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Question

A circular dial is divided into 10 equal parts from the origin toward the circumference. The parts are labeled clockwise from 1 to 28 in increments of 3. Beginning directly to the right, they are marked as 1, 4, 7, 10, 13, 16, 19, 22, 25, and 28. A needle is fixed vertically at the center, with the arrow pointing directly upward to the line between 22 and 25 at the top center of the dial.

If this spinner is to be spun once, how likely will it stop on a multiple of 9?

(1 point)
Responses

impossible
impossible

unlikely
unlikely

likely
likely

certain
certain
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Answer 1

To determine the likelihood of the spinner stopping on a multiple of 9, we first need to identify the numbers marked on the dial:

1, 4, 7, 10, 13, 16, 19, 22, 25, 28

Next, we check which of these numbers are multiples of 9:

• 1 is not a multiple of 9.
• 4 is not a multiple of 9.
• 7 is not a multiple of 9.
• 10 is not a multiple of 9.
• 13 is not a multiple of 9.
• 16 is not a multiple of 9.
• 19 is not a multiple of 9.
• 22 is not a multiple of 9.
• 25 is not a multiple of 9.
• 28 is not a multiple of 9.

As we can see, none of the numbers on the dial are multiples of 9.

Therefore, the probability of the spinner stopping on a multiple of 9 is impossible.