Homework Answers
4) answer :- Option(e)
5)Answer :- Option(b)
6)Answer :- Option(a)
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please answer 4-6
4. Which of the following relationships between compound interest factors is not correct?...
Question 1 If you know the future value or worth of something and would like to...
Question 1 If you know the future value or worth of something and would like to know what its present value or worth is, which interest factor could you use? Present worth factor for a uniform series Capital recovery factor Present worth factor for a single payment Compound amount factor Question 2 If you are given a series of payments into the future and want to know their present value or worth, what is the best interest factor to use?...
Find P given F is often represented by the following term: which letter is correct a...
Find P given F is often represented by the following term: which letter is correct a Present worth factor b Capital recovery factor c Compound amount factor d. Sinking fund factor
Question 1: (6 points) A company having a cost of capital rate of 8% purchases a...
Question 1: (6 points) A company having a cost of capital rate of 8% purchases a $200,000 track loader. This machine has an expected service life of 5 years and will be used 2,000 hrs per year. The estimated salvage value at the end of 5 years is $40,000. Fuel, oil, grease, and minor maintenance are estimated to cost $35.00 per operating hour. No tire replacement since this is a track machine. TABLE 2.1 Economic analysis relationships Name Symbol Converts...
please show all steps, and answer all of the question. The second pictures states how it...
please show all steps, and answer all of the question.
The second pictures states how it should be solved.
thanks!
1-S. A large engineering company is considering the following three major capital investments (each investment is considered to have zero salvage value at the end of its life) (b) For a MARR of 15%, which investment, if any, would you select? Why? (How do you handle the unequal lives?) InitialEstimated capital annual receipts Estimated annual disbursements Estimated life required Investment...
Consider the accompanying cash flow diagram. Compute the equivalent annual worth at i= 10 % 6....
Consider the accompanying cash flow diagram. Compute the equivalent annual worth at i= 10 % 6. $5,000 $6,000 $4,000 $3,000 2 4 56 Years $3,000 Click the icon to view the interest factors for discrete compounding when i 10% per year The equivalent annual worth is $ (Round to the nearest dollar.) 8: More Info Equal Payment Series Single Payment Gradient Series Gradient Present Compound Present Compound Amount Sinking Present Capital Recovery Gradient Worth Fund Worth Uniform Amount Factor Factor...
Using equivalence calculations involving multiple formulas (section 4.10) and interest rate of 8%, calculate the present...
Using equivalence calculations involving multiple formulas (section 4.10) and interest rate of 8%, calculate the present equivalent value in year 0 of the cash flows shown in the diagram. Note that your calculations will include both positive and negative cash flows. $2,000 $5,310 $5,310 $5,310 $5,310 $5,310 SEN $3,000 $3,000 $3,000 $3,000 $3,000 Years $10,000 8% 8 % Compound Interest Factors Uniform Payment Series Single Payment Compound Present Amount Worth Factor Factor Find F Find P Given P Given F...
please show all steps, I need help on part D. Thanks! row detailed and appropriate cash...
please show all steps, I need help on part D.
Thanks!
row detailed and appropriate cash flow diagrams for each problem, and use quations to solve each problem. Show and explain all work Facter Name ormula Converts to given P to P given F to A given F to A given fP to F given A to P given A Single Payment Compound Amount (F/P, ma, n) P/F, P%, n) AF.%, n) (AP, ins, n) Single Payment Present Worth Uniform...
Problem 6-2 (book/static) Question Help Years SS.OCC $6,000 58.000 511000 s14000 517000 Click the icon to...
Problem 6-2 (book/static) Question Help Years SS.OCC $6,000 58.000 511000 s14000 517000 Click the icon to view the interest factors for discrete compounding when i = 13% per year. i More Info The equivalent annual worth is $ . (Round to the nearest doller.) Single Payment Equal Payment Series Gradient Series N Compound Amount Factor (F/P, I, N) 1.1300 1.2769 1.4429 1.6305 1.8424 Present Worth Factor (P/F, I, NJ 0.8850 0.7831 0.6931 0.6133 0.5428 Compound Amount Factor (F/A,,N) 1.0000 2.1300...
1 year= $3000 2 year= $3000 3 year= $6000 4 year= $9000 5 year= $12000 6 year= $15000 X Problem 6-2 (algorithmic) Questi...
