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LMAO......!!!!

Yes, you beat me fair and square    LOL   :)))

AND I DON'T CARE - TAKE THAT !!!

already answered

z-2(1-z)

= z-2--2z

= z-2+2z

= 3z-2

FV=1000(1+0.09/12)^9

I hope that you are having a really good day Alan 

I thought 'Up' was a fabulous movie too and I hardly ever watch movies.

I will admit that I can be partial to a good kid flick though :))

12ysquared(y/3+9x/y squared)4yto the 3rd-72

$$\\12y^2(\frac{y}{3}+\frac{9x}{y^2})*4y^{1/3}-72\\\\
(4y^3+108x)*4y^{1/3}-72\\\\
4y^{1/3}(4y^3+108x)-72\\\\
16y^{3.\dot{3}}+432xy^{0.\dot{3}}-72\\\\$$

Many thanks all!  I'm 21 (plus or minus an error bar) today! 

(I have seen the film Up - I think it's really good)

1 followed by 8 zeros

100,000,000

$$(Cos(x) + iSin(x))^9 = e^{ix}^9=e^{9xi}$$

Using Euler's formula       

I think Geno's answer is correct too.  :)

Maybe this will help

I will interprete this to mean

1300-10% of 1300

=1300-     130

=1170

That is not a problem Rosala :)

Hi Geno, I was looking at what you did.

If you choose the non US prize winners like you did that is 16*15 then you are saying that there are 240 permutaions of those 2 places.

then 1 us person must be chosen.  this can be done in 3 ways

now that US person can be 1st 2nd or 3rd     So 3 ways

240*3*3 = 2160   

You should not multiply this by 2 as all the possible permutations of the non US prize winners in relation to each other have already been accounted for.

To find "x," we just take the square root of the the other side......so we have

√[9^2 + 7^2] =

√[81 + 49] =

√130    .....this is the hypotenuse....

Yes that might help too - might help me too LOL  :))

....Or a bit more wine......

I do not think that anyone on this forum is really strong when it comes to probability.

I do however think that your knowledge has been improving very rapidly in this subject area.  Now all you really need is a bit more confidence.  

Sorry....I misread that one.......maybe I'm still dizzy....LOL!!!

Why are you talking like I wrote that last annonymous comment Chris?

I did not write that.    

However,

I just looked at the question and I agree with you and anon.  That is how I would have done it.

I am not saying it is necessarily correct but it does make sense to me.     

I gave Sorasyn 3 points for a good try.....and since I'm a mod....I have the power to giveth ....or taketh away.....

OK...thanks, Melody....you know....I'm NEVER too sure about my answers to any probability question....!!!!

I beat you, Melody........!!!!!

P(no more thatn 4 fours) = P(no fours)+P(1four)+P(2 fours)+P(3 fours)+P(4 fours)

$$\\\left(\frac{5}{6}\right)^{12}
\;\;+\;\;^{12}C_1\left(\frac{1}{6}\right)^{1}\left(\frac{5}{6}\right)^{11}
\;\;+\;\;^{12}C_2 \left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{10}\\\\
\;\;+\;\;^{12}C_3\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right)^{9}
\;\;+\;\;^{12}C_4\left(\frac{1}{6}\right)^{4}\left(\frac{5}{6}\right)^{8}$$

100 pesos = $2.26

So

2.26 / 100 = x / 2,000,000      cross-multiply

x = about $45, 200    {actually, about $45, 212}