many people prepare for retirement by making monthly contributions to a savings program. suppose that $2,500…

many people prepare for retirement by making monthly contributions to a savings program. suppose that $2,500 is set aside each year and invested in a savings account that pays 8% interest per year, compounded continuously.\na. determine the accumulated savings in this account at the end of 25 years.\nb. in part (a), suppose that an annuity will be withdrawn from savings that have been accumulated at the eoy 25. the annuity will extend from the eoy 26 to the eoy 33. what is the value of this annuity if the interest rate and compounding frequency in part (a) do not change?\nclick the icon to view the interest and annuity table for continuous compounding when i = 8% per year\na. the accumulated savings amount at the end of 25 years will be $ (round to the nearest dollar.)
Answer
Explanation:
Step1: Recall the formula for continuous - compounding of an annuity
The formula for the future value of a continuous - compounding annuity is $F = A\frac{e^{rt}-1}{e^{r}-1}$, where $A$ is the annual payment, $r$ is the annual interest rate, and $t$ is the number of years. Here, $A = 2500$, $r=0.08$, and $t = 25$.
Step2: Calculate the exponent values
First, calculate $e^{rt}$ and $e^{r}$. We know that $e^{rt}=e^{0.08\times25}=e^{2}$ and $e^{r}=e^{0.08}$. Using the approximation $e\approx2.71828$, we have $e^{2}\approx7.38906$ and $e^{0.08}\approx1.08329$.
Step3: Substitute values into the formula
$F = 2500\times\frac{e^{2}-1}{e^{0.08}-1}=2500\times\frac{7.38906 - 1}{1.08329-1}=2500\times\frac{6.38906}{0.08329}$.
Step4: Perform the calculation
$F = 2500\times76.719\approx191798$.
Answer:
$191798$
For part b:
Explanation:
Step1: Recall the present - worth formula for an annuity
The present - worth of an annuity $P$ with continuous compounding is related to the future - worth $F$ (from part a) by the formula for the present value of an annuity. The present value of an annuity $A$ for $n$ years with continuous compounding is $P = A\frac{1 - e^{-rn}}{e^{r}-1}$. We know the present value $P$ is the amount accumulated at the end of year 25 ($F$ from part a), $r = 0.08$, and $n=8$ (from EOY 26 to EOY 33). First, we need to re - arrange the formula to solve for $A$. We get $A = P\frac{(e^{r}-1)}{1 - e^{-rn}}$.
Step2: Calculate $e^{-rn}$
$e^{-rn}=e^{-0.08\times8}=e^{-0.64}\approx0.52729$.
Step3: Substitute values
We know $P = 191798$, $e^{r}\approx1.08329$, and $e^{-rn}\approx0.52729$. Then $A=191798\times\frac{1.08329 - 1}{1 - 0.52729}=191798\times\frac{0.08329}{0.47271}$.
Step4: Perform the calculation
$A=191798\times0.1762=33794$.
Answer:
$33794$