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lutik1710 [3]
3 years ago
7

Suppose you draw two cards from a standard deck of 52 playing cards. What is the probability that they are both sixes? Keep at l

east 3 significant digits in your answer. (For example, 0.00012345 could be entered as 0.000123.)
Computers and Technology
1 answer:
GaryK [48]3 years ago
7 0

Answer:

0.00452

Explanation:

There are 4 sixes in a deck. So the chance that the first card you draw is a six, is 4/52. Then there are only 3 sixes left and 51 cards. So the chance that the second one is also a six is 3/51.

The combined chance is the multiplication, i.e., 4/52 * 3/51 = 0.00452

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(Palindrome integer) Write the methods with the following headers // Return the reversal of an integer, i.e., reverse(456) retur
Debora [2.8K]

Answer:

import java.util.Scanner;

public class PalindromeInteger {

public static void main(String[] args) {

 // Create an object of the Scanner class to allow for user's inputs

 Scanner input = new Scanner(System.in);

 // Create a prompt to display the aim of the program.

 System.out.println("Program to check whether or not a number is a palindrome");

 // Prompt the user to enter an integer number.

 System.out.println("Please enter an integer : ");

 // Receive user's input and store in an int variable, number.

 int number = input.nextInt();

 // Then, call the isPalindrome() method to check if or not the

 // number is a  palindrome integer.

 // Display the necessary output.

 if (isPalindrome(number)) {

  System.out.println(number + " is a palindrome integer");

 }

 else {

  System.out.println(number + " is a not a palindrome integer");

 }

}

// Method to return the reversal of an integer.

// It receives the integer as a parameter.

public static int reverse(int num) {

 // First convert the number into a string as follows:

 // Concatenate it with an empty quote.

 // Store the result in a String variable, str.

 String str = "" + num;

 // Create and initialize a String variable to hold the reversal of the  

 // string str. i.e rev_str

 String rev_str = "";

 // Create a loop to cycle through each character in the string str,

 // beginning at  the last character down to the first character.

 // At every cycle, append the character to the rev_str variable.

 // At the end of the loop, rev_str will contain the reversed of str

 for (int i = str.length() - 1; i >= 0; i--) {

  rev_str += str.charAt(i);

 }

 // Convert the rev_str to an integer using the Integer.parseInt()

 // method.  Store the result in an integer variable reversed_num.

 int reversed_num = Integer.parseInt(rev_str);

 // Return the reversed_num

 return reversed_num;

}

// Method to check whether or not a number is a palindrome integer.

// It takes in the number as parameter and returns a true or false.

// A number is a palindrome integer if reversing the number does not

//  change the  number itself.

public static boolean isPalindrome(int number) {

 // check if the number is the same as its reversal by calling the

 //reverse()  method declared earlier. Return true if yes, otherwise,

               // return false.

 if (number == reverse(number)) {

  return true;

 }

 return false;

}

}

Explanation:

The source code file for the program has also been attached to this response for readability. Please download the file and go through the comments in the code carefully as it explains every segment of the code.

Hope this helps!

Download java
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What guidelines should you follow when adding graphics to your presentations?
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B, C, and D are the correct answers.
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2 years ago
Input 10 integers and display the following:
LekaFEV [45]

Answer:

// code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;

   int largest=INT_MIN;

   int smallest=INT_MAX;

   int n;

   cout<<"Enter 10 Integers:";

   // read 10 Integers

   for(int a=0;a<10;a++)

   {

       cin>>n;

       // find largest

       if(n>largest)

       largest=n;

       // find smallest

       if(n<smallest)

       smallest=n;

       // if input is even

       if(n%2==0)

       {  

           // sum of even

           sum_even+=n;

           // even count

           eve_count++;

       }

       else

       {

           // sum of odd    

          sum_odd+=n;

          // odd count

          odd_count++;

       }

   }

   

   // print sum of even

   cout<<"Sum of all even numbers is: "<<sum_even<<endl;

   // print sum of odd

   cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;

   // print largest

   cout<<"largest Integer is: "<<largest<<endl;

   // print smallest

   cout<<"smallest Integer is: "<<smallest<<endl;

   // print even count

   cout<<"count of even number is: "<<eve_count<<endl;

   // print odd cout

   cout<<"count of odd number is: "<<odd_count<<endl;

return 0;

}

Explanation:

Read an integer from user.If the input is greater that largest then update the  largest.If the input is smaller than smallest then update the smallest.Then check  if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.

Output:

Enter 10 Integers:1 3 4  2 10 11 12 44 5 20                                                                                

Sum of all even numbers is: 92                                                                                            

Sum of all odd numbers is: 20                                                                                              

largest Integer is: 44                                                                                                    

smallest Integer is: 1                                                                                                    

count of even number is: 6                                                                                                

count of odd number is: 4

3 0
3 years ago
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