Wikipedia:Reference desk/Archives/Mathematics/2007 October 20

= October 20 =

math
i need halp to do this can some one halp me place.thank you 1) -5p-7=-27 2)4n-12=28 3)2s+6=0 —Preceding unsigned comment added by 4.236.132.174 (talk) 00:40, 20 October 2007 (UTC)


 * Isolate the variables. By performing the same actions on both sides of the equation, try to get the numbers on just one side and the variable on the other. Strad 01:27, 20 October 2007 (UTC)

Furthermore, do you remember the order of operations??? Keep in mind that when solving an equation you use the reverse order, Addition/Subtraction first, Multiplication/Division 2nd, and so on. parenthesis are worked on outside to inside vs the usual inside-out method. This is because you are altering both expressions that are on either side of the equals sign. Think about it this way, when trying to solve for a variable you must undo what the expression (on the left in this case) is doing to that variable. So what happens last?? Addition/Subtraction, then next to last... etc. So add 7 to both sides, in problem 1) to eliminate the -7 from the left side, remember to add 7 to both sides or you lose your equality and thus you would not get a correct solution. We add, not subtract because we must use an opposite to eliminate the -7 term. Then you have -5p=-20, next divide both side by -5. So now you have p=4. And it is generally a good idea to check your answer by substituting it in place of p (use parenthesis so as, to insure proper order of operations) and show the the left and right sides are equal. That is not necessary but if you're not compete in your answer being correct that can verify whether or not it is. A math-wiki 05:45, 20 October 2007 (UTC)

how much a person in this profession make veterinary
i need halp plase —Preceding unsigned comment added by 4.236.132.174 (talk) 00:50, 20 October 2007 (UTC)


 * Somewhere in the region of $$\sqrt{france}\,$$, I would expect. 86.141.145.45 01:45, 20 October 2007 (UTC)


 * I find it hard to believe someone can properly spell "profession" and "veterinary" but misspell "help" and "please". --KSmrqT 06:28, 20 October 2007 (UTC)

They make $$\sqrt{-i}$$ dollars a week plus tips. 210.49.155.124 07:29, 20 October 2007 (UTC)


 * ok - this person is asking on the science desk similar questions - so why here? Anyway just out of respect please don't take the piss. Please.87.102.17.46 10:07, 20 October 2007 (UTC)

The maths desk is not really for financial advice - such as estimated wages - the answer to your question depends very much on which country you are thinking about - which you need to tell us.87.102.17.46 10:09, 20 October 2007 (UTC)

Typically a vetinarian earns above average wages or is well paid.87.102.17.46 10:48, 20 October 2007 (UTC)

Vysochanskiï-Petunin inequality
I read the article on the Vysochanskiï-Petunin inequality, and am trying to understand it, and its implications on the interpretation of control charts using 3-sigma limits, i.e. lambda=3. The article states: "The sole restriction the distribution is that it be unimodal and have finite variance. (This implies that it is a continuous probability distribution except at the mode, which may have a non-zero probability.)" I am having problems with the above, because up until now I have always thought that a probability distribution is either continuous or discrete, but the statement in parenthesis seems to imply that it can be both (continuous for some arguments, discrete for others). To me, this raises several questions:
 * Is there such a thing as a unimodal discrete probability distribution? The statement cited above seems to indicate that the answer is "no", but oviously many discrete probability distributions have only one mode.
 * If the answer to the preceding question is "yes", does the Vysochanskiï-Petunin inequality apply to a unimodal discrete probability distribution?
 * What is the probability density of a "continuous" probability distribution at the mode, if the probability is non-zero? Undefined?
 * Could someone give an example of a reasonably occurring probability density with mode=0, nonzero probability at 0, and arguments that are positive or zero, which would either satisfy nor not satisfy the requirements of the inequality?

Thank you! --NorwegianBluetalk 12:24, 20 October 2007 (UTC)
 * Probability distributions can be quite general. Given your sample space, say $$\mathcal{X}=\mathbb{R}$$ and any σ-algebra on it, you can define any function from that  σ-algebra to [0, 1] which satisfies a few axioms, and you get a probability distribution. Continuous and discrete are just two narrow, yet common, kinds. The distribution kind alluded to is not really one of them, but resembles one or the other in different regions.
 * No (unless there is just one point where the probability is positive), and this becomes clear if you take a look at the definition of a Unimodal function. The probability density for a discrete distribution is 0 for most real numbers. Hence, for any number where the probability is positive, you have a local maximum, and thus the distribution is not unimodal.
 * At the basic level it is undefined, but the Dirac delta function can be introduced to deal with it.
 * That "reasonably occurring" bit is tricky... Can't think of any right now.
 * -- Meni Rosenfeld (talk) 13:06, 20 October 2007 (UTC)
 * -- Meni Rosenfeld (talk) 13:06, 20 October 2007 (UTC)


