Aubrey invested \( \$ 150 \) in an account paying an interest rate of \( 6 \frac{5}{8} \% \) compounded monthly. Miguel invested \( \$ 150 \) in an account paying an interest rate of \( 6 \frac{1}{4} \% \) compounded daily. After years, how much more money would Aubrey have in her account than Miguel, to the nearest dollar?

To find out how much more money Aubrey has compared to Miguel after 10 years, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] **For Aubrey:** - Principal (P) = \$150 - Annual interest rate (r) = \(6 \frac{5}{8}\% = 6.625\% = 0.06625\) - Compounded monthly (n = 12) - Time (t) = 10 years \[ A_{\text{Aubrey}} = 150 \left(1 + \frac{0.06625}{12}\right)^{12 \times 10} \] \[ A_{\text{Aubrey}} = 150 \left(1 + 0.00552083\right)^{120} \] \[ A_{\text{Aubrey}} = 150 (1.00552083)^{120} \] \[ A_{\text{Aubrey}} \approx 150 \times 1.7137 \] \[ A_{\text{Aubrey}} \approx 257.06 \] **For Miguel:** - Principal (P) = \$150 - Annual interest rate (r) = \(6 \frac{1}{4}\% = 6.25\% = 0.0625\) - Compounded daily (n = 365) - Time (t) = 10 years \[ A_{\text{Miguel}} = 150 \left(1 + \frac{0.0625}{365}\right)^{365 \times 10} \] \[ A_{\text{Miguel}} = 150 \left(1 + 0.00017123\right)^{3650} \] \[ A_{\text{Miguel}} = 150 (1.00017123)^{3650} \] \[ A_{\text{Miguel}} \approx 150 \times 1.7203 \] \[ A_{\text{Miguel}} \approx 258.05 \] **Difference:** \[ \text{Difference} = A_{\text{Aubrey}} - A_{\text{Miguel}} \] \[ \text{Difference} \approx 257.06 - 258.05 \approx -0.99 \] Thus, after rounding to the nearest dollar, the difference is approximately **\$1.** **Answer:** After ten years, Aubrey has about \$1 less than Miguel, after rounding to the nearest dollar.