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.) Suppose we have three alternatives, and the least common multiple of lives equal to 10 years and MARR-12%. If the annual worth for a life cycles is equal to $50.000 for alternative A. what will be the present werth ever a least common multiple of A) 5340,700 B) 5282.511 C) $309.718 D) More information is needed to calculate the present worth 7) Four investment alternatives (A, B, C and D) are being considered by a firm. There are five factors used to compare between the proposals as shown in the following table. A D Factors Weight 20 Rating 7 N 6 7 20 30 7 6 10 7 10 Which of the alternative is better using Weighted Factor Comparison c. dD *) Determine the equivalence uniform serious ever (1.5) for the below cash flow at 4 year 1 CF (SR) 2000 2000 2. (2000 (PA8%, 6) + 3000 (PF8%. 5) + 5000 (PF8%. (AP 88,5) b.2000 PIA 4) 1000 (PF8%, 5) 5000PF 8%, 6)) (P85,6) C. (2000 (PA8%, 6) +1000 (PF8%, 5) 3000 (PF846) P8,5) d.(2000 (PA8% 4) 1000 (PF 8%, 53000 (PF8%, 6P855) 9) Consider the following cash flow series. At 6% per year compounded annually. Calculate the present worth of the below cash flow series? year 0 CF (SR) -10.000 6.900 6.000 5.500 5.000 4.50 4.000 3.500 3.000 a. 20415.8 6,781.5 d Cannot be measured 10) Considera cash flow and interest profile as shown below: year CF at EOY (SR) -10,000 6.500 6.000 5.500 Interest rate (*) 10 The formula to be used for finding the equivalent anual werth for the above cash flow from the present worth 2. APF%, 1)MFP8%)PF 6%, 1)A(FP 10% INFP8%, 1PF 6%.1) MAP%DLAP8%, HAP% TXAP 10%, 1)) CALAF%. 1)+ACAP 8%, IKAF6% 1+ACAP 10% IVAP 8%. TXAF 6%. 1) d. AP/F 6% 1) (PF8%, IPF8% DPF 10%, 1)