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ivanzaharov [21]
3 years ago
11

Technologies can offer safety and protection for people

Computers and Technology
1 answer:
inn [45]3 years ago
6 0

Answer:

True

Explanation:

This is certainly true. If the technologies are being used correctly and ethically then they are certainly going to safeguard the people and make their life easier and better. However, if the technology is not used ethically, and technology is given in the wrong hands, it can be very lethal. Like we can consider the nuclear bombs. If a nuclear bomb technology is being given to the wrong hands then they can use it to do the destruction, and that will not be good for us. And all should understand this. Proper and ethical use is a must for technology.

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Care should be taken whenever you post comments or photos to any social media or other websites, because each of these online ac
REY [17]

Answer:

True

Explanation:

Building your personal brand requires extreme care when it comes to social media presence. In the age of internet there is nothing hidden, so whatever you post or comment will be seen by your audience and it can go in either way. It can either make your brand or break it. It is of utmost importance to leave a pleasant and positive impact on your audience. One should not indulge in inappropriate or controversial debates and try to avoid being too personal on social media.  

6 0
3 years ago
Is there an ethically acceptable reason to study and use the various attack methods described in this chapter? Can you think of
avanturin [10]

Answer:

Please see below  

Explanation:

Yes, there indeed is ethical justification for hacking certain computer systems. Since computer scientists are required to keep the system secure from external threats, so they make use of it when testing the network for potential loopholes that could make it vulnerable. It is beneficial in that it can help manifest the weaknesses present in the system, which can then be corrected for.

6 0
3 years ago
. Which one of the following is not a JAD participant? 1. Managers 2. System analyst 3. Business development manager 4. Users
asambeis [7]

Answer: 3) Business development manager

Explanation:

 JAD stand for joint application development and it is the methodology which basically involved users and client in the development and designing of the application.

It is the process which basically accelerate the designing of the information technology system. JAD involve managers , system analyst and users to develop the dynamic solution. But it does not involve the business development manager as the JAD basically focused o the individual approach in the development.

6 0
3 years ago
Which of the following ethical guidelines is best described as refraining from using granted authorizations for personal gain or
Artyom0805 [142]

Answer:

A. Do not use computer access or the Internet for unprofessional or inappropriate purposes.

7 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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