Melody

It is good to see you back Max 

Ok, thanks, I think I get it.

I hope I can use it appropriately in the future. :)

I would use probablility contour mapping to answer this question.

A line segment of length 5 is broken at two random points along its length. What is the probability that the shortest of the three new segments has length longer than 1?

Let the lengths be x,   5-y and  y-x

Note that when you add all these you get a lenght of 5  

so

0

Lets graph that.

The darkest area represents the entire sample space.   How much area does it have?

Area = 0.5*5*5 = 12.5 units squared

Now how much of this sample space will have the shortest side bigger than than 1 unit?

mm

I think it is a lot easier to graph the sample space where one side is less than 1.

These are mutually exclusive and exhaustive events so their probabilities will be add up to 1

The lengths are  x,   5-y and  y-x

When will a length be LESS than 1 unit

\(0

The dark shaded area is where there IS a side that is less then one unit.

SO the triangle in the middle is where there is NO SIDE less than 1 unit.

Area of trianlge in middle = 0.5*2*2 = 2 units squared.

So the probability that the shortest length will be longer than 1 unit is  \(\frac{2}{12.5}= \frac{4}{25}=\frac{16}{100}=16\%\)

Thanks guest.

I don't understand the 4!

They do not have to be all the same ???

\(f(x) = \begin{cases} k(x) &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases}\)

Mmm.

The first thing I notice is that f(2)=2  so this point will be its own inverse (it lies on the line y=x)

That helps becasue it gives me less things I have to think about.

\(\text{This will work if k(x) is the inverse of   } \\f(x)=2+(x-2)^2 \;\; where\;\ x\le 2\)

So I need to find the inverse:

let\(y=2+(x-2)^2    \qquad    x\le 2, \quad y\ge 2\\ \text{I need to make x the subject}\\ y-2=(x-2)^2\\ \pm\sqrt{y-2}=x-2\\ \pm\sqrt{y-2}+2=x\\ \text{But x has to be less then 2 so it must be the neg square root}\\ x=-\sqrt{y-2}+2 \qquad    x\le 2, \quad y\ge 2\\ \text{Now for the inverse swap the x and y values}\\ y=-\sqrt{x-2}+2 \qquad x\ge2, \quad y\le 2 \)

so

\(f(x) = \begin{cases} -\sqrt{x-2}+2 &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases}\\ \text{will be its own inverse}\)

so

\(g(x)=-\sqrt{x-2}+2 \qquad\text{where}\;\;x>2\)

Here is the graph:

Thanks Ginger.

I am thinking about the logic behind Ginger's answer.

We have 10 different colored dice.

How many ways can I chose 2 (these will be 1)  10C2

Now there are 8 dice left, how many ways can i choose 2 (these will be 3)  8C2

Now there are 6 dice left, how many ways can I choose 2 (these will be 5)  6C2

Now there are 4 dice left. Each can be 2,4 or 6 so there are  3^4 posible combinations

So total number of possible combinations is   10C2 * 8C2 * 6C2 * 3^4 

nCr(10,2)*nCr(8,2)*nCr(6,2)*3^4 = 1530900    Cool !!  

This works and that is great.

I am always reluctant to use logic like this because sometimes it works and sometimes it doesn't and I can never understand why that is.  Maybe I understand now idk but historically I had a lot of problems with this. (historically for me it often led to double counting)

Thanks very much Heureka,

Max has asked me if I can help him to understand your answer so that is why I am posting here.

Firstly max, every normal door has two sides. 

Here is a very simple example.  Every room has 2 doors.  Here, each room has 2 doors. So the sum of the doors in the rooms is 4. But as you can see, there are 3 doors.  If I want to count both sides of all doors then I can do that by thinking of the outside of the house as one big room.  There are 2 doors to the grand outdoors. So there must be 4+2=6 door sides.

That is there are  6/2 = 3 doors altogether in the house.

The number of door sides in any house is the sum of all the doors in all the rooms + the number of doors that are on the outside of the house.  (you can think of the outside of the house as one big room)

And, the number of doors in any house is half the number of door sides (since every door has 2 sides)

So 

the number of doors on the outside + the number of doors in each room = 2 * the total number of doors that the house has.

We are told that the number of doors in each room is even. And a positive even number can be expressed as 2N (where N is a positive integer) so

the number of doors on the outside +  2N = 2 * the total number of doors that the house has.

rearranging we get...

the number of doors on the outside = 2 * the total number of doors that the house has - 2N

the number of doors on the outside = 2 *(the total number of doors that the house has - N)

So if every room has an even number of doors the number of doors on the outside of the house is even.

As Heureka stated, closet doors do not count :)

Daniel, writing the answer 24 does not teach anyone anything. 

We are here to teach people :)

John is 3/4 the age of James. Three years ago, John was 5/7 the age of James. How old is James?

First thing I am going to do is change the names cause they are too confusing.

James becomes his twin sister Amy.

John is 3/4 the age of Amy. Three years ago, John was 5/7 the age of Amy. How old is Amy?

Let John's age now be J.  Three years ago his age was J-3

Let Amy's age now be A.        Three years ago her age was A-3

John is 3/4 the age of Amy.

so

\(J=\frac{3}{4}*A\\ J=\frac{3A}{4}\)

Three years ago, John was 5/7 the age of Amy.

\(J-3=\frac{5}{7}*(A-3)\\ J-3=\frac{5A-15}{7}\\ J=\frac{5A-15}{7}+3\\ \)

so

\(\frac{3A}{4}=\frac{5A-15}{7}+3\\ 28*\frac{3A}{4}=28*[\frac{5A-15}{7}+3]\\ 21A=4(5A-15)+84\\ 21A=20A-60+84\\ A=24\\ \)

Amy and therefore James, is 24 years old.