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erica [24]
3 years ago
8

Jason is computer professional who has access to sensitive information. He shares this information with another organization in

the same industry. Which professional code of conduct did Jason violate?
Computers and Technology
1 answer:
Temka [501]3 years ago
4 0

<em>Jason violated the </em><em>confidentiality of information. </em>

<em> </em>

<em>Part of information security is "confidentiality of information" wherein an employee must kept all the information he may acquire when he is working in a company, confidential. When we say confidential, he is not allowed to disclose anything that he may know about company's operation, sensitive information to anybody who is outside the company (like in the example, another organization). Because data that he might disclose may get on hands by company's competitors and it can be used against the company. They might steal ideas and strategies. </em>

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Answer:

See explaination

Explanation:

dateType.h

#ifndef dateType_H

#define dateType_H

class dateType

{

public:

void setDate(int month, int day, int year);

//Function to set the date.

//The member variables dMonth, dDay, and dYear are set

//according to the parameters.

//Postcondition: dMonth = month; dDay = day;

// dYear = year

int getDay() const;

//Function to return the day.

//Postcondition: The value of dDay is returned.

int getMonth() const;

//Function to return the month.

//Postcondition: The value of dMonth is returned.

int getYear() const;

//Function to return the year.

//Postcondition: The value of dYear is returned.

void printDate() const;

//Function to output the date in the form mm-dd-yyyy.

void isLeapYear() const;

dateType(int month = 1, int day = 1, int year = 1900);

//Constructor to set the date

//The member variables dMonth, dDay, and dYear are set

//according to the parameters.

//Postcondition: dMonth = month; dDay = day; dYear = year;

// If no values are specified, the default

// values are used to initialize the member

// variables.

private:

int dMonth; //variable to store the month

int dDay; //variable to store the day

int dYear; //variable to store the year

};

#endif

dateType.cpp

#include <iostream>

#include "dateType.h"

using namespace std;

void dateType::setDate(int month, int day, int year)

{

// Checking month is valid

while(month<1 || month>12)

{

cout << "Enterd month "<<month<< " is wrong"<<endl;

cout << "Enter correct month"<<endl;

cin>>month;

}

dMonth = month;

// Checking date is valid

while(day<1 || day>31)

{

cout << "Enterd date "<<day<<" is wrong"<<endl;

cout<<"Enter correct date"<<endl;

cin>>day;

}

dDay = day;

int count_digits = 0;

int flag=0;

int year1;

// Counting number of digits in year

while(flag==0)

{

year1=year;

count_digits=0;

while (year) {

year /= 10;

count_digits++;

}

if(count_digits != 4)

{

cout << "Enterd year "<<year1<<" is wrong"<<endl;

cout<<"Enter correct year"<<endl;

cin>>year;

flag=0;

}

else

flag=1;

}

dYear = year1;

}

int dateType::getDay() const

{

return dDay;

}

int dateType::getMonth() const

{

return dMonth;

}

int dateType::getYear() const

{

return dYear;

}

void dateType::printDate() const

{

cout << dMonth << "-" << dDay << "-" << dYear;

}

void dateType::isLeapYear() const

{

if ( dYear%400 == 0)

cout<<endl<<dYear<< " is leap year.\n";

else if ( dYear%100 == 0)

cout<<endl<<dYear<< " is leap year.\n";

else if ( dYear%4 == 0 )

cout<<endl<<dYear<< " is leap year.\n";

else

cout<<endl<<dYear<< " is not leap year.\n";

}

//Constructor with parameters

dateType::dateType(int month, int day, int year)

{

// Checking month is valid

while(month<1 || month>12)

{

cout << "Enterd month "<<month<< " is wrong"<<endl;

cout << "Enter correct month"<<endl;

cin>>month;

}

dMonth = month;

// Checking date is valid

while(day<1 || day>31)

{

cout << "Enterd date "<<day<<" is wrong"<<endl;

cout<<"Enter correct date"<<endl;

cin>>day;

}

dDay = day;

int count_digits = 0;

int flag=0;

int year1;

// Counting number of digits in year

while(flag==0)

{

year1=year;

count_digits=0;

while (year) {

year /= 10;

count_digits++;

}

if(count_digits != 4)

{

cout << "Enterd year "<<year1<<" is wrong"<<endl;

cout<<"Enter correct year"<<endl;

cin>>year;

flag=0;

}

else

flag=1;

}

dYear = year1;

}

main.cpp

#include<iostream>

#include "dateType.h"

using namespace std;

int main()

{

dateType *dt1=new dateType();

cout<<"Date is "<<endl;

dt1->printDate();

cout<<endl;

dt1->isLeapYear();

cout<<endl;

dateType *dt2=new dateType(11,14,2019);

cout<<"Date is "<<endl;

dt2->printDate();

cout<<endl;

dt2->isLeapYear();

cout<<endl;

dt2->setDate(13,32,2016);

cout<<"Date is "<<endl;

dt2->printDate();

cout<<endl;

dt2->isLeapYear();

cout<<endl;

dt1->setDate(10,10,198);

cout<<"Date is "<<endl;

dt1->printDate();

cout<<endl;

dt1->isLeapYear();

cout<<endl;

system("pause");

return 0;

}

6 0
3 years ago
Suppose we compute a depth-first search tree rooted at u and obtain a tree t that includes all nodes of g.
Temka [501]

G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.

<h3>What are DFS and BFS?</h3>

An algorithm for navigating or examining tree or graph data structures is called depth-first search. The algorithm moves as far as it can along each branch before turning around, starting at the root node.

The breadth-first search strategy can be used to look for a node in a tree data structure that has a specific property. Before moving on to the nodes at the next depth level, it begins at the root of the tree and investigates every node there.

First, we reveal that G exists a tree when both BFS-tree and DFS-tree are exact.

If G and T are not exact, then there should exist a border e(u, v) in G, that does not belong to T.

In such a case:

- in the DFS tree, one of u or v, should be a prototype of the other.

- in the BFS tree, u and v can differ by only one level.

Since, both DFS-tree and BFS-tree are the very tree T,

it follows that one of u and v should be a prototype of the other and they can discuss by only one party.

This means that the border joining them must be in T.

So, there can not be any limits in G which are not in T.

In the two-part of evidence:

Since G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.

The complete question is:

We have a connected graph G = (V, E), and a specific vertex u ∈ V.

Suppose we compute a depth-first search tree rooted at u, and obtain a tree T that includes all nodes of G.

Suppose we then compute a breadth-first search tree rooted at u, and obtain the same tree T.

Prove that G = T. (In other words, if T is both a depth-first search tree and a breadth-first search tree rooted at u, then G cannot contain any edges that do not belong to T.)

To learn more about  DFS and BFS, refer to:

brainly.com/question/13014003

#SPJ4

5 0
2 years ago
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