Answer:
The answer is action eventlistener
Explanation:
It is an event handler and it is easy to implement.In java we call them even listeners it is function or a sub routine or a procedure that waits for an event to occur and respond to an event from a GUI component.
Answer:
// here is code in C++.
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int n;
double average,sum=0,x;
cout<<"enter the Value of N:";
// read the value of N
cin>>n;
cout<<"enter "<<n<<" Numbers:";
// read n Numbers
for(int a=0;a<n;a++)
{
cin>>x;
// calculate total sum of all numbers
sum=sum+x;
}
// calculate average
average=sum/n;
// print average
cout<<"average of "<<n<<" Numbers is: "<<average<<endl;
return 0;
}
Explanation:
Read the total number from user i.e "n".Then read "n" numbers from user with for loop and sum them all.Find there average by dividing the sum with n.And print the average.
Output:
enter the Value of N:5
enter 5 Numbers:20.5 19.7 21.3 18.6 22.1
average of 5 Numbers is: 20.44
Answer:
ESC
Explanation:
It allows the user to abort, cancel, or close an operation.
The information that would be the next to be processed by the receiving device is: IP at the internet layer.
<h3>Internet protocol (IP)</h3>
Internet protocol is a network protocol that help to transmit data to user device across the internet or network which inturn make it possible for user device to connect and communicate over the network.
Internet protocol at the internet layer is a TCP/IP software protocol which sole purpose is to transfer data packets across the internet after receiving and processing the data.
Inconclusion the information that would be the next to be processed by the receiving device is: IP at the internet layer.
Learn more about internet protocol here:brainly.com/question/17820678
Answer:
<u>C program to find the sum of the series( 1/2 + 2/3 + ... + i/i+1)</u>
#include <stdio.h>
double m(int i);//function declaration
//driver function
int main() {
int i;
printf("Enter number of item in the series-\n");//Taking input from user
scanf("%d",&i);
double a= m(i);//Calling function
printf("sum=%lf",a);
return 0;
}
double m(int i)//Defining function
{
double j,k;
double sum=0;
for(j=1;j<i+1;j++)//Loop for the sum
{
k=j+1;
sum=sum+(j/k);
}
return sum;
}
<u>Output:</u>
Enter number of item in the series-5
sum=3.550000
