Since SolarFlare is missing in action bmacri or Tom_Frohman may ask the next question. I see no reason to wait an eternity.
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Since SolarFlare is missing in action bmacri or Tom_Frohman may ask the next question. I see no reason to wait an eternity.
Should be someone other than me. Right now the only question I can think of is :
Why is man born only to suffer, die and eat at Taco Bell? There is no wrong answer.
Tom, here is my serious answer for you:
We live in a fallen world but God has been made right with us through our Christ Jesus. You need to look at the history of the world in the Bible (Old Testament) to understand the world and why it is the way it is. Everyone suffers. This is how the world is. And everyone dies. Unless you repent and have faith, then you will have eternal life in the Kingdom of God. And all the suffering which Christ Jesus helped you endure while on this world will be gone forever. If you don't, you face death and **** where God will turn his back on you forever. Repent and believe that Jesus died for our sins. And you can then have all the taco bell in the world. :)
Actually, world domination takes its tolls, if you know what I mean. Progress is dandy, by the way. Although they did steal my weightlifting Bronze :(.Quote:
Originally Posted by Tom Frohman
I don't have a question... if I tried to find one it would be a stretch...
I told you that doping inflicts great harm to your body and mind... But you're not listening.Quote:
Originally Posted by SolarFlare
Easy one, just to make it go ...
SolarFlare is sun burning at the top of a 200m cliff and cilu at the sand beneath it (assume a vertical climb) ... how long will SolarFlare be enjoying sun when cilu gets into the shadow? (they are facing West, obviously)
Well, you did not say what is the time, but I will enjoy the sun until 12:03.
Or until SolarFlare jumps over me and we both die in a carnage... :D
Nope, you are sooooo lazy that you will enjoy the sun until the sun dissapears under the horizon at what time SolarFlare will still be drinking some kind of alcoholic mix, with solar glasses on his eyes, a couple of nice girls/boys (is he/she a he or a she? :rolleyes: ) at his/her side, looking the sun going down slowly ... for how long?
I am so lazzy to figure out a good answer, but perhaps over the week-end it will come to me... The shadow I mean. :D
What is croakey?
Quote:
Originally Posted by YourSurrogateGod
Is it a German built motorbike?
An encrypting method used by some native americans, based on the singing rithm of the frogs of a particular river.
I be dmm... :ehh:Quote:
Originally Posted by DeepButi
I remember hearing the word a while ago, I don't remember what it means. I figured that some of you gents would know.Quote:
Originally Posted by Deniz
Someone should ask a question and resurect this thread.
Okay here's one. How much would the earth's rotation have to increase for the people to float in the air? Assume you are at the equator and gravity is 9.8 meters/sec. A teacher told her second grade students (my neice) this was possible. Give an aproximate answer so ignore things like air friction and stuff like that. So what I'm saying is keep it simple.
-Ben
Assumes:
Earth Radius: 6,378 km
Equatorial circumference: 40,000 km
Rotation speed: 40,000 / 24 h = 1670 km/h = 464 m/s
Centrifugal force: Fc = mv^2/R
Gravity force: G = mg
To begin fly Fc must (at least) equal G.
Thus: mv^2/R = mg
v = sqrt(gR);
v = 7910 m/s = 25884 km/h
To make you float in the air, the earth rotation speed must increase 17 times.
Unfortunatelly for you, Earth rotation speed is actually decreasing. That makes the day longer (than 24 hours) and the year shorter (that 365 days).
Please look here Be sure to read the full thread.
What's your point? I'm not following... :confused:
All my life's a circle....
It seems like we've been here before
I can't rmember when (oh yes I can)
But i've got this funny feeling that we'll all be together again....
But if you're floating, then there's no static friction to pull you around with the Earth... meaning you'll stop orbiting... meaning you'll fall to the Earth... meaning friction will pull you back to orbiting velocity... meaning... I've got a headache...
Some people don't need to move at all....
I'm sorry if this question was asked before. I'm not really up to rereading the whole thread each time I have a question to make sure it was asked before.
As for the answer I did it a little differently. I'm not sure which is right but here's how I did it.
The radius of the earth is 6378 km. Ahh I found my problem. I'll make it in red and contiune with my answer even though it's wrong. For you to float (basically be in orbit) you need to displace as much distance as you fall each second. So I made a right triangle. The hypotinus(sp?) is 6378km and one side is 6378km - .0098km (the distance you fall in one second). Calcing the last isde you get 11.18 km. So you need to travell that far is one sec. So 11180 m/sec - 461m/sec(current velocity not sure where I found that either) you get and increase of 10719 m/sec is need to float.
So I messed up with how much a person falls in one second. If I change that to what it should be 4.9 m then the answer I get is 7906 m/sec which is basically the same as yours. Good job.
So what's the next question?
-Ben
I have no idea whatsoever if this question was asked and answered before.
