f.3 maths ~ 15marks...

1) Let the radius of the base of the cone be r cm, h be the height of the cone :
(4/3)πr^3 = πr^2 * h
(4/3) r = h
The cone can be with a base with radius = r cm and the height = 4/3r cm.
2)We can cut the cone // to the base at x% of the height to get it.
The volume of the frustum
= The volume of the cone - The volume of the small cone with the height .
Let the volume of the cone = V
87.5%V = V - (1-x%)^3 * V
0.875 = 1 - (1-x%)^3
(1-x%)^3 = 0.125
1-x% = 0.5
x% = 50%
So cut the cone // to the base at the mid point of the height.
3)
Let G be the original number of green balls in the box :
P(getting a blue ball and a green ball)
= P(G , B) + P(B , G)
= [G/(5+G)] * [5/(4+G)] + [5/(5+G)] * [G/(4+G)]
= 10G / [(5+G)(4+G)] > 0.5
20G >(5+G)(4+G)
20G > 20 + 9G + G^2
G^2 - 11G + 20 < 0
(G - 8.7...)(G - 2.298...) < 0
2.298...< G < 8.7...
3 <= G <= 8 (G is a integer)
The original number of green balls in the box can be 3 or 4 or 5 or 6 or 7 or 8.






2009-08-01 00:32:06 補充：
Let the radius of the base of the cone be r cm, h be the height of the cone :

(4/3)πr^3 = (1/3)πr^2 * h

4r = h

The cone can be with a base with radius = r cm and the height = 4r cm.

2009-08-01 00:33:23 補充：
Corrections of Q1)

Let the radius of the base of the cone be r cm, h be the height of the cone :

(4/3)πr^3 = (1/3)πr^2 * h

4r = h

The cone can be with a base with radius = r cm and the height = 4r cm.