Question

Use the compound interest formula, A=P(1+rn)nt , to find the following amount of money after the given amount of time: Blake opens a savings account in a bank with an annual interest rate of 2.7%. If he deposits $3,500.00 in a savings account and the interest is compounded annually, how much will he have in the account after 3 years? (1 point) Responses $3,783.50 $3,783.50 $3,791.22 $3,791.22 $7,169.34 $7,169.34 $7,285.50

Answer 1

To solve the problem using the compound interest formula, we first need to clarify the variables in the given formula:

The formula for compound interest is:

\[
A = P(1 + r/n)^{nt}
\]

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 (in decimal).
- \( n \) is the number of times interest is compounded per year.
- \( t \) is the number of years the money is invested or borrowed.

In Blake's scenario:
- \( P = 3500 \)
- \( r = 2.7\% = 0.027 \)
- Since the interest is compounded annually, \( n = 1 \).
- \( t = 3 \) years.

Now we can plug these values into the formula:

\[
A = 3500(1 + \frac{0.027}{1})^{1 \times 3}
\]

Calculating inside the parentheses first:

\[
A = 3500(1 + 0.027)^{3}
\]
\[
A = 3500(1.027)^{3}
\]

Now calculate \( (1.027)^{3} \):

\[
(1.027)^{3} \approx 1.08364
\]

Next, multiply this by 3500:

\[
A \approx 3500 \times 1.08364 \approx 3791.74
\]

So, rounding to two decimal places:

\[
A \approx 3791.22
\]

Thus, after 3 years, Blake will have approximately **$3,791.22** in his account.

The correct response is **$3,791.22**.