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In-s [12.5K]
3 years ago
9

List the six parts in order, for the format of a two-column proof.

Mathematics
2 answers:
Alecsey [184]3 years ago
3 0
<span>A formal two-column proof contains the following components:

</span><span> Statement of the </span><span>original problem
</span> Diagram, marked with "Given<span>" information
</span>Re-statement<span> of the "Given" information in the proof
</span>Complete supporting reasons<span> for each step in the proof
</span><span> The </span>"Prove"<span> statement as the last statement
</span>Conclusion

Hope this answers the question. Have a nice day.
fgiga [73]3 years ago
3 0

Answer:

Statement of the theorem

Figure

Given Information

Conclusion to prove

Plan of proof

​Proof

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(5, -5)

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Question 6 The mineral content of a particular brand of supplement pills is normally distributed with mean 490 mg and variance o
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0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

Step-by-step explanation:

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.

Mean 490 mg and variance of 400.

This means that \mu = 490, \sigma = \sqrt{400} = 20

What is the probability that a randomly selected pill contains at least 500 mg of minerals?

This is 1 subtracted by the p-value of Z when X = 500. So

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

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Z = 0.5 has a p-value of 0.6915.

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0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

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