FortuneFaded
Member
Looks fine to me but just leave the answer as a sin function. No need for a decimal expansion.
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The Math Help Thread
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Looks fine to me but just leave the answer as a sin function. No need for a decimal expansion.
triplestation
Member
if the answer is correct i think i just perfect scored my calc exam 
thanks for taking a look, guys!
thanks for taking a look, guys!
notoriousKTR
Member
The answer is right, however that integral is an improper integral (since the function is undefined when x is 0). There are a couple extra steps you have to show work for to get that answer.
spelltropy
Member
Someone mind attempting this question? I got 2pi/9 but I'm not sure if I got the region right - if you get something different please explain how you got the region, its a bit of a weird one (I think)...
Particulam
Neo Member
Anyone got any knowledge about k-NN (nearest neighbor) algorithm? I'm trying to do some classification (speech).
Currently this is what I do: After generating a set of vectors, I compare these to other sets of vectors stored in separate matrices. Each matrix represent a different class. I use an algorithm called Dynamic Time Warping for comparing the different matrices. For simplification we can assume I'm just using the euclidean distance between the sets. This gives me the distance between my generated set of vectors and each class.
I found a text on the web suggesting that one should apply a k-NN algorithm to these distances, and this I don't understand. Would you also believe that this would improve the result? I want to try it, but I do not know how to use k-NN. If anyone is familiar with this I would be very happy for some hints.
Cheers
Currently this is what I do: After generating a set of vectors, I compare these to other sets of vectors stored in separate matrices. Each matrix represent a different class. I use an algorithm called Dynamic Time Warping for comparing the different matrices. For simplification we can assume I'm just using the euclidean distance between the sets. This gives me the distance between my generated set of vectors and each class.
I found a text on the web suggesting that one should apply a k-NN algorithm to these distances, and this I don't understand. Would you also believe that this would improve the result? I want to try it, but I do not know how to use k-NN. If anyone is familiar with this I would be very happy for some hints.
Cheers
Someone mind attempting this question? I got 2pi/9 but I'm not sure if I got the region right - if you get something different please explain how you got the region, its a bit of a weird one (I think)...
I'm away from my computer, so I can't draw the solid and show you the bounds of r, theta, and phi by drawing, sorry. I am, however, going by the picture on wikipedia under the spherical coordinates article, so please note the picture there.
To me, it's easy to see that the solid is an ice cream cone with an spherical ice cream on top. Your inequality says that, for a fixed z, the (x, y) loci form a circle of radius z, and that the triple (x, y, z) is inside the sphere of radius of 1, which gives the shape of the ice cream on top.
I used r in (0, 1), theta in (0, pi/4), and phi in (0, 2pi) to get pi/8 as the answer.
Someone mind attempting this question? I got 2pi/9 but I'm not sure if I got the region right - if you get something different please explain how you got the region, its a bit of a weird one (I think)...
(Edit: Just noticed that the american system is different from the one used here, I use Thetas and Phi switched, so Theta is the one around the Z axis and Phi is the other one
)Well, lets see if I can help:
the region is defined as z >= 0, x^2 + y^2 <= z^2 <= 1 - x^2 - y^2
Then, the region is inside the cone x^2 + y^2 = z^2, and inside the sphere x^2 + y^2 + z^2 = 1
This region looks like an ice cream cone
Now, to get the Spherical Coordinates:
(theta, phi, r)
Theta: 0 <= theta <= 2*pi, as its symmetrical around the Z axis.
Phi: 0 <= Phi <= pi/4, as its limited by the cone (and this cone is the rotation of the z^2 = x^2)
Radius: 0 <= r <= 1, all lines starting from the origin inside the cone are limited by the sphere.
