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azamat
3 years ago
12

After 6 months the simple interest earned annually on an investment of $6,000 was $463. Find the interest rate to the nearest te

nth of a percent.
Mathematics
1 answer:
daser333 [38]3 years ago
3 0
Keeping in mind that 6 months is not even a year, since there are 12 months in a year, then 6 months is 6/12, or 1/2 a year, thus

\bf ~~~~~~ \textit{Simple Interest Earned}
\\\\
I = Prt\qquad 
\begin{cases}
I=\textit{interest earned}\to &\$463\\
P=\textit{original amount deposited}\to& \$6000\\
r=rate\to r\%\to \frac{r}{100}\\
t=years\to &\frac{1}{2}
\end{cases}
\\\\\\
463=(6000)(r)\left( \frac{1}{2} \right)\implies \cfrac{463}{(6000)\left( \frac{1}{2} \right)}=r\implies 0.154\approx r
\\\\\\
r\%\approx 100\cdot 0.154\implies r\approx \stackrel{\%}{15.4}
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3 years ago
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olga55 [171]

Greetings.

The answer is (5,4)

Explanation:

Solving by elimination can be done by adding or subtracting of first and two equation.

Find any terms that make themselves = 0.

And that is the y-term.

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Therefore, we eliminate the y-term first.

\left \{ {{-x+2y=3} \atop {3x-2y=7}} \right. \\2x=10

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