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https://pwntestprep.com/2021/08/could-you-please-explain-number-7-on-page-29/
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Could you please explain number 7 on page 29?
This is in the plug in chapter because you can get it simply by making up numbers that add up to 180° inside the triangle. Say, for example, that the angles inside the triangle are 50°, 60°, and 70°. Now use supplemental angle rules to fill in the exterior angles of the triangle. For example, the 50° angle will have 130° angles on either side of it.
Now add them all up and you’ll get 900°.
If you want to know why that works, note that a full circle is 360°, and each of the marked angles is almost a full circle, but missing one piece. And it turns out, the missing pieces are all vertical angles of the angles inside the triangle. Since you know a triangle has 180°, you know that the sum of all the marked angles is 3(360°)-180°.
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# Inferences
Comprehension is one of the key parts of learning English Grammar. The word comprehension has arrived from ‘comprehend’ which means ‘to understand’. Inference or Infer in reading comprehension is the ability to understand the meaning of a passage of text by reading in between the lines when all the information is not provided. Inferences might be difficult for a few students but they can be simplified for better understanding.
## Understanding Inferences
There are a few ways on how to begin figuring out what the infer of a given passage is:
• We need to look out for clues, some may be obvious and some may be subtle
• Look for keywords
• There can be more than one answer or point to a particular aspect of the passage.
• Whatever inferences we draw from the passage, we must have substantial reasons to support it.
Source: Pinterest
When we are posed with a question about the inference of a given passage, we need to ask the most obvious questions to be able to derive an infer.
• What is the inference of this passage?
• From the given passage can I use that will help me get to an infer.
• What points can I use to support the inference I have derived?
Let us now look at an example with questions and answers:
### Example 1
Every day after work Paul took his muddy boots off on the steps of the front porch. Alice would have a fit if the boots made it so far as the welcome mat. He then took off his dusty overalls and threw them into a plastic garbage bag; Alice left a new garbage bag tied to the porch railing for him every morning.
On his way in the house, he dropped the garbage bag off at the washing machine and went straight up the stairs to the shower as he was instructed. He would eat dinner with her after he was “presentable,” as Alice had often said.
• What type of job does Paul work?
Ans: Paul does some kind of a job as a laborer, miner, digger etc. that requires him to get dirty.
How did we land at this inference? The passage mentions “muddy boots’ and ‘took off his dusty overalls’, both of which indicate the nature of his job.
• Describe Alice.
Ans: From the passage, we come to know that Alice preferred to keep her house clean and worked around to make sure no dirt from Paul’s shoes or clothes would dirty the house. She had a good control of her house.
How did we reach this inference? The passage gives us clues such as “Alice would have a fit if the boots made it so far as the welcome mat” and ” Alice left a new garbage bag tied to the porch railing for him every morning”.
## Solved Example for You on Infer
Passage:
Crack! Thunderstruck and rain poured. Max stared blankly out the window, trying to contain his emotions that raged like the weather. He was beginning to lose it. Dropping the kite from his hand, Max broke out into a full sob. His mother comforted him, “There, there, Max. We’ll just find something else to do.” She began to unpack the picnic basket that was on the counter and offered him a sandwich. Max snapped, “I don’t wanna sand-mich!” A flash from the sky lit up the living room. Boom! Mom sighed.
Question: Why was Max upset?
1. Lightning light up the room with a loud noise
2. He didn’t want to eat the sandwich
3. It is raining and he cannot go outside to play
4. He wanted to go on a picnic.
Sol. (c) It is raining and he cannot go outside to play. The passage mentions ‘ Dropping the kite from his hand, Max broke out into a full sob’ which indicates that Max was ready to go fly his kite until the thunderstruck and rain poured.
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Q.1
Q.2
In a triangle, sides are in A.P. and biggest angle is twice of smallest, then the ratio of sides are
Q.3
The value of $\mathrm{cos}52°+\mathrm{cos}68°+\mathrm{cos}172°$ is
Q.4
$\mathrm{tan}70°$ is equals to
Q.5
The angle of elevation of top of a tower from a point P on a plane field is $30°$ . If the height of the tower is 100m, then the distance of point P from the bottom of the tower wil be
Q.6
Angle of elevtion of top of a tower from a point on earth is $30°$ . After walking 30m towards the building, the angle of elevation of top of the building becomes $60°$ . What is the height of the building?
Q.7
$\sqrt{\frac{1-\mathrm{sin}A}{1+\mathrm{sin}A}}=?$
Q.8
The value of $\mathrm{sin}30°+\mathrm{tan}45°-\mathrm{cos}60°$ is
Q.9
If tan $\mathrm{tan}\theta =\frac{1}{2}$ and $\mathrm{tan}\phi =\frac{1}{3}$ then the value of $\theta +\phi$ will be
Q.10
$\mathrm{tan}1°.\mathrm{tan}2°.\mathrm{tan}3°.\mathrm{tan}4°........\mathrm{tan}80°$ is equals to
Q.11
The value of $\mathrm{cos}20°.\mathrm{cos}40°.\mathrm{cos}60°.\mathrm{cos}80°$ is
Q.12
The value of $\mathrm{sin}10°.\mathrm{sin}30°.\mathrm{sin}50°.\mathrm{sin}70°$ is
Q.13
The value of $\mathrm{tan}20°.\mathrm{tan}40°.\mathrm{tan}60°.\mathrm{tan}80°$ is
Q.14
Maximum and minimum value of $7\mathrm{cos}\theta +24\mathrm{sin}\theta$ is
Q.15
The value of (tanA+secA-1) cos A is
Q.16
The value of $\frac{\mathrm{sec}x-1+\mathrm{tan}x}{\mathrm{tan}x-\mathrm{sec}x+1}$ will be
Q.17
In a triangle ABC if a = 4, b = 3 and $\mathrm{sin}A=\frac{4}{3}$ then the value of $\angle B$ is
Q.18
$\frac{\sqrt{1+\mathrm{sin}x}+\sqrt{1-\mathrm{sin}x}}{\sqrt{1+\mathrm{sin}x}-\sqrt{1-\mathrm{sin}x}}$ is equal to
Q.19
The value of $\mathrm{cot}41°.\mathrm{cot}42°.\mathrm{cot}43°.\mathrm{cot}44°.\mathrm{cot}45°.\mathrm{cot}46°.\mathrm{cot}47°.\mathrm{cot}48°.\mathrm{cot}49°$ is equals to
Q.20
If ${\mathrm{tan}}^{2}B=\frac{1-\mathrm{sin}A}{1+\mathrm{sin}A}$ ,the what will be the value A + 2B
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The Solar Panel Requirements Calculator
Mar 16, · How Many Solar Panels do you Need for kWh per Month? The first step is calculating the kilowatts needed. You must simply divide the average daily kWh by the peak sun hours. Assuming a day month, an electricity generation of 1, kWh is equivalent to kWh per day. If the site gets 6. Now calculate what size solar panel you will need. Based on a ten hour day of light, the calculation is simple: 1, Watt hours / 10 hours sunlight = Watt solar panel. The reality is that most summer days give about 15 hours of sunlight and winter you get about hours of sunlight.
We estimate that a typical home needs between 20 and 25 solar panels to cover percent of its electricity usage. The formula we used to estimate the number of solar panels you need to power your home depends on three key factors: annual energy usagepanel wattageand production ratios.
What does that mean exactly? Here are the assumptions we made, and how we did our math:. Annual electricity usage : Your annual electricity usage is the amount of electricity you use in your home over a full year. Measured in kilowatt-hours kWhthis number is influenced by the appliances in your home that use electricity and how often you use them. Refrigerators, air conditioning units, small kitchen appliances, lights, chargers, and more all use electricity.
According to the U. Wattage is measured in watts Wand most solar panels fall in the range of — watts of power. These numbers are almost never — depending on how much sunlight your system will get which is primarily based on your geographic locationyour production ratio will change accordingly. For example, a 10 kW system that produces 14 kWh of electricity in a year has a production ratio of 1. In the U. We have our three main assumptions energy use, solar panel wattage, and production ratios — now how do those numbers translate to an estimated number of solar panels for your home?
The formula looks like this:. The amount of power kWh your solar energy system can produce depends on how much sunlight exposure your roof receives, which in turn how to determine solar panel needs your production ratio.
The amount of sunlight you get in a year depends on both where you are in how to determine solar panel needs country, and what time of year it is. For instance, California has more sunny days annually than New England.
The California household needs about a seven kW system to cover percent of their energy needs. By comparison, the comparable household in Massachusetts needs about a nine kW system to cover their energy needs. Homeowners in less sunny areas, like Massachusetts, can make up for this disparity by simply using more efficient panels or increasing the what apple cider vinegar should i drink of their solar energy system, resulting in slightly more solar panels on their rooftop.
In our long example at the beginning of this piece, we determined that an 8 kW system would probably cover the average energy use for an American household if you live in an area with a production ratio of 1. For California shoppers, this might actually be realistic, but for folks in the Northeast or areas with less sun, these estimates might be a bit high on the production end and low on the number of panels needed. Below is a table that will give you a sense of how much space your system will take up on your roof, depending on the power output of the solar panels you select.
Perhaps one of the most difficult aspects of sizing a solar panel system is estimating annual energy usage for your household.
A number of larger consumer products or add-ons can significantly change your annual kWh requirements and greatly impact how many panels you will need. By reviewing the various kWh requirements for everyday household appliances and products, one thing is clear: certain add-ons will dramatically change monthly energy use, and can have an outsized impact on the size solar panel system you should install.
For example, pairing your electric vehicle with solar panels is a great way to reduce carbon emissions and improve energy efficiency; however, it should be planned accordingly considering it could potentially double the size of your PV system. Though it is certainly possible to install a solar system and then add more panels later to accommodate increased energy needs, the most pragmatic option is to size your system as accurately as possible based on your expected purchases—such as an how to prune a dieffenbachia vehicle, swimming pool or central air system.
There are multiple variables to consider when seeking how to implement stack in java the best solar panels on the market. For any homeowner in the early what do jehovah witness believe of shopping for solar that would just like a ballpark estimate for an installation, try our Solar Calculator that offers upfront cost and long-term savings estimates based on your location and roof type.
For those looking to get quotes from local contractors today, check out our quote comparison platform. I am considering solar for a commercial project comprised of residential units and 24 retail spaces that I am designing. I am trying to assess the aesthetic impact. Your email address will not be published. Enter your zip code to find out what solar panels cost from installers near you.
2. How many watts do you currently use?
For now, calculate 30% of your homemade solar panel's rating. This takes into consideration the cheaper glass or plexiglass used in DIY panels, heating of the panels and lack of expensive electronics. For example if your solar panel is rated at watts, figure on about 30 watts output. Do not panic, we will raise these numbers later. Calculate Solar Panel Costs for Your Home. Determine your average monthly electricity usage. You probably consume more electricity during certain months of the year. Refer to your electricity Calculate your daily kWh usage. Estimate the amount of sunlight your solar panels will receive. Factor. Feb 18, · Solar panel wattage: Also known as a solar panel’s power rating, panel wattage is the electricity output of a specific solar panel under ideal conditions. Wattage is measured in watts (W), and most solar panels fall in the range of – watts of power. We’re using watts as an average panel in these calculations.
Follow the Off Grid Project and learn how to calculate your own solar energy requirements. The following article will guide you step by step through calculating how many solar panels you need, how large a battery bank, what sized power inverter and what size solar charge controller. This website will be constantly updated as the project continues, so check back often. Making your own solar panels is the first step.
The next step is to calculate just how much power you really need. You can always make more solar panels as your energy consumption increases. When you start out off the grid, there will be a lot of changes in your household. One of these is watching your power use. In the Off Grid Project, You will learn how much energy the average household appliance consumes.
Learn how to make the calculations needed to figure out your total household energy requirements. Then learn how to put it all together. Check out the other projects covered on the list to the right. Solar panel output is actually lower than the manufacturer rated numbers under normal circumstances.
The Off Grid Project will show you how to increase those numbers. This takes into consideration the cheaper glass or plexiglass used in DIY panels, heating of the panels and lack of expensive electronics. For example if your solar panel is rated at watts, figure on about 30 watts output. Do not panic, we will raise these numbers later. But when you first install a homamade solar panel, do not be surprised at a much lower charging current than you had expected.
This also allows for clouds and partially sunny conditions. The sun does not shine The sun is also not shining directly on your solar panels at all times. Depending on where you live and where you have the solar panels installed, you will get varying hours of direct sunlight. In the setting of The Off Grid Project the sun rises just after 5am in the summer, but is not hitting the solar panels until about am due to mountains and trees in the way.
The same goes for sunset. The sun sets around but has already left the solar panels after pm. That gives about 8 hours of full sunlight on the panels. You can use the Free Solar Panel Requirements Calculator to calculate how many solar panels and what equipment you will need.
Take note that this calculator shows the full manufacturers rated solar panel output. The projects on the pages listed below will be constantly updated throughout the coming months. Please bookmark this site and check back often. Instructions, photos and video will be added regularly as the project progresses.
List of Projects How to make a solar panel How to calculate your solar power requirements How to install your solar panels How to make a windmill on a budget How to use off grid power efficiently How to make passive solar heaters How to make a solar water heater How to provide fresh water How to provide running water How to dispose of solid and liquid waste How to make a rocket stove How to make a geothermal cooling system How to move to natural cleaners.
#### 3 thoughts on “How to determine solar panel needs”
• ###### Akinoshicage
04.07.2021 in 20:12
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# ๐ก Temperature model (codim 2)
This is a classical example from the Trilinos library.
This is a simple example in which we aim at solving $\Delta T+\alpha N(T,\beta)=0$ with boundary conditions $T(0) = T(1)=\beta$. This example is coded in examples/chan.jl. We start with some imports:
using Revise, BifurcationKit, LinearAlgebra, Plots, Parameters
const BK = BifurcationKit
N(x; a = 0.5, b = 0.01) = 1 + (x + a*x^2)/(1 + b*x^2)
We then write our functional:
function F_chan(x, p)
@unpack ฮฑ, ฮฒ = p
f = similar(x)
n = length(x)
f[1] = x[1] - ฮฒ
f[n] = x[n] - ฮฒ
for i=2:n-1
f[i] = (x[i-1] - 2 * x[i] + x[i+1]) * (n-1)^2 + ฮฑ * N(x[i], b = ฮฒ)
end
return f
end
We want to call a Newton solver. We first need an initial guess:
n = 101
sol0 = [(i-1)*(n-i)/n^2+0.1 for i=1:n]
# set of parameters
par = (ฮฑ = 3.3, ฮฒ = 0.01)
Finally, we need to provide some parameters for the Newton iterations. This is done by calling
optnewton = NewtonPar(tol = 1e-11, verbose = true)
We call the Newton solver:
prob = BifurcationProblem(F_chan, sol0, par, (@lens _.ฮฑ),
# function to plot the solution
plot_solution = (x, p; k...) -> plot!(x; ylabel="solution", label="", k...))
sol = @time newton( prob, optnewton)
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
โ Newton step residual linear iterations โ
โโโโโโโโโโโโโโโฌโโโโโโโโโโโโโโโโโโโโโโโฌโโโโโโโโโโโโโโโโโค
โ 0 โ 2.3440e+01 โ 0 โ
โ 1 โ 1.3774e+00 โ 1 โ
โ 2 โ 1.6267e-02 โ 1 โ
โ 3 โ 2.4336e-06 โ 1 โ
โ 4 โ 6.7452e-12 โ 1 โ
โโโโโโโโโโโโโโโดโโโโโโโโโโโโโโโโโโโโโโโดโโโโโโโโโโโโโโโโโ
0.000908 seconds (361 allocations: 2.302 MiB)
Note that, in this case, we did not give the Jacobian. It was computed internally using Automatic Differentiation.
We can perform numerical continuation w.r.t. the parameter $\alpha$. This time, we need to provide additional parameters, but now for the continuation method:
optcont = ContinuationPar(dsmin = 0.01, dsmax = 0.2, ds= 0.1, p_min = 0., p_max = 4.2,
newton_options = NewtonPar(max_iterations = 10, tol = 1e-9))
Next, we call the continuation routine as follows.
br = continuation(prob, PALC(), optcont; plot = true)
nothing #hide
The parameter axis lens = @lens _.ฮฑ is used to extract the component of par corresponding to ฮฑ. Internally, it is used as get(par, lens) which returns 3.3.
Tip
We don't need to call newton first in order to use continuation.
You should see
scene = title!("") #hide
The left figure is the norm of the solution as function of the parameter $p=\alpha$, the y-axis can be changed by passing a different recordFromSolution to BifurcationProblem. The top right figure is the value of $\alpha$ as function of the iteration number. The bottom right is the solution for the current value of the parameter. This last plot can be modified by changing the argument plotSolution to BifurcationProblem.
Bif. point detection
Two Fold points were detected. This can be seen by looking at br.specialpoint, by the black dots on the continuation plots when doing plot(br, plotfold=true) or by typing br in the REPL. Note that the bifurcation points are located in br.specialpoint.
## Continuation of Fold points
We get a summary of the branch by doing
br
โโ Curve type: EquilibriumCont
โโ Number of points: 58
โโ Type of vectors: Vector{Float64}
โโ Parameter ฮฑ starts at 3.3, ends at 4.2
โโ Algo: PALC
โโ Special points:
- # 1, bp at ฮฑ โ +4.04103799 โ (+4.04103799, +4.04115545), |ฮดp|=1e-04, [converged], ฮด = ( 1, 0), step = 5
- # 2, bp at ฮฑ โ +3.15565319 โ (+3.15565221, +3.15565319), |ฮดp|=1e-06, [converged], ฮด = (-1, 0), step = 23
- # 3, endpoint at ฮฑ โ +4.20000000, step = 57
We can take the first Fold point, which has been guessed during the previous continuation run and locate it precisely. However, this only works well when the jacobian is computed analytically. We use automatic differentiation for that
# index of the Fold bifurcation point in br.specialpoint
indfold = 2
outfold = newton(
#index of the fold point
br, indfold)
BK.converged(outfold) && printstyled(color=:red, "--> We found a Fold Point at ฮฑ = ", outfold.u.p, ", ฮฒ = 0.01, from ", br.specialpoint[indfold].param,"\n")
--> We found a Fold Point at ฮฑ = 3.155650731611167, ฮฒ = 0.01, from 3.1556531887068466
We can finally continue this fold point in the plane $(ฮฑ, ฮฒ)$ by performing a Fold Point continuation. In the present case, we find a Cusp point.
Tip
We don't need to call newton first in order to use continuation for the codim 2 curve of bifurcation points.
outfoldco = continuation(br, indfold,
# second parameter axis to use for codim 2 curve
(@lens _.ฮฒ),
# we disable the computation of eigenvalues, it makes little sense here
ContinuationPar(optcont, detect_bifurcation = 0))
scene = plot(outfoldco, plotfold = true, legend = :bottomright)
Tip
The performances for computing the curve of Fold is not that great. It is because we use the default solver tailored for ODE. If you pass jacobian_ma = :minaug to the last continuation call, you should see a great improvement in performances.
## Using GMRES or another linear solver
We continue the previous example but now using Matrix Free methods. The user can pass its own solver by implementing a version of LinearSolver. Some linear solvers have been implemented from KrylovKit.jl and IterativeSolvers.jl (see Linear solvers (LS) for more information), we can use them here. Note that we can also use preconditioners as shown below. The same functionality is present for the eigensolver.