 * Thanks a lot, Meni, I think I understood. To verify, I'll try to give examples of "reasonably occuring" probability distributions with nonzero probability at 0, and which satisfy and don't satisfy the requirement. Am I correct in thinking that the distribution

p(x<0) = 0 p(x=0) = 0.3 p(x>0) = 0.7*exp(-x)


 * satisfies the requirement, whereas the distribution

p(x<0) = 0 p(x=0) = 0.2 p(x>0) = 0.7*exp(-x)+0.1*g(x)
 * where g(x) has the properties

g(x) = 0 when x<0 The integral of g from -inf to +inf is 1 g(5) > exp(-5)


 * does not? --NorwegianBluetalk 15:34, 20 October 2007 (UTC)
 * The idea for the first looks good, but the notation isn't. The distribution you mean can be described by:
 * $$P(X\le a) = \left\{\begin{array}{ll}0&a<0\\1 - 0.7\exp(-a)&a\ge0\end{array}\right.$$
 * As for the second, same problem with the notation, and I'm not sure we know enough about g to deduce anything. If, on the other hand, we know that $$g'(5)>7\exp(-5)\;\!$$, I think we can conclude the distribution isn't unimodal. -- Meni Rosenfeld (talk) 16:49, 20 October 2007 (UTC)


 * Thanks again. I'm sorry about the notation, I knew it was awful and should have apologized beforehand. I also realize that I failed to make a clear distinction between densities and probabilities, and that I failed to take into account the factors 0.7 and 0.1 that I was multiplying exp(-x) and g(x) with. By cutting and pasting from the wiki math notation in your reply, I'll have a second go at rephrasing what I was trying to express:


 * The first distribution was intended to be:
 * $$P(a \le X \le b) = \left\{\begin{array}{ll}0&a\le b < 0\\0.3&a=b=0\\0.7\int_{a}^{b}\exp(-t)\,dt &0< a\le b\end{array}\right.$$


 * The second distribution was intended to be:


 * $$P(a \le X \le b) = \left\{\begin{array}{ll}0&a\le b < 0\\0.2&a=b=0\\0.7\int_{a}^{b}\exp(-t)\,dt + 0.1\int_{a}^{b}g(t)\,dt &0{\color{red}\mathbf{7}}exp(-5)$$


 * Sorry about persevering, I just want to make sure I understand:
 * Did I get the notation right?
 * If so, is my first distribution just a more awkward way of expressing what you did in two lines?
 * Does the first distribution satisfy the requirements of the Vysochanskiï-Petunin inequality?
 * Does the second distribution violate the requirements of the Vysochanskiï-Petunin inequality?
 * -NorwegianBluetalk 19:53, 20 October 2007 (UTC)
 * Much better. It seems a bit unconventional, but it gets the necessary information across.
 * Pretty much, yes. Basically, by calculating $$P(X \le b) - P(X \le a)$$ using my formula, you will get the same results as your formula (there are some technical details involved).
 * Yes.
 * I think not. The factor of 7 wasn't the only thing I have changed - I have given a condition on the derivative of g rather than g itself. To ensure being non-unimodal, there must be some point where the derivative of the density function (which is also the second derivative of the cumulative density function) is positive. Ensuring that with a condition on g is trickier.
 * -- Meni Rosenfeld (talk) 22:23, 20 October 2007 (UTC)
 * I understand. Your replies have been very helpful. Thank you! --NorwegianBluetalk 11:16, 21 October 2007 (UTC)

list of numbers
could you please give me a comprehensive list of all types of numbers, ( names only ) i mean to say not inly the kinds of numbers like natural, whole.integers,.etc but also the others like prime, perfect, abundant, kapreskar,ghosh number,etc..... 59.93.102.39 23:55, 20 October 2007 (UTC)


 * Try Category:Integer sequences. —Keenan Pepper 00:09, 21 October 2007 (UTC)
 * Or Category:Numbers and the sub-categories therein.87.102.16.28 09:00, 21 October 2007 (UTC)


 * This link starts to answer your question http://mathworld.wolfram.com/search/?query=number there's over 100- pages to go through though..87.102.16.28 09:00, 21 October 2007 (UTC)


 * http://www.research.att.com/~njas/sequences/Seis.html is also pretty usefull. 48v 23:15, 21 October 2007 (UTC)