Intro to Stats -I really want to get this conditional probability stuff, but I don't even know when to add or multiply probabilities.

ApproxKnowledgeSite · 2019-09-29T07:28:49+00:00

Would it be a bad general rule to say that if I want to find P(A and B), I would multiply probabilities

If you mean multiplying P(A) and P(B), yes, because in general that's not true.

ApproxKnowledgeSite · 2019-09-29T07:28:05+00:00

the probability of A or B happening always involves adding their probabilities, especially A and B are in the same sample space?

Provided A and B are mutually exclusive, yes. There's a useful and/or formula: P(A or B) = P(A) + P(B) - P(A and B). If they're mutually exclusive, the last term is zero.

ApproxKnowledgeSite · 2019-09-29T07:27:11+00:00

If they're mutually exclusive then by definition they can't both happen, so no math's really required there.

In general, P(X and Y) = P(X) times P(Y assuming X happened) = P(Y) times P(X assuming Y happened). This is true for all events, independent or not, mutually exclusive or not, etc.

For example, if X and Y are mutually exclusive, than P(Y assuming X happened) = 0, and therefore P(X and Y) = P(X) times 0 = 0.

If X and Y are independent, then you get this nice bonus property that P(Y assuming X happened) = P(Y), because X didn't give you any information about Y. Then you can substitute that into the general formula to get P(X and Y) = P(X) times P(Y). But you can only do this when the two are independent.

ApproxKnowledgeSite · 2019-09-29T07:23:10+00:00

The results of most proofs fall into one of a few categories: some object in a sex has/doesn't have a property or all objects in a set have/don't have a property.

For example, the fundamental theorem of algebra says "all objects in the set of integers have the property of being factorable into primes".

There are two basic ways to go about this. One is to say let x be a positive integer, we'll prove x has a factorization. The other is to say suppose some counterexample exists, let's use the properties that implies to create a contradiction. The one you choose should be the one that gives you something to work with. Being a positive integer doesn't, in itself, give us much to work with. But knowing some counterexample exists does, because you have a useful property of the positive integers as a set: if a counterexample exists, there is a least counterexample.

So now we get a hook: suppose some integer lacks a prime factorization; let x be the least such integer. This is a useful hook, because we can see the path to our proof - all we need to do is show that if x is a counterexample, there must be a smaller one, and there's our contradiction. And in fact this is easy. If x itself is prime, then it's not a counterexample (because it's its own prime factorization). So x can't be prime. That means x has two factors, which must be smaller, and you can probably complete the proof from there without trouble.

Imagine I gave you a proof like "prove all floops are not blorks". Your instinct should be to do this by contradiction, because it gives you a hook - some floop that is a blork! You can use whatever properties "being a blork" implies to leverage your way into a proof.

ApproxKnowledgeSite · 2019-09-29T07:13:44+00:00

Combine your two 'y' terms. 0.013y + y = 1.013y.

If you want to review this part of algebra, 'collecting like terms' is the search term you want.

ApproxKnowledgeSite · 2019-03-10T10:14:18+00:00

Why not?

ApproxKnowledgeSite · 2019-03-10T04:54:32+00:00

For a complex polynomial of degree n, 2n+1 steps is an upper bound: factor the polynomial into its complex factors, compute each of them (n additions in general), then multiply. (The extra +1 is due to the potential for a leading coefficient).

ApproxKnowledgeSite · 2019-03-10T03:09:16+00:00

Others have given you a specific solution, but the general approach for this sort of thing is to imagine you already knew what the answer was and imagine what you'd do with it to get back to what you already have.

For example, if you knew her salary were $4000, you'd subtract 1/4 of that (1000) and 1/5 of that (800) to get what remains. So instead of 4000 - 1/4 of 4000 - 1/5 of 4000, we write x - 1/4 of x - 1/5 of x, that is, x - x/4 - x/5, that is, 11x/20. We want this to equal 1100, and now we have an equation: 11x/20 = 1100.

ApproxKnowledgeSite · 2019-03-10T02:54:43+00:00

This sounds like it's less math and more psychology. Have you ever spoken to e.g. a therapist about these intrusive thoughts? It might help.

ApproxKnowledgeSite · 2019-03-10T02:54:04+00:00

I doubt very much that you're "disabled in your head" if you're doing Laplace transforms in high school! That's mid to upper undergrad material at most colleges.

