The Math Help Thread

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Mar 15, 2010
18,875
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Indianapolis, IN
Well, there is a quartic formula, but it is absolutely horrifying, so what kgtrep said, use softwares.
As it so happens, a beam where w_a = w_b has a max deflection at L/2. A beam where w_a = 0 and w_b > 0 has a max deflection at ~0.52L . So in the case where 0 < w_a < w_b, the max deflection will be between 0.50 and 0.52L . Sooooo I'm going to assume 0.51L and call it a day, haha.

Thanks for all the help and input, though. Coming up with a simple general solution for that expression simply wasn't worth my time or effort. That's engineering for you! ;)
 
Jul 14, 2012
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You need some motivation to find the right answer (you can do it):

The Sleipner A platform produces oil and gas in the North Sea and is supported on the seabed at a water depth of 82 m. It is a Condeep type platform with a concrete gravity base structure consisting of 24 cells and with a total base area of 16 000 m2. Four cells are elongated to shafts supporting the platform deck. The first concrete base structure for Sleipner A sprang a leak and sank under a controlled ballasting operation during preparation for deck mating in Gandsfjorden outside Stavanger, Norway on 23 August 1991.
[...]
The conclusion of the investigation was that the loss was caused by a failure in a cell wall, resulting in a serious crack and a leakage that the pumps were not able to cope with. The wall failed as a result of a combination of a serious error in the finite element analysis and insufficient anchorage of the reinforcement in a critical zone.

[...] The top deck weighs 57,000 tons, and provides accommodation for about 200 people and support for drilling equipment weighing about 40,000 tons. When the first model sank in August 1991, the crash caused a seismic event registering 3.0 on the Richter scale, and left nothing but a pile of debris at 220m of depth. The failure involved a total economic loss of about $700 million.
[...]
The post accident investigation traced the error to inaccurate finite element approximation of the linear elastic model of the tricell (using the popular finite element program NASTRAN). The shear stresses were underestimated by 47%, leading to insufficient design. In particular, certain concrete walls were not thick enough. More careful finite element analysis, made after the accident, predicted that failure would occur with this design at a depth of 62m, which matches well with the actual occurrence at 65m.
http://ta.twi.tudelft.nl/users/vuik/wi211/disasters.html
If nothing else just try pushing it through wolfram alpha with chacters it can read, I tried it quick and the underscores fudged the input.
 
Nov 24, 2013
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Well, there is a quartic formula, but it is absolutely horrifying, so what kgtrep said, use softwares.
It is indeed. :)

As it so happens, a beam where w_a = w_b has a max deflection at L/2. A beam where w_a = 0 and w_b > 0 has a max deflection at ~0.52L . So in the case where 0 < w_a < w_b, the max deflection will be between 0.50 and 0.52L . Sooooo I'm going to assume 0.51L and call it a day, haha.

Thanks for all the help and input, though. Coming up with a simple general solution for that expression simply wasn't worth my time or effort. That's engineering for you! ;)
You may want to be careful with the approximation. By superposition, the deflection function can be found by taking the sum of two deflections. However, the global maximum of the sum isn't necessarily the sum of the global maxima.

As you may know, we would normally use finite element method to approximate the deflection function. We know the deflection at the nodes (for standard FEM), so we can easily approximate the maximum deflection by finding the maximum among the nodal deflections. Then, we would refine our answer by using more elements.

----

The failure studies that Partial Gamification posted are quite real, in that there have been many instances where a failure by mathematicians and engineers to properly characterize the physical behavior and use the correct elements (i.e. understand the theory behind FEM) resulted in damages of varying degree. It's important for both groups to understand both fields.
 
Mar 15, 2010
18,875
16
805
Indianapolis, IN
Don't take this the wrong way, but I wouldn't be putting my PE on the line if I wasn't confident in my assumption. Yes, I have run STAAD models and written Excel sheets, MATLAB codes, and Mathcad routines to verify my assumption...But those things take a fraction of the time required to actually solve and verify the quartic function.

Hell, I just put together an Excel file here that does the superposition of a uniformly distributed load and a uniformly increasing (from 0) load. It comes out to approximately 0.52L in every single variable case I can come up with. This makes perfect sense since the worst case scenario is when w_a = 0 and w_b = infinity (max deflection is still at 0.52L in this case). So while it's nice to have a magic solution available, it's not always practical in the real world. Understanding the implications of the equations and what beam theory suggests is what leads me to my final assumption of 0.51L.

