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mariarad [96]
2 years ago
13

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

er at his​ school, with an ERA of 3.02. ​Also, Alice had the lowest ERA of any female pitcher at the school with an ERA of 3.16. For the​ males, the mean ERA was 4.206 and the standard deviation was 0.846. For the​ females, the mean ERA was 4.519 and the standard deviation was 0.789. Find their respective​ z-scores. Which player had the better year relative to their​ peers, Perry or Alice​
Mathematics
1 answer:
Blizzard [7]2 years ago
7 0

Answer:

z score Perry z=-1.402

z score Alice z=-1.722

Alice had better year in comparison with Perry.

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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In a right triangle, the measure of one acute angle is 12 more than twice the measure of the other acute angle. Find the measure
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Given:

In a right triangle, the measure of one acute angle is 12 more than twice the measure of the other acute angle.

To find:

The measures of the 2 acute angles of the triangle.

Solution:

Let x be the measure of one acute angle. Then the measure of another acute is (2x+12).

According to the angle sum property, the sum of all interior angles of a triangle is 180 degrees. So,

x+(2x+12)+90=180

3x+102=180

3x=180-102

3x=78

Divide both sides by 3.

x=\dfrac{78}{3}

x=26

The measure of one acute angle is 26 degrees. So, the measure of another acute angle is:

2x+12=2(26)+12

2x+12=52+12

2x+12=64

Therefore, the measures of two acute angles are 26° and 64° respectively.

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Answer:

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Step-by-step explanation:

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Identify the percent of change as an increase or a decrease 20 to 30.
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Answer:

50% increase.

Step-by-step explanation:

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If 55% of a number is 132 and 25% of the same number is 60, find 80% of that number.
UNO [17]

Answer:

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Step-by-step explanation:

It is always wise to look carefully at the question to see what information is given, and what information you are asked for. You are given two different fractions of the same number, and you want the value of a third fraction of that number. That third fraction is the sum of the other two.

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The value of 80% of the number will be the sum of 55% and 25% of the number.

  n×80% = n×55% +n×25%

  n×80% = 132 +60

  n×80% = 192

80% of the number is 192.

_____

<em>Additional comment</em>

Your first clue that there may be something special about this problem lies in the fact that one fraction of a number is all you need in order to find any other fraction. Here, we're given two fractions (55% and 25%), so that is more information than we need. Why? Because it simplifies the problem to one of addition. No proportions need to be solved.

__

If you were interested, you could determine 100% of the number to be ...

  132/n = 55/100   ⇒   n = 132(100/55) = 240

or

  60/n = 25/100   ⇒   n = 60(100/25) = 240

Then ...

  0.80n = 0.80(240) = 192 . . . . 80% of the number

Or, you could write proportions for the 80% value:

  80%/55% = n/132   ⇒   n = 132(80/55) = 192 . . . . 80% of the number

  80%/25% = n/60   ⇒   n = 60(80/25) = 192 . . . . 80% of the number

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"Look carefully" includes thinking about the ways the <em>givens</em> and the <em>finds</em> are related.

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