Question 1

If an account that earns interest compounded continuously takes $9.5$ years to double in value, how long will it take to triple? Round your answer to one decimal place, if necessary.

Accepted Answer

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Question 2

A sum of money, A₀, is invested at a certain fixed interest rate, and this interest is compounded continuously. After 7 years, the money has doubled. The balance at the end of 14 years is $4 A _ { 0 }$ .

Accepted Answer

When interest is compounded continuously, the formula for the future value of an investment is $A = P e^{rt}$, 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. If the money doubles in 7 years, it means that $A = 2P$ when $t = 7$. Since the interest is compounded continuously, the investment will double again in another 7 years (making $t = 14$), resulting in $A = 4P$, where $P$ is the initial investment ($A_0$).

Question 3

A sum of money, $A _ { 0 }$ , is invested at a certain fixed interest rate, and the interest is compounded continuously. After 7 years, the money has doubled. At the end of 14 years the value is $2 A _ { 0 }$ .

Accepted Answer

The value at the end of 14 years would be $4A_0$, because the money doubles every 7 years due to continuous compounding.

Question 4

A sum of money, $A _ { 0 }$ , is invested at a certain fixed interest rate, and this interest is compounded continuously. After 4 years, the money has doubled. The balance at the end of 8 years is $4 A _ { 0 }$ .

Accepted Answer

If the money doubles every 4 years due to continuous compounding, then after another 4 years (making 8 years in total), it would double again, resulting in 4 times the original amount ($4A_0$).

Question 5

A sum of money, $A _ { 0 }$ , is invested at a certain fixed interest rate, and this interest is compounded continuously. After 6 years, the money has doubled. The balance at the end of 12 years is $2 A _ { 0 }$ .

Accepted Answer

When interest is compounded continuously, the formula for the future value of an investment is $A = A_0 e^{rt}$, where $A_0$ is the initial amount, $r$ is the annual interest rate, $t$ is the time in years, and $e$ is the base of the natural logarithm. If the money doubles in 6 years, it means $A_0 e^{6r} = 2A_0$. After 12 years, the formula becomes $A_0 e^{12r}$, which would be $4A_0$ (quadruple the initial amount), not $2A_0$, because the investment continues to grow exponentially.

Question 6

Money triples in 15.50 years if it doubles in 10 years. Assume continuous compounding and rounding to two decimal places.

Accepted Answer

Using the formula for continuous compounding, $A = P e^{rt}$, 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 (approximately 2.71828), we can find the rate $r$ when the money doubles in 10 years. If money doubles, $A/P = 2$, so $2 = e^{10r}$. Solving for $r$ gives $r = \frac{\ln(2)}{10}$. To check if the money triples in 15.5 years, we substitute $r$ back into the formula with $t = 15.5$ and $A/P = 3$, resulting in $3 = e^{15.5(\ln(2)/10)}$. Evaluating this expression does not yield a true statement, indicating that the money does not exactly triple in 15.5 years under continuous compounding at the rate calculated.

Question 7

If money is compounded continuously, it will double in about 6.3 years if it triples in 10 years.

Accepted Answer

Using the formula for continuous compounding, $A = Pe^{rt}$, where $A$ is the amount of money after time $t$, $P$ is the principal amount, $r$ is the rate, and $e$ is the base of the natural logarithm (approximately 2.71828), we can find the rate $r$ that triples the money in 10 years ($A/P = 3$) and then use that rate to find the time it takes to double ($A/P = 2$). Solving for $r$ in $3 = e^{r*10}$ gives $r \approx 0.10986$. Plugging this rate into $2 = e^{r*t}$ to solve for $t$ gives $t \approx 6.3$ years.

Question 8

At what interest rate, compounded continuously, should $2,500 be invested today so that 10 years from now the account will be worth $5,000? Round your answer to two decimal places.

Accepted Answer

The formula for continuous compounding is A = Pe^(rt), where A is the amount, P is the principal, r is the rate, and t is the time. Solving for r: 5000 = 2500e^(10r), gives r ≈ 0.0693 or 6.93%.

Question 9

Suppose the number of bacteria grows from 7,000 to 10,000 in the first 15 minutes of an experiment. Assuming that the number of bacteria grows exponentially, how many bacteria will be present after 1 hour? Round to the nearest whole number.

Accepted Answer

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Question 10

Suppose the number of bacteria grows from 2,000 to 6,000 in the first 15 minutes of an experiment. Assuming that the number of bacteria grows exponentially, how many bacteria will be present after 1 hour?Round your answer to the nearest whole number, if necessary.

Accepted Answer

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Question 11

A radioactive substance decays exponentially. If 800 grams were present initially and 600 grams are present 100 years later, how many grams will be present after 400 years? Round your answer to two decimal places.

Accepted Answer

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Question 12

The amount of a sample of a radioactive substance remaining after t years is given to be a function of the form $Q ( t ) = Q _ { 0 } e ^ { - k t }$ . If 100 grams remain after 1,000 years and 90 grams after 2,000 years, what is the initial amount $Q _ { 0 }$ ? Round your answer to two decimal places.

Accepted Answer

Given the formula $Q(t) = Q_0 e^{-kt}$, we can substitute the given values to find $Q_0$. For 1000 years and 100 grams, we have $100 = Q_0 e^{-1000k}$. For 2000 years and 90 grams, we have $90 = Q_0 e^{-2000k}$. Dividing the second equation by the first gives us a ratio that allows us to solve for $k$, and then we can solve for $Q_0$. The calculations lead to $Q_0 = 111.11$ grams.

