Question

Matthew invested $5000 in an account that earns 3.8% interest, compounded annually.
The formula for compound interest is A = P(1 + r/n)^nt
How much did Matthew have in the account after 3 years?
$5591.93
$5594.37
$5570.00
$6900.00

Answer 1

To find out how much Matthew has in his account after 3 years with an investment of $5000, an interest rate of 3.8%, compounded annually, we can use the compound interest formula:

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

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 (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:
- \( P = 5000 \)
- \( r = 3.8\% = 0.038 \)
- \( n = 1 \) (since it is compounded annually)
- \( t = 3 \)

Now, substituting the values into the formula:

\[
A = 5000(1 + 0.038/1)^{1 \cdot 3}
\]

This simplifies to:

\[
A = 5000(1 + 0.038)^{3}
\]
\[
A = 5000(1.038)^{3}
\]

Now we calculate \( (1.038)^{3} \):

\[
(1.038)^{3} \approx 1.117649
\]

Now substituting it back:

\[
A \approx 5000 \times 1.117649 \approx 5588.25
\]

So, rounding to two decimal places:

\[
A \approx 5588.25
\]

Thus, after 3 years, Matthew had approximately \( 5588.25 \) in the account.

None of the provided options directly match \( 5588.25 \), but it seems likely that a calculation or rounding difference could account for this as it gets close to the choices provided. Among your options, the closest value is \( 5591.93. \)

So, the answer to your question is \( \text{\$5591.93} \).