Answer:
See explaination
Explanation:
import java.util.Scanner;
public class Word
{
public static void main(String args[])
{
Scanner read=new Scanner(System.in);
char repeat='Y';
String phrase=null;
int index=0;
while(repeat=='Y')
{
System.out.println("enter a phrase :");
phrase=read.nextLine();
while(index<=0)
{
System.out.println("enter an index greater than 0");
index=Integer.parseInt(read.nextLine());
}
String s;
int spaces = phrase == null ? 0 : phrase.length() - phrase.replace(" ", "").length();
int numofwords=spaces+1;
if(index>numofwords)
{
index=numofwords;
}
System.out.println("word is: "+ getWord(phrase,index));
System.out.println("do you want to repeat (Y/N)");
repeat=read.nextLine().charAt(0);
index=0;
}
read.close();
}
private static String getWord(String phrase, int index) {
// TODO Auto-generated method stub
Scanner in =new Scanner(phrase);
String word=null;
int wordindex=0;
while(wordindex!=index)
{
word=in.next();
wordindex++;
}
in.close();
return word;
}
}
Check attachment screenshot
Answer:
True
Explanation:
Not(False) OR False = True OR False = True
Answer:

Explanation:
The Hexa-decimal numbers have base 16 and includes numbers:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F
The given steps are performed to convert a decimal number to hexa-decimal number, here to convert decimal number 35 to hexa-decimal number:
- Divide 35 by 16
- Note the remainder, r which is 3 here and quotient which is 2
- Again divide 2 (quotient) by 16 and note the remainder, r' which is 2 and quotient is 0
- We will stop here as the quotient is now 0. Usually division by 16 is repeated until we get quotient = 0
- Now arrange the remainder in reverse order to get the hexa-decimal number as r'r
- The hexa-decimal number is

Yes it is D like the other person said
Answer:
4 bits
Explanation:
With 4 bits you can count to 15, because 2⁴=16. The maximum number you can express is always one less, i.e., 16-1=15.
In general, with n bits you can count to 2ⁿ-1.