# derivative of N
dN(x; a = 0.5, b = 0.01) = (1-b*x^2+2*a*x)/(1+b*x^2)^2
# Matrix Free version of the differential of F_chan
# Very easy to write since we have F_chan.
# We could use Automatic Differentiation as well
function dF_chan(x, dx, p)
@unpack ฮฑ, ฮฒ = p
out = similar(x)
n = length(x)
out[1] = dx[1]
out[n] = dx[n]
for i=2:n-1
out[i] = (dx[i-1] - 2 * dx[i] + dx[i+1]) * (n-1)^2 + ฮฑ * dN(x[i], b = ฮฒ) * dx[i]
end
return out
end
# we create a new linear solver
ls = GMRESKrylovKit(dim = 100)
# and pass it to the newton parameters
optnewton_mf = NewtonPar(verbose = true, linsolver = ls, tol = 1e-10)
# we change the problem with the new jacobian
prob = re_make(prob;
# we pass the differential a x,
# which is a linear operator in dx
J = (x, p) -> (dx -> dF_chan(x, dx, p))
)
# we can then call the newton solver
out_mf = @time newton(prob, optnewton_mf)
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
โ Newton step residual linear iterations โ
โโโโโโโโโโโโโโโฌโโโโโโโโโโโโโโโโโโโโโโโฌโโโโโโโโโโโโโโโโโค
โ 0 โ 2.3440e+01 โ 0 โ
โ 1 โ 1.3774e+00 โ 68 โ
โ 2 โ 1.6267e-02 โ 98 โ
โ 3 โ 2.4336e-06 โ 74 โ
โ 4 โ 5.6275e-12 โ 63 โ
โโโโโโโโโโโโโโโดโโโโโโโโโโโโโโโโโโโโโโโดโโโโโโโโโโโโโโโโโ
0.980502 seconds (1.41 M allocations: 95.330 MiB, 99.63% compilation time)
We can improve this computation, i.e. reduce the number of Linear-Iterations, by using a preconditioner
using SparseArrays
# define preconditioner which is basically ฮ
P = spdiagm(0 => -2 * (n-1)^2 * ones(n), -1 => (n-1)^2 * ones(n-1), 1 => (n-1)^2 * ones(n-1))
P[1,1:2] .= [1, 0.];P[end,end-1:end] .= [0, 1.]
# define gmres solver with left preconditioner
ls = GMRESIterativeSolvers(reltol = 1e-4, N = length(sol.u), restart = 10, maxiter = 10, Pl = lu(P))
optnewton_mf = NewtonPar(verbose = true, linsolver = ls, tol = 1e-10)
out_mf = @time newton(prob, optnewton_mf)
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
โ Newton step residual linear iterations โ
โโโโโโโโโโโโโโโฌโโโโโโโโโโโโโโโโโโโโโโโฌโโโโโโโโโโโโโโโโโค
โ 0 โ 2.3440e+01 โ 0 โ
โ 1 โ 1.3777e+00 โ 3 โ
โ 2 โ 1.6266e-02 โ 3 โ
โ 3 โ 2.3699e-05 โ 2 โ
โ 4 โ 4.8929e-09 โ 3 โ
โ 5 โ 5.6333e-12 โ 4 โ
โโโโโโโโโโโโโโโดโโโโโโโโโโโโโโโโโโโโโโโดโโโโโโโโโโโโโโโโโ
1.127177 seconds (2.38 M allocations: 163.401 MiB, 99.93% compilation time)
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https://www.netlib.org/lapack/explore-html-3.6.1/da/d0b/group__complex16__lin_ga18ef0428f1704f00a93eb3fad38c9151.html
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LAPACK 3.6.1 LAPACK: Linear Algebra PACKage
subroutine zget03 ( integer N, complex*16, dimension( lda, * ) A, integer LDA, complex*16, dimension( ldainv, * ) AINV, integer LDAINV, complex*16, dimension( ldwork, * ) WORK, integer LDWORK, double precision, dimension( * ) RWORK, double precision RCOND, double precision RESID )
ZGET03
Purpose:
ZGET03 computes the residual for a general matrix times its inverse:
norm( I - AINV*A ) / ( N * norm(A) * norm(AINV) * EPS ),
where EPS is the machine epsilon.
Parameters
[in] N N is INTEGER The number of rows and columns of the matrix A. N >= 0. [in] A A is COMPLEX*16 array, dimension (LDA,N) The original N x N matrix A. [in] LDA LDA is INTEGER The leading dimension of the array A. LDA >= max(1,N). [in] AINV AINV is COMPLEX*16 array, dimension (LDAINV,N) The inverse of the matrix A. [in] LDAINV LDAINV is INTEGER The leading dimension of the array AINV. LDAINV >= max(1,N). [out] WORK WORK is COMPLEX*16 array, dimension (LDWORK,N) [in] LDWORK LDWORK is INTEGER The leading dimension of the array WORK. LDWORK >= max(1,N). [out] RWORK RWORK is DOUBLE PRECISION array, dimension (N) [out] RCOND RCOND is DOUBLE PRECISION The reciprocal of the condition number of A, computed as ( 1/norm(A) ) / norm(AINV). [out] RESID RESID is DOUBLE PRECISION norm(I - AINV*A) / ( N * norm(A) * norm(AINV) * EPS )
Date
November 2011
Definition at line 112 of file zget03.f.
112 *
113 * -- LAPACK test routine (version 3.4.0) --
114 * -- LAPACK is a software package provided by Univ. of Tennessee, --
115 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
116 * November 2011
117 *
118 * .. Scalar Arguments ..
119 INTEGER lda, ldainv, ldwork, n
120 DOUBLE PRECISION rcond, resid
121 * ..
122 * .. Array Arguments ..
123 DOUBLE PRECISION rwork( * )
124 COMPLEX*16 a( lda, * ), ainv( ldainv, * ),
125 \$ work( ldwork, * )
126 * ..
127 *
128 * =====================================================================
129 *
130 * .. Parameters ..
131 DOUBLE PRECISION zero, one
132 parameter ( zero = 0.0d+0, one = 1.0d+0 )
133 COMPLEX*16 czero, cone
134 parameter ( czero = ( 0.0d+0, 0.0d+0 ),
135 \$ cone = ( 1.0d+0, 0.0d+0 ) )
136 * ..
137 * .. Local Scalars ..
138 INTEGER i
139 DOUBLE PRECISION ainvnm, anorm, eps
140 * ..
141 * .. External Functions ..
142 DOUBLE PRECISION dlamch, zlange
143 EXTERNAL dlamch, zlange
144 * ..
145 * .. External Subroutines ..
146 EXTERNAL zgemm
147 * ..
148 * .. Intrinsic Functions ..
149 INTRINSIC dble
150 * ..
151 * .. Executable Statements ..
152 *
153 * Quick exit if N = 0.
154 *
155 IF( n.LE.0 ) THEN
156 rcond = one
157 resid = zero
158 RETURN
159 END IF
160 *
161 * Exit with RESID = 1/EPS if ANORM = 0 or AINVNM = 0.
162 *
163 eps = dlamch( 'Epsilon' )
164 anorm = zlange( '1', n, n, a, lda, rwork )
165 ainvnm = zlange( '1', n, n, ainv, ldainv, rwork )
166 IF( anorm.LE.zero .OR. ainvnm.LE.zero ) THEN
167 rcond = zero
168 resid = one / eps
169 RETURN
170 END IF
171 rcond = ( one / anorm ) / ainvnm
172 *
173 * Compute I - A * AINV
174 *
175 CALL zgemm( 'No transpose', 'No transpose', n, n, n, -cone, ainv,
176 \$ ldainv, a, lda, czero, work, ldwork )
177 DO 10 i = 1, n
178 work( i, i ) = cone + work( i, i )
179 10 CONTINUE
180 *
181 * Compute norm(I - AINV*A) / (N * norm(A) * norm(AINV) * EPS)
182 *
183 resid = zlange( '1', n, n, work, ldwork, rwork )
184 *
185 resid = ( ( resid*rcond ) / eps ) / dble( n )
186 *
187 RETURN
188 *
189 * End of ZGET03
190 *
double precision function dlamch(CMACH)
DLAMCH
Definition: dlamch.f:65
subroutine zgemm(TRANSA, TRANSB, M, N, K, ALPHA, A, LDA, B, LDB, BETA, C, LDC)
ZGEMM
Definition: zgemm.f:189
double precision function zlange(NORM, M, N, A, LDA, WORK)
ZLANGE returns the value of the 1-norm, Frobenius norm, infinity-norm, or the largest absolute value ...
Definition: zlange.f:117
Here is the call graph for this function:
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http://math.stackexchange.com/questions/202532/gaining-insight-into-the-inverse-image-sheaf
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# Gaining insight into the Inverse Image Sheaf
Let $f: X \rightarrow Y$ be a continuous map of topological spaces and let $G$ be a sheaf of sets on $Y$. I am trying to understand the definition of the inverse image sheaf $f^{-1}G$ on $X$. This is the sheaf associated to the preshief $U \rightarrow \varinjlim_{V \supseteq f(U)} G(V)$ where $V$ is open in $Y$ and $U$ open in $X$. In particular i am trying to understand the quantity $\varinjlim_{V \supseteq f(U)} G(V)$. As i understand, this is a colimit in the category of sets. I read the category-theoretic definition of the colimit using co-cones, but i am having a hard time making a connection/interpretation. Also, is this a filtered colimit or just a colimit?
-
It's a filtered colimit – if two open sets contain $f(U)$, then so does their intersection. – Zhen Lin Sep 26 '12 at 1:36
@ZhenLin: Thanks, i am gonna work with this hint... – Manos Sep 26 '12 at 2:15
1) First of all you have the crucial property $(f^{-1}G)_x=G_{f(x)}$ for all $x\in X$.
In particular, applying this to the inclusion $i:\lbrace x\rbrace \hookrightarrow X$, of a point, you get the interesting equality $(i^{-1}G)_x=G_x$ which shows that taking an inverse image is a generalization of taking the stalk of a sheaf at a point.
2) This is a situation where the pre-Grothendieck interpretation of a sheaf as an étalé space is quite illuminating:
If $Et(G)\to Y$ is the étalé space corresponding to $G$, take the fiber product $Z\stackrel {def}{=}X\times _Y Et(G)\to X$ in the category of topological spaces.
This is an étalé space over $X$ and the corresponding sheaf of sections is the sheaf on $X$ you are looking for: $Sh_Z=f^{-1}G$.
3) Beware that in the category of schemes (or in any other geometric category like that of analytic spaces) the important pull-back for sheaves of $\mathcal O_Y$-modules over $Y$ is not the topological one we are discussing but the algebraic one, namely $$f^{*}G =f^{-1}G \otimes _{f^{-1}(\mathcal O_Y)} \mathcal O_X$$
The simplest contrasting example is that of a scheme $X$ over a field $k$ and the inclusion of a rational point $i:\lbrace x\rbrace \hookrightarrow X$. There is a huge difference between both pull-backs of $\mathcal O_X$:
$$(i^{-1}\mathcal O_X)_x = \mathcal O_{X,x}\neq (i^{*}\mathcal O_X)_x =k$$
-
+1 Georges. I only realised the power of passing to the stalk when trying to prove that sheafification commutes with restriction of a sheaf to an open set! – user38268 Jul 8 '13 at 11:20
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A274369 Let the starting square of Langton's ant have coordinates (0, 0), with the ant looking in negative x-direction. a(n) is the x-coordinate of the ant after n moves. 4
0, 0, 1, 1, 0, 0, -1, -1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 2, 2, 1, 1, 2, 2, 3, 3, 2, 2, 1, 1, 2, 2, 3, 3, 4, 4, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 1, 1, 0, 0, -1, -1, 0, 0, -1, -1, 0, 0, -1, -1, -2, -2, -1, -1, -2, -2, -3, -3, -2, -2, -1, -1, -2, -2, -3 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,21 LINKS Rémy Sigrist, Table of n, a(n) for n = 0..15000 Felix Fröhlich, Coordinates of Langton's ant. Wikipedia, Langton's ant. FORMULA a(n+104) = a(n) + 2 for n > 9975. - Andrey Zabolotskiy, Jul 05 2016 PROG (Python) # A274369: Langton's ant by Andrey Zabolotskiy, Jul 05 2016 def ant(n): steps = [(1, 0), (0, 1), (-1, 0), (0, -1)] black = set() x = y = 0 position = [(x, y)] direction = 2 for _ in range(n): if (x, y) in black: black.remove((x, y)) direction += 1 else: black.add((x, y)) direction -= 1 (dx, dy) = steps[direction%4] x += dx y += dy position.append((x, y)) return position print ([p[0] for p in ant(100)]) # change p[0] to p[1] to get y-coordinates CROSSREFS Cf. A274370 (y-coordinate). Cf. A102358, A102369, A204810, A255938, A261990, A269757. Sequence in context: A165123 A318439 A106180 * A055091 A014678 A164516 Adjacent sequences: A274366 A274367 A274368 * A274370 A274371 A274372 KEYWORD sign,look AUTHOR Felix Fröhlich, Jun 19 2016 STATUS approved
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Last modified January 25 01:49 EST 2020. Contains 331229 sequences. (Running on oeis4.)
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## Elementary Technical Mathematics
Published by Brooks Cole
# Chapter 2 - Section 2.1 - Addition of Signed Numbers - Signed Numbers Drill 1 - Page 109: 6
-14
#### Work Step by Step
Add the absolute value of (-7) and (-7) which is 7 + 7 = 14 and place a negative sign in front of the result for -14
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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Highlighted
Frequent Visitor
## Max slicer date as input for dateadd
I am trying to make a measure wich always calculates an average over the past 5 months, based on the max selected date in a slicer
The problem i have is that the dateadd function doenst work with a calculation like, does someone have a solution.
Parts of the Function at the moment (there are some more variables but they work fine, problem is the bold parts):
var periode = DATESBETWEEN(V_PeriodeMaand[EersteDagVdMaand], DATEADD(max(V_PeriodeMaand[EersteDagVdMaand]), -4, MONTH) , [Laatstedatumselectie]
RETURN
CALCULATE((Verzuimdagen/Beschikbaredagen), FILTER(V_PeriodeMaand, V_PeriodeMaand[EersteDagVdMaand] in(periode)))
In case the language causes any problems, in essence this is the var:
DATESBETWEEN(Period, DATEADD(max(Period), -4, MONTH) , max(Period)
2 ACCEPTED SOLUTIONS
Accepted Solutions
Highlighted
Resolver III
Hello, @MM_NL
The Syntax of DATEADD function is
DATEADD(<dates>,<number_of_intervals>,<interval>)
<dates> is column, not scalar value. MAX
(Period) isn't work. You must set column, if you need only one date, you can use construction something like this:
DATEADD ( FILTER ( Table[Period], Table[Period] = MAX (Table[Period] ) ), .......)
Highlighted
Frequent Visitor
I solved it, by replacing the table input in the filter expression by lastdate() on the date column
Old:
DATEADD(FILTER(Period, Period[Datecolumn] = max(Period[Datecolumn]))
New
DATEADD(FILTER(LASTDATE(Period[Datecolumn]), Period[Datecolumn] = max(Period[Datecolumn]))
this way the filter expression accepted a column input instead of whole table
Thank you Popov!
4 REPLIES 4
Highlighted
Resolver III
Hello, @MM_NL
The Syntax of DATEADD function is
DATEADD(<dates>,<number_of_intervals>,<interval>)
<dates> is column, not scalar value. MAX
(Period) isn't work. You must set column, if you need only one date, you can use construction something like this:
DATEADD ( FILTER ( Table[Period], Table[Period] = MAX (Table[Period] ) ), .......)
Highlighted
Frequent Visitor
I tried this and the error i get now is:
"A table expression containing more than one column was specified in the call to function 'dateadd'. this is not supported"
Highlighted
Resolver III
Can you share your formula?
Highlighted
Frequent Visitor
I solved it, by replacing the table input in the filter expression by lastdate() on the date column
Old:
DATEADD(FILTER(Period, Period[Datecolumn] = max(Period[Datecolumn]))
New
DATEADD(FILTER(LASTDATE(Period[Datecolumn]), Period[Datecolumn] = max(Period[Datecolumn]))
this way the filter expression accepted a column input instead of whole table
Thank you Popov!
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https://byjus.com/question-answer/let-a-1-2-3-b-1-3-5-if-relation-r-from-a-to-1/
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Question
# Let A = [1, 2, 3], B = [1, 3, 5]. If relation R from A to B is given by = {(1, 3), (2, 5), (3, 3)}, Then R−1 is (a) {(3, 3), (3, 1), (5, 2)} (b) {(1, 3), (2, 5), (3, 3)} (c) {(1, 3), (5, 2)} (d) None of these
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Solution
## (a) {(3, 3), (3, 1), (5, 2)} A = {1, 2, 3}, B ={1, 3, 5} R = {(1, 3), (2, 5), (3, 3)} ∴ R−1 = {(3,1),(5,2),(3,3)}
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Main
# Blackbody problem
""
#### Contents
Code
set term svg
set termoption enhanced
set encoding utf8
set output 'Black_body.svg'
unset key
set tics nomirror out
set border 3
set xrange [0:6]
set yrange [0:15]
set ylabel "Spectral radiance (kW • sr⁻¹ • m⁻² • nm⁻¹)"
set xlabel "Wavelength (μm)"
set label "5000 K" at 0.45,13.1
set label "4000 K" at 0.6,4.55
set label "3000 K" at 0.8,1.4
set label "Classical theory (5000 K)" at 1.55,11
set grid lc rgb "light-blue"
# length unit is micrometre
c=3e14 # speed of light
h=6.626e-22 # Planck constant
k=1.38e-11 # Boltzmann constant
# Planck curves
p1(x)=1e-6*2*h*c**2/(x**5*(exp(h*c/(x*k*3000))-1))
p2(x)=1e-6*2*h*c**2/(x**5*(exp(h*c/(x*k*4000))-1))
p3(x)=1e-6*2*h*c**2/(x**5*(exp(h*c/(x*k*5000))-1))
p4(x)=1e-6*2*h*c**2/(x**5*(exp(h*c/(x*k*6000))-1))
# Rayleigh-Jeans curve
rj(x)=1e-6*2*c*k*5000/(x**4)
plot p1(x) lw 2, p2(x) lw 2, p3(x) lw 2, p4(x) lw 2
plot rj(x) lw 2 lc rgb "black"
Sources:
Darth Kule. Produced by GNUPLOT 4.4 patchlevel 0, background spectrum based on http://commons.wikimedia.org/wiki/File:Linear_visible_spectrum.svg Dedicated to the public domain by the author.
gnuplot 4.2 Copyright (C) 1986 - 1993, 1998, 2004, 2007 Thomas Williams, Colin Kelley
Permission to use, copy, and distribute this software and its documentation for any purpose with or without fee is hereby granted, provided that the above copyright notice appear in all copies and that both that copyright notice and this permission notice appear in supporting documentation.
Recommended:
Categories: Gunplot Recipes
Tags:
This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.
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https://www.topperlearning.com/answer/the-perimeter-of-an-equilate/sj27ig8uu
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Request a call back
The perimeter of an equilateral triangle is 18 cm. What is the length of a side?
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
Equilateral triangles are those which have all the sides of equal measure. It means 3 equal sides. Let s be the side of the triangle.
Here, perimeter = 18 cm
So, Perimeter = 3 x s
18 = 3s [swap both sides]
3s = 18 s = = 6
Hence, the length of each side of the triangle is 6 cm.
