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3 years ago

Describe all the main stress causal agents

Computers and Technology

1 answer:

LekaFEV [45]3 years ago

5 0

Answer:

Causes of Stress

Being unhappy in your job.

Having a heavy workload or too much responsibility.

Working long hours.

Having poor management, unclear expectations of your work, or no say in the decision-making process.

Working under dangerous conditions.

Being insecure about your chance for advancement or risk of termination.

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Please design a Java GUI application with two JTextField for user to enter the first name and last name. Add two JButtons with a

Ede4ka [16]

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.

5 0

3 years ago

Suppose you are organizing a party for a large group of your friends. Your friends are pretty opinionated, though, and you don't

Iteru [2.4K]

Answer:

Relative greedy algorithm is not optimal

Explanation:

Proving that Relative’s greedy algorithm is not optimal.

This can be further proved by the following example.

Let us assumethe friends to be invited be Ali, Bill, David, Dennis, Grace, Eemi, and Sam.

The enemy list of each friend is shown below:

• Ali: Bill. David, Dennis

• Bill: Ali, Grace, Eemi, Sam

• David: Ali, Grace, Eemi, Sam

• Dennis: Ali, Grace, Eemi, Sam

• Grace: Bill. David, Dennis. Eemi . Sam

• Eemi: Bill, David, Dennis, Grace, Sam

• Sam: Bill, David, Dennis, Eemi, Grace

From the enemy list, the one with fewer enemies is Ali.

If we invite Ali, then Bill, David, and Dennis cannot be invited.

Next person that can be invited is one from Grace, Eemi, and Sam

If we choose any one of them we cannot add any other person.

For example, if we choose Grace, all other members are enemies of Ali and Grace. So only Ali

and Grace can only be invited.

So we get a list of two members using relative’s greedy algorithm.

This is not the optimal solution.

The optimal solution is a list of 3 members

• Bill, David, and Dennis can be invited to the party.

Hence, it is proved that relative’s greedy algorithm is not optimal, where the maximum number of friends is not invited to the party.

3 0

4 years ago

Define a method pyramidVolume with double parameters baseLength, baseWidth, and pyramidHeight, that returns as a double the volu

marysya [2.9K]

<h2>Question:</h2>

Define a method pyramidVolume with double parameters baseLength, baseWidth, and pyramidHeight, that returns as a double the volume of a pyramid with a rectangular base. Relevant geometry equations:

Volume = base area x height x 1/3

Base area = base length x base width.

(Watch out for integer division)

import java.util.Scanner;

public class CalcPyramidVolume {

/* Your solution goes here */

public static void main (String [] args) {

System.out.println("Volume for 1.0, 1.0, 1.0 is: " + pyramidVolume(1.0, 1.0, 1.0));

return;

}

<h2>Answer:</h2><h2></h2>

import java.util.Scanner;

public class CalcPyramidVolume {

/* Your solution goes here */

public static void main (String [] args) {

System.out.println("Volume for 1.0, 1.0, 1.0 is: " + pyramidVolume(1.0, 1.0, 1.0));

return;

}

//Begin method definition

public static double pyramidVolume(double baseLength, double baseWidth, double pyramidHeight){

//First, calculate the base area of the pyramid

//store the result in a double variable

double baseArea = baseLength * baseWidth;

//Then, calculate the volume of the pyramid

//using the base area and the base width

double volume = 1 / 3.0 * baseArea * pyramidHeight;

//return the volume

return volume;

}

<h2>Output:</h2>

Volume for 1.0, 1.0, 1.0 is: 0.3333333333333333

<h2>Explanation:</h2>

The code above contains comments explaining the important lines of the code. The output of the code has also been provided above.

The parts of the code that are worth noting are:

(i) The return type of the method pyramidVolume should be a <em>double</em> since calculations are done using <em>double</em> values. i.e the method header should be written as

<em>public static double pyramidVolume(){</em>

<em />

(ii) The method requires three(3) parameters of type double: baseLength, baseWidth and pyramidHeight. These should be included in the method. Therefore, the complete header definition should be:

<em>public static double pyramidVolume(double baseLength, double baseWidth, double pyramidHeight){</em>

<em />

(iii) The formula for calculating the volume of the pyramid should be:

volume = 1 / 3.0 * baseArea * baseWidth;

Rather than;

volume = 1 / 3 * baseArea * baseWidth;

This is because integer division yields integer result. If 1 / 3 is evaluated, the result will be 0 since the decimal part will be truncated thereby making the result of the volume = 1 / 3 * baseArea * baseWidth will be 0.

Therefore, the work around for that should be to write 1 / 3.0 or 1.0 / 3.0 or 1.0 / 3.

(iv) After the calculation, the result of volume should be returned by the method. This will enable a proper call by the main method for execution.

7 0

3 years ago

How many bits must be “flipped” (i.e., changed from 0 to 1 or from 1 to 0) in order to capitalize a lowercase ‘a’ that’s represe

Neporo4naja [7]

<span>To capitalize lowercase “a” which is 0110001 which is “A” you will need to flip the following bites 01000001<span> as represented in ASCII. Since we are only looking at 2bit digit which is 0 and 1 which has a 256 possible combinations from 0 up to 255. </span></span>

6 0

3 years ago

Write a for loop that displays the following set of numbers: 0,10,20,30,40,50....1000

ELEN [110]

C# program code:

int i = 0
while (i<=1000)
{
console.Writeline("{0}",i);
i = i + 10;
}

Explanation:
First we set variable to initial value. In this example it is 0. Then we enter into while loop. This type of loop executes the code until the condition is fulfilled. In our case while loop checks if i <=1000. It is and then it writes it on the screen. Next step is to increase it by 10. Then it does the same code again.
Last number that will be printed is 1000. After that it will increase i to 1010 and it will exit the loop.

3 0

3 years ago

Describe all the main stress causal agents​

Describe all the main stress causal agents