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Author Topic: Mathematics Help Thread  (Read 227030 times)

Vector

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Re: Mathematics Help Thread
« Reply #1140 on: August 28, 2013, 11:59:31 pm »

Yeah, I'm just recommending it because he seems to need practice with the formula.

Arithmetic isn't terrible for everyone, though.  It's not true that "all REAL mathematicians are bad at computation" or something like that.  Especially if you pay attention to what you're doing when you do it, making it meaningful, you can become quite fast.
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Jim Groovester

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Re: Mathematics Help Thread
« Reply #1141 on: August 29, 2013, 12:32:08 am »


I'm not sure how to set out the answer OR what they're really looking for...

This is a straightforward application of Bayes' Theorem.

The problem tells you to find P(G|Dc) which by Bayes' Theorem is P(G|Dc) = P(Dc|G) P(G) / P(Dc).

We know the probabilities of P(Dc|G) and P(G), but we don't know the straight-up chance of the customer not defaulting, so we need to figure this out. We use the law of total probability:

P(Dc) = P(Dc|G) P(G) + P(Dc|Gc) P(Gc) = .94 * .7 + .65 * .3 = 0.853

Substituting this back in:

P(G|Dc) = P(Dc|G) P(G) / P(Dc) = .94 * .7 / 0.853 ≈ 0.7713
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Parsely

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Re: Mathematics Help Thread
« Reply #1142 on: August 29, 2013, 06:31:33 pm »

PTW.
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Gamerlord

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Re: Mathematics Help Thread
« Reply #1143 on: August 29, 2013, 07:39:27 pm »


I'm not sure how to set out the answer OR what they're really looking for...

This is a straightforward application of Bayes' Theorem.

The problem tells you to find P(G|Dc) which by Bayes' Theorem is P(G|Dc) = P(Dc|G) P(G) / P(Dc).

We know the probabilities of P(Dc|G) and P(G), but we don't know the straight-up chance of the customer not defaulting, so we need to figure this out. We use the law of total probability:

P(Dc) = P(Dc|G) P(G) + P(Dc|Gc) P(Gc) = .94 * .7 + .65 * .3 = 0.853

Substituting this back in:

P(G|Dc) = P(Dc|G) P(G) / P(Dc) = .94 * .7 / 0.853 ≈ 0.7713
Okay, so by that working the probability of them being a 'good credit risk' is approx 75%. Right, that means MY working is definitely wrong. Back to the drawing board I suppose.

MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1144 on: August 30, 2013, 01:49:29 pm »

So I found this geometry game on Vector's tumblr. It took a while to solve all the problems, but they were tons of fun.

Spoiler: My solutions (click to show/hide)
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RedWarrior0

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Re: Mathematics Help Thread
« Reply #1145 on: September 02, 2013, 11:39:48 am »

Wait what vector has a tumblr?
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quinnr

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Re: Mathematics Help Thread
« Reply #1146 on: September 02, 2013, 01:17:18 pm »

So there is likely a very easy solution to this problem that I'm missing, but I am terrible with logarithms and haven't used them in quite a while...
The equation is just 10^(4log3), and the instructions are just to simplify the equation. I'm totally stumped though..
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Another

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Re: Mathematics Help Thread
« Reply #1147 on: September 02, 2013, 01:27:38 pm »

Is that decimal (commonly written as lg) or natural (commonly written as ln) logarithm?

Assuming decimal - 10^(4*lg(3))=10^(lg(3)*4)=(10^lg(3))^4=3^4.
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quinnr

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Re: Mathematics Help Thread
« Reply #1148 on: September 02, 2013, 01:29:32 pm »

Is that decimal (commonly written as lg) or natural (commonly written as ln) logarithm?

Assuming decimal - 10^(4*lg(3))=10^(lg(3)*4)=(10^lg(3))^4=3^4.

It's decimal.

Also I feel like a giant idiot now thanks a ton. It's just been so long since I've had to use logarithms at all, totally forgot.
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Gamerlord

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Re: Mathematics Help Thread
« Reply #1149 on: September 03, 2013, 09:42:39 pm »

Okay, in a bit I may post up some math questions. Is there anyone here who wold be willing to help me understand HOW to do them? I'm asking because I can be a bit annoying at times and that only gets worse when I don't know something.

Vector

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Re: Mathematics Help Thread
« Reply #1150 on: September 03, 2013, 10:00:58 pm »

Okay, in a bit I may post up some math questions. Is there anyone here who wold be willing to help me understand HOW to do them? I'm asking because I can be a bit annoying at times and that only gets worse when I don't know something.

If you're okay with me being annoying back, yes.
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"The question of the usefulness of poetry arises only in periods of its decline, while in periods of its flowering, no one doubts its total uselessness." - Boris Pasternak

nonbinary/genderfluid/genderqueer renegade mathematician and mafia subforum limpet. please avoid quoting me.

pronouns: prefer neutral ones, others are fine. height: 5'3".

Pnx

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Re: Mathematics Help Thread
« Reply #1151 on: September 04, 2013, 01:39:23 pm »

Find the exact value of:
Sin[ 2Cos-1(3/5) ]

I know that the inverse cosine will give me an angle value between 45 degrees, and 0 degrees, and that doubling it will give me something in the same quadrant. But I'm not sure how to get an exact value out of this algebraically.

I feel sort of embarrassed by how my brain just refuses to do math right now.
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Another

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Re: Mathematics Help Thread
« Reply #1152 on: September 04, 2013, 02:16:02 pm »

sin(2*x)=2*sin(x)*cos(x)
sin(acos(x))=sqrt(1-cos(acos(x))^2)=sqrt(1-x^2)
The sign of the above square root is always non-negative.
Spoiler: answer (click to show/hide)
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1153 on: September 04, 2013, 02:22:27 pm »

sin(acos(3/5)) = 4/5
sin(2*acos(3/5)) = 2*sin(acos(3/5))*cos(acos(3/5)) = 2*(4/5)*(3/5) = 24/25.

EDIT:

Okay, in a bit I may post up some math questions. Is there anyone here who wold be willing to help me understand HOW to do them? I'm asking because I can be a bit annoying at times and that only gets worse when I don't know something.
I'm very resistant to being annoyed. Post ahead.
« Last Edit: September 04, 2013, 02:24:02 pm by MagmaMcFry »
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Pnx

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Re: Mathematics Help Thread
« Reply #1154 on: September 04, 2013, 02:57:45 pm »

Ok, I feel like kind of a moron for not realising the double angle thingemy could be used here. Thanks.
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