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发表于 13-7-2009 07:08 PM
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19.
Β= 3α 代入
α + Β = -b/a,
αΒ= c/a,
得出 4α=-b/a----1
3α²=c/a-----2
让 方程式 1 平方,
得,
16α²=b²/a²-----3
3/2 :
16/3=b2/ac
3b²=16ac (证毕)
--------------------------------
ac²x² + (b3 -3abc)x + a²c=0
设 此方程式为
ux²+vx+z=0,
x²+vx/u+z/u=0.
α/Β²+Β/α²= -v/u----[1]
(α/Β² ) (Β/α² ) =z/u----[2]
[2]: 1/αB=z/u
a/c=z/u 代入方程式.
[1]: α³/α²B² + B³/α²B² =-v/u
(α³+B³ ) /α²B² = -v/u
(α+B ) (α²-αB+B² ) =-v/u
(-b/a)/[(α+B)²-3αB] = -v/u
(-b/a)/[(-b/a)²-3c/a]=-v/u
(b³-3abc)/a=v/u 代入方程式
接下来应该懂了吧?
[ 本帖最后由 飘逸手语 于 13-7-2009 07:13 PM 编辑 ] |
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发表于 13-7-2009 09:03 PM
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大家怎样读Probability的??
我form5时底麻麻地
现在PC!傻傻分不清楚
+-*/也傻傻分不清楚 |
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发表于 20-7-2009 08:56 PM
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回复 63# 芭樂 的帖子
p is "pie.".have order.c is choose order not important. |
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发表于 20-7-2009 09:02 PM
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发表于 8-8-2009 12:39 PM
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发表于 13-8-2009 05:08 PM
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1-(1+x)^-12 = 12x
solve x 等于多少? |
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发表于 13-8-2009 05:26 PM
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1. Let a and b be equal non-zero quantities
a = b
2. Multiply through by a
a^2 = ab
3. Subtract b^2
a^2 - b^2 = ab - b^2
4. Factor both sides
(a - b)(a + b) = b(a - b)
5. Divide out (a - b)
a + b = b
6. Observing that a = b
b + b = b
7. Combine like terms on the left
2b = b
8. Divide by the non-zero b
2 = 1
数学真神奇。。。 2 = 1 |
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发表于 13-8-2009 11:25 PM
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5. Divide out (a - b) <<< 这个有问题 |
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发表于 14-8-2009 09:24 PM
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原帖由 Dicardo 于 13-8-2009 05:26 PM 发表
1. Let a and b be equal non-zero quantities
a = b
2. Multiply through by a
a^2 = ab
3. Subtract b^2
a^2 - b^2 = ab - b^2
4. Factor both sides
(a - b)(a + b) = b(a ...
大家看到以上的矛盾了吗?1 + 1 竟然等于2!!
如笨蛋一个所说的,问题出在除以(a - b)的这个行动。
注意到
a = b,
所以
a - b = 0。
这就是为什么0不能做除数,
因为这会造成上列的矛盾。 |
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发表于 15-8-2009 07:15 PM
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回复 66# 龍弟 的帖子
恩... 我们老师是推荐oxford 的书。
longman 的也不错,example 比较多。
我觉得还是oxford的比较好。 |
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发表于 14-9-2009 05:48 PM
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how to solve this question?
A continuous random variable X is such that P(X<x)=mx+c for 1<x<5.
Calculate the values of m and c.
[the < is greater and equal to...so sorry because i dun know how 2 find that simbol] |
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发表于 15-9-2009 08:57 AM
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P(X<=x)已经是cumulative density function。
所以有两个properties,when x=1, P(X<=1)=0,m(1)+c=0
when x=5, P(X<=5)=1, m(5)+c=1
接下来的应该没问题了吧? |
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发表于 15-9-2009 04:06 PM
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哦..我明白了...谢谢你!! 下次不会再请教!! |
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发表于 22-9-2009 10:32 PM
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请问有人可以帮我解一下吗??谢谢
X is a continuous random variable with probability density function as follows.
f(x)= { 3/4 (x-1)(3-x) , 1<x<3
{0 , otherwise}
Two independent observations , X1 and X2
for the variable X were taken.Find
E(X1 +2X2)
ans:6 |
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发表于 23-9-2009 11:16 AM
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原帖由 premier66 于 2009/9/22 10:32 PM 发表
请问有人可以帮我解一下吗??谢谢
X is a continuous random variable with probability density function as follows.
f(x)= { 3/4 (x-1)(3-x) , 1
Since they are independent random variables,
E( x1 + 2x2) = E(x1) + 2E(x2) [ You can prove this equation using E(x) = xP(X=x)]
Some eg , E(x1 + x2) = E(x1) + E(x2) , E( x1 - x2) = E(x1) - E(x2) , E(3x1 - 4x2) = 3E(x1) - 4E(x2)
Var (x1 + 4x2) = Var (x1) + 16 Var (x2) ] use Var (x) = E(x^2) - [E(x)]^2 to prove.
Var (x1 -3x2) = Var (x1) + 9 Var (x2) ] |
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发表于 23-9-2009 11:17 AM
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原帖由 premier66 于 2009/9/22 10:32 PM 发表
请问有人可以帮我解一下吗??谢谢
X is a continuous random variable with probability density function as follows.
f(x)= { 3/4 (x-1)(3-x) , 1
E(x1 + 2x2) = E(x1) + 2E(x2) |
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发表于 23-9-2009 01:31 PM
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发表于 23-9-2009 06:40 PM
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发表于 23-9-2009 10:46 PM
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E(X)=integrate 1 to 3 3/4 x(x-1)(3-x) = 2
X1=X2=X
E(X1 +2X2) =E(X+2X)=E(3X)=3E(X)=6
问题很奇怪,唯有这样才能做到答案。
为什么突然跑出X1和X2? |
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发表于 23-9-2009 11:50 PM
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E(X)={ x f(x) dx
= 3/4{ x(x-1)(3-x) dx
= 3/4 { 4x^2-3x-x^3 dx
= 3/4[ 4x^3/3 - 3x^2/2 - x^4/4]
= 3/4 [8/3]
= 2
E(X1 + 2X2) = E(X1) + 2E(X2)
= 2 + 2(2)
= 2+4
=6
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