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algol13
3 years ago
9

Find the volume of the given figure.*

Mathematics
2 answers:
enyata [817]3 years ago
8 0
117.286 , that’s what i got
balu736 [363]3 years ago
4 0

Answer:

117.286

Step-by-step explanation:

Volume of a cone formula:  v=\pi r^{2} \frac{h}{3}

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A 12 ounce bottle of beer contains 355 mL. The Random Brewing Company programs its machinery so that the amount of beer in a bot
erica [24]

Answer:

a) 79.67% probability that a randomly selected bottle has at least 355 ml

b) 99.29% probability that the bottles of a randomly selected 6-pack of beer have an average of at least 355 ml.

c) 88.88% probability that the total amount of beer in the 6-pack is less than 2131.5 ml

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 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.

In this question:

\mu = 355.5, \sigma = 0.6

a. What is the probability that a randomly selected bottle has at least 355 ml?

This is 1 subtracted by the pvalue of Z when X = 355. So

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

Z = \frac{355 - 355.5}{0.6}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033

1 - 0.2033 = 0.7967

79.67% probability that a randomly selected bottle has at least 355 ml.

b. What is the probability that the bottles of a randomly selected 6-pack of beer have an average of at least 355 ml?

Now n = 6, s = \frac{0.5}{\sqrt{6}} = 0.2041

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

By the Central Limit Theorem

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

Z = \frac{355 - 355.5}{0.2041}

Z = -2.45

Z = -2.45 has a pvalue of 0.0071

1 - 0.0071 = 0.9929

99.29% probability that the bottles of a randomly selected 6-pack of beer have an average of at least 355 ml.

c. What is the probability that the total amount of beer in the 6-pack is less than 2131.5 ml?

2131.5/6 = 355.25

This is the pvalue of Z when X = 355.25.

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

Z = \frac{355.25 - 355.5}{0.2041}

Z = -1.22

Z = -1.22 has a pvalue of 0.1112

1 - 0.1112 = 0.8888

88.88% probability that the total amount of beer in the 6-pack is less than 2131.5 ml

5 0
3 years ago
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