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 &\$3000\\ r=rate\to 5.2\%\to \frac{5.2}{100}\to &0.052\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\to &4\\ t=years\to &10 \end{cases} \\\\\\ A=3000\left(1+\frac{0.052}{4}\right)^{4\cdot 10}\implies A=3000(1.013)^{40} \\\\\\ A\approx 5029.201878874637[/tex]
bayleecase2002
Future Value = 3,000(1 +(0.052/4))^4*10

= 3,000(1.013)^40

= $5,029.20 after 10 years

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