The nice thing about application versus theory is that you don't need anything more than repeatable examples to consider it good. :)

E: Also, that failure article posted is not related to simple beam theory. That's more shell membrane theory (of which I also consider myself pretty competent at given what I do for a living). In my line of work, FEM is something we use to verify our hand calculations. We fully expect steel elements to come out ~1/32-1/16" thinner via FEM. When we see 1/2"+, we know something is up. People who put all of their eggs in the FEM basket are asking for trouble.
 
Nov 24, 2013
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Don't take this the wrong way, but I wouldn't be putting my PE on the line if I wasn't confident in my assumption. Yes, I have run STAAD models and written Excel sheets, MATLAB codes, and Mathcad routines to verify my assumption...But those things take a fraction of the time required to actually solve and verify the quartic function.

Hell, I just put together an Excel file here that does the superposition of a uniformly distributed load and a uniformly increasing (from 0) load. It comes out to approximately 0.52L in every single variable case I can come up with.

The nice thing about application versus theory is that you don't need anything more than repeatable examples to consider it good. :)

E: Also, that failure article posted is not related to simple beam theory. That's more shell membrane theory (of which I also consider myself pretty competent at given what I do for a living). In my line of work, FEM is something we use to verify our hand calculations. We fully expect steel elements to come out ~1/32-1/16" thinner via FEM. When we see 1/2"+, we know something is up. People who put all of their eggs in the FEM basket are asking for trouble.
Nah, no offense taken. In fact, it was my bad to assume that you considered only two cases and were interpolating these two. I'm glad to see you did much more (and was really happy to see a structure-related question here).
 
Apr 12, 2010
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Sketch the region of integration and change the order of integration

double integral f(x,y) dx dy

bounds for dy [-2, 2]

bounds for dx [0, sqrt(4-y^2)]


I sketched it and it seems like they are integrating half a circle.

I don't really get how to change the order.

Would the new bounds just be
for dy [-2, sqrt(4-x^2)]
for dx [0, 2]
 
Jul 18, 2014
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Sketch the region of integration and change the order of integration

double integral f(x,y) dx dy

bounds for dy [-2, 2]

bounds for dx [0, sqrt(4-y^2)]


I sketched it and it seems like they are integrating half a circle.

I don't really get how to change the order.

Would the new bounds just be
for dy [-2, sqrt(4-x^2)]
for dx [0, 2]
Close. The -2 for dy is wrong.
 
Mar 9, 2010
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Not sure if this is the right thread, but i recently bought a book called "Engineering mathematics 7th edition K.A Stroud" to relearn math from scratch. Is this the right book to start with, it has 5 star rating on amazon?
So far the 20 first pages has been alright still in the arithmetic section.

I want to relearn math to complement my programming skills so mostly practical
usage.
 
Jul 14, 2012
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Not sure if this is the right thread, but i recently bought a book called "Engineering mathematics 7th edition K.A Stroud" to relearn math from scratch. Is this the right book to start with, it has 5 star rating on amazon?
So far the 20 first pages has been alright still in the arithmetic section.

I want to relearn math to complement my programming skills so mostly practical
usage.
How are you at reading textbooks on your own? It will have a preamble about "how to read this book" that should give an overview of all the little hintboxes, layout of the chapters, and sometimes even different approaches for going through the book (might suggest to skip a chapter or two depending on the cirriculm its being used for).

Does something like a textbook inspire you to study, and I mean owning one? Honestly, if its a sort of magic trinket that gets you into it more then keep it but if not return it. It looks like a great book but there is no single source to goto for mathematics.

Counting, arithmetic, geometry, trigonometry, algebra, calculus, probability and statistics, differential equations... "higher maths." The internet has a lot of great free resources. Your local library has math books.

For the cost of the book, you could buy a AAA-title or two 30$ Italian meals.

I am biased against the extortion racket that is academic publishing.
 
Mar 9, 2010
12,132
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How are you at reading textbooks on your own? It will have a preamble about "how to read this book" that should give an overview of all the little hintboxes, layout of the chapters, and sometimes even different approaches for going through the book (might suggest to skip a chapter or two depending on the cirriculm its being used for).

Does something like a textbook inspire you to study, and I mean owning one? Honestly, if its a sort of magic trinket that gets you into it more then keep it but if not return it. It looks like a great book but there is no single source to goto for mathematics.

