Answer:
(1) $11955.38
(2) $12228.62
(3) $12293.527
(4) $12326.6
Step-by-step explanation:
Compound interest is given by
A= P( 1 + r/n)^n*t
A= Final amount
P = initial amount = 6000
r = interest rate = 9% = 0.09
n = number of times interest applied per time period
t = number of time periods elapsed
(1) Compounded annually
n = 1 , t = 8
A = 6000( 1+ 0.09/1) ^1*8
A = 6000( 1.09) ^8 = $11955.38
(2) compounded quarterly
n = 4 , t = 8
A = 6000( 1+ 0.09/4) ^4*8
= 6000( 1.0225)^32 = $12228.62
(3) compounded monthly
n = 12 , t = 8
A = 6000( 1+0.09/12)^12*8 = $12293.527
(4) compounded continuously
A = P* e^rt
r = 0.09 , t = 8
= 6000* e^0.09*8
= 6000* e^0.72 = $12326.6
The Amount according to the given scenario will be:
(1) $11955.38
(2) $12228.62
(3) $12293.527
(4) $12326.6
The given values are:
Initial amount,
Interest rate,
(a)
If interest compounded annually then,
→ [tex]A = P(\frac{1+r}{n} )^n\times t[/tex]
[tex]= 6000(\frac{1+0.09}{1} )^1\times 8[/tex]
[tex]= 11955.38[/tex] ($)
(b)
If interest compounded quarterly then,
→ [tex]A = P(\frac{1+r}{n} )^n\times t[/tex]
[tex]= 6000(\frac{1+0.09}{4} )^4\times 8[/tex]
[tex]= 12228.62[/tex] ($)
(c)
If interest compounded monthly then,
→ [tex]A = P(\frac{1+r}{n} )^n\times t[/tex]
[tex]=P(\frac{1+0.09}{12} )^{12}\times 8[/tex]
[tex]= 12293.527[/tex] ($)
(d)
If interest compounded continuously then,
→ [tex]A = P\times e^{rt}[/tex]
[tex]= 6000\times e^{0.09\times 8}[/tex]
[tex]= 6000\times e^{0.72}[/tex]
[tex]= 12326.6[/tex] ($)
Thus the answers above are appropriate.
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