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faust18 [17]
3 years ago
10

Jason is working on a project that requires him to manage a huge amount of data. The spreadsheet he is working on has data relat

ed to the official budget from cell A1 through C9, and data related to actual expenditures from cell L1 through N19. How will Jason track and compare the data as he works on the spreadsheet?
A. use split worksheet view to break the worksheet into different visible sections
B. use page break view to break the worksheet into different visible sections
C. use full screen view to break the worksheet into different visible sections
D. use normal view to break the worksheet into different visible sections
E. use page layout view to break the worksheet into different visible sections
Computers and Technology
1 answer:
Alinara [238K]3 years ago
6 0

A. use split worksheet view to break the worksheet into different visible sections

<u>Explanation:</u>

Jason is working on a project that requires him to manage a huge amount of data. There are numerous instances when a user has to manage a huge amount of data within a single spreadsheet. It will cause trouble for the user to simultaneously access the data from more than one table in the same sheet.

To deal with this problem, Jason can track and compare this data as he works on the spreadsheet by using the split worksheet view to break the worksheet into different visible sections. The different sections provide a better view of the tables and data extraction and manipulation become easy.

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Write a Java class to perform the following: 1. Write a method to search the following array using a linear search, ( target ele
Alina [70]

Answer:

Check the explanation

Explanation:

Linear search in JAVA:-

import java.util.Scanner;

class linearsearch

{

  public static void main(String args[])

  {

     int count, number, item, arr[];

     

     Scanner console = new Scanner(System.in);

     System.out.println("Enter numbers:");

     number = console.nextInt();

   

     arr = new int[number];

     System.out.println("Enter " + number + " ");

     

     for (count = 0; count < number; count++)

       arr[count] = console.nextInt();

     System.out.println("Enter search value:");

     item = console.nextInt();

     for (count = 0; count < number; count++)

     {

        if (arr[count] == item)

        {

          System.out.println(item+" present at "+(count+1));

         

          break;

        }

     }

     if (count == number)

       System.out.println(item + " doesn't found in array.");

  }

}

Kindly check the first attached image below for the code output.

Selection Sort in JAVA:-

public class selectionsort {

   public static void selectionsort(int[] array){

       for (int i = 0; i < array.length - 1; i++)

       {

           int ind = i;

           for (int j = i + 1; j < array.length; j++){

               if (array[j] < array[ind]){

                   ind = j;

               }

           }

           int smaller_number = array[ind];  

           array[ind] = array[i];

           array[i] = smaller_number;

       }

   }

     

   public static void main(String a[]){

       int[] arr = {9,94,4,2,43,18,32,12};

       System.out.println("Before Selection Sort");

       for(int i:arr){

           System.out.print(i+" ");

       }

       System.out.println();

         

       selectionsort(arr);

       

       System.out.println("After Selection Sort");

       for(int i:arr){

           System.out.print(i+" ");

       }

   }

}  

Kindly check the second attached image below for the code output.

Bubble Sort in JAVA:-

public class bubblesort {

   static void bubblesort(int[] array) {

       int num = array.length;

       int temp = 0;

        for(int i=0; i < num; i++){

                for(int j=1; j < (num-i); j++){

                         if(array[j-1] > array[j]){

                           

                                temp = array[j-1];

                                array[j-1] = array[j];

                                array[j] = temp;

                        }

                         

                }

        }

   }

   public static void main(String[] args) {

               int arr1[] ={3333,60,25,32,55,620,85};

               

               System.out.println("Before Bubble Sort");

               for(int i=0; i < arr1.length; i++){

                       System.out.print(arr1[i] + " ");

               }

               System.out.println();

                 

               bubblesort(arr1);

               

               System.out.println("After Bubble Sort");

               for(int i=0; i < arr1.length; i++){

                       System.out.print(arr1[i] + " ");

               }

 

       }

}  

Kindly check the third attached image below for the code output.

Binary search in JAVA:-

public class binarysearch {

  public int binarySearch(int[] array, int x) {

     return binarySearch(array, x, 0, array.length - 1);

  }

  private int binarySearch(int[ ] arr, int x,

        int lw, int hg) {

     if (lw > hg) return -1;

     int middle = (lw + hg)/2;

     if (arr[middle] == x) return middle;

     else if (arr[middle] < x)

        return binarySearch(arr, x, middle+1, hg);

     else

        return binarySearch(arr, x, lw, middle-1);

  }

  public static void main(String[] args) {

     binarysearch obj = new binarysearch();

     int[] ar =

       { 22, 18,12,14,36,59,74,98,41,23,

        34,50,45,49,31,53,74,56,57,80,

        61,68,37,12,58,79,904,56,99};

     for (int i = 0; i < ar.length; i++)

        System.out.print(obj.binarySearch(ar,

           ar[i]) + " ");

     System.out.println();

     System.out.print(obj.binarySearch(ar,19) +" ");

     System.out.print(obj.binarySearch(ar,25)+" ");

     System.out.print(obj.binarySearch(ar,82)+" ");

     System.out.print(obj.binarySearch(ar,19)+" ");

     System.out.println();

  }

}

Kindly check the fourth attached image below for the code output

7 0
3 years ago
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