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elena55 [62]
3 years ago
8

On a coordinate plane, a line goes from point (negative 3, negative 2) to (2, 4). The distance between the two points pictured i

s d = Use the distance formula to find n. d = StartRoot (x 2 minus x 1) squared + (y 2 minus y 1) squared EndRoot n =
Mathematics
2 answers:
GalinKa [24]3 years ago
8 0

Answer:

N=61

Step-by-step explanation:

Mnenie [13.5K]3 years ago
7 0

Answer:

n =

Step-by-step explanation:

Although the image is not shown, I imagine that what they are asking (n) is the length of the segment that joins both points. If that is the case, we first proceed to identify each of the given points with the general form of coordinate pairs on the plane. For example, we identify:

(-3,-2) = (x_1,y_1)\\(2,4)=(x_2,y_2)\\

Now we use the distance formula replacing the general coordinates by the given values as follows:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\d=\sqrt{(2-(-3))^2+(4-(-2))^2}\\d=\sqrt{5^2+6^2} \\d=\sqrt{25+36} \\d=\sqrt{61}

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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

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