Question 116300
{{{A = P (1 + r/n)^(nt)}}}

{{{A}}} is the amount of returned

{{{P}}} is the principal amount deposited

{{{r}}} is the annual interest rate (expressed as a decimal)

{{{n}}} is the compound period

{{{t}}} is the number of years

a)

{{{n = 1}}} 

{{{P =  4000 }}}$

{{{r}}} = {{{7}}}%

{{{t = 8}}}


{{{A = P (1 + r/n)^(nt)}}}


{{{A = 4000 (1 + .07/1)^(1*8)}}}


{{{A =  4000 (1 + .07)^8}}}


{{{A = 4000 (1.07)^8}}}


{{{A =  4000 (1.72)}}}


{{{A = $ 4000 (1.72)}}}


{{{A = $ 6880}}}…….$


b)

given:

{{{n = 12}}} 

{{{P = 4000 }}}

{{{r}}} = {{{.07}}}

{{{t = 8}}}

{{{A = P (1 + r/n)^(nt)}}}


{{{A =  4000 (1 + .07/12)^(12*8)}}}


{{{A = 4000 (1 + .0058)^96}}}


{{{A = 4000 (1. 0058)^96}}}


{{{A = 4000 (1.74)}}}


{{{A = 4000 (1.74)}}}


{{{A =  6960}}}…….$

 

c)

compounding monthly yield more interest; the more frequently an account pays interest, the faster you can start earning interest on interest

Compounded annually:

{{{A = P (1 + r/n)^(nt)}}}


{{{A = 4000 (1 + .07/1)^1}}}


{{{A = 4000 (1.07)^1}}}


{{{A = 4000 (1.07)}}}


{{{A = 4280}}}	…….$

	
Compounded monthly: 

{{{A = P (1 + r/n)^(nt)}}}


{{{A = 4000 (1 + .07/12)^12}}}


{{{A = 4000 (1.07)^12}}}


{{{A = 4000 (2.52)}}}


{{{A = 10080}}}	…….$


d)


let

{{{P = 4000 }}}


{{{r}}} = {{{.07}}}


{{{t = 8}}}

answer:

{{{A=Pe^(rt)}}}


{{{A=4000(2.7183)^(.07*8)}}}


{{{A=4000(2.7183)^( 0.56)}}}


{{{A=4000(1.75)}}}


{{{A= 7000}}}… …….$


e)


let
{{{P =  4000 }}}

if {{{A= 7000}}}… …….$.....double is {{{2A= 14000}}}… 


{{{r}}} = {{{.07}}}


{{{t = t}}}

answer:


{{{A=Pe^(rt)}}}


{{{e^(rt) = A/P}}}


{{{(e)^(.07t) = 14000/4000}}}


{{{(e)^(.07t) = 14/4}}}


{{{(e)^(.07t) = 3.5}}}

	
{{{ln(e^.07t) = ln(3.5)}}}	


{{{.07t = 1.253}}}


{{{t = 1.253/.07}}}


{{{t = 17.90}}}..years