Answered by | 04 Jun, 2014, 03:23: PM
## Application Videos
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https://www.gradesaver.com/textbooks/engineering/computer-science/computer-science-an-overview-global-edition/chapter-1-data-storage-section-1-6-storing-integers-questions-exercises-page-64/10
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## Computer Science: An Overview: Global Edition (12th Edition)
$$\begin{array}{|c|c|}\hline & {\text { Answer }} \\ \hline a & {1101} \\ \hline b & {0011} \\ \hline c & {1011} \\ \hline d & {1000} \\ \hline e & {1111} \\ \hline f & {0000} \\ \hline\end{array}$$
$$\begin{array}{|c|c|}\hline & {\text { Answer }} & {\text { Reason }} \\ \hline a & {1101} & { because\ 5 + 8 = 13 → 1101\ } \\ \hline b & {0011} & { because\ -5 + 8 = 3 → 0011\ } \\ \hline c & {1011} & { because\ 3 + 8 = 11 → 1011\ } \\ \hline d & {1000} & { because\ 0 + 8 = 8 → 1000\ } \\ \hline e & {1111} & { because\ 7 + 8 = 15 → 1111\ } \\ \hline f & {0000} & { because\ -8 + 8 = 0 → 0000\ } \\ \hline\end{array}$$ ----- you can see the image below:
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http://aptitudetests4me.com/Basic_Numeracy_253.html
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Aptitude Tests 4 Me
Basic Numeracy/Quantitative Aptitude
1055. A and B together have Sterling 1210. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount does B have?
(a) Sterling 460 (b) Sterling 484 (c) Sterling 550 (d) Sterling 664
1056. Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is
(a) 2 : 5 (b) 3 : 5 (c) 4 : 5 (d) 6 : 7
1057. A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Euro 1000 more than D, what is B's share?
(a) Euro 500 (b) Euro 1500 (c) Euro 2000 (d) None of these
1058. Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
(a) 2 : 3 : 4 (b) 6 : 7 : 8 (c) 6 : 8 : 9 (d) None of these
TOTAL
Detailed Solution
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https://newpkjobs.com/qa/question-what-kind-of-point-is-0.html
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# Question: What Kind Of Point Is 0?
## Can energy negative?
When a system does work on its surroundings, energy is lost, therefore E is negative..
## What is origin math?
In mathematics, the origin of a Euclidean space is a special point, usually denoted by the letter O, used as a fixed point of reference for the geometry of the surrounding space. In physical problems, the choice of origin is often arbitrary, meaning any choice of origin will ultimately give the same answer.
## What does Origin mean?
origin, source, inception, root mean the point at which something begins its course or existence. origin applies to the things or persons from which something is ultimately derived and often to the causes operating before the thing itself comes into being.
## Is zero point energy infinite?
In the standard quantum field theory, not only does the vacuum (zero-point) energy have an absolute infinite value, but also all the real excited states have such an irregular value; this is because these energies correspond to the zero-point energy of an infinite number of harmonic oscillators ( ).
## Why zero point energy is not zero?
Actually, every oscillator or quantum system has a residual energy even at absolute zero, such non zero-energy is, (½) νh , which is the residual energy of the quantum system and later called zero point energy. … The “vacuum”, thus has zero point fluctuations as well as the zero point field and hence zero point energy.
## Is zero divided by zero infinity?
Infinity is not a real number, and even if it were, it wouldn’t be the answer to dividing something by zero. There is no number that you can multiply by 0 to get a non-zero number. There is NO solution, so any non-zero number divided by 0 is undefined.
## What is another name for the point located at 0 0?
The horizontal axis in the coordinate plane is called the x-axis. The vertical axis is called the y-axis. The point at which the two axes intersect is called the origin. The origin is at 0 on the x-axis and 0 on the y-axis.
## What is a zero dimensional object?
Basic Dimensions A 0-dimensional object is a point, which has no length, height, or depth. A 1-dimensional object is a line, or line segment, which has length, but no other characteristics. A 2-dimensional object has length and height, but no depth. Examples of 2D objects are planes and polygons.
## Is a dot zero dimensional?
A dot is defined as a figure on a three-dimensional plane having no length, no breadth, and no height. That means it has no dimension.
## Is zero point energy dark energy?
The steady and dynamic states are studied for a spherical cloud of zero point energy photons. The “antigravitational” force due to its pressure gradient then represents dark energy, and its gravitational force due to the energy density represents dark matter.
1 : one-fourth of a circle. 2 : any of the four parts into which something is divided by two imaginary or real lines that intersect each other at right angles. quadrant. noun. quad·rant | \ ˈkwäd-rənt \
## Why does a point has zero dimension?
A point has Hausdorff dimension 0 because it can be covered by a single ball of arbitrarily small radius.
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https://us.metamath.org/mpeuni/gneispacef.html
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Mathbox for Richard Penner < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > gneispacef Structured version Visualization version GIF version
Theorem gneispacef 40825
Description: A generic neighborhood space is a function with a range that is a subset of the powerset of the powerset of its domain. (Contributed by RP, 15-Apr-2021.)
Hypothesis
Ref Expression
gneispace.a 𝐴 = {𝑓 ∣ (𝑓:dom 𝑓⟶(𝒫 (𝒫 dom 𝑓 ∖ {∅}) ∖ {∅}) ∧ ∀𝑝 ∈ dom 𝑓𝑛 ∈ (𝑓𝑝)(𝑝𝑛 ∧ ∀𝑠 ∈ 𝒫 dom 𝑓(𝑛𝑠𝑠 ∈ (𝑓𝑝))))}
Assertion
Ref Expression
gneispacef (𝐹𝐴𝐹:dom 𝐹⟶(𝒫 (𝒫 dom 𝐹 ∖ {∅}) ∖ {∅}))
Distinct variable groups: 𝑛,𝐹,𝑝,𝑓 𝐹,𝑠,𝑓 𝑓,𝑛,𝑝
Allowed substitution hints: 𝐴(𝑓,𝑛,𝑠,𝑝)
Proof of Theorem gneispacef
StepHypRef Expression
1 gneispace.a . . . 4 𝐴 = {𝑓 ∣ (𝑓:dom 𝑓⟶(𝒫 (𝒫 dom 𝑓 ∖ {∅}) ∖ {∅}) ∧ ∀𝑝 ∈ dom 𝑓𝑛 ∈ (𝑓𝑝)(𝑝𝑛 ∧ ∀𝑠 ∈ 𝒫 dom 𝑓(𝑛𝑠𝑠 ∈ (𝑓𝑝))))}
21gneispace2 40822 . . 3 (𝐹𝐴 → (𝐹𝐴 ↔ (𝐹:dom 𝐹⟶(𝒫 (𝒫 dom 𝐹 ∖ {∅}) ∖ {∅}) ∧ ∀𝑝 ∈ dom 𝐹𝑛 ∈ (𝐹𝑝)(𝑝𝑛 ∧ ∀𝑠 ∈ 𝒫 dom 𝐹(𝑛𝑠𝑠 ∈ (𝐹𝑝))))))
32ibi 270 . 2 (𝐹𝐴 → (𝐹:dom 𝐹⟶(𝒫 (𝒫 dom 𝐹 ∖ {∅}) ∖ {∅}) ∧ ∀𝑝 ∈ dom 𝐹𝑛 ∈ (𝐹𝑝)(𝑝𝑛 ∧ ∀𝑠 ∈ 𝒫 dom 𝐹(𝑛𝑠𝑠 ∈ (𝐹𝑝)))))
43simpld 498 1 (𝐹𝐴𝐹:dom 𝐹⟶(𝒫 (𝒫 dom 𝐹 ∖ {∅}) ∖ {∅}))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 399 = wceq 1538 ∈ wcel 2112 {cab 2779 ∀wral 3109 ∖ cdif 3881 ⊆ wss 3884 ∅c0 4246 𝒫 cpw 4500 {csn 4528 dom cdm 5523 ⟶wf 6324 ‘cfv 6328 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2114 ax-9 2122 ax-10 2143 ax-11 2159 ax-12 2176 ax-ext 2773 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-clab 2780 df-cleq 2794 df-clel 2873 df-nfc 2941 df-ral 3114 df-rab 3118 df-v 3446 df-dif 3887 df-un 3889 df-in 3891 df-ss 3901 df-pw 4502 df-sn 4529 df-pr 4531 df-op 4535 df-uni 4804 df-br 5034 df-opab 5096 df-rel 5530 df-cnv 5531 df-co 5532 df-dm 5533 df-rn 5534 df-iota 6287 df-fun 6330 df-fn 6331 df-f 6332 df-fv 6336 This theorem is referenced by: gneispacefun 40827 gneispacern 40828
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https://oeis.org/A182904
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A182904 Number of weighted lattice paths in L_n having no peaks. The members of L_n are paths of weight n that start at (0,0), end on the horizontal axis and whose steps are of the following four kinds: an (1,0)-step with weight 1, an (1,0)-step with weight 2, a (1,1)-step with weight 2, and a (1,-1)-step with weight 1. The weight of a path is the sum of the weights of its steps. A peak is a (1,1)-step followed by a (1,-1)-step. 1
1, 1, 2, 4, 9, 21, 48, 112, 263, 623, 1484, 3550, 8525, 20537, 49612, 120136, 291519, 708699, 1725714, 4208364, 10276173, 25122829, 61486180, 150632012, 369361757, 906462529, 2226297008, 5471757126, 13457326605, 33117622245, 81547372396 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS a(n)=A182903(n,0). REFERENCES M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306. E. Munarini, N. Zagaglia Salvi, On the rank polynomial of the lattice of order ideals of fences and crowns, Discrete Mathematics 259 (2002), 163-177. LINKS Table of n, a(n) for n=0..30. FORMULA G.f.: g=1/sqrt(1-2z-z^2-z^4-2z^5+z^6). Conjecture: n*a(n) +(n-2)*a(n-1) +(-7*n+10)*a(n-2) +3*(-n+2)*a(n-3) +(-n+2)*a(n-4) +(-5*n+14)*a(n-5) +(-5*n+18)*a(n-6) +3*(n-4)*a(n-7)=0. - R. J. Mathar, Jun 14 2016 EXAMPLE a(3)=4. Indeed, denoting by h (H) the (1,0)-step of weight 1 (2), and U=(1,1), D=(1,-1), we have hhh, hH, Hh, and DU. MAPLE g := 1/sqrt(1-2*z-z^2-z^4-2*z^5+z^6): gser := series(g, z = 0, 35):seq(coeff(gser, z, n), n = 0 .. 30); CROSSREFS Cf. A182903. Sequence in context: A084634 A137256 A051164 * A281425 A101891 A119967 Adjacent sequences: A182901 A182902 A182903 * A182905 A182906 A182907 KEYWORD nonn AUTHOR Emeric Deutsch, Dec 16 2010 STATUS approved
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Last modified September 21 07:12 EDT 2023. Contains 365494 sequences. (Running on oeis4.)
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We all want to get back to our lives—go out for a drink, see a movie, hug our parents and grandparents, and bring our cities back to life. When will the virus be under control? And how do we measure that?
Much of the media coverage of COVID-19 talks about R0 or “R-nought,” the average number of people each sick person infects. If R0 is bigger than 1, cases grow exponentially, and an epidemic spreads across the population. But if we can keep R0 below 1, we can limit the disease to isolated outbreaks and keep it under control. This has produced a flood of websites that estimate R0 in each country or each state, treating R0=1 as the definition of success.
But R0 is only an average. Your ability to practice social distancing depends on whether you are a first responder or health care worker, whether you have to work in close quarters, or whether you can work comfortably from home. (I’m one of the lucky few who get paid for writing this from my garden.) It depends on how seriously you take your government’s warnings, and how seriously your government takes the warnings of public health experts. And it depends on the structure of your family and your home.
As a result, R0 varies wildly, not just from region to region, but across social space as well. In New Mexico, Santa Fe has very few new cases, but there is an explosion of cases in rural, low-income areas. As of May 12, 57 percent of our confirmed cases are Native American, even though that community forms just 11 percent of the state’s population. Clearly R0 is larger in some parts of the state and of society than others.
And even if R0 is less than 1, outbreaks can be surprisingly large. Suppose you meet 10 people while you are infectious, and that you infect each one with probability 8 percent. The average number of people you infect is 10×0.08 = 0.8, less than 1. But those you infect may infect others in turn, and so on. If an outbreak starts with you, how many “descendants” will you have? A classic calculation shows that, if R0=0.8, then the average number of people in this chain reaction is 1/(1–0.8)=1/0.2=5. But like R0 itself, this is only an average. Like earthquakes and forest fires, outbreaks have a “heavy tail” where large events are common.
Here is a visualization of 100 random outbreaks. The average size is indeed 5, and most outbreaks are small. But about 1 percent have size 50 or more, 10 times the average, and in this simulation the largest of these 100 outbreaks has size 82. This tail gets heavier if R0 is just below the critical point R0 =1. If R0=0.9, the average outbreak size is 10, but 1 percent have size 140 or more.
This tail has real effects. Imagine 100 small towns, each with a hospital that can handle 10 cases. If every town has the average number of cases, they can all ride out the storm. But there’s a good chance that one of them will have 50 or 100, creating a “hot spot” beyond their ability to respond.
The tail of large events gets even heavier if we add superspreading. We often talk of “superspreaders” as individuals with higher viral loads, or who by choice or necessity interact with many others. But it’s more accurate to talk about superspreading events and situations—like the Biogen meeting, the chorus rehearsal, or the pork processing plant, as well as prisons and nursing homes—where the virus spreads to many of those present.
Suppose that 20 percent of cases generate one new case, 10 percent generate two, 4 percent generate five, and 1 percent “superspread” and generate 20 (and the remaining 65 percent infect no one). The average number of new cases is again R0=0.8. Let’s generate 100 random outbreaks with this new scenario.
The average outbreak size is still five, but now the tail is much heavier. If just one of the 100 original cases is involved in superspreading, we get a large outbreak. If there are several generations of superspreading, the size multiplies. As a result, large outbreaks are quite common, and the largest one in this simulation has 663 people in it.
What does all this mean? First, it can be misleading to look at statewide or national averages and celebrate if R0 seems to be falling below 1. The epidemic could still be raging in particular places or among particular groups.
Second, even if we can hold R0 below 1, we need to prepare for hot spots. Even if the average outbreak is small, large outbreaks will occur, due to superspreading or simply by chance. If we do a fantastic job at testing and contact tracing—using both technology and human effort—we will get this pandemic under control, but for the foreseeable future there will be times and places where it flares up and strains local resources. Through these flare-ups, we have to do our best to help each other, and hope that intelligent, generous voices prevail.
Cristopher Moore is a professor at the Santa Fe Institute. He has published over 150 papers at the boundary between physics and computer science, including the theory of social networks and algorithms for analyzing their structure. He is an elected fellow of the American Physical Society, the American Mathematical Society, and the American Association for the Advancement of Science. With Stephan Mertens, he is the author of The Nature of Computation.
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https://www.neetprep.com/question/5661-Given-systemAsBgCgIf-concentration-C-eqilibrium-increased-factor--willcause-eqilibrium-concentration-B-change-two-times-its-original-value-one-half-its-original-value-times-its-original-valuetimes-its-original-value/54-Chemistry/651-Equilibrium?courseId=8
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# Given a system: If the concentration of C at eqilibrium is increased by a factor 2, it will cause the eqilibrium concentration of B to change to: 1. two times of its original value 2. one half of its original value 3. 2$\sqrt{2}$ times of its original value 4. $\frac{1}{2\sqrt{2}}$ times of its original value
Subtopic: Kp, Kc & Factors Affecting them |
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A weak acid, HA has a Ka of 1.00 x 10-5. If 0.100 mole of this acid is dissolved in one litre of water, the percentage of acid dissociated at equibrium is closest to
1. 99.9%
2. 1.00%
3. 99.9%
4. 0.100%
Subtopic: Ionisation Constant of Acid, Base & Salt |
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If little heat is added to ice $⇌$ liquid equilibrium in a sealed container, then:
1. Pressure will rise
2. Temperature will rise
3. Temperature will fall
4. No change in P and T
Subtopic: Kp, Kc & Factors Affecting them |
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In the equilibrium,
2SO2(g) + O2 (g) $⇌$2SO3(g), the partial pressure of SO2, O2 and SO3 are 0.662, 0.101 and 0.331 atm respectively. What should be the partial pressure of oxygen so that the equilibrium concentration of SO2 and SO3 are equal.
1. 0.4 atm
2. 1.0 atm
3. 0.8 atm
4. 0.25 atm
Subtopic: Kp, Kc & Factors Affecting them |
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Ionisation constant of CH3COOH is 1.7 X 10-5 and concentration of H+ ions is 3.4 X 10-4.Then, find out initial concentration of CH3COOH molecules.
1. 3.4 X 10-4
2. 3.4 X 10-3
3. 6.8 X 10-4
4. 6.8 X 10-3
Subtopic: Ionisation Constant of Acid, Base & Salt |
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The formation of phosgene is represented as,
CO + Cl2 $⇌$COCl2
The reaction is carried out in 500 mL flask. At equilibrium o.3 mole of phosgene, 0.1 mole of CO and 0.1 mole of Cl2 are present. The equilibrium constant of the reaction is:
1. 30
2. 15
3. 5
4. 3
Subtopic: Introduction To Equilibrium |
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For the equilibrium, PCl5 $⇌$ PCl3 + Cl2,
${\mathrm{K}}_{\mathrm{C}}={\mathrm{\alpha }}^{2}/\left(1-\mathrm{\alpha }\right)\mathrm{V}$, temperature remaining constant:
1. Kc will increase with the increase in volume
2. Kc will increase with the decrease in volume
3. Kc will not change with the change in volume
4. Kc may increase or decrease with the change in volume depending upon its numerical value.
Subtopic: Kp, Kc & Factors Affecting them |
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For the reaction,
PCl5(g) PCl3(g) + Cl2 (g)
The forward reaction at constant temperature is favoured by:
1. Introducing an inert gas at a constant volume
2. Introducing chlorine gas at a constant volume
3. Introducing an inert gas at constant pressure
4. None of the above
Subtopic: Le Chatelier's principle |
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In the dissociation of 2HI $⇌$H2 + I2, the degree of dissociation will be influenced by the:
1. Addition of inert gas at constant volume
2. Addition of inert gas at constant pressure
3. Increase of temperature
4. Increase of pressure
Subtopic: Kp, Kc & Factors Affecting them |
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The correct order of increasing [H3O+] in the following aqueous solutions is:
1. 0.01 M H2S < 0.01 M H2SO4<0.01 M NaCl<0.01M NaNO2
2. 0.01 M NaCl< 0.01 M NaNO2<0.01 M H2S< 0.01 M H2SO4
3. 0.01 M NaNO2<0.01 M NaCl<0.01 M H2S<0.01 M H2SO4
4. 0.01 M H2S<0.01 M NaNO2<0.01 M NaCl<0.01 M H2SO4
Subtopic: Kp, Kc & Factors Affecting them |
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https://eanswers.in/math/question5392169
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Agold bracelet is sold for rs. 14,500 at a loss of 20%. what is the cost price of the gold bracelet?
, 18.10.2019 11:00, yogeshsahu40
# Agold bracelet is sold for rs. 14,500 at a loss of 20%. what is the cost price of the gold bracelet?
### Other questions on the subject: Math
Math, 19.08.2019 22:00, helper6462
How to solve non terminal decimal number when bar is before decimal?
Two different dice are tossed together find the probability of getting a sum 10 of the numbers on the two dice
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https://www.w3resource.com/javascript-exercises/javascript-date-exercise-25.php
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# JavaScript: Get a full textual representation of a month
## JavaScript Datetime: Exercise-25 with Solution
Write a JavaScript function to get a full textual representation of a month, such as January or June.
Test Data:
dt = new Date(2015, 10, 1);
console.log(full_month(dt));
"November"
Sample Solution:-
HTML Code:
``````<!DOCTYPE html>
<html>
<meta charset="utf-8">
<title>JavaScript function to get a full textual representation of a month, such as January or June.</title>
<body>
</body>
</html>
```
```
JavaScript Code:
``````Date.longMonths = ['January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December'];
function full_month(dt)
{
return Date.longMonths[dt.getMonth()];
}
dt = new Date();
console.log(full_month(dt));
dt = new Date(2015, 10, 1);
console.log(full_month(dt));
```
```
Sample Output:
```June
November
```
Flowchart:
Live Demo:
See the Pen JavaScript - Get a full textual representation of a month-date-ex-25 by w3resource (@w3resource) on CodePen.
