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aleksandr82 [10.1K]
3 years ago
11

The length of a rectangle is 20 units more than its width. The area of the rectangle is x4−100.

Mathematics
2 answers:
attashe74 [19]3 years ago
7 0

Answer:

<h2>x2−10 because the area expression can be rewritten as (x2−10)(x2+10) which equals (x2−10)((x2−10)+20).</h2>

Step-by-step explanation:

The area of the rectangle is x^{4} -100, which can be factored as (x^{2} +10)(x^{2} -10), because it's the difference of two perfect squares.

But, we know that l=20+w, where l is the length and w is the width.

Additionally, l=x^{2} -10+20 and w=x^{2} -10, which means l=x^{2} +10.

Therefore, the right answer is A.

nadezda [96]3 years ago
4 0

Answer:

1. x^2-10 because the area expression can be rewritten as (x^2-10)(x^2+10)which equals (x^2-10)((x^2-10)+20).

Step-by-step explanation:

Area of the rectangle =(x^4-100)

x^4-100=(x^2)^2-10^2\\$Applying difference of two squares: a^2-b^2=(a-b)(a+b)\\(x^2)^2-10^2=(x^2-10)(x^2+10)

Since the length of a rectangle is 20 units more than its width.

Width: x^2-10\\Length=x^2+10=x^2-10+20

The correct option is therefore 1.

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Step-by-step explanation:

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An SRS of 25 recent birth records at the local hospital was selected. In the sample, the average birth weight was x = 119.6 ounc
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Answer:

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

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\mu_{\bar X}= \mu = 119.6

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\mu_{\bar X}= \mu = 119.6

And now for the deviation we have this:

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So then the correct answer for this caee would be:

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Step-by-step explanation:

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