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Eva8 [605]
3 years ago
15

Can someone help me on this please

Mathematics
2 answers:
DENIUS [597]3 years ago
8 0

I believe the answer is C because if you add the other numbers to 1/2 witch in a decimal is 0.5 the other choices all are continues and 0.314 doesn't continue forever when added 0.5

Hope this helps

elixir [45]3 years ago
3 0
The answer is C) 0.314
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g Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distrib
algol13

Answer:

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

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

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125

This means that n = 125, s = \frac{7320}{\sqrt{125}}

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

Z = \frac{39000 - 39725}{\frac{7320}{\sqrt{125}}}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

4 0
2 years ago
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I believe x equals one
3 0
3 years ago
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Answer:

It would be 6.8

8 0
3 years ago
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Lena had $24 at the beginning
the movie ticket was $6
it's just basic algebra
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3 years ago
!! HELP ASAP, I ONLY HAVE A FEW MINUTES LEFT BEFORE MY HOMEWORK IS DUE !!
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Y/3=12 should be y=63 because 63/3=12
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