Melody

Is Mellie allowed to do that ????

I have never seen a binomial used the way Mellie has used it - are  you allowed to do that ??

I've only  ever used it where p+q=1     

 If you can that would be a really good thing to know.  ://

I am not convinced that you are allowed to do what you did either Alan.

You have multiplied them like as if the 2 probabilities are independant of each other but they are not independent of each other. ://

I was just trying to work it out on my own and I had the same problem as you Alan.

The meaning is not at all clear.

Mmm

Thanks Zac  

Mmm,

I was thinking about this and I think one of the triangles has to be the biggest and they are all equally likely do the probability of ABP being the biggest is 1/3

I also decided to follow Chris and try a visual approach.

You want the probability of triangle 1 being the biggest.

If P is in 1b or 2a then   triangle 3 is the biggest

If P is in  2b or 3a then triangle 1 is the biggest

If P is in 3b or 1a then triangle 2 is the biggest.

So

The probability that triangle 1 is the biggest is 1/3

That is what I think anyway :)

Could another mathematician please comment on this :)

-2.8=-1(c-0.65)     mult bot sides by =1

2.8=c-0.65

now just add 0.65 to both sides 

$$\\a^2+b(b+1)a+b^3\\
a^2+ab(b+1)+b^3\\
a^2+ab^2+ab+b^3\\$$

Thanks Alan,

Wouldn't you have to say as    $$x\rightarrow 0$$     since h is the denominator and so it cannot equal 0 ?

You have presented this beautifully Zac :))

There is a little thing I would like you to consider. 

If you multiply both sides by 3 right at the very beginning then the fraction disappears immediately.

With some equations this makes the problem much simpler :)

If you mean [sqrt3(48)] / [sqrt3(10)]

then the sqrt3 cancels out leaving you with 48/10=4.8

Thanks Badinage,

I usually do it that way too.   But be warned ..... :)

I entered   sin(acos( 0.6372))

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.637\: \!2}}\right)}\right)} = {\mathtt{0.770\: \!698\: \!488\: \!386}}$$

Only thing you need to be careful of is the sign.

Cos is pos in the 1st and 4th quad.   Sin is pos in the 1st but neg in the 4th 

so

the answer can be     + or -   0.7707