And, the (x,y,z) in Spherical Coordinates are the following:
x = r * sin(phi) * cos(theta)
y = r * sin(phi) * sin (theta)
z = r * cos(phi)
And, the Jacobian is r^2 * sin(phi)
So, z * jacobian = r * cos(phi) * r^2 * sin(phi) = r^3 * sin(phi) * cos(phi)
∫(0 -> 2*pi) d(theta) ∫(0 -> pi/4) sin(phi) * cos (phi) d(phi) ∫(0 -> 1) r^3 d(r)
∫ d(theta) = theta
=> ∫(0 -> 2*pi) d(theta) = 2*pi - 0 = 2*pi
∫sin(phi) * cos (phi) d(phi) = -1/2 * cos^2(phi)
=> ∫(0 -> pi/4) sin(phi) * cos (phi) d(phi) = -1/2 * [cos^2(pi/4) - cos^2(0)] =
= -1/2 * [1/2 - 1] = 1/4
∫r^3 d(r) = r^4 / 4
=> ∫(0 -> 1) r^3 d(r) = 1/4 * (1 - 0) = 1/4
2*pi * 1/4 * 1/4 = pi * 2/16 = pi/8
I believe this is the answer, if nothing is wrong
spelltropy
Member
Cheers both of you, I see it now, I had the region drawn correctly, justt for some reason I decided 0<r<sqrt2
Need some help understanding this (possible) math joke. In the latest episode of Big Bang Theory the vanity card at the end of the episode ended with this:
And I think I need help.
1+(1/(2+(1/(3+(1/(4+(1/(5))))))))*(e^(i*pi))*det{{8, 47},{14, 43}}*((95 mod 12)/log2(1024))+V
Is it just some random thing he made up or is there some insidey math joke hidden in it?
And I think I need help.
1+(1/(2+(1/(3+(1/(4+(1/(5))))))))*(e^(i*pi))*det{{8, 47},{14, 43}}*((95 mod 12)/log2(1024))+V
Is it just some random thing he made up or is there some insidey math joke hidden in it?
Stumpokapow
listen to the mad man
Note: Much of machine learning is domain-specific techniques and I know very little about speech classification, except that I remember an instructor saying that ICA is canonically used in noisy speech situations. So I can't speak to what you're reading about the use of k-NN in your field. But the consensus in machine learning is generally prediction-driven, so they mostly use what works. There is generally not any theoretical or subject-matter grounding in why one technique is chosen over another technique. Typically when someone approaches an ML problem, they just seek out other domain-specific experts and see what has worked for them. Everything I'm going to say is a general description of k-NN. I'm also, as my tag suggests, going to be offline for the next week or so so I can't do any follow-up here. So, buyer beware.
k-NN is an algorithm where you have a number of well-classified observations, and you have an observation you wish to classify. k-NN takes a parameter k. The algorithm takes the k nearest observations to your test observation (minimizing distance using any algorithm; Euclidean is one choice) and looks at their classes. The modal class among the k nearest neighbors is chosen as the class for classifying the result. Typically if the quantity of training data is sufficient and the classes are sufficiently separable in the dimensional space you're working, you would choose a low k > 1. If k=1, then the algorithm simply classifies any observation as the nearest well-classified neighbor's classification.
I actually just used k-NN on a 135-dimensional dataset with no feature selection. With ~8000 training points, kNN was about to correctly classify about ~80% of my test set, which was comparable to more computationally complex methods of classification (kernel SVM, for example). I was able to do k-NN in essentially real time compared with 3-5 minute convex optimization processing on the same problem using kernel SVM.
Real empirical classification accuracy from the work I just did:
I think for the sake of your sanity you should probably stop thinking in terms of "matrices" even if that's how you factually store your data for processing. The unit of analysis is an observation (which is a point in r-dimensional space, where r is the number of rows in your column-vector for the observation).
P.S. standardize your data before you run k-NN (scale each dimension to mean 0, unit variance). Otherwise if your dimensions have significantly differing orders of magnitude--say height in inches in one dimension and annual income in another dimension--any distance formula is going to be massively distorted.
