ࡱ> WYVM Nbjbj== "WWJlTTTTTTTh####\L$h>$%*N)N)N)N)N)N)N)>>>>>>>$T? tAC>TN)N)N)N)N)C>@.TTN)N)X>@.@.@.N)"TN)TN)>@.N)>@.@.&69TT:N)% :CDh!#p)~::Dn>0>:4B-RB:@.hhTTTTModule 8 Notes and Solutions You need to be able to use specific formulas to solve the problems in this module. The formulas have all been given to you in other documents that are posted on the module 8 website and the problems have been divided by type so identifying the needed formula shouldnt be difficult. It is essential that you learn how to use your calculator. There are notes provided that will walk you through solutions on the calculator but each brand of calculator has its quirks. Each set of practice problems (except set 3) are divided into 2 parts. The first part is a table. You only have to take those numbers and put them into the correct formula and use your calculator to get the answer. The second part has several word problems. You have to find the information in the problem, decide which piece of information is missing, use the correct formula to find the missing piece and then answer the question. These problems are all set up in this document and a few of them are worked through the solution. Set 1 Practice Problems for Module 8, Unit 1 Simple Interest Formula T = P(1 + I) = P + PI 1) How much money will Joe have at the end of a year if he invests $5,000 at 6.75% simple interest? P = 5000 I = .0675 T = 5000(1 + .0675)2) How much money will Therese have at the end of a year if she invests $12,500 at 7.25% simple interest? P = 12500 I = .0725 T = 12500(1 + .0725)3) How much money will $8,500 earn if it is invested at 7.15% simple interest? This problem only asks for the amount of interest that the investment will earn. It only asks for PI. P = 8500 I = .0715 Interest = PI = (8500)(.0715)4) How much money will $4750 earn if it is invested at 6.85% simple interest? P = 4750 I = .0685 Interest = PI = (4750)(.0685)5) A man invested $5,000 in two plans, part at 6% and the rest at 8% simple interest. The total income the first year is $330. How much did he invest in each plan? This problem divides an investment sum into two plans. He invests x amount at 6% and 5000 x is invested at 8%. The income is $330. Therefore, we have the equation: .06x + .08(5000 x) = 330. Solve and answer the question.6) Connie invested $720 partly at 3.5% and partly at 5%. After a year, both investments together yield the same income as if the whole were invested at 4.5%. How much did she invest in each rate? The first part is set up the same way as #5 but the yearly income is (.045)(720). Therefore, we have the equation: .035x + .05(720 x) = (.045)(720) Solve and answer the question. 7) Jennifer invested $2,000 more in plan A at 5% than plan B at 9% simple interest. The total income the first year is $1,070. How much did she invest in each plan? (By now, you might be asking why would someone invest so much more at the lower interest rate.) 2000 more means x + 2000 so we have the equation: .05(x + 2000) + .09x = 1070 Solve and answer the question.8) John invested $3,000 at 7% and $4,000 at 8% simple interest that is calculated once a year. What additional amount must he invest at 7% to give him a total income of $600 after one year? We have .07(3000) + .08(4000) = 530 but we want 600 income. The problem wants us to add x amount to the 3000 invested at 8%. Therefore, we have the equation: .07(3000 + x) + .08(4000) = 600 Solve and answer the question.9) How much money does Rebecca need to invest if she wants to earn $1,200 yearly income off an 8.5% simple interest investment? Were only looking