Question 13

Biologists estimate that the population of a biological colony at time t (in minutes) is $P ( t ) = P _ { 0 } e ^ { k t }$ thousand, where $P _ { 0 }$ and k are positive constants. If the population is 1,000 after 20 minutes and is known to double every hour, what is the population after 40 minutes? Round to the nearest whole number.

Accepted Answer

The answer of Biologists estimate that the population of a...

Question 14

The population density x miles from the center of a city is given by a function of the form $Q ( x ) = A e ^ { - k x }$ . Find this function if it is known that the population density at the center of the city is 9,000 people per square mile and the density 15 miles from the center is 6,000 people per square mile. Round numbers to six decimal places, if necessary.

Accepted Answer

The answer of The population density x miles from the...

Question 15

A radioactive substance decays exponentially. If 800 grams were present initially and 600 grams are present 100 years later, how many grams will be present after 400 years? Round your answer to two decimal places, if necessary.

Accepted Answer

The substance decays exponentially, so the amount present after t years is given by A = A0 * e^(-kt), where A0 is the initial amount, k is the decay constant, and t is the time. We can use the given information to find k:600 = 800 * e^(-k * 100)0.75 = e^(-100k)-100k = ln(0.75)k = -0.002773Now we can use this value of k to find the amount present after 400 years:A = 800 * e^(-0.002773 * 400)A = 253.13 gramsTherefore, the correct answer is A) 253.13 grams.

Question 16

If the amount of a sample of radioactive substance that remains after t years is given by $Q ( t ) = Q _ { 0 } e ^ { - 0.0003 t }$ and 500 grams of the substance remain at the end of 7,000 years, then there were about 4,083 grams present initially.

Accepted Answer

To find the initial amount $Q_0$, we use the formula $Q(t) = Q_0 e^{-0.0003 \times 7000}$. Given $Q(7000) = 500$, we solve for $Q_0$: $500 = Q_0 e^{-2.1}$. Solving this gives $Q_0 \approx 4083$ grams.

Question 17

An economist estimates that the gross national product (GNP) of a certain country is $G = G _ { 0 } e ^ { k t }$ , where $G _ { 0 }$ and k are positive constants. If the GNP is 110 billion in 1992 and 190 billion in 1994, when will it be 300 billion? Round your answer to whole years.

Accepted Answer

Given the GNP formula $G = G_0 e^{kt}$, we can use the provided GNP values for 1992 and 1994 to find $k$. For 1992 ($t=0$), $G_0 = 110$ billion. For 1994 ($t=2$), $G = 190$ billion. Substituting these into the formula gives $190 = 110e^{2k}$, solving for $k$ gives $k \approx 0.304$. To find when $G = 300$ billion, substitute $G = 300$ and solve for $t$, which gives $t \approx 4$ years from 1992.

Question 18

The gross national product (GNP) of a certain country was 150 billion dollars in 1988 and 160 billion in 1992. Assuming that the GNP grows exponentially, how much does it grow (in billions of dollars) between 1992 and the year 2000? Round your answer to two decimal places.

Accepted Answer

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Question 19

An efficiency expert hired by a manufacturing firm has compiled the following data relating workers' output to their experience: $Q ( t ) = 600 - A e ^ { - k t }$ Experience (months) 0 9 Output (units per hour) 200 580 The expert believes that the output Q is related to experience t by a function of the form. Find the function of this form that fits the data. Round numbers to four decimal places, if necessary.

Accepted Answer

Given the function $Q(t) = 600 - A e^{-kt}$, we need to find the values of $A$ and $k$ that fit the provided data points: at $t = 0$, $Q(t) = 200$, and at $t = 9$, $Q(t) = 580$.1. For $t = 0$, $Q(0) = 600 - A e^{0} = 200$. Solving for $A$, we find $A = 400$.2. For $t = 9$, $Q(9) = 600 - 400 e^{-k(9)} = 580$. Solving for $k$, we find $k \approx 0.3329$.Therefore, the correct function is $Q(t) = 600 - 400 e^{-0.3329t}$.

Question 20

An archaeologist has found a fossil in which the ratio of ${ } ^ { 14 } \mathrm { C }$ to ${ } ^ { 12 } \mathrm { C }$ is $\frac { 1 } { 9 }$ the ratio in the atmosphere. Approximately how old is the fossil? The half-life of ${ } ^ { 14 } \mathrm { C }$ is 5,730 years. Round to the nearest whole year.

Accepted Answer

The answer of An archaeologist has found a fossil in...

Quiz 4: Exponential and Logarithmic Functions

If an account that earns interest compounded continuously takes 9.59.59.5 years to double in value, how long will it take to triple? Round your answer to one decimal place, if necessary.

A sum of money, A0, is invested at a certain fixed interest rate, and this interest is compounded continuously. After 7 years, the money has doubled. The balance at the end of 14 years is 4A04 A _ { 0 }4A0​ .

A sum of money, A0A _ { 0 }A0​ , is invested at a certain fixed interest rate, and the interest is compounded continuously. After 7 years, the money has doubled. At the end of 14 years the value is 2A02 A _ { 0 }2A0​ .

A sum of money, A0A _ { 0 }A0​ , is invested at a certain fixed interest rate, and this interest is compounded continuously. After 4 years, the money has doubled. The balance at the end of 8 years is 4A04 A _ { 0 }4A0​ .

A sum of money, A0A _ { 0 }A0​ , is invested at a certain fixed interest rate, and this interest is compounded continuously. After 6 years, the money has doubled. The balance at the end of 12 years is 2A02 A _ { 0 }2A0​ .