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**coolwind**- Replies: 1

Find F(n) for n=1,2,3...

F(n)=

det[1 2 .... n]

[ n+1 n+2 .... 2n ]

[ . ................. . ]

[ . ............... . ]

[ . ................. . ]

[n(n-1)+1 n(n-1)+2 ...n^2]

Sloved!!

**coolwind**- Replies: 0

If the pdf of f(x1)=(x1)e^[(-x1^2)/2] , x1>=0

f(x2)=1/2*pi , 0<x2<2*pi

(a)Find the pdf of Y1=X1cos(X2) and Y2=X1sin(X2)

(b)Is Y1 and Y2 independent of each other?

Ricky wrote:

This is a fairly simple integral. Do you know integration by parts?

Thanks~I see!!

**coolwind**- Replies: 2

∫te^(-st)dt where 0<t<∞

Thank you!!:cool:

**coolwind**- Replies: 4

I(f)= ∫(sin2pi100t)e^(j2pift)dt where range 0<t<∞

THANKS

Hi fgarb, this is my communication systems homework.

**coolwind**- Replies: 5

Find Fourier transform

x(t)=1/(1+(t/3)^2)

Thank you

**coolwind**- Replies: 2

∫(sinc^2)4f df ,where integral range -∞ to +∞

Thanks~~!!

**coolwind**- Replies: 2

If a signal X(t)=5cos(2pi60t)-3sin(2pi180t)

let Cn=(1/T)∫X(t)e^(-jwnt) dt where T=1/60 , w=2(pi)f f=1/T

find the Cn ?

Thank you~~

**coolwind**- Replies: 0

Use reduction of order ,to fund a second solution y2(x).

1. y''+16y=0 ; y1(x)=cos4x

2. xy''+y'=0 ; y1(x)=lnx

Thanks~~!!

luca-deltodesco wrote:

1.a)

you have 15 places, for one sister can be. for the other sister, there are 14 more places, and given that sister A isnt on the edge, 2 places out of the 14 she could be in, if A is on the edge, there is only 1 out of 14 places, so the answer isP(0<A<15) = 13/15, P(B next to A | 0<A<15) = 2/14

P(A=0 or A=15) = 2/15, P(B next to A | A=0 or A=15) = 1/14

P(B next to A) = 13/15*2/14 + 2/15*1/14 = 28/210 = 2/15 = 0.1333...

Good!

pi man wrote:

I come up with a different solution.

A: Consider all of the different pairings the 2 could be in. There are (15 choose 2)possibilities. That's 105 different pairings. Now count how many pairings are next to each other: (1,2), (2,1), (2,3), (3,2), ... (14,15) and (15, 14). That's 28 pairings out of the possible 105. That's .2666, which is double Luca's answer. And that explains my mistake. There are 15 * 14 possible pairings, not (15 choose 2). So that makes it 28 possibilities out of 210 which is .133333.

B: You still have the 210 different pairings but the number of pairings that meet the requirements is smaller: (1,7), (7,1), (2, 8), (8,2),..., (9,15) & (15,9). There are 18 good pairings out of the 210, or .0857

2. I have no idea what you're asking for here.

Thank you,and the second problem is a combinations with repetition problem.

**coolwind**- Replies: 4

1.At tht Mary sorority the 15 sisters who are seniors line up in a random manner for a graduation picture.

Two of these sisters are Columba and Piret.What is the probability that this graduation picture will find

(a)Piret and Columba standing next to each other?(b)exact five sisters standing between Columba and

Piret?

2. Consider the 2^19 compositions of 20.(a)How many have each summand

even?(b)How many have each summand a multiple of 4?

Thank you very much~~:cool:

Ricky wrote:

coolwind, are you sure she doesn't have 1 dozen of each?

Ricky,this problem is from my textbook(written by Ralph P.Grimaldi)

Ricky wrote:

Something is wrong in b.

If she has 2 dozen of each (24), and she is only picking 20 beads total, then it doesn't matter how many of each she has. It could be infinite.

So she has n choices for the first bead, n choices for the second bead, n choices for the.... which is n*n*n....*n = n^20. So n^20 = 230230 which comes out to 1.85399, which must be wrong.

Unless I'm missing something.

Hi,Ricky

the ans is n=7.

krassi_holmz wrote:

(a)

If a1!=a2!=a3!=a4, then: 3258

If not: 4475

If you want a1<=a2<=a3<=a4, you will get 242 different solutions.

Right,the ans is 4475.

How did you count?

luca-deltodesco wrote:

im thinking b = n

but also, are the beads put back in the mix, or kept seperate?

What 's the different?:D

luca-deltodesco wrote:

for (a) can they be the same?

Hi,luca-deltodesco

the ans is yes.

**coolwind**- Replies: 12

(a)Determine the number of integer solutions of

a1+a2+a3+a4=32

where a1,a2,a3>0, 0<a4<=25

(b)Mary has two dozen each of n different colored beads.

If she can select 20 beads(with repetition of colors allowed)

in 230,230 ways,what is the value of n?

Thanks

ryos wrote:

a) This is a combination problem. Basically, you are choosing three vertices from a set of n, where order doesn't matter. So, the formula is:

This will give you the number of triangles you can make from the vertices of

anyregular polygon!b) In order for this to work, no two vertices may be adjacent to another. Unfortunately, I don't have time right now to figure out that formula.

ok,thank you

**coolwind**- Replies: 2

How many triangles are determined by the vertices of a regular polygon of n sides?

How many if no side of the polygon is to be a side of any triangle?

Thanks~~

**coolwind**- Replies: 3

70=T+c

110=T+ce^(K/2)

145=T+ce^K

solve T = ?

**coolwind**- Replies: 2

2P(n,2)+50=P(2n,2)

P:"Permutation"

ganesh wrote:

-1²=1²

(1-2)² = (2-1)²

Taking square root on both the sides,

1-2=2-1

2=4

Dividing both the sides by 2,

1=2.

But we know 1+1=2,

Therefore,

1=1+1

or

1+1=1

cool...:cool:

**coolwind**- Replies: 4

∫(t+1)e^(-jwkt)dt

Thanks...