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Rudiy27
4 years ago
6

A(n) _____ occurs when a visitor requests a page from a web site.'

Computers and Technology
1 answer:
lawyer [7]4 years ago
8 0
If everything goes right, A file transfer occurs.

The question is so broad, I don't know what you're looking for. How does httpd work? How does HTTP work? I don't know.
You might be interested in
A vast global network that is made up of many smaller interconnected networks is known as:
Galina-37 [17]

The answer is The Internet.   It is a vast global network that is made up of many smaller interconnected networks. It connects to millions of computer units world wide, in which any computer can communicate with any other computer as long as they are both connected to the Internet. It also made access to information and communication easier.

6 0
4 years ago
Read 2 more answers
A start-up employs interns. The following details of interns are stored
Rufina [12.5K]

Answer: See below.

class Intern:

   def __init__(self):

       self.first_name = input("Enter first name: ")

       self.last_name = input("Enter last name: ")

       self.address = input("Enter address: ")

       self.mobile_number = input("Enter mobile number: ")

       self.e_mail = input("Enter e-mail: ")

   def getdata(self):

       print("First name: ", self.first_name)

       print("Last name: ", self.last_name)

       print("Address: ", self.address)

       print("Mobile number: ", self.mobile_number)

       print("E-mail: ", self.e_mail)

   def putdata(self):

       print("First name: ", self.first_name)

       print("Last name: ", self.last_name)

       print("Address: ", self.address)

       print("Mobile number: ", self.mobile_number)

       print("E-mail: ", self.e_mail)

Explanation:

7 0
2 years ago
Return true if the given non-negative number is a multiple of 3 or 5, but not both. Use the % "mod" operator.
Pani-rosa [81]

the following C++ function will return true if the given non - negative number is multiple of 5 or 3 else it will return false.

bool function( int x ){

// variable to check if it is multiple of both or not

int number =0;

if(3%x == 0){

number++;

}

if(5%x == 0){

number++;

}

// Now returning by deciding

if( number <=1)

return true;

else

return false

}

4 0
4 years ago
Scrabble is a word game in which words are constructed from letter tiles, each letter tile containing a point value. The value o
Fantom [35]

Complete question:

Scrabble is a word game in which words are constructed from letter tiles, each letter tile containing a point value. The value of a word is the sum of each tile's points added to any points provided by the word's placement on the game board. Write a program using the given dictionary of letters and point values that takes a word as input and outputs the base total value of the word (before being put onto a board). Ex:  If the input is:  PYTHON

the output is: 14

part of the code:

tile_dict = { 'A': 1, 'B': 3, 'C': 3, 'D': 2, 'E': 1, 'F': 4, 'G': 2, 'H': 4, 'I': 1, 'J': 8,  'K': 5, 'L': 1, 'M': 3, 'N': 1, 'O': 1, 'P': 3, 'Q': 10, 'R': 1, 'S': 1, 'T': 1,  'U': 1, 'V': 4, 'W': 4, 'X': 8, 'Y': 4, 'Z': 10 }

Answer:

Complete the program as thus:

word = input("Word: ").upper()

points = 0

for i in range(len(word)):

   for key, value in tile_dict.items():

       if key == word[i]:

           points+=value

           break

print("Points: "+str(points))

Explanation:

This gets input from the user in capital letters

word = input("Word: ").upper()

This initializes the number of points to 0

points = 0

This iterates through the letters of the input word

for i in range(len(word)):

For every letter, this iterates through the dictionary

   for key, value in tile_dict.items():

This locates each letters

       if key == word[i]:

This adds the point

           points+=value

The inner loop is exited

           break

This prints the total points

print("Points: "+str(points))

6 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
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