Counting, arithmetic, geometry, trigonometry, algebra, calculus, probability and statistics, differential equations... "higher maths." The internet has a lot of great free resources. Your local library has math books.

For the cost of the book, you could buy a AAA-title or two 30$ Italian meals.

I am biased against the extortion racket that is academic publishing.
The money is not a issue, i'm relearning the math for hobby programming projects
before i had to spend evening time for programming homework assignments.
Now that i work i have a couple of evening hours free i can fill with hobby stuff.
A good math base opens up a lot of opportunity for interesting hobby projects
that is my main motivation to relearn from scratch.

I also work better with paper format then with videos or the sites, i get distracted
fast. So with a physical medium i just shut off the computer and sit through it.

Given its mostly for hobby projects i don't really see the need to get a proof heavy book
maybe later on.
 
Jul 14, 2012
3,397
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The money is not a issue, i'm relearning the math for hobby programming projects
before i had to spend evening time for programming homework assignments.
Now that i work i have a couple of evening hours free i can fill with hobby stuff.
A good math base opens up a lot of opportunity for interesting hobby projects
that is my main motivation to relearn from scratch.

I also work better with paper format then with videos or the sites, i get distracted
fast. So with a physical medium i just shut off the computer and sit through it.

Given its mostly for hobby projects i don't really see the need to get a proof heavy book
maybe later on.
Something to consider is that this book's technology assignments look like they are going to be on Excel and handheld calculator. There are more program-specific texts. Its about what you think is right because to an extent it doesn't matter* as long as you want to learn from whatever source you are using (if you just want to read chapters and do exercises).

If you can do a little coding then any of the Excel or handheld assignments should be doable with another program but the book will be referencing that program and those devices.

If money doens't matter; later on if this is your thing, look into mathematica or get a raspberry pi (I think rasbian [os] comes with mathematica still and the ras.pis are a little more powerful now).

I'm sure others know of even more options and better programs.

*assuming reputable publisher
 
Nov 24, 2013
745
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Sketch the region of integration and change the order of integration

double integral f(x,y) dx dy

bounds for dy [-2, 2]

bounds for dx [0, sqrt(4-y^2)]


I sketched it and it seems like they are integrating half a circle.

I don't really get how to change the order.

Would the new bounds just be
for dy [-2, sqrt(4-x^2)]
for dx [0, 2]
It helps to draw a picture:



On the left side of the picture, the integral corresponds to the infinitesimal volume under the curve of f between y (some fixed value) and (y + dy). Once we figure out that integral, we would let y vary between -2 and 2 to find the total volume (double integral).

Similarly, the integral on the right corresponds to the infinitesimal volume between x (some fixed value) and (x + dx). Then, we would let x vary between 0 and 2.

Note, the domain of integration that you came up with would look like this:

 
Sep 18, 2012
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Hi guys, having a really hard time with this statistics question, b in particular. Am I supposed to cube the function and integrate from 0 to y? That won't give me what the question is looking for though. Thanks in advance for the help!



Suppose that the probability density function (p.d.f.) of the lifetime
(in days) of a particular type of component is

fX(x) = 5 x^4/(500)^5 , 0 &#8804; x < 500.


(a) Compute the probability a component will fail in less than 365
days.

(b) Redundancy is an engineering principle to increase the reliability
of a system by assembling components in parallel. To decrease the
probability in part (a), three independent components are placed
in parallel. So the system will fail only when all three components
fail. Let Y = max{X1, X2, X3} denote the lifetime of such a
system, where Xi denotes the lifetime of the ith component. Show
that:

fY (y) = 15 y^14/(500)^15 , y > 0.

Hint: First construct FY (y) = P(Y &#8804; y), by noticing that
{Y &#8804; y} = {X1 &#8804; y} &#8745; {X2 &#8804; y} &#8745; {X3 &#8804; y}.
 

Stumpokapow

listen to the mad man
May 21, 2006
17,232
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Hi guys, having a really hard time with this statistics question, b in particular. Am I supposed to cube the function and integrate from 0 to y? That won't give me what the question is looking for though. Thanks in advance for the help!
Suppose that the probability density function (p.d.f.) of the lifetime
(in days) of a particular type of component is

fX(x) = 5 x^4/(500)^5 , 0 &#8804; x < 500.

