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Morgan is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 0.75%. If Morgan would like to end up with $99,000 after 10 years, 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:

Morgan is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 0.75% 0.75 \% . If Morgan would like to end up with $99,000 \$ 99,000 after 1010 years, 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)n−1i) 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:

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Q. Morgan is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 0.75% 0.75 \% . If Morgan would like to end up with $99,000 \$ 99,000 after 1010 years, 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)n−1i) 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) = $99,000\$99,000\newlineii (monthly interest rate) = 0.75%0.75\% or 0.00750.0075 when converted to decimal\newlinenn (number of periods in 1010 years, with monthly contributions) = 1010 years ∗* 1212 months/year = 120120 months
  2. Plug into Formula: Plug the given values into the formula to solve for dd (the amount invested at the end of each period).A=d×((1+i)n−1i)A = d \times \left(\frac{(1 + i)^{n} - 1}{i}\right)(\newline\)$99,000=d×((1+0.0075)120−10.0075)\$99,000 = d \times \left(\frac{(1 + 0.0075)^{120} - 1}{0.0075}\right)
  3. Calculate (1+i)n(1 + i)^{n}: Calculate the value of (1+i)n(1 + i)^{n}.(1+0.0075)120=(1.0075)120(1 + 0.0075)^{120} = (1.0075)^{120}
  4. Calculate ((1+i)n−1)((1 + i)^{n} - 1): Calculate the value of ((1+i)n−1)((1 + i)^{n} - 1).
    (1.0075)120−1(1.0075)^{120} - 1
  5. Calculate Denominator: Calculate the denominator of the formula ii. 0.00750.0075
  6. Calculate Entire Right Side: Calculate the entire right side of the equation, which is the expression for dd.d=inlinelatex199,000((1.0075)120−1)/0.0075d = \frac{inline_latex_199,000}{\left(\left(1.0075\right)^{120} - 1\right) / 0.0075}
  7. Compute Values: Use a calculator to compute the values.\newlined=$99,000((1.0075)120−1)/0.0075d = \frac{\$99,000}{((1.0075)^{120} - 1) / 0.0075}\newlined≈$99,000(2.039887307…−1)/0.0075d \approx \frac{\$99,000}{(2.039887307\ldots - 1) / 0.0075}\newlined≈$99,0001.039887307…/0.0075d \approx \frac{\$99,000}{1.039887307\ldots / 0.0075}\newlined≈$99,000138.6516409…d \approx \frac{\$99,000}{138.6516409\ldots}\newlined≈$714.14d \approx \$714.14
  8. Round Monthly Contribution: Round the monthly contribution to the nearest dollar. d≈$(714)d \approx \$(714)

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