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Nezavi [6.7K]
3 years ago
5

Please help urgent!!

Mathematics
1 answer:
trapecia [35]3 years ago
8 0

Answer:

<em>The area of the rectangle is</em> A=12b^6+6b^3-18b^2

Step-by-step explanation:

<u>Area of a rectangle</u>

A rectangle of height H and width W has an area of:

A=H*W

The rectangle shown in the figure has a height of:

H=3b^2

And a width of

W=4b^4+2b-6

The area is

A=(3b^2)(4b^4+2b-6)

Operating:

A=(3b^2)*4b^4+(3b^2)*2b-(3b^2)6

A=12b^6+6b^3-18b^2

The area of the rectangle is A=12b^6+6b^3-18b^2

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We assume the lunch prices we observe are drawn from a normal distribution with true mean \mu and standard deviation 0.68 in dollars.


We average n=45 samples to get \bar{x}.


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Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains \mu.


Our interval takes the form of ( \bar{x} - z \sigma, \bar{x} + z \sigma ) as \bar{x} is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".


Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.


Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.


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We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is


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in other words a margin of error of


\pm 1.65(.101) = \pm 0.167 dollars


That's around plus or minus 17 cents.




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