The answer to the question is: encapsulation
Answers- My computer, My Documents and Recyle bin.
2,3,4 option
Answer:
Descriptives is the correct answer of this question.
Explanation:
Some software packages provide data column definitions that include qualitative summaries such as control averages, mean, average, minimum, standard deviation, number of zero values, number of empty records, etc.
- A descriptives summary is a sentence that gives someone information or something.
- Description is the style of narration creation aimed at making a location, an event, a character or a community vivid.
There are 2 types of Descriptives :-
- Narrative type.
- Argumentative type.
Answer:
import java.io.*;
import java.util.Scanner;
class divide {
public static void main (String[] args) {
Scanner num=new Scanner(System.in);//scanner object.
int userNum=num.nextInt();
while(userNum>1)//while loop.
{
userNum/=2;//dividing the userNum.
System.out.print(userNum+" ");//printing the userNum.
}
}
}
Input:-
40
Output:-
20 10 5 2 1
Input:-
2
Output:-
1
Input:-
0
Output:-
No Output
Input:-
-1
Output:-
No Output.
Explanation:
In the program While loop is used.In the while loop it divides the userNum by 2 in each iteration and prints the value of userNum.The inputs and corresponding outputs are written in the answer.
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
