Math thread! Come in and have some pi...

[IronProdigy]

No its not a math question. Im programming a system that keep tracks of people visiting places. So I want to find the probability that at least x users will visit a place during 1 hour intervals. I want to base this of a dataset of 1 year.

What id did first was counting unique people per hour and averaged that over the entire dataset and got some averages per hour.

That what I did, prolly not the right thing to do. That why i ask.

I thought this maybe was a math question so I asked here.

I'm not familiar with programming but the problem with the probability is that you didn't give a set value for x. Because of that, you're answer will have to depend on x and the probability will be a discrete sum. You'll also have to give two cases, one with x= 0 and one with x > 0. Because the number of people entering a store is a random variable, which we don't have a maximum, the values can range from 0 to infinity. The distribution will also be a poisson, and if you're not familiar with a poisson distribution, then the problem will be much harder. I think you can use the Normal distribution but that depends on the size of your data set.

[noisybrah]

I'm not familiar with programming but the problem with the probability is that you didn't give a set value for x. Because of that, you're answer will have to depend on x and the probability will be a discrete sum. You'll also have to give two cases, one with x= 0 and one with x > 0. Because the number of people entering a store is a random variable, which we don't have a maximum, the values can range from 0 to infinity. The distribution will also be a poisson, and if you're not familiar with a poisson distribution, then the problem will be much harder. I think you can use the Normal distribution but that depends on the size of your data set.

Lets say x = 5. I want to find the probability that at least 5 people out of lets say 1000 people will enter a store.

And I could do some logic if there is 80%probability that 5 people enter a store the next hour.

Note to self:sign up for stat course next year lol

[IronProdigy]

Lets say x = 5. I want to find the probability that at least 5 people out of lets say 1000 people will enter a store.

And I could do some logic if there is 80%probability that 5 people enter a store the next hour.

Note to self:sign up for stat course next year lol

You're question is off. You're saying whats the probability of at least 5 people showing up in the store if 1000 people show up. The probability would be 1 because you KNOW 1000 people will show up.

[IronProdigy]

i think he's saying there are 1000 visits to a set of stores. what is the probability at least 5 of those visits were at a particular store.

This sounds like finding the number of solutions

x_1+...+x_n = 1000

where there are n stores, x_i is the number of visits to store i, and we want x_1 >=5 (Without loss of generality, the store in question is store 1)

If he phrased the question like that, then i would approach it as a binomial with n = 1000 and p = (1/m) where m is the total number of stores. P(at least 5) = 1 - P(x < 5) and then evaluate the binomials for x = 0,1,2,3,4. You're way could be more effective and quicker though.

[IronProdigy]

Im not sure if you're answer is correct, although i'm worried about why you didn't include the probability of choosing the given store, but you can check by using a smaller sample size. Say n=10 and we want to know P(x>1). By the binomial distribution f(x) = (n choose k)[(p)^k][(1-p)^n-k]
Suppose the probability of a customer entering the given store is 0.4 (p = 0.4)
Then P(x>1) =
1 - [ ((10 choose 0)(.6)^10 + ( 10 choose 1)(.4)(.6)^9)]
Im not educated enough( or smart enough) to tell if youre logic is right but if you can get the same answer using your method, then i dont see why it should be wrong.

[Chrysippus]

Bump for awareness, we still getting math questions in misc

[DrinkingBull]

Yeah, there is 5 Real Analysis classes at my school, a two part sequence at 300 level and then a 3 part sequence 400/500 level mixed undergrad/grad.

It's hard man. I think I could have scraped by with a poor grade relatively easily, however, getting in A was very difficult. I probably spent in the range of 20 hours a week, I'm sure mathematician and slacker can verify that I posted an analysis question in here roughly everyday for the 8 weeks.

I also took it accelerated in the summer so it was just really quick and I followed it up w/ number theory which was a bit much analysis and number theory back to back Monday through Friday.

I just took the first in the 300 level, but taking the second of the 300 level next term. Not looking forward to it. I'll probably be blowing this thread up again.

**** I can't believe it's been already over a year since I was inquiring about this. Well it turns out I only need one extra math class to get the minor so I'm planning on taking real analysis in the fall.

Background:
- Calc 1 - 3
- Diff Eqs
- Linear Algebra
- Discrete Math (the intro to proofs class at my school)
- 300 level Probability Theory (ECE dept)
- Signals and Systems (ECE again but basically an applied math class, RIP'd from convolution integrals)
- Complexity Theory (ECE, part of the Data Structures and Algorithms class)

I've gotten a B in every single 300 level and above math class I've taken at this school lmao. Hoping to maybe go out with a bang and get the A in analysis.

[Didlid]

Hey maths brahs wanna help me out with this question?

Suppose that f: D → R, where |D| > |R|. Can you conclude any of the following? Explain.
(i) There are at least two distinct elements d1 and d2 of D such that f (d₁) = f (d₂)
(ii) Every element r in R is the image of at least two distinct elements d₁ and d₂ of D such that f (d1₁) = f (d2₂) = r

No idea where to start to be honest with it, none of our material even covers what that means so i think its just extra reading

[Chrysippus]

MM how do you feel about this? I'm going to do it tonight, any tips?

