How did you find this out?!
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The Math Help Thread
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How did you find this out?!
nincompoop
Banned
By looking at the probability tree. The probability of selecting a red ball from the first urn is 2/5 and the probability of selecting a white ball is 3/5. So you know the first urn has some multiple of 5 balls. Maybe it has 2 red and 3 white, or maybe it has 4 red and 6 white, or maybe it has 120 red and 180 white, who knows. (Actually I don't think you can conclude exactly how many balls the first urn has with the information they give you, just that it has a multiple of five balls.)
Anyway, after you put the first ball in the second urn, either your probability of picking red is 5/7 and your probability of picking white is 2/7, or your probability of picking red is 4/7 and your probability of picking white is 3/7, depending on which ball you picked from the first urn. So it must have started out with 4 red balls and 2 white balls, because then if you put in a red ball from the first urn you would have 5 red and 2 white and if you put in a white ball from the first urn you would have 4 red and 3 white.
Anyway, after you put the first ball in the second urn, either your probability of picking red is 5/7 and your probability of picking white is 2/7, or your probability of picking red is 4/7 and your probability of picking white is 3/7, depending on which ball you picked from the first urn. So it must have started out with 4 red balls and 2 white balls, because then if you put in a red ball from the first urn you would have 5 red and 2 white and if you put in a white ball from the first urn you would have 4 red and 3 white.
You just blew my mind. I was looking at the whole tree the wrong way. Thanks a million!!
Genesis Knight
Member
I need some help. Does anyone have a clue of how to do an inverse Laplace transform of this equation:
You should try to do a partial fraction expansion. Wolfram Alpha might be able to help. If not, just try to do the partial fraction expansion yourself.I need some help. Does anyone have a clue of how to do an inverse Laplace transform of this equation:
Genesis Knight
Member
Wolfram Alpha gives me a solution, but I'm just learning this concept so I'm having difficulty doing a partial fraction expansion. Would it look like this:
I need some help with probability again. The question is : Find the probability that a 5-card hand dealt from a standard 52-card deck will have 5 spades or 5
hearts.
I did 52C5 to find all possible hands and got 2,598,960. After that I'm just straight lost. Any help would be amazing.
edit: got it, (13/53)(12/52)(11/52)(10/52)(9/52)+(13/53)(12/52)(11/52)(10/52)(9/52) = .00099
hearts.
I did 52C5 to find all possible hands and got 2,598,960. After that I'm just straight lost. Any help would be amazing.
edit: got it, (13/53)(12/52)(11/52)(10/52)(9/52)+(13/53)(12/52)(11/52)(10/52)(9/52) = .00099
Wolfram Alpha gives me a solution, but I'm just learning this concept so I'm having difficulty doing a partial fraction expansion. Would it look like this:
If you replace A/(s+1) with (As + D)/(s+1)^2 then it should be fine. You might want to look up how to do it properly though, it gets a bit more complicated the more powers of s you have in the denominator.
Polydeuces
Member
Really struggling on optimization, I wasn't really good with geometry and word problems. Is there any websites that will help me on this? If anything ill post the problem here and if I can get any hints, it will be greatly appreciated. This is for calculus btw. Im primarily using derivatives which Im good at.
nincompoop
Banned
No, you would use (As + D)/(...) if you had an irreducible quadratic factor in the denominator. If you have a linear term raised to a higher power then your partial fraction decomposition will consist of one term for each power, each with a constant as the numerator. So if (s + 1)^2 is a factor of the denominator then the partial fraction decomposition will have A/(s+1) + B/(s+1)^2.
Have you seen this function? f(x) = [1/2a when -a < x < a, 0 if |x| > a].
You can think of the impulse function as this f(x) when a approaches 0. The area below this function is always 1.
Thanks. I was just going by memory, and it has been a while since I performed partial fraction expansion.
Horse Detective
Why the long case?
Can anyone on Gaf help me with Geometric sequences?
Sure I'll give it a go, what's the problem?
Oh and
Thanks for this. I later found it in the textbook too (wasn't mentioned in our lecture notes).
systemfehler
Member
Hi!
I am just doing some calculations on electric circuits (switches, RLC) and need some help with complex numbers. It is too long ago and honestly I never liked them
phi0 = arg(I) + arg(R) - arg(R + j*w*L) (comes from the equation: abs(I * R/(R+j*w*L)))
translates to phi + 0 - tan^-1(w*L/R) if I am correct with I = Î * exp^(j*phi) because
z = tan^-1(b/a) and in that case a = R and b = j*w*L.
