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Kipish [7]
2 years ago
14

Krustyburger just paid a dividend of $2 and has a required return of 15%. Which of the following equations represent's today's v

alue of this stock if Krustyburger expects a 10 percent constant growth rate in dividends? a. $2(1.10)/0.15 b. $2/[0.15 - 0.10] c. $2/0.15 d. $2(1.10)/[0.15-0.10]
Business
1 answer:
Neko [114]2 years ago
8 0

Answer:

d. $2(1.10)/[0.15-0.10]

Explanation:

The formula to compute the today value of the stock by using the Gordon model is shown below:

= Next year dividend ÷ (Required rate of return - growth rate)

where,

Next year dividend is

= $2 + $2 × 10%

= $2 + 0.2

= $2.2

And, the required rate of return is 15%

Plus the growth rate of return is 10%

So, the today value of the stock is

= $2.2 ÷ (15% - 10%

= $44

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An all-equity firm is considering the following projects:
FromTheMoon [43]

Answer:

Projects Y and Z

b. Projects W and Z

c. Projects W and Y

Explanation:

CAPM equation : Expected return = Risk free rate + Beta x (Expected market return - Risk free rate)

W = 4% + [0.85 x (11% - 4%)] = 9.95%

X = 4% + (0.92 x 7%) = 10.44%

Y = 4% + (1.09 x 7%) = 11.63%

Z = 4% + (1.35 x 7%) = 13.45%

Projects Y and Z have an expected return greater than 11%

b. Projects W and Z should be accepted because its expected return is higher than the IRR

c. Project W would be incorrectly rejected because the expected rate of return is less than the overall cost of capital (i.e. 9.95 is less than 11). But its expected rate of return is greater than the IRR

Y would be incorrectly accepted because its expected rate of return is greater  than the overall cost of capital but its expected rate of return is less than the IRR

4 0
2 years ago
You are evaluating two different silicon wafer milling machines. The Techron I costs $285,000, has a three-year life, and has pr
KonstantinChe [14]

Answer:

EAC Techron I = -$141,050

EAC Techron II = -$138,181

Explanation:

Techron I costs $285,000, has a three-year life, and has pretax operating costs of $78,000 per year. Salvage value $55,000, use straight line depreciation.

annuity factor = [1 - 1/(1 + r)ⁿ] / r = [1 - 1/(1 + 0.11)³] / 0.11 = 2.4437

depreciation expense per year = ($285,000 - $55,000) / 3 = $76,667

cash outflow years 1 and 2 = [($78,000 + $76,667) x (1 - 24%)] - $76,667 = ($154,667 x 0.76) - $76,667 = $40,880

cash outflow year 3 = [($78,000 + $76,667) x (1 - 24%)] - $76,667 - $55,000 = ($154,667 x 0.76) - $76,667 - $55,000 = -$14,120

NPV = -285,000 - 40,880/1.11 - 40,880/1.11² + 14,120/1.11³ = -285,000 - 36,829 - 33,179 + 10,324 = -344,684

EAC = NPV / annuity factor = -344,684 / 2.4437 = -$141,050

Techron II costs $495,000, has a five-year life, and has pretax operating costs of $45,000 per year. Salvage value $55,000, use straight line depreciation.

annuity factor = [1 - 1/(1 + r)ⁿ] / r = [1 - 1/(1 + 0.11)⁵] / 0.11 = 3.6959

depreciation expense per year = ($495,000 - $55,000) / 5 = $88,000

cash outflow years 1 through 4 = [($45,000 + $88,000) x (1 - 24%)] - $88,000 = ($133,000 x 0.76) - $88,000 = $13,080

cash outflow year 5 = [($45,000 + $88,000) x (1 - 24%)] - $88,000 - $55,000 = ($133,000 x 0.76) - $88,000 - $55,000 = -$41,920

NPV = -495,000 - 13,080/1.11 - 13,080/1.11² - 13,080/1.11³ - 13,080/1.11⁴ + 41,920/1.11⁵ = -495,000 - 11,784 - 10,616 - 9,564 - 8,616 + 24,877 = -510,703

EAC = NPV / annuity factor = -510,703 / 3.6959 = -$138,181

4 0
2 years ago
A population gathers plants and animals for survival. They need at least 360 units of​ energy, 300 units of​ protein, and 8 hide
stellarik [79]

Answer:

0 units of plants and 18 animals.

Explanation:

From the information provided, we know that this population must hunt at least 8 animals, to obtain the hides necessary to survive. When hunting such a quantity of animals, they would also obtain 160 units of energy and 200 units of protein. In terms of costs, this equals 80 hours of labor.

How can the population obtain the additional 200 units of energy and 100 of protein they need to live? They have two options: gather plants or keep hunting animals, of course, respecting the available quantities.

In the former case, at least 10 units of a plant would have to be gathered, this is the only way to obtain the additional units of energy and protein they are looking for. With 8 animals and 10 units of plants, they would have 460 units of energy, 300 of protein and 8 hides, and would have to invest 280 hours of labor.

If the population decided to reduce labor by gathering fewer plants (which is the activity that requires more time) but hunting the same amount of animals, it would obtain 430 units of energy, 290 units of protein and 8 hides. That means that it would not survive because it would not meet the requirement of 300 proteins.

Therefore, the only way to reduce hours of labor without risking survival would be gathering fewer plants but hunting more animals. Following this logic, the optimal amount would be 18 animals and 0 units of plants, as this would allow people to obtain the necessary resources to survive at the minimum cost in terms of labor. Furthermore, it would fit the physical limit of 27 units of plants and 23 animals. This combination would allow to obtain 360 units of energy, 450 of protein and 18 hides by investing only 180 hours of labor.

The attached image shows what happens when the quantity of animals is increased and that of plants is decreased until reaching the optimal combination of 18 animals and no units of plants.

4 0
2 years ago
Kid's world industries has projected sales of 67,000 machines for the current year. the estimated january 1 inventory is 6,000 u
Valentin [98]
So, let us see the facts. The company needs to sell 67000 units throughout the year. We also need to have 15000 units in the storage so that we have 15000 in December. Hence, we need 82000 totally. But there are also 6000 already in storage. Hence we only need to produce 82000-6000=76000 units. If anything is unclear just comment.
4 0
2 years ago
Use the following information to answer next three questions: IO PI IRR LIFEProject 1 $300,000 1.12 14.38% 15 yearsProject 2 $15
Evgesh-ka [11]

Answer:

Project 1

Explanation:

                    IO          PI    IRR       LIFE

Project 1 $300,000 1.12 14.38% 15 years

Project 2 $150,000 1.08 13.32% 6 years

Project 3 $100,000 1.20 16.46% 3 years

Assume that the cost of capital is 12%.

We should invest in  the projects that have the highest profitability index (PI) first.

PI = present value of project's cash flows / initial outlay

Projects with a high PI should also have high IRRs and this applies to this situation:

  1. Project 3 has a PI of 1.2 and an IRR of 16.46%
  2. Project 1 has a PI of 1.12 and an IRR of 14.38%
  3. Project 2 has a PI of 1.08 and an IRR of 13.32%

If the protects weren't mutually exclusive and the company had enough money for the 3 of them, then it should invest in all of them. But that is not the case, here, since the company has to decide in which project it will invest (only 1 project). The first option should be project 3, but since it cannot be repeated, and its life is short, I would go for project 1.

Besides, it is the only possible answer since you have to choose only 1 project (remember projects are mutually exclusive).

6 0
1 year ago
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