Answer:
The code is witten in Java and given in the explanation section.
Using a Scanner object we receive a the user's input and with the System.out.println we output the variable
Explanation:
import java.util.Scanner;
public class VariableOutput{
public static void main (String [] args){
Scanner input = new Scanner(System.in);
System.out.println("Enter an Integer 15 or 40");
int numObjects = input.nextInt();
System.out.println("You entered the number "+numObjects);
}
}
Answer:
for (int h = k; h >= 0; h--)
Explanation:
From the list of given options, option C answers the question.
In the outer loop
Initially, k = 0
In the inner loop,
h = k = 0
The value of h will be printed once because h>=0 means 0>=0 and this implies once
To the outer loop
k = 1
The inner loop will always assume value of k;
So,
h = 1
This will be printed twice because of the condition h>=0 means 1>=0.
Since 1 and 0 are >=0; 1 will be printed twice
To the outer loop
k = 2
The inner loop
h = 2
This will be printed thrice because of the condition h>=0 means 2>=0.
Since 2, 1 and 0 are >=0; 2 will be printed thrice
To the outer loop
k = 3
The inner loop
h = 3
This will be printed four times because of the condition h>=0 means 3>=0.
Since 3, 2, 1 and 0 are >=0; 3 will be printed four times
Answer:
Option b, Option c, and Option d are the correct options.
Explanation:
While Sherri is replacing the laptops processor because her's laptop working slow as compared to the past performance then, she decided to ensure to use the right processors for her laptop's motherboard. Then, she verifies that not under the warranty, and she also determines that the teardown procedure is available or not.
Answer:
static int checkSymbol(char ch)
{
switch (ch)
{
case '+':
case '-':
return 1;
case '*':
case '/':
return 2;
case '^':
return 3;
}
return -1;
}
static String convertInfixToPostfix(String expression)
{
String calculation = new String("");
Stack<Character> operands = new Stack<>();
Stack<Character> operators = new Stack<>();
for (int i = 0; i<expression.length(); ++i)
{
char c = expression.charAt(i);
if (Character.isLetterOrDigit(c))
operands.push(c);
else if (c == '(')
operators.push(c);
else if (c == ')')
{
while (!operators.isEmpty() && operators.peek() != '(')
operands.push(operators.pop());
if (!operators.isEmpty() && operators.peek() != '(')
return NULL;
else
operators.pop();
}
else
{
while (!operators.isEmpty() && checkSymbol(c) <= checkSymbol(operators.peek()))
operands.push(operators.pop());
operators.push(c);
}
}
while (!operators.isEmpty())
operands.push(operators.pop());
while (!operands.isEmpty())
calculation+=operands.pop();
calculation=calculation.reverse();
return calculation;
}
Explanation:
- Create the checkSymbol function to see what symbol is being passed to the stack.
- Create the convertInfixToPostfix function that keeps track of the operands and the operators stack.
- Use conditional statements to check whether the character being passed is a letter, digit, symbol or a bracket.
- While the operators is not empty, keep pushing the character to the operators stack.
- At last reverse and return the calculation which has all the results.
