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Rashon is saving money and plans on making monthly contributions into an account earning an annual interest rate of 
3.9% compounded monthly. If Rashon would like to end up with 
$144,000 after 10 years, how much does he need to contribute to the account every month, to the nearest dollar? Use the following formula to determine your answer.

A=d(((1+i)^(n)-1)/(i))

A= the future value of the account after 
n periods

d= the amount invested at the end of each period

i= the interest rate per period

n= the number of periods
Answer:

Rashon is saving money and plans on making monthly contributions into an account earning an annual interest rate of 3.9% 3.9 \% compounded monthly. If Rashon would like to end up with $144,000 \$ 144,000 after 1010 years, how much does he need to contribute to the account every month, to the nearest dollar? Use the following formula to determine your answer.\newlineA=d((1+i)n1i) A=d\left(\frac{(1+i)^{n}-1}{i}\right) \newlineA= A= the future value of the account after n n periods\newlined= d= the amount invested at the end of each period\newlinei= i= the interest rate per period\newlinen= n= the number of periods\newlineAnswer:

Full solution

Q. Rashon is saving money and plans on making monthly contributions into an account earning an annual interest rate of 3.9% 3.9 \% compounded monthly. If Rashon would like to end up with $144,000 \$ 144,000 after 1010 years, how much does he need to contribute to the account every month, to the nearest dollar? Use the following formula to determine your answer.\newlineA=d((1+i)n1i) A=d\left(\frac{(1+i)^{n}-1}{i}\right) \newlineA= A= the future value of the account after n n periods\newlined= d= the amount invested at the end of each period\newlinei= i= the interest rate per period\newlinen= n= the number of periods\newlineAnswer:
  1. Identify Given Values: Identify the given values from the problem.\newlineA=$144,000A = \$144,000 (future value of the account)\newlineannual interest rate = 3.9%3.9\%\newlinen = 1010 years 12* 12 months/year = 120120 months (number of periods)\newlinei = (3.9%3.9\% annual interest rate) / 1212 months = 0.003250.00325 (interest rate per period)
  2. Plug into Formula: Plug the identified values into the compound interest formula.\newlineA=d×((1+i)n1)/iA = d \times \left(\left(1 + i\right)^{n} - 1\right) / i\newline$144,000=d×((1+0.00325)1201)/0.00325\$144,000 = d \times \left(\left(1 + 0.00325\right)^{120} - 1\right) / 0.00325
  3. Calculate Exponentiation: Calculate the value inside the parentheses (1+i)n(1 + i)^{n}.(1+0.00325)120=(1.00325)120(1 + 0.00325)^{120} = (1.00325)^{120}
  4. Substitute Calculated Value: Calculate the exponentiation from the previous step.\newline(1.00325)1201.457689(1.00325)^{120} \approx 1.457689
  5. Calculate Numerator: Substitute the calculated value back into the formula.\newline$144,000=d×(1.45768910.00325)\$144,000 = d \times \left(\frac{1.457689 - 1}{0.00325}\right)
  6. Divide by Interest Rate: Calculate the numerator of the fraction.\newline1.4576891=0.4576891.457689 - 1 = 0.457689
  7. Solve for d: Divide the numerator by the interest rate per period.\newline0.4576890.00325140.8264615\frac{0.457689}{0.00325} \approx 140.8264615
  8. Round Monthly Contribution: Solve for dd by dividing the future value AA by the result from the previous step.\newline$144,000/140.82646151022.55\$144,000 / 140.8264615 \approx 1022.55
  9. Round Monthly Contribution: Solve for dd by dividing the future value AA by the result from the previous step.\newline$144,000/140.82646151022.55\$144,000 / 140.8264615 \approx 1022.55Round the monthly contribution to the nearest dollar.\newlined$1023d \approx \$1023

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