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It is indicated by experimention that the outcome of this algorithm is very efficient,especially in those problem who fulfil triangle inequation.

实验表明,此类算法求解的近似度很高,其是在满足三角不等式的问题中,误差

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ageless, agelike, agelong, Agenais, agency, agenda, agendas, agendum, Agene, agenesia,

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3G, 401(K), a,

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耶鲁开课:博弈论

It's because if you look at those inequalities, it's monotone.

这是因为如果你审视那些,它们是单调的。

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耶鲁开课:博弈论

So once you cross this inequality, you can't go back.

因此,一旦你跨越了这个,就无法回头。

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可汗院:统计(视频版)

But we can operate on all of them at the same time, this entire inequality.

但我们能够同时对所有这些进行操作,即整个

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耶鲁开课:博弈论

I've just added something to both sides of an inequality.

我刚刚在的两边都加上了一些东西。

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之美

Well, he and I just made a video on a certain quantum mechanical topic - " Bell's inequalities" .

好吧,他和我刚刚做了一个关于某个量子力主题的视频——“贝尔”。

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可汗院:统计(视频版)

So what we want to do is multiply this entire inequality by this value right over here.

因此,我们希望将整个乘以这个值。

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可汗院:统计(视频版)

Just be clear, I'm just multiplying all three sides of this inequality by this number right over here.

只是为了清楚起见,我只是将这个的三边都乘以了这里的这个

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可汗院:统计(视频版)

So what I'm going to do is multiply this entire inequality by negative 1.

因此,我打算将这个的两边同时乘以负1。

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耶鲁开课:博弈论

So, in here, the inequality is met: the sum is bigger than 1; and out here it's less than 1.

因此,在这里,得到满足:总和于 1;而在这里它小于 1。

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可汗院:统计(视频版)

Have this inequality expressed in terms of mu.

将这个用μ表示。

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可汗院:统计(视频版)

But since we are multiplying an inequality by a negative number you have to swap the inequality sign.

但由于我们是在两边乘以一个负,你必须调换的符号。

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可汗院:统计(视频版)

And then we still have our inequalities-- is going to be less than 17.17 minus the mean, which is less than 2.13.

然后我们仍然有这些——它将小于17.17减去平均值,这又小于2.13。

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耶鲁开课:博弈论

So at the beginning of the game, when people are out here, if we ask the question, is P1D plus P2D-1, is it bigger than 1?

因此,在游戏初期, 当人们还在场时,如果我们提出疑问:P1D 加上 P2D-1 是否于 1? 在场的情况下,这个是否成立呢?

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耶鲁开课:博弈论

There's going to be a critical point when, for the first time, the inequality switches from being incorrect to correct, from being false to being true.

首次从错误变为正确,从假变为真时, 将出现一个关键点。

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可汗院:统计(视频版)

If you do that, if you multiply the entire thing times negative 1, this quantity right here, this negative 2.13 will become a positive 2.13.

如果你这样做, 将整个乘以负1,这里的这个量, 即负2.13, 将会变成正2.13。

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耶鲁开课:博弈论

We're saying this inequality, which I'm claiming is going to be important, but none of you know why yet, I'm claiming is going to be important.

我们正在讨论这个, 我断言它将非常重要,尽管你们现在还不知道原因, 但我坚持认为它至关重要。

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之美

But of course, for those of you who want some more passive viewing, don't forget that Henry and I just put out a video on Bell's inequalities over on MinutePhysics.

但是当然, 对于那些想要一些被动观看的朋友们,别忘了亨利和我刚刚在分钟物理频道上发布了一段关于贝尔的视频。

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可汗院:统计(视频版)

So if I do that-- so let me just do that right over-- so if I multiply this entire-- this is really two equations or two inequalities I should say.

所以,如果我这么做——让我就在这里直接进行——如果我将整个子——这实际上是两个方程或两个,我应该说。即这个量于那个量,而那个量又于另一个量。

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耶鲁开课:博弈论

This inequality is not met, not met, not met, not met, not met, not met, not met, not met, not met, and then it's met, met, met, met, met, met.

这个未被满足,未被满足,未被满足,未被满足,未被满足,未被满足,未被满足,未被满足,未被满足,然后它被满足了,满足了,满足了,满足了,满足了,满足了。

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agglomerant, agglomerate, agglomerates, agglomeratic, agglomerating, agglomeration, agglomerative, agglomerator, agglutinability, agglutinable,

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3G, 401(K), a,
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