Calculate the theoretical probability of choosing an even-numbered card from a traditional deck of 52 cards. For this question, face cards are not counted as odd or even. Simplify.(1 point)

Responses

713
Start Fraction 7 over 13 End Fraction

513
Start Fraction 5 over 13 End Fraction

2052
Start Fraction 20 over 52 End Fraction

413

5 over 13

why not 20/52

The theoretical probability of choosing an even-numbered card can be calculated by first determining how many even-numbered cards are in a deck of 52 cards. In a traditional deck of 52 cards, there are 26 red cards and 26 black cards. Half of these cards are even-numbered, so there are 26 / 2 = 13 even-numbered cards in total.

Therefore, out of 52 cards, 13 are even-numbered. The probability of choosing an even-numbered card is given by the fraction:

13/52

This fraction can be simplified to:

1/4

So, the correct answer is 1/4 or 5/13.

bot they said not counted as odd or even

Apologies for the confusion. If face cards are not counted as odd or even, and we are only considering the numbered cards (2-10), then there are 20 numbered cards in each suite. Since there are four suits in a deck, there are 20 x 4 = 80 numbered cards in total.

Out of these numbered cards, half of them are even-numbered, which means there are 80 / 2 = 40 even-numbered cards.

Therefore, the probability of choosing an even-numbered card from a traditional deck of 52 cards would be:

40 even-numbered cards / 52 total cards = 40/52 = 10/13

Thus, the theoretical probability of choosing an even-numbered card (when face cards are not counted as odd or even) is 10/13.

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Use the image to answer the 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

certain
certain

likely
likely
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Since the spinner is divided into 10 equal parts from the origin toward the circumference and labeled in increments of 3, we can see that the numbers that are a multiple of 9 are 9 and 27 (as they fall on the "9" markers when counted in increments of 3).

Out of the numbers labeled on the spinner, only one of them is a multiple of 9, which is 9 (since 27 is not labeled).

Therefore, the probability of the spinner stopping on a multiple of 9 is 1 out of 10, which means it is unlikely to stop on a multiple of 9 when spun once.

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