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2 years ago

Classify the figure. Identify its vertices, edges and bases.

Mathematics

1 answer:

vodka [1.7K]2 years ago

7 0

The direct variation function is y = 0.3x

<h3>How to determine the direct variation function?</h3>

The direct variation from y to x is represented as:

y = kx

When y = 1.5 and x = 5, we have

1.5 = 5k

Divide by 5

k = 0.3

Substitute k = 0.3 in y = kx

y = 0.3x

Hence, the direct variation function is y = 0.3x

Read more about direct variation at:

brainly.com/question/14254277

#SPJ1

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3 years ago

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3 years ago

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Using the z-distribution, it is found that the 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

<h3>z-distribution interval:</h3>

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

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

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

For this problem:

1215 samples, hence $n = 1215$ .

33% was mislabeled or misidentified, hence $p = 0.33$ .

95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

<h3>The lower limit of this interval is:</h3>

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 - 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3036$

<h3>The upper limit of this interval is:</h3>

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 + 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3564$

The 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

You can learn more about the use of the z-distribution to build a confidence interval at brainly.com/question/25730047

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3 years ago

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3 years ago

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-1

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