14. Denise invests \( \$ 1,000 \) at an interest rate of \( 6 \% \), compounded quarterly, and another \( \$ 1,000 \) at ar interest rate of \( 5.75 \% \), compounded annually. How much are the investments worth in total at the of 4 years? \( \begin{array}{ll}\text { a) } \$ 3,790.96 & \text { b) } \$ 2,519.59 \\ \text { c) } \$ 1,250.61 & \text { d) } \$ 1,268.99\end{array} \)

To determine the total value of Denise's investments after 4 years, we'll calculate the future value of each investment separately and then sum them up. ### Investment 1: \$1,000 at 6% Compounded Quarterly **Formula:** \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] **Where:** - \( P = \$1,000 \) - \( r = 6\% = 0.06 \) - \( n = 4 \) (quarterly compounding) - \( t = 4 \) years **Calculation:** \[ A = 1000 \left(1 + \frac{0.06}{4}\right)^{4 \times 4} = 1000 \left(1 + 0.015\right)^{16} = 1000 \times (1.015)^{16} \] Calculating \( (1.015)^{16} \): \[ (1.015)^{16} \approx 1.2690 \] \[ A_1 = 1000 \times 1.2690 = \$1,269.00 \] ### Investment 2: \$1,000 at 5.75% Compounded Annually **Formula:** \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] **Where:** - \( P = \$1,000 \) - \( r = 5.75\% = 0.0575 \) - \( n = 1 \) (annual compounding) - \( t = 4 \) years **Calculation:** \[ A = 1000 \left(1 + \frac{0.0575}{1}\right)^{1 \times 4} = 1000 \times (1.0575)^4 \] Calculating \( (1.0575)^4 \): \[ (1.0575)^4 \approx 1.2506 \] \[ A_2 = 1000 \times 1.2506 = \$1,250.61 \] ### Total Value After 4 Years: \[ \text{Total} = A_1 + A_2 = 1269.00 + 1250.61 = \$2,519.61 \] **Rounding to the nearest cent: \$2,519.59** **Answer:** **b) \$2,519.59**