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 \( \$ 3,800 \) ?

We start with the compound interest formula: \[ A = P\left(1+\frac{r}{m}\right)^{mt} \] where: - \( A \) is the final amount, - \( P \) is the principal, - \( r \) is the annual interest rate (in decimal form), - \( m \) is the number of compounding periods per year, - \( t \) is the time in years. For this problem, we have: \[ P = 2400,\quad A = 3800,\quad m = 4,\quad t = 13 \] Substitute these values into the formula: \[ 3800 = 2400\left(1+\frac{r}{4}\right)^{52} \] Divide both sides by \(2400\): \[ \frac{3800}{2400} = \left(1+\frac{r}{4}\right)^{52} \] Simplify the left side: \[ \frac{3800}{2400} = 1.58333\ldots \] Thus: \[ 1.58333\ldots = \left(1+\frac{r}{4}\right)^{52} \] Take the \(52^\text{nd}\) root of both sides: \[ 1+\frac{r}{4} = \left(1.58333\ldots\right)^{\frac{1}{52}} \] Now, solve for \( r \): \[ \frac{r}{4} = \left(1.58333\ldots\right)^{\frac{1}{52}} - 1 \] \[ r = 4\left[\left(1.58333\ldots\right)^{\frac{1}{52}} - 1\right] \] We now compute the value: 1. Compute the natural logarithm: \[ \ln(1.58333\ldots) \approx 0.45953 \] 2. Divide by 52: \[ \frac{0.45953}{52} \approx 0.00884 \] 3. Exponentiate: \[ \left(1.58333\ldots\right)^{\frac{1}{52}} = e^{0.00884} \approx 1.00888 \] 4. Subtract 1: \[ 1.00888 - 1 = 0.00888 \] 5. Multiply by 4: \[ r \approx 4(0.00888) = 0.03552 \] Expressing \(r\) as a percentage: \[ 0.03552 \times 100 \approx 3.55\% \] Thus, the required annual interest rate is approximately \(3.55\%\).