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manuel deposits $10000 for 12 yr in an account paying 4% compounded annually.He then puts this total amount on deposit in another account paying 5% compounded semiannually for another 9yr. Find the total amount on deposit after the entire 14 yr period.

Respuesta :

[tex]\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$10000\\ r=rate\to 4\%\to \frac{4}{100}\to &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &12 \end{cases} \\\\\\ A=10000\left(1+\frac{0.04}{1}\right)^{1\cdot 12}\implies A=1000(1.04)^{12}\\\\\\ A\approx 16010.32[/tex]

he then turns around and grabs that money and sticks it for another 9 years,

[tex]\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} ~~ \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$16010.32\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semi-annually, thus twice} \end{array}\to &2\\ t=years\to &9 \end{cases} \\\\\\ A=16010.32\left(1+\frac{0.05}{2}\right)^{2\cdot 9}\implies A=16010.32(1.025)^{18} \\\\\\ A\approx 24970.64[/tex]

add both amounts, and that's how much is for the whole 21 years.

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