(a) Compute the probability a component will fail in less than 365
days.
This is a PDF, not a PMF, and so we know that "days" is a continuous number. To transform PDF into CDF, integrate from -infinity (in this case 0 is fine, since it's the lowest value for which the PDF is defined) to the upper number of days; in this case, 364 (since the problem is asking "less" than 365). 365

Integral of a constant times a function = constant times the integral of a function, so take (5/(500^5)) out of the function.

Edit: Fixed.
Int(0 to 365) x^4 should be trivial to you, then take the constant back in to cancel out the 5s, and you get the following for the definite integral:

365^5 / 500^5 - 0 = ~0.2073

(b) Redundancy is an engineering principle to increase the reliability
of a system by assembling components in parallel. To decrease the
probability in part (a), three independent components are placed
in parallel. So the system will fail only when all three components
fail. Let Y = max{X1, X2, X3} denote the lifetime of such a
system, where Xi denotes the lifetime of the ith component. Show
that:

fY (y) = 15 y^14/(500)^15 , y > 0.

Hint: First construct FY (y) = P(Y &#8804; y), by noticing that
{Y &#8804; y} = {X1 &#8804; y} &#8745; {X2 &#8804; y} &#8745; {X3 &#8804; y}.
I think the hint gives you what you need.

f(x) = 5x^4/(500^5) <-- original PDF for single component failure
F(X <= x) = x^5/(500^5) <--- integration of f(x) to go from PDF to CDF
F(Y <= y) = (y^5/(500^5))^3 = (y^15)/(500^15) <-- transform from single component failure to three-in-parallel failure considering their hint
f(y)= d/dy F(y) = 1/(500^15) * d/dy y^15 = 15y^14 / (500^15) <-- recovering three-in-parallel PDF given three-in-parallel CDF

I believe your error is that you were trying to cube the PDF when you should have been cubing the CDF and then recovering the resulting PDF, am I correct?
 
Sep 18, 2012
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This is a PDF, not a PMF, and so we know that "days" is a continuous number. To transform PDF into CDF, integrate from -infinity (in this case 0 is fine, since it's the lowest value for which the PDF is defined) to the upper number of days; in this case, 364 (since the problem is asking "less" than 365).

Integral of a constant times a function = constant times the integral of a function, so take (5/(500^5)) out of the function.

Int(0 to 364) x^4 should be trivial to you, then take the constant back in to cancel out the 5s, and you get the following for the definite integral:

364^5 / 500^5 - 0 = ~0.2045



I think the hint gives you what you need.

f(x) = 5x^4/(500^5) <-- original PDF for single component failure
F(X <= x) = x^5/(500^5) <--- integration of f(x) to go from PDF to CDF
F(Y <= y) = (y^5/(500^5))^3 = (y^15)/(500^15) <-- transform from single component failure to three-in-parallel failure considering their hint
f(y)= d/dy F(y) = 1/(500^15) * d/dy y^15 = 15y^14 / (500^15) <-- recovering three-in-parallel PDF given three-in-parallel CDF

I believe your error is that you were trying to cube the PDF when you should have been cubing the CDF and then recovering the resulting PDF, am I correct?
Ok for the last part, I see what I did wrong now. Yes I was cubing the PDF instead. Thank you very much for the help again!!!
 
Sep 18, 2012
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This is a PDF, not a PMF, and so we know that "days" is a continuous number. To transform PDF into CDF, integrate from -infinity (in this case 0 is fine, since it's the lowest value for which the PDF is defined) to the upper number of days; in this case, 364 (since the problem is asking "less" than 365).
One small question about what you are saying here. Why is it 364 and not 365? I know its asking for less than 365, but wouldn't it be the integral from 0 to 365 anyways since days is continuous? Thanks
 

Stumpokapow

listen to the mad man
May 21, 2006
17,232
3
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One small question about what you are saying here. Why is it 364 and not 365? I know its asking for less than 365, but wouldn't it be the integral from 0 to 365 anyways since days is continuous? Thanks
Yes you are correct, that's my bad. I've edited my original post. Since it's continuous, you want 0 to 365. Were it discrete, you'd want 0 to 364.
 
Sep 18, 2012
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Another one I'm working on:

The surface flaws in plastic panels used in the interior of automobiles
can be modeled as a Poisson process with a rate of 0.057 flaws per
square foot of plastic panel. Assume an automobile interior contains
10 square feet of plastic panel.

