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Allushta [10]
3 years ago
14

The template code provided is intended to take two inputs, x and y, from the user and print "pass" if one or more of the followi

ng is true:
x is not less than 4
y is not greater than 5 and x + y is less than 7
However, when using De Morgan's law to simplify this code, the programmer has made some mistakes. Can you correct the errors so the code functions as intended?
/* Lesson 6 Coding Activity Question 2 */
import java.util.Scanner;
public class U3_L6_Activity_Two{
public static void main(String[] args){
Scanner scan = new Scanner(System.in);
int x = scan.nextInt();
int y = scan.nextInt();
if(!((x 5) || x+y > 7))
System.out.println("pass");
}
}
Computers and Technology
1 answer:
just olya [345]3 years ago
7 0

import java.util.Scanner;

public class U3_L6_Activity_Two{

   public static void main(String[] args){

       Scanner scan = new Scanner(System.in);

       int x = scan.nextInt();

       int y = scan.nextInt();

       if(x>=4|| ((y < 5) && ((x+y) < 7))){

           System.out.println("pass");

       }

   }

}

I'm pretty sure this is what you're looking for. Best of luck.

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Answer:

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a) Machine's Netmask = /27 : therefore no. of remaining bits for hosts =32-27 = 5 bits.

b) No. of usable addresses in the child network = 25 -2 = 32-2 =30 [Since first(network ID of the machine) and last ip (broadcast address of the machine ) addresses are not used ]

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First, find out the network id of machine can be found out by doing bitwise AND machine's IP and Machine's subnet mask :

01111110. 01111111.01010101.10101010 (IP)

11111111. 11111111 .11111111 .11100000 (Subnet Mask)

01111110. 01111111. 01010101.10100000 (Network ID )

first usable address is on this child network (subnet) :   01111110. 01111111. 01010101.10100001

: 126.127.85.161

d) what the last usable address is on this child network (subnet) :  01111110. 01111111. 01010101.10111110

: 126.127.85.190

e) what the child network’s (subnet's) broadcast address is :

In the directed broadcast address , all the host bits are 1. Therefore, broadcast address :

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01111110. 01111111. 01010101.10100000 (126.127.85.160)

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