If 20600 dollars is invested at an interest rate of 10 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:a) The value of the investment is $33176.506b) The value of the investment is $33555.53c) The value of the investment is $33893.36d) The value of the investment is $33961.33Step-by-step explanation:This is a compound interest problemCompound interest formula:The compound interest formula is given by:[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]A: Amount of money(Balance)P: Principal(Initial deposit)r: interest rate(as a decimal value)n: number of times that interest is compounded per unit t t: time the money is invested or borrowed for.In our problem, we have:A: the value we want to findP = 20600(the value invested)r = 0.1n: Will change for each lettert = 5.a) If the interest is compounded anually, n = 1. So.[tex]A = 20600(1 + \frac{0.1}{1})^{1*5}[/tex][tex]A = 33176.506[/tex]The value of the investment is $33176.506b) If the interest is compounded semianually, it happens twice a year, which means n = 2. So:[tex]A = 20600(1 + \frac{0.1}{2})^{2*5}[/tex][tex]A = 33555.23[/tex]The value of the investment is $33555.53c) If the interest is compounded monthly, it happens 12 times a year, which means n = 12. So:[tex]A = 20600(1 + \frac{0.1}{12})^{12*5}[/tex][tex]A = 33893.36[/tex]The value of the investment is $33893.36d) If the interest is compounded monthly, it happens 365 times a year, which means n = 365. So:[tex]A = 20600(1 + \frac{0.1}{365})^{365*5}[/tex][tex]A = 33961.33[/tex]The value of the investment is $33961.33