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iogann1982 [59]
3 years ago
13

Move down one screen​

Computers and Technology
2 answers:
netineya [11]3 years ago
6 0

Answer:

k(kkkkkkkkkkklkkkkkkkkkk(

Musya8 [376]3 years ago
5 0
Okay , I will


step by step explanation:
You might be interested in
Which of the following is the best way to make sure you understand what someone has said?
aivan3 [116]

Answer:

I would choose either paraphrase the info or take notes.

Explanation:

8 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

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1 year ago
What is the primary reason that routing on the internet is redundant
Elena-2011 [213]

Most of the time rerouting on the internet is redundant because a router already does it automatically. For example if one path has a lot of traffic it with put the user on another path with less traffic but same destination.

<span />
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3 years ago
A computer uses a memory unit with 256k words of 32 bits each. A bina.ry
iogann1982 [59]

Answer:

Touch and hold a clip to pin it. Unpinned clips will be deleted after 1 hour.

4 0
3 years ago
If a schema is not given, you may assume a table structure that makes sense in the context of the question. (using sql queries)
adell [148]

Answer:

The correct query is;

select * from EMPLOYEE where Employee_Name = 'JOE' or Employee_Name = 'Joe' or Employee_Name = 'joe';

where EMPLOYEE refer to the table name and type attribute name is Employee_Name

Explanation:

Here, the first thing we will do is to assume the name of the table.

Let’s assume the table name is EMPLOYEE, where the column i.e attribute from which we will be extracting our information is Employee_Name

The correct query to get the piece of information we are looking for will be;

select * from EMPLOYEE where Employee_Name = 'JOE' or Employee_Name = 'Joe' or Employee_Name = 'joe';

6 0
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