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bija089 [108]
2 years ago
5

A mother invests $7000 in a bank account at the time of her daughter's birth. The interest is 18) compounded quarterly at a rate

of 8%. What will be the value of the daughter's account on her twentieth birthday, assuming no other deposits or withdrawals are made during this period?
Mathematics
1 answer:
Alex2 years ago
5 0

well, from birth to your twentieth birthday that'll just be 20 years, so

~~~~~~ \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 &\$7000\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years\dotfill &20 \end{cases} \\\\\\ A=7000\left(1+\frac{0.08}{4}\right)^{4\cdot 20}\implies A=7000(1.02)^{80}\implies A\approx 34128.07

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Answer:

t=2.5\ years  

Step-by-step explanation:

we know that

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A=P(e)^{rt}  

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we have  

t=?\ years\\ P=\$12,500\\P=\$14,000\\ r=4.5\%=4.5/100=0.045  

substitute in the formula above   and solve for t

14,000=12,500(e)^{0.045t}  

Simplify

14,000/12,500=(e)^{0.045t}  

1.12=(e)^{0.045t}  

Apply ln both sides

ln(1.12)=ln[(e)^{0.045t}]  

ln(1.12)=(0.045t)ln(e)  

Remember that

ln(e)=1  

so

ln(1.12)=(0.045t)  

t=ln(1.12)/(0.045)  

t=2.5\ years  

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