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1) True.
2) True.
3) False.
4) False.
5) True.
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Type A B $8,650.00 $13,900.00 First cost, $ Maintenance cost, $/year Salvage value Life, years $2,200.00...
Some proffesional from here request more information but below is all I have, There is something...
Some proffesional from here request more information but below
is all I have, There is something missing? if not please help me
here
Objective: This activity has the purpose of helping students compute an equivalent annual worth quantity using the software Microsoft Excel. Also to perform a sensitivity analysis. (Objective 7) Student Instructions: Read example to be used to compute an equivalent annual worth cash flow. Example Equivalent Annual Worth with different lives: A mechanical engineer is considering two types...
an annual maintenance cost of $72,000, and a salvage value of $78,000 after its 10-year life....
an annual maintenance cost of $72,000, and a salvage value of $78,000 after its 10-year life. Use an interest rate of 6% a. b) $871,000 b.c) $1,256,890 c. a) $187,142 d. d) $1,452 367
engineering economics Two mutually exclusive alternatives The minimum with a 20-year life and no salvage value....
engineering economics
Two mutually exclusive alternatives The minimum with a 20-year life and no salvage value. 4. attractive rate of return is 6%. 4000 5000 639 We would like to know how sensitive the the initial cost of ho w much higher than 400) can ce the preferred alternative? (Hint: find the present worth of decision is to our estimate of initial cost be and still have A alternative? . both alternative to start with) (15 pts) a. 4100 b....
α is the probability of a Type I error, which occurs when we accept the alternative...
α is the probability of a Type I error, which occurs when we accept the alternative H1 when the null hypothesis Ho is true. True False A Type II error occurs when when a false null hypothesis is rejected. True False If a null hypothesis is rejected at the 5% significance level but not at the 1% significance level, then the p-value of the test is less than 1%. True False The power of a test is the probability of...
Nules. Enter the information exactly as below. Factor Annuity (05"/1.ES/"E71/((1+5)*E7-1) =3*G3 30,000.00 Investment Save: 300000 0.10...
Nules. Enter the information exactly as below. Factor Annuity (05"/1.ES/"E71/((1+5)*E7-1) =3*G3 30,000.00 Investment Save: 300000 0.10 MARR Market value $120,000.00 (A/P =300000*0.1 0.15 $8,000.00 AFI, E5/((1+5)*87-1) 1666 Benefit *H4/H3-H6) (do not forget to write the = sign) To format columns with numbers: right click the mouse in the cell you whic Cancel S 1. When the MARR is 15% and the life of the heating system asset is 6 years is the alternative acceptable? a. True b. False 2. When...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Main...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Maintenance costs, $ 20.000 15,000 ear Overhaul cots in vear 5 10.000 Salvage values, S 8.000 20,000 200,000 Life, years Match the closest correct answers for the below questions: Calculate the Present worth PW of A A. -86,748.66 B. -97,743.33] # Calculate the Annual worth AW of B Calculate...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Main...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Maintenance costs, $ 20.000 15,000 ear Overhaul cots in vear 5 10.000 Salvage values, S 8.000 20,000 200,000 Life, years Match the closest correct answers for the below questions: Calculate the Present worth PW of A A. -86,748.66 B. -97,743.33] # Calculate the Annual worth AW of B Calculate...
please add the cash flow diagram as well b Problem 1. The maintenance cost for a...
please add the cash flow diagram as well
b Problem 1. The maintenance cost for a certain machine is $1000 per year for the first 5 years and S2000 per year for the next 5 years. At an interest rate of 10% per year, the annual worth in years 1 through 10 of the maintenance cost is closest to Answer: CFD: Equation:
3. Determine the present worth of a maintenance contract that has a cost of $50,000 in...
3. Determine the present worth of a maintenance contract that has a cost of $50,000 in year 1 and annual increases of 8% per year for 10 years. Use an interest rate of 8% per year. (10 points) Pg=. 2 - 8v. Sop VVVV. 72 70 Now $s < 4. The equivalent present worth of a geometric gradient series of cash flows for 10 years was found to be $19,776. If the interest rate was 15% per year and the...
