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olga55 [171]
3 years ago
10

c724 wgu True or false. A storage device consists of all the components that work together to process data into useful informati

on
Computers and Technology
1 answer:
Law Incorporation [45]3 years ago
8 0

Answer:

False

Explanation:

A output device consists of all the components that work together to process data into useful information

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galben [10]

Hey Litz06,

Sorry that happened to you.

Try changing your password to something complicated so it’s more secure.

5 0
3 years ago
Plz help, correct answer will get brainliest (if i can, if i can't ill still rate and give thanks)
Elena-2011 [213]

Answer:

E-book : online edition of a new novel .

e-zine:online issue of today’s newspaper.

Online reference: online encyclopedia

blog: online website that posts restaurants reviews

Explanation:

Are you doing edge?

Also pleaseeeeeeeeeeeeeeeeeee mark me brainliest.

6 0
3 years ago
If you are asked to design an optimal workflow using technology for a car accident claim process, what factors would you conside
yan [13]

Answer:

The factors to consider when building a solution for optical workflow for car accident are the type of claim to be modelled, also to examine the level of automation in vehicles.

Some of the software solutions that meet the requirements are the A1 Tracker  which is very effective to the enterprise business and small business working with insurance. we also have the Mutual expert, Ninja Quote.

Explanation:

Solution

In suggesting an optical workflow for the car accident claim process it is of importance to understand the following factors such as, the type of claim to be modelled, it will be determined by the policy that the clients have for their cars. this will assist in classifying the workflow and giving a clear path of the design.

The number of expected claims is also a very important factor that should be added in the design. Another important factor to examine  the level of automation of the vehicles, this can be very useful in collecting the accident data in a more efficient way and also making it simple to design the model.

Presently there are several workflow softwares in the insurance that have encouraged a great deal and have made the workflow in the insurance companies more greater.

Some of the soft wares are the A1 Tracker  which are very effective and efficient in are highly applicable to the enterprise business and small business working with insurance, the software is convenient in monitoring the work progress in the company.

The other soft wares that have been quite useful in the industry for example the Insly, ISI enterprise, Mutual expert, Ninja Quoter among others. These software are some of the best soft wares available in the market that have met the workflow requirement in the insurance sector.

7 0
3 years ago
Which of the following need NOT be completed separately if a worksheet is prepared?
Anika [276]
C because you already have you sheet prepared
3 0
3 years ago
Define a class called TreeNode containing three data fields: element, left and right. The element is a generic type. Create cons
m_a_m_a [10]

Answer:

See explaination

Explanation:

// Class for BinarySearchTreeNode

class TreeNode

{

// To store key value

public int key;

// To point to left child

public TreeNode left;

// To point to right child

public TreeNode right;

// Default constructor to initialize instance variables

public TreeNode(int key)

{

this.key = key;

left = null;

right = null;

key = 0;

}// End of default constructor

}// End of class

// Defines a class to crate binary tree

class BinaryTree

{

// Creates root node

TreeNode root;

int numberElement;

// Default constructor to initialize root

public BinaryTree()

{

this.root = null;

numberElement = 0;

}// End of default constructor

// Method to insert key

public void insert(int key)

{

// Creates a node using parameterized constructor

TreeNode newNode = new TreeNode(key);

numberElement++;

// Checks if root is null then this node is the first node

if(root == null)

{

// root is pointing to new node

root = newNode;

return;

}// End of if condition

// Otherwise at least one node available

// Declares current node points to root

TreeNode currentNode = root;

// Declares parent node assigns null

TreeNode parentNode = null;

// Loops till node inserted

while(true)

{

// Parent node points to current node

parentNode = currentNode;

// Checks if parameter key is less than the current node key

if(key < currentNode.key)

{

// Current node points to current node left

currentNode = currentNode.left;

// Checks if current node is null

if(currentNode == null)

{

// Parent node left points to new node

parentNode.left = newNode;

return;

}// End of inner if condition

}// End of outer if condition

// Otherwise parameter key is greater than the current node key

else

{

// Current node points to current node right

currentNode = currentNode.right;

// Checks if current node is null

if(currentNode == null)

{

// Parent node right points to new node

parentNode.right = newNode;

return;

}// End of inner if condition

}// End of outer if condition

}// End of while

}// End of method

// Method to check tree is balanced or not

private int checkBalance(TreeNode currentNode)

{

// Checks if current node is null then return 0 for balanced

if (currentNode == null)

return 0;

// Recursively calls the method with left child and

// stores the return value as height of left sub tree

int leftSubtreeHeight = checkBalance(currentNode.left);

// Checks if left sub tree height is -1 then return -1

// for not balanced

if (leftSubtreeHeight == -1)

return -1;

// Recursively calls the method with right child and

// stores the return value as height of right sub tree

int rightSubtreeHeight = checkBalance(currentNode.right);

// Checks if right sub tree height is -1 then return -1

// for not balanced

if (rightSubtreeHeight == -1) return -1;

// Checks if left and right sub tree difference is greater than 1

// then return -1 for not balanced

if (Math.abs(leftSubtreeHeight - rightSubtreeHeight) > 1)

return -1;

// Returns the maximum value of left and right subtree plus one

return (Math.max(leftSubtreeHeight, rightSubtreeHeight) + 1);

}// End of method

// Method to calls the check balance method

// returns false for not balanced if check balance method returns -1

// otherwise return true for balanced

public boolean balanceCheck()

{

// Calls the method to check balance

// Returns false for not balanced if method returns -1

if (checkBalance(root) == -1)

return false;

// Otherwise returns true

return true;

}//End of method

// Method for In Order traversal

public void inorder()

{

inorder(root);

}//End of method

// Method for In Order traversal recursively

private void inorder(TreeNode root)

{

// Checks if root is not null

if (root != null)

{

// Recursively calls the method with left child

inorder(root.left);

// Displays current node value

System.out.print(root.key + " ");

// Recursively calls the method with right child

inorder(root.right);

}// End of if condition

}// End of method

}// End of class BinaryTree

// Driver class definition

class BalancedBinaryTreeCheck

{

// main method definition

public static void main(String args[])

{

// Creates an object of class BinaryTree

BinaryTree treeOne = new BinaryTree();

// Calls the method to insert node

treeOne.insert(1);

treeOne.insert(2);

treeOne.insert(3);

treeOne.insert(4);

treeOne.insert(5);

treeOne.insert(8);

// Calls the method to display in order traversal

System.out.print("\n In order traversal of Tree One: ");

treeOne.inorder();

if (treeOne.balanceCheck())

System.out.println("\n Tree One is balanced");

else

System.out.println("\n Tree One is not balanced");

BinaryTree

BinaryTree treeTwo = new BinaryTree();

treeTwo.insert(10);

treeTwo.insert(18);

treeTwo.insert(8);

treeTwo.insert(14);

treeTwo.insert(25);

treeTwo.insert(9);

treeTwo.insert(5);

System.out.print("\n\n In order traversal of Tree Two: ");

treeTwo.inorder();

if (treeTwo.balanceCheck())

System.out.println("\n Tree Two is balanced");

else

System.out.println("\n Tree Two is not balanced");

}// End of main method

}// End of driver class

5 0
3 years ago
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