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http://forthmath.blogspot.com/2016/05/words-for-testing-conjectures.html
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## Sunday, May 15, 2016
### Words for testing conjectures
Making code for testing conjectures can be cumbersome even in the case of easy programming. So far in Zet there is the possibility of make and calculate with sets of numbers interactively.
{ 1 1000 | prime } ok
{ 1 1000 | pairprime } { 1 1000 | notpairprime }
union ok
zet= . -1 ok
Conditions so far are
: all dup = ;
: odd 1 and ;
: 1mod4 4 mod 1 = ;
: 3mod4 4 mod 3 = ;
: sqr dup sqrtf dup * = ;
: sqrfree dup radical = ;
: pairprime dup prime over 2 + prime rot 2 - prime or and ;
: notpairprime dup prime swap pairprime 0= and ;
: semiprime bigomega 2 = ; \ A product of two primes?
: uniprime smallomega 1 = ; \ Only divisional by one prime?
: biprime smallomega 2 = ; \ Exact two different primes?
The construction { 1 10000 | pairprime } is fancy but slow and risk overflow in data stack. All the pairprimes in the intervall will first be created on the stack and then be moved to the zst-stack. It's better to check number for number and create the set directly on the zst-stack.
: intcond \ low hi xt -- | -- s "intervall condition"
loc{ xt }
swap 0 -rot
do i xt execute
if i >zst 1+ then
loop 2* negate >zst ;
utime 1 100000 ' pairprime intcond utime cr d- d. cardinality .
-35954 2447 ok
A set of 2447 primes is created in about 0.04 seconds. This construction is also possible to use in definitions, then using ['] instead of '.
To filtrate a set on the zst-stack:
: setcond \ xt -- | s -- s' "set condition"
loc{ xt } 0
foreach
do zst> dup xt execute
if >xst 1+ else drop then
loop dup 0
do xst> >zst
loop 2* negate >zst ;
{ 1 100 | prime } ' 1mod4 setcond cr zet.
{5,13,17,29,37,41,53,61,73,89,97} ok
It's also nice to be able to create the image of a function:
: intimage \ low hi xt -- | -- s "intervall image"
loc{ xt }
swap 2dup
do i xt execute >zst
loop - 2* negate >zst
set-sort reduce ;
: setimage \ xt -- | s -- s' "set image"
loc{ xt } 0
foreach
do zst> xt execute >xst 1+
loop dup 0
do xst> >zst
loop 2* negate >zst
set-sort reduce ;
Functions so far are:
log~ ( n -- nr ) where nr=1+²log n
random ( u1 -- u2 ) where 0≤u2<u1
nextprime ( numb -- prime )
prevprime ( numb -- prime )
sqrtf ( m -- n ) "floor"
sqrtc ( m -- n ) "ceiling"
radical ( n -- r )
totients ( n -- t )
bigomega ( n -- b )
smallomega ( n -- s )
ufaculty ( u -- u! )
pnr@ ( n -- p ) prime number n
pi ( x -- n ) number of primes ≤ x
Functions and conditions both must have the stackdiagram ( m -- n ), but the concept will be generalized.
1 20 ' radical intimage zet. {1,2,3,5,6,7,10,11,13,14,15,17,19} ok
Some test functions:
: square dup * ; \ x → x²
: sqr>prime square nextprime ; \ x → nextprime(x²)
: sqr<prime square prevprime ; \ x → prevprime(x²)
: foo dup totients mod ; \ x → x(mod ϕ(x)) Euler's totient.
{ 1 100 | all } ' foo setimage cr zet.
{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,23,25,27,31,33,35,39} ok
1 100 ' square intimage ' foo setimage cr zet.
{0,3,5,7,11,13,17,19,20,23,27,28,29,31,37,41,43,44,47,52,53,59,61,67,68,71,73,76,79,80,83,89,92,97,105,112,116,124,125,148,164,172,176,180,188,189,208,243,252,272,304,320,343,368,385,396,429,448,468,500,585,704,720,825,945,969,1008,1105,1197,1280,1309,1372,1540,1620,1701,1725,1729,1785,2185,2187,2625,2697,3069,3861} ok
Hmm, it seems like all odd primes less than 100 belongs to the image...
1 10000 ' square intimage ' foo setimage ok
1 10000 ' prime intcond ok
zswap diff zet. {2} ok
Well, it might be sound to expect non dramatic explanations to conjectures, especially conjectures concerning primes.
To check relations R m there is a need for testing subsets of Cartesian products, sets of pairs of integers.
: paircond \ xt -- | s -- s'
loc{ xt } 0
foreach
do zdup zet> drop xt execute
if zst xst setmove 1+ else zdrop then
loop 6 * negate >xst
xst zst setmove ;
{ 1 10 | all } zdup cartprod ' = paircond cr zet.
{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9)} ok
: pairimage \ xt -- | s -- s'
loc{ xt } 0
foreach
do 1+ zet> drop xt execute >xst
loop dup 0
do xst> >zst
loop 2* negate >zst
set-sort reduce ;
{ 2 10 | all } zdup cartprod ' * pairimage cr zet.
{4,6,8,9,10,12,14,15,16,18,20,21,24,25,27,28,30,32,35,36,40,42,45,48,49,54,56,63,64,72,81} ok
Some conditions and functions N²→N are =, <>, <, >, >=, <=, +, *, /, mod, **, ugcd, -, invmod, legendre, jacobi, kronecker, gnorm, choose, where m ≥ n for m n - and m n must be coprime for m n invmod. the gnorm of two integers m n is the norm of the gaussian integer m+in, that is the number m²+n².
: coprime ugcd 1 = ;
: divide swap mod 0= ;
{ 1 10 | all } zdup cartprod ' coprime paircond cr zet.
{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8),(1,9),(2,1),(2,3),(2,5),(2,7),(2,9),(3,1),(3,2),(3,4),(3,5),(3,7),(3,8),(4,1),(4,3),(4,5),(4,7),(4,9),(5,1),(5,2),(5,3),(5,4),(5,6),(5,7),(5,8),(5,9),(6,1),(6,5),(6,7),(7,1),(7,2),(7,3),(7,4),(7,5),(7,6),(7,8),(7,9),(8,1),(8,3),(8,5),(8,7),(8,9),(9,1),(9,2),(9,4),(9,5),(9,7),(9,8)} ok
{ 1 10 | all } zdup cartprod ' divide paircond cr zet.
{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8),(1,9),(2,2),(2,4),(2,6),(2,8),(3,3),(3,6),(3,9),(4,4),(4,8),(5,5),(6,6),(7,7),(8,8),(9,9)} ok
{ 1 10 | all } zdup cartprod ' coprime paircond ' gnorm pairimage cr zet.
{2,5,10,13,17,25,26,29,34,37,41,50,53,58,61,65,73,74,82,85,89,97,106,113,130,145} ok
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# Is regression the same as correlation?
The difference between these two statistical measurements is that correlation measures the degree of a relationship between two variables (x and y), whereas regression is how one variable affects another.
## Is regression the same as correlation?
The difference between these two statistical measurements is that correlation measures the degree of a relationship between two variables (x and y), whereas regression is how one variable affects another.
### What does β mean in regression?
The beta coefficient is the degree of change in the outcome variable for every 1-unit of change in the predictor variable.
#### Is standardized beta equal to correlation?
A beta weight is a standardized regression coefficient (the slope of a line in a regression equation). They are used when both the criterion and predictor variables are standardized (i.e. converted to z-scores). A beta weight will equal the correlation coefficient when there is a single predictor variable.
What is the difference between beta and beta coefficient?
In other words, standardized beta coefficients are the coefficients that you would get if the variables in the regression were all converted to z-scores before running the analysis. 2. Beta is the correlation coefficient range from 0-1, higher the value of beta stronger the association between variables.
Why is regression better than correlation?
Regression simply means that the average value of y is a function of x, i.e. it changes with x. Regression equation is often more useful than the correlation coefficient. It enables us to predict y from x and gives us a better summary of the relationship between the two variables.
## What is the difference between correlation and regression in statistics?
‘Correlation’ as the name says it determines the interconnection or a co-relationship between the variables. ‘Regression’ explains how an independent variable is numerically associated with the dependent variable. In Correlation, both the independent and dependent values have no difference.
### What is β in statistics?
Beta (β) refers to the probability of Type II error in a statistical hypothesis test. Frequently, the power of a test, equal to 1–β rather than β itself, is referred to as a measure of quality for a hypothesis test.
#### What is the difference between B and beta in multiple regression?
According to my knowledge if you are using the regression model, β is generally used for denoting population regression coefficient and B or b is used for denoting realisation (value of) regression coefficient in sample.
Is beta the same as R-Squared?
Beta is an estimate of the marginal effect of a unit change in the return on a market index on the return of the chose security. R-squared (R2) is an estimate of how much beta and alpha together help to explain the return on a security, versus how much is random variation.
What is the difference between correlation and regression?
Correlation and regression are two terms in statistics that are related, but not quite the same. In this tutorial, we’ll provide a brief explanation of both terms and explain how they’re similar and different. What is Correlation? Correlation measures the linear association between two variables, x and y. It has a value between -1 and 1 where:
## What is the difference between beta and correlation coefficient?
All Answers (13) The beta values, or b coefficients, are estimates of the parameters of the straight line equation underlying your data set. The absolute value of the correlation coefficient is a measure of the alignment of the points in your data set.
### What is the correlation between these two variables?
In other words, we can visually see that there is a positive correlation between the two variables. Using a calculator, we can find that the correlation between these two variables is r = 0.915. Since this value is close to 1, it confirms that there is a strong positive correlation between the two variables. What is Regression?
#### What are beta values in regression analysis?
The beta values in regression are the estimated coeficients of the explanatory variables indicating a change on response variable caused by a unit change of respective explanatory variable keeping all the other explanatory variables constant/unchanged.
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Number 46420422
Properties of number 46420422
Cross Sum:
Factorization:
2 * 3 * 37 * 211 * 991
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
2c451c6
Base 32:
1c8ke6
sin(46420422)
-0.69234767065485
cos(46420422)
-0.72156406710618
tan(46420422)
0.95950962945189
ln(46420422)
17.653250049678
lg(46420422)
7.6667090842336
sqrt(46420422)
6813.2534078808
Square(46420422)
Number Look Up
Look Up
46420422 (forty-six million four hundred twenty thousand four hundred twenty-two) is a amazing number. The cross sum of 46420422 is 24. If you factorisate 46420422 you will get these result 2 * 3 * 37 * 211 * 991. The number 46420422 has 32 divisors ( 1, 2, 3, 6, 37, 74, 111, 211, 222, 422, 633, 991, 1266, 1982, 2973, 5946, 7807, 15614, 23421, 36667, 46842, 73334, 110001, 209101, 220002, 418202, 627303, 1254606, 7736737, 15473474, 23210211, 46420422 ) whith a sum of 95898624. 46420422 is not a prime number. The figure 46420422 is not a fibonacci number. The number 46420422 is not a Bell Number. The number 46420422 is not a Catalan Number. The convertion of 46420422 to base 2 (Binary) is 10110001000101000111000110. The convertion of 46420422 to base 3 (Ternary) is 10020100101211220. The convertion of 46420422 to base 4 (Quaternary) is 2301011013012. The convertion of 46420422 to base 5 (Quintal) is 43340423142. The convertion of 46420422 to base 8 (Octal) is 261050706. The convertion of 46420422 to base 16 (Hexadecimal) is 2c451c6. The convertion of 46420422 to base 32 is 1c8ke6. The sine of the figure 46420422 is -0.69234767065485. The cosine of the figure 46420422 is -0.72156406710618. The tangent of 46420422 is 0.95950962945189. The square root of 46420422 is 6813.2534078808.
If you square 46420422 you will get the following result 2154855578658084. The natural logarithm of 46420422 is 17.653250049678 and the decimal logarithm is 7.6667090842336. You should now know that 46420422 is very special number!
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Solar Panels and Solar Radiation Flux Density Help - Very Confused
AN630078
Homework Statement:
Hello, I have a question which I am really struggling with attached below. I have tried to answer all sections, and areas in which there is an absence of solutions I have tried to suggest possible methods or approaches I could take. I am very uncertain though and would thus be very grateful of any help. I know there are quite a few questions here but they are related to one another, therefore, where I am uncertain about aspects this may be because I have made a mistake earlier, so if anyone could offer possible suggestions to help I would be very grateful. I am most stuck on question 1 b and c, and the same for 2 b and c which is essentially repeating the method but for the winter reading. This is supposedly an A-Level question I am completing for revision purposes yet I have never come across solar radiation flux density, which is causing me great confusion.
A company invests in 14 solar panels to generate electricity in the north-east of England. Each panel has an area of 1.38 m^2.
The average energy produced per day during the summer is 16.1 kWh and the average energy produced per day during the winter is 2.1 kWh. The panels are all oriented South and the roof is inclined at 35° (which gives an angle of 55° between the panel and the radiation).
Question 1:
a) Calculate the power generated in the summer
b) The solar panels are 100 % efficient, find the solar radiation flux density? Consider the angle of inclination in your calculation.
c) It is found that the solar panels are only 20.8 % efficient, meaning the calculated value of the solar radiation flux density in part c is an underestimate. Find the actual solar radiation flux density in north-eastern England during the summer?
Question 2: Repeat for the winter reading.
Question 3: It is often very cold and cloudy in the north-east of England, considering this is the textbook’s value of 300 Wm^-2 a reasonable example?
Relevant Equations:
P=E/t
Intensity (Wm^-2)= Power (W)/ Area (m^2)
Question 1;
a) P=E/t
E=5.796*10^7 J energy produced per day during the summer
However, I am not certain how to calculate the time period, since although this concerns the energy produced per day, the sun does not shine for the entire duration of this 24 hour period. Also, I am unsure of the average length of daylight hours in the north-east of England in the summer, and I do not want to insert such a value as it has not been specified by the question. Consequently, I have proceeded taking the time as one day but I do not think this calculation will produce the correct result for the power generated in the day during the summer.
t= 24 hours * 60 minutes * 60 seconds = 86400 s
P=5.796*10^7/86400
P=4025/6=640.8333… ~ 641 W to three significant figures
b) This is where I am having significant difficulty as I am utterly unfamiliar with the concept of solar radiation flux density. I have researched the term further but have had little fortune in finding anything helpful. I do not have anyone else to ask therefore I would be very grateful for any help or governance to assist my understanding.
The concepts I have come across neglect the angle of inclination also which is why I do not think I have arrived at the correct conclusion.
I did however come across the equation for the Solar Flux Density (Sd), which is the amount of solar energy per unit area on a sphere centred at the Sun with a distance d:
Sd =L/(4πd^2) W/m^2
Or perhaps the formula for the Solar Constant (S), which is defined as the solar energy density at the mean distance of Earth from the sun (1.5 x 1011 m):
S = L / (4 π d^2)
S=(3.9*10^26 W)/[4*3.14*(1.5 *10^11 m)^2] = 1370 W/m^2
I do not know whether either of these are applicable here?
c) Again, I am not certain how to calculate the solar radiation flux density, regardless of the efficiency of the solar panels, which is why I have not answered this section.
Question 2;
a) P=E/t
E=7.56*10^6 J per day
t=? Again, I am uncertain of the time period, as the sun does not shine for the entire duration of a 24 hour period, and the average length of daylight hours in the north-east of England in the winter is shorter than the summer. However, taking the energy reading as per day I shall also take the time period as one day;
t= 86400 s
P=7.56*10^6/86400
P=175/2=87.5 W
b) and c) I do not know how to find the solar radiation flux density of the solar panels in the winter when the solar panels are 100% efficient or 20.8% efficient.
Question 3;
I am very confused here, firstly I have no idea which textbook the question is referring to. Secondly, I assume the value of 300 Wm^-2 is a measure of intensity calculated by; Intensity (Wm^-2)= Power (W)/ Area (m^2)
Consequently, to evaluate the reading of 300 Wm^-2 would I input the values for the power in the summer and for the winter, to see whether this is a reasonable example? I do not think my values found for the power are correct but I will proceed nonetheless to show how I may continue.
In the summer;
Intensity= 641/ 1.38
Intensity = 464.4927.. ~ 464 Wm^-2 to three significant figures
In the winter;
Intensity= 87.5/ 1.38
Intensity=63.40579… ~ 63.4 Wm^-2 to three significant figures
I do not think my calculated values for power in the summer or winter are correct, however, if they were evaluating the calculations for the intensity I would assume the textbook’s reading of 300 Wm^-2 is not a reasonable example, as this figure is too low for the summer and too high for the winter.
Thank you to anyone who replies 👍
Gold Member
Congratulations, you seemed to have overthunk the entire problem... well, mostly.
I find it reasonable to assume that, since the author has felt the need to include reasonably obvious statements, like "consider the angle of inclination in your calculation" and "...meaning the calculated value of the solar radiation flux density in part c b is an underestimate", the student really isn't expected to do much (or anything, really) in the way of independent research in order to answer the problem.
(except maybe part c, where it looks like they want you to either do a calculation or look something up, elsewhere)
As far as "solar radiation flux density" is concerned, all the words are pretty self-explanatory : what do you think it means within the context of the problem ? What units would be appropriate to use ?
Last edited:
AN630078
Hello, upon further reearch I have found that the terms intensity and radiation flux can be used interchangeably,
Congratulations, you seemed to have overthunk the entire problem.
I find it reasonable to assume that, since the author has felt the need to include reasonably obvious statements, like "consider the angle of inclination in your calculation" and "...meaning the calculated value of the solar radiation flux density in part c b is an underestimate", you really aren't being expected to do much(or anything, really) in the way of independent research in order to answer the problem.
As far as "solar radiation flux density" is concerned, all the words are pretty self-explanatory : what do you think it means within the context of the problem ? What units would be appropriate to use ?
@hmmm27 Thank you for your reply. Would "solar radiation flux density" be another term for intensity, which has units of Wm^-2?
So in calculating the solar radiation flux density could I use the equation for intensity?
Intensity (Wm^-2)= Power (W)/ Area (m^2)
But I do not know how to include the angle of inclination here?
Gold Member
Hello, upon further reearch I have found that the terms intensity and radiation flux can be used interchangeably,
@hmmm27 Thank you for your reply. Would "solar radiation flux density" be another term for intensity, which has units of Wm^-2?
It's an actual term, which I found out by typing the term into an Internet search engine.
But I do not know how to include the angle of inclination here?
How do you think the relationship between a) the incoming sunlight, and b) the angle of the panel with respect to that, affects the amount of sunlight received by the panel.
AN630078
It's an actual term, which I found out by typing the term into an Internet search engine.
How do you think the relationship between a) the incoming sunlight, and b) the angle of the panel with respect to that, affects the amount of sunlight received by the panel.
Thank you for your reply. Really? I have not found the term specifically related to solar radiation flux density and I have searched online and in my textbooks, which show nothing to do with it.
I think the amount of incoming sunlgiht, and thus the count of photons, is going to increase more rapidly when the solar panel is facing the sun (at 90 degrees) and will be lowest when it is edge on (at 0 degrees) to the sun. However, I do not know how to relate this angle within the equation I=P/A?
Gold Member
use ""s around the term ...
re: the latter, have you done any trigonometry ? sines, cosines, etc.
If there's a flat piece of paper between you and something you're trying to look at, how does the angle of the paper determine how much of your vision is being blocked ? Obviously if it's flat-on to your line of sight, it's the area of the paper, and if it's edge-on, it doesn't block at all. But, what if it's at an angle ? How much does it block, then ? If you shine a flashlight at it, how does the angle determine how much light hits the paper ?
Last edited:
AN630078
use ""s around the term ...
re: the latter, have you done any trigonometry ? sines, cosines, etc.
If there's a flat piece of paper between you and something you're trying to look at, how does the angle of the paper determine how much of your vision is being blocked ? Obviously if it's flat-on to your line of sight, it's the area of the paper, and if it's edge-on, it doesn't block at all. But, what if it's at an angle ? How much does it block, then ? If you shine a flashlight at it, how does the angle determine how much light hits the paper ?
Thank you for your reply. I tried googling "solar radiation flux density" in quotation marks as you suggested but I have either just come across academic research or companies selling solar panels.
Yes, I have done trigonometry. Also, in reference to your analogy as the angle of the paer increases from being edge-on to flat-on it blocks more of my line of sight. Similarly, as the angle increases if light is shone onto the paper the amount of light hitting the paper increases and more is absorbed.
Gold Member
Thank you for your reply. I tried googling "solar radiation flux density" in quotation marks as you suggested but I have either just come across academic research or companies selling solar panels.
<sigh> got to love "targeted search results" (...not). Sorry about that : my first few hits concern a device called a "pyranometer", which measures the aforementioned "solar radiance flux density" in ##kW/m^2## . Actual photon counts not required, nor relevant.
Yes, I have done trigonometry. Also, in reference to your analogy as the angle of the paer increases from being edge-on to flat-on it blocks more of my line of sight. Similarly, as the angle increases if light is shone onto the paper the amount of light hitting the paper increases and more is absorbed.
Okay, so given that you now know you're dealing with units in square - not cubic - metres, you should be good with the calculation, yes ?
AN630078
<sigh> got to love "targeted search results" (...not). Sorry about that : my first few hits concern a device called a "pyranometer", which measures the aforementioned "solar radiance flux density" in ##kW/m^2## . Actual photon counts not required, nor relevant.
Okay, so given that you now know you're dealing with units in square - not cubic - metres, you should be good with the calculation, yes ?
I am still not sure of the calculation. Would I use Intensity (Wms^-2)=Power (W)/ area(m^2)
Power in the summer was calculated to be 671 W
Then take the total area = area of panel * sin θ
Total area=1.38 m^2 * sin 55
Total area = 1.1304... ~ 1.13m^2
Solar radiation flux = 593.5795.. ~ 594 Wm^-2 to 3.s.f
I am still rather confused here, also would my calaulated value for the summer be correct?
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The problem statement has several doubtful parts.
As you note, the total output over 24 hours is not directly useful in finding the solar flux. For one thing, the angle changes, so the stated angle of incidence only applies at one instant.
Assuming that is midday, we really want the peak output. Now, we could assume the output follows a sine curve from dawn to dusk (different intervals in summer and winter, of course) and deduce the peak kW from the total kWh.
Then again, the given 55 degrees is curious. This is at a latitude of around 55N. If the panels are tilted up at 35 degrees then at midday in midsummer the sun's angle to the normal will be 55-23.5-35 degrees, almost square-on. In midwinter, 55+23.5-35 degrees, around 45 degrees.
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The problem statement has several doubtful parts.
As you note, the total output over 24 hours is not directly useful in finding the solar flux. For one thing, the angle changes, so the stated angle of incidence only applies at one instant.
Assuming that is midday, we really want the peak output. Now, we could assume the output follows a sine curve from dawn to dusk (different intervals in summer and winter, of course) and deduce the peak kW from the total kWh.
Then again, the given 55 degrees is curious. This is at a latitude of around 55N. If the panels are tilted up at 35 degrees then at midday in midsummer the sun's angle to the normal will be 55-23.5-35 degrees, almost square-on. In midwinter, 55+23.5-35 degrees, around 45 degrees.
Thank you for your reply. Yes I agree with you and was thinking surely the angle of incidence applies at one instant.
How would I find the actaul power output then without overcomplicating matters and using the information in the question?
How could I find the peak kW from the total kWh also?
Thank you for explaining that 55 degrees is an unsual angle also, how this would be almost square-on in the summer. I think the entire question is a little peculiar.
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How could I find the peak kW from the total kWh also?
AS I wrote, you can model it as a sine curve from 0 to pi.
If the peak power is A then at time t it is A sin(t). The output over time T is ##\int_0^T A\sin(\omega t).dt=[-\frac A{\omega}\cos(\omega t)]_0^T=\frac{2A}{\omega}##, where ##\omega=\frac{\pi }{12h}##.
So, given this is 16.1kWh we deduce ##A=\frac{\pi 16.1}{24}## kW, or about 2kW.
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AS I wrote, you can model it as a sine curve from 0 to pi.
If the peak power is A then at time t it is A sin(t). The output over time T is ##\int_0^T A\sin(\omega t).dt=[-\frac A{\omega}\cos(\omega t)]_0^T=\frac{2A}{\omega}##, where ##\omega=\frac{\pi }{12h}##.
So, given this is 16.1kWh we deduce ##A=\frac{\pi 16.1}{24}## kW, or about 2kW.
Thank you for your reply, I truly appreciate it, although I do not understand it fully and think it offers greater depth than required by the question. Would the peak kW then be used to find the power output using
I have been thinking how to find the solar radiation flux density. If the power generated in the summer is 641W then for question 1 b could I proceed as follows;
solar flux = 1/sin(θ) *(power output of the panels / area of panels)
solar flux = 1/sin(55) *(671 W / 1.38 m^2)
solar flux = 593.57952 ~ 594 Wm^-2
1.c When it is found that the solar panels are only 20.8 % efficient, would I calculate the actual solar radiation flux density using the formula:
efficiency= useful power output/ total power input * 100%
useful power output=efficiency/100%*total power input
useful power output=0.208 * 671 W
useful power output=139.568 ~ 140 W to 3.s.f
Question 2; Repeating the previous method for the winter:
P=E/t
E=7.56*10^6 J energy produced per day during the summer
To find the solar radiation flux density in the winter;
solar flux = 1/sin(θ) *(power output of the panels / area of panels)
solar flux = 1/sin(55) *(87.5 W / 1.38 m^2)
solar flux = 77.40418... ~ 77.4 Wm^-2
If it is found the solar panels are only 20.8% efficient;
useful power output=efficiency/100%*total power input
useful power output=0.208 * 87.5 W
useful power output= 18.2 W to 3.s.f
Question 3;is 300 Wm^-2 a reasonable example?
In hindsight I believe that the values I have calculated for the power generated are correct, though I am still a little uncertain of them. Therefore, I do not think the example of 300 Wm^-2 is reasonable since this is 294 Wm^-2 too low for the summer and 222.6 Wm^-2 too high for the winter. Even though it is often overcast and cloudy in northern England, the value of 300 Wm^-2 is too general and midrange for the contrast between the summer and winter months.
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solar flux = 1/sin(55) *(671 W / 1.38 m^2)
It doesn’t make sense to use an average power over 24 hours (671 or 641?) and a fixed angle of 55 degrees. That's why you should be using the peak output.
No need to do the calculus, just argue that most of the energy will be generated over 6 to 8 hours, so multiply the average power by 3 or 4 to get peak power.
But you are forgetting this is the total over 14 panels, each with the given area.
efficiency= useful power output/ total power input * 100%
useful power output=efficiency/100%*total power input
Think that through again.
And bear in mind that it ought to be giving you a bigger answer for the solar flux.
Question 3;is 300 Wm^-2 a reasonable example?
Reasonable for what? Solar flux? Irradiation of the panels? Output from the panels? Peak or average over 24h?
Btw, doesn't matter if it is cold in the NE, that actually improves efficiency. Just need the sunshine.
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@haruspex Thank you for your reply. I am still really confused here.
a) So in the summer, to find the power generated multiply the average power per day by 4 (since the peak power is generated in 8 hours for example). Also, you mention that I forgot that this is the total over 14 panels, each with the given area.
5.796*10^7 J * 4=2.32*10^8 J to 3.s.f
Area = 1.38*14=19.32 m^2
P=2.32*10^8/19.32=11904761.9 ~ 1.19*10^7 W
And for the winter the peak power would be;
7.56*10^6 J*4=3.02*10^7 J to 3.s.f
Then P=3.02*10^7/19.32=22878787.88 ~ 2.30* 10^7 W
b) Could I still use (1/sin θ) *(power output of the panels / area of panels) to find the solar radiation flux?
c) You said to rethink the formula for efficiency, so would my answer have been wrong ? How can I correct this?
Question 3; I have no idea what "reasonable" is here referring to. I think by the units of 300 Wm^-2 I would assume the intensity?
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a) So in the summer, to find the power generated multiply the average power per day by 4 (since the peak power is generated in 8 hours for example)
I see you have tagged this "a)". Does that mean you are using this to answer part a)?
I didn't pay much attention to that part before... not sure what it is asking, but my best guess is that they just want the average power over 24 hours, which is a simple matter of dividing the daily energy output by the length of a day.
You do need it for part b).
Suppose the peak output power is Ppeak. If most of the energy is generated over h hours each day then the total for the day will be about hPpeak, so the average power will be hPpeak/24.
In the present case, you are (unusually) given the average and want to find the peak, so we reverse this to get Ppeak=Pavg*24/h.
In Sydney, the standard is that the daily total is only equal to about 4 hours of peak output, so h=4, and you would multiply by 6 to get peak from average. But if you try to derive it from simply assuming a sine curve you get h=24/π, and a multiplier of only π. This discrepancy is partly because in the early morning and late afternoon it is not just that the angle of incidence is low; there is also the issue of all that extra atmosphere the light has to pass through. So the sine curve overestimates the power at those times. Another reason is that the peak is only achievable when the sun is square on to the panel, and with typical installations that only happens at the equinoxes.
Go with a multiplier of four.
b) Could I still use (1/sin θ) *(power output of the panels / area of panels) to find the solar radiation flux?
Yes, if that means peak output, theta is measured as angle to the panel (not angle to the normal), and we are assuming 100% efficiency here. The difficulty is knowing what the question setter intends to be used for theta. If we assume the panels are oriented to maximise output at the equinoxes, theta would be 23.5 degrees at the solstices.
You said to rethink the formula for efficiency
I think the lack of parentheses led me to misinterpret your formula.
Question 3; I have no idea what "reasonable" is here referring to. I think by the units of 300 Wm^-2 I would assume the intensity?
The units would also be appropriate for power output per unit area of panel, but given the context of the immediately preceding question I would think it means solar irradiance per unit area of Earth's surface in NE England. Question is, when?
For square-on irradiance, it is about 1000W/m2, so 300 would be with the sun at only 18 degrees above the horizon. It is 35 degrees above midday at an equinox, 12 above in midwinter. So I don't have a good guess as to what the 300 represents.
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I see you have tagged this "a)". Does that mean you are using this to answer part a)?
I didn't pay much attention to that part before... not sure what it is asking, but my best guess is that they just want the average power over 24 hours, which is a simple matter of dividing the daily energy output by the length of a day.
You do need it for part b).
Suppose the peak output power is Ppeak. If most of the energy is generated over h hours each day then the total for the day will be about hPpeak, so the average power will be hPpeak/24.
In the present case, you are (unusually) given the average and want to find the peak, so we reverse this to get Ppeak=Pavg*24/h.
In Sydney, the standard is that the daily total is only equal to about 4 hours of peak output, so h=4, and you would multiply by 6 to get peak from average. But if you try to derive it from simply assuming a sine curve you get h=24/π, and a multiplier of only π. This discrepancy is partly because in the early morning and late afternoon it is not just that the angle of incidence is low; there is also the issue of all that extra atmosphere the light has to pass through. So the sine curve overestimates the power at those times. Another reason is that the peak is only achievable when the sun is square on to the panel, and with typical installations that only happens at the equinoxes.
Go with a multiplier of four.
Yes, if that means peak output, theta is measured as angle to the panel (not angle to the normal), and we are assuming 100% efficiency here. The difficulty is knowing what the question setter intends to be used for theta. If we assume the panels are oriented to maximise output at the equinoxes, theta would be 23.5 degrees at the solstices.
I think the lack of parentheses led me to misinterpret your formula.
The units would also be appropriate for power output per unit area of panel, but given the context of the immediately preceding question I would think it means solar irradiance per unit area of Earth's surface in NE England. Question is, when?
For square-on irradiance, it is about 1000W/m2, so 300 would be with the sun at only 18 degrees above the horizon. It is 35 degrees above midday at an equinox, 12 above in midwinter. So I don't have a good guess as to what the 300 represents.
Thank you for your reply. I know that you stated it would not make sense to use average power over 24 hours and that I should use the peak output, I am just unsure about doing so as no specific information has been given regarding the peak daylight hours? I think it is negligible on the part of the way the question is written but I do think it is implying that one should find the average over 24 hours, since this can be found with the information given. Even the questions like "calculate the power generated in the summer" are a little ambiguous, is this per day, or the average across the whole of the summer, or the total adding the power each day in summer?
a) Yes, I was trying to approach the problem in order. So is this asking for the average power generated across the whole of the summer? i.e if summer is approximately 91 days?
Ok, so if the peak power occurs over 4 hours;
t=4*60=240 s
Average Power=(240*5.796*10^7)/(24*3600)
Average Power=161,000 W
b) I think the angle I am supposed to use is 55 degrees between the solar panels and the radiation.
Solar radiation flux= (1/sin θ) *(power output of the panels / area of panels)
Solar radiation flux=10173.12157 ~ 102000 Wm^-2 to 3.s.f.
c) If the panels are 20.8% efficient
useful power output=(efficiency/100%)*total power input
useful power output=0.208*161,000
actual solar radiation flux = 33488 W
Question 2; For the winter reading:
iI the peak power occurs over 4 hours;
t=4*60=240 s
Average Power=(240*7.56*10^6)/(24*3600)
Average Power=21000 W
Solar radiation flux= (1/sin θ) *(power output of the panels / area of panels)
Solar radiation flux=1326.92890~ 1327 Wm^-2 to 3.s.f
If the panels are 20.8% efficient:
useful power output=(efficiency/100%)*total power input
useful power output=0.208*21000 W
actual solar radiation flux = 4368 W
Would this be correct? The readings are far greater than my previous calculations though.
Question 3;
If 300 Wm^-2 is the sun at only 18 degrees above the horizon, would this mean that this occurs only at a specific time of day at this angle, therefore it would not be a reasonable example as this is limited to a specific point of solar radiation flux output, compared to the averages previously calculated. Honestly I do not know and I am grasping at straws here 😂
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is this asking for the average power generated across the whole of the summer? i.e if summer is approximately 91 days?
Ok, so if the peak power occurs over 4 hours;
t=4*60=240 s
Average Power=(240*5.796*10^7)/(24*3600)
Average Power=161,000 W
The question confuses me. It asks for the "power generated in the summer". If it asked for the energy generated it would be a matter of multiplying 16.1 kWh x however many days you think summer is.
But it asks for power... is that average power over a summer's day? If so, 16.1 kWh/24h.
Anyway, I cannot make sense of your calculation. You cannot multiply by 91 days and end up with a number of Watts. Joules, possibly.
b) I think the angle I am supposed to use is 55 degrees between the solar panels and the radiation.
Solar radiation flux= (1/sin θ) *(power output of the panels / area of panels)
Solar radiation flux=10173.12157 ~ 102000 Wm^-2 to 3.s.f.
The peak combined power output from the panels cannot be more than 4kW at most (16.1kWh/4h). That solar flux is 100 times reality.
c) If the panels are 20.8% efficient
useful power output=(efficiency/100%)*total power input
useful power output=0.208*161,000
actual solar radiation flux = 33488 W
You are confused between input and output. Actual solar radiation would be input, so = output *100%/efficiency.
Question 3;
I cannot figure out what this 300W/m2 refers to, so I cannot help with this one.
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I cannot figure out what this 300W/m2 refers to, so I cannot help with this one.
Hi. Can I chip in? The question is poorly written. I believe all references to "solar radiation flux density"
(intensity) are intended to be *average value over 24 hours*.
If I'm correct, then 300W/m2 is meant to be the generally accepted average intensity over a full day. (Obviously during the daytime, the intensity is much larger, and at night time it is effectively zero.)
However, it is not entirely clear if 300W/m2 is meant to be the summer or winter value.
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Hi. Can I chip in? The question is poorly written. I believe all references to "solar radiation flux density"
(intensity) are intended to be *average value over 24 hours*.
If I'm correct, then 300W/m2 is meant to be the generally accepted average intensity over a full day. (Obviously during the daytime, the intensity is much larger, and at night time it is effectively zero.)
However, it is not entirely clear if 300W/m2 is meant to be the summer or winter value.
Yes, that looks reasonable. So a) is asking for the average power over 24h on summer's day.
Since it asks the same questions re summer and winter, the 300W/m2 is probably the average of the two.
Any similarly brilliant insight re the 35 degree and 55 degree info?
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… a) is asking for the average power over 24h on summer's day.
Since it asks the same questions re summer and winter, the 300W/m2 is probably the average of the two.
1a) and 2a) are the average output powers of the panels in summer and winter.
1b) and 2b) are a the average intensities of the sunlight in summer and winter.
If 300W/m2 is meant to be an annual average, then it should broadly match the average of the answers to 1b)and 2b). But that’s just a guess.
[Any similarly brilliant insight re the 35 degree and 55 degree info
I hope you're not being sarcastic! Yes, the wording in the question is poor. But I think
“an angle of 55° between the panel and the radiation”
means that the angle of incidence (to thepanel's normal) is 90° – 55° = 35°. So if the intensity is I, the useful component of intensity is Icos(35°). But again, that’s just a guess.
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So to find 1a) The average power output in the panels in the summer;
P=5.796*10^7/86400
P=670.8333… ~ 671 W to three significant figures
b) Would the total area be per panel or per 14 panels? i.e. 1.38 m^-2 or (1.38*14)=19.32 m^2
Also, should the angle used here be 35 degrees? If so why would it not be 55 degrees?
Solar flux = 1/sin(θ) *(power output of the panels / area of panels)
solar flux = 1/sin(35) *(671 W / 19.32 m^2)
solar radiation flux = 60.5513... ~ 60.6 Wm^-2
Or would it be with the area per panel: 1/sin(35) *(671 W / 1.38 m^2)=847.719~848 Wm^-2
c)Have I used the efficiency formula incorrectly here? Would it be efficiency = (useful power output/total power input)*100%
useful power output = (efficiency/100%)*total power input
useful power output = 0.208 *671
useful power output =139.568 ~ 140 W to 3.s.f.
2. a) The average power output in the panels in the winter;
E=7.56*10^6 J per day
P=E/t
P=7.56*10^6/86400 = 87.5 W
b) Again, this depends on the area and angle so I am still uncertain.
Solar flux = 1/sin(θ) *(power output of the panels / area of panels)
solar flux = 1/sin(35) *(87.5 W / 19.32 m^2)=7.8960~7.89 Wm^-2
or
solar flux =1/sin(35) *(87.5 W / 1.38m^2)=110.5446 ~ 111 Wm^-2
I think the are must be per panel because otherwise the value for the intensity appears far too low for the area of the 14 panels.
c)useful power output = 0.208 *87.5
useful power output =18.2 W to 3.s.f.
3. If 300 Wm^-2 is supposed to be the average of the summer and winter reading, would this be 111 Wm^-2+848 Wm^-2 =479.5 Wm^-2
This far surpasses the value of 300 Wm^-2, meaning the textbook's example is an underestimate of the average intensity.
Would my calculations be correct, and if not how and where can I amend them? Thank you very much for your help. 😁👍
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So to find 1a) The average power output in the panels in the summer;
P=5.796*10^7/86400
P=670.8333… ~ 671 W to three significant figures
Agreed. But if you understand what power means and how to use units you can much more simply:
P = 16.1kWh/day = 16100W h/(24h) = 16100/24 W = 671W (the answer to 1a))
b) Would the total area be per panel or per 14 panels? i.e. 1.38 m^-2 or (1.38*14)=19.32 m^2
The total area A = 1.38*14=19.32 2 m2 How else could you interpret the word 'total'?!
Also, should the angle used here be 35 degrees? If so why would it not be 55 degrees?
The original question is so badly written we can't be certain what angle to use. I believe the angle of incidence (relative to the normal) is 35º for the reasons I already explained in post #21. Note that cos(35º) = sin(55º) so you can use either. But your guess is as good as mine.
Solar flux = 1/sin(θ) *(power output of the panels / area of panels)
solar flux = 1/sin(35) *(671 W / 19.32 m^2)
solar radiation flux = 60.5513... ~ 60.6 Wm^-2
I don't like the term solar flux. Intensity (I) is the correct term. I would write:
I = Power/effective area = 670.8/(19.32cos(35º)} = 42.4W/m²
So that's the answer to 1b).
Or would it be with the area per panel: 1/sin(35) *(671 W / 1.38 m^2)=847.719~848 Wm^-2
No.
c)Have I used the efficiency formula incorrectly here? Would it be efficiency = (useful power output/total power input)*100%
No! Useful power output is NOT actual solar radiation flux. Solar radiation flux is what is going IN, Useful power is what is coming OUT. And there is a simpler way to do the calculation if you can think mathematically.
Useful power output = 671W (as calculated in part 1a). That hasn't changed. But now we realize the efficiency was much less than 100%. This means the intensity must have been much higher than 42.4W/m². Since the efficicency was 20.8%, by proportion intensity must have actually been 42.4 * 100/20.8 = 203.8 = 204W/m² (so that's the answer to 1c)
If you check, you will see I've actually done 1a), 1b) and 1c) for you! 2a), 2b) and 2c) are exactly the same calculation but using 2.1kWh rather than 16.1kWh. If you think carefully, the answerS to 2a, 2b) and 2c) are the same as for 1a), 1b) and 1c) but multipled by 2.1/16.1.
So you should be abe to complete Q2.
For Q3, 300W/m² is well above the answer to 1c) and it will be even more the answer to 2c). This suggests 300W/m² is much too high.
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Agreed. But if you understand what power means and how to use units you can much more simply:
P = 16.1kWh/day = 16100W h/(24h) = 16100/24 W = 671W (the answer to 1a))
The total area A = 1.38*14=19.32 2 m2 How else could you interpret the word 'total'?!
The original question is so badly written we can't be certain what angle to use. I believe the angle of incidence (relative to the normal) is 35º for the reasons I already explained in post #21. Note that cos(35º) = sin(55º) so you can use either. But your guess is as good as mine.
I don't like the term solar flux. Intensity (I) is the correct term. I would write:
I = Power/effective area = 670.8/(19.32cos(35º)} = 42.4W/m²
So that's the answer to 1b).
No.
No! Useful power output is NOT actual solar radiation flux. Solar radiation flux is what is going IN, Useful power is what is coming OUT. And there is a simpler way to do the calculation if you can think mathematically.
Useful power output = 671W (as calculated in part 1a). That hasn't changed. But now we realize the efficiency was much less than 100%. This means the intensity must have been much higher than 42.4W/m². Since the efficiency was 20.8%, by proportion intensity must have actually been 42.4 * 100/20.8 = 203.8 = 204W/m² (so that's the answer to 1c)
If you check, you will see I've actually done 1a), 1b) and 1c) for you! 2a), 2b) and 2c) are exactly the same calculation but using 2.1kWh rather than 16.1kWh. If you think carefully, the answerS to 2a, 2b) and 2c) are the same as for 1a), 1b) and 1c) but multiple by 2.1/16.1.
So you should be abe to complete Q2.
For Q3, 300W/m² is well above the answer to 1c) and it will be even more the answer to 2c). This suggests 300W/m² is much too high.
Repeating the procedure you have exhibited in question 1;
2. a) P=2.1 kWh/day = 2100 h/(24h) = 87.5W
b) No I agree, I had not actually come across the term "solar radiation flux density" prior to this question and am far more familiar with using intensity.
I = Power/effective area = 87.5/(19.32cos(35º)} = 5.5288 ~ 5.53W/m²
c) Oh ok so would 5.53W/m² be the intensity of the solar panel at 20.8% efficiency?
5.53*100/20.8=26.5811 Wm^-2 (the actual solar radiation flux)
Question 3) This is not a reasonable example as 300Wm^-2 is a tremendous overestimate of the average intensity per day in both the summer and winter. Even if this is assumed to be the average intensity over the year (summer and winter), that would be equal to (42.4+5.53)/2=13.29055 Wm^-2 which is still far lower than the example given.
Thank you again, do I need to amned anything here? 👍
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I hope you're not being sarcastic!
I worried you might think that, but I assure you not. Hence the Like.
But I think “an angle of 55° between the panel and the radiation”
means that the angle of incidence (to thepanel's normal) is 90° – 55° = 35°.
Yes, it feels like it is trying to say something like that, but it says:
"the roof is inclined at 35° (which gives an angle of 55° between the panel and the radiation)."
No, that gives an angle of 55° to the vertical. The angle to the radiation depends also on the angle of the rays to the vertical.
As I noted, the location is around 55°N. Angling the panel at 35° to the horizontal gives an average angle between the normal and the midday sun of 20°. At the solstices we can add and subtract 23.5°, so nearly square on in midsummer but around 45° in midwinter.
