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3 years ago

Which intermediate sanction programs take form of residential facilities that provide supervision to offenders?

1 answer:

4 0

Community Supervision and Corrections Departments are residential facilities run by the Department of Corrections at which offenders on probation and post-parole release receive supervision, vocational training, and counseling.

Beginner

My bad if i'm not wrong or i am

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A forecast that addresses the business cycle by predicting planning indicators is

zloy xaker [14]

Answer:

B. a demand forecast.

Explanation:

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3 years ago

Compare and contrast the role of management, as reflected by management theory, in the 20thcentury organization with the role of

Alexxx [7]

Do it yourself this gets you no where im sorry

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3 years ago

In a local market, the monthly price of internet access service decreases from $40 to $30, and the total quantity of monthly

deff fn [24]

The answer is : The demand is elastic.

Elasticity =

[(80,000 - 180,000)/((80,000+180,000)/2)]/[($40 - $30)/(($40 + $30)/2)]|

[(-100,000/130,000)]/[(10/55)] = -.7692/.1818= -4.23

The answer is -4.23, however when considering own price elasticity of demand, we ignore the negative sign and look at the absolute value to determine whether it is elastic or inelastic.

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3 years ago

No trespassing signs are an example of owners enforcing their right to

Ilia_Sergeevich [38]

Answer:

I guess the ans is their right to exclude people from your property.

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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.

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3 years ago