1 year= $3000
2 year= $3000
3 year= $6000
4 year= $9000
5 year= $12000
6 year= $15000
X Problem 6-2 (algorithmic) Question Help O $10,000 OOO Years 0 ! 53.00 53.000 $.000 $2,000 $12.000 315.000 Click the icon to view the interest factors for discrete compounding when = 7% per year. i More Info The equivalent annual worth is S . (Round to the nearest dollar) Single Payment Compound Present Amount Worth Factor Factor (F/P.I, NJ (P/F, I, NJ...
Please show all work, the directions is on the first pictures. Please show all steps, thanks! ...
Please show all work, the directions is on the first
pictures. Please show all steps, thanks!
Draw detailed and appropriate cash flow diagrams for each problem, and use the EE Equations to solve each problem. Show and explain all work. Factor Name Formula Converts to Fgiven P to P given F to A given F to A given F to F given A to P given A Symbol (F/P, i%, n) (P/F, i%, n) (A/F, i%, n) Single Payment Compound...
Question 1: (6 points) A company having a cost of capital rate of 8% purchases a $200,000 track loader. This machine has an expected service life of 5 years and will be used 2,000 hrs per year. The estimated salvage value at the end of 5 years is $40,000. Fuel, oil, grease, and minor maintenance are estimated to cost $35.00 per operating hour. No tire replacement since this is a track machine. TABLE 2.1 Economic analysis relationships Name Symbol Converts...
please show all steps, and answer all of the question.
The second pictures states how it should be solved.
thanks!
1-S. A large engineering company is considering the following three major capital investments (each investment is considered to have zero salvage value at the end of its life) (b) For a MARR of 15%, which investment, if any, would you select? Why? (How do you handle the unequal lives?) InitialEstimated capital annual receipts Estimated annual disbursements Estimated life required Investment...
Consider the accompanying cash flow diagram. Compute the equivalent annual worth at i= 10 % 6. $5,000 $6,000 $4,000 $3,000 2 4 56 Years $3,000 Click the icon to view the interest factors for discrete compounding when i 10% per year The equivalent annual worth is $ (Round to the nearest dollar.) 8: More Info Equal Payment Series Single Payment Gradient Series Gradient Present Compound Present Compound Amount Sinking Present Capital Recovery Gradient Worth Fund Worth Uniform Amount Factor Factor...
Using equivalence calculations involving multiple formulas (section 4.10) and interest rate of 8%, calculate the present equivalent value in year 0 of the cash flows shown in the diagram. Note that your calculations will include both positive and negative cash flows. $2,000 $5,310 $5,310 $5,310 $5,310 $5,310 SEN $3,000 $3,000 $3,000 $3,000 $3,000 Years $10,000 8% 8 % Compound Interest Factors Uniform Payment Series Single Payment Compound Present Amount Worth Factor Factor Find F Find P Given P Given F...
please show all steps, I need help on part D.
Thanks!
row detailed and appropriate cash flow diagrams for each problem, and use quations to solve each problem. Show and explain all work Facter Name ormula Converts to given P to P given F to A given F to A given fP to F given A to P given A Single Payment Compound Amount (F/P, ma, n) P/F, P%, n) AF.%, n) (AP, ins, n) Single Payment Present Worth Uniform...
Problem 6-2 (book/static) Question Help Years SS.OCC $6,000 58.000 511000 s14000 517000 Click the icon to view the interest factors for discrete compounding when i = 13% per year. i More Info The equivalent annual worth is $ . (Round to the nearest doller.) Single Payment Equal Payment Series Gradient Series N Compound Amount Factor (F/P, I, N) 1.1300 1.2769 1.4429 1.6305 1.8424 Present Worth Factor (P/F, I, NJ 0.8850 0.7831 0.6931 0.6133 0.5428 Compound Amount Factor (F/A,,N) 1.0000 2.1300...
1 year= $3000
2 year= $3000
3 year= $6000
4 year= $9000
5 year= $12000
6 year= $15000
X Problem 6-2 (algorithmic) Question Help O $10,000 OOO Years 0 ! 53.00 53.000 $.000 $2,000 $12.000 315.000 Click the icon to view the interest factors for discrete compounding when = 7% per year. i More Info The equivalent annual worth is S . (Round to the nearest dollar) Single Payment Compound Present Amount Worth Factor Factor (F/P.I, NJ (P/F, I, NJ...
Please show all work, the directions is on the first
pictures. Please show all steps, thanks!
Draw detailed and appropriate cash flow diagrams for each problem, and use the EE Equations to solve each problem. Show and explain all work. Factor Name Formula Converts to Fgiven P to P given F to A given F to A given F to F given A to P given A Symbol (F/P, i%, n) (P/F, i%, n) (A/F, i%, n) Single Payment Compound...
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