You told me to use the KISS principle. (Kepp It Simple, Stupid). So I did. I don't understand your calculations with 1 second fall, etc. When you fly you don't fall and fly again. :D
I will come with a question tomorrow.
/Physics laws must have mathematical beauty.
The idea is the same as how something stays in orbit. It is always falling to earth but it stays in orbit because it is moving in a direction perpendicular to gravity. So in one second something falls to the earth 4.9 meters assuming gravity at 9.8 m/sec. So that forms the right triangle and using a*a + b*b = c*c you get the horizontal side as 7906 meters. Since the earth is so close to flat I just say the earth needs to rotated that much.
See attached picture. I hope that helps.
I think your way is better though.
-Ben
A very easy one:
An oblong garden, half a meter longer than wide, consists entirely of a gravel-walk, spirally arranged, a meter wide and 3630 meters long. What are the dimensions of the garden?
is this the topic of riddle freaks?
i love riddles, but not mathematic ones :p
ps.. anyone seen nemesis game? its an interesting movie on this subject.
cilu, i didn't understand your problem very well. I drew some scratch in M$ Paint to see if you agree with my understanding:
You got it pretty well.
Hmmmm! Ok I'll try to solve this one. I've seen it before in school, but long time ago hehe.
Long time, no answer... :confused: If this one is too hard (though it's so simple) I will post another... :sick:
Oh well, I seached the forums and found this long thread by chance, I am not boasting but I truly love humors and fun stuff,
I now have a tricky funny problem that I think would tease your mind sometimes especially on weekends like this. You can post anything about my problem, give me any ideas that you like to,
I have a multiset S (S1....Sn) and another set dS (S21, S31,....,Sn1,S32, S42,....Sn2,.....,Snn-1)
A multiset is a set whose elements can be repeated whereas in a set,its elemest cant.
For example, S={2,1,4,3} => dS={-1,2,1,3}
From dS, tell me all of the ways you think up to build up S again ???
Quote:
Originally Posted by cilu
half a meter longer then wide
That's not enough. What are the dimensions?
Well, what does S21, S31 etc. stand for? I.e. there seems to be a way of deriving dS from S, but how? From your example, it doesn't seem to be the multiplication of the items in S, so what is it?Quote:
Originally Posted by Hediea
you problem is wrong:
multiset S(S1....Sn)
set dS(S21, S31,....,Sn1,S32, S42,....Sn2,.....,Snn-1)
multiset S has N elements
set dS has N-1 + N-2 +... + 1 elements, that is SUM[i=1, N-1] (i) = N*(N-1)/2.
If N = 4, then dS should have 6 elements, but in your example both S and dS have 4.
If dS has 4 elements, there is no integer solution for N.
So, give us a valid example.
the valid dS seems pretty obvious to me:
For example, S={2,1,4,3} => dS={-1,2,1,3, 2, -1}
As it seems that Smn = Sn - Sm
that's how it starts anyway, until he has forgotten or removed the duplicated elements (that he should not have done given dS is a multi set).
Smn = Sn - Sm was obvious from the very first moment. However, I was expecting a valid example. He did not say enything avout removing duplicate entries...
Well, dS is not a multiset, so you can't have duplicates, so his example is valid.
A first observation is that you won't be able to get the exact multiset S back, since you can always add or remove a constant. Another thing is that since S is a multiset, you can have lots of different solutions in S for a given dS.
For example, say you are given the dS of {-1, 0, 1}
S could be: {2, 1, 2}
Or it could be: {2, 1, 2, 1, 2}
Or it could be: {3, 2, 3, 2, 3, 2, 3}
etc.
Here's a pretty interesting problem. The reason I pose it here is that both my friend and I solved it independently: I solved through a long, mathetmatical route, while he solved via a comparatively quick logic-based route.
Suppose there is a row of k lightswitches, each wired to its own lightbulb, and initially off. Someone turns on one of the lightswitches randomly. Then they toggle another lightswitch randomly (if it's the same one as the first one, the light goes off. Otherwise a second light comes on). This process is repeated n times (so that n total toggles are made). What is the expected value for the number of lights that will be on when the process is complete, in terms of n and k?
(I have tried to make this as clear as possible; unfortunately I won't be around until Sunday to clarify ambiguities I have overlooked)
hmm here is my shot ...
starting is n=0 and num of lights on is also 0, both even,
n=1, numLightsOn = 1
n=2, numLightsOn = 0 / 2
n=3, numLightsOn = 1 / 3
n=4, numLightsOn = 0 / 2 / 4
so,
if n is even, number of lightss on is all even numbers less than k and less than n
if n is odd, number of lightss on is all oddnumbers less than k and less than n
so..
numLightsOn = 0+(n%2) , 2+(n%2) , 4+(n%2) ... k+(n%2)
so, how? is it right..?
Ever heard of Pascal pyramid? It looks like this:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
....
