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3 years ago

Subtracting 3xy2 from 8xy2 gives the same result as the expression

Mathematics

1 answer:

denpristay [2]3 years ago

5 0

The last one 7xy^2 - 2xy^2.

They both give you 5xy^2

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Question is in the picture:

cupoosta [38]

It’s 25 bc every 2 it goes up 5 dollars

4 0

2 years ago

Read 2 more answers

When 19 is added to one-quarter of a number, the result is 40. What is the number?

Kaylis [27]

Answer:

5.25

Step-by-step explanation:

1. 19 + 1/4x =40

2. 1/4x=40-19

3. 1/4x=21

4. x= 21/1/4

5. x= 5.25

6 0

3 years ago

Read 2 more answers

If MN = PQ AND PQ = RS then MN = RS

shtirl [24]

The answer for this problem would be : true

8 0

3 years ago

A quality control expert wants to estimate the proportion of defective components that are being manufactured by his company. A

tigry1 [53]

Answer:

A sample of 1032 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

A sample of 300 components showed that 20 were defective.

This means that $\pi = \frac{20}{300} = 0.0667$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

How large a sample is needed to estimate the true proportion of defective components to within 2.5 percentage points with 99% confidence?

A sample of n is needed.

n is found when M = 0.025. So

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

$0.025 = 2.575\sqrt{\frac{0.0667*0.9333}{n}}$

$0.02\sqrt{n} = 2.575\sqrt{0.0667*0.9333}$

$\sqrt{n} = \frac{2.575\sqrt{0.0667*0.9333}}{0.02}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.0667*0.9333}}{0.02})^2$

$n = 1031.9$

Rounding up

A sample of 1032 is needed.

6 0

2 years ago

Can cointegrated variables have different differencing functions

Bess [88]

If there are 2 variables and they are I(1) then u can apply enger and granger ... Recently I came across the 1st difference OLS model in a thesis

hope this helps!

8 0

3 years ago