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zalisa [80]
1 year ago
5

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.Refer to

Exhibit 6-3. What is the minimum weight of the middle 95% of the players?
Mathematics
1 answer:
Dafna11 [192]1 year ago
3 0

Given that

the weight of football players is distributed with a mean of 200 pounds and a standard deviation of 25 pounds.

And we need to find What is the minimum weight of the middle 95% of the players?

Explanation -

Using the Empirical Rule, 95% of the distribution will fall within 2 times of the standard deviation from the mean.

Two standard deviations = 2 x 25 pounds = 50 pounds

So the minimum weight = 200 pounds - 50 pounds = 150 pounds

Hence the final answer is 150 pounds.

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Find a compact form for generating functions of the sequence 1, 8,27,... , k^3
pantera1 [17]

This sequence has generating function

F(x)=\displaystyle\sum_{k\ge0}k^3x^k

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Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k

Take the derivative to get

\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k

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Take the derivative again:

\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k

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Take the derivative one more time:

\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k

\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k

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\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}

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iris [78.8K]

Answer:

a) y =2/3 x + 7.

b) x = -1

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Step-by-step explanation:

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b) y can take any value but x is always -1.

c) Slope m = (-1-3)/ (1 - -4)

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When x = -1 , y = 4 so using the slope intercept form of a line

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4 = -3/2 * -1 + b  where b = the y-intercept.

4 = 3/2 + b

b = 4 - 3/2 = 5/2.

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