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Oduvanchick [21]
3 years ago
10

Helppppppppppppppppppppp

Mathematics
2 answers:
katrin2010 [14]3 years ago
8 0

Answer:

E

Step-by-step explanation:

You use subtraction property of equality to get rid of that 4.

Subtract 4 from both sides basically.

You get 2x > 6

Then you divide both sides by 2 to get x > 3

Then you graph it.

So it starts on point 3, and since x is greater than 3, it goes to points greater than 3, so 4, 5, 6, ect. but it never goes below or on 3.

The graph starts on three, as 3.00000000000000001 is a viable solution, but  since it says > instead of the underlined >, it is an open circle like o---->

So it is E.

babymother [125]3 years ago
5 0

Answer: its B

Step-by-step explanation:

because the circle is open and its the greater than sign

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In college​ basketball, a shot made from beyond a designated arc radiating about 20 ft from the basket is worth three points ins
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a) By the Central Limit Theorem, the distribution would be approximately normal, with mean \mu = 0.45 and standard deviation s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497

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Step-by-step explanation:

To solve this question, we are going to need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

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In this problem, we have that:

p = 0.45, n = 100

a)By the Central Limit Theorem, the distribution would be approximately normal, with mean \mu = 0.45 and standard deviation s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.45*0.55}{100}} = 0.0497

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Z = \frac{X - \mu}{\sigma}

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Z = \frac{0.25 – 0.45}{0.0497}

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c) Making 25 or less shots has a pvalue of -4.02. Z-scores lower than -2 are considered

surprising, so yes, it would be surprising for him to make only 25 of these shots.

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