STPM-学校功课討論区

Leong13

a husband and wife couple is taking driving tests. if the probability that either the husband or the wife passes the test, each time the test is taken, is 0.8,
i) find the probability that the husband of the wife passes the test after taking the test exactly twice,
ii) find the probability that both the husband and wife pass the test after taking the test more than two times.

mathlim

原帖由 Leong13 于 23-8-2007 09:23 PM 发表
這樣什麼時候才知道題目要用Ｐor Ｃ？

基本上跟次序有关就用P,只考虑组合就用C咯！

Leong13

我還有問題
1. the temperature of a body, measured in degrees, at t minutes, is x. newton's law of cooling states that the rate of fall in the temperature of the body placed in a room with a constant temperature of Xo where X0 < x, is proportional to the difference of the temperature of the body over the room temperature. write a differential equation that connects the variables x and t. the room temperature X0 is 27 degrees. fiven that x = 63 when t = 0, x = 45 when t= 6ln2. prove that x= 27 + 36e^-t/6

find,
a. the fall in the temperature of the body after being left in the room for 7 minute,
b. the time that lapses before the body cools down to within one degree from the room temperature.

proving 我會so 不用作出來，我要的是a and b

2. let p(t) be the  world populaion at the time t ( in terms of years), and k is a positive constant such that p(t) < k for each t. if the rate of change in the population is proportional to the product of the population within the excess of k over the population, write a differential equation to describe this, taking a as the constant of proportion.
if p(o) = Po, show that
p(t) = ( kPo )/ [Po + ( k-Po) e^ -akt]

if a = 4.64 x 10^-13 , k = 50 x 10^9, and Po =  4.3 x 10^9, determine the time t when the population reaches 8.6 billion

[ 本帖最后由 Leong13 于 26-8-2007 08:42 AM 编辑 ]

jhwong_alen

原帖由 Leong13 于 25-8-2007 10:13 PM 发表
我還有問題
1. the temperature of a body, measured in degrees, at t minutes, is x. newton's law of cooling states that the rate of fall in the temperature of the body placed in a room with a cons ...

我试试看。。
a)subtitude t=7 into X=27+36e^(-t/6)
X=27+36e^(-7/6)
X=38.2105
when t=0 Xo=63 [given in question]
fall in temp.=Xo-X=63-38.2705=24.7895=24.8 (1decimalpoint)

b)use -ln|Xo-X|=t/6-ln36
since Xo-X=1 (given in question)
-ln1=t/6-ln36
t/6=ln36 (ln1=0)
t=6ln36
t=21.5011 minutes

[ 本帖最后由 jhwong_alen 于 26-8-2007 11:05 PM 编辑 ]

jhwong_alen

原帖由 Leong13 于 25-8-2007 10:13 PM 发表
我還有問題
1. the temperature of a body, measured in degrees, at t minutes, is x. newton's law of cooling states that the rate of fall in the temperature of the body placed in a room with a cons ...

dP(t)/dt=aP(t)[k-P(t)]
Int dP(t)/P(t)[k-P(t)] =Int a dt
Int kdP(t)/P(t)[k-P(t)]=Int ak dt
Int dP(t)/P(t) + Int dP(t)/[k-P(t)] = Int ak dt
ln|P(t)| - ln|k-P(t)|=akt + c
ln|[k-P(t)]/P(t)|=-akt + c....(1)
e^(-akt + c)=[k-P(t)]/P(t)
when t=0,P(t)=Po
e^c=(k-Po)/Po....(2)
then,
[(k-Po)/Po]e^(-akt)=[k-P(t)]/P(t)
(k-Po)e^(-akt)=Po[k-P(t)]/P(t)
Po+(k-Po)e^(-akt)=kPo/P(t)
P(t)=kPo/{Po=(k-Po)e^(-akt)}   [proved]

From(1)&(2),
ln|[k-P(t)]/P(t)|=-akt + ln|(k-Po)/Po|
ln|[k-P(t)]Po/(k-Po)P(t)|=-akt
ln|P(t)(k-Po)/Po[k-P(t)]|=akt
t=ln|P(t)(k-Po)/Po[k-P(t)|/ak
然后subtitute,
P(t)=8.6billion
Po=4.3x10^9
a=4.64x10^-13
k=50x10^9
就能找到答案了。。。
因为我不懂billion是10^多少。。所以到这里就不会作了。。

[ 本帖最后由 jhwong_alen 于 26-8-2007 03:10 PM 编辑 ]

Leong13

a的答安是２４。８
b對

jhwong_alen

原帖由 Leong13 于 26-8-2007 08:22 PM 发表
a的答安是２４。８
b對

sorry..没看清楚问题。。。我已经编辑好了。。

Leong13

billion是１０^９，我找到答安了，thank you。

Leong13

show that

( 1 + 2x )^3/2  d [ y / sqr ( 1 + 2x ) ]/dx =
( 1 + 2x )dy/dx - y
where y is a function of x. hence solve the diffential equation

( 1 + 2x )dy/dx - y = x ( 1 + 2x )
giving that y = 0 when x = 1.

jhwong_alen

原帖由 Leong13 于 27-8-2007 06:03 PM 发表
show that

( 1 + 2x )^3/2  d [ y / sqr ( 1 + 2x ) ]/dx =
( 1 + 2x )dy/dx - y
where y is a function of x. hence solve the diffential equation

( 1 + 2x )dy/dx - y = x ( 1 + 2x )
giving t ...

