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I have 7 black and 3 white marbles. I will randomly pick 2 of them.

What is the probability that both of them are white?

How to solve this?

\(\begin{array}{|rcll|} \hline && \frac{3}{10} \cdot \frac{2}{9} \\ &=& \frac{1}{15} \qquad (6.\bar{6}\ \%) \\ \hline \end{array}\)

What is the mean of 80, 92, 77, 83, 72, 91, 84?

\(\begin{array}{|rcll|} \hline && 80 + \frac{0+12-3+3-8+11+4}{7} \\ &=& 80 + \frac{12-8+11+4}{7} \\ &=& 80 + \frac{27-8}{7} \\ &=& 80 + \frac{19}{7} \\ &=& 80 + 2\frac{5}{7} \\ &=& 82\frac{5}{7} \\ &=& \frac{579}{7} \\ &=& 82.7142857143 \\ \hline \end{array}\)

prove 
sec2x = (sec^2x*csc^2x)/ (csc^2x - sec^2x)

\(\begin{array}{|rcll|} \hline && \frac{\sec^2(x)\cdot \csc^2(x)}{\csc^2(x) - \sec^2(x)} \\ &=& \frac{1}{ \frac{\csc^2(x) - \sec^2(x)}{\sec^2(x)\cdot \csc^2(x)} } \\ &=& \frac{1}{ \frac{\csc^2(x)}{\sec^2(x)\cdot \csc^2(x)} -\frac{\sec^2(x)}{\sec^2(x)\cdot \csc^2(x)} } \\ &=& \frac{1}{ \frac{1}{\sec^2(x)} -\frac{1}{ \csc^2(x)} } \quad & | \quad \frac{1}{\sec(x)}=\cos(x) \quad \frac{1}{ \csc(x)}=\sin(x) \\ &=& \frac{1}{ \cos^2(x) - \sin^2(x) } \quad & | \quad \cos(2x) = \cos^2(x) - \sin^2(x) \\ &=& \frac{1}{ \cos(2x) } \quad & | \quad \frac{1}{\cos(x)}=\sec(x)\\ &=& \sec(2x) \\ \hline \end{array} \)

sec2x = (sec^2x*csc^2x)/ (csc^2x - sec^2x)    simplify the right side

(sec^2x*csc^2x)/ (csc^2x - sec^2x)   =

( 1/cos^2x * 1 / sin^2x)  / ( 1/sin^2x -  1/cos^2x)  =

Get a common denominator for the fractions in the denominator  = sin^2xcos^2x

[ 1/(cos^2x sin^2x)] / [ ( cos^2x - sin^2x) / ( sin^2x cos^2x) ]  =

Invert the fraction in the denominator and multiply by the numerator

[ 1/(cos^2x sin^2x) ] * (sin^2x cos^2x)  / [cos^2x - sin^2x]  =

1 / [ cos^2x  - sin^2x ] =

1/ cos2x    =

sec2x

Guest
1 hour ago     

is 69 96 upsidedown?

Well if the question was 3*3223 then you are definitely reading the calculator upside down.  :)

Don't forget, this is a calculator website. It can do this. But, if you do it...22/7 = 3.1428571428571429. Has the first three digits of Pi.

Here you go :   https://en.wikipedia.org/wiki/Fractal

The number of bacteria, y, present after x hours  =

y  =  2500*(2)^{(x / 3)}

On the calculator here you can just type... tan^-1(x) or you could put arctan. They both mean the same exact thing.

no one can answer that

Yes

What * 9   = 63.6       divide both sides by 9

What  =  63.6 / 9  =  106 / 15

Find the diameter and radius, multiply by pi factor

Thanks a lot.

(1/4) * (2/3) * (1/2) * (1/3)  =    2 / 72  =   1 / 36

20 * 70%  =

20 * .7  =

14  points to get a "C"

2√3  * 3   =

3 * 2√3  =

3 * 2 * √3  =

6√3

Base times Hight

21   yolo

Who are you

We have an infinite set of points that would work......the set of points would be all those on a circle with a radius of 13 and with a center  of (2,3)

8.5%   means that, every hour, there are  91.5% remaining from the previous hour.........so the function is  :

y =  650(.915)^x       where y is the number of bacteria present after x hours

The domain is [0, infinity (theoretically) )

The range is (0, 650]

Here's the graph : https://www.desmos.com/calculator/pwxvqmil5o

Half of the bacteria remains when y = 325........this occurs at about 7.8 hours from the start

Volume = Base times Height

Base = pi time radius squared