Find the value of each of these investments at the end of the period specified below. (a) \( \$ 3500 \) invested at \( 5 \frac{2}{5} \% \) compounded annually for five years (b) \( \$ 6000 \) invested at \( 7 \% \) compounded semiannually for ten years (a) The amount in the account after five years is \( \$ \square \) (Round to the nearest cent as needed.)

To compute the values of these investments, we can use the formula for compound interest: \[ A = P (1 + r/n)^{nt} \] where: - \( A \) = the amount of money accumulated after n years, including interest. - \( P \) = principal amount (the initial amount of money). - \( r \) = annual interest rate (decimal). - \( n \) = number of times that interest is compounded per year. - \( t \) = number of years the money is invested or borrowed. Now, let's solve each investment. ### (a) For \( \$ 3500 \) invested at \( 5 \frac{2}{5} \% = 5.4\% = 0.054 \) compounded annually for five years: - \( P = 3500 \) - \( r = 0.054 \) - \( n = 1 \) (compounded annually) - \( t = 5 \) Using the formula: \[ A = 3500 \left(1 + \frac{0.054}{1}\right)^{1 \times 5} \] \[ A = 3500 \left(1 + 0.054\right)^{5} \] \[ A = 3500 \left(1.054\right)^{5} \] Calculating \( (1.054)^{5} \approx 1.295618 \): \[ A \approx 3500 \times 1.295618 \approx 4525.66 \] The amount in the account after five years is **\$4525.66**. ### (b) For \( \$ 6000 \) invested at \( 7 \% = 0.07 \) compounded semiannually for ten years: - \( P = 6000 \) - \( r = 0.07 \) - \( n = 2 \) (compounded semiannually) - \( t = 10 \) Using the formula: \[ A = 6000 \left(1 + \frac{0.07}{2}\right)^{2 \times 10} \] \[ A = 6000 \left(1 + 0.035\right)^{20} \] \[ A = 6000 \left(1.035\right)^{20} \] Calculating \( (1.035)^{20} \approx 1.806111 \): \[ A \approx 6000 \times 1.806111 \approx 10836.67 \] The amount in the account after ten years is **\$10836.67**.