Jeriel is going to invest \( \$ 2,400 \) and leave it in an account for 13 years. Assuming the interest is compounded quarterly, what interest rate, to the nearest hundredth of a percent, would be required in order for Jeriel to end up with \( \$ 3,800 \) ?

To solve this problem, we can use the compound interest formula: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] 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 (in decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the number of years the money is invested. In this instance, we have: - \( A = 3800 \) - \( P = 2400 \) - \( n = 4 \) (quarterly) - \( t = 13 \) We need to solve for \( r \). Rearranging the formula gives us: \[ 3800 = 2400 \left(1 + \frac{r}{4}\right)^{4 \times 13} \] We can simplify and isolate the term with \( r \): \[ \frac{3800}{2400} = \left(1 + \frac{r}{4}\right)^{52} \] Calculating the left side: \[ \frac{3800}{2400} \approx 1.5833 \] Now taking the 52nd root: \[ 1 + \frac{r}{4} = 1.5833^{\frac{1}{52}} \] Calculating \( 1.5833^{\frac{1}{52}} \) gives approximately: \[ 1 + \frac{r}{4} \approx 1.01026 \] Subtracting 1 and then multiplying by 4 to solve for \( r \): \[ \frac{r}{4} \approx 0.01026 \] \[ r \approx 0.04104 \] To convert it into a percentage, multiply by 100, giving us: \[ r \approx 4.10\% \] Thus, the interest rate required for Jeriel to end up with \( \$3,800 \) after 13 years, compounded quarterly, is approximately **4.10%**.