(a) Let W be the surface area in square feet required to observe a
flaw. Give the mean and the standard deviation of W.
(b) What is the probability that there are no surface flaws in an automobile&#8217;s
interior?
(c) We inspect the cars one at a time for flaws. What is the probability
that the 10th inspected car is the 3rd car containing at least 2
flaws?


(a) - For the mean of W I did the reciprocal of the mean of x and got 17.54. And for the standard deviation I took the root of 17.54. Should I have done the reciprocal of the standard deviation of x instead, I am not sure here.

(b) - I did b(0;10,0.057) ~ p(0;10x0.057) = 0.566
To me this does not sound right, if there is expected to be 1 flaw every 17.54 square feet than shouldn't the probability of being no flaws in 10 square meters be just below 50%

(c) - This one I'm not sure at all


Thank you so much for the help in advance!!
 
Jun 5, 2014
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Another one I'm working on:

The surface flaws in plastic panels used in the interior of automobiles
can be modeled as a Poisson process with a rate of 0.057 flaws per
square foot of plastic panel. Assume an automobile interior contains
10 square feet of plastic panel.

(a) Let W be the surface area in square feet required to observe a
flaw. Give the mean and the standard deviation of W.
(b) What is the probability that there are no surface flaws in an automobile’s
interior?
(c) We inspect the cars one at a time for flaws. What is the probability
that the 10th inspected car is the 3rd car containing at least 2
flaws?


(a) - For the mean of W I did the reciprocal of the mean of x and got 17.54. And for the standard deviation I took the root of 17.54. Should I have done the reciprocal of the standard deviation of x instead, I am not sure here.
In general, the mean of a reciprocal is not the reciprocal of the mean, though it may be in specific cases. In order to know whether this is one of those cases or not, you need to determine the actual distribution of the random variable you're interested in. (This would also answer your question about what the standard deviation should be.) Have you learned about the distribution of waiting times for a Poisson process? Though this question is in 2 dimensions instead of 1, it can be handled in a very similar fashion.

(b) - I did b(0;10,0.057) ~ p(0;10x0.057) = 0.566
To me this does not sound right, if there is expected to be 1 flaw every 17.54 square feet than shouldn't the probability of being no flaws in 10 square meters be just below 50%

(c) - This one I'm not sure at all
(b) is correct. Your intuition may not be accounting for the possibility of having multiple flaws in 10 square meters, which makes the expectation higher than it otherwise would be.

For (c), let p be the probability that a car has fewer than 2 defects, and let q=1-p be the probability that a car has 2 or more defects. You should be able to determine exact expressions for p and q based on the Poisson distribution. For the 10th car examined to be the 3rd with 2 or more defects, we have to have exactly 2 of the first 9 cars having 2 or more defects, which can be done in C(9,2) ways. Thus the probability desired is

( C(9,2) p^7 q^2 ) * q = C(9,2) p^7 q^3
 
May 8, 2014
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I got an A in Descrete Math 1 last semester and have been given the opportunity to apply for a tutor for DM1 during the Fall. Has anybody here ever been a college tutor before? To what extent do you need to understand the subject? I feel like I could study during the summer so that I haven't forgotten any of the material, but I worry I may not be able to work out more challenging problems. So to those that have done it before, what is generally the expectation?
 
Sep 18, 2012
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In general, the mean of a reciprocal is not the reciprocal of the mean, though it may be in specific cases. In order to know whether this is one of those cases or not, you need to determine the actual distribution of the random variable you're interested in. (This would also answer your question about what the standard deviation should be.) Have you learned about the distribution of waiting times for a Poisson process? Though this question is in 2 dimensions instead of 1, it can be handled in a very similar fashion.
Thanks for replying. I understand the explanation for part c. But for part a are you saying what I am doing is wrong? I have learned the mean of a poisson distribution, which is lamda*length but where lamda is the average flaws per unit length . This gives me the expected number of flaws in the 10 squares meters which is not the mean I am looking for. So the mean area for there to be on flaw is not 10/(10*0.057) = 1/0.057 ? Thank you again
 
Jun 5, 2014
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Thanks for replying. I understand the explanation for part c. But for part a are you saying what I am doing is wrong? I have learned the mean of a poisson distribution, which is lamda*length but where lamda is the average flaws per unit length . This gives me the expected number of flaws in the 10 squares meters which is not the mean I am looking for. So the mean area for there to be on flaw is not 10/(10*0.057) = 1/0.057 ? Thank you again
I wasn't directly saying whether the numerical value of your answer was right or wrong, just pointing out that the intuitive reasoning by which you arrive at 1/.057 actually requires justification, as well as the name of the concept you should look up.
 
