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Delvig [45]
4 years ago
9

Clunker Motors Inc. is recalling all vehicles in its Extravagant line from model years 1999-2002 as well all vehicles in its Guz

zler line from model years 2004-2007. A boolean variable named recalled has been declared. Given a variable modelYear and a String modelName write a statement that assigns true to recalled if the values of modelYear and modelName match the recall details and assigns false otherwise. I am doing this in JAVA Programming
Computers and Technology
1 answer:
Aleksandr [31]4 years ago
5 0

Answer:

The following code is written in java programming language:

//set if statement

if (((modelYear >= 1999) && (modelYear <= 2002) && (modelName == "Extravagant")) || ((modelYear >= 2004) && (modelYear <= 2007) && (modelName == "Guzzler")))

{

recalled = true;    //initialized Boolean value

}

else      //set else statement

{

recalled = false; ////initialized Boolean value

}

Explanation:

Here, we set the if statement and set condition, if the value of modelYear is greater than equal to 1999 and less that equal to 2002 and modelName is equal to "Extravagant" or the value of modelYear is greater than equal to 2004 and less than equal to 2007 and the model year is equal to "Guzzler", than "recalled" initialized to "true".

Otherwise "recalled" initialized to "true".

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6 0
4 years ago
explain the following joke: “There are 10 types of people in the world: those who understand binary and those who don’t.”
sasho [114]

It means that there are people  who understand binary and those that do not understand it. That is, binary is said to be the way that computers are known to express numbers.

<h3>What is the joke about?</h3>

This is known to be a popular  joke that is often  used by people who are known to be great savvy in the field of mathematics.

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The binary system is one that relies on numbers that are known to be  in groups of 2.

Therefore, If the speaker of the above phrase is one who is able to understand binary, that person  would  be able to say that that the phrase  is correctly written as  "there are 2 types of people that understand binary".

Learn more about binary from

brainly.com/question/21475482

#SPJ1

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2 years ago
To open the find and replace dialog box with shortcut keys, hold down ctrl and press ________
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Hope this helps.
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Before you leave for a trip, you should check ___________.
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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
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