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by: Mr. Dayna Spencer


Mr. Dayna Spencer
GPA 3.88


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Class Notes
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This 5 page Class Notes was uploaded by Mr. Dayna Spencer on Saturday September 12, 2015. The Class Notes belongs to ENGR 2110 at University of Georgia taught by Sellers in Fall. Since its upload, it has received 27 views. For similar materials see /class/202415/engr-2110-university-of-georgia in Engineering and Tech at University of Georgia.

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Date Created: 09/12/15
42 A man borrowed 500 from a bank on October 15th payments due on the 15th of each month beginning Novembe how much must he pay per month A 500AP116 50000679 3395 PMT 3397 43 A local finance company will loan 10000 to a home of 49900 each The first payment is due thirty days after the month are they charging A PAPi24 or APi24 AP 49910000 00499 fron month or 18 per year RATE 150 018 414 A loan company has been advertising on television a plat 1087 per month This payment is for interest only and includes n interest rate that they are charging P 1000 interest payment 1087 nominal interest 10871 415 What effective interest rate per annum corresponds t I 1month39 effective rate 1Im1 101121 0127 or 12quot EFFECT 126825 424 A young engineer wishes to become a millionaire byt investment he can obtain a 15 rate of return He plans to add year beginning on his 20th birthday and continuing through his in this project each year We will draw the cash flows on the board but the number of invr have enough money in the bank to yield 1000000 one year late F1000000FP151 100 Then A 869600AF1540 869600000056 48698 From calculator 48878 PMT S 48877 From the file additional Chapter4 problems Problem 48 Mary Smith took a car loan of 12000 to pay back in 60 montl understanding that the interest rate may be changed sometim the monthly payment for Mary the loan balance immediately after the 2 the monthly payment for the remainder peroid RDOOVOHU39IbWNH MMMMMMMMMMMMMMMMMMMMMMMMm on lt 1200000 1185307 1170466 1155478 1140339 1125049 1109606 1094009 1078256 1062345 1046275 1030045 1013652 997095 980372 963483 946424 929195 911794 894219 876467 858539 840431 822142 803670 390 a a MMMMMMMMMMMMMMMMMMMMMMMM 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 26693 25556 12000 11853 11705 11555 11403 11250 11096 10940 10783 10623 10463 10300 10137 9971 9804 9635 9464 9292 9118 8942 8765 8585 8404 8221 quotD quotU MMMMMMMMMMMMMMMMMMMMMMMM 14693 14840 14989 15139 15290 15443 15597 15753 15911 16070 16231 16393 16557 16722 16890 17059 17229 17401 17575 17751 17929 18108 18289 18472 He must repay the loan in 16 equal monthly r 15th If interest is computed at 1 per month eowner It is to be repaid in 24 monthly payments 3 10000 is received What interest rate per 1 the compound interest tablesexactly 15 per 1 where one may borrow 1000 and make a payment of 3 payment to the principal What is the nominal annual 21000 13 or 13 o a nominal rate if 12 compounded monthly 7 he time he is sixty years old He believes that a careful a uniform sum of money to his investment program each 59th birthday How much money must the engineer set aside estments is 40 59201 and on the 59th birthday he wants to r at 15 per year 00008696 5869600


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