Answer:
The program is as follows:
#include <iostream>
#include <iomanip>
using namespace std;
int main(){
int num1, num2, num3;
cin>>num1>>num2>>num3;
cout << fixed << setprecision(2);
cout<<(num1 + num2 + num3)/3<<" ";
cout<<num1 * num2 * num3<<" ";
return 0;
}
Explanation:
This declares three integer variables
int num1, num2, num3;
This gets input for the three integers
cin>>num1>>num2>>num3;
This is used to set the precision to 2
cout << fixed << setprecision(2);
This prints the average
cout<<(num1 + num2 + num3)/3<<" ";
This prints the product
cout<<num1 * num2 * num3<<" ";
Answer:
Code is completed below
Explanation:
Source Code in Java:
class Parenthesis
{
boolean hasBalancedParentheses(String eq) //method to check and return if a set of parenthesis is balanced or not
{
int count=0; //to store count of current unclosed opening brackets
for(int i=0;i<eq.length();i++)
{
if(eq.charAt(i)=='(')
count++;
else if(eq.charAt(i)==')')
count--;
if(count<0) //if count falls below zero, there were more closing brackets than opening brackets till this point
return false;
}
if(count>0) //if count is still more than zero, there were less closing brackets than opening brackets
return false;
return true; //if program reaches this point, it has passed all the tests
}
public static void main(String args[])
{
//testing the function
Parenthesis ob=new Parenthesis();
System.out.println(ob.hasBalancedParentheses("()(()())((())())"));
System.out.println(ob.hasBalancedParentheses(")((()())(()))())"));
}
}
Answer:
See explanation below
Explanation:
Previos concepts
First Come First Serve (FCFS) "is an operating system scheduling algorithm that automatically executes queued requests and processes in order of their arrival".
Shortest job next (SJN), or the shortest job first (SJF) or shortest "is a scheduling policy that selects for execution the waiting process with the smallest execution time".
Shortest remaining time (SRF) "is a scheduling method that is a preemptive version of shortest job next scheduling'".
Round robin (RR) is an algorithm where the time parts "are assigned to each process in equal portions and in circular order, handling all processes without priority"
Solution for the problem
Assuming the dataset given on the plot attached.
Part a
For this algorithm the result would be:
Job A 0-6
Job B 6-(6+3) = 6-9
Job C 9-(9+1) = 9-10
Job D 10-(10+4) = 10-14
Part b
For this algorithm the result would be:
Job A 0-6
Job C 6-(6+1) = 6-7
Job B 7-(7+3) = 7-10
Job D 10-(10+4) = 10-14
Part c
For this algorithm the result would be:
Job A 0-1 until 14
Job B 2-(2+3) = 2-5
Job C 3-(3+2) = 3-5
Job D 9-(9+5) = 9-14
Part d
For this algorithm the result would be:
Job A 0-2 , 7-9, 12-14
Job B 2-4, 9-10
Job C 4-(4+1) = 4-5
Job D 5-7, 10-12
what the answer to this question....
