Question 6, 8.4.19 HW Score: \( 38.6 \%, 22 \) of 57 points Points:0 of 4 Find the monthly payment \( R \) needed to have a sinking fund accumulate the future value \( A \). The yearly interest rate \( r \) and the time \( f \) in years are given below. Interest is compo monthly. \[ A=\$ 4,000 ; r=6.5 \%, t=4 \] \[ R=S \] \( \square \) (Round up to the nearest cent.)

To find the monthly payment \( R \) needed to accumulate a future value \( A \) in a sinking fund, we can use the formula for the future value of a sinking fund: \[ A = R \cdot \frac{(1 + i)^{nt} - 1}{i} \] Where: - \( A \) is the future value of the sinking fund. - \( R \) is the monthly payment. - \( i \) is the monthly interest rate (annual interest rate divided by 12). - \( n \) is the number of times interest is compounded per year (12 for monthly). - \( t \) is the number of years. ### Step 1: Extract known values - \( A = 4000 \) - \( r = 6.5\% = 0.065 \) - \( t = 4 \) ### Step 2: Calculate the monthly interest rate \( i \) \[ i = \frac{r}{12} = \frac{0.065}{12} \] ### Step 3: Calculate the total number of payments \( nt \) \[ n = 12 \quad \text{(monthly payments)} \] \[ nt = 12 \cdot 4 = 48 \] ### Step 4: Substitute values into the future value formula and solve for \( R \) Rearranging the formula to solve for \( R \): \[ R = A \cdot \frac{i}{(1 + i)^{nt} - 1} \] Now we will calculate \( i \) and then substitute all values into the formula to find \( R \). Let's perform the calculations. 1. Calculate \( i \): \[ i = \frac{0.065}{12} \] 2. Substitute \( i \) and other values into the formula for \( R \): \[ R = 4000 \cdot \frac{i}{(1 + i)^{48} - 1} \] Now, I will calculate these values. Calculate the value by following steps: - step0: Calculate: \(\frac{0.065}{12}\) - step1: Convert the expressions: \(\frac{\frac{13}{200}}{12}\) - step2: Multiply by the reciprocal: \(\frac{13}{200}\times \frac{1}{12}\) - step3: Multiply the fractions: \(\frac{13}{200\times 12}\) - step4: Multiply: \(\frac{13}{2400}\) Calculate or simplify the expression \( 4000*(0.065/12)/((1+(0.065/12))^48-1) \). Calculate the value by following steps: - step0: Calculate: \(\frac{4000\left(\frac{0.065}{12}\right)}{\left(\left(1+\left(\frac{0.065}{12}\right)\right)^{48}-1\right)}\) - step1: Remove the parentheses: \(\frac{4000\left(\frac{0.065}{12}\right)}{\left(1+\left(\frac{0.065}{12}\right)\right)^{48}-1}\) - step2: Divide the terms: \(\frac{4000\left(\frac{0.065}{12}\right)}{\left(1+\frac{13}{2400}\right)^{48}-1}\) - step3: Add the numbers: \(\frac{4000\left(\frac{0.065}{12}\right)}{\left(\frac{2413}{2400}\right)^{48}-1}\) - step4: Divide the terms: \(\frac{4000\times \frac{13}{2400}}{\left(\frac{2413}{2400}\right)^{48}-1}\) - step5: Multiply the numbers: \(\frac{\frac{65}{3}}{\left(\frac{2413}{2400}\right)^{48}-1}\) - step6: Subtract the numbers: \(\frac{\frac{65}{3}}{\frac{2413^{48}-2400^{48}}{2400^{48}}}\) - step7: Multiply by the reciprocal: \(\frac{65}{3}\times \frac{2400^{48}}{2413^{48}-2400^{48}}\) - step8: Rewrite the expression: \(\frac{65}{3}\times \frac{3^{48}\times 800^{48}}{2413^{48}-2400^{48}}\) - step9: Reduce the numbers: \(65\times \frac{3^{47}\times 800^{48}}{2413^{48}-2400^{48}}\) - step10: Multiply: \(\frac{65\times 3^{47}\times 800^{48}}{2413^{48}-2400^{48}}\) ### Step 1: Calculate the monthly interest rate \( i \) The monthly interest rate \( i \) is calculated as follows: \[ i = \frac{0.065}{12} \approx 0.00541667 \] ### Step 2: Calculate the monthly payment \( R \) Now, substituting \( i \) into the formula for \( R \): \[ R = 4000 \cdot \frac{0.00541667}{(1 + 0.00541667)^{48} - 1} \] Calculating the denominator: \[ (1 + 0.00541667)^{48} \approx 1.296 \] Thus, \[ (1 + 0.00541667)^{48} - 1 \approx 1.296 - 1 = 0.296 \] Now substituting back into the formula for \( R \): \[ R \approx 4000 \cdot \frac{0.00541667}{0.296} \] Calculating \( R \): \[ R \approx 4000 \cdot 0.0183 \approx 73.193145 \] ### Final Step: Round up to the nearest cent Rounding \( R \) to the nearest cent gives: \[ R \approx 73.19 \] Thus, the monthly payment \( R \) needed to accumulate the future value of \$4000 in 4 years at an interest rate of 6.5% compounded monthly is approximately **\$73.19**.