Women outscore men in IQ tests for first time since records began

Well men, looks like they are hotter and smarter. I kinda suspect this isnt the first time really but the first time a man hasn't been able to tamper with the results...

http://www.telegraph.co.uk/education/educationnews/9401241/IQ-tests-women-score-higher-than-men.html

I think women probably always knew deep down that they were the more intelligent ones but as the gentler sex we were quiet about it and let men continue to believe they ruled the world.

Fascinating...

So who invented the car? The answer escapes me my dear chaps.

That's nice.

i read the article. what the hell was the average? i read that in the past women lagged by as much as 5 pts. so what was the new average according to these exams?

I've never been one to take to heart an exam that is supposed to measure your intelligence, as a means of showing how inteligent someone was. It usually almost always means to me that your good at taking exams.

That said, first time women are in the lead, and i read a bunch of generalizations and theories of latent female domination as if saying "if only they realized".

if they were all that smart then they would all still be in the kitchen only speaking when spoken too :lol:

I kid i kid, just wanted to say it before someone else did ;)

Well men, looks like they are hotter and smarter.Sedin44

That has yet to be decided.

I don't buy it. The IQ score difference between men and women is very small, and the IQ advantage or disadvantage of women relative to men varies from country to country. This is probably just some result blown out of proportion.

Also, the comments on that article are unbelievably stupid - what a waste of brain power. That article is a painful reminder of why I never like to read comments on Telegraph articles in general.

Okay then.

Our society used to be biased against women, now it's biased against men. dkdk999

Our society used to be biased against women, now it's biased against men.dkdk999

I also heard recently that water is the driest substance on the planet.

Also, the comments on that article are unbelievably stupid - what a waste of brain power. That article is a painful reminder of why I never like to read comments on Telegraph articles in general.ghoklebutter

I am guessing a woman did this study. If not it was one of those men that come to the aid of feminists thinking he is going to thin out the herd of men for himself and get laid. Either scenario, it doesn't matter. It's the government who helped women steal everything from men and most likely the same who hosted these tests and then graded them. Nothing but the same favoritism, biased, helping women win every divorce, stealing every man's child government who has been doing it all alone. I think ANYONE is smarter when the Government does all of your thinking and fighting for you

I think women probably always knew deep down that they were the more intelligent ones but as the gentler sex we were quiet about it and let men continue to believe they ruled the world.

They can't be much worse than some of the stuff that will be and has already been posted in this thre-

[quote=""]I am guessing a woman did this study. If not it was one of those men that come to the aid of feminists thinking he is going to thin out the herd of men for himself and get laid. Either scenario, it doesn't matter. It's the government who helped women steal everything from men and most likely the same who hosted these tests and then graded them. Nothing but the same favoritism, biased, helping women win every divorce, stealing every man's child government who has been doing it all alone. I think ANYONE is smarter when the Government does all of your thinking and fighting for youBlood-Scribe

I love how that commenter went on to post that while ignoring the fact that the person who did the study is male and that he isn't even sure if this result is a definite trend.

[QUOTE="Blood-Scribe"]

They can't be much worse than some of the stuff that will be and has already been posted in this thre-

[quote=""]I am guessing a woman did this study. If not it was one of those men that come to the aid of feminists thinking he is going to thin out the herd of men for himself and get laid. Either scenario, it doesn't matter. It's the government who helped women steal everything from men and most likely the same who hosted these tests and then graded them. Nothing but the same favoritism, biased, helping women win every divorce, stealing every man's child government who has been doing it all alone. I think ANYONE is smarter when the Government does all of your thinking and fighting for youghoklebutter

I love how that commenter went on to post that while ignoring the fact that the person who did the study is male and that he isn't even sure if this result is a definite trend.

Oh, I'm sorry, which gender has had a bigger impact on science, history, and all the important elements of society? Yeah, I thought so. Honestly, everyone is stupider these days. Not just men. Reed_Bowie

Oh, I'm sorry, which gender has had a bigger impact on science, history, and all the important elements of society? Yeah, I thought so.Reed_Bowie

Ok.

