A sales transaction was coded with an invalid customer account code (XXX-XX-XXX rather than XXX-XXX-XXX). The error was not dete
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Answer:
Check the explanation
Explanation:
Code :
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
public class nameFrame {
private JFrame mainFrame;
public nameFrame(){
prepareGUI();
}
public static void main(String[] args){
nameFrame nameFrame = new nameFrame();
}
private void prepareGUI(){
mainFrame = new JFrame("Name Frame");
mainFrame.setSize(450,250);
mainFrame.setLayout(new GridLayout(5, 2));
mainFrame.addWindowListener(new WindowAdapter() {
public void windowClosing(WindowEvent windowEvent){
System.exit(0);
}
});
JLabel fname= new JLabel("First Name : ");
JLabel lname = new JLabel("Last Name : ");
JLabel fullname = new JLabel("Full Name : ");
JTextField fnameText = new JTextField(15);
JTextField lnameText = new JTextField(15);
JTextField fullnameText = new JTextField(30);
fullnameText.setEditable(false);
JButton submit = new JButton("submit");
submit.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent e) {
String data = fnameText.getText() + " " + lnameText.getText();
fullnameText.setText(data);
}
});
JButton clear = new JButton("clear");
clear.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent e) {
fnameText.setText("");
lnameText.setText("");
fullnameText.setText("");
}
});
mainFrame.add(fname);
mainFrame.add(fnameText);
mainFrame.add(lname);
mainFrame.add(lnameText);
mainFrame.add(fullname);
mainFrame.add(fullnameText);
mainFrame.add(submit);
mainFrame.add(clear);
mainFrame.setVisible(true);
}
}
Kindly check the attached output image below.
Answer:
Following are the solution to the given question:
Explanation:
Please find the complete and correct question in the attachment file.
For Point a:
For the second round,
A is selects kA(2) either 0 or 1, so for each of them, that is .
B selects for each choice with the probability of
.
If wins the second rear race.
For Point b:
Throughout this example, also selects to be either 0 or 1 with such a
probability. So, although B chooses
from
the probabilities each are
:
For point c:
Assume that B tries again 16 times (typical value), and it destroys. In addition, throughout the exponential background n is obtained at 10 when choosing k between 0 to 2n−1. The probability of A winning all 13 backoff events is:
Let the k value kA(i) be A for the backoff race I select. Because A retains the breed
For point d:
Two stations A and B are supposed. They assume that B will try 16 times afterward. Even so, for A, 16 races were likely to also be won at a rate of 0.82 For Just higher expectations of three A, B, and C stations. For Station A, possibility to win all backoffs
