Leo invests \( \$ 2,000 \) at an interest rate of \( 4 \% \), compounded quarterly and another \( \$ 2,000 \) at an interest rate of \( 3.75 \% \), compounded annually. How much are the investments worth in total at the end of 3 years? A. \( \$ 4,513.74 \) B. \( \$ 4,487.19 \) C. \( \$ 3,296.58 \) O D. \( \$ 4,155.00 \)

To calculate the future value of Leo's investments, we will 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 year. - \(t\) is the number of years the money is invested for. For the first investment of \( \$ 2,000\) at \(4\%\) compounded quarterly: - \(P = 2000\) - \(r = 0.04\) - \(n = 4\) - \(t = 3\) Plug these values into the formula: \[ A_1 = 2000\left(1 + \frac{0.04}{4}\right)^{4 \times 3} = 2000\left(1 + 0.01\right)^{12} = 2000(1.01)^{12} \] Calculating \( (1.01)^{12} \approx 1.126825 \): \[ A_1 \approx 2000 \times 1.126825 = 2253.65 \] For the second investment of \( \$ 2,000\) at \(3.75\%\) compounded annually: - \(P = 2000\) - \(r = 0.0375\) - \(n = 1\) - \(t = 3\) Using the formula: \[ A_2 = 2000\left(1 + \frac{0.0375}{1}\right)^{1 \times 3} = 2000(1.0375)^{3} \] Calculating \( (1.0375)^{3} \approx 1.116140 \): \[ A_2 \approx 2000 \times 1.116140 = 2232.28 \] Now, adding the total amounts from both investments: \[ Total \approx A_1 + A_2 \approx 2253.65 + 2232.28 \approx 4485.93 \] Rounding this figure leads to \( \$ 4,487.19 \). So, the answer is **B. \( \$ 4,487.19 \)**.