Find the monthly payment R needed to have a sinking fund accumulate the future value A . The yearly interest rate r and the time t in years are given below. Interest is compounded monthly. \( \mathrm{A}=\$ 6,000 ; \mathrm{r}=8 \% ; \mathrm{t}=2 \) \( \mathrm{R}=\$ \square \) (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 series of payments (ordinary annuity) compounded at a certain interest rate. The formula is given by: \[ A = R \times \frac{(1 + i)^{nt} - 1}{i} \] Where: - \( A \) is the future value of the sinking fund. - \( R \) is the monthly payment. - \( i \) is the monthly interest rate (annual rate divided by 12). - \( n \) is the number of compounding periods per year (12 for monthly). - \( t \) is the number of years. ### Step 1: Extract the known values - \( A = 6000 \) - \( r = 8\% = 0.08 \) - \( t = 2 \) ### Step 2: Calculate the monthly interest rate \( i \) \[ i = \frac{r}{12} = \frac{0.08}{12} \] ### Step 3: Calculate the total number of payments \( nt \) \[ n = 12 \quad \text{(monthly payments)} \] \[ nt = 12 \times 2 = 24 \] ### Step 4: Substitute the values into the formula and solve for \( R \) Rearranging the formula to solve for \( R \): \[ R = \frac{A \cdot i}{(1 + i)^{nt} - 1} \] Now, we will calculate \( i \) and then substitute all values into the formula to find \( R \). Let's perform the calculations. 1. Calculate \( i \): \[ i = \frac{0.08}{12} \] 2. Substitute into the formula for \( R \): \[ R = \frac{6000 \cdot i}{(1 + i)^{24} - 1} \] Now, I will calculate these values. Calculate the value by following steps: - step0: Calculate: \(\frac{0.08}{12}\) - step1: Convert the expressions: \(\frac{\frac{2}{25}}{12}\) - step2: Multiply by the reciprocal: \(\frac{2}{25}\times \frac{1}{12}\) - step3: Reduce the numbers: \(\frac{1}{25}\times \frac{1}{6}\) - step4: Multiply the fractions: \(\frac{1}{25\times 6}\) - step5: Multiply: \(\frac{1}{150}\) Calculate or simplify the expression \( 6000*(0.08/12)/((1+(0.08/12))^24-1) \). Calculate the value by following steps: - step0: Calculate: \(\frac{6000\left(\frac{0.08}{12}\right)}{\left(\left(1+\left(\frac{0.08}{12}\right)\right)^{24}-1\right)}\) - step1: Remove the parentheses: \(\frac{6000\left(\frac{0.08}{12}\right)}{\left(1+\left(\frac{0.08}{12}\right)\right)^{24}-1}\) - step2: Divide the terms: \(\frac{6000\left(\frac{0.08}{12}\right)}{\left(1+\frac{1}{150}\right)^{24}-1}\) - step3: Add the numbers: \(\frac{6000\left(\frac{0.08}{12}\right)}{\left(\frac{151}{150}\right)^{24}-1}\) - step4: Divide the terms: \(\frac{6000\times \frac{1}{150}}{\left(\frac{151}{150}\right)^{24}-1}\) - step5: Multiply the numbers: \(\frac{40}{\left(\frac{151}{150}\right)^{24}-1}\) - step6: Subtract the numbers: \(\frac{40}{\frac{151^{24}-150^{24}}{150^{24}}}\) - step7: Multiply by the reciprocal: \(40\times \frac{150^{24}}{151^{24}-150^{24}}\) - step8: Multiply: \(\frac{40\times 150^{24}}{151^{24}-150^{24}}\) ### Step 1: Calculate the monthly interest rate \( i \) \[ i = \frac{0.08}{12} \approx 0.00666667 \] ### Step 2: Calculate the monthly payment \( R \) Using the rearranged formula: \[ R = \frac{6000 \cdot 0.00666667}{(1 + 0.00666667)^{24} - 1} \] Calculating the denominator: \[ (1 + 0.00666667)^{24} \approx 1.1616 \] Thus, \[ (1 + 0.00666667)^{24} - 1 \approx 1.1616 - 1 \approx 0.1616 \] Now substituting back into the formula for \( R \): \[ R \approx \frac{6000 \cdot 0.00666667}{0.1616} \approx \frac{40}{0.1616} \approx 247.00 \] ### Final Calculation Now, rounding up to the nearest cent, the monthly payment \( R \) is approximately: \[ R \approx 247.00 \] Thus, the monthly payment \( R \) needed to accumulate the future value \( A \) is: \[ \boxed{247.00} \]