Example
So, say you are operating in 3-dimensional space for the sake of geometric comprehension and the fact that I'm waiting at an airport and can't sanity check my work because I don't have a copy of R near me). And say your well classified training points are:
Class 1: [1, 1, 1] [1, 3, 2] [1, 0, 0] [0, 0, 0] [2, 5, 3] [10, 11, 8]
Class -1: [10, 10, 14] [14, 8, 7] [9, 10, 12] [22, 24, 17] [18, 18, 18]
Say your test point is:
[12, 12, 12]
If you run k-NN, k = 3, the algorithm will identify the three nearest points--assume for this example that whatever distance calculation you use gives you these results:
[10, 10, 14] [9, 10, 12] [10, 11, 8]
What are their classes:
-1, -1, 1
The modal class among the k=3 nearest neighbors is -1, and so we classify the test point as class -1.
MisterLuffy
Member
This is for my programming homework. I want to ask on how to calculate base salary and current annual salary?
Need some help with a problem as I can't understand the steps taken:
10^X - 0.2 * 10^X = 40
Somehow here you're supposed to factor it out and get ( 0.8 * 10^X = 40 )
Then divide 40/0.8 ending up with the logarithm ( log(50) = X )
Now, I'm thinking here, that ( 1*10 )^X = 10^X and if I take 1 - 0.2 I do get 0.8.
As far as I can tell that's the only way I'll end up with ( 0.8 * 10^X ).
But what rule of mathematics says that I can randomly take out a 1 from ( 1*10 )^X, add it to another term, and then just ignore the ( *10 )^X that's left?
Maybe this seems like a silly question but I just don't get the connection.
10^X - 0.2 * 10^X = 40
Somehow here you're supposed to factor it out and get ( 0.8 * 10^X = 40 )
Then divide 40/0.8 ending up with the logarithm ( log(50) = X )
Now, I'm thinking here, that ( 1*10 )^X = 10^X and if I take 1 - 0.2 I do get 0.8.
As far as I can tell that's the only way I'll end up with ( 0.8 * 10^X ).
But what rule of mathematics says that I can randomly take out a 1 from ( 1*10 )^X, add it to another term, and then just ignore the ( *10 )^X that's left?
Maybe this seems like a silly question but I just don't get the connection.
FortuneFaded
Member
It's more like 10^X = 1*(10^X) because multiplying by 1 does nothing.
Thanks for some clarification but I still don't get it.
I mean, obviously you take the 1 from ( 1*(10^X) ) minus the 0.2, but why and how?
I've been looking in a lot of places but I just can't get to terms with the steps taken.
FortuneFaded
Member
Thanks for some clarification but I still don't get it.
I mean, obviously you take the 1 from ( 1*(10^X) ) minus the 0.2, but why and how?
I've been looking in a lot of places but I just can't get to terms with the steps taken.
10^X - 0.2*(10^X)
= 1*(10^X) - 0.2*(10^X)
= (1-0.2)*10^X
=0.8*(10^X)
Once you've done the first step, then you can take out 10^X as a common factor.
We are using the distributive law, which says that ac+bc = (a+b)c. (This might be backwards from the way you've thought about it before,i.e. (a+b)c=ab+ac, where we start with what's on the left and end up with what's on the right, but the equal sign lets us go the other way just as easily.) Here a=1, b=-0.2, and c=10^X, soThanks for some clarification but I still don't get it.
I mean, obviously you take the 1 from ( 1*(10^X) ) minus the 0.2, but why and how?
I've been looking in a lot of places but I just can't get to terms with the steps taken.
1(10^X)-0.2(10^X) = (1-0.2)10^X.
Thanks for some clarification but I still don't get it.
I mean, obviously you take the 1 from ( 1*(10^X) ) minus the 0.2, but why and how?
I've been looking in a lot of places but I just can't get to terms with the steps taken.
Neph1x, it is because of the distributive property, which governs how real numbers are to behave when both multiplication and addition are present. In particular, the distributive property tells us that,
a * (b + c) = a * b + a * c, for all real numbers a, b, and c.
In your case (assume that x is a real number),
1 * 10^x - 0.2 * 10^x
= (1 - 0.2) * 10^x
= 0.8 * 10^x.
