What is the difference between the = operator and the == operator in Java?

In Python please.

SOVA2 [1]

Answer:

negatives = []

zeros = []

positives = []

while True:

number = input("Enter a number: ")

if number == "":

break

else:

number = int(number)

if number < 0:

negatives.append(number)

elif number == 0:

zeros.append(number)

else:

positives.append(number)

for n in negatives:

print(n)

for z in zeros:

print(z)

for p in positives:

print(p)

Explanation:

Initialize three lists to hold the numbers

Create a while loop that iterates until the user enters a blank line

Inside the loop:

If the number is smaller than 0, put it in the negatives list

If the number is 0, put it in the zeros list

Otherwise, put the number in the negatives list

When the while loop is done, create three for loops to print the numbers inside the lists

5 0

3 years ago

Select the correct answer from each drop-down menu. Which two factors do ergonomic principles consider a part of a job? Ergonomi

Rudiy27

Answer:mental stress

Explanation:

4 0

3 years ago

Suppose that the format for license plates in a certain state is two letters followed by four numbers. (a) How many different pl

Ymorist [56]

Answer:

(a) 6,760,000 plates

(b) 3,407,040 plates

(c) 6,084,000 plates

Explanation:

The very first thing to note about this question is the number of characters involved in the license plate format (6 characters in this case; 2 letters and 4 numbers). The letters come first and then the numbers follow.

There are a total of 26 possible letters (A-Z) and 10 possible numbers (0 - 9) that can be chosen. We can then proceed to the first question;

(a) Here, the total number of possible plates is to be determined. This is done as follows:

Character 1 (Letter): There are 26 possible letters

Character 2 (Letter): There are 26 possible letters

Character 3 (number): There are 10 possible numbers

Character 4 (number): There are 10 possible numbers

Character 5 (number): There are 10 possible numbers

Character 6 (number): There are 10 possible numbers

So, total number of different plates will be obtained by multiplying all the possibilities: 26 × 26 × 10 × 10 × 10 × 10 = 6,760,000 plates

(b) This second part puts a constraint on the usage of the numbers, unlike the question (a), where there was no constraint at all.

Since there is no constraint on the letters, we can write that:

Character 1 (Letter): There are 26 possible letters

Character 2 (Letter): There are 26 possible letters

For the first number as well, we can write:

Character 3 (number): There are 10 possible numbers

However, for the remaining characters, the possibilities will continually reduce by a value of 1, since we can not use a number that has been used before. So,

Character 4 (number): There are 9 possible numbers

Character 5 (number): There are 8 possible numbers

Character 6 (number): There are 7 possible numbers

So, total number of different plates will be: 26 × 26 × 10 × 9 × 8 × 7 = 3,407,040 plates

(c) Here, repetitions are allowed as in questions (a), but there can not be four zeros. This implies that the maximum number of zeros in any plate will be three. Thus, there will be maximum possibilities on all characters until the last one which will be constrained.

Character 1 (Letter): There are 26 possible letters

Character 2 (Letter): There are 26 possible letters

Character 3 (number): There are 10 possible numbers

Character 4 (number): There are 10 possible numbers

Character 5 (number): There are 10 possible numbers

Character 6 (number): There are 9 possible numbers

Total number of plates will therefore be: 26 × 26 × 10 × 10 × 10 × 9 = 6,084,000 plates.

7 0

3 years ago

The return value is a one-dimensional array that contains two elements. These two elements indicate the rows and column indices

erma4kov [3.2K]

Answer:

In Java:

import java.util.Scanner;

import java.util.Arrays;

public class Main {

public static void main(String args[]) {

Scanner input = new Scanner(System.in);

int row, col;

System.out.print("Rows: "); row = input.nextInt();

System.out.print("Cols: "); col = input.nextInt();

int[][] Array2D = new int[row][col];

System.out.print("Enter array elements: ");

for(int i =0;i<row;i++){

for(int j =0;j<col;j++){

Array2D[i][j] = input.nextInt();

} }

int[] maxarray = findmax(Array2D,row,col);

System.out.println("Row: "+(maxarray[0]+1));

System.out.println("Column: "+(maxarray[1]+1));

}

public static int[] findmax(int[][] Array2D,int row, int col) {

int max = Array2D[0][0];

int []maxitem = new int [2];

for(int i =0;i<row;i++){

for(int j =0;j<col;j++){

if(Array2D[i][j] > max){

maxitem[0] = i;

maxitem[1] = j; } } }

return maxitem;

}

Explanation:

The next two lines import the scanner and array libraries

import java.util.Scanner;

import java.util.Arrays;

The class of the program

public class Main {

The main method begins here

public static void main(String args[]) {

The scanner function is called in the program

Scanner input = new Scanner(System.in);

This declares the array row and column

int row, col;

This next instructions prompt the user for rows and get the row input

System.out.print("Rows: "); row = input.nextInt();

This next instructions prompt the user for columns and get the column input

System.out.print("Cols: "); col = input.nextInt();

This declares the 2D array

int[][] Array2D = new int[row][col];

This prompts the user for array elements

System.out.print("Enter array elements: ");

The following iteration populates the 2D array

for(int i =0;i<row;i++){

for(int j =0;j<col;j++){

Array2D[i][j] = input.nextInt();

} }

This calls the findmax function. The returned array of the function is saved in maxarray

int[] maxarray = findmax(Array2D,row,col);

This prints the row position

System.out.println("Row: "+(maxarray[0]+1));

This prints the column position

System.out.println("Column: "+(maxarray[1]+1));

The main method ends here

}

The findmax function begins here

public static int[] findmax(int[][] Array2D,int row, int col) {

This initializes the maximum to the first element of the array

int max = Array2D[0][0];

This declares maxitem. The array gets the position of the maximum element

int []maxitem = new int [2];

The following iteration gets the position of the maximum element

for(int i =0;i<row;i++){

for(int j =0;j<col;j++){

if(Array2D[i][j] > max){

maxitem[0] = i; -- The row position is saved in index 0

maxitem[1] = j; -- The column position is saved in index 1} } }

This returns the 1 d array

return maxitem;

}

7 0

3 years ago

RAM is usually a. secondary storage b. a static type of memory, used for permanent storage c. a volatile type of memory, used fo

PolarNik [594]

Answer:

The correct answer to the following question will be Option C (Volatile type of memory).

Explanation:

Random Access Memory, a variety of computer's memory which could be retrieved accidentally i.e., some memory can be obtained without any of the previous bytes being touched.

Volatile memory is a type of a storage in computer that only retains its information or data while powering the appliance.
Almost all of the RAM used in Pc's for the primary storage is unstable storage.

So, Option C is the right answer.

8 0

3 years ago