Nov 24, 2013
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I got an A in Descrete Math 1 last semester and have been given the opportunity to apply for a tutor for DM1 during the Fall. Has anybody here ever been a college tutor before? To what extent do you need to understand the subject? I feel like I could study during the summer so that I haven't forgotten any of the material, but I worry I may not be able to work out more challenging problems. So to those that have done it before, what is generally the expectation?
I was, for the math department and for a federally-supported program in my college. What I did for the former was what you may do, e.g. grade homework/tests, tutor students seeking help for an hour or two a week. For that, you will want to be aware of what your department expects of the students to have learned by the end of a semester.

Over the summer you could look at different books/websites to see what are different intuitive ways to solve a problem, as I imagine some students will get one approach, while some another. I wouldn't worry about not getting the hard problems; you will find that you learn a lot more as you tutor, and if there is still a problem you can't solve... nothing wrong with that. Tutors are humans too, and there's also the option of asking your professor or other tutors for help.
 
Sep 18, 2012
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Sorry for asking so many questions, but I have a midterm next week and I want to make sure I get all of this. My question (which relates a bit to the last question I asked):

For a poisson ratio, I know the mean ratio of X is the average # of outcomes per unit length. What if I am not asked for the mean of X but instead the mean of length for one outcome.

Example:
In a communication system, many messages arriving at a node are
grouped into a packet before being transmitted in the network. Suppose
that the message arrive at the node according to a Poisson process at a
rate of 36 messages per minute. A packet is a grouping of 3 messages.

So here the mean ratio is 12 packets (36 messages), and if I were asked for the mean # messages for a given time it would be mean = mean ratio x time , but this gives the mean of X.

What if I were for example:
A packet has been formed and transmitted. What is the mean
waiting time (in minutes) until the next packet is formed and
transmitted?

This is not the mean of X. So in my head I would do 1/mean of X. First I am not sure if this is even correct, to me it makes sense. But also, should there not be a given formula for this, if so I cannot find it in my book. If there is no formula for this given situation, what would be the reasoning as to why it turns out to be the reciprocal of the mean of X?

I know this question is longer than it needs to be, sorry, I am just really looking to get clarification on this. Thank you to who ever responds again :)
 
Mar 30, 2013
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If you have a Poisson process, the time between events (i.e. arrival of messages) is exponential with a mean of 1/36 minutes.

For a packet to be formed you need three messages, so the time for it to be formed is the sum of three independent expontentials with a mean of 1/36 minutes. The mean, then, is easy to calculate.

Since they are independent, the mean is just the sum of the means.
 
Sep 18, 2012
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If you have a Poisson process, the time between events (i.e. arrival of messages) is exponential with a mean of 1/36 minutes.

For a packet to be formed you need three messages, so the time for it to be formed is the sum of three independent expontentials with a mean of 1/36 minutes. The mean, then, is easy to calculate.

Since they are independent, the mean is just the sum of the means.
Thank you! So is this like the general rule for mean of length/time for an outcome in poisson process?
 

Duji

Member
Jul 4, 2012
1,435
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Canada
Been a long ass time since I've factored a cubic polynomial.

x^3 - 5x^2 - 2x + 24 = 0.

So far I plugged in -2 and got (x+2) as a factor. No idea how to figure out the rest. Pls help.
 
May 10, 2014
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Been a long ass time since I've factored a cubic polynomial.

x^3 - 5x^2 - 2x + 24 = 0.

So far I plugged in -2 and got (x+2) as a factor. No idea how to figure out the rest. Pls help.
Divide "x^3 - 5x^2 - 2x + 24" by (x+2) (polynomial long division). That should give you the answer (you may need to factor afterwards). Or synthetic division, that works just as well.
 
Nov 30, 2010
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Been a long ass time since I've factored a cubic polynomial.

x^3 - 5x^2 - 2x + 24 = 0.

So far I plugged in -2 and got (x+2) as a factor. No idea how to figure out the rest. Pls help.
If you know that x+2 is a factor, then you can always use long division to simplify the expression down. In this case, you'll get a quadratic equation that you should be able to factor easily just be inspection or with the quadratic formula.

Of course, this technique requires determining one factor first by plugging and checking. Two Words' suggestion of using Pascal's Triangle is a lot more versatile.
 
