If \( \$ 500 \) are deposited into an account with a \( 4.5 \% \) interest rate, compounded annually, what is the balance after 6 years? \[ F=\$[?] \] \( F=P\left(1+\frac{r}{n}\right)^{n t} \) Round to the nearest cent.

To calculate the future balance using the formula \( F = P\left(1 + \frac{r}{n}\right)^{nt} \), where \( P = 500 \), \( r = 0.045 \) (4.5%), \( n = 1 \) (compounded annually), and \( t = 6 \): \[ F = 500\left(1 + \frac{0.045}{1}\right)^{1 \cdot 6} = 500\left(1 + 0.045\right)^{6} = 500\left(1.045\right)^{6} \] Calculating \( (1.045)^{6} \) gives approximately \( 1.28368 \). Now, multiplying this by \( 500 \): \[ F \approx 500 \cdot 1.28368 \approx 641.84 \] Thus, the balance after 6 years will be approximately \( F = \$641.84 \).