Melody

you made a mistake anon  

2x-5=17

You have to ADD 5 to both sides (to get rid of the -5)

$$\\(5xy^3)^2\\
=5xy^3\times 5xy^3\\
=25x^2y^6$$

now my original interpretation was 2 red lights in a row and the others can be anything, including red.

I am going to have another go at this one.

If exactly 2 red ones there is  32 posibilities.

If there are 3 red ones then it gets harder  Mmm

I don't know - I'll think about it later     

Alan, what interpretation have you answered?

I have no idea what you are talking about :/

Hi Zac,

your answer looks great.  Did you use GeoGebra?

Thanks Shadow bolt

I think it is more helpful if you do these by hand.

Students have to know how to do these without a calculator.

Perhaps you could use the calc to check your answer and to show people how it can be used :)

There are several answers to this.

I could say it is by convention.

or I could say because it works.

Think about this.    3! is the number of ways you can line 3 people up in a queue

2! is the number of ways that you can line up 2 people    A then b  or B then A

1! is the number of ways you can line up 1 person.

How many ways can you line up 0 people.  Well 1 way.  There is only one way that you can put nobody in a queque

0!=1

I am sure that there are many more explanations but I guess the most important one is that it makes probablility questions work.  

Lame perhaps but true :)

I am not familiar with the monte-carlo simulations.

And like many of these questions this is open to different interpretations.  

Note: this answer has been changed so has my interpretation :)

I am going to interprete it to mean - Exactly 2 red lights in a row and none of the others are red

now they can be

1&2    or    2&3    or    3&4    or     4&5     So that is 4 ways

the number of ways that the 2 reds can come first is    1*1*2*2*2 = 8

So altogether there are    8*4 = 32 ways the you can get exactly 2 red lights in a rowand no other red lights

Now the total number of possibilities is    $${{\mathtt{3}}}^{{\mathtt{5}}} = {\mathtt{243}}$$

So I think that the answer is     $${\frac{{\mathtt{32}}}{{\mathtt{243}}}} = {\mathtt{0.131\: \!687\: \!242\: \!798\: \!353\: \!9}}$$

you canif you use the formula :)

I will give you both some points ;)