Melody

That would depend on the terrain.

I think your question is incomplete

If the whole trip is flat then the answer is 120m/hr

Very nice Tiggsy.

It is good to see you here :)

Thanks Alan,   

I want to find the smallest distance between  $(x,\frac{x^2-8}{2}) \qquad and \qquad (0,0)$

The distance between these is

 $d=\sqrt{(x-0)^2+(\frac{x^2-8}{2}-0)^2} \\d= \sqrt{x^2+(\frac{x^2-8}{2})^2}$

We need to minimize that expression and solve for x

I will get the same result if I minimize 

 $x^2+(\frac{x^2-8}{2})^2 \qquad or \qquad 4x^2+(x^2-8)^2$

Let

$T=4x^2+(x^2-8)^2\\ \frac{dT}{dx}=8x+2(x^2-8)^1*2x\\ \frac{dT}{dx}=8x+4x(x^2-8)\\ \frac{dT}{dx}=8x+4x^3-32x\\ \frac{dT}{dx}=4x^3-24x\\ \frac{dT}{dx}=4x(x^2-6)\\ \frac{dT}{dx}=4x(x-\sqrt 6)(x+\sqrt 6)\\~\\ \frac{dT}{dx}=0 \;\;\;\;when\;\;\;\;\ x=0,\;\;x=\pm\sqrt6$

When x=0 

     $d= \sqrt{x^2+(\frac{x^2-8}{2})^2}\\ d= \sqrt{16}\\ d=4$

When   $x=\pm\sqrt6$

$d= \sqrt{x^2+(\frac{x^2-8}{2})^2}\\ d= \sqrt{6+(\frac{6-8}{2})^2}\\ d= \sqrt{6+1}\\ d=\sqrt7\\ d\approx 2.64$

So it looks to me like the min distance is  when x= $\sqrt7 \quad \text{and it occurs when }\quad x=\pm\sqrt6$

I'll look at the second derivative for verification.

$\frac{dT}{dx}=4x^3-24x\\ \frac{d^2T}{dx^2}=12x^2-24\\ \frac{d^2T}{dx^2}=12(x^2-2)\\ $

When    $x=\pm \sqrt 6\\ \frac{d^xT}{dx^2}= 48>0 \quad \text{So minimum}$

When    $x= 0\\ \frac{d^xT}{dx^2}= -24<0 \quad \text{So maximum}$

So I get the minimum distance to be    sqrt7

And here is the pic

LaTex:

T=4x^2+(x^2-8)^2\\
\frac{dT}{dx}=8x+2(x^2-8)^1*2x\\
\frac{dT}{dx}=8x+4x(x^2-8)\\
\frac{dT}{dx}=8x+4x^3-32x\\
\frac{dT}{dx}=4x^3-24x\\
\frac{dT}{dx}=4x(x^2-6)\\
\frac{dT}{dx}=4x(x-\sqrt 6)(x+\sqrt 6)\\~\\
\frac{dT}{dx}=0 \;\;\;\;when\;\;\;\;\ x=0,\;\;x=\pm\sqrt6

a)  (4C2)^4 = 3^4 * 4^4 =  81*256 = 

b)  (4C3)^4 = 4^4 = 256

That depend on which child in the queue you are talking about.

The first 2 children can choose any that they want.

After that it dpends on what the ealier children have.

what you have here is 2 scoops of each flavour, distributed one each to 6 different shildren.

AABBCC   that is the scoops.

How many ways can that be arranged.

6!/(2!2!2!) = 6!/8 = 3*5*6 = 90 ways that the icecream can be given out.

I'm sorry but I do not understand what you have done here.

You are welcome :)

Of course I haven't proven anything.

It is just an assumption that the pattern will continue.

I get 4

Miyu is giving out 9 identical chocolates to her 7 friends, including Dhruv.  

Put the friends in order Dhruv can go first but it really doesnet matter

There will be  (9+7-1) C8 = 15C8  ways to distribute the chocolates

Now let Dvruv have her 2 chocs.

There are 6 friends left and 7 choc let so that makes     12C6 ways

So the likelyhood that Dvruv gets exactly 2 chocs is   

12C6  divided by  15C8  = 924/6435 = 28/195  which is approx   14.35% chance