What happens in this case:
phi0 = arg(I) + arg(R) - arg(r + j*w*l + 1/(j*w*C))? I just added up j*w*l + 1/j*w*c to (w^2*L*C - 1)/(w*C) and said now this is "b" and a = R again. Is that correct?
Thanks in advance, didn't know where else to post this - too bad there is no electrical engineering help OT
I am just doing some calculations on electric circuits (switches, RLC) and need some help with complex numbers. It is too long ago and honestly I never liked them
phi0 = arg(I) + arg(R) - arg(R + j*w*L) (comes from the equation: abs(I * R/(R+j*w*L)))
translates to phi + 0 - tan^-1(w*L/R) if I am correct with I = Î * exp^(j*phi) because
z = tan^-1(b/a) and in that case a = R and b = j*w*L.
What happens in this case:
phi0 = arg(I) + arg(R) - arg(r + j*w*l + 1/(j*w*C))? I just added up j*w*l + 1/j*w*c to (w^2*L*C - 1)/(w*C) and said now this is "b" and a = R again. Is that correct?
Thanks in advance, didn't know where else to post this - too bad there is no electrical engineering help OT
Sorry to bump this old thread, but I'm practicing for a college placement exam and was hoping someone on here would explain something for me.
I'm working with the equation"(-4^2 - 8 * 4)/-4", in fraction form. Naturally, I squared my -4 to +16 and then completed my arithmetic. But, for whatever reason, the test I'm taking is insisting that my -4 remains a negative rather than becoming a positive. It has been a while since I studied exponents, but I can't remember a single example of a negative remaining a negative when squared. Am I forgetting an exception to the rule here?
I'm working with the equation"(-4^2 - 8 * 4)/-4", in fraction form. Naturally, I squared my -4 to +16 and then completed my arithmetic. But, for whatever reason, the test I'm taking is insisting that my -4 remains a negative rather than becoming a positive. It has been a while since I studied exponents, but I can't remember a single example of a negative remaining a negative when squared. Am I forgetting an exception to the rule here?
It's just about precedence. (-4)^2 is different from -4^2
Yup, this is correct. Remember -4^2 is like saying -1*4^2.
It isn't a law; it's order of operations. PEMDAS, right? E, for exponents, comes before M, for multiplication (which is what a negative sign is). In -4^2, the 4^2 comes first. Thus, -16.
Parentheses takes precedence over exponents, so that's why (-4)^2 = 16.
I need to take my college placement test for math. Around here MATH 1010 is the basic level math class for college, the minimum you'll need for a degree. I'm not sure what it is called. 1050 is College Algebra and below 1010 are 900-level remedial classes.
Anyway, I'm hoping to test into 1010. Can anyone who knows be able to tell me what I would need to study for the test? I'm thinking that it would including factoring, but I'm not sure what else I would need to study.
Anyway, I'm hoping to test into 1010. Can anyone who knows be able to tell me what I would need to study for the test? I'm thinking that it would including factoring, but I'm not sure what else I would need to study.
Axalon
Member
They probably never explicitly taught it, but just think of the negative sign as implicit multiplication of -1 to whatever number it's in front of, and just use the standard order of operations. Generally though, -x^y is the negative of x^y unless -x is surrounded by parentheses.
When it comes to this kind of thing, most colleges actually sell prep books for their respective tests. They're generally cheap and well worth it if all you need is a refresher.
As always:
www.khanacademy.org and www.patrickjmt.com
(What you'll want to focus on for 1010 is beginning functions, graphing, factoring, finding zeroes, solving systems of linear equations, and a few things here and there. Just check out those websites if you choose to not purchase a study guide)
edit: Do yourself a favor. Try to test into 1050. If you don't make it, no big deal, but if you prepare as if you were attempting to get into 1050, then your chance of getting into 1010 is MUCH higher. Otherwise you may wind up in remedial math :\
Do you know if you'll be taking the standard accuplacer or a school specific test?
Yeah, I love Khan Academy and have PatrickJMT bookmarked, I'm just not sure what I need to study. The bookstore only has generic study guides, nothing specific to the placement test, which is why I'm not quite sure what I should focus on. I just don't want to get stuck having to take a remedial class.
I believe it is the standard Accuplacer, but I may be wrong.