Improve this sample solution and post your code through Disqus
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
## JavaScript: Tips of the Day
Sort Number Arrays
JavaScript arrays come with a built-in sort method. This sort method converts the array elements into strings and performs a lexicographical sort on it by default. This can cause issues when sorting number arrays. Hence here is a simple solution to overcome this problem.
```[0,10,4,9,123,54,1].sort((a,b) => a-b);
>>> [0, 1, 4, 9, 10, 54, 123]
```
You are providing a function to compare two elements in the number array to the sort method. This function helps us the receive the correct output.
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http://www.openwetware.org/index.php?title=Holcombe:ProgrammingInR&oldid=410258
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# Holcombe:ProgrammingInR
### Members
Alex Holcombe
Sarah McIntyre
Fahed Jbarah
• Shih-Yu Lo
• Patrick Goodbourn
Lizzy Nguyen
Alumni
### Other
R is an interactive programming language for statistics. The syntax is very idiosyncratic, and not really in a good way. Try R for programmers for a description. However it may have menu-driven versions maybe available R commander we haven't tried that and another one is pmg GTK maybe here
In the lab we have the book Using R for Introductory Statistics. R_Statistics introduces you to R
Dani has posted some example code and graphs on his personal website.
R reference cheatsheet, also a file here Media:Matlab-python-xref.pdf that gives equivalent code for doing array operations in MATLAB, Python, and R plot parameters
There is a wiki with some good tips here. Also Data frame tips, list of R websites
Functions in R can only return one parameter. Delete nearly everything in memory: rm(list = ls())
### dataframe stuff
Examining your data frame or object, let's say it's called datos
typeof(datos) #returns "list!"
str(datos) #tells you it's a dataframe, number of observations, columns, etc
str(datos)
summary(datos) #good for ggplot objects also
df\$varWithExtraLevels = factor(df\$varWithExtraLevels)
length(df) #number of columns of dataframe
names(df) #names of columns of dataframe
library(Hmisc);
describe(df)
#Calling typeof() on a dataframe returns "list"
rm(objectToBeDeleted)
Don't use the function attach. It seems to leave lots of data in the 'environment' that can cause problems later. Also it makes the code harder to understand.
table(dataRaw\$speed,dataRaw\$relPhaseOuterRing)
## Creating Graphs (usu. ggplot2)
In the lab we usually use the package, ggplot2, for graphs. Ask Sarah about the ggplot2 book.
For custom colour schemes, the Chart of R Colors is a helpful resource. The table with named colours is most useful.
ggplot2 tips:
#where 'g' is your ggplot object
str(g) #gives you everything!
summary(g) #gives a summary
last_plot() #refreshes the last plot and returns the struct
p=p+scale_y_continuous(breaks=c(0,0.5,1),minor_breaks=c(.25,.75)) #changing axis breaks
p=p+opts(title="one-behind errors")
p=p+xlab('relative phase')
opts(panel.grid.minor = theme_blank())+
last_plot() + opts(panel.grid.minor = theme_line(colour = "black") )
facet_grid(.~subject) # rows~columns
g<-g+stat_summary(fun.y=mean,geom="point",position="jitter") #getting jitter to work when you're collapsing across other variables with stat_summary
g=g+ stat_smooth(method="glm",family="binomial",fullrange=TRUE) #adds logistic smoother to plot
#it's impossible to extract the function fits however because they're fit on the fly
#how can I fit an arbitrary function, like a psychometric function, e.g. cumulative gaussian with chance rate and lapse rate? Logistic is restricted to 0->1
order used for scale mapping (color, etc.) is perhaps the order of the levels property of the vector. This gives bizarre results because R's sort(unique(x)) default does not alphabetize strings.
## Debugging in R
How to examine and try things with a questionable variable within a function?
ee <<- resultsMeans #make global, violating all principles of good coding #DEBUG
STOP
## fitting psychometric functions
Malte Kuss hosts the R library PsychoFun on his personal webpage rather than c-ran server. So you must download is package, unzip it, and install it by inside R going to Packages&Data->Package Installer->Local Package Directory->Install, go inside the PsychoFun directory you've unzipped, and click Open.
I needed to constrain width of psychometric function to be quite narrow. Prior I was using followed lognormal distribution. Then if want mode to be say .1, have to feed it a mean parameter of -2.3 because ln(.1) = -2. Unfortunately the PsychoFun code doesn't allow using a negative parameter for that prior, so I had to change the code. To do so, you go into the downloaded version of PsychoFun folder before you install it, where you can find PsychoFun.R in the R subdirectory. I commented out line 56. Then have to reinstall with Package Installer inside R, "Local Package Directory" option, after in my case first deleting original PsychoFun installation in /Library/Frameworks/R.framework/Versions/2.10/Resources/library/
Technical Report explains many more terms than JoV article: Acceptance rate:Next sample in chain only accepted if quantity on p.481 of JoV paper is good Kinetic, Potential energy from Hamiltonian algorithm
## doing ANOVAs etc
I think I had too many error terms reducing error terms
## Dealing with circular data
von Mises vs. wrapped Gaussian,
see Swindale, N. V. (1998). Orientation tuning curves: empirical description and estimation of parameters. Biol Cybern, 78(1), 45-56.
## Setting up a proxy in R on a Mac
The easiest way to set up a proxy is simply to create a file called ".Rprofile" in your user directory (~ or Users/username/) with the line:
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## Wavefolding?
DSP, Plug-in and Host development discussion.
joshb
KVRist
85 posts since 13 Apr, 2016
Can anyone give me any insight into a wavefolding algorithm? I believe it's also referred to as foldover?
Thx
valhallasound
KVRAF
3426 posts since 15 Nov, 2006 from Pacific NW
joshb wrote:Can anyone give me any insight into a wavefolding algorithm? I believe it's also referred to as foldover?
Thx
Try this:
out = sinf(gain*in);
As gain gets larger, the output will start to fold back on itself. Very similar to the Serge wave multiplier, but with some nice sine curving, instead of the triangle wave folding found in the Serge circuit.
joshb
KVRist
85 posts since 13 Apr, 2016
Thanks, but that doesn't do what I was expecting, which was mirroring around a threshold. Maybe that's what you're referring to with the Serge?
2DaT
KVRist
316 posts since 21 Jun, 2013
joshb
KVRist
85 posts since 13 Apr, 2016
Wow, thank you. That's very cool. Both the demo animation and the actual function.
"apply triangle"...is that a known function? Never heard of it and it really beats the hell out of the 18 ugly lines of code I threw together:
Code: Select all
``````float fold(float x)
{
float sign = 1.0f;
if(x < 0.0f)
sign = -1.0;
x *= sign;
if(x > threshold)
{
const float remainder = std::fmod(x, threshold);
const int numFolds = (int)std::floor(x / threshold);
float y;
if(numFolds % 2 == 0)
y = remainder;
else
y = threshold - remainder;
return y * sign;
}
return x * sign;
}``````
mystran
KVRAF
5032 posts since 12 Feb, 2006 from Helsinki, Finland
2DaT wrote:Take input, apply triangle.
https://www.desmos.com/calculator/ge2wvg2wgj
Oh... that round() trick is really neat, saves some remapping when compared to using fmod(). Have to try to remember it.
If you'd like Signaldust to return, please ask Katinka Tuisku to resign.
2DaT
KVRist
316 posts since 21 Jun, 2013
mystran wrote: Oh... that round() trick is really neat, saves some remapping when compared to using fmod(). Have to try to remember it.
Code: Select all
``````//Range (-0.5,0.5)
__m128 fold(__m128 x)
{
__m128i int_round = _mm_cvtps_epi32(x);
__m128 frac = _mm_sub_ps(x,_mm_cvtepi32_ps(int_round));
return _mm_xor_ps(frac,_mm_castsi128_ps(_mm_slli_epi32(int_round,31)));
}
``````
Here's another. Not plain C, but i guess you know sse2
2DaT
KVRist
316 posts since 21 Jun, 2013
joshb wrote: "apply triangle"...is that a known function? Never heard of it and it really beats the hell out of the 18 ugly lines of code I threw together:
Have a look at:
https://en.wikipedia.org/wiki/Triangle_wave
Smashed Transistors
KVRist
132 posts since 10 Oct, 2014
Polynomial integration/differentiation can be used to limit aliasing.
With this method, instead of using a punctual sample, you get a mean value over a sampling interval.
This method can be combined with x2 oversampling.
Here is a simplified code snippet from an Axoloti object.
Code: Select all
``````//init code
float x0 = 0, x1 = 0, y0 = 0, y1 = 0;
//sample rate code
x1 = x0; y1 = y0;
x0 = input * drive; // input drive
float f0 = x0+16.5f;
int i0 = (int)f0; //rounding
if(i0 & 1){
f0 = 2 * (f0 - i0) - 1.0f; //wavefold segment
y0 = 0.25f*(f0*f0-1); // and its smooth integral
} else {
f0 = -2 * (f0 - i0) + 1.0f; //wavefold segment
y0 = -0.25f*(f0*f0-1); // and its smooth integral
}
float x1_x0 = x1 - x0;
if(fabs(x1_x0) > 0.001f){ // if the interval is large enough,
out = (y1 - y0) / (x1_x0); // we differentiate
}else{ // else we take the
out = f0; // direct value
}``````
aciddose
KVRAF
12071 posts since 7 Dec, 2004
While there is a very narrow focus on abs(), it is important to recognize that abs() is merely one possible non-linear function, while any non-linear function can be used to create this effect.
http://xhip.net/effects/?p=Multiplier
https://soundcloud.com/xhip/multiplier
abs() is an infinite order non-linearity while it is also possible to use low order (2nd, 3rd) to approximate it very well. These can be trivially anti-aliased because they generate a limited number of harmonics. (N^2 = at most 2x the bandwidth.)
The famous "Serge wave multiplier" used a diode clamp (NOT APPROXIMATED BY TANH()!) as the non-linear function.
Search for past threads on the topic, I'm not going to go into detail just repeating what has already been said.
Free plug-ins for Windows, MacOS and Linux. Xhip Synthesizer v8.0 and Xhip Effects Bundle v6.7.
joshb
KVRist
85 posts since 13 Apr, 2016
The algorithm that 2DaT most generously gave:
Code: Select all
``out = 4.0 * (std::abs(0.25 * in + 0.25 - std::round(0.25 * in + 0.25)) - 0.25);``
is really close to what I'm trying to do. But I played around with this in Logic:
bitcrusher.jpg
and was hoping to achieve that. I played around with that really elegant algorithm, but couldn't get there.
Any help would be greatly appreciated.
You do not have the required permissions to view the files attached to this post.
aciddose
KVRAF
12071 posts since 7 Dec, 2004
Code: Select all
``````float sign = x > 0 ? 1.0f : -1.0f;
float y = abs(x);
while (y >= 1.0f || y < 0.0f) {
if (y > 0.0f) {
y = 1.0f - y;
} else {
y = -y;
}
}
return sign * y;
``````
Just off the top of my head. If it doesn't work right, figure it out and fix it.
Free plug-ins for Windows, MacOS and Linux. Xhip Synthesizer v8.0 and Xhip Effects Bundle v6.7.
aciddose
KVRAF
12071 posts since 7 Dec, 2004
Just a tip: give up on DSP and do something else with your time.
Free plug-ins for Windows, MacOS and Linux. Xhip Synthesizer v8.0 and Xhip Effects Bundle v6.7.
2DaT
KVRist
316 posts since 21 Jun, 2013
aciddose
KVRAF
12071 posts since 7 Dec, 2004
Although that wraps at 1/2 which is a major problem.
A simple solution is to scale the input/output like so:
https://www.desmos.com/calculator/f9cedeyybp
Noting of course that these operations can be applied to the bits of the float value far more efficiently using some mask magic. I know it can be done trivially as an optimization but I won't get in to it: anyone in need of the performance boost can invest that time themselves.
Unfortunately nothing will cure the atrocious aliasing resulting from such an implementation; you're limited to over-sampling and 6 dB per power of 2.
Better functions are like this:
https://www.desmos.com/calculator/yay1rwvk2w
Which is only third order and so can be perfectly anti-aliased with only 3x over-sampling.
Since this example uses 3 stages it requires a 3^3 (27x) over-sample, but it may make sense to go for 8x, 12x or other values for optimization purposes. Practically speaking such a wave-multiplier has little benefit on super-sonic partials anyway so pre-filtering the input can make a lot of sense to decrease over-sampling needs.
It's also possible to apply dynamic range processing (limiter or compressor) to set the maximum depth of the effect so as to ensure the bandwidth is limited far below the absolute maximum of the function. In many cases an expander and limiter combination (inverted compansion) can make the depth effect more exponential (DX series FM-like) and limit the maximum bandwidth with minimal processing expense compared to over-sampling methods.
Free plug-ins for Windows, MacOS and Linux. Xhip Synthesizer v8.0 and Xhip Effects Bundle v6.7.
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# How Long is…?
We understand that it takes about a second to say “about a second” or that a minute is a relatively small amount of time consisting of sixty seconds. We are taught that a foot is 12 inches, but it is much harder to imagine that 5,280 of them would make up a mile. Yet there are other phrases we have heard that we may have even less of an idea of what they mean. Here are some examples.
A Generation
We have heard of “a once in a generation athlete” or “once in a generation solar phenomenon.” But how long is a generation? Should it be taken literally, as if how long it takes a complete generation to pass on? Actually, while not exact, a generation is usually understood to be about 20-30 years, the amount of time it takes one generation to replicate into another. A six generation family would include a living Great-Grandparent, Grandparent, Parent, Child, a Grandchild and Great-Grandchild.
A Season
If you think all of our seasons are exactly the same length, you are mistaken. In fact, they change in length slightly as to when they start and end each year. Summer is actually our longest season at 93 days and 15 hours, followed by Spring at 92 days and 19 hours. Fall is 89 days and 20 hours in length and winter is 89 days in length.
Score
A score is a period of 20 years. When Lincoln gave his Gettysburg address saying “Four score and seven years ago” he meant 87 years ago.
Fortnight
Before being associated with a video game, a fortnight was a definition of a period of time meaning two weeks. “I’ll be back in a fortnight” meant I will see you in two weeks. Fortnite, the game, uses a slightly different spelling.
While a business day generally lasts 8 hours, if a vendor says your package will be delivered in three business days that does not mean 24 hours (3×8). A business day refers to a day business is conducted through the week, Monday through Friday. If an item is promised for delivery in 3 business days and is ordered on a Tuesday, you should have it by Thursday or Friday. If ordered on a Friday, however, a three business day delivery means the package should arrive by the following Tuesday or Wednesday.
An Eternity
No one know how long this is, so it generally refers to “never gonna happen.”
The average lifespan for a woman in the U.S. is about 78 years while the life expectancy for a male in the U.S. is about 75 years.
Unlike eternity, we know our lifespan has limits. This is why so many people buy life insurance, to continue to support those left behind in the event of their death. Life insurance can pay old bills, account for final funeral expenses and even provide for a future college education of a child or grandchild. If you would like to know more, contact one of our independent insurance agents with your questions and to get a no obligation quote.
May 28, 2024
May 27, 2024
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# Most populated range of floating point numbers in array
6 views (last 30 days)
lvn on 8 Sep 2014
Commented: lvn on 9 Sep 2014
histc can be used on a floating point array to find the bin with the largest number of elements. However these bins are fixed, and for a fixed width of bin, might not be optimal.
Example
a=[0 0.01 0.4 0.45 0.55 0.56 0.60]
histc(a,[0 0.5 1])
ans =
4 3 0
So the most frequent bin is [0, 0.5]. However, I am interested in a function that finds the range of at most 0.5 wide, with the most elements, so in this case [0.4, 0.6] which contains 5 elements. Does anybody know an elegant way of doing this?
Roger Stafford on 8 Sep 2014
Perhaps you won't consider this for-loop solution elegant, but it should be computationally efficient. I will assume that your array 'a' is already in ascending order, as in your example. If not, you should sort it first before using the following code:
d = 0.5; % <-- or whatever you choose
n = length(a);
i1 = 1;
for i2 = 1:n
if a(i2)-a(i1) <= d
m = i2;
else
i1 = i1+1;
end
end
The interval [a(m-n+i1),a(m)] is of width less than or equal to d and contains the maximum number of points among such intervals. If there are other such intervals with the same number of points, this is the first one encountered.
lvn on 9 Sep 2014
Fantastic! My solution was a very slow double for loop over all begin and end combinations, and for my purposes just too slow. Yours works like a charm. Thank you very much!
Honglei Chen on 8 Sep 2014
You can just do
histc(a,[0 0.4 0.6 1])
if you know for sure you want the interval [0.4 0.6]. Otherwise, you can use hist
y = hist(a,[0 0.5 1])
which specifies the center
lvn on 8 Sep 2014
Thanks for the answer, the problem is that I do not know in advance the boundaries of the interval (from one run to another, the vector attains different values). Only the maximum width of the interval is known.
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# geometry
posted by .
The radius R of a sphere is 5.6cm. Calculate the sphere's surface area A.
Use the value 3.14, and round your answer to the nearest tenth.
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# 77.7 minutes in hours and minutes
## Result
77.7 minutes equals 1 hour and 17.7 minutes
You can also convert 77.7 minutes to hours.
## How to convert 77.7 minutes to hours and minutes?
In order to convert 77.7 minutes to hours and minutes we first need to convert 77.7 minutes into hours.
We know that 1 minute is equal to 1/60 hours, therefore to convert 77.7 minutes to hours we simply multiply 77.7 minutes by 1/60 hours:
77.7 minutes × 1/60 hours = 1.295 hours
We already know the amount of hours is 1. Now we have to find out the amount of minutes, to do so we take the decimal part of 1.295 hours and convert it into minutes. In this case we need to convert 0.295 hours into minutes. To convert 0.295 hours to minutes we simply multiply 0.295 hours by 60 minutes.
0.295 hours × 60 minutes = 17.7 minutes
Finally, we can say that 77.7 minutes in hours and minutes is equivalent to 1 hour and 17.7 minutes:
77.7 minutes = 1 hour and 17.7 minutes
Seventy-seven point seven minutes is equal to one hour and seventeen point seven minutes.
## Conversion table
For quick reference purposes, below is the minutes and hours to minutes conversion table:
minutes(min) hours(hr) minutes(min)
78.7 minutes 1 hours 18.7 minutes
79.7 minutes 1 hours 19.7 minutes
80.7 minutes 1 hours 20.7 minutes
81.7 minutes 1 hours 21.7 minutes
82.7 minutes 1 hours 22.7 minutes
83.7 minutes 1 hours 23.7 minutes
84.7 minutes 1 hours 24.7 minutes
85.7 minutes 1 hours 25.7 minutes
86.7 minutes 1 hours 26.7 minutes
87.7 minutes 1 hours 27.7 minutes
## Units definitions
The units involved in this conversion are hours and minutes. This is how they are defined:
### Hours
The minute is a unit of time usually equal to 1⁄60 (the first sexagesimal fraction) of an hour, or 60 seconds. In the UTC time standard, a minute on rare occasions has 61 seconds, a consequence of leap seconds (there is a provision to insert a negative leap second, which would result in a 59-second minute, but this has never happened in more than 40 years under this system). Although not an SI unit, the minute is accepted for use with SI units. The SI symbol for minute or minutes is min (without a dot).
### Minutes
An hour (symbol: h, also abbreviated hr) is a unit of time conventionally reckoned as 1⁄24 of a day and scientifically reckoned between 3,599 and 3,601 seconds. In the modern metric system, hours are an accepted unit of time defined as 3,600 atomic seconds. There are 60 minutes in an hour, and 24 hours in a day.
