Question

Faith invested $1,100 in an account paying an interest rate of 5.1% compounded continuously. Assuming no deposits or withdrawals are made, how much money, to the nearest ten dollars, would be in the account after 18 years?

Answer 1

To calculate the future value of an investment with continuously compounded interest, we can use the formula:

\[
A = Pe^{rt}
\]

where:
- \( A \) is the amount of money accumulated after time \( t \),
- \( P \) is the principal amount (the initial investment),
- \( r \) is the annual interest rate (in decimal),
- \( t \) is the time the money is invested for (in years),
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).

In this case:
- \( P = 1100 \)
- \( r = 5.1\% = 0.051 \)
- \( t = 18 \)

Now, substituting these values into the formula:

\[
A = 1100 \times e^{(0.051 \times 18)}
\]

First, calculate \( 0.051 \times 18 \):

\[
0.051 \times 18 = 0.918
\]

Now calculate \( e^{0.918} \):

\[
e^{0.918} \approx 2.499
\]

Now substitute this back into the formula to find \( A \):

\[
A \approx 1100 \times 2.499 \approx 2748.90
\]

Rounding this to the nearest ten dollars, we find that the amount in the account after 18 years is approximately:

\[
\boxed{2750}
\]