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3 years ago

Which set of coordinates will provide the verticals for the translation of HJK 4 units down

Mathematics

1 answer:

Ierofanga [76]3 years ago

4 0

Answer:

oh ___ dam thats the ____ that I want

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The segments shown below could form a triangle.<br><br> O A. True<br> O B. False

andreyandreev [35.5K]

The answer would be A

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3 years ago

Read 2 more answers

Bob's Gift Shop sold 700 cards for Mother's Day. One salesman, Sebastian, sold 4% of the cards sold for Mother's Day. How many c

Nikolay [14]

Answer:

Step-by-step explanation:

700 x 4% = 700 x 0.04 = 28

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3 years ago

Please help solve for RN

sp2606 [1]

Answer:

I guess...

Step-by-step explanation:

Since QR and PN are parallels, I think what happens in MN and MP is proportional, so:

RN = x

8 is to 10 as

5 is to x

8 * x = 5 * 10

8x = 50

x = 50 / 8

x = 6.25

RN = 6.25

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2 years ago

A student participates in a Coke versus Pepsi taste test. She correctly identifies which soda is which four times out of six tri

Marysya12 [62]

Answer:

Step-by-step explanation:

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3 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