Question
How much money will there be in an account at the end of 9 years if $17000 is deposited at 3% interest compounded semiannually question mark (Assume no withdrawals are made.)
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Question content area right
Part 1
The amount after 9 years will be $
enter your response here.
(Round to the nearest cent as needed.)
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.
.
Question content area right
Part 1
The amount after 9 years will be $
enter your response here.
(Round to the nearest cent as needed.)
Answers
GPT-4o mini
To calculate the future value of an investment with compound interest, we can use the formula:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \(A\) = the amount of money accumulated after n years, including interest.
- \(P\) = the principal amount (the initial amount of money).
- \(r\) = annual interest rate (decimal).
- \(n\) = number of times that interest is compounded per year.
- \(t\) = the number of years the money is invested for.
Given:
- \(P = 17000\)
- \(r = 0.03\) (3% as a decimal)
- \(n = 2\) (because interest is compounded semiannually)
- \(t = 9\)
Now plug the values into the formula:
\[
A = 17000 \left(1 + \frac{0.03}{2}\right)^{2 \cdot 9}
\]
Calculating step by step:
1. Calculate \(\frac{r}{n}\):
\[
\frac{0.03}{2} = 0.015
\]
2. Calculate \(nt\):
\[
2 \cdot 9 = 18
\]
3. Substitute these values back into the formula:
\[
A = 17000 \left(1 + 0.015\right)^{18}
\]
4. Calculate \(1 + 0.015\):
\[
1 + 0.015 = 1.015
\]
5. Now calculate \(A\):
\[
A = 17000 \cdot (1.015)^{18}
\]
First, calculate \((1.015)^{18}\):
\[
(1.015)^{18} \approx 1.346855007
\]
6. Now, multiply by the principal:
\[
A \approx 17000 \cdot 1.346855007 \approx 22995.53512
\]
7. Round to the nearest cent:
\[
A \approx 22995.54
\]
Thus, the amount after 9 years will be approximately **$22995.54**.
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \(A\) = the amount of money accumulated after n years, including interest.
- \(P\) = the principal amount (the initial amount of money).
- \(r\) = annual interest rate (decimal).
- \(n\) = number of times that interest is compounded per year.
- \(t\) = the number of years the money is invested for.
Given:
- \(P = 17000\)
- \(r = 0.03\) (3% as a decimal)
- \(n = 2\) (because interest is compounded semiannually)
- \(t = 9\)
Now plug the values into the formula:
\[
A = 17000 \left(1 + \frac{0.03}{2}\right)^{2 \cdot 9}
\]
Calculating step by step:
1. Calculate \(\frac{r}{n}\):
\[
\frac{0.03}{2} = 0.015
\]
2. Calculate \(nt\):
\[
2 \cdot 9 = 18
\]
3. Substitute these values back into the formula:
\[
A = 17000 \left(1 + 0.015\right)^{18}
\]
4. Calculate \(1 + 0.015\):
\[
1 + 0.015 = 1.015
\]
5. Now calculate \(A\):
\[
A = 17000 \cdot (1.015)^{18}
\]
First, calculate \((1.015)^{18}\):
\[
(1.015)^{18} \approx 1.346855007
\]
6. Now, multiply by the principal:
\[
A \approx 17000 \cdot 1.346855007 \approx 22995.53512
\]
7. Round to the nearest cent:
\[
A \approx 22995.54
\]
Thus, the amount after 9 years will be approximately **$22995.54**.