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shusha [124]
2 years ago
7

Which measure of central tendency is the most affected by outliers

Mathematics
1 answer:
aliina [53]2 years ago
7 0

Mean measure of central tendency is the most affected by outliers. Then the correct option is B.

<h3>What is Mean?</h3>

The mean is the straightforward meaning of the normal of a lot of numbers. In measurements, one of the markers of focal propensity is the mean. The normal is alluded to as the number-crunching mean. It's the proportion of the number of genuine perceptions to the absolute number of perceptions.

Mean measure of central tendency is the most affected by outliers.

Then the correct option is B.

More about the mean link is given below.

brainly.com/question/521501

#SPJ1

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Answer:

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Step-by-step explanation:

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3 years ago
Find the exact length of the curve. 36y2 = (x2 − 4)3, 5 ≤ x ≤ 9, y ≥ 0
IrinaK [193]
We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.

We are given: 36 y^{2} =( x^{2} -4)^3
We divide by 36 and take the root of both sides to obtain: y = \sqrt{ \frac{( x^{2} -4)^3}{36} }

Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: y =  \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}

Let's leave that for the moment and look at the formula for arc length. The formula is L= \int\limits^c_d {ds} where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.

Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows: ds= \sqrt{1+( \frac{dy}{dx})^2 } dx

As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.

As you can see from the formula we need to find dy/dx and square it. Let's do that now.

We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here x^2-4). More formally, we can let u=x^{2} -4 and then consider the derivative of u^{3/2}du. Either way, we obtain,

\frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2}

Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4}

This means that in our case:
ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx
ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx
ds=  \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx

Recall, the formula for arc length: L= \int\limits^c_d {ds}
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:

L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.

That is, [(\frac{1}{2}) ( \frac{9^3}{3}) -9]-([(\frac{1}{2}) ( \frac{5^3}{3}) -5]=( \frac{9^3}{6}-9)-( \frac{5^3}{6}-5})=\frac{290}{3}


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The answer to that question would be 15
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