This matters because the student has to make use of the angle in the calculation.
I tried looking up UK websites advising on mounting angle to see what is done in practice. The results are worrying. More than one says to elevate the panel at 35 to 40, on the basis of 90-latitude (!). http://www.reuk.co.uk/wordpress/solar/solar-panel-mounting-angle/ says latitude+10 to 15 degrees, but offers no explanation. My guess is so as to favour wintertime output, but it will cut total output over the year.
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Update.. maybe the UK advice of 35 to 40 elevation maximises the annual output. Trouble is, peak demand will be winter.
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I don't like the term solar flux. Intensity (I) is the correct term
Intensity is another word for energy flux density. But in common (maybe sloppy) parlance, people like to drop the word "density". So you just have to keep your wits about you as to whether someone is talking about the evaluated surface integral (formally flux), or the integrand (formally flux density)!
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I worried you might think that, but I assure you not. Hence the Like.
Yes, it feels like it is trying to say something like that, but it says:
"the roof is inclined at 35° (which gives an angle of 55° between the panel and the radiation)."
No, that gives an angle of 55° to the vertical. The angle to the radiation depends also on the angle of the rays to the vertical.
As I noted, the location is around 55°N. Angling the panel at 35° to the horizontal gives an average angle between the normal and the midday sun of 20°. At the solstices we can add and subtract 23.5°, so nearly square on in midsummer but around 45° in midwinter.
This matters because the student has to make use of the angle in the calculation.
I tried looking up UK websites advising on mounting angle to see what is done in practice. The results are worrying. More than one says to elevate the panel at 35 to 40, on the basis of 90-latitude (!). http://www.reuk.co.uk/wordpress/solar/solar-panel-mounting-angle/ says latitude+10 to 15 degrees, but offers no explanation. My guess is so as to favour wintertime output, but it will cut total output over the year.
Thank you very much for your reply and for pursuing this further. My apologies but I am a little confused what the correct angle to use would be when finding the intensity?
AN630078
Update.. maybe the UK advice of 35 to 40 elevation maximises the annual output. Trouble is, peak demand will be winter.
I do not know whether this will help but I found this regarding radiation flux (intensity) in a textbook online 👍
Here is a link to it also, https://www.pearsonschoolsandfecolleges.co.uk/AssetsLibrary/SECTORS/Secondary/PDFs/Science/EdexcelScience/ALevelRevisionGuides/EdexcelASPhysicsRG_9781846905957_pg70-78_web.pdf
It also mentions efficiency, as required by part c.
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2a) P=2.1 kWh/day = 2100 h/(24h) = 87.5W
2b) I = Power/effective area = 87.5/(19.32cos(35º)} = 5.5288 ~ 5.53W/m²
Agreed.
2c) Oh ok so would 5.53W/m² be the intensity of the solar panel at 20.8% efficiency?
No. It is meaningless to refer to the "intensity of the solar panel". The solar panel does not have an intensity. The intensity is the amount of power from the sun per square metre. 5.53W/m² is the intensity *calculated assuming the solar panels are 100% efficient*. So 5.53W/m² is not an accurate value of intensity (because efficiency is not 100%).
But we know the panels are only 20.8% efficient. So intensity must be (roughly 5 times) bigger than 5.53W/m².
5.53*100/20.8=26.5811 Wm^-2 (the actual solar radiation flux)
Yes but round to 3 sig. figs.
Question 3) This is not a reasonable example as 300Wm^-2 is a tremendous overestimate of the average intensity per day in both the summer and winter. Even if this is assumed to be the average intensity over the year (summer and winter), that would be equal to (42.4+5.53)/2=13.29055 Wm^-2 which is still far lower than the example given.
No. First, 'example' is the wrong word. And we physicists don't use words like 'tremendous' in this context! Just say (for example) "300W/m² is an overestimate ...
'per day' is wrong. You = mean averaged over a day.
13.29055 is too many significant figures.
The most *accurate* values of intensity are the ones calculated allowing for efficiency. So you should be averaging the answers from 1c) and 2c), not from 1b) and 2b).
(204 + 26.6)/2 = 115Wm/s²
So the 300W/m² (assumed to be an annual average), compared to 115W/m², is roughly between two and three times too big.
AN630078
AN630078
Agreed.
No. It is meaningless to refer to the "intensity of the solar panel". The solar panel does not have an intensity. The intensity is the amount of power from the sun per square metre. 5.53W/m² is the intensity *calculated assuming the solar panels are 100% efficient*. So 5.53W/m² is not an accurate value of intensity (because efficiency is not 100%).
But we know the panels are only 20.8% efficient. So intensity must be (roughly 5 times) bigger than 5.53W/m².
Yes but round to 3 sig. figs.
No. First, 'example' is the wrong word. And we physicists don't use words like 'tremendous' in this context! Just say (for example) "300W/m² is an overestimate ...
'per day' is wrong. You = mean averaged over a day.
13.29055 is too many significant figures.
The most *accurate* values of intensity are the ones calculated allowing for efficiency. So you should be averaging the answers from 1c) and 2c), not from 1b) and 2b).
(204 + 26.6)/2 = 115Wm/s²
So the 300W/m² (assumed to be an annual average), compared to 115W/m², is roughly between two and three times too big.
2. c) Sorry my mistake, clearly I misused intensity here.
Question 3) "example" was the wording from the original question. Oh right, I will correct my average value also, thank you for remarking on that.
Can I verify that it would be correct to use an angle of 35 degrees when finding the intensity, as I have become a little uncertain?
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Intensity is another word for energy flux density. But in common (maybe sloppy) parlance, people like to drop the word "density". So you just have to keep your wits about you as to whether someone is talking about the evaluated surface integral (formally flux), or the integrand (formally flux density)!
Yes indeed. In general, we have to be prepared to interpret loosely used terminology in context. It's very difficult for the younger students who have had little exposure to different terminologies.
etotheipi and AN630078
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Can I verify that it would be correct to use an angle of 35 degrees when finding the intensity, as I have become a little uncertain?
I can't verify that. The wording in the question is unclear. 35º is is simply my best guess using the available information, as I explained in message #21. Is this a written piece of work you have to hand-in? If so precede it with a statement such as:
"The direction of solar radiation relative to the panel is unclear. I have assumed an angle of incidence of 35º".
You will not be penalised if the angle is not 35º because the question is so obviously poorly-worded.
AN630078
AN630078
I can't verify that. The wording in the question is unclear. 35º is is simply my best guess using the available information, as I explained in message #21. Is this a written piece of work you have to hand-in? If so precede it with a statement such as:
"The direction of solar radiation relative to the panel is unclear. I have assumed an angle of incidence of 35º".
You will not be penalised if the angle is not 35º because the question is so obviously poorly-worded.
Thank you very much for your reply and for your advice, I admit I will probably use your statement. It is a question my physics tutor sent to me in a quantum physics revision package of questions, but I had not come across a homogenous problem to this before, at least not within the context. If you like, once I have feeback from them I can inform you of the angle demanded by the question. Thank you again 👍😁
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Gold Member
2022 Award
I worried you might think that, but I assure you not. Hence the Like.
Thankyou!
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# How to plot confidence interval in Python?
Hello Guys, How are you all? Hope You all Are Fine. Today We Are Going To learn about How to plot confidence interval in Python in Python. So Here I am Explain to you all the possible Methods here.
## How to plot confidence interval in Python?
1. How to plot confidence interval in Python?
Let's assume that we have three categories and lower and upper bounds of confidence intervals of a certain estimator across these three categories:
2. plot confidence interval in Python
Let's assume that we have three categories and lower and upper bounds of confidence intervals of a certain estimator across these three categories:
## Method 1
There are several ways to accomplish what you asking for:
Using only matplotlib
from matplotlib import pyplot as plt
import numpy as np
#some example data
x = np.linspace(0.1, 9.9, 20)
y = 3.0 * x
#some confidence interval
ci = 1.96 * np.std(y)/np.sqrt(len(x))
fig, ax = plt.subplots()
ax.plot(x,y)
ax.fill_between(x, (y-ci), (y+ci), color='b', alpha=.1)
fill_between does what you are looking for. For more information on how to use this function
Output
Alternatively, go for seaborn, which supports this using lineplot or regplot
## Method 2
Let’s assume that we have three categories and lower and upper bounds of confidence intervals of a certain estimator across these three categories:
data_dict = {}
data_dict['category'] = ['category 1','category 2','category 3']
data_dict['lower'] = [0.1,0.2,0.15]
data_dict['upper'] = [0.22,0.3,0.21]
dataset = pd.DataFrame(data_dict)
You can plot the confidence interval for each of these categories using the following code:
for lower,upper,y in zip(dataset['lower'],dataset['upper'],range(len(dataset))):
plt.plot((lower,upper),(y,y),'ro-',color='orange')
plt.yticks(range(len(dataset)),list(dataset['category']))
Resulting with the following graph:
## Summery
It’s all About this issue. Hope all Methods helped you a lot. Comment below Your thoughts and your queries. Also, Comment below which Method worked for you? Thank You.
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Kernel density estimation (KDE) is a procedure that provides an alternative to the use of histograms as a means of generating frequency distributions. The use of the kernel function for lines is adapted from the quartic kernel function for point densities as described in Silverman (1986, p. 76, equation 4.5). If Gaussian kernel functions are used to approximate a set of discrete data points, the optimal choice for bandwidth is: h = ( 4 σ ^ 5 3 n) 1 5 ≈ 1.06 σ ^ n − 1 / 5. where σ ^ is the standard deviation of the samples. 9/20/2018 Kernel density estimation - Wikipedia 1/8 Kernel density estimation In statistics, kernel density estimation ( KDE ) is a non-parametric way to estimate the probability density function of a random variable. The density at each output raster cell is calculated by adding the values of all the kernel surfaces where they overlay the raster cell center. For instance, … Kernel density estimation is a way to estimate the probability density function (PDF) of a random variable in a non-parametric way. It includes … However, there are situations where these conditions do not hold. In this section, we will explore the motivation and uses of KDE. The Kernel Density Estimation is a mathematic process of finding an estimate probability density function of a random variable. Setting the hist flag to False in distplot will yield the kernel density estimation plot. The estimation attempts to infer characteristics of a population, based on a finite data set. The kernel density estimation task involves the estimation of the probability density function $$f$$ at a given point $$\vx$$. Let {x1, x2, …, xn} be a random sample from some distribution whose pdf f(x) is not known. The data smoothing problem often is used in signal processing and data science, as it is a powerful … This idea is simplest to understand by looking at the example in the diagrams below. Kernel density estimate is an integral part of the statistical tool box. Later we’ll see how changing bandwidth affects the overall appearance of a kernel density estimate. A kernel density estimation (KDE) is a non-parametric method for estimating the pdf of a random variable based on a random sample using some kernel K and some smoothing parameter (aka bandwidth) h > 0. gaussian_kde works for both uni-variate and multi-variate data. We estimate f(x) as follows: Kernel Density Estimation (KDE) is a way to estimate the probability density function of a continuous random variable. Motivation A simple local estimate could just count the number of training examples $$\dash{\vx} \in \unlabeledset$$ in the neighborhood of the given data point $$\vx$$. 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Population are, there are situations where these conditions do not hold ) is a fundamental data smoothing problem inferences. Section, we will explore the motivation and uses of KDE density..
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# ACT® Math
Free Version
Easy
ACTMAT-JYJQ4N
In the figure below, the combined diameters of the four congruent smaller semicircles make up the diameter of the larger semicircle.
If the diameter of the larger semicircle is 8 cm, what is the area of the shaded region, in square centimeters?
A
${4\pi}$
B
${6\pi}$
C
${12\pi}$
D
${24\pi}$
E
${48\pi}$
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https://www.ipl.org/essay/Advantages-And-Disadvantages-Of-Fuzzy-Clustering-FKYUQJS3RJ48R
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# Advantages And Disadvantages Of Fuzzy Clustering
3280 Words14 Pages
OPTIMAL SENTENCE CLUSTERING USING HIERARCHICAL FUZZY RELATIONAL CLUSTERING INTEGRATED WITH ARTIFICIAL BEE COLONY ALGORITHM
Miss.Prasanthini.R1 Dr.Santhi.V2
PG Student Associate professor
PSG College of Technology Department of Computer Science
Department of Computer Science PSG College of Technology
Coimbatore Coimbatore
E-mail id: prasanthini962@gmail.com E-mail id: sannthi79@gmail.com
Abstract
Clustering is the process of aggregating or grouping data items. In many text processing activities the role of data clustering is inevitable. Sentence clustering plays a vital role in text mining and text processing activities. In real world, same idea can be conveyed using different sentences and for this reason
The prototypes or mixtures of Gaussians based conventional Fuzzy clustering approaches do not represent sentence similarity measures in a common metric space as well as it requires the data to be Euclidean. This becomes a disadvantage and for this reason relational fuzzy clustering algorithm is used, which takes relational data as input. But the documents usually represent information in a hierarchical structure. The fuzzy relational algorithm forms simple or flat partition clustering which gives us a single set of clusters. Hence there is no particular organization or structure within them. But there are cases where one cluster may be a sub cluster of other cluster. (At the time of clustering images, flowers should form the super cluster with roses and marigolds as the sub cluster) So for this reason hierarchical clustering is proposed, which is depicted by a tree diagram or…show more content…
Basic flow diagram
Clustering famous quotations
The famous quotation data set is used in this paper. The context provided in this data set is rich and challenging since it contains a lot of semantic content and are often couched in poetic use of language.
Hierarchical clustering
In every clustering technique the data items are classified based on some properties. Hierarchical clustering uses distance matrix as the clustering criteria to build the hierarchical structure. A hierarchical tree is a nested set of partitions represented by a tree diagram or dendrogram. Its working is based on the union between two closest clusters or to generate sub cluster of a cluster.
Hierarchical Fuzzy Relational Clustering (HFRC)
HFRCA algorithm is a recent renowned algorithm for sentence clustering and is capable of identifying sub clusters. The algorithm proceeds with the similarity measure calculation between the sentences. After which the PageRank is calculated, using which the sentences are clustered.
Artificial Bee colony
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https://www.physicsforums.com/threads/fourier-series.60310/
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Fourier Series
Okay. so i have a problem that i'm finding tricky to work backwards from.
I'm given the coefficients
Period 4 and x(t) is continuous.
a_{k} = 0 if k=0
a_{k} = j^k*sin(k*Pi/4)/(k*Pi) if k!=0
Supposed to find x(t).
Now i've tried a few things but have come up with many different answers. Mostly I am confused as they all depend on the value of k. I am obviously lost as to how to reverse this.
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http://www.gurufocus.com/term/NCAV/CINF/Net%2BCurrent%2BAsset%2BValue/Cincinnati%2BFinancial%2BCorporation
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Switch to:
GuruFocus has detected 1 Warning Sign with Cincinnati Financial Corp \$CINF.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Cincinnati Financial Corp (NAS:CINF)
Net-Net Working Capital Per Share
\$-69.56 (As of Sep. 2016)
In calculating the Net-Net Working Capital (NNWC), Benjamin Graham assumed that a companys accounts receivable is only worth 75% its value, its inventory is only worth 50% of its value, but its liabilities have to be paid in full. In addition, Graham believed that preferred stock belongs on the liability side of the balance sheet, not as part of capital and surplus. This is a conservative way of estimating the companys value.
Cincinnati Financial Corp's net-net working capital per share for the quarter that ended in Sep. 2016 was \$-69.56.
Definition
Cincinnati Financial Corp's Net-Net Working Capital (NNWC) per share for the fiscal year that ended in Dec. 2015 is calculated as
Net-Net Working Capital Per Share (A: Dec. 2015 ) = (Cash And Cash Equivalents + 0.75 * Acct. Receivable + 0.5 * Inventory - Total Liabilities - Preferred Stock) / Shares Outstanding = (544 + 0.75 * 1493 + 0.5 * 0 - 12461 - 0) / 163.90 = -65.88
Cincinnati Financial Corp's Net-Net Working Capital (NNWC) per share for the quarter that ended in Sep. 2016 is calculated as
Net-Net Working Capital Per Share (Q: Sep. 2016 ) = (Cash And Cash Equivalents + 0.75 * Acct. Receivable + 0.5 * Inventory - Total Liabilities - Preferred Stock) / Shares Outstanding = (700 + 0.75 * 1571 + 0.5 * 0 - 13334 - 0) / 164.70 = -69.56
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
In calculating the Net-Net Working Capital (NNWC), Benjamin Graham assumed that a companys accounts receivable is only worth 75% its value, its inventory is only worth 50% of its value, but its liabilities have to be paid in full.
In addition, Graham believed that preferred stock belongs on the liability side of the balance sheet, not as part of capital and surplus. In "Security Analysis", preferred stock is dubbed "an imperfect creditorship position" that is best placed on the balance sheet alongside funded debt.
This is a conservative way of estimating the companys value.
Explanation
One research study, covering the years 1970 through 1983 showed that portfolios picked at the beginning of each year, and held for one year, returned 29.4 percent, on average, over the 13-year period, compared to 11.5 percent for the S&P 500 Index. Other studies of Grahams strategy produced similar results.
Benjamin Graham looked for companies whose market values were less than two-thirds of their net-net value. They are collected under our Net-Net screener. GuruFocus also publishes a monthly Net-Net newsletter.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Cincinnati Financial Corp Annual Data
Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 NNWC -53.12 -57.12 -44.63 -51.32 -54.70 -57.36 -59.15 -50.94 -63.98 -65.88
Cincinnati Financial Corp Quarterly Data
Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 NNWC -63.25 -64.68 -63.98 -63.90 -64.92 -65.55 -65.88 -66.62 -69.56 -69.56
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Genetic Algorithm to Draw a Graph? Position assignment problem
I have an assignment problem at hand and am wondering how suitable it would be to apply local search techniques to reach a desirable solution (the search space is quite large).
I have a directed graph (a flow-chart) that I would like to visualize on 2-D plane in a way that it is very clear, understandable and easy to read by human-eye. Therefore; I will be assigning (x,y) positions to each vertex. I'm thinking of solving this problem using simulated annealing, genetic algorithms, or any such method you can suggest
Input: A graph G = (V,E)
Output: A set of assignments, `{(xi, yi) for each vi in V}`. In other words, each vertex will be assigned a position (x, y) where the coordinates are all integers and >= 0.
These are the criteria that I will use to judge a solution (I welcome any suggestions):
• Number of intersecting edges should be minimal,
• All edges flow in one direction (i.e from left to right),
• High angular resolution (the smallest angle formed by two edges incident on the same vertex),
• Small area - least important.
Furthermore; I have an initial configuration (assignment of positions to vertices), made by hand. It is very messy and that's why I'm trying to automate the process.
My questions are,
• How wise would it be to go with local search techniques? How likely would it produce a desired outcome?
• And what should I start with? Simulated annealing, genetic algorithms or something else?
• Should I seed randomly at the beginning or use the initial configuration to start with?
• Or, if you already know of a similar implementation/pseudo-code/thing, please point me to it.
Any help will be greatly appreciated. Thanks.
EDIT: It doesn't need to be fast - not in real-time. Furthermore; |V|=~200 and each vertex has about 1.5 outgoing edges on average. The graph has no disconnected components. It does involve cycles.
-
Is the main task drawing graphs? There's an awful lot of research already on graph drawing, you should probably read up on that first ... and if this is not your main task, then just use some existing library to do the graph drawing part. – Szabolcs Jul 12 '11 at 12:41
Have you found a solution how to store coordinates and edges in gene to create not so complex genetic algorithm? I mean you decided to store it as a String with 1,0 values or you have found a solution how to store coordinates and edges? – kuldarim Mar 25 at 17:15
I would suggest looking at http://www.graphviz.org/Theory.php since graphviz is one of the leading open source graph visualizers.
Depending on what the assignment is, maybe it would make sense to use graphviz for the visualization altogether.
-
This paper is a pretty good overview of the various approaches. Roberto Tomassia's book is also a good bet.
-
Thanks MPG. The paper is really good but it doesn't point me to a solid algorithm. It is only an overview. Furthermore; the reviews on the book say that it is mostly full of theory but not practical implementations. – Murat Jul 12 '11 at 12:49
http://oreilly.com/catalog/9780596529321 - In this book you might find implementation of genetic algorithm for fine visualization of 2D graph.
In similar situations I'm prefer using genetic algorithm. Also you might start with random initialized population - according to my experience after few iterations, you'll find quite good (but also not the best) solution.
Also, using java you're may paralell this algorithm (isolated islands strategy) - it is rather efficient improvement.
Also I'd like to advice you Differential evolution algorithm. From my experience - it finds solution much more quickly than genetic optimization.
-
``````function String generateGenetic()
String genetic = "";
for each vertex in your graph
Generate random x and y;
String xy = Transform x and y to a fixed-length bit string;
genetic + = xy;
endfor
return genetic;
``````
write a function double evaluate(String genetic) which will give you a level of statisfaction. (probably based on the how many edges intersect and edges direction.
``````int population = 1000;
int max_iterations = 1000;
double satisfaction = 0;
String[] genetics = new String[population]; //this is ur population;
while((satisfaction<0.8)&&(count<max_iterations)){
for (int i=0;i<population;i++){
if(evaluate(genetics[i])>satisfaction)
satisfaction = evaluate(genetics[i]);
else
manipulate(genetics[i]);
}
}
``````
funciton manipulate can flip some bit of the string or multiple bits or a portion that encodes x and y of a vertex or maybe generate completely a new genetic string or try to solve a problem inside it(direct an edge).
-
In plenty of problems transformation from numeric vector to bit vector is unnecessary. But if you want do that, I'll advice to use Gray codes – stemm Jul 12 '11 at 13:33
To answer your first question, I must say it depends. It depends on a number of different factors such as:
• How fast it needs to be (does it need to be done in real-time?)
• How many vertices there are
• How many edges there are compared to the number of vertices (i.e. is it a dense or sparse graph?)
If it needs to be done in a real-time, then local search techniques would not be best as they can take a while to run before getting a good result. They would only be fast enough if the size of the graph was small. And if it's small to begin with, you shouldn't have to use local search to begin with.
There are already algorithms out there for rendering graphs as you describe. The question is, at which point does the problem grow too big for them to be effective? I don't know the answer to that question, but I'm sure you could do some research to find out.
Now going on to your questions about implementation of a local search.
From my personal experience, simulated annealing is easier to implement than a genetic algorithm. However I think this problem translates nicely into both settings. I would start with SA though.
For simulated annealing, you would start out with a random configuration. Then you can randomly perturb the configuration by moving one or more vertices some random distance. I'm sure you can complete the details of the algorithm.
For a genetic algorithm approach, you can also start out with a random population (each graph has random coordinates for the vertices). A mutation can be like the perturbation in SA algorithm I described. Recombination can simply be taking random vertices from the parents and using them in the child graph. Again, I'm sure you can fill in the blanks.
The sum up: Use local search only if your graph is big enough to warrant it and if you don't need it to be done super quickly (say less than a few seconds). Otherwise use a different algorithm.
EDIT: In light of your graph parameters, I think you can do just use whichever algorithm is easiest to code. With V=200, even an O(V^3) algorithm would be sufficient. Personally I feel like simulated annealing would be easiest and the best route.
-
Thanks tskuzzy for the answer. I've answered your questions by editing the original post. It doesn't need to be fast - not in real-time. Furthermore; |V|=~200 and each vertex has about 1.5 outgoing edges on average. I will into other existing methods to see how long they take... – Murat Jul 12 '11 at 12:56
Edited my answer accordingly. – tskuzzy Jul 12 '11 at 13:04
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Community Profile
# amoda
Last seen: 3 maanden ago Active since 2021
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https://abstract-polytopes.com/chiral/1792/1083553/8.html
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Questions?
See the FAQ
or other info.
# Chiral Polytope of Type {14,7}
This page is part of the Atlas of Small Chiral Polytopes
Regular Cover :{14,7}*1792d with group SmallGroup(1792,1083553) = (C26 ⋊ D7) x C2of order 1792
Rank : 3
Schlafli Type : {14,7}
Rotation Group : SmallGroup(112,41) = (C23 ⋊ C7) x C2 of order 112
Number of vertices, edges, etc : 16, 56, 8
If Aut({14,7}*1792d)=<s0, s1, s2>, then this chiral polytope is ({14,7}*1792d)/N, where
N=<s0*s1*s2*s1*s0*s1*s2*s1*s2*s1*s0*s1*s0*s1, s1*s0*s1*s0*s1*s0*s1*s2*s1*s2*s1*s0*s1*s2, s1*s2*s1*s0*s1*s2*s1*s0*s1*s0*s1*s0*s1*s2> of order 8
Facet : (Regular) 14-gon
Vertex Figure : (Regular) 7-gon
Finitely Presented Group Representation of the Rotation Group(GAP) :
```F := FreeGroup("sig1","sig2");;
sig1 := F.1;; sig2 := F.2;;
rels := [ sig1^-1*sig2^-1*sig1^-1*sig2^-1, sig1*sig1*sig1*sig2^-1*sig1*sig2^-1*sig2^-1 ];;
rotpoly := F / rels;;
```
Finitely Presented Group Representation of the Rotation Group (Magma) :
```rotpoly<sig1,sig2> := Group< sig1,sig2 | sig1^-1*sig2^-1*sig1^-1*sig2^-1, sig1*sig1*sig1*sig2^-1*sig1*sig2^-1*sig2^-1 >;
```
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https://calculator.academy/total-rent-calculator/
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Enter the rent ($), the utilities ($), and the rental insurance ($) into the calculator to determine the Total Rent. ## Total Rent Formula The following formula is used to calculate the Total Rent. Rtotal = R + U + I • Where Rtotal is the Total Rent ($)
• R is the rent ($) • U is the utilities ($)
• I is the rental insurance ($) ## How to Calculate Total Rent? The following example problems outline how to calculate Total Rent. Example Problem #1 1. First, determine the rent ($). In this example, the rent ($) is given as 1500 . 2. Next, determine the utilities ($). For this problem, the utilities ($) is given as 100 . 3. Next, determine the rental insurance ($). In this case, the rental insurance ($) is found to be 30. 4. Finally, calculate the Total Rent using the formula above: Rtotal = R + U + I Inserting the values from above yields: Rtotal = 1500 + 100 + 30 = 1630 ($)
Example Problem #2
The variables needed for this problem are provided below:
rent ($) = 1600 utilities ($) = 200
rental insurance ($) = 50 Entering these values and solving gives: Rtotal = 1600 +200 + 50 = 1850 ($)
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# Thread: Help! What do i do???
1. ## Help! What do i do???
Question: You decide you would like to concrete the alfresco area.
Due to budget restrictions you can only afford 1.5m cubed of concrete. Assuming, for aesthetic purposes, the concrete will need to line up with at least one of either the western external wall or the eastern external wall of the house. Council regulations require that the depth of concrete to be 100mm deep.
Determine all the possible ways the concrete slab could satisfy these requirments by calculating the dimensions for each. State and justify which slab is the most practical and why. Include a labelled diagram that clearly shows the shape and dimensions (in mm) your final concrete slab
My initial thoughts are:
V= AH
1.5 = A X 100mm
A = 1500/100
A = 15
.... idk how that helps
or
1.5cubed = 6500 x W x 100mm
or
1.5cubed = L x 3300 x 100mm
ANY IDEAS WOULD BE GREATLY APPRECIATED
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http://www.gilbertandgeorge.co.uk/ypah6b/6a54fd-long-division-with-complex-numbers
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https://www.chilimath.com/lessons/advanced-algebra/dividing-complex-numbers/, http://www.mesacc.edu/~scotz47781/mat120/notes/complex/dividing/dividing_complex.html, http://tutorial.math.lamar.edu/Classes/CalcII/PolarCoordinates.aspx, consider supporting our work with a contribution to wikiHow. worksheet This way, a complex number is defined as a polynomial with real coefficients in the single indeterminate i, for which the relation i 2 + 1 = 0 is imposed. File: Lesson 4 Division with Complex Numbers . So the root of negative number √-n can be solved as √-1 * n = √n i, where n is a positive real number. \frac{ 9 + 4 }{ -4 - 9 } All tip submissions are carefully reviewed before being published, This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. \frac{ \red 3 - \blue{ 2i}}{\blue{ 2i} - \red { 3} } Algebraic long division is very similar to traditional long division (which you may have come across earlier in your education). Up Next. Look carefully at the problems 1.5 and 1.6 below. Using synthetic division to factor a polynomial with imaginary zeros. Keep reading to learn how to divide complex numbers using polar coordinates! By using our site, you agree to our. Another step is to find the conjugate of the denominator. Free Complex Number Calculator for division, multiplication, Addition, and Subtraction $\big( \frac{ 4 -5i}{ 5i -4 } \big) \big( \frac { 5i \red + 4 }{ 5i \red + 4 } \big)$, \$ Figure 1.18 Division of the complex numbers z1/z2. Step 1: To divide complex numbers, you must multiply by the conjugate. Search. Multiply \big( \frac{ 3 -2i}{ 2i -3 } \big) \big( \frac { 2i \red + 3 }{ 2i \red + 3 } \big) \frac{ 35 + 14i -20i \red - 8 }{ 49 \blue{-28i + 28i} - \red - 16 } 5 + 2 i 7 + 4 i. Your support helps wikiHow to create more in-depth illustrated articles and videos and to share our trusted brand of instructional content with millions of people all over the world. $$(7 + 4i)$$ is $$(7 \red - 4i)$$. Multiply \\ $$3 + 2i$$ is $$(3 \red -2i)$$. Please consider making a contribution to wikiHow today. \frac{\blue{20i} + 16 -25\red{i^2} -\blue{20i}} Main content. 0 Downloads. NB: If the polynomial/ expression that you are dividing has a term in x missing, add such a term by placing a zero in front of it. \frac{ 9 \blue{ -6i -6i } + 4 \red{i^2 } }{ 9 \blue{ -6i +6i } - 4 \red{i^2 }} \text{ } _{ \small{ \red { [1] }}} Let's divide the following 2 complex numbers. \\ \boxed{-1} This is termed the algebra of complex numbers. Search for courses, skills, and videos. Keep reading to learn how to divide complex numbers using polar coordinates! \\ \\ File: Lesson 4 Division with Complex Numbers . Courses. \\ \frac{ 6 -8i \red + 30 }{ 4 \red + 36}= \frac{ 36 -8i }{ 40 } {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/d\/d7\/Complex_number_illustration.svg.png\/460px-Complex_number_illustration.svg.png","bigUrl":"\/images\/thumb\/d\/d7\/Complex_number_illustration.svg.png\/519px-Complex_number_illustration.svg.png","smallWidth":460,"smallHeight":495,"bigWidth":520,"bigHeight":560,"licensing":"
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Algebra Tutorials!
Friday 12th of July
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# Dividing Mixed Numbers
Dividing mixed numbers is similar to dividing fractions, except that there is an additional step.
To Divide Mixed Numbers
• change the mixed numbers to their equivalent improper fractions,
• follow the steps for dividing fractions, and
EXAMPLE 1
Find .
Solution
EXAMPLE 2
What is ?
Solution
EXAMPLE 3
There are yards of silk in a roll. If it takes yards to make one designer tie, how many ties can be made from the roll?
Solution
The question is: How many ’s fit into . It tells us that we must divide.
So 9 ties can be made from the roll of silk.
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# More Mathematics for 3D Modelling
This will be the last session on 3D Modelling, designed to tie up loose ends. It contents will therefore depend on the challenges encountered in the three first sessions.
# Learning Outcomes
• Be able to use homogeneous coordinates to model and manipulate motion in 3D
# Briefing and questions
• Quaternions. Ma (2004) Appendix 2.A.
# Exercises
1. Review the stage turntable exercise from last week. Use homogeneous co-ordinates to find the global co-ordinates of the actress (item 5). Check that your calculations match with what you did with heterogeneous co-ordinates.
2. Given a rotational matrix $\begin{bmatrix} \cos(\pi/6) & -\sin(\pi/6) & 0 \\ \sin(\pi/6) & \cos(\pi/6) & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and a translation $$\vec{v}=[1,0,2]$$.
What are the homogenous matrices describing each of the following operations:
• rotate by $$R$$ and then translate by $$\vec{v}$$
• translate by $$\vec{v}$$ and then rotate by $$R$$
3. Review the crane exercise from last week and redo it using homogenous co-ordinates. Check that your calculations match regardless of the method used.
4. Suppose you have rotated by $$R$$ and then translated by $$\vec{v}$$ as given in the previous exercise. What is the homogeneous matrix to undo this operation?
## Exercises from Ma (2004)
1. Exercise 2.11. To calculate eigenvalues and -vectors in Python, you can use numpy.linalg.eig.
2. Exercise 2.7.
3. Exercise 2.10.
4. Ma (2004:40) Exercise 2.14. Hint: start by drawing
5. Ma (2004:40) Exercise 2.13
# Debrief
I showed this sample code in class. It shows one way of dealing with homogeneous co-ordinates together with the plotting tools that we have used.
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Continue to Site
# Connecting a piezoelectric transducer
Status
Not open for further replies.
#### Eight
##### Member level 2
Hello.
I've been looking at some piezo elements with their driving circuits online, and I'm puzzled by a few things. A piezo element is supposedly a capacitive load, and, according to some forum posts, the polarity doesn't really matter. The manufacturers normally specify their rated peak-to-peak voltage. Now, suppose I have a driving circuit, which generates a 16Vp-p 50% duty cycle square waveform at 1 kHz. Assuming the piezo transducer is rated for this voltage, I'd like to feed the given waveform to the element, but I'm mystified how to correctly connect the device.
Should I connect one end to the driving circuit and the other to ground? Or should I create a low-impedance virtual ground at +8V in place of the normal 0V ground (or rather use a square wave generator i.e. a full-bridge that generates output from -8 to +8V in respect to the ground rather than 0V to 16V range)? See the attached picture. In both cases the element will be driven by a 16Vp-p range, but the mean voltage will shifted by 8V down to ground, allowing negative voltage to appear on the piezo element.
What is the difference in performance when driving a piezoelectric transducer in the two given ways?
#### Attachments
• piezo.gif
23.6 KB · Views: 84
There's no difference as long as the applied voltage is considerably below the maximal permitted DC voltage level. Otherwise depolarisation of the piezo can happen.
A piezo transducer strongly resonates at a frequency between 3kHz and 6kHz. If you feed it with a 1kHz squarewave then most of its output will be at its odd harmonic number of 3kHz or 5kHz.
#### Attachments
• piezo speaker.PNG
52.6 KB · Views: 75
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An analysis used for the overall amplitude of a signal is called the root-mean-square (RMS) amplitude or level. Conceptually, it describes the average signal amplitude. However, it is different than simply measuring the arithmetic mean of a signal.
An audio signal can have both positive and negative amplitude values. If we took the arithmetic mean of a sine wave, the negative values would offset the positive values and the result would be zero. This approach is not informative about the average signal level.
This is where the RMS level can be useful. It is based on the magnitude of a signal as a measure of signal strength, regardless of whether the amplitude is positive or negative. The magnitude is calculated by squaring each sample value (so they are all positive), then the signal average is calculated, eventually followed by the square root operation. More completely, the RMS level is, “the square root of the arithmetic mean of the signal squared.”
We can use the RMS level as a reference while changing the amplitude of a signal. This is called RMS Normalization.
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# Is the result always n+1?
I'm reading a book on algorithms by Kurt Mehlhorn and Peter Sanders. On page 2, the following Theorem is stated:
The addition of two n-digit integers requires exactly n primitive operations. The result is an n+1-digit integer.
Is the result always n+1. What if you have 2 2-digit integers like 11 and 22 or others like that?
-
At most $n+1$ digits (also, maybe the book considers $033$ as a $3$ digit number), and I don't think the last two tags are appropriate for this question – Belgi Aug 5 '12 at 4:32
To use the decimal system as an example, consider the case when you're adding $\underbrace{99\dots99}_{n\text{ times}}$ to itself... – J. M. Aug 5 '12 at 4:33
In fact, the text here makes this same point right before starting the section on addition: "We can artificially turn any n-digit integer into an m-digit integer for any m≥n by adding additional leading zeros. Concretely, “425” and “000425” represent the same integer. We shall use a and b for the two operands of an addition or multiplication and assume throughout this section that a and b are n-digit integers. The assumption that both operands have the same length simplifies the presentation without changing the key message of the chapter. We shall come back to this remark at the end of the chapter.We refer to the digits of a as an−1 to a0, with an−1 being the most significant digit (also called leading digit) and a0 being the least significant digit, and write a = ($a_{n−1} . . . a_0$). The leading digit may be zero."
To see why n+1 digits are sufficient here, if you are working in base b, then any n-digit number is less than $b^n$, so the sum of two such numbers is less than $2b^n$, which is less than or equal to $b^{n+1}$ if b is at least 2. Being less than $b^{n+1}$ means that n+1 digits are sufficient to represent it.
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# What types of algorithms are used in AI contract analysis?
Understanding artificial intelligence (AI) contract analysis can be daunting for the uninitiated. Not only does one need to understand the fundamentals of AI, but also what types of algorithms are used in AI contract analysis. In this blog post, we will explore some of the most commonly used algorithms for AI contract analysis and how they work in practice. We’ll discuss deep learning, natural language processing (NLP), neural networks, and more to help you gain a better understanding of how AI is being used in contract analysis today.
## Artificial intelligence (AI)
Artificial intelligence (AI) is a field of computer science that deals with the design and development of intelligent computer systems. AI contract analysis algorithms are used to automatically analyze and interpret contracts. These algorithms can be used to identify important terms and clauses, determine the relationships between parties, and predict the outcomes of negotiations.
## Contract analysis
There are a few different types of algorithms that can be used for contract analysis. The most common type is called a neural network algorithm. This type of algorithm is inspired by the way the human brain works and can learn from data. Another type of algorithm that is often used is called a Bayesian algorithm. This type of algorithm uses probability to make predictions based on data.
## Algorithms
In artificial intelligence, there are many different types of algorithms that can be used for contract analysis. Some of the most common include:
-Rule-based systems: These algorithms use a set of rules to analyze contracts and identify relevant information.
-Statistical methods: These algorithms use statistical techniques to find patterns in data sets.
-Machine learning: These algorithms learn from data and get better over time at identifying patterns.
## How algorithms are used in contract analysis
Contract analysis is a process of reviewing and extracting key information from contracts. This can be done manually, but is often done using algorithms.
There are different types of algorithms that can be used for contract analysis, each with its own strengths and weaknesses. Some of the most common algorithms include:
1. Regular expression-based extraction: This type of algorithm uses regular expressions to identify and extract key information from contracts. It is fast and efficient, but can be limited in its accuracy.
2. Semantic parsing: This type of algorithm uses semantic knowledge to identify and extract key information from contracts. It is more accurate than regular expression-based extraction, but can be slower and more resource intensive.
3. Neural network-based extraction: This type of algorithm uses neural networks to identify and extract key information from contracts. It is highly accurate, but can be slow and resource intensive.
## Benefits of using algorithms in contract analysis
Contract analysis can be a time-consuming and error-prone process, especially when done manually. Algorithms can be used to automate and speed up the contract analysis process, and can also help to improve accuracy by reducing human error.
There are many different algorithms that can be used for contract analysis, each with its own strengths and weaknesses. Some of the more popular algorithms include:
• Rule-based systems: These systems use a set of rules to identify clauses in a contract. The rules can be written by humans or generated automatically from a sample of contracts.
• Statistical models: These models use statistical methods to analyze contracts. They can be used to identify patterns in contracts or to predict future events.
• Neural networks: These are similar to statistical models, but they use artificial neural networks instead of traditional statistical methods. Neural networks can learn from data and improve their predictions over time.
Algorithms can be used to automate many tasks in the contract analysis process, including clause identification, classification, and extraction. They can also help to improve accuracy by reducing human error. In some cases, algorithms may even be able to outperform human analysts.
## Challenges of using algorithms in contract analysis
The use of algorithms in contract analysis is not without its challenges. One challenge is the potential for bias in the results of the analysis. Algorithms are often designed with a particular purpose or goal in mind, and this can lead to results that favor one party over another. Another challenge is the lack of transparency in how the algorithms work. This can make it difficult for parties to understand why certain results were generated, and makes it more difficult to identify errors or potential areas of improvement. Additionally, algorithms can be resource-intensive, and may require significant computing power to run effectively. This can make them impractical for use in many settings, particularly those with limited resources.
## Conclusion
AI contract analysis requires advanced algorithms to help organizations and businesses understand the complexities of legal contracts. There are a variety of different algorithms that can be used in AI contract analysis, such as natural language processing, document summarization, keyword extraction, and topic modeling. Each algorithm has its own strengths and weaknesses so it is important to determine which one is best suited for your specific use case. With the right combination of algorithms, you can make sure you get the most accurate results when performing AI contract analysis.
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# Maths 2U
posted by .
how do you find the horizontal and vertical assymptotes of the curve
y= x/ x+3 and also the intercepts.
so that you can approximate the area under the curve from x=1 to x=5 using 5 function values?
THANKYOU,
• Maths 2U -
There is a vertical asymptote at x = -3 and a horizontal asymptote of y = 1, as x approaches + or = infinity.
The x-intercept is where x = 0, which would be y = (0,0). The y intercept would be where y = 0, or (also) (0,0). There is ony one intercept.
To approximate the area uner the curve, use the trapezoidal or Simpson's rule. and computed values of f(x) at x = 1,2,3,4 and 5. These values are 1/4, 2/5, 1/2, 4/7 and 5/8.
• Maths 2U -
What happens to the function as x>>-3 ?
What happens to the function as x>>inf
What happens to the function as x>>-inf
appx area. I am not certain of the technique you have been taught.
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Cody
# Problem 60. The Goldbach Conjecture
Solution 782811
Submitted on 23 Nov 2015 by Grégory Cristino
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% nList = 28:6:76; for i = 1:length(nList) n = nList(i); [p1,p2] = goldbach(n) assert(isprime(p1) && isprime(p2) && (p1+p2==n)); end
p1 = 23 p2 = 5 p1 = 31 p2 = 3 p1 = 37 p2 = 3 p1 = 43 p2 = 3 p1 = 47 p2 = 5 p1 = 53 p2 = 5 p1 = 61 p2 = 3 p1 = 67 p2 = 3 p1 = 73 p2 = 3
2 Pass
%% nList = [18 20 22 100 102 114 1000 2000 36 3600]; for i = 1:length(nList) n = nList(i); [p1,p2] = goldbach(n) assert(isprime(p1) && isprime(p2) && (p1+p2==n)); end
p1 = 13 p2 = 5 p1 = 17 p2 = 3 p1 = 19 p2 = 3 p1 = 97 p2 = 3 p1 = 97 p2 = 5 p1 = 109 p2 = 5 p1 = 997 p2 = 3 p1 = 1997 p2 = 3 p1 = 31 p2 = 5 p1 = 3593 p2 = 7
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# Knowingpains Financial FormulasKnowingpains Financial Formulas
Finance Formulas / July 18, 2018 / Chanel Cleveland
## Marginal Revenue Product FormulaMarginal Revenue Product Formula
If a corporation has issued only one type, or class, of stock it will be common stock. ("Preferred stock" is discussed later.) While "common" sounds...
Finance Formulas / July 18, 2018 / Natalia Atkins
## National Income FormulaNational Income Formula
Bond valuation is a technique for determining the theoretical fair value of a particular bond. Bond valuation includes calculating the present value of the bond's...
Finance Formulas / July 18, 2018 / Tatiana Douglas
### Marginal Cost FormulaMarginal Cost Formula
The cost of equity is the return a company requires to decide if an investment meets capital return requirements. It is often used as a...
Finance Formulas / July 17, 2018 / Rory Wise
#### Magic Formula InvestingMagic Formula Investing
The topic of business valuation is frequently discussed in corporate finance. Business valuation is typically conducted when a company is looking to sell all or...
Finance Formulas / July 17, 2018 / Chanel Cleveland
##### Net Working Capital FormulaNet Working Capital Formula
An accounting ratio compares two line items in a company’s financial statements, namely made up of its income statement, balance sheet and cash flow statement....
Finance Formulas / July 17, 2018 / Alia Marquez
###### Increasing Annuity FormulaIncreasing Annuity Formula
Current assets are generally listed first on a company's balance sheet and will be presented in the order of liquidity. That means they will appear...