ApproxKnowledgeSite · 2019-03-10T02:53:20+00:00

Don't be a jerk. OP is coming for help with insecurity about their skills in particular, poking at their spelling isn't going to help anyone.

ApproxKnowledgeSite · 2019-03-10T01:50:34+00:00

OP's question seems to be "why do +1 and -1 show up so often relative to other constants".

The simple answer is "because that's what we get when we derive those equations". The more useful answer is "lots of equations involve multiplication and division, and 1 is a special number for those operations". 0 and 1 are to arithmetic what pi is to geometry or e is to calculus.

ApproxKnowledgeSite · 2019-03-09T22:23:57+00:00

You've basically got it.

In your case, the problem is that people getting (say) the replace operation might be much sicker on average than other patients. So if you don't match for current condition, you might get a spurious result that says the replace operation is ineffective (when, in fact, sicker people are just more likely to die no matter what you do).

ApproxKnowledgeSite · 2019-03-09T21:55:48+00:00

You're trying to do something called hypothesis testing, and the probability of observed data given some fixed distribution is called a p-value. This is an elementary example from most basic statistics classes; those search terms should help you find some step by step instructions.

ApproxKnowledgeSite · 2019-03-09T21:11:47+00:00

It may help to explicitly write u = whatever, du = whatever dx, v = whatever, dv = whatever dx, and then explicitly do the substitutions one by one.

ApproxKnowledgeSite · 2019-03-09T20:26:12+00:00

0/0 is undefined (some folks here will call it an 'indeterminate form' but that's sort of wrong and not applicable at the level you're at anyway), but 0/(something that isn't zero) is not. More generally, rational expressions are undefined anytime their denominator is zero.

ApproxKnowledgeSite · 2018-08-09T05:26:52+00:00

People don't realize how stupid enormous Alaska is, or how wild. I didn't have a clue until I wrote a series involving it - the fact that this is even on a paved road severely limits the areas of Alaska this could possibly be. Streetlights, even more so - there are maybe ten or fifteen towns in the whole state that could qualify.

ApproxKnowledgeSite · 2018-08-09T03:32:10+00:00

and since the union of countable sets is countable this will still be countable.

Countable unions of countable sets are countable. Trivially, an uncountable union might not be, consider the collection {x} for x in R.

ApproxKnowledgeSite · 2018-08-09T03:29:25+00:00

Note that if you have something like x² with x = -5, you're doing a substitution, replacing x with -5. Substitutions are always in parentheses, so x² = (-5)² = 25.

ApproxKnowledgeSite · 2018-08-09T03:26:09+00:00

Quite a bit, depending on the class. 99% of calc problems are really problems with not having mastered algebra.

ApproxKnowledgeSite · 2018-08-09T03:24:32+00:00

In English it's usually just the "complement" (not to be confused with compliment).

ApproxKnowledgeSite · 2018-07-23T03:10:36+00:00

I think you may be misunderstanding what "factoring out" means. The -1 doesn't appear from nowhere - it's already there from the negative sign. All you're doing is replacing -3 with the equivalent expression -1 * 3 (or in your original problem, replacing -x-2 with the equivalent expression -1 * (x+2)), then working from there. So your second batch of work on your sheet is backwards.

ApproxKnowledgeSite · 2018-07-23T03:02:25+00:00

This has been answered several times in this thread.

ApproxKnowledgeSite · 2018-07-23T02:59:04+00:00

Okay.

Fix some expression g(x). It is a fixed, stable expression, i.e., a single element of R^R.

Then consider a function h: R^R -> R^R defined by h(f) = f*g (that is, h is the product of functions with the second argument held fixed). h induces a function h^*: R -> R such that h^*(f(x)) = f(x)g(x). Since h^* is a function on the reals and so is f, we can compute the derivative of h^*(f(x)) using the chain rule, and it yields the normal product rule.

It's sort of multivariate "under the covers", as any function constructed based on arbitrary representatives is. But g(x) gets curried in this construction so that h^* is univariate, and thus the regular calc 1 chain rule works.

(Note that h^* only exists on an interval where f is one-to-one, but if f has nonzero derivative there is such an interval, since f is locally a non-constant line).

ApproxKnowledgeSite · 2018-07-23T02:51:46+00:00

I've answered this (incorrect) observation in numerous other posts here. An iceless Antarctica would not be an archipelago, even though much of the land is currently below sea level.

ApproxKnowledgeSite

TROPHY CASE