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DochEvi [55]
3 years ago
9

Please help me solve this

Mathematics
1 answer:
damaskus [11]3 years ago
8 0
N=8
x= -19

Let me know if you need to see how to solve it
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Anybody good at geometry I need help with this worksheet
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Number 1: x≈4.9412824

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Number 3: x≈−5.04938819

Number 4: x=−1+2√6,−1−2√6x=-1+26,-1-26

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6 0
3 years ago
Teen obesity:
Law Incorporation [45]

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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3 years ago
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EleoNora [17]
B, since 13 ppl have 2 or less TVs while 12 have 3 or more
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