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3 years ago

11

A spinner has regions numbered 1 through 21. what is the probability that the spinner will stop on an even number or a multiple

of 3

2 answers:

MAXImum [283]3 years ago

6 0

Answer: $\dfrac{2}{3}$

Step-by-step explanation:

Given : The total number of regions in the spinner = 21

The outcomes for even number =2,4,6,8,10,12,14,16,18,20

The outcomes for multiple of 3 = 3,6,9,12,15,18,21

The number of outcomes of getting an even number or a multiple of 3 = 14

Then, the probability that the spinner will stop on an even number or a multiple of 3 will be :-

$\dfrac{14}{21}=\dfrac{2}{3}$

Hence, the probability that the spinner will stop on an even number or a multiple of 3 is $\dfrac{2}{3}$

sertanlavr [38]3 years ago

5 0

10 of the numbers are even. 4 more are odd multiples of 3. The probability of landing on even or a multiple of 3 is (10+4)/21 = 2/3.

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Answer:

1. $xy = 63$

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Step-by-step explanation:

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- Find y when x = 21

Given that thee variation is inversely proportional;

This implies that

$y\ \alpha\ \frac{1}{x}$

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svp [43]

Answer:

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Option A is correct.

Step-by-step explanation:

We will solve to find out Which expression has a positive quotient.

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We will use these rules

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