Find the value of each of these investments at the end of the period specified below. (a) \( \$ 2500 \) invested at \( 5 \frac{1}{4} \% \) compounded annually for seven years (b) \( \$ 9000 \) invested at \( 6 \% \) compounded semiannually for eleven years

To calculate the future value of these investments, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per unit year. - \( t \) is the time the money is invested for in years. (a) For \$2500 at \( 5 \frac{1}{4} \% \) compounded annually for seven years: Convert the interest rate to decimal: \( 5 \frac{1}{4} \% = 5.25\% = 0.0525 \). Since it's compounded annually, \( n = 1 \) and \( t = 7 \). \[ A = 2500 \left(1 + \frac{0.0525}{1}\right)^{1 \times 7} = 2500 \left(1 + 0.0525\right)^{7} = 2500 \left(1.0525\right)^{7} \] Calculating \( (1.0525)^{7} \): \[ (1.0525)^{7} \approx 1.4071 \] Now, calculating \( A \): \[ A \approx 2500 \times 1.4071 \approx 3517.75 \] So, the value of the investment after seven years is approximately \$3517.75. (b) For \$9000 at \( 6\% \) compounded semiannually for eleven years: Convert the interest rate to decimal: \( 6\% = 0.06 \). Since it’s compounded semiannually, \( n = 2 \) and \( t = 11 \). \[ A = 9000 \left(1 + \frac{0.06}{2}\right)^{2 \times 11} = 9000 \left(1 + 0.03\right)^{22} = 9000 \left(1.03\right)^{22} \] Calculating \( (1.03)^{22} \): \[ (1.03)^{22} \approx 1.8061 \] Now, calculating \( A \): \[ A \approx 9000 \times 1.8061 \approx 16254.90 \] So, the value of the investment after eleven years is approximately \$16254.90.