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Umnica
6 days ago
11

The UF athletic department wanted to determine if students came to more than 4 home football games on average per season. They r

andomly surveyed 10 students. Assume all responses were independent of each other. Can we assume normal distribution of the sampling distribution of the sample mean?
A. No, not enough students were sampled
B. No, the randomization condition was not met
C. No the independence assumption was not met
D.Yes
Mathematics
1 answer:
Svet_ta [9.5K]6 days ago
5 0
The correct response is D; we can treat the survey as one involving a single variable, responding with either "yes, I attended more than 4 games" or "no." Given the random survey of 10 students meets the randomization criterion and that their responses are independent, options B and C can be dismissed. However, since only 10 students were surveyed, the confidence interval will not be narrow. As per Statistical Solutions, a minimum of 10 subjects per variable is essential for regression analysis. If the query concerns the number of games each student attended, the potential variables increase; conversely, if it solely asks, “Did you attend more than 4 games?”, then we only consider a single variable, making 10 students sufficient.
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David is performing the following construction. Based on the markings that are present, what construction is he performing?
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Answer:

He is dividing the angle BAC into two equal parts.

Step-by-step explanation:

Initially, he places the compass at point A and draws two small arcs intersecting points D and E. Next, setting the compass at D and then at E, he draws two arcs that intersect between the line segments AB and AC.

The bisecting line is drawn from point A through the intersection of these arcs.

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1 month ago
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Zayed is helping his classmates get ready for their math test by making them identical packages of pencils and calculators. He h
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To find the maximum number of identical packs we see we have 72 pencils and 24 calculators.

This involves discovering the largest number that divides both 72 and 24 evenly,
which is known as the GCM or greatest common multiplier.

To determine the GCM, factor 72 into primes and group them:
72=2 times 2 times 2 times 3 times 3
24=2 times 2 times 2 times 3
Thus, the common grouping is 2 times 2 times 2 times 3, equating to 24.
Therefore, the maximum number of packs is 24.

For pencils:
72 divided by 24=3
Resulting in 3 pencils per pack.

For calculators:
24 divided by 24=1
So, 1 calculator per pack.

The outcome is 3 pencils and 1 calculator in each pack.
6 0
23 days ago
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Based on daily measurements, Bob's weight has a mean of 200 pounds with a standard deviation of 16 pounds, while Mary's weight h
babunello [8423]

Answer:

(C) They have the same coefficient of variation

Step-by-step explanation:

The coefficient of variation (CV) is calculated using the formula:

CV = \frac{\sigma}{\mu}

Where \sigma represents standard deviation and \mu represents the mean.

Bob's average weight is 200 pounds with a standard deviation of 16 pounds

This indicates that \sigma = 16, \mu = 200.

Thus, his coefficient of variation is

CV = \frac{16}{200} = 0.08

Mary's average weight is 125 pounds, with a standard deviation of 10 pounds.

This implies \sigma = 10, \mu = 125

Therefore, her coefficient of variation is

CV = \frac{10}{125} = 0.08

Since both have the same coefficient of variation, the accurate response is.

(C) They have the same coefficient of variation

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24 days ago
Myrtle has a credit card that uses the average daily balance method. For the first 21 days of one of her billing cycles, her bal
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A

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19 days ago
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Construct an interval estimate for the given parameter using the given sample statistic and margin of error. For μ1-μ2, using x¯
zzz [9093]

Answer: CI = (0, 8)

Step-by-step explanation: The confidence interval for the difference in means is given as

Lower limit

= (x1 - x2) + margin of error

Upper limit

= (x1 - x2) - margin of error

Where x1 - x2 = 8 and the margin of error = 8

For the lower limit,

= 8 - 8 = 0

For the upper limit

= 8 + 8 = 16

Thus, CI = (0, 8)

4 0
22 days ago
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