Respuesta :
Let's use accumulation factor-compound formula:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Where:
A = Future value or savings
P = Principal or initial investment
n = Number of times interest is compounded per year
t = time
(a) If they begin the deposits at the end of each month when their child is a newborn, so that they have 18 years of deposits, how large must the deposit be?
In this case:
A = 250000
P = ?
n = 12 (because the interest is compounded monthly)
t = 18
r=9% = 0.09
[tex]\begin{gathered} 250000=P(1+\frac{0.09}{12})^{12\cdot18} \\ 250000=P(1.0075)^{216} \\ 250000=P(5.022637555) \\ \text{Solving for P:} \\ P=\frac{250000}{5.022637555}\approx49774.64 \end{gathered}[/tex]If they do not begin making deposits until their child is 10 years old, so that they have only 8 years of deposits, how large must the deposit be? Round your final answer to two decimal places.
In this case t changes from 18 to 8, therefore:
[tex]\begin{gathered} 250000=P(1+\frac{0.09}{12})^{12\cdot8} \\ 250000=P(1.0075)^{96} \\ 250000=P(2.048921228) \\ \text{Solving for P:} \\ P=\frac{250000}{2.048921228}\approx122015.43 \end{gathered}[/tex]