Finance Formulas / July 17, 2018 / Cecelia Weiss
## Operating Profit FormulaOperating Profit Formula
DebtEquity (DE) Ratio, calculated by dividing a company’s total liabilities by its stockholders' equity, is a debt ratio used to measure a company's financial leverage....
Finance Formulas / July 17, 2018 / Rory Wise
## Interest Earned FormulaInterest Earned Formula
The cash flow statement (CFS) measures how well a company manages its cash position, meaning how well the company generates cash to pay its debt...
Finance Formulas / July 17, 2018 / Cecelia Weiss
## Interest Rate FormulaInterest Rate Formula
Given that the debtequity ratio measures a company’s debt relative to the total value of its stock, it is most often used to gauge the...
Finance Formulas / July 17, 2018 / Cecelia Weiss
## Net Margin FormulaNet Margin Formula
Total debt to total assets is a leverage ratio that defines the total amount of debt relative to assets. This metric enables comparisons of leverage...
Knowingpains
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In Free Printable Worksheets252 views
4.32 / 5 ( 171votes )
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# Trading With VWAP and MVWAP
## What Is VWAP and MVWAP?
Volume-weighted average price (VWAP) and moving volume-weighted average price (MVWAP) are trading tools that can be used by all traders to ensure they are getting the best price. However, these tools are used most frequently by short-term traders and in algorithm-based trading programs.
### Key Takeaways
• Volume-weighted average price (VWAP) and moving volume-weighted average price (MVWAP) are trading tools that can be used by all traders to ensure they are getting the best price.
• VWAP is the average price a security has traded at throughout the day, based on both volume and price.
• MVWAP is a user defined average of VWAP calculations and has no final value as it can run fluidly from one day to the next.
## Understanding VWAP and MVWAP
MVWAP may be used by longer-term traders, but VWAP only looks at one day at a time due to its intraday calculation. Both indicators are a special type of price average that takes into account volume which provides a much more accurate snapshot of price action. The indicators also act as benchmarks for individuals and institutions that wish to gauge if they had good execution or poor execution on their order.
## Calculating VWAP
VWAP is the average price a security has traded at throughout the day, based on both volume and price and is important because it provides traders with insight into both the trend and value of a security.
The VWAP calculation is performed by charting software and displays an overlay on the chart representing the calculations. This display takes the form of a line, similar to other moving averages. How that line is calculated is as follows:
• Choose your time frame (tick chart, 1 minute, 5 minutes, etc.)
• Calculate the typical price for the first period (and all periods in the day following). Typical price is attained by taking adding the high, low and close, and dividing by three: (H+L+C)/3
• Multiply this typical price by the volume for that period. This will give you a value called TPV.
• Keep a running total of the TPV values, called cumulative-TPV. This is attained by continually adding the most recent TPV to the prior values (except for the first period, since there will be no prior value). This figure should get larger as the day progresses.
• Keep a running total of cumulative volume. Do this by continually adding the most recent volume to the prior volume. This number should also get larger as the day progresses.
• Calculate VWAP with your information: [cumulative TPV ÷ cumulative volume]. This will provide a volume-weighted average price for each period and will provide the data to create the flowing line that overlays the price data on the chart.
It is likely best to use a spreadsheet program to track the data if you are doing this manually. A spreadsheet can be easily set up with column headings as shown in the picture below. The appropriate calculations would need to be inputted.
Attaining the MVWAP is quite simple after VWAP has been calculated. An MVWAP is basically an average of the VWAP values. VWAP is only calculated per day, but MVWAP can move from day to day because it is an average of an average. This provides longer-term traders with a moving average volume-weighted price.
If a trader wanted a 10-period MVWAP, they would simply wait for the first 10 periods to elapse, then average the first 10 VWAP calculations. This would provide the trader with the MVWAP that starts being plotted at period 10. To continue getting the MVWAP calculation, average the most recent 10 VWAP figures, include a new a VWAP from the most recent period, and drop the VWAP from 11 periods earlier.
## Application to Charts
While understanding the indicators and the associated calculations is important, charting software can do the calculations for us. On software that does not include VWAP or MVWAP, it may still be possible to program the indicator into the software using the calculations above.
By selecting the VWAP indicator, it will appear on the chart. Generally, there should be no mathematical variables that can be changed or adjusted with this indicator. If a trader wishes to use the moving MVWAP indicator, they can adjust how many periods to average in the calculation. This can be done by adjusting the variable in the charting platform. Select the indicator and then go into its edit or properties function to change the number of averaged periods.
## VWAP vs. MVWAP
There are a few major differences between the indicators that need to be understood.
VWAP will provide a running total throughout the day. Thus, the final value of the day is the volume-weighted average price for the day. For example, if using a one-minute chart for a particular stock, there are 390 (6.5 hours X 60 minutes) calculations that will be made for the day, with the last one providing the day's VWAP.
MVWAP, on the other hand, will provide an average of the number of VWAP calculations to analyze. This means there is no final value for MVWAP, as it can run fluidly from one day to the next, providing an average of the VWAP value over time. This makes the MVWAP much more customizable. It can be tailored to suit specific needs. It can also be made much more responsive to market moves for short-term trades and strategies, or it can smooth out market noise if a longer period is chosen.
VWAP provides valuable information to buy-and-hold traders, especially post execution (or end of day). It lets traders know if they received a better-than-average price that day or a worse price. MVWAP does not necessarily provide this same information.
VWAP will start fresh every day. Volume is heavy in the first period after the markets open, therefore, this action usually weighs heavily into the VWAP calculation. MVWAP can be carried from day to day, as it will always average the most recent periods (10 for example), is less susceptible to any individual period and becomes progressively less so the more periods that are averaged.
## General Strategies
When a security is trending, we can use VWAP and MVWAP to gain information from the market. If the price is above VWAP, it is a good intraday price to sell. If the price is below VWAP, it is a good intraday price to buy. However, there is a caveat to using this intraday. Prices are dynamic and what appears to be a good price at one point in the day may not be by day's end.
On upward trending days, traders can attempt to buy as prices bounce off MVWAP or VWAP. Alternatively, they can sell in a downtrend as price pushes up toward the line. The figure below shows three days of price action in the iShares Silver Trust ETF (SLV). As the price rose, it stayed largely above the VWAP and MVWAP, and declines toward the lines provided buying opportunities. As the price fell, it stayed largely below the indicators, and rallies toward the lines were selling opportunities.
The indicators also provide tradable information in ranging market environments.
On ranging days, traders can buy as price crosses above VWAP/MVWAP and sell as price crosses below VWAP/MVWAP for quick trades. This method runs the risk of being caught in whipsaw action. Alternatively, a trader can use other indicators, including support and resistance, to attempt to buy when the price is below the VWAP and MVWAP and sell when the price is above the two indicators.
At the end of the day, if securities were bought below the VWAP, the price attained was better than average. If the security was sold above the VWAP, it was a better-than-average sale price.
## The Bottom Line
VWAP and MVWAP are useful indicators that have some differences between them. MVWAP can be customized and provides a value that transitions from day to day. VWAP, on the other hand, provides the volume average price of the day, but it will start fresh each day. MVWAP can be used to smooth data and reduce market noise, or tweaked to be more responsive to price changes. If a trader sells above the daily VWAP, they get a better-than-average sale price. Similarly, traders that buy below the VWAP get a better-than-average purchase price. On trending days, attempting to capture pullbacks toward the VWAP and MVWAP can produce a profitable result if the trend continues.
Article Sources
Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy.
1. Optuma. "Volume Weighted Average Price Overview."
2. Schwab. "StreetSmart Edge® User Guide," Pages 91-92, 293, 361, 378-379.
3. Zhou, Hao, et al. "Algorithmic Trading in Turbulent Markets." Pacific Basin Finance Journal, Vol. 62, 2020, p. 101358
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What is $10$? If this is your first time learning about the binary number system, then this question may seem odd. Of course it’s ten, right?
Let’s try something different. Have you ever heard this joke?
There are $10$ types of people: those who understand binary and those who don’t.
Unless you’re familiar with binary numbers, this probably doesn’t make much sense. But by the end of this article, you’ll understand this awful joke!
In this beginner’s tutorial, we’ll look at everything you need to know about the binary number system, but we’ll also take a quick look at decimal and hexadecimal, as they’re closely related. I’ll include relevant bits of code and real-life examples to help you appreciate the beauty of binary.
## What Is a Number System?
Before we look at binary, let’s take a step back and discuss number systems more generally.
It may seem strange to think of number systems in the plural if this is your first time learning about them. That’s because the majority of the world is familiar with just one system: the decimal number system, also known as the Arabic number system. This number system uses the digits $0–9$ to represent numbers symbolically, based on their position in a string.
For example, in the decimal number system, $579$ expands to this:
$579 = 5(10^2) + 7(10^1) + 9(10^0) = 500 + 70 + 9$
In school, you were taught that the $5$ in $579$ is in the hundredths place, the $7$ is in the tens place, and the $9$ is in the ones place. Notice that the $5$ is multiplied by one hundred ($10^2$), the $7$ by ten ($10^1$), and the $9$ by one ($10^0$) to form the decimal number $579$. We say that the number $579$ is positional because the digits, from left to right, correspond to a specific power of ten based on the position of the digit in the number.
Here, the number $10$ is what we call the base (aka radix) of our number system. Notice the powers of $10$ in the expanded expression above: $10^2$, $10^1$, and $10^0$. For this reason, the terms decimal and base ten are interchangeable.
In the decimal number system, a number is represented by placing digits into “buckets” that represent increasing powers of ten, starting with $10^0$ in the rightmost “bucket,” followed by $10^1$ to its immediate left, and so on infinitely:
Any unused buckets to the far left have an implicit value of $0$ in them. We usually trim leading zeros because there is no use in saying $00579$ when that’s mathematically identical to $579$.
Why did humans pick $10$ to be the base of their preferred number system? Likely because most people are born with ten fingers and ten toes, and we’re used to counting with our fingers when we’re young. So it’s natural for us to have adopted ten as the base of our number system.
### Bases, Exponents, and Digits
As I’ve already hinted, the decimal number system (base $10$) isn’t the only one in existence. Let’s use a more general notation to represent number systems beyond just our familiar one.
In a number system with a fixed base of $b$, the available digits range from $0$ to $b - 1$. For example, in the decimal number system ($b = 10$), we can only use the digits $0, 1, 2, ..., 9$. When you run out of digits to stuff into a single bucket, you carry over a one to the next power of the base. For example, to get to the number after $99$, you carry a one to the bucket representing the next power of ten ($100$).
Now, suppose that we have a string of digits $d_{n-1} d_{n-2} ... d_0$ (where $n$ is the number of digits). Maybe that’s $d_2 d_1 d_0 = 579$ from our earlier example. That string expands like this:
$d_{n-1} b^{n-1} + d_{n-2} b^{n-2} + ... + d_{0} b^0$
And you can visualize it like this:
Using our same example, $d_{n-1} b^{n-1} + d_{n-2} b^{n-2} + ... + d_{0} b^0 = 5(10^2) + 7(10^1) + 9(10^0)$. Again, we have buckets from right to left in increasing powers of our base ($10$), as depicted below:
Now, in reality, you can have a number system that uses a base of $2$, $3$, $4$, $120$, and so on. Some of these have special names because they’re used more often than others:
Base Name Description
1 Unary Also known as tallying. A number n is represented by picking an arbitrary character and repeating it n times (e.g., xxxx would be 4).
2 Binary Only two digits: zero and one. Most commonly used in computing. Everything on a computer is, at the lowest possible level, stored using the binary number system.
8 Octal Only eight digits are available: 0–7.
16 Hexadecimal Fifteen digits: 0–9 and a–f. Often used to express binary strings more compactly.
60 Sexagesimal How many seconds are in a minute? How many minutes in an hour? This is the basis of the modern circular coordinate system (degrees, minutes, and seconds).
For this reason, when discussing number systems, we usually subscript a number with its base to clarify its value. Alternatively, you can prepend a number with a certain string (usually 0b for binary or 0x/# for hexadecimal). So we’d write $579$ as $579_{10}$, or the binary number $1001$ as $1001_2$ (or $\text{0b}1001$). Otherwise, if we were to merely write the number $1001$ without providing any context, nobody would know whether that’s in binary, octal, decimal, hexadecimal, and so on because the digits $0$ and $1$ are valid in all of those number systems, too!
## The Binary Number System
We’re all familiar with decimal numbers because we use them everyday. But what about the binary number system?
By definition, the binary number system has a base of $2$, and thus we can only work with two digits to compose numbers: $0$ and $1$. Technically speaking, we don’t call these digits—they’re called bits in binary lingo. Each “bucket” in a binary string represents an increasing power of two: $2^0$, $2^1$, $2^2$, and so on.
The leftmost bit is called the most significant bit (MSB), while the rightmost bit is called the least significant bit (LSB).
Here are some examples of representing decimal numbers in the binary number system:
• Zero: $0_{10} = 0_2$. Expansion: $0 (2^0)$
• One: $1_{10} = 1_2$. Expansion: $1(2^0)$
• Two: $2_{10} = 10_2$. Expansion: $1(2^1) + 0(2^0)$
• Three: $3_{10} = 11_2$. Expansion: $1(2^1) + 1(2^0)$
• Four: $4_{10} = 100_2$. Expansion: $1(2^2) + 0(2^1) + 0(2^0)$
• Five: $5_{10} = 101_2$. Expansion: $1(2^2) + 0(2^1) + 1(2^0)$
Having learned the binary number system, you should now understand the joke from earlier:
There are $10$ types of people: those who understand binary and those who don’t.
Here, we really mean the binary equivalent of two, which looks like ten to our eyes when it’s not properly subscripted: $10_2 = 1 × 2^1 = 2_{10}$.
### Binary Is Close to the Hardware of a Computer
Why do we bother with using the binary number system in the first place? Doesn’t it seem like a whole lot of extra work to represent numbers in this manner when we could instead use the decimal number system? Well, yes—if you’re writing these out by hand, it’s certainly more work to represent (and manipulate) binary numbers.
You may not see any point in using binary if you haven’t learned about computer architecture at a low level. Internally, computers are nothing more than electrical circuits tied to hardware. Current either flows through a wire or doesn’t—a binary state. Likewise, computers use logic gates (AND/OR/NOR/XOR) to control the flow of a program’s execution, and these take binary inputs (true/false). The best way to represent these low-level interactions is to use the binary number system: $0$ means “off” (or false in its boolean form) and $1$ means “on” (true).
Everything on your computer—the files you save and the software you install—is represented as nothing more than zeros and ones. But how is this possible?
### The Unicode Standard
Suppose you create a file on your computer and store some basic text in it:
echo Hello, Binary > file
At the end of the day, your computer can’t store a character like H, e, l, or o (or even the space between two words) literally. Computers only know how to work with binary. Thus, we need some way to convert these characters to numbers. And that’s why the Unicode standard was introduced.
Unicode is the most widely accepted character encoding standard: a method of representing human-readable characters like H, e, ,, ?, and 9 numerically so that computers can understand and use them like we do. Each character maps to a unique number known as a code point.
For example, the chart below shows a very limited subset of Unicode characters (known as the ASCII standard) and their corresponding code points:
For the sake of brevity, we’ll focus on just the ASCII standard for now, even though it doesn’t capture the full range of characters in the Unicode standard and the complexities that come with needing to support hundreds of thousands of characters.
The ASCII standard supports only 128 characters, each mapped to a unique number:
• Arabic digits: $0-9$ (10)
• Uppercase Latin letters: $A-Z$ (26)
• Lowercase Latin letters: $a-z$ (26)
• Punctuation and special characters (66)
Again, note that while the ASCII standard only allows us to represent a tiny fraction of Unicode characters, it’s simple enough that it can help us better understand how characters are stored on computers.
### 1 ASCII Character = 1 Byte
In the decimal number system, we’re used to working with digits. In binary, as we already saw, we’re used to working with bits. There’s another special group of digits in binary that’s worth mentioning: A sequence of eight bits is called a byte.
Here are some examples of valid bytes:
00000000
10000000
11101011
11111111
… and any other valid permutation of eight $0$s and $1$s that you can think of.
Why is this relevant? Because on modern computers, characters are represented using bytes.
Recall that the ASCII encoding format needs to support a total of 128 characters. So how many unique number can we represent with $8$ bits (a byte)?
Well, using the product rule from combinatorics, we have eight “buckets,” each with two possible values: either a $0$ or a $1$. Thus, we have $2 × 2 × ... × 2 = 2^8$ possible values.
In decimal, this is $2^8 = 256$ possible values. By comparison, $2^7 = 128$. And $128$ happens to be the number of characters that we want to represent.
So… That’s weird, and seemingly wasteful, right? Why do we use $8$ bits (one byte) to represent a character when we could use $7$ bits instead and meet the precise character count that we need?
Good question! We use bytes because it’s not possible to evenly divide a group of $7$ bits, making certain low-level computations difficult if we decide to use $7$ bits to represent a character. In contrast, a byte can be evenly split into powers of two:
11101011
[1110][1011]
[11][10][10][11]
The key takeaway here is that we only need one byte to store one character on a computer. This means that a string of five characters—like Hello—occupies five bytes of space, with each byte being the numerical representation of the corresponding character per the ASCII format.
Remember the file we created earlier? Let’s view its binary representation using the xxd Unix tool:
xxd -b file
The -b flag stands for binary. Here’s the output that you’ll get:
00000000: 01001000 01100101 01101100 01101100 01101111 00101100 Hello,
00000006: 00100000 01000010 01101001 01101110 01100001 01110010 Binar
0000000c: 01111001 00001010 y.
The first line shows a sequence of six bytes, each corresponding to one character in Hello,.
Let’s decode the first two bytes using our knowledge of the binary number system and ASCII:
• $01001000 = 1(2^6) + 1(2^3) = 72_{10}$. Per our ASCII table, this corresponds to $H$.
• $01100101 = 1(2^6) + 1(2^5) + 1(2^2) + 1(2^0) = 101_{10}$, which is $e$ in ASCII.
Cool! Looks like the logic pans out. You can repeat this for all of the other bytes as well. Notice that on the second line, we have a leading space (from Hello, Binary), represented as $2^5 = 32_{10}$ in ASCII (which is indeed Space per the table).
By the way, what’s up with the numbers along the left-hand side of the output? What does $0000000c$ even mean? Time to explore another important number system!
As I mentioned in the table from earlier, the hexadecimal number system is closely related to binary because it’s often used to express binary numbers more compactly, instead of writing out a whole bunch of zeros and ones.
The hexadecimal number system has a base of $16$, meaning its digits range from $0–15$.
This is our first time encountering a number system whose digits are made up of more than two characters. How do we squeeze $10$, $11$, or $15$ into a single “bucket” or “slot” for a digit? To be clear, this is perfectly doable if you have clear delimiters between digits, like vertical lines—without which you wouldn’t know if $15$ is a one followed by a five or a single digit of $15$ in the ones place. But in reality, using delimiters isn’t practical.
Let’s take a step back and consider a simple hexadecimal number:
$0x42$
What does this mean to us humans in our decimal number system? Well, all we have to do is multiply each digit by its corresponding power of $16$:
$0x42 = 4(16^1) + 2(16^0) = 64_{10} + 2_{10} = 66_{10}$
Okay, so that’s a simple hex number. Back to the problem at hand: How do we represent the hex digits $10$, $11$, and so on? Here’s an example that’s pretty confusing unless we introduce some alternative notation:
$0x15$
Is this a $15$ in a single slot or a $1$ and a $5$ in two separate slots? One way to make this less ambiguous is to use some kind of delimiter between slots, but again, that’s not very practical:
$0x8[15]29$
The better solution that people came up with is to map $10–15$ to the the English letters $a–f$. Note that we could’ve also used any other symbols to represent these digits. As long as we agree on a convention and stick with it, there’s no ambiguity as to what a number represents.
Here’s an example of a hexadecimal number that uses one of these digits:
$0xf4$
And here’s its expansion:
$0xf4 = 15(16^1) + 4(16^0) = 240_{10} + 4_{10} = 244_{10}$
There’s nothing magical about the hexadecimal number system—it works just like unary, binary, decimal, and others. All that’s different is the base!
Before we move on, let’s revisit the output from earlier when we used xxd on our sample file:
00000000: 01001000 01100101 01101100 01101100 01101111 00101100 Hello,
00000006: 00100000 01000010 01101001 01101110 01100001 01110010 Binar
0000000c: 01111001 00001010 y.
The numbers along the left-hand side mark the starting byte for each line of text on the far right. For example, the first line of text (Hello,) ranges from byte #0 (H) to byte #5 (,). The next line is marked as $00000006$, meaning we’re now looking at bytes #6 through 11 (B to r). Finally, the last label should make sense now that you know the hexadecimal number system: c maps to $12$, meaning the byte that follows corresponds to the twelfth character in our file.
### How to Convert Between Binary and Hexadecimal
Now that we know a bit about binary and hexadecimal, let’s look at how we can convert between the two systems.
Say you’re given this binary string and you’d like to represent it in hexadecimal:
$011011100101$
While at first this may seem like a pretty difficult task, it’s actually straightforward!
Let’s do a bit of a thought exercise: In the hexadecimal number system, we have $16$ digits from $0$ to $15$. Over in binary land, how many bits do we need to represent these $16$ values?
The answer is four because $2^4 = 16$. With four “buckets,” we can create the numbers zero ($0000$), one ($0001$), ten ($1010$), all the way up to fifteen ($1111$). This means that when you’re given a binary string, all you have to do is split it into groups of four bits and evaluate them to convert binary to hexadecimal!
011011100101
[0110][1110][0101]
6 14 5
Now we just replace $10–15$ with $a-f$ and we’re done: $0x6e5$.
What about the reverse process? How do you convert a hexadecimal number to binary? Say you’re given the hexadecimal number $0xad$. What do we know about each hexadecimal digit?
Well, from our earlier exercise, we know that four bits comprise one hex digit. So we can convert each individual hex digit to its $4$-bit representation and then stick each group together!
$a_{16} = 10_{10} = 1010_{2} \\ d_{16} = 13_{10} = 1101_{2} \\ ad_{16} = 10101101_{2}$
### Real-World Application: Colors in RGB/Hex
While we’re on the topic of binary and hexadecimal, it’s worth taking a look at one real-world use case for the things we’ve learned so far: RGB and hex colors.
Colors have three components: red, green, and blue (RGB). With LED (light-emitting diode) displays, each pixel is really split into these three components using a color diode. If a color component is set to $0$, then it’s effectively turned off. Otherwise, its intensity is modulated between $0$ and $255$, giving us a color format like rgb(0-255, 0-255, 0-255).
Let’s consider this hex color: #4287f5. What is it in the RGB format?
Well, we need to split this hex string evenly between red, green, and blue. That’s two digits per color:
$[42][87][f5]$
Now, we interpret the decimal equivalent for each part:
• Red: $42_{16} = 4(16^1) + 2(16^0) = 66$
• Green: $87_{16} = 8(16^1) + 7(16^0) = 135$
• Blue: $f5_{16} = 15(16^1) + 5(16^0) = 245$
That means #4287f5 is really rgb(66, 135, 245)! You can verify this using a Color Converter:
For practice, let’s convert this to binary as well. I’ll mark the groups of four bits to make it easier to see how I did this (you could also convert from the decimal RGB representation if you want to):
$0x4287f5 = 0b[0100][0010][1000][0111][1111][0101]$
Now, two groups of four bits will represent one component of the color (red/green/blue):
$0b[01000010][10000111][11110101]$
Notice that each color component takes up a byte ($8$ bits) of space.
#### How Many Colors Are There?
As an additional exercise, how many unique colors can you possibly have in the modern RGB format?
We know that each component (red/green/blue) is represented using one byte ($8$ bits). So the colors we’re used to are really $24$-bit colors.
That means there are a whopping $2^{24} = 16,777,216$ possible unique colors that you can generate using hex/rgb! The $24$-bit color system is known as truecolor, and it’s capable of representing millions of colors.
Note that you could just as well have performed this calculation using hex: #4287f5. There are six slots, each capable of taking on a value from $0$ to $f$. That gives us a total of $16 × 16 × ... × 16 = 16^6 = 16,777,216$ values—the same result as before.
Or, if you’re using the decimal RGB format, the math still pans out:
$256 × 256 × 256 = 16,777,216$
#### What Are 8-Bit Colors?
On older systems with limited memory, colors were represented using just eight bits (one byte). These 8-bit colors had a very limited palette, which meant that most computer graphics didn’t have gradual color transitions (so images looked very pixelated/grainy). With only $8$ bits to work with, you are limited to just $2^8 = 256$ colors!
Naturally, you may be wondering: How did they split $8$ bits evenly among red, green, and blue? After all, $8$ isn’t divisible by three!
Well, the answer is that they didn’t. The process of splitting these bits among the color components is called color quantization, and the most common method (known as 8-bit truecolor) split the bits as 3-3-2 red-green-blue. Apparently, this is because the human eye is less sensitive to blue light than the other two, and thus it simply made sense to distribute the bits heavily in favor of red and green and leave blue with one less bit to work with.
## Signed Binary Number System: Two’s Complement
Now that we’ve covered decimal, binary, and hexadecimal, I’d like us to revisit the binary number system and learn how to represent negative numbers. Because so far, we’ve only looked at positive numbers. How do we store the negative sign?
To give us some context, I’ll assume that we’re working with standard $32$-bit integers that most computers support. We could just as well look at $64$-bit or $N$-bit integers, but it’s good to have a simple basis for a discussion.
If we have $32$ bits to fiddle with, that means we can represent a total of $2^{32} = 4,294,967,296$ (4 billion) numbers. More generally, if you have $N$ bits to work with, you can represent $2^N$ values. But we’d like to split this number range evenly between negatives and positives.
Positive or negative… positive or negative. One thing or another thing—ring a bell? That sounds like it’s binary in nature. And hey—we’re already using binary to store our numbers! Why not reserve just a single bit to represent the sign? We can have the most significant (leading) bit be a $0$ when our number is positive and a $1$ when it’s negative!
Earlier, when we were first looking at the binary number systems, I mentioned that you can strip leading zeros because they are meaningless. This is true except when you actually care about distinguishing between positive and negative numbers in binary. Now, we need to be careful—if you strip all leading zeros, you my be left with a leading $1$, and that would imply that your number is negative (in a signed number system).
You can think of two’s complement as a new perspective or lens through which we look at binary numbers. The number $100_2$ ordinarily means $4_{10}$ if we don’t care about its sign (i.e., we assume it’s unsigned). But if we do care, then we have to ask ourselves (or whoever provided us this number) whether it’s a signed number.
### How Does Two’s Complement Work?
What does a leading $1$ actually represent when you expand a signed binary number, and how do we convert a positive number to a negative one, and vice versa? For example, suppose we’re looking at the number $22_{10}$, which is represented like this in unsigned binary:
$10110_2$
Since we’re looking at signed binary, we need to pad this number with an extra $0$ out in front (or else a leading $1$ would imply that it’s negative):
$010110_2$
Okay, so this is positive $22_{10}$. How do we represent $-22_{10}$ in binary?
There are two ways we can do this: the intuitive (longer) approach and the “shortcut” approach. I’ll show you both, but I’ll start with the more intuitive one.
#### The Intuitive Approach: What Does a Leading 1 Denote?
Given an $N$-bit binary string, a leading $1$ in two’s complement represents $-1$ multiplied by its corresponding power of two ($2^{n-1}$). A digit of $1$ in any other slot represents $+1$ times its corresponding power of two.
For example, the signed number $11010_2$ has this expansion:
$11010_2 = -1(2^4) + 1(2^3) + 1(2^1) = -16_{10} + 8_{10} + 2_{10} = -6_{10}$
We simply treat the leading $1$ as a negative, and that changes the resulting sum in our expansion.
#### Two’s Complement Shortcut: Flip the Bits and Add 1
To convert a number represented in two’s complement binary to its opposite sign, follow these two simple steps:
1. Flip all of the bits ($0$ becomes $1$ and vice versa).
2. Add $1$ to the result.
For example, let’s convert $43_{10}$ to $-43_{10}$ in binary:
+43 in binary: 0101011
Flipped: 1010100
What is this number? It should be $-43_{10}$, so let’s expand it by hand to verify:
$-1(2^6) + 1(2^4) + 1(2^2) + 1(2^0) = -64_{10} + 16_{10} + 4_{10} + 1_{10} = -43$
Sure enough, the process works!
#### How Many Signed Binary Numbers Are There?
We’ve seen that in a signed binary system, the most significant bit is reserved for the sign. What does this do to our number range? Effectively, it halves it!
Let’s consider $32$-bit integers again. Whereas before we had $32$ bits to work with for the magnitude of an unsigned number, we now have only $31$ for the magnitude of a signed number (because the 32nd bit is reserved for the sign):
Unsigned magnitude bits: [31 30 29 ... 0]
Signed magnitude bits: 31 [30 29 ... 0]
We went from having $2^{32}$ numbers to $2^{31}$ positive and negative numbers, which is precisely half of what we started with ($\frac{2^{32}}{2} = 2^{31}$).
More generally, if you have an $N$-bit signed binary string, there are going to be $2^N$ values, split evenly between $2^{n-1}$ positives and $2^{n-1}$ negatives.
Notice that the number zero gets bunched in with the positives and not the negatives:
Signed zero: 0 0 0 0 ... 0 0 0 0
Bits: 31 30 29 28 ... 3 2 1 0
As we’re about to see, this has an interesting consequence.
#### What Is the Largest Signed 32-bit Integer?
The largest signed 32-bit integer is positive, meaning its leading bit is a zero. So we just need to maximize the remaining bits to get the largest possible value:
Num: 0 1 1 1 ... 1
Bits: 31 30 29 28 ... 0
This is $2^{31} - 1$, which is $2,147,483,647$. In Java, this number is stored in Integer.MAX_VALUE, and in C++, it’s std::numeric_limits<int>::max().
More generally, for an $N$-bit system, the largest signed integer is $2^{n-1}-1$.
Why did we subtract a one at the end? Because we start counting at one, but computers start at zero. As I mentioned in the previous section, the number zero gets grouped along with the positives when we split our number range (by convention):
Signed zero: 0 0 0 0 ... 0 0 0 0
Bits: 31 30 29 28 ... 3 2 1 0
So to get the largest signed integer, we need to subtract one.
##### Real-World Application: Video Game Currency
In video games like RuneScape that use $32$-bit signed integers to represent in-game currency, the max “cash stack” that you can have caps out at exactly $2^{31} - 1$, which is roughly 2.1 billion.
Now you know why! If you’re wondering why they don’t just use unsigned ints, it’s because RuneScape runs on Java, and Java doesn’t support unsigned ints (except in SE 8+).
#### What Is the Smallest Signed 32-bit Integer?
This occurs when we set the leading bit to be a $1$ and set all remaining bits to be a $0$:
Num: 1 0 0 0 ... 0
Bits: 31 30 29 28 ... 0
Why? Because recall that in the expansion of negative numbers in two’s complement binary, the leading $1$ is a $-1$ times $2^{n-1}$, and a $1$ in any other position will be treated as $+1$ times its corresponding power of two. Since we want the smallest negative number, we don’t want any positive terms, as those take away from our magnitude. So we set all remaining bits to be $0$.
Answer: $-2^{31}$
In Java, this value is stored in Integer.MIN_VALUE. In C++, it’s in std::numeric_limits<int>::min().
More generally, if we have an $N$-bit system, the smallest representable signed int is $-2^{n-1}$.
Notice that the magnitude of the smallest signed $32$-bit integer is exactly one greater than the magnitude of the largest signed $32$-bit integer. As mentioned previously, this is because of where we chose to group the number zero itself, which “steals” one magnitude from that group’s available bits.
## Binary Arithmetic
Spoiler: Adding, subtracting, multiplying, and dividing numbers in the binary number system is exactly the same as it is in decimal!
We’ll first revisit what we learned in elementary school for decimal numbers and then look at how to add two binary numbers.
To add two numbers in the decimal number system, you stack them on top of one another visually and work your way from right to left, adding two digits and “carrying the one” as needed.
Now you should know what carrying the one really means: When you run out of digits to represent something in your fixed-base number system (e.g., $13$ isn’t a digit in base $10$), you represent the part that you can in the current digits place and move over to the next power of your base (the “column” to the left of your current one).
For example, let’s add $24$ and $18$ in decimal:
24
+ 18
————
42
We first add the $4$ and $8$ to get $12$, which is not a digit we support in the decimal number system. So we represent the part that we can ($2$) and carry the remaining value (ten) over to the next column as a $1$ ($1 × 10^1 = 10_{10}$). In that column, we have $1_{10} + 2_{10} + 1_{10} = 4_{10}$:
1 <-- carried
24
+ 18
————————
42
Now, let’s add these same two numbers ($24_{10}$ and $18_{10}$) using the binary number system:
11000
+ 10010
———————
101010
We work from right to left:
• Ones place: $0 + 0 = 0$
• Twos place: $0 + 1 = 1$
• Fours place: $0 + 0 = 0$
• Eighths place: $1 + 0 = 1$
• Sixteens place: $1 + 1 = 10_2$ (two)
That last step deserves some clarification: When we try to add the two ones, we get $1_2 + 1_2 = 10_2$ (two), so we put a $0$ in the current column and carry over the $1$ to the next power of two, where we have a bunch of implicit leading zeros:
1 <-- carry bits
0000 ... 00011000
0000 ... + 00010010
—————————————————————
0000 ... 00101010
In that column, $1 (carried) + 0(implicit) = 1$.
If we expand the result, we’ll find that it’s the same answer we got over in decimal:
$1(2^5) + 1(2^3) + 1(2^1) = 32 + 8 + 2 = 42_{10}$
Let’s look at one more example to get comfortable with carrying bits in binary addition: $22_{10} + 14_{10}$, which we know to be $36_{10}$:
10110
+ 01110
———————
100100
Something interesting happens when we look at the twos place (the $2^1$ column): We add $1_2$ to $1_2$, giving us two ($10_2$), so we put a zero in the $2^1$ column and carry the remaining one.
Now we have three ones in the $2^2$ column: $1_2(carried) + 1_2(operand1) + 1_2(operand2) = 11_2$ (three). So we put a one in the $2^2$ column and carry a one yet again. Rinse and repeat!
1111 <-- carry bits
0000 ... 00010110
0000 ... + 00001110
—————————————————————
0000 ... 00100100
Once again, it’s a good practice to expand the result so you can verify your work:
$1(2^5) + 1(2^2) = 32_{10} + 4_{10} = 36_{10}$
### Subtracting Binary Numbers
Subtraction is addition with a negative operand: $a - b = a + (-b)$. Now that we know how to represent negative numbers in the binary system thanks to two’s complement, this should be a piece of cake: negate the second operand and perform addition.
For example, what’s $12_{10} - 26_{10}$? In decimal, we know this to be $-14_{10}$. Over in binary, we know that $12_{10}$ is $01100$. What about $-26_{10}$? We’ll represent that using two’s complement.
We start by first representing $26_{10}$ in binary:
$+26_{10} = 011010_2$
Now we negate it by flipping the bits and adding one:
26 in binary: 011010
Flipped: 100101
Then, stack up the operands and add them like before:
11 <-- carry bits
001100
+ 100110
——————————
110010
Notice that the result has a leading one, which we know denotes a negative number in signed binary. So we at least got the sign part right! Let’s check the magnitude:
$-1(2^5) + 1(2^4) + 1(2^1) = -32_{10} + 16_{10} + 2_{10} = -14_{10}$
Adding and subtracting numbers in the binary number system is no different than in the decimal system! We’re just working with bits instead of digits.
### Multiplying Binary Numbers
Let’s remind ourselves how we multiply numbers in decimal:
21
x 12
————
Remember the process? We multiply the $2$ by each digit in the first multiplicand and write out the result under the bar:
21
x 12
————
42
Then we move on to the $1$ in $12$ and repeat the process, but adding a $0$ in the right column of the result. Add the two intermediate products to get the answer:
21
x 12
—————
42
+ 210
—————
252
Guess what? The process is exactly the same in the binary number system!
Let’s multiply these same two numbers in binary. They are $21_{10} = 010101$ and $12_{10} = 01100$:
010101
x 01100
—————————
Obviously, this is going to be more involved in binary since we’re working with bits (and thus longer strings), but the logic is still the same. In fact, beyond having to write out so many intermediate results, we actually have it much easier over in binary. Whenever a digit is $1$, you simply copy down the first multiplicand, padded with zeros. Whenever it’s a zero times the first multiplicand, the result is zero!
010101
x 01100
————————————
000000
0000000
01010100
010101000
+ 0000000000
————————————
0011111100
Expanding this in binary, we get:
$0011111100_2 = 1(2^7) + 1(2^6) + 1(2^5) + 1(2^4) + 1(2^3) + 1(2^2) = 252_{10}$
Easy peasy. The same process applies regardless of whether your multiplicands are signed or unsigned.
### Dividing Binary Numbers
Let’s divide $126_{10}$ by $12_{10}$ using long division:
0 1 0 . 5
_______
12 |1 2 6
- 1 2
————
0 6
- 0
——————
6 0
- 6 0
—————
0
Answer: $10.5$.
Now let’s repeat the process over in the binary number system. Note that I’m going to strip leading zeros to make my life easier since we’re working with two unsigned numbers:
_______
1100 |1111110
Take things one digit at a time, and reference this useful YouTube video if you get stuck:
0 0 0 1 0 1 0 . 1
______________
1 1 0 0 |1 1 1 1 1 1 0 . 0
-0
——
1 1
- 0
————
1 1 1
- 0
——————
1 1 1 1
- 1 1 0 0
————————
1 1 1
- 0
——————————
1 1 1 1
- 1 1 0 0
—————————
0 0 1 1 0
- 0
—————————
1 1 0
- 0
—————
1 1 0 0
- 1 1 0 0
———————
0 0 0 0
Answer: $01010.1$.
What does the $1$ to the right of the decimal point represent? Well, in the decimal number system, anything to the right of the decimal point represents a negative power of ten: $10^{-1}$, $10^{-2}$, and so on.
As you may have guessed, in the binary number system, these are $2^{-1}$, $2^{-2}$, and so on. So $.1$ above really means $1(2^{-1})$, which is $\frac{1}{2} = 0.5_{10}$ in decimal. And of course, the part in front of the decimal point evaluates to $10_{10}$.
That gives us $10_{10} + 0.5_{10} = 10.5$. So our answer using binary long division is exactly the same as the one we got over in decimal!
### Integer Overflow and Underflow in Binary
What happens if you try to add one to the largest representable $N$-bit signed integer?
For example, if $N = 32$, we’re really asking what happens if we try adding one to the largest representable $32$-bit signed int.
Let’s give it a shot:
0111...11111
+ 0000...00001
————————————————
In the rightmost column, we’ll get $1_2 + 1_2 = 10_2$, so that’s a zero carry a one. But as a result, all of the remaining additions will be $1_2 + 1_2$ since we’ll always carry a one until we get to the leading bit:
11111111111 <-- carry bits
0111...11111 (2^{N-1} - 1)
+ 0000...00001 (1)
————————————————
1000...00000 (-2^{N-1})
And what number is that in signed binary? Hmm… Looks like it’s the smallest representable negative number! What we’ve observed here is called integer overflow. When you try to go past the largest representable signed integer in a given $N$-bit system, the result overflows or wraps around.
What if we try to subtract one from the smallest representable $N$-bit signed integer? First, we’ll represent $-1_{10}$ as a signed integer in binary:
1 in binary: 0000...00001
Flipped: 1111...11110
Now let’s add this to the smallest representable signed integer:
1 <-- carry bits
1000...00000 (-2^{N-1})
+ 1111...11111 (-1)
————————————————
1|0111...11111 (2^{N-1} - 1)
Notice that the result carries an additional bit over, yielding a result that has $N+1$ bits. But our system only supports $N$ bits, so that leading $1$ is actually discarded. The result is the largest representable $N$-bit signed integer, and this is known as integer underflow.
Overflow and underflow are things you should be mindful of in programs that are performing lots of computations, as you may end up getting unexpected results.
## The Binary Number System: Additional Topics for Exploration
That about does it for this introduction to the binary number system! We took a pretty in-depth look at decimal, binary, and hexadecimal, and I hope you now have a greater appreciation for the binary number system and the role that it plays in computing.
In reality, there’s much more to learn beyond what we covered here. If you’re curious, I encourage you to look into representing floating point numbers in binary using the IEE754 format.
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Question: The Giant Wheel at Cedar Point Amusement Park is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height 136 feet. It completes two revolutions in 2 minutes and 7 seconds. Assuming the riders are at the edge of the circle, how fast are they traveling in miles per hour?
This problem is all sort's of FUN! I was working on this problem trying to figure it out. Everything that I tried I kept always getting the wrong answer. Here is the big problem that I was facing.
I was trying to convert two revolutions into 1 revolution and cut the time in half. My thought here was since a circle is $2\pi$ is a complete revolution so should 1 revolution and 1 minute and 3.5 seconds. This is where my logic was WRONG. I cleaned this up by representing two revolutions in seconds instead of trying to convert them to a complete revolution around the circle, in other words, the circumference of the circle.
$$speed=\frac { 2 }{ 127 } (\frac { rev }{ sec } )$$
After doing this the math works out perfectly!
If you have a hard time reading the solution, it's probably a good idea to review how to find the solution
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# Re: st: How to store marginal effect value after using margin command?
From Steve Samuels <[email protected]> To [email protected] Subject Re: st: How to store marginal effect value after using margin command? Date Mon, 10 Jun 2013 10:14:10 -0400
``` Vũ Võ :
If you've now studied the -help- for -margins-, you'll know that as with
-mfx-, the marginal effects are contained in the returned matrix r(b)
and, with the post option, in e(b), as Kit stated. You can extract
values from these matrices, e.g.:
. scalar m1 = el(r(b),1,1)
But you really cannot do much with scalars; they are just numbers
unconnected to the command that generated them. The real advantage of
the post option is that it makes the estimated effects available as
Stata system "underscore variables" (type "help system variables"). To
find the names of those variables, you run a second -margins- command
with the "coeflegend" option, immediately after the first.
. sysuse auto, clear
. gen wtk = weight/1000
. reg price rep78##c.wtk
. margins, dydx(wtk rep78) post
. margins, coeflegend
. nlcom _b[wtk] + _b[2.rep78]/_b[3.rep78] //silly manipulation
. nlcom _b[w] + _b[2.r]/_b[3.r] // abbreviate names
Your original question was whether -margins- could show the partial
derivatives that you had derived "by hand". The answer is "Yes". These
partial derivatives are contained in r(Jacobian) and e(Jacobian):
. reg price c.wtk##c.wtk
. margins, dydx(wtk) at(wtk =(1(1)4))
. matrix list r(Jacobian)
For a constructive use of Jacobians, see Jeff Pitblado's post at:
http://www.stata.com/statalist/archive/2010-04/msg00860.html
Steve
> On Jun 8, 2013, at 8:20 PM, William Buchanan wrote:
>
> If you're using the - margins - command, the - post - option is clearly documented in the help file. Additionally, the Statalist FAQ suggests looking through the documentation as a good first step prior to querying the list. If this isn't the answer that you were looking for it may be a good idea to be more explicit about your end goal.
>
> HTH,
> Billy
>
Sent from my iPhone
On Jun 8, 2013, at 16:54, Vũ Võ <[email protected]> wrote:
> Dear Sergiy,
>
> Thank you for your message. Probably, you misunderstand my question
> nor I am clear in raising a question.
