Question

5. John invested $4000 into a savings account that pays 1.75% interest compounded quarterly.

   1. (15 points) Write an exponential function that models the value of John’s account after t

      years have passed.

   2. (10 points) How much money will John have in the account after 54 months? Round to 2 decimal places and include units in your answer.

6. The half-life of cobalt-56 is 77.27 days.

   1. (15 points) If a sample contains 10 grams of cobalt-56, write an exponential function

      that models the amount of cobalt-56 left after t days have passed.

   2. (15 points) How long will it take for the sample of cobalt-56 to decay to 8 grams? Round to 2 decimal places and include units in your answer.

Answer 1

in one year het Po be the initial amount 8 is the date of interest n is the number of compoundings on to is the total number of years Then the amount accumulated in to years is PCt) = P. (1+x)

Since the compound he compounding us done quarterly, the number of compoundings ber year is n = 4 Yate of interest = 1.75% = 0.0175 Initial amount = $4000 The value of John's account after to years us P(+) = 4000 (1+ 0.0175 ) * * P(+) = $4000 (1.004375)** P(+) = $4000 (1.017615179) 2 54 months = 54 years = 4.5 years t = 4.5 years Amount in John's account after 54 months of 4.5 years is P(4,5) = $4000 (1.017615179) = $4326.99 $43211 4.5

I Equation modelling the dee decay of radioactive materials is Aft) = A, ēkt Whele Act) is the amount remaining after to days A is the initial amount k is decay late Given unitial amount of Cobalt – 56 is A = 10 grams Half life =77.27 days → 77.27 A (t) = 5 geams at 5= 10 (77.27) ► ok (17.272 2 K= 7 1 27 In (2) day! K = 8.9704 x 103 day! Amount of Cobalt - 56 after to days A ft) = lo ē 89704x178+ glame ③ we have to find it when A (t) = 8 geams → lo=8.9704x16+ = 8 8.9704x163 += In ( ) += 24.8755 days It will take [24.88 days applociamately