$16000 is invested at an APR of 3.5% compounded daily. Write a numerical expression that would compute the valueof the investment after 30 years.Preview

Answer:The value of the investment after 30 years is [tex]\$45,720.12[/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 in this problem we have  [tex]t=30\ years\\ P=\$16,000\\ r=3.5\%=3.5/100=0.035\\n=365[/tex]  substitute in the formula above  [tex]A=16,000(1+\frac{0.035}{365})^{365*30}[/tex]  [tex]A=16,000(1+\frac{0.035}{365})^{10,950}[/tex]  [tex]A=\$45,720.12[/tex]