Question

-.161 The amount of a certain radioactive material (in grams) in a storage facility at time t is given by c(t) = 17 e where time is measured in years. (a) How much of the radioactive material was present initially? (b) What is the hall-life of the radioactive material? (Hint: For what value oft is c(t) = 8.5?) (a) Initially, there were grams of the radioactive material. (Round to two decimal places as needed.)

Answer 1

classmate Date Page (1) 1. -16t The amount of a centain radioactive material in a storage facility at time t is given by cut) = 17. e where time is measured in years (a) How much of the radioactive material was present initially? (6) what is the half-life of the radioactive material Solution (a) If we consider the initial stage the ime will be t=o. - 16t then from the given formulat alt=17ē we get O -16.0 c(t=o) = 17. e 17e 10 ct) = clo) = 17 . The amount of material was present initially is 17 grams the radioactive matemal is the time t when (6) Half life of s.s [ie, the amount the onginal initial quantity] of maknal becomes the half of Then CLt) = 8.5 -16t 17e 8.5 -16t e 8.5/17 -16€ e 42 taking natural logarithm on boleth sides we get inean ince 16t) = ln (42) In(a/b)= Ina-lnb In 2=0.693147 -16 t = In Cyz) t= -4/ In (42) 7 / ( 1n) -ln 2) + n2 = 0.04332 0.693147

Date Page Hence the half-life of the given radio active madenial is t=0.04332 Find the interest on the $1850 at 6.5 for 9 months. following loan. For a pnnciple amount & the Interest for a time pened to years rate Ry is. I= P.NR 100 Here P = $ 1850 N = 9 months = 3 2 years R = 6.5%. Then issox 3 x 6.5 다. 100 555 X 6.5 40 7215 80 90.1875 Hence the Interest on the amount $1850 is ~ $90.1875 As we round to the nearest cents it will be $90 3. Find the interest the following loan Assume 365 days in a year $ 35,300 at 15:25 y. for 161 days the nterest for a time penod of For a panaple amout P, years at a rate of R. PNR 100 given by Here P= $ 35, 300 = 161 days = 161 years 365 R: 15.25 7.

classmate Date Page (3) hen. Interest I a 35,300X 161 x 15.25 100 365 568 33 x 15.25 365 2374 529 2375 [Rounded to nearest cent] Hence the interest on the amount $35300 is, $ 2375 4. For the given corporate bond, whose annual simple interest rate is Provided find semi annual simple interest payment and the total nestest earned over the life of the bond. Assume 365 in a year. $5500 company A, 30-years bond, 6.059 y. (a) when interest is computed for semiannually it means that the is Sin months computing Demod gwen the pondple amount P = $5500 Rate of interest R = 6.059 4. Time duration 12 years 6 12. Hence the semiannual interest on the corporate bond is T = POR 100 I 85 500 X 1 x 6.059 2 X 100 a 166.6225 Z 167 CRounded to nearest cent) semi-inerest on the corporate bond is $167

classmate Date Page the bond is given (6 Total interest earned over the life of by POUR 100 where P= $5500 30 years R= 6.059 % T = 5500X BOX 6.059 9997.35 Hence the total interest canned over the tre of the bond is Is $ 9997. [Rounded to nearest Cent] Find the future value of this loan. $ 11,756 at 4.5% for 16 months Here the Procple amount P= $ 1756 and rate of interest R= 4.57. N = 16 months. - 16 4 years and time period future value fa Then the PCLRN) 100 fv = $ 11, 756 (1+ 4.4.5) 11756( 1 + 1813) 11756 (1+0.06) 11a56( 40.06) 11756 (1.06) 12461.36 Hence future value is $ 1246L [Rounded to nearest cent?

classmate Date Page Find the future value $9591 of the loan. Assume that Bbs days is a year 6.05% annual simple interest for 12 months at Future value of the loan. fv = P(1H AN Tous 9591 (lt 6.os GOSX) asal (1.8605) lol21. 255 Hence the future value of the loan is, fx = $10171 (Rounded to nearest cent Note: All the interests are calculated as simple interest as not menhoned any of the above need as specifically compound or Simple. If compound interest please comment-

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