Q:

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?

Accepted Solution

A:
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 decimalt is Number of Time Periods n is the number of times interest is compounded per yearPart 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]