Answer: d. the time quantum should be as small as possible.
Explanation: Round robin is a CPU scheduling algorithm. If the time quantum is extremely small, number of context switches is very high, round robin scheduling will be same as processor sharing algorithm. Then the algorithm is good. Very large time quantum will make response time of processes too much which may not be tolerated in interactive environment.
Answer:
PART ONE
- import java.util.Scanner;
- public class CountToLimit {
- public static void main(String[] args) {
- Scanner scnr = new Scanner(System.in);
- int countLimit = 0;
- int printVal = 0;
- // Get user input
- System.out.println("Enter Count Limit");
- countLimit = scnr.nextInt();
- do {
- System.out.print(printVal + " ");
- printVal = printVal + 1;
- } while ( printVal<=countLimit );
- System.out.println("");
- return;
- }
- }
PART TWO
- import java.util.Scanner;
- public class NumberPrompt {
- public static void main (String [] args) {
- Scanner scnr = new Scanner(System.in);
- System.out.print("Your number < 100: ");
- int userInput = scnr.nextInt();
- do {
- System.out.print("Your number < 100: ");
- userInput = scnr.nextInt();
- }while (userInput>=100);
- System.out.println("Your number < 100 is: " + userInput);
- return;
- }
- }
Explanation:
In Part one of the question, The condition for the do...while loop had to be stated this is stated on line 14
In part 2, A do....while loop that will repeatedly prompt user to enter a number less than 100 is created. from line 7 to line 10

Convert
0.625 to binary

Translate 0.625 into a fraction. We all know that 0.5 is ½. We know that the remainder, 0.125, is ⅛. Add them together, and you get ½ + ⅛ = ⅝.
Now, in binary, the positions to the right of the point are , which is ½, ¼, and ⅛ respectively.
⅝ is 5 × ⅛. 5 in binary is 101. So, ⅝ is
= 0.101
Answer:
Grass
Explanation:
When the wind blows grass or small plants start dancing (?)
<em>#</em><em>S</em><em>p</em><em>r</em><em>e</em><em>a</em><em>d</em><em>T</em><em>h</em><em>e</em><em>K</em><em>n</em><em>o</em><em>w</em><em>l</em><em>e</em><em>d</em><em>g</em><em>e</em>