(1+2x)^(3/2) d[y (1+2x)^(-1/2)]/dx
=(1+2x)^(3/2) [(1+2x)^(-1/2) dy/dx - y (1+2x)^(-3/2)]
=(1+2x) dy/dx - y (proved).....(1)

from (1) y=(1+2x) dy/dx - (1+2x)^(3/2) d[y (1+2x)^(-1/2)]/dx
subtitude into (1+2x) dy/dx - y = x (1+2x)

(1+2x) dy/dx -{(1+2x) dy/dx - (1+2x)^(3/2) d[y (1+2x)^(-1/2)]/dx}= x (1+2x)

(1+2x)^(3/2) d[y (1+2x)^(-1/2)]/dx = x (1+2x)

d[y (1+2x)^(-1/2)]/dx =x (1+2x) (1+2x)^(-3/2)

d[y (1+2x)^(-1/2)]/dx =x (1+2x)^(-1/2)

∫ d[y (1+2x)^(-1/2)]=  ∫ x (1+2x)^(-1/2)ｄｘ

let u^2=1+2x,u du/dx=1

[y (1+2x)^(-1/2)] = 1/2∫(u^2-1)du

[y (1+2x)^(-1/2)] = [(u^3)/6 - u/2] + c

[y (1+2x)^(-1/2)] =[(1+2x)^(3/2)/6 - (1+2x)^(1/2)/2] + c

y = [(1+2x)^(1/2)][(1+2x)^(3/2)/6 - (1+2x)^(1/2)/2] + d

y = [(1+2x)^(2)/6 - 3(1+2x)/6] + d

y = (1+2x)[(1+2x-3)/6] + d

y = (1+2x) [(2x-2)/6] + d

y = (1+2x)[(x-1)/3] + d

when y=0,x=1

0 = 3(0) + d

d=0

hence,
y = (1+2x)[(x-1)/3]+0
y = (1+2x)[(x-1)/3]   (proved)

[ 本帖最后由 jhwong_alen 于 29-8-2007 08:58 PM 编辑 ]

Leong13

ans= y = (1+2x)(x-1)/3

jhwong_alen

原帖由 Leong13 于 28-8-2007 08:53 PM 发表
ans= y = (1+2x)(x-1)/3

我编辑上面的答案了，做到了。。谢谢哦。。。

img3nius

帮解....

1. state 1/r(r+1)(r+3) as the sum of 3 partial fraction , hence , or other wise find
SUM 1/r(r+1)(r+3) , expressing your answer in the form a + b/(n+1) +c/(n+2) + d/(n+3)
where a b c d are constandt to be found

img3nius

2. find the number A , B and C sch that
    1+r^2 =  A(r+2)(r+1) + B (r+1) + C for all values of r
hence , or other wise , prove that
SUM ( 1+r^2)r! = n (n+1)!

jhwong_alen

原帖由 img3nius 于 29-8-2007 11:51 PM 发表
帮解....

1. state 1/r(r+1)(r+3) as the sum of 3 partial fraction , hence , or other wise find
SUM 1/r(r+1)(r+3) , expressing your answer in the form a + b/(n+1) +c/(n+2) + d/(n+3)
where a b ...

我试试看阿。。。

A(r+1)(r+3)+Br(r+3)+Cr(r+1)=1
when r=0, A=1/3
when r=-1, B=-1/2
when r=-3, C=1/6

1/r(r+1)(r+3)=1/r - 1/2(r+1) + 1/6(r+3)

sum 1/r(r+1)(r+3)
= sum r+2/r(r+1)(r+2)(r+3)
= sum 1/(r+1)(r+2)(r+3) + sum 2/r(r+1)(r+2)(r+3)

let Un=1/(n+2)(n+3)
Un-&#8321;- Un = 1/(n+1)(n+2) - 1/(n+2)(n+3)
          = 2/(n+1)(n+2)(n+3)

let f(n)=1/(n+1)(n+2)(n+3)
f(n-1)-f(n) = 1/n(n+1)(n+2) - 1/(n+1)(n+2)(n+3)
            = 3/n(n+1)(n+2)(n+3)

hence,
sum 1/r(r+1)(r+3)
= sum r+2/r(r+1)(r+2)(r+3)
= sum 1/(r+1)(r+2)(r+3) + sum 2/r(r+1)(r+2)(r+3)
= 1/2 sum 2/(r+1)(r+2)(r+3) + 2/3 sum 3/r(r+1)(r+2)(r+3)
= 1/2 x [Uo-Un] + 2/3 x [f(0)-f(n)]
= 1/2 x [1/6 - 1/(n+2)(n+3)] + 2/3 x [1/6 - 1/(n+1)(n+2)(n+3)]

