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)

We start with the compound‐interest formula \[ A = P\left(1+\frac{r}{n}\right)^{n t}, \] where \( A \) is the accumulated amount, \( P \) is the principal (initial investment), \( r \) is the annual interest rate (in decimal form), \( n \) is the number of compounding periods per year, and \( t \) is the time in years. In this problem, we have: - \( A = 6494.00 \) dollars, - \( r = 0.04057 \) (since \( 4.057\% = 0.04057 \)), - Daily compounding, so \( n = 365 \), - \( t = \frac{31}{365} \) years (since 31 days). Notice that \[ n t = 365 \times \frac{31}{365} = 31. \] Thus, the formula becomes \[ 6494.00 = P\left(1+\frac{0.04057}{365}\right)^{31}. \] We solve for \( P \) by rearranging the equation: \[ P = \frac{6494.00}{\left(1+\frac{0.04057}{365}\right)^{31}}. \] Now, calculating the daily interest factor: \[ 1+\frac{0.04057}{365} \approx 1 + 0.000111178 \] Raise this factor to the 31st power. Since for small \( x \), \( \ln(1+x) \approx x \), we have \[ \left(1+0.000111178\right)^{31} \approx e^{31 \times 0.000111178} = e^{0.0034465} \approx 1.003451. \] Now, substitute back to find \( P \): \[ P \approx \frac{6494.00}{1.003451} \approx 6468.07. \] Thus, you would need to invest approximately \(\$6,468.07\).