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Sloths, HELP!

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Mar 26 '11
Question from pegghetti.
Anyone?

[ALSO, SIDE NOTE: Sorry it took like 2 days to get these up—I’ve been on holiday and just completely forgot to keep checking the inbox! Apologies!!!]

Question from pegghetti.

Anyone?

[ALSO, SIDE NOTE: Sorry it took like 2 days to get these up—I’ve been on holiday and just completely forgot to keep checking the inbox! Apologies!!!]

Mar 26 '11

find the number of symmetric relations on a set of 3 elements.

so say I have A={a,b,c}
so we have relation R on A
which gives us R includes {aRa, aRb, aRc, bRa, bRb, bRc, cRc, cRa, cRb}

then…all of them are symmetric, right? which means we have 9 different symmetric relations on A?
just checking my answer haha. not too sure what to do P:

- ima-gina-ation

Anybody out there know the answer? :)

Mar 17 '11
So, I want to note that I used WolframAlpha for this—and for the same reason you should if you don’t know how to evaluate an integral (or if you’re just lazy like me because you realized that knowing how to integrate a hundred different types of integrals is generally wasted brain space): it gives you all intermediary steps for one possible solution to integrating the integrand. 
Hope this helped, whoa-bro!

So, I want to note that I used WolframAlpha for this—and for the same reason you should if you don’t know how to evaluate an integral (or if you’re just lazy like me because you realized that knowing how to integrate a hundred different types of integrals is generally wasted brain space): it gives you all intermediary steps for one possible solution to integrating the integrand. 

Hope this helped, whoa-bro!

Mar 16 '11

“For #s B-D, do you know if they’re asking for marginal or conditional probability (or something else)?” - blank-infinity

For B through D, they’re asking for conditional probabilities. This is because they’re asking for the probability of one event given that the other has already happened.
Just in case you need further steps into solving these problems, here’s my attempt at explaining:
So, for example, in B, the question is asking for P(Ulcer | None). First, you find the marginal total for None. Like you wrote in the margins, it’s 0.01+0.22 = 0.23. I find that it’s also helpful to circle/mark the relevant cells in the table (the ones you added up to get to P(None).
So since P(None) = 0.23 by adding up the probabilities in the “None” row, P(Ulcer | None) can be found by looking for the cell corresponding to P(Ulcer AND None). That’s 0.01.
Just think of this like a fraction, right? 0.01 out of the total 0.23 relevant to the question.
Finally, then, our answer is:
P(Ulcer | None) = 0.01/0.23 = bout 0.04
Hope that helped! :)

For #s B-D, do you know if they’re asking for marginal or conditional probability (or something else)?” - blank-infinity

For B through D, they’re asking for conditional probabilities. This is because they’re asking for the probability of one event given that the other has already happened.

Just in case you need further steps into solving these problems, here’s my attempt at explaining:

So, for example, in B, the question is asking for P(Ulcer | None). First, you find the marginal total for None. Like you wrote in the margins, it’s 0.01+0.22 = 0.23. I find that it’s also helpful to circle/mark the relevant cells in the table (the ones you added up to get to P(None).

So since P(None) = 0.23 by adding up the probabilities in the “None” row, P(Ulcer | None) can be found by looking for the cell corresponding to P(Ulcer AND None). That’s 0.01.

Just think of this like a fraction, right? 0.01 out of the total 0.23 relevant to the question.

Finally, then, our answer is:

P(Ulcer | None) = 0.01/0.23 = bout 0.04

Hope that helped! :)

Mar 14 '11
Another solution! I’m not sure who posted this question since in the picture it says anonymous.
Thank you to intothecontinuum for answering another question!

Another solution! I’m not sure who posted this question since in the picture it says anonymous.

Thank you to intothecontinuum for answering another question!

Mar 14 '11
Looks like we’ve got an answer to sigma-notation's question!

Thank you intothecontinuum for your answer!

There is still one unanswered question below for anyone willing to help

Looks like we’ve got an answer to sigma-notation's question!

Thank you intothecontinuum for your answer!

There is still one unanswered question below for anyone willing to help

Mar 14 '11
Mar 14 '11
1 note Tags: anyone?
Mar 13 '11

Hello, World!

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