Question An account is opened with an initial deposit of \( \$ 700 \) and earns \( 3.3 \% \) interest compounded semi-annually. What will the account be worth in 3 years? Round your answer to the nearest cent. Do NOT round until you have calculated the final answer.

To determine the future value of an account with compound interest, we can use the **compound interest formula**: \[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) = the amount of money accumulated after **n** years, including interest. - \( P \) = the principal investment amount (\$700 in this case). - \( r \) = the annual interest rate (decimal). - \( n \) = the number of times that interest is compounded per year. - \( t \) = the time the money is invested for in years. **Given:** - Principal (\( P \)) = \$700 - Annual interest rate (\( r \)) = 3.3% = 0.033 (as a decimal) - Compounded semi-annually, so \( n = 2 \) - Time (\( t \)) = 3 years **Plugging the values into the formula:** \[ A = 700 \times \left(1 + \frac{0.033}{2}\right)^{2 \times 3} \] \[ A = 700 \times \left(1 + 0.0165\right)^{6} \] \[ A = 700 \times \left(1.0165\right)^{6} \] **Calculating \( \left(1.0165\right)^{6} \):** 1.0165 raised to the 6th power can be calculated step-by-step: \[ 1.0165^2 = 1.0165 \times 1.0165 = 1.03327225 \] \[ 1.03327225 \times 1.0165 = 1.05031597 \] \[ 1.05031597 \times 1.0165 = 1.067596116 \] \[ 1.067596116 \times 1.0165 = 1.08523246 \] \[ 1.08523246 \times 1.0165 = 1.10308434 \] \[ 1.10308434 \times 1.0165 = 1.121229 \] So, \[ \left(1.0165\right)^{6} \approx 1.10308434 \] **Final Calculation:** \[ A = 700 \times 1.10308434 \approx 772.158038 \] **Rounding to the nearest cent:** \[ A \approx \$772.16 \] **Conclusion:** After **3 years**, the account will be worth approximately **\$772.16**.