[QUOTE="Reed_Bowie"]Oh, I'm sorry, which gender has had a bigger impact on science, history, and all the important elements of society? Yeah, I thought so.Blood-Scribe

what r u doing

[QUOTE="Reed_Bowie"]Oh, I'm sorry, which gender has had a bigger impact on science, history, and all the important elements of society? Yeah, I thought so.Blood-Scribe

Also Davey power series solutions to linear DEs are killing me

fvcking recursion formulas :{

Because there wasn't really anything to elaborate upon?Blood-Scribe

[QUOTE="Blood-Scribe"]Because there wasn't really anything to elaborate upon?Sajo7

[QUOTE="Blood-Scribe"]Because there wasn't really anything to elaborate upon?Sajo7

[QUOTE="Blood-Scribe"]Because there wasn't really anything to elaborate upon?Sajo7

[QUOTE="Sajo7"][QUOTE="Blood-Scribe"]Because there wasn't really anything to elaborate upon?Blood-Scribe

Give the Kardashians the test. That'll even the odds.

[QUOTE="Reed_Bowie"]Oh, I'm sorry, which gender has had a bigger impact on science, history, and all the important elements of society? Yeah, I thought so.Blood-Scribe

[QUOTE="Blood-Scribe"][QUOTE="Reed_Bowie"]Oh, I'm sorry, which gender has had a bigger impact on science, history, and all the important elements of society? Yeah, I thought so.TheWalkingGhost

[QUOTE="ghoklebutter"][QUOTE="Blood-Scribe"] Eugh, hate anything that involves series solutions. You just plug and chug but the whole process takes up a whole goddamn page. Just wait until you get to Laplace transforms, IT'LL BE WORTH IT.Blood-Scribe

It's pretty nice actually - here's a typical example:

y'' + 5y' + 6y = e^x

[D^2]y + 5Dy + 6y = e^x

[D^2 + 5D + 6](y) = e^x

Because [D - 1]e^x = De^x - e^x = e^x - e^x = 0

Multiply each side by [D - 1] to annihilate the right side of the DE:

[D - 1][D^2 + 5D + 6](y) = [D - 1][D + 2][D + 3](y) = 0

D = -3, -2, 1, where D = 1 is a root for the nonhomogenous solution and D = -3, -2 are roots for the homogenous one.

The complete solution is y(x) = ae^(-3x) + be^(-2x) + ce^x

The only real downside is that it only works for nonhomogenous equations that involve exponential, polynomial, and/or trigonometric functions on the right side.

[QUOTE="Blood-Scribe"][QUOTE="ghoklebutter"] I've studied Laplace transforms a bit - they're absolutely lovely! Also, I may be voicing a minority opinion here, but I truly adore the annihilator method (the one that involves "factoring out" the dependent variable out of the derivatives of a constant coefficient DE to find a characteristic polynomial consisting of differential operators).ghoklebutter

It's pretty nice actually - here's a typical example:

y'' + 5y' + 6y = e^x

[D^2]y + 5Dy + 6y = e^x

[D^2 + 5D + 6](y) = e^x

Because [D - 1]e^x = De^x - e^x = e^x - e^x = 0

Multiply each side by [D - 1] to annihilate the right side of the DE:

[D - 1][D^2 + 5D + 6](y) = [D - 1][D + 2][D + 3](y) = 0

D = -3, -2, 1, where D = 1 is a root for the nonhomogenous solution and D = -3, -2 are roots for the homogenous one.

The complete solution is y(x) = ae^(-3x) + be^(-2x) + ce^x

The only real downside is that it only works for nonhomogenous equations that involve exponential, polynomial, and/or trigonometric functions on the right side.

e: Would've been nice to have known that instead of having to guess with undeterminied coefficients ;[

[QUOTE="Blood-Scribe"][QUOTE="ghoklebutter"] I've studied Laplace transforms a bit - they're absolutely lovely! Also, I may be voicing a minority opinion here, but I truly adore the annihilator method (the one that involves "factoring out" the dependent variable out of the derivatives of a constant coefficient DE to find a characteristic polynomial consisting of differential operators).ghoklebutter

It's pretty nice actually - here's a typical example:

y'' + 5y' + 6y = e^x

[D^2]y + 5Dy + 6y = e^x

[D^2 + 5D + 6](y) = e^x

Because [D - 1]e^x = De^x - e^x = e^x - e^x = 0

Multiply each side by [D - 1] to annihilate the right side of the DE:

[D - 1][D^2 + 5D + 6](y) = [D - 1][D + 2][D + 3](y) = 0

D = -3, -2, 1, where D = 1 is a root for the nonhomogenous solution and D = -3, -2 are roots for the homogenous one.

