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Sliva [168]
3 years ago
9

A stateless firewall inspects each incoming packet to determine whether it belongs to a currently active connection.

Computers and Technology
1 answer:
satela [25.4K]3 years ago
3 0

Answer:

False

Explanation:

Stateless firewall uses characteristics of the incoming packet to determine whether the packet should be permitted or rejected. Examples of such packets characteristics include:

- Source IP address

- Destination IP address

- Protocol

- Port

Stateless firewall does not keep track of active connection (state information) to be able to filter packets by that criteria.

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Create a Boolean function odd_number_digits(n) that returns True when a positive integer has an odd number of digits. (You may a
Snowcat [4.5K]

Answer:

Following are the code to this question:

import java.util.*;//import package for user input

public class Main//defining class main

{

   public static boolean odd_number_digits(int n)//defining boolean method odd_number_digits

   {

       if(n>0 && n%2!=0)//defining if block that check value is positive and odd number

       {

       return true;//return value true

       }

       else//defining else block

       {

           return false;//return false value

       }

   }

   public static void  sum_odd_digits(int n)//defining a method sum_odd_digits

   {

       int sum=0,i;//defining integer variable

       for(i=0;i<=n;i++)//defining for loop

       {

           if(i%2!=0)//defining if block for odd number

           {

               sum=sum+i;//add odd number

           }

       }

      System.out.print(sum);//use print method to print sum value

   }

public static void main(String[] args) //defining main method

   {

       Scanner ox=new Scanner(System.in);//creating Scanner object

       int n= ox.nextInt();//defining integer variable for input value

       System.out.print(odd_number_digits(n)+ "\n");//use print method to call method

       System.out.println("Sum of odd numbers: ");//print message

       sum_odd_digits(n);//calling method

   }

}

Output:

please find the attachment.

Explanation:

In the above code, two methods "odd_number_digits and sum_odd_digits" are defined in which the first method return type is boolean because it will true or false value, and the second method returns the sum of odd numbers.

In the "odd_number_digits" method,  an integer variable passes as an argument and inside the method, if block is used that check value is a positive and odd number then it will return a true value.

In the "sum_odd_digits" method, it accepts an integer parameter "n", and define integer variable "sum" inside the method, which uses the for loop, inside the loop if block is used that counts odd numbers and adds its in sum and print its value.

8 0
3 years ago
What outline feature can the Navigation pane browse the document by?
sattari [20]
It can browse the document by headings.
5 0
3 years ago
Read 2 more answers
What is the new file system available in windows server 2012?
exis [7]
Windows 2012 introduced a proprietary File system to success NTFS with ReSF (Resilient File System). There are many improvements made over NTFS as larger disks sizes, better quota management, additional attributes etc. also, the speed has significantly increased. There are many more improvement that you can simply Google.
6 0
4 years ago
Write a class called MagicSquare, which contains a single method called check that accepts a two-dimensional array as an argumen
ira [324]

Answer:

public class MagicSquare {

   public static void main(String[] args) {

       int[][] square = {

               { 8, 11, 14, 1},

               {13, 2, 7,12},

               { 3, 16, 9, 6},

               {10, 5, 4, 15}

       };

       System.out.printf("The square %s a magic square. %n",

               (isMagicSquare(square) ? "is" : "is not"));

   }

   public static boolean isMagicSquare(int[][] square) {

       if(square.length != square[0].length) {

           return false;

       }

       int sum = 0;

       for(int i = 0; i < square[0].length; ++i) {

           sum += square[0][i];

       }

       int d1 = 0, d2 = 0;

       for(int i = 0; i < square.length; ++i) {

           int row_sum = 0;

           int col_sum = 0;

           for(int j = 0; j < square[0].length; ++j) {

               if(i == j) {

                   d1 += square[i][j];

               }

               if(j == square.length-i-1) {

                   d2 += square[i][j];

               }

               row_sum += square[i][j];

               col_sum += square[j][i];

           }

           if(row_sum != sum || col_sum != sum) {

               return false;

           }

       }

       return d1 == sum && d2 == sum;

   }

}

5 0
3 years ago
Which of the following is true of lossy and lossless compression algorithms?
taurus [48]

B. Lossy compression algorithms are typically better than lossless compression algorithms at reducing the number of bits needed to represent a piece of data.

7 0
3 years ago
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