CPhill

a)

       2   4   6    8

1     3   5   7   9   

3     5   7   9  11

5     7   9   11 13

b)  The probability of a "7" is   3/12  = 1/4

This means that if 120 play the game, the school will collect   120 * 20p  = 2400p

But...1/4 of the people will win  =  30 people...and the school will have to spend 30 * 30p =

900p  for lollipops

So...the school will collect   2400 - 900  =   1500p

We can find the height using the Pythagorean Theorem

height  =  sqrt (10^2 - 6^2 )  =  sqrt (100 - 36)  = sqrt (64)  = 8 ft

Volume is

3.14/ 3  *  (6)^2 * 8  =

3.14 / 3  *  36 * 8 =

3.14 * 8 * 36/3  =

3.14 * 8 * 12  = 

3.14 * 96  =

301  ft^3

2 3/8  =   19/ 8  =  the diameter

The radius is (1/2) of this  =  19/8 * 1/2  =   19/16

The volume of the cone is

(1/3) pi * (19/16)^2 * 6  =

2 pi * (19/16)^2  =

2.82 pi   in^3

Call the side of the cube, S

The volume of the cube  is  S^3  =  1S^3

The volume of the cylinder is  pi * (S/2)^2 * S  =

pi * S^2 / 4  * S  =

(pi / 4) S^3

pi/4  < 1    

So....the cube has the greater volume

Circumference  =  2 * pi * radius

So we have

8.5 pi  =  2 * pi * r      divide both sides by 2 pi

8.5/2  =  r  =  4.25 in

Volume_cone  =  (1/3) pi * (radius)^2 * height     =

(1/3) * pi * (4.25)^2 * 15  =

5 * (4.25)^2 * pi  = 

90.31 pi   in^3

Volume_cylinder  =  3.14 * radius^2 * height  =

[Note that we are given a diameter of 5....the radius is 1/2 of this ]

3.14 * (5/2)^2 *12  =

3.14 * 25/4  * 12  =

3.14 * 75  =  

235.5 ft^3

First 6 primes  =

2, 3, 5, 7, 11, 13

Notice that each can choose any one of six numbers....so...the total number of possibilities is  6 * 6  = 36

But  the only odd sums occur when one of the chooses a 2 and the other chooses an odd prime....

So for   (Paul, Jessie)....these will occur  as  ( 2, odd)  or ( odd, 2) 

And there are 10 total sets where an odd sum occurs...so....the probability of an even sum is :

26/36  =    13/18

Here's my best shot.....

"Anchor"  A  anywhere on the circle

Now.....going clockwise from A  we have the following possibilities

A - B  - C - D

A - B  - D - C

A - C - D -  B

A - C - B - D

A - D - B - C

A - D - C - B

Note that they will intersect  only when A and B are separated by another point

So...the  probability is    2/6   =  1/3

Since order is not a concern, we have 5 ways to pick a vowel....21 ways to pick one consonant.... and 20 ways to pick the second consonant...so....

5 * 21 * 20   =

2100 ways

g(x)  = f(x^2)

g(-1)  = f( [-1]^2)  =  f(1)

And

g(9)  = f(9^2)  = f(81)

The interval  is  [ 81   - 1 ]   =   80 units wide