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Nana76 [90]
4 years ago
12

Create a C# GUI Windows Form application named JobDemo that declares and uses Job objects. The Job class holds job information f

or a home repair service. The class has five properties that include a job number, customer name, job description, estimated hours, and price for the job. Create a constructor that requires parameters for all the data except price. Include auto-implemented properties for the job number, customer name, and job description, but not for hours or price; the price field value is calculated as estimated hours times $45.00 whenever the hours value is set.
Also create the following for the class:

An Equals() method that determines two Jobs are equal if they have the same job number A ToString() method that returns a string containing all job information

The JobDemo Windows Form declares a few Job objects, sets their values, and demonstrates that all the methods work as expected.

Using the Job class you created in (a), write a new application named JobDemo2 that creates an array of five Job objects. Prompt the user for values for each Job. Do not allow duplicate job numbers; force the user to reenter the job when a duplicate job number is entered. When five valid objects have been entered (use List), display them all, plus a total of all prices.

Create a RushJob class that derives from Job. A RushJob has a $150.00 premium that is added to the normal price of the job. Override any methods in the parent class as necessary. Create a new Windows Form named JobDemo3 that creates an a List of five RushJobs. Prompt the user for values for each, and do not allow duplicates

Create a RushJob class that derives from Job. A RushJob has a $150.00 premium that is added to the normal price of the job. Override any methods in the parent class as necessary. Write a new Windows Form named JobDemo3 that creates a List of five RushJobs. Prompt the user for values for each, and do not allow duplicate job numbers. When five valid RushJob objects have been entered, display them all, plus a total of all prices. Make any necessary modifications to the RushJob class so that it can be sorted by job number.
Computers and Technology
1 answer:
murzikaleks [220]4 years ago
6 0

Answer:

The explanation of the given question is described in the below section.

Explanation:

This would undoubtedly be inappropriate to expose the organization where low ratings were issued but it would still be acceptable to do the same with prior authorization and approval from either the senior management.

  • Doing something really without top management expertise is immoral and refusing to obey the rules of the organization.
  • Furthermore, one may not be convinced of the supervisor's purpose for collecting this evidence or data, therefore, a documented proof about why this info is necessary might serve as a proof documentation for either the prosecution of that same immediate future.

So, it's the right answer.

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Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

^nC_r = \frac{n!}{(n-r)!r!}

Where n = 15\ and\ r = 4

^{15}C_4 = \frac{15!}{(15-4)!4!}

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^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}

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^{15}C_4 = 1365

<em>Hence, there are 1365 ways </em>

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

p = 3/15

p = 0.2

q = 1 - p

q = 1 - 0.2

q = 0.8

Solving further using binomial;

(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

Probability =  ^nC_3pq^3

Substitute 4 for n, 0.2 for p and 0.8 for q

Probability =  ^4C_3 * 0.2 * 0.8^3

Probability =  \frac{4!}{3!1!} * 0.2 * 0.8^3

Probability = 4 * 0.2 * 0.8^3

Probability = 0.4096

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

Probability =  q^n

Substitute 4 for n and 0.8 for q

Probability =  0.8^4

Probability = 0.4096

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

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