Final answer:
The probability that a spinner lands on an even number or a multiple of 3 when divided into 18 equal sections is the sum of unique even numbers, unique multiples of 3, and numbers that are both, giving us a total of 12 favorable outcomes out of 18, or 2/3.
Step-by-step explanation:
The question is asking for the probability that a spinner lands on an even number or a multiple of 3 when the spinner is divided into 18 equal sections, numbered 1 to 18. To find this, we must first identify the outcomes that satisfy the given conditions. Even numbers between 1 and 18 are: 2, 4, 6, 8, 10, 12, 14, 16, and 18. Multiples of 3 are: 3, 6, 9, 12, 15, and 18. Notice that some numbers are both even and multiples of three, specifically 6, 12, and 18. We should not count these twice.
Even numbers that are not multiples of 3: 2, 4, 8, 10, 14, 16 (6 numbers)
Multiples of 3 that are not even: 3, 9, 15 (3 numbers)
Numbers that are both: 6, 12, 18 (3 numbers)
Adding these together, we have 6 + 3 + 3 = 12 outcomes that fit the criteria out of a total of 18 possible outcomes. Therefore, the probability that the spinner lands on either an even number or a multiple of 3 is 12/18, which simplifies to 2/3 or 0.6667. So, the correct answer is D. 12/18.