STPM-学校功课討論区

lavendar_o5

有人帮忙解第457楼吗？

flash

原帖由 lavendar_o5 于 23-10-2007 10:34 AM 发表



请问如何算呢？

1. 7C5(0.25)^5(0.75)^2 + 7C6(0.25)^6(0.75) + (0.25)^7 = 0.013

2. 0.2*0.013 + 0.8 [7C3(0.75)^3(0.25)^4 + 7C4(0.75)^4(0.25)^3 + ....(0.75)^7] = 0.792

[ 本帖最后由 flash 于 24-10-2007 06:43 PM 编辑 ]

dunwan2tellu

原帖由 lavendar_o5 于 21-10-2007 09:23 AM 发表
A family consisting of the father and seven other members. Each time a holiday trip is suggested, the father and the family members must accept or reject it. The probability that the father will accept a holiday trip is 4 / 5 . The other members make their decision independently, but the probability that each member agrees with the father is 3 / 4 . A holiday trip is agreed by the family members when the number of members who accept it exceeds the number of members who oppose it, or if the number of members who accept it is equal to the number who oppose it and the father is among those who accept it.

1        Find the probability that a holiday trip is agreed by the family members if the father oppose it
2        Find the probability that a holiday trip is agreed by the family members.

1. P(agree | father opose) = P(agree n father opose)/P(father opose)

P(father opose) = 1/5 ;
P(agree n father opose) = P(5 to 7 members agree but father opose)
                        = Sum_{k=5 to 7} (1/5)*((7Ck * (1/4)^k * (3/4)^(7-k))  
                        = ....
                        = A

P(agree|father opose) = 5A ...

2 .P(agree) = P(agree n father opose) + P(agree n father agree)

P(agree n father agree) = P(3 to 7 member agree and father agree)
                        = sum_{k=4 to 7} (4/5)*((7Ck * (3/4)^k * (1/4)^(7-k))
                        = B

=> P(agree) = A + B

**flash 兄快了一步

Leong13

為什麼沒有人幫我解４５０樓的問題

Leong13

1. the assembly of the student of a school
begins 15 minutes after 8 o'clock every
school day.  a male student changes his
shoes once he reaches the school. the
time taken in minutes, from the time he
arrives at school, to changes his shoes,
can be taken to have a normal distribution
with mean 2 and standard deviation 1.5.
the time, in minutes, after eight o'clock,
the student arrivess at school can be
taken to have a normal distribution with
mean and standard deviation 2. the time
taken to change his shoes can be asumed
to be independent of th etime he arrives
at the school.

a. show that , on a day selected at random,
   the probability that the student has
   changed his hsoes before assembly begins
   is 0.885

lavendar_o5

At time 0700 hours, the position vectors of boats A, B and C are 2i - 3j, 6i + 5j and -2i respectively. The velocities of boats A and B are 7i + j and i-7j respectively. Boat C moves in such a way that its velocity relative to boat A is parallel to the vector -5i and its velocity relative to boat B is parallel to the vector i + 8j. Distances are given in km and veloicities in km / h.

Find the time when boats A and B collide, and the positions of boats A and B at this instant.

lavendar_o5

有人帮忙解上题吗？??

zfc

高手们请帮我解决这一题：
Factorise 2-x+2x^2-x^3 completely. Hence solve the inequality 2(1+x^2) < x(1+x)^2.

请检查我的答案：
我factorise到(x-2)(-x^2-1)
inequality 就做到
-x^3-x+2 < 0  过后我发现他跟上面要factorise的cubic equation有点关系，所以
(x-2)(-x^2-1)-2x^2 < 0  过后就不会做了，请问我是否做错了？

img3nius

我想问
statistik里面的 khi kuasa dua (x^2)的
它的v(darjah kebebasan ) = bilangan kelas - bilangan sekatan
这里的sekatan 是什么意思?
他的数值取决于什么?

还有 顺便问
今年还有马来文的考卷吗?
还是只有英文了?

dunwan2tellu

At time 0700 hours, the position vectors of boats A, B and C are 2i - 3j, 6i + 5j and -2i respectively. The velocities of boats A and B are 7i + j and i-7j respectively. Boat C moves in such a way that its velocity relative to boat A is parallel to the vector -5i and its velocity relative to boat B is parallel to the vector i + 8j. Distances are given in km and veloicities in km / h.

Find the time when boats A and B collide, and the positions of boats A and B at this instant.

