Math question

1a)
(x-5)^2+(8-x-5)^2 = 4
(x-5)^2+(3-x)^2 = 4
x^2-10x+25+ (9-6x+x^2) = 4
2x^2 -16x + 30 = 0
x^2 - 8x + 15 = 0
(x-5)(x-3) = 0
x = 5 or 3


2a)
there are total 90 no. 2-digit number (from 10 - 99)
and there are 9 from them can be divided by 11 (ie. 11,22,33,44,55,66,77,88,99)
P = 9/90
=1/10

b)(i)
Let x be the no. of red bands, 3x be the no. of green bands
x + 3x + 100 = 156
4x = 56
x = 14
red 14 no., green = 14*3 = 42 no.

(ii)
P(red) = 14/156
= 7/78

(iii)
P(red) = 10/81
ie. 10/81 = 14+y/ 156+y
1560+10y = 1134+81y
426 = 71y
y = 6

c)
P(open) + P(lock & open by 2 keys)
= 1/3 + [2/3*(1/5+ 4/5*1/5)]
= 1/3 + (2/3 * 9/25)
= 1/3 + 6/25
= 25+18/75
= 43/75

2007-06-21 23:48:49 補充：
c) should beP(open) P(lock & open by 2 keys)= 1/3 [2/3*(1/5 4/5*1/4)]= 1/3 (2/3 * 2/5)= 1/3 4/15= (5 4)/15= 3/5

1a)
唔知你有無打錯野,我跟你條式做
(x-5)^2+(8-x-5)^2 = 4
(x-5)^2+(-x+3)^2 = 4
x^2 - 10x + 25 + x^2 - 6x + 9 = 4
2x^2 - 16x + 30 = 0
x^2 - 8x + 15 = 0
(x - 5)(x - 3) = 0
x = 3 or x = 5

2a)
慢慢數都計到
2-digit number 有 10 至 99
其中 11, 22, 33 ... 99 divisible by 11

p(2-digit number chosen at random is divisible by 11)
= 9/90
= 1/10

b)
i)
(156 - 100) (3/4) = 42

ii)
number of red bands = 156 - 42 - 100 = 14
p(red) = 14/156 = 7/78

iii)
let y be number of red bands added
(14+y)/(156+y) = 10/81
81 (14+y) = 10 (156+y)
y = 6

c)
p(door unlocked) + p(門鎖左,第一次選錯 key,第二次選岩)
+ p(門鎖左,第一次就選岩 key)
= 1/3 + (2/3)(4/5)(1/4) + (2/3)(1/5)
= 9/15