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Dima020 [189]
3 years ago
7

PLEASE HELP ME HELP ME NO IM NOT OKAY AND ITS NOT EASY FOR ME.....Vladimir is reworking an old slideshow. He needs to delete som

e of the text, combine or summarize other parts, and create a few new slides as well. He wants to only look at the text. Is there any easy way to do that?
A.
Yes, he can use the Outline view to see all the text in one place.
B.
Yes, he can use the Slide Master to view all the text in one place.
C.
No, he'll have to use the Slide pane to navigate the slides one by one.
D.
No, he'll have to copy all the text out into a word processor.
Computers and Technology
1 answer:
Mandarinka [93]3 years ago
7 0

Answer:

Outline view

Explanation:

This is a formatted window that shows all text, headers, and titles in a popup on the left side of the screen. Slide master shows all the slides but includes slide details such as inserted images and smart art.

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The birthday paradox says that the probability that two people in a room will have the same birthday is more than half, provided
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Answer:

The Java code is given below with appropriate comments for explanation

Explanation:

// java code to contradict birth day paradox

import java.util.Random;

public class BirthDayParadox

{

public static void main(String[] args)

{

   Random randNum = new Random();

   int people = 5;

   int[] birth_Day = new int[365+1];

   // setting up birthsdays

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

       birth_Day[i] = i + 1;

 

   int iteration;

   // varying number n

   while (people <= 100)

   {

       System.out.println("Number of people: " + people);

       // creating new birth day array

       int[] newbirth_Day = new int[people];

       int count = 0;

       iteration = 100000;

       while(iteration != 0)

       {

           count = 0;

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

           {

               // generating random birth day

               int day = randNum.nextInt(365);

               newbirth_Day[i] = birth_Day[day];

           }

           // check for same birthdays

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

           {

               int bday = newbirth_Day[i];

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

               {

                   if (bday == newbirth_Day[j])

                   {

                       count++;

                       break;

                   }

               }

           }

           iteration = iteration - 1;

       }

       System.out.println("Probability: " + count + "/" + 100000);

       System.out.println();

       people += 5;

   }

}

}

4 0
3 years ago
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