[Didlid]

The first one is obvious. It's a restatement of the pigeonhole principle.

The second one is not true. For example every element in D can be sent to the same element in R (suppose R has at least two elements for this counter -example to work).

Thanks a lot dude, just looked up the pigeonhole principle and that is exactly what i was looking for. You made me actually understand it too which was nice of you instead of just straight up giving the answer.

[Chrysippus]

Yeah, gonna need a translation of this. I can sort of make it out, but want to be certain.

[Slacker23]

[img]https://i.imgur.com/cV2Qwxrh.jpg[img]

Yeah, gonna need a translation of this. I can sort of make it out, but want to be certain.

If a function f(x) is convex in the interval [a,b] and discontinuous at the interior point x0 of the interval, than for any positive k and any neighbourhood of x0 there exists a point E in that neighbourhood such that f(E)>=k.

[TrueBallerBrah]

Hey guys, I'm wondering if you can help me out with this trig problem. It's been a while since I've done any math and I can't remember where to go from here.

X = Asin(Bt+C)+D

I want to solve for t.

So far I'm at:

(X-D)/A=sin(Bt+C)

(X-D)/A=sinBtcosC+sinCcosBt

Don't know where to go from here.

[charity4thepoor]

(X-D)/A=sin(Bt+C)

You're going to want to take the inverse sine of each side here. Note: inverse sine has a limited domain, so you may need to do something to get all your values of t.

[TrueBallerBrah]

You're going to want to take the inverse sine of each side here. Note: inverse sine has a limited domain, so you may need to do something to get all your values of t.

I thought you couldn't do that because the sine is in the form of sin (a+b)?

Edit nvm, you're right.

[MrJensenn]

Can someone just quickly remind me of how these kinds of problems are solved:
Two factories produce something each produces 50% of the total production, factory A have 2% defects and factory B have 3%.
What is the combined defect rate?

[MrJensenn]

2% of A will be defective, 3% of B will defective. each of A and B produce 0.5*total production

total percentage of defects: 0.02*0.5+0.03*0.5

Thanks man, highly appreciated.

[InfoBrah]

Any help with these two brahs?

[numberguy12]

Informally:

I would use proof by contradiction and use the unique factorization theorem, but then again my proofs tend to be inelegant.

For the first, assume for all prime factors p of n, that p^t is congruent to either 0 or 1 (mod r) and get your contradiction (if any of p^t are congruent to 0, then n^t is congruent to 0, if all are congruent to 1, then n^t congruent to 1 (mod r) - remembering n is a product of powers of p's by unique factorization, so n^t a product of p^t's)...arrive at contradiction.

For the second, assume there is no positive integer a such that n=p^a...thus n has a unique prime factor not equal to p..in the original assumption n^k1p^l1 = n^k2p^l2, with k1 not equal k2, l1 not equal l2, isolating n and p will show you that the other prime factor of n (mentioned above), divides a power of p, a contradiction. This would need formalized.

Edit: also in the first problem, to show that there indeed exists a prime factor of n, as all that is given is n>0, note that n cannot be 1, as then n^t would be congruent to 1 (mod r). Thus n is 2 or greater, and has a prime factor.

[InfoBrah]

Informally:

I would use proof by contradiction and use the unique factorization theorem, but then again my proofs tend to be inelegant.

For the first, assume for all prime factors p of n, that p^t is congruent to either 0 or 1 (mod r) and get your contradiction (if any of p^t are congruent to 0, then n^t is congruent to 0, if all are congruent to 1, then n^t congruent to 1 (mod r) - remembering n is a product of powers of p's by unique factorization, so n^t a product of p^t's)...arrive at contradiction.

For the second, assume there is no positive integer a such that n=p^a...thus n has a unique prime factor not equal to p..in the original assumption n^k1p^l1 = n^k2p^l2, with k1 not equal k2, l1 not equal l2, isolating n and p will show you that the other prime factor of n (mentioned above), divides a power of p, a contradiction. This would need formalized.

Edit: also in the first problem, to show that there indeed exists a prime factor of n, as all that is given is n>0, note that n cannot be 1, as then n^t would be congruent to 1 (mod r). Thus n is 2 or greater, and has a prime factor.

Simple enough, thanx brah.

[Miscer42069srs]

Answer the following questions. ​(a) The random variable x is distributed​ normally, with x~N(80,100) Find the probability that x is greater than 90. x >90. ​(b) Find P(x<85). ​(c) Find P(76<x<84).

[Miscer42069srs]

Answer the following questions. ​(a) The random variable x is distributed​ normally, with x~N(80,100) Find the probability that x is greater than 90. x >90. ​(b) Find P(x<85). ​(c) Find P(76<x<84).

BUMP, i really need then answer preferably brahs il learn the stuff tmoro (normal distribution)

[dat]

Answer the following questions. ​(a) The random variable x is distributed​ normally, with x~N(80,100) Find the probability that x is greater than 90. x >90. ​(b) Find P(x<85). ​(c) Find P(76<x<84).

