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3 years ago

Which command compiles the Java source code file Welcome.java?

Computers and Technology

1 answer:

GarryVolchara [31]3 years ago

3 0

Answer:

javac Welcome.java

Explanation:

In order to the file to compile with the default compiler in the JDK, the instruction must be of the form:

javac filename

(Note: the filename with the .java extension)

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Write a program to calculate the total salary of the employees in an office. The program will first prompt the user to enter the

raketka [301]

Answer:

import java.util.Scanner;

public class Salary

{

public static void main(String[] args) {

int numberOfEmployees;

double payRate, numberOfHours, totalPay = 0;

Scanner input1 = new Scanner(System.in);

Scanner input2 = new Scanner(System.in);

System.out.print("Enter the number of employees: ");

numberOfEmployees = input1.nextInt();

for(int i=1; i<=numberOfEmployees; i++) {

System.out.print("Enter the pay rate for employee " + i + ": ");

payRate = input2.nextDouble();

System.out.print("Enter the number of hours worked: ");

numberOfHours = input2.nextDouble();

totalPay += (payRate * numberOfHours);

}

System.out.println("Total payment is: "+ totalPay);

}

Explanation:

- Declare the variables

- Ask the user the for the number of employees

- Inside the for loop, ask the user for pay rate and hours worked for each employee

- Calculate the total pay

- <u>Outside of the loop</u>, print the total pay

6 0

3 years ago

Read 2 more answers

Which process is used to protect transmitted data in a vpn?

wlad13 [49]

Here is the answer: <span>Tunneling</span>

3 0

3 years ago

nder a Huffman encoding ofnsymbols with frequenciesf1, f2, . . . , fn, what is the longest a codewordcould possibly be

Natasha2012 [34]

Answer:

The longest codeword that could possibly be for "n" symbol is n-1 bits.

Explanation:

From the given information:

Suppose we are to consider a set of frequencies $\mathtt{f_1,f_2,f_3 ...,f_n}$ , for which f is a symbol for the length n. Therefore, the longest codeword that could possibly be for "n" symbol is n-1 bits.

However, during the encoding for "n" in conjunction with n-2, then the possibilities for n are; $\mathtt{\dfrac{1}{2}, \dfrac{1}{4}, ... \dfrac{1}{2^{n-2}}}$