at interest earned so Income = PI. Thus, 1200 = P(.085)How much money does Jim need to invest if he wants to earn $800 yearly income off a 6.25% simple interest investment? Were only looking at interest earned so Income = PI. Thus, 800 = P(.0625)11) Laura has $132,000 to invest. She wants to earn $20,000 a year from this investment. What simple interest rate would she need to invest at to earn this yearly income? Were only looking at interest earned so Income = PI. Thus, 20000 = 132000(I)12) Kristina has $150,000 to invest. She wants to live off the interest of this investment and needs to earn $12,500 a year to live comfortably. What simple interest rate would she need to invest at to reach this yearly income? Were only looking at interest earned so Income = PI. Thus, 12500 = 150000(I)13) How much money needs to be invested at 5.5% simple interest to make $6,000 yearly income? Were only looking at interest earned so Income = PI. Thus, 6000 = P(.055)14) Howard wants to earn $25,000 interest on a $325,000 investment. What simple interest rate would earn this income? Were only looking at interest earned so Income = PI. Thus, 25000 = 325000(I)15) DePaul University wants to establish an endowment fund. The interest from this fund will provide scholarship money for adult students admitted to the School for New Learning. The university wants to provide fifteen $5000 annual scholarships and has $1 million to start the fund. What simple interest rate will earn enough money to provide the funds for these scholarships? 15(5000) = 75000 = Income. Were only looking at interest earned so Income = PI. Thus, 75000 = 1000000(I) Set 2 Practice Problems for Module 8, Unit 2 Compound Interest Formula  EMBED Equation.3  1) How much money will a 5 year CD be worth if $1,500 is invested and it earns 6.5% interest that compounds daily? Y = 5, P = 1500, I = .065 and n = 365.  EMBED Equation.3 2) Jane invests $2800 in a 3 year CD that earns 5.8% interest which compounds daily. How much money will she have at the end of three years? P = 2800, Y = 3, I = .058 and n = 365.  EMBED Equation.3 3) Jackie invests $4,800 in a CD that earns 6.5% interest that compounds daily. What is the value of the CD after 4 years? after 6 years? after 8 years? P = 4800, I = .065, n = .065 and Y = 4, 6 & 8.  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3 4) Tom invests $3,500 in an account that earns 5.75% interest that compounds daily and $3,500 in an account that earns 6.25% interest that compounds every 2 weeks. Which account will have more money in 5 years? Investment 1: P = 3500, I = .0575, n = 365 and Y = 5.  EMBED Equation.3  Investment 2: P = 3500, I = .0625, n = 26 and Y = 5.  EMBED Equation.3  Compare Ts and answer the question.5) Jennifer invests $5,000 in a fixed income IRA that is guaranteed to earn 7.85% interest that compounds monthly. She also invests $5,000 in another fixed income IRA that earns 6.5% interest that compounds daily. She is 35 years old now and plans to retire when she is 63. What will each IRA be worth when she retires? Investment 1: P = 5000, I = .0785, n = 12 and Y = 63 35 = 28.  EMBED Equation.3  Investment 2: P = 5000, I = .065, n = 365 & Y = 28.  EMBED Equation.3  Solve and answer the question. 6) Which of the following interest rates will improve a $1,500 investment the most over a 10 year period? a) 6.5% interest compounded daily. b) 6.9% interest compounded weekly. c) 7.5% interest compounded every two weeks(biweekly). d) 8% interest compounded monthly. a) P = 1500, Y = 10, I = .065 and n = 365.  EMBED Equation.3  b) P = 1500, Y = 10, I = .069 and n = 52.  