Answer:
midpoint i'm pretty sure. :)
Explanation:
Answer:
- import java.util.Arrays;
- import java.util.Scanner;
-
- public class Main {
- public static void main(String[] args) {
- Scanner input = new Scanner(System.in);
- double distances [] = new double[7];
-
- for(int i=0; i < distances.length; i++){
- System.out.print("Input running distance for day " + (i+1) + ": ");
- distances[i] = input.nextDouble();
- }
-
- System.out.println(Arrays.toString(distances));
- }
- }
Explanation:
The solution code is written in Java.
Firstly, create a Scanner object to get user input for running distance (Line 6). Next, declare a distances array and set the array size to 7 because we intend to use this array to hold running distance for 7 days per week.
Next, use the for loop that run for 7 times to repeatedly prompt user to input running distance for each day and store each value to the array distances (Line 9 -12).
At last, display the array to console terminal (Line 14).
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
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the relation and the table names.
Answer:
Heyy I'm feeling sad and stressed. Wbu?
Explanation:
