How much would [tex]$500 invested at 9% interest compounded annually be worth after 4 years? Round your answer to the nearest cent.

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

A. $[/tex]705.79
B. [tex]$716.66
C. $[/tex]680.00
D. $546.54



Answer :

To determine the future value of a [tex]$\$[/tex]500[tex]$ investment compounded annually at an annual interest rate of $[/tex]9\%[tex]$ over $[/tex]4[tex]$ years, we can use the compound interest formula: \[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( P \) is the principal amount ($[/tex]500).
- [tex]\( r \)[/tex] is the annual interest rate (9% or 0.09 as a decimal).
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year (annually, so [tex]\( n = 1 \)[/tex]).
- [tex]\( t \)[/tex] is the number of years the money is invested (4 years).

Let's plug in the values:

1. Substitute [tex]\( P = 500 \)[/tex], [tex]\( r = 0.09 \)[/tex], [tex]\( n = 1 \)[/tex], and [tex]\( t = 4 \)[/tex] into the formula:

[tex]\[ A(t) = 500 \left(1 + \frac{0.09}{1}\right)^{1 \times 4} \][/tex]

2. Simplify the equation inside the parentheses:

[tex]\[ A(t) = 500 \left(1 + 0.09\right)^4 \][/tex]

3. Add the values inside the parentheses:

[tex]\[ A(t) = 500 \left(1.09\right)^4 \][/tex]

4. Calculate the exponentiation:

[tex]\[ 1.09^4 \approx 1.41158 \][/tex]

5. Multiply the principal amount by this result:

[tex]\[ A(t) = 500 \times 1.41158 = 705.79 \][/tex]

6. Round to the nearest cent:

[tex]\[ A(t) \approx 705.79 \][/tex]

Thus, after 4 years, [tex]$\$[/tex]500[tex]$ invested at $[/tex]9\%[tex]$ interest compounded annually would be worth approximately $[/tex]\[tex]$705.79$[/tex].

The correct answer is:
A. [tex]$\$[/tex]705.79$