CPhill

Here's the way I approached this one.......

In the square BEDA, the diagonal AE  represents all the x = y values from 0 to 1.  The segment FG will hold all the y values that are .25 greater than their associated x values at any point on this segment. So, since the square holds all possible (x,y)  values from 0 to 1 where x is the first  number chosen and y is the second one chosen, the only y values of interest lie in the triangle GBF. All the y values in here will exceed all their associated x values by more than .25. Note that the x value at H = 7.5. And, at point G, the associated y value will be the largest possible, i.e., 1.

And the area of this truangle = (1/2)(.75)(.75)sin90  = .28125  = 9/32

So,  the total area of BEDA encompasses all possible (x,y) values from 0 to 1. And the area of this square is just 1.

And triangle GBF is 9/32 of this area = probability that the second number chosen will exceed the first by more than .25 ..... (1/4).

I'm not great at these, Mellie....so Alan or Melody might correct my answer if it's incorrect, but anyway.....

The expected value  = (Amount won) x (probability of winning that amount)

So we have

($2)(1/6) + ($4)(1/6) + ($6)(1/6)  =

(1/6) ($2 + $4 + $6)  =

(1/6) ($12)  =

$2

We have:

4                                  

∫ x + 1  dx   =    (1/2)[16 - 4] + [4 - 2]  =  6 + 2  = 8 square units

2

OK, rosala.....will do....!!!

Rosala......strange coincidence.....the author, Markus Zusak, was born near where Melody lives - Sydney, Australia.......!!!!!

I havent heard of this book before, rosala.......after doing some research, it looks like a very good one......I'll have to get it from my local library  !!!!

Thanks for the info......

The "child" function is given by :

y = (8)cos3(x - pi/3)

See both here: 

Note......

(x - 5) / [ x^2 - 6x + 5] =

(x - 5) / [ (x - 1) (x -5) ] =

1 / (x - 1)

There will be a "hole" at x = 5......but the discontinuity at x = 1  isn't "removable".....there's a vertical asymptote at that point......

Here's the graph.......

Oops.... I made a slight error

It should be

(57/tan15 - 57/tan25) =   about 90.49m

I'll give you a pic in a second......

Remember that  we have the co-function identity....

csc(u)  = sec(pi/2 - u)   ....so....

If csc(x)  = 4

Then....sec(pi/2 - x)  also = 4