Madison invested \( \$ 3,300 \) in an account paying an interest rate of \( 6.7 \% \) compounded daily. Assuming no deposits or withdrawals are made, how much money, to the nearest ten dollars, would be in the account after 5 years?

To calculate the amount in Madison's account after 5 years with daily compounding interest, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) is the amount in the account after time \( t \) - \( P \) is the principal amount (initial investment, \( \$ 3,300 \)) - \( r \) is the annual interest rate (as a decimal, \( 0.067 \)) - \( n \) is the number of times that interest is compounded per year (daily, so \( 365 \)) - \( t \) is the number of years the money is invested or borrowed (5 years) Plugging in the values: \[ A = 3300 \left(1 + \frac{0.067}{365}\right)^{365 \times 5} \] Calculating this step by step: 1. Calculate \( \frac{0.067}{365} \), which is approximately \( 0.0001835616438356 \). 2. Calculate \( 1 + 0.0001835616438356 \), which is approximately \( 1.0001835616438356 \). 3. Calculate \( 365 \times 5 = 1825 \). 4. Calculate \( A = 3300 \left(1.0001835616438356\right)^{1825} \). Using a calculator for the exponentiation, we find: \[ (1.0001835616438356)^{1825} \approx 1.375794679803 \] Finally, calculate \( A \): \[ A \approx 3300 \times 1.375794679803 \approx 4549.12 \] Rounding this to the nearest ten dollars gives us approximately \( \$ 4,550 \).