>
> For example:
>
> sysuse auto, clear
>
> reg price price wgt mpg c.wgt#c.wgt c.mpg#c.mpg c.wgt#c.mpg
>
>
> Source | SS df MS Number of obs = 74
> -------------+------------------------------ F( 5, 68) = 12.84
> Model | 308384833 5 61676966.6 Prob > F = 0.0000
> Residual | 326680563 68 4804125.93 R-squared = 0.4856
> -------------+------------------------------ Adj R-squared = 0.4478
> Total | 635065396 73 8699525.97 Root MSE = 2191.8
>
> ------------------------------------------------------------------------------
> price | Coef. Std. Err. t P>|t| [95% Conf. Interval]
> -------------+----------------------------------------------------------------
> weight | -31.88985 9.148215 -3.49 0.001 -50.14483 -13.63487
> mpg | -3549.495 1126.464 -3.15 0.002 -5797.318 -1301.672
> |
> c.weight#|
> c.weight | .0034574 .0008629 4.01 0.000 .0017355 .0051792
> |
> c.mpg#c.mpg | 38.74472 12.62339 3.07 0.003 13.55514 63.93431
> |
> c.weight#|
> c.mpg | .5421927 .1971854 2.75 0.008 .1487154 .9356701
> |
> _cons | 92690.55 25520.53 3.63 0.001 41765.12 143616
> ------------------------------------------------------------------------------
>
> So, I would like to calculate the marginal effects of weight and mpg
> on price by using command:
> margins, dydx(weight mpg)
>
> and get ->
>
> Average marginal effects Number of obs = 74
> Model VCE : OLS
>
> Expression : Linear prediction, predict()
> dy/dx w.r.t. : weight mpg
>
> ------------------------------------------------------------------------------
> | Delta-method
> | dy/dx Std. Err. z P>|z| [95% Conf. Interval]
> -------------+----------------------------------------------------------------
> weight | .5360966 .6516553 0.82 0.411 -.7411243 1.813318
> mpg | -262.0502 103.681 -2.53 0.011 -465.2612 -58.83919
> ------------------------------------------------------------------------------
>
>
> My question is how to obtain or store in to scalar or whatever in
> stata values of 0.5360966 and -262.0502 in the above table?
>
> If i use mfx , after that using matrix list e(b) as you suggested, it
> will give me coefficients of the regression. To some extent, it can
> say that these are marginal effets, but not what I would like to
> obtain.
>
> Any suggestion are highly appreciated. Thanks
>
> Vu
>
> On 9 June 2013 00:07, Sergiy Radyakin <[email protected]> wrote:
>> type either 'return list' or 'ereturn list' (depending on the command)
>> to see which results are saved after these commands. From what you
>> describe you are interested in matrices r(b) and e(b)
>>
>> sysuse auto, clear
>> generate lp=log(price)
>> regress lp weight length headroom mpg
>> mfx
>> matrix list e(b)
>>
>> Best, Sergiy
>>
>> On Sat, Jun 8, 2013 at 4:50 PM, Vũ Võ <[email protected]> wrote:
>>> Hi everyone,
>>>
>>> I am writing a code in do file to calculate the marginal effect of x
>>> on y (log-linear model)
>>>
>>> The function form as following:
>>>
>>> Ln(Y) = a0 + a1*X + a2*X^2 + e ----> Y = exp(a0 + a1*X + a2*X^2)
>>>
>>> So, dY/dX = (a1 + 2*a2*X)*exp(a0 + a1*X + a2*X^2) eq(1)
>>>
>>> If I run the regression by command line as:
>>>
>>> reg lnY x c.x#c.x
>>>
>>> after that using margins command:
>>>
>>> margins, dydx(x).
>>>
>>> I will obtain the marginal effect of x on Ln(Y): (a1 + 2*a2*X).
>>>
>>> How can I calculate the dY/dX as in eq(1) by using the do file, not
>>> calculating manually?
>>>
>>>
>>> 2. Second question is:
>>>
>>> If I use command margins, predict(p) and obtain the results as following:
>>>
>>> Marginal effects after hetprob
>>> y = Pr(y) (predict, p)
>>> = .54284206
>>> ------------------------------------------------------------------------------
>>> variable | dy/dx Std. Err. z p>|z| [ 95% C.I. ] X
>>> ---------+--------------------------------------------------------------------
>>> x | .3704576 .03237 11.44 0.000 .307015 .4339 .020114
>>> xhet | -.0736092 .02423 -3.04 0.002 -.121095 -.026124 .502716
>>> off1 | (offset1) 1.00516
>>> off2 | (offset2) 1.09709
>>> ------------------------------------------------------------------------------
>>>
>>> The question is how to store the marginal effects value, say in this
>>> case is 0.3704576,
>>>
>>> in the scalar in stata?
>>>
>>>
>>> Thank you so much.
>>>
>>> Vo Duc Hoang Vu
>>> *
>>> * For searches and help try:
>>> * http://www.stata.com/help.cgi?search
>>> * http://www.stata.com/support/faqs/resources/statalist-faq/
>>> * http://www.ats.ucla.edu/stat/stata/
>>
>> *
>> * For searches and help try:
>> * http://www.stata.com/help.cgi?search
>> * http://www.stata.com/support/faqs/resources/statalist-faq/
>> * http://www.ats.ucla.edu/stat/stata/
>
>
>
> --
>
> Vo Duc Hoang Vu
> Faculty of Development Economics
> University of Economics HCMC
> 1 A Hoang Dieu Street
> Phu Nhuan Dist. Ho Chi Minh
> Vietnam
> Tel: +84 94 550 22 77 (VN); +31 624 63 29 45 (NL)
> Fax: +84 8 38477948 (VN)
> E-mail:[email protected]
> http://www.voduchoangvu.net
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/faqs/resources/statalist-faq/
> * http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/faqs/resources/statalist-faq/
* http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/faqs/resources/statalist-faq/
* http://www.ats.ucla.edu/stat/stata/
```
• References:
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# Two semiconductor physics problems with similar data but different solution methods
Prob 1: Find the conductivity of n-type Ge at room temperature. Assuming one donor atom in each 108 atoms. The density of Ge is 5.32 x 103 kg/m3 and the atomic weight is 72.66 kg/k-mol. Given e=1.6 x 10-19 C, μe=0.38 m2/V-s and μh=0.18 m2/V-s.
Hint: No of Ge atoms per cm3 = 6.023 x 1023 x (atoms/mole) x (1 mole/72.6 g) x (5.32g/cm3) = 4.41 x 1022.
Prob 2: A specimen of germanium at 300 K for which the density of carriers is 2.5 x 1013, is doped with impurity atoms such that there is one impurity atom for 106 germanium atoms. All the impurity atoms may be assumed ionized. The resistivity of doped material is 0.039 Ω-cm. Carreir mobility for germanium at 300 K is 3600 cm2/V-s. For the doped material find the electron and hole concentration.
## .....................................................................................
In order to solve these two problems, we need to get the donor atom concentration. Please tell me how should I find the value of donor concentration in each problem.
• Is this a homework question? – Daniel Dec 21 '17 at 6:05
• No, this is a solved problem from a text book where 2 different methods to find donor concentration is used when the problem data is similar. Which is what I am confused about and asking the experts here to put some light on it...Not posting the solution here on purpose to not affect your judgement. – John deo Dec 21 '17 at 6:09
• Do you want me to post the solution for each in the comment here? – John deo Dec 21 '17 at 6:15
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http://www.jacobsilterra.com/2013/04/08/theoretical-maximum-of-the-bechdel-test/?replytocom=3876
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# Theoretical maximum of the Bechdel Test
Background:
For those that may not know, the Bechdel test has 3 criteria. A movie/TV show/book/whatever must have at least 1 scene where there exist:
0. 2 female characters
1. talking to each other
2. about something other than a man
A rather depressing amount of movies/TV shows fail this relatively simple test. Most characters seem to be male, and the female ones exist just so that men will have love interests. Since I’m a numbers guy, I wondered what fraction of media would pass the test if there were actually no sexism.
Parameters:
N = The number of characters
C = The number of distinct conversational pairings in a given item of media. A “conversation pairing” means any 2 characters talking to each other, and distinct means at least 1 of the characters is different than any other pairing.
p_g = Fraction of characters of an acceptable gender (ie female)[1]
p_t = Probability that a conversation is about an acceptable topic (ie something other than a man)
For each conversation that happens, the probability it will cause the item to pass a Bechdel-type test is:
Pr(1 convo passes) = (Probability the convo is between two acceptable people) x (Probability the topic is acceptable) ~= p_g^2 x p_t.
Since we aren’t counting conversations one has with oneself, technically that’s a slight overestimate. In general there will be N choose 2 = N * (N-1) / 2 possible conversations. However, only (p_g x N) choose 2 = (p_g x N)(p_g x N-1)/2 are within a given gender. For N = 10 and p_g = 0.5, this is 20/90 = 2/9, a touch less than p_g^2 = 1/4th. So I’m gonna make this approximation, it’s the least of our troubles.
For an item to pass it only needs 1 passing conversation, that probability is:
Pr(whole item passes) = 1 – (1 – Pr(1 convo passes) )^C
It’s rather difficult to estimate how many distinct conversations might exist. In a 40 minute TV show, figuring a scene lasts 2 minutes, there would be time for 20. Given episodes tend to match characters up together, so one would expect multiple scenes with the same people, reducing that number. There also need to be enough characters for this number of distinct conversations to be possible, but that’s usually not an issue. Even a 6-character show would have 15 possible pairings, and that’s assuming no guest stars.
Guessing at p_t is even more difficult. The simpest assumption would be that all conversations are about another character. However, it’s extremely likely that a conversation between two characters of one gender will be about a character of a different gender, since it’s likely to be romantic. That isn’t sexist, just heteronormative. One might also assume that two characters are talking about some other character not on screen, which would be more likely to be the opposite gender (given that we’ve already taken up two characters to have the convo).
So since I can’t come up with a solid number, let’s explore the parameter space a bit. Here are some plausible (re: made up) numbers for each parameter, and the corresponding odds of passing the test
Graph:
Chart:
p_g C p_t Pr(passes) 0.5 22 0.3 1 0.2 5 0.05 0.1 0.2 5 0.1 0.18 0.2 5 0.15 0.27 0.2 5 0.2 0.34 0.2 5 0.25 0.41 0.2 5 0.3 0.47 0.2 5 0.35 0.53 0.2 5 0.4 0.58 0.2 5 0.45 0.63 0.2 5 0.5 0.67 0.5 5 0.05 0.23 0.5 5 0.1 0.41 0.5 5 0.15 0.56 0.5 5 0.2 0.67 0.5 5 0.25 0.76 0.5 5 0.3 0.83 0.5 5 0.35 0.88 0.5 5 0.4 0.92 0.5 5 0.45 0.95 0.5 5 0.5 0.97 0.5 10 0.05 0.4 0.5 10 0.05 0.4 0.5 10 0.1 0.65 0.5 10 0.15 0.8 0.5 10 0.2 0.89 0.5 10 0.25 0.94 0.5 10 0.3 0.97 0.5 10 0.35 0.99 0.5 10 0.4 0.99 0.5 10 0.45 1 0.5 10 0.5 1
The first row is my guess at a realistic value for C, and idealistic values for p_g and p_t. Since it’s a rounding-error from 1.0, I tweaked the numbers to see some variability. We see that in the sexism-free condition (p_g = 0.5), for even a modestly large number of distinct conversation pairings, the probability of passing pretty rapidly approaches 1.
So keep on the lookout. Because until damn near everything you watch passes the Bechdel test, sexism still exists.
-Jacob
by
1. [1]For simplicity of language I’m going to assume that all characters’ gender is equal to their biological sex, and that all characters are either male or female.
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Имя: Пароль: Зарегистрироваться Восстановить пароль
28/01/2023 15:02:48
# Рабочее место участника
## Задачи
694. The Teacher's Side of Math
Ограничения: время – 5s/10s, память – 64MiB Ввод: input.txt или стандартный ввод Вывод: output.txt или стандартный вывод
Послать решение Blockly Посылки Темы Где Обсудить (0)
One of the tasks students routinely carry out in their mathematics classes is to solve a polynomial equation. It is, given a polynomial, say X^2\ -\ 4X\ +\ 1, to find its roots (2\ ±\ sqrt(3)).
If the students’ task is to find the roots of a given polynomial, the teacher’s task is then to find a polynomial that has a given root. Ms. Galsone is an enthusiastic mathematics teacher who is bored with finding solutions of quadratic equations that are as simple as a\ +\ b\ sqrt(c). She wanted to make higher-degree equations whose solutions are a little more complicated. As usual in problems in mathematics classes, she wants to maintain all coefficients to be integers and keep the degree of the polynomial as small as possible (provided it has the specified root). Please help her by writing a program that carries out the task of the teacher’s side.
You are given a number t of the form:
t\ =\ root{m}(a)\ +\ root{n}(b)
where a and b are distinct prime numbers, and m and n are integers greater than 1.
In this problem, you are asked to find t’s minimal polynomial on integers, which is the polynomial F(X)\ =\ a_d\ X^d\ +\ a_{d-1}X^{d-1}\ +\ …\ +\ a_1\ X\ +\ a_0 satisfying the following conditions.
• Coefficients a_0,\ …,\ a_d are integers and a_d\ >\ 0.
• F(t)\ =\ 0.
• The degree d is minimum among polynomials satisfying the above two conditions.
• F(X) is primitive. That is, coefficients a_0,\ …,\ a_d have no common divisors greater than one.
For example, the minimal polynomial of sqrt(3)+sqrt(2) on integers is F(X)\ =\ X^4-10X^2+1. Verifying F(t)\ =\ 0 is as simple as the following (α\ =\ sqrt(3), β\ =\ sqrt(2)).
F(t)\ =\ (α\ +\ β)^4\ -\ 10(α\ +\ β)^2\ +\ 1
=\ (α^4\ +\ 4α^3β\ +\ 6α^2β^2\ +\ 4αβ^3\ +\ β^4)\ -\ 10(α^2\ +\ 2αβ\ +\ β^2)\ +\ 1
=\ 9\ +\ 12αβ\ +\ 36\ +\ 8αβ\ +\ 4\ -\ 10(3\ +\ 2αβ\ +\ 2)\ +\ 1
=\ (9\ +\ 36\ +\ 4\ -\ 50\ +\ 1)\ +\ (12\ +\ 8\ -\ 20)αβ
=\ 0
Verifying that the degree of F(t) is in fact minimum is a bit more difficult. Fortunately, under the condition given in this problem, which is that a and b are distinct prime numbers and m and n greater than one, the degree of the minimal polynomial is always mn. Moreover, it is always monic. That is, the coefficient of its highest-order term (a_d) is one.
Input
The input consists of multiple datasets, each in the following format.
a m b n
This line represents root{m}(a)\ +\ root{n}(b). The last dataset is followed by a single line consisting of four zeros. Numbers in a single line are separated by a single space. Every dataset satisfies the following conditions.
• root{m}(a)\ +\ root{n}(b) ≤ 4.
• "mn"\ ≤\ 20.
• The coefficients of the answer a_0,\ …,\ a_d are between (-2^31\ +\ 1) and (2^{-31}\ -\ 1), inclusive.
Output
For each dataset, output the coefficients of its minimal polynomial on integers F(X)\ =\ a_d\ X^d\ +\ a_{d-1}X^{d-1}\ +\ …\ +\ a_1\ X\ +\ a_0, in the following format.
a_d a_{d-1}a_1 a_0
Non-negative integers must be printed without a sign (+ or -). Numbers in a single line must be separated by a single space and no other characters or extra spaces may appear in the output.
Sample Input
3 2 2 2
3 2 2 3
2 2 3 4
31 4 2 3
3 2 2 7
0 0 0 0
Output for the Sample Input
1 0 -10 0 1
1 0 -9 -4 27 -36 -23
1 0 -8 0 18 0 -104 0 1
1 0 0 -8 -93 0 24 -2976 2883 -32 -3720 -23064 -29775
1 0 -21 0 189 0 -945 -4 2835 -252 -5103 -1260 5103 -756 -2183
Source: ACM ICPC Asia RC, Tokyo, 2007
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# Understanding numbers in your bills and payslips
Managing money can mean looking at a lot of numbers, sometimes too many. It can all become confusing and tempting not to bother at all.
If this sounds like you, here are some ideas that might help you.
## Council tax bill
### Step 1
One bill that it's important to understand is for council tax.
It looks something like this.
There area lot of numbers on this bill. It is easy to feel a bit lost when you see it.
### Step 2
When you get your council tax bill, you're likely to be trying to find out how much you need to pay and when.
Lots of this information doesn't help you do that. Everything in the blue box can be ignored.
It's not that the information isn't useful, you just don't need it right now.
### Step 3
Once you have got just the numbers you need - it can look much clearer.
The total is in the blue box, and below that it shows how much you have to pay and when.
Finding the numbers that matter most, and ignoring the rest, can help you feel less anxious.
## Energy bill
### Step 1
Let's look at another example. Your energy bill will usually look something like this.
It's full of numbers!
The numbers in the blue box are good to have in case you want to check your bill.
But if you just want to know how much you need to pay, you can ignore them for now.
### Step 2
If you just want to know how much you need to pay so you can plan your budget, you only need one number from the bill.
The others can cause confusion and make you feel anxious, so ignore them for now.
## Payslip
### Step 1
Payslips always show a lot of numbers.
Depending on what you're trying to find out, only some of the numbers will help.
Often you just want to check how much you're being paid - so we'll use this as an example.
### Step 2
To check how much you're getting paid for the month, you can ignore most of the numbers.
Once you know that - it's easier not to be put off by the information overload.
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https://socratic.org/questions/does-the-series-sum-n-1-oo-5n-3n-5-n-3-n-converge-or-diverge
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# Does the series sum_{n=1}^oo (5n)^(3n)/(5^n+3)^n converge or diverge?
Mar 18, 2018
See below.
#### Explanation:
Considering
${a}_{n} = {\left(5 n\right)}^{3 n} / {\left({5}^{n} + 3\right)}^{n}$
we have
${a}_{n} < {\left(5 n\right)}^{3 n} / {\left({5}^{n}\right)}^{n}$
now we will compare ${a}_{n}$ with $\frac{1}{n} ^ 2$ that we know to converge.
Taking $\ln$
$\ln {a}_{n} < \left(3 n\right) \ln \left(5 + \ln n\right) - {n}^{2} \ln 5$ and as we can easily verify exists a ${n}_{0}$ such that $n > {n}_{0} \Rightarrow \ln {a}_{n} < \ln \left(\frac{1}{n} ^ 2\right)$
As we know, $\ln$ is a strictly increasing function so considering $n > {n}_{0}$
$\ln {a}_{n} < \ln \left(\frac{1}{n} ^ 2\right) \Rightarrow {a}_{n} < \frac{1}{n} ^ 2$
hence the series
${\sum}_{n = 1}^{\infty} {\left(5 n\right)}^{3 n} / {\left({5}^{n} + 3\right)}^{n}$
converges.
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https://www.cs.montana.edu/paxton/classes/spring-2016/csci338/lectures/ticket/Ticket4.html
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# Golden Ticket: Chapter 4
## Optional Get-Together
• When: Tonight at 7:00 p.m.
• Where: 721 South 6th Avenue
• What: Watch CodeBreaker, a movie about Alan Turing.
• What: Play games such as CSCI 338 Charades. If you have a fun game that works in a group setting, bring it along!
• Munchies: Pizza and soda!
• There is an annual competition to identify new benchmarks and promote new approaches to the satisfiability problem.
## The Hardest Problems in NP
• In 1971, Steve Cook proved that the satisfiability problem is the hardest type of NP problem.
• Since then, many other NP problems have been shown to be equivalent to satisfiability.
• These problems are called NP-Complete (NPC) problems.
• To demonstrate that Problem X is NPC, (1) show that Problem X is in NP and (2) show a polynomial reduction from a known NPC problem to Problem X. (We will learn more about this in section 7.4 of the Sipser book.)
• Sudoku is NP-complete.
• Minesweeper is NP-complete.
• Tetris is NP-complete.
• The Traveling Salesperson Problem is NP-complete.
• The Bin Packing Problem is NP-complete.
• The status of some problems is unknown. For example, graph isomorphism and factoring are both believed to be harder than P but easier than NPC.
## Active Learning Problem
Model the following statements as a satisfiability problem. What is the solution?
• Qing can only meet on Monday, Wednesday or Thursday
• Upulee can't meet on Wednesday
• Hunter can't meet on either Tuesday or Friday
## Active Learning Problem
Model the following Frenemy statements as a satisfiability problem that enables friendship cliques to be identified.
• Mike and Binhai are friends
• Brittany and Brendan are friends
• Binhai and Brendan are friends
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https://socratic.org/questions/556ebe65581e2a43976fabcb#150324
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# Question fabcb
Jun 3, 2015
Consider a sample of 182 grams of such substance (approximately one mole). It contains:
$\text{182 grams · 0.40" / "12.01 g/mol" = 6.06" mol of C atoms}$;
$\text{182 grams · 0.067" / "1.008 g/mol" = 12.1" mol of H atoms}$;
$\text{182 grams · 0.533" / "16.00 g/mol" = 6.06" mol of C atoms}$;
Given that in a molecule there are integer number of atoms, that equal numbers of moles means equal numbers of atoms, we deduce that in each molecule there are:
a) the same number of carbon and oxygen atoms, i.e. 6
b) double number of H atoms, that is 12.
The molecular formula is therefore: ${C}_{6} {H}_{12} {O}_{6}$.
The true molecular mass should be:
C·6 + H·12 + O·6 =
12,0·6 + 1,0·12 + 16,0·6 = 180#
A value of 182 deduced from osmosis experiments is just one percent bigger, as due to experimental errors.
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https://stats.stackexchange.com/questions/319651/determining-probability-of-smallest-departure-delay
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# Determining Probability of Smallest Departure Delay
I am working with library(nycflights13)dataset in R and I'm trying to provide an answer to this specific question:
If I am leaving before noon, what are my top two airline options at each airport (JFK, LGA, EWR) that will have the least amount of delay time?
At first I created a summarised table that displayed each airport's top two airline departure times (smallest departure delay) on average.
flights %>%
filter(sched_dep_time < 1200, origin %in% c("JFK", "LGA", "EWR")) %>%
aggregate(dep_delay ~ carrier + origin, ., mean) %>%
group_by(origin) %>%
top_n(n=-2, wt=dep_delay)
But then I thought, means aren't really a good summary stat I should base my decision off of. So I started plotting my data
I can see there are quite a few outliers but my thought is to keep them in the dataset as they provide valuable information on how badly a departure can be delayed at times. After getting a glimpse of the entire dataset, I wanted to look closer at departure times that are negative (meaning departed early) or around zero.
These plots are valuable but don't really make it obvious which airlines and airport would be the best for me to take given all the information I have. My next thought is to estimate the CDF for each airlines departure delay and then compute $P(X \leq 0)$. At that point I can then select the airlines that have the highest probability of having a departure delay of at most zero.
My question is two parts:
1. Is this sound statistical reasoning? Is there a test I can perform that would be better? I Really would just like some guidance on thinking this problem through.
2. If this is sound statistical thinking, are there any resources you can direct me towards that would teach me how to implement it in R?
• Time intervals can be naturally modeled using the Poisson distribution where the smallest mean would hint the smallest delay airport. On top of that, you want to do ranking since you wish to select top 2 (which Poisson distributions have the smallest means). My 2 cents. – Vladislavs Dovgalecs Dec 20 '17 at 19:09
• @VladislavsDovgalecs Thanks for the comment. Do you have any suggestions on resources I could look to do this and/or do it in R? – dylanjm Dec 20 '17 at 19:26
• I see a hierarchical Bayes model where each airline delay time is modeled using a Poisson distribution. The respective airport is modeled by a top level parameter \mu (Gaussian? Gamma?) that controls shrinkages for each airline and enables better parameter estimates. You can go a level higher for multiple airports. You can model that in Stan (see the 8 schools example). The Stan community on discourse is very helpful. John Kruschke's "puppies" book "Doing Bayesian Data Analysis" goes into great lengths explaining this type of models and offers many examples (R and JAGS). – Vladislavs Dovgalecs Dec 20 '17 at 19:32
## 1 Answer
Data visualization is always a great way to start. Here is how I would approach it.
Since you are only interested in before noon - limit all data from 6am to noon.
Run three different ANOVAs (delay time as dependent variable, aircraft carrier as independent variable) for each airport. Or you could run one model including airport as another main effect along with the interactions.
Examine residuals for severe violations of model assumptions.
Investigate the output to see which airlines are significantly shorter in flight delay. Will need to do some multiplicity adjustments.
Googling ANOVA and R will give you plenty of material.
There are other questions to consider such as do I want to almost guarantee a delay time of less than 30 minutes compared to no delay time at all? This, I believe, requires more advanced models. To answer this I would use a Bayesian approach in JAGS and compute theses posterior probabilities. However, for your sake, I think you can just focus on least square means from the ANOVA models.
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https://www.puzzleprime.com/brain-teasers/mathematics/worm-in-an-apple/
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Worm in an Apple
There is a perfectly spherical apple with radius 50mm. A worm has entered the apple, made a tunnel of length 99mm through it and left. Prove that we can slice the apple in two pieces through the center, so that one of them is untouched by the worm.
Let the entering point is A, the leaving point is B and the center of the apple is C. Consider the plane P containing the points A, B and C and project the worm’s tunnel on it. Since 99 < 2×50, the convex hull of the tunnel’s projection will not contain the center C. Therefore we can find a line L through C, such that the tunnel’s projection is entirely in one of the semi-planes of P with respect to L. Now cut the apple with a slice orthogonal to P passing through the line L and you are done.
Unknown Author
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http://www.notesmela.com/work/
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# Work
by • 04/07/2012 • GeneralComments (0)595
## Work
Work is said to be done when a force causes a displacement to a body on which it acts.
Work is a scalar quantity. It is the dot product of applied force F and displacement d.
Diagram Coming Soon
W = F . d
W = F d cos θ ………………………… (1)
Where θ is the angle between F and d.
Equation (1) can be written as
W = (F cos θ) d
i.e., work done is the product of the component of force (F cos θ) in the direction of displacement and the magnitude of displacement d.
equation (1) can also be written as
W = F (d cos θ)
i.e., work done is the product of magnitude of force F and the component of the displacement (d cos θ) in the direction of force.
Unit of Work
M.K.S system → Joule, BTU, eV
C.G.S system → Erg
F.P.S system → Foot Pound
1 BTU = 1055 joule
1 eV = 1.60 x 10(-19)
Of the knots online essay and ruggedness of poor human nature than there lay
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http://electronics-electrical-engineering.blogspot.com/2011/09/steps-for-design-binary-to-gray-code.html
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## Subscribe
Skin Design:
Free Blogger Skins
## Friday, September 23, 2011
### Steps for Design binary to gray code converter
Design binary to gray code converter
It is easy to convert the binary number to gray number. First, for example take a binary number i.e ,101101.
Step 1: 101101 first write the MSB as it is i.e, darken bit i.e, 1
Step 2: add MSB and bit next to the MSB i.e, 1+0=1
Step 3: again add 0 and 1 we get i.e, 0+1=1
Step 4: again add 1 and 1 we get 1+1=0
Step 5: in previous step carry is occurred so neglect that carry . note that don't add carry to next add numbers
step 6: add 0+1 0+1= 1
Answer: Gray Code is : 111011
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# Rearrange vectors so partial sums are all non-negative
Consider we are given a collection of $$n$$ vectors in $$d$$ dimensions, we want to decide if they can be rearranged into $$v_1,\ldots,v_n$$ such that $$\sum_{i=1}^j v_i\geq \textbf{0}$$ for all $$j\in [n]$$. Here $$a \geq \textbf{0}$$ if and only if all coordinates is at least $$0$$.
Have this problem been studied before?
I don't know if it has an official name, but it is NP complete; I give a reduction idea from Exact Cover from 3-Sets:
Given $$n = 3q$$, $$X = \{x_1,...,x_{n}\}$$ and $$C_1,C_2,...,C_m$$ a collection of 3 elements subsets.
Use the following set of $$3q + 2$$ dimensions vectors:
$$v_{fill} = [1_1, 1_2, ... ,1_{3q}, \; q_z ,\; 0_b]$$ (the subscripts are just labels to show the element "column" position)
For each $$C_i = \{ x_{i1}, x_{i2}, x_{i3} \}$$
$$v_{C_i} = [ 0,...,0, -1_{i1},0,...,0,-1_{i2},0,-1_{i3},0,...,0, \; -1_{z}, \; 1_{b} ]$$
$$v_{clean} = [ m, m, ..., m, m_{z}, -q_{b} ]$$
Starting from $$[0,0,...,0,\; 0_{z}, \; 0_{b}]$$ the only vector that doesn't contain negative elements is $$v_{fill}$$ so it must be the first of the sum.
Then in order to be able to add $$v_{clean}$$ that has a $$-q$$ in column $$b$$; we must add exactly $$q$$ vectors $$v_{C_i}$$ that contains $$[....,1_{b}]$$; we cannot add more than $$q$$ otherwisee the column $$z$$ becomes negative.
Furthermore we cannot add two $$C_i$$ that has the same element in the first $$3q$$ columns, otherwise it would become negative.
At this point $$v_{clean}$$ can be added; and, finally, the remaining $$v_{C_i}$$ can be added ("cleaned")
Example:
n = 3q = 6; q=2; m = 3
C_1={1,2,3}, C_2={ 4,5,6}, C_3={1,2,4}
v_fill = [ 1, 1, 1, 1, 1, 1, 2, 0 ]
v_C_1 = [ 1, 1, 1, 0, 0, 0,-1, 1 ]
v_C_2 = [ 0, 0, 0, 1, 1, 1,-1, 1 ]
v_C_3 = [ 1, 1, 0, 1, 0, 0,-1, 1 ]
v_clear= [ 3, 3, 3, 3, 3, 3, 3,-2 ]
Sum: v_fill + v_C_1 + v_C_2 + v_clear + v_C_3
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Home
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Evaluate: inte^(x)(x^(3)-x+2)/...
# Evaluate: inte^(x)(x^(3)-x+2)/(x^(2)+1)^(2)dx
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
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Solution :
e^(x)(x+1)/(x^(2)+1)+C
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# How do I get the output of this text in latex with sagetex?
Using Sage\TeX, one can use Sage to compute things and put them into your \LaTeX{} document. For example, there are $\sage{number_of_partitions(1269)}$ integer partitions of $1269$. You don't need to compute the number yourself, or even cut and paste it from somewhere.
Here's some Sage code:
\begin{sageblock} f(x) = exp(x) * sin(2*x) \end{sageblock}
The second derivative of $f$ is
[ \frac{\mathrm{d}^{2}}{\mathrm{d}x^{2}} \sage{f(x)} = \sage{diff(f, x, 2)(x)}. ]
Here's a plot of $f$ from $-1$ to $1$:
\sageplot{plot(f, -1, 1)}
\end{document}
edit retag close merge delete
( 2020-06-12 18:20:36 +0200 )edit
I don't know how to run a code in latex with sagetex. I have got some calculations in sage and I want latex to print them out.
( 2020-06-12 18:25:42 +0200 )edit
Sort by ยป oldest newest most voted
EDIT: a good starting point is this documentation.
Very roughly :
• create your text in LaTeX, interspeding Sage fragments as needed : this will be, say, yourtext.tex.
• pdflatex yourtext.tex : this will create yourtext.sagetex.sage, and give warninges about undefined references.
• sage yourtext.sagetex.sage. This will, among other effects, update yourtext.tex.
• pdflatex yourtext.tex (usually, you have to do it twice).
yourtext.pdf is now the sought document...
EDIT : The documenntation compiled in PDF form an an example tec=xt are available in $SAGE_ROOT/local/share/doc/sagetex/. To recompile it (and get the indices) : in your installation, you have: (sage-sh) charpent@zen-book-flip:~$ ls \$SAGE_ROOT/local/share/texmf/tex/latex/sagetex/
CONTRIBUTORS py-and-sty.dtx sagetex.dtx scripts.dtx
example.tex remote-sagetex.dtx sagetex.ins
extractsagecode.py remote-sagetex.py sagetexparse.py
makestatic.py run-sagetex-if-necessary.py sagetex.sty
Compiling sagetex.dtx with pdflatex (in a scratch directory) will give you instructions on how to compile both Sagetex and its documentation. More details here.
This cannot be done automatically in Sage's installation, because it requires a (not inconsequential) functional installation of LaTeX, which is not a Sage dependency.
The fie example.tex is also quite instructive... Fiddle with it a bit.
HTH,
more
Thanks for your help. I do not really understand the third point. Now latex is telling me the following:"Run Sage on sage.sagetex.sage, and then run LaTeX on sage.tex again.".
( 2020-06-12 19:42:32 +0200 )edit
I suppose you named your LaTeX file sage.tex, right ? So just do what you're instructed to do :
• Step 1 : pdflatex sage.tex gives you :
• asage.pdf file with [??] marks in place of your Sage results, which are supposed to exist in files yet nonexistant.
• a sage.sagetex.sage file (as well as other files), which contain your Sage code (wrapped in execution instructions).
• Step 2 : sage sage.sagetex.sage will run this file through Sage, generating a bunch of .tex files containing your results.
• Step 3 pdflatex sage.tex will reconpile your tex file, where the references can now be obtained from the files created at step 2.
The new material added in step 2 may well change page, figure and table numbering, thus entailing further recompilation.
( 2020-06-12 23:26:33 +0200 )edit
Now I have the file .sagetex.sout: % This file was autogenerated from 2020-06-12-230428.sagetex.sage with % sagetex.py version 2019/01/09 v3.3 %b138f8a4fc7b23f763cd933974ea7272% md5sum of corresponding .sage file (minus "goboom", "current_tex_line", and pause/unpause lines)
( 2020-06-12 23:32:07 +0200 )edit
That has been generated by Step 2, and contains (references to) your Sage results. They are referenced by the sagetex.tex file (via the \sagexxx macroes), and ill be read during Step 3. Go ahead !
( 2020-06-12 23:46:26 +0200 )edit
Thank you for your patience. I do not really know how to do step 3. If I recompile the latex file nothing changes.
( 2020-06-13 00:19:53 +0200 )edit
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Question 11: When a tractor moves with uniform velocity, its heavier wheel rotates slowly than its lighter wheel. Explain. ANSWER A tractor has larger rear wheels and smaller front wheels. When a tractor moves with uniform velocity, the larger rear wheel rotates slowly than the smaller front wheels. Suppose ‘R’ is the radius of the larger wheel and ‘r’ is the radius of the smaller wheel. The linear distance covered by one rotation is given by, For larger wheel, So the linear distance covered by the larger wheel s1 is R/r times more than the distance s2 covered by the smaller wheel. Since ‘R’ is greater than ‘r’, their quotient is also greater than 1 and, hence, s1 is greater than s2. This means in the same number of rotations, the larger wheel will cover more distance than the smaller one. When the tractor is moving with uniform speed, both wheels cover same distance in same time, therefore, s1 = s2. In order s1 = s2, we should have (R/r = 1). Now for (R/r = 1), either the larger wheel rotate slowly or the smaller wheel faster. Anyways, the heavier wheel should rotate slowly than the lighter one.
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# Plane to Ship Displacement (using component method)
1. Sep 27, 2009
### joeseppe
1. The problem statement, all variables and given/known data
A coastguard station locates a ship at range 15.4 km and bearing 123° clockwise from north.
From the same station a plane is at horizontal range 19.4 km, 150° clockwise from north, with elevation 2.06 km.
What is the vector displacement from plane to ship, let i represent east, j north, and k up.
2. Relevant equations
3. The attempt at a solution
Shipx=15.4cos123 = -8.39i km
Shipy=15.4sin123 = 12.92j km
Planex=19.4cos150 = -16.8i km
Planey=19.4sin150 = 9.7j km
Planez=2.06 k km
Therefore Displacement PtoS = (P-S)
=(-8.41i , -3.22j , 2.06k) km
Last edited: Sep 27, 2009
2. Sep 27, 2009
### Delphi51
You seem to have these reversed; the sine gives the i (east) value.
3. Sep 27, 2009
### joeseppe
Really? So Sin is for the X direction, and cos is for the Y? That's not what my textbook says?
Last edited: Sep 27, 2009
4. Sep 27, 2009
### Delphi51
But if you draw the diagram and note that the ship is 57 degrees away from the south line, then you would naturally say that sin(57) = x/15.4 so x = 15.4*sin(57) = 12.9 to the east.
5. Sep 27, 2009
### joeseppe
Yeah I did draw a diagram, I just don't have a scanner to upload it.
I never thought to look at it as two triangles though.
So what I actually should have is:
Shipx=15.4sin57 = 12.9i
Shipy=15.4cos57 = -8.39j
Planex=19.4sin30 = 9.7i km
Planey=19.4cos30 = -16.8j km
Planez=2.06k km
Therefore Displacement PtoS = (P-S)
=(-3.20i , -8.41j , 2.06k) km
Right??
Last edited: Sep 27, 2009
6. Sep 27, 2009
### Delphi51
Hmm, from the plane you would have to go east, north and down to get to the ship.
Therefore I think it should be (3.2i, 8.41j, -2.06k).
7. Sep 27, 2009
### joeseppe
Makes sense. Thanks again for the help! :)
8. Sep 27, 2009
### Delphi51
Most welcome.
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# Questions for the more advanced players
#### diabloblanco
##### Legend
Silver Level
I have a couple questions for those that are actually students of hold 'em and are at the more advanced levels with regards to knowledge of the game.
First, I was rummaging around in some old threads last night, just looking for something interesting and I came upon a thread where a player was bitching about an opponent in an online casino being dealt pocket aces 2 consecutive hands. From what I have seen so far this person complains a lot so the thread didn't really strike me until I got to a reply where another poster spat out some numbers and came up with odds of 48,000 and change to 1 of this actually happening. This number as I recall was based on the odds of it happening in one game then doubled or something similar. This confused me and here's why.
Statistically, isn't the chance of being dealt a particular hand, be it AA or 88, the same for each hand dealt. The act of gathering played cards, shuffling, cutting, and dealing are all factors that are completely detatched from the game and are indvidual random events that contribute to the unpredictability of the dealt cards. Correct me if I'm wrong, but doesn't that mean, when simplified, that the odds of being dealt a particular hand no matter the rank don't fluctuate with number of consecutive hands? Doesn't that also mean that pocket aces have an equal statistical probability of being in the hole on every single hand that is dealt? Given of course that no other mitigating factors are changed such as number of players at the table.
If what I am thinking is correct then why all the complaining from people when opponents are dealt consecutive pocket pairs in online play where RNG's are used? The only factor that seems to me would affect the probability of pairs would be the number of players in the game. Am I completely wrong or right? Thanks for reading this long post and in advance for your thoughts on this.
V
#### VegasGrinder
##### Rock Star
Silver Level
The odds of gettin a specific hand 2 times in a row will always be larger than just getting it once. This is with anything in life that have mathmatical odds.
To make it simple. If you have 10-1 odds on something. The odds of you getting it 2 times is 20-1. The odds of getting it 2 times in a row would be even larger.
#### diabloblanco
##### Legend
Silver Level
But isn't that "statistically speaking", not true...I mean I understand that it is improbbable...but each dealing of the cards is a single solitary act that irrespective of the previous hands can come up any 2 cards. I get the concept that it may "seem" less likely that it will happen, but when dealing with an inanimate, manipulated object, like a deck of cards it seems to me like there are too many variables for any outcome to be accurately predicted. Its all realative....
I am off to the net in search of what I am looking for, I will post back. Maybe I am just making this more complex than in actually is.
BTW, VegasGrinder, thanks for the reply. I would like to know, however, how one would come to this conclusion regarding this specific example. I'm not looking for a simplification of it, or someone to say "you're wrong." I want to be shown how I'm wrong and how to correctly figure the odds the right way.
Last edited:
#### buckster436
##### Cardschat Hall of Famer - RIP Buck
Silver Level
in our noble poker freeroll last week i got pocket aces two hands in a row at the beginning of the tourney ( and i showed them ) then the cards went down hill from there. i would rather get pocket AA towards the end of the tourney rather than the beginning. ( i went all-in with them both times and no one called,i did that cause everytime i play slo-play them someone catches something and beats me) 65%-- 35% percent of the time.
buckster436
#### IrishDave
##### A Member
Silver Level
I got pocket Q's last night back to back in a ring game - lost on both to a river though. I agree that I'd rather have the good cards come later as if you get the nuts early the cards tend to fall off quickly. In a SNG I played last week the first hand I got was 2-2, flop was 2-2-A. Quads to start out and it was the last real hand I got...
#### diabloblanco
##### Legend
Silver Level
I really appreciate the comments, but please don't hi-jack this thread, I'm really looking for the solution to this and I don't want it to get miles off-track. Thanks.
W
#### WesCharge
##### Enthusiast
Silver Level
this will accure exactly 0.00204746% of the time
#### HoldemChamp
##### Rock Star
Silver Level
Well,
I read a post where some person sad the got them 3 times in a row.
I myself have gotten them two times in a row more than once.
The other day playing Omaha. I got AK 10 times in like 340 hands and QJ almost as many times during that 40 hands as well. Obviously not as rare as getting AA 2 or 3 times in a row. But, still very odd. Worse part is none of them won.
The next day I doubled up from what I had lost. I felt better.
#### Grumbledook
##### Legend
Silver Level
It was me that got them 3 times in a row, only had to show them the last time, won the first one preflop and the other on the flop.
Sure its long odds of it happening, that doesn't mean it can't happen though.
#### bpazjr13
##### Rock Star
Silver Level
Knowledge for you:
The odds of drawing AA on any one hand is 1/221. It is a single event. The odds of drawing AA on the next hand are also 1/221, I agree. It is also a single event.
However, when you look at the odds of drawing AA 2 times in a row, then you are looking at a single event, that event being AA coming up 2 times in a row, and the odds of that is 1/221 times 1/221, or 1/48841. If you want to look at the single event of AA coming up 3 times in a row, the it is 1/10,793,861,
Last edited:
#### The_Missnary
##### Enthusiast
Silver Level
WesCharge said:
this will accure exactly 0.00204746% of the time
Nice job on the math! I did it on paper and I was close.. Here's how it looked:
Chance of any one ace is 4/52.
Chance of getting another ace is 3/51.
Chance of getting pocket aces (or any other) is 4/52 X 3/51, or 0.4%
Chance of getting aces two hands in a row is 4/52 X 3/51 X 4/52 X 3/51, or .002%.
X
#### xdmanx007
##### Legend
Bronze Level
diabloblanco said:
I have a couple questions for those that are actually students of hold 'em and are at the more advanced levels with regards to knowledge of the game.
First, I was rummaging around in some old threads last night, just looking for something interesting and I came upon a thread where a player was bitching about an opponent in an online casino being dealt pocket aces 2 consecutive hands. From what I have seen so far this person complains a lot so the thread didn't really strike me until I got to a reply where another poster spat out some numbers and came up with odds of 48,000 and change to 1 of this actually happening. This number as I recall was based on the odds of it happening in one game then doubled or something similar. This confused me and here's why.
Statistically, isn't the chance of being dealt a particular hand, be it AA or 88, the same for each hand dealt. The act of gathering played cards, shuffling, cutting, and dealing are all factors that are completely detatched from the game and are indvidual random events that contribute to the unpredictability of the dealt cards. Correct me if I'm wrong, but doesn't that mean, when simplified, that the odds of being dealt a particular hand no matter the rank don't fluctuate with number of consecutive hands? Doesn't that also mean that pocket aces have an equal statistical probability of being in the hole on every single hand that is dealt? Given of course that no other mitigating factors are changed such as number of players at the table.
If what I am thinking is correct then why all the complaining from people when opponents are dealt consecutive pocket pairs in online play where RNG's are used? The only factor that seems to me would affect the probability of pairs would be the number of players in the game. Am I completely wrong or right? Thanks for reading this long post and in advance for your thoughts on this.
You are correct... For example you will not hit a 4 flush draw on the turn or river 35% of the time you have a 35% chace of hitting everyhand! I believe that is simplifing what you are getting at. You can't look at a hand like well I have a 1 in 4 chance and missed the last 3 so the next should fall. The percentages are the percentages for an individual hand. Yes you are correct MOST players don't look at it this way but it the correct way to look each hand. Although over 100,000 hands or so the numbers should be close! I hope that is what you were looking for if I need to clarify I am happy to try :hmmmm2:
Last edited:
#### bpazjr13
##### Rock Star
Silver Level
xdman what the hell are you not smoking?
M
#### mrbrightside
##### Freeroll #16 Winner
Bronze Level
diablo is right. the events are independent of each other. its like rolling a dice 1/6 odds of hitting a 6. if you roll it again you have the exact same odds of hitting a 6, 1/6. If i could remember my statistics terminology it would be a bit clearer.
X
#### xdmanx007
##### Legend
Bronze Level
bpazjr13 said:
xdman what the hell are you not smoking?
