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Lisa [10]
3 years ago
8

Coach Elyson claims that 90% of her volleyball serves are good serves. The captain of the team thinks Coach Elyson has a lower r

ate of good serves. To test this, 50 of the coach's serves are randomly selected and 41 of them are good. To determine if these data provide convincing evidence that the proportion of Coach Elyson's serves that are good is less than 90%, 100 trials of a simulation are conducted. The team captain's hypotheses are: H o: p = 90% and H a: p < 90%, where p = the true proportion of Coach Elyson's serves that are good. Based on the results of the simulation, what is an estimate of the P-value of the test? ​
Mathematics
1 answer:
Mrrafil [7]3 years ago
8 0

Answer:

The correct answer is: 2%

Step-by-step explanation:

Took the quiz

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A bag contains 6 blue marbles 10 red marbles and 9 green marbles. If 2 marbles are drawn at random without replacement what is t
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Answer:

P(1 and 2 are green) = 3/25

Step-by-step explanation:

We need to know the total number of marbles

6 Blue + 10 red + 9 green = 25 marbles

P ( 1st marble is green) = green/ total

                                       = 9/25

Since we don't put it back, there are  

6 Blue + 10 red + 8 green = 24 marbles

P ( 2st marble is green) = green/ total

                                       = 8/24  = 1/3

P(1 and 2 are green) = P(1st green) * P(2nd green)

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4 years ago
What are the two consecutive integers and 56
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3 0
4 years ago
What is a triangle with a right angle between two 6cm sides?​
Kobotan [32]

Answer:

That will be a 45°-45°-90° triangle with sides 6, 6, 6√2 cm

Step-by-step explanation:

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3 years ago
Nicole had already run 17 miles on her own and expects to run 1 mile during each track practice. How many miles would nicole hav
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Explanation:

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gavmur [86]

Answer:

The first and third quartiles were calculated incorrectly. Also, this means the inter quartile range is incorrect.

Step-by-step explanation:

The first quartile is going to be the median of the lower half of the data set {89, 93, 99, 110}  this median is (93+99)/2, or 96.

The third quartile can be calculated the same way with the data set {135, 144, 152, 159} This gives us 148 as the third quartile.

Lastly, the interquartile range is just the difference between the first and third quartiles, which is 52.

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3 years ago
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