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Xelga [282]
3 years ago
9

Which relationship is possible when two tables share the same primary key?

Computers and Technology
1 answer:
klasskru [66]3 years ago
6 0

Answer:

Many-to-one

Explanation:

Many-to-one relationships is possible when two tables share the same primary key it is because one entity contains values that refer to another entity that has unique values. It often enforced by primary key relationships, and the relationships typically are between fact and dimension tables and between levels in a hierarchy.

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A DFA is equivalent in power to an NFA. True False
sattari [20]

Answer: True

Explanation:

  Yes, the given statement is true that a DFA is equivalent to NFA in terms of power. For any type of NFA we can easily build an equal DFA so, the NFA are not much powerful as compared to DFA. Both NFA and DFA are characterized by a similar type of class.

DFA is a special case of NFA and They both defined in the same class of language. Each condition in the DFA get summarized by all the condition that the NFA has itself.

4 0
3 years ago
Write a program in C which will open a text file named Story.txt. You can create the file using any text editor like notepad etc
ddd [48]

Answer:

See explaination

Explanation:

#include <stdio.h>

#include <stdlib.h>

int main()

{

FILE * file_object;

char file_name[100];

char ch;

int characters=0, words=0;

printf("Enter source file name: ");

scanf("%s", file_name); //asking user to enter the file name

file_object = fopen(file_name, "r"); //open file in read mode

if (file_object == NULL)

{

printf("\nUnable to open file.file not exist\n"); //check if the file is present or not

}

while ((ch = fgetc(file_object)) != EOF) //read each character till the end of the file

{

if (ch == ' ' || ch == '\t' || ch == '\n' || ch == '\0') //if character is space or tab or new line or null character increment word count

words++;

else

characters++; //else increment character count this assures that there is no spaces count

}

printf("The file story.txt has the following Statistics:\n"); //finally print the final statistics

if (characters > 0)

{

printf("Words: %d\n", words+1); //for last word purpose just increment the count of words

printf("Characters (no spaces): %d\n", characters);

}

fclose(file_object); //close the file object

return 0;

}

6 0
4 years ago
Maeve has created the website, but she is not able to include more than one blank space at a time. Even if she hits the SPACE bu
GenaCL600 [577]

Answer:

For the browser to display multiple spaces, Maeve must use &nbsp; in her code to create more spaces.

Explanation:

&nbsp; is known as hard space or fixed space. NBSP stands for Non-Breaking Space. The statement will continue without breaking into another line.

Maeve can easily use the &nbsp; many times to get the required space she needs.

5 0
4 years ago
Did the Z3 computer invented by Konrad Zuse have a negative effect on society?
musickatia [10]

Answer:

Explanation:

Không

4 0
2 years ago
Read 2 more answers
The birthday paradox says that the probability that two people in a room will have the same birthday is more than half, provided
poizon [28]

Answer:

The Java code is given below with appropriate comments for explanation

Explanation:

// java code to contradict birth day paradox

import java.util.Random;

public class BirthDayParadox

{

public static void main(String[] args)

{

   Random randNum = new Random();

   int people = 5;

   int[] birth_Day = new int[365+1];

   // setting up birthsdays

   for (int i = 0; i < birth_Day.length; i++)

       birth_Day[i] = i + 1;

 

   int iteration;

   // varying number n

   while (people <= 100)

   {

       System.out.println("Number of people: " + people);

       // creating new birth day array

       int[] newbirth_Day = new int[people];

       int count = 0;

       iteration = 100000;

       while(iteration != 0)

       {

           count = 0;

           for (int i = 0; i < newbirth_Day.length; i++)

           {

               // generating random birth day

               int day = randNum.nextInt(365);

               newbirth_Day[i] = birth_Day[day];

           }

           // check for same birthdays

           for (int i = 0; i < newbirth_Day.length; i++)

           {

               int bday = newbirth_Day[i];

               for (int j = i+1; j < newbirth_Day.length; j++)

               {

                   if (bday == newbirth_Day[j])

                   {

                       count++;

                       break;

                   }

               }

           }

           iteration = iteration - 1;

       }

       System.out.println("Probability: " + count + "/" + 100000);

       System.out.println();

       people += 5;

   }

}

}

4 0
4 years ago
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