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Sagot :
Okay, here we have this:
According with the provided information, we are going to replace in the following formula:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Replacing we obtain:
[tex]\begin{gathered} 3000=P(1+\frac{0.05}{4})^{4\cdot15} \\ \end{gathered}[/tex]Now, let's solve for P, that is the principal investment:
[tex]\begin{gathered} P\mleft(1+\frac{0.05}{4}\mright)^{4\cdot\: 15}=3000 \\ P\frac{4.05^{60}}{4^{60}}=3000 \\ P=3000\cdot\frac{4^{60}}{4.05^{60}} \\ P\approx1,423.70 \end{gathered}[/tex]Finally we obtain that the deposit wold be aproximately of $1,423.70.
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