Apparently not anything good enough.... I still remember last night not start remembering last week a month from now
#### bpazjr13
##### Rock Star
Silver Level
yea, but the odds of hitting two 6's in a row is not 1/6 it is 1/36 and you dont need to be a genius to figure it out
#### MsStyque
##### Rock Star
Silver Level
I have never been able to calculate odds as math is not my forte'. However, from the above posts, it does not seem that this is unlikely but not toooo unlikely.
#### diabloblanco
##### Legend
Silver Level
Thank god people finally got what I was saying. And no bpazjr13, you don't have to be a genius to figure out this particular problem, which is good because you got it wrong so there's still hope for ya. Unless you're smarter than Albert Einstein, your dice throwing analogy is 100% wrong. When a dice is tossed, the chance of it hitting any one particular number is 1 in 6. On the second, third, fourth, all the way through 3,000,000th toss, the chance of it hitting the same number doesn't change. Same with cards.
BTW, I would have left that first sentence out of my post if you hadn't made the inference that other posters were less than intelligent for disagreeing, especially since they were correct.
#### bpazjr13
##### Rock Star
Silver Level
so..you are saying that rolling a 6 once has the same chance as rolling two 6's in a row does?
that makes absolutely no sense at all...
roll number 1 - you roll a six 1/6 of the time
roll number 2 - you roll another six 1/36 of the time
roll number 3 - you roll another six 1/216 of the time
not...
rolling 3 sixes in a row has a 1/6 chance of happening
#### diabloblanco
##### Legend
Silver Level
The odds of it happening are 1 in 6 on each individual roll. They are all seperate occourances independant of one another. How can the probability change? Does the dice still have six sides? If it does the answer is 1 in 6. On 6000 rolls, the 6 will come up close to 1000 times. Thus, a 1 in 6 chance.
#### bpazjr13
##### Rock Star
Silver Level
yes in 6000 rolls it will come around 1,000 times but the odds of it coming up 1,000 times in a row is not 1000/6000 (1/6)..that is close to impossible
you are wrong....go get an education in probabilities
#### diabloblanco
##### Legend
Silver Level
Are you going to apologize when you are proved wrong? Why do you assume I am uneducated? So far, you have proven that you are just an asshole, not an educated man. My goal now is to prove positive that you aren't as sharp as you think you are. If, and that is a huge IF, I am wrong (which I am all but positive I'm not), I will admit it. Will you be a big enough person to do the same? Probably not.
S
#### SwgCrazy
##### Rising Star
Bronze Level
:icon_boun I hate to say it, but the original poster of this thread is wrong. Sorry, not trying to be offensive but bpazjr13 has it correct. The odds of hitting a 6 is 1/6 the first time and every 6000 rolls you will roll close to 1000 6's, but it wouldn't be in a row. The odds of hitting 1000 straight 6's is in the billions/trillions:1
Sorry, you could probably look it up on the internet under like learn probabilities or something.
#### Four Dogs
##### Legend
Silver Level
VegasGrinder said:
The odds of gettin a specific hand 2 times in a row will always be larger than just getting it once. This is with anything in life that have mathmatical odds.
To make it simple. If you have 10-1 odds on something. The odds of you getting it 2 times is 20-1. The odds of getting it 2 times in a row would be even larger.
Not quite. If the odds of something happening are 10:1, the odds of it happening again are 100%. The odds of it happening consecutively are 10:1 x 10:1 = 100:1
#### bubbasbestbabe
##### Suckout Queen
Silver Level
the odds that this is boring me...100%
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View more editions
Solutions
Introduction to Algorithms
# TEXTBOOK SOLUTIONS FOR Introduction to Algorithms 3rd Edition
• 954 step-by-step solutions
• Solved by publishers, professors & experts
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Chapter: Problem:
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SAMPLE SOLUTION
Chapter: Problem:
100% (9 ratings)
• Step 1 of 2
Maximum sub-array problem on array of negative values
The maximum sub-array problem is to find the sub-array in an array of positive and negative integers, where the sum of the elements of the sub-array is the maximum sum.
For example, consider the array of values. The contiguous sub-array is with the largest sum as 7.
• Step 2 of 2
• Assume that the array A contains all negative values.
• In each recursive call, FIND-MAXIMUM-SUBARRAY calls FIND-MAX-CROSSING SUBARRAY.
• FIND-MAX-CROSSING SUBARRAY returns the sum of two sub arrays(cross- sum).
• Since all the values in A are negative, FIND-MAX-CROSSING SUBARRAY always returns a cross-sum (a negative value) that is less than the largest negative value in A.
• In lines 7-11, FIND-MAXIMUM-SUBARRAY compares the left-sum and right-sum with cross-sum and returns corresponding three values.
• Since the largest negative value of A is always greater than the cross-sum, ultimately FIND-MAXIMUM-SUBARRAY returns the largest negative value of A and index of the largest negative value in A.
Therefore, if all the elements of A are negative, then FIND-MAXIMUM-SUBARRAY returns the largest negative value in A and its index.
Corresponding Textbook
Introduction to Algorithms | 3rd Edition
9780262033848ISBN-13: 0262033844ISBN:
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capacitance measurement of the
# capacitance measurement of the
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• ### Capacitance and Inductance Measurements with PXI DMMs ...
Jan 11, 2018· Capacitor (C) and Inductor (L) Measurements and Their Preferred Models Back to Top. 3. Capacitance and Inductance Measurement Considerations Capacitors. A capacitor is an electronic component that is capable of storing energy as charge.
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• ### Circuit measures capacitance or inductance | EDN
the basic equation for capacitance is: To measure the value of a capacitor, measure the input voltage and then adjust the frequency of the signal generator to make the output voltage one-half of the input voltage. You need not use a 2-to-1 ratio for VIN/VOUT.
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• ### C Fundamentals of Semiconductor C-V Measurements
• Leaky (high D) capacitance measurements • Using hardware and software to acquire the data • Parameter extractions Overcoming these challenges requires careful attention to the techniques used along with appropriate hardware and software. Low Capacitance Measurements. If C is .
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• ### Electronics Measurements: Capacitance - dummies
Capacitance is measured in units called farads (abbreviated F). The definition of one farad is deceptively simple. A one-farad capacitor holds a voltage across the plates of exactly one volt when it''s charged with exactly one ampere per second of current.
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• ### What is the capacitance of the capacitor and how is it ...
Capacitance is the ratio of the change in an electric charge in a system to the corresponding change in its electric potential. It can be measured in the following ways: 1. You can measure the time it takes to charge capacitor to one time constant...
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• ### Capacitors & Capacitance Calculations Formulas Equations ...
Equations for combining capacitors in series and parallel are given below. Additional equations are given for capacitors of various configurations. As these figures and formulas indicate, capacitance is a measure of the ability of two surfaces to store an electric charge.
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• ### Capacitance measurement | Article about Capacitance ...
Capacitance measurement. The measurement of the ratio of the charge induced on a conductor to the change in potential with respect to a neighboring conductor which induces the charge. In a multiconductor system there are capacitances between each pair of conductors.
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• ### Instruction of Parallel Plate Capacitor Experiments - SMU
measuring the capacitance with and without the insulator one can calculate the value of ε r of this material. The second type of experiments employs the fixed capacitor. Measure the capacitance of this capacitor in air, and then immerse the capacitor in the (non-conducting) liquid of which you want to measure its dielectric constant.
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• ### Capacitance and Charge on a Capacitors Plates
Capacitance can also be determined from the dimensions or area, A of the plates and the properties of the dielectric material between the plates. A measure of the dielectric material is given by the permittivity, ( ε ), or the dielectric constant.
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• ### Challenges and solutions for Impedance measurements
Measurement Method Agilent Product Frequency Range Auto-Balancing Bridge 4263B LCR Meter E4981A C Meter E4980A Precision LCR Meter 4285A Precision LCR Meter 4294A Precision Impedance Analyzer 100 Hz to 100 kHz spot 120 Hz/1 kHz/1 MHz spot 20 Hz to 2 MHz 75 kHz to 30 MHz 40 Hz to 110 MHz E4982A RF LCR Meter E4991A Impedance/Material Analyzer
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• ### Capacitance and Tan- δ Measurement Of Winding: – A B C of ...
2 Capacitance and Tan- δ Measurement Of Winding: Objective: Dissipation factor/Loss factor and capacitance measurement of winding is carried out to ascertain the general condition of the ground and inter-winding insulation of transformer and reactors. Instrument Required: 1. Portable Automatic Capacitance & tan-δ kit. 2. Volt Meter 3.
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• ### ELECTROCHEMICAL CAPACITANCE MEASUREMENTS TO .
capacitance measurements of a Platinum working electrode before and after glucose adsorption in buffered solutions. We noticed that the obtained results explain the
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• ### How to Read a Multimeter''s Capacitance | Hunker
A farad is a unit of capacitance defined by the International System of Units. Capacitors are usually measured in units of micro-farad. Displayed as μF F, mfd, MFD, MF and UF, micro-farad measurement on a multimeter is usually indicated by "MFD." Compare the measured MFD with the capacitors expected MFD measurement.
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• ### Measure Capacitance With Arduino: 6 Steps (with Pictures)
Capacitance is an object''s ability to store an electric charge. Reasonably, this object is referred to as a capacitor. A capacitor that stores this charge in an electric field between two conductive plates is known as a parallel plate capacitor.
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• ### What is Cable Capacitance? Definition & Capacitance in a ...
4. Two conductors, B and C are connected to the sheath and the capacitance is measured between these and the third conductor A. The capacitance arrangement of the system, then reduces. The capacitance measured is this case = C s + C c + C c = 2C c + C s. From the above test, the value of C c and C s can also be determined separately.
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• ### How to Make an Arduino Capacitance Meter
Each capacitance meter has an RC circuit with known resistor values and an unknown capacitor value. The Arduino will measure the voltage at the capacitor and record the time it takes to reach 63.2% of it''s voltage when fully charged (the time constant).
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• ### 4.5 The p-n junction capacitance - University of Colorado ...
4.5.4 Capacitance measurement and parameter extraction. A capacitance versus voltage measurement can be used to obtain the built-in voltage and the doping density of a one-sided p-n diode. Plotting one over the capacitance squared one expects a linear dependence as expressed by:
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• ### Capacitance Measurement - YouTube
Aug 24, 2015· Video on testing capacitance of capacitors installed in NEPSI''s medium-voltage metal-enclosed power capacitor banks and harmonic filter banks.
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• ### Keysight Technologies Direct Power MOSFET Capacitance ...
High-Voltage MOSFET Capacitance Measurement Basics (continued) Let us now consider what happens when we try to measure any single one of these three capacitances using a capacitance meter.
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• ### Testing Electrochemical Capacitor: CV-EIS-Leakage Current
In the plot above, impedance in this region depends on DC voltage, so the capacitance must also depend on DC voltage. Measurement of Leakage Current. Leakage current can be measured in at least two ways: Apply a DC voltage to a capacitor and measure the current required to maintain that voltage.
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• ### Capacitance meter - Wikipedia
A capacitance meter is a piece of electronic test equipment used to measure capacitance, mainly of discrete capacitors. Depending on the sophistication of the meter, it may display the capacitance only, or it may also measure a number of other parameters such .
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• ### Why is it necessary to measure capacitance at different ...
Generally, the 10uF and under have applied voltage of 1.0 ± 0.2 Vrms. But over 10uF, the applied voltage is 0.5 ± 0.1 Vrms. High capacitance capacitors have very low impedance, therefore to supply enough current to make the measurement, the power supply needs more current than that supplied at 1.0 ± .
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• ### How to Make an Arduino Capacitance Meter
How to Make an Arduino Capacitance Meter. One meter would accurately measure values in the 1000 μF range, but it would fail in the nF and pF range. Another capacitance meter was accurate in the nF and pF ranges, but failed in the μF range. So, I''ll show you three different meters that together will cover a range of about 10 pF to 3900 μF.
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• ### Capacitance - Wikipedia
https://en.wikipedia/wiki/CapacitanceOverviewGet Price
• ### FDC1004: Basics of Capacitive Sensing and Applications
Capacitance Measurement Basics 1 Capacitance Measurement Basics Capacitance is the ability of a capacitor to store an electrical charge. A common form – a parallel plate capacitor – the capacitance is calculated by C = Q / V, where C is the capacitance related by the stored charge Q .
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• ### Capacitance meters | Amazon
Some meters may measure parameters in addition to capacitance, such as continuity, resistance, or inductance. Considerations when choosing a capacitance meter include the capacitance range to be tested. Capacitance meters measure capacitance within a specified range, such as 0.01 to 10,000 microfarads or 200 picofarads to 20 millifarads.
### Capacitance measurement techniques and solutions from ...
The capacitance measurement is a fundamental and important measurement to understand the characteristics of the devices such as semiconductors, components, materials .
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• ### Capacitance Unit Converter - Measurement conversion A-I
Capacitance - the amount of electrical charges an isolated conductor can hold. The SI unit used to describe capacitance is the farad, the symbol is C.
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• ### Capacitance Meter (with Pictures)
Capacitance Meter. Capacitors are vital components in electronics, but sometimes they are broken, or the value printed on the cap has become unreadable. Because my multi-meter does not have a capacitance measurement, I decided to make one! The principle of measuring capacitance is quite simple.
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• ### distributed capacitance paper
7 MEASURING TRANSFORMER DISTRIBUTED CAPACITANCE by Kirby Creel Figure 6 shows the phase angle of current plotted against frequency, using a Hewlett Packard HP-4194A. The plot is the same coil as discussed in (2). The resonant frequency is 499 KHz (zero in this case is actually +0.00084°).
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• ### Capacitance-voltage profiling techniques for ...
The method involves the measurement of the junction barrier. capacitance at different reverse voltage biases. From C(V) data then N(x) is. inferred – N being the doping concentration, usually measured in cm-3 and x is. the spatial coordinate (cm or µm).
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CC-MAIN-2020-34
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https://www.numbersaplenty.com/31223333
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Search a number
31223333 is a prime number
BaseRepresentation
bin111011100011…
…0111000100101
32011202022100202
41313012320211
530443121313
63033120245
7526252061
oct167067045
964668322
1031223333
1116696609
12a559085
136612a5b
14420aaa1
152b1b558
hex1dc6e25
31223333 has 2 divisors, whose sum is σ = 31223334. Its totient is φ = 31223332.
The previous prime is 31223303. The next prime is 31223359. The reversal of 31223333 is 33332213.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 25381444 + 5841889 = 5038^2 + 2417^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-31223333 is a prime.
31223333 is a modest number, since divided by 3333 gives 3122 as remainder.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (31223303) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 15611666 + 15611667.
It is an arithmetic number, because the mean of its divisors is an integer number (15611667).
Almost surely, 231223333 is an apocalyptic number.
It is an amenable number.
31223333 is a deficient number, since it is larger than the sum of its proper divisors (1).
31223333 is an equidigital number, since it uses as much as digits as its factorization.
31223333 is an evil number, because the sum of its binary digits is even.
The product of its digits is 972, while the sum is 20.
The square root of 31223333 is about 5587.7842656996. The cubic root of 31223333 is about 314.8906414707.
Adding to 31223333 its reverse (33332213), we get a palindrome (64555546).
The spelling of 31223333 in words is "thirty-one million, two hundred twenty-three thousand, three hundred thirty-three".
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https://www.teacherspayteachers.com/Product/Area-Worksheets-Mini-Bundle-3334445
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# Area Worksheets Mini Bundle
Subject
Resource Type
Common Core Standards
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2 MB|21 pages
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Product Description
These 4 Color by Solution Activities cover everything to do with AREA!
Includes Area of Circles, Area of Triangles, Area of Special Quadrilaterals and Area of Composite Figures. You will find these are the perfect activity to supplement your lesson on Area! Use as a whole group activity, small group even centers! Great component to academic practice (homework) and checking for comprehension.
These activities are all sold separately or purchase together to save money \$\$\$
They all include 9 to 12 problems using a variety of questions from pictorial, word problems, coordinate planes, decomposing whole figures and finding missing components to a formula. All include a quick reference answer key and the Composite Figures activity includes a detailed answer key if needed due to the complexity!
Common Core Aligned: 6.G.A.1 and 7.G.B.4
All four Color by Solution Activities also have a whimsy color page for quick assessment that tells the students what they will be covering next! On the last activity, I've incorporated a Self-Reflection component since this is the ending of the Area Unit. Students will gauge their comprehension, identify how hard they worked, what they could do improve on and any extra help they could receive from the teacher. This makes for a perfect exit ticket or to form intervention groups.
Enjoy!
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Other Math Games and Center Ideas
I Have…Who Has…? Equivalent Expressions
3 Versions of a classic math game (easy, medium & difficult) perfect for all your learners!
Scavenger Hunt – Integer Edition
Highly engaging Integer activity with 10 word problems and extension questions for each!
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© Personal Copyright: The purchase of this product allows the teacher to use in their personal classroom. Please do not share with other educators unless additional licenses have been purchased. Site and District licenses are also available.
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https://scicomp.stackexchange.com/questions/36724/verification-of-order-of-convergence-of-implicit-euler-method-to-numerically-sol
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# Verification of order of convergence of Implicit Euler Method to numerically solve Black-Scholes PDE
I'm trying to verify the order of convergence for implicit Euler method to numerically solve Black-Scholes PDE. Theory says that it should be $$O(\Delta t + \Delta S^2).$$ My code is working absolutely fine. I'm getting almost zero error when I compare my solution to the one obtained by Black-Scholes formula. However, I'm getting incorrect order of convergence. Does it have something to do with how I choose parameters? Could someone have a look at my code and tell me where I'm going wrong if there is a mistake somewhere? Or perhaps convey to me what option parameters need to be chosen so that I get the correct order of convergence?
Here is the MATLAB code:
function [V, t, S] = bspdeimp(pc, K, T, r, sigma, Smax, Nt, NS)
% [V, t, S] = bspdeimp(pc, K, T, r, sigma, Smax, Nt, NS)
% Prices European put or call option using implicit finite difference method.
% -- Input arguments --
% pc = 'put' or 'call'
% K = strike price
% T = expiry time
% r = risk-free interest rate
% sigma = volatility
% Smax = max value of S used in finite difference grid
% Nt = number of time steps
% NS = number of grid points along S axis
% -- Output arguments --
% V = values of option at asset values in S
% t = vector of time points
% S = vector of asset prices
% Time discretization
Dt = T / Nt;
t = linspace(0, T, Nt+1);
% Asset price discretization
DS = Smax / (NS);
S = linspace(0, Smax, NS+1)';
% Storage for option values
V = zeros(length(S), length(t));
% Set initial condition at expiry and boundary values
switch lower(pc(1))
case 'p' % put option
V(1,:) = K * exp(-r*(T-t));
V(end,:) = 0;
V(1:NS-1,end) = max( K - S(2:end-1), 0);
case 'c' % call option
V(1,:) = 0;
V(end,:) = Smax - K * exp(-r*(T-t));
V(2:end-1,end) = max( S(2:end-1) - K, 0);
otherwise
error(['unknown option type ' pc])
end
% Define index vector of interior asset prices
J = 2:NS;
c1 = sigma^2 * S(J).^2 * Dt / ( 2 * DS^2 );
c2 = r * S(J) * Dt / ( 2 * DS );
alpha = -c1 + c2;
beta = 1 + r*Dt + 2*c1;
gamma = -c1 - c2;
% Create and factor the sparse matrix for the linear system.
A = spdiags([[alpha(2:NS-1); 0], beta, [0;gamma(1:NS-2)]], [-1, 0, 1], NS-1, NS-1);
[L,U,p] = lu(A,'vector');
% Time step backwards from expiry
for n = Nt:-1:1
% Set up RHS vector from values at time step n+1
b = V(J,n+1);
% Adjust for boundary values at S = 0 and S = Smax
b(1) = b(1) - alpha(1)*V(1,n);
b(NS-1) = b(NS-1) - gamma(NS-1)*V(NS+1,n);
% Solve linear system A * V(J,n) = b using existing LU factorization
% V(J,n) = A \ b; would recalculate factorization at each time step
V(J,n) = U \ ( L \ b(p) );
end
function [c, dcds] = blackscholes(S, K, r, sigma, Tmt)
% [c, dcds] = blackscholes(S, K, r, sigma, Tmt)
% Black and Scholes formula for the value of a call option
% and its derivative with respect to volatility sigma
% S = underlying asset price
% K = strike price
% r = risk-free interest rate
% Tmt = time to maturity = T - t where T = expiry
% If sigma is a vector of volatilities, then both the
% call value and its derivatives are vectors of the same size.
%
% Uses normpdf and normcdf from Statistics toolbox.
s = sigma * sqrt(Tmt);
d1 = ( log(S/K) + ( r + sigma.^2/2)*(Tmt) ) ./ s;
d2 = d1 - s;
% Use normpdf and normcdf from Statistics toolbox
c = S .* normcdf(d1) - K * exp(-r*Tmt) * normcdf(d2);
% Derivative of call value w.r.t. volatility sigma
dcds = S .* normpdf(d1) * sqrt(Tmt);
% Test script for the functions to solve Black - Scholes PDE
% for the value of a call or put option.
% Define option parameters
r = 0.1;
sigma = 0.4;
T = 5/12;
K = 50;
pc = 'call';
% Define discretization parameters
% Asset price S goes from 0 to Smax
Smax = 100;
% Compare with Black and Scholes formula for call
Tmt = T;
[c, dcds] = blackscholes(S, K, r, sigma, Tmt);
Err = c - V(:,1);
fprintf("\nConvergence with respect to Dt\n")
fprintf("\n%6s %10s %10s %8s\n\n","NS", "Nt", "max error", "rate")
nrows = 8;
NS = 100;
Nt = 200;
err = zeros(nrows,1);
for row = 1:nrows
Nt = 2 * Nt;
[V, t, S] = bspdeimp(pc, K, T, r, sigma, Smax, Nt, NS);
for n = 1:Nt+1
err_n = norm( V(:,n) - blackscholes(S, K, r, sigma, T - t(n)), Inf);
if err_n > err(row)
err(row) = err_n;
end
end
if row == 1
fprintf("%6d %10d %10.2e\n", NS, Nt, err(row))
else
ratio = err(row-1) / err(row);
rate = log2(ratio);
fprintf("%6d %10d %10.2e %8.3f\n", NS, Nt, err(row), rate)
end
end
fprintf("\nConvergence with respect to Ds\n")
fprintf("\n%6s %10s %10s %8s\n\n","Ns", "Nt", "max error", "rate")
nrows = 5;
NS = 50;
Nt = 2000;
err = zeros(nrows,1);
for row = 1:nrows
NS = 2 * NS;
%[U, x, t] = iEuler(a, L, T, f, gamma0, gammaL, u0, NS, Nt);
[V, t, S] = bspdeimp(pc, K, T, r, sigma, Smax, Nt, NS);
for n = 1:Nt+1
err_n = norm( V(:,n) - blackscholes(S, K, r, sigma, T - t(n)), Inf);
if err_n > err(row)
err(row) = err_n;
end
end
if row == 1
fprintf("%6d %10d %10.2e\n", NS, Nt, err(row))
else
ratio = err(row-1) / err(row);
rate = log2(ratio);
fprintf("%6d %10d %10.2e %8.3f\n", NS, Nt, err(row), rate)
end
end
$$$$
You forgot to multiply the norm of the difference between the numerical solution and the exact one by the discretization step. In your case it is enough to divide the err_n by Nt when computing the order of time discretization, and by NS when computing the order of space discretization. I got a perfect first order accuracy for the time discretization, in the case of space discretization the order seems to approach the first one from above as the initial function is not differentiable in the space.
Here is the output of the corrected code:
Convergence with respect to Dt
NS Nt max error rate
100 400 2.59e-04
100 800 1.19e-04 1.120
100 1600 5.79e-05 1.042
100 3200 2.83e-05 1.032
100 6400 1.40e-05 1.017
100 12800 6.95e-06 1.009
100 25600 3.46e-06 1.005
100 51200 1.73e-06 1.002
Convergence with respect to Ds
Ns Nt max error rate
100 2000 9.17e-04
200 2000 2.53e-04 1.858
400 2000 7.03e-05 1.848
800 2000 2.20e-05 1.678
1600 2000 8.92e-06 1.300
`
• Thanks!! It works now but I'm not sure as to why we need to divide err_n by Nt when computing the order of time discretization, and so forth by NS . We know that order of convergence $\rho = \frac{log(E_1/E_2)}{log(h_1/h_2)}$ where $E_1 = u_{h_1} - u*$ and $E_2 = u_{h_2} - u*.$ My code calculates rate = log2(ratio) which is the same as the formula. So I don't understand why we need to do the divisions that you have suggested? Also, not sure why you say multiply in the first line when you are actually dividing and how did you deduce that initial function is not differentiable in the space
– user36184
Jan 24 '21 at 20:01
• I am answering on a phone. Have a look to definitions of errors, your one is wrong. You have to e.g. multiply the difference with e.g. DeltaS that is same as divided by NS. Because DeltaS=Smax/NS, it is the same. The initial condition is not differentiable, because it contains the function max that is not differentiable in zero. Jan 25 '21 at 6:16
• Thanks!!! Could you please refer me to some relevant literature where the definition of error is given. I'm having a hard time finding out relevant resources. Also, in my code I'm using infinity norm to compute error , what will happen if I use L2 norm and do I have to make any changes?
– user36184
Jan 25 '21 at 9:53
• A simple explanation for a "correct way" to compute the error is that you should do a fair comparison for different discretizations steps (i.e. different NS and Nt). Then an "averaged error", e.g. the sum of absolute differences in each grid node divided by the number of nodes, can do the job. More theoretically based approach is that you should place (extend) all numerical solutions to one functional space, e.g. using piecewise linear functions by interpolating values in nodes, and compare errors with a norm of that space (e.g. an integral norm) using enough precise numerical quadrature. Jan 26 '21 at 15:29
• A good reference to be found also on the web is the book of Randy LeVeque on "Finite Difference Methods for Ordinary and Partial Differential Equations", where there is a Chapter (Appendix) on Measuring errors. Jan 26 '21 at 15:39
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# How To 3, Light Switch Wiring Perfect Wiring Diagram, Two, Switch Uk, Wire 3, Switch Uk, 3 Way Galleries
## Other recommended diagram ideas:
13 Practical How To 3, Light Switch Wiring Collections - So lets say the wires you've got in container one are all of them. A 14/2 is power in, 14/2 is to the light and the 3 twine of route is to the opposite three manner. Take the black cord from the opposite 3 way and join that to the black cord to the light. And the twine from the mild to the white from the electricity and the black from the energy to the three way inside the box with all of the wires. Hi dominick, how might i twine 2 lighting to 2 three manner switches? I've a mild at the pinnacle of the steps and a mild at the lowest of the steps. So i want to put a transfer at the bottom of the steps and at the top of the stairs. Thank you.
Sure three methods switches on both ends. There are some approaches to cord a 3 manner. Now not just the way i display. This is the comment section of the page you’re on. Use the question phase to where you can submit photos. Top proper “questions”. I need extra element on what you're asking. In case you're simplest looking to add a extra mild then you definately might just join your wires from the prevailing light and not the switch. If i have you ever right here.
Mr. Dominick, i apprehend the fundamental wiring for a three-way transfer and your video helps to enhance my knowledge. My wife asked me to install new led furniture within the hallway. One is on a separate 2-way transfer (works satisfactory) and the opposite is hooked up to three-way switches. But, i appear to be getting residual/inductive current; the fixture glows while switched-off. The power appears to be coming into the light and jogging into the switches. I'm able to’t appear to segregate the residual contemporary to hold the mild from glowing. Any thoughts?. Why do i ask to sign up to ask a question? Properly easy definitely. One, i want to ensure you are actual and which you do get my answer. Seems fair to me. And yes your statistics is very safe with me.
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1. ## Help reading functions with denominator
I've been having issues seeing functions that have denominators. I have few issues understanding most regular functions and how they are manipulated but in my effort to return to school I have missed out on understanding functions when combined with a denominator.
For example: $\displaystyle f(x)=\frac{|x-1|}{2-2x}$
Is there any method to seeing that the lines are at +1/2 and -1/2? Is there a good website that can teach me methods of visualizing various fraction functions?
2. ## Re: Help reading functions with denominator
$\displaystyle f(x)=\frac{|x-1|}{2-2x}$
Maybe rewrite it as $\displaystyle f(x)=\frac{|1-x|}{2(1-x)}$ then it's easier to see that f(x)=1/2 when 1-x>0 and f(x)=-1/2 when 1-x<0.
I'd also recommend f(x)=|1-x|/(2(1-x)) - Wolfram|Alpha and you could download a graph plotter to play with.
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Explore BrainMass
# Accounting for Corporations
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# Calculate the potential
1. Jan 18, 2006
### stunner5000pt
the trajectory of a particle moving in a cnetral potential V(r) is given by $r = a + b \sin(\eta \phi)$ where a b and eta are constants
Compute the potential V(r) in which the particel moves (for arbitrary a, b, and eta) and sketch $V_{eff} (r) = V(r) + \frac{L^2}{2mr^2}$ for cases a =b
sketching is not hte problem here.. i just need to find out V(r)
well i know that i need to compute the force f first and then integrate f w.r.t. r to get V(r)
find find $$u = \frac{1}{r} = \frac{1}{a + b \sin(\eta \phi)}$$
then using $$f = -\frac{L^2 u^2}{m} (u'' + u)$$
and $$u '' = \frac{2b^2 \eta^2 \cos^2 (\eta \phi)}{r^3} + \frac{b \eta^2 \sin(\eta \phi)}{r^2}$$ (phew!)
once i sub expressions for u'' and u i get
$$f(r) = - \frac{L^2}{mr^2} \left( \frac{2b^2 \eta^2 \cos^2 (\eta \phi)}{r^3} + \frac{b \eta^2 \sin(\eta \phi)}{r^2} \right)+ \frac{1}{r}$$
is that fine? Can i simplify that any more in terms of r?
are the steps correct? Any problems with the derivative? Just need to know if i can simplify any further.
Your help is greatly appreciated!
Thank you
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## BPUT Syllabus
BPUT Syllabus of MBA
Module-I Statistical Methods: Measures of central tendency and dispersion: StandardDeviation, moments, measures of skewness and kurtosis. Simple Correlation, calculation ofcorrelation coefficient, probable error, Rank correlation.Regression: Linear regression,calculation of regression coefficients, Time series Model (Component, Uses, Moving AverageMethod, Least Square Method) Module II Probability: Concept, Addition, Conditional Probability Baye’s theorem,Probability Distributions: Binomial, Poisson and Normal Module III Decision Sciences & role of quantitative techniques. Linear Programming: Concept, Formulation & Graphical and Simplex SolutionAssignment Models: Concept, Flood’s Technique / Hungarian Method, applications including restricted & multiple assignments.Transportation Models: Concept, Formulation, Problem types: Balanced, Unbalanced, Minimization, Maximization Basic initial solution using North West Corner, Least Cost & VAM, and Optimal Solution using MODI. Module-IV Queuing Theory: Concept, Single Server (M/M/I,)Markov Chains & Simulation Techniques: Markov chains: Applications related to management functional areas, Decision Theory: Concept, Decision under risk (EMV) & uncertainty GameTheory : Concept, 2 zero sum game with dominance, Pure & Mixed Strategy. Module – V The concerned faculty shall have the liberty to define the course contents under this module and teach students accordingly
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M.Tech BPUT Syllabus
Module I Introduction to RM:Meaning and significance of research.Importance of scientific research in decision making.Types of research and research process.Identification of research problem and formulation of hypothesis.Research Designs. Module II Measurement and Data Collection. Primary data, Secondary data, Design of questionnaire ; Sampling fundamentals and sample designs. Measurement and Scaling Techniques, Data Processing. Module III Data Analysis – I:Hypothesis testing; Z-test, t-test, F-test, Chi-square test.Analysis of variance.Non-parametric Test – Sign Test, Run test, Krushall – Wallis test Module IV Data Analysis – II:Factor analysis, Multiple Regressions Analysis.Discriminant Analysis, Use of SPS Package.
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BPUT Syllabus for M.Sc
FPYC-101 CLASSICAL MECHANICS & SPECIAL THEORY OF RELATIVITY-I
UNIT I
Constrained Motion Constraints, Classification of constraints, Principle of virtual work, D’Alembert’s principle and its application. Lagrangian formulation limitations of Newtonian formulation, degrees of freedom , generalized coordinates and velocities, Derivation of Lagrange’s equation, Calculus of variation, Euler – Lagrange equation, derivation of Lagrange’s equation from Hamilton’s principle , simple application of Lagrange’s equation, Cyclic coordinates , Symmetry and conservation theorems.
Unit-II
Phase Space and the motion of the system Hamiltonian Hamilton’s Canonical Equation of motion Physical significance of H Advantage of Hamiltonian approach Deduction of canonical equation from variational principle Hamiltonian canonical equation of motion in different co ordinate system Application of Hamilton’s equation of motion in different co ordinate system Application of hamiltonian’s equation of motion Hamiltonian for charged particle in an electromagnetic field Principle of least action.
Unit-III
Canonical or Contact transformations, Advantage of Canonical transformation, example of canonical transformation Solution of simple harmonic oscillator problem Condition for transformation to be canonical. Infinitesimal contact transformation. Hamiltonian Jacobi Method, Solution of harmonic oscillator problem by Hamilton Jacobi Method. Particle Falling Freely, Hamilton Jacobi equation for Hamilton Characteristic function.
Unit-IV
Poisson Bracket-Definition Invariance of poission bracket with respect to canonical transformation Equation of motion in poisson bracket form Jacobi’s identity Infinitesimal contact transformation interpretation in terms of poision’s bracket The angular momentum and poisson bracket Poisson bracket in quantum mechanics Lagrange bracket Relation in Lagrange and poisson bracket Liouvilies Theorem. Galilean transformation (GT) Invariance of Newton’s laws and laws of conservation of linear momentum and kinetic energy under GT. Michelson Morley experiment postulates of special theory of relativity Lorentz transformation length contraction time dilation Velocity transformation relativistic mass and momentum mass energy relation.
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FPYC-102 HEAT AND THERMODYNAMICS
Unit-I Thermodynamical system , Principles of thermodynamics , concept of thermodynamic state , Zeroth law of thermodynamics , work done in isothermal and isobaric processes, Heat and work ,Free energy and their application, internal energy function and the first law of thermodynamics , application to various processes , CP – CV , Equation of state for adiabatic process, work done in adiabatic process , Equations of state. Ideal gases and their PVT relations, Gas mixtures. Unit-II Maxwell- Boltzmann formula for distribution of molecular speed (statement of formula and discussion), Average RMS and most probable speed, Mean free path, Degrees of freedom,The principle of equipartition of energy, The Vanderwaals equation of state, Evaluation ofcritical constants Unit-III Zeroeth law of thermodynamics, Heat capacity, Second law of thermodynamics, Carnot’s engine, Carnot theorem, The thermodynamic scale of temperature, Entropy, entropy change in reversible and irreversible processes, mathematical formulation of second law, Maxwell’s relations, first TdS equation, second TdS equation, Phase change, Clausius-Clapeyron equation Unit-IV Thermal conductivity, Conduction along a uniform bar, rectilinear flow of heat, Experimental determination of Thermal conductivity (Ingen-Hausz’s method).Blackbody radiation, Emissive and absorptive power, Kirchoff’s law, Stefan-Boltzmann’s law,Energy distribution in the blackbody spectrum, Wein’s law and Rayleigh-Jean’s law (Statement of formula and discussion). Planck’s radiation formula, derivation of Rayleigh-Jean’s formula, Wein’s formula and Stefan-Boltzmann law using Planck’s formula.
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Module-I : (14 Hours) Basic Concepts of Differential EquationDegree EquationEquations of First order but of Higher Degree Module-II : (14 Hours) Linear Equations with Constant coefficientExistence and Uniqueness of solution Module-III : (12 Hours) Series Solution and special functionGaseous stateLiquid stateAtomic structureLiquid stateAtomic structurePeriodic propertiesChemical bonding
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# Lecture Notes for EE133A L. Vandenberghe Fall Quarter 2014-15
```Fall Quarter 2014-15
Lecture Notes for EE133A
L. Vandenberghe
46
Chapter 4
Exercises on least-norm
problems
4.1 Minimum-energy optimal control. A simple model of a vehicle moving in one dimension
is given by
s1 (t + 1)
s2 (t + 1)
=
1
0
1
0.95
s1 (t)
s2 (t)
+
0
0.1
u(t),
t = 0, 1, 2, . . . .
s1 (t) is the position at time t, s2 (t) is the velocity at time t, and u(t) is the actuator input.
Roughly speaking, the equations state that the actuator input affects the velocity, which
in turn affects the position. The coefficient 0.95 means that the velocity decays by 5% in
one sample period (for example, because of friction), if no actuator signal is applied. We
assume that the vehicle is initially at rest at position 0: s1 (0) = s2 (0) = 0.
We will solve the minimum energy optimal control problem: for a given time horizon N ,
choose inputs u(0), . . . , u(N − 1) so as to minimize the total energy consumed, which we
assume is given by
N −1
E=
X
u(t)2 .
t=0
In addition, the input sequence must satisfy the constraint s1 (N ) = 10, s2 (N ) = 0. Our
task therefore is to bring the vehicle to the final position s1 (N ) = 10 with final velocity
s2 (N ) = 0, as efficiently as possible.
(a) Formulate the minimum energy optimal control problem as a least-norm problem
minimize
subject to
kxk2
Cx = d.
Clearly state what the variables x, and the problem data C and d are.
(b) Solve the problem for N = 30. Plot the optimal u(t), the resulting position s1 (t),
and velocity s2 (t).
(c) Solve the problem for N = 2, 3, . . . , 29. For each N calculate the energy E consumed
by the optimal input sequence. Plot E versus N . (The plot looks best if you use a
logarithmic scale for E, i.e., semilogy instead of plot.)
(d) Suppose we allow the final position to deviate from 10. However, if s1 (N ) 6= 10,
we have to pay a penalty, equal to (s1 (N ) − 10)2 . The problem is to find the input
48
4
Exercises on least-norm problems
sequence that minimizes the sum of the energy E consumed by the input and the
terminal position penalty,
N −1
X
u(t)2 + (s1 (N ) − 10)2 ,
t=0
subject to the constraint s2 (N ) = 0.
Formulate this problem as a least-norm problem, and solve it for N = 30. Plot
the optimal input signals u(t), the resulting position s1 (t) and the resulting velocity
s2 (t).
Remark. If C is right invertible, then the MATLAB command x = C \ d computes a
solution to Cx = d, but it is not the least-norm solution.
We can use the command x = C’ * ((C*C’) \ d) to compute the least-norm solution
using the Cholesky factorization method. (MATLAB will recognize that CC T is positive
definite and use the Cholesky factorization to solve CC T z = d.) We can also use the QR
factorization method, using the code
[Q, R] = qr(C’, 0);
x = Q * (R’ \ d);
4.2 Two vehicles are moving along a straight line. For the first vehicle we use the same model
as in exercise 4.1:
s1 (t + 1)
s2 (t + 1)
=
1
0
1
0.95
s1 (t)
s2 (t)
+
0
0.1
u(t),
t = 0, 1, 2, . . . ,
s1 (t) is the position at time t, s2 (t) is the velocity at time t, and u(t) is the actuator
input. We assume that the vehicle is initially at rest at position 0: s1 (0) = s2 (0) = 0.
The model for the second vehicle is
p1 (t + 1)
p2 (t + 1)
=
1
0
1
0.8
p1 (t)
p2 (t)
+
0
0.2
v(t),
t = 0, 1, 2, . . . ,
p1 (t) is the position at time t, p2 (t) is the velocity at time t, and v(t) is the actuator input.
We assume that the second vehicle is initially at rest at position 1: p1 (0) = 1, p2 (0) = 0.
Formulate the following problem as a least-norm problem, and solve it in MATLAB (see
the remark at the end of exercise 4.1). Find the control inputs u(0), u(1), . . . , u(19) and
v(0), v(1), . . . , v(19) that minimize the total energy
19
X
u(t)2 +
t=0
19
X
v(t)2
t=0
and satisfy the following three conditions:
s1 (20) = p1 (20),
s2 (20) = 0,
p2 (20) = 0.
(4.1)
In other words, at time t = 20 the two vehicles must have velocity zero, and be at the
same position. (The final position itself is not specified, i.e., you are free to choose any
value as long as s1 (20) = p1 (20).)
Plot the positions s1 (t) and p1 (t) of the two vehicles, for t = 1, 2, . . . , 20.
4.3 Explain how you would solve the following problems using the QR factorization.
(a) Find the solution of Cx = d with the smallest value of
minimize
subject to
Pn
i=1
wi x2i :
Pn
wi x2i
Cx = d.
i=1
The problem data are the p × n matrix C, the p-vector d, and the n vector w. We
assume that A is right invertible, and wi > 0 for all i.
Exercises
49
(b) Find the solution of Cx = d with the smallest value of kxk2 − cT x:
kxk2 − cT x
Cx = d.
minimize
subject to
The problem data are the n-vector c, the p × n matrix C, and the p-vector d. We
assume that C is right-invertible.
4.4 Show how to solve the following problems using the QR factorization of A. In each
problem A is a left invertible m × n matrix. Clearly state the different steps in your
method. Also give a flop count, including all the terms that are quadratic (order m2 , mn,
or n2 ), or cubic (order m3 , m2 n, mn2 , n3 ).
If you know several methods, give the most efficient one.
(a) Solve the set of linear equations
0
A
AT
I
x
y
=
b
c
.
The variables are the n-vector x and the m-vector y.
(b) Solve the least-squares problem
minimize
2kAx − bk2 + 3kAx − ck2 .
The variable is the n-vector x.
(c) Solve the least-norm problem
kxk2 + kyk2
AT x − 2AT y = b.
minimize
subject to
The variables are the m-vectors x and y.
(d) Solve the quadratic minimization problem
minimize
xT AT Ax + bT x + c.
The variable is the n-vector x.
4.5 If A is a left invertible m × n matrix, and D is an m × m diagonal matrix with positive
diagonal elements, then the coefficient matrix of the equation
D2
AT
A
0
x
ˆ
yˆ
=
b
c
is nonsingular. Therefore the equation has a unique solution x
ˆ, yˆ.
(a) Show that x
ˆ is the solution of the optimization problem
minimize
subject to
kDx − D−1 bk2
AT x = c.
(b) Show that yˆ is the solution of the optimization problem
minimize
kD−1 (Ay − b)k2 + 2cT y.
(Hint: set the gradient of the cost function to zero.)
(c) Describe an efficient method, based on the QR factorization of D−1 A, for computing
x
ˆ and yˆ. Clearly state the different steps in your algorithm, the complexity of each
step (number of flops for large m, n), and the total complexity.
50
4
Exercises on least-norm problems
4.6 Let A be an m × n matrix, b an n-vector, and suppose the QR factorization
AT
b
=
Q1
Q2
R11
0
R12
R22
exists. The matrix Q1 has size n × m, Q2 has size n × 1, R11 has size m × m, R12 is m × 1,
and R22 is a scalar. Show that x
ˆ = Q2 R22 solves the optimization problem
minimize
subject to
kx − bk2
Ax = 0.
4.7 Suppose A is a left invertible matrix of size m×n. Let x
ˆ be the solution of the optimization
problem
minimize kAx − bk2 + 2cT x
with b ∈ Rm and c ∈ Rn , and let yˆ be the solution of
minimize
subject to
ky − bk2
AT y = c.
(a) Show that yˆ = b − Aˆ
x. (Hint. To find an expression for x
kAx − bk2 + 2cT x equal to zero.)
(b) Describe an efficient method for calculating x
ˆ and yˆ using the QR factorization of A.
Clearly state the different steps in your algorithm and give a flop count, including
all terms that are quadratic (m2 , mn, n2 ) or cubic (m3 , m2 n, mn2 , n3 ) in m and n.
4.8 Consider the underdetermined set of linear equations
Ax + By = b
where the p-vector b, the p × p matrix A, and the p × q matrix B are given. The variables
are the p-vector x and the q-vector y. We assume that A is nonsingular, and that B is
left invertible (which implies q ≤ p). The equations are underdetermined, so there are
infinitely many solutions. For example, we can pick any y, and solve the set of linear
equations Ax = b − By to find x.