The complete solution is y(x) = ae^(-3x) + be^(-2x) + ce^x

The only real downside is that it only works for nonhomogenous equations that involve exponential, polynomial, and/or trigonometric functions on the right side.

Get that moonspeak off my OT.

[QUOTE="ghoklebutter"]

[QUOTE="Blood-Scribe"] We never really went over the annihilator method in depth, sadly. The professor just mentioned it off-hand and showed us the notation and that was pretty much it. Didn't even do an example with it :< Although it sounds a lot like undetermined coefficients (which was balls and completely inferior to variation of parameters).CHOASXIII

It's pretty nice actually - here's a typical example:

y'' + 5y' + 6y = e^x

[D^2]y + 5Dy + 6y = e^x

[D^2 + 5D + 6](y) = e^x

Because [D - 1]e^x = De^x - e^x = e^x - e^x = 0

Multiply each side by [D - 1] to annihilate the right side of the DE:

[D - 1][D^2 + 5D + 6](y) = [D - 1][D + 2][D + 3](y) = 0

D = -3, -2, 1, where D = 1 is a root for the nonhomogenous solution and D = -3, -2 are roots for the homogenous one.

The complete solution is y(x) = ae^(-3x) + be^(-2x) + ce^x

The only real downside is that it only works for nonhomogenous equations that involve exponential, polynomial, and/or trigonometric functions on the right side.

Get that moonspeak off my OT.

[QUOTE="ghoklebutter"]

[QUOTE="Blood-Scribe"] We never really went over the annihilator method in depth, sadly. The professor just mentioned it off-hand and showed us the notation and that was pretty much it. Didn't even do an example with it :< Although it sounds a lot like undetermined coefficients (which was balls and completely inferior to variation of parameters).CHOASXIII

It's pretty nice actually - here's a typical example:

y'' + 5y' + 6y = e^x

[D^2]y + 5Dy + 6y = e^x

[D^2 + 5D + 6](y) = e^x

Because [D - 1]e^x = De^x - e^x = e^x - e^x = 0

Multiply each side by [D - 1] to annihilate the right side of the DE:

[D - 1][D^2 + 5D + 6](y) = [D - 1][D + 2][D + 3](y) = 0

D = -3, -2, 1, where D = 1 is a root for the nonhomogenous solution and D = -3, -2 are roots for the homogenous one.

The complete solution is y(x) = ae^(-3x) + be^(-2x) + ce^x

The only real downside is that it only works for nonhomogenous equations that involve exponential, polynomial, and/or trigonometric functions on the right side.

Get that moonspeak off my OT.

[QUOTE="ghoklebutter"]

[QUOTE="Blood-Scribe"] We never really went over the annihilator method in depth, sadly. The professor just mentioned it off-hand and showed us the notation and that was pretty much it. Didn't even do an example with it :< Although it sounds a lot like undetermined coefficients (which was balls and completely inferior to variation of parameters).Blood-Scribe

It's pretty nice actually - here's a typical example:

y'' + 5y' + 6y = e^x

[D^2]y + 5Dy + 6y = e^x

[D^2 + 5D + 6](y) = e^x

Because [D - 1]e^x = De^x - e^x = e^x - e^x = 0

Multiply each side by [D - 1] to annihilate the right side of the DE:

[D - 1][D^2 + 5D + 6](y) = [D - 1][D + 2][D + 3](y) = 0

D = -3, -2, 1, where D = 1 is a root for the nonhomogenous solution and D = -3, -2 are roots for the homogenous one.

The complete solution is y(x) = ae^(-3x) + be^(-2x) + ce^x

The only real downside is that it only works for nonhomogenous equations that involve exponential, polynomial, and/or trigonometric functions on the right side.

e: Would've been nice to have known that instead of having to guess with undeterminied coefficients ;[