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# PROFILE
Check out all the problems this user has already solved.
Problem Problem Name Ranking Submission Language Runtime Submission Date
1756 Genetic Algorithm 00003º 19295874 Python 3 0.000 8/23/20, 3:33:53 PM
1142 PUM 05582º 7515618 C 0.008 7/12/17, 1:54:57 AM
1038 Snack 13688º 7515038 C 0.000 7/12/17, 12:46:58 AM
1037 Interval 12620º 7514906 C 0.000 7/12/17, 12:31:13 AM
1036 Bhaskara's Formula 12736º 7514718 C 0.000 7/12/17, 12:09:38 AM
1035 Selection Test 1 13960º 7514279 C 0.000 7/11/17, 11:32:09 PM
1020 Age in Days 15103º 7509232 C 0.000 7/11/17, 5:59:22 AM
1019 Time Conversion 15183º 7509177 C 0.000 7/11/17, 5:30:50 AM
1018 Banknotes 14044º 7509062 C 0.000 7/11/17, 4:15:10 AM
1017 Fuel Spent 15889º 7508788 C 0.000 7/11/17, 2:52:16 AM
1016 Distance 15267º 7507619 C 0.000 7/10/17, 11:13:06 PM
1015 Distance Between Two Points 16871º 7507196 C 0.000 7/10/17, 10:32:33 PM
1014 Consumption 17220º 7506993 C 0.000 7/10/17, 10:11:58 PM
1013 The Greatest 16363º 7506838 C 0.000 7/10/17, 8:26:20 PM
1012 Area 17309º 7506046 C 0.000 7/10/17, 8:04:54 PM
1010 Simple Calculate 19471º 7505135 C 0.000 7/10/17, 6:32:39 PM
1008 Salary 22241º 7503927 C 0.000 7/10/17, 3:33:27 PM
1007 Difference 23597º 7502297 C 0.000 7/10/17, 4:02:36 AM
1005 Average 1 23655º 7502283 C 0.000 7/10/17, 3:58:35 AM
1004 Simple Product 26755º 7502262 C 0.000 7/10/17, 3:52:14 AM
1003 Simple Sum 27656º 7502242 C 0.000 7/10/17, 3:46:47 AM
1002 Area of a Circle 70842º 18969669 C 0.000 7/10/17, 3:38:12 AM
1001 Extremely Basic 34016º 7501835 C 0.000 7/10/17, 1:54:06 AM
1006 Average 2 15673º 4474809 C++ 0.000 5/25/16, 8:15:29 PM
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https://math.stackexchange.com/questions/2903310/question-about-the-definition-of-the-least-upper-bound-property
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# Question about the definition of the least upper bound property
Definition: Let $A$ the set with order relation. We say that the set $A$ has least upper bound property if any $A_0\subset A$, $A_0\neq \varnothing$ which has upper bound has the least upper bound.
Question 1: When we say "has upper bound..." do we mean that its upper bound is in $A$?
Question 2: When we say "has the least upper bound..." do we mean that its least upper bound is in $A$?
Example: Consider the set $A=(-1,1)$ of real numbers in the usual order. Assuming the fact that the real numbers have least upper bound property, it follows that the set $A$ has the least upper bound property (why?). For given any subset of $A$ having an upper bound in $A$ , it follows that its least upper bound must be in $A$. For example, the subset $\{-1/2n: n\in \mathbb{N}\}$ of $A$, thought it has no largest element, does have a least upper bound in $A$, the number $0$.
$\quad$ On the other hand, the set $B=(-1,0)\cup (0,1)$ does not have th least upper bound property . The subset $\{-1/2n: n\in > \mathbb{N}\}$ of $B$ is bounded above by any element of $(0,1)$, but it has no least upper bound in $B$.
I have read this example very carefully and I guess that it provides an example of subsets of reals which has LUB-property and has not, respectively.
Do I correctly interpreted the meaning of above example?
• Q1&2. Yes.${}{}$ Commented Sep 2, 2018 at 23:24
Your point is that if a set $A$ has the least upper bound property, it does not imply that every subset of A also has the least upper bound property.
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https://www.physicsforums.com/threads/deflection-of-tapered-beam-cantilever.353149/
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# Deflection of tapered beam cantilever
1. Nov 9, 2009
### SJB
Hi and thank you for taking a look at my problem..
I'm very familiar with doing deflection calculations for beams of constant cross-section but I find myself needing to now do calculations on a beam of varying cross-section and I'm completely lost.
It's a relatively simple case in that the beam is a round/tubular, constant/symmetrical taper. It's supported in cantilever at the large end and the load is a single point load at the tip. I've attached a simple sketch if it helps.
As a starting point, is it even possible to derive an equation for the deflection of the beam or can this only be calculated by FEA? (excuse my ignorance!)
If anyone can offer any help/pointers/equations I would be very grateful.
Simon
#### Attached Files:
• ###### Cantilever tapered beam.pdf
File size:
10 KB
Views:
2,262
2. Nov 12, 2009
### piygar
This is certainly possible to derive eq'n of delection.
You will need to derive the deflection equation for you case, from equation:
E*I*d2y/dx2 = -M,where I is varying with position of section, and will be a function of x.
Therfore
E*d2y/dx2 = -M/Ix, and integrate this equation twice.
you will need to do this carefully and in the end you will get a big,scary expression of deflection.
You can definitely varify your result with FEA.
Take a look at Roark also for some empirical formula,depending upon ratio of both end area of inertias.
3. Dec 2, 2009
### SJB
Hello Piygar,
Thank you for your reply, I'll give this a go and if successful will post back the equation in case it is useful to others.
Thanks
Simon
4. Dec 21, 2009
### nvn
SJB: The cantilever tip deflection would be as follows, with x = 0 at the fixed support.
$$y_{\,\mathrm{max}}=\frac{1}{E}\int_{0}^{L} \int_{0}^{x}\frac{M(x)}{I(x)}\,dx\ dx$$
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Many idiom/grammer questions. Pls help! : GMAT Verbal Section
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# Many idiom/grammer questions. Pls help!
Author Message
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Joined: 01 Sep 2003
Posts: 2
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Many idiom/grammer questions. Pls help! [#permalink]
### Show Tags
01 Sep 2003, 22:32
Please help me. I am not sure if the following usages are idomatic. Thank you so much!
1. to do A, instead of to do B
2. propose doing
propose to do
3. difference between "excepting" and "except"
4. 1) The reason of somthing is because....
2) The reason of somthing is that....
3) for the reason that....
5. A proved itself to be...
6. warrant doing something
7. differences among "second", "secondly" and "secondarily" as adverbs.
8. difference between "less" and "lesser"
9. Is "one registered" less idomatic than "one that is registered"?
10. paper of all kinds [is | are] ??
All kinds of paper [is | are] ??
11. good enough that...
good enough so that..
12. (OG 251) its number are greater than... (why use "numbers", rather than "number"?)
13. a) seem as if ...
b) she seems an old woman.
14 "Even as" = "Just as"?
15 "Not all their members meet..." = "All their members do not meet"?
16. requirement for something
require of somebody to do somthing
I think it is ok to say "require something of somebody to do somthing" (OG181)
17 "while once" = "because"?
18. too good for somebody to do something
19. as an instance
20. Can "Also" be put at the beginning of a sentence? OG 241 ("also the sun" in B)
21 call for something to do
22 invest into ...
23 A proves less popular than B. (I think there should be a "does" before or after B. Why is it omitted?)
24 Is is right to say "as though during the formation of the universe"? I know it is OK to say "although accouting 5% of the population".
If you have any questions
New!
Many idiom/grammer questions. Pls help! [#permalink] 01 Sep 2003, 22:32
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# Many idiom/grammer questions. Pls help!
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# Distribution after adding noise to a Gaussian Distribution
I have a random variable $$X \sim \mathcal{N}(\mu_X, \sigma_X^2)$$. Now, I add a noise $$N \sim \mathcal{N}(0, \sigma_N^2)$$ to $$X$$ to get $$Y$$ ($$Y = X + N$$). Thus, $$Y \sim \mathcal{N}(\mu_X, \sigma_X^2 + \sigma_N^2)$$. Are $$Y$$ and $$X$$ bivariate normal now? If so, what would be the density function $$f_{Y|X}$$?
• If $N$ is independent of $X$ then $(X,Y)=(X,X+N)$ has a bivariate normal distribution. Aug 19, 2020 at 5:34
• Right. So I can assume some value $\rho$ as correlation coefficient and use the bivaraiate normal formula to represent $f_{y|x}$, right? Aug 19, 2020 at 5:36
We know that $$X+N\: | \: X=x \sim x+N \: | \: X=x.$$ If we also know that $$X$$ and $$N$$ are independent, then we can infer that $$x+N \: | \: X=x \sim x+N \sim \mathcal{N}(x, \sigma_N^2),$$ which means that $$Y|X \sim \mathcal{N}(X,\sigma_N^2)$$.
For the part about bivariate normal distribution, you may use that linear transformations of a bivariate normal vector is again bivariate normal. So if $$(X,N)$$ is bivariate normal (which it is if we assume independence), then $$(X,X+N)$$ is also bivariate normal.
• Does it mean that the mean of $Y|X$ is $\mu_X$ and variance is $\sigma_N^2$? Aug 19, 2020 at 14:56
• The mean of $Y|X$ is $X$ (not $\mu_X$) and variance is $\sigma_N^2$. The mean of $Y$ is however $$\mathbb{E}[Y] = \mathbb{E}[\mathbb{E}[Y \:|\: X]] = \mathbb{E}[X]=\mu_X$$ Aug 19, 2020 at 17:27
• So, if I use $X = x_1$, then the mean of $Y|X$ would be $x_1$. Correct? Aug 19, 2020 at 17:32
• Yes, although when a particular value of $X$ is considered, you should write $Y|X=x_1$, as in "The mean of $Y|X=x_1$ is $x_1$". Aug 19, 2020 at 17:35
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# Significance of isospin and hypercharge
by robousy
Tags: hypercharge, isospin, significance
P: 335 This may sounds like a dumb question but I want to figure out the 'point' of them. I know that isospin was an attempt to describe the proton and nucleon as an isospin doublet and that hypercharge seems to me to be a nifty little relation between the electric charge and the isospin - but what is the point of them now that we know that we cannot describe the nucleon as an isospin doublet and that it is in fact constructed of quarks. Is it that: a) Before we new about quarks it seemed like a good idea. or b) We can indeed build a model based on isospin. I am learning about GUTS at the moment and specifically $$SU(2)_L \times SU(1)_Y$$ What is the significance of the hypercharge Y subscript in the $$SU(1)_Y$$? It seems (to my admittedly ignorant mind) that hypercharge is nothing more than a nifty relation to charge so why does it get elevated to the subscript level $$SU(1)_Y$$ in GUTS??? Please help me understand the significance!
P: 335 Ok - I know now so I don't want to waste anyone time!! :)
P: 19
Quote by robousy Ok - I know now so I don't want to waste anyone time!! :)
well tell me what u have understood...i 'll b glad 2 read what u 've interpreted of it.....
P: 21,637
## Significance of isospin and hypercharge
A collection of comments on hypercharge and isospin.
Conservation of strangeness is not in fact an independent conservation law, but can be viewed as a combination of the conservation of charge, isospin, and baryon number. It is often expressed in terms of hypercharge Y, defined by: Y = S + B = 2(Q-I), where S = Strangeness B = Baryon number Q = charge I = isospin Isospin and either hypercharge or strangeness are the quantum numbers often used to draw particle diagrams for the hadrons.
However - it seems it can more complicated - http://en.wikipedia.org/wiki/Hypercharge
http://en.wikipedia.org/wiki/Isospin
As for strangeness -
A property of hadrons which is conserved in particle reactions caused by the strong force and which has in a weak interaction.
Eric Weisstein's Scienceworld at Wolfram.com - http://scienceworld.wolfram.com/phys...rangeness.html
http://en.wikipedia.org/wiki/Strangeness - See the definition here, which then states
The reason for this unintuitive definition is that the concept of strangeness was defined before the existence of quarks was discovered, and for consistency with the original definition the strange quark must have strangeness -1, and the anti-strange quark must have strangeness +1.
Particle physics and supporting theories do appear as strange.
Sometimes one is faced with the need for a theory (understanding) without having all the necessary information. Remember back to the 1800's before quantum physics - the physicists had glimpses of atomic structure, but just didn't have all the information.
A neat bit of trivia -
Unlike his brother Maurice, who was primarily an experimental physicist, Louis de Broglie had the mind of a theoretician rather than that of an experimenter or engineer. His 1924 doctoral thesis, Recherches sur la théorie des quanta (Research on Quantum Theory), introduced his theory of electron waves.
- from http://en.wikipedia.org/wiki/Louis_de_Broglie
and http://en.wikipedia.org/wiki/Murray_Gell-Mann
P: 335
Quote by preet0283 well tell me what u have understood...i 'll b glad 2 read what u 've interpreted of it.....
Well, due to the almost identical masses of the proton and neutron we hypothesized the existence of a nucleon isospin doublet. Clearly the p and n have differing electric charges so a formula was created by Gellman and Nishijima linking the isospin with the electric charge: Q=Y+H where H is the hypercharge and Y the isospin. It turns out that hypercharge is actually composed more fundamentally of B-L where B is Baryon and L Letpton number.
Thats the OLD interpretation. I am starting to realize that there is a new interpretation of Isospin which is a bit more complicated and the name is just a carry on from the old days. Something to do with the generators of the groups of the standard model and the Cartan Subalgebra eigenvalues being used to construct the charge generator...(maybe)
If you can expose more of the 'new' isospin idea that would be good.
Thanks!
P: 817
I am learning about GUTS at the moment and specifically $$SU(2)_L \times SU(1)_Y$$
What is SU(1)_Y?
Look, do you mean the electroweak gauge group SU(2)XU(1) or chiral symmetry group SU(2)XSU(2)?
THE 1ST GROUP HAS SUBSCRIPTS (W) OR (L) ON SU(2) TO MEAN WEAK OR LEFTHANDED FIELDS, AND Y ON U(1) TO MEAN THE WEAK HYPERCHARGE(THE GENERATOR OF U(1)).
THE CHIRAL GROUP SU(2)XSU(2) OFTEN WRITTEN AS SU(2)_LXSU(2)_R.
FROM NOW ON, BEFORE I ANSWER ANY OF YOUR QUESTIONS, I NEED TO KNOW WETHER OR NOT YOU ARE LEARNING SOME THING FROM US. TELL ME; ARE WE WASTING OUR TIME?
P: 335
Quote by samalkhaiat What is SU(1)_Y? Look, do you mean the electroweak gauge group SU(2)XU(1) or chiral symmetry group SU(2)XSU(2)? THE 1ST GROUP HAS SUBSCRIPTS (W) OR (L) ON SU(2) TO MEAN WEAK OR LEFTHANDED FIELDS, AND Y ON U(1) TO MEAN THE WEAK HYPERCHARGE(THE GENERATOR OF U(1)). THE CHIRAL GROUP SU(2)XSU(2) OFTEN WRITTEN AS SU(2)_LXSU(2)_R. FROM NOW ON, BEFORE I ANSWER ANY OF YOUR QUESTIONS, I NEED TO KNOW WETHER OR NOT YOU ARE LEARNING SOME THING FROM US. TELL ME; ARE WE WASTING OUR TIME?
$$SU(3) \times SU(2) \times U(1)_Y$$
Yes, the U(1)_Y weak hypercharge generator.
Yes, I am learning - some aspects faster than others, but please, if you think you are wasting your time then please do not answer. I am sure I would not be offended if noone left an answer.
HW Helper
P: 1,204
Quote by robousy If you can expose more of the 'new' isospin idea that would be good.
Perhaps it would be good to remember that the fundamental particles are not actually little indivisible things that cannot be divided. A better description of them is to think of the wave functions one dealt with in quantum mechanics.
With wave functions, you can take the sum of two wave functions and, because the equations are linear, the result is another wave function with properties somehow sort of midway between the other two wave functions. For example:
$$\psi = \psi_A + \psi_B$$
where the two wave functions on the right satisfy an operator equation like:
$$\mathcal{O} \psi_A = A \psi_A$$ and same for B.
The sum of the A and B wave functions is a valid wave function but is not likely to be an eigenfunction for the operator like A and B were. So if you define "particle" as the things that are eigenfunctions for that operator, then the sum is not a "particle".
In the case of the elementary particles, we consider charge, Q, to be one of the operators that define what are the elementary particles. For example, if "e" is the electron, and \psi_e is an electron wave function, we have the operator equation:
$$\mathcal{Q} \psi_e = -e \psi_e$$
since the charge of the electron is -e. Similarly,
$$\mathcal{Q} \psi_\nu = 0$$
since the charge of the neutrino is zero.
So we classify the elementary particles in a way that makes them eigenfunctions of (electric) charge and mass (and parity or whatever).
But just because we classify elementary particles in this way does not mean that we cannot reclasify them in the same we can reclassify wave functions by taking linear combinations of them. And some of the alternative methods of classifying them would make more sense in certain circumstances.
An example is the Cabibbo angle. The up and down quarks (of the "electron family") are eigenstates of the electric charge operator and are eigenstates of mass, but they are not eigenstates of the "weak charge" (when I was in grad school it was called "neutral charge") operator in the sense that when you change an up quark into a down quark by emitting a W-, you don't actually get a pure down quark. Instead, you have to mix in some of the other families of quarks, the "muon family" and "tau family". W+ and W- interactions for quarks involve changing from a +2/3 to a -1/3 (or -2/3 to a +1/3), so you can fix the weak charges by either mixing the (u,c,t), or by mixing the (d,s,b). Nowadays, it is done by mixing the (down,strange,bottom) and it is called the "CKM" matrix.
As an alternative, when you have a pure down quark and arrange for it to emit a W-, it doesn't become a pure up quark but instead ends up with a mixture of top and charm. Thus you could instead define a CKM type matrix as mixing the (u,t,c).
With the leptons, there is also a mixing between the neutral leptons (neutrinos) as compared to the charged leptons. As with the quarks, the mixing appears when you define the particles according to their masses (and therefore into the families or "flavors"). The electron, muon and tau are the mass eigenstates of the charged leptons. When one of these particles emits a W-, the charged lepton changes to a mixture of neutrinos.
Now with the quarks, the effect of the emission of a W+ or W- is a relatively small probability of a change in the family. Therefore we collect the quarks into pairs, (up,down), etc. But with the neutrinos, the mixtures are more democratic so it's hard to say which of the neutrino eigenstates corresponds to the electron and which to the muon and tau. So the neutrinos are generally defined according to what you get when you take a W+/- out of the corresponding lepton mass eigenstate. That is, we talk about an electron neutrino, a muon neutrino and a tau neutrino. This means that the lepton analog of the quark CKM matrix is defined in sort of reverse, an extra opportunity for confusion. It is my belief that the quark mixing is more pure than the lepton mixing arises naturally from the way they are produced from subparticles but that's another story; in the standard model, these are all fairly arbitrary parameters.
If you want to always define "particle" as things that are eigenstates of electric charge and mass, then you will have the weak interactions mixing particle types. But you could instead define "particle" as things that are eigenstates of weak charge and mass, and that would leave you with electric interactions that mixed particle types.
It is also possible to define particles as the things that are eigenstates of both electric and weak interactions. If you do this, then you will have particles that are of mixed mass eigenstates. I suspect that things will be simplest in this basis. In any of these three cases, it is important to note that the mixing is over corresponding particles in the different families, and that the families differ only in their masses.
Now what does this have to do with weak isospin and weak hypercharge? Weak isospin and hypercharge, together, give electric charge and weak charge. Weak isospin is the SU(2) type symmetry between the two objects that a weak force W+/- converts between. Once you've defined weak isospin and electric charge, the difference between these is also defined and (twice the difference) is called weak hypercharge.
