Chords have numerical values and their notes quarters, eights, sixteens, they're all just fraction variables.

And now I'm going to be really demanding.

And the how was harmony, stacking up the pitches to form chords.

Dissonance. Look...

We'll let R do the work.

And the basic chords were the ones we still have with us, the triads, either the major one, which we think is happy, or the minor one, which we perceive as sad.

The chorus that goes with the highlight Seems like supporting the high low note should do it

So you are, in fact, a chord each of you an individual chord.

Dispersion

Variance

We're almost at the correlation matrix.

So that leads me to the final concept in this, in this segment and that's just variants and co variants.

The answer will be 0.025.

Then we squared them and then we summed the square deviation scores.

If I just put the microphone down for a bit,

That chord doesn't work. Now, he tries the E again. That chord doesn't work.

A monitor for an Icecream compilation network

Distributed Process Launcher

Dispersion samples

Field variance?

Quarter note. Half note.

Then we can sum the squared deviations scores and divide by n. So if we sum this column we get 922.1 divided by ten is 92.21.

Remember the notation, the sum of squares and mean squares.

The most transformative experience I've ever had in that single moment, hearing dissonance and harmony and people singing, people together, the shared vision.

So James Surowieki has this idea of the wisdom of crowds, that says that a whole bunch of people are smarter than any individual.

That's just dividing by N.

And now I have a variance co variance matrix.

So 7 will be the correct mean.

Dispersion of the normal distribution

The sudden noise scattered the birds.

He gets to E, but it's the wrong chord. Now, he tries E again.

It's pretty cool.

Because those correlations were based on samples that had a broader range of cognitive ability.

With this formula in mind, we find that the variance of two coin flips is 0.5, and the variance of the sum of 10 coin flips which is 2.5, which is 10 times the number over here.

We gotta split them up!

Whereas over here, they're 121, 400, 196, 25, 0.

Not only can you say how many notes, it's being able to play them all at the same time.

The ee eve ning hangs be neath ... there's the potential for little lovely sounds on every one of those beats.

You already learned that the means add up. So we get 2µ, which is 1 for the µ .

So those would be the high peaks of the density field.

(Noise) So even if you're in a noisy environment like this and we all spend a lot of time in places like this

Right.

Let's go

The mean is now a vector with 1 element for each of the variance.

The way you get that is divide the 6618 by 83. How about the variance?