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olchik [2.2K]
3 years ago
5

6 2/3−4 4/9=? six and two over three minus four and four over nine equals?

Mathematics
2 answers:
In-s [12.5K]3 years ago
8 0
The answer is 2 2/9.
tino4ka555 [31]3 years ago
5 0
<span>2.22222222222 is the answer
</span>
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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 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 375 minutes and standard deviation 68 minutes. So \mu = 375, \sigma = 68

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So n = 6, s = \frac{68}{\sqrt{6}} = 27.76

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

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

Z = \frac{410 - 375}{27.76}

Z = 1.26

Z = 1.26 has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So \mu = 522, \sigma = 106

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So n = 6, s = \frac{106}{\sqrt{6}} = 43.27

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

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

Z = \frac{410 - 523}{43.27}

Z = -2.61

Z = -2.61 has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

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3 years ago
Millie has 5 yards of blue fabric and 7 yards of pink fabric.how many quilt squares can she make with the fabric she has if both
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Millie can only make 5 quilt squares. If she needs 1 yard of blue fabric and 1 yard of pink fabric to make 1 quilt square and only has 5 yards of blue fabric and 7 yards of pink fabric, she can only make 5 quilt squares.
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a carpenter estimates that it will take one person 54 hours to complete a job he plans to have three people work on the job for
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 2 days = 54/2 = 27 hours a day
Your welcome
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In his free time. Gary spends 8 hours per week on the internet. and 11 hours per week playing video games. If Gary has 5 hours o
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