Answer:
Part a) [tex]\$17,958.56[/tex]
Part b) [tex]\$32,251.00[/tex]
Part c) [tex]\$57,918.16[/tex]
Step-by-step explanation:
The complete question is
If $10,000 is invested at an interest rate of 10% per year, compounded semiannually, find the value of the investment after the given number of years. (Round your answers to the nearest cent.)
a)6 years
b)12 years
c)18 years
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) 6 years
we have
[tex]t=6\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2[/tex]
substitute in the formula above
[tex]A=10,000(1+\frac{0.10}{2})^{2*6}[/tex]
[tex]A=10,000(1.05)^{12}[/tex]
[tex]A=\$17,958.56[/tex]
Part b) 12 years
we have
[tex]t=12\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2[/tex]
substitute in the formula above
[tex]A=10,000(1+\frac{0.10}{2})^{2*12}[/tex]
[tex]A=10,000(1.05)^{24}[/tex]
[tex]A=\$32,251.00[/tex]
Part c) 18 years
we have
[tex]t=18\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2[/tex]
substitute in the formula above
[tex]A=10,000(1+\frac{0.10}{2})^{2*18}[/tex]
[tex]A=10,000(1.05)^{36}[/tex]
[tex]A=\$57,918.16[/tex]
