Which of these is an example of a system?

A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

Dima020 [189]

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 $n = 15\ and\ r = 4$

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

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

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

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

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)

From the given parameters; 3 out of 15 is detective;

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;

Probability that at least one is defective + Probability that at none is defective = 1

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

8 0

3 years ago

Write a copy assignment operator for CarCounter that assigns objToCopy.carCount to the new objects's carCount, then returns *thi

Olin [163]

The following cose will be used to copy assignment operator for CarCounter

Explanation:

Complete Program:

#include <iostream>

using namespace std;

class CarCounter

{

public:

CarCounter();

CarCounter& operator=(const CarCounter& objToCopy);

void SetCarCount(const int setVal)

{

carCount = setVal;

}

int GetCarCount() const

{

return carCount;

}

private:

int carCount;

};

CarCounter::CarCounter()

{

carCount = 0;

return;

}

// FIXME write copy assignment operator

/* Your solution goes here */

CarCounter& CarCounter::operator=(const CarCounter& objToCopy)

{

if(this != &objToCopy)

carCount = objToCopy.carCount;

return *this;

}

int main()

{

CarCounter frontParkingLot;

CarCounter backParkingLot;

frontParkingLot.SetCarCount(12);

backParkingLot = frontParkingLot;

cout << "Cars counted: " << backParkingLot.GetCarCount();

cout << endl << endl;

system("pause");

return 0;

}

5 0

3 years ago

What is the use of html in websites

Alborosie

HTML is a very basic markup language and requires memorization of a few dozen HTML commands that structure the look and layout of a web page. Before writing any HTML code or designing your first web page, you must decide on an HTML editor or text editor, such as Notepad or Word Pad.

4 0

3 years ago

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Dam naj za pierwszej osobie która odpowie dobrze!

Rama09 [41]

I don’t understand that language

8 0

3 years ago

You don’t design new software. Instead, you work with software that already exists to be sure that it keeps running smoothly. Wh

Arada [10]

Answer:

Explanation:

If you are developing software products, run complex projects, you should ... All the same, to make the right choice of an SDLC methodology, you ... Generally, there are stages involved in all the different methodologies. ... You will therefore have to keep both the old and new systems running for some time.

5 0

2 years ago

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