When the electric and weak interactions are combined into an electroweak interaction, the appropriate currents (i.e. moving charges) are weak isospin (which is a vector) and weak hypercharge (a scalar). These two are mixed by the Weinberg angle so that instead of interacting directly with gauge bosons according to weak isospin and weak hypercharge, they instead use weak charge and electric charge. Thus the photon is a mixture of a weak isospin and weak hypercharge interaction.
It's late. I hope I haven't made too many serious mistakes.
Carl
P: 4,008 CarlB, Just wanted to let you know that i really like your last post. You have the ability to explain (conceptually) difficult stuff in an easy language. I like your style, man. I wanna ask you to contribute to the "elementary particles presented thread" if you want. For example, you could make a reference to this post into that thread. Or, if you feel that certain aspects/topics of theoretical physics need to be explained, i invite you to post them there and thus expand our elementary particles-library. I would be very greatful regards marlon
HW Helper
P: 1,204
Quote by marlon CarlB, Just wanted to let you know that i really like your last post. You have the ability to explain (conceptually) difficult stuff in an easy language. I like your style, man. I wanna ask you to contribute to the "elementary particles presented thread" if you want. For example, you could make a reference to this post into that thread. Or, if you feel that certain aspects/topics of theoretical physics need to be explained, i invite you to post them there and thus expand our elementary particles-library. I would be very greatful regards marlon
Whoa! Before you conclude that I know what I'm talking about, you should be aware that (a) I am not associated with any physics (or math) departments and in fact make a living by doing stuff like driving forklifts, using a nail gun, and soldering; (b) I started grad school back when Carter was president and never finished a PhD; (c) I don't believe in Einstein's relativity; (d) I believe that using symmetry to define elementary particles is fundamentally misguided; (e) I don't believe in the quantum mechanical vacuum (along with Schwinger as it turns out); (f) I believe that the symmetry breaking seen in the standard model is actually a part of spacetime itself rather than an attribute of the particles per se; (g) I believe that standard matter is condensed from tachyons; and (h) I have a nasty habit of going to physics conferences and being ignored for supporting points (c-g). In short, I really can't think of anyone you'd want less to be telling you about the standard model.
On the other hand, I do believe that what I wrote above is a fairly accurate portrayal of what both I and the standard model believe about the elementary particles and you're certainly welcome to copy it where you wish.
The mixing matrix for the weak force as applied to the quarks is called the "CKM" matrix. The same thing applied to the leptons is called the "MNS"
matrix. Naturally I'm busily attempting to unify all this with my own version of particle theory, but if you value your sanity (or your standing in the standard physics community, I advise you to stay far away).
Carl
P: 4,008 CarlB at what university did you study ? Besides, you can have whatever personal opinion that you want, i do not care. I read your "personal caracterization" very thouroughly and i admit i would be the exact opposite of you. However, this does not take away the fact that lot's of your post are of high quality and that is my honest opinion. So, YES, i would like to have you on board when it comes to explaining difficult concepts to laymen, which one of the primary intentions of this great forum. For the rest, you are forgiven :) marlon ps : why is it that only the really good and talented people are so very honest and direct ? :)
P: 335 CarlB. I've been meaning to thank you for ages now since you kindly posted a long and no doubt lucid explanation to my question. TBH I haven't had a chance to go through it yet - but I plan to next week (when a 'long lost friend' I am entertaining returns home) - but thanks again!
Related Discussions High Energy, Nuclear, Particle Physics 6 High Energy, Nuclear, Particle Physics 2 High Energy, Nuclear, Particle Physics 14 High Energy, Nuclear, Particle Physics 4 Advanced Physics Homework 13
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Search a number
33321 = 329383
BaseRepresentation
bin1000001000101001
31200201010
420020221
52031241
6414133
7166101
oct101051
950633
1033321
1123042
1217349
1312222
14c201
159d16
hex8229
33321 has 8 divisors (see below), whose sum is σ = 46080. Its totient is φ = 21392.
The previous prime is 33317. The next prime is 33329. The reversal of 33321 is 12333.
Adding to 33321 its sum of digits (12), we get a palindrome (33333).
Adding to 33321 its reverse (12333), we get a palindrome (45654).
It can be divided in two parts, 3332 and 1, that added together give a palindrome (3333).
It is a happy number.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 33321 - 22 = 33317 is a prime.
It is a super-2 number, since 2×333212 = 2220578082, which contains 22 as substring.
It is a Curzon number.
It is a plaindrome in base 13.
It is a nialpdrome in base 10.
It is a junction number, because it is equal to n+sod(n) for n = 33297 and 33306.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (33329) 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 7 ways as a sum of consecutive naturals, for example, 105 + ... + 278.
It is an arithmetic number, because the mean of its divisors is an integer number (5760).
233321 is an apocalyptic number.
It is an amenable number.
33321 is a deficient number, since it is larger than the sum of its proper divisors (12759).
33321 is a wasteful number, since it uses less digits than its factorization.
33321 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 415.
The product of its digits is 54, while the sum is 12.
The square root of 33321 is about 182.5404064858. The cubic root of 33321 is about 32.1790097631.
The spelling of 33321 in words is "thirty-three thousand, three hundred twenty-one".
Divisors: 1 3 29 87 383 1149 11107 33321
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# Tricky one
1. Oct 29, 2007
### BMW25
A fireworks rocket is launched vertically upward at 40 m/s. At the peak of its trajectory, it explodes into two equal-mass fragments. One reaches the ground t1= 2.51s after the explosion.
When does the second reach the ground?
t= ......??
logiclly, it is 2.51 s also ....but I don't know. so could you pls guys help with that out ??
2. Oct 29, 2007
### qspeechc
Ok, find where it is at its peak. At its peak its velocity is zero. Use conservation of momentum to get the initial directions and speeds. That is, let the first fragment travel in some initial direction with an angle theta to the horizontal (or whatever you like), then get the other fragment initial direction and speed, and solve the equations of motion.
3. Oct 29, 2007
### BMW25
thnx dude for your post. but could you pls explain a little bit more for me?
4. Oct 29, 2007
### qspeechc
Ok, use the equations of motion to find where (at what height) the rocket reaches its peak. Then define a set of axes: to the right horizontally is you +x axis, and, say, vertically upwards is your +y axis. Then say immediately after the explosion, the first fragment leaves at an angle theta relative to the +x axis (in the usual way we measure angles, in the anti-clockwise direction), with a speed v.
Then you can use the equations of motion to find this angle theta and the speed v, since we know it took 2.51 seconds to reach the ground.
Now we have the initial direction and speed the first fragment had. Also, initially, linear momentum was conserved, so the other fragment flew off in the opposite direction but with the same speed (because their masses are the same). You know the angle the first one flew off in, so you can find the angle the other one flew off in, because the total momentum initially must have been zero.
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# How far is Bimini from Arthur's Town?
The distance between Arthur's Town (Arthur's Town Airport) and Bimini (South Bimini Airport) is 237 miles / 381 kilometers / 206 nautical miles.
237
Miles
381
Kilometers
206
Nautical miles
## Distance from Arthur's Town to Bimini
There are several ways to calculate the distance from Arthur's Town to Bimini. Here are two standard methods:
Vincenty's formula (applied above)
• 236.697 miles
• 380.926 kilometers
• 205.684 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 236.416 miles
• 380.474 kilometers
• 205.440 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Arthur's Town to Bimini?
The estimated flight time from Arthur's Town Airport to South Bimini Airport is 56 minutes.
## Flight carbon footprint between Arthur's Town Airport (ATC) and South Bimini Airport (BIM)
On average, flying from Arthur's Town to Bimini generates about 60 kg of CO2 per passenger, and 60 kilograms equals 132 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path from Arthur's Town to Bimini
See the map of the shortest flight path between Arthur's Town Airport (ATC) and South Bimini Airport (BIM).
## Airport information
Origin Arthur's Town Airport
City: Arthur's Town
Country: Bahamas
IATA Code: ATC
ICAO Code: MYCA
Coordinates: 24°37′45″N, 75°40′25″W
Destination South Bimini Airport
City: Bimini
Country: Bahamas
IATA Code: BIM
ICAO Code: MYBS
Coordinates: 25°41′59″N, 79°15′52″W
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# Transformations Level 7
## Learn how to solve mixed transformation problems and answer exam-style questions.
##### MenuLevel 1Level 2Level 3Level 4Level 5Level 6Level 7Exam-StyleHelpMore...
This is level 7: describe the transformation from the diagrams. You can earn a trophy if you get at least 9 questions correct and you do this activity online.
1. Shape B can be transformed to shape C with a reflection. What is the equation of the line of reflection? 2. Shape F can be transformed to shape C with a reflection. What is the equation of the line of reflection? 3. Shape F can be transformed to shape E with a reflection. What is the equation of the line of reflection? 4. Shape F can be transformed to shape A with a reflection. What is the equation of the line of reflection? 5. Shape E can be transformed to shape G with a rotation of 90° clockwise. What are the coordinates of the centre of rotation? 6. Shape A is rotated 180° about the point (5,10). Over which shape will it now be? 7. By how many degrees must shape H be rotated around the point (5,10) in a clockwise direction to fit over shape F? 8. Shape A is translated by the vector $$\begin{pmatrix} 0 \\ -2 \\ \end{pmatrix}$$ then rotated 90° clockwise about the point (6,9). Over which shape will it now be? 9. Shape F is translated by the vector $$\begin{pmatrix} 2 \\ 0 \\ \end{pmatrix}$$ then rotated 180° clockwise to fit over shape B. What are the coordinates of the centre of rotation? 10. Shape B is translated by the vector $$\begin{pmatrix} -4 \\ -4 \\ \end{pmatrix}$$ then reflected to fit over shape E. What is the equation of the line of reflection? 11. Shape A is reflected in the line x = 4. It can then be transformed to shape F with a rotation of 90° anticlockwise. What are the coordinates of the centre of rotation? 12. Shape H is enlarged with centre of enlargement (5,10) to fit over shape D. What is the scale factor of this enlargement?
Check
This is Transformations level 7. You can also try:
Level 1 Level 2 Level 3 Level 4 Level 5 Level 6
## Instructions
Try your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help.
When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file.
## More Activities:
Mathematicians are not the people who find Maths easy; they are the people who enjoy how mystifying, puzzling and hard it is. Are you a mathematician?
Comment recorded on the 1 February 'Starter of the Day' page by Terry Shaw, Beaulieu Convent School:
"Really good site. Lots of good ideas for starters. Use it most of the time in KS3."
Comment recorded on the s /Indice 'Starter of the Day' page by Busolla, Australia:
"Thank you very much for providing these resources for free for teachers and students. It has been engaging for the students - all trying to reach their highest level and competing with their peers while also learning. Thank you very much!"
Each month a newsletter is published containing details of the new additions to the Transum website and a new puzzle of the month.
The newsletter is then duplicated as a podcast which is available on the major delivery networks. You can listen to the podcast while you are commuting, exercising or relaxing.
Transum breaking news is available on Twitter @Transum and if that's not enough there is also a Transum Facebook page.
#### Numskull
Interactive, randomly-generated, number-based logic puzzle based on the Latin square designed to develop numeracy skills. These puzzles are drag and drop and can earn you a Transum Trophy.
There are answers to this exercise but they are available in this space to teachers, tutors and parents who have logged in to their Transum subscription on this computer.
A Transum subscription unlocks the answers to the online exercises, quizzes and puzzles. It also provides the teacher with access to quality external links on each of the Transum Topic pages and the facility to add to the collection themselves.
Subscribers can manage class lists, lesson plans and assessment data in the Class Admin application and have access to reports of the Transum Trophies earned by class members.
Subscribe
## Go Maths
Learning and understanding Mathematics, at every level, requires learner engagement. Mathematics is not a spectator sport. Sometimes traditional teaching fails to actively involve students. One way to address the problem is through the use of interactive activities and this web site provides many of those. The Go Maths page is an alphabetical list of free activities designed for students in Secondary/High school.
## Maths Map
Are you looking for something specific? An exercise to supplement the topic you are studying at school at the moment perhaps. Navigate using our Maths Map to find exercises, puzzles and Maths lesson starters grouped by topic.
## Teachers
If you found this activity useful don't forget to record it in your scheme of work or learning management system. The short URL, ready to be copied and pasted, is as follows:
Alternatively, if you use Google Classroom, all you have to do is click on the green icon below in order to add this activity to one of your classes.
It may be worth remembering that if Transum.org should go offline for whatever reason, there is a mirror site at Transum.info that contains most of the resources that are available here on Transum.org.
When planning to use technology in your lesson always have a plan B!
Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.
For Students:
For All:
## Description of Levels
Close
Level 1 - Reflect a shape in a given line
Level 2 - Translate a shape by a given vector
Level 3 - Rotate a shape about a given point by a given angle
Level 4 - Enlarge a shape from a given point by a given scale factor
Level 5 - Transform a shape by a given matrix
Level 6 - Mixed questions without diagrams
Level 7 - Describe the transformation from the diagrams
Exam Style questions are in the style of GCSE or IB/A-level exam paper questions and worked solutions are available for Transum subscribers.
There is a printable worksheet to go with this activity.
Finally try Blow Up, a clallenge to find all the points that could be the centre of enlargement of a shape if the image does not go off the grid. Great fun!
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
## Help
You may want to open the Graph Plotter in a new tab to help you answer these questions.
Alternatively you could print some Graph Paper, have it laminated then use a non-permanent pen as a handy working resource.
Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly. You can double-click the 'Check' button to make it float at the bottom of your screen.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
Close
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#### Search Results for ages 11-15
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In this activity, design and construct an NXT car or snail which is capable of traveling extremely slowly. The cars compete in a snail race with the last car to cross the finish line crowned as the winner.
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Construct a rotation sensor with the highest degree of accuracy and repeatability.
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Relay Race - RCX (Lesson Plan/Activity)
In this activity, design and construct three RCX cars and program them to run a relay race. Each car is programmed to start the next car on its team by using the Mail function in ROBOLAB.
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Program a car equipped with a rotation sensor to return to its original place no matter how far it is rolled away from the origin. Attain this goal by using both on-off control and proportional control.
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Get an NXT car to follow a square path while driving.
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Is it possible for humans to survive on Mars? Help mankind take a giant step in this information gathering process by constructing a rover that has a wide array of remote sensing capabilities and sending it to the red planet to collect data.
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Students will construct a sturdy chair for a stuffed-animal bear. The chair must be able to support the bear from the front, sides and back. Ultimately, the chair must keep Mr. Bear in a sitting position without him falling out of the chair.
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Students will investigate the properties of sturdy structures by constructing a LEGO wall. The wall must be about 6 bricks high and withstand the "Flick Test". The Flick Test involves flicking the top of the wall to see if the wall fails or tips over.
Ages: 5-11
Strongest Shape (Lesson Plan/Activity)
Students will investigate different shapes to determine which shape is the strongest. Students will construct a roof to be placed on a previously built house. The roof must reflect their shape choice for the strongest shape. To test their structure, they will push on the top of the roof to check its...
Ages: 5-11
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https://www.physicsforums.com/threads/a-steel-rod-and-an-iron-rod-is-placed-inside-an-ac-current-soleoid.576422/
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# Homework Help: A steel rod and an iron rod is placed inside an ac current soleoid
1. Feb 10, 2012
### sgstudent
What happens? I think nothing should happen since the rods will not get magnetised at all due to the ac current..
Thanks for the help!
2. Feb 11, 2012
### technician
If you have low frequency AC then the bars will become magnetised with changing magnetic polarity. The bars could repel each other although the fact that one bar is steel and the other is iron will produce different strengths so .....? Not sure !!
At higher ( mains) frequencies I think you are maybe correct!.......I know that mains AC solenoids can be used to demagnetize steel bars by withdrawing them slowly.
I imagine there is not one simple answer.
3. Feb 11, 2012
### sgstudent
Hi thanks for the help. I'm Singaporean so ky syllabus is a lot different, we only touch the basics of magnetism so I'm unsure about this. May I ask what is the relationship between magnetic domain and magnetic field? And why when magnetic shielding occurs, the shielding object does not get magnetised and have its own field?
This question came out in a test. I'll let you know the answer my teacher gives. Thanks!
4. Feb 11, 2012
### gsal
That's probably not true, right? I mean nothing does not happen...I am sure something happens, it is a matter of whether what happens is of interest to you or not or whether it affects you or not...so, maybe some context would have been nice.
Maybe the rods do not get magnetized because of the changing magnetic field, but something happens...because the relative motion between the magnetic field and the rods, eddy currents will be induced into the rods and they will heat up due to their internal resistance...certainly the iron rod, I don't know about the steel one, if by making it out of steel, they meant non-magnetic steel?
5. Feb 11, 2012
### sgstudent
The question just said, if the dc current was changed to a low frequency ac current, describe what will happen to the rods and explain why.
I know that steel is a hard magnetic material so its harder to get magnetised, but I thought the field will still pass through it so they will still repel no matter what. Thanks for the help
6. Feb 11, 2012
### gsal
Nope, the original posting said absolutely nothing about dc.
7. Feb 11, 2012
### sgstudent
8. Feb 13, 2012
### sgstudent
Can I get more help in this
9. Mar 2, 2012
### sgstudent
This is what my school teacher said: there are 2 scenarios that can happen
1) the iron will keep changing polarity with the changing current. While steel will retain its polarity from the first current direction. This is because steel is a hard magnetic material so it retains it magnetism longer. Thus, it will continuously attract and repel.(this is when the frequency is very very low)
2) the steel and iron will keep repelling each other. This is because the iron is a soft magnetic material so it is able to get magnetised easily. Even though steel is a hard magnetic material, it is able to switch polarity so it will repel from the iron rod.
But won't the steel be unmagnetised?
10. Mar 2, 2012
### Staff: Mentor
To demagnetise a steel object you slowly withdraw it from within a strong AC field. Emphasis on the "slowly".
11. Mar 2, 2012
### sgstudent
I don't quite get it, slowly is meant to demagnetise it, but I'm still unsure about what if the observations in this case.
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https://www.doubtnut.com/question-answer/a-manufacturing-company-makes-two-models-a-and-b-of-a-product-each-piece-of-mode-a-requires-9-labour-642584608
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# A manufacturing company makes two models A and B of a product. Each piece of mode a requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of Mode B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricati2 and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of Rs. 8000 on each piece of model A and Rs. 12000 on each piece of Model B. How many pieces of Model A and Model B should be manufactured per week to realise a maximum profit? What is the maximum profit per week?
Updated On: 17-04-2022
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Transcript
in this version support experience of model a&y pieces of modern tree are very effective for each model a request 9 labour and a model Tere class 12 letter of for fabricating their features of model a&y 1982 9 X + 12 x + Y equal to 180
model a required for labour and software required by labour finishing but not equal to 30 now the profit from model 8008 to talk about other model be that for model be jealous of the total profit from of model a features of model 350
4 x pieces how a y pieces of the that will be 8000 plus 8008 + 12000 so that is over maximize now subject to the constraint subject to the conference + 4 x +
3 Y equal to 30 X and Y equal to 4 the region Defined by the can be diagrammatically you can say you can say on graph that will be represented as Zakhmi ke pass Tantra diagram of the 2020 1226 a and c is common table and diagram
we can write all the components that is and zero comma now we will put the equation there as that equals to 800 of 1000 + 12 agent of wine equal to 1000 x 0 1000 x 0
2012 1000 x 0 12000 X 0 28000 X 20000 now what well commerce six 38000 x 12 2012 1000 x 6 16800 now aur last date 1000
X 0 + 12000 x 10 20000 now this is our maximum value so the value of x = 2 158000 at X equal to 12 and Y = 26 to now has a manufacturing companies could produce 12 of model and 653 to realise maximum profit and the profit 180000 MPOnline
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Master the 7 pillars of school success
What is a vertical line ? Here are several ways to describe a vertical line.
