George wishes to have \( \$ 12,500 \) in 12 years. How much he should invest now at \( 12 \% \) interest, compounded quarterly in order to have \( \$ 12,500,12 \) years from now. select one: Oa. \( \$ 6,623.25 \) O.b. \( \$ 3,208.50 \) Oc. \( \$ 3,024.94 \) od. \( \$ 7,000.00 \)

To solve this problem, 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 of money). - \( r \) is the annual interest rate (in decimal form). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for, in years. Given: - \( A = \$12,500 \) - \( r = 12\% = 0.12 \) - \( n = 4 \) (compounded quarterly) - \( t = 12 \) years We need to find the principal amount \( P \) that George should invest now. Substitute the given values into the formula: \[ 12500 = P \left(1 + \frac{0.12}{4}\right)^{4 \times 12} \] Now, we can solve for \( P \) using the formula for compound interest. Solve the equation \( 12500 = P * (1 + 0.12/4)^{4*12} \). Solve the equation by following steps: - step0: Solve for \(P\): \(12500=P\left(1+\frac{0.12}{4}\right)^{4\times 12}\) - step1: Simplify: \(12500=\frac{103^{48}}{100^{48}}\times P\) - step2: Swap the sides: \(\frac{103^{48}}{100^{48}}\times P=12500\) - step3: Rewrite the expression: \(\frac{103^{48}P}{100^{48}}=12500\) - step4: Cross multiply: \(103^{48}P=100^{48}\times 12500\) - step5: Simplify the equation: \(103^{48}P=12500\times 100^{48}\) - step6: Divide both sides: \(\frac{103^{48}P}{103^{48}}=\frac{12500\times 100^{48}}{103^{48}}\) - step7: Divide the numbers: \(P=\frac{12500\times 100^{48}}{103^{48}}\) The amount George should invest now to have $12,500 in 12 years at 12% interest, compounded quarterly is: \[ P = \frac{12500 \times 100^{48}}{103^{48}} \] Let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(\frac{12500\times 100^{48}}{103^{48}}\) Therefore, George should invest approximately $3,024.99 now to have $12,500 in 12 years at 12% interest, compounded quarterly.