P$ is the principal amount, $r$ is the rate, and $e$ is the base of the natural logarithm. Here, $A = 4 \times 3.40 = 13.60$, $P = 3.40$, and $r = 0.06$. Solving for $t$, we get $t = \frac{\ln(A/P)}{r} = \frac{\ln(4)}{0.06} \approx 23.1$ years.

Question 4

The half-life of an element is $4.5 \times 10 ^ { 11 } \text { yr. }$
How long does it take a sample of the element to decay to $\frac { 2 } { 3 }$ of its original mass? Use $\mathrm { A } = \mathrm { A } _ { 0 } ( 0.5 ) ^ { ( \mathrm { t } / \mathrm { T } ) }$ where $\mathrm { A } 0$ is the initial amount,  T  is the half-life, and  t  is the time. (Express results in scientific notation, rounded to the nearest hundredth.)
A) $2.63 \times 10 ^ { 11 } \mathrm { yr }$
B) $5.83 \times 10 ^ { 11 } \mathrm { yr }$
C) $1.29 \times 10 ^ { 11 } \mathrm { yr }$
D) $1.26 \times 10 ^ { 11 } \mathrm { yr }$

Accepted Answer

To find the time it takes for a sample to decay to $\frac{2}{3}$ of its original mass, we use the given formula with $A/A_0 = \frac{2}{3}$ and $T = 4.5 \times 10^{11}$ years. Solving for $t$, we get $t = T \times \frac{\log(2/3)}{\log(0.5)} \approx 2.63 \times 10^{11}$ years.

Question 5

How long will it take a sample of radioactive substance to decay to half of its original amount, if it decays according to the functi $\mathrm { A } ( \mathrm { t } ) = 250 \mathrm { e } ^ { - .223 \mathrm { t } }$ , where t is the time in years? Round your answer to the nearest
Hundredth year.
A) 27.87 yr
B) 55.75 yr
C) 3.11 yr
D) 24.76 yr

Accepted Answer

To find the half-life, we set $A(t) = \frac{250}{2} = 125$ and solve for $t$. Substituting into the equation gives $125 = 250e^{-0.223t}$. Dividing both sides by 250 gives $0.5 = e^{-0.223t}$. Taking the natural logarithm of both sides gives $\ln(0.5) = -0.223t$. Solving for $t$ gives $t = \frac{\ln(0.5)}{-0.223} \approx 3.11$ years.

Question 6

The purchasing power of a dollar is decreasing at the rate of 4.9% annually, compounded continuously. How long will it take for the purchasing power of $1.00 to be worth $0.74? Round answers to the nearest hundredth.
A) 0.06 yr
B) 0.61 yr
C) 6.15 yr
D) 15.10 yr

Accepted Answer

To find the time it takes for the purchasing power of $1.00 to decrease to $0.74 with a continuous annual decrease rate of 4.9%, we use the formula for continuous compounding: $A = Pe^{rt}$, where $A$ is the amount after time $t$, $P$ is the principal amount, $r$ is the rate of decrease, and $t$ is the time in years. Here, $A = 0.74$, $P = 1$, and $r = -0.049$ (since it's a decrease). Rearranging the formula to solve for $t$, we get $t = \frac{\ln(A/P)}{r}$. Plugging in the values, $t = \frac{\ln(0.74)}{-0.049}$, which approximately equals 6.15 years.

Question 7

A sample of 400 grams of radioactive substance decays according to the function $\mathrm { A } ( \mathrm { t } ) = 400 \mathrm { e } ^ { - .035 \mathrm { t } }$ , where t is the time in years. How much of the substance will be left in the sample after 10 years? Round your answer to the
Nearest whole gram.
A) 0 g
B) 12 g
C) 1 g
D) 282 g

Accepted Answer

To find the amount of substance left after 10 years, substitute $t = 10$ into the given function: $A(10) = 400e^{-0.035 \times 10}$. Simplifying, $A(10) = 400e^{-0.35}$. Calculating this gives approximately 282 grams.

Question 8

Suppose the consumption of electricity grows at 4.4% per year, compounded continuously. Find the number of years before the use of electricity has tripled. Round the answer to the nearest hundredth.
A) 0.25 yr
B) 68.18 yr
C) 2.50 yr
D) 24.97 yr

Accepted Answer

To find the number of years before the use of electricity has tripled when growing at a rate of 4.4% per year compounded continuously, we use the formula for continuous compounding: $A = Pe^{rt}$, where $A$ is the amount after time $t$, $P$ is the initial amount, $r$ is the rate of growth, and $e$ is the base of the natural logarithm. Here, we want $A$ to be 3 times $P$ (tripled), so we set up the equation $3 = e^{0.044t}$. Solving for $t$, we take the natural logarithm of both sides to get $ln(3) = 0.044t$, which gives $t = \frac{ln(3)}{0.044}$, resulting in approximately 24.97 years.

Question 9

In the formula A(t) = A0ekt, A is the amount of radioactive material remaining from an initial amount A0 at a given time t, and k is a negative constant determined by the nature of the material. A certain radioactive isotope
Decays at a rate of 0.3% annually. Determine the half-life of this isotope, to the nearest year.
A) 167 yr
B) 231 yr
C) 2 yr
D) 100 yr

Accepted Answer

The answer of In the formula A(t) = A0ekt, A...

