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Komok [63]
4 years ago
6

Java Programming home > 1.14: zylab training: Interleaved input/output Н zyBooks catalog Try submitting it for grading (click

"Submit mode", then "Submit for grading"). Notice that the test cases fail. The first test case's highlighting indicates that output 3 and newline were not expected. In the second test case, the-5 and newline were not expected 2 Remove the code that echoes the user's input back to the output, and submit again. Now the test cases should all pass 0/2 ACTIVITY 1.14.1: zylab training: Interleaved input/output DoubleNum.java Load default template 1 import java.util.Scanner; 3 public class DoubleNum public static void main(String args) { Scorner sehr = new Scanner(System.in); int x; 7 System.out.println("Enter x: "); 9 - scnr .nextInto: 10 11 System.out.println(); // Student mistakenly is echoing the input to output to match example 12 System.out.println("x doubled is".02 Develop mode Submit mode Run your program as often as you'd like, before submitting for grading. Below, type any needed Input values in the first box, then click Run program and observe the programs cutout in the second be Enter program input (optional) of your code requires input values, provide them here Ad Cat
Computers and Technology
1 answer:
Andru [333]4 years ago
5 0

Answer:

The following are the code to this question:

code:

System.out.println(x);  //use print method to print value.

Explanation:

In the given question, it simplifies or deletes code in the 11th line. This line has a large major statement.  

In the case, the tests fail because only 1 line of output is required by the tester, but two lines are obtained instead.

It's to demonstrate that the input/output or testing requirements function throughout the model.

You might be interested in
Write the definition of a function named printpoweroftwostars that receives a non-negative integer n and prints a line consistin
Aleksandr-060686 [28]
To accomplish this without using a loop,
we can use math on a string.

Example:
print("apple" * 8)

Output:
appleappleappleappleappleappleappleapple

In this example,
the multiplication by 8 actually creates 8 copies of the string.

So that's the type of logic we want to apply to our problem.

<span>def powersOfTwo(number):
if number >= 0:
return print("*" * 2**number)
else:
<span>return

Hmm I can't make indentations in this box,
so it's doesn't format correctly.
Hopefully you get the idea though.

We're taking the string containing an asterisk and copying it 2^(number) times.

Beyond that you will need to call the function below.
Test it with some different values.

powersOfTwo(4) should print 2^4 asterisks: ****************</span></span>
4 0
4 years ago
What is adobe photoshop?
Hitman42 [59]
Its a photoshop that u need to pay for
4 0
3 years ago
Read 2 more answers
I just need the flowchart and pseudocode.No need program.
nevsk [136]

The pseudocode algorithm for the given program is:

  1. PROCESS saving account transactions
  2. REQUEST for previous account balance
  3. FOR every deposit made, (+)
  4. FOR every withdrawal made (-)

<h3>What is a Pseudocode?</h3>

This refers to the use of plain language to describe the sequence of steps for solving a problem in human language.

Hence, we can see that the complete step is given below:

5. DISPLAY "Withdrawal" when (-) is used.

6. DISPLAY "Deposit" when (+) is used

7. ELSE

8. PRINT "Error"

#SPJ1

Read more about flowcharts and pseudocodes here:

brainly.com/question/24735155

3 0
2 years ago
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
Why are mobile apps often easier to develop than desktop apps?
Karo-lina-s [1.5K]

Answer:

Mobile app creation platforms are available that reduce the need to code.

Explanation:

Simple app creation softwares and tools are all over the place for mobile app creation. To create a mobile app, no coding knowledge or experience is needed unlike desktops apps that requires the knowledge of some coding languages like python, JavaScript etc.

The availability of these softwares and platforms have made mobile apps creation easy.

6 0
3 years ago
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