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Joseph is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 
0.75%. If Joseph would like to end up with 
$259,000 after 13 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:

Joseph is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 0.75% 0.75 \% . If Joseph would like to end up with $259,000 \$ 259,000 after 1313 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. Joseph is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 0.75% 0.75 \% . If Joseph would like to end up with $259,000 \$ 259,000 after 1313 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.\newlineAA (future value of the account) = $259,000\$259,000\newlineii (monthly interest rate) = 0.75%0.75\% or 0.00750.0075 when converted to decimal\newlinenn (total number of periods in months) = 1313 years 12* 12 months/year = 156156 months\newlineWe will use the formula A=d((1+i)n1i)A=d\left(\frac{(1+i)^{n}-1}{i}\right) to find $259,000\$259,00000, the monthly contribution.
  2. Substitute Values into Formula: Substitute the given values into the formula.\newlineA=$259,000A = \$259,000\newlinei=0.0075i = 0.0075\newlinen=156n = 156\newlineNow, plug these values into the formula to solve for dd.
  3. Calculate Exponent: Calculate the value inside the parentheses and the exponent.\newlineFirst, calculate (1+i)n(1+i)^n which is (1+0.0075)156(1+0.0075)^{156}.
  4. Compute (1+0.0075)156(1+0.0075)^{156}: Use a calculator to compute (1+0.0075)156(1+0.0075)^{156}.(1+0.0075)1563.4449(1+0.0075)^{156} \approx 3.4449 (rounded to four decimal places for simplicity)
  5. Subtract 11: Subtract 11 from the result obtained in Step 44.\newline3.444912.44493.4449 - 1 \approx 2.4449
  6. Divide by ii: Divide the result from Step 55 by ii.2.44490.0075325.9867\frac{2.4449}{0.0075} \approx 325.9867
  7. Substitute into Formula: Substitute the result from Step 66 into the formula to solve for dd.
    $259,000=d×325.9867\$259,000 = d \times 325.9867
    Now, solve for dd by dividing both sides of the equation by 325.9867325.9867.
  8. Calculate Monthly Contribution: Calculate the monthly contribution dd. \newlined=$259,000325.9867$794.55d = \frac{\$259,000}{325.9867} \approx \$794.55\newlineSince the question asks for the nearest dollar, we round this to $795\$795.

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