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Sagot :
well, keeping in mind that a year has 365 years, so let's see
[tex]\stackrel{Oct}{10}~~ + ~~\stackrel{Nov}{30}~~ + ~~\stackrel{Dec}{31}~~ + ~~\stackrel{Jan}{31}~~ + ~~\stackrel{Feb}{21}\implies 123~days[/tex]
so the 3600 were borrowed for only 123 days of a year, that'd be 123/365 of a year, thus
[tex]~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$3694.63\\ P=\textit{original amount deposited}\dotfill & \$3600\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &\frac{123}{365} \end{cases}[/tex]
[tex]3694.63=3600[1+(\frac{r}{100})(\frac{123}{365})]\implies \cfrac{3694.63}{3600}=1+\cfrac{123r}{36500} \\\\\\ \cfrac{3694.63}{3600}-1=\cfrac{123r}{36500}\implies \left( \cfrac{36500}{123} \right)\left( \cfrac{3694.63}{3600}-1 \right)=r\implies \stackrel{\%}{7.8}\approx r[/tex]
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