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If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years for the following compounding methods, to the nearest cent.

(a) Annual: $
(b) Semiannual: $
(c) Monthly: $
(d) Daily: $

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Respuesta :

calculista

Answer:

Part A) Annual [tex]\$66,480.95[/tex]  

Part B) Semiannual [tex]\$66,862.38[/tex]  

Part C) Monthly [tex]\$67,195.44[/tex]  

Part D) Daily [tex]\$67,261.54[/tex]  

Step-by-step explanation:

we know that

The compound interest formula is equal to  

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

so

Part A) Annual

in this problem we have  

[tex]t=5\ years\\ P=\$47,400\\ r=0.07\\n=1[/tex]  

substitute in the formula above  

[tex]A=\$47,400(1+\frac{0.07}{1})^{1*5}[/tex]  

[tex]A=\$47,400(1.07)^{5}[/tex]  

[tex]A=\$66,480.95[/tex]  

Part B) Semiannual

in this problem we have  

[tex]t=5\ years\\ P=\$47,400\\ r=0.07\\n=2[/tex]  

substitute in the formula above  

[tex]A=\$47,400(1+\frac{0.07}{2})^{2*5}[/tex]  

[tex]A=\$47,400(1.035)^{10}[/tex]  

[tex]A=\$66,862.38[/tex]  

Part C) Monthly

in this problem we have  

[tex]t=5\ years\\ P=\$47,400\\ r=0.07\\n=12[/tex]  

substitute in the formula above  

[tex]A=\$47,400(1+\frac{0.07}{12})^{12*5}[/tex]  

[tex]A=\$47,400(1.0058)^{60}[/tex]  

[tex]A=\$67,195.44[/tex]  

Part D) Daily

in this problem we have  

[tex]t=5\ years\\ P=\$47,400\\ r=0.07\\n=365[/tex]  

substitute in the formula above  

[tex]A=\$47,400(1+\frac{0.07}{365})^{365*5}[/tex]  

[tex]A=\$47,400(1.0002)^{1,825}[/tex]  

[tex]A=\$67,261.54[/tex]  

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