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!(4 %2  == 0) && ((8*2) > (3*3)) , is it true or false?Group of answer choicesTrueFalse

Question

!(4 % 2 == 0) && ((8 * 2) > (3 * 3)), is it true or false?

Group of answer choices

  • True
  • False
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Solution

Let's break down the expression step by step:

  1. 4 % 2 == 0: The modulus operator (%) gives the remainder of the division of two numbers. In this case, 4 divided by 2 has a remainder of 0, so 4 % 2 equals 0. Therefore, 4 % 2 == 0 is true.

  2. (82) > (33): This is comparing whether 16 is greater than 9, which is true.

  3. !(4 %2 == 0): The exclamation mark (!) is a logical NOT operator. It inverts the truth value of the expression that follows it. So, !(4 % 2 == 0) is false because 4 % 2 == 0 is true.

  4. Finally, we have false && true. The logical AND operator (&&) returns true if both operands are true and false otherwise. In this case, it returns false.

So, the whole expression "!(4 %2 == 0) && ((82) > (33))" is false.

This problem has been solved

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