Question 6, 8.4.19 HW Score: \( 38.6 \%, 22 \) of 57 points Points: 0 of 4 Save 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. \[ A=\$ 4,000 ; r=2.5 \% ; t=1 \] \[ R=S \] \( \square \) (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: \[ A = R \times \frac{(1 + i)^{nt} - 1}{i} \] 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 - \( A = 4000 \) - \( r = 2.5\% = 0.025 \) - \( t = 1 \) year ### Step 2: Calculate the monthly interest rate \( i \) \[ i = \frac{r}{12} = \frac{0.025}{12} \] ### Step 3: Calculate the number of compounding periods \( nt \) \[ n = 12 \quad \text{(monthly)} \] \[ nt = 12 \times 1 = 12 \] ### Step 4: Substitute the values into the formula and solve for \( R \) Rearranging the formula to solve for \( R \): \[ R = A \times \frac{i}{(1 + i)^{nt} - 1} \] Now, we will calculate \( i \) and then substitute all values into the formula to find \( R \). Let's perform the calculations. 1. Calculate \( i \): \[ i = \frac{0.025}{12} \] 2. Substitute into the formula for \( R \): \[ R = 4000 \times \frac{i}{(1 + i)^{12} - 1} \] Now, I will calculate these values. Calculate the value by following steps: - step0: Calculate: \(\frac{0.025}{12}\) - step1: Convert the expressions: \(\frac{\frac{1}{40}}{12}\) - step2: Multiply by the reciprocal: \(\frac{1}{40}\times \frac{1}{12}\) - step3: Multiply the fractions: \(\frac{1}{40\times 12}\) - step4: Multiply: \(\frac{1}{480}\) Calculate or simplify the expression \( 4000*(0.025/12)/((1+(0.025/12))^12-1) \). Calculate the value by following steps: - step0: Calculate: \(\frac{4000\left(\frac{0.025}{12}\right)}{\left(\left(1+\left(\frac{0.025}{12}\right)\right)^{12}-1\right)}\) - step1: Remove the parentheses: \(\frac{4000\left(\frac{0.025}{12}\right)}{\left(1+\left(\frac{0.025}{12}\right)\right)^{12}-1}\) - step2: Divide the terms: \(\frac{4000\left(\frac{0.025}{12}\right)}{\left(1+\frac{1}{480}\right)^{12}-1}\) - step3: Add the numbers: \(\frac{4000\left(\frac{0.025}{12}\right)}{\left(\frac{481}{480}\right)^{12}-1}\) - step4: Divide the terms: \(\frac{4000\times \frac{1}{480}}{\left(\frac{481}{480}\right)^{12}-1}\) - step5: Multiply the numbers: \(\frac{\frac{25}{3}}{\left(\frac{481}{480}\right)^{12}-1}\) - step6: Subtract the numbers: \(\frac{\frac{25}{3}}{\frac{481^{12}-480^{12}}{480^{12}}}\) - step7: Multiply by the reciprocal: \(\frac{25}{3}\times \frac{480^{12}}{481^{12}-480^{12}}\) - step8: Rewrite the expression: \(\frac{25}{3}\times \frac{3^{12}\times 160^{12}}{481^{12}-480^{12}}\) - step9: Reduce the numbers: \(25\times \frac{3^{11}\times 160^{12}}{481^{12}-480^{12}}\) - step10: Multiply: \(\frac{25\times 3^{11}\times 160^{12}}{481^{12}-480^{12}}\) ### Step 1: Calculate the monthly interest rate \( i \) The monthly interest rate \( i \) is calculated as follows: \[ i = \frac{0.025}{12} \approx 0.00208333 \] ### Step 2: Substitute into the formula for \( R \) Now, substituting \( i \) into the formula for \( R \): \[ R = 4000 \times \frac{0.00208333}{(1 + 0.00208333)^{12} - 1} \] Calculating the denominator: \[ (1 + 0.00208333)^{12} \approx 1.0254 \] Thus, \[ (1 + 0.00208333)^{12} - 1 \approx 0.0254 \] Now substituting back into the formula for \( R \): \[ R = 4000 \times \frac{0.00208333}{0.0254} \] ### Step 3: Calculate \( R \) Now, let's calculate \( R \). Calculate the value by following steps: - step0: Calculate: \(4000\left(\frac{0.00208333}{0.0254}\right)\) - step1: Divide the terms: \(4000\times \frac{208333}{2540000}\) - step2: Reduce the numbers: \(1\times \frac{208333}{635}\) - step3: Multiply: \(\frac{208333}{635}\) ### Final Calculation of \( R \) After performing the calculations, we find: \[ R \approx 328.083465 \] ### Step 4: Round up to the nearest cent Rounding \( R \) to the nearest cent gives: \[ R \approx 328.08 \] ### Conclusion The monthly payment \( R \) needed to accumulate the future value of \$4,000 in a sinking fund, with an interest rate of 2.5% compounded monthly over 1 year, is approximately: \[ \boxed{328.08} \]