Melody

Well it is wrtitten in LaTex, so I opened the [LaTex Formula] box and copied it in.

This is what it is supposed to look like.   

$$4\pi r^2$$     is the surface area of a sphere.

BUT I really don't know what this is about.

One of our higher level mathematicians might be able to help you more.  

$$L_s=\int_o^{\infty} L_{\lambda} d\lambda = 4\pi R_s^2 \int_0^{\infty} F_{\lambda} d\lambda$$

Hi Chris,

If I have to graph by hand this is the way I always do - as a build up from a parent graph.

I am still not as good at it as I would like to be but thinking about what should happen and then testing it with Desmos graphing calculator is an easy, efficient way to practice and to check one's logic.

Hi Mellie,

I have answered you one your original post.

Thanks for reposting as requested :)

Hi Mellie,

It is very late and I am not going to get a final answer tonight but I will show you where my thoughts are taking me.

I'm going to let 1 and 6 be the biased sides so 2,3,4 and 5 all have a chance of 1/6 of being rolled.

so the chance of throwing a 1,2,3 or 4 is      $${\frac{{\mathtt{4}}}{{\mathtt{6}}}}$$  =  $${\frac{{\mathtt{2}}}{{\mathtt{3}}}}$$

the chance of throwing a 1 is    $${\frac{{\mathtt{m}}}{{\mathtt{n}}}}$$       and the chance of throwing a 6 is      $${\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{m}}}{{\mathtt{n}}}}$$

now to roll a 7 these are the possibilities

$$\\P(1,6) =\frac{m}{n}\times\frac{n-3m}{3n}=\frac{m(n-3m)}{3n^2}\\\\
P(2,5) = \frac{1}{6}\times\frac{1}{6}=\frac{1}{36}\\\\
P(3,4) = \frac{1}{6}\times\frac{1}{6}=\frac{1}{36}\\\\
P(4,3) = \frac{1}{6}\times\frac{1}{6}=\frac{1}{36}\\\\
P(5,2) = \frac{1}{6}\times\frac{1}{6}=\frac{1}{36}\\\\
P(6,1) = \frac{n-3m}{3n\times}\frac{m}{n}=\frac{m(n-3m)}{3n^2}\\\\
NOW\\\\$$

$$\\\frac{4}{36}+\frac{2m(n-3m)}{3n^2}=\frac{47}{288}\\\\
\frac{32}{288}+\frac{2m(n-3m)}{3n^2}=\frac{47}{288}\\\\
\frac{2m(n-3m)}{3n^2}=\frac{15}{288}\\\\
\frac{m(n-3m)}{n^2}=\frac{15*3}{288*2}\\\\
\frac{m(n-3m)}{n^2}=\frac{5}{64}\\\\
$Now if we are lucky n will equal 8$\\\\
m(8-3m)=5\\\\
8m-3m^2=5\\\\
$Just by sight I can see that m=1 will work.$\\\\
$So one possibility, maybe the only possibility is $ n=8 \;and\; m=1\\\\
m+n=9$$

There is probably a much quicker way of doing this.  Plus I have not shown that this is the ONLY solutions. 

What is $F$  ?

The sides are numbered 1,2,3,4,5, and 6   So what is F  ?

I write a 'wrap' most nights.

It just wraps up the day as I see it.  I thank people for their answers and point out posts that may have specific interest for people.  I try to choose a variety of posts so that there is sometihing for most people who are interested in mathematics. 

I encouage people to recommend posts to me.  If you want to nominate a post for the wrap please copy the address and paste it into a private message for me.  I'd like that :)

The Lantern lights the way to the wrap - that is why it is a lantern.  :)

I put out the lantern post immediately after I write the wrap :)

If there is anything that I want to communicate to the forum as a whole I will put it in my wrap. 

So it is a communication/information post. :)

@@ End of Day Wrap   Thurs 9/4/15    Sydney, Australia    Time 12:30am  (really Friday) ♪ ♫

Hello everyone,

Fri 10/4/15

FTJ means "For the Juniors"

Thanks Natthada,

If you do these using the [calculation] button then it is easier for you and we can all see that the answer is correct.

It looks like this

$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{666\,432\,885\,543}} = {\mathtt{3\,332\,164\,427\,715}}$$

Yes, I made a stupid error.  Thanks Alan

I knew that there could easily be mistakes which is why I told the asker to check my answer.

I am sure your answer is correct.  I am not going to bother with correcting mine