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How much in dollars) is needed in an account that earns 9.6% compounded monthly to withdraw $1,000 at the end of each month for 15 years?
(Round your answer to two decimal places.)
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saakazplayz

Answer:

$4196.58

Step-by-step explanation:

Compound formula =

A = P(1+r)^n

P = Primary

R = Rate

N = Time

Amount = $1000(1+0.096/12)^(15*12)

=

$4196.58263959721

=

$4196.58

Using compound interest, it is found that $42,892.04 needs to be invested.

Compound interest:

[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]

  • A(t) is the amount of money after t years.  
  • P is the principal(the initial sum of money).  
  • r is the interest rate(as a decimal value).  
  • n is the number of times that interest is compounded per year.  
  • t is the time in years for which the money is invested or borrowed.

In this problem:

  • Interest of 9.6%, hence [tex]r = 0.096[/tex].
  • Compounded monthly, hence [tex]n = 12[/tex].
  • 15 years, hence [tex]t = 15[/tex].
  • $1,000 withdrawn monthly for these entire 15 years, hence [tex]A(t) = 1000(15)(12) = 180000[/tex]

To find the amount that need to be invested, we solve for P, hence:

[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]

[tex]180000 = P\left(1 + \frac{0.096}{12}\right)^{12(15)}[/tex]

[tex](1.008)^{180}P = 180000[/tex]

[tex]P = \frac{180000}{(1.008)^{180}}[/tex]

[tex]P = 42892.04[/tex]

$42,892.04 needs to be invested.

A similar problem is given at https://brainly.com/question/24507395

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