Question

Find the palance in the account after the given period.
$5000 deposit earning 1.5% compounded quarterly after 3 years
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2024-0...48.40 A
Screen Shot
2024-0...40.01 A
3
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2024-0...40.30 A
$5,229.70
$7,777.27
$7,604.38
$5,538.86

Answer 1

To find the balance in the account after 3 years for a deposit of $5000 earning 1.5% interest compounded quarterly, we can use the compound interest formula:

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

where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial deposit or investment).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time the money is invested for in years.

In this case:
- \( P = 5000 \)
- \( r = 0.015 \) (1.5% as a decimal)
- \( n = 4 \) (quarterly)
- \( t = 3 \)

Plugging in these values:

\[
A = 5000 \left(1 + \frac{0.015}{4}\right)^{4 \times 3}
\]

Calculating the interest rate per compounding period:

\[
\frac{0.015}{4} = 0.00375
\]

Now substituting this into the formula:

\[
A = 5000 \left(1 + 0.00375\right)^{12}
\]

Calculating inside the parentheses:

\[
1 + 0.00375 = 1.00375
\]

Now, raise this to the power of 12:

\[
A = 5000 \left(1.00375^{12}\right)
\]

Calculating \(1.00375^{12}\):

\[
1.00375^{12} \approx 1.045682 (can vary slightly based on calculator precision)
\]

Now multiplying by the principal:

\[
A = 5000 \times 1.045682 \approx 5228.41
\]

So, the balance in the account after 3 years is approximately:

\[
\boxed{5228.41}
\]

Looking at the provided options, the closest answer to our calculation of approximately $5228.41 is **$5,229.70**.