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GrogVix [38]
2 years ago
6

In 2007, this wireless security algorithm was rendered useless by capturing packets and discovering the passkey in a matter of s

econds. This security flaw led to a network invasion of TJ Maxx and data theft through a technique known as wardriving. Which Algorithm is this referring to?
A. Wired Equivalent Privacy (WEP)
B. Wi-Fi Protected Access (WPA)
C. Wi-Fi Protected Access 2 (WPA2)
D. Temporal Key Integrity Protocol (TKIP)
Computers and Technology
2 answers:
Basile [38]2 years ago
7 0

Answer:

The correct answer to the following question is option A. Wired Equivalent Privacy (WEP).

Explanation:

WEP (Wired Equivalent Privacy) is the security protocol which is specified in IEEE Wi-Fi (Wireless Fidelity) standard 802.11b, which is designed for providing a WLAN (Wireless Local Area Network) with the level of privacy and security.

Wardriving is act of searching for the WI-FI (Wireless Fidelity) networks by the person in the moving vehicle by using PDA (Personal Digital Assistant), smartphones or portable computer.

sammy [17]2 years ago
6 0

Answer:

The answer to this question is the option "A".

Explanation:

The  WEP (Wired Equivalent Privacy) is a security protocol suite. Which is defined in the IEEE(Institute of Electrical and Electronics Engineers) Wi-Fi (Wireless Fidelity) standard 802.11b. This project invented to produce a wireless local area network (WLAN) with a level of security and secrecy. Which is normally demanded from a wired LAN. In WEP two devices need to share their password to communicate with one or more devices. The difficulty with WEP includes the use of a 24-bit initialization vector, that will sometimes repeat itself during transmission. So the answer to this question is WEP.

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Write a class for a Cat that is a subclass of Pet. In addition to a name and owner, a cat will have a breed and will say "meow"
fenix001 [56]

Answer:

The Java class is given below with appropriate tags for better understanding

Explanation:

public class Cat extends Pet{

  private String breed;

 public Cat(String name, String owner, String breed){

      /* implementation not shown */

      super(name, owner);

      this.breed = breed;

  }

  public String getBreed() {

      return breed;

  }

  public void setBreed(String breed) {

      this.breed = breed;

  }

  public String speak(){ /* implementation not shown */  

      return "Purring…";

  }

}

8 0
2 years ago
Summary: Given integer values for red, green, and blue, subtract the gray from each value. Computers represent color by combinin
Dmitry_Shevchenko [17]

Answer:

Follows are the code to this question:

#include<iostream>//defining header file

using namespace std;// use package

int main()//main method

{

int red,green,blue,x;//declaring integer variable

cin>> red >>green>>blue;//use input method to input value

if(red<green && red<blue)//defining if block that check red value is greater then green and blue  

{

x = red;//use x variable to store red value

}

else if(green<blue)//defining else if block that check green value greater then blue  

{

x= green; //use x variable to store green value

}

else//defining else block

{

x=blue;//use x variable to store blue value

}

red -= x;//subtract input integer value from x  

green -=x; //subtract input integer value from x

blue -= x;//subtract input integer value from x

cout<<red<<" "<<green<<" "<<blue;//print value

return 0;

}

Output:

130 50 130

80 0 80

Explanation:

In the given code, inside the main method, four integers "red, green, blue, and x" are defined, in which "red, green, and blue" is used for input the value from the user end. In the next step, a conditional statement is used, in the if block, it checks red variable value is greater than then "green and blue" variable. If the condition is true, it will store red variable value in "x", otherwise, it will goto else if block.

  • In this block, it checks the green variable value greater than the blue variable value. if the condition is true it will store the green variable value in x variable.
  • In the next step, else block is defined, that store blue variable value in x variable, at the last step input variable, is used that subtracts the value from x and print its value.      
5 0
2 years ago
Import java.util.scanner; public class sumofmax { public double findmax(double num1, double num2) { double maxval; // note: if-e
jeyben [28]

Here you go,


Import java.util.scanner

public class SumOfMax {

   public static double findMax(double num1, double num2) {

       double maxVal = 0.0;

       // Note: if-else statements need not be understood to

       // complete this activity

       if (num1 > num2) { // if num1 is greater than num2,

           maxVal = num1; // then num1 is the maxVal.

       }

       else { // Otherwise,

           maxVal = num2; // num2 is the maxVal.

       }

       return maxVal;

   }

   public static void main(String[] args) {

       double numA = 5.0;

       double numB = 10.0;

       double numY = 3.0;

       double numZ = 7.0;

       double maxSum = 0.0;

       /* Your solution goes here */

       maxSum = findMax(numA, numB); // first call of findMax

       maxSum = maxSum + findMax(numY, numZ); // second call

       System.out.print("maxSum is: " + maxSum);

       return;

   }

}

/*

Output:

maxSum is: 17.0

*/

6 0
2 years ago
Write a program to declare a matrix A[][] of order (MXN) where ‘M’ is the number of rows and ‘N’ is the
Liula [17]

Answer:

import java.io.*;

import java.util.Arrays;

class Main {

   public static void main(String args[])

   throws IOException{

       // Set up keyboard input

       InputStreamReader in = new InputStreamReader(System.in);

       BufferedReader br = new BufferedReader(in);

 

       // Prompt for dimensions MxN of the matrix

       System.out.print("M = ");

       int m = Integer.parseInt(br.readLine());

       System.out.print("N = ");

       int n = Integer.parseInt(br.readLine());

       // Check if input is within bounds, exit if not

       if(m <= 2 || m >= 10 || n <= 2 || n >= 10){

           System.out.println("Matrix size out of range.");

           return;

       }

       // Declare the matrix as two-dimensional int array

       int a[][] = new int[m][n];

 

       // Prompt for values of the matrix elements

       System.out.println("Enter elements of matrix:");

       for(int i = 0; i < m; i++){

           for(int j = 0; j < n; j++){

               a[i][j] = Integer.parseInt(br.readLine());

           }

       }

       // Output the original matrix

       System.out.println("Original Matrix:");

       printMatrix(a);

       // Sort each row

       for(int i = 0; i < m; i++){

         Arrays.sort(a[i]);

       }

       // Print sorted matrix

       System.out.println("Matrix after sorting rows:");

       printMatrix(a);

   }

   // Print the matrix elements separated by tabs

   public static void printMatrix(int[][] a) {

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

           for(int j = 0; j < a[i].length; j++)

               System.out.print(a[i][j] + "\t");

           System.out.println();

       }

   }

}

Explanation:

I fixed the mistake in the original code and put comments in to describe each section. The mistake was that the entire matrix was sorted, while only the individual rows needed to be sorted. This even simplifies the program. I also factored out a printMatrix() method because it is used twice.

4 0
1 year ago
The processing of data in a computer involves the interplay between its various hardware components.
Ronch [10]

True.

Data processing involves the conversion of raw data and the flow of data through the Central Processing Unit and Memory to output devices. Each CPU in a computer contains two primary elements: the Arithmetic Logic Unit (ALU) and the control unit. The Arithmetic Logic Unit performs complex mathematical calculations and logical comparisons. On the other hand, the control unit accesses computer instructions, decodes them, and controls the flow of data in and out of the Memory, ALU, primary and secondary storage, and various other output devices.  


8 0
2 years ago
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