How much do you need to invest in an account earning an annual interest rate of \( 4.057 \% \) compounded daily, so hat your money will grow to \( \$ 6,494.00 \) in 31 days? Principal: Note: Your answers should include a dollar sign and be accurate to two decimal places)

To determine the principal amount needed to invest, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time in years. Given: - \( A = 6494 \) - \( r = 4.057\% = 0.04057 \) - \( n = 365 \) (daily compounding) - \( t = \frac{31}{365} \) (31 days in years) Now we solve for \( P \): \[ 6494 = P \left(1 + \frac{0.04057}{365}\right)^{365 \times \frac{31}{365}} \] Calculating \( \left(1 + \frac{0.04057}{365}\right)^{31} : \) First, calculate \( \frac{0.04057}{365} \approx 0.0001112 \). Then, \( (1 + 0.0001112)^{31} \) gives approximately \( 1.00351 \). Now we solve for \( P \): \[ 6494 = P \cdot 1.00351 \] So: \[ P = \frac{6494}{1.00351} \approx 6475.57 \] Thus, the principal amount needed to invest is approximately **$6,475.57**.