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Firdavs [7]
1 year ago
13

The layer of the ISO/OSI responsible for source to destination delivery.

Computers and Technology
1 answer:
Andrej [43]1 year ago
7 0

Answer:

(b) network

Explanation:

The Network Layer is the OSI Model's third layer. It is in charge of packet transport from source to destination or host to host across various networks. The layer receives data from the transport layer, adds a header to it, and sends it to the data link layer.

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Imagine you are responsible for making a presentation that includes a representation of the logic flow through a process. You un
svetoff [14.1K]

Answer:

find reliable resources

Explanation:

if you find reliable resources, than you might able to have more help i know im not answering..but im trynna help

3 0
2 years ago
Which two climates have moderate rainfall and experience warm summers and cold winters due to their position relative to mountai
77julia77 [94]
Most probably grassland and steppes
6 0
2 years ago
Write a program that converts a number entered in Roman numerals to decimal. your program should consist of a class, say, Roman.
arlik [135]

Answer:

The code is given below in Java with appropriate comments

Explanation:

import java.util.*;

public class RomantoDecimal {

    public static void main(String[] args)

    {

         Scanner SC = new Scanner(System.in);

         RomantoDecimal r = new RomantoDecimal();

         System.out.println("Enter a Roman number :");

         // INPUT A ROMAN NUMBER

         String rNum = SC.next();

         // CALL convertToDecimal FOR CONVERSION OF ROMAN TO DECIMAL

         r.convertToDecimal(rNum);

    }

    // M=1000, D=500, C=100, L=50, X=10, V=5, I=1

    public void convertToDecimal(String roamNo)

    {

         int number = 0;

         // TEACK EACH DIGIT IN THE GIVEN NUMBER IN REVERSE ORDER

         for (int i = roamNo.length() - 1; i >= 0; i--)

         {

             // FIND OUT WHETHER IT IS 'M' OR NOT

             if (roamNo.charAt(i) == 'M')

             {

                  if (i != 0)

                  { // CHECK WHETHER THERE IS C BEFORE M

                       if (roamNo.charAt(i - 1) == 'C')

                       {

                            number = number + 900;

                            i--;

                            continue;

                       }

                  }

                  number = number + 1000;

             }

             // FIND OUT WHETHER IT IS 'D' OR NOT

             else if (roamNo.charAt(i) == 'D')

             {

                  if (i != 0)

                  {

                  // CHECK WHETHER THERE IS C BEFORE D

                       if (roamNo.charAt(i - 1) == 'C')

                       {

                            number = number + 400;

                            i--;

                            continue;

                       }

                 }

                  number = number + 500;

             }

            // FIND OUT WHETHER IT IS 'C' OR NOT

             else if (roamNo.charAt(i) == 'C')

             {

                  if (i != 0)

                  {

                       if (roamNo.charAt(i - 1) == 'X')

                       {

                            number = number + 90;

                            i--;

                            continue;

                      }

                  }

                  number = number + 100;

             }

             else if (roamNo.charAt(i) == 'L')

             {

                  if (i != 0)

                  {

                       if (roamNo.charAt(i) == 'X')

                       {

                            number = number + 40;

                            i--;

                            continue;

                       }

                  }

                  number = number + 50;

             }

             // FIND OUT WHETHER IT IS 'X' OR NOT

             else if (roamNo.charAt(i) == 'X')

             {

                  if (i != 0)

                  {

                       if (roamNo.charAt(i - 1) == 'I')

                       {

                            number = number + 9;

                            i--;

                            continue;

                       }

                  }

                  number = number + 10;

             }

             // FIND OUT WHETHER IT IS 'V' OR NOT

             else if (roamNo.charAt(i) == 'V')

             {

                  if (i != 0)

                  {

                       if (roamNo.charAt(i - 1) == 'I')

                       {

                            number = number + 4;

                            i--;

                            continue;

                       }

                  }

                  number = number + 5;

             }

            else // FIND OUT WHETHER IT IS 'I' OR NOT

             {

                  number = number + 1;

             }

         }// end for loop

         System.out.println("Roman number: " + roamNo + "\nIts Decimal Equivalent: " + number);

    }

}

3 0
3 years ago
Write a python program to read four numbers (representing the four octets of an IP) and print the next five IP addresses. Be sur
sdas [7]

Answer:

first_octet = int(input("Enter the first octet: "))

second_octet = int(input("Enter the second octet: "))

third_octet = int(input("Enter the third octet: "))

forth_octet = int(input("Enter the forth octet: "))

octet_start = forth_octet + 1

octet_end = forth_octet + 6

if (1 <= first_octet <= 255) and (0 <= second_octet <= 255) and (0 <= third_octet <= 255) and (0 <= forth_octet <= 255):

   for ip in range(octet_start, octet_end):

       forth_octet = forth_octet + 1

       if forth_octet > 255:

           forth_octet = (forth_octet % 255) - 1

           third_octet = third_octet + 1

           if third_octet > 255:

               third_octet = (third_octet % 255) - 1

               second_octet = second_octet + 1

               if second_octet > 255:

                   second_octet = (second_octet % 255) - 1

                   first_octet = first_octet + 1

                   if first_octet > 255:

                       print("No more available IP!")

                       break

       print(str(first_octet) + "." + str(second_octet) + "." + str(third_octet) + "." + str(forth_octet))

else:

   print("Invalid input!")

Explanation:

- Ask the user for the octets

- Initialize the start and end points of the loop, since we will be printing next 5 IP range is set accordingly

- Check if the octets meet the restrictions

- Inside the loop, increase the forth octet by 1 on each iteration

- Check if octets reach the limit - 255, if they are greater than 255, calculate the mod and subtract 1. Then increase the previous octet by 1.

For example, if the input is: 1. 1. 20. 255, next ones will be:

1. 1. 21. 0

1. 1. 21. 1

1. 1. 21. 2

1. 1. 21. 3

1. 1. 21. 4

There is an exception for the first octet, if it reaches 255 and others also reach 255, this means there are no IP available.

- Print the result

8 0
3 years ago
Plz hurry !!!
lakkis [162]

Answer: The <u>Ordering Tag List </u>defines the numbering styles of lists in HTML pages.

<ol>

</ol>

Inside this tag, you have <li></li> which will contain individual lists.

For example:

<ol>

   <li> This is my first point. </li>

   <li> This is my second point. </li>

   <li> This is my second point. </li>

</ol>

is displayed as:

1. This is my first point. 

2. This is my second point. 

3. This is my second point.

Read more on Brainly.com - brainly.com/question/9724249#readmore

Explanation:

8 0
3 years ago
Read 2 more answers
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