stan has made a $125.30 monthly deposit into an account that pays 1.5% interest, compounded monthly, for 35…

Explanation:

Step1: Calculate the future - value of the annuity of deposits

The formula for the future - value of an ordinary annuity is $F = A\times\frac{(1 + r)^{n}-1}{r}$, where $A$ is the annuity payment, $r$ is the interest rate per period, and $n$ is the number of periods. The monthly deposit $A=$125.30$, the monthly interest rate $r=\frac{0.015}{12}=0.00125$, and the number of periods $n = 35\times12=420$ months. $F = 125.30\times\frac{(1 + 0.00125)^{420}-1}{0.00125}$ First, calculate $(1 + 0.00125)^{420}$. Let $x=(1 + 0.00125)^{420}$, then $\ln(x)=420\times\ln(1.00125)\approx420\times0.001249\approx0.5246$. So, $x = e^{0.5246}\approx1.6907$. $F = 125.30\times\frac{1.6907 - 1}{0.00125}=125.30\times\frac{0.6907}{0.00125}=125.30\times552.56=$69155.768$.

Step2: Calculate the monthly withdrawal amount

Now, we use the present - value of an ordinary annuity formula $P = W\times\frac{1-(1 + r)^{-m}}{r}$ to find the monthly withdrawal amount $W$, where $P$ is the present value (the amount in the account from step 1), $r$ is the monthly interest rate, and $m$ is the number of withdrawal periods. We know $P = 69155.768$, $r = 0.00125$, and $m=20\times12 = 240$ months. $69155.768=W\times\frac{1-(1 + 0.00125)^{-240}}{0.00125}$ First, calculate $(1 + 0.00125)^{-240}$. Let $y=(1 + 0.00125)^{-240}$, then $\ln(y)=- 240\times\ln(1.00125)\approx-240\times0.001249\approx - 0.29976$. So, $y = e^{-0.29976}\approx0.7413$. $1-(1 + 0.00125)^{-240}=1 - 0.7413 = 0.2587$. $W=\frac{69155.768\times0.00125}{0.2587}=\frac{86.44471}{0.2587}\approx333.71$.

Answer:

d. $$333.71$