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a. Suppose we had $15,192 cash and invested it in the bank at 16 percent interest, how much would you have at the end of 1, 2, 3, 4 years, assuming annual compounding?

Respuesta :

Answer:

Part a) [tex]\$17,622.72[/tex]

Part b) [tex]\$20,442.36[/tex]

Part c) [tex]\$23,713.13[/tex]

Part d) [tex]\$27,507.23[/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

Part a) How much would you have at the end of 1 year?

in this problem we have

[tex]t=1\ years\\ P=\$15,192\\ r=0.16\\n=1[/tex]

substitute in the formula above

[tex]A=15,192(1+\frac{0.16}{1})^{1*1}=\$17,622.72[/tex]

Part b) How much would you have at the end of 2 year?

in this problem we have

[tex]t=2\ years\\ P=\$15,192\\ r=0.16\\n=1[/tex]

substitute in the formula above

[tex]A=15,192(1+\frac{0.16}{1})^{1*2}=\$20,442.36[/tex]

Part c) How much would you have at the end of 3 year?

in this problem we have

[tex]t=3\ years\\ P=\$15,192\\ r=0.16\\n=1[/tex]

substitute in the formula above

[tex]A=15,192(1+\frac{0.16}{1})^{1*3}=\$23,713.13[/tex]

Part d) How much would you have at the end of 4 year?

in this problem we have

[tex]t=4\ years\\ P=\$15,192\\ r=0.16\\n=1[/tex]

substitute in the formula above

[tex]A=15,192(1+\frac{0.16}{1})^{1*4}=\$27,507.23[/tex]

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