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Sagot :
SOLUTION
We will apply the compound interest formula
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \text{Where } \\ A=\text{ amount aft}er\text{ 4 years }=\text{?} \\ P=\text{ principal, that is money invested = }6000\text{ dollars } \\ r=\text{ interest rate = }\frac{3.1}{100}=0.031 \\ n=n\text{umber of times compounding, that is quarterly = 4} \\ t\text{ = time in years = 4 years } \end{gathered}[/tex]Substituting the values into the equation we have
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ A=6000(1+\frac{0.031}{4})^{4\times4} \\ A=6000(1+0.00775)^{16} \\ A=6000(1.00775)^{16} \\ A=6000\times1.1314748 \\ A=6788.84916 \end{gathered}[/tex]Hence the answer is $6788.85
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