D(n+3)+E(n+2)=1             F(n+2)(n+3)+G(n+1)(n+3)+H(n+1)(n+2)=1
when n=-2, D=1              when n=-1, F=1/2
when n=-3, E=-1             when n=-2, G=-1
                            when n=-3, H=1/2

= 1/2 x [1/6 - 1/(n+2) + 1/(n+3)] + 2/3 x [1/6 - 1/2(n+1) + 1/(n+2) -1/2(n+3)]
= 1/12 + 2/18 - 1/3(n+1) - 1/2(n+2) + 2/3(n+2) + 1/2(n+3) - 1/3(n+3)
= 7/36 - 1/3(n+1) + 1/6(n+2) + 1/6(n+3)

solution; a=7/36,b=-1/3,c=d=1/6

[ 本帖最后由 jhwong_alen 于 30-8-2007 03:53 PM 编辑 ]

jhwong_alen

原帖由 img3nius 于 30-8-2007 12:02 AM 发表
2. find the number A , B and C sch that
    1+r^2 =  A(r+2)(r+1) + B (r+1) + C for all values of r
hence , or other wise , prove that
SUM ( 1+r^2)r! = n (n+1)!

by equation coefficient,
A=1
3A+B=0,B=-3
2A+B+C=1,C=2
solution; A=1,B=-3,C=2

不懂将做对吗？
let Un=n(n+1)!
Un-&#8321;- Un = (n-1)n! - n(n+1)!
          = (n-1)n! - n(n+1)(n!)
          = n![n-1-n(n+1)]
          = n![n - 1 - n^2 - n]
          = n![-n^2-1]
          = -n!(n^2+1)

sum(1+r^2)r! = -sum -r!(r^2+1)
             = -(Uo - Un)
             = -[0 - n(n+1)!]
             = n(n+1)!  [proved]

Leong13

你好厲害，thank you

Leong13

each supabrek packet contains a card from a set of n picture cards. a packet selected at random has the same probability of containing any one of the n cards.
the smith family already possess k ( < n ) different picture cards. they purchase a few more packets, selected one by one randomly, until they find a card that is different from the k cards they already have. if this happens when they have bought R packets, show that


P ( R = r ) = [( n-k )/n ][ (k/n)^r-1 ]
given the result ∞
                 ∑  rx^r-1 = (1-x)^-2, for 0<X<1, show
                 r=1
that E (R)= n/ ( n-k ).
for the case n = 4, deduce that the expected total number of packes that a family must buy if they do not possess any yet, and picks the packets one by one until they have a complete set is 25/3.

晨天

Matrixxxxxxxxxxxx


１）Assume all is square matrix
当   AX=B
SO,  A^-1*AX=A^-1*B
      IX=A^-1*B
       X=A^-1*B

当 AXC＝B
X？？？？？
是不是     X＝A＾-1 * B * C＾-1
or         X＝C＾-1 * B * A＾-1
or 其他？？？？？ 如何知道A＾-1 的位置， 是不是当A在domain的左边， A＾-1 也会在左边？？？？？？

可以在这方面解释吗...........


２）
4X-7Y＝8
3X-5Y＝0
use XA＝B to solve
*是XA＝B， 不是 AX＝B
请各位大大帮忙

jhwong_alen

原帖由 晨天 于 31-8-2007 09:01 PM 发表
Matrixxxxxxxxxxxx


１）Assume all is square matrix
当   AX=B
SO,  A^-1*AX=A^-1*B
      IX=A^-1*B
       X=A^-1*B

当 AXC＝B
X？？？？？
是不是     X＝A＾-1 * B * C＾-1
or         X＝ ...

AXC=B
A&#8254;&sup1;AXC=A&#8254;&sup1;B
IXC=A&#8254;&sup1;B
XCC&#8254;&sup1;=A&#8254;&sup1;BC&#8254;&sup1;
XI=A&#8254;&sup1;BC&#8254;&sup1;
X=A&#8254;&sup1;BC&#8254;&sup1;
如果LHS是把A&#8254;&sup1;乘在左边，那么RHS也就在左边。。这好像叫premultiply
如果右边，好像是post multiply...