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Step2247 [10]
3 years ago
10

Object-oriented systems have three general types of cohesion: _____, _____, and _____. A. method, class, inheritance B. method,

generalization/specialization, inheritance C. generalization/specialization, class, object D. method, class, generalization/specialization E. functional, sequential, procedural
Computers and Technology
1 answer:
Verdich [7]3 years ago
4 0

Answer:

D. method, class, generalization/specialization

Explanation:

The Cohesion is said to be the level to what an element of a module is related to others. Generally, the cohesion can be understood as an internal adhesive that holds together the modules. If the software is good then it is going to have high cohesion. Method and class are type of cohesion, as they hold the modules together. And generalization/specialization also is a general type of cohesion as generalization tracks the common features among the class, and binds them into one superclass. However, the specialization means creating subclasses out of the classes, and which is the meaning of the specialization. In an ideal situation, there should be high cohesion or single responsibility. However, in the case of inheritance, there can be multiple responsibilities. And hence composition like a generalization, specialization, association, aggregation are used to make use of the composition for reuse rather than inheritance. Have you seen language supporting multiple inheritances. You will never see it though there is some way. And its never allowed to ensure high cohesion. And hence three general types of cohesion are method, class and generalization/specialization. And this is option D.

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Suppose the daytime processing load consists of 65% CPU activity and 35% disk activity. Your customers are complaining that the
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The answer to this question can be described as follows:

Explanation:

Given data:

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S_{DISK} = (\frac{1}{(1-f_2)+\frac{f_2}{k_2}}) \\\\ S_{DISK} = (\frac{1}{(1-0.35)+\frac{0.35}{3}}) \\\\ = -0.5\% .... \rightarrow (2)

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equation (3) and (4),

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From (3) and (4) result,

The disc choice is therefore the best choice for a quicker system if you ever don't care about the cost.

c)

The break-event point for the upgrades:

=4702.19 x-0.5

= -2351.095

From (2) and (3))

Therefore, when you pay the sum for disc upgrades, all is equal $ -2351.095

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