Below we define four solutions that minimize some measure of the magnitude of x, or y,
or both. For each of these solutions, describe the factorizations (QR, Cholesky, or LU)
that you would use to calculate x and y. Clearly specify the matrices that you factor,
and the type of factorization. If you know several methods, give the most efficient one.
(a) The solution x, y with the smallest value of kxk2 + kyk2
(b) The solution x, y with the smallest value of kxk2 + 2kyk2 .
(c) The solution x, y with the smallest value of kyk2 .
(d) The solution x, y with the smallest value of kxk2 .
```
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| 4.03125
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CC-MAIN-2018-47
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https://www.numbersaplenty.com/812
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|
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| 706,859,470
| 3,511
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Search a number
812 = 22729
BaseRepresentation
bin1100101100
31010002
430230
511222
63432
72240
oct1454
91102
10812
11679
12578
134a6
14420
15392
hex32c
812 has 12 divisors (see below), whose sum is σ = 1680. Its totient is φ = 336.
The previous prime is 811. The next prime is 821. The reversal of 812 is 218.
812 = T3 + T4 + ... + T16.
812 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
812 is an esthetic number in base 6, because in such base its adjacent digits differ by 1.
It is an alternating number because its digits alternate between even and odd.
812 is a modest number, since divided by 12 gives 8 as remainder.
It is a plaindrome in base 5, base 11 and base 12.
It is a nialpdrome in base 14.
It is a zygodrome in base 5.
It is a junction number, because it is equal to n+sod(n) for n = 793 and 802.
It is not an unprimeable number, because it can be changed into a prime (811) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (5) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 14 + ... + 42.
It is an arithmetic number, because the mean of its divisors is an integer number (140).
It is a pronic number, being equal to 28×29.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 812, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (840).
812 is a primitive abundant number, since it is smaller than the sum of its proper divisors, none of which is abundant.
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
812 is a wasteful number, since it uses less digits than its factorization.
812 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 40 (or 38 counting only the distinct ones).
The product of its digits is 16, while the sum is 11.
The square root of 812 is about 28.4956136976. The cubic root of 812 is about 9.3293633910.
Adding to 812 its product of digits (16), we get a palindrome (828).
The spelling of 812 in words is "eight hundred twelve", and thus it is an aban number and an oban number.
Divisors: 1 2 4 7 14 28 29 58 116 203 406 812
| 668
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|
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https://www.mathconverse.com/en/Definitions/Perimeter/
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Home / All Definitions / Geometry / Perimeter Definition
# Perimeter Definition
A perimeter is a path that encompasses or surrounds a two-dimensional shape. The term perimeter refers either to the curve constituting the boundary of a lamina or else to the length of this boundary. The term may be used either for the path, or its length in one dimension. It can be thought of as the length of the outline of a shape. For a polygon, the perimeter is the sum of the lengths of the sides. The perimeter of a circle or ellipse is called its circumference although that term is used by some authors to refer to the perimeter of an arbitrary curved geometric figure.
## Overview
Calculating the perimeter has several practical applications. A calculated perimeter is the length of fence required to surround a yard or garden. The perimeter of a wheel/circle (its circumference) describes how far it will roll in one revolution. Similarly, the amount of string wound around a spool is related to the spool's perimeter. If the length of the string was exact, it would equal the perimeter.
## Perimeter & Area Formulas:
Volume formulas for various figures including lateral surface area, area of base(s), and total surface area:
Geometric Shape Perimeter Formula Area Formula Symbols Rectangle P = 2l + 2w A = lw l = length, w = width. Square P = 4a A = a2 a = length of a side. Triangle P = a + b + c A = 1⁄2bh a, b, c = side lengths of the triangle. b = base of triangle, h = height of triangle. Parallelogram P = 2b + 2h A = bh b = base, h = height. Trapezoid P = a + b + c + d A = 1⁄2(a + b)h a, b, c, d = sides of trapezoid. a, b = two bases, h = vertical height. Ellipse P = 2π√ (a2⁄2 + b2⁄2) A = πab a = radius of the major axis, b = radius of the minor axis. Kite P = 2a + 2b A = 1⁄2d1d2 a = length of the first pair of equal sides, b = length of the second pair of equal sides. d1 = long diagonal of kite, d2 = short diagonal of kite. Rhombus P = 4a A = 1⁄2d1d2 a = length of a side. d1 = length of a diagonal, d2 = length of the other diagonal. Pentagon P = 5a A = 5⁄2sa a = length of a side. s = side of the pentagon, a = apothem length. Hexagon P = 6a A = 3√ 3 ⁄2a2 a = length of a side. Heptagon P = 7a A = pa⁄2 a = length of a side. p = perimeter value, a = apothem length. Octagon P = 8a A = 2a2(1 + √ 2 ) a = length of a side.
### Sources
“Perimeter.” Wikipedia, Wikimedia Foundation, 31 Mar. 2020, en.wikipedia.org/wiki/Perimeter.
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| 889
| 3,245
|
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| 4.3125
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Welcome! Check out our free B-School Guides to learn how you compare with other applicants.
## exponents.....
This topic has 1 expert reply and 2 member replies
vaswani.sharan Just gettin' started!
Joined
10 Apr 2012
Posted:
3 messages
exponents..... Sun Apr 15, 2012 11:53 am
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
Hey, just getting my math feet back under me, I am not so bad at exponents, but this question is bugging me. Not sure how I could quickly solve it.
can some one show me a fast way?
(5^21)(4^11)=2x10^n
n=21
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shubham_k Just gettin' started!
Joined
08 Apr 2012
Posted:
27 messages
Thanked:
20 times
Sun Apr 15, 2012 12:00 pm
We see "5" and "4" on the left side and "10" on the right side.. somehow we want to make the bases the same.
So, let's start by turning the "4" into "2^2":
(5^21)((2^2)^11) = 2*10^n
(5^21)(2^22) = 2*10^n
Now, we want to be able to simplify the left side, but we can only do so if the exponents are equal, so:
(5^21)(2^21)(2^1) = 2*10^n
if we divide both sides by 2, we get:
(5^21)(2^21) = 10^n
and since the exponents are equal on the left side, we can say:
(5*2)^21 = 10^21
and finally
10^21 = 10^n
so 21 = n.
Not a difficult one i guess.....
Thanked by: vaswani.sharan
Shalabh's Quants Really wants to Beat The GMAT!
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134 messages
Followed by:
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Sun Apr 15, 2012 1:10 pm
vaswani.sharan wrote:
Hey, just getting my math feet back under me, I am not so bad at exponents, but this question is bugging me. Not sure how I could quickly solve it.
can some one show me a fast way?
(5^21)(4^11)=2x10^n
n=21
Lets break it into prime factors...
(5^21)(4^11)=2x10^n
5^21*(2^2)^11=2*(2.5)^n
5^21*2^22 = 2*2^n.5^n
5^21*2^22 = 2^(n+1).5^n
Either equate exponents of 2 or 5. Both gives n = 21.
_________________
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e-GMAT Instructor
### GMAT/MBA Expert
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GMAT Score:
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Sun Apr 15, 2012 6:21 pm
vaswani.sharan wrote:
Hey, just getting my math feet back under me, I am not so bad at exponents, but this question is bugging me. Not sure how I could quickly solve it.
can some one show me a fast way?
(5^21)(4^11)=2x10^n
n=21
(5^21)(4^11) = 2 x 10^n
(5^21)(2^2)^11 = 2 x (2 * 5)^n
(5^21)(2^22) = 2 x 2^n * 5^n
(5^21)(2^22) = 2^(n + 1) * 5^n
Now since the bases are the same, so exponents will also be the same on both sides.
So, n = 21
_________________
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GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)
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# Help with this DE problem
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Jan 31st 2010, 03:22 PM
whitepenguin
Help with this DE problem
Hi
I got a problem in DE too
I checked all the possible methods that I've learned so far includes Linear, Homogeneous,Bernoulli, and exact equation however none of them worked for me
$y' = \frac{y}{3x - y^2}$
Please give me just a little hint on this.
Thank you
• Jan 31st 2010, 04:50 PM
Krizalid
Put $y=\sqrt xt$ and that'd turn the ODE into a separable one.
• Jan 31st 2010, 04:56 PM
whitepenguin
Hi, can you tell me how you know to put it that way? Is there any general method for this kind of problem?
Thank you
• Jan 31st 2010, 05:29 PM
Krizalid
mm, i didn't know, i just made that substitution and it worked out! :)
• Jan 31st 2010, 05:48 PM
whitepenguin
So I guess I'm pretty dumb, coz I've been thinking about that since yesterday...(Crying)
• Jan 31st 2010, 05:49 PM
Krizalid
no, don't say that.
apply the substitution, what do you get?
• Jan 31st 2010, 06:43 PM
whitepenguin
Yes, I was be able to solve the problem using your substitution.
so.. the seprable form is this...
$\frac {6 - 2v^2}{v-3+v^2}dv = \frac {dx}{x}$
Thank you.
BTW.
I got another I.V.P where I found $c = \sqrt4$ , I end up with c= +-2
which one should I choose?
• Jan 31st 2010, 06:45 PM
Krizalid
$\sqrt4=2,$ and not $-2.$
did you stop to think how a positive function could deliver a negative value?
• Jan 31st 2010, 06:51 PM
whitepenguin
Quote:
Originally Posted by Krizalid
did you stop to think how a positive function could deliver a negative value?
Im not sure what you're asking
an you give me a little explain? ,why is it not negative?
I found the SOLN , and plug back to orginal problem, with both case c=+-2
,looks like both satify the problem.....
I dont really understand...
And what about uniqueness ... ???? I though they said IVP is unique soln, but... I found 2 C's....(Thinking)
• Feb 1st 2010, 09:30 PM
snaes
general solution
Hi, i read over this post i think i know how to find a "general solution" for this probelm. Its actually a specific case for this problem, but it'll help you find an integrating factor in the form $x^ay^b$. where "a" and "b" are constants.
Get equation in exact form:
$(M)dx+(N)dy=0$
$(-y)dx+(3x-y)dy=0$
Make this fraction =1
$\dfrac{M_y-N_x}{N\dfrac{a}{x}-M\dfrac{b}{y}}=1$
By making "a" and "b" appropriot values this should make the fraction 1. Thereby giving you the values of "a" and "b"to fill in the integrating factor $x^ay^b$.
This should make the equation turn into an "exact differential equation"
Hope this helps!
*Note: I havent tried this for your specific problem, but this is the solution my professor has taught us and has worked for me on other problems.
• Feb 1st 2010, 09:32 PM
Krizalid
that's funny, you got the same result as me and i've never studied well exact equations. :D
• Feb 2nd 2010, 05:00 AM
Calculus26
For another approach
http://www.mathhelpforum.com/math-he...c06ef73a-1.gif
rewrite
dx/dy = (3x-y^2)/y
dx/dy -(3/y)x = -y
use integrating factor 1/y^3
x/y^3 = 1/y + c
x = y^2 + cy^3
• Feb 2nd 2010, 10:36 AM
whitepenguin
Quote:
Originally Posted by snaes
Hi, i read over this post i think i know how to find a "general solution" for this probelm. Its actually a specific case for this problem, but it'll help you find an integrating factor in the form $x^ay^b$. where "a" and "b" are constants.
Get equation in exact form:
$(M)dx+(N)dy=0$
$(-y)dx+(3x-y)dy=0$
Make this fraction =1
$\dfrac{M_y-N_x}{N\dfrac{a}{x}-M\dfrac{b}{y}}=1$
By making "a" and "b" appropriot values this should make the fraction 1. Thereby giving you the values of "a" and "b"to fill in the integrating factor $x^ay^b$.
This should make the equation turn into an "exact differential equation"
Hope this helps!
*Note: I havent tried this for your specific problem, but this is the solution my professor has taught us and has worked for me on other problems.
Hi, can you put it in a little detail? How do you get $3x-y^2$ to $3x - y$
• Feb 2nd 2010, 10:44 AM
whitepenguin
Quote:
Originally Posted by Calculus26
For another approach
http://www.mathhelpforum.com/math-he...c06ef73a-1.gif
rewrite
dx/dy = (3x-y^2)/y
dx/dy -(3/y)x = -y
use integrating factor 1/y^3
x/y^3 = 1/y + c
x = y^2 + cy^3
So you make this look like a linear eqn?
I thought it should be dy/dx?
Im confused tho, coz y is a function has x .... or may be i'm wrong
Can you give me an explain
Thanks
• Feb 2nd 2010, 02:13 PM
Calculus26
I just inverted the equation to get x as a fn of y to make the solution a simple linear ODE.
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# Thread: Finding a subgroup
1. ## Finding a subgroup
Hi,
started a topic on subgroups this week and I can't seem to find any examples to help me with a couple of questions.
The first one is 'Find all the subgroups of Z30' I'm not to sure if this is just the factors of 30 am I right?
The second part of the question I'm stuck on is not necessarily about subgroups (or maybe it is, not to sure!) but it asks to determine the order of every element in (i) D8 and (ii). D12 .
Would appreciate very much if anyone could help me, thanks!
2. ## Re: Finding a subgroup
some things you should be aware of (and might try to prove, if you feel ambitious):
1) every subgroup of a cyclic group is cyclic.
2) if G is cyclic of order n, and k is a divisor of n, G has EXACTLY one subgroup of order k.
in particular, (2) indicates your hunch is correct. the subgroups of Z30 are:
<0> = {0}
<1> = Z30 (and also equals <7>, <11>, <13>, <17>, <19>, <23>, and <29>)
<2> = {0,2,4,6,8,10,12,14,16,18,20,22,24,26,28} (this is also equal to <4>, <8>, <14>, <16>, <22>, 26>, and <28>)
<3> = {0,3,6,9,12,15,18,21,24,27} (this is also equal to <9>, <21>, and <27>)
<5> = {0,5,10,15,20,25} (this is also equal to <25>)
<6> = {0,6,12,18,24} (this is also equal to <12>, <18>, and <24>)
<10> = {0,10,20} (this also equals <20>)
<15> = {0,15}
we can prove some results about D2n, the dihedral group of order 2n. note that this can be presented as <r,s> where:
rn = e
s2 = e
sr = r-1s.
you can prove by induction on |k| that:
srk = r-ks for all integers k.
this means all elements of the form rks have order 2:
(rks)2 = (rks)(rks) = rk(srk)s = rk(r-ks)s = s2 = e.
all the other elements are in <r>, and can be analyzed like elements of any cyclic group.
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# physicslab7 - La b 7 Voltage Current and Resistance I Staci...
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La b 7: Voltage, Current, and Resistance I Staci Williams Nicole Harty, Kevin Schesing, Caitling Kubota
Section 015
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Theory: Voltage Two points have a voltage between them if work is done by the electric field as a charge is moved from one of the points to another. Voltage is the work done on a unit positive charge by the electric field as the unit positive charge is moved from one of the two points to the other point, with units of joules per coulomb, or volt. Wires have resistance and are carrying currents, and the wire is not at equipotential. The earth is not a good conductor. Voltage can be supplied by batteries, power supplies, and signal generators. Ideal voltage sources maintains the prescribed voltage across the terminals no matter how much current is supplied, but these do not exist. Voltage can be seen as a function of time and the voltage across the terminals of a signal generator may repeat itself in a length of time called the period. The voltage for one period is called one oscillation or one cycle. The inverse of the period is frequency, which is the number of periods or oscillations per second. The shape of the voltage vs. time graph is called the waveform, which may vary in shapes such as sine, square, triangular, and ramp. Voltage can be
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## Wednesday, June 26, 2013
### Problem 893: Triangle, Circumcircle, Incenter, Circle, Tangent, Collinear Points
Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the dynamic geometry demonstration of problem 892.
1. Consider the line DE and try to prove that I lies on DE.
Let P be a point on CE on the other side of E.
Let ∠BAC be 2A and ∠DEC = X.
∠IBC = (1/2)∠ABC = (1/2)∠DCP = (1/2)*2*∠DEC = ∠DEC, so I,B,E,C concylic.
Therefore ∠BIE = ∠BCE = 2A.
Now consider ∠BDE = (1/2)∠BCE = (1/2)*2A = A.
Hence,
∠DBI = ∠DBC - ∠IBC
= ∠BDC - X
= ∠BDE + ∠EDC - X
= A + ∠DEC - X
= A + X - X
= A
Combining all the results, we have ∠DBI = A, ∠BDE = A, and ∠BIE = 2A
By the idea of exterior angle of triangle, D,I,E is collinear.
1. To William
Your statement " ∠IBC = (1/2)∠ABC = (1/2)∠DCP = (1/2)*2*∠DEC = ∠DEC, so I,B,E,C concylic." may not correct.
∠IBC =∠DEC doesn't imply I,B,E,C concylic .I,B,E,C concylic only if D, I, E are collinear .
Peter Tran
2. https://goo.gl/photos/Kmxu4USsFCZZEZXG8
Connect IC and IB
Triangle DIC congruent to BIC ..( case SAS)
So ∠ (IDC)= ∠ (IBC)= ∠ (B/2)
Triangle DCE is isosceles and ∠ (DCE)= ∠ (DCB)+ ∠ (BCE)= ∠ ( C) +∠ (A)= 180- ∠ (B)
So ∠ (CDE)=90-1/2. ∠ (DCE)= 90-(90-∠ (B)/2))= ∠ (B/2)
So ∠ (CDE)= ∠ (IDC) => D, I, E are collinear
3. Problem 893
Let DE intersects the bisector the angle ACB at point C.Is <BEK=<BED=<BAD/2=<KCB so KBEC is cyclic.But <KBC=<KEC=<DEC=90-<ECD/2=90-(<ECB+<BCA)/2=90-(<CAB+<BCA)/2=<ABC/2=<IBC.So the points K,I conside .Therefore the points D,I and E are
collinear.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
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##### Trigonometry Homework Help
Mathematics Tutor: None Selected Time limit: 1 Day
A right triangle has one leg that measures 8 feet and another that measures 15 feet . What are the measures of the acute angles?
Oct 16th, 2015
Thank you for the opportunity to help you with your question!
=8/15
Angle = TAN inverse 8/15
=28.07
90+28.07+x=180
x=180-118.7
=61.93
therefore
angles
=28.07
=61.93
Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 16th, 2015
...
Oct 16th, 2015
...
Oct 16th, 2015
May 30th, 2017
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Course Outline
• segmentGetting Started (Don't Skip This Part)
• segmentStatistics and Data Science: A Modeling Approach
• segmentPART I: EXPLORING VARIATION
• segmentChapter 1 - Welcome to Statistics: A Modeling Approach
• segmentChapter 2 - Understanding Data
• segmentChapter 3 - Examining Distributions
• segmentChapter 4 - Explaining Variation
• segmentPART II: MODELING VARIATION
• segmentChapter 5 - A Simple Model
• segmentChapter 6 - Quantifying Error
• segmentChapter 7 - Adding an Explanatory Variable to the Model
• segmentChapter 8 - Models with a Quantitative Explanatory Variable
• segmentPART III: EVALUATING MODELS
• segmentChapter 9 - The Logic of Inference
• segmentChapter 10 - Model Comparison with F
• segmentChapter 11 - Parameter Estimation and Confidence Intervals
• segmentPART IV: MULTIVARIATE MODELS
• segmentChapter 12 - Introduction to Multivariate Models
• segmentChapter 13 - Multivariate Model Comparisons
• segmentFinishing Up (Don't Skip This Part!)
• segmentResources
list Statistics and Data Science: A Modeling Approach
Book
• College / Advanced Statistics and Data Science (ABCD)
• College / Statistics and Data Science (ABC)
• High School / Advanced Statistics and Data Science I (ABC)
• High School / Statistics and Data Science I (AB)
• High School / Statistics and Data Science II (XCD)
12.2 Visualizing Price = Home Size + Neighborhood
Let’s explore this idea with some visualizations. We will start with a graph of the home size model, plotting PriceK by HomeSizeK, with this code: gf_point(PriceK ~ HomeSizeK, data = Smallville). We will then explore some ways we could visualize the effect of Neighborhood above and beyond that of HomeSizeK.
Using Facet Grids
Here’s a scatter plot of PriceK by HomeSizeK for the 32 homes in Smallville.
One way to integrate Neighborhood into the same visualization is to make a grid of scatter plots, each one representing a different neighborhood. We can do this by chaining on gf_facet_grid(Neighborhood ~ .) on top of the scatter plot.
Because we put Neighborhood before the tilde (Neighborhood ~ .) the two graphs will be stacked vertically (i.e., along the y-axis). To put the graphs side-by-side (i.e., in a grid along the x-axis), we would put the variable after the tilde: . ~ Neighborhood. Notice that in R, as in GLM notation, we usually follow the form Y ~ X.
In the code block below, try putting the two scatter plots, one for each Neighborhood, side by side in a horizontal grid.
require(coursekata) # delete when coursekata-r updated Smallville <- read.csv("https://docs.google.com/spreadsheets/d/e/2PACX-1vTUey0jLO87REoQRRGJeG43iN1lkds_lmcnke1fuvS7BTb62jLucJ4WeIt7RW4mfRpk8n5iYvNmgf5l/pub?gid=1024959265&single=true&output=csv") Smallville$Neighborhood <- factor(Smallville$Neighborhood) Smallville$HasFireplace <- factor(Smallville$HasFireplace) # Make a horizontal grid of scatter plots using Neighborhood gf_point(PriceK~ HomeSizeK, data = Smallville) gf_point(PriceK~ HomeSizeK, data = Smallville) %>% gf_facet_grid(. ~ Neighborhood) # temporary SCT ex() %>% check_error()
CK Code: D1_Code_Visualizing_01
Based on these plots, you can see that knowing both neighborhood and home size would improve your predictions. One way to see this is to look, within each neighborhood, at the prices of homes that are between 1000 and 1500 square feet (i.e., HomeSizeK between 1.0 and 1.5). We have colored them differently in the faceted plot below. You can see that even for homes the same size, there still are higher prices in Downtown than in Eastside.
Using Color
Another approach to adding neighborhood to the scatter plot of PriceK by HomeSizeK is to assign different colors to points representing homes from the different neighborhoods. You can do this by adding color = ~Neighborhood to the scatter plot. (The ~ tilde tells R that Neighborhood is a variable.) Try it in the code block below.
require(coursekata) # delete when coursekata-r updated Smallville <- read.csv("https://docs.google.com/spreadsheets/d/e/2PACX-1vTUey0jLO87REoQRRGJeG43iN1lkds_lmcnke1fuvS7BTb62jLucJ4WeIt7RW4mfRpk8n5iYvNmgf5l/pub?gid=1024959265&single=true&output=csv") Smallville$Neighborhood <- factor(Smallville$Neighborhood) Smallville$HasFireplace <- factor(Smallville$HasFireplace) # Add in the color argument gf_point(PriceK ~ HomeSizeK, data = Smallville)gf_point(PriceK ~ HomeSizeK, data = Smallville, color = ~Neighborhood) # temporary SCT ex() %>% check_error()
CK Code: D1_Code_Visualizing_02
We used this code (also overlaying the HomeSizeK regression line on the scatter plot) to get the graph below.
HomeSizeK_model <- lm(PriceK ~ HomeSizeK, data = Smallville)
gf_point(PriceK ~ HomeSizeK, data = Smallville, color= ~ Neighborhood) %>%
gf_model(HomeSizeK_model, color = "black")
Adding the regression line makes it easier to see the error (or residuals) leftover from the HomeSizeK model. Notice that the teal dots (homes from Downtown) are mostly above the regression line (i.e., with positive residuals from the HomeSizeK model) while the purple dots (from Eastside) are mostly below the line (negative residuals).
This indicates that Downtown homes are generally more expensive than what the home size model would predict, while Eastside homes are less expensive. This pattern is a clue that tells us that adding Neighborhood into the HomeSizeK model will explain additional variation in PriceK above and beyond that explained by just the home size model alone.
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# chemistry
posted by on .
Balance this equation by following equations of the REDOX Method.
Cu + HNO3 --->Cu(NO3)2 + NO + H2O
• chemistry - ,
Cu changes from zero on the left to +2 on the right.
N changes from +5 on the left to + 2 on the right (for NO--don't concern yourself yet with the nitrate on the right since it doesn't change.
So Cu to Cu^2+ is a loss of 2e and N is a gain of 3e. Multiply the Cu half by 3 and the HNO3/NO half by 2.
That gives us
3Cu + 2HNO3 ==> 3Cu^2+ + 2NO + 4H2O
That balances everything but H and the charge. Since we have a Cu^2+ here it will use nitrate to balance it and form Cu(NO3)2 so we add six more HNO3 on the left to form 3Cu(NO3)2 which makes 6 more H^+ added on the left at the same time.
3Cu + 8HNO3 ==> 3Cu(NO3)2 + 2NO + 4H2O
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## How to Find a Critical Value for a Left-Tailed Test
Finding a critical value for a left-tailed test will have a familiar feel to it if you are used to looking up areas in a z-table (which should have been covered by this point in any elementary statistics class). This article shows you how to find α=.012 for a left-tailed test. If you like our easy to follow explanations of statistics, check out our easy to follow book, which has hundreds more examples, just like this one.
Area under a normal distribution curve (alpha is shaded yellow)
Step 1: Draw a diagram, like the one above. Shade in the area in the left tail.
This area represents α.
Step 2:Subtract alpha (α) from 0.5.
0.5-0.012=0.488.
Step 3: Find the result from step 2 in the center part of the z-table.
The closest area to 0.488 is at z=2.26. If you can’t find the exact area, just find the closest number and read the z value for that number.
Step 4:Add a negative sign to Step 3.
-2.26.
That’s it! Like the explanation? Check out our statistics how-to book, with a how-to for every elementary statistics problem type.
Feel like Cheating at Statistics? This is the Statistics Handbook that your professor doesn't want you to see. So easy, it's Practically Cheating. Find out more »
## 28 Responses to “How to Find a Critical Value for a Left-Tailed Test”
1. ### Angie Widdows said:
Oct 15, 09 at 9:40 am
This was also a helpful example. One problem that I seem to have with these types of problems is figuring out what value on the table is closest to the number we are to look for. I tend to get in a hurry and pick the wrong number. That is something I have been trying to work on.
2. ### Jennifer Thomas said:
Oct 17, 09 at 7:21 am
So if it’s in between, we could use 3.03 or 3.04 and both would be considered correct?
3. ### Stephanie said:
Oct 17, 09 at 8:45 am
Both would be considered correct in my book. Usually Mathzone will pick problems that aren’t ambiguous like that but if you are marked wrong because of that type of error–let me know and I’ll fix it.
Stephanie
4. ### How to Do Everything Statistics » How to Support or Reject a Null Hypothesis said:
Oct 17, 09 at 10:01 am
[...] critical value. You should have already covered that topic by this point, but here is a how-to on finding the critical value for a left-tailed test as a reminder. You may need to find a right-tailed or two-tailed [...]
5. ### Lisa Barcomb said:
Oct 18, 09 at 10:17 am
yeah at first I was reading this and did not really understand it but then all of a sudden it came to me and then once I saw the bell and figured out the critical area then it was really simple. I noticed that we had some of these problems on our homework and they started to become easy
6. ### Vanessa DuBarry said:
Oct 27, 09 at 11:00 am
This also helped me because I didnt know we had to put a negative sign fot the problems with left tailed which I should of caught because its the left side and its always negative, so it helps to have examples like these to help with those mistakes.
7. ### Donna Allen said:
Oct 27, 09 at 2:13 pm
I have had a few instances where mathzone has marked my question incorrect because of variation in the number that could be used from the z-table. This is good to know for future reference.
8. ### Vanessa DuBarry said:
Oct 29, 09 at 11:19 am
I dont know why I dont understand getting the critical value. in this example it says to subtract 0.5-0.012 but here you said the answer is 0.4988 but when I subtract it gives me 0.488 so I dont know what IM I doing wrong?.. like for example I have a=0.05 so I subtract that from 0.5 and it gives me 0.45 but that its wrong because the answer is 0.4505 how? please help what Im I doing wrong
9. ### Stephanie said:
Oct 29, 09 at 1:29 pm
You just caught a type. 0.5-0.012 is indeed 0.488.
You are looking for the closest value in the z-table. It doesn’t have every number. So if you are looking for .45, the closest value may be .4505. That isn’t an “exact” answer, but it will work for your calculations. The TI-83 will give you a more exact answer, but even that isn’t perfect :)
Stephanie
10. ### Donna Allen said:
Nov 02, 09 at 3:03 pm
I understand how to arrive at the answer. I’m just not sure how to tell if a word problem is right-tailed or left-tailed or two-tailed.
11. ### Vanessa said:
Nov 03, 09 at 6:08 pm
How do I determine if it the test is two tailed left tailed or right tailed?
12. ### Stephanie said:
Nov 05, 09 at 4:55 am
Vanessa,
This should help!
13. ### How to Do Everything Statistics » How to Support or Reject a Null Hypothesis (Using a P-Value) said:
Nov 05, 09 at 9:26 am
[...] critical value. You should have already covered that topic by this point, but here is a how-to on finding the critical value for a left-tailed test as a reminder. You may need to find a right-tailed or two-tailed [...]
14. ### Catherine Flanagan said:
Nov 08, 09 at 8:11 am
This website is a life-saver! I spent about ten minutes trying to understand these kinds of problems. This website explains it so easily! Thank you.
15. ### cherr said:
Nov 12, 09 at 9:49 am
why do you have to substract it from 0.5?
16. ### Stephanie said:
Nov 12, 09 at 11:24 am
The are under the curve is 1. So the area in the left-hand side of the curve is 0.5. You’re looking for the area 0.5-alpha.
Stephanie
17. ### Arturo said:
Nov 22, 09 at 9:10 am
Hi, this site seems to be very helpful to me too. I had the same problem as Vanessa dubary, I was confuse, but I think I got it all now. In my book, I have astatement to find a left tailed test with 95% confidence. that is, alpha = 1-.95=0.05. Right?? but now I just look for 0.05 in the midle of the table, I dont have to substract again 0.5-0.05 in this case. am I right??. then the answer would be -1.64
18. ### Arturo said:
Nov 22, 09 at 9:10 am
Can we make interpolations in the Z table to find a number that is not in the table?
19. ### Stephanie said:
Nov 24, 09 at 2:41 pm
Arturo,
You do have to subtract. 0.5-0.05, before you look it up in the table…
Stephanie
20. ### Stephanie said:
Dec 01, 09 at 2:40 am
Absolutely.
21. ### Statistics How To» Blog Archive » How to Support or Reject a Null Hypothesis (for a Proportion) said:
Jan 13, 10 at 12:08 pm
[...] 2: Find the critical value. Here is a how-to on finding the critical value for a left-tailed test as a reminder. You may need to find a right-tailed or two-tailed [...]
22. ### crystal lydick said:
Mar 27, 10 at 11:41 am
This page really helped explain how to find a critical value for a Left-tailed test. I couldn’t figure out how to do these problems until I found this on the statistics how to forum. If it wasn’t for this forum, I would be completely lost. At least with this forum and the practically cheating book we have something to help guide us in the right direction.
23. ### Cheryl said:
Oct 25, 10 at 11:31 am
How do I find the critical value for α=.025 for a left-tailed test? Is there a zscore table on this site?
24. ### Sam said:
Apr 14, 12 at 10:27 am
YOU SAVED MY LIFE!!
25. ### Andale said:
Apr 14, 12 at 5:31 pm
:)
26. ### Meena said:
May 22, 12 at 2:55 am
That was a very simple and easy to understand example. Thank you :)
27. ### Wanda said:
Jul 22, 12 at 8:55 am
-2.26 takes me to .0119 in the z table ??? what am I doing wrong?
28. ### Andale said:
Jul 22, 12 at 9:47 am
Are you using this z-table?
Stephanie
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# Basic Time Dilation Question
1. Feb 9, 2013
### i.h87
Forgive me if this has already been asked as I'm trying to understand this, but I can't seem to find the right info..
Basically, what I want to know is if it is possible to chase something so fast that you wont catch it due to time dilation. If an object is travelling at a low sublight speed and another object is following the same path at a near light speed can time dilation make it so that the faster object never catches up to the slower object, or is this nullified because traveling on the same path gives them the same frame of reference?
2. Feb 9, 2013
### Staff: Mentor
No. As long as both objects are traveling at less than the speed of light, the faster object will always catch up to the slower one eventually.
I'm not sure what you mean by this, but the reason the faster object will catch up is that, if you look at things in the frame in which that object is at rest, the slower object is moving *towards* the faster object in that frame.
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# 30 Day look back
✭✭✭
Hello,
I am using the formula below to count all instances of rows that contain "ADCVD" in my "2022 Archive RFI Reason" with my "2022 Archive Created Date Range".
My problem is that my Jan 21 data is located in in my "2021 Archive Created Date Range" sheet.
How can I modify this formula to look back 30 days in both archives?
=IF(MONTH(TODAY()) = 1, COUNTIFS({2022 Archive Created Date Range}, IFERROR(MONTH(@cell), 0) = 12, {2022 Archive Created Date Range}, IFERROR(YEAR(@cell), 0) = YEAR(TODAY()) - 1), COUNTIFS({2022 Archive Created Date Range}, IFERROR(MONTH(@cell), 0) = MONTH(TODAY()) - 1, {2022 Archive Created Date Range}, IFERROR(YEAR(@cell), 0) = YEAR(TODAY()), {2022 Archive RFI Reason}, CONTAINS("ADCVD", @cell)))
Tags:
• ✭✭✭✭✭✭
Sorry. I missed grabbing that bit.
=COUNTIFS({2022 Archive Created Date Range}, AND(IFERROR(MONTH(@cell), 0) = IF(MONTH(TODAY()) = 1, 12, MONTH(TODAY()) - 1), IFERROR(YEAR(@cell), 0) = YEAR(TODAY()) - IF(MONTH(TODAY()) = 1, 1, 0)), {2022 Archive RFI Reason}, CONTAINS("ADCVD", @cell)) + COUNTIFS({2021 Archive Created Date Range}, AND(IFERROR(MONTH(@cell), 0) = IF(MONTH(TODAY()) = 1, 12, MONTH(TODAY()) - 1), IFERROR(YEAR(@cell), 0) = YEAR(TODAY()) - IF(MONTH(TODAY()) = 1, 1, 0)), {2021 Archive RFI Reason}, CONTAINS("ADCVD", @cell))
• ✭✭✭✭✭✭
Hi @JP Pedicino ,
I may be oversimplifying it, but could it be as simple as copying the current formula, changing the range name to the 2021 one, and adding it to the 2022 formula? You may have to tweak the month/year calculations a bit, too. So (edited items in bold below):
=IF(MONTH(TODAY()) = 1, COUNTIFS({2022 Archive Created Date Range}, IFERROR(MONTH(@cell), 0) = 12, {2022 Archive Created Date Range}, IFERROR(YEAR(@cell), 0) = YEAR(TODAY()) - 1), COUNTIFS({2022 Archive Created Date Range}, IFERROR(MONTH(@cell), 0) = MONTH(TODAY()) - 1, {2022 Archive Created Date Range}, IFERROR(YEAR(@cell), 0) = YEAR(TODAY()), {2022 Archive RFI Reason}, CONTAINS("ADCVD", @cell))) + IF(MONTH(TODAY()) = 1, COUNTIFS({2021 Archive Created Date Range}, IFERROR(MONTH(@cell), 0) = 12, {2021 Archive Created Date Range}, IFERROR(YEAR(@cell), 0) = YEAR(TODAY()) - 1), COUNTIFS({2021 Archive Created Date Range}, IFERROR(MONTH(@cell), 0) = MONTH(TODAY()) + 11, {2021 Archive Created Date Range}, IFERROR(YEAR(@cell), 0) = YEAR(TODAY() - 1), {2021 Archive RFI Reason}, CONTAINS("ADCVD", @cell)))
Let me know if it works.
• ✭✭✭
Thank you, but my issue is that is has to look at both 2021 and 2022 sheets in the formula. If it finds nothing in 2021 then is needs to go to 2022.
• ✭✭✭✭✭✭
My first suggestion would be to shorten your existing formula up a bit by putting the IF statement inside of the COUNTIFS instead of using two COUNTIFS. From there you would replicate the formula to point at the second sheet and then add them together.
Simplified:
=COUNTIFS({2022 Archive Created Date Range}, AND(IFERROR(MONTH(@cell), 0) = IF(MONTH(TODAY()) = 1, 12, MONTH(TODAY()) - 1), IFERROR(YEAR(@cell), 0) = YEAR(TODAY()) - IF(MONTH(TODAY()) = 1, 1, 0)))
For both sheets:
=COUNTIFS({2022 Archive Created Date Range}, AND(IFERROR(MONTH(@cell), 0) = IF(MONTH(TODAY()) = 1, 12, MONTH(TODAY()) - 1), IFERROR(YEAR(@cell), 0) = YEAR(TODAY()) - IF(MONTH(TODAY()) = 1, 1, 0))) + COUNTIFS({2021 Archive Created Date Range}, AND(IFERROR(MONTH(@cell), 0) = IF(MONTH(TODAY()) = 1, 12, MONTH(TODAY()) - 1), IFERROR(YEAR(@cell), 0) = YEAR(TODAY()) - IF(MONTH(TODAY()) = 1, 1, 0)))
The idea behind this is that the COUNTIFS would generate a zero on one sheet and a count from the second since the data for the month you are wanting to count shouldn't be in both sheets.
• ✭✭✭
Thank you Paul, but how about the CONTAINS portion of the formula? What I do with it since it has to look within both 2021 and 2022?
• ✭✭✭✭✭✭
Sorry. I missed grabbing that bit.
=COUNTIFS({2022 Archive Created Date Range}, AND(IFERROR(MONTH(@cell), 0) = IF(MONTH(TODAY()) = 1, 12, MONTH(TODAY()) - 1), IFERROR(YEAR(@cell), 0) = YEAR(TODAY()) - IF(MONTH(TODAY()) = 1, 1, 0)), {2022 Archive RFI Reason}, CONTAINS("ADCVD", @cell)) + COUNTIFS({2021 Archive Created Date Range}, AND(IFERROR(MONTH(@cell), 0) = IF(MONTH(TODAY()) = 1, 12, MONTH(TODAY()) - 1), IFERROR(YEAR(@cell), 0) = YEAR(TODAY()) - IF(MONTH(TODAY()) = 1, 1, 0)), {2021 Archive RFI Reason}, CONTAINS("ADCVD", @cell))
• ✭✭✭
Paul,
Thank you so much for your assistance!
You're absolutely correct, regarding your clean-up of the formula, it worked like a charm. I also really appreciate you taking the time to teach me a better way to approach the issue.
• ✭✭✭✭✭✭
Happy to help. 👍️
If you find yourself repeating the same range/criteria sets within a formula (or in the case of IF statements the same output), then I always suggest seeing if you can consolidate somehow.
In this case, the IF is shorter than the COUNTIFS, so we can save keystrokes by repeating the IF and only doing the COUNTIFS once. Saving keystrokes means shorter formulas which (usually) means more efficient formulas. More efficient formulas can make a huge impact on sheet performance when you start getting into larger and/or more complex sheets.
## Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template!
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# Circle Theorem
## Circle Theorem 1
The angle at the center is twice the angle at the circumference.
## Circle Theorem 2
The angle in a semi-cicle is 90°.
## Circle Theorem 3
Angles in the same segment are equal.
## Circle Theorem 5
The lengths of the two tangents from a point to a circle are equal.
## Circle Theorem 6
The angle between a tangent and a radius in a circle is 90°.
## Circle Theorem 7
Alternate segment theorem:
The angle (α) between the tangent and the chord at the point of contact (D) is equal to the angle (β) in the alternate segment*.
## Circle Theorem 8
Perpendicular from the centre bisects the chord:
Circle Theorems for Arcs and Chords:
• If two chords are congruent, then their corresponding arcs are congruent.
• If the diameter or radius is perpendicular to a chord, then it bisects the chord and its arc.
• In the same circle or congruent circles, two chords are congruent if and only if they are equidistant from the center.
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http://mathcentral.uregina.ca/QQ/database/QQ.09.99/buboltz1.html
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My name is michelle and I just need to convert 25 ft across a circle into how many square feet. Hi Michelle, The area of a circle is given by pi times the square of the radius. Your circle has a diameter of 25 ft so the radius is 25/2 = 12.5 ft. Thus the area is pi x12.52 sq. ft. Cheers, Penny Go to Math Central
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https://www.sololearn.com/Discuss/2860905/data-science-last-project-data-science-pandas-pandas-pandas
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Data Science Last Project: Data Science - Pandas Pandas Pandas | Sololearn: Learn to code for FREE!
0
# Data Science Last Project: Data Science - Pandas Pandas Pandas
My solution gives 4 correct out of 5 test cases, can someone help me out? Assume that there are two clusters among the given two-dimensional data points and two random points (0, 0), and (2, 2) are the initial cluster centroids. Calculate the euclidean distance between each data point and each of the centroid, assign each data point to its nearest centroid, then calculate the new centroid. If there's a tie, assign the data point to the cluster with centroid (0, 0). If none of the data points were assigned to the given centroid, return None.
8/19/2021 12:13:36 AM
Shivam Panchal
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'
# Search results
Found 1945 matches
Electromagnetic mass ( longitudinal mass) by Lorentz
Due to the self-induction effect, electrostatic energy behaves as having some sort of momentum and “apparent” electromagnetic mass, which can ... more
Q factor for a series resonant circuit (RL circuits)
It is defined as the peak energy stored in the circuit divided by the average energy dissipated in it per cycle at resonance; Q factor is directly ... more
Doppler Effect - for stationary observer
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Cycloid ( parametric equation Y-coordinate)
A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slippage. It is an example of a ... more
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# Convert Yottahertz to Cycles Per Day
### Kyle's Converter > Frequency > Yottahertz > Yottahertz to Cycles Per Day
Yottahertz (YHz) Cycles Per Day (cpd)* Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Cycles Per Day to Yottahertz
(or just enter a value in the "to" field)
#### Please share if you found this tool useful:
Unit Descriptions
1 Yottahertz:
1 Yottahertz is exactly one septillion Hertz. 1 YHz = 1 x 1024 Hz. 1 YHz = 1000000000000000000000000 Hz.
1 Cycle per Day:
A period of 1 day is equal to 1/86400 Hertz frequency. Period is the inverse of frequency: 1 Hz = 1/86400 cpd or approximately 0.000011574074074074 cpd.
Conversions Table
1 Yottahertz to Cycles Per Day = 8.64E+2870 Yottahertz to Cycles Per Day = 6.048E+30
2 Yottahertz to Cycles Per Day = 1.728E+2980 Yottahertz to Cycles Per Day = 6.912E+30
3 Yottahertz to Cycles Per Day = 2.592E+2990 Yottahertz to Cycles Per Day = 7.776E+30
4 Yottahertz to Cycles Per Day = 3.456E+29100 Yottahertz to Cycles Per Day = 8.64E+30
5 Yottahertz to Cycles Per Day = 4.32E+29200 Yottahertz to Cycles Per Day = 1.728E+31
6 Yottahertz to Cycles Per Day = 5.184E+29300 Yottahertz to Cycles Per Day = 2.592E+31
7 Yottahertz to Cycles Per Day = 6.048E+29400 Yottahertz to Cycles Per Day = 3.456E+31
8 Yottahertz to Cycles Per Day = 6.912E+29500 Yottahertz to Cycles Per Day = 4.32E+31
9 Yottahertz to Cycles Per Day = 7.776E+29600 Yottahertz to Cycles Per Day = 5.184E+31
10 Yottahertz to Cycles Per Day = 8.64E+29800 Yottahertz to Cycles Per Day = 6.912E+31
20 Yottahertz to Cycles Per Day = 1.728E+30900 Yottahertz to Cycles Per Day = 7.776E+31
30 Yottahertz to Cycles Per Day = 2.592E+301,000 Yottahertz to Cycles Per Day = 8.64E+31
40 Yottahertz to Cycles Per Day = 3.456E+3010,000 Yottahertz to Cycles Per Day = 8.64E+32
50 Yottahertz to Cycles Per Day = 4.32E+30100,000 Yottahertz to Cycles Per Day = 8.64E+33
60 Yottahertz to Cycles Per Day = 5.184E+301,000,000 Yottahertz to Cycles Per Day = 8.64E+34
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https://alperezescudero.wordpress.com/2009/07/07/first-results-for-the-dynamics-of-c-elegans/
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## First results for the dynamics of C elegans
Eigenvectors and eigenvalues MUST be complex. Still, real solution for the voltage is possible (each eigenvector/value has another one which is complex conjugate. If the modulus of the coefficient of both is the same, the final linear combination is real).
