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maw [93]
3 years ago
15

A common stock pays an annual dividend per share of $1.80. The risk-free rate is 5%, and the risk premium for this stock is 4%.

If the annual dividend is expected to remain at $1.80 per share, what is the value of the stock
Business
1 answer:
ArbitrLikvidat [17]3 years ago
3 0

Answer:

The value of the stock today is $20

Explanation:

Using the CAPM equation, we first calculate the required rate of retunr on the stock.

The equation for CAPM is,

r = rRF + Beta * rpM

Where,

  • rRF is the risk free rate
  • rpM is the risk premium on market
  • Beta * rpM is the risk premium on stock

r = 0.05 + 0.04

r = 0.09 or 9%

The value of the stock can be calculated using the zero growth model of DDM. The DDM values the stock based on the present value of the expected future dividends from the stock. As the dividend from the stock is expected to remain constant through out to an indefinite period, the value of the stock today is,

P0 = Dividend / r

P0 = 1.8 / 0.09

P0 = $20

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Delta Construction Corporation, a general contractor, hires Eagle Electrical Company, a subcontractor, to wire a new office buil
nignag [31]

Answer: excused by Delta's failure to pay.

Explanation:

Delta Construction Corporation hires Eagle Electrical Company, as a subcontractor, to wire its new office building. After the completion of the work, Eagles is owed more than $50000.

Eagle's suspension of work is most likely due to the excuse by Delta's failure to pay. Delta has a right to pay up the money owed to Eagle. Lack of payment can lead to court cases.

3 0
3 years ago
Becca is a manager in the credit department for 3STAR Motors. Joe Greene is a new employee in her department. While Joe has been
Dmitriy789 [7]

Answer: coaching Joe rather than helping him.

Explanation:

Based on the scenario in the question, Becca's approach to getting Joe up to speed indicates that she is coaching rather Joe than helping him.

A coach is someone who guides someone and makes them better and believe in themselves. Since Becca has provided Joe with guidance by offering advice, encouragement, and instructions, this shows that Becca has been coaching him.

4 0
3 years ago
Constable Co. reported the following information at December 31, Year 1:
zhenek [66]

Answer:

The amount of Current liabilities is $7,710

Explanation:

The amount of current liabilities on the classified balance sheet is seen below;

Constable Corp.

Balance sheet as at December 31, year 1.

Current liabilities

Accounts payable $4,540

Wages payable $3,170

Total $7,710

5 0
3 years ago
Your company wants to set aside a fixed amount every year to a sinking fund to replace a piece of industrial equipment costing $
olga2289 [7]

Answer:

$4,5243.63

Explanation:

Data provided in the question:

Future value = $250,000

Interest rate = 5% = 0.05

Time = 5 years

Now,

Future value = C\times\left[ \frac{(1+i)^{n}-1}{i} \right]

here,

C = Regular deposit amount

i = Interest rate per period

n = number of periods

Future value = C\times\left[ \frac{(1+i)^{n}-1}{i} \right]

or

$250,000 = C\times\left[ \frac{(1+0.05)^{ 5}-1}{ 0.05} \right]

$250,000 = C\times\left[ \frac{ 1.05^{ 5}-1}{ 0.05} \right]

$250,000 = C\times\left[ \frac{ 1.276282 - 1}{ 0.05} \right]

$250,000 = C × 5.52564

or

C = \frac{ 250000 }{ 5.52564 }

C = $4,5243.63

7 0
4 years ago
The new Fore and Aft Marina is to be located on the Ohio River near Madison, Indiana. Assume that Fore and Aft decides to build
Nataly [62]

Po = 0.5385, Lq = 0.0593 boats, Wq = 0.5930 minutes, W = 6.5930 minutes.

<u>Explanation:</u>

The problem is that of Multiple-server Queuing Model.

Number of servers, M = 2.

Arrival rate, \lambda= 6 boats per hour.

Service rate, \mu= 10 boats per hour.

Probability of zero boats in the system,\mathrm{PO}=1 /\{[(1 / 0 !) \times(6 / 10) 0+(1 / 1 !) \times(6 / 10) 1]+[(6 / 10) 2 /(2 ! \times(1-(6 /(2 \times 10)))]\} = 0.5385

<u>Average number of boats waiting in line for service:</u>

Lq =[\lambda.\mu.( \lambda / \mu )M / {(M – 1)! (M. \mu – \lambda )2}] x P0

= [\{6 \times 10 \times(6 / 10) 2\} /\{(2-1) ! \times((2 \times 10)-6) 2\}] \times 0.5385 = 0.0593 boats.

The average time a boat will spend waiting for service, Wq  =  0.0593 divide by 6 = 0.009883 hours = 0.5930 minutes.

The average time a boat will spend at the dock, W =  0.009883 plus (1 divide 10) = 0.109883 hours = 6.5930 minutes.

4 0
3 years ago
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