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3 years ago

Guys Help please asapppp

Mathematics

1 answer:

Zigmanuir [339]3 years ago

3 0

1. Isolate radical on one side of the equation.

2. Square both sides of the equation

( usually the first sentence is different but I think that’s the answer, hope that helped :) )

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Which variable did you plot on the x-axis, and which variable did you plot on the y-axis? Explain why you assigned the variables

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4 years ago

Mislabeled seafood In 2013 the environmental group Oceana (usa.oceana.org) analyzed 1215 samples of seafood purchased across the

Step2247 [10]

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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2 years ago

Is 0.07 x 10 equivalent 0.07

Nezavi [6.7K]

Answer:

It is the equivalent dose at 10 mm below a specified point of the body. H p (0.07) is the shallow dose equivalent, also referred to as skin dose equivalent or surface dose equivalent. It is the equivalent dose at 0.07 mm below the surface of the body. Currently, RadTarge II measures Hp (10).

Step-by-step explanation:

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alexgriva [62]

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