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Charlotte is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 
0.7%. If Charlotte would like to end up with 
$13,000 after 35 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:

Charlotte is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 0.7% 0.7 \% . If Charlotte would like to end up with $13,000 \$ 13,000 after 3535 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. Charlotte is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 0.7% 0.7 \% . If Charlotte would like to end up with $13,000 \$ 13,000 after 3535 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 Given Values: Identify the given values from the problem.\newlineAA (future value of the account) = $13,000\$13,000\newlineii (interest rate per period) = 0.7%0.7\% per month, which is 0.0070.007 in decimal form\newlinenn (number of periods) = 3535 months\newlineWe need to find dd (the amount invested at the end of each period).
  2. Plug Values into Formula: Plug the given values into the formula to solve for dd. The formula is A=d×((1+i)n1)/iA = d \times \left(\left(1 + i\right)^{n} - 1\right) / i. We have A=$13,000A = \$13,000, i=0.007i = 0.007, and n=35n = 35.
  3. Calculate (1+i)n(1 + i)^n: Calculate the value of (1+i)n(1 + i)^n.(1+i)n=(1+0.007)35(1 + i)^n = (1 + 0.007)^{35}Use a calculator to find the value.(1+0.007)351.276(1 + 0.007)^{35} \approx 1.276
  4. Calculate Numerator: Calculate the numerator of the formula: ((1+i)n1)((1 + i)^{n} - 1). ((1+i)n1)=1.2761((1 + i)^{n} - 1) = 1.276 - 1 ((1+i)n1)0.276((1 + i)^{n} - 1) \approx 0.276
  5. Calculate Denominator: Calculate the denominator of the formula: i. The denominator is simply the interest rate per period in decimal form, which is 0.0070.007.
  6. Calculate Fraction: Calculate the entire fraction part of the formula: (((1+i)(n)1)/i)(((1 + i)^{(n)} - 1) / i). (((1 + i)^{(n)} - 1) / i) = rac{0.276}{0.007} (((1 + i)^{(n)} - 1) / i) \ hickapprox 39.429
  7. Calculate d: Calculate d using the formula A=d×((1+i)n1)/iA = d \times \left(\left(1 + i\right)^{n} - 1\right) / i.
    $13,000=d×39.429\$13,000 = d \times 39.429
    To find d, divide $13,000\$13,000 by 39.42939.429.
    d$13,000/39.429d \approx \$13,000 / 39.429
    d$329.49d \approx \$329.49
  8. Round Monthly Contribution: Round the monthly contribution to the nearest dollar. d$(329.49)d \approx \$(329.49) is approximately $(329)\$(329) when rounded to the nearest dollar.

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