With no membrane conductance:
>> V0=zeros(279,1);
>> solucion=matrizsistema2solucion(M,V0);
>> plot(solucion.autovalores,’.’)
This figure matches with Figure 8 of Mitya’s paper (y axis is imaginary part)
We see that there are positive eigenvalues. This has two consequences:
– Diverging voltage.
– Apparently some computations are not performed correctly. I get non-negligible imaginary parts in the voltages.
I see that I need about 30 in the membrane conductance to get all eigenvalues negative:
>> solucion=matrizsistema2solucion(M,V0);
>> plot(solucion.autovalores,’.’)
FIRST EXAMPLE: Excitation of neuron 1.
>> V0=zeros(279,1);
>> V0(1)=1;
>> solucion=matrizsistema2solucion(M,V0);
>> V=solucion2voltajes(solucion,0:.001:.5);
>> plot(V(1,:))
>> plot(V(2,:))
And so on…
SECOND EXAMPLE: Random excitation
>> V0=rand(279,1);
>> solucion=matrizsistema2solucion(M,V0);
>> V=solucion2voltajes(solucion,0:.001:.5);
>> imagesc(V)
What are those ‘rough’ zones?
>> plot(V(149,:))
What the…?
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https://mapdldocs.pyansys.com/api/inline.html
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# Inline Functions#
These are wrapped versions of inline APDL functions that perform operations like finding the x-coordinate of a node given its number (`Query.nx`).
class ansys.mapdl.core.inline_functions.Query(mapdl)#
Class containing all the inline functions of APDL.
Most of the results of these methods are shortcuts for specific combinations of arguments supplied to `ansys.mapdl.core.Mapdl.get()`.
Currently implemented functions:
• `centrx(e)` - get the centroid x-coordinate of element e
• `centry(e)` - get the centroid y-coordinate of element e
• `centrz(e)` - get the centroid z-coordinate of element e
• `nx(n)` - get the x-coordinate of node n
• `ny(n)` - get the y-coordinate of node n
• `nz(n)` - get the z-coordinate of node n
• `kx(k)` - get the x-coordinate of keypoint k
• `ky(k)` - get the y-coordinate of keypoint k
• `kz(k)` - get the z-coordinate of keypoint k
• `lx(n, lfrac)` - X-coordinate of line `n` at length fraction `lfrac`
• `ly(n, lfrac)` - Y-coordinate of line `n` at length fraction `lfrac`
• `lz(n, lfrac)` - Z-coordinate of line `n` at length fraction `lfrac`
• `lsx(n, lfrac)` - X-slope of line `n` at length fraction `lfrac`
• `lsy(n, lfrac)` - Y-slope of line `n` at length fraction `lfrac`
• `lsz(n, lfrac)` - Z-slope of line `n` at length fraction `lfrac`
• `ux(n)` - get the structural displacement at node n in x
• `uy(n)` - get the structural displacement at node n in y
• `uz(n)` - get the structural displacement at node n in z
• `rotx(n)` - get the rotational displacement at node n in x
• `roty(n)` - get the rotational displacement at node n in y
• `rotz(n)` - get the rotational displacement at node n in z
• `nsel(n)` - get the selection status of node n
• `ksel(k)` - get the selection status of keypoint k
• `lsel(n)` - get the selection status of line n
• `asel(a)` - get the selection status of area a
• `esel(n)` - get the selection status of element e
• `vsel(v)` - get the selection status of volume v
• `ndnext(n)` - get the next selected node with a number greater than n.
• `kpnext(k)` - get the next selected keypoint with a number greater than k.
• `lsnext(n)` - get the next selected line with a number greater than n.
• `arnext(a)` - get the next selected area with a number greater than a.
• `elnext(e)` - get the next selected element with a number greater than e.
• `vlnext(v)` - get the next selected volume with a number greater than v.
• `nnear(n)` - get the selected node nearest node n.
• `knear(k)` - get the selected keypoint nearest keypoint k.
• `enearn(n)` - get the selected element nearest node n.
• `node(x, y, z)` - get the node closest to coordinate (x, y, z)
• `kp(x, y, z)` - get the keypoint closest to coordinate (x, y, z)
Examples
In this example we construct a solid box and mesh it. Then we use the `Query` methods `nx`, `ny`, and `nz` to find the cartesian coordinates of the first node. We can access these through the `mapdl.queries` property.
```>>> from ansys.mapdl.core import launch_mapdl
>>> mapdl = launch_mapdl()
>>> mapdl.prep7()
>>> mapdl.et(1, 'SOLID5')
>>> mapdl.block(0, 10, 0, 20, 0, 30)
>>> mapdl.esize(2)
>>> mapdl.vmesh('ALL')
>>> q = mapdl.queries
>>> q.nx(1), q.ny(1), q.nz(1)
0.0 20.0 0.0
```
`Query.node`(x, y, z) Return node closest to coordinate `(x, y, z)`. `Query.kp`(x, y, z) Return keypoint closest to coordinate `(x, y, z)`. Return the x coordinate of the element centroid. Return the y coordinate of the element centroid. Return the z coordinate of the element centroid. Return the x coordinate of a keypont. Return the y coordinate of a keypont. Return the z coordinate of a keypont. Return the x coordinate of a node. Return the y coordinate of a node. Return the z coordinate of a node. Returns x-component of structural displacement at a node. Returns y-component of structural displacement at a node. Returns z-component of structural displacement at a node. Returns x-component of rotational displacement at a node. Returns y-component of rotational displacement at a node. Returns z-component of rotational displacement at a node. `Query.lx`(n, lfrac) X-coordinate of line `n` at length fraction `lfrac`. `Query.ly`(n, lfrac) Y-coordinate of line `n` at length fraction `lfrac`. `Query.lz`(n, lfrac) Z-coordinate of line `n` at length fraction `lfrac`. `Query.lsx`(n, lfrac) X-slope of line `n` at length fraction `lfrac`. `Query.lsy`(n, lfrac) Y-slope of line `n` at length fraction `lfrac`. `Query.lsz`(n, lfrac) Z-slope of line `n` at length fraction `lfrac`. Returns selection status of a node. Returns selection status of a keypoint. Returns selection status of a line. Returns selection status of an area. Returns selection status of an element. Returns selection status of a volume.
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en
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https://www.diagramelectric.co/how-to-calculate-power-in-dc-parallel-circuits/
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# How To Calculate Power In Dc Parallel Circuits
## How to Calculate Power in DC Parallel Circuits
In a parallel circuit, the voltage across each resistor is the same, but the current through each resistor is different. The total current in the circuit is the sum of the currents through each resistor. The power dissipated by a resistor is equal to the square of the current through the resistor multiplied by the resistance of the resistor. $$P = I^2R$$ where: * $P$ is the power in watts * $I$ is the current in amps * $R$ is the resistance in ohms In a parallel circuit, the total power dissipated is equal to the sum of the powers dissipated by each resistor. $$P_{total} = \sum_{i=1}^n P_i$$ where: * $P_{total}$ is the total power in watts * $P_i$ is the power dissipated by resistor $i$ in watts * $n$ is the number of resistors in the circuit To calculate the power in a parallel circuit, you can use the following steps: 1. Find the total current in the circuit. 2. Find the resistance of each resistor. 3. Calculate the power dissipated by each resistor. 4. Add the powers dissipated by each resistor to find the total power in the circuit. Here is an example of how to calculate the power in a parallel circuit. A parallel circuit has three resistors: $R_1 = 2\Omega$, $R_2 = 4\Omega$, and $R_3 = 6\Omega$. The voltage across the circuit is $V = 12\volts$. To find the total current in the circuit, we use the following formula: $$I = \frac{V}{R}$$ In this case, we have: $$I = \frac{12\volts}{2\Omega + 4\Omega + 6\Omega} = 2\amps$$ To find the resistance of each resistor, we use the following formula: $$R = \frac{V}{I}$$ In this case, we have: * $R_1 = \frac{12\volts}{2\amps} = 6\Omega$ * $R_2 = \frac{12\volts}{4\amps} = 3\Omega$ * $R_3 = \frac{12\volts}{6\amps} = 2\Omega$ To calculate the power dissipated by each resistor, we use the following formula: $$P = I^2R$$ In this case, we have: * $P_1 = (2\amps)^2(6\Omega) = 24\watts$ * $P_2 = (2\amps)^2(3\Omega) = 12\watts$ * $P_3 = (2\amps)^2(2\Omega) = 8\watts$ To find the total power in the circuit, we add the powers dissipated by each resistor: $$P_{total} = 24\watts + 12\watts + 8\watts = 44\watts$$ Therefore, the total power dissipated in the circuit is 44 watts. ## Conclusion In this article, we have shown how to calculate power in a DC parallel circuit. We have used a step-by-step example to show how to find the total current in the circuit, the resistance of each resistor, and the power dissipated by each resistor. We have also shown how to find the total power in the circuit.
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https://hoogle.haskell.org/?hoogle=map%20-package%3Abase%20-is%3Aexact%20-is%3Aexact%20-package%3Acase-insensitive%20-package%3Atext%20-package%3Aunordered-containers%20-package%3Aaeson
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# map -package:base -is:exact -is:exact -package:case-insensitive -package:text -package:unordered-containers -package:aeson
O(n) map f xs is the ByteString obtained by applying f to each element of xs.
O(n) map f xs is the ByteString obtained by applying f to each element of xs
O(n) map f xs is the ShortByteString obtained by applying f to each element of xs.
O(n). Map a function over all values in the map.
```map (++ "x") (fromList [(5,"a"), (3,"b")]) == fromList [(3, "bx"), (5, "ax")]
```
O(n*min(n,W)). map f s is the set obtained by applying f to each element of s. It's worth noting that the size of the result may be smaller if, for some (x,y), x /= y && f x == f y
O(n). Map a function over all values in the map.
```map (++ "x") (fromList [(5,"a"), (3,"b")]) == fromList [(3, "bx"), (5, "ax")]
```
O(n*log n). map f s is the set obtained by applying f to each element of s. It's worth noting that the size of the result may be smaller if, for some (x,y), x /= y && f x == f y
O(n) Map a function over a vector.
Map a function over a Bundle
O(n) Map a function over a vector.
O(n) Map a function over a vector.
O(n) Map a function over a vector.
O(n) Map a function over a vector.
map f xs is the list obtained by applying f to each element of xs, i.e.,
```map f [x1, x2, ..., xn] == [f x1, f x2, ..., f xn]
map f [x1, x2, ...] == [f x1, f x2, ...]
```
```>>> map (+1) [1, 2, 3]
```
Apply a function to each element of a Stream, lazily
map f xs is the list obtained by applying f to each element of xs, i.e.,
```map f [x1, x2, ..., xn] == [f x1, f x2, ..., f xn]
map f [x1, x2, ...] == [f x1, f x2, ...]
```
```>>> map (+1) [1, 2, 3]
[2,3,4]
```
Note: You should use Data.Map.Strict instead of this module if:
• You will eventually need all the values stored.
• The stored values don't represent large virtual data structures to be lazily computed.
An efficient implementation of ordered maps from keys to values (dictionaries). These modules are intended to be imported qualified, to avoid name clashes with Prelude functions, e.g.
```import qualified Data.Map as Map
```
The implementation of Map is based on size balanced binary trees (or trees of bounded balance) as described by:
• Stephen Adams, "Efficient sets: a balancing act", Journal of Functional Programming 3(4):553-562, October 1993, http://www.swiss.ai.mit.edu/~adams/BB/.
• J. Nievergelt and E.M. Reingold, "Binary search trees of bounded balance", SIAM journal of computing 2(1), March 1973.
Bounds for union, intersection, and difference are as given by
Note that the implementation is left-biased -- the elements of a first argument are always preferred to the second, for example in union or insert. Warning: The size of the map must not exceed maxBound::Int. Violation of this condition is not detected and if the size limit is exceeded, its behaviour is undefined. Operation comments contain the operation time complexity in the Big-O notation (http://en.wikipedia.org/wiki/Big_O_notation).
A Map from keys k to values a. The Semigroup operation for Map is union, which prefers values from the left operand. If m1 maps a key k to a value a1, and m2 maps the same key to a different value a2, then their union m1 <> m2 maps k to a1.
Not on Stackage, so not searched. Class of key-value maps
The mapAccumL function behaves like a combination of map and foldl; it applies a function to each element of a ByteString, passing an accumulating parameter from left to right, and returning a final value of this accumulator together with the new ByteString.
The mapAccumR function behaves like a combination of map and foldr; it applies a function to each element of a ByteString, passing an accumulating parameter from right to left, and returning a final value of this accumulator together with the new ByteString.
The mapAccumL function behaves like a combination of map and foldl; it applies a function to each element of a ByteString, passing an accumulating parameter from left to right, and returning a final value of this accumulator together with the new ByteString.
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Community Profile
# Romesh
### University of Sydney
51 2011 年以降の合計貢献数
#### Romesh's バッジ
How to enlarge legend marker size in r2014b?
Confirming on R2017a that the second output of legend has the necessary handles e.g. [~,b] = legend('foo','bar','baz') K...
3年以上 前 | 2
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*± ± ± ± ± ± ± ± ± ± ±* * Imagine you need to subtract one...
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A GUI to make it easy to manually count cells
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How to create an empty array of structs?
If you know the fields in advance, you can create a 0x0 struct array WITH fields using the syntax s = struct('a',{},'b',{})...
7年以上 前 | 20
Estimating the error of a trapezoid method integral
I'm going to revive this just because I think this problem is fairly widespread and there isn't much available by way of resourc...
8年弱 前 | 3
you just measured its surface area, that is the input.
Nearest Numbers
Given a row vector of numbers, find the indices of the two nearest numbers. Examples: [index1 index2] = nearestNumbers([2 5 3...
Which values occur exactly three times?
Return a list of all values (sorted smallest to largest) that appear exactly three times in the input vector x. So if x = [1 2...
Increment a number, given its digits
Take as input an array of digits (e.g. x = [1 2 3]) and output an array of digits that is that number "incremented" properly, (i...
Detect a number and replace with two NaN's
Write code which replaces the number 1 with two NaNs. Example X = [ 1 2 NaN 4 1 3 7 NaN 1 4 NaN 2] ...
Keyboard replacement with stack trace
Display a stack trace showing where keyboard was used
Storing persistent variable in a standalone GUI
There is a likely reason that you have this problem. When the program is compiled, you might have included the XML file inside t...
8年以上 前 | 0
Find best placement for ordered dominoes (harder)
Given a list of ordered pairs, find the order they should be placed in a line, such that the sum of the absolute values of the d...
Scoring for oriented dominoes
Given a list of ordered pairs, and the order they should be placed in a line, find the sum of the absolute values of the differe...
Love triangles
Given a vector of lengths [a b c], determines whether a triangle with non-zero area (in two-dimensional Euclidean space, smarty!...
Too mean-spirited
Find the mean of each consecutive pair of numbers in the input row vector. For example, x=[1 2 3] ----> y = [1.5 2.5] x=[1...
Target sorting
Sort the given list of numbers |a| according to how far away each element is from the target value |t|. The result should return...
Which doors are open?
There are n doors in an alley. Initially they are all shut. You have been tasked to go down the alley n times, and open/shut the...
The Hitchhiker's Guide to MATLAB
Output logical "true" if the input is the answer to life, the universe and everything. Otherwise, output logical "false".
Swap the input arguments
Write a two-input, two-output function that swaps its two input arguments. For example: [q,r] = swap(5,10) returns q = ...
surface of a spherical planet
you just discovered its circumference, that is the input.
Solve the Sudoku Row
*Description* A simple yet tedious task occurs near the end of most Sudoku-solving algorithms, computerized or manual. The ta...
The Goldbach Conjecture
The <http://en.wikipedia.org/wiki/Goldbach's_conjecture Goldbach conjecture> asserts that every even integer greater than 2 can ...
How to find the position of an element in a vector without using the find function
Write a function posX=findPosition(x,y) where x is a vector and y is the number that you are searching for. Examples: fin...
De-dupe
Remove all the redundant elements in a vector, but keep the first occurrence of each value in its original location. So if a =...
Function Iterator
Given a handle fh to a function which takes a scalar input and returns a scalar output and an integer n >= 1, return a handle f...
Return the 3n+1 sequence for n
A Collatz sequence is the sequence where, for a given number n, the next number in the sequence is either n/2 if the number is e...
<http://en.wikipedia.org/wiki/Benford%27s_law Benford's Law> states that the distribution of leading digits is not random. This...
Sort a list of complex numbers based on far they are from the origin.
Given a list of complex numbers z, return a list zSorted such that the numbers that are farthest from the origin (0+0i) appear f...
Weighted average
Given two lists of numbers, determine the weighted average. Example [1 2 3] and [10 15 20] should result in 33.333...
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latest
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en
| 0.782104
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https://brilliant.org/discussions/thread/rule-of-sum-and-rule-of-product/
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# This note has been used to help create the Rule of Sum and Rule of Product Problem Solving wiki
The Rule of Sum (Addition Principle) and Rule of Product (Multiplication Principle) are principles of counting that are used to build up the theory of enumerative combinatorics.
Rule of Sum / Addition principle: If there are $n$ ways of doings something, and $m$ ways of doing another thing, both of which cannot be done at the same time, then there are $n+m$ ways to choose one of these actions.
Rule of Product / Multiplication principle: If there are $n$ ways of doing something, and $m$ ways of doing another thing after that, then there are $n\times m$ ways to perform both of these actions.
## Worked Examples
### 1. Calvin wants to go to Milwaukee. He can choose from $3$ bus services or $2$ train services to head from home to downtown Chicago. From there, he can choose from 2 bus services or 3 train services to head to Milwaukee. How many ways are there for him to get to Milwaukee?
Solution: Since Calvin can either take a bus or a train downtown , he has $3+2 =5$ ways to head downtown (Rule of Sum). After which, he can either take a bus or a train to Milwaukee, hence he has another $2+3=5$ ways to head to Milwaukee (Rule of Sum). Thus in total, he has $5 \times 5 = 25$ ways to head from home to Milwaukee (Rule of Product).
### 2. Calvin wants to go to Milwaukee (see previous question). This time, he has to purchase a bus concession (which will only allow him to take buses), or a train concession (which will only allow him to take trains). If he only has money for $1$ of these concessions, how many ways are there for him to get to Milwaukee?
Solution: If Calvin purchases a bus concession, he has $3 \times 2=6$ ways to get to Milwaukee (Rule of Product). If Calvin purchases a train concession, he has $2\times3=6$ ways to get to Milwaukee (Rule of Product). Hence, he has $6+6=12$ ways to get to Milwaukee in total (Rule of Sum).
### 3. Six friends Andy, Bandy, Candy, Dandy, Endy and Fandy want to sit in a row at the cinema. If there are only six seats available, how many ways can we seat these friends?
Solution: For the first seat, we have a choice of any of the 6 friends. After seating the first person, for the second seat, we have a choice of any of the remaining 5 friends. After seating the second person, for the third seat, we have a choice of any of the remaining 4 friends. After seating the third person, for the fourth seat, we have a choice of any of the remaining 3 friends. After seating the fourth person, for the fifth seat, we have a choice of any of the remaining 2 friends. After seating the fifth person, for the sixth seat, we have a choice of only 1 of the remaining friends. Hence, by the Rule of Product, there are $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ ways to seat these 6 people. More generally, this problem is known as a Permutation. There are $n! = n \times (n-1) \times (n-2) \times \ldots \times 1$ ways to seat $n$ people in a row.
### 4. How many positive divisors does $2000 = 2^4 5^3$ have?
Solution: Any positive divisor of 2000 must have the form $2^a 5^b$, where $a$ and $b$ are integers satisfying $0 \leq a \leq 4, 0 \leq b \leq 3$. There are 5 possibilities for $a$ and 4 possibilities for $b$, hence there are $5 \times 4 = 20$ (Rule of Product) positive divisors of 2000 in all.
Note by Arron Kau
7 years, 4 months ago
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The last two of them really helped thank you!
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# Teacher Gameroom
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Engage students with this fun, interactive community helpers game. In this "Who am I" game, students must guess the community worker. There are 10 questions and you just click on each question to go to it. Each question has 3 clues and student have
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Here is a powerpoint game specifically for 3rd grade. I made this for my after school students who are struggling. They enjoy it very much and your students will too. It is for fractions and covers just about everything we cover at our school in
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Engage students with this fun, interactive fractions game. Students must determine what the fraction is when given a number line and a single point. This is a student vs. students game so students are competing against one another. There's also a
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Engage 4th grade students with this fun, interactive equivalent fraction game. In this write and wipe powerpoint game, students determine the equivalent for the fraction provided. Students are given the numerator of the fraction and have to
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5 worksheets of money word problems. There are 2 types of worksheets. The first set of worksheets are basic addition and subtraction word problems. The second set is a short story of kids finding money and student must figure out how much they
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The powerpoint game is just like Jeopardy. Click on a question and students answer it correctly to get the points. This game caters to 4th grade geometry. Questions include identifying lines, angles, and symmetry. Students will love it. Great for
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Engage 1st grade students with this fun, interactive time game. In this teacher vs student powerpoint game, its students against the teacher as they review analog time of clocks. Students are given a clock with time to the hour or half hour. There
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Engage students with this fun, interactive pictograph game. Students will answer questions about two pictographs. Questions include most/least favorites, totals for each group, and comparing groups. There are 2 graphs in this game, each with 10
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Engage students with this fun, interactive pronouns game. This mini game reviews pronouns by giving students a sentences with a noun. The students must choose the best pronoun to replace the given noun. . Nouns are identified in yellow and after
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Reinforce fractions skills with this equivalent fractions worksheets. In this set of worksheets, students are making equivalent fractions. There are two fractions bars in each problem for students to use as a visual and fractions below each set.
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Engage students with this fun, interactive money game. You students will love playing the exciting Jeopardy - style PowerPoint game. Students compete against one another as they answer questions about money. Questions include counting coins and
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215 ratings
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Engage 2nd grade students with this fun, interactive time game. In this powerpoint game, students review various topics on time. This include telling time, elapsed time, and more. You can play against your students or in teams. There are 25
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Engage students with this fun, interactive Numbers game. In this reviewpowerpoint game, students demonstrate their math skills by practicing their math fluency with factors, multiples, prime, and composite numbers. There are 6 categories for a
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This game reviews basic information about the 3 branches of government: Executive, Legislative, and Judicial. Students are give a clue about one of the three branches and they must decided which branch it is. There are 21 questions and you just
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165 ratings
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Engage students with this fun, interactive ELA game. In this tic-tac-toe powerpoint game, students read the sight word and place their "X" or "O". There are 3 boards, each with a different set of words. There are eight (8) words and a free space per
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This ppt game reviews comparing fractions up to 12ths. There are 20 questions and you can have up to 4 teams. The question disappears after you've clicked on it so you know you've answered it. There is a type-in scoreboard. The scoreboard can be
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186 ratings
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Check for student understanding with this fun, interactive powerpoint game. Students are given a 2d shape and they must figure out how many lines of symmetry it contains. Includes triangles, squares, rectangles, and many more! There are 10 questions
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173 ratings
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Engage students with this fun, interactive adding and subtraction game. In this Jeopardy powerpoint game, students are practicing adding and subtracting both like and unlike fractions. This game also includes word problems. Your students will enjoy
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Engage students with this fun, interactive punctuation game. In this teacher vs. student powerpoint game, students are adding the correct punctuation to sentences. There are 3 options: period (.), exclamation mark (!), and question mark (?). There
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209 ratings
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This worksheets helps students practice expanded form with 3 digit numbers. Students are given a number and must write the value of each number. There is a guide at the top to help struggling students.Great for a guided math center or rainy day
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153 ratings
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TEACHING EXPERIENCE
I have taught 5th grade math and science. I did my student teaching in Kindergarten and 5th grade. I have observed several different classrooms during my college years. I was a math tutor for grades 3rd, 4th, and 5th grade at Northwoods Elementary as well as a 2nd grade teacher.
MY TEACHING STYLE
I love using technology, especially the Activboard/Smartboard. I love direct teaching as well as cooperative groups.
HONORS/AWARDS/SHINING TEACHER MOMENT
Many high school awards, especially in Math. I took all honors classes in High school including Geometry honors, Algebra 2 Honors, and Algebra 3 Honors.
MY OWN EDUCATIONAL HISTORY
Graduated from NC Wesleyan College. I did my student teaching in 5th grade as well as kindergarten. Began my career as a 5th grade math and science teacher. I then moved onto become a math tutor for 3rd, 4th, and 5th grade. A year later, I moved to 2nd grader teacher.
Teachers have so much to worry about. From paper work, to assessment, to the curriculum, and beyond. When I decided to make a game, I think about what the students likes, what the teacher likes, and what could I create to make the classroom a better place. Not just for the teacher, but for the teacher as well. Making games that everyone loves will get the students excited, the teacher relaxed, and the administration pleased. Check out what others are saying about these wonderfully amazing powerpoint games. Store logo and banners designed by A Little Piece of Africa
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Friday
May 6, 2016
# Homework Help: Calculus
Posted by Brian on Saturday, March 10, 2012 at 2:46pm.
For the parabola given by (y – 3)^2 = 8(x + 5), find the following:
(a) direction parabola opens (to the left, right, up, or down) Answer: ___________
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http://physics.stackexchange.com/questions/12035/why-does-light-reflect-more-intensely-when-it-hits-a-surface-at-a-large-angle/12077
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# Why does light reflect more intensely when it hits a surface at a large angle?
I mean, what is happening at a microscopic level to cause this behavior? Here's what I got from Wikipedia:
1. On Reflection (physics)#Reflection of light it says that "solving Maxwell's equations for a light ray striking a boundary allows the derivation of the Fresnel equations, which can be used to predict how much of the light reflected, and how much is refracted in a given situation."
2. On Specular reflection#Explanation it says that "for most interfaces between materials, the fraction of the light that is reflected increases with increasing angle of incidence $\theta_i$" (but doesn't explain why)
3. Finally, on Reflection coefficient#Optics, it says basically nothing, redirecting the reader to the Fresnel equations article.
What I'm trying to find, instead, is a basic level explanation that could provide an intuition on why this happens, rather than analytic formulations or equations to calculate these values. Is there a good analogy that explains this behavior?
-
Not only is it reflected more intensely, but sometimes there is no smooth (non-scattered) reflected image at all at a straight angle; then when you're observing at a grazing angle the surface could appear mirror-like. This is noticed when you're making optics (e.g. telescope mirrors) and you're polishing the glass: when you're working with fine abrasives such as 5 micron Al2O3, the mirror surface seems matte at a straight angle, but appears smooth and reflective at a grazing angle. – Florin Andrei Jul 8 '11 at 22:30
First, I just want to remind readers that it is NOT true that "more glancing angle always means more reflection". For p-polarized light, as the angle goes away from the normal, it gets less and less reflective, then at the Brewster angle it's not reflective at all, and then beyond the Brewster angle it becomes more reflective again:
Nevertheless, it's certainly true that as the angle approaches perfectly glancing, the reflection approaches 100%. Even though the question asks for non-mathematical answers, the math is pretty simple and understandable in my opinion...here it is for reference. (I don't have any non-mathematical answer that's better than other peoples'.)
The Maxwell's equations boundary conditions say that certain components of the electric and magnetic fields have to be continuous across the boundary. The situation at almost-glancing angle is that the incoming and reflected light waves almost perfectly cancel each other out (opposite phase, almost-equal magnitude), leaving almost no fields on one side of the boundary; and since there's almost no transmitted light, there's almost no fields on the other side of the boundary too. So everything is continuous, "zero equals zero".
The reason this cannot work at other angles is that two waves cannot destructively interfere unless they point the same direction. (If two waves have equal and opposite electric fields and equal and opposite magnetic fields, then they have to point the same direction, there's a "right-hand rule" about this.) At glancing angle, the incident and reflected waves are pointing almost the same direction, so they can destructively interfere. At other angles, the incident and reflected waves are pointing different directions, so they cannot destructively interfere, so there has to be a transmitted wave to make the boundary conditions work. :-)
-
Thanks, Steve! I'd accept this answer, since it seems to have received the blessing of the community as the most correct explanation of the phenomenon. However, I must admit that I cannot, with my current knowledge, understand everything in it. Then again, perhaps explaining this at a layman level would require a much longer text, to introduce all the relevant concepts. I'll try to become more familiarized with the theory behind this and maybe later if I can understand it better, I'll select it as the best answer. I feel that's would be more honest and aligned with the scientific spirit. – waldyrious Jul 11 '11 at 13:55
Update: revisiting the question without all those thoughts in my head (I was preparing for an exam at the time) allowed me to realize that your explanation was actually quite intuitive and simple. I'll accept it now. Thanks, and sorry for the delay! :) – waldyrious Jan 7 '12 at 15:22
Well, imagine shooting a piece of glass with a gun. If you shoot at a glancing angle, it is more likely to ricochet off the glass without damaging it. This is because the impulse required to reflect the bullet is smaller for shallow grazing angles, as most of the bullets momentum is parallel to the interface. Now obviously the physics of reflecting a photon is very different than reflecting a bullet, but the analogy is that the impulse required to reflect the photon becomes smaller, and it becomes relatively "easier" for the medium to supply that small impulse than to let the photon go through.
-
This makes sense and would be my first intuition, but I expected that while the result is similar to macroscopic bouncing, the principles behind it would be potentially very different. I'd love if you could expand this analogy with a bit more detail on what happens microscopically (if that's possible when adopting this metaphor). – waldyrious Jul 8 '11 at 17:48
user1631's analogy holds. You can think of the reflection as the surface generating an electromagnetic wave which cancels the incoming wave on the non-reflecting side and reflects the wave on the reflecting side.
At a high angle of incidence this wave needs to have a high intensity. At a low angle of incidence, the wave needs a lower intensity.
At higher intensities, more of this "reflecting wave" is going to be transmitted through the surface or turned into wasted energy.
In other words, at low angles, you only need to change the direction of the EM vector a little bit. At high angles, it needs to change direction completely.
How does reflection work? might help.
-
The linked question was very interesting. Thanks for the pointer! – waldyrious Jul 9 '11 at 0:01
Perhaps it is because in the case of grazing incidence a lot more scatterer are in the path of the photon. At least this is why I think we observe X-ray reflection at all and were able to build Chandra and Rosat.
But I would like to hear a better explanation as well.
-
The area of the intersection of the ray with the surface at A is greater than the area of intersection at B.
To a greater area correspond more atoms to reflect the light.
A justification:
The equations related to graphs on Steve answer (the reflected coefficients $R_s and R_p$) are the Fresnel equations when I read there $(n_1/n_2\cdot\sin\theta_i)^2$ I see a proportionality to the area.
The angles are in relation to the normal and sin² traces an area. A sensivity of the equations in relation to this factor is apparent in the graphs ($\theta_i$), because $\sin\theta_i, \sin\theta_i^2$ will 'follow' the shape of ($\theta_i$)
I think that my viewpoint is justified.
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How to add customized distance to ComplexHeatmap
1
0
Entering edit mode
17 months ago
Hi there, I constructed a co-expression network using WGCNA. As you could know, we identify co-expressed gene modules by performing hierarchical clustering method based on TOM-based dissimilarity. So, a distance matrix was computed here to draw a dendrogram of eigen-genes is TOM-based dissimilarity.
Now, I want to draw a heatmap that visualizes the overlap of eigen-genes among inter-modules using the function heatmap by the package ComplexHeatmap. But the problem is, the function heatmap does not provide the above TOM-based dissimilarity distance method. I wonder how to do it? The code lines I used to draw heatmap as follows:
> plotTOM = dissTOM^7; #Transform dissTOM with a power to make moderately strong connections more visible in the heatmap
> diag(plotTOM) = NA; #Set diagonal to NA for a nicer plot
> sizeGrWindow(9,9)
> heatmap(plotTOM, show_row_names = TRUE, show_column_names = FALSE,
row_dend_reorder = TRUE, column_dend_reorder = TRUE,
clustering_distance_rows = ???,
clustering_distance_columns = ???,
clustering_method_rows = "ward.D2",
clustering_method_columns = "ward.D2")
I define my own distance in R like:
> softPower = 6;
> adjacency = adjacency(df, power = softPower); #df is my gene expression data
> dissTOM = 1-TOM #distance matrix
dendrogram ComplexHeatmap • 1.3k views
2
Entering edit mode
Hey,
You can add any function that you want via:
Heatmap(...,
clustering_distance_columns = function(x) as.dist(1-cor(t(x))),
clustering_method_columns = 'ward.D2',
clustering_distance_rows = function(x) as.dist(1-cor(t(x))),
clustering_method_rows = 'ward.D2',
...)
Kevin
0
Entering edit mode
> Heatmap(plotTOM,
> show_row_names = TRUE, show_column_names = TRUE,
> row_dend_reorder = TRUE, column_dend_reorder = TRUE,
> clustering_method_rows = "ward.D2",
> clustering_method_columns = "ward.D2",
> show_heatmap_legend = FALSE,
> row_names_gp = gpar(fontsize = 8),
> column_names_gp = gpar(fontsize = 8))
But the thing is, the dendrogram height show on the top and the left of heatmap don't look right.
0
Entering edit mode
I do not believe you are implementing this in the right way. Basically, in order to use my solution in the way that you want, you have to understand really well how the WGCNA functions work.
I am not sure, but you may need something like this:
clustering_distance_columns = function(x) 1 - TOMsimilarity(adjacency(x, power = softPower))
1
Entering edit mode
17 months ago
jbalberge ▴ 160
As plotTOM is already a distance-based matrix, you probably don't want to compute the distance between individuals again. You may rather use the dendrogram of your choice in the Heatmap arguments:
clustTOM <- hclust(as.dist(dissTOM), method="average")
Heatmap(dissTOM, cluster_rows = clustTOM, cluster_columns = clustTOM)
0
Entering edit mode
Many thanks for your helps. It is kind of a fast way to visualize what I want. But I've faced with a new problem above. Can you help me to solve it?
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# Problem Centered problems
Date conversion 05.01.2017 Size 4.06 Kb.
## Problem Centered problems
1. Frank Rd and James Rd. make a perpendicular intersection. The state wants to build a new road. The new road will intersect 3 miles north of the intersection on Frank Rd. and 4 miles west of the intersection on James Rd. How long will the new road be that intersects Frank and James Rd? The new road would cost \$10 per foot to pave. What would be the cost of the new road?
2. The mobile phone company is anchoring wires to the top of their 1200 ft high communication towers. The cable for the support wire comes in a roll that is 3900 ft long. The company requires you to use the entire roll. The cable can only be cut twice to ensure its strength. All cables need to be equal. How long will each cable be and how far from the base of the tower do they need to be anchored?
3. In the city planning meeting, a scale drawing of a park was drawn. The park fills inside a square city block. The scale was 3 inches equal 3/10 miles. One side of the city blocks was 4 inches in the drawing. One member of the city planners said, " Their needs to be a short cut through the park from the corners." How long in miles will the short cut be? Round answers to the nearest tenth of a mile.
4. You are planning to put a new digital flat TV on a wall that is 12 ft long and 9 ft high. The digital TV has a diagonal of 72 inches. The length of the TV is twice the width of the TV. How much of the wall will still need to be decorated around the TV?
1. 5 miles long cost \$264,000
2. 1300 ft cables, 500 ft from the tower
4. 32.2 sq in width rounded to 3 ft
64.3 sq in length rounded to 5ft
total area for TV is 15 sq ft
remaining area around TV is 93 sq ft.
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Úvod Výuka Studium Odkazy
## NMST434 Modern statistical methods (official information)
Requirements:
The knowledge of the statistics and probability theory at the level of courses Mathematical Statistics 1 and 2 (NMSA331 and NMSA332), Probability Theory 1 (NMSA333) and Linear regression (NMSA407).
Among others we will use the following concepts: almost sure convergence, convergence in probability, convergence in distribution, law of large numbers and central limit theorem for independent and identically distributed random vectors. A nice overview (in Czech language) of most of the results that are used in the course is available here.
(Assumed) content of the course:
• Asymptotic methods - Delta Theorem
• Theory of maximum likelihood
• Profile, conditional and marginal likelihood
• M-estimators and Z-estimators, Quasi-likelihood, Robust estimation
• Bootstrap
• Quantile regression
• EM-algorithm
• Methods for missing data
• Kernel density estimation
• Kernel nonparametric regression
Exercise classes
Exercise classes accompany the lecture with both theoretical as well as practical examples and illustrations.
Lecture material and actual information
Exercise class 3 (7. 3. 2017)
Exercise class 6 (28. 3. 2017)
Exercise class 7 (4. 4. 2017)
Exercise class 8 (11. 4. 2017)
• Working script - nmst434-E08.R
• Datasets - IQ-en.RData, (the code supposes that datasets are placed in the subdirectory `data' of the working directory)
Exercise class 9 (18. 4. 2017)
Exercise class 10 (25. 4. 2017)
• Working script - nmst434-E10.R
• AAUP.RData (the working script assumes that datasets are stored in the directory 'data' that is placed in the working directory)
Exercise class 11 (2. 5. 2017)
Exercise class 12 (9. 5. 2017)
Exercise class 13 (16. 5. 2017)
Exercise class 14 (23. 5. 2017)
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# Search by Topic
#### Resources tagged with Working systematically similar to Intersection Sudoku 2:
Filter by: Content type:
Age range:
Challenge level:
### There are 127 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Intersection Sudoku 1
##### Age 11 to 16 Challenge Level:
A Sudoku with a twist.
### Intersection Sudoku 2
##### Age 11 to 16 Challenge Level:
A Sudoku with a twist.
##### Age 11 to 16 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
### Pole Star Sudoku
##### Age 14 to 18 Challenge Level:
A Sudoku based on clues that give the differences between adjacent cells.
### Twin Corresponding Sudoku III
##### Age 11 to 16 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Seasonal Twin Sudokus
##### Age 11 to 16 Challenge Level:
This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?
### Twin Corresponding Sudoku
##### Age 11 to 18 Challenge Level:
This sudoku requires you to have "double vision" - two Sudoku's for the price of one
### Latin Lilies
##### Age 7 to 16 Challenge Level:
Have a go at this game which has been inspired by the Big Internet Math-Off 2019. Can you gain more columns of lily pads than your opponent?
### Diagonal Sums Sudoku
##### Age 7 to 16 Challenge Level:
Solve this Sudoku puzzle whose clues are in the form of sums of the numbers which should appear in diagonal opposite cells.
### Corresponding Sudokus
##### Age 11 to 18
This second Sudoku article discusses "Corresponding Sudokus" which are pairs of Sudokus with terms that can be matched using a substitution rule.
### Ratio Sudoku 2
##### Age 11 to 16 Challenge Level:
A Sudoku with clues as ratios.
### Wallpaper Sudoku
##### Age 11 to 16 Challenge Level:
A Sudoku that uses transformations as supporting clues.
### Twin Corresponding Sudokus II
##### Age 11 to 16 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Colour in the Square
##### Age 7 to 16 Challenge Level:
Can you put the 25 coloured tiles into the 5 x 5 square so that no column, no row and no diagonal line have tiles of the same colour in them?
##### Age 11 to 16 Challenge Level:
This is a variation of sudoku which contains a set of special clue-numbers. Each set of 4 small digits stands for the numbers in the four cells of the grid adjacent to this set.
### Ratio Sudoku 3
##### Age 11 to 16 Challenge Level:
A Sudoku with clues as ratios or fractions.
### Games Related to Nim
##### Age 5 to 16
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Ratio Sudoku 1
##### Age 11 to 16 Challenge Level:
A Sudoku with clues as ratios.
### Intersection Sums Sudoku
##### Age 7 to 16 Challenge Level:
A Sudoku with clues given as sums of entries.
### The Naked Pair in Sudoku
##### Age 7 to 16
A particular technique for solving Sudoku puzzles, known as "naked pair", is explained in this easy-to-read article.
### Twin Line-swapping Sudoku
##### Age 14 to 16 Challenge Level:
A pair of Sudoku puzzles that together lead to a complete solution.
### Simultaneous Equations Sudoku
##### Age 11 to 16 Challenge Level:
Solve the equations to identify the clue numbers in this Sudoku problem.
### Diagonal Product Sudoku
##### Age 11 to 16 Challenge Level:
Given the products of diagonally opposite cells - can you complete this Sudoku?
### Integrated Sums Sudoku
##### Age 11 to 16 Challenge Level:
The puzzle can be solved with the help of small clue-numbers which are either placed on the border lines between selected pairs of neighbouring squares of the grid or placed after slash marks on. . . .
### Pole Star Sudoku 2
##### Age 11 to 16 Challenge Level:
This Sudoku, based on differences. Using the one clue number can you find the solution?
### Integrated Product Sudoku
##### Age 11 to 16 Challenge Level:
This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid.
##### Age 11 to 16 Challenge Level:
Four small numbers give the clue to the contents of the four surrounding cells.
### LCM Sudoku II
##### Age 11 to 18 Challenge Level:
You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.
### Bochap Sudoku
##### Age 11 to 16 Challenge Level:
This Sudoku combines all four arithmetic operations.
### Colour Islands Sudoku 2
##### Age 11 to 18 Challenge Level:
In this Sudoku, there are three coloured "islands" in the 9x9 grid. Within each "island" EVERY group of nine cells that form a 3x3 square must contain the numbers 1 through 9.
### Constellation Sudoku
##### Age 14 to 18 Challenge Level:
Special clue numbers related to the difference between numbers in two adjacent cells and values of the stars in the "constellation" make this a doubly interesting problem.
### Magnetic Personality
##### Age 7 to 16 Challenge Level:
60 pieces and a challenge. What can you make and how many of the pieces can you use creating skeleton polyhedra?
### Smith and Jones
##### Age 14 to 16 Challenge Level:
Mr Smith and Mr Jones are two maths teachers. By asking questions, the answers to which may be right or wrong, Mr Jones is able to find the number of the house Mr Smith lives in... Or not!
### Magic W
##### Age 14 to 16 Challenge Level:
Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.
### Instant Insanity
##### Age 11 to 18 Challenge Level:
Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.
### All-variables Sudoku
##### Age 11 to 18 Challenge Level:
The challenge is to find the values of the variables if you are to solve this Sudoku.
### Alphabetti Sudoku
##### Age 11 to 16 Challenge Level:
This Sudoku requires you to do some working backwards before working forwards.
### Warmsnug Double Glazing
##### Age 14 to 16 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### Twin Chute-swapping Sudoku
##### Age 14 to 18 Challenge Level:
A pair of Sudokus with lots in common. In fact they are the same problem but rearranged. Can you find how they relate to solve them both?
### Difference Dynamics
##### Age 14 to 18 Challenge Level:
Take three whole numbers. The differences between them give you three new numbers. Find the differences between the new numbers and keep repeating this. What happens?
### Rainstorm Sudoku
##### Age 14 to 16 Challenge Level:
Use the clues about the shaded areas to help solve this sudoku
### Rectangle Outline Sudoku
##### Age 11 to 16 Challenge Level:
Each of the main diagonals of this sudoku must contain the numbers 1 to 9 and each rectangle width the numbers 1 to 4.
### One Out One Under
##### Age 14 to 16 Challenge Level:
Imagine a stack of numbered cards with one on top. Discard the top, put the next card to the bottom and repeat continuously. Can you predict the last card?
### Plum Tree
##### Age 14 to 18 Challenge Level:
Label this plum tree graph to make it totally magic!
### Magic Caterpillars
##### Age 14 to 18 Challenge Level:
Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.
### Olympic Logic
##### Age 11 to 16 Challenge Level:
Can you use your powers of logic and deduction to work out the missing information in these sporty situations?
### Latin Squares
##### Age 11 to 18
A Latin square of order n is an array of n symbols in which each symbol occurs exactly once in each row and exactly once in each column.
### Difference Sudoku
##### Age 14 to 16 Challenge Level:
Use the differences to find the solution to this Sudoku.
### Star Product Sudoku
##### Age 11 to 16 Challenge Level:
The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid.
### LCM Sudoku
##### Age 14 to 16 Challenge Level:
Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.
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https://pvillage.org/archives/5362
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Pvillage.org
Informative blog about fresh lifehacks
What is electric current in simple words?
What is electric current in simple words?
