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Sagot :
let me answer the last one first
compound interest is just an exponential sequence really, where the next value is some exponential amount of the previous.
hmmm that can happen in say, a bouncing ball, the 1st bounce is high, let it keep on boucing by itself and the next bounce is usually a compounded value of the previous one, since it's smaller it'd be a Decay type of equation.
hmmm it also happens in say population growth, of any organism, humans, bees, amoebas.
now let's do the others
[tex]~~~~~~ \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}\dotfill &\$3000\\ r=rate\to 2.6\%\to \frac{2.6}{100}\dotfill &0.026\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years \end{cases} \\\\\\ A=3000\left(1+\frac{0.026}{4}\right)^{4\cdot t}\implies \boxed{A=3000(1.0065)^t} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{\textit{after 15 years}}{t=15}\hspace{5em} A=3000(1.0065)^{15}\implies A\approx 3306.19 \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$3000\\ r=rate\to 2.6\%\to \frac{2.6}{100}\dotfill &0.026\\ t=years\dotfill &15 \end{cases} \\\\\\ A=3000e^{0.026\cdot 15}\implies A=3000e^{0.39}\implies A\approx 4430.94[/tex]
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