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Allushta [10]
2 years ago
12

A store is located 10 miles from my house, it takes the average female 20 minutes to talk from the store back to her house. Why

has it been 12 years and my mom still isn't back?
Mathematics
2 answers:
grin007 [14]2 years ago
8 0

Answer:

Because she may have been abducted by aliens and there time way is way slower than ours so 20 mins for them is like 15 years for ua, ao you should have her back in 3 years. Lemme get brainliest too

Step-by-step explanation:

White raven [17]2 years ago
6 0

Answer:

well the reason is she found the store but she saw the news paper that had a better deal across the country so she is walking across the world to get the better deal because she is crazy for a deal shell do anything for the deal because she loves to save money

Step-by-step explanation:

IM CORRECT OK MY G

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skelet666 [1.2K]
Remember y=mx+b

Since y=2 that means it is a horizontal line at 2 on the y axis which means it does not have a slope. The slope would be zero.
6 0
3 years ago
It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

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}}.

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.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So \mu = 373, \sigma = 67

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So n = 5, s = \frac{67}{\sqrt{5}} = 29.96

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 373}{29.96}

Z = 1.23

Z = 1.23 has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So \mu = 526, \sigma = 107

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So n = 5, s = \frac{107}{\sqrt{5}} = 47.86

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 526}{47.86}

Z = -2.42

Z = -2.42 has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

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Answer:

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Step-by-step explanation:

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Answer:

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Then reorder and see what their original values were

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