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February 6

Sterlings

How much is 4,596,411 pounds of sterling in U.S dollars.

[1] Black Carrot 00:56, 6 February 2007 (UTC)[reply]

9.04389828 million U.S. dollars, at whatever rate of exchange Google is using currently.…86.146.174.174 01:50, 6 February 2007 (UTC)[reply]

Sum of Normal distributions

It looks to me that if f(x) and g(x) are normal PDFs, then (f(x)+g(x))/2 is also a normal PDF. Is this so? What are the parameters of this new PDF? --대조 | Talk 09:22, 6 February 2007 (UTC)[reply]

Unfortunately, no. I think this is never the case (unless f and g are the same), but it is easiest to consider distributions which have very different means and very low variance. Then the average PDF will have two peaks, so the distribution will be bimodal which is clearly not normal. Compare taking the average (or any linear combination) of the variables which is normal, with the obvious mean and a variance which depends on the covariance of the variables. -- Meni Rosenfeld (talk) 09:45, 6 February 2007 (UTC)[reply]

And, in the latter case of linear combination of the variables, the pdf should be a convolution of the two original pdf. --Lemontea 01:37, 27 February 2007 (UTC)[reply]

You are, of course, right. Thank you. Once again, I am reminded that to get the right answer one must ask the right question. --대조 | Talk 14:07, 6 February 2007 (UTC)[reply]

Kurtosis

What is the difference between kurtosis, as defined in our article, and Karl Pearson's measure of kurtosis, as defined by the R package 'moments'? I realise that the former subtracts 3 while the latter does not, but is that all, or are there other differences? --NorwegianBlue^talk 09:59, 6 February 2007 (UTC)[reply]

Found this while browsing for an answer, and I realise that the concept of kurtosis may be more complex than I thought when posting the question. Grateful for any input, though. --NorwegianBlue^talk 11:47, 6 February 2007 (UTC)[reply]

If it takes 20 men 6 hours to dig a hole, how long will it take 10 men to dig half a hole?

Haha --AMorris (talk)●(contribs) 10:09, 6 February 2007 (UTC)[reply]

3 hours. Plus some time to figure out how they define half a hole. Since holes are usually cylindrical prisms, I would argue that half a hole would be shaped like a hemispherical prism. Depth would have to be standardized to some reference hole, so one would have to add some time to find a suitable reference hole. Manholes wouldn't be suitable because you'd need access to the sewrage system, unless one of the men was working for the city and was part of a hole-digging convention.

So what have we learned from all this? It's three hours and a bit, depending what the men work as when they're not digging holes and some thinking time. Hope that helps!

Half a hole is a hole. Proto::► 12:12, 6 February 2007 (UTC)[reply]

No, it isn't, because you have found a suitable reference hole.

Treating this as a serious question, the standard method is as follows. Work out man-hours per hole. Divide by 2 to find man-hours for half a hole. Divide by 10 to find hours it takes team of 10 to put in this many man-hours. And the result is not 3 hours. But it really depends on how you define half a hole, as everyone else has pointed out. Gandalf61 14:58, 6 February 2007 (UTC)[reply]

If you mean a hole with half the volume, then the answer is 6 hours, not 3. --Wirbelwindヴィルヴェルヴィント (talk) 18:45, 6 February 2007 (UTC)[reply]

If this is a serious question, we have a serious problem. For 20 men to take 6 hours to dig a hole, that must be one awesome hole! Is it the size of the Chunnel? Is it dug in solid granite? Are they paid by the hour? Do they spend all their time stepping on each others' toes? Recommended reading: The Mythical Man-Month. Chances are in real life the 10 men can dig half a hole much more quickly, because they'll spend less time in meetings, and they won't have to invest the 80% of effort required for the final 20% of full completion. --KSmrq^T 19:08, 6 February 2007 (UTC)[reply]

In real life I've seen 20 'men' take six hours to remove one half wheelbarrow of turf - I was one of those men.87.102.77.140 19:32, 6 February 2007 (UTC)[reply]

Whoops. So it's 6 hours and a bit, with provisos.

It's a trick question, as there's no such thing as half a hole. I've seen this before. (yes, if you used a 'reference hole', there'd be an answer, if we assume that the hole is sufficiently large for its circumference to allow 10 men to dig it simultaneously). Proto::► 15:40, 9 February 2007 (UTC)[reply]

I found this question in a puzzle book. It says that there's no such thing as half a hole, since half a hole is a hole.

