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>> No.12142459 [View]
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12142459

>>12142410
>Keep getting/seeing 4r/3pi

That's the centroid of a solid/filled semicircle. You need to use the centroid of a semicircular arc to find the surface area of a sphere.

>> No.12126539 [View]
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12126539

>>12126424
>BEMDAS

Nah, we call it BODMAS (O stands for order) or BIDMAS (I stands for indices/index).

>> No.12095309 [View]
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12095309

>>12095000
You need to figure out how many rolls yield a given value [math]k[/math] and divide this by the total number of possible rolls ([math]400[/math] in this case, which I hope is obvious).

When trying to calculate the number of rolls which output a given value, start small. Clearly, [math](1,1)[/math] is the only roll which will yield [math]1[/math]. Then [math](1,2)[/math], [math](2,1)[/math] and [math](2,2)[/math] will output [math]2[/math]. If you imagine the problem visually (as a [math]20 × 20[/math] grid), a pattern is already forming. [math]k=1[/math] gave a [math]1 × 1[/math] square, then [math]k=2[/math] extended this to a [math]2 × 2[/math] square. As we increase [math]k[/math] we just keep adding L-shapes to extend this square. For any value of [math]k[/math], this L-shape consists of [math]2k - 1[/math] outcomes (I'll let you verify this yourself). This gives us a probability distribution defined by [eqn]P(Y = k) = \frac{2k - 1}{400}[/eqn]

>> No.12088456 [View]
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12088456

>>12088145
Yes. You can group the $1 bills into piles of 20 and there would be exactly as many* piles of $1 bills as individual $20 bills. This quantity is denoted [math]\aleph _0[/math]

*formally, this means you can create a bijection between them https://en.wikipedia.org/wiki/Bijection

>> No.12066367 [View]
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12066367

>>12065721
>>12065721
Maybe check out the Khan Academy: www.khanacademy.org/math

There's quite a lot of algebra on there and the Precalculus section has complex numbers

>> No.12042076 [View]
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12042076

>>12041890

Write the playback speed as a fraction (e.g. [math]1.5 = \frac{3}{2}[/math]). Flip over this fraction and multiply it by the video length. The result is the time it will take to watch the video at the given speed.

>> No.12001195 [View]
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12001195

>>12001105
>>12001147
https://www.bbc.co.uk/bitesize/guides/zg6vcj6/revision/6

Just gotta keep track of the minus signs bud

>> No.11645489 [View]
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11645489

>>11645120

Yep



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