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MrRissso [65]
3 years ago
6

Qual parâmetro é usado é usado na query sql para a presentar os dados em ordem decrescente?Leitura Avançada

Computers and Technology
1 answer:
nikitadnepr [17]3 years ago
4 0

Answer:

gg

Explanation:

gg

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Using filtering as a strategy to deal with information overload requires Group of answer choices reviewing all unsolicited infor
STALIN [3.7K]

Answer:

The answer is "Choice B"

Explanation:

Please find the complete question in the attached file.

A filter is a tool for removing undesirable parts. Eliminate solid from a fluid, for example. The filter may mean the filter action: it could be used as a verb. When the filter is mentioned, various branches of science and technology often refer to a certain type of device. Filtering tries to decide the data you need and the data you have to address the overload.

4 0
3 years ago
What are rules that enforce basic and fundamental information-based constraints?
irga5000 [103]
The
answer should be C.
8 0
3 years ago
Write a c++ program to print even numbers from 1 to 20​
Novay_Z [31]

Answer:

#include <bits/stdc++.h>

using namespace std;

// Function to print even numbers

void printEvenNumbers(int N)

{

cout << "Even: ";

for (int i = 1; i <= 2 * N; i++) {

// Numbers that are divisible by 2

if (i % 2 == 0)

cout << i << " ";

}

}

// Function to print odd numbers

void printOddNumbers(int N)

{

cout << "\nOdd: ";

for (int i = 1; i <= 2 * N; i++) {

// Numbers that are not divisible by 2

if (i % 2 != 0)

cout << i << " ";

}

}

// Driver code

int main()

{

int N = 20;

printEvenNumbers(N);

printOddNumbers(N);

return 0;

}

Explanation:

Note: This will find both odd and even numbers, you have to change the number above to the number of your choice

For even numbers

Even number are numbers that are divisible by 2.

To print even numbers from 1 to N, traverse each number from 1.

Check if these numbers are divisible by 2.

If true, print that number.

For odd numbers

Odd number are numbers that are not divisible by 2.

To print Odd numbers from 1 to N, traverse each number from 1.

Check if these numbers are not divisible by 2.

If true, print that number

8 0
3 years ago
CLS
Ainat [17]

Answer:

INPUT "Enter your marks in computer"<u><em>;</em></u>CS

<u><em>IF</em></u> C>40 THEN

PRINT "You are passed. <u><em>"</em></u>

ELSE

PRINT "You are failed."

END <u><em>IF</em></u>

Explanation:

see corrections above.

5 0
4 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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