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DerKrebs [107]
3 years ago
8

The memory management layer is the part of the kernel that serves out all memory allocation requests.

Computers and Technology
1 answer:
svetlana [45]3 years ago
4 0
Most likely it’s False . But look it up just in case
You might be interested in
Negative numbers are encoded using the __________ technique. a. two’s complement b. floating point c. ASCII d. Unicode
OLEGan [10]

Negative numbers are encoded using the two’s complement technique. Two's complement is used to encode negative numbers.

Option  A is correct .

<h3>What method does the in data type use to store negative numbers?</h3>

Most implementations you're likely to encounter store negative signed integers in a form known as two's complement. Another important way of storing negative signed numbers is called one's complement. The one's complement of an N-bit number x is basically defined as x with all bits inverted.

<h3>What is encoding method?</h3>

An encoding technique is the application of established industry conventions to a coded character set to create a coded character scheme. Such rules determine the number of bits required to store the numeric representation of a given character and its code position in the encoding.

Learn more about two complement technique encoding :

brainly.com/question/26668609

#SPJ4

5 0
1 year ago
I am not a living being, I am a cylindrical shape that has three to eight sides. I never die. I can build anything again. What i
Setler79 [48]

Answer:

A shape

Explanation:

4 0
3 years ago
What happens if two functions are defined with the same name, even if they are in different arguments within the same .py files?
yan [13]

Answer:

Following are the code to this question:

def data(a):#defining method data that accepts parameter

   print (a)#print parameter value

def data(b):#defining method data that accepts parameter

   print (b)#print parameter value

x=input("enter value: ")#defining variable x that5 input value from user

print(data(x))#call method data

Output:

enter value: hello..

hello..

None

Explanation:

  • As the above code, it is clear defines that python doesn't support the method overloading because More than one method can't be specified in a python class with the same name and python method arguments have no type.
  • The single argument method may be named using an integer, a series, or a double value, that's why we can say that it is not allowed.
7 0
3 years ago
. It has been said that technology will be the end of management. Maybe. How about artificial intelligence
Dvinal [7]

Answer:

Yes

Explanation:

Artificial Intelligence is just a subcategory of Technology. That being said, if any type of technology has the ability to do the job of a human being in the management sector of a company it would be Artificial Intelligence. This is because AI is designed to be able to analyze data, discover patterns, and make decisions based on those patterns. These decisions are incredibly sophisticated, efficient, and made incredibly fast. It also learns the more that it makes decisions, therefore increasing its efficiency the more that it does a specific task. This would represent the same tasks that management is responsible for getting done, but the AI is able to do it faster, cheaper, and more efficiently. So, yes, AI is very capable of bringing the end of management.

6 0
3 years ago
Write a removeDuplicates() method for the LinkedList class we saw in lecture. The method will remove all duplicate elements from
oksano4ka [1.4K]

Answer:

removeDuplicates() function:-

//removeDuplicates() function removes duplicate elements form linked list.

   void removeDuplicates() {

     

       //declare 3 ListNode pointers ptr1,ptr2 and duplicate.

       //initially, all points to null.

       ListNode ptr1 = null, ptr2 = null, duplicate = null;

       

       //make ptr1 equals to head.

       ptr1 = head;

        //run while loop till ptr1 points to second last node.

       //pick elements one by one..

       while (ptr1 != null && ptr1.next != null)

       {

               // make ptr2 equals to ptr1.

               //or make ptr2 points to same node as ptr1.

           ptr2 = ptr1;

           //run second while loop to compare all elements with above selected element(ptr1->val).

           while (ptr2.next != null)

           {

              //if element pointed by ptr1 is same as element pointed by ptr2.next.

               //Then, we have found duplicate element.

               //Now , we have to remove this duplicate element.

               if (ptr1.val == ptr2.next.val)

               {

                  //make duplicate pointer points to node where ptr2.next points(duplicate node).

                       duplicate = ptr2.next;

                       //change links to remove duplicate node from linked list.

                       //make ptr2.next points to duplicate.next.

                   ptr2.next = duplicate.next;

               }

               

             //if element pointed by ptr1 is different from element pointed by ptr2.next.

