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Victoria is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 
0.475%. If Victoria would like to end up with 
$15,000 after 26 months, how much does she 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:

Victoria is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 0.475% 0.475 \% . If Victoria would like to end up with $15,000 \$ 15,000 after 2626 months, how much does she 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. Victoria is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 0.475% 0.475 \% . If Victoria would like to end up with $15,000 \$ 15,000 after 2626 months, how much does she need to contribute to the account every month, to the nearest dollar? Use the following formula to determine your answer.\newlineA=d((1+i)n1i) A=d\left(\frac{(1+i)^{n}-1}{i}\right) \newlineA= A= the future value of the account after n n periods\newlined= d= the amount invested at the end of each period\newlinei= i= the interest rate per period\newlinen= n= the number of periods\newlineAnswer:
  1. Identify variables: Identify the variables from the problem to use in the formula.\newlineFuture value of the account, A=$(15,000)A = \$(15,000)\newlineMonthly interest rate, i=0.475%i = 0.475\% or 0.004750.00475 (as a decimal)\newlineNumber of periods, n=26n = 26 months\newlineWe need to find the monthly contribution, dd.
  2. Plug values into formula: Plug the known values into the formula to solve for dd.A=d×((1+i)n1i)A = d \times \left(\frac{(1 + i)^{n} - 1}{i}\right)(\newline\)$15,000=d×((1+0.00475)2610.00475)\$15,000 = d \times \left(\frac{(1 + 0.00475)^{26} - 1}{0.00475}\right)
  3. Calculate compound factor: Calculate the compound factor (1+i)n(1 + i)^{n}.(1+0.00475)26=(1.00475)26(1 + 0.00475)^{26} = (1.00475)^{26}
  4. Evaluate compound factor: Evaluate the compound factor.\newline(1.00475)261.13047(1.00475)^{26} \approx 1.13047 (rounded to five decimal places for precision)
  5. Substitute compound factor: Substitute the compound factor back into the formula.\newline$15,000=d×(1.1304710.00475)\$15,000 = d \times \left(\frac{1.13047 - 1}{0.00475}\right)
  6. Calculate numerator: Calculate the numerator of the fraction. 1.130471=0.130471.13047 - 1 = 0.13047
  7. Divide numerator by rate: Divide the numerator by the interest rate to find the value of the fraction.\newline0.13047/0.0047527.467370.13047 / 0.00475 \approx 27.46737 (rounded to five decimal places for precision)
  8. Solve for d: Solve for d by dividing the future value of the account by the value of the fraction.\newline$15,000/27.46737546.039\$15,000 / 27.46737 \approx 546.039 (rounded to three decimal places for precision)
  9. Round monthly contribution: Round the monthly contribution to the nearest dollar. \newlined$546d \approx \$546 (to the nearest dollar)

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