Question

Hw21 4.1: Problem 7 Previous Problem List Next (1 point) Get hele entering answer See a similar example (PDF) andA-Pe to solve the exercises below. Use the compound interest formulas A Round your answer to the nearest cent and make sure to include the dollar sign Find the accumulated value of an investment of $11070 for 7 years at an annual interest rate of 5.5% if the money is compounded seminannually: compounded monthly: compounded quarterly compounded continuously: Note: You can earn partial credit on this problem.

Answer 1

Solution :

Let

P- $11070

$\mathrm{t = 7\: years}$

$\mathrm{r\:=\:0.055\:or\:5.5\%\:annual\:interest\:rate}$

$\mathrm{A\:=\:future\:value\:\left(accumulated\:value\right)}$

$\mathrm{Now,}$

$\mathrm{For\:compounded\:semiannually\::}$

$\mathrm{n = 2}$

$\mathrm{A=P\left(1+\frac{r}{n}\right)^{nt}}=11070\left(1+\frac{0.055}{2}\right)^{2*7}$

$=11070\left(1+0.0275\right)^{14}$

$=11070*1.0275^{14}$

$=11070*1.46199$

$=16184.23$

$\mathrm{and}$

$\mathrm{For\:compounded\:monthly\::}$

$\mathrm{n = 12}$

$\mathrm{A=P\left(1+\frac{r}{n}\right)^{nt}}$

$=11070\left(1+\frac{0.055}{12}\right)^{12*7}$

$=11070\left(1+0.00458 \right)^{84}$

$=11070*(1.00458)^{84}$

$=11070*1.46791$

$=16249.76$

$\mathrm{and}$

$\mathrm{For\:compounded\:quaterly\::}$

$\mathrm{n = 4}$

$\mathrm{A=P\left(1+\frac{r}{n}\right)^{nt}}$

$=11070\left(1+\frac{0.055}{4}\right)^{4*7}$

11070 (1 0.01375)2

11070(1.01375)

$=11070*1.46576$

$=16225.96$

$\mathrm{and}$

$\mathrm{For\:compounded\:continuously\::}$

$\mathrm{A=Pe^{rt}}=11070e^{0.055*7}=11070e^{0.385}=11070*1.46961=16268.58$