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A sum of $1000 was invested for 4 years, and the interest was compounded semiannually. If this sum amounted to $1459.54 in the given time, what was the interest rate? (Round your answer to two decimal places.)

Verified Answer :

\(A=P(1+ \frac{r}{n})^{nt}\)
A=furuter aount
P=present amount
r=rate in decimal
n=number of times per year it is compounded
t=time in years


A=1459.54
P=1000
r=r
n=2 (semianually is 2 times per year)
t=4

\(1459.54=1000(1+ \frac{r}{2})^{(2)(4)}\)
\(1459.54=1000(1+ \frac{r}{2})^{8}\)
divide both sides by 1000
\(1.45954=(1+ \frac{r}{2})^{8}\)
\(1.45954=(\frac{2+r}{2})^{8}\)
\(1.45954=\frac{(2+r)^8}{2^8}\)
\(1.45954=\frac{(2+r)^8}{256}\)
times both sides by 256
373.64224=\((2+r)^8\)
take the 8th root of both sides
\(\sqrt[8]{373.64224} =2+r\)
minus 2 from both sides
\(\sqrt[8]{373.64224} -2=r\)
aprox
0.0968=r
9.67%

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