Jun 5, 2014
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The problem with paper is that I never have enough space. It's also impractical to edit or to cut and paste (unless I literally cut and paste...) Loose sheets easily get disorganized and lost, while bound sheets can be cumbersome when referring to disparate pieces of information in scattered locations. All that said, I still haven't found anything better than pen and paper. I was hoping a Surface Pro 3 with OneNote would do it, but the quality of the inking experience wasn't up to par.
 
Jul 18, 2014
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I'd love some help with these, if possible!




For the first one, well, it's seven sided. Don't overthink it, and don't let the algebraic equation confuse you. They outright tell you what the perimeter is by giving you x=8. If it's seven sided, what operation do you need to split the perimeter into seven pieces?

For the second one, think of the equation that will find the area. Draw one if you can't think of it. For the hexagon in particular, it is made of 6 isosceles triangles.

For the third one, think of a way to partition 140 into a 3:4 ratio (hint: think of 140 as if it were divided into 7 pieces). Then use the pythagorean theorem since the diagonal makes a right triangle.

For the fourth one, since it is a square, you should easily find what the length of each side is since the problem gives you a perimeter. What is the area of the square then? Equate that with the area of the circle and solve for the radius.
 
May 23, 2006
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For the first one, well, it's seven sided. Don't overthink it, and don't let the algebraic equation confuse you. They outright tell you what the perimeter is by giving you x=8. If it's seven sided, what operation do you need to split the perimeter into seven pieces?

For the second one, think of the equation that will find the area. Draw one if you can't think of it. For the hexagon in particular, it is made of 6 isosceles triangles.

For the third one, think of a way to partition 140 into a 3:4 ratio (hint: think of 140 as if it were divided into 7 pieces). Then use the pythagorean theorem since the diagonal makes a right triangle.

For the fourth one, since it is a square, you should easily find what the length of each side is since the problem gives you a perimeter. What is the area of the square then? Equate that with the area of the circle and solve for the radius.
Your response is appreciated! I think I mistakenly pasted the first one, as I'm good on that.

Second: how do I find the area of D, specifically? I'm not sure what to do if I only have the diagonal.

Third: Still stumped!

Fourth: The area of the square/circle is (pi)^2/4, right? If so, I still can't get the right answer for whatever reason.
 
Jul 18, 2014
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Your response is appreciated! I think I mistakenly pasted the first one, as I'm good on that.

Second: how do I find the area of D, specifically? I'm not sure what to do if I only have the diagonal.

Third: Still stumped!

Fourth: The area of the square/circle is (pi)^2/4, right? If so, I still can't get the right answer for whatever reason.
For the second one, think about how c) can help you find d)

For the fourth one, that's actually only the area of the square. Now, you know that it is equal to the area of the circle. Do the algebra.
 
Jun 5, 2014
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Your response is appreciated! I think I mistakenly pasted the first one, as I'm good on that.

Second: how do I find the area of D, specifically? I'm not sure what to do if I only have the diagonal.

Third: Still stumped!

Fourth: The area of the square/circle is (pi)^2/4, right? If so, I still can't get the right answer for whatever reason.
For the second, just knowing the diagonal doesn't uniquely determine the area. If the rectangle is long and skinny, then the area can be arbitrarily close to 0. If the rectangle is a square, then the area can be 12.5 square inches. You can also get any area in between those two extremes by adjusting the proportions of the rectangle. That said, the question is asking for the item with the largest area. Since the rectangle can't be any bigger than 12.5 in^2, and some of the other items have a larger area, it doesn't matter that the exact area of the rectangle cannot be determined.
 

Stumpokapow

listen to the mad man
May 21, 2006
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Third: Still stumped!
Given:
2l + 2w = 140
w = 4/3 l

Substitute the second equation into the first
2l + 8/3 l = 140
14/3 l = 140
14l = 420
l = 30

Substitute the value of l into the equation to get w
2(30) + 2(w) = 140
2w = 80
w = 40

Verify width and length have the correct ratio (they do)

Length of a diagonal is gotten through pythogorean theorem.
a^2 + b^2 = c^2
(30^2) + (40^2) = c^2
900 + 1600 = c^2
2500 = c^2
c = 50

I mean, this is not a difficult question, the big thing you need to do is sit down and think about what you're being given, what you're being asked for, and what things like "ratio" actually mean. What's the formula for the perimeter of a rectangle? p = 2l + 2w. And then it's just solving basic linear equations.
 
