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ricky bobby wants to buy a new automobile for $23000 in 4 years. How much money must ricky’s original investment be if he makes a single deposit into an account monthly compounding and annual interest rate of 3.70% in order to reach his goal ?

Sagot :

[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\$23000\\ P=\textit{original amount deposited}\\ r=rate\to 3.70\%\to \frac{3.70}{100}\dotfill &0.037\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &4 \end{cases}[/tex]

[tex]23000=P\left(1+\frac{0.037}{12}\right)^{12\cdot 4}\implies 23000=P\left( \frac{12037}{12000} \right)^{48}\implies \cfrac{23000}{~~ \left( \frac{12037}{12000} \right)^{48}~~}=P \\\\\\ \cfrac{23000}{~~ \frac{12037^{48}}{12000^{48}}~~}=P\implies \cfrac{23000\cdot 12000^{48}}{12037^{48}}=P\implies 19840.43\approx P[/tex]

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