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ziro4ka [17]
3 years ago
15

Which expression is equivalent to 200? 210 10/2 10./20 100V

Mathematics
2 answers:
elena-s [515]3 years ago
8 0

Answer:

100v

Step-by-step explanation:

Can i get Brainliest??

Anastasy [175]3 years ago
4 0

Answer:

100V

Step-by-step explanation:

210 is obviously higher

10/2 is only 5 when simplified

10/20 is only 1/2 when simplified

100V if, and I'm guessing, V is going to equal 2

100V f(x)=2

200

Hope this helps!

Brainliest??

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The amount of potato chips an 18-ounce bag contains follows a normal distribution with a mean of 18.5 ounces and a standard devi
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Answer:

100% probability that the sample mean weight of these 100 bags is less than 18.6 ounces

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 normally distributed samples are 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 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, we have that:

\mu = 18.5, \sigma = 0.2, n = 100, s = \frac{0.2}{\sqrt{100}} = 0.02

What is the probability that the sample mean weight of these 100 bags is less than 18.6 ounces

This is the pvalue of Z when X = 18.6. So

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

By the Central Limit Theorem

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

Z = \frac{18.6 - 18.5}{0.02}

Z = 5 has a pvalue of 1

100% probability that the sample mean weight of these 100 bags is less than 18.6 ounces

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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group like terms

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