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Olegator [25]
2 years ago
13

Koi jinda hei kya hello​

Computers and Technology
2 answers:
PSYCHO15rus [73]2 years ago
5 0

Answer:

hi I m here

Explanation:

jinda bandi hu

Bezzdna [24]2 years ago
5 0

Which type of question is this ?

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Which is the least technically experienced technical support group?
Nezavi [6.7K]

Answer:

Tier 1 support

Explanation:

Of the given options, tier 1 support technical group is the least experienced group. This group of technicians is also referred to as level 1 technical group.

The tier 1 or level support are made up of junior technician, and they have few technical understandings.

Their roles include email response and basic troubleshooting, attending to phone calls, among others.

When a problem cannot be solved by tier 1 support  technicians, they pass the problem to tier 2 support technicians,  

5 0
2 years ago
Match the different sources of payments to their descriptions. an electronic device that aids consumers to make online payments
LekaFEV [45]

<u>Automatic Draft-</u> A convenient payment method where funds are automatically debited from your account.

<u>Cash-</u> The physical form of currency.

<u>Credit Card-</u> Allows you to make a purchase and pay for it later.

<u>Digital Wallet-</u> An electronic device that aids consumers to make online payments through various websites.

For anyone still looking, hope this helps.

6 0
3 years ago
In 1-2 sentences, describe how to use the thesaurus in the Word Processor you have used.
jasenka [17]
Thesaurus is a tool use to find the synonym of the of a word.
Here's the way to use the thesaurus in the word processor.
=> highlight the word, then right-click, Navigate to synonyms and the words will  display.
3 0
3 years ago
Read 2 more answers
A perfect binary tree is a complete binary tree with all levels fully filled. Add a method in the BST class to return true if th
algol [13]

Answer:

class BST {

static class Node

{

int val;

Node left, right;

}

static boolean checkPerfectBT(Node node, int h, int d)

{

if (node == null)

return true;

 

if (node.left == null && node.right == null)

{

if(h==d+1)

return true;

else

return false;

}

 

if (node.left == null || node.right== null)

return false;

 

return checkPerfectBT(node.left, h, d+1) && checkPerfectBT(node.right, h, d+1);

}

static int height(Node node)

{

int dep = 0;

while (node != null)

{

node = node.right;

dep=dep+1;

}

return dep;

}

static boolean isPerfectBT(Node node)

{

int h = height(node);

return checkPerfectBT(node, h, 0);

}

 

static Node addNode(int x)

{

Node node = new Node();

node.val= x;

node.right = null;

node.left = null;

return node;

}

public static void main(String args[])

{

int i,j,k;

Node node= null;

node = addNode(34);

node.left = addNode(2);

node.right = addNode(322);

 

node.left.left = addNode(21);

node.left.right = addNode(23);

node.right.left = addNode(37);

node.right.right = addNode(54);

 

if (isPerfectBT(node) == true)

System.out.println("This is a Perfect Binary tree");

else

System.out.println("This is Not a perfect Binary Tree");

}

}

Explanation:

  • Calculate the depth of BST by using a while loop until it reaches a null value.
  • In the addNode method, make an object of Node class.
  • In the main method, pass the values to addNode method.
  • Finally display the relevant message to show if it's Perfect or not.
6 0
3 years ago
The 7-bit ASCII code for the character ‘&amp;’ is: 0100110 An odd parity check bit is now added to this code so 8 bits are trans
Alex787 [66]

Answer

First part:

The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.

Second part:

The invalid bit sequence are option a. 01001000 and d. 11100111

Explanation:

Explanation for first part:

In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.

If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.

If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.  

Explanation for second part:

A valid odd parity bit sequence will always have odd number of 1s.

Since in option a and d,  total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.

And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.

7 0
3 years ago
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