• A line perpendicular to the horizon.
• A line that runs up and down on a page.
• A line on a graph that will not cross the y intercept and does not have a y intercept.
In Geometry a horizontal line is a line that is parallel to the horizon. Here are a couple of other ways to describe a horizontal line.
• A line that is perpendicular to a vertical line.
• On a page, a horizontal line is a straight line that runs from left to right .
• A line that has a slope value of zero.
• A horizontal line has a rise which is zero.
• A line that will not cross the x axis, and does not have a x intercept.
• A horizontal line is parallel to the x axis on a coordinate plane.
Common Core Standard 8.E.E.6 8th Grade Math
### What makes a vertical line and horizontal line different?
A horizontal line has a slope of zero and is written y= c c=the constant
Write the equation for a horizontal line that goes through the point (7,3)
The line crosses the y axis at 3, so the equation is y=3
• vertical line has an undefined slope and the equation is written x=c c=the constant.
Write the equation for a vertical line passing through (4,-5)
x= 4
Horizontal lines have a slope of zero, and run parallel to the x axis.
Vertical lines have a undefined slope, and run parallel to the y axis.
The equation for a horizontal line equals y=c , and the equation for a vertical line equals x=c.
If a horizontal line crosses a vertical line the two lines would be perpendicular to one another.
### Parallel,Perpendicular,Horizontal, and Vertical Lines
A line that is parallel to a horizontal line has a slope of zero. (rise/run, 0/1).
A line perpendicular to a horizontal line has a slope that is the negative reciprocal of zero (rise/run ,-1/0).
Find the equation of a line parallel to y=5, and passes through the point (5,3)
Remember, two parallel never intersect.
The equation y=5 tells you that the line is a horizontal line, so the second line has a slope of zero so the equation would equal y=3.
Write the equation for a line that is parallel to x=6, and passes through the points (12,-4).
Both lines are vertical lines therefore the equation would equal x=12.
Write the equation for a line that is perpendicular to the line x=5 and passes through the points (6,8).
Remember perpendicular lines intersect and form a 90 degree angle.
The original line is a horizontal line, therefore the equation will be y=8.
Vertical Line
Horizontal Line
The Horizontal line D has a slope equal to 0 and the equation is:
y = 2 ( because this is where the line crosses the y axis)
Write the equation of a vertical passing through (-3,3)
x = - 3 ( because this is where the line crosses the x axis)
The equations and slopes of vertical and horizontal lines are slightly different than a regular line, and can be described as " Special. "
### Examples of Horizontal lines on a coordinate plane.
Transcript
Hi welcome to MooMooMath. Today we are going to look at special slopes of horizontal and vertical lines. I’m going to review the guidelines of slopes. This first example has a slope that is going up when you look from left to right so it is a positive slope and we always count our slope as rise over run. So in the first example we will rise positive 4 and run positive 3 so the slope is positive 4⁄3.Let’s look at the second example, if you move from left to right notice the line is going downward so it will have a negative slope. When we go to count it we will go down 2 and over 1, 2 so it will have a slope of negative 2 over 2 so it will have a slope of negative 1. Again the slope is negative and going down from left to right. Now let’s look at our two special cases. The first one does not go up or down but is horizontal. So any time you have a horizontal lie you have a slope of 0 why is that? You have a rise of 0 and a run of any value so our slope will just be 0. Let’s look at this last one. The equation for this line is y= positive 2. Whenever you have a horizontal line it is always y = the constant of where line crosses the y axis. In this case it crosses the y axis at 2. Now in the last case you have a vertical line. Now the vertical line does not have a slope, or sometimes they call it an undefined slope. Why is it undefined or no slope? We have a rise of any value but it has a run of 0 what do we know when we divide any value by 0? You can’t do that so that is why the value is undefined. So this crosses the X axis at two so the equation of the line is x equals two and no Y value. Let’s look at a summary of our slopes. So here is our Cartesian coordinate plane to find the slope from a to b we count our rise is 3 and our run is 4 so this line is ¾ our slope of our next line is down one over four so it is negative ¼. The slope of our special lines is: the vertical lines (C) are undefined slope or no slope and the slope of the horizontal line (d) is 0. The equation for line d is negative 2 y because that is where the line crosses the y axis a so it is y= negative 2 the vertical line (C) has an equation of x = negative 3 whenever you have a vertical line it is x equals and a horizontal line is y equals that constant. Hope this was helpful.
## Horizontal Lines
You may also enjoy .......
Transversal angles
Perpendicular Lines
D
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According to the American Time Use Survey conducted by the
According to the American Time Use Survey conducted by the Bureau of Labor Statistics (www.bls.gov/atus/), Americans spent an average of 985.50 hours watching television in 2010. Suppose that the standard deviation of the distribution of times that Americans spent watching television in 2010 is 285.20 hours.
a. Using Chebyshev’s theorem, find at least what percentage of Americans watched television in
2010 for
i. 272.50 to 1698.50 hours
ii. 129.90 to 1841.10 hours
b. Using Chebyshev’s theorem, find the interval that contains the time (in hours) that at least 75% of Americans spent watching television in 2010.
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# Math All sessions first category second school third school fourth quality 5th level 6th grade 7th class 8th class 9th grade 10th standard eleventh class 12th level 13th standard.
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In this article learn what numbers are and you will learn about the different monthly payment methods. Initially, you’ll have the figures to 10 know. Then this phone numbers are more substantial. Discover how you may with amounts to 20 and later on even get around to 100th
If you want to share you gummy bears with your friend, you can also divide numbers, for example. You might have 20 gummy bears. What is it necessary to do? Right, you write about the jelly children on justice. This means that every single exactly the www.edutopia.org same range of jelly toddlers get. You can then have the two 10 gummy bears.
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„As well as“ and „minus“ are referred to as simple arithmetic. In school, you start to add up to 20 with plus and minus. If you can do that all right, we continue learned with plus and minus to 100th
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About most of these issues, one can learn considerably right here around. Evaluate you only to your video lessons. Have a good time!
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# What is the minimum acceleration that stops the rod from sliding in the frictionless surface? [closed]
What should be the block's acceleration 'a' such that the rod present inside, inclined at at the angle 'theta' and mass 'm' will be obstructed from sliding down. From car's frame of reference, a pseudoforce 'ma' must indeed act opposite to the direction of acceleration to stop it from sliding, but my calculations are not bringing the actual answer according to my book(a=gcot(theta)). The given picture is about what I tried with all the forces acting.
You've basically drawn the diagram right; for a steady-state the net force must be 0 and hence $|\vec N_2| = m~g$ and $|\vec N_1| = m~|\vec a|.$ The remaining relation is a torque balance, because it's also in a steady-state rotationally. Let's take the torques around the point that $\theta$ is being measured around; then $\vec N_2$ and $m \vec a$ provide no torque and it's all about $\vec N_1$ contributing a torque $|\vec N_1| \sin\theta$ and the gravitational force contributing a torque $-m g \cos\theta$.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2014-04-01 22:56:41
iLloyd054
Member
Registered: 2014-04-01
Posts: 10
### Probability Problem
1.
a.) Your instructor gives the class a surprise four-question with 3-choice quiz. You have not studied the material being quizzed and therefore decide to answer the four questions by randomly guessing the answers without reading the questions or the answers.
Calculate
i) what is the probability that you have more than half of the answers correct?
ii) what do you think the class average number of correct answers will be, if an entire class answers the quiz by guessing ?
b) The incomes of senior executive in a large corporation are normally distributed with a standard deviation of RM 1200. A cutback is pending, at which time those who earn less RM 28,000 will be discharged. If such a cut represents 5% of the junior executives, what is the current mean salary of the group of senior executives?
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## #2 2014-04-01 23:28:56
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: Probability Problem
Hi;
a)
b)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #3 2014-04-02 08:59:07
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: Probability Problem
Hi
I am actually getting 7/81 for a)i), bobbym. Are you sure you are counting only the cases with 3 and 4 correct answers?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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## #4 2014-04-02 09:02:43
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: Probability Problem
Yes, I am counting 3, and 4 correct.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #5 2014-04-02 09:08:48
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: Probability Problem
What do you get for 3?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
## #6 2014-04-02 09:11:32
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: Probability Problem
8 / 81
If you want to do this in M:
``````d = BinomialDistribution[4, 1/3];
Probability[x > 2, x \[Distributed] d]``````
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #7 2014-04-02 09:17:04
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: Probability Problem
How are you getting that?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
## #8 2014-04-02 09:19:07
ShivamS
Member
Registered: 2011-02-07
Posts: 3,648
### Re: Probability Problem
Bobbym, your code does not work by the way.
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## #9 2014-04-02 09:21:39
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: Probability Problem
Did you copy and paste? It works fine here.
Hi anonimnystefy, this is a straight binomial distribution problem. The distributions of statistics are even more powerful then gf's for solving problems and M knows all about them.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #10 2014-04-02 09:31:28
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: Probability Problem
Hi bobbym
I made the mistake in taking the Binomial[4,3] to be 3 when doing it by hand.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
## #11 2014-04-02 09:37:58
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: Probability Problem
But of course my young friend. If someone's answer disagrees with mine then I automatically know it is incorrect.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #12 2014-04-02 10:00:00
ElainaVW
Member
Registered: 2013-04-29
Posts: 580
### Re: Probability Problem
Keep joking with him like that and he isn't going to be your friend long. Why must you always tease the people you like?
The multivariable generating function can also be used.
The probability is 1 / 9.
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## #13 2014-04-02 10:01:45
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: Probability Problem
Well, I never teased you...
You are correct about the gf though.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #14 2014-04-02 10:06:13
ElainaVW
Member
Registered: 2013-04-29
Posts: 580
### Re: Probability Problem
Sure you wanted to go into that in the other thread? I mean about the artificial intelligence work. You are correct and all but young people should not be exposed to different viewpoints. It will confuse them.
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## #15 2014-04-02 10:09:43
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: Probability Problem
I could not disagree more. pappym and grandpappyd exposed me to much of what I know to be true now when I was much younger than him. I expect his reaction will be disbelief and dismissal same as mine was but young people can handle differing opinions easier than you think.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #16 2014-04-02 10:13:23
ElainaVW
Member
Registered: 2013-04-29
Posts: 580
### Re: Probability Problem
How will you handle the rest of the posted problems?
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## #17 2014-04-02 10:14:49
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: Probability Problem
They look like homework so I will start to stress to the OP the importance of at least trying to solve a problem before asking for help.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #18 2014-04-02 11:10:15
eigenguy
Member
Registered: 2014-03-18
Posts: 78
### Re: Probability Problem
A different approach.
"Having thus refreshed ourselves in the oasis of a proof, we now turn again into the desert of definitions." - Bröcker & Jänich
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NEWS ARCHIVE
# WHERE THINGS HARDER?
The force with which a body is attracted to earth, decreases as the elevation above the earth's surface. If we raised kilo weight to a height of 6400 km, i.e. remove it from the center of the globe on two of its radius, the force of gravity would fail in 22, i.e. 4 times, and a weight on a spring steelyard would pull just 250 g instead of 1000. According to the law of gravity globe attracts external body as if all its mass was concentrated in the center, and the force of this attraction decreases inversely as the square of the distance. In our case, the distance of the weights from the center of the Earth has doubled, and because of the weak attraction 22, i.e., four times. Removing the weight on 12800 km from the earth surface, i.e. at triple the distance from the center of the Earth, we would weaken the attraction 32, i.e. 9 times.
1000-gram weight would weigh only 111 grams, etc.
There is the idea that, venturing with weight in the bowels of the Earth, i.e., bringing the body to the center of our planet, we should see an increase in gravity: the weight in the depths of the Earth should weigh more. This conjecture is incorrect: with a hollow in the Earth body does not increase in weight, but on the contrary, decreases.
The reason is that in this case, the magnetic particles of the Earth are located on one side of the body, but on different sides. Look at the picture.
Why with a hollow in the Earth the force of gravity weakens.
You can see that the weight placed deep in the Earth, drawn down by the particles located below the weights, but at the same time is drawn up of those particles that lie above it. You can prove that ultimately matters is capturing only a ball whose radius is the distance from the centre of the Earth to the location of the body. therefore, the weight of the body as depressions in the Ground would quickly decrease. Reaching the center of the Earth, the body will completely lose weight, become weightless as the surrounding particles cause its there in all directions with equal force.
So, just more body weighs on the surface of the Ground; remove from her high into the weight decreases [As would occur if the globe was quite homogeneous density: in fact, the density of the Earth increases with approaching to the center; therefore, the force of gravity when going deep into the Earth first, at some distance, growing, and only then begins to weaken.].
Entertaining physics J. Perelman
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## Hannah and her sweets
Apparently students in the UK have been protesting against the following question on a GCSE math exam (see e. g. coverage at The Guardian):
There are n sweets in a bag. Six of the sweets are orange. The rest of the sweets are yellow. Hannah takes a random sweet from the bag. She eats the sweet. Hannah then takes at random another sweet from the bag. She eats the sweet. The probability that Hannah eats two orange sweets is 1/3. Show that n²-n-90=0.
The probability that the first sweet is orange is $6/n$. Now there are five orange sweets left out of $n-1$, so the probability that the second sweet is orange, assuming that the first one is, is $5/(n-1)$. Therefore we need to solve $(6/n) \times (5/(n-1)) = 1/3$. Multiplying it out gives
${30 \over n(n-1)} = {1 \over 3}$
and we can easily rearrange to get $90 = n(n-1)$. So $n = 10$; there are 10 sweets. (I guarantee you that a bunch of students went straight for the quadratic formula at this point – but you don’t have to, it’s easy to find two consecutive numbers that multiply to 90.) According to the BBC, setters of the exam point out that this was supposed to be one of the more difficult questions, “targeted at students aiming for A and A* grades”.
There’s a question this reminds me of, of which I don’t recall the original source: there are r raspberry sweets and b blueberry sweets in a bag. You take two of them at random; the probability that they have the same flavor is exactly 1/2. What are possible values for r and b? (Okay, so I’ve heard it with “red” and “blue”, but let’s go with fruit flavors.) (Okay, so I’ve heard it with “red” and “blue”, but let’s go with fruit flavors.) See for example this Quora question. This one is a bit trickier, and depends on getting lucky and choosing the right parametrization of the problem. If the number of reds/raspberries is $r$ and the number of blue(berries) is $b$, then we have
${r \over r+b} {r-1 \over r+b-1} + {b \over r+b} {b-1 \over r+b-1} = {1 \over 2};$
the first term is the probability of drawing red-red and the second is the probability of drawing blue-blue. We can rewrite as
$2(r(r-1) + b(b-1)) = (r+b)(r+b-1)$
but that isn’t much of a help, to be honest. Solving for $r$ in terms of $b$ gives
$r = {1 \over 2} \left( 2b \pm \sqrt{8b+1} + 1 \right)$
and if you happen to know the obscure piece of trivia that for integer $b$, $8b+1$ is a perfect square if and only if $b$ is triangular, then you can show that $r$ and $b$ must be two consecutive triangular numbers. For example $b = 15$ leads to the solutions $r = 21$ and $r = 10$.
But if you use n = r + b, then is as pointed out at the Quora answer, $r = (n \pm \sqrt{n})/2$ (with different notation) and this doesn’t require knowing anything obscure. Setting $n = 36$ leads immediately to the solutions $r = 21, r = 15$, for example. In general, to make this work out the number of sweets needs to be a perfect square.
What if there are three colors of sweets? How can we choose the number of sweets of each color to make the probability of getting a match equal to one-third?
Posted in Uncategorized | 6 Comments
## The most mathematical flag: Nepal
Do you like geometrical constructions? Then you should like the flag of Nepal, which is the only non-rectangular flag of a nation, is actually defined by a Euclidean ruler and compass construction in their constitution. See a step by step drawing of the construction or this video from Numberphile.
The flag of the state of Ohio is also non-rectangular, but sadly their laws only describe it as “burgee-shaped”. It might be possible to extract a construction from the picture in this brochure from the Ohio secretary of state, but it looks like many of the points are specified but their position in a larger rectangle. Ohio’s flag is analytic geometry; Nepal’s is synthetic.
Posted in Uncategorized | 5 Comments
## Today in p-values
From Nature, by Jeffrey T. Leek and Roger D. Peng : p-values are just the tip of the iceberg (that is, of ways statistics are misused). Here’s Leek’s post at Simply Statistics giving some examples of subcultures of data analysis.
Andrew Gelman also posted today on good, mediocre, and bad p-values, quoting an article he wrote in 2012, P-values and statistical practice.
Posted in Uncategorized | 10 Comments
## Stark and Freishtat on course evaluations
Philip Stark (UC Berkeley statistics) and Richard Freishtat (UC Berkeley Center for Teaching and Learning, which supports undergraduate education) have written An evaluation of course evaluations.
Stark and Freishtat observe, among other things:
• that nonresponse bias is a serious problem;
• that averaging ordinal variables doesn’t make sense (are a 3 and a 7 on a seven-point scale the same as two 5s?);
• that students can more effectively comment on some aspects of pedagogy than others;
• That student evaluations are influenced by student grade expectations, and by instructor gender, age, ethnicity, and attractiveness…
If anybody should know how hard measurement is, it’s statisticians.
With this in mind, I’m amused to remember that when I was there, teaching evaluation averages by instructor and course were actually posted on a bulletin board, with other information relevant to students, outside the departmental office on the third floor of Evans Hall. Apparently the department has a more “holistic” procedure in place now for evaluating teaching; I was not at Berkeley long enough to comment on the old process. (Two academic years, as a lecturer.)
To be honest, I often found student comments more useful than grades – but it is difficult to read those comments. The format of the evaluations and the fact that they’re usually given at the end of a class period seems designed to discourage thorough comments (and Stark and Freishtat point out that comments in evaluations of technical courses tend to be less discursive). And the most critical comments tend to stick in one’s craw, which is only human nature.
Posted in Uncategorized | 1 Comment
## Coo, shiver my sceptre!
A couple weeks ago James Grime linked on Twitter to a puzzle in the January 8, 1981 issue of New Scientist, which runs as follows:
“Beauty? Courage? Generosity? Patience? Wisdom? Which do you wish for the new-born Princess?” asked the Good Fairy.
“Beauty and Wisdom will do nicely, thank you,” replied the King, not wanting to seem greedy.
“Wait! For each gift you name, I shall bestow on her two of the other gifts instead. Each name of a gift triggers a different pair. Each gift is triggered by two of the names. But if you mention both names, they cancel out and she will not get that gift at all.”
“Coo, shiver my sceptre!” exclaimed His Majesty.
“Quite simple! For instance if you ask for Beauty and Courage, she will receive Generosity and Patience. If you ask for Beauty, Generosity and Patience, she will receive those three and Wisdom too. You in fact wished for Beauty and Wisdom. She shall have them, provided you ask for them in the simplest way.”
So we can associate each of the five gifts — let’s denote them by $B, C, G, P, W$ — with two of the others. Let’s denote this by $f(B) = X + Y$, for example, where $X$ and $Y$ are the two gifts triggered by $B$. So what we know is
$f(B) + f(C) = G + P$
and
$f(B) + f(G) + f(P) = B + G + P + W.$
Formally, $B, C, G, P, W$ are generators of $(\mathbb{Z}/2\mathbb{Z})^5$; informally they’re symbols that when added to themselves cancel out. Now, this function $f$ is a homomorphism from $(\mathbb{Z}/2\mathbb{Z})^5$ to itself – that is, $f(x+y) = f(x) + f(y)$. (This means we can do the cancelling either before or after translating from the language of what was wished for to what actually happens.) So therefore we know
$f(B+C) = G+P, f(B+G+P) = B+G+P+W$.
