Answer Happy

The value of the present worth for alternative A at an interest rate of 8% per year is closest to: A B First cost, $ -23,000-30,000 Annual operating cost, $ -4000 -2,500 Salvage value, $ 3000 1000 3 6 O $-55,588 O $-54,474 $-55,478 Life, years $-59,788
Machine A Machine B First cost, SR 100,000 -700,000 Annual Operation Cost, SR/ year -50,000 -5,000 Salvage value, SR 9000 1,000,000 Life (years) 10 The capitlized cost for machine B using an interest rate of 10% per year is: $-790.000 $-950.000 $-850.000 $-750.000
2 points Initial cost $ Annual operating cost, $ per year Salvage value, $ Life, years Machine X $-80,000 $-20,000 $10,000 2 Machine Y 3-95,000 $-15,000 $10,000 4 The interest rate is 10% per year. The present worth of machine Yis: O $-175,980 S-122,060 $-112,320 OS-163,040 Question 6 of 10
Initial cost, $ Annual operating cost, $ per year Salvage value, $ Life, years Machine X $-80,000 $-20,000 $10,000 2 Machine Y $-95,000 $-15,000 $10,000 4 The interest rate is 10% per year. The equation that will calculate the present worth of machine X is: O PW X =-80,000-20,000( PIA 10%,4)-80,000(P/F,10% 2) +10,000(P/F,10%,4) PW X =-80,000-15,000(PIA 10%, 4)+30,000( P/F, 10%,4) PW X =-80,000-20,000(PIA 10%,2)+10,000( P/F 10% 2) O PW X =-80,000-20,000(PIA 10% 4) -70,000(P/F 10%,2)+10,000(P/F 10%,4) millive this response
Z points Alternative X has a first cost of $30,000. an operating cost of $9,000 per year, and a $5,000 salvage value after 5 years Altomate Y il con $45,000 at n=0 with an operating cost of $4,000 per year after that and a salvage value of $7,000 after 5 years. At an MARR of 125 per year, what is the present worth for alternative Y? -$58,557 -$55,447 -$54,778 -$55.998
The value of the present worth for alternative B at an interest rate of 8% per year is closest to: А B First cost, $ -23,000-30,000 Annual operating cost, $ -4000 -2,500 Salvage value, $ 3000 1000 Life, years 3 6 -$39,352 O -$33,277 -$35,767 -$34,244 L Click Submit to complete this assessment.