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Snezhnost [94]
3 years ago
9

Sequence encryption is a series of encryptions and decryptions between a number of systems, wherein each system in a network dec

rypts the message sent to it and then reencrypts it using different keys and sends it to the next neighbor. This process continues until the message reaches the final destination.
a. True
b. False
Computers and Technology
1 answer:
valentina_108 [34]3 years ago
3 0

Answer:

The answer is True.

Explanation:

In a Network data transfer, it is essential for the security purpose that data should be in encrypted form which an unreadable form and it will be decrypted when it reaches the destination system.

It is important for the organization to encrypt data to make it safe from hackers and cryptographic technology is used in data Sequence encryption.

There is a key that is used to decrypt data in the network system and encryption and decryption is a sequential process for the data starting from the initial system and ending in the destination system.

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Which step is first in changing the proofing language of an entire document?
Genrish500 [490]
Select the whole document by pressing Ctrl+a.
7 0
3 years ago
Mezcla de una señal con otra para facilitar su transporte o transmisión.
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Answer:

telecomunicaciones

Explanation:

6 0
4 years ago
Define a function SetTime, with int parameters hoursVal and minutesVal, that returns a struct of type TimeHrMin. The function sh
Lapatulllka [165]

Answer:

#include <stdio.h>

typedef struct TimeHrMin_struct //struct

{

int hours;

int minutes;

} TimeHrMin;

struct TimeHrMin_struct SetTime(int hoursVal,int minutesVal) //SetTime function

{

struct TimeHrMin_struct str;

str.hours=hoursVal; //assigning the values

str.minutes=minutesVal;

return str; //returning the struct

}

int main(void)

{

TimeHrMin studentLateness;

int hours;

int minutes;

scanf("%d %d", &hours, &minutes);

studentLateness = SetTime(hours, minutes); //calling the function

printf("The student is %d hours and %d minutes late.\n", studentLateness.hours, studentLateness.minutes);

return 0;

}

Explanation:

4 0
3 years ago
What is the decimal form of the following binary ip address? 11001101.00111001.10101001.01000010
jasenka [17]

The decimal form of the 11001101.00111001.10101001.01000010 binary ip address is 205.57.169.66.

An IP address is a binary number with 32 bits. The 32 bits are divided into four octets, which are groupings of 8 bits each. An IP address, however, is shown as a dotted decimal number (for example: 205.57. 32.9).

Memory regions are given binary addresses by the computer system. But in order to access a memory location, the system utilises a certain number of bits. We can address two memory regions with 1 bit. We can address 4 memory locations with 2 bits and 8 memory locations with 3 bits.

The 4 sets of 8 bits in each of the 4 directions are simply written down to create the 32-bit binary IP address.

Learn more about binary ip address:

brainly.com/question/14541891

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3 0
2 years ago
1. Write a telephone lookup program. Read a data set of 1, 000 names and telephone numbers from a file that contains the numbers
Fed [463]

Answer:

Kindly note that, you're to replace "at" with shift 2 as the brainly text editor can't take the symbol

Explanation:

PhoneLookup.java

import java.io.FileReader;

import java.io.IOException;

import java.util.Scanner;

public class PhoneLookup

{

  public static void main(String[] args) throws IOException

  {

     Scanner in = new Scanner(System.in);

     System.out.println("Enter the name of the phonebook file: ");

     String fileName = in.nextLine();

     LookupTable table = new LookupTable();

     FileReader reader = new FileReader(fileName);

     table.read(new Scanner(reader));

   

     boolean more = true;

     while (more)

     {

        System.out.println("Lookup N)ame, P)hone number, Q)uit?");

        String cmd = in.nextLine();

       

        if (cmd.equalsIgnoreCase("Q"))

           more = false;

        else if (cmd.equalsIgnoreCase("N"))

        {

           System.out.println("Enter name:");

           String n = in.nextLine();

           System.out.println("Phone number: " + table.lookup(n));

        }

        else if (cmd.equalsIgnoreCase("P"))

        {

           System.out.println("Enter phone number:");

           String n = in.nextLine();

           System.out.println("Name: " + table.reverseLookup(n));

        }

     }

  }

}

LookupTable.java

import java.util.ArrayList;

import java.util.Collections;

import java.util.Scanner;

/**

  A table for lookups and reverse lookups

*/

public class LookupTable

{

  private ArrayList<Item> people;

  /**

     Constructs a LookupTable object.

  */

  public LookupTable()

  {

      people = new ArrayList<Item>();

  }

  /**

     Reads key/value pairs.

     "at"param in the scanner for reading the input

  */

  public void read(Scanner in)

  {

     while(in.hasNext()){

         String name = in.nextLine();

         String number = in.nextLine();

         people.add(new Item(name, number));

     }

  }

  /**

     Looks up an item in the table.

     "at"param k the key to find

     "at"return the value with the given key, or null if no

     such item was found.

  */

  public String lookup(String k)

  {

     String output = null;

     for(Item item: people){

         if(k.equals(item.getName())){

             output = item.getNumber();

         }

     }

     return output;

  }

  /**

     Looks up an item in the table.

     "at"param v the value to find

     "at"return the key with the given value, or null if no

     such item was found.

  */

  public String reverseLookup(String v)

  {

      String output = null;

         for(Item item: people){

             if(v.equals(item.getNumber())){

                 output = item.getName();

             }

         }

         return output;

  }

}

Item.java

public class Item {

  private String name, number;

 

  public Item(String aName, String aNumber){

      name = aName;

      number = aNumber;

  }

 

  public String getName(){

      return name;

  }

 

  public String getNumber(){

      return number;

  }

}

input.txt

Abbott, Amy

408-924-1669

Abeyta, Ric

408-924-2185

Abrams, Arthur

408-924-6120

Abriam-Yago, Kathy

408-924-3159

Accardo, Dan

408-924-2236

Acevedo, Elvira

408-924-5200

Acevedo, Gloria

408-924-6556

Achtenhagen, Stephen

408-924-3522

Kindly check the attached output image below.

3 0
3 years ago
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