C(n,0) C(n,1) C(n,2) ... C(n,n)
First colum and last are always 1. The others are computed this way
a1
a21 a22
a31 a32 a33
a41 a42 a43 a44
a32 = a21 + a22
a42 = a31 + a32 ; a43 = a32 + a33
...
a[i][j] = a[i-1][j-1] + a[i-1][j]
where C(n,k) mean k combination taken from n. (well, I'm not very good at math in english but I hope I'll make it as clear as possible)
if you want to see how you can get the pyramid, compute (x+y)^n. Here ^ means power.
(x + y)^0 = 1
(x + y)^1 = 1*x + 1*y
(x + y)^2 = 1*x^2 + 2*xy + 1*y^2
(x + y)^3 = 1*x^3 + 3*xy^2 + 3*yx^2 + 1*y^3
...
and so on
What does it have to do with our problem? Well, it is the answer. At least almost.
Let me explain. we have k switches, all OFF. That means at the beginning the number of ON switches is 0. After each step you can either have increase the number of ON swtiches with 2 or have it unmodified. In the first phase consist of a turning one switch ON. That only applies to the OFF switches, ON switches are not affected in this phase. In the second phase you can either turn another OFF switch to ON, or toggle and ON switch to OFF, in which case you have returned to a previous configuration (only the number of ON switches counts, not their position in the row).
So, after step 1 you can either have 2 switches ON or none. That is
1:0 1:2
No, we go to the second step (n=2). If after the first step we had 0 ONs, after step 2 we can have either 2 ONs or 0. If after the first step we had 2 ONs, after the second we can have 2 ONs or 4ONs. To summarize:
1:0 2:2 1:4
Getting back to Pascal's pyramid:
step = 0 1:0
step = 1 1:0 1:2
step = 2 1:0 2:2 1:4
step = 3 1:0 3:2 3:4 1:6
step = 4 1:0 4:2 6:4 4:6 1:8
....
step = n C(n,0):0 C(n,1):2 C(n,k')
if k is odd, than k' = k-1
if k is even than k' = k
Note that the fact k is odd or even does not matter. Proof:
k = 5
n = 0
00000
n=1
00000 , 11000
n = 2
00000 , 11000 , 11000 , 11110
n = 3
00000 , 11000, 11000 , 11110 , 11000 , 11110 , 11110
end
11110 from step 2, can only generate 11110, because there is only 1 OFF switch that will be set to ON in phase1. in phase 2 any of the 5 ON switches (position doesn't matter) will be turned OFF.
k = 4
n = 0
0000
n = 1
0000 , 1100
1:0 1:2
n = 2
0000 , 1100 , 1100 , 1111
1:0 2:2 2:4
n = 3
0000 , 1100 , 1100 , 1111 , 1100 , 1111
1:0 3:2 2:4
n = 4
0000 , 1100 , 1100 , 1111 , 1100 , 1111 , 1100 , 1111
1:0 4:2 3:4
If you experiment with k=6, k=8, etc (I told you that the parity is unimportant) you will observe that Pascal's pyramid must be cut after (k+2)/2 colums.
Here is an example for K=6:
1
1 1
1 2 1
1 3 3 1
1 4 6 3
1 5 10 6
I hope you noticed that when the pyramid is cut, the last element is not 1, but is a copy of the upper-left element.
So the answer is that Pascal's Pyramid (sometimes cut) gives the probability.
Cheers! :wave:
read more about Pascal's pyramid
long post hurt brain.
Find a single formula - no summation or Pascal's triangle references (although it can probably be achieved that way - what a great triangle). It also turns out that the formula is independent of whether k is odd or even.
Is that so?Quote:
Originally Posted by Joe Nellis
I'll give it a shot and say that the answer is 63 2/11.
:D
let
- P(i) be the probability that 2*i switches are ON (there can be only even number of ON switches)
- k' be k/2 if k = 2*j (i.e. k even) and (k-1)/2 is k = 2*j+1 (i.e. k odd)
- C(n,i) i combinations of n
P(i) = C(n, i), if n<=k'
P(i) = C(n, i), if n>k' and i<k'
P(i) = C(n-1, i-1), if n>k' and i=k'
Of course you can put it simplier as
P(i) = C(n, i), if i<k'
P(i) = C(n-1, i-1), if i=k'
[QUOTE=cilu]there can be only even number of ON switches[\QUOTE]
???
Bad english? :o Sorry. Or you don't understand why there can only be 0,2,4, etc. switches turned ON? If so, please read my first post. It's obvious. :cool:
k=1, n=3. guarantees one switch on after your "step 1"Quote:
Originally Posted by cilu
Cilu, your method explores possibility instead of probability and if I'm not mistaken, assumes k>>n.
To clarify, the solution is an equation for the expected value (defined as "The weighted average of a probability distribution"), and the assumption that k>n is invalid.