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svet-max [94.6K]
3 years ago
15

The elevation E, in meters, above sea level at which the boiling point of a certain liquid is t degrees Celsius is given by the

function shown below. At what elevation is the boiling point 119.5? 120? E(t)= 1200(120-t)+520(120-t)^2 At what elevation is the boiling point 119.5?
E(119.5)= (the answer) meters
Mathematics
1 answer:
Anna007 [38]3 years ago
5 0

Answer:

E(119.5)= 730m

Step-by-step explanation:

Given

E(t)= 1200(120-t)+520(120-t)^2

Required

Determine E(119.5)

E(t)= 1200(120-t)+520(120-t)^2

Simply substitute 119.5 for t

E(119.5)= 1200(120-119.5)+520(120-119.5)^2

Evaluate the expressions in bracket

E(119.5)= 1200(0.5)+520(0.5)^2

Solve 0.5²

E(119.5)= 1200 * 0.5+520*0.25

E(119.5)= 600+130

E(119.5)= 730

Hence;

E(119.5)= 730m

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Answer:

185

Step-by-step explanation:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,2021.22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40

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The Statistical Abstract of the United States published by the U.S. Census Bureau reports that the average annual consumption of
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Answer:

a) P(\bar X

b)P(98

c) P(\bar X

d) P(93

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 heights of a population, and for this case we know the distribution for X is given by:

X \sim N(99.9,30)  

Where \mu=99.9 and \sigma=30

We select a random sample of n=36. And from the central limit theorem we know that the distribution for the sample is given by:

\bar X \sim N(\mu =99.9 , \frac{\sigma}{\sqrt{n}}= \frac{30}{\sqrt{38}}=4.87)

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

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

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

P(\bar X

=P(Z

Part b

For this case we want this probability:

P(98

=P(\frac{98-99.9}{\frac{30}{\sqrt{38}}}

And we can find this probability on this way:

P(-0.39

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(-0.39

Part c

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

P(\bar X

P(\bar X

Part d

For this case we want this probability:

P(93

=P(\frac{93-99.9}{\frac{30}{\sqrt{38}}}

And we can find this probability on this way:

P(-1.42

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(-1.42

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