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dezoksy [38]
2 years ago
14

Marie made the box plot below to display a set of data. Which set of data could the box plot represent?

Mathematics
1 answer:
Assoli18 [71]2 years ago
7 0

Answer:

  see below

Step-by-step explanation:

First of all, you want to find the data set the matches the extreme values of 5 and 35. That eliminates the 2nd and 4th choices.

Then you want to find the data set that has a median of 15. The first data set has a middle value (median) of 20, so that choice is eliminated.

The data set of the 3rd choice matches the box plot extremes, median, and quartile values.

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On a scale drawing, a bookshelf is 8 inches tall. The scale factor is 1/8. What is the height of the bookshelf?
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8 inches = 1/8 of the real height of the book shelf
Multiply 8 by 8 because the question shows 1/8 of the book shelf
8x8=64
The book shelf is 64 inches.
Or 5 feet and 4 inches.
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Savannah recorded the average rainfall amount, in inches, for two cities over the course of 6 months.
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2 years ago
Which of the following is either a median or an altitude?
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PR is an altitude

QT is a median.

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What is the next term in the pattern? 1,-1,2,-2,3...
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3 years ago
Read 2 more answers
Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.5 and a mean diameter of 205
Crank

Answer:

P(205-0.3=204.7

z=\frac{204.7-205}{\frac{1.5}{\sqrt{79}}}=-1.778

z=\frac{205.3-205}{\frac{1.5}{\sqrt{79}}}=1.778

So we can find this probability:

P(-1.778

And then since the interest is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches using the complement rule we got:

P = 1-0.9243 = 0.0757

Step-by-step explanation:

Let X the random variable that represent the diamters of interest for this case, and for this case we know the following info

Where \mu=205 and \sigma=1.5

We can begin finding this probability this probability

P(205-0.3=204.7

For this case they select a sample of n=79>30, so then we have enough evidence to use the central limit theorem and the distirbution for the sample mean can be approximated with:

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

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

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

And we can find the z scores for each limit and we got:

z=\frac{204.7-205}{\frac{1.5}{\sqrt{79}}}=-1.778

z=\frac{205.3-205}{\frac{1.5}{\sqrt{79}}}=1.778

So we can find this probability:

P(-1.778

And then since the interest is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches using the complement rule we got:

P = 1-0.9243 = 0.0757

6 0
3 years ago
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