EMBED Equation.3  c) P = 1500, Y = 10, I = .075 and n = 26.  EMBED Equation.3  d) P = 1500, Y = 10, I = .08 and n = 12.  EMBED Equation.3 Sue has $1,000 to invest. She wants it to earn $1,000 in 12 years. What interest rate would she have to invest at if the interest compounded daily? P = 1000, Y = 12 and n = 365. She wants to earn $1000 in 12 years so T = 2000. I is missing so we use the formula:  EMBED Equation.3  She needs to find a 5.78% interest rate.Henry has an 8 year old daughter and has $10,000 to invest so she can go to college when she is 18. He wants the money to double by the time she is ready to go. What interest rate does he need to invest at if the interest compounds daily? P = 10000, Y = 18 8 = 10, n = 365 and T = 2(10000) = 20000. I is missing. Thus,  EMBED Equation.3 Jay has $5,000 to invest. He wants to buy a really nice computer in 5 years with the income from this investment and hell need about $2,000. What interest rate does he need to invest at if the interest compounds daily? P = 5000, Y = 5, n = 365 and T = 5000 + 2000 = 7000. I is missing.  EMBED Equation.3 Elena invests money in two accounts. She invests $1,000 at 6.75% that compounds daily. She invests $1,500 at 7.5% that compounds weekly. How much does she have in each account at the end of 10 years? Plan a) P = 1000, I = .0675, n = 365 and Y = 10.  EMBED Equation.3  Plan b) P = 1500, I = .075, n = 52 and Y = 10.  EMBED Equation.3 Set 4 Practice Problems for Module 8, Unit 4 Compound Interest Formula Solved for Y:  EMBED Equation.3  APR Formula APR = (1 + I/n)n - 1 How long will it take $10,000 to generate $5,000 of income if it is invested at 5.5% that compounds daily? T = 10000 + 5000 = 15000, P = 10000, I = .055, n = 365.  EMBED Equation.3  Thus, it would take about 7.4 years for $10,000 to earn $5000 in interest at 5.5%. How long will it take $2,500 to generate $6,000 of income if it is invested at 7.5% that compounds daily? T = 2500 + 6000 = 8500, P = 2500, I = .075, n = 365.  EMBED Equation.3 How long will it take $3,400 to earn $2,000 of interest if it is invested in a CD that earns 6.4% that compounds weekly? T = 3400 + 2000 = 5400, P = 3400, I = .064, n = 52.  EMBED Equation.3  Bob and Sue are going to have a baby. They have $10,000 to start a college fund. They invest the money in a CD that earns 7.2% interest that compounds weekly. How long will it take to earn $5,000? $8,000? $10,000? How much money will the account have when the child turns 18? P = 10000, I = .072, n = 52. When T = 10000 + 5000 = 15000,  EMBED Equation.3  When T = 10000 + 8000 = 18000,  EMBED Equation.3  When T = 10000 + 10000 = 20000,  EMBED Equation.3  Which of the following is the best investment? (Hint: Find the APR of each investment) a) $2,000 @ 6.5% interest compounded daily. b) $2,000 @ 6.9% interest compounded weekly. c) $2,000 @ 7.5% interest compounded every two weeks(biweekly). d) $2,000 @ 8% interest compounded monthly. a) APR = (1 + .065/365)365 - 1 b) APR = (1 + .069/52)52 - 1 c) APR = (1 + .075/26)26 - 1 d) APR = (1 + .08/12)12 - 1 Tim wants to invest $2,500. He is comparing two investments. The first has a 6.5% interest rate that compounds daily and the other has a 7.3% interest rate that compounds weekly. Compare the APR of each and determine which is the better investment. Prove your conclusion by calculating the income for each investment for 5 years. How long will it take each investment to triple? a) Investment 1: APR = (1 + .065/365)365 1 Investment 2: APR = (1 + .073/52)52 1 b) Investment 1: P = 2500, I = .065, Y = 5, n = 