Please note that, the commutative property of multiplication and the inverse property of addition are also at work above.
----
You will want to go back to an algebra textbook and study what properties the real numbers have. I think, we often take these properties for granted and don't emphasize their importance enough, and as a result, many students come out of an algebra class not knowing what algebra is.
Algebra is not about solving linear, quadratic, or exponential equations in your case, but really concerns the question of, given a set (collection) of objects and some operations (rules) with which we can manipulate these objects, what we are able to (or unable to) accomplish. Hence, understanding what properties these objects (e.g. real numbers) and operations (addition and multiplication) satisfy is crucial to understanding how to solve the equation that you were given.
YES! That's the missing link!
No idea why I didn't think about that before, guess I got confused by all the variables.
No matter, I'll read up on the distributive law once again just to fill any gaps.
Thanks a bunch!
thesaucetastic
Unconfirmed Member
Back again for a quick question, but it's more of a finance thing I guess.
There's something called a binomial lattice that's used for determining the price of an option, but this method takes O(n^2) so it's not great for large n. In that case, we use the Black-Scholes formula because it has O(1), since it's not dependent on n at all (i.e., time is assumed to be continuous instead of having discrete periods like in the binomial model).
The problem I have is transitioning from the binomial lattice to the Black Scholes formula (since for large n, it approaches the Black Scholes value). The only hint I got was "calculate the expected value of expected values", but I'm not sure what that means.
Maybe I should look at the derivation for the Black Scholes formula again when I get some sleep.
edit: Okay, so Work = Force * distance right? I'm assuming the force is applied consistently/constantly over the distance (otherwise you'd have to use Work = integral of F(d) * ds where * is the dot product, I think.) ((It's been a while since I took physics/multivariate calculus so correct me if I'm wrong.))
To pump the water up, you have to counteract gravity, so the force applied is F = mg, where gravity is taken to be in the negative direction (so the work you're doing is positive). To calculate the mass, you have to figure out the volume of water being pumped, which is where the integral comes from. The volume is length * width * depth/height of the water = 8 * integral of x^2 from 0 to 4 * (8-2) (because the center of the tank is 4 feet deep, so to get to the bottom of the tank, you'd have to go down another 4 feet, but don't forget that we're only lowering the water level to 2 feet and not to 0.). I get 48 * 64/3 = 1024 feet^3. Now you need to convert from feet^3 to kilograms, just look it up the conversion factors/make use of the fact that 1 mL = cm^3 = 1 g or something.
The distance the force is applied is the distance the water spout is from the ground. So that's the height of the trough + the distance from the trough to the spout = 8 feet + 5 feet = 13 feet.
Disclaimer: I havne't slept in 2 days, trust me at your own risk.
There's something called a binomial lattice that's used for determining the price of an option, but this method takes O(n^2) so it's not great for large n. In that case, we use the Black-Scholes formula because it has O(1), since it's not dependent on n at all (i.e., time is assumed to be continuous instead of having discrete periods like in the binomial model).
The problem I have is transitioning from the binomial lattice to the Black Scholes formula (since for large n, it approaches the Black Scholes value). The only hint I got was "calculate the expected value of expected values", but I'm not sure what that means.
Maybe I should look at the derivation for the Black Scholes formula again when I get some sleep.
I don't have much time to figure this one out right now 'cos I'm about to leave (and I'm sleep deprived but what else is new), but I can give you a hint to the shape if that helps. Do you know what a trough is? Like something piggies eat out of? It looks like that, with round (parabolic) ends.
edit: Okay, so Work = Force * distance right? I'm assuming the force is applied consistently/constantly over the distance (otherwise you'd have to use Work = integral of F(d) * ds where * is the dot product, I think.) ((It's been a while since I took physics/multivariate calculus so correct me if I'm wrong.))