Questions that pretend to be clever when they are not are annoying! If any partial progress towards the first "hole" mentioned in the question counts as "a hole", why does it "take" 20 men 6 hours to dig it? Based on the book's answer, when the men are half way into the digging (actually even much earlier), they'd have already got "a hole". It wouldn't "take" the rest of the work to "dig a hole". --72.78.237.95 21:41, 14 February 2007 (UTC)[reply]

Quadratic equations

In school, we learnt about quadratic equations:

1. Their form: ax² + bx + c = 0

2. The three methods of solving them (factorisation, completing the square and the quadratic formula)

3. How to use b² - 4ac to find the nature of the roots of a quadratic equation.

I find quadratic equations fascinating. We're going to learn more about them next year, but I don't have next year's textbook yet. So, to help me prepare for next year, and my future, could you please tell me: 2x+4y=10,x+5y=11 1. Differentiation is to calculus as quadratic equations are to...what? Which fields of mathematics do quadratic equations play a major role in?

2. Some important applications of quadratic equations, especially in computing. —Preceding unsigned comment added by 218.186.9.3 (talk • contribs)

Sounds as if you have understood the relationship between roots and factors, and you have covered the different methods of solving quadratics. Do you understand the link between completing the square and the quadratic formula ? Can you show that they both give the same result ? Not sure if you have covered complex numbers and so complex roots yet i.e. what you do when b² - 4ac is negative - maybe you will cover that next year. To answer your Q1, quadratic equations are a special case of polynomial equations - but they come up in many different areas of mathematics, just because they are the next simplest case after linear functions. Gandalf61 12:44, 6 February 2007 (UTC)[reply]

Since x²+y²=r² is the equation of a circle - r is the radius - so for a given x , y is the solution of a simple quadratic - means that quadratic equations turn up all the time when considering circles.. and in fact ellipses, parabolas, and spheres to name a few. Quadratic equations turn up almost all the time in further mathematics - one reason being that pythagorus's theorem is quadratic in form (it uses powers of 2 eg x squared) - So you'll be seeing a lot of them.. especially in relation to triangles, angles, also movement under the effect of gravity.

For the reasons give above quadratic equations often come up in computer graphics - typically when drawing simple curves.

They also turn up in least squares statistical analysis (line fitting) which you may have or may have not done yet.

There are many more examples which I'm sure others will be able to give you.87.102.13.26 14:21, 6 February 2007 (UTC)[reply]

This all depends on which year you are in now! I can assume that you are in year 100/11 and doing GCSEs. Quadratics are in many branches of maths. they fit into differentiation/intergration, the main topic of Polynomials, Graph work and many more.MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 21:58, 12 February 2007 (UTC)[reply]

Hello, and thanks for the replies, everyone. I'm taking my O Levels this year, so I'll be studying the A Level curriculum in the next 2 years. I understand that the quadratic formula is derived from the completing the square formula. We have not covered complex roots yet, so I'd appreciate any information on them. Thanks for telling me the formulas which usually become quadratic equations when linear functions are substituted into them. Since cubic equations and quadratic equations are somewhat similar, do cubic equations also have a discriminant and formula?

Cubics do have a formula, but it is very long and messy, and you don't really need it ever, higher grade polymers (than quadratic) don't have discriminants, and with odd functions, the graph will always have at least one route. MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 19:01, 14 February 2007 (UTC)[reply]

Actually, discriminants can be defined for any polynomial (see discriminant) but it gets very complicated. For example, the cubic

ax^{3}+bx^{2}+cx+d

has discriminant

b^{2}c^{2}-4ac^{3}-4b^{3}d-27a^{2}d^{2}+18abcd

. These higher-order discriminants are first taught to third-year undergraduates (at least where I am) so you probably don't need to know about them unless you're feeling especially keen. Algebraist 15:10, 17 February 2007 (UTC)[reply]

What's x and y

If $0,2x-0,1y=0,03$ and $0,4x+1,2y=3$ , what is the values of $x$ and $y$ ? I've figured out that it is $0,255$ and $2,1$ respectively, but I do not think that is correct...? Lol. Please help? This is kinda a homework question, yes... ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 13:29, 6 February 2007 (UTC)[reply]

