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inysia [295]
3 years ago
10

What legal punishment can people face for plagerism

Computers and Technology
1 answer:
Maru [420]3 years ago
5 0
They can get a fine.
You might be interested in
Write a program that grades arithmetic quizzes as follows: Ask the user how many questions are in the quiz. Ask the user to ente
kirill115 [55]

Answer:

The program in C++ is as follows:

#include <iostream>

using namespace std;

int main(){

   int questions, answer;

   cout<<"Questions: ";

   cin>>questions;

   int answerkey[questions];

   cout<<"Enter answer keys: ";

   for(int i = 0; i< questions; i++){

       cin>>answerkey[i];    }

   int correct = 0;

   cout<<"Enter answers: ";

   for(int i = 0; i< questions; i++){

       cin>>answer;

       if(answer == answerkey[i]){

           correct++;        }    }

   cout<<"Correct answers: "<<correct<<endl;

   cout<<"Percentage correct : "<<(100 * correct)/questions<<"%";

   return 0;

}

Explanation:

This declares the number of questions and the answers submitted to each equation

   int questions, answer;

Prompt to get the number of questions

   cout<<"Questions: ";

This gets input for the number of questions

   cin>>questions;

This declares the answerkey as an array

   int answerkey[questions];

Prompt to get the answer key

   cout<<"Enter answer keys: ";

This iteration gets the answer key for each question

<em>    for(int i = 0; i< questions; i++){</em>

<em>        cin>>answerkey[i];    }</em>

This initializes the number of correct answers to 0

   int correct = 0;

Prompt to get the enter the answers

   cout<<"Enter answers: ";

This iterates through the answer keys

   for(int i = 0; i< questions; i++){

This gets the answer to each question

       cin>>answer;

This compares the answer to the answer key of the question

       if(answer == answerkey[i]){

If they are the same, correct is incremented by 1

           correct++;        }    }

Print the number of correct answers

   cout<<"Correct answers: "<<correct<<endl;

Print the percentage of correct answers

   cout<<"Percentage correct : "<<(100 * correct)/questions<<"%";

7 0
3 years ago
I have six nuts and six bolts. Exactly one nut goes with each bolt. The nuts are all different sizes, but it’s hard to compare t
juin [17]

Answer:

Explanation:

In order to arrange the corresponding nuts and bolts in order using quicksort algorithm, we need to first create two arrays , one for nuts and another for bolts namely nutsArr[6] and boltsArr[6]. Now, using one of the bolts as pivot, we can rearrange the nuts in the nuts array such that the nuts on left side of the element chosen (i.e, the ith element indexed as nutArr[i]) are smaller than the nut at ith position and nuts to the right side of nutsArr[i] are larger than the nut at position "I". We implement this strategy recursively to sort the nuts array. The reason that we need to use bolts for sorting nuts is that nuts are not comparable among themselves and bolts are not comparable among themselves(as mentioned in the question)

The pseudocode for the given problem goes as follows:

// method to quick sort the elements in the two arrays

quickSort(nutsArr[start...end], boltsArr[start...end]): if start < end: // choose a nut from nutsArr at random randElement = nutsArr[random(start, end+1)] // partition the boltsArr using the randElement random pivot pivot = partition(boltsArr[start...end], randElement) // partition nutsArr around the bolt at the pivot position partition(nutsArr[start...end], boltsArr[pivot]) // call quickSort by passing first partition quickSort(nutsArr[start...pivot-1], boltsArr[start...pivot-1]) // call quickSort by passing second partition quickSort(nutsArr[pivot+1...end], boltsArr[pivot+1...end])

// method to partition the array passed as parameter, it also takes pivot as parameter

partition(character array, integer start, integer end, character pivot)

{

       integer i = start;

loop from j = start to j < end

       {

check if array[j] < pivot

{

swap (array[i],array[j])

               increase i by 1;

           }

 else check if array[j] = pivot

{

               swap (array[end],array[j])

               decrease i by 1;

           }

       }

swap (array[i] , array[end])

       return partition index i;

}

7 0
3 years ago
For our computer club in school, we are making a discord server and seeing who can get the most members. I am making an Anime Se
Gwar [14]

Answer:

misteragua#1874

Explanation:

5 0
2 years ago
The process of ensuring that web pages coded with new or advanced techniques still are usable in browsers that do not offer supp
Leya [2.2K]

Answer:

(C) progressive enhancement

Explanation:

Progressive enhancement is a web design technique that first underlines the main functionality web page and then incorporates increasingly enhanced and advanced features and design levels.

This process enables the users to use and access the basic features of the web pages. These users can use any browser to access the main contents of the web page. Progressive Enhancement also offers an advanced version of the website for those with more advanced and sophisticated computer browsers or faster internet connection.

So the user can access basic features of a web sites as well as the complex features. This process first makes sure the basic intended purpose and main contents of the web site is accessible to all users before adding complex features which are supported by different browsers and devices.

Validation is a process which checks that code of the website is as per the world wide web standards and also checks that content of web pages and design of websites is being properly displayed and the site is accessible. So this is not the correct option.

Technique of altering a web site in a way that it can appear higher in the search engine results is called optimization so this is not a correct option.

7 0
3 years ago
What protocol suite below is the most commonly used protocol for local area network (lan) communication?
grandymaker [24]
Answer TCP/IP is the most common protocol in Local Area Networks.
3 0
3 years ago
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