DIA Dow Jones 30 (ETF)

[0zK]

http://el-marketero.blogspot.com/

muy bueno che !

[mendezfederico]

Ibupirak, el blog es tuyo?

[el_sobrino]

Estimados osos fieles,no desaprovechen este regalo del cielo que les presenta el mercado hoy para salir un poquito mas caro que ayer....no quedara nada de esta suba fantasma a fin de la rueda...yo se porque se los digo.

Afortunado el que escucha la palabra

[0zK]

hip hip urra ... hip hip URRE !!!


onda 3 va hasta 2.25 vas a disfrutar mas de lo que pensas

ven bien a URRE?

pipo, a la larga tiene que ir a cerrar el gap en usd 2.25 .. va a ser un pelpa que habra que tenerle paciencia , ... 2,25 si los rompe a usd 3 derechito .. si las noticias vienen favorables, puede ser un driver para que vaya a buscar maximos por el impulso que pueda tener ..

[Phantom]

Tomo una balanza de platos. Separo las bolas en 3 grupos de 4 bolas c/u y los llamo A - B - C. Reservo C.

1er Pesada: Peso A y B --> Son iguales? --> La bola distintiva está en C - Si A y B pesan distinto, la bola distintiva está en uno de los dos grupos. Supongo que la bola más pesada está en A.

Al grupo A de 4 bolas lo separo en 2 grupos de 2 bolas cada uno y los llamo D y E.

2da Pesada: Peso D y E --> No pueden ser iguales. Tomo el más pesado. Supongo que es D.

3er Pesada: Pongo en cada plato de la balanza una de las bolas de D y veo cual es la más pesada.

Alguno que lo resuelva.....pero pensando que la bola distintiva es más liviana que el resto!

[mendezfederico]

Hola Johnwayne avisame cuando apareces asi vemos como te paso la data , a PAC se la envío por mail, 1000 disculpas por no contestar ayer, que estaba en el curso (muuuuuuuuuuuuy bueno)

[Aviador]

siguen los temblores en japon y no pueden enfriar el reactor !!

[Luca]

si, se ve en los markets.
es increible que le driver sea ese.
tengo CNN en vivo y futures y ver como se correlaciona todo.

TEPCO dice que el agua vertida está siendo efectiva en el enfriamiento de las piscinas de fuel.

[Poo]

TEPCO dice que el agua vertida está siendo efectiva en el enfriamiento de las piscinas de fuel.

[0zK]

1+1= ?

[virrey]

12 Balls (D10)
SOLUTION

We found a good solution to this puzzle on the Cut The Knot website. They show a solution taken from the book Mathematical Spectrum, by Brian D. Bundy. Bundy shows two solutions: the first one requires different courses of action depending on the outcome of previous weighings, so it is not particularly elegant or easy to remember. The second solution involves a fixed course of action in all circumstances, and is the one that follows.

In this method we weigh four specified balls against four other specified balls in each of the three weighings and note the result. If we observe say the left hand side of the balance, then for an individual weighing there are three possible alternatives: the left hand side is heavy (>), light (<) or equal (=) as compared with the right hand side of the balance.

Since three weighings are allowed, the number of different results that can be obtained is just the number of arrangements (with repetitions allowed) of the three symbols >, <, =, i.e. 3 3 = 27. If we use all twelve balls in the three weighings, and ensure that no particular ball appears on the same side of the balance in all three weighings, the outcomes >>>, <<<, === are not possible. We thus have only 24 possible outcomes and we shall show that it is possible to set up a one to one correspondence between these 24 outcomes and the conclusion that a particular ball among the twelve is heavy or light.

The 24 outcomes can be divided into two groups of twelve in each group. If we call the reverse of an outcome the outcome obtained by replacing > by <, < by > and leaving = unchanged, one group of twelve will be the reverses of the other group and vice versa. We can thus write the 24 outcomes in the form of two arrays, each array having three rows (the three weighings) and twelve columns (the twelve balls), so that each row contains four >'s, four <'s and four ='s. Thus we have, for example,

> > > > < < < < = = = =
< = = = < < > > > < > =
= > < = > = < > > = < <
A B C D E F G H I J K L
< < < < > > > > = = = =
> = = = > > < < < > < =
= < > = < = > < < = > >

We consider just the top array, and for each weighing (row) place the balls corresponding to a > in the left, and the balls corresponding to a < in the right of the balance. Thus we would weigh A, B, C, D against E, F, G, H; then G, H, I, K against A, E, F, J; and finally B, E, H, I against C, G, K, L. The results of these three weighings as observed on the left of the balance are noted. If the outcome is >, <, = we conclude that ball A is heavy; if >, =, > ball B is heavy, etc; if =, =, < ball L is heavy. If we obtain an outcome that appears in the lower array, we conclude that the corresponding ball is light. Thus for <, >, = ball A is light, etc; =, =, > means ball L is light.