Q:

How much would you invest today to have $9500 in 8 years if the effective annual rate of interest is 4%?Suppose that an investment of $5750 accumulates to $11533.20 at the end of 13 years, then the effective annual interest rate is i= ?At an effective annual rate of interest of 5.3%, the present value of $7425.70 due in t years is $3250. Determine t

Accepted Solution

A:
Answer:a) You should invest $6941.90 today.b) The effective annual interest rate is 11%.c) t is approximately 6.Step-by-step explanation:These are compound interest problems. The compound interest formula is given by: [tex]A = P(1 + \frac{r}{n})^{nt}[/tex]Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.a) How much would you invest today to have $9500 in 8 years if the effective annual rate of interest is 4%?Here, we want to find the value of P when [tex]A = 9500, t = 8, n = 1, r = 0.04[/tex].[tex]9500 = P(1 + \frac{0.04}{1})^{8}[/tex][tex]P = \frac{9500}{1.3685}[/tex][tex]P = 6941.90[/tex].You should invest $6941.90 today.b) Suppose that an investment of $5750 accumulates to $11533.20 at the end of 13 years, then the effective annual interest rate is i= ?Here, we have that [tex]A = 11533.20, P = 5750, t = 13, n = 1[/tex], and we want to find the value of i, that is r on the formula above the solutions.[tex]11533.20 = 5750(1 + r)^{13}[/tex][tex]\sqrt[13]{11533.20} = \sqrt[13]{5750(1 + r)^{13}}[/tex][tex]2.05 = 1.94(1 + r)[/tex][tex]r = 0.11[/tex]The effective annual interest rate is 11%.c) At an effective annual rate of interest of 5.3%, the present value of $7425.70 due in t years is $3250. Determine t.Here, we have that [tex]A = 7425.70 + 3250 = 10675.7, P = 7425.7, r = 0.053, n = 1[/tex] and we have to find t. So[tex]10675.7 = 7425.7(1 + \frac{0.053}{1})^{t}[/tex][tex](1 + 0.053)^{t} = \frac{10675.7}{7425.7}[/tex] [tex](1.0553)^{t} = 1.4377[/tex]We have that:log_{a}a^{n} = nSo[tex]log_{1.0553} (1.0553)^{t}  = log_{1.0553} 1.4377[/tex][tex]t = 5.95[/tex]t is approximately 6.