To pump the water up, you have to counteract gravity, so the force applied is F = mg, where gravity is taken to be in the negative direction (so the work you're doing is positive). To calculate the mass, you have to figure out the volume of water being pumped, which is where the integral comes from. The volume is length * width * depth/height of the water = 8 * integral of x^2 from 0 to 4 * (8-2) (because the center of the tank is 4 feet deep, so to get to the bottom of the tank, you'd have to go down another 4 feet, but don't forget that we're only lowering the water level to 2 feet and not to 0.). I get 48 * 64/3 = 1024 feet^3. Now you need to convert from feet^3 to kilograms, just look it up the conversion factors/make use of the fact that 1 mL = cm^3 = 1 g or something.
The distance the force is applied is the distance the water spout is from the ground. So that's the height of the trough + the distance from the trough to the spout = 8 feet + 5 feet = 13 feet.
Disclaimer: I havne't slept in 2 days, trust me at your own risk.
Thought I'd make another post to reply to yet another awesome explanation.
Yeah I know what you mean; I remember all my old math classes where we were taught a lot of formulas.
Funny thing is, the teacher never once bothered to explain why the formula works and why and when you can or cannot apply it.
So I have quite an uphill struggle here, but I'm making good progress!
Well, any progress is good to be honest, it just takes time to "unlearn" some very bad habits.
Can add as a footnote that I've always hated maths with a passion.
But seeing all the proper connections and how and why to apply formulas makes it interesting and ultimately fun and rewarding.
thesaucetastic
Unconfirmed Member
Also, does anyone have any good resources for abstract algebra? I want to touch up on rings, but I forget the name of the textbook I had on it.
I eventually found the integral to this as
integral from 2 to 4 of : (62.5) 8 (2sqrt
) (9-y) dy8 (2sqrt
) dy = Volume of rectangular slice of solid (length * width * height). The only thing I'm confused about is why the width was 2sqrt instead of just sqrt, or rather, why were we trying to find 2x instead of just x?Not really sure if I have the same answer given I didn't simplify mine at all. I did have different limits of integration (2 to 4). Thanks anyway, though.
Dummit & Foote, Hungerford, and Lang are all standard references, and I would rank them in that order from most to least readable. My personal preference is Rotman's Advanced Modern Algebra.
thesaucetastic
Unconfirmed Member
Welp, this is embarrassing. You're totally right, and this is why I shouldn't do math on no sleep. I should go fix that right now.integral from 2 to 4 of : (62.5) 8 (2sqrt
Thanks, I'll check all of them out.
I'm taking Physics 1 in Summer 1, Physics 2 in Summer 2 and Discrete Math 2 in the fall. How should I prep for these classes? I guess I won't worry about Physics 2 prep until I complete Physics 1. What is the material for Physics 1 and Discrete 2 that I should make sure I'm strong on before the classes start?
notoriousKTR
Member
The iLectureOnline videos on Youtube (https://www.youtube.com/channel/UCiGxYawhEp4QyFcX0R60YdQ) really helped me a lot for Physics. Most colleges uses the "University Physics" book and these videos follows the book.
Orbis Tabula
Member
I learned with Fraleigh's "A First Course in Abstract Algebra." I enjoyed it well enough, although I haven't read enough to really say how it stacks up to other texts. But it was good enough for me.
Thanks. I subscribed to it. Also, I fixed my post, I'm taking Discrete Math 2 in the Fall.
spelltropy
Member
Given some initial value problems, what do I do to prove they have/don't have a unique solution? the question states i don't have to prove continuity, and it is just sufficient to state where it fails if it does - i'm not quite sure what I should be doing here at all. How do i find where it fails/whether it fails?
You're right in that you need trig substitution.
The first step is to use rewrite the exponent in the denominator.
Using the laws of exponents, you can re-write that as (sqrt(9-y^2))^3
At this point, you should be noticing a familiar formula [sqrt(a^2 - y^2)]
For that configuration we use the identity 1-sin^2(theta) = cos^2(theta).
So you want to choose an appropriate substitution for y that will allow you to make use of this identity.
If you need more help let me know, it's a fairly lengthy problem.