Sorry to bump this again, but I've run into something odd. In my continuing practice for a placement exam I ran into a quadratic factor that conflicted with the laws of quadratics as I was taught. When I was first taught to factor quadratics I was told that, if your third term is negative, the larger of your two factors takes the same sign as your second term and your smaller term takes the opposite. The practice tests I'm taking seem to disregard this and sign their terms however they need in order to get the sum or difference of the second term.
For example: 5x^2 -7x -6 factors into (5x + 3)(x - 2). This solution is correct, but it contradicts the law I was taught.
Now, I've mostly been using the quadratic formula since I noticed this, but I'm still curious if the law about factoring quadratics that I was taught was simply wrong, or if I forgot a different law.
For example: 5x^2 -7x -6 factors into (5x + 3)(x - 2). This solution is correct, but it contradicts the law I was taught.
Now, I've mostly been using the quadratic formula since I noticed this, but I'm still curious if the law about factoring quadratics that I was taught was simply wrong, or if I forgot a different law.
Well, if you have a correct equation that contradicts a "law", the law must be incorrect. Maths deal in absolutes.
That was my first though as well. What is odd though, is that this law I was taught has worked in countless of practice questions up until this one.
I'm pretty sure this is true if a=1. Doesn't apply in this case where a=5.
Seem possible, but other equations I've worked with a coefficient higher than 1 have factored using this method. Could just be a coincidence though. Regardless, I generally prefer the use the quadratic formula so it doesn't really matter that much.
So the only "practice" I know of goes like this: factor by grouping.
For 5x^2 - 7x - 6, multiply A and C giving you 30. Now, find factors of 30, which, when added together give you B (-7 in this case)
These factors are 10 and 3.
So:
5x^2 - 10x + 3x - 6
Factor out common terms: 5x(x - 2) + 3(x - 2)
(x - 2) is common, and thus is a factor. The remaining factor is then (5x + 3).
edit: You need to be very careful when people give you advice or methods which are just handy tricks for learning how to solve the problem. If you learn a shortcut, great, but always "know" the long way, or the hard way, because it is generally the *only* way which will always yield the proper answer.
If your "law" is breaking, consider just the method instead.
I was never taught this method oddly enough. Just basic factoring and the quadratic formula. I have come to the conclusion though that this "law" I was told about when I was taught quadratics taught quadratics was a load of crap. Very strange that it took this long for the method to break though.
systemfehler
Member
Why don't you just solve the equation for 0 and then use x1 and x2 as your factors?
If you have something like a*x^2 + b*x +c = 0 then use the formula
x1/x2 = (-b +/- sqrt(b^2 - 4*a*c)) / 2*a
After that you can just write: a*x^2 + b*x + c = a * (x - x1) * (x - x2)
For equations where a = 1 you can also use a simpler solution but I think it is best to remember this formula - it solves "every" quadratic equation. Just be carefull if b^2 - 4*a*c is negative - you have to use complex numbers in that case.
In this special case:
5x^2 - 7x - 6 = 0
a = 5, b = -7, c = -6
==> x1/x2 = (-(-7) +/- sqrt((-7)^2 - 4*5*(-6)) / 2 * 5 = (7 +/- sqrt(49 + 120)) / 10
= (7 +/- sqrt(169)9 / 10 = (7 +/- 13) / 10
==> x1 = 20/10 = 2 and x2 = 6/10 = 3/5
and that leads to 5x^2 - 7x -6 = 5 * (x - 2) * (x + 3/5) = (x-2) * (5x+3)
Memento_Mori15
Member
Nvm, solved it. 
systemfehler
Member
A bit embarrasing but I can't figure it out so far:
How can I describe the following graph:
So far I would say I use a step response/heaviside for - infinite until t0 to make the graph zero. After that I guess it should be something like U * ...
Guess it should be pretty easy but right now I can't see it
How can I describe the following graph:
So far I would say I use a step response/heaviside for - infinite until t0 to make the graph zero. After that I guess it should be something like U * ...
Guess it should be pretty easy but right now I can't see it
A bit embarrasing but I can't figure it out so far:
How can I describe the following graph:
So far I would say I use a step response/heaviside for - infinite until t0 to make the graph zero. After that I guess it should be something like U * ...
Guess it should be pretty easy but right now I can't see it
I would imagine it is something like:
Code:
f(t)= 0 for -inf < t < t0
U(t1 - t)/(t1-t0) for t0<=t <= t1
0 for t1 < t < infsystemfehler
Member
Snarfington
Member
Okay, really easy question for you all I'm sure but my mind's gone blank and I haven't touched on this in months and have forgotten it entirely. 
