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Tom [10]
2 years ago
8

Frequently used _____________ can be saved as _____________ for use in analysis, dashboards, reports, tickets, and alerts.

Computers and Technology
1 answer:
Ilia_Sergeevich [38]2 years ago
4 0

Frequently used filters can be saved as queries for use in analysis, dashboards, reports, tickets, and alerts.

<h3>What are the types of filters in a computer?</h3>

The low-pass filter, high-pass filter, band-pass filter, and notch filter are the four main categories of filters (or the band-reject or band-stop filter).

The following is a list of the three filtration kinds.

Vacuum Filtration: In order to quickly draw fluid through a filter, vacuum filtration uses a vacuum pump.

Centrifugal Filtration: In this technique, the material to be filtered is spun at a very high speed.

You can use the filter tool to identify the crucial elements you require by filtering a column of data within a table. You have the option to sort by date, number, alphabetical order, and more using the sorting tool. The use of sorting and filtering is explored in the example that follows, along with some sophisticated sorting methods.

It is possible to save frequently used filters as filters, queries for analysis, dashboards, reports, tickets, and alerts.

To learn more about filter tool refer to:

brainly.com/question/10169776

#SPJ4

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8 0
4 years ago
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Refer to the method f:
Goshia [24]
<h2>This function will land up in infinite function call</h2>

Explanation:

first time when the function gets invoked,

f(6,8), so k=6 & n=8, inside the function it checks k==n, ie. 6==8, returns false, then one more if is available, so 6>8 is check for , once again it is false and else loop is executed, the function is called recursively using f(k-n,n), that is f(6-8,8), it means f(-2,8) is passed.

Second time,

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So this goes recursively and ends in an infinite function call.

5 0
4 years ago
When you use the Bing Image Search for online pictures, you will be searching the Internet for pictures that have been filtered
just olya [345]

Answer:

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Explanation:

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7 0
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When you are using the internet to research a school project, and you ignore the banner ads on websites even though you see them
FinnZ [79.3K]

Answer:

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Data Structure in C++
agasfer [191]

The code .cpp is available bellow

#include<iostream>

using namespace std;

//declaring variables

void merge(int* ip, int sz, int* opt, bool opt_asc); //merging

int* mergesort(int* ip, int sz);

void mergesort(int *ip, int sz, int* opt, bool opt_asc);

void merge(int* ip, int sz, int* opt, bool opt_asc)

{

  int s1 = 0;

  int mid_sz = sz / 2;

  int s2 = mid_sz;

  int e2 = sz;

  int s3 = 0;

  int end3 = sz;

  int i, j;

   

  if (opt_asc==true)

  {

      i = s1;

      j = e2 - 1;

      while (i < mid_sz && j >= s2)

      {

          if (*(ip + i) > *(ip + j))

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j--;

          }

          else if (*(ip + i) <= *(ip + j))

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i++;

          }

      }

      if (i != mid_sz)

      {

          while (i < mid_sz)

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i++;

          }

      }

      if (j >= s2)

      {

          while (j >= s2)

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j--;

          }

      }

  }

  else

  {

      i = mid_sz - 1;

      j = s2;

      while (i >= s1 && j <e2)

      {

          if (*(ip + i) > *(ip + j))

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i--;

          }

          else if (*(ip + i) <= *(ip + j))

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j++;

          }

      }

      if (i >= s1)

      {

          while (i >= s1)

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i--;

          }

      }

      if (j != e2)

      {

          while (j < e2)

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j++;

          }

      }

  }

   

  for (i = 0; i < sz; i++)

      *(ip + i) = *(opt + i);

}

int* mergesort(int* ip, int sz)

{

  int* opt = new int[sz];

   

  mergesort(ip, sz, opt, true);

  return opt;

}

void mergesort(int *ip, int sz, int* opt, bool opt_asc)

{

  if (sz > 1)

  {

      int q = sz / 2;

      mergesort(ip, sz / 2, opt, true);

      mergesort(ip + sz / 2, sz - sz / 2, opt + sz / 2, false);

      merge(ip, sz, opt, opt_asc);

  }

}

int main()

{

  int arr1[12] = { 5, 6, 9, 8,25,36, 3, 2, 5, 16, 87, 12 };

  int arr2[14] = { 2, 3, 4, 5, 1, 20,15,30, 2, 3, 4, 6, 9,12 };

  int arr3[10] = { 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 };

  int *opt;

  cout << "Arays after sorting:\n";

  cout << "Array 1 : ";

  opt = mergesort(arr1, 12);

  for (int i = 0; i < 12; i++)

      cout << opt[i] << " ";

  cout << endl;

  cout << "Array 2 : ";

  opt = mergesort(arr2, 14);

  for (int i = 0; i < 14; i++)

      cout << opt[i] << " ";

  cout << endl;

  cout << "Array 3 : ";

  opt = mergesort(arr3, 10);

  for (int i = 0; i < 10; i++)

      cout << opt[i] << " ";

  cout << endl;

  return 0;

}

4 0
4 years ago
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