A 和  B 不会  collide , .
Let position of A ,B after time t be P_a,P_b, 那么
P_a = (2,-3) + t(7,1)
P_b = (6,5) + t(1,-7)

如果会  collide , 那么 P_a = P_b
=> t(6,8) = (4,8)
无解

高手们请帮我解决这一题：
Factorise 2-x+2x^2-x^3 completely. Hence solve the inequality 2(1+x^2) < x(1+x)^2.

请检查我的答案：
我factorise到(x-2)(-x^2-1)
inequality 就做到
-x^3-x+2 < 0  过后我发现他跟上面要factorise的cubic equation有点关系，所以
(x-2)(-x^2-1)-2x^2 < 0  过后就不会做了，请问我是否做错了？

直接 factorise -x^3 - x + 2 = - (x-1)(x^2 + x +2)
所以 -(x-1)<0 => x>1

我想问
statistik里面的 khi kuasa dua (x^2)的
它的v(darjah kebebasan ) = bilangan kelas - bilangan sekatan
这里的sekatan 是什么意思?
他的数值取决于什么?

还有 顺便问
今年还有马来文的考卷吗?
还是只有英文了?

bilangan sekatan = 你在做 chi-square test 时有几个限制

chi-square test 里我们假设 Sum of observed data = Sum of expected data , 这就是一个限制，所以是一个 bilangan sekatan.所以 degree of freedom 要 -1

在 m x n contigency table 里，degree of freedom = (m-1)(n-1)

因为每一个 row 和每一个 column 的 sum 都要 fix 着，这就是限制，有几个限制？row 有  m 个所以 有 m 个 的 限制. Column 有  n 个 所以 n 个 限制. 他们之间有一个共同的 sum , 所以要 -1
total restriction = m+n-1

degree of freedom = mn - [m + n -1] = (m-1)(n-1)

STPM further math 到去年为止还是双语出题。今年则不知

zfc

原帖由 dunwan2tellu 于 5-11-2007 04:12 PM 发表
直接 factorise -x^3 - x + 2 = - (x-1)(x^2 + x +2)
所以 -(x-1)<0 => x>1

那 x^2 + x + 2不用管它了吗？为什么呢？

hamilan911

因为 x^2+x+2 = (x+0.5)^2 + 1.75  > 0 for all values of x

Leong13

請幫我solve這題，
(modulus x + 3) / ( modulus x - 3 ) >5

TCLGT

原帖由 Leong13 于 9-11-2007 11:01 PM 发表
請幫我solve這題，
(modulus x + 3) / ( modulus x - 3 ) >5

(modulus x + 3) / ( modulus x - 3 ) >5
(modulus x + 3-5modulus x+15) / ( modulus x - 3 )>0
(-4modulus x + 18) / ( modulus x - 3 )>0

-4modulus x+18>0
x<18/4
x>-18/4

( modulus x - 3 )>0
x>3
x<3

therefore {x: -18/4<x<-3 , 3<x<18/4}

Leong13

如果我的答案這樣些能嗎？

{ x:  -18/4 < x <18/4 u x > 3 u x < -3 }

TCLGT

不能。。你可以检查答案。。
你放x=0时，(modulus x + 3) / ( modulus x - 3 )<5
这不符合他要的eqn
同样的，你放x>3时，你会发现(modulus x + 3) / ( modulus x - 3 )<5。。

dunwan2tellu

原帖由 Leong13 于 10-11-2007 11:24 AM 发表
如果我的答案這樣些能嗎？

{ x:  -18/4 < x  3 u x < -3 }

U 只是用在 set 里的写法
应该是

{x: -18/4 < x < -3} U {x:3<x<18/4} 比较真确

或

{x:3<|x|<18/4}

[ 本帖最后由 dunwan2tellu 于 10-11-2007 03:20 PM 编辑 ]

Leong13

我找到values可是我不知到怎樣放，可一解釋嗎？

TCLGT

找到value后，画number line来决定+-

                                                                  ——|————|————|————|——
                                                                       -9/2        -3               3              9/2
-4modulus x+18                                        -         +                +             +            -
( modulus x - 3 )                                        +        +                -              +            +
(-4modulus x + 18) / ( modulus x - 3 )    -          +                -              +            -

因为(-4modulus x + 18) / ( modulus x - 3 )>0
所以{x: -18/4<x<-3 , 3<x<18/4} 或{x: -18/4 < x < -3} U {x:3<x<18/4}

chwk87

请问要怎样kamir (1/(x^3+1))?谢谢

[ 本帖最后由 chwk87 于 17-11-2007 03:19 PM 编辑 ]