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vivado [14]
2 years ago
5

I don't understand this question can someone help me with this please, please explain!! and thank you for the help!! If u dont k

now try your best. Please do it quick, please do all :D

Mathematics
1 answer:
mestny [16]2 years ago
6 0

Answer:

#1=d

#2=a

Step-by-step explanation:

i am not sure about #1 though good luck

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7+2 Officalbrotherz Tik Tok
iVinArrow [24]

Answer:

9

Step-by-step explanation:

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3 years ago
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A 500-page book contains 250 sheets of paper. The thickness of the paper used to manufacture the book has mean 0.08 mm and stand
LenaWriter [7]

Answer:

10.20% probability that a randomly chosen book is more than 20.2 mm thick

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

250 sheets, each sheet has mean 0.08 mm and standard deviation 0.01 mm.

So for the book.

\mu = 250*0.08 = 20, \sigma = \sqrt{250}*0.01 = 0.158

What is the probability that a randomly chosen book is more than 20.2 mm thick (not including the covers)

This is 1 subtracted by the pvalue of Z when X = 20.2. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{20.2 - 20}{0.158}

Z = 1.27

Z = 1.27 has a pvalue of 0.8980

1 - 0.8980 = 0.1020

10.20% probability that a randomly chosen book is more than 20.2 mm thick

7 0
3 years ago
Find the area of the triangle​
attashe74 [19]
1/2 base x height

That’s it
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3 years ago
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Using an ordered alphabet of 26 letters, how many ways are there to choose a set of six letters such that no two letters in the
mestny [16]
To find our solution, we can start off by creating a string of 27 boxes, all followed by the letters of the alphabet. Underneath the boxes, we can place 6 pairs of boxes and 15 empty boxes.The stars represent the six letters we pick. The empty boxes to the left of the stars provide the "padding" needed to ensure that no two adjacent letters are chosen. We can create this - 
\binom {21} {6}= \frac{21*20*19*18*17*16}{6*5*4*3*2*1} =21*19*17*8=54264
Thus, the answer is that there are \boxed{54264} ways to choose a set of six letters such that no two letters in the set are adjacent in the alphabet. Hope this helped and have a phenomenal New Year! <em>2018</em>
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What is the number of acute obtuse and right angle of a rectangle
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The number of a acute angle in a rectangle is 89 through 1 the obtuse angle in a rectangle is 91 through inf and a right angle of a rectangle is 90 degrees 
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