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quester [9]
3 years ago
9

The first thing you must consider in any type of communication is ______.

Computers and Technology
2 answers:
vivado [14]3 years ago
6 0

The correct answer to this questio is option letter C

Wittaler [7]3 years ago
5 0
As for this problem together with the options as answers presented with it, the most probable and the most likely answer would be C. the purpose of the communication.

The first thing you must consider in any type of communication is the purpose of the communication itself. Determining the purpose behind the communication would, in turn, pave way to determining the other things that is needed to be determined such as the type of communication, the style of communication, the timing of the communication, and so on and so forth.
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After selecting a cell in the data range, what are some ways to create a table? Check all that apply.
lana66690 [7]

Answer: the answer is B /E

3 0
3 years ago
Read 2 more answers
A city government is attempting to reduce the digital divide between groups with differing access to computing and the Internet.
sergeinik [125]

Answer:

C

Explanation:

Putting all government forms on the city web site is the least activity likely to be effective in the purpose of reducing digital divide.

Holding basic computer classes at the community centers will very much help to reduce the digital divide.

Providing free wireless internet connections at locations in low-income neighborhood will also reduce the gap of digital divide

Requiring that every city school has computers that meet a minimum hardware and software will made computing resources available to users thereby reducing digital divide.

5 0
3 years ago
You are gong to buy a computer but first you want to do some research to help you select the best model for your needs
DerKrebs [107]

Answer:

Research what kind of computer you want.

Explanation:

I don't know exactly what you are asking here but when doing research you have got to know what you want like how much ram you want, how much storage you want, what brand of computer you want, what core processor you want, and what graphics card you want. You also have to decide if you want a laptop or a full sized desk top computer.

8 0
3 years ago
All margins of a report are _____ inch. *<br><br> 1<br> 2<br> 3<br> 4
MA_775_DIABLO [31]

Answer:

The answer is "1 inch"

Explanation:

The margin is also known as space, it is the distance between the document 's text and edge. In default, its distance is 1, and it also allows you to customize, the new document, which is set to standard.

  • In other words, we can say that it implies the one-inch gap between the text and each edge, which is based on your specifications.
  • It also allows you to customize the margin size of the document, and other choices were wrong because it is the customize size.

6 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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