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4 years ago

I need help asap will mark you brain list. Show your work

Mathematics

1 answer:

Anna11 [10]4 years ago

3 0

Answer:

1/25
1/9

Step-by-step explanation:

<u>Solving exponents</u>

a. 5^-2 = 1/5^2 = 1/25
b. (3^2)^3/3^8 = 3^6/3^8 = 1/3^2 = 1/9

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A parallelogram<span> is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. Plotting the points given above would give as a parallelogram where line made by AB is parallel and congruent with line CD. Hope this answers the question.</span>

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3 years ago

Given the equation 2x + 4 = 4x − 2, select the reasoning that correctly solves for x.

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Subtract 4, subtract 2x, then divide by -2

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4 years ago

Margo traveled 528.8 miles on her trip. Write a statement that compares the values of the ones and tenths digits

gavmur [86]

Answer:

In the number 528.8 the value of 8 in the ones place is 10 times the value of 0.8 in the tenths place

Step-by-step explanation:

we have

The number 528.8 miles

Rewrite the number in digits value

$528.8=500+20+8+0.8$

The value of the ones digit is 8 ones or 8

The value of the tenths digit is 0.8 tenths or 0.8

Find the quotient

8/0.8=10

therefore

In the number 528.8 the value of 8 in the ones place is 10 times the value of 0.8 in the tenths place

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3 years ago

30-60-90 special right triangles

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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