I think it really depends on your differential equation and what you want to do with it (i.e. it's a case by case situation), but if there is nonlinearity or non-differentiability involved, you would suspect that there can be multiple solutions. I don't know if there are sufficient conditions for non-uniqueness, but you can certainly try to guess two solutions that satisfy the equation and initial condition.
spelltropy
Member
Okay, there were three different examples where I had to check for a unique solution. The first was a simple separation of variables where I could just solve it for the unique solution. The second was non linear, y y' = x for y(1)=0. This would give two solutions, +- of a square root or whatever, so its not unique? The final question was straight forward too, so I guess its just the second one that's the issue - the exact wording of the question is "You do not need to prove continuity, it is sufficient to just state where it fails, if it does so." How would I do it for this? Is just writing the two solutions right or am I missing a beat here?
I'm not sure what the person who wrote the question was exactly looking for, but we do see that yy' = x is equivalent to (y^2)' = 2x.
So, by our usual existence/uniqueness theorem, the IVP has a unique solution near x = 1 in terms of the variable y^2, but not in terms of y like you said.
AzureSkies
Member
So GAF, I'm getting entirely thrown off by this question, the only one I haven't finished for this assignment:
I'm already dead.
notoriousKTR
Member
To simplify a radical, you want to breakup the radicand into perfect squares multiplied to some non-perfect squares.
For example, sqrt(48x^2) can be written as sqrt(16x^2 * 3), where 16x^2 is a perfect square and 3 is not.
By the distributive property, sqrt(16x^2 * 3) can be written as sqrt(16x^2) * sqrt(3).
The sqrt(16x^2) is 4x, and nothing can be done to the sqrt(3).
Therefore, sqrt(48x^2) is simplified to 4x * sqrt(3).
For example, sqrt(48x^2) can be written as sqrt(16x^2 * 3), where 16x^2 is a perfect square and 3 is not.
By the distributive property, sqrt(16x^2 * 3) can be written as sqrt(16x^2) * sqrt(3).
The sqrt(16x^2) is 4x, and nothing can be done to the sqrt(3).
Therefore, sqrt(48x^2) is simplified to 4x * sqrt(3).
Hello there ! Any body here willing to help this poor student with some probability stuff ? Here is the problem: We have 2 jar A and B. Jar A contains 3 white ball and 2 red ball. Jar B contains 1 white ball and 4 red ball. We choose randomly one jar and drawn with replacement 2 ball. We have Ua the event "Jar A is chosen" , Ub "Jar B is chosen" and X the number of white ball drawn.
With this come 4 questions:
1) Calculate P(Ua) and P(Ub)
2) Calculate P(X = k |Ua ) et P(X=k | Ub) for 0 <= k <= 2
3) Deduce X probability law and calculate E(X)
4) Knowing we have drawn 2 red ball, what are the probability we have drawn from Jar B ?
I know that P(Ua) and P(Ub) = 1/2 but for the others questions I'm a little lost.
With this come 4 questions:
1) Calculate P(Ua) and P(Ub)
2) Calculate P(X = k |Ua ) et P(X=k | Ub) for 0 <= k <= 2
3) Deduce X probability law and calculate E(X)
4) Knowing we have drawn 2 red ball, what are the probability we have drawn from Jar B ?
I know that P(Ua) and P(Ub) = 1/2 but for the others questions I'm a little lost.
I'm really bad
I'm assuming by replacement, you are saying that you pick one ball, put it back, and then redraw:
2) Think what P(X = k |Ua ) means. It's telling you I've picked Jar A. Which has 3 white and 2 red. Since the balls are picked with replacement (i.e. I pick one then put it back) we can imagine that the outcome of each trial (picking a ball) are independent. Which means that:
k = 0 (Red then Red) : (2/5)*(2/5) or
k = 1 (Red then White or White then Red) : 2 * (3/5*2/5)
k = 2 ( White then White): (3/5)*3/5
or to generalise, you can think of it as bernoulli trial with success probability p =3/5 and n = 2 trials(k) = (n choose k) * (p)^k*(1-p)^(n-k)
You can use the same reasoning the for Ub.
