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defon
3 years ago
13

Baby Boomers are an example of what type of trend?

Computers and Technology
2 answers:
Bond [772]3 years ago
8 0
The baby boomers is a type of demographic group. The demographic groups are usually categorized by age, income bracket, social class and so on. Baby boomers are classified as a generation of a group of people and also therefore, by their age range. 
FromTheMoon [43]3 years ago
3 0

Baby Boomers are an example of the demographic type of trend. Correct answer: A

The demographic group baby boomers includes those people born worldwide between 1946 and 1964. The term "baby boomer" is also used in a cultural context, associated with a rejection or redefinition of traditional values.

You might be interested in
What were the advantages and disadvantages of agricultural technology to us.
Digiron [165]

Advantages of technology in agriculture include expediting crop production rate and crop quantity, which in turn reduces costs of production for farmers and food costs for consumers, and even makes crops more nutritious and livestock bigger and meatier.

The excessive use of chemicals by the help of machines reduces the fertility of the land.Lack of practical knowledge the farmers cant handle the machines properly.While the cost of maintenance is very high.Overuse of machines may lead to environmental damage.It is efficient but has many side effects and drawbacks.

5 0
3 years ago
Read 2 more answers
When you connect several home devices using a wireless router, what network topology are you using? A. bus B. mesh C. star D. ri
bonufazy [111]

Answer:

Star Topology

Explanation:

Because the definition of Star Topoplogy is: In star topology each device in the network is connected to a central device called hub. Unlike Mesh topology, star topology doesn’t allow direct communication between devices, a device must have to communicate through hub. If one device wants to send data to other device, it has to first send the data to hub and then the hub transmit that data to the designated device.

3 0
3 years ago
Given a floating-point formal with a k-bit exponent and an n-bit (fraction, write formulas for the exponent E, significant M, th
ANEK [815]

Answer:

A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112  ≈ 7/4

fractional value ( F )  = 0.112

And, numeric value of the quantity ( V )  = 7

The exponent bits will be represented  as :  100----01.

while The fraction bits will be represented  as : 1100---0.

<u>B) The largest odd integer that can be represented exactly </u>

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

=  1.11------12 = 2 - 2-n

also the fractional value ( F ) =  

0.11------12 = 1 – 2-n

Also, Value, V = 2n+1 – 1

The exponent bits  will be represented  as follows:  n + 2k-1 – 1.

while The bit representation for the fraction will be as follows: 11---11.

<u>C) The reciprocal of the smallest positive normalized value </u>

The numerical value of the equity ( V ) = 22k-1-2

The exponential value ( E )  = 2k-1 – 2

While the significand value ( M )  = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

Explanation:

E = integer value of exponent

M = significand value

F = fractional value

V = numeric value of quantity

A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112  ≈ 7/4

fractional value ( F )  = 0.112

And, numeric value of the quantity ( V )  = 7

The exponent bits will be represented  as :  100----01.

while The fraction bits will be represented  as : 1100---0.

<u>B) The largest odd integer that can be represented exactly </u>

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

=  1.11------12 = 2 - 2-n

also the fractional value ( F ) =  

0.11------12 = 1 – 2-n

Also, Value, V = 2n+1 – 1

The exponent bits  will be represented  as follows:  n + 2k-1 – 1.

while The bit representation for the fraction will be as follows: 11---11.

<u>C) The reciprocal of the smallest positive normalized value </u>

The numerical value of the equity ( V ) = 22k-1-2

The exponential value ( E )  = 2k-1 – 2

While the significand value ( M )  = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

8 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
4 years ago
Which of the following are examples of packages you can import and use classes from in java?
Svetlanka [38]
It’s D java. int i think
8 0
3 years ago
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