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: $

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 thatThe 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 soPart A) Annualin 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) Semiannualin 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) Monthlyin 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) Dailyin 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]