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sladkih [1.3K]
3 years ago
10

Which statement describes the behavior of the function f (x) = StartFraction 2 x Over 1 minus x squared EndFraction?

Mathematics
2 answers:
Ghella [55]3 years ago
6 0

Answer:

It's B on edge

Step-by-step explanation:

Anastaziya [24]3 years ago
3 0

Answer:

Option B.

Step-by-step explanation:

The given function is

f(x)=\dfrac{2x}{1-x^2}

We have find the behavior of the function f (x) as  x approaches infinity.

The given function can be rewritten as

f(x)=\dfrac{2x}{x^2(\dfrac{1}{x^2}-1)}

f(x)=\dfrac{2}{x(\dfrac{1}{x^2}-1)}

lim_{x\rightarrow \infty}f(x)=lim_{x\rightarrow \infty}\dfrac{2}{x(\dfrac{1}{x^2}-1)}

Apply limit.

lim_{x\rightarrow \infty}f(x)=\dfrac{2}{\infty(\dfrac{1}{\infty^2}-1)}

lim_{x\rightarrow \infty}f(x)=0

The graph approaches 0 as x approaches infinity.

Therefore, he correct option is B.

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Answer:

95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

Step-by-step explanation:

We are given that 15% of a random sample of 300 U.S. public high school students were obese.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                        P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample % of U.S. public high school students who were obese = 15%

           n = sample of U.S. public high school students = 300

           p = population percentage of all U.S. public high school students

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

<u></u>

<u>So, 95% confidence interval for the population proportion, p is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                             of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u> = [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

  = [ 0.15-1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } } , 0.15+1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } } ]

  = [0.110 , 0.190]

Therefore, 95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

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