"Solve the simultaneous equations x-y=1 and (2^x)(3^y)=432"
Any tips as to where to start? Thanks.
"Solve the simultaneous equations x-y=1 and (2^x)(3^y)=432"
Any tips as to where to start? Thanks.
Okay, really easy question for you all I'm sure but my mind's gone blank and I haven't touched on this in months and have forgotten it entirely.
"Solve the simultaneous equations x-y=1 and (2^x)(3^y)=432"
Any tips as to where to start? Thanks.
x -1=y
(2^x)(3^(x-1)) =432
(2^x)(3^x)(3^(-1)) =432
(2^x)(3^x)(1/3) =432
6^x = 1296
x = ln(1296)/ln(6) = 4 therefore y = 3
x = ln(1296)/ln(6) = 4 therefore y = 3
Okay, really easy question for you all I'm sure but my mind's gone blank and I haven't touched on this in months and have forgotten it entirely.
"Solve the simultaneous equations x-y=1 and (2^x)(3^y)=432"
Any tips as to where to start? Thanks.
First equation tells us x=1+y
Sub this into our second equation:
(2^x)(3^y)=(2^(1+y))(3^y)=432
Take logs of both sides:
ln((2^1+y)(3^y))=ln(2^(1+y))+ln(3^y)=(1+y)ln(2)+yln(3)=ln(432)
rearrange and we have:
y=ln(432)-ln(2)/(ln(2)+ln(3))
This can be simplified down.
Therefore we get
y=3
x=4
Snarfington
Member
Got it. Thanks everyone! That took much longer than it should have.
SniperHunter
Banned
Learning calculus at the moment (just started) need some help with this problem: The mass M in grams of undissolved sugar left in a teacup after t seconds is given by M = 10.5-.4t^2
I need to find the average rate of change in the interval 0<_t <_1
This is what I got:
http://i.imgur.com/AOcag.jpg
is this right?
I need to find the average rate of change in the interval 0<_t <_1
This is what I got:
http://i.imgur.com/AOcag.jpg
is this right?
-0.4 grams/second, technically, but yeah, looks fine.
Metroid Hunter
Banned
Help me GAF!! I have a final coming up and I have no idea how to do the end of this problem. I feel like there must be a really simple solution, but I just can't find it. It's the part B.
Wouldn't this be expected value? Correct me if I'm wrong please.
SniperHunter
Banned
thanks, I really appreciate it. I am second guessing myself cause I got 78% on the polynomial unit :/
So GAF I suck using trig identities, and would like to improve my skills because I know I will need this later on in my upcoming Calculus classes. Is there a book or some websites that anyone could recommend for me. I'm also looking for some good Math books/Calculus books to help me improve.
SniperHunter
Banned
I need help with chain rule, the power rule and the combination of rules for exponents with polynomial expressions. I went to Khan academy and watched his videos but his examples were really simple.
These are the type of questions I am having trouble with: y = xe^(2x) or y = e^-(2x+5). Is there a website with a lot of problems? I need to practice! It gets really confusing when you are trying to find second or third derivatives....
These are the type of questions I am having trouble with: y = xe^(2x) or y = e^-(2x+5). Is there a website with a lot of problems? I need to practice! It gets really confusing when you are trying to find second or third derivatives....
Partial Gamification
Banned
Save your money on materials, the web has what you need. Consider investing in a decent graphing calculator, keep in mind some are not allowed on standardized tests but the symbolic calculators are great for checking one's work. The fundamental knowledge to have is the Pythagorean Theorem. Other identities require a familiarity but typically the more complicated ones can usually be referenced. Become intimate with the graphs of the trigonometric functions. Most university calculus books open with a chapter on functions and rehash applicable trigonometric concepts in a designate section. List of Trigonometric Identities
I recommend MIT's Lectures on youtube. The recitations are the most explicit demos. If web-searchs are not turning up what you need to practice (search "chain rule" "problem set" ~derivative site:.edu), stop by the local library and snap some pics of the problem-set(s) you want. Problem sets from UC Davis
SniperHunter
Banned
Partial [URL="http://www.youtube.com/watch?v=wezQdmwolMU" said:
That's exactly what I was looking for, thanks a lot bro. And speaking of trig identities, there was this question that appeared on my exam back earlier this year, it really left me totally stumped. I will post it if I manage to find it.
SniperHunter
Banned
Ok, anyone what to have a go at this?
tanѲ+cscѲcosѲ = 1/(sinѲcosѲ
tanѲ+cscѲcosѲ = 1/(sinѲcosѲ
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