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Anastasy [175]
2 years ago
7

List the seven basic internal components found in a computer tower

Computers and Technology
1 answer:
Softa [21]2 years ago
6 0
The basic internal components found in a computer tower also called as computer case are the following:

1. The motherboard
2. The power supply
3. Hard drives
4. Fan
5. Random Access Memory - RAM
6. CD - Rom's and DVD - Rom's
7. Ports and Hubs

These are found inside your PC case.

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Create a new Die object. (Refer to Die.html for documentation.)Create a loop that will iterate at least 100 times. In the loop b
hjlf

Answer:

Java code is given below

Explanation:

import java.util.Random;

class Die{

private int sides;

public Die(){

sides = 6;

}

public Die(int s){

sides = s;

}

public int Roll(){

Random r = new Random();

return r.nextInt(6)+1;

}

}

class DieRoll{

public static void main(String[] args) {

Die die = new Die();

int arr[] = new int[6];

for(int i=0; i<6; i++)

arr[i] = 0;

for(int i=0; i<100; i++){

int r = die.Roll();

arr[r-1]++;

}

for(int i=0; i<6; i++)

System.out.println((i+1)+" was rolled "+arr[i]+" times.");

}

}

8 0
3 years ago
In Scratch, you have to choose backdrops from a limited number in the Scratch image library.
forsale [732]
I would say it would be True
5 0
3 years ago
Input 10 integers and display the following:
LekaFEV [45]

Answer:

// code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;

   int largest=INT_MIN;

   int smallest=INT_MAX;

   int n;

   cout<<"Enter 10 Integers:";

   // read 10 Integers

   for(int a=0;a<10;a++)

   {

       cin>>n;

       // find largest

       if(n>largest)

       largest=n;

       // find smallest

       if(n<smallest)

       smallest=n;

       // if input is even

       if(n%2==0)

       {  

           // sum of even

           sum_even+=n;

           // even count

           eve_count++;

       }

       else

       {

           // sum of odd    

          sum_odd+=n;

          // odd count

          odd_count++;

       }

   }

   

   // print sum of even

   cout<<"Sum of all even numbers is: "<<sum_even<<endl;

   // print sum of odd

   cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;

   // print largest

   cout<<"largest Integer is: "<<largest<<endl;

   // print smallest

   cout<<"smallest Integer is: "<<smallest<<endl;

   // print even count

   cout<<"count of even number is: "<<eve_count<<endl;

   // print odd cout

   cout<<"count of odd number is: "<<odd_count<<endl;

return 0;

}

Explanation:

Read an integer from user.If the input is greater that largest then update the  largest.If the input is smaller than smallest then update the smallest.Then check  if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.

Output:

Enter 10 Integers:1 3 4  2 10 11 12 44 5 20                                                                                

Sum of all even numbers is: 92                                                                                            

Sum of all odd numbers is: 20                                                                                              

largest Integer is: 44                                                                                                    

smallest Integer is: 1                                                                                                    

count of even number is: 6                                                                                                

count of odd number is: 4

3 0
3 years ago
CODING IN C++ URGENT HELP PLEASE
forsale [732]

Answer:

this is hard i thought i knew it wow u have hard stuff in your school

Explanation:

7 0
3 years ago
gven an IP address and mask of 192.168.12.0/24, design an IP addressing scheme that satisfies the following requirements. Subnet
Reika [66]
Uh 69 +23= 92
Hope this helps :)
8 0
3 years ago
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