This is about as basic as it gets with normal distributions... convert to z variable where z ~ N(0,1) by z = (x-80)/100:

(a) P(x>90) = P(z>(90-80)/100) = P(z>0.1) = 1 - P(z<0.1) = 1 - cdf(0.1) (you should be able to calculate this with your calculator/matlab/whatever you use)

(b) You can do this one on your own you lazy git

(c) P(76 < x < 84) = P(x<84) - P(x<76) = you can do the rest

[numberguy12]

This is about as basic as it gets with normal distributions... convert to z variable where z ~ N(0,1) by z = (x-80)/100:

To the original poster of this problem, make sure you are noting whether 100 is the standard deviation or variance. Looking at the numbers, if it's the variance, it will give standard deviation sqrt(100) = 10, which will change z score to (x-80)/10 in the above calculations.

[Miscer42069srs]

Am aware this stuff is basic but I only chose a stats module recently so only been doing for like a month..I haven't learnt normal distribution yet (will be sure to do so, just needed help on something due tmoro)

[shortythree]

anyone have a fast way of factoring cubic polynomials besides grouping?

[numberguy12]

anyone have a fast way of factoring cubic polynomials besides grouping?

generally I try to find a root of the cubic by inspection, and then use polynomial long division to find the resulting quadratic factor. Example:

factor x^3-15x-4

By inspection we see x=4 is a root, so divide (x^3-15x-4) by (x-4) using long division to get (x^2+4x+1). Thus the final factored form is (x-4)(x^2+4x+1).

Note on finding the first root. Textbook problems will typically have cherry picked cubics which have a root evident by inspection, like the one you listed. If the root is not immediately evident, there are several ways to go about this. There are various approximation methods to hone down upon a root, as well as graphing the cubic. Cardan's cubic formula will find exact roots (not recommended).

The best answer here is "if" all coefficients in the cubic are integers (and this is generally the case), can use the rational root theorem to try and find a root, which states a rational root of a cubic must have form (factor of last term/factor of coefficient of x^3). To use your example...root would be 1/1, or -1/1. Check each option until you indeed find one is the root (in your case -1). So the factor is x+1.

To apply the rational root theorem to my example above, the possible rational roots would be 4/1, -4/1, 2/1, -2/1, 1/1, or -1/1. Check until you find which is actually a root (4 here). So factor is x-4.

[shortythree]

All factoring is really done by grouping. Its all based on isolating a common expression and "un-multiplying". Only faster way is to look for patterns.

There is no governing principle for factoring (in general, 5th degree and higher cannot be "factored" in the normal sense), but a place to start is pascal's triangle (binomial coefficients) which gives you your result instantly.

If you wanted to can consider the multiplied-out forms of:

(ax+b)^3
(ax+b)^2(cx+d)
(ax^2+bx+c)(dx+e)

All cubics are factored as one of the above. The first is given pascal's triangle and is easy to recognize.

a^3x^3 + 3a^2bx^2 + 3ab^2x + b^3

ex

8t^3-60t^2 + 150t - 125

First thing to notice: 2^3=8, (-5)^3=-125, so you know a and b, if it factors this way

Now verify the other coefficients to get (2t-5)^3

as with anything practice will help you recognize these things

Obviously know the sum/difference of cubes formulas too.

generally I try to find a root of the cubic by inspection, and then use polynomial long division to find the resulting quadratic factor. Example:

factor x^3-15x-4

By inspection we see x=4 is a root, so divide (x^3-15x-4) by (x-4) using long division to get (x^2+4x+1). Thus the final factored form is (x-4)(x^2+4x+1).

Note on finding the first root. Textbook problems will typically have cherry picked cubics which have a root evident by inspection, like the one you listed. If the root is not immediately evident, there are several ways to go about this. There are various approximation methods to hone down upon a root, as well as graphing the cubic. Cardan's cubic formula will find exact roots (not recommended).

The best answer here is "if" all coefficients in the cubic are integers (and this is generally the case), can use the rational root theorem to try and find a root, which states a rational root of a cubic must have form (factor of last term/factor of coefficient of x^3). To use your example...root would be 1/1, or -1/1. Check each option until you indeed find one is the root (in your case -1). So the factor is x+1.

To apply the rational root theorem to my example above, the possible rational roots would be 4/1, -4/1, 2/1, -2/1, 1/1, or -1/1. Check until you find which is actually a root (4 here). So factor is x-4.

cool thanks for the answers

[TrueBallerBrah]

Question for you brahs.

If h(x) = x + sqrt(x), then h-inverse(6) = ?.

So I figured I would just have the first equation and set it to 6 = x + sqrt(x) and solve for x.

So

6 = x + sqrt(x)

6x = x^2 + x

0 = x^2 -5x

0 = x(x-5)

x= 5 or 0.

but the book says it's 4? I don't know what I did wrong.

[TrueBallerBrah]

^ error here.

Let u=sqrt(x).

Then

6 = u^2 + u

u^2+u-6=0

(u-2)(u+3)=0

u=2.. (since u>0)

sqrt(x)=2

x=4

Ah that makes sense. But why can I not multiply x to both sides to get rid of the sqrt and then solve for x? The answer shouldn't change if I do it to both sides.