If \( \$ 550 \) are deposited into an account with a \( 7.5 \% \) interest rate, compounded nonthly, what is the balance after 14 years? \[ F=\$[?] \] \( F=P\left(1+\frac{t}{n}\right)^{\mathrm{nt}} \)

To find the future balance of an investment compounded monthly, you can use the formula: \[ F = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( P = 550 \) - \( r = 0.075 \) (which is \( 7.5\% \)) - \( n = 12 \) (monthly compounding) - \( t = 14 \) Now, substituting these values into the formula: 1. Calculate \( \frac{r}{n} = \frac{0.075}{12} = 0.00625 \). 2. Calculate \( nt = 12 \times 14 = 168 \). 3. Finally, substitute and calculate: \[ F = 550 \left(1 + 0.00625\right)^{168} = 550 \left(1.00625\right)^{168} \] Using a calculator to compute \( (1.00625)^{168} \) gives approximately \( 2.833157 \). Now, calculate: \[ F \approx 550 \times 2.833157 \approx 1558.24 \] Thus, the balance after 14 years will be approximately: \[ F \approx \$1558.24 \] The balance after 14 years would be approximately \( \$1558.24 \).