Question 10

The amount of particulate matter left in solution during a filtering process is given by the equation $P = 700 ( 2 ) ^ { - 0.8 n }$, where $\mathrm { n }$ is the number of filtering steps. Find the amounts left for $\mathrm { n } = 0$ and $\mathrm { n } = 5$. (Round to the nearest whole number.)
A) 1400, 44
B) 700, 22
C) 700, 44
D) 700, 11,200

Accepted Answer

For $n = 0$, $P = 700(2)^{-0.8(0)} = 700(1) = 700$. For $n = 5$, $P = 700(2)^{-0.8(5)} \approx 700(0.0625) \approx 44$.

Question 11

How long will it take for prices in the economy to double at a 7% annual inflation rate?
A) 16.24 yr
B) 23.45 yr
C) 9.01 yr
D) 10.24 yr

Accepted Answer

The answer of How long will it take for prices...

Question 12

Coffee is best enjoyed at a temperature of 118°F. A restaurant owner wants to discover the temperature T at which he should serve his coffee so that it will have cooled to this ideal temperature in 4 minutes. He discovers
That a cup of coffee served at 197°F cools to 184°F in one minute when his restaurant is at 68°F. If he maintains
The restaurant temperature at 68°F, at what temperature should he serve the coffee to meet his goal?
A) 144°F
B) 139°F
C) 164°F
D) 197°F

Accepted Answer

The answer of Coffee is best enjoyed at a temperature...

Question 13

An economist predicts that the buying power B(x) of a dollar x years from now will decrease according to the form $B ( x ) = 0.51 ^ { x }$ . How much will today's dollar be worth in 7 years? Round the answer to the nearest cent.
A) $3.57
B) $2.70
C) $0.62
D) $0.01

Accepted Answer

The answer of An economist predicts that the buying power...

Question 14

In year 0 , the number of female athletes participating in Summer Olympic-Type Games was 450 . In year 11 , about 3550 participated in the Summer Olympics in Atlanta. Assuming that  P(0)=500  and that the exponential model applies, find the value of k rounded to the hundredths place, and write the function.
A) $\mathrm { k } = 0.17 ; \mathrm { P } ( \mathrm { t } ) = 500 \mathrm { e } ^ { 0.17 \mathrm { t } }$
B) $k = 0.19 ; P ( t ) = 500 e ^ { 0.29 t }$
C) $\mathrm { k } = 0.21 ; \mathrm { P } ( \mathrm { t } ) = 500 \mathrm { e } ^ { 0.21 \mathrm { t } }$
D) $\mathrm { k } = 0.19 ; \mathrm { P } ( \mathrm { t } ) = 500 \mathrm { e } ^ { 0.19 \mathrm { t } }$

Accepted Answer

The exponential growth model is $P(t) = P_0e^{kt}$, where $P_0$ is the initial amount, $P(t)$ is the amount at time $t$, and $k$ is the growth rate. Given $P_0 = 500$ and $P(11) = 3550$, we solve for $k$ using the equation $3550 = 500e^{11k}$. Solving for $k$ gives $k \approx 0.19$, so the correct function is $P(t) = 500e^{0.19t}$.

Question 15

Suppose the number of Quickie hamburgers (in millions) served yearly can be modeled by $f ( x ) = 35.8 e ^ { 0.17 x }$ Approximate how many years it takes for the number of hamburgers served to reach 54 million. Round to the nearest year.
A) 2 yr
B) 3 yr
C) 1 yr
D) 4 yr

Accepted Answer

To find the number of years it takes for the hamburgers served to reach 54 million, set $f(x) = 54$ and solve for $x$:  $54 = 35.8 e^{0.17x}$  Divide both sides by 35.8:  $1.5084 \approx e^{0.17x}$  Take the natural logarithm of both sides:  $\ln(1.5084) \approx 0.17x$  Solve for $x$:  $x \approx \frac{\ln(1.5084)}{0.17} \approx 2$  So, it takes approximately 2 years.

Question 16

A certain country's Consumer Price Index is approximated by A( $\mathrm { A } ( \mathrm { t } ) = 100 \mathrm { e } ^ { 0.024 \mathrm { t } }$ , where t represents the number of years. Use the function to determine the year in which costs will be 50% higher than in year 0.
A) year 13
B) year 15
C) year 19
D) year 17

Accepted Answer

The answer of A certain country's Consumer Price Index is...

Question 17

A certain radioactive isotope has a half-life of approximately 1500 years. How many years to the nearest year would be required for a given amount of this isotope to decay to 70% of that amount?
A) 2605 yr
B) 772 yr
C) 702 yr
D) 450 yr

Accepted Answer

The amount of radioactive isotope remaining after t years is given by the formula A = A0 * (1/2)^(t/h), where A0 is the initial amount, h is the half-life, and t is the time elapsed. We want to find t such that A = 0.7A0. Substituting these values into the formula, we get 0.7A0 = A0 * (1/2)^(t/h). Dividing both sides by A0, we get 0.7 = (1/2)^(t/h). Taking the logarithm of both sides, we get log(0.7) = (t/h) * log(1/2). Solving for t, we get t = h * log(0.7) / log(1/2) = 1500 * log(0.7) / log(1/2) = 772 years.

