Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

#4. In this problem, you will nod to use the interest formulas stated in my less

ID: 3018613 • Letter: #

Question

#4. In this problem, you will nod to use the interest formulas stated in my lesson plans for Is on 3.2 (which are in the Helpful Stuff section on Canvas) and which you used in Lesson 3.2 in the Study Guide. In some parts, you may need to solve an equation in order to get the answer. Don't forget that you can solve epuations using the calculator (a.k.a., the Intersection of Graphs method). Also, in each problem, we assume that each investment is a "one time" investment, meaning money is only invested at the beginning. (a) Joe investol $2500 in an account with annual interest rate of 3.5% compounded monthly, what was the value of this investment after 7 years? Round your answer to the nearest cent. 4a (b) Sally invested some money in an account with annual interest rate of 7.4% compounded con- tinuously. After 10 years, the account contained $25, 000. How much money did Sally invest in the beginning? Round your answer to the nearest cent. 4b (c) Ten years ago, Morgan invested S13,000 in an account in which the interest was compounded quarterly. Now, ten years later, the account contains $20,000. What was the annual interest rate, expressed as a percentage? Round your answer to two decimal places. 4c

Explanation / Answer

(a)    P = 2500

         r = 3.5% = 0.035

         n = 12

         t = 7 years

      Now   A = P (1 + r/n)^(nt) = 2500*(1 + 0.035/12)^(12*7) = $3192.9

(b)    A = 25000

         r = 7.4% = 0.074

         t = 10 years

      Now   A = P*e^(rt) =======> P = A/ e^(rt) = 25000/e^(0.074*10) = $11927.8

(c) P = 13000

        A = 20000

         r = ?

         n = 4

         t = 10 years

      Now   A = P (1 + r/n)^(nt)

           20000 = 13000*(1 + r/4)^(4*10)

           (1 + r/4)^40 = 20/13

            1 + r/4 = 1.010828

            r = 0.0433 = 4.33%