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Sphinxa [80]
3 years ago
7

AI is not embraced everywhere in every industry because _______.

Computers and Technology
1 answer:
Nitella [24]3 years ago
7 0

Answer:

See below:

Explanation:

AI is not embraced in every industry because of the type of industry and what it requires. In some situations, a human is needed since an AI cannot determine what to do and can sometimes even cause death.

AI isn't embraced everywhere because for some things we need a person, lets say for example, we need an AI to respond to 911 calls, that simply won't work due to the many situations and training the bot too will be a nightmare. For example, if a caller says a word that normal humans can understand, or a "code word" that every human knows but they haven't bothered to tell the bot, that will be the end of the line for them.

There are mutliple reasons to not use AI depending on the situation whether its life and death or just plain impractical.

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What is the significant feature of computer capabilities?​
Pepsi [2]
<h2>Hey mate </h2><h2>Here is ur answer..! ⬇️⬇️</h2>

Explanation:

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4 0
3 years ago
Question # 4
arlik [135]

Answer:

Explanatio Morse code

4 0
3 years ago
Let's implement a classic algorithm: binary search on an array. Implement a class named BinarySearcher that provides one static
yKpoI14uk [10]

Answer:

Hope this helped you, and if it did , do consider giving brainliest.

Explanation:

import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

 

//   main method

  public static void main(String[] args) {

     

//   create a list of Comparable type

     

      List<Comparable> list = new ArrayList<>();

     

//       add elements

     

      list.add(1);

      list.add(2);

      list.add(3);

      list.add(4);

      list.add(5);

      list.add(6);

      list.add(7);

     

//       print list

     

      System.out.println("\nList : "+list);

     

//       test search method

     

      Comparable a = 7;

      System.out.println("\nSearch for 7 : "+search(list,a));

     

      Comparable b = 3;

      System.out.println("\nSearch for 3 : "+search(list,b));

     

      Comparable c = 9;

      System.out.println("\nSearch for 9 : "+search(list,c));

     

      Comparable d = 1;

      System.out.println("\nSearch for 1 : "+search(list,d));

     

      Comparable e = 12;

      System.out.println("\nSearch for 12 : "+search(list,e));

     

      Comparable f = 0;

      System.out.println("\nSearch for 0 : "+search(list,f));

     

  }

 

 

 

//   static method named search takes arguments Comparable list and Comparable parameter

  public static boolean search(List<Comparable> list, Comparable par) {

     

//       if list is empty or parameter is null the throw IllegalArgumentException

     

      if(list.isEmpty() || par == null ) {

         

          throw new IllegalArgumentException();

         

      }

     

//       binary search

     

//       declare variables

     

      int start=0;

     

      int end =list.size()-1;

     

//       using while loop

     

      while(start<=end) {

         

//           mid element

         

          int mid =(start+end)/2;

         

//           if par equal to mid element then return

         

          if(list.get(mid).equals(par) )

          {

              return true ;

             

          }  

         

//           if mid is less than parameter

         

          else if (list.get(mid).compareTo(par) < 0 ) {

                 

              start=mid+1;

          }

         

//           if mid is greater than parameter

         

          else {

              end=mid-1;

          }

      }

     

//       if not found then retuen false

     

      return false;

     

  }

 

 

}import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

 

//   main method

  public static void main(String[] args) {

     

//   create a list of Comparable type

     

      List<Comparable> list = new ArrayList<>();

     

//       add elements

     

      list.add(1);

      list.add(2);

      list.add(3);

      list.add(4);

      list.add(5);

      list.add(6);

      list.add(7);

     

//       print list

     

      System.out.println("\nList : "+list);

     

//       test search method

     

      Comparable a = 7;

      System.out.println("\nSearch for 7 : "+search(list,a));

     

      Comparable b = 3;

      System.out.println("\nSearch for 3 : "+search(list,b));

     

      Comparable c = 9;

      System.out.println("\nSearch for 9 : "+search(list,c));

     

      Comparable d = 1;

      System.out.println("\nSearch for 1 : "+search(list,d));

     

      Comparable e = 12;

      System.out.println("\nSearch for 12 : "+search(list,e));

     

      Comparable f = 0;

      System.out.println("\nSearch for 0 : "+search(list,f));

     

  }

 

 

 

//   static method named search takes arguments Comparable list and Comparable parameter

  public static boolean search(List<Comparable> list, Comparable par) {

     

//       if list is empty or parameter is null the throw IllegalArgumentException

     

      if(list.isEmpty() || par == null ) {

         

          throw new IllegalArgumentException();

         

      }

     

//       binary search

     

//       declare variables

     

      int start=0;

     

      int end =list.size()-1;

     

//       using while loop

     

      while(start<=end) {

         

//           mid element

         

          int mid =(start+end)/2;

         

//           if par equal to mid element then return

         

          if(list.get(mid).equals(par) )

          {

              return true ;

             

          }  

         

//           if mid is less than parameter

         

          else if (list.get(mid).compareTo(par) < 0 ) {

                 

              start=mid+1;

          }

         

//           if mid is greater than parameter

         

          else {

              end=mid-1;

          }

      }

     

//       if not found then retuen false

     

      return false;

     

  }

 

 

}

7 0
3 years ago
Once a manual is written, it may change over time as procedures change. What's a recommended format for a manual to incorporate
zhenek [66]

Answer:

Print the manual in a loose leaf binder so even small changes can be replaced

Explanation:

Procedure  manual is the document that contains the business policies and strategies. These strategies can be changed time to time in the interest of company.

As these policies or procedures changes, it is necessary to communicate with all employees. To communicate this updated information to all employees it is necessary to print all that pages, where the changes have been made to replace them in the current manual. If we print whole manual it may increase the cost of printing. To reduce printing cost and wastage of pages only print those particular pages that have been changed. This will help to communicate to the existing employees as well  as to newly hired employees.

3 0
3 years ago
Given a Fully Associative cache with 4 lines using LRU replacement. The word size is one byte, there is one word per block, and
uranmaximum [27]

Answer:

3 bits

Explanation:

Capacity of main memory=16 Bytes=24

The number of address bits= 4 bits.

The size of the word= 1 Byte=20

The word bits=0.

Number of lines =4

Number of sets required=21

The sets bits is =1

The number of offset bits=20=0

Number of tag bits= total number of address bits - (word bits + offset bits + set bits)

= 4 - 0 -0- 1

= 3 bits

8 0
3 years ago
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