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3 years ago

Find the median 23 ,15, 38 ,34 ,22, 2, 21 ,27 ,26 ,14, 32, 30, 19

2 answers:

3 0

First put them in ascending numerical order
2, 14, 15, 19, 22, 21, 26, 27, 23, 30, 32, 34, 38
Then start to eliminate numbers by pair (the farthest number from each side) until you get to 1 or two numbers left (in the very middle)
in this case you will be left with 26, which means that the median is 26

NikAS [45]3 years ago

3 0

23. You out the numbers in numerical order, once done you will start crossing them out from each side. If odd you will have a number from there, if not you will have a central number.

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2 years ago

Zia was driving his truck on the superhighway. His speed was recorded by the motorway camera between 6:00 am to 8:30 am. He cove

kodGreya [7K]

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2 years ago

Can someone solve this?

ladessa [460]

First question:

Recall that $\cos^2x+\sin^2x=1$ and $\sqrt{x^2}=|x|$ for all $x$ . So

$\sqrt{1-\cos^2x}=\sqrt{\sin^2x}=|\sin x|$

$\sqrt{1-\sin^2x}=\sqrt{\cos^2x}=|\cos x|$

For $0$ , we expect both $\cos x>0$ and $\sin x>0$ (i.e. the sine and cosine of any angle that lies in the first quadrant must be positive). By definition of absolute value, $|x|=x$ if $x>0$ .

So we have

$\dfrac{\sqrt{1-\cos^2x}}{\sin x}+\dfrac{\sqrt{1-\sin^2x}}{\cos x}=\dfrac{\sin x}{\sin x}+\dfrac{\cos x}{\cos x}=1+1=\boxed{2}$

making H the answer.

Second question:

C is always true, because the inequality reduces to x > y.

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3 years ago