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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

likely
likely

unlikely
unlikely

certain
certain

impossible

Answer 1

To determine the likelihood of the spinner stopping on a multiple of 9, we need to identify the numbers on the dial and see which of them are multiples of 9.

The numbers on the dial are: 1, 4, 7, 10, 13, 16, 19, 22, 25, and 28.

Now, let's check for the multiples of 9 within this set:

• The multiples of 9 are: 0, 9, 18, 27, etc.
• Among these, we can see that 9 (the first multiple of 9) is not in the set.
• The next multiples of 9 (18 and 27) are also not present.

Since there are no multiples of 9 on the dial, the chances of the spinner stopping on a multiple of 9 are zero.

Therefore, the correct answer is impossible.