Melody

Hi Shaniab :)

Well I think that Guildenstein should stop going into Horatio's phone and that there should be no bet!!! LOL

When will  the last 2 digits be greater than 49

Well, the second last digit can be 5,6,7,8 or 9    That is 5 choices

and the last digit can be anything at all.

P(last 2 digits are great than 49)= (5/10)*1 = 0.5

So yes this has the same odds as a fair coin toss.   

Does that answer the question?

Or if you use the input box type in 

nCr(10,3)

Thanks Chris :)

Oh - it is really good for private messages because you cannot upload a file there :)

I snipped it straight into Gyazo.  It is quicker that way.  

Have you used Gyazo?  It is free and it works the same as a snip only instead of saving it onto your computer is  justs gets immediately uploaded. :)

All you can do is leave out the brackets - it is as simple as it gets already  :)

Well it is nice to know that I am capable of choosing a simple method sometimes.

This is very rare for me.    LOL

That is ok Chris - we were probably answering at the same time :)

$$\\x^2 + 5/9x=1/9\\$$

multiply both sides by 9 to get rid of the fractions

$$\\9x^2+5x=1\\
9x^2+5x-1=0\\
$now you can use the quadratic formula to solve it.$$$

just do it on the sit calc  (or do you need to do it by hand?)

Enter it as

(7+1/4)*(9 +3/4)*2