               //then it is not duplicate element.

               //So, move ptr2 = ptr2.next.

               else

               {

                   ptr2 = ptr2.next;

               }

           }

           

           //move ptr1 = ptr1.next, after check duplicate elements for first node.

           //Now, we check duplicacy for second node and so on.

           //so, move ptr1  by one node.

           ptr1 = ptr1.next;

       }

   }

Explanation:

Complete Code:-

//Create Linked List Class.

class LinkedList {

       //Create head pointer.

       static ListNode head;

       //define structure of ListNode.

       //it has int val(data) and pointer to ListNode i.e, next.

   static class ListNode {

       int val;

       ListNode next;

       //constructor to  create and initialize a node.

       ListNode(int d) {

               val = d;

           next = null;

       }

   }

//removeDuplicates() function removes duplicate elements form linked list.

   void removeDuplicates() {

       

       //declare 3 ListNode pointers ptr1,ptr2 and duplicate.

       //initially, all points to null.

       ListNode ptr1 = null, ptr2 = null, duplicate = null;

       

       //make ptr1 equals to head.

       ptr1 = head;

       

       

       //run while loop till ptr1 points to second last node.

       //pick elements one by one..

       while (ptr1 != null && ptr1.next != null)

       {

               // make ptr2 equals to ptr1.

               //or make ptr2 points to same node as ptr1.

           ptr2 = ptr1;

           //run second while loop to compare all elements with above selected element(ptr1->val).

           while (ptr2.next != null)

           {

              //if element pointed by ptr1 is same as element pointed by ptr2.next.

               //Then, we have found duplicate element.

               //Now , we have to remove this duplicate element.

               if (ptr1.val == ptr2.next.val)

               {

                  //make duplicate pointer points to node where ptr2.next points(duplicate node).

                       duplicate = ptr2.next;

                       

                       //change links to remove duplicate node from linked list.

                       //make ptr2.next points to duplicate.next.

                   ptr2.next = duplicate.next;

               }

               

             //if element pointed by ptr1 is different from element pointed by ptr2.next.

               //then it is not duplicate element.

               //So, move ptr2 = ptr2.next.

               else

               {

                   ptr2 = ptr2.next;

               }

           }

           

           //move ptr1 = ptr1.next, after check duplicate elements for first node.

           //Now, we check duplicacy for second node and so on.

           //so, move ptr1  by one node.

           ptr1 = ptr1.next;

       }

   }

   //display() function prints linked list.

   void display(ListNode node)

   {

       //run while loop till last node.

       while (node != null)

       {

               //print node value of current node.

           System.out.print(node.val + " ");

           

           //move node pointer by one node.

           node = node.next;

       }

   }

   public static void main(String[] args) {

       

       //Create object of Linked List class.

       LinkedList list = new LinkedList();

       

       //first we create nodes and connect them to form a linked list.

       //Create Linked List 1-> 2-> 3-> 2-> 4-> 2-> 5-> 2.

       

       //Create a Node having node data = 1 and assign head pointer to it.

       //As head is listNode of static type. so, we call head pointer using class Name instead of object name.

       LinkedList.head = new ListNode(1);

       

       //Create a Node having node data = 2 and assign head.next to it.

       LinkedList.head.next = new ListNode(2);

       LinkedList.head.next.next = new ListNode(3);

       LinkedList.head.next.next.next = new ListNode(2);

       LinkedList.head.next.next.next.next = new ListNode(4);

       LinkedList.head.next.next.next.next.next = new ListNode(2);

       LinkedList.head.next.next.next.next.next.next = new ListNode(5);

       LinkedList.head.next.next.next.next.next.next.next = new ListNode(2);

       //display linked list before Removing duplicates.

       System.out.println("Linked List before removing duplicates : ");

       list.display(head);

       //call removeDuplicates() function to remove duplicates from linked list.

       list.removeDuplicates();

       System.out.println("")

       //display linked list after Removing duplicates.

       System.out.println("Linked List after removing duplicates :  ");

       list.display(head);

   }

}

Output:-

6 0
3 years ago
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