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kykrilka [37]
3 years ago
5

As an interviewer, it is not necessary to be at eye level with interviewee True False

Computers and Technology
2 answers:
ANTONII [103]3 years ago
7 0

Answer:

True

Hope this helps

Tamiku [17]3 years ago
3 0

Answer:

True

Explanation:

Have a great day

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3. What do you need to do before you can sort and filter data in a database?
Digiron [165]

3. format the table

4. idk

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3 years ago
Practice Home Survey
Ksivusya [100]

Answer:

Since this is a survey without preparation, you can move ahead as you like. However, still, you need to figure out the attributes. Here you need to register plugged in hours for each of your family members. And the same thing is abbreviated as screen time as well. You need to make a table with two columns, a family member's name, and the screen time or plugged in time. And you are not required to do anything else, as the survey is the method of collecting the data. And it does not deal with the analysis. The analysis will be done, when you will sit as an analyst. And in fact, you are making this impromptu survey for the family head, who is in charge of your family which can be your father or mother or your grand father or grand mother. And they can use these details for various findings, like how can they reduce cable TV cost per month, and so on. Your job is to collect the accurate data. And same thing, you need to repeat for all the survey that you are asked to conduct. Always remember, you need to know what data you need to collect through a survey. Plan for the survey only after finding the complete details about this. And even when asked to conduct as unprepared, this knowledge is essential.  

Explanation:

Please check the attachment.

5 0
4 years ago
Select the correct answer.
IrinaK [193]
The answer is the letter (D)
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
Which shortcut key aligns to the center of a page
riadik2000 [5.3K]

Answer:

To make text centered, select and highlight the text first, then hold down Ctrl (the control key) on the keyboard and press E. To make text right aligned, select and highlight the text first, then hold down Ctrl (the control key) on the keyboard and then press R.

Explanation:

4 0
3 years ago
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