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vovikov84 [41]
3 years ago
7

Use substitution to solve the system. X = 2y +6 2x + 5y = 3

Mathematics
1 answer:
joja [24]3 years ago
8 0
Point Form:

(4,−1)(4,-1)

Equation Form

X = 4, y = -1
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Omg thank you
It equals 4 OR this could be a trick question and it equals 22 :)
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Step-by-step explanation:

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The distribution of weights of United States pennies is approximately normal with a mean of 2.5 grams and a standard deviation o
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Answer:

a) P(X

b) \bar X \sim N(2.5, \frac{0.03}{\sqrt{10}}=0.00948)

c) P(\bar X

And using a calculator, excel or the normal standard table we have that:

P(Z

d) Figure attached

e) If we don't know the distribution then we can't ensure that the sample mean would be distributed like this:

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

And we can't estimate the probabilities on a easy way.

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".  

Part a

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

X \sim N(2.5,0.03)  

Where \mu=2.5 and \sigma=0.03

We are interested on this probability

P(X

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X

And we can find this probability on this way:

P(-0.50

Part b

Since the distribution for X is normal then the distribution for the sample mean is:

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

\bar X \sim N(2.5, \frac{0.03}{\sqrt{10}}=0.00948)

Part c

P(\bar X

And using a calculator, excel or the normal standard table we have that:

P(Z

Part d

See the figure attached the deviation for the sample mean is lower for this reason we have the pattern in the graph attached.

Part e

If we don't know the distribution then we can't ensure that the sample mean would be distributed like this:

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

And we can't estimate the probabilities on a easy way.

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Which of the equations given below would generate a straight line? Help :)
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