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Leto [7]
3 years ago
11

When you print ____, the presentation is printed with one or more slides on each piece of paper?

Computers and Technology
1 answer:
Artist 52 [7]3 years ago
8 0
When you print handouts, the presentation is printed with one or more slides on each piece of paper. When you print handouts, it's best to have the presentation printed on them for those who can not view the presentation clearly to follow along with. This also serves as a great tool for people who missed the presentation to receive a handout later or for those there to take home and review. 
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Explanation:

if you allow cookie site for fake websites. It can absolutely be possible.

5 0
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What do the power key and refresh key do when pressed together
9966 [12]

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shuts down the laptop.

Explanation:

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3 0
3 years ago
Which presidential race has been called “the dirtiest in American history”
sdas [7]
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3 0
4 years ago
Implement a method called bubbleSort, that takes an ArrayList, sorts it using bubble sort algorithm, and returns a sorted list;
boyakko [2]

Answer:

Java algorithm is given below

Explanation:

import java.util.ArrayList;

import java.util.Arrays;

import java.util.Random;

public class Temp {

   static void bubbleSort(ArrayList<Integer> list) {

       int n = list.size();

       for (int p = 0; p < n - 1; p++)

           for (int q = 0; q < n - p - 1; q++)

               if (list.get(q) > list.get(q + 1)) {

                   int temp = list.get(q);

                   list.set(q, list.get(q + 1));

                   list.set(q + 1, temp);

               }

   }

   static void selectionSort(ArrayList<Integer> list) {

       int n = list.size();

       for (int p = 0; p < n - 1; p++) {

           int minimumIndex = p;

           for (int q = p + 1; q < n; q++)

               if (list.get(q) < list.get(minimumIndex))

                   minimumIndex = q;

           int temp = list.get(p);

           list.set(p, list.get(minimumIndex));

           list.set(minimumIndex, temp);

       }

   }

   static void insertionSort(ArrayList<Integer> list) {

       int size = list.size();

       int p, val, q;

       for (p = 1; p < size; p++) {

           val = list.get(p);

           q = p - 1;

           while (q >= 0 && list.get(q) > val) {

               list.set(q + 1, list.get(q));

               q = q - 1;

           }

           list.set(q + 1, val);

       }

   }

   static void mergeSort(ArrayList<Integer> list, int low, int high) {

       if (low < high && (high - low) >= 1) {

           int mid = (high + low) / 2;

           mergeSort(list, low, mid);

           mergeSort(list, mid + 1, high);

           merger(list, low, mid, high);

       }

   }

   static void merger(ArrayList<Integer> list, int low, int mid, int high) {

       ArrayList<Integer> mergedArray = new ArrayList<Integer>();

       int left = low;

       int right = mid + 1;

       while (left <= mid && right <= high) {

           if (list.get(left) <= list.get(right)) {

               mergedArray.add(list.get(left));

               left++;

           } else {

               mergedArray.add(list.get(right));

               right++;

           }

       }

       while (left <= mid) {

           mergedArray.add(list.get(left));

           left++;

       }

       while (right <= high) {

           mergedArray.add(list.get(right));

           right++;

       }

       int i = 0;

       int j = low;

       while (i < mergedArray.size()) {

           list.set(j, mergedArray.get(i++));

           j++;

       }

   }

   public static void main(String[] args) throws Exception {

       ArrayList<Integer> list = new ArrayList<>();

       Random rand = new Random(System.currentTimeMillis());

       for (int i = 0; i < 10000; i++) {

           list.add(rand.nextInt(256));

       }

       long start = System.currentTimeMillis();

       selectionSort(list);

       int sel = (int) (System.currentTimeMillis() - start);

       System.out.println("Selection Sort time (in ms): " + sel);

       list.clear();

       // ------------------------

       rand = new Random(System.currentTimeMillis());

       for (int i = 0; i < 10000; i++) {

           list.add(rand.nextInt(256));

       }

       start = System.currentTimeMillis();

       insertionSort(list);

       int ins = (int) (System.currentTimeMillis() - start);

       System.out.println("Insertion Sort time (in ms): " + ins);

       list.clear();

       // ------------------------

       rand = new Random(System.currentTimeMillis());

       for (int i = 0; i < 10000; i++) {

           list.add(rand.nextInt(256));

       }

       start = System.currentTimeMillis();

       bubbleSort(list);

       int bub = (int) (System.currentTimeMillis() - start);

       System.out.println("Bubble Sort time (in ms): " + bub);

       list.clear();

       // ---------------------------

       rand = new Random(System.currentTimeMillis());

       for (int i = 0; i < 10000; i++) {

           list.add(rand.nextInt(256));

       }

       start = System.currentTimeMillis();

       mergeSort(list, 0, list.size() - 1);

       int mer = (int) (System.currentTimeMillis() - start);

       System.out.println("Merge Sort time (in ms): " + mer);

       long m = Math.min(Math.min(sel, ins), Math.min(bub, mer));

       if (m == sel)

           System.out.println("Selection Sort is fastest");

       else if (m == ins)

           System.out.println("insertion Sort is fastest");

       else if (mer == m)

           System.out.println("Merge Sort is fastest");

       else if (m == bub)

           System.out.println("Bubble Sort is fastest");

   }

}

6 0
3 years ago
Consider a sequence of method invocations as follows: main calls m1, m1 calls m2, m2 calls m3 and then m2 calls m4, m3 calls m5.
Minchanka [31]

Answer:

M2 is the answer for the above question.

Explanation:

  • The above question states the scenario, where firstly main function calls the M1 function.
  • Then M1 function calls the function M2.
  • Then M2 function calls the function M3.
  • Then M3 function calls the function M5.
  • Then Again M2 function will resume and calls the function M4.
  • Then Again M2 function will resume because when the M4 will ends it return to the statement of the M2 function from which the M4 function has been called.
  • It is because when any function ends its execution it returns to that line of a statement from which it has been called. This is the property of programming language to call the function.
  • So when the M4 will terminate its execution then the M2 function will resume again.
4 0
4 years ago
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