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3 years ago

Divide 6/13 by 6/12 . A. 9/16 B. 1/12 C. 12/13 D. 13/12

Mathematics

2 answers:

astra-53 [7]3 years ago

8 0

$\frac{6}{13}:\frac{6}{12}=\frac{6}{13}*\frac{12}{6}= \boxed{\frac{12}{13}}$

cluponka [151]3 years ago

8 0

$\frac{6}{13} / \frac{6}{12} = \frac{6}{13}* \frac{12}{6} = \frac{72}{78} = \frac{12}{13}$
Therefore your answer is C. 12/13

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Supposing a normal distribution, we find that:

The diameter of the smallest tree that is an outlier is of 16.36 inches.

--------------------------------

We suppose that tree diameters are normally distributed with mean 8.8 inches and standard deviation 2.8 inches.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score 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, which is the percentile of X.

In this problem:

Mean of 8.8 inches, thus $\mu = 8.8$ .
Standard deviation of 2.8 inches, thus $\sigma = 2.8$ .

The interquartile range(IQR) is the difference between the 75th and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

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

$-0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = -0.675(2.8)$

$X = 6.91$

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

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

$0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = 0.675(2.8)$

$X = 10.69$

The IQR is:

$IQR = 10.69 - 6.91 = 3.78$

What is the diameter, in inches, of the smallest tree that is an outlier?

The diameter is 1.5IQR above the 75th percentile, thus:

$10.69 + 1.5(3.78) = 16.36$

The diameter of the smallest tree that is an outlier is of 16.36 inches.

A similar problem is given at brainly.com/question/15683591

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2 years ago

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Answer:

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Step-by-step explanation:

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