Cost of new machinery $200,000 Machinery estimated useful life 10 years Estimated salvage value $0 Straight-line...
Cost of new machinery $200,000 Machinery estimated useful life 10 years Estimated salvage value $0 Straight-line depreciation Annual labor hours reduction 10,000 Annual operating costs reduction $4,000 Average hourly labor rate $5.50 Income tax rate 40% Discount rate 10% Part 1: Calculate the annual incremental income after taxes and the annual net cash flow for each year Part 2: Calculate the payback period Part 3: Calculate the rate of return on average investment Part 4: Calculate the net present value...
Some proffesional from here request more information but below
is all I have, There is something missing? if not please help me
here
Objective: This activity has the purpose of helping students compute an equivalent annual worth quantity using the software Microsoft Excel. Also to perform a sensitivity analysis. (Objective 7) Student Instructions: Read example to be used to compute an equivalent annual worth cash flow. Example Equivalent Annual Worth with different lives: A mechanical engineer is considering two types...
an annual maintenance cost of $72,000, and a salvage value of $78,000 after its 10-year life. Use an interest rate of 6% a. b) $871,000 b.c) $1,256,890 c. a) $187,142 d. d) $1,452 367
engineering economics
Two mutually exclusive alternatives The minimum with a 20-year life and no salvage value. 4. attractive rate of return is 6%. 4000 5000 639 We would like to know how sensitive the the initial cost of ho w much higher than 400) can ce the preferred alternative? (Hint: find the present worth of decision is to our estimate of initial cost be and still have A alternative? . both alternative to start with) (15 pts) a. 4100 b....
Nules. Enter the information exactly as below. Factor Annuity (05"/1.ES/"E71/((1+5)*E7-1) =3*G3 30,000.00 Investment Save: 300000 0.10 MARR Market value $120,000.00 (A/P =300000*0.1 0.15 $8,000.00 AFI, E5/((1+5)*87-1) 1666 Benefit *H4/H3-H6) (do not forget to write the = sign) To format columns with numbers: right click the mouse in the cell you whic Cancel S 1. When the MARR is 15% and the life of the heating system asset is 6 years is the alternative acceptable? a. True b. False 2. When...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Maintenance costs, $ 20.000 15,000 ear Overhaul cots in vear 5 10.000 Salvage values, S 8.000 20,000 200,000 Life, years Match the closest correct answers for the below questions: Calculate the Present worth PW of A A. -86,748.66 B. -97,743.33] # Calculate the Annual worth AW of B Calculate...
Question 11 For the cash flows provided hereunder, answer the questions and determine which alternative is best at an interest rate of 10% Alternative 350,000 150,000 0,000 450,000 First cost, S Maintenance costs, $ 20.000 15,000 ear Overhaul cots in vear 5 10.000 Salvage values, S 8.000 20,000 200,000 Life, years Match the closest correct answers for the below questions: Calculate the Present worth PW of A A. -86,748.66 B. -97,743.33] # Calculate the Annual worth AW of B Calculate...
please add the cash flow diagram as well
b Problem 1. The maintenance cost for a certain machine is $1000 per year for the first 5 years and S2000 per year for the next 5 years. At an interest rate of 10% per year, the annual worth in years 1 through 10 of the maintenance cost is closest to Answer: CFD: Equation:
3. Determine the present worth of a maintenance contract that has a cost of $50,000 in year 1 and annual increases of 8% per year for 10 years. Use an interest rate of 8% per year. (10 points) Pg=. 2 - 8v. Sop VVVV. 72 70 Now $s < 4. The equivalent present worth of a geometric gradient series of cash flows for 10 years was found to be $19,776. If the interest rate was 15% per year and the...
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