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Kryger [21]
3 years ago
15

2 plus 2 minus 6 plus 9 times 4

Mathematics
2 answers:
RideAnS [48]3 years ago
7 0
Answer: 28
2+2=4 4-6= -2+ 9= 7•4= 28
Dvinal [7]3 years ago
3 0

Answer:

28

Step-by-step explanation:

2+2=4

4-6=-2

-2+9=7

7x4=28

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Kim wants to determine a 90 percent confidence interval for the true proportion of high school students in the area who attend t
PolarNik [594]

Answer:

We need a sample of at least 752 students.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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The margin of error of the interval is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

90% confidence level

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

How large of a sample must she have to get a margin of error less than 0.03

We need a sample of at least n students.

n is found when M = 0.03.

We have no information about the true proportion, so we use \pi = 0.5.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.645*0.5

\sqrt{n} = \frac{1.645*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.645*0.5}{0.03})^{2}

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Rounding up

We need a sample of at least 752 students.

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4 years ago
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3 years ago
A. Use the​ one-mean t-interval procedure with the sample​ mean, sample​ size, sample standard​ deviation, and confidence level
PtichkaEL [24]

Answer:

a) (17.227, 22.773)

b) 2.773

c) 2.773

Step-by-step explanation:

Given:

Sample size, n = 25

Standard deviation, s = 4

Sample mean, x' = 20

Level of significance, a = 0.98 = 1 - 0.98 = 0.02

The degrees of freedom, df, for a t-distribution = n - 1 = 25 - 1 = 24

Using the t table, the Critical value = t_\alpha _/_2, _d_f = t_0_._0_2_/_2, _2_4 = t_0_._0_1_, _2_4 = 3.4668

Margin of error, E = t_\alpha _/_2, _d_f * \frac{\sigma}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

Limits of 98% confidence interval, we have:

Lower limit : x' - M.E = 20 - 2.773 = 17.227

Upper limit: x' + M.E = 20 + 2.773 = 22.773

Therefore, (17.227, 22.773) is 98% confidence interval.

b) Let's the margin of error by taking half the length of the confidence interval.

Since we are to use half the length of CI, we have:

M.E = \frac{22.773 - 17.227}{2} = 2.773

c)M.E = t_\alpha _/_2, _d_f *s/\sqrt{n}

= t * \frac{s}{\sqrt{n}} = 3.4668 * \frac{4}{\sqrt{25}} = 2.773

4 0
3 years ago
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