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2 years ago

Hannah is mixing paints to create a green color. Her mixture for green requires 2 parts blue, 1 part purple, and 4 parts yellow.

Mathematics

1 answer:

alexdok [17]2 years ago

4 0

Answer: A

Step-by-step explanation:

Divide each part in 2.

2/2 parts blue, 1/2 part purple, and 4/2 parts yellow

Which equals l:

1 part blue, 0.5 part purple, and 2 parts yellow

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an outlier may be defined as a data point that is more than 1.5 times the interquartile range below the lower quartile or is mor

Ira Lisetskai [31]

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

3 0

2 years ago

Find the magnitude and direction (in degrees) of the vector. (Assume 0° ≤ < 360°.)

ddd [48]

$6i~~ + ~~2\sqrt{3}j\implies < \stackrel{a}{6}~~,~~\stackrel{b}{2\sqrt{3}} > \\\\[-0.35em] ~\dotfill\\\\ \stackrel{magnitude}{\sqrt{a^2+b^2}}\implies \sqrt{6^2+(2\sqrt{3})^2}\implies \sqrt{36+(2^2\cdot 3)}\implies \sqrt{36+12}\implies \sqrt{48} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{direction}{tan^{-1}\left( \cfrac{b}{a}\right)}\implies tan^{-1}\left( \cfrac{2\sqrt{3}}{6} \right)\implies tan^{-1}\left( \cfrac{\sqrt{3}}{3} \right) \\\\\\ tan^{-1}\left( \cfrac{1}{\sqrt{3}} \right)\implies 30^o$