Question
In 1851, a person sold a house to a lady for $24. If the lady had put the $24 into a bank account paying 6% interest, how much would the investment have been worth in the year 2011 if interest were compounded in the following ways?
a. monthly b. continuously
Question content area bottom
Part 1
a. If compounded monthly, the investment would be worth $
enter your response here in 2011.
(Round to the nearest dollar as needed.)
a. monthly b. continuously
Question content area bottom
Part 1
a. If compounded monthly, the investment would be worth $
enter your response here in 2011.
(Round to the nearest dollar as needed.)
Answers
GPT-4o mini
To calculate how much the investment would be worth in the year 2011 if the $24 were put into a bank account paying 6% interest compounded monthly, we can use the formula for compound interest:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \(A\) = the amount of money accumulated after n years, including interest.
- \(P\) = the principal amount (the initial amount of money, which is $24 in this case).
- \(r\) = the annual interest rate (decimal) (which is 6% or 0.06).
- \(n\) = the number of times that interest is compounded per year (which is 12 for monthly).
- \(t\) = the number of years the money is invested (from 1851 to 2011, which is 160 years).
Let’s plug in the values:
- \(P = 24\)
- \(r = 0.06\)
- \(n = 12\)
- \(t = 2011 - 1851 = 160\)
Now calculate:
\[
A = 24 \left(1 + \frac{0.06}{12}\right)^{12 \times 160}
\]
Calculating the inside of the parentheses:
\[
1 + \frac{0.06}{12} = 1 + 0.005 = 1.005
\]
Now calculating the exponent:
\[
12 \times 160 = 1920
\]
Now we can calculate \(A\):
\[
A = 24 \times (1.005)^{1920}
\]
First calculate \( (1.005)^{1920} \):
Using a calculator,
\[
(1.005)^{1920} \approx 7.313286
\]
Now calculate \(A\):
\[
A \approx 24 \times 7.313286 \approx 175.52
\]
Rounding to the nearest dollar, the investment would be worth approximately:
\[
\boxed{176}
\]
So, if compounded monthly, the investment would be worth **$176 in 2011**.
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \(A\) = the amount of money accumulated after n years, including interest.
- \(P\) = the principal amount (the initial amount of money, which is $24 in this case).
- \(r\) = the annual interest rate (decimal) (which is 6% or 0.06).
- \(n\) = the number of times that interest is compounded per year (which is 12 for monthly).
- \(t\) = the number of years the money is invested (from 1851 to 2011, which is 160 years).
Let’s plug in the values:
- \(P = 24\)
- \(r = 0.06\)
- \(n = 12\)
- \(t = 2011 - 1851 = 160\)
Now calculate:
\[
A = 24 \left(1 + \frac{0.06}{12}\right)^{12 \times 160}
\]
Calculating the inside of the parentheses:
\[
1 + \frac{0.06}{12} = 1 + 0.005 = 1.005
\]
Now calculating the exponent:
\[
12 \times 160 = 1920
\]
Now we can calculate \(A\):
\[
A = 24 \times (1.005)^{1920}
\]
First calculate \( (1.005)^{1920} \):
Using a calculator,
\[
(1.005)^{1920} \approx 7.313286
\]
Now calculate \(A\):
\[
A \approx 24 \times 7.313286 \approx 175.52
\]
Rounding to the nearest dollar, the investment would be worth approximately:
\[
\boxed{176}
\]
So, if compounded monthly, the investment would be worth **$176 in 2011**.