365. T = 2500(1 + .065/365)365(5) Investment2: P = 2500, I = .073, Y = 5, n = 52. T = 2500(1 + .073/52)52(5) c) T = 3(2500) = 7500 Investment 1:  EMBED Equation.3  Investment 2:  EMBED Equation.3  Cheryl has $15,000 to invest. She wants to earn $2,500 yearly income. How long will it take the $15,000 dollars to increase to this level if it is invested at 6.5% that compounds daily? The first question is, How much money will earn $2,500 yearly income at 6.5% interest that compounds daily. This is a simple interest calculation, Income = PI, if we have the right interest rate. Fortunately, we can calculate the APR. APR = (1 + .065/365)365 1 = .06715. Therefore, P = 2500/.06715 = 37228.5. This is T we need to answer the second question, How long will it take $15,000 to increase? We have: T = 37228.5, P = 15000, I = .065, n = 365.  EMBED Equation.3  8) Terry wants to retire someday and he needs an income of $22,000 a year from his investments. He currently has $112,000 in his retirement fund that is increasing at 7.2 % compounding daily. How long will it take his retirement fund to grow to this level? Use the same strategy as #7. I = .072, n = 365. APR = (1 + .072/365)365 1 = .07465 Thus, P = 22000/.07465 = 294717.67 T = 294717.67, P = 112000, I = .072, n = 365.  EMBED Equation.3   Set 5 Practice Problems for Module 8, Unit 5. The future value of an annuity.  EMBED Equation.3  1) Peggy wants to save $100 a month in an annuity that earns 5.5% interest that compounds monthly. If she deposited the first $100 today, how much will the annuity be worth in 5 years? 10 years? 20 years? P = 100, I = .055, n = 12. When Y = 5,  EMBED Equation.3  The annuity will be worth $6888.13 in 5 years. When Y = 10,  EMBED Equation.3  and when Y = 20,  EMBED Equation.3  2) James wants to take $10 out of his weekly check and invest that money in an annuity that earns 6.25% compounded weekly. He will be 35 soon and wants to retire at 55. What would the annuity be worth at age 55 if he started it on his 35th birthday? At age 65? P = 10, I = .0625, n = 52. Y = 55 35 = 20 OR Y = 65 35 = 30. When Y = 20,  EMBED Equation.3  and when Y = 30,  EMBED Equation.3 3) How long will it take an annuity to grow to $10,000 dollars if you invest $200 a month at 6.5% interest compounded monthly. T = 10000, P = 200, I = .065 and n = 12. We are looking for Y.  EMBED Equation.3  is used when T, I, P and n are given.  EMBED Equation.3  Thus, it will take about 3.7 years for this annuity to grow to $10,000.4) Sandra wants to invest in an annuity that earns 7.75% that compounds monthly. She wants to save $25,000. She wants to know how long it will take her to reach her goal if she saves at the rate of $100 per month and at $200 per month. T = 25000, I = .0775, n = 12, P = 100 OR 200. We are looking for Y. When P = 100,  EMBED Equation.3  and when P = 200,  EMBED Equation.3  5) Mickey wants to save $28,000 in 10 years. He is thinking about investing in an annuity that earns 5.5% interest that compounds monthly. What would his monthly payment have to be in order to reach his goal? T = 28000, Y = 10, I = .055, n = 12.  EMBED Equation.3  is used when T, I, Y and n are given.  EMBED Equation.3  Mickey should save about $175.5 a month for 10 years if the interest rate is 5.5%.6) Jesse wants to save $150,000 for retirement. She can retire in 15 years but also has the option of working 25 more years. She is considering an annuity that earns 6.15% interest that compounds monthly. What would her monthly payment have to be in order to reach her financial goal in 15 years? 