Melody

Excellent answer Mellie :)

This answer is going in the wrap :))

Hi Rosala :)

Don't worry Brittany, Chris and i won't bite you.  

We have seen your teeth and we know when we are out-toothed!

You should use your teeth on Zegroes, that is what I think !!

Thanks Alan :)

Thanks Alan,

I had to think about that one, but yes of course any x where sin x =0 will be a solution.

This is why it is best to use facorization to make sure you get ALL the solutions.

$$\\2sinxcosx=2sinx\\
2sinxcosx-2sinx=0\\
2sinx(cosx-1)=0\\
sinx=0\;\;OR\;\;cosx=1\\
x=0,\pi\;\; OR\;\;x=0\\
$The 2 solution are $x=0\:\;and\;\;x=\pi$$

Thanks Alan, I did it exactly the same way BUT

can you see any way of getting the total number of paths possible?

I suppose you could do it backwards.

Any individual path has the probability of   $${\frac{{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{6}}}}}$$      

So

$${\frac{{\mathtt{1}}}{{{\mathtt{2}}}^{{\mathtt{6}}}}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{1}} \Rightarrow {\mathtt{x}} = {\mathtt{64}}$$

So there must be 64 possible paths taken, 22 of these are favourable outcomes

so   P(ending at startpoint) =      $${\frac{{\mathtt{22}}}{{\mathtt{64}}}} = {\frac{{\mathtt{11}}}{{\mathtt{32}}}} = {\mathtt{0.343\: \!75}}$$

This is interesting -

I cannot remember ever using probability backwards like this to work out the number of possible outcomes.  

you just type it in.

for some you may need to put the x  (or whatever) in brackets.

ok thanks Alan :)

I do not understand what i am being asked to do :/

Well you could read the old posts I guess.

I put one or two of them in the wrap I think :)