Files 6 Engineering Econo x M y.pdf pcs/10pcs Gold C. O 7 amazonprime has... @ 3 Best Pra 35,000(P/F,15%,6) (C) PW, = -250,000 – 26,000(P/A,15%,6) + 35,000(P/E15%,6) @ PW, * - 250,000 - 26,000(P/A,15%,6) - 35,000(P/E.15%,6) 5.48 In comparing alternatives I and J by the present worth method, the equation that yields the present worth of alternative I is: (a) PW, = - 150.000 + 11,000(P/1.15%,3) + 25,000(P/E 15%,3) (b) PW, = - 150,000 + 11,000(P/A,15%,6) + 1 25.000(P/E 15%,6) (0) PW, = - 150.000 + 11.000(P/A.15%,6) + 175.000(P/E.15%,3) + 25,000(P/E15%,6) (d) PW, = 150.000 + 11.000(P/A.15%,6) – 125,000(P/E 15%,3)+25.0000P F 15%,6) Problems 5.49 and 5.50 are based on the following information. Machine X Machine Y Initial cost, s 80.000 95.000 Annual operating cost. $ per ved 0 000 15.000 Salvare values 0000 3000 (b) ( C (d FW,= 3-86,020 FW, = $-81,274 FW, = $-70.178 5.44 The present worth of $50,000 now, $10,000 per year In years 1 through 15, and $20,000 per year in years 16 through infinity at 10% per year is closest to: (a) Less than $-169,000 (b) S-169,580 (C) $-173,940 (d) $-195,730 Init Ann Salu Life The 5.45 5.45 A donor (you) wishes to start an endowment that will provide scholarship money of $40,000 per year beginning in year 5 and continuing indefinitely. If the university earns 10% per year on the endow- ment, the amount you must donate now is closest to: (a) S-225,470 (b) S-248,360 (c) S-273,200 (d) S-293,820 Problems 5.46 through 5.48 are based on the following information. Alternative! Alternative Initial cost. S 150,000 250.000 Annual income, S per year 20.000 40.000 Annual expenses. S per year 9.000 14,000 Salvage value, s 25,000 35.000 Life: years The interest rate is 15% per year.