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AnnyKZ [126]
1 year ago
6

Julie is working on a spreadsheet with data about the company's profits. They will be sharing this data with customers. She chec

ks her e-mail and reads a memo that certain data should be kept confidential. She looks over her spreadsheet and sees that she needs to eliminate row 2 and row 3. Julie should _____.
highlight row 2 and row 3 and use the delete command
select B2 and press Delete
highlight row 2 and use the clear command
highlight row 2 and row 3 and make the background black
Computers and Technology
2 answers:
Alika [10]1 year ago
7 0
Julie should highlight row 2 and row 3 and use the delete command.
grigory [225]1 year ago
5 0

Answer: Highlight row 2 and 3 and use the delete command.

Anything that is confidential should not be included in a document that will be released to the customers of a company.

Select B2 and press delete, would just remove the contents of the cell B2.

Highlight row 2 and use the clear command will only remove the contents of the selected rows and leave a blank space.

Highlight row 2 and row 3 and make the background black. This will be one of the worst things you can do, as the customers may have accessibility to edit the contents and simply remove the formatting to see the data hidden in the rows.

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2 years ago
If s=abcd is a string defined over Σ = {a,bc,d}then reverse of s is dcba.<br> Δ True<br> Δ False
Minchanka [31]

Answer:

True

Explanation:

If s=abcd is a string defined over {a,b,c,d}, it corresponds to a regular expression which can be represented using a finite automata. Then the reverse of the string essentially corresponds to another finite automata where the starting state becomes the accepting state and vice versa.Moreover all the directions of state transitions will be reversed for each of the transitions in the original automata.

With these modifications, the new finite automata will accept a string which is reverse of the original string ,namely, dcba and this string will ne part of the reverse language.

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2 years ago
You have several pictures of different sizes that you would like to frame.A local picture framing store offers two types of fram
exis [7]

Answer:

written in java

Note :  Package name was excluded

 

import java.util.Scanner;

public class Main {

   public static void main(String[] args) {

       //declared 8 uninitialised variable  

       String input;

       double cost, colorC, frameC, paperC, glassC, crownC;

       char crowns;

       int type, color, crown, length, width, area, perimeter;

       

       //creating an instance of the Scanner class to accept value from user

       Scanner scanner = new Scanner(System.in);

       System.out.print("Enter length of frame: ");

       length = scanner.nextInt();

       System.out.print("Enter width of frame: ");

       width = scanner.nextInt();

       System.out.print("Do you want a fancy frame or a regular frame\nEnter 1 for fancy, 2 for regular: ");

       type = scanner.nextInt();

       if (type == 1) {

           frameC = 0.25;

       } else {

           frameC = 0.15;

       }

       System.out.print("Do you want a white frame or a colored frame\nEnter 1 for white, 2 for colored: ");

       color = scanner.nextInt();

       if (color == 2) {

           colorC = .10;

       } else{

           colorC = 0;

       }

       System.out.print("Do you want crowns \nEnter y for yes, n for no ");

       input = scanner.next();

       crowns = input.charAt(0);

       if (crowns == 'y' || crowns == 'Y') {

           System.out.print("How many crowns?");

           crown = scanner.nextInt();

       } else

           crown = 0;

       if (crown > 0)

           crownC = crown * .35;

       else

           crownC = 0;

       area = length * width;

       perimeter = length * 2 + width * 2;

       paperC = area * .02;

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       cost = glassC + paperC + crownC + colorC * perimeter + frameC * perimeter;

       System.out.println("The cost of your frame is: $" + cost);

   }

}

5 0
2 years ago
We have said that the average number of comparisons need to find a target value in an n-element list using sequential search is
bija089 [108]

Answer:

Part a: If the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b: If the list contains n elements (where n is even) the middle terms are  at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c: The average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

Explanation:

Suppose the list is such that the starting index is 0.

Part a

If list has 15 elements, the middle item would be given at 7th index i.e.

there are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(8,9,10,11,12,13,14) above it. It will have to run 8 comparisons  to find the middle term.

If list has 17 elements, the middle item would be given at 8th index i.e.

there are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(9,10,11,12,13,14,15,16) above it.It will have to run 9 comparisons  to find the middle term.

If list has 21 elements, the middle item would be given at 10th index i.e.

there are 10 indices (0,1,2,3,4,5,6,7,8,9) below it and 10 indices (11,12,13,14,15,16,17,18,19,20) above it.It will have to run 11 comparisons  to find the middle term.

Now this indicates that if the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b

If list has 16 elements, there are two middle terms as  one at would be at 7th index and the one at 8th index .There are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(9,10,11,12,13,14,15) above it. It will have to run 9 comparisons  to find the middle terms.

If list has 18 elements, there are two middle terms as  one at would be at 8th index and the one at 9th index .There are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(10,11,12,13,14,15,16,17) above it. It will have to run 10 comparisons  to find the middle terms.

If list has 20 elements, there are two middle terms as  one at would be at 9th index and the one at 10th index .There are 9 indices(0,1,2,3,4,5,6,7,8) below it and 9 indices(11,12,13,14,15,16,17,18,19) above it. It will have to run 11 comparisons  to find the middle terms.

Now this indicates that if the list contains n elements (where n is even) the middle terms are  at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c

So the average number of comparisons is given as

((n+1)/2+(n+2)/2)/2=(2n+3)/4

So the average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

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