2條f.3 maths題 exam用20分!

哇，好深wor! 我學校考試很少出這類題目。

62a.

Area of Region I = pi * x^2
Area of Region II = pi * 12^2 - pi * x^2 = pi (144 - x^2)
Area of Region III = pi * z^2 - pi * 12^2 = pi * (z^2 - 144)

pi * x^2 * 1 + pi (144 - x^2) * 0 +pi * (z^2 - 144) * (-4) = 0
pi * x^2 - 4 * pi * (z^2 - 144) = 0
pi * (x^2 - 4 * (z^2 - 144)) = 0
x^2 - 4 * (z^2 - 144) = 0
x^2 = 4 * (z^2 - 144)
x = root(4 * (z^2 - 144))
x = 2 root (z^2 - 144)

62b.
2 root (z^2 - 144) < 12
root (z^2 - 144) < 6
z^2 - 144 < 36
z^2 < 180
z < root 180
z < 13.4 (to the nearest tenth)

because z > 12 and z < 13.4
therefore 12 < z < 13.4
therefore The only possible integral value of z is 13.

x = 2 root (13^2 - 144) = 10

63a.
Possible area
= [5 - (2/2) - (2/2)]cm * [5 - (2/2) - (2/2)]cm
= 9 cm^2

63b.
Expected value of the number of tokens given
= 10 * (9cm^2 * 4 / 20^2) + 0 * [(20^2 - 9cm^2 * 4) / 20^2]
= 9/10

63c.
Expected value of the number of tokens gained or lost at the end of each game
= (-1 + 10) * (9cm^2 * 4 / 20^2) + -1 * [(20^2 - 9cm^2 * 4) / 20^2]
= 0.81 - 0.91
= -0.1

therefore No, I will not play this game.