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horsena [70]
3 years ago
15

Which relationship is possible when two tables share the same primary key? one-to-one one-to-many many-to-one many-to-many

Computers and Technology
2 answers:
Finger [1]3 years ago
8 0
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.
disa [49]3 years ago
3 0

for plato many-to-one is not the answer

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What does a compiler do, and why is it necessary when using higher-level languages?
UkoKoshka [18]

Answer:

A compiler converts human readable instructions into machine code (machine readable instructions). Without it, a computer will not be able to understand the code that was written and execute it. Since higher programmer languages are easier for humans to read and write and effective compiler is needed. This has to do with how the compiler does much of the work when it comes to programming.

A great example is a drag and drop programming language. The compiler does all the work in the background before the machine can actually execute the code, but the language itself is incredibly easy to read and write by human standards. Without the compiler, it would be impossible for the machine to execute any code.

Explanation:

High Level programming languages like Python has a system that turns our easy to read human code into something the computer can actually read. Python for example, doesn't compile the code as it was design to do it as it runs.

Low level programming languages like Objective C uses a compiler to change the human readable code to machine code. You can tell a compiler was used when there is a 2nd file, one that can't be read by humans. This is the compiled code that the machine actually runs.

8 0
4 years ago
Which of the following statement is true?
Alexxx [7]

Answer:

b. the IP address can be spoofed, so if you want to read response from the deceived party, you can use IP spoofing to hide yourself.

Explanation:

3 0
3 years ago
Create a loop that will output all the numbers less than 200 that are evenly divisible (meaning remainder is zero) by both 5 and
Readme [11.4K]

Answer:

public class num7 {

   public static void main(String[] args) {

       int n =1;

       while(n<200){

           if(n%5==0 && n%7==0){

               System.out.print(n);

               System.out.print(",");

           }

           n++;

       }

   }

}

Explanation:

  • In Java programming Language
  • Create and initialize an int variable (n=1)
  • Create a while loop with the condition while (n<200)
  • Within the while loop use the modulo operator % to check for divisibility by 5 and 7
  • Print the numbers divisible by 5 and 7
4 0
3 years ago
Write a program to sort an array of 100,000 random elements using quicksort as follows: Sort the arrays using pivot as the middl
Stels [109]

Answer:

Check the explanation

Explanation:

#include<iostream.h>

#include<algorithm.h>

#include<climits.h>

#include<bits/stdc++.h>

#include<cstring.h>

using namespace std;

int partition(int arr[], int l, int r, int k);

int kthSmallest(int arr[], int l, int r, int k);

void quickSort(int arr[], int l, int h)

{

if (l < h)

{

// Find size of current subarray

int n = h-l+1;

 

// Find median of arr[].

int med = kthSmallest(arr, l, h, n/2);

 

// Partition the array around median

int p = partition(arr, l, h, med);

 

// Recur for left and right of partition

quickSort(arr, l, p - 1);

quickSort(arr, p + 1, h);

}

int findMedian(int arr[], int n)

{

sort(arr, arr+n); // Sort the array

return arr[n/2]; // Return middle element

}

int kthSmallest(int arr[], int l, int r, int k)

{

// If k is smaller than number of elements in array

if (k > 0 && k <= r - l + 1)

{

int n = r-l+1; // Number of elements in arr[l..r]

 

// Divide arr[] in groups of size 5, calculate median

// of every group and store it in median[] array.

int i, median[(n+4)/5]; // There will be floor((n+4)/5) groups;

for (i=0; i<n/5; i++)

median[i] = findMedian(arr+l+i*5, 5);

if (i*5 < n) //For last group with less than 5 elements

{

median[i] = findMedian(arr+l+i*5, n%5);

i++;

}

int medOfMed = (i == 1)? median[i-1]:

kthSmallest(median, 0, i-1, i/2);

int pos = partition(arr, l, r, medOfMed);

if (pos-l == k-1)

return arr[pos];

if (pos-l > k-1) // If position is more, recur for left

return kthSmallest(arr, l, pos-1, k);

return kthSmallest(arr, pos+1, r, k-pos+l-1);

}

return INT_MAX;

}

void swap(int *a, int *b)

{

int temp = *a;

*a = *b;

*b = temp;

}

int partition(int arr[], int l, int r, int x)

{

// Search for x in arr[l..r] and move it to end

int i;

for (i=l; i<r; i++)

if (arr[i] == x)

break;

swap(&arr[i], &arr[r]);

 

// Standard partition algorithm

i = l;

for (int j = l; j <= r - 1; j++)

{

if (arr[j] <= x)

{

swap(&arr[i], &arr[j]);

i++;

}

}

swap(&arr[i], &arr[r]);

return i;

}

 

/* Function to print an array */

void printArray(int arr[], int size)

{

int i;

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

cout << arr[i] << " ";

cout << endl;

}

 

// Driver program to test above functions

int main()

{

float a;

clock_t time_req;

int arr[] = {1000, 10, 7, 8, 9, 30, 900, 1, 5, 6, 20};

int n = sizeof(arr)/sizeof(arr[0]);

quickSort(arr, 0, n-1);

cout << "Sorted array is\n";

printArray(arr, n);

time_req = clock();

for(int i=0; i<200000; i++)

{

a = log(i*i*i*i);

}

time_req = clock()- time_req;

cout << "Processor time taken for multiplication: "

<< (float)time_req/CLOCKS_PER_SEC << " seconds" << endl;

 

// Using pow function

time_req = clock();

for(int i=0; i<200000; i++)

{

a = log(pow(i, 4));

}

time_req = clock() - time_req;

cout << "Processor time taken in pow function: "

<< (float)time_req/CLOCKS_PER_S

return 0;

}

..................................................................................................................................................................................................................................................................................................................................

OR

.......................

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

 

// Swap utility

void swap(long int* a, long int* b)

{

int tmp = *a;

*a = *b;

*b = tmp;

}

 

// Bubble sort

void bubbleSort(long int a[], long int n)

{

for (long int i = 0; i < n - 1; i++) {

for (long int j = 0; j < n - 1 - i; j++) {

if (a[j] > a[j + 1]) {

swap(&a[j], &a[j + 1]);

}

}

}

}

 

// Insertion sort

void insertionSort(long int arr[], long int n)

{

long int i, key, j;

for (i = 1; i < n; i++) {

key = arr[i];

j = i - 1;

 

// Move elements of arr[0..i-1], that are

// greater than key, to one position ahead

// of their current position

while (j >= 0 && arr[j] > key) {

arr[j + 1] = arr[j];

j = j - 1;

}

arr[j + 1] = key;

}

}

 

// Selection sort

void selectionSort(long int arr[], long int n)

{

long int i, j, midx;

 

for (i = 0; i < n - 1; i++) {

 

// Find the minimum element in unsorted array

midx = i;

 

for (j = i + 1; j < n; j++)

if (arr[j] < arr[min_idx])

midx = j;

 

// for plotting graph with integer values

printf("%li, %li, %li, %li\n",

n,

(long int)tim1[it],

(long int)tim2[it],

(long int)tim3[it]);

 

// increases the size of array by 10000

n += 10000;

}

 

return 0;

}

8 0
4 years ago
Read 2 more answers
Pls can anyone be so kind and answer this question.....i need the answer urgently
mixas84 [53]

Answer:

so u have to be smaet

Explanation:

8 0
3 years ago
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