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abruzzese [7]
3 years ago
6

Green field country is planning to conduct a cricket match between two teams A and B. a large crowd is expected in the stadium a

s 10000 tickets have already been sold. To fulfill the viewer's requirements, the match administrators are willing to install large display screens inside the stadium
Which should display the match Live through several cameras installed on different angles in the stadium.
After carefully reading and analyzing the given scenario, you are required to suggest the most suitable serial transmission mode with two strong reasons.
Computers and Technology
1 answer:
AnnZ [28]3 years ago
5 0

Answer:

For this requirement, we first need the Network Interface Card which supports the serial transmission, and two such can be RS-485 and EIA RS-422A. And you need to remember that parallel transmission is speedier than serial, however, they are more prone to the noise. And in both of the above cases, we can have multiple receivers and one transmitter and all that over a single line. And this line can be as long as 6000 feet and the speed can be as high as 10 MBPS, which is quite awesome and certainly apt for the above requirement.

Explanation:

Please check the answer.

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Hyper-inflation is where prices rise dramatically over a short period of time. What is the cause given for cases of hyper-inflat
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8 0
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Read 2 more answers
Create a Binary Expressions Tree Class and create a menu driven programyour program should be able to read multiple expressions
solong [7]

Answer:

Explanation:

Program:

#include<iostream>

#include <bits/stdc++.h>

using namespace std;

//check for operator

bool isOperator(char c)

{

switch(c)

{

case '+': case '-': case '/': case '*': case '^':

return true;

}

return false;

}

//Converter class

class Converter

{

private:

string str;

public:

//constructor

Converter(string s):str(s){}

//convert from infix to postfix expression

string toPostFix(string str)

{

stack <char> as;

int i, pre1, pre2;

string result="";

as.push('(');

str = str + ")";

for (i = 0; i < str.size(); i++)

{

char ch = str[i];

if(ch==' ') continue;

if (ch == '(')

as.push(ch);

else if (ch == ')')

{

while (as.size() != 0 && as.top() != '('){

result = result + as.top() + " ";

as.pop();

}

as.pop();

}

else if(isOperator(ch))

{

while (as.size() != 0 && as.top() != '(')

{

pre1 = precedence(ch);

pre2 = precedence(as.top());

if (pre2 >= pre1){

result = result + as.top() + " ";

as.pop();

}

else break;

}

as.push(ch);

}

else

{

result = result + ch;

}

}

while(as.size() != 0 && as.top() != '(') {

result += as.top() + " ";

as.pop();

}

return result;

}

//return the precedence of an operator

int precedence(char ch)

{

int choice = 0;

switch (ch) {

case '+':

choice = 0;

break;

case '-':

choice = 0;

break;

case '*':

choice = 1;

break;

case '/':

choice = 1;

break;

case '^':

choice = 2;

default:

choice = -999;

}

return choice;

}

};

//Node class

class Node

{

public:

string element;

Node *leftChild;

Node *rightChild;

//constructors

Node (string s):element(s),leftChild(nullptr),rightChild(nullptr) {}

Node (string s, Node* l, Node* r):element(s),leftChild(l),rightChild(r) {}

};

//ExpressionTree class

class ExpressionTree

{

public:

//expression tree construction

Node* covert(string postfix)

{

stack <Node*> stk;

Node *t = nullptr;

for(int i=0; i<postfix.size(); i++)

{

if(postfix[i]==' ') continue;

string s(1, postfix[i]);

t = new Node(s);

if(!isOperator(postfix[i]))

{

stk.push(t);

}

else

{

Node *r = nullptr, *l = nullptr;

if(!stk.empty()){

r = stk.top();

stk.pop();

}

if(!stk.empty()){

l = stk.top();

stk.pop();

}

t->leftChild = l;

t->rightChild = r;

stk.push(t);

}

}

return stk.top();

}

//inorder traversal

void infix(Node *root)

{

if(root!=nullptr)

{

cout<< "(";

infix(root->leftChild);

cout<<root->element;

infix(root->rightChild);

cout<<")";

}

}

//postorder traversal

void postfix(Node *root)

{

if(root!=nullptr)

{

postfix(root->leftChild);

postfix(root->rightChild);

cout << root->element << " ";

}

}

//preorder traversal

void prefix(Node *root)

{

if(root!=nullptr)

{

cout<< root->element << " ";

prefix(root->leftChild);

prefix(root->rightChild);

}

}

};

//main method

int main()

{

string infix;

cout<<"Enter the expression: ";

cin >> infix;

Converter conv(infix);

string postfix = conv.toPostFix(infix);

cout<<"Postfix Expression: " << postfix<<endl;

if(postfix == "")

{

cout<<"Invalid expression";

return 1;

}

ExpressionTree etree;

Node *root = etree.covert(postfix);

cout<<"Infix: ";

etree.infix(root);

cout<<endl;

cout<<"Prefix: ";

etree.prefix(root);

cout<<endl;

cout<< "Postfix: ";

etree.postfix(root);

cout<<endl;

return 0;

}

3 0
3 years ago
People who connect one network to another within the company or even across organizations are A. central connectors. B. boundary
Lunna [17]

People who connect one network to another within the company or even across organizations are boundary spanners.  

People who serve as ambassadors to linking information and communications on informal networks within different department in an organisation and with other organisations or personnel  far and near for the good of their organisation are called Boundary Spanners.

The importance of Boundary  spanning in organisation includes

  • Exchange of expertise information
  • Nurturing of connections with people from different part s of the world.
  • Improved innovation in businesses

See more here: brainly.com/question/14728967

5 0
3 years ago
Careers and information technology deal with
son4ous [18]

Answer:

computers and telecommunications

Explanation:

3 0
2 years ago
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