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Madeline deposits 
$5,800 every year into an account earning an annual interest rate of 
5.2% compounded annually. How much would she have in the account after 15 years, 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:

Madeline deposits $5,800 \$ 5,800 every year into an account earning an annual interest rate of 5.2% 5.2 \% compounded annually. How much would she have in the account after 1515 years, 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. Madeline deposits $5,800 \$ 5,800 every year into an account earning an annual interest rate of 5.2% 5.2 \% compounded annually. How much would she have in the account after 1515 years, 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.\newlineWe are given:\newlined=$5,800d = \$5,800 (the amount invested at the end of each period)\newlinei=5.2%i = 5.2\% or 0.0520.052 (the interest rate per period)\newlinen=15n = 15 (the number of periods)\newlineWe need to find AA, the future value of the account after nn periods.
  2. Convert interest rate: Convert the percentage interest rate to a decimal. i=5.2%=5.2100=0.052i = 5.2\% = \frac{5.2}{100} = 0.052
  3. Plug values into formula: Plug the values into the compound interest formula.\newlineA=d×((1+i)n1)/iA = d \times \left(\left(1 + i\right)^n - 1\right) / i\newlineA=$(5,800)×((1+0.052)151)/0.052A = \$(5,800) \times \left(\left(1 + 0.052\right)^{15} - 1\right) / 0.052
  4. Calculate compound factor: Calculate the compound factor (1+i)n(1 + i)^n.(1+i)n=(1+0.052)15(1 + i)^n = (1 + 0.052)^{15}(1+i)n1.05215(1 + i)^n \approx 1.052^{15}(1+i)n2.1133137(1 + i)^n \approx 2.1133137 (rounded to 77 decimal places for precision)
  5. Calculate numerator: Calculate the numerator of the formula: ((1+i)n1)((1 + i)^n - 1). ((1+i)n1)2.11331371((1 + i)^n - 1) \approx 2.1133137 - 1 ((1+i)n1)1.1133137((1 + i)^n - 1) \approx 1.1133137
  6. Divide by interest rate: Divide the result from Step 55 by the interest rate ii. \newline1.11331370.05221.4094952\frac{1.1133137}{0.052} \approx 21.4094952 (rounded to 77 decimal places for precision)
  7. Multiply by amount: Multiply the result from Step 66 by the amount invested at the end of each period dd.A$(5,800×21.4094952)A \approx \$(5,800 \times 21.4094952)A$124,174.8716A \approx \$124,174.8716
  8. Round future value: Round the future value of the account to the nearest dollar. A$124,175A \approx \$124,175

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