3) E(X) = P(Ub) * E(X|Ua) + P(Ua)*E(X|Ub)
E(X) = 0.5 * (0*4/25 +1*2 * (3/5*2/5) + 2*9/25) + 0.5*(1*2*(1/5))
To get the above, think what E(X|Ua) means: On average how many white balls will I get if i pick Urn A, well: I have 4/25 probability of getting a 0, 16/25 probability of getting 1, and 6/25 probability of getting 2, so my expected value is 0*4/25+1*16/25+2*6/25
4) The question is asking, what is P(B|X=2), which from Bayes's rule is: P(X=2|B)*P(B)/P(X=2)
P(X=2|B) = (4/5)*(4/5)
P(B) = 1/2
P(X=2) = (0.5*(3/5)*3/5+0.5*2/5*2/5)
2) Think what P(X = k |Ua ) means. It's telling you I've picked Jar A. Which has 3 white and 2 red. Since the balls are picked with replacement (i.e. I pick one then put it back) we can imagine that the outcome of each trial (picking a ball) are independent. Which means that:
k = 0 (Red then Red) : (2/5)*(2/5) or
k = 1 (Red then White or White then Red) : 2 * (3/5*2/5)
k = 2 ( White then White): (3/5)*3/5
or to generalise, you can think of it as bernoulli trial with success probability p =3/5 and n = 2 trials
(k) = (n choose k) * (p)^k*(1-p)^(n-k)You can use the same reasoning the for Ub.
3) E(X) = P(Ub) * E(X|Ua) + P(Ua)*E(X|Ub)
E(X) = 0.5 * (0*4/25 +1*2 * (3/5*2/5) + 2*9/25) + 0.5*(1*2*(1/5))
To get the above, think what E(X|Ua) means: On average how many white balls will I get if i pick Urn A, well: I have 4/25 probability of getting a 0, 16/25 probability of getting 1, and 6/25 probability of getting 2, so my expected value is 0*4/25+1*16/25+2*6/25
4) The question is asking, what is P(B|X=2), which from Bayes's rule is: P(X=2|B)*P(B)/P(X=2)
P(X=2|B) = (4/5)*(4/5)
P(B) = 1/2
P(X=2) = (0.5*(3/5)*3/5+0.5*2/5*2/5)
I am retaking a college algebra class and got dinged on an exam for keeping my denominator as part of a subtraction problem involving fractions. All expressions were given the correct LCM.
I have the problem listed in email tags on this post. Don't want to post the screenshot directly because the test is still open, and I don't want to be accused of sharing answers if by chance someone recognizes the exam.
I did not complete the problem (ran out of time), but you can see I kept my fraction while the provided answer clears it.
I admit I am very tired right now and am probably misunderstanding or missing something obvious.
Your arguments leading to your third equation are correct (good job on specifying what x cannot be equal to in the beginning). I'm not sure why you came up with your fourth equation, though.
To answer your earlier question, we are told to make the denominators of two fractions the same before adding or subtracting the numerators, because we want to use the distributive property of real numbers.
We see that,
a/c + b/c
= a * (1/c) + b * (1/c)
= (a + b) * (1/c)
= (a + b)/c.
If two fractions had different denominators, e.g. a/c and b/d (with c and d being different numbers), then the third line above wouldn't hold. In other words, we could not add the numerators.
I understand why denominators must be common in order to add and subtract. After getting a full night's sleep I think I see what the problem was. Earlier on the exam there was an addition problem that kept the answer as a fraction. This subtraction problem wanted a whole number as the answer. I see now the directions for the addition problem were "solve by simplifying" and this one is simply "solve," which must mean "solve for x." I did not find x in the addition problem, I only reduced and simplified the addition as far as possible. If we were only to simplify, would the negative form of my answer to this subtraction problem be correct?
To put it another way, I think I started by solving and ended by simplifying.
Assuming linearity it's just 100 divided by 16.
You want to know how much "percentage" you can use per hour.
So you're looking for percent per hour, or percent/hour, or 100/16.
So you can use 6.25% per hour.
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