May 23, 2006
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Truthfully, I should know how to do all of this stuff. The problem is that it's been almost a decade since I've had to worry about it. I'm sincerely grateful to everyone for the help so far!

Given:
2l + 2w = 140
w = 4/3 l

Substitute the second equation into the first
2l + 8/3 l = 140
14/3 l = 140
14l = 420
l = 30

Substitute the value of l into the equation to get w
2(30) + 2(w) = 140
2w = 80
w = 40

Verify width and length have the correct ratio (they do)

Length of a diagonal is gotten through pythogorean theorem.
a^2 + b^2 = c^2
(30^2) + (40^2) = c^2
900 + 1600 = c^2
2500 = c^2
c = 50

I mean, this is not a difficult question, the big thing you need to do is sit down and think about what you're being given, what you're being asked for, and what things like "ratio" actually mean. What's the formula for the perimeter of a rectangle? p = 2l + 2w. And then it's just solving basic linear equations.
Yes, of course--thank you. I feel like an idiot!

For the second, just knowing the diagonal doesn't uniquely determine the area. If the rectangle is long and skinny, then the area can be arbitrarily close to 0. If the rectangle is a square, then the area can be 12.5 square inches. You can also get any area in between those two extremes by adjusting the proportions of the rectangle. That said, the question is asking for the item with the largest area. Since the rectangle can't be any bigger than 12.5 in^2, and some of the other items have a larger area, it doesn't matter that the exact area of the rectangle cannot be determined.
That makes sense! That should not have been confusing to me.

For the second one, think about how c) can help you find d)

For the fourth one, that's actually only the area of the square. Now, you know that it is equal to the area of the circle. Do the algebra.
In regard to the fourth problem, I was consistently making one careless mistake. The answer is C, right?
 
Oct 8, 2009
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Does anyone know a good website that can make one better at using their calculator? My exams actually allows you to have a calculator when you are fixing problems in the oral exam, but it's more like the input that is intimidating for me. I'll read about fixing things on paper and with head calculations, but I find it difficult to use some of the correct input - the order which to calculate. Things like Sinus and linear equations and stuff like that. Khan doesn't really have that so I am a bit lost on this:(
 
Nov 24, 2013
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In regard to the fourth problem, I was consistently making one careless mistake. The answer is C, right?
Nah, the answer is B. Let's walk through the problem.

We are told that a circle and a square have the same area, and we are interested in the perimeter (circumference) of the circle. Recall that,

area of circle = pi * r^2,
area of square = s^2,
perimeter of circle = 2 * pi * r,

where I used r to denote the radius of the circle, and s the square's side length.

We already have one equation:

(1) pi * r^2 = s^2.

----

Now, we are told the perimeter of the square. Recall that, in terms of the square's side length, the perimeter is given by,

perimeter of square = 4 * s.

Hence, we have the equation 4 * s = 2 * pi, which means the side length s must be,

(2) s = pi / 2.

----

From equations (1) and (2), we can now find the radius of the circle:

(3) r = s / sqrt(pi) = sqrt(pi) / 2.

Note that r = -s / sqrt(pi) is also a solution to equation (1). However, physically, r represents the radius of the circle, which we assume to be non-negative. So we discard the solution r = -s / sqrt(pi), only consider r = +s / sqrt(pi).

Now that we have the radius, we can finally calculate the perimeter of the circle:

perimeter of circle = 2 * pi * r = pi * sqrt(pi).
 
May 23, 2006
6,141
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1,115
SF Bay Area, CA
From equations (1) and (2), we can now find the radius of the circle:

(3) r = s / sqrt(pi) = sqrt(pi) / 2.

Note that r = -s / sqrt(pi) is also a solution to equation (1). However, physically, r represents the radius of the circle, which we assume to be non-negative. So we discard the solution r = -s / sqrt(pi), only consider r = +s / sqrt(pi).

Now that we have the radius, we can finally calculate the perimeter of the circle:

perimeter of circle = 2 * pi * r = pi * sqrt(pi).
This is where I'm totally losing you. I also found that a side length of the square is pi/2, but wouldn't we then square that to find the area of the square? And since we know the area of the square is equal to the circle, would we then set that equal to pi(r)^2 to find the radius?

I'm assuming here that the area of the circle/square is pi/4.
 
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