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Maru [420]
4 years ago
5

The process of providing and denying access to objects is called:

Computers and Technology
1 answer:
tatiyna4 years ago
5 0

The correct answer is access control. It is because this has the function of providing a user an access when the individual is considered or belong to the authorized users and if not, they are likely to be denied to the access that they requested if they belong to the unauthorized users.

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The purpose of this programming project is to demonstrate a significant culmination of most constructs learned thus far in the c
Oksana_A [137]

Answer:

Java program explained below

Explanation:

here is your files : ----------------------

Word.java : --------------

import java.util.*;

public class Word implements Comparable<Word>{

private String normal;

private String canonical;

public Word(){

 normal = "";

 canonical = "";

}

public Word(String norm){

 setNormal(norm);

}

public void setNormal(String norm){

 normal = norm;

 char[] arr = norm.toCharArray();

 Arrays.sort(arr);

 canonical = new String(arr);

}

public String getNormal(){

 return normal;

}

public String getCanonical(){

 return canonical;

}

public int compareTo(Word word){

 return canonical.compareTo(word.getCanonical());

}

public String toString(){

 return "("+normal+", "+canonical+")";

}

}

AnagramFamily.java : ------------------------------

import java.util.*;

public class AnagramFamily implements Comparable<AnagramFamily>{

private LinkedList<Word> words;

private int size;

private class WordComp implements Comparator{

 public int compare(Object o1,Object o2){

  Word w1 = (Word)o1;

  Word w2 = (Word)o2;

  return w2.getNormal().compareTo(w1.getNormal());

 }

}

public AnagramFamily(){

 words = new LinkedList<>();

 size = 0;

}

public LinkedList<Word> getAnagrams(){

 return words;

}

public void addAnagram(Word word){

 words.add(word);

 size++;

}

public void sort(){

 Collections.sort(words,new WordComp());

}

public int getSize(){

 return size;

}

public int compareTo(AnagramFamily anag){

 Integer i1 = new Integer(size);

 Integer i2 = new Integer(anag.getSize());

 return i2.compareTo(i1);

}

public String toString(){

 return "{ Anagrams Family Size : "+size+" "+words.toString()+"}";

}

}

WordMain.java : ------------------------------

import java.util.*;

import java.io.File;

import java.io.PrintWriter;

public class WordMain{

public static void readFile(LinkedList<Word> words){

 try{

  Scanner sc = new Scanner(new File("words.txt"));

  while(sc.hasNext()){

   words.add(new Word(sc.next()));

  }

  sc.close();

 }catch(Exception e){

  e.printStackTrace();

  System.exit(-1);

 }

}

public static void findAnagrams(LinkedList<AnagramFamily> anagrams,LinkedList<Word> words){

 Iterator<Word> itr = words.iterator();

 while(itr.hasNext()){

  Iterator<AnagramFamily> aitr = anagrams.iterator();

  Word temp = itr.next();

  System.out.println(temp);

  boolean st = true;

  while(aitr.hasNext()){

   AnagramFamily anag = aitr.next();

   Iterator<Word> anags = anag.getAnagrams().iterator();

   while(anags.hasNext()){

    Word t1 = anags.next();

    if(t1.compareTo(temp) == 0 && !t1.getNormal().equals(temp.getNormal())){

     anag.addAnagram(temp);

     st = false;

     break;

    }

   }

   anag.sort();

  }

  if(st){

   AnagramFamily anag = new AnagramFamily();

   anag.addAnagram(temp);

   anagrams.add(anag);

   System.out.println(anag);

  }

 }

}

public static void writeOutput(LinkedList<AnagramFamily> anagrams){

 try{

  PrintWriter pw = new PrintWriter(new File("out6.txt"));

  Collections.sort(anagrams);

  int i = 0;

  Iterator<AnagramFamily> aitr = anagrams.iterator();

  while(i < 5 && aitr.hasNext()){

   AnagramFamily anag = aitr.next();

   anag.sort();

   pw.println(anag);

   i++;

