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seraphim [82]
4 years ago
12

Priscilla is providing the junior analysts in her firm with some real-world illustrations to explain some of the recommendations

that they must be prepared to make to clients, based on what they have studied in their coursework, in order to solidify their understanding.Which is a reason why Priscilla would offer in-house software development?A) The client has a number of very distinctive security requirements.B) The client wants to be able to use fewer technical development staff members.C) The client is looking to spend the least possible time in implementation.D) The client wishes to receive future upgrades from the vendors.
Computers and Technology
1 answer:
Vlad [161]4 years ago
7 0

Answer:

The answer is letter A.

Explanation:

The client has a number of very distinctive security requirements.

Building a software  product using internal company's resources and inside the company is a good option if the client has the required technology and professionals available.

You might be interested in
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
Ask what kind of pet the user has. If they enter cat, print "Too bad...", if they enter dog, print "Lucky you!" (You can change
nirvana33 [79]

Answer:

Explanation:

a = input("what kind of pet the user has")

if a == 'cat':

  print("Too bad")

elif a == 'dog':

   print("Lucky you!")

8 0
3 years ago
What is hypertext? Whata search engine? What is IT?
Verizon [17]

Answer:

Hypertext is a text in a document which references some other link within the same page or another page.

Search engine is an information retrieval system from larger pool of resources.

IT is for information technology whereby with the help of systems and communication medium we are able to send, retrieve information from one place to another.

Explanation:

Some website page can contain hypertext which is a link to another page. These are simple text upon clicking it lands us to another page or a pop up activity.

Examples of search engine is google with which we can search and retrieve information from a large database servers.

IT governs the ways these information exchanges taking place between client and sender with help of systems(computers) and communicating medium(internet).

7 0
3 years ago
Read 2 more answers
A small workgroup inherits a second-hand printer without networking capabilities. Which of the following is the BEST method of s
butalik [34]
Answer: c

Explanation: connect the printer to a users work station and share the device
5 0
3 years ago
In this assignment, you will write a program that will contain two functions, setlsbs() and getlsbs(). These functions will use
kicyunya [14]

Answer:

#include "Lab11_jlilly3_204.h"

#include <stdio.h>

#include <stdlib.h>

#define BYTETOBINARYPATTERN "%d%d%d%d%d%d%d%d"

#define BYTETOBINARY(byte) \

(byte & 0x80 ? 1 : 0), \

(byte & 0x40 ? 1 : 0), \

(byte & 0x20 ? 1 : 0), \

(byte & 0x10 ? 1 : 0), \

(byte & 0x08 ? 1 : 0), \

(byte & 0x04 ? 1 : 0), \

(byte & 0x02 ? 1 : 0), \

(byte & 0x01 ? 1 : 0)

#define PRINTBIN(x) printf(BYTETOBINARYPATTERN, BYTETOBINARY(x));

//! Set the least significant bit for all values in an array to the coorrisponding bit pos. in b_o

void setlsbs(unsigned char* p, unsigned char b_0)

{

int i;

for (i = 0; i < 8; i++)

{

int x = b_0 & 1;

// If x is one set the low order bit to 1 using or

if (x)

{

p[i] |= x;

}

// If the value is a 0 and with one's complement to set lo to 0

else

{

p[i] &= ~1;

}

// shift our bit to the right to get the next value in the bit

b_0 = b_0 >> 1;

}

}

//! Get the least significant bit from each position in an array and add it to a byte

unsigned char getlsbs(unsigned char *p)

{

int i;

unsigned char c = 0;

for (i = 0; i < 8; i++)

{

// find out what the lsb is

int x = p[i] & 1;

if (x)

{

// if the lsb was a one insert a one and shift it into position

c |= 1 << i;

}

}

return c;

}

//! Print a given array in decimal

void print_decimal(unsigned char* p)

{

int k;

for (k = 0; k < 8; k++)

{

printf("Num at %d: %d\n", k, p[k]);

}

}

//! Print a given array in in bianary representation

void print_bin(unsigned char* p)

{

int c;

for (c = 0; c < 8; c++)

{

printf("Num at %d ", c);

PRINTBIN(p[c]);

printf("\n");

}

}

6 0
3 years ago
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