Answered

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5. Sabrina put $3,000 into a Money Market (high-yield savings)
account with an interest rate of 2.6% compounded quarterly.

a. Write an equation to model the amount in the account over
time.

b. Assuming no deposits or withdrawals are made, how much
money would be in the account after 15 years? Show your
calculation.

c. How would this problem be different if it was compounded
"continuously" instead? Calculate the amount after 15 years.

d. Name 2 or more other situations (other than a savings
account) where you might encounter compound interest later in
your life.

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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