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lesantik [10]
3 years ago
6

vector is flying his helicopter at 200 miles per hour. how long has he been flying if he has traveled 1,200 miles? A. 2 hours B.

60 hours C. 6 hours D. 10 hours??? please help
Mathematics
2 answers:
seropon [69]3 years ago
6 0
C 6hours because the helicopter can cover 200miles in an hour so if you want to find out 1200 miles all you need to do is divide 1200 by 200 which gives you an answer if 6 xx
Ierofanga [76]3 years ago
4 0
6 hours, because 1,200 divided by 6 is 200. Think about it at 200 mph it will take you 1 hr to go 200 miles two hours will be 400 miles and so on. Does that make sense?
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Answer:

a) We are interested on this probability

P(X>13.17)

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>13.17)=P(\frac{X-\mu}{\sigma}>\frac{13.17-\mu}{\sigma})=P(Z>\frac{13.17-13}{0.5})=P(z>0.34)=0.367

b) P(\bar X >13.17)

The new z score is given by:

z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

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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(13,0.5)  

Where \mu=13 and \sigma=0.5

We are interested on this probability

P(X>13.17)

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>13.17)=P(\frac{X-\mu}{\sigma}>\frac{13.17-\mu}{\sigma})=P(Z>\frac{13.17-13}{0.5})=P(z>0.34)=0.367

Part b

Since the distribution for X is normal then we can conclude that the distribution for the sample mean \bar X is given by:

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

And we want this probability:

P(\bar X >13.17)

The new z score is given by:

z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P(\bar X >13.17)=P(Z> \frac{13.17-13}{\frac{0.5}{\sqrt{36}}})= P(Z>2.04) =0.0207

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