25 years? T = 150000, I = .0615, n = 12, Y = 15 OR 25. We are looking for P. When Y = 15,  EMBED Equation.3  and when Y = 25,  EMBED Equation.3 7) How much would an annuity be worth if you initially deposited $1000 and then made monthly payments of $100 for 10 years if the interest rate was 5.85% compounded monthly? The $1000 deposit will gain interest as a compound interest investment. Just add the $100 monthly annuity to the other investment. P = 1000, I = .0585, Y = 10 and n = 12. T$1000 = 1000(1 + .0585/12)10(12) P = 100, I = .0585, Y = 10 and n = 12.  EMBED Equation.3  The investment will be worth T$1000 + T$100annuity 8) Kathy invested $2,500 in an annuity that earns 6.45% interest compounded weekly and she will also deposit $50 a week. How much will the annuity be worth in 10 years? in 20 years? P = 2500, I = .0645, n = 52, Y = 10 OR 20 When Y = 10, T$2500 = 2500(1 + .0645/52)10(52) and when Y = 20, T$2500 = 2500(1 + .0645/52)20(52) P = 50, I = .0645, n = 52, Y = 10 OR 20. When Y = 10,  EMBED Equation.3  and when Y = 20,  EMBED Equation.3  Set 6 Practice Problems for Module 8, Unit 6. Monthly payment for a conventional loan.  EMBED Equation.3  Interest is compounding monthly unless otherwise noted. Jim wants to buy a $22,000 dollar car. He has 10% to put down and he is borrowing the rest at 4.5% over 5 years. What is the monthly payment? P = 22000 22000(.1) = 22000 2200 = 19800. Y = 5, I = .045 and n = 12.  EMBED Equation.3  Jim will have to make monthly payments of 369.13 for 5 years to pay off the $19,800 loan.Tina wants to by a condominium that is $65,000. She is putting 5% down and will take out a 30 year loan at 8.5% interest to cover the rest. What is her monthly payment? P = 65,000 65000(.05) = 65000 3250 = 61750. Y = 30, I = .085, n = 12.  EMBED Equation.3 John is buying a $80,000 car. He will finance the entire package and pay 2.8% on the 4 year loan. The yearly insurance will be $3,200 and will be rolled into the loan. What is his monthly payment? P = 80,000, Y = 4, I = .028, n = 12.  EMBED Equation.3  Divide the $3200 by 12 and add to the monthly payment.Joan has $25,000 of credit card debt. She has cut up the cards and wants to pay off the debt in 5 years. The interest rate is 13.9% that compounds daily. What will the monthly payment have to be in order for her to reach her goal? P = 25000, Y = 5, I = .139, n = 12.  EMBED Equation.3 Gerard owes $12,000 to his credit union. He can negotiate the interest rate and repayment schedule. He can choose 6.5% for 5 years, 6.85% for 6 years and 6.95% for 7 years. What is the monthly payment for each plan? What is the total amount that Gerard would pay for each plan? Case a) P = 12000, n = 12, I = .065, Y = 5.  EMBED Equation.3  Case b) P = 12000, n = 12, I = .0685, Y = 6.  EMBED Equation.3  Case c) P = 12000, n = 12, I = .0695, Y = 7.  EMBED Equation.3  Multiply the monthly payment by (Yn) to get the total amount paid. Tammy wants to buy a car. She can finance $20,000 for 5 years at 2.3% or 3 years at 2.5 %. What will the monthly payment be for each plan? What is the total amount she will pay in each plan? Case a) P = 20000, I = .023, Y = 5, n = 12.  EMBED Equation.3  Case b) P = 20000, I = .025, Y = 3, n = 12.  EMBED Equation.3  Multiply the monthly payment by (Yn) to get the total amount paid. Bob and Julie are buying a house for $225,000. They have a 30% down payment and the $8,500 closing costs are being rolled into the loan. Property taxes are $3,800 per year and homeowners insurance $2,400 a year. They have a choice between a 30-year mortgage @ 7.2% or a 15-year mortgage @ 7.5%. What are the monthly payments for each mortgage? What is the total cost of the house for each plan based on these monthly payment calculations? 