STPM-学校功课討論区

邵逸夫

哦~~原來是這樣！。。
慘咯~！！這三題都是今天測驗其中的題目。。
我好像錯了

原帖由 dunwan2tellu 于 7-8-2006 04:40 PM 发表


a)找 x 吗？
首先必须满足 x+2>0 , 和 x-2> 0 ==> x>2 ...(1)
之后用log 的rule， log2 (x+2)/(x-2) > 1 ==> (x+2)/(x-2) > 2
==> (x+2)/(x-2) - 2 > 0 ==> (6-x)/(x-2) > 0
==> (x-6)(x-2) > 0
==> x > 6 , x < 2 ....(2)

为什么我找到的是相反的orz，x<6,x>2

这一步(6-x)/(x-2) > 0
怎么去到==> (x-6)(x-2) > 0

dunwan2tellu

原帖由 yks 于 9-8-2006 09:41 AM 发表

为什么我找到的是相反的orz，x<6,x>2

这一步(6-x)/(x-2) > 0
怎么去到==> (x-6)(x-2) > 0

不好意思，我打错了。正确的答案是 2<x<6

a school offers japanese and arabic as two foreign languages.the probability that astudent who studies japanese also studies arabic is 1/4 and1/5 of the students who study arabic also studie japanese too. among the students, 1/5 of them do not study any of the two languages. what is the probability that a studebt in the school studies both languages? are the events of studying japanese and arabic independent?

dunwan2tellu

原帖由 ~ciyun~ 于 9-8-2006 09:04 PM 发表
a school offers japanese and arabic as two foreign languages.the probability that astudent who studies japanese also studies arabic is 1/4 and1/5 of the students who study arabic also studie japane ...

提示：
J=japanese , A=arabic , 根据题目，

P(A|J) = 1/4
P(J|A) = 1/5
P(A'n J') = 1/5

要我们找 P(A U J) 和 P(A) , P(J)

这里 P(A|J) = P(A n J)/P(J) ; P(J|A) = P(A n J)/P(A)

可以继续吗？

Dexer

the functions f,g,h are defined as

f(x)=  X+modulus X
       -----------
           2

g(x)=  { 2 ,     x < 0
       { 1+x     x >_ 0

h(x)=  {  x  ,    x < 0
       {  x^3  ,  x >_ 0

( >_ is greater or equal)


a) without using graph,show tat the composite function  fg is not continuous at x=0

b)  show that the composite function fh is differentiable at x=0

help,thanks

dunwan2tellu

a)
when x<0 , g(x) = 2 => f(g(x))=f(2)=2
when x>=0 , g(x)=x+1 => f(g(x))=f(x+1)=(x + |x+1|)/2 = (2x+1)/2 (因为x+1>0)

when x -> 0+ , f(g(x)) -> 1/2
when x-> 0- , f(g(x)) -> 2

so f(g(x)) not continuos at x=0

b)
when x<0 , h(x)=x =>f(h(x)) = f(x) = (x +|x|)/2 = (x-x)/2 = 0 (因为 x<0 => |x| = -x )
so (f(h(x))' = 0 (at x->0- )

when x>=0 , h(x)=x^3 , f(h(x)) = f(x^3) = (x^3+|x^3|)/2 = (x^3 +x^3)/2=x^3
so (f(h(x))'=3x^2 = 0 (at x->0+)

and (f(h(x))' = 0 (at x=0)
so (f(h(x)) is differentiable at x=0

Dexer

another one....

A competition between 2 boats,A &B,consists of a series of independent races and the competition will be won by the 1st boat that wins 3races.Each race will be won by either A or B,and their probability of winning are influenced by the weather.In bad weather,P A wins is 0.9,while in fine weather,P of A wins is 0.4 .For each race,the weather is either fine or bad,and the(P) of bad weather is 0.2.Given the 1st race is won by A,find the conditional (P) that:

a) A will win the competition

(P)=probability

dunwan2tellu

原帖由 Dexer 于 10-8-2006 07:12 PM 发表
another one....

