Melanie is going to invest in an account paying an interest rate of 3.1% compounded
daily. How much would Melanie need to invest, to the nearest dollar, for the value of
the account to reach $11,900 in 14 years?


Answer :

Using compound interest, it is found that Melanie would need to invest a value of $7,710.

What is compound interest?

The amount of money earned, in compound interest, after t years, is given by:

[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, we have that:

  • Rate of 3.1%, hence [tex]r = 0.031[/tex].
  • Compounded daily, hence [tex]n = 365[/tex].
  • Value of $11,900 in 14 years, hence [tex]t = 14, A(t) = 11900[/tex].
  • The amount she needs to invest is P.

Hence:

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

[tex]11900 = P\left(1 + \frac{0.031}{365}\right)^{365(14)}[/tex]

[tex]1.54339P = 11900[/tex]

[tex]P = \frac{11900}{1.54339}[/tex]

[tex]P = 7710[/tex]

Melanie would need to invest a value of $7,710.

To learn more about compound interest, you can take a look at https://brainly.com/question/25781328