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Gnesinka [82]
4 years ago
8

Help, pls!!!

Mathematics
1 answer:
alina1380 [7]4 years ago
6 0

Answer:

Step-by-step explanation:

Let's start with this.  When do you think the mean would NOT be an appropriate measure of center?  Well, maybe your  data points are 1,2,4,1,5,3, 1,000,000.  If you took the mean of that it wouldn't be anywhere near any of the numbers, and would just be in between two.  So not in the center at all.  

The point is you want your data set well balanced.  You want about an equal number on one side as the other.  So let's look at yours.  

The furthest away from the middle has 11 on the left side and 12 on the right.  If youw ere weighing them on a scale and took the 11 and 12 as the weight, they would be pretty close to equal.  The next two are 21 and 23.  Less close, but still only two away.  The middle is the middle, so nothing to balance it out with.  If you look at it as a whole, the right side is 3 more than the left.  I would say 3 is still pretty close when you are looking at "weights" above 10.  So I would say mean is a pretty appropriate measure of center.  

Stats takes a lot of "gut feelings" like this.  Thinking, "yeah, these are pretty close" so you'll get the hang of it pretty soon.  

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If the denominator does not equal 0, the equivalent expression of \frac{14x^4y^6}{7x^8y^2} is \frac{2y^{4}}{x^{4}}\frac{x^{10} y^{14}}{729}

<h3>How to determine the equivalent expression?</h3>

The expression is given as:

\frac{14x^4y^6}{7x^8y^2}

Divide 14 by 7

\frac{14x^4y^6}{7x^8y^2} = \frac{2x^4y^6}{x^8y^2}

Apply the law of indices

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Evaluate the differences in the exponents

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