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ANEK [815]
3 years ago
9

A college football coach has decided to recruit only the heaviest 15% of high school football players. He knows that high school

players’ weights are normally distributed and that this year, the mean weight is 225 pounds with a standard deviation of 43 pounds. Calculate the weight at which the coach should start recruiting players.
Mathematics
1 answer:
Alla [95]3 years ago
3 0

Answer:

The coach should start recruiting players with weight 269.55 pounds or more.                                                    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 225 pounds

Standard Deviation, σ = 43 pounds

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.15

P( X > x) = P( z > \displaystyle\frac{x - 225}{43})=0.15  

= 1 -P( z \leq \displaystyle\frac{x - 225}{43})=0.15  

=P( z \leq \displaystyle\frac{x - 225}{43})=0.85  

Calculation the value from standard normal z table, we have,  

P(z < 1.036) = 0.85

\displaystyle\frac{x - 225}{43} = 1.036\\\\x = 269.548 \approx 269.55

Thus, the coach should start recruiting players with weight 269.55 pounds or more.

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explain step by step how to solve. please point out any similarities in the steps used for each question
Mamont248 [21]
#1
m∠A=180 - 115 - 24 = 41°

By the law of sines:

b/sinB = a/sinA  ⇒  
b = (a*sinB)/sinA = (21*sin24°)/sin41° = (21*0.4067)/0.656 ≈ 13

c/sinC = a/sinA  ⇒  
c = (a*sinC)/sinA = (21*sin115°)/sin41° = (21*0.9063)/0.656 ≈ 29


#2
m∠C=180 - 119 - 27 = 34°

By the law of sines:

b/sinB = a/sinA  ⇒  
b = (a*sinB)/sinA = (13*sin119°)/sin27° = (13*0.8746)/0.454 ≈ 25

c/sinC = a/sinA  ⇒  
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#3
m∠C=180 - 57 - 37 = 86°

By the law of sines:

c/sinC = a/sinA  ⇒  
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Solve the following equation. Remember to check for extraneous solutions 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).
xxMikexx [17]

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-1, 2, 6

Step-by-step explanation:

We have to solve the equation as follows: 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).

Now, we have, \frac{1}{x-6} +\frac{x}{x-2} = \frac{4}{x^{2}-8x+12 }

⇒\frac{(x-2)+x(x-6)}{(x-2)(x-6)} = \frac{4}{x^{2}-8x+12 }

⇒\frac{x-2+x^{2}-6x }{(x-2)(x-6)} =\frac{4}{(x-2)(x-6)}

⇒\frac{(x-2)(x-6)}{x^{2}-5x-2 }=\frac{(x-2)(x-6)}{4}

⇒(x-2)(x-6)[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0

⇒ (x-2)(x-6) =0 or, [\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0

If, (x-2)(x-6) =0, then x=2 or x=6

If, [\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0, then x^{2} -5x-2=4

and (x-6)(x+1) =0

Therefore, x=6 or -1

So the solutions for x are -1, 2 6. (Answer)

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Answer:

(x, y) = (0, 4)

Step-by-step explanation:

The two lines intersect at their y-intercept: (x, y) = (0, 4).

6 0
3 years ago
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