A competition between 2 boats,A &B,consists of a series of independent races and the competition will be won by the 1st boat that wins 3races.Each race will be won by either  ...

不是很确定，应该是 0.25 .
A 在 bad weather 胜利是 0.9 . 那么在 bad weather 输就是 0.1 .
A = A win , X = bad weather , G = good weather ,
那么已知 P(A|X) = 0.9 , P(A|G) = 0.4 , P(X) = 0.2
那么 P(G) = 1-P(X) = 0.8

P(A n X) = P(A|X) * P(X) = 0.18
P(A n G) = P(A|G) * P(G) = 0.32
P(A win in bad OR good weather) = P(A n X) + P(A n G) = 0.50

表示每场胜利机会是 0.50 .
那么连胜三场 P(win win win| 1st win) = (0.5*0.5*0.5)/(0.5) = 0.25

total number of candidates in STPM is 58
mathematics 30
economics 48
history 34

mathematics and economics 23
economics and history 30
mathematics and history 13
(a) find the probability that randomly selected candidate takes
(i)only mathematics
(ii) mathematics and economics but not history
(iii)all three subjects
(b) what is the probability that a candidate takes history but not economics if he takes mathematics?

dunwan2tellu

原帖由 ~ciyun~ 于 11-8-2006 03:44 PM 发表
total number of candidates in STPM is 58
mathematics 30
economics 48
history 34

mathematics and economics 23
economics and history 30
mathematics and history 13
(a) find the probability t ...

我建议你用 venn diagram .

是不是这样？
P(AUAUC)=P(A)+P(B)+P(C)-P(ANB)-P(BNC)-P(CNA)+P(ANBNC)
用这个formula

dunwan2tellu

原帖由 ~ciyun~ 于 11-8-2006 04:03 PM 发表
是不是这样？
P(AUAUC)=P(A)+P(B)+P(C)-P(ANB)-P(BNC)-P(CNA)+P(ANBNC)
用这个formula

你要用这个 formula 也行。不然的话用 venn diagram 也行。

原帖由 dunwan2tellu 于 11/8/2006 16:05 发表


你要用这个 formula 也行。不然的话用 venn diagram 也行。

好理解谢谢

Dexer

原帖由 dunwan2tellu 于 11-8-2006 03:32 PM 发表


不是很确定，应该是 0.25 .
A 在 bad weather 胜利是 0.9 . 那么在 bad weather 输就是 0.1 .
A = A win , X = bad weather , G = good weather ,
那么已知 P(A|X) = 0.9 , P(A|G) = 0.4 , P(X) = 0.2
那 ...

答案是0.6875耶，这好难

Dexer

原帖由 dunwan2tellu 于 10-8-2006 07:06 PM 发表
a)
when x<0 , g(x) = 2 => f(g(x))=f(2)=2
when x>=0 , g(x)=x+1 => f(g(x))=f(x+1)=(x + |x+1|)/2 = (2x+1)/2 (因为x+1>0)

when x -> 0+ , f(g(x)) -> 1/2
when x-> 0- , f(g(x) ...

不是很明白(b) leh....
可否解释？？谢谢

dunwan2tellu

原帖由 Dexer 于 11-8-2006 05:04 PM 发表




不是很明白(b) leh....
可否解释？？谢谢

a function ,f(x) is differentiable at x=a if

(i) lim_(x->a+) f'(x) = lim_(x->a-) f'(x) = f'(a)

Dexer

噢！！谢谢你，明白了！！

可不可以在帮忙
integrate{( x^2)*(e^-x^3)}

dunwan2tellu

d/dx [ e^(-x^3)] = -3x^2 * e^(-x^3)

=> integrate [ x^2 * e^(-x^3) ] = -1/3 integrate [ -3x^2 * e^(-x^3)]
= -1/3 [e^(-x^3)] + C

Dexer

噢噢？？！！这么简单哦，哈哈。。。谢谢