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Misha Larkins [42]
3 years ago
14

The ability of services to grow or shrink based on need is called __________.

Computers and Technology
1 answer:
Talja [164]3 years ago
3 0
This kind of ability is called rapid elasticity. The resources can be maximized by a consumer to provide wider webs and applications.It can also make the server stop, just to reduce the cost in the process.  The flexibility in adapting to needs of a company is an advantage. Rapid elasticity can be done in a public cloud.
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Is(are) input value(s) to use with a program to test the accuracy of the output.
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test cases

Explanation:

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This image shows a web designer's grids for different pages on a website. The uppermost box on each page is the website's identi
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A keyboard shortcut is a key or combination of keys that you press to access a feature or perform a command. ___________________
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8 0
3 years ago
The birthday paradox says that the probability that two people in a room will have the same birthday is more than half, provided
poizon [28]

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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