An electric current is a flow of an electric charge between two points. The electric charge in the electrons causes resistors in light bulbs and electric fires to heat up. It also creates magnetic fields in motors, inductors, and generators.
How is electricity defined for kids?
Electricity is the flow of tiny particles called electrons and protons. It can also mean the energy you get when electrons flow from place to place. Electricity can be seen in nature in a bolt of lightning.
What is electric current explain with diagram?
An electric current is a flow of particles (electrons) flowing through wires and components. It is the rate of flow of charge. If the electric charge flows through a conductor, we say that there is an electric current in the conductor. In the circuits using metallic wires, electrons constitute a flow of charges.
How do you explain electricity to a 6 year old?
To help a child understand the properties of static electricity, rub a balloon on your hair or a wool blanket or sweater, and watch how it sticks to the wall for a few moments. Explain that this is an example of negatively and positively charged atoms reacting to each other.
How many volts are in an ohm?
Ohm to Volt/ampere Conversion Table
Ohm Volt/ampere [V/A]
1 ohm 1 V/A
2 ohm 2 V/A
3 ohm 3 V/A
5 ohm 5 V/A
What is ohm full form?
Ohm, abbreviation Ω, unit of electrical resistance in the metre-kilogram-second system, named in honour of the 19th-century German physicist Georg Simon Ohm.
What is called electric pressure?
Voltage is another term for electrical pressure. Voltage is measured in volts. Voltage or electrical pressure is measured in volts. One volt is the amount of pressure required to flow one ampere of electricity through one ohm of resistance.
How does electromagnet work for kids?
Electromagnet facts for kids. Kids Encyclopedia Facts. Electromagnet attracts paper clips when current is applied creating a magnetic field, loses them when current and magnetic field are removed. When current flows through any wire it makes a magnetic field around the wire.
What are circuits for kids?
Parallel Circuit Definition for Kids. There are different meanings of the term circuit. In a literal sense, a circuit is a diagram or route that begins and ends at the same point or place. The path may be circular, a square, rectangular, triangle , polygonal or any kind of geometric shape as long as the start and the end points are the same.
What four things are needed to complete an electric circuit?
calculators and other small electronic devices.
• Scissors – Mostly available at home.
• Two Sided Tape – This tape has adhesive on both sides.
• Cardboard- You can find these in form of cartons and boxes at home.
• electronic devices.
• What are facts about electric current?
Electric current is the name for the flow of electrons that makes up the movement of electric charge. Current flows when the voltage on one end of a conductor differs from the voltage on the other end of a conductor. A force that most people deal with nearly every day, flowing current includes lighting,…
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https://goprep.co/ex-8.2-q6-the-radius-and-height-of-a-cylinder-are-in-the-i-1nkh1y
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GIVEN : radius of solid cylinder = “r”
height of solid cylinder = “h”
=
volume of the cylinder “V” = 4400cm2
we take π =
TO FIND : radius of a solid cylinder = ?
PROCEDURE :
As we know, that
Volume of a cylinder “V” = π r2h
Also, we have, =
= h = r ×
So we now put the values of V and h in the above formula for V,
So we get, 4400 = π ×(r)2×( r × )
4400 = × r3×
4400 × × = r3
200 × 5 = r3
1000 = r3
So, r = 1000 = 10
radius of the given cylinder = 10cm.
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http://owenduffy.net/transmissionline/VSWR/Zs50.htm
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# Is Zs of a HF ham tx typically 50+j0?
It is often assumed that a ham transmitters designed for a nominal 50+j0Ω load can be accurately represented generally by a Thevenin equivalent circuit with a source impedance of 50+j0Ω. This assumption may then be used for instance to infer the reduction in power delivered to an antenna of certain VSWR using the concept of Mismatch Loss.
Thevenin’s theorem for linear networks states that any combination of voltage sources, current sources and resistors with two terminals can be represented by an electrically equivalent circuit of a single voltage source Vs and a single series resistance Rs. In the frequency domain, the theorem can also be applied to impedance, not just resistance. A Thevenin equivalent circuit can be used to determine how much power would be developed in a load of any given impedance.
An interesting property of a Thevenin source (constant Vs and with Zs=50+j0Ω) is that the ‘forward power’ indicated by a quality directional wattmeter calibrated for 50+j0Ω will be independent of load impedance.
The following shows that Vf=Vs/2, ie Vf is independent of Zl, and therefore ‘forward power’ (Vf2/Z0) is independent of Zl. The proof starts by calculating the voltage V at the terminals of load Zl where Vs is the Thevenin equivalent source voltage and the Thevenin equivalent source impedance (Zs) in matched system is taken to be Z0 , Γ is the complex reflection coefficient due to Zl.
(Hint: to simplify the second last expression, multiply top and bottom of the right hand expression by Zl+Z0.)
A simple and easily performed test for practical hams therefore is to connect a transmitter via a directional wattmeter to an ATU and dummy load. If one can vary the ATU settings to vary the load impedance seen by the transmitter within VSWR(50) limits of say 1.5:1, the indicated ‘forward power’ should not vary. (If you swing the load VSWR too far, you may trigger some further non-linear effects like VSWR protection, power control, voltage or current saturation of the PA, which are further reasons why the assumed Zs may be invalid.)
Try the experiment at different frequencies, different power levels, you are likely to find that the results are frequency dependent and power level dependent.
Of course, nothing is perfect so an acceptable tolerance for depending on the assumed Zs=50+j0Ω for Mismatch Loss calculations might be that ‘forward power’ doesn’t vary by more than 10% of the ‘reflected power’ at any point. The worst case ‘reflected power’ for the experiment as described is 4%, so a variation of ‘forward power’ by more than 0.4% (ie 0.4W in 100W) at any point would indicate significant error in any Mismatch calculations based on an assumed Zs=50+j0Ω.
Now, it is very hard to discern such a small variation in ‘forward power’, but for most transmitters that I have ever tested, the variation is more like 5% and readily seen on a meter.
If you perform this test on a range of transmitters, at a range of frequencies, you will probably be convinced that whilst it is possible that Zs=50+j0Ω, it is not always, or even often the case, and that calculations such as Mismatch Loss that depend on that assumption are in error.
# Changes
Version Date Description 1.01 20/06/2010 Initial. 1.02 1.03
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https://community.teradata.com/t5/Analytics/Extract-numeric-values-from-string/m-p/19304/highlight/true
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Analytics
Fan
## Extract numeric values from string
Hi!
I can't find solution, may be somebody can help me.
Example:
`SELECT'ABC123D666656-522815EF!@55#' AS COL1`
I need to extract 666656-52281. But other string chartes can be different. Also can vary their number. I need XXXXXX-XXXXX (where X=[0-9]) from any kind of text.
I stoped on this step:
`SELECT'ABC123D666656-522815EF!@55#' AS COL1,REGEXP_REPLACE(COL1,'[^0-9]+',' ',1,0,'i') AS COL2`
P.S. Sorry for my English.
Tags (3)
4 REPLIES 4
Highlighted
## Re: Extract numeric values from string
Use a REGEXP_SUBSTR to extract a group of digits` [0-9]+` followed by a dash` - `followed by a group of digits` [0-9]+`
`REGEXP_SUBSTR(COL1,'[0-9]+-[0-9]+') AS COL2`
Fan
## Re: Extract numeric values from string
Thank you! But is it posible to count numeric values in string? Because I need strong number count before and after '-'. Situation when string is:
`SELECT'ABC123D45345345666656-5228155668EF!@55#' AS COL1`
I need the same result XXXXXX-XXXXX.
Thank you one more time!
Fan
## Re: Extract numeric values from string
thank I find anwer!
REGEXP_SUBSTR(COL1,'[0-9][0-9][0-9][0-9][0-9][0-9]-[0-9][0-9][0-9][0-9][0-9]')
I didn't see that it's an exact number of digits, no need to repeat` [0-9] `multiple times, you can specify the number directly:
`REGEXP_SUBSTR(COL1,'[0-9]{6}-[0-9]{5}')`
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https://feet-to-cm.appspot.com/3580-feet-to-cm.html
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Feet To Cm
# 3580 ft to cm3580 Feet to Centimeters
ft
=
cm
## How to convert 3580 feet to centimeters?
3580 ft * 30.48 cm = 109118.4 cm 1 ft
A common question is How many foot in 3580 centimeter? And the answer is 117.454068241 ft in 3580 cm. Likewise the question how many centimeter in 3580 foot has the answer of 109118.4 cm in 3580 ft.
## How much are 3580 feet in centimeters?
3580 feet equal 109118.4 centimeters (3580ft = 109118.4cm). Converting 3580 ft to cm is easy. Simply use our calculator above, or apply the formula to change the length 3580 ft to cm.
## Convert 3580 ft to common lengths
UnitLength
Nanometer1.091184e+12 nm
Micrometer1091184000.0 µm
Millimeter1091184.0 mm
Centimeter109118.4 cm
Inch42960.0 in
Foot3580.0 ft
Yard1193.33333333 yd
Meter1091.184 m
Kilometer1.091184 km
Mile0.678030303 mi
Nautical mile0.5891922246 nmi
## What is 3580 feet in cm?
To convert 3580 ft to cm multiply the length in feet by 30.48. The 3580 ft in cm formula is [cm] = 3580 * 30.48. Thus, for 3580 feet in centimeter we get 109118.4 cm.
## Alternative spelling
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# Lesson 19 Make Two-digit Numbers
• Let’s make two-digit numbers with tens and ones in different ways.
## Warm-up Which One Doesn’t Belong: Different Ways to Show a Number
Which one doesn’t belong?
## Activity 1 Make 65 Using Tens and Ones
Create a collection of 65.
You may not break apart any towers.
You may not make any new towers.
Show your collection in a way that others will understand.
If you have time, think of another way to make 65 using the cubes in the bag.
## Activity 2 Make 37 in Different Ways
How many ways can you make 37?
Show your thinking using drawings, numbers, or words.
## Activity 3 Centers: Choice Time
Choose a center.
Greatest of Them All
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ERROR: type should be string, got "https://stoneswww.academickids.com/encyclopedia/index.php/H%E9non_map"
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Missing image
HenonMapImage.png
Hénon attractor for a = 1.4 and b = 0.3
The Hénon map is a discrete-time dynamical system. It is one of the most studied examples of dynamical systems that exhibit chaotic behavior. The Hénon map takes a point (xy) in the plane and maps it to a new point
[itex]x_{n+1} = y_n+1-a x_n^2\,[itex]
[itex]y_{n+1} = b x_n\,[itex].
The map depends on two constants a and b, which have the canonical values of a = 1.4 and b = 0.3.
The map was introduced by Michel Hénon as a simplified model of the Poincaré section of the Lorenz model. For the canonical map (a = 1.4 and b = 0.3) an initial point of the plane will either approach a set of points known as the Hénon strange attractor, or diverge to infinity. The Hénon attractor is a fractal, smooth in one direction and a Cantor set in another. Numerical estimates yield a correlation dimension of 1.42 ± 0.02 (Grassberger, 1983) and a Hausdorff dimension of 1.261 ± 0.003 (Russel 1980) when a = 1.4 and b = 0.3.
As a dynamical system, the canonical Hénon map is interesting because, unlike the logistic map, its orbits defy a simple description.
## Decomposition
The Hénon map may be decomposed into an area-preserving bend:
[itex](x_1,y_1) = (x, 1 - ax^2 + y)[itex]
a contraction in the "x" direction:
[itex](x_2,y_2) = (bx_1, y_1)[itex]
and a reflection in the line y = x:
[itex](x_3, y_3) = (y_2, x_2)[itex]
## References
• Art and Cultures
• Musical Instruments (http://academickids.com/encyclopedia/index.php/List_of_musical_instruments)
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Ancient Civilizations (http://www.academickids.com/encyclopedia/index.php/Ancient_Civilizations)
• Industrial Revolution (http://www.academickids.com/encyclopedia/index.php/Industrial_Revolution)
• Middle Ages (http://www.academickids.com/encyclopedia/index.php/Middle_Ages)
• United States (http://www.academickids.com/encyclopedia/index.php/United_States)
• World History (http://www.academickids.com/encyclopedia/index.php/History_of_the_world)
• Human Body (http://www.academickids.com/encyclopedia/index.php/Human_Body)
• Physical Science (http://www.academickids.com/encyclopedia/index.php/Physical_Science)
• Social Studies (http://www.academickids.com/encyclopedia/index.php/Social_Studies)
• Space and Astronomy
• Solar System (http://www.academickids.com/encyclopedia/index.php/Solar_System)
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Free sample Financial statement analysis of flight centre group
## Financial Statement Analysis Of Flight Centre Group
Question
Task: Part A Reformatted Financial Statements and Financial Analysis (15 %)
•Based on the same company chosen in the group assignment, reformat the company’s financial statements for the past five years in an excel spreadsheet. Produce an individual professional report where ratio and cash flow analysis is performed to evaluate the current and past performance of the company and its sustainability in the future. The report should discuss in details what happened and why the ratios/cash flows changed. Specifically, the report should include the following discussions:
Ratio analysis (6 marks):
•Calculate and discuss key ratios such as ROE, RNOA, PM, ATO, FLEV and NBC.
•Break down and analyse PM and ATO ratios in further details. Identify and discuss three significant expense items that have caused major changes in profit margin. Identify and discuss three major assets or liabilities whose turnover ratios have contributed to the overall change in assets efficiency.
•Briefly describe the ratios trend. The analysis should elaborate on the economic, industry and business factors that drive the changes in ratios. The discussion should consistently reflect the same firm fundamentals identified in the group report.
Cash flow analysis (3 marks):
•Calculate liquidity, solvency and cash flow ratios.
•Analyse financial risk and cash flow management of the company based on ratios.
Overall Report Quality (1 mark):
•The report should be readily comprehensible, condensed and within the word limit. Information should be collected from various reliable sources to inform analysis and references are properly cited. Tables and graphs should be used to effectively present information.
Ratio analysis
In the financial statement analysis, this ratio is measured by dividing the net profit after tax from the turnover of the company. The profit margin is the return the company is getting over its revenue for the period (Davydov 2016).
The percent of gain on the sales of the Flight Centre is measured. In the given diagram the starting ratio belongs to 2019 going backward to 2015 (Flight Centre Group Ltd 2019). Thus we can see from the financial statement analysis that the company margin on its total sales is constant and with the given rising sales every year the margin is constant. Thus for making the additional revenue, the company is not required to decrease its profit margin. This shows that the demands for the company products are increasing without any decrease in the price by reducing the margin. The price depends on other factors also being a reduction in the cost of production via more efficient use of resources (Deegan 2016).
This ratio is measured by dividing the total revenue from the assets invested in the company. This ratio measures how efficiently the assets of the company are being used to generate revenue for the company (Sherman 2015). In the given case of financial statement analysis we can see the net return from assets has reduced from 2015 and 2016 to recent years but the consistency is maintained and much fluctuation is not present. The assets are used efficiently for the generation of sales. Thus the increase in the assets of the company over the years has proportionately increased its revenue to cover the same (Hitchner 2016).
The Return on Net Assets is the measurement of the use of the assets of the company concerning the net profit generated. Since all the assets and income are operational the figures for the above ratio and this ratio will show the same results (Sherman 2015). The difference between the above ratio and the ratio discussed earlier in the financial statement analysis was that in the last ratio we discussed about the generation of the revenue in the current ratio we discuss the profit on such sales. Since the profit margin is constant thus as the sales increased the operating profit also increased (Tosun 2019). Thus this ratio also shows that the company is consistent in its performance.
The Return on Equity in the financial statement analysis defines the net income available for the equity investment for the year. This is computed by dividing the income available for equity shareholders by total equity investment including equity share capital, reserves and surplus, etc. The financial data used to conduct financial statement analysis shows the return on equity was decreasing till 2017 thereafter the recovery has started taking place. The shareholders are getting more gain on their invested capital in the company.
The financial leverage used in financial statement analysis shows the amount of borrowings of the company as compared to its equity. The capital structures of external funds to its internal funds are compared. The leverage is very variable and is highest in the year 2019. The company has increased its dependence in the external borrowing. As per the financial statement analysis, the benefit from the use of external borrowing can be realised only of Cost of Borrowing is less than the return on Investment of the company. This will increase the shareholder’s wealth.
Ratio trend
In case of financial statement analysis, these are the major expenses which have lead to the change in the profit margin and changed the most in the recent years. The cost of sales which have incurred from the year 2017 has increased up to 70% and thus is the major reason for decrease in profits from the year 2015. Second major expense is the employee’s benefits expense. It is not only large in volume that is quantity but also in its variations (Flight Centre Group Ltd 2019). In 5 years the expense has increased by 24% and this high amount of expenditure has huge impact on profitability of the company. The other miscellaneous expenses which are considered minor have also varied the most. It has increased by almost 50% over the 5 years (Flight Centre Group Ltd 2019). The given are essential part of the operations and controlling this would not be an easy task but the proper methods should be taken to reduce cost and increase the profit margin.
The above chart used for financial statement analysis shows the major assets and liabilities of the company which have largely impacted the asset efficiency of the company. The main assets representing the liquidity of the company that is cash and cash equivalents are showing a decreasing trend which is alarming and if not maintained may lead to defaults in the payments of the company. As against the cash and cash equivalents, the payables and borrowings of the company are showing the increasing trend. The profitability of the company declined and the leverage increased. The company has more payables and even have taken more external borrowings (Morecroft 2015). The financial statement analysis signifies that the high borrowings will increase the burden on the profit of the company and even the interest fixed payments would further impact the liquidity position of the company.
Thus from the given financial statement analysis, the company is maintaining its margin as per the industrial standards but the dependence on the borrowing is impacting the capital structure. Moreover, the trend of the past five years indicates a declining trend in most of the ratio. The required return of the equity shareholders will rise as the leverage of the company increases as compared to other companies in the same industry.
Break down and financial statement analysis
Profitability ratios are a measure of the profitability of the company generated through its operating activities. It displays the capacity of the company in the transformation of the sales into profit generation for the company (Shuli 2011). The profitability ratios also give an idea about the company’s control over its expenses and taxes. The profitability ratio refers to the percentage of the revenue or turnover that is left behind after the company has paid for all its taxes and expenses. Flight Center’s profitability ratios for the last five years are worked out as below:
The net profit after tax has been used for working out this ratio during financial statement analysis. It can be observed that the company had a profitability margin of 11% in the year 2015 which fell to 9% in the succeeding year and then to 8% (Flight Centre Group Ltd 2019).
Over the last two years, it has remained more or less constant at 9%. This ratio is used by the creditors and bankers for judging the capability of the company in repayment of their loans and dues. If this ratio is falling, it is an indicator that the expenses of the company are increasing and the company needs to keep over the same to finance its capital and other requirements (Sherman 2015). In the case of Flight Australia, as the profits margin has fallen from 11% to 9% in the last five years, the financial statement analysis of the significant expenses that have contributed to this fall is done.
Considering the financial statement analysis, it can be stated that there has been a significant increase in the employee benefits expenses in the last 5 years. It works out to a 24% increase for the year 2019 in comparison to the year 2015. The reason for the same can be attributed to the general price rise of commodities and services over the world in the last few years. The second significant expense item is the cost of sales which has been incorporated as a separate expense item from the year 2017 and onwards (Flight Centre Group Ltd 2019). In the financial statement analysis of last three years itself it has witnessed an increase of 70% which is pretty high in this arena. The item of other expenses has increased by 49% in the year 2019 when compared to the year 2015. The company has certainly added new items under this category and it has also led to a significant price increase in total. Thus the combined impact of these expenses has brought down the net profit margin of the company.
Asset Turnover Ratios
This is another significant ratio considered in the financial statement analysis to analyze how effectively the company has been able to utilize its assets and lead to profit generation for the company. It is the comparison of the revenues of the company with the value of the assets to understand the amount of revenue generated for every dollar of asset owned by the company (Pucheta-Martiinez & Garcia-Meca 2019). Companies with a high asset turnover ratio enjoy a good financial standing and reputation in the market. Based on the financial statement analysis of annual reports of Flight Center for the last five years, the asset turnover ratio has been worked out as below:
The results imply that for every \$1 of the assets, the company is able to generate 89 cents of revenue. It would be significant to note that over the last five years, the assets turnover ratio has remained more or less constant with just a little decrease. The decrease is also not very significant in comparison to the other market conditions and the company performance. Yet financial statement analysis is done on the significant items that have led to this decline in the assets turnover ratio over years.
The movement in a few items of assets and liabilities has been analyzed below:
Cash and cash equivalents are the most liquid form of current assets available with any company. It is always necessary to maintain a high quick ratio while performing financial statement analysis to ensure the liquidity of the company. At Flight Center, it can be observed that over the years, there has been a decline in the balance of cash and cash equivalents held by the company.
The decline is calculated at 15% in the last five years. The reason for the same can be attributed to the company’s increasing obligations to meet its operating expenses and the cash required for funding its working capital requirements (Flight Centre Group Ltd 2019). The company should still strive towards maintaining a high amount of cash and cash equivalents by proper financial statement analysis.
Trade and other payables refer to the current liabilities that are repayable in less than a year’s time. There has been an increase of 53% in the balance of trade and other payables in the last five years. This also contributes towards the decrease in the current assets and subsequently has an impact on the assets turnover ratio.
Borrowings are usually short term (current) and long term (non-current). The table above shows the comparison of the current borrowings that refers to short term liabilities. According to the financial statement analysis, there has been a significant increase in this balance of borrowings over the last five years and this increase is worked out as 158%. This item has a significant impact on the decrease in the assets turnover ratio.
Financial statement analysis: Cash flow analysis
Liquidity ratios: An ideal current ratio is considered to be around 1.33, As per the ratios calculated, we see that the current ratio is appropriate and sufficient enough to cover its current liabilities (Peirson et al 2015). The company current assets are sufficient and the working capital in the company is good. The company is perfectly using its working capital. A 1:1 or slightly higher ratio is considered to be an ideal quick ratio. The company has a less than 1:1 ratio which shows that the company liquid assets are not sufficient to cover its entire current liabilities. There is a lack of liquid assets in the company. Lastly, the cash ratio represents the ratio of cash and marketable securities and current liabilities. This ratio shows the company’s ability to repay its current liabilities using cash and its equivalents like marketable securities. In this case of financial statement analysis, the cash ratio is below one which shows that company cash and its equivalents are not sufficient to pay off current liabilities.
Solvency ratios: These ratios depict the company's financial statement analysis and ability to pay off its debts. Debts include short term and long term debts. The first ratio is the ratio of long term debts and total equity. This ratio is 0.07 which shows that the company Long term debts are negligible in comparison to the total equity of the company. Next is the ratio of Total Debt and Total equity which is 0.15. This ratio indicates that the company’s capital is healthy and the company does not require long or short term debts for business operations (Mersland & Urgeghe 2013). Next ratio in this segment is the ratio of Total debt and total assets which is a paltry ratio of 0.05. This indicates that the company assets are owned by the company and there are hardly any loans against it. Next ratio is financial leverage ratio which is Total assets and total equity. This ratio based on the financial statement analysis is approx 2.39 which shows that the company has more than twice the capital/equity of the company. The Company’s financial health is very good the company has got quite a good amount of retained earnings as well. The proprietary ratio is the opposite of financial leverage and is the ratio of total equity upon total assets. This ratio shows that equity is 42 % of total assets.
Cash flow ratios: These ratios for financial statement analysis show the cash flows of the company which are equally important in business. The current liability coverage ratio which is cash flow from operating activities upon average current liabilities is quite low at 0.16 which shows that the operating activities are not producing sufficient cash to cover its current liabilities, Next is Share price upon cash flows per share which comes around 15.05 which shows that the company has better value. Even the top S&P 500 companies have this ratio around 14-15. Next is cash flow margin ratio which suggests the company ability to convert its sales in cash flows (Carlon 2019). The company is not able to convert its sales into cash flows; its ratio is quite less at around 0.06 which is very less. The last ratio used in financial statement analysis is Cash flow coverage ratio which is suggestive the company's capacity to use its cash flows from operations to pay off its debts (Carlon 2019). The company does not have much of debts while its cash flows are sufficient and cover its debts by 1.5 times approx. This indicates a healthy financial situation for the company.
References
Carlon, S., 2019, Financial accounting: reporting, analysis and decision making. Financial statement analysis 6th ed. Milton, QLD John Wiley and Sons Australia, Ltd
Davydov, D 2016, Debt structure and corporate performance in emerging markets. Research in International Business and Finance, vol. 38,pp. 299-311.
Deegan, C. M 2016, Financial accounting, McGraw-Hill Education.
Flight Centre Group Ltd 2015, Flight Centre Group Ltd 2015annual report & accounts, viewed 15 May 2020, https://www.fctgl.com/wp-content/uploads/2017/08/2015-Annual-Report.pdf
Flight Centre Group Ltd 2016, Flight Centre Group Ltd 2016 annual report & accounts, viewed 15 May 2020, http://www.annualreports.com/Company/flight-centre-travel-group-ltd
Flight Centre Group Ltd 2017, Flight Centre Group Ltd 2017 annual report & accounts, viewed 15 May 2020, https://www.fctgl.com/wp-content/uploads/2017/09/Flight-Centre-Travel-Group-Annual-Report-2017.pdf
Flight Centre Group Ltd 2018, Flight Centre Group Ltd 2018 annual report & accounts, viewed 15 May 2020, https://www.fctgl.com/wp-content/uploads/2018/09/Computershare-FLT-Final-Annual-Report.pdf
Flight Centre Group Ltd 2019, Flight Centre Group Ltd 2019 annual report & accounts, viewed 15 May 2020, https://www.fctgl.com/wp-content/uploads/2019/10/FLT-Annual-Report-FY19.pdf
Hitchner, J. R 2016, Financial valuation, John Wiley & Sons, Incorporated.
Mersland, R., & Urgeghe, L 2013, International Debt Financing and Performance of Microfinance Institutions, Strategic Change vol. financial statement analysis 22, pp. 36-47,
Morecroft, JD 2015, Strategic modelling and business dynamics: A feedback systems approach, John Wiley & Sons, Hoboken,
Peirson, G, Brown, R., Easton, S, Howard, P. & Pinder, S 2015, Finance, 12th ed. North Ryde: McGraw-Hill Australia.
Pucheta-Martiinez, M. & Garcia-Meca, E 2019, Monitoring, corporate performance and institutional directors. Australian Accounting Review, vol. 29, no. 1, pp. 208-219. doi:10.1111/auar.12262
Sherman, E 2015, A manager's guide to financial analysis : Powerful tools for analyzing the numbers and making the best decisions for your business, Ama Self-Study
Shuli, I 2011, Earnings management and the quality of the financial reporting. Perspective of Innovation in Economics and Buisness (PIEB), vol. 8, no. 2, pp. 45- 48.
Tosun, O 2019, Why do large shareholders adopt a short-term versus a long-term investment horizon in different firms? Financial statement analysis Financial Review, vol. 53, no. 4, pp. 763-800.
Appendix
Balance Sheet
Assets Fiscal year is July-June. All values AUD Millions. 2019 2018 2017 2016 2015 Cash & Short Term Investments 352 553 535 1,466 1,458 Cash Only 337 445 426 1,316 1,378 Short-Term Investments 16 108 109 150 80 Cash & Short Term Investments Growth -36.30% 3.42% -63.53% 0.55% - Cash & ST Investments / Total Assets 10.00% 16.14% 16.59% 48.40% 51.96% Total Accounts Receivable 932 868 809 757 661 Accounts Receivables, Net 559 525 762 672 635 Accounts Receivables, Gross 578 540 775 683 642 Bad Debt/Doubtful Accounts -18 -15 -13 -11 -7 Other Receivables 373 343 46 84 26 Accounts Receivable Growth 7.38% 7.38% 6.88% 14.44% - Accounts Receivable Turnover 3.28 3.37 3.39 3.45 3.57 Inventories 2 2 1 2 2 Finished Goods - - - - 2 Progress Payments & Other - - - 2 - Other Current Assets 1,022 - - - 32 Prepaid Expenses 54 50 42 39 32 Miscellaneous Current Assets 968 972 951 - - Total Current Assets 2,308 2,446 2,338 2,263 2,153 Net Property, Plant & Equipment 240 248 256 216 196 Property, Plant & Equipment - Gross 626 585 562 533 466 Buildings 34 33 33 32 45 Machinery & Equipment 592 551 528 502 421 Accumulated Depreciation 386 337 305 317 269 Buildings 10 9 8 6 8 Machinery & Equipment 376 328 297 311 261 Total Investments and Advances 86 - - - 17 LT Investment - Affiliate Companies 86 39 69 19 15 Other Long-Term Investments - - - 2 2 Long-Term Note Receivable 0 6 5 - - Intangible Assets 769 586 471 448 386 Net Goodwill 599 460 372 367 340 Net Other Intangibles 170 127 100 81 47 Other Assets 19 - - - - Tangible Other Assets 19 5 2 - - Total Assets 3,524 3,425 3,223 3,029 2,806 Assets - Total - Growth 2.89% 6.28% 6.41% 7.95% - Asset Turnover 0.88 - - - - Return On Average Assets 7.59% - - - - Liabilities & Shareholders' Equity All values AUD Millions. 2019 2018 2017 2016 2015 ST Debt & Current Portion LT Debt 85 35 56 77 33 Short Term Debt 12 10 56 77 17 Current Portion of Long Term Debt 72 25 - - 16 Accounts Payable 1,458 1,448 1,386 1,311 943 Accounts Payable Growth 0.67% 4.48% 5.76% 38.96% - Income Tax Payable 11 20 8 9 6 Other Current Liabilities 202 - - - 470 Accrued Payroll 101 93 83 79 64 Miscellaneous Current Liabilities 100 102 102 91 406 Total Current Liabilities 1,755 1,699 1,635 1,567 1,453 Current Ratio 1.31 1.44 1.43 1.44 1.48 Quick Ratio 1.31 1.44 1.43 1.44 1.48 Cash Ratio 0.2 0.33 0.33 0.94 1 Long-Term Debt 100 1 - - - Long-Term Debt excl. Capitalized Leases 100 1 - - - Non-Convertible Debt 100 1 - - - Provision for Risks & Charges 48 41 37 31 36 Deferred Taxes -56 -58 -42 -46 -36 Deferred Taxes - Credit 47 38 39 36 18 Deferred Taxes - Debit 103 96 82 81 54 Other Liabilities 111 - - - 29 Other Liabilities (excl. Deferred Income) 63 76 60 34 29 Deferred Income 48 56 23 16 - Total Liabilities 2,062 - - - 1,536 Total Liabilities / Total Assets 58.51% 0.00% 0.00% 0.00% 54.74% Common Equity (Total) 1,462 1,510 1,429 1,346 1,270 Common Stock Par/Carry Value 406 404 403 399 396 Retained Earnings 1,053 1,109 1,015 923 837 Cumulative Translation Adjustment/Unrealized For. Exch. Gain 29 -8 4 22 28 Unrealized Gain/Loss Marketable Securities 0 1 0 0 - Revaluation Reserves - - - - 0 Other Appropriated Reserves -14 15 8 3 9 Treasury Stock -12 -11 -2 - - Common Equity / Total Assets 41.48% 44.08% 44.33% 44.44% 45.26% Total Shareholders' Equity 1,462 1,510 1,429 1,346 1,270 Total Shareholders' Equity / Total Assets 41.48% 44.08% 44.33% 44.44% 45.26% Accumulated Minority Interest 0 5 - - - Total Equity 1,462 1,515 1,429 1,346 1,270 Liabilities & Shareholders' Equity 3,524 3,425 3,223 3,029 2,806
Income Statement
Fiscal year is July-June. All values AUD Millions. 2019 2018 2017 2016 2015 Sales/Revenue 3,055 2,923 2,740 2,612 2,363 Sales Growth 4.53% 6.69% 4.89% 10.53% - Cost of Goods Sold (COGS) incl. D&A 1,832 - - - - COGS excluding D&A 1,749 1,639 1,543 - - Depreciation & Amortization Expense 83 78 75 67 54 Depreciation 64 65 61 56 47 Amortization of Intangibles 19 14 14 10 7 COGS Growth 0.00% - - - - Gross Income 1,223 1,206 1,121 - - Gross Income Growth 1.43% 7.57% 0.00% - - Gross Profit Margin 40.03% - - - - SG&A Expense 880 - - - 1,781 Other SG&A 880 828 795 1,997 1,781 SGA Growth 0.00% - - 0.00% - EBIT 344 - - - 528 Unusual Expense 28 - - - 0 Non Operating Income/Expense 27 2 -2 -178 -165 Non-Operating Interest Income 24 25 27 26 30 Interest Expense 25 - - - 27 Interest Expense Growth 0.00% - - 0.00% - Gross Interest Expense 25 25 28 28 27 Pretax Income 342 362 323 344 366 Pretax Income Growth -5.45% 12.05% -6.07% -6.03% - Pretax Margin 11.20% - - - - Income Tax 79 99 95 100 110 Income Tax - Current Domestic 77 108 90 117 104 Income Tax - Deferred Domestic 2 -8 5 -17 5 Equity in Affiliates 1 2 2 1 0 Consolidated Net Income 264 265 231 245 257 Minority Interest Expense 0 1 - - - Net Income 264 - - - 257 Net Income Growth 0.00% - - 0.00% - Net Margin 8.64% - - - - Net Income After Extraordinaries 264 0 0 0 257 Net Income Available to Common 264 - - - 257 EPS (Basic) 2.36 2.35 2.06 2.18 2.29 EPS (Basic) Growth 0.20% 14.27% -5.73% -4.83% - Basic Shares Outstanding 112 112 112 112 112 EPS (Diluted) 2.34 2.34 2.05 2.18 2.29 EPS (Diluted) Growth 0.12% 14.13% -6.00% -4.72% - Diluted Shares Outstanding 113 113 113 112 112 EBITDA 427 - - - 582 EBITDA Growth 0.00% - - 0.00% - EBITDA Margin 13.96% - - - - EBIT 344 - - - 528
Cash flow Statement
Operating activities Fiscal year is July-June. All values AUD Thousands. 2019 2018 2017 2016 2015 Net Income before Extra ordinaries 2,64,174.00 2,64,782.00 2,30,773.00 2,44,556.00 2,56,553.00 Net Income Growth -0.23% 14.74% -5.64% -4.68% - Depreciation, Depletion & Amortization 82,370.00 77,802.00 74,975.00 66,091.00 54,103.00 Depreciation and Depletion 63,069.00 64,222.00 61,030.00 55,893.00 46,703.00 Amortization of Intangible Assets 19,301.00 13,580.00 13,945.00 10,198.00 7,400.00 Deferred Taxes & Investment Tax Credit -106 -12,861.00 2,045.00 -16,578.00 4,823.00 Deferred Taxes -106 -12,861.00 2,045.00 -16,578.00 4,823.00 Other Funds -1,577.00 -3,043.00 730 37,195.00 -16,063.00 Funds from Operations 3,44,861.00 3,26,680.00 3,08,523.00 3,31,264.00 2,99,416.00 Changes in Working Capital -65,410.00 -11,810.00 -13,169.00 25,969.00 63,786.00 Receivables - - - -12,033.00 -72,708.00 Inventories -34 -407 490 138 -739 Accounts Payable -26,470.00 41,070.00 - 23,470.00 1,44,047.00 Income Taxes Payable -13,575.00 12,657.00 -14,180.00 7,308.00 -11,891.00 Other Assets/Liabilities -25,331.00 -65,130.00 521 7,086.00 5,077.00 Net Operating Cash Flow 2,79,451.00 - - - 3,63,202.00 Net Operating Cash Flow Growth 0.00% - - 0.00% - Net Operating Cash Flow / Sales 9.15% 0.00% 0.00% 0.00% 15.37% Investing Activities All values AUD Thousands. 2019 2018 2017 2016 2015 Capital Expenditures -1,00,982.00 - - - -82,850.00 Capital Expenditures (Fixed Assets) -53,352.00 -50,957.00 -75,786.00 -93,852.00 -72,496.00 Capital Expenditures (Other Assets) -47,630.00 -36,326.00 -28,338.00 -27,136.00 -10,354.00 Capital Expenditures Growth 0.00% - - 0.00% - Capital Expenditures / Sales -3.31% 0.00% 0.00% 0.00% -3.51% Net Assets from Acquisitions -1,44,937.00 -60,019.00 -9,646.00 -55,895.00 -527 Sale of Fixed Assets & Businesses - - - 17,201.00 - Purchase/Sale of Investments 36,932.00 - - - -31,577.00 Purchase of Investments -75,946.00 -2,692.00 -54,100.00 -1,41,175.00 -39,342.00 Sale/Maturity of Investments 1,12,878.00 3,017.00 11,780.00 10,029.00 7,765.00 Other Uses -9,883.00 - - - - Net Investing Cash Flow -2,18,870.00 - - - -1,14,954.00 Net Investing Cash Flow Growth 0.00% - - 0.00% - Net Investing Cash Flow / Sales -7.16% 0.00% 0.00% 0.00% -4.86% Financing Activities All values AUD Thousands. 2019 2018 2017 2016 2015 Cash Dividends Paid - Total -3,19,441.00 -1,55,629.00 -1,38,339.00 -1,58,354.00 -1,53,108.00 Common Dividends -3,19,441.00 -1,55,629.00 -1,38,339.00 -1,58,354.00 -1,53,108.00 Change in Capital Stock -3,967.00 - - - 4,697.00 Repurchase of Common & Preferred Stk. -9,837.00 -13,449.00 -2,816.00 - - Sale of Common & Preferred Stock 5,870.00 5,147.00 5,614.00 3,534.00 4,697.00 Proceeds from Stock Options 5,870.00 5,147.00 5,614.00 3,534.00 4,697.00 Issuance/Reduction of Debt, Net 1,48,686.00 - - - -17,426.00 Other Funds -346 - - - - Other Uses -346 -35 - - - Net Financing Cash Flow -1,75,068.00 - - - -1,65,837.00 Net Financing Cash Flow Growth 0.00% - - 0.00% - Net Financing Cash Flow / Sales -5.73% 0.00% 0.00% 0.00% -7.02% Exchange Rate Effect 13,747.00 9,648.00 -11,853.00 -18,541.00 33,892.00 Net Change in Cash -1,00,740.00 -8,656.00 -33,738.00 -62,599.00 1,16,303.00 Free Cash Flow 2,26,099.00 - - - 2,90,706.00 Free Cash Flow Growth 0.00% - - 0.00% - Free Cash Flow Yield -2.22% - - - -
Ratio computation
(in \$'000) Year ended 30 June 2019 2018 2017 2016 2015 Profit after tax 2,64,174 2,64,782 2,30,773 2,44,556 2,56,553 Revenue 30,55,268 29,22,985 27,39,734 26,66,176 23,96,989 Profit Margin 9% 9% 8% 9% 11% (in \$'000) Year ended 30 June 2019 2018 2017 2016 2015 Employee Benefits 15,91,965 15,10,897 14,51,041 14,32,796 12,83,661 Cost of Sales 1,57,231 1,27,843 92,370 - - Other expenses 5,21,537 4,84,756 4,37,953 4,00,624 3,49,853 (in \$'000) Year ended 30 June 2019 2018 2017 2016 2015 Revenue 30,55,268 29,22,985 27,39,734 26,66,176 23,96,989 Average Assets 34,46,207 33,00,354 30,98,402 28,94,641 25,99,177 Asset Turnover Ratio 0.89 0.89 0.88 0.92 0.92 (in \$'000) Year ended 30 June 2019 2018 2017 2016 2015 Cash and Cash equivalents 11,72,252 12,72,992 12,81,648 13,15,984 13,77,985 Trade and other payables 15,17,845 15,65,288 15,14,210 14,29,572 9,90,500 Borrowings 84,710 34,846 55,866 76,845 32,806 (in \$'000) Year ended 30 June 2019 2018 2017 2016 2015 Profit after tax 2,64,174 2,64,782 2,30,773 2,44,556 2,56,553 Average Assets 34,46,207 33,00,354 30,98,402 28,94,641 25,99,177 Return on Net Assets 8% 8% 7% 8% 10%
Year ended 30 June 2019 2018 2017 2016 2015 Profit after tax 2,64,174 2,64,782 2,30,773 2,44,556 2,56,553 Revenue 30,55,268 29,22,985 27,39,734 26,66,176 23,96,989 Profit Margin 9% 9% 8% 9% 11% (in \$'000) Year ended 30 June 2019 2018 2017 2016 2015 Employee Benefits 15,91,965 15,10,897 14,51,041 14,32,796 12,83,661 Cost of Sales 1,57,231 1,27,843 92,370 - - Other expenses 5,21,537 4,84,756 4,37,953 4,00,624 3,49,853 (in \$'000) Year ended 30 June 2019 2018 2017 2016 2015 Revenue 30,55,268 29,22,985 27,39,734 26,66,176 23,96,989 Average Assets 34,46,207 33,00,354 30,98,402 28,94,641 25,99,177 Asset Turnover Ratio 0.89 0.89 0.88 0.92 0.92 (in \$'000) Year ended 30 June 2019 2018 2017 2016 2015 Cash and Cash equivalents 11,72,252 12,72,992 12,81,648 13,15,984 13,77,985 Trade and other payables 15,17,845 15,65,288 15,14,210 14,29,572 9,90,500 Borrowings 84,710 34,846 55,866 76,845 32,806
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Anonymous
Not applicable
## Help with estimating
Hello
I have a simple file which is based on a regulatory returns a company submits to a regulatory body. The company submits this file at the end of each financial year and the revenue amongst other things is then used to determine regulatory fees. If a company does not submit on time then their fees are based on their previous year's revenue. I wanted to be able to calculate the company's (c in the below example) estimated revenue for FY19/20 as they have yet to submit. I also wanted to calculate the whole revenue for all companies for FY19/20 including the estimated values however I am stuck on how to start both points as it means trying to create a value for company c for FY19/20 when it doesn't exist. I'm sure there must be a solution and I hope you can help.
Jo
Company Financial Year Revenue A FY17/18 10 A FY18/19 20 A FY19/20 30 B FY17/18 15 B FY18/19 32 B FY19/20 45 C FY17/18 20 C FY18/19 40
4 REPLIES 4
Community Support
@Anonymous
I create a sample with blank revenue for FY19/20. You need to create an Index Column to get the revenue from previous row (previews FY) as Est Revenue.
BTW, I am not sure if this is your requirement, but it is not possible to add a row from no where in to the current table, you need to have the Company or FY in the first place.
``Est Revenue = IF('Table'[Revenue]=BLANK(),CALCULATE(SUM('Table'[Revenue]),FILTER('Table',[Index]=EARLIER('Table'[Index])-1),ALLEXCEPT('Table','Table'[Company])))``
Paul Zheng _ Community Support Team
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Super User
So what is your formula for calculating estimated revenue? Average of the previous years? maximum, minimum?
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Anonymous
Not applicable
Hi Greg
Apologies for not mentioning in the original post. It's just basically the revenue from the previous financial year.
Jo
Super User
I would use an Enter Data query to enter your estimates. Then you could use UNION to union the two tables together and perform operations on them.
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https://visualfractions.com/unit-converter/convert-10-arcsec-to-arcmin/
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# Convert 10 arcsec to arcmin
So you want to convert 10 arcseconds into arcminutes? If you're in a rush and just need the answer, the calculator below is all you need. The answer is 0.16666646666707 arcminutes.
## How to convert arcseconds to arcminutes
We all use different units of measurement every day. Whether you're in a foreign country and need to convert the local imperial units to metric, or you're baking a cake and need to convert to a unit you are more familiar with.
Luckily, converting most units is very, very simple. In this case, all you need to know is that 1 arcsec is equal to 0.016666646666707 arcmin.
Once you know what 1 arcsec is in arcminutes, you can simply multiply 0.016666646666707 by the total arcseconds you want to calculate.
So for our example here we have 10 arcseconds. So all we do is multiply 10 by 0.016666646666707:
10 x 0.016666646666707 = 0.16666646666707
## What is the best conversion unit for 10 arcsec?
As an added little bonus conversion for you, we can also calculate the best unit of measurement for 10 arcsec.
What is the "best" unit of measurement? To keep it simple, let's say that the best unit of measure is the one that is the lowest possible without going below 1. The reason for this is that the lowest number generally makes it easier to understand the measurement.
For 10 arcsec the best unit of measurement is arcseconds, and the amount is 10 arcsec.
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