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Finger [1]
3 years ago
10

Duval Inc. uses only equity capital, and it has two equally-sized divisions. Division A’s cost of capital is 10.0%, Division B’s

cost is 14.0%, and the corporate (composite) WACC is 12.0%. All of Division A’s projects are equally risky, as are all of Division B's projects. However, the projects of Division A are less risky than those of Division B. Which of the following projects should the firm accept?1. A Division B project with a 13% return.2. A Division B project with a 12% return.3. A Division A project with an 11% return.4. A Division A project with a 9% return.5.A Division B project with an 11% return.
Business
1 answer:
Lorico [155]3 years ago
8 0

Answer:

The answer is 3. A Division A project with an 11%.

Explanation:

The projects Duval Inc. should accept is the projects generating an expected return higher than its Cost of Capital.

Because all of Division A’s projects are equally risky, as are all of Division B's projects, each project of Division A will have Cost of capital of 10.0% and each project of Division B will have a Cost of capital of 14%.

For 5 projects given, only project described in (3) has expected return higher than its cost of capital ( 11% in comparison to 10%).

As a result, (3) is the correct answer.

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Violations of security policies are considered to be a(n) __________ issue upon which proper disciplinary actions must be taken.
kari74 [83]

<u>Answer: </u>

Violations of security policies are considered to be a law enforcement issue upon which proper disciplinary actions must be taken.

<u>Explanation: </u>

  • The security policies put into place by the organization actually serve as the first wall of protection for the organization.
  • The organization makes it clear through its security policies what actions would be deemed as an attempt to breach the security and what actions are prohibited within and outside the premises of the organization as an employee of the organization:
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3 years ago
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Arisa [49]

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8 0
2 years ago
What can cause the market demand curve for day-old bread -- an inferior good -- to shift rightward (an increase in demand)
Sedbober [7]

A factor that can cause the market demand curve for day-old bread to shift rightward is a decrease in income.

<h3>What is an inferior good?</h3>

An inferior good is a good whose demand falls when income rises and increases when income falls. An inferior good is a good whose demand increases when income falls.

<h3>What is an increase in demand?</h3>

An increase in demand occurs when the demand for a good increase as a result of factors other than a change in the price of a good. When there is an increase in demand, there would be a shift to the right of the demand curve.

For more information about the increase in demand, please check: brainly.com/question/25871620

5 0
3 years ago
The inventory of Bramble Company on December 31, 2020, consists of the following items. Part Quantity Cost per Unit Net Realizab
Brrunno [24]

Answer:

$462,562

Explanation:

Lower of Cost and Net Realizable Value (LCNRV) records ending inventory at the lowest between purchase costs or net realizable value.

Part         Q             Cost per Unit          Net RV            Total

110        620              <u>$121.00</u>                 $127.00          $75,020

111        1,080             $76.20                  <u>$66.00</u>            $71,280        

112         540             $101.60                  <u>$97.00</u>           $52,380

113         220            <u>$215.90</u>                $228.60           $47,498

120        440           <u>$260.00</u>                $264.00          $114,400

121      1,400              $20.00                     <u>$1.00</u>             $1,400

122       330             $304.80               <u>$298.00</u>         $100,584

Total                                                                             $462,562

6 0
3 years ago
For a particular flight from Dulles to SF, an airline uses wide-body jets with a capacity of 440 passengers. It costs the airlin
BARSIC [14]

Answer:

$378.75

Explanation:

Data provided:

Capacity = 440 passengers

Operating cost = $4,000 + $70(Number of passengers)

Expected number of passengers = 440 - 0.64T

Ticket price = T

Total operational cost = $4000 + $70( 440-0.64T )

Total operational cost = $34,800 - 44.8T

Thus,

Total revenue = Number of passengers × Ticket price

= (440 - 0.64T)T

= 440T - 0.64T²

also,

Total profit ,P(T) = Total revenue - Total operational cost

P(T) = ( 440T - 0.64T²) - (34,800 - 44.8T)

P(T) = - 0.64T² - 34,800 + 484.8T

Now,

Differentiating with respect to ticket price T

P'(T) = -0.64(2)T - 0 + 484.8(1)

or

P'(T) = - 1.28T + 484.8 ..............(1)

For point of maxima or minima

P'(T) = 0

or

 - 1.28T + 484.8 = 0

or

1.28T = 484.8

or

T = $378.75

now,

again differentiating (1) to check for maxima or minima

P''(T)= -1.26(1) + 0

P''(T) = -1.26

Since,

P"(T)  < 0

Hence,

T = $378.75 will maximise the profit

6 0
4 years ago
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