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Blizzard [7]
2 years ago
9

3х + 9y - 8x - 2y +7​

Mathematics
1 answer:
Sonja [21]2 years ago
7 0

Answer:

-5x+7y+7

Step-by-step explanation:

3х + 9y - 8x - 2y +7​ can be written like the equation below to help solve it

3x-8x+9y-2y+7

=-5x+7y+7

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A kilobyte is 210 bytes and a megabyte is 220 bytes. How many kilobytes are in a megabyte?
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Around 1.04. Exactly 1.047619048

Step-by-step explanation:

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Alex_Xolod [135]
6x - 7 = 4x + 7
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50 ÷ (-5) =<br><br><br> Please simplify
mafiozo [28]

Answer:

-10

Step-by-step explanation:

50 ÷ (-5)

First ignore the brackets and the negative sign.

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What is the probability that a five-card poker hand contains the ace of hearts?
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2 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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