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il63 [147K]
3 years ago
12

A ____ object is used to hold data that is retrieved from a database via the OleDbDataAdapter connection. a. DataRecord b. DataS

ource c. DataFile d. DataTable
Computers and Technology
2 answers:
Mumz [18]3 years ago
4 0

Answer:

the answer is

data table

Explanation:

i think so

mariarad [96]3 years ago
3 0

Answer:

datatable

Explanation:

You might be interested in
Anyone know how to translate this 1100111111110100000110 pls n ty!4!:$;
marysya [2.9K]

Answer:

1100111111110100000110 = �

Binary -> UTF-16

3,407,110 in decimal form

Additionally, it also translates to the color green in hexadecimal.

Not really sure what you are trying to translate this in to though.

7 0
3 years ago
Read 2 more answers
and assuming main memory is initially unloaded, show the page faulting behavior using the following page replacement policies. h
Svet_ta [14]

FIFO

// C++ implementation of FIFO page replacement

// in Operating Systems.

#include<bits/stdc++.h>

using namespace std;

// Function to find page faults using FIFO

int pageFaults(int pages[], int n, int capacity)

{

   // To represent set of current pages. We use

   // an unordered_set so that we quickly check

   // if a page is present in set or not

   unordered_set<int> s;

   // To store the pages in FIFO manner

   queue<int> indexes;

   // Start from initial page

   int page_faults = 0;

   for (int i=0; i<n; i++)

   {

       // Check if the set can hold more pages

       if (s.size() < capacity)

       {

           // Insert it into set if not present

           // already which represents page fault

           if (s.find(pages[i])==s.end())

           {

               // Insert the current page into the set

               s.insert(pages[i]);

               // increment page fault

               page_faults++;

               // Push the current page into the queue

               indexes.push(pages[i]);

           }

       }

       // If the set is full then need to perform FIFO

       // i.e. remove the first page of the queue from

       // set and queue both and insert the current page

       else

       {

           // Check if current page is not already

           // present in the set

           if (s.find(pages[i]) == s.end())

           {

               // Store the first page in the

               // queue to be used to find and

               // erase the page from the set

               int val = indexes.front();

               

               // Pop the first page from the queue

               indexes.pop();

               // Remove the indexes page from the set

               s.erase(val);

               // insert the current page in the set

               s.insert(pages[i]);

               // push the current page into

               // the queue

               indexes.push(pages[i]);

               // Increment page faults

               page_faults++;

           }

       }

   }

   return page_faults;

}

// Driver code

int main()

{

   int pages[] = {7, 0, 1, 2, 0, 3, 0, 4,

               2, 3, 0, 3, 2};

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

   int capacity = 4;

   cout << pageFaults(pages, n, capacity);

   return 0;

}

LRU

//C++ implementation of above algorithm

#include<bits/stdc++.h>

using namespace std;

// Function to find page faults using indexes

int pageFaults(int pages[], int n, int capacity)

{

   // To represent set of current pages. We use

   // an unordered_set so that we quickly check

   // if a page is present in set or not

   unordered_set<int> s;

   // To store least recently used indexes

   // of pages.

   unordered_map<int, int> indexes;

   // Start from initial page

   int page_faults = 0;

   for (int i=0; i<n; i++)

   {

       // Check if the set can hold more pages

       if (s.size() < capacity)

       {

           // Insert it into set if not present

           // already which represents page fault

           if (s.find(pages[i])==s.end())

           {

               s.insert(pages[i]);

               // increment page fault

               page_faults++;

           }

           // Store the recently used index of

           // each page

           indexes[pages[i]] = i;

       }

       // If the set is full then need to perform lru

       // i.e. remove the least recently used page

       // and insert the current page

       else

       {

           // Check if current page is not already

           // present in the set

           if (s.find(pages[i]) == s.end())

           {

               // Find the least recently used pages

               // that is present in the set

               int lru = INT_MAX, val;

               for (auto it=s.begin(); it!=s.end(); it++)

               {

                   if (indexes[*it] < lru)

                   {

                       lru = indexes[*it];

                       val = *it;

                   }

               }

               // Remove the indexes page

               s.erase(val);

               // insert the current page

               s.insert(pages[i]);

               // Increment page faults

               page_faults++;

           }

           // Update the current page index

           indexes[pages[i]] = i;

       }

   }

   return page_faults;

}

// Driver code

int main()

{

   int pages[] = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2};

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

   int capacity = 4;

   cout << pageFaults(pages, n, capacity);

   return 0;

}

You can learn more about this at:

brainly.com/question/13013958#SPJ4

4 0
1 year ago
What are the 5 font/typography families
larisa [96]

Answer:

Arial

Helvetica

Verdana

Calibri

Noto

Explanation:

8 0
3 years ago
What is the capacity of a disk with two platters, 10,000 cylinders, an average of 400 sectors per track, and 512 bytes per secto
natita [175]

Answer:

The capacity of the disk is "40 GB".

Explanation:

Given value:

\to \text{cylinder}= 10,000\\\\\to \frac{sector}{track} = 400\\\\\to \frac{bytes}{sector} = 512\\

\text{Calculating the size of track} =  \frac{bytes}{sector} \times  \frac{sector}{track}

                                            = 512 \times  400 \\\\ = 204,800 \ \ Or  \ \ 200 \ K

\text{Calculating the size of surface} = \frac{byte}{track} \times \ cylinder

                                               = 2000 \ K \times  10,000 \\\\ = 20, 000,000 \ K

\text{Calculating the capacity of a disk} = \frac{byte}{surface} \times \frac{surface}{disk}

                                                     = 20, 000,000 \ K \times  2 \\\\ = 40, 000,000 \ K \\\\= 40 \ GB

Cache memory is often used to speed up the runtime. It will enable us to improve performance unless we can store information, that is retrieved constantly in memory space.

  • Hit Cache: when the Processor relates to representing database data, Cache Hit results.
  • Miss Cache: when this Processor responds to a non-present cache data, this will trigger its miss of cache.
  • It doesn't have to retrieve the Main Memory through secondary storage.
3 0
3 years ago
What is a program that, when installed on a computer, records every keystroke and mouse click?
Mrac [35]
"Key logger" This could be software or hardware that does this.
6 0
3 years ago
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