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Luda [366]
3 years ago
5

How to write in inqvality

Mathematics
1 answer:
Kitty [74]3 years ago
8 0

Answer:

If you mean "How do I write an inequality?", the answer is COEFFICIENTxVARIABLE {any symbol} COEFFICIENTxVARIABLE >/< INTEGER

In other words, you should have something like this: -3x + 4v < 63

Or this: 51x - 16v > 11

Or anything along those lines.

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Help please :) I'd greatly appreciate it
azamat
= 4i (2i) <span>√6
= 8i^2 </span><span>√6 but i^2 = -1
= - 8</span><span>√6</span><span>

</span>
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
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

6 0
3 years ago
I need to know how to solve it and get the answer right to check my work
Ede4ka [16]
Solve it by finding out how many times 22 can be put into 22, which is once, so the answer is 1.
3 0
3 years ago
Read 2 more answers
An equation of the line that passes through the point (0, -1)<br> and whose slope is 2 is
BARSIC [14]
I believe that it is y=2x-1
7 0
2 years ago
Read 2 more answers
Nadia owes her friend $8 for a movie ticket, $3 for popcorn, and $2 for a soda. Which of the following represents how much money
Novay_Z [31]
8+2+3= 13. So Nadia owes her friend $13
8 0
3 years ago
Read 2 more answers
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