Multiply the first equation by -2 and add it to the second to get y. Then put your value for y back into one of the equations (your choice) to get x. –King Bee ^{(T • C)} 14:43, 6 February 2007 (UTC)[reply]

Here's a longer explanation..(try the above first it probably is a better way to learn how to do this)

The standard way to solve these linear equations is to first eliminate either x or y (it's your choice which) eg you had

0.2x - 0.1y = 0,03

adding 0.1y to both sides, the equation becomes:

0.2x = 0.03 + 0.1y  (multiply by five is the same as divide by 0.2 to give just x..)
x = 0.15 + 0.5y

The second equation was

0.4x + 1.2y = 3 (replacing x by 0.15 + 0.5y  ..)
0.4(0.15 + 0.5y) + 1.2y = 3 (expand the multiplication in the brackets)
0.6 + 0.2y + 1.2y = 3
0.6 + 1.4y = 3
1.4y = 2.4
14y = 24
y = 24/14 = 12/7 =1 5/7

You can do the same for y to get a value for x - unfortunately I can't find a simple explanation for this on wikipedia.

However you could look at http://www.themathpage.com/alg/simultaneous-equations.htm , this page is also quite clear http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-simultaneous1.pdf (neads adobe acrobat). The equations you have are called "simultaneous linear equations". There 'simultaneous' because there are two or more variables in one equation. (linear because there are no powers like x squared or x cubed..). If you want to search for other answers - be warned a lot of the pages cover examples where there are many unknowns eg 4a+7b+1.3c+0.5d = 4 ... the method is the same but might seem confusing.

Try the links or look at the method above and see if it can make sense. If you didn't understand the steps I used above then ask.87.102.13.26 14:46, 6 February 2007 (UTC)[reply]

I think I more or less understand, thanks. I had trouble multiplying the co-efficients(?) of the x en y (whatever you call that in English lol (changeable?)) so that I can add or subtract the one equation from the other in order to eliminate one of the variables, yes that's the word lol. Anyway, thanks for the help. ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 18:11, 6 February 2007 (UTC)[reply]

OK good - as for multiplying the coefficients - what I'd do is first divide so that the coefficient is 1 eg:

4x+3y=15  divide by 4 gives x+0.75y = 15/4
or 0.2x + 7y = 1 divide by 0.2 gives x + 7/0.2y = 1/0.2 ie x+35y=5

and then rearrange to get the equation in terms of x eg x= 15/4 - 0.75y

then you can replace x in the other equation... (that's two steps - as you get better at it you might be able to do it in just one step)87.102.13.26 18:19, 6 February 2007 (UTC)[reply]

K cool, thanks. ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 10:17, 7 February 2007 (UTC)[reply]

???

4³?

4³ is a way of writing 4 × 4 × 4 = 64. (In the same way, 4⁵ = 4 × 4 × 4 × 4 × 4, and 4ⁿ = 4 × 4 × ... × 4, n times). Darkhorse06 17:51, 6 February 2007 (UTC)[reply]

I think this was a yes or no question… Okay, I kid, but more generally, and for the glorious sake of redundancy:

a^{n}=\underbrace {a\times \cdots \times a} _{n},

− Twas Now ( talk • contribs • e-mail ) 06:08, 8 February 2007 (UTC)[reply]

Or you can use a calculator. —Preceding unsigned comment added by Johnnyb2963 (talk • contribs)

...which is 4 [x^y button] 3 --h2g2bob 15:42, 17 February 2007 (UTC)[reply]

Independent events

Easy question. By the definition of two independent events A and B, Pr(A ∩ B) = Pr(A)Pr(B). It can also be shown that A is independent of B′ (the complement event of B) and vice-versa. What I'd like to know is whether A′ and B′ are independent, given that A and B are independent. Thanks in advance. -- mattb @ 2007-02-06T19:31Z

Yes. If A and B are independent then, as you said, A is independent of B′. Now, using the exact same result, with B′ as the first event and A as the second, this implies the A′ is independent of B′. -- Meni Rosenfeld (talk) 19:43, 6 February 2007 (UTC)[reply]

Ah, so simple. I'm embarrassed to have missed it. Thank you. -- mattb @ 2007-02-06T20:16Z