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Kay [80]
2 years ago
10

I Want to Test Somebody, Can you answer this?

Mathematics
1 answer:
Ratling [72]2 years ago
7 0
Answer 1/3

Explanation:

Write all numbers above common denominator

Then subtract the numbers
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I need help<br> please and thank you
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I believe it’s
D.51is it B ??
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f the original price of a table is p and the sale price of the table is p−0.2p, which other expression represents the sale price
Sphinxa [80]

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0.2p is the correct

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If sin x = -1/2, and 270° &lt; x &lt; 360°, what is cos(x - 30°)?
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Step-by-step explanation:

Look at the picture

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A queen ant lays 1.83 x 10^6 eggs over a period of 30 days. Assuming she lays the same number of eggs each day, about how many e
Leya [2.2K]
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3 years ago
A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator.
AVprozaik [17]

Answer:

X \sim N(61,14)  

Where \mu=61 and \sigma=14

Since the distribution for X is normal then the distribution for the sample mean  is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\mu_{\bar X} = 61

\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715

So then is appropiate use the normal distribution to find the probabilities for \bar X

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  The letter \phi(b) is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: \phi(b)=P(z

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(61,14)  

Where \mu=61 and \sigma=14

Since the distribution for X is normal then the distribution for the sample mean \bar X is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\mu_{\bar X} = 61

\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715

So then is appropiate use the normal distribution to find the probabilities for \bar X

8 0
3 years ago
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