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3 years ago

Simplify:

Mathematics

1 answer:

OverLord2011 [107]3 years ago

8 0

Step-by-step explanation:

Simplify :

x4−6x3+2x−7+7x3−x+5x2+2−x4

Rearranging and collecting the like terms, we get:

=(x4−x4)+(−6x3+7x3)+5x2+(2x−x)+(−7+2)

=(1−1)x4+(−6+7)x3+5x2+(2−1)x+(−7+2)

=(1)x3+5x2+(1)x−7+2

=x3+5x2+x−5

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3 years ago

One year Perry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitch

Blizzard [7]

Answer:

z score Perry $z=-1.402$

z score Alice $z=-1.722$

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Step-by-step explanation:

Consider the provided information.

One year Perry had the lowest ERA of any male pitcher at his school, with an ERA of 3.02. For the males, the mean ERA was 4.206 and the standard deviation was 0.846.

To find z score use the formula.

$z=\frac{x-\mu}{\sigma}$

Here μ=4.206 and σ=0.846

$z=\frac{3.02-4.206}{0.846}$

$z=\frac{-1.186}{0.846}$

$z=-1.402$

Alice had the lowest ERA of any female pitcher at the school with an ERA of 3.16. For the females, the mean ERA was 4.519 and the standard deviation was 0.789.

Find the z score

where μ=4.519 and σ=0.789

$z=\frac{3.16-4.519 }{0.789}$

$z=\frac{-1.359}{0.789}$

$z=-1.722$

The Perry had an ERA with a z-score is –1.402. The Alice had an ERA with a z-score is –1.722.

It is clear that the z-score value for Perry is greater than the z-score value for Alice. This indicates that Alice had better year in comparison with Perry.

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