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"  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMOPQRSTUX[\]_^`abdcefgihjklmnpoqrsutvwxyz{|~}Root Entry, F7WDZData NWordDocument+"ObjectPool.OeD7WD_1075982489cFeDeDOle CompObjfObjInfo  !$'(+./2569<=@CDGJKNQRUXY\_`cfghijknqrsvyz} FMicrosoft Equation 3.0 DS Equation Equation.39qN\ T=P1+In() Yn FMicrosoft Equation 3.0 DS EqEquation Native j_1076572518 F VDDOle CompObj fuation Equation.39qm0L T=15001+.065365() (5)(365) FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo Equation Native  _1076572635FDDOle  CompObjfObjInfoEquation Native _1076572777 FpDpDm~ T=28001+.058365() 3(365) FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObjfObjInfoEquation Native mȞ T Y=4 =48001+.065365() 4(365) FMicrosoft Equation 3.0 DS Equation Equation.39qmXa  T Y=6 =_1076572831FPSDPSDOle CompObjfObjInfoEquation Native _1076572877"F0D0DOle "CompObj #f48001+.065365() 6(365) FMicrosoft Equation 3.0 DS Equation Equation.39qm,b T Y=8 =48001+.065365ObjInfo!%Equation Native &_1076573047$F{D{DOle )() 8(365) FMicrosoft Equation 3.0 DS Equation Equation.39qm T=35001+.0575365() 5(365)CompObj#%*fObjInfo&,Equation Native -_1076573147;)F$D$DOle 0CompObj(*1fObjInfo+3Equation Native 4 FMicrosoft Equation 3.0 DS Equation Equation.39qmz T=35001+.057526() 5(26)_1076573333.FТ(DТ(DOle 7CompObj-/8fObjInfo0: FMicrosoft Equation 3.0 DS Equation Equation.39qm~ T=50001+.078512() 28(12) FMicrosoft Equation 3.0 DS EqEquation Native ;_1076573509,63F6-D6-DOle >CompObj24?fuation Equation.39qmpl T=50001+.065365() 28(365) FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo5AEquation Native B_10765736968FC0DC0DOle ECompObj79FfObjInfo:HEquation Native I_10765737731E=F4D4Dm T=15001+.065365() 10(365) FMicrosoft Equation 3.0 DS Equation Equation.39qOle LCompObj<>MfObjInfo?OEquation Native Pmz T=15001+.06952() 10(52) FMicrosoft Equation 3.0 DS Equation Equation.39qmz< T=1500_1076573842BFk9Dk9DOle SCompObjACTfObjInfoDVEquation Native W_1076573903@JGF=D=DOle ZCompObjFH[f1+.07526() 10(26) FMicrosoft Equation 3.0 DS Equation Equation.39qmvD T=15001+.0812() 10ObjInfoI]Equation Native ^_1076574109LFpBDpBDOle a(12) FMicrosoft Equation 3.0 DS Equation Equation.39qmm I=36520001000() 1(12)(365) "1()CompObjKMbfObjInfoNdEquation Native e_1076574681'QFP'GDP'GD=3652() ) 14380 "1[]=365(1.000158265"1)=.0578 FMicrosoft Equation 3.0 DS Equation Equation.39qOle lCompObjPRmfObjInfoSoEquation Native pm I=3652000010000() 110(365) "1() FMicrosoft Equation 3.0 DS Equation Equation.39q_1076574885VF0KD0KDOle tCompObjUWufObjInfoXwEquation Native x_1076696783[FpNDpNDOle {CompObjZ\|fm I=36570005000() 15(365) "1() FMicrosoft Equation 3.0 DS Equation Equation.39qoL  e@T=1000ObjInfo]~Equation Native _1076696782TY`FpZDpZDOle 8  e@1+  e@.0675  e@365 () e@  e@10(365) FMicrosoft Equation 3.0 DS Equation Equation.39qo8  e@T=1500  e@1+  e@.075  e@52 CompObj_afObjInfobEquation Native _1075989359eF ^D ^D() e@  e@10(52) FMicrosoft Equation 3.0 DS Equation Equation.39qȫG Y=logTP()nlog1+In()[]Ole CompObjdffObjInfogEquation Native _1076696997^rjFbDbDOle CompObjikfObjInfol FMicrosoft Equation 3.0 DS Equation Equation.39qo:L  e@Y=  e@log  e@  e@15000  e@10000 () e@365  e@log  e@1+  e@.055  e@365 Equation Native V_1076697380oFp2gDp2gDOle CompObjnpf()[] e@=log(1.5)365[log(1.000150685)=.17609365(.0000654367)=.17609.02388=7.3726 FMicrosoft Equation 3.0 DS Equation Equation.39qo  e@Y=  e@log  e@  e@8500  e@2500 () e@365  e@log  e@1+  e@.075  