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3 years ago

For which distributions is the median the best measure of center? select each correct answer.

Mathematics

1 answer:

emmainna [20.7K]3 years ago

7 0

Step-by-step explanation:

you have to add all of the together so like graph number 1 add all those numbers together and then divide by how much numbers there are and that is the median

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PLEASE HELP ME ASAP!

AleksandrR [38]

Ok so first we find the equation that equals one variable.

2y = -x + 9

3x - 6y = -15

We solve for y.

2y = -x + 9

y = -x/2 + 9/2

Then we plug in this y value into the other equation to keep only one variable so we can solve for it.

3x - 6y = -15

3(-x + 9/2) - 6y = -15

-3x + 27/2 - 6y = -15

-9y + 27/2 = -15

-9y = 3/2

-y = 3/18

y = -3/18

Then we plug in this numerical y-value into the first equation which we found out by solving an equation for y.

y = -x/2 + 9/2

-3/18 = -x/2 + 9/2

-84/18 = -x/2

-x = 9 1/3

x = -28/3

Your answer would be (-28/3, -3/18)

Hope this helps!

6 0

3 years ago

Read 2 more answers

Experts please please PLEASE answer this question ASAP! Due today.

quester [9]

The height of the given trapezoid is 7.5 m.

Step-by-step explanation:

Step 1:

The trapezoid's area is calculated by averaging the base lengths and multiplying it with the trapezoid's height.

The trapezoid's area, $A = \frac{b_{1}+b_{2}}{2} (h).$

Here $b_{1}$ is the lower base length and $b_{2}$ is the upper base length while h is the height.

Step 2:

In the given problem, $b_{1}=5 \mathrm{m}$ and $b_{2}=3 \mathrm{m}$ . Assume the height is h m.

The trapezoid's area $= 30.$

$30 = \frac{5+3}{2} (h), 30 = (4)(h).$

$h = \frac{30}{4} = 7.5.$

So the height of the given trapezoid is 7.5 m.

8 0

3 years ago

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$