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GuDViN [60]
3 years ago
13

A primary job of team management is to manage player salaries so that productive players make more than others. This serves the

purpose of rewarding players and providing incentives. For defensive backs, bonuses are given for interceptions. Each interception earns a player $25,000. If Ken makes $600,000 in salary and gets seven interceptions per year, what is the percentage increase in his overall pay? (Round to the nearest percentage.)
Mathematics
2 answers:
Art [367]3 years ago
8 0

Answer: 29%

Step-by-step explanation:

Given : The annual salary of Ken = $600,000

The number of interceptions earned by her = 7

Each interception value = $25,000

Therefore, total money added to her income =7\times25,000=\$175,000

Now, the percentage increase in his overall pay is given by :-

P=\frac{\text{change in income}}{\text{Salary}}\times100\\\\\Rightarrow P=\frac{175000}{600000}\times100=29.1666666667\approx29\%

Hence,  the percentage increase in his overall pay is 29%.

kirza4 [7]3 years ago
3 0

Answer:

29%

Step-by-step explanation:


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Answer:

(1-\sqrt{2})a^2

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

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Thus,

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Consider isosceles triangle MSC. In this triangle

MS=MC=(\sqrt{2}-1)a.

The area of this triangle is

A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}

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A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2

5 0
3 years ago
Which of the following expressions could be used to represent "the difference between -12 and 10"?
pshichka [43]

Answer:

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Step-by-step explanation:

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A pretzel maker was interested in knowing the number of pretzels in each bag is sold. The results of the research are shown in t
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Answer:

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Step-by-step explanation:

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From the box and whisker plot attached ; the maximum value = 68

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bazaltina [42]

Answer:

The 95% confidence interval for the population mean is between 61.5 and 68.5.

Step-by-step explanation:

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The upper end of the interval is the sample mean added to M. So it is 65 + 3.5 = 68.5

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