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Radda [10]
2 years ago
6

Sheffield Corp. owns the following assets: Asset Cost Salvage Estimated Useful Life A $540000 $42000 10 years B 201000 23500 5 y

ears C 490000 22000 12 years What is the composite life of Sheffield's assets?
Business
1 answer:
rusak2 [61]2 years ago
4 0

Answer:

The composite life is 9.19.

Explanation:

Below is the calculation for composite life of assets:

Composite life = Total Depreciable Cost ÷ Total Annual Depreciation

Composite life = 1143500 ÷ 124300

Composite life = 9.19

The composite life is 9.19.

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3 0
3 years ago
if, after one year, the yield to maturity on a multiyear coupon bond that was issued at par is lower than the coupon rate, what
maw [93]

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<h3>What is the cost of a $1,000 par value, three year, zero-coupon bond?</h3>

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To Know more about coupon rate

brainly.com/question/16913107

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5 0
1 year ago
Angela Fox and Zooey Caulfield were food and nutrition majors at State University, as well as close friends and roommates. Upon
S_A_V [24]

Answer:

  a) maximize 12x+16y subject to x+y≤60; x+2y≤80; 2x-3y≥0; x-9y≤0.

  b) 40 fish and 20 beef dinners

  c) $800

Explanation:

Let x and y represent the numbers of fish and beef dinners to prepare, respectively. Then the relations these values must satisfy are ...

  x + y ≤ 60 . . . . . a maximum of 60 dinners will be sold

  0.25x + 0.50y ≤ 20 . . . . . kitchen hours cannot exceed 20

  x/y ≥ 3/2 . . . . . . at least 3 fish dinners for each 2 beef dinners will be sold

  y ≥ 0.10(x +y) . . . . at least 10% of dinners sold will be beef

While satisfying these relations, we want to maximize the profit function:

  p = 12x +16y

a) The linear programming problem can be formulated as ...

  Maximize 12x +16y, subject to ...

  • x + y ≤ 60
  • x + 2y ≤ 80
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  • x - 9y ≤ 0

__

b) The graph shows the constraint inequalities with the comparison symbol reversed. The effect of that is to shade the area that is NOT part of the solution set, leaving the feasible region white. The vertex of the (white) feasible region that makes the profit line farthest from the origin is the solution we're looking for. Once the profit line is plotted so we can compare its slope to the lines bounding the feasible region, it becomes clear which vertex is the one that maximizes profit.

The solution is (x, y) = (40, 20).

  • 40 fish dinners
  • 20 beef dinners

__

c) The maximum earnings are estimated to be ...

  ($12)(40) +($16)(20) = $800

5 0
3 years ago
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