What is a similarity between the empirical rule and chebychev's theorem?

1 answer:

4 0

The Empirical Rule is used when data distribution is bell shaped, whereas Chebyshev's theorem is used for all distribution shapes

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How do i solve this

laila [671]

The ratio is 2:3 for the side lengths of the triangles.

so 6:9 is one side and 5:x would be the other side

2/3 of five is 3.33 and multiplying that by 2 gives you 6.67

4 0

3 years ago

Find the distance between the points (-2, 4) and (5, 4). 3 7 8

Lina20 [59]

Try this solution:
distance= $\sqrt{(5-(-2)^2+(4-4)^2} = \sqrt{49} =7.$

answer: 7

8 0

3 years ago

It's said to say my question, but it's in the pic, so I don't know what to say. But thanks

BARSIC [14]

Answer: A, or 0.2x

Step-by-step explanation:

To figure how expressions and percent's match up, replace the x with an easy number. For example, I use 4.

20% of 4 is 0.8, and 4 x 0.2 is also 0.8

Therefore, the expression to represent that percentage is A.

Also, the other answers were WAYYY off.

B, or x - 20 is flawed since x could be anything.

C, is also flawed in the same way.

And D makes -76.

A is the only sensible one and is proven to work.

Hope this helped!

-Nick

3 0

3 years ago

HELPPP !! LOOK AT PICTURE

WINSTONCH [101]

Answer:

A) (-infinity,-2] and D) [0,2]

Step-by-step explanation:

7 0

3 years ago

Find the critical numbers sin^2x+cosx 0 < x< 2pi

ryzh [129]

$\bf sin^2(x)+cos(x)\iff [sin(x)]^2+cos(x)\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=2sin(x)cos(x)-sin(x)\implies 0=2sin(x)cos(x)-sin(x)
\\\\\\
0=sin(x)[2cos(x)-1]\implies 
\begin{cases}
0=sin(x)\\
sin^{-1}(0)=\measuredangle x\\
\pi \\
----------\\
0=2cos(x)-1\\\\ \frac{1}{2}=cos(x)\\\\ cos^{-1}\left( \frac{1}{2} \right)=\measuredangle x\\\\ \frac{\pi }{3},\frac{5\pi }{3}
\end{cases}$

5 0

3 years ago

What is a similarity between the empirical rule and​ chebychev's theorem?

What is a similarity between the empirical rule and chebychev's theorem?