  }

  aitr = anagrams.iterator();

  pw.println("\n\nAnagramsFamily size 8 datas : ");

  while(aitr.hasNext()){

   AnagramFamily anag = aitr.next();

   anag.sort();

   if(anag.getSize() == 8){

    pw.println(anag);

   }

  }

  pw.close();

 }catch(Exception e){

  e.printStackTrace();

  System.exit(-1);

 }

}

public static void main(String[] args) {

 LinkedList<Word> words = new LinkedList<>();

 readFile(words);

 Collections.sort(words);

 //System.out.println(words.toString());

 LinkedList<AnagramFamily> anagrams = new LinkedList<>();

 findAnagrams(anagrams,words);

 //System.out.println(anagrams.toString());

 writeOutput(anagrams);

}

}

8 0
3 years ago
For dinner, a restaurant allows you to choose either Menu Option A: five appetizers and three main dishes or Menu Option B: thre
alukav5142 [94]

Answer:

The formula used in this question is called the probability of combinations or combination formula.

Explanation:

Solution

Given that:

Formula applied is stated as follows:

nCr = no of ways to choose r objects from n objects

  = n!/(r!*(n-r)!)

The Data given :

Menu A : 5 appetizers and 3 main dishes

Menu B : 3 appetizers and 4 main dishes

Total appetizers - 6

Total main dishes - 5

Now,

Part A :

Total ways = No of ways to select menu A + no of ways to select menu B

          = (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

          = 6C5*5C3 + 6C3*5C4

          = 6*10 + 20*5

          = 160

Part B :

Since, we can select the same number of appetizers/main dish again so the number of ways to select appetizers/main dishes will be = (total appetizers/main dishes)^(no of appetizers/main dishes to be selected)

Total ways = No of ways to select menu A + no of ways to select menu B  

          = (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

          = (6^5)*(5^3) + (6^3)*(5^4)

          = 7776*125 + 216*625

          = 1107000

Part C :

No of ways to select same appetizers and main dish for all the options

= No of ways to select menu A + no of ways to select menu B  

= (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

=(6*5) + (6*5)

= 60

Total ways = Part B - (same appetizers and main dish selected)      

= 1107000 - 60

= 1106940

3 0
4 years ago
In your own words, describe what Internet Protocols are. Why is it important to have agreed upon protocols?
Anni [7]

Answer:

Internet Protocol refers to a set of rules that govern how data packets are transmitted over a network. Internet protocol describes how data packets move through a network. Its important to have agreed upon protocols because Computers make use of protocols as well, to enable them to communicate. Devices need to communicate. When two devices want to successfully communicate, they must agree to follow some rules about the way they will do it.

8 0
3 years ago
Which of the following statements is TRUE about startup capital?
8_murik_8 [283]
Only the first statement is true. "Startup capital" is merely defined as the financial resources invested in a business in order to get the business started, hence the term startup. Money spent for the purposes of balancing personal finances is nonessential to the initiation of a new business's activities, and thus can not be considered startup capital.
3 0
3 years ago
Create a file using any word processing program or text editor. Write an application that displays the files, name, containing f
MArishka [77]

Answer:

below is a Shell I have ;

1.Import java.nio.file.*;

2.Import java.nio.file attribute.*;

3.Import java.10 Exception;

4.public class FileStatistics

5.{

6.public static void main(string []args)

7.{

8.path file=

9.

paths.get("C:\\Java\\chapter.13\\TestData.txt")

10.try

11.{

12.\\declare count and then display path, file, name and folder name.

13.

14.

15.

16.\\declare a BasicFileAttributes object, then add statements to display the file's size and creation time.

17.

18.}

19.

20.catch (10 Exception e)

21.{

22.\\add display 10 Exception

23.

24.}

25.}

4 0
2 years ago
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