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Feliz [49]
3 years ago
11

A diagnostic program built into a computer to test the hardware components before the computer boots up

Computers and Technology
1 answer:
Allushta [10]3 years ago
8 0
BIOS thank me later
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A desktop computer (named workstation22) can’t connect to the network. A network card was purchased without documentation or dri
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Answer:

<em>The first step is to download the driver file from the Internet using  another Computer system, Copy the driver to a flash or CD then run the installation file on the Desktop Computer.</em>

Explanation:

<em>Driver files can be easily gotten from software sites online or from other secure websites, most times you have to pay for these driver files in some sites for secure and authentic download.</em>

<em>it is very necessary  that the user takes note of the exact system name, model and system architecture with respect to when downloading the driver file.</em>

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All of the following are aspects of the search process except?
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<span>d. select information is sent to the search engines database to be indexed. </span>
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Write a destructor for the CarCounter class that outputs the following. End with newline.
ycow [4]

Answer:

Following are the code to this question:

CarCounter::~CarCounter()//Defining destructor CarCounter

{

cout << "Destroying CarCounter\n";//print message Destroying CarCounter

}

Explanation:

Following are the full program to this question:

#include <iostream>//Defining header file

using namespace std;

class CarCounter //Defining class CarCounter

{

public:

CarCounter();//Defining constructor CarCounter

~CarCounter();//Defining destructor CarCounter

private:

int carCount;//Defining integer variable carCount

};

CarCounter::CarCounter()//declaring constructor  

{

carCount = 0;//assign value in carCount variable

return;//using return keyword

}

CarCounter::~CarCounter()//Defining destructor CarCounter

{

cout << "Destroying CarCounter\n";//print message Destroying CarCounter

}

int main() //Defining main method

{

CarCounter* parkingLot = new CarCounter();//Defining class object parkingLot

delete parkingLot;//

return 0;

}

  • In the given C++ language code, a class "CarCounter" is defined, and inside the class, a "constructor, Destructors, and an integer variable" is defined.  
  • Outside the class, the scope resolution operator is used to define the constructor and assign value "0" in the integer variable.  
  • In the above-given code, the scope resolution operator, to define destructor and inside this cout function is used, which prints a message.  
  • In the main method, the class object is created, which automatically calls its class constructor and destructors.  
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What is the meaning of web browser
anastassius [24]
Is that a joke... if ain't cheatin you repeatin
8 0
3 years ago
1. How many bits would you need to address a 2M × 32 memory if:
Dominik [7]

Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0
3 years ago
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