Kyoko has \( \$ 15,000 \) that she wants to invest. Her bank has several accounts to choose from. Her goal is to have \( \$ 20,000 \) by the time she finishes graduate school in 7 years. To the nearest hundredth of a percent, what should her minimum annual interest rate be in order to reach her goal assuming they compound daily? (Hint: solve the compound interest formula for the intrerest rate. Also, assume there are 365 days in a year) \%

To find the minimum annual interest rate Kyoko needs, 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 (in this case, \( A = 20,000 \)), - \( P \) is the principal amount (the initial money, \( P = 15,000 \)), - \( r \) is the annual interest rate (in decimal), - \( n \) is the number of times that interest is compounded per unit year (for daily compounding, \( n = 365 \)), - \( t \) is the time the money is invested for (in years, \( t = 7 \)). Rearranging the formula to solve for \( r \): 1. Start with the formula: \[ 20,000 = 15,000 \left(1 + \frac{r}{365}\right)^{365 \times 7} \] 2. Divide both sides by 15,000: \[ \frac{20,000}{15,000} = \left(1 + \frac{r}{365}\right)^{2555} \] 3. This simplifies to: \[ \frac{4}{3} = \left(1 + \frac{r}{365}\right)^{2555} \] 4. Take the natural logarithm of both sides: \[ \ln\left(\frac{4}{3}\right) = 2555 \cdot \ln\left(1 + \frac{r}{365}\right) \] 5. Solve for \( \frac{r}{365} \): \[ \frac{r}{365} = e^{\frac{\ln\left(\frac{4}{3}\right)}{2555}} - 1 \] 6. Multiply through by 365 to solve for \( r \): 7. Converting it back to a percentage by multiplying by 100. Upon calculating \( r \), it results in approximately \( 0.037309 \) or \( 3.73\% \). To the nearest hundredth of a percent, Kyoko needs a minimum annual interest rate of approximately **3.73%**.