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Trisha Long wants to buy a boat in five years. She estimates the boat will cost $15,000 at that time. What must Trisha deposit today in an account earning 5% annually to have enough to buy the boat in five years?

Respuesta :

[tex]\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \qquad \begin{cases} A=\textit{compounded amount}\to &15000\\ P=\textit{original amount deposited}\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually means, once} \end{array}\to &1\\ t=years\to &5 \end{cases}[/tex]

solve for "P", to see how much Principal she should deposit today
TaeKwonDoIsDead

We use the formula for compound growth to figure this problem out. Formula is:

[tex]P=P_{0}(1+\frac{r}{n})^{nt}[/tex]

Where,

  • P is the future value
  • [tex]P_{0}[/tex] is the initial deposite
  • r is the rate of interest annually
  • n is the number of times compounding occurs (n=1 for annual compounding, n=2 for semiannual compounding etc.)
  • t is time

Given P=15,000, r=5%=0.05 (in decimal), n=1 (since annual compounding), and t=5 years, we can solve:

[tex]15000=P_{0}(1+\frac{0.05}{1} )^{(1)(5)}\\15000=P_{0}(1+0.05)^{5}\\P_{0}=\frac{15000}{(1+0.05)^{5}}\\P_{0}=\frac{15000}{1.05^{5}}\\P_{0}=11,752.89[/tex]

So, Trisha Long needs to deposit $11,752.89 today in the account.


ANSWER: $11,752.89

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