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katrin2010 [14]
3 years ago
9

What is a common indicator of a phishing attempt cyber awareness 2022

Computers and Technology
1 answer:
vivado [14]3 years ago
7 0

There are several types of malicious document. A common indicator of a phishing attempt cyber awareness 2022 is that It includes a threat of dire circumstances.

  • Phishing attacks often makes use of email or malicious websites to infect the machine with malware and viruses so as to collect personal and financial information.

Cybercriminals often uses different means to lure users to click on a link or open an attachment that infects their computers thereby producing vulnerabilities for criminals to use to attack.

Learn more from

brainly.com/question/24069038

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How desktop case or chassis designed?, material and steps?​
MariettaO [177]

They may require things such as plastic cut outs and glass depending if you want to be able to have a window to look into the pc. Some cases are 3D Printed although they have to be joined by screws at the end due to the fact that one whole print can take over a week and then it has to be well doesn't but looks better when sanded.

3 0
2 years ago
This program will store roster and rating information for a soccer team. Coaches rate players during tryouts to ensure a balance
aksik [14]

Answer:

import java.util.Scanner;

public class TeamRoster {

  public static void main(String[] args) {

     Scanner scan = new Scanner(System.in);

     int[][] players = new int[5][2];

     boolean keepAlive = true;

     char input;

     

     for (int i = 0; i < 5; i++) {

        System.out.println("Enter player " + (i+1) + "'s jersey number: ");

        players[i][0] = scan.nextInt();

        if( players[i][0] >= 0 && players[i][0] < 100 ) {

             System.out.println("Enter player number less than 100");

               System.out.println("Enter player " + (i+1) + "'s rating: ");

               players[i][1] = scan.nextInt();

               if ( players[i][1] >= 1 && players[i][1] <10 ) {

                     System.out.println();

               }    else {

                     System.out.println("Enter player " + (i+1) + "'s rating: ");

                     players[i][1] = scan.nextInt();

               }

     }

     }

     System.out.println();

     outputRoster(players, 0);

     

     while (keepAlive) {

        outputMenu();

        input = scan.next().charAt(0);

        if (input == 'q') {

           keepAlive = false;

           break;

        } else if (input == 'o') {

           outputRoster(players, 0);

        } else if (input == 'u') {

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

           int jerseyNum = scan.nextInt();

           System.out.println("Enter a new rating for the player: ");

           int newRating = scan.nextInt();

           for (int l = 0; l < 5; l++) {

              if (players[l][0] == jerseyNum) {

                 players[l][1] = newRating;

              }

           }

        } else if (input == 'a') {

           System.out.println("Enter a rating: ");

           int rating = scan.nextInt();

           outputRoster(players, rating);

        } else if (input == 'r') {

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

           int jerseyNum = scan.nextInt();

           boolean exists = true;

           for (int l = 0; l < 5; l++) {

              if (players[l][0] == jerseyNum) {

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

                 players[l][0] = scan.nextInt();

                 System.out.println("Enter a rating for the new player: ");

                 players[l][1] = scan.nextInt();

              }

           }

           

        }

     }

     

     return;

  }

   

  public static void outputRoster(int[][] players, int min) {

     System.out.println(((min>0) ? ("ABOVE " + min) : ("ROSTER")));

     int item = 1;

     for (int[] player : players) {

        if (player[1] > min) {

           System.out.println("Player " + item + " -- Jersey number: " + player[0] + ", Rating: " + player[1]);

        }

        item++;

     }

     System.out.println();

  }

   

  public static void outputMenu() {

     System.out.println("MENU");

     System.out.println("u - Update player rating");

     System.out.println("a - Output players above a rating");

     System.out.println("r - Replace player");

     System.out.println("o - Output roster");

     System.out.println("q - Quit\n");

     System.out.println("Choose an option: ");  

  }

}

Explanation:

The program is written in Java.

It is a bit straightforward.

The Scanner module is declared and used to collect input from the user.

The program prompts the user for each player jersey then rating.

All five players details are collected.

The program shows the output of all the entries.

5 0
3 years ago
A laptop computer communicates with a router wirelessly, by means of radio signals. the router is connected by cable directly to
Hoochie [10]
The answer would be 2500 Bits
4 0
3 years ago
2. Consider a 2 GHz processor where two important programs, A and B, take one second each to execute. Each program has a CPI of
allsm [11]

Answer:

program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

Explanation:

Finding number of instructions in A and B using time taken by the original processor :

The clock speed of the original processor is 2 GHz.

which means each clock takes, 1/clockspeed

= 1 / 2GH = 0.5ns

Now, the CPI for this processor is 1.5 for both programs A and B. therefore each instruction takes 1.5 clock cycles.

Let's say there are n instructions in each program.

therefore time taken to execute n instructions

= n * CPI * cycletime = n * 1.5 * 0.5ns

from the question, each program takes 1 sec to execute in the original processor.

therefore

n * 1.5 * 0.5ns = 1sec

n = 1.3333 * 10^9

So, number of instructions in each program is 1.3333 * 10^9

the new processor :

The cycle time for the new processor is 0.6 ns.

Time taken by program A = time taken to execute n instructions

=  n * CPI * cycletime

= 1.3333 * 10^9 * 1.1 * 0.6ns

= 0.88 sec

Time taken by program B = time taken to execute n instructions

= n * CPI * cycletime

= 1.3333 * 10^9 * 1.4 * 0.6

= 1.12 sec

Now, program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

5 0
3 years ago
What setting in Word keeps single lines of a new
Gre4nikov [31]

Answer:

C (Window/Orphan control)

Explanation:

An "orphan" in formatting is a single line of text that is left alone/separated from the rest of the paragraph. Orphan control prevents this from happening by keeping lines together.

3 0
4 years ago
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