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Licemer1 [7]
3 years ago
5

PLEASE PLEASE PlEASE I NEED HELP!!!

Mathematics
1 answer:
Eva8 [605]3 years ago
6 0

Answer:

As the property of angle on circle:

Arc DF = 2 x angle DEF = 2 x 72 = 144 deg

=> Option B is correct

Hope this helps!

:)

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Answer:

Step-by-step explanation:

Given that:

the sample proportion p = 0.39

sample size = 100

Then np = 39

Using normal approximation

The sampling distribution from the sample proportion is approximately normal.

Thus, mean \mu _{\hat p} = p = 0.39

The standard deviation;

\sigma = \sqrt{\dfrac{p(1-p)}{n} }

\sigma = \sqrt{\dfrac{0.39(1-0.39)}{100} }

\sigma = 0.048

The test statistics can be computed as:

Z = \dfrac{{\hat _{p}} - \mu_{_ {\hat p}} }{\sigma_{\hat p}}

Z = \dfrac{0.3 - 0.39 }{0.0488}

Z = -1. 8 4

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(b)

Here;

the sample proportion = 0.39

the sample size n = 400

Since np = 400 * 0.39 = 156

Thus, using normal approximation.

From the sample proportion, the sampling distribution is approximate to the mean \mu_{\hat p} =  p = 0.39

the standard deviation \sigma_{\hat p} = \sqrt{\dfrac{p(1-p)}{n} }

\sigma_{\hat p} = \sqrt{\dfrac{0.39 (1-0.39)}{400} }

\sigma_{\hat p} =0.0244

The test statistics can be computed as:

Z = \dfrac{{\hat _{p}} - \mu_{_ {\hat p}} }{\sigma_{\hat p}}

Z = \dfrac{0.3 - 0.39 }{0.0244}

Z = -3.69

From the z - tables;

P (\hat p \le 0.3 ) = P(z \le -3.69)

\mathbf{P (\hat p \le 0.3 ) = 0.0001}

(c) The effect of the sample size on the sampling distribution is that:

As sample size builds up, the standard deviation of the sampling distribution decreases.

In addition to that, reduction in the standard deviation resulted in increases in the Z score, and the probability of having a sample proportion  that is less than 30% also decreases.

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1.37x + 1.08y = 16.36
x + y = 13

x = 13 - y
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