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Over [174]
3 years ago
14

A user’s laptop hard drive contains sensitive information. The user often plugs the laptop into the corporate network. A sensiti

ve file from the laptop has been found on another users laptop. How could the user have prevented this breach?
Computers and Technology
1 answer:
jekas [21]3 years ago
6 0

Answer:

the user should disable file and print sharing on the laptop

Explanation:

According to my research on information technology and cyber security, I can say that based on the information provided within the question in order to prevent this the user should disable file and print sharing on the laptop. This is because this file and printer sharing allows users to see your files that you have available as shared and copy or print those files. Which if they are sensitive data you should just disable this option so that people on the same network cannot see or access these files.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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How was wifi made? bro
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Explanation: ...

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If a user's input string matches a known text message abbreviation, output the unabbreviated form, else output: Unknown. Support
Molodets [167]

Answer:

Here is the JAVA program. I  have added a few more abbreviations apart from LOL and IDK.

import java.util.Scanner; // to take input from user

public class MessageAbbreviation {

public static void main(String[] args) { //start of main() function body

   Scanner input = new Scanner(System.in); // create object of Scanner

       String abbreviation= ""; //stores the text message abbreviation

/* In the following lines the abbreviation string type variables contain the un-abbreviated forms */

           String LOL = "laughing out loud";

           String IDK = "i don't know";

           String BFF = "best friends forever";

           String IMHO = "in my humble opinion";

           String TMI = "too much information";

           String TYT = "take your time";

           String IDC = "I don't care";

           String FYI = "for your information";

           String BTW = "by the way";

           String OMG = "oh my God";

//prompts the user to enter an abbreviation for example LOL

           System.out.println("Input an abbreviation:" + " ");

           if(input.hasNext()) { // reads the abbreviation form the user

           abbreviation= input.next(); }

// scans and reads the abbreviation from user in the abbreviation variable

/*switch statement checks the input abbreviation with the cases and prints the un abbreviated form in output accordingly. */

    switch(abbreviation) {

       case "LOL" : System.out.println(LOL);

                    break;

       case "IDK" : System.out.println(IDK);

                    break;

       case "BFF" : System.out.println(BFF);

                    break;

       case "IMHO": System.out.println(IMHO);

                     break;

       case "TMI": System.out.println(TMI);

                     break;

       case "TYT": System.out.println(TYT);

                     break;

       case "IDC": System.out.println(IDC);

                     break;

       case "FYI": System.out.println(FYI);

                     break;

       case "BTW": System.out.println(BTW);

                     break;

       case "OMG": System.out.println(OMG);

                     break;

                     

       default    : System.out.println("Unknown"); } } }

Explanation:

The program first prompts the user to enter an abbreviation. Lets say the user enters LOL. Now the switch statement has some cases which are the conditions that are checked against the input abbreviation. For example if the user enters LOL then the following statement is executed:

case "LOL" : System.out.println(LOL);

This means if the case is LOL then the message ( un abbreviated form) stored in the LOL variable is displayed on output screen So the output is laughing out loud.

Anything else entered other than the given abbreviations will display the message: Unknown

The output is attached as screen shot.

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