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lana66690 [7]
3 years ago
7

Security A: Probability 30% of Return 7%; Probability 50% of Return 12%; Probability of 20% of Return 17%. Security B: Probabili

ty 35% of Return 6%; Probability 65% of Return 16% What is the expected return for Security A
Business
1 answer:
earnstyle [38]3 years ago
5 0

Answer:

the expected return for security A is 11.50%

Explanation:

The computation of the expected return for security A is shown below:

The  Expected return for security A is

= Respective Return × respective probability

= (0.3 × 7) + (0.5 × 12) + (0.2 × 17)

= 11.50%

hence, the expected return for security A is 11.50%

We simply applied the above formula so that the correct value could come

And, the same is to be considered

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Why do people need to buy and sell products or services answers?
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A company purchases inventory on terms of net 30 days and resells to its customers on terms of net 15 days. The inventory conver
Tanzania [10]

Answer:

The company's cash conversion cycle is 75 days

Explanation:

The conversion cycle is the number of days that a business takes to convert its investment in inventory into cash flowing from the purchase and sales of the business.

Conversion cycle = Payable days + Inventory in Stock days + Receivable days

where

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6 0
3 years ago
A mining company owns two mines, each of which produces three grades (high, medium, and low) of ore. The company has a contract
sashaice [31]

Answer:

They should operate Mine 1 for 1 hour and Mine 2 for 3 hours to meet the contractual obligations and minimize cost.

Explanation:

The formulation of the linear programming is:

Objective function:

C=200M_1+160M_2

Restrictions:

- High-grade ore: 6M_1+2M_2\geq12

- Medium-grade ore: 2M_1+2M_2\geq8

- Low-grade ore: 4M_1+12M_2\geq24

- No negative hours: M_1,M_2\geq0

We start graphing the restrictions in a M1-M2 plane.

In the figure attached, we have the feasible region, where all the restrictions are validated, and the four points of intersection of 2 restrictions.

In one of this four points lies the minimum cost.

Graphically, we can graph the cost function over this feasible region, with different cost levels. When the line cost intersects one of the four points with the lowest level of cost, this is the optimum combination.

(NOTE: it is best to start with a low guessing of the cost and going up until it reaches one point in the feasible region).

The solution is for the point (M1=1, M2=3), with a cost of C=$680.

The cost function graph is attached.

7 0
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