Joe needs \( \$ 12,000 \) in 4 years. What amount can he deposit at the end of each month at \( 5 \% \) annual rate of interest compounded monthly, so he will have his \( \$ 12,000 \) ? Round to the nearest cent. A. \( \$ 226.35 \) B. \( \$ 314.78 \) C. \( \$ 325.16 \) D. \( \$ 287.62 \)

To find the monthly deposit Joe needs to make, we can use the future value of a series formula, which is: \[ FV = P \times \frac{(1 + r)^n - 1}{r} \] Where: - \( FV \) is the future value (\$12,000) - \( P \) is the monthly deposit - \( r \) is the monthly interest rate (annual rate / 12) - \( n \) is the total number of deposits (months) Given: - \( FV = 12000 \) - Annual interest rate = \( 5\% \) or \( 0.05 \) - Monthly interest rate \( r = \frac{0.05}{12} = 0.0041667 \) - Total months \( n = 4 \times 12 = 48 \) We need to solve for \( P \): Rearranging the formula gives: \[ P = \frac{FV \cdot r}{(1 + r)^n - 1} \] Substituting the values we have: \[ P = \frac{12000 \cdot 0.0041667}{(1 + 0.0041667)^{48} - 1} \] Calculating: 1. Calculate \( (1 + 0.0041667)^{48} \): \[ (1 + 0.0041667)^{48} \approx 1.221386 \] 2. Subtract 1: \[ 1.221386 - 1 \approx 0.221386 \] 3. Calculate the numerator: \[ 12000 \cdot 0.0041667 \approx 50 \] 4. Calculate \( P \): \[ P \approx \frac{50}{0.221386} \approx 226.35 \] Thus, the amount Joe needs to deposit at the end of each month is \( \boxed{\$226.35} \). Joe will indeed have exactly $12,000 in 4 years if he follows this plan!