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8090 [49]
2 years ago
5

The parallel boxplots below display the weights of bags of two different brands, A and B, of granola.

Mathematics
1 answer:
KiRa [710]2 years ago
6 0

Answer:

A. The bag weights for brand A have less variability than the bag weights for brand B.

Step-by-step explanation:

In a box plot display, measure of variability can be determined by the length of the rectangular box or/and by the length of the whiskers.

The longer or greater the length, the more the variability the data set has. The shorter or smaller the length, the lesser the variability.

The box plot display of Brand A has shorter rectangular box and a shorter whisker length compared to the box plot display of Brand B. Therefore, it can be concluded that: bag weights for Brand A have less variability compared to bag weights for Brand B.

The correct statement of comparison is:

"A. The bag weights for brand A have less variability than the bag weights for brand B."

Option B is incorrect. Bag weights for Brand A do not have more variability than those of Brand B.

Option C and option D are both incorrect. Neither an outlier nor range can be used to represent or describe "typical value" for a given data set.

Typical bag weights can be well represented or described by average bag weights or median weight of the data set.

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Answer:

A) Yes, because P (F∩S) = 0

Step-by-step explanation:

Hello!

50 customers of a store were asked to choose between two discounts:

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Discount 2: 10% off all purchases for the week.

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S: the selected person choose discount 2.

Two events are mutually exclusive when the occurrence of one of them prevents the other from occurring in one repetition of the trial and the intersection between these two events is void with zero probability of happening.

In this case, since the customers were asked to choose one out of the two events, if the customer chooses the first one, then he couldn't have chosen the second one and vice-versa. Then the intersection between these two events has zero probability, symbolically:

P(F∩S)=0

I hope it helps!

8 0
2 years ago
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tatiyna

Answer:

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Step-by-step explanation:

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5 0
2 years ago
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Alejandro surveyed his classmates to determine who has ever gone surfing and who has ever gone snowboarding. Let A be the event
Gekata [30.6K]

Answer:

\text{A and B are independent events}, P(A|B)=P(A)=0.16

Step-by-step explanation:

First of all we need to know when does two events become independent:

For the two events to be independent, P(A|B)=P(A) that is if condition on one does not effect the probability of other event.

Here, in our case the only option that satisfies the condition for the events to be independent is P(A|B)=P(A)=0.16. Rest are not in accordance with the definition of independent events.

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2 years ago
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andriy [413]
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x = 10
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20 and 22




3 0
2 years ago
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