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Ratling [72]
3 years ago
15

What is the output after running the following code snippet? int number = 600; if (number < 200) { System.out.println("Low sp

ender"); } else if (number < 500) { System.out.println("Spending in moderation"); } else if (number < 1000) { System.out.println("Above average!"); } else { System.out.println("High Roller!"); }
Computers and Technology
1 answer:
Nutka1998 [239]3 years ago
8 0

Answer:

Above average!

Explanation:

The code snippet is testing different values of the integer variable number.

If number is less than 200 It prints Low Spender

else If number is more than 200 but less than 500 It prints Spending in moderation

else if number is more than 500 but less than 100 It prints Above Average!

In the code snippet given, number is set to 600 So it prints Above Average!

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Answer: i just learned something new.

Explanation:

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A __________ note is a private note that you leave for yourself or for other people who might use the presentation file
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Answer:

The answer is that it is a speaker note.

Explanation:

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Select the true statement from the choices below. Group of answer choices Invalid code may cause browsers to render the pages sl
IrinaK [193]

Answer:

(A) A web page will not display in a browser unless it passes syntax validation testing.

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5 0
3 years ago
Determine whether the compound condition is True or False.
Vlada [557]

The  compound condition are:

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<h3>What is compound condition?</h3>

A compound statement is known to be one that shows up as the body of another statement, e.g. as in if statement.

The  compound condition are:

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8 0
2 years ago
Convert infix to postfix
kvv77 [185]

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.
4 0
3 years ago
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