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 that    The compound interest formula is equal to  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) 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} \\A=47,400(1.07)^{5}\\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} \\A=47,400(1.035)^{10}\\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}\\A=47,400(1.0058)^{60}\\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}\\A=47,400(1.0002)^{1,825}\\A=\$67,261.54[/tex]