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anastassius [24]
3 years ago
6

Pls, answer!!!!. Please answer the Geometry worksheet. 50 points, and brainliest

Mathematics
2 answers:
dimaraw [331]3 years ago
7 0
Can i get more information or a picture so i can help
crimeas [40]3 years ago
4 0

Answer:

you didnt put the picture out !!!!!!!!

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Suppose a random sample of 36 is selected from a population with a standard deviation of 12. if the sample mean is 98, the 95% c
kvasek [131]
<span>36.......................................................</span>
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3 years ago
Help me please I’ll give 5 stars
baherus [9]

But I want Brainliest

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3 years ago
Please help with this
Evgen [1.6K]

( 20 - 16 / 16 ) × 100 =

( 4 / 16 ) × 100 =

( 1 / 4 ) × 100 =

100/4 =

% 25

7 0
3 years ago
The Dow Jones Industrial Average has had a mean gain of 432 pear year with a standard deviation of 722. A random sample of 40 ye
trasher [3.6K]

Answer:

66.98% probability that the mean gain for the sample was between 250 and 500.

Step-by-step explanation:

To solve this problem, it is important to know the Normal probability distribution and the Central limit theorem.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 432, \sigma = 722, n = 40, s = \frac{722}{\sqrt{40}} = 114.16

What is the probability that the mean gain for the sample was between 250 and 500?

This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 250.

So

X = 500

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

By the Central Limit Theorem

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

Z = \frac{500 - 432}{114.16}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257.

X = 250

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

By the Central Limit Theorem

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

Z = \frac{250 - 432}{114.16}

Z = -1.59

Z = -1.59 has a pvalue of 0.0559.

So there is a 0.7257 - 0.0559 = 0.6698 = 66.98% probability that the mean gain for the sample was between 250 and 500.

8 0
3 years ago
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