e@365 ()[]ObjInfoqEquation Native _1076697513mwtFPkDPkDOle CompObjsufObjInfovEquation Native _1076697623yF0ZpD0ZpD FMicrosoft Equation 3.0 DS Equation Equation.39qo8  e@Y=  e@log  e@  e@5400  e@3400 () e@52  e@log  e@1+  e@.064  e@52 ()[]Ole CompObjxzfObjInfo{Equation Native  FMicrosoft Equation 3.0 DS Equation Equation.39qo  e@Y=  e@log  e@  e@15000  e@10000 () e@52  e@log  e@1+  e@.072  e@52 ()[] FMicrosoft Equation 3.0 DS Equation Equation.39qo  e@Y=  e@log  e@  e@18000  e@10000 () e@52  e@log  _1076697730h~FtDtDOle CompObj}fObjInfoEquation Native _1076697802F`"xDyDOle CompObjfe@1+  e@.072  e@52 ()[] FMicrosoft Equation 3.0 DS Equation Equation.39qo  e@Y=  e@log  e@  e@20000  e@100ObjInfoEquation Native _1076698462F@|D@|DOle 00 () e@52  e@log  e@1+  e@.072  e@52 ()[] FMicrosoft Equation 3.0 DS Equation Equation.39qo  e@Y=  e@loCompObjfObjInfoEquation Native _1076698531F JD JDg  e@  e@7500  e@2500 () e@365  e@log  e@1+  e@.065  e@365 ()[] FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObjfObjInfoEquation Native oX$  e@Y=  e@log  e@  e@7500  e@2500 () e@52  e@log  e@1+  e@.073  e@52 ()[] FMicrosoft Equation 3.0 DS Equation Equation.39q_1076699241FޅDޅDOle CompObjfObjInfool  e@Y=  e@log  e@  e@37228.5  e@15000 () e@365  e@log  e@1+  e@.065  e@365 ()[] FMicrosoft Equation 3.0 DS EqEquation Native _1076699540FqDqDOle CompObjfuation Equation.39qol  e@Y=  e@log  e@  e@294717.67  e@112000 () e@365  e@log  e@1+  e@.072  e@365 ()[]ObjInfoEquation Native +_1076699661FDDOle CompObjfObjInfoEquation Native _1076699987F D D FMicrosoft Equation 3.0 DS Equation Equation.39qoo  T=P(1+I/n) Yn "1I/n[] FMicrosoft Equation 3.0 DS EqOle CompObjfObjInfoEquation Native 1uation Equation.39qoX T=100(1+.055/12) 5(12) "1.055/12[]=100(1.0045833) 60 "1.0045833[]=1001.3157    !"%()*-0123456789<?@ABEHIJKNQRUXYZ[\]^adefilmnqtuvwz}~"1.0045833[]=100.3157.0045833()=6888.13 FMicrosoft Equation 3.0 DS Equation Equation.39qo T=100_1076700386|FHDHDOle CompObjfObjInfoEquation Native _1076700467FۢDۢDOle  CompObj f(1+.055/12) 10(12) "1.055/12[] FMicrosoft Equation 3.0 DS Equation Equation.39qo T=100ObjInfoEquation Native _1076700646F D DOle (1+.055/12) 20(12) "1.055/12[] FMicrosoft Equation 3.0 DS Equation Equation.39qo8 T=10(CompObjfObjInfoEquation Native _1076700709F}D}D1+.0625/52) 20(52) "1.0625/52[] FMicrosoft Equation 3.0 DS Equation Equation.39qo$ T=10(Ole CompObjfObjInfoEquation Native 1+.0625/52) 30(52) "1.0625/52[] FMicrosoft Equation 3.0 DS Equation Equation.39qo  e@Y=  e@lo_1076700843FDDOle #CompObj$fObjInfo&Equation Native '_1076700892FDDOle +CompObj,fg  e@  e@  e@TI  e@Pn () e@+1 [] e@nlog(1+I/n) FMicrosoft Equation 3.0 DS Equation Equation.39qo8  e@Y=  e@loObjInfo.Equation Native /_1076701442F8D8DOle :g  e@  e@  e@10000(.065)  e@200(12) () e@+1 [] e@12log(1+.065/12)=log.2708333+1 [] e@12log(1.00541667) [] e@=log(1.270833)12(.0023461)=.104088.028153=3.697 FMicrosoft Equation 3.0 DS Equation Equation.39qo   e@Y=  e@log  e@  e@  e@25000(.CompObj;fObjInfo=Equation Native >(_1076701499F̼D̼D0775)  e@100(12) () e@+1 [] e@12log(1+.0775/12) FMicrosoft Equation 3.0 DS Equation Equation.39qo    e@Y=  e@loOle CCompObjDfObjInfoFEquation Native G(g  e@  e@  e@25000(.0775)  e@200(12) () e@+1 [] e@12log(1+.0775/12) FMicrosoft Equation 3.0 DS Equation Equation.39q_1076701680FٿD``DOle LCompObjMfObjInfoOEquation Native P_1076701725FmDmDOle SCompObjTfoq P=TIn1+In() Yn "1 FMicrosoft Equation 3.0 DS Equation Equation.39qo, P=280ObjInfoVEquation Native W_1076702496FDDOle _00(.055)121+.05512() 10(12) "1=128.33(1.0045833) 120 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