A computer network is a group of computers that use a set of common communication protocols over digital interconnections for the purpose of sharing resources located on or provided by the network nodes.
Answer:
By Using the Greedy- Activity- Selection algorithm
Explanation:
The Greedy- Activity- Selection algorithm in this case involves
First finding a maximum size set S1, of compatible activities from S for the first lecture hall.
Then using it again to find a maximum size set S2 of compatible activities from S - S1 for the second hall.
This is repeated till all the activities are assigned.
It requires θ(n2) time in its worse .
Answer:
import java.util.Scanner;
public class num8 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("Enter month's budget");
double monthBudget = in.nextDouble();
double totalExpenses = 0.0;
double n;
do{
System.out.println("Enter expenses Enter zero to stop");
n = in.nextDouble();
totalExpenses += n;
}while(n>0);
System.out.println("Total expenses is "+totalExpenses);
System.out.println("The amount over your budget is "+ Math.abs(monthBudget-totalExpenses));
}
}
Explanation:
- Using Java programming language
- Prompt user for month's budget
- Use Scanner class to receive and store the amount entered in a variable
- Use a do while loop to continuously request user to enter amount of expenses
- Use a variable totalExpenses to add up all the expenses inside the do while loop
- Terminate the loop when user enters 0 as amount.
- Subtract totalExpenses from monthBudget and display the difference as the amount over the budget
The answer a. record in the table.
Answer:
The following are the description of storing bits into Hard-disk.
Explanation:
As we know computer only know the binary language, that is in the form of zero's and once's "0's and 1's". In the hard drive, it requires a magnetically covered rotating disc, that's "head" passes over its platter.
- It marked 0's and 1's on the platter as tiny electronic areas in the north.
- Its head goes to the very same location to read its information again, the north and south places pass there and assume from the 0s and 1s which are contained.
