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Sagot :
We can model this problem by an exponential growth:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]where A is the amount accumulated, P is the principal, r is the interest rate, n is the number of times per year and t is the time. By substituting our given data, we get
[tex]\begin{gathered} A=1000(1+\frac{0.036}{12})^{12t} \\ A=1000(1+0.003)^{12t} \end{gathered}[/tex]therefore, the model is
[tex]A=1000(1.003)^{12t}[/tex]Now, by substituting t=10 years, we have
[tex]A=1000(1.003)^{120}[/tex]then, the amount will be
[tex]A=1432.55\text{ dollars}[/tex]
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