Question
John has a bank account with \( \$ 7,362 \). He decides to invest the money at \( 4.85 \% \) interest, compounded annually. How much will the investment be worth after 9 years? Round to the nearest dollar. Type your numerical answer below.
Ask by Turnbull Wells. in the United States
Mar 14,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The investment will be worth \$11,273 after 9 years.
Solution
We start with the compound interest formula:
\[
A = P(1 + r)^n
\]
where:
- \( P = 7362 \) is the principal,
- \( r = 0.0485 \) is the annual interest rate (in decimal), and
- \( n = 9 \) is the number of years.
Substituting the values into the formula:
\[
A = 7362 \left(1 + 0.0485\right)^9 = 7362 \left(1.0485\right)^9
\]
Next, we calculate \( \left(1.0485\right)^9 \). Using a calculator, we find:
\[
\left(1.0485\right)^9 \approx 1.5313
\]
Now, multiplying by the principal:
\[
A \approx 7362 \times 1.5313 \approx 11273.43
\]
Rounding to the nearest dollar gives:
\[
A \approx 11273
\]
11273
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Simplify this solution Extra Insights
To find the future value of John's investment, we use the formula for compound interest: \[ A = P (1 + r)^n \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (\$7,362). - \( r \) is the annual interest rate (decimal) (4.85% = 0.0485). - \( n \) is the number of years the money is invested (9). Plugging in the values: \[ A = 7362 (1 + 0.0485)^9 \] Now calculating: \[ A = 7362 (1.0485)^9 \approx 7362 \times 1.487215 \approx 10901.03 \] Rounding to the nearest dollar, the investment will be worth approximately \( \$10,901 \).