Question 18

How long will it take for $8200 to grow to $42,000 at an interest rate of 4.7% if the interest is compounded continuously? Round the number of years to the nearest hundredth.
A) 34.76 yr
B) 1.11 yr
C) 3475.61 yr
D) 3.48 yr

Accepted Answer

To solve this, we use the formula for continuous compounding interest, $A = Pe^{rt}$, where $A$ is the amount of money accumulated after n years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (decimal), and $t$ is the time the money is invested for in years. Rearranging the formula to solve for $t$, we get $t = \frac{\ln(A/P)}{r}$. Plugging in the given values: $A = 42000$, $P = 8200$, and $r = 0.047$, we find $t = \frac{\ln(42000/8200)}{0.047}$, which calculates to approximately 34.76 years.

Question 19

The decay of 921 mg of an isotope is given by A(t) = 921e-0.031t, where t is time in years. Find the amount left after 41 years.
A) 258 mg
B) 129 mg
C) 251 mg
D) 893 mg

Accepted Answer

The amount left after 41 years is calculated by substituting t = 41 into the given decay formula: A(41) = 921e^(-0.031*41). This calculation results in approximately 258 mg.

Question 20

Assume the cost of a car is $21,000. With continuous compounding in effect, find the number of years it would take to double the cost of the car at an annual inflation rate of 6.2%. Round the answer to the nearest hundredth.
A) 11.18 yr
B) 160.52 yr
C) 1.61 yr
D) 171.70 yr

Accepted Answer

The formula for continuous compounding is $A = Pe^{rt}$, where $A$ is the amount after time $t$, $P$ is the principal amount, $r$ is the rate, and $e$ is the base of the natural logarithm. To double the cost, $A = 2P$, so $2 = e^{0.062t}$. Taking the natural logarithm of both sides gives $\ln(2) = 0.062t$, solving for $t$ gives $t = \frac{\ln(2)}{0.062} \approx 11.18$ years.

Quiz 5: Inverse, Exponential, and Logarithmic Functions

A certain radioactive isotope decays at a rate of 0.175% annually. Determine the half-life of this isotope, to the nearest year.

An artifact is discovered at a certain site. If it has 53% of the carbon-14 it originally contained, what is the approximate age of the artifact to the nearest year? (carbon-14 decays at the rate of 0.0125% annually.)

Assume the cost of a gallon of milk is $3.40. With continuous compounding, find the time it would take the cost to be 4 times as much (to the nearest tenth of a year) , at an annual inflation rate of 6%.

How long will it take a sample of radioactive substance to decay to half of its original amount, if it decays according to the functi $\mathrm { A } ( \mathrm { t } ) = 250 \mathrm { e } ^ { - .223 \mathrm { t } }$ , where t is the time in years? Round your answer to the nearest Hundredth year.

The purchasing power of a dollar is decreasing at the rate of 4.9% annually, compounded continuously. How long will it take for the purchasing power of $1.00 to be worth $0.74? Round answers to the nearest hundredth.

Suppose the consumption of electricity grows at 4.4% per year, compounded continuously. Find the number of years before the use of electricity has tripled. Round the answer to the nearest hundredth.

The amount of particulate matter left in solution during a filtering process is given by the equation $P = 700 ( 2 ) ^ { - 0.8 n }$ , where $\mathrm { n }$ is the number of filtering steps. Find the amounts left for $\mathrm { n } = 0$ and $\mathrm { n } = 5$ . (Round to the nearest whole number.)

How long will it take for prices in the economy to double at a 7% annual inflation rate?

An economist predicts that the buying power B(x) of a dollar x years from now will decrease according to the form $B ( x ) = 0.51 ^ { x }$ . How much will today's dollar be worth in 7 years? Round the answer to the nearest cent.

Suppose the number of Quickie hamburgers (in millions) served yearly can be modeled by $f ( x ) = 35.8 e ^ { 0.17 x }$ Approximate how many years it takes for the number of hamburgers served to reach 54 million. Round to the nearest year.

A certain country's Consumer Price Index is approximated by A( $\mathrm { A } ( \mathrm { t } ) = 100 \mathrm { e } ^ { 0.024 \mathrm { t } }$ , where t represents the number of years. Use the function to determine the year in which costs will be 50% higher than in year 0.

A certain radioactive isotope has a half-life of approximately 1500 years. How many years to the nearest year would be required for a given amount of this isotope to decay to 70% of that amount?

How long will it take for $8200 to grow to $42,000 at an interest rate of 4.7% if the interest is compounded continuously? Round the number of years to the nearest hundredth.

The decay of 921 mg of an isotope is given by A(t) = 921e-0.031t, where t is time in years. Find the amount left after 41 years.

Assume the cost of a car is $21,000. With continuous compounding in effect, find the number of years it would take to double the cost of the car at an annual inflation rate of 6.2%. Round the answer to the nearest hundredth.