$f(C+G+P) = B+W$
which gives a way to get both beauty and wisdom — namely, asking for courage, generosity, and patience.
But what’s $f(B+C+G+P+W)$ — that is, what do you get if you ask for everything? We have that “each name of a gift triggers a different pair, and each gift is triggered by two of the names”. So we have
$f(B+C+G+P+W) = 2B+2C+2G+2P+2W$
since each gift is triggered twice. And the right-hand side there is just zero. So
$f(B+C+G+P+W) = 0$
and we can rewrite this:
$f((C+G+P) + (B+W)) = 0.$
But since $f$ is a homomorphism that’s just
$f(C+G+P) + f(B+W) = 0$
which is another way of saying $f(C+G+P) = f(B+W)$. So in fact $f(B+W) = B+W$ — that is, to get beauty and wisdom, just ask for them.
To see that this is the simplest possible solution, we need to show that no single gift triggers both beauty and wisdom. I can’t come up with a “clean” way to do this, but we can go brute force. Beauty must trigger wisdom, wisdom must trigger beauty, and beauty and wisdom both trigger the same other gift. This gift can’t be courage – if it were, then we’d have $f(B) = W+C$, but we know $f(B+C) = G+P$, so adding these we’d have $f(C) = W+C+G+P$, which is impossible. So the “other” gift is either generosity or patience. Repeatedly applying the constraints that every gift can occur twice and the facts we already know, those lead to the solutions
$f(B) = W+G, f(C) = W+P , f(G) = C+P, f(P) = B+C, f(W) = B+G$
and
$f(B) = W+P, f(C) = W+G, f(G) = B+C, f(P) = C+G, f(W) = B+P$
respectively, which differ by exchanging $G$ and $P$ wherever they appear. In no case is there a single gift $X$ with $f(X) = B+W$, so the solution $f(B+W) = B+W$ is indeed the simplest one.
Note: Jim Randell has been doing programmatic solutions to these puzzles for quite a while now.
Posted in Uncategorized | 2 Comments
## Help with social science about science
Perhaps of interest to many of my readers: Melanie Sinche, a career counselor and consultant, and research associate at Harvard Law’s Labor and Worklife Program, is working on a study of the career paths of PhD receipients in the sciences. If you:
– Earned a PhD in any of the physical, computational, social, life sciences or engineering between 2004 and 2014
– Has ever worked, trained, or studied in the U.S.
then you should fill out this survey and help to paint a more accurate picture of what PhD recipients actually do.
Edited, April 16: turns out I forgot to link to the actual survey.
(H/T Tamara Broderick)
Posted in Uncategorized | 2 Comments
## Various computations of the odds of a perfect March Madness bracket
Joseph Nebus has written a series of posts on the entropy in basketball results: for a single team, for both teams, in the win-loss results for a 64-team tournament like March Madness. For the final question he gets an answer of about 48 bits. That is, the probability of guessing the winners and losers of a tournament correctly is on the order of 248.
From blind guessing one gets 1 in 263, quoted for example in this USA Today story, with the caveat that:
Even so, allowing for some knowledge of college basketball and taking it account the norms of the NCAA tournament, the odds of a perfect bracket are still about 1 in 128 billion, according to DePaul math professor Jay Bergen.
This refers to Jeff Bergen’s video, “where does 1 in 128 billion come from”. Note 128 billion is roughly 237. Bergen’s strategy is to assume that the top seeds always win, since this is the most likely outcome.
The fact that two reasonable people gave two such different answers is an example of just how hard it is to estimate small probabilities. But both of these models gave up on using empirical data after the first round. Yet matchups between the “favorite” seeds should happen fairly often, and there will be data! Let’s look at win probabilities by seed as compiled at mcubed.net. In a tournament where all the favorites win, we’ll have:
• in the first round, four matches of 1 vs. 16, 2 vs. 15, …, 8 vs. 9, one in each region
• four matches of 1 vs. 8, 2 vs. 7, 3 vs. 6, 4 vs. 5, one in each region
• four matches of 1 vs. 4, 2 vs. 3, one in each region
• four matches of 1 vs. 2, one in each region
• three matches of 1 vs. 1, from different regions, of course
Historically, 1 seeds are 124-0 against 16 seeds, 2 seeds are 117-7 against 15 seeds, and so on until 8 seeds are 79-69 against 9 seeds. So the probability of picking all eight first-round games in one region perfectly is
$(124/124)(117/124)(104/124)(99/123)(97/144)(95/144)(90/148)(79/148) = 0.09188$
and the probability of getting all 32 first-round games right is the fourth power of this, about $7.12 \times 10^{-5}$ or one in 14,000. (The different denominators correspond to different numbers of times each matchup has occurred, presumably due to changes in the tournament structure; the 64-team field only dates back to 1985. Oddly enough, the Washington Post reports that nobody ever seems to pick a perfect first round. This isn’t a contradiction – nobody is boring enough to pick the strategy with the highest expected value for that bet, when the bet most people are interested in is trying to win their pool.
The probability of picking a perfect second round in any given region is $(65/81)(64/88)(46/84)(49/88) = 0.17796$; for all four regions it’s the fourth power of this, about $1.00 \times 10^{-3}$.
The third round in each region consists of a 1-vs-4 game and a 2-vs-3 game, where the favorites win with probability 46/68 and 36/59 respectively; the probability of picking all eight third round games correctly is $((46/68)(36/59))^4 = 0.0290$.
The fourth round in each region is a 1-vs-2 game, where the 1 seed has historically won with probability 38/69; the probability of picking all four correctly is $(38/69)^4 = 0.0919$.
Finally, the probability of picking all three Final Four games correctly is 1/8 – the model knows nothing beyond seeding.
Multiplying this all out, I get that the probability of picking all 63 games correctly is
$(7.12 \times 10^{-5}) (1.00 \times 10^{-3}) (0.0290) (0.0919) (1/8) = 2.38 \times 10^{-11}$
or about one in 42 billion, in a generic tournament. For what it’s worth, FiveThirtyEight gave 1 in 1.6 billion this year and 1 in 7.4 billion last year, using a model that actually knew something about basketball.
Posted in Uncategorized | 2 Comments
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# Thread: Sampling Question - Help!!!
1. ## Sampling Question - Help!!!
A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. One each page the mean area devoted to display ads was measured ( a display ad is a large block of multicolered illustrations, maps and text). The data (in square millimeters) are shown below:
0 260 356 403 536 0 268 369 428 536
268 396 469 536 162 338 403 536 536 130
(a)Construct a 95 percent confidence interval for the true mean. (b)Why might normality be an issue here? (c)What sample size would be needed to obtain an error of +/- square millimeters with 99 percent confidence? (d)If this is not a reasonable requirement suggest one that is.
I need help understanding what the questions are asking and how to input info into excel or minitab. Please help!!!!!!!!!!!
2. Originally Posted by chocolat771
A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. One each page the mean area devoted to display ads was measured ( a display ad is a large block of multicolered illustrations, maps and text). The data (in square millimeters) are shown below:
0 260 356 403 536 0 268 369 428 536
268 396 469 536 162 338 403 536 536 130
(a)Construct a 95 percent confidence interval for the true mean. (b)Why might normality be an issue here? (c)What sample size would be needed to obtain an error of +/- square millimeters with 99 percent confidence? (d)If this is not a reasonable requirement suggest one that is.
I need help understanding what the questions are asking and how to input info into excel or minitab. Please help!!!!!!!!!!!
First let's take this one step at a time. You need a sample mean and standard deviation. Your number set is:
0,260,356,403,536,0,268,369,428,536,268,396,469,53 6,162,338,403,536,536,130
Enter that here --> Mean, Variance, and Standard Deviation
I get $\displaystyle \bar{X} = 346.5$
$\displaystyle \sigma = 166.0643$
Now, since your sample size is less than 30, we should not do the normal test, but rather, the Student t-test. To see how that works, go here:
Student t-Test Confidence Interval for Mean
Using our mean and standard deviation that we calculated, n=20, and 95 for our confidence interval, I get:
$\displaystyle 268.7804 < \mu < 424.2196$
PS - Scroll down in the math for a message on how to calculate this using Excel.
For your questions c and d, you can calculate a 99% confidence interval for your current sample size, but I'm not sure I understand what you mean by what sample size needed to ensure 99% confidence.
3. Originally Posted by chocolat771
A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. One each page the mean area devoted to display ads was measured ( a display ad is a large block of multicolered illustrations, maps and text). The data (in square millimeters) are shown below:
0 260 356 403 536 0 268 369 428 536
268 396 469 536 162 338 403 536 536 130
(a)Construct a 95 percent confidence interval for the true mean. (b)Why might normality be an issue here? (c)What sample size would be needed to obtain an error of +/- ...... square millimeters with 99 percent confidence? (d)If this is not a reasonable requirement suggest one that is.
I need help understanding what the questions are asking and how to input info into excel or minitab. Mr F says: I'd suggest that you go to a minitab or excel reference book or manual.
There is a value missing where I've inserted the red dots. Part (c) cannot be done without the missing value.
(a) Use the t-distribution because population variance is unknown and sample size is small (n < 50).
df = 20 - 1 = 19. $\displaystyle t_{\alpha/2} = t_{0.025} = 2.093$ (using http://www.anu.edu.au/nceph/surfstat...e/tables/t.php, for example).
$\displaystyle \bar{x} - t_{\alpha/2} \frac{s}{\sqrt{n}} < \mu < \bar{x} + t_{\alpha/2} \frac{s}{\sqrt{n}}$
where $\displaystyle \bar{x}$ is the sample mean and s is the sample sd.
(b) For sample sizes less than ~15: The t-distribution can be used if the population is close to normal. The t-distribution should not be used if the population is non-normal or if outliers are present in the sample data set.
For sample sizes ~15 or greater and less than ~40: t-distribution can be safely used whether or not the population is normal, except in the presence of outliers or strong skewness in the data set.
Your data set has outliers .....
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# Real number multiplicative inverses expressed in another form
I've been asked to express the multiplicative inverse of $3 + \sqrt{5}$ in the form $c + d\sqrt{5}$, where $c,d$ are rational numbers.
I understand that for some rational numbers $c,d$ we must have: $$1 = (3 + \sqrt{5})(c + d\sqrt{5}).$$
I was able to answer for the multiplicative inverse of $2 +\sqrt{3}$. We find that $1 = (2 +\sqrt{3})(c + d\sqrt{3})$ where $c = 2$ and $d = -1$. However this seems to be related to the original $2 +\sqrt{3}$; and in the problem at hand this is not the case. So I'm rather confused.
How could i go about solving this? I'd like steps without the answer IF possible; moreover, how could i prepare for more general questions in this form?
-
Let's generalise this: you want to write $\dfrac{1}{x+y\sqrt{z}}$ in the form $b+c\sqrt{z}$:
$$\dfrac{1}{x+y\sqrt{z}} = \dfrac{(x-y\sqrt{z})}{(x+y\sqrt{z})(x-y\sqrt{z})} = \dfrac{x-y\sqrt{z}}{x^2-y^2z} = \dfrac{x}{x^2-y^2z} + \dfrac{-y}{x^2-y^2z} \sqrt{z}.$$
So $b= \frac{x}{x^2-y^2z}$ and $c= \frac{-y}{x^2-y^2z}$.
It works for complex numbers too.
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\begin{align} Answer &= \frac{1}{3+\sqrt{5}}\\ \mbox{Multiply and Divide by Conjugate}&\\ &= \frac{1}{3+\sqrt{5}}\cdot \frac{3-\sqrt{5}}{3-\sqrt{5}}\\ &= \frac{3-\sqrt{5}}{(3)^2-(\sqrt{5})^2}\\ \end{align}
$$\frac{3}{4}-\frac{\sqrt{5}}{4}$$
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We have $a+b\sqrt c$, where $a,b,c$ are known, and $c$ is not a perfect square, and we want to find $x$ and $y$ to make $$(x +y\sqrt c)(a + b\sqrt c) = 1.$$
Multiplying out, we get $$(ax + bcy) + (ay+bx)\sqrt c = 1.$$
The first term, $ax+bcy$, is an integer and its sum with $(ay+bx)\sqrt c$ is $1$, another integer, so $(ay+bx)\sqrt c$ must be an integer as well. The only way this can happen is if $ay+bx=0$, in which case we must have $ax+bcy = 1$. So we have two equations in $x$ and $y$:
\begin{align} ay&+bx&=0 \\ ax&+bcy&=1 \end{align}
We can solve these by any of the usual methods and get \begin{align} x&=\frac{a}{a^2-b^2c} \\ y&=\frac{-b}{a^2-b^2c} \end{align}
so the solution is $$\frac{a}{a^2-b^2c} + \frac{-b}{a^2-b^2c}\sqrt c.$$
Checking with your $2+\sqrt3$ example, we have $a=2, b=1, c=3$, so $a^2-b^2c = 1$ and the inverse should be simply $a-b\sqrt 3 = 2 -\sqrt 3$, as you said.
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Point from 2 points given distances 1-20 21
From: George (GKSL4) 27 Oct 2010 (1 of 21)
Hi Michael, Is there a way (shortcut) to have a point with known distances from 2 given points on a standard view? Thank you, George Image Attachments:
From: Michael Gibson 27 Oct 2010 (2 of 21)
3832.2 In reply to 3832.1 Hi George, > Is there a way (shortcut) to have a point with known > distances from 2 given points on a standard view? If you create 2 circles you can then find that point using intersection snap. I'm not really sure if there is much possibility to make it any shorter than that, because you need to identify the location of P1, and the distance of L1, and then the location of P2 and the distance of L2. Creating 2 circles is the same number of picks as that, pick the location of P1 for the center of the circle, and type in L1 for the radius of the circle, pick the location of P2 for the radius of the 2nd circle, and type in L2 for the radius of the 2nd circle... - Michael
From: George (GKSL4) 27 Oct 2010 (3 of 21)
3832.3 In reply to 3832.2 Hi Michael, Thank you for the immediate answer. Right now I use this solution, as you can see in the attachment, but I ask for an automated way because most times I have many points estimating triangles from field measurements. George
From: Michael Gibson 27 Oct 2010 (4 of 21)
3832.4 In reply to 3832.3 Hi George, could you describe how you would want an automated method to work? I can't see how an automated method will be able to do it in fewer picks - don't you need to pick 2 points and enter 2 distances? That's the same number of picks for creating 2 circles... - Michael
From: George (GKSL4) 27 Oct 2010 (5 of 21)
3832.5 In reply to 3832.4 Hi Michael, The most annoying step is that you have to delete 2 circles and then go to the next pair of points and so on. Thanks again. George PS. Wish I have construction circles..
From: Frenchy Pilou (PILOU) 28 Oct 2010 (6 of 21)
circles "helper lines" will be fine ;) --- Pilou Is beautiful that please without concept! My Gallery
From: Michael Gibson 28 Oct 2010 (7 of 21)
3832.7 In reply to 3832.5 Hi George, > The most annoying step is that you have to delete 2 > circles and then go to the next pair of points and so on. Well, the only alternative short of a whole "construction circle" mechanism (which is not really possible to do as a plug-in, it would need quite a bit of internal support stuff for that), would seem to be to make a plug-in that creates 2 point objects, since there are actually 2 possible solutions where the 2 circles intersect. So it's either delete 2 circles or delete 2 points... Would it help any if there was a command to delete the last 2 created curves so you could do that with 1 keypress without needing to actually select them? - Michael
From: coi (MARCO) 28 Oct 2010 (8 of 21)
3832.8 In reply to 3832.6 hey george you could use the SelectClosedCurves for a tinytiny workflow-enhancement http://kyticka.webzdarma.cz/3d/moi/#SelectClosedCurves it selects all closed curves, so it doesn't bother with points and lines.. your work flow would be something like this 1. draw circle 1 -> radius 2. draw circle 2 -> radius 3. pick point -> intersection 1 or 2 4. SelectClosedCurves -> delete cheers, marco EDITED: 28 Oct 2010 by MARCO
From: George (GKSL4) 28 Oct 2010 (9 of 21)
Hi all, Thank you for your solutions. Michael, I think is better to have 2 points , since one will be used (the remaining point(s) does not complicate the plan). Probably a selection between them, about the side, will solve it. Thanks, George
From: Frenchy Pilou (PILOU) 28 Oct 2010 (10 of 21)
3832.10 In reply to 3832.8 < it selects all closed curves, so it doesn't bother with points and lines.. yes but it's not dangerous? If another closed curves than circles are existing anywhere in the screen? They will be killed also ;)
From: coi (MARCO) 28 Oct 2010 (11 of 21)
3832.11 In reply to 3832.10 it's dangerous..that's for sure ;]
From: Frenchy Pilou (PILOU) 28 Oct 2010 (12 of 21)
3832.12 In reply to 3832.11 so just a little selection by rectangle right to left is maybe speedy and sufficient before the circles helpers lines :) (if they come a day in the v3 ;)
From: Michael Gibson 28 Oct 2010 (13 of 21)
3832.13 In reply to 3832.9 Hi George, > I think is better to have 2 points , since one will be used > (the remaining point(s) does not complicate the plan). > Probably a selection between them, about the side, will solve it. I should be able to set something for that up. Would a command that had a sequence like this work: Pick point P1 Enter distance L1 Pick point P2 Enter distance L2 Pick point, intersection point closest to the picked point is taken. Or would you want a different sequence, like pick both points first and then enter the distances? - Michael
From: George (GKSL4) 28 Oct 2010 (14 of 21)
3832.14 In reply to 3832.13 Hi Michael, Your sequence >Pick point P1 Enter distance L1 Pick point P2 Enter distance L2 Pick point, intersection point closest to the picked point is taken.< will be the best. Thanks, George
From: Michael Gibson 28 Oct 2010 (15 of 21)
3832.15 In reply to 3832.14 Hi George, I've attached a plug-in here that you can try. To install it, download the attached .zip file and unzip it, then copy the 2 files to the \commands sub-folder inside of MoI's main installation folder. That will then make a new command called Intersect2Circles available. To trigger the command, set up a keyboard shortcut under Options > Shortcut keys, and for the command part of the shortcut put in: Intersect2Circles Then when you push that key, this custom command should start and it works as described above. It lets you create 2 circles and then at the last step you pick a point to tell it which intersection you want to take. For the last pick you just need to pick a point closer to one of the 2 possibilities, you don't need to actually put it at the intersection although you can if you want. The circles will be displayed only while you are inside of the command and when you are finished only the single point will stick around and the circles will disappear. I hope that might help for processing a lot of these! - Michael Attachments:
From: Frenchy Pilou (PILOU) 28 Oct 2010 (16 of 21)
Works like a charm! bravo! A little trick if you dont make a another shortcut and don't want re type the command Type one time, copy it then just copy past it for the other time ;) PS I obtain an error message when I draw first point in Top view, second point in Front view Seems your script works only in the same window and not with circles not coplanar ? EDITED: 28 Oct 2010 by PILOU Attachments:
From: Michael Gibson 28 Oct 2010 (17 of 21)
3832.17 In reply to 3832.16 Hi Pilou, > A little trick if you dont make a another shortcut and > don't want re type the command > Type one time, copy it then just copy past it for > the other time ;) Also if you just want to repeat the last command that was run you can do that by right-clicking in a view port or pushing the Enter key after the command has finished. > Seems your script works only in the same window > and not with circles not coplanar ? Yes, the circles must be coplanar and also intersecting for this command to work properly. - Michael
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{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
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https://stackoverflow.com/questions/15887885/converting-a-one-item-list-to-an-integer
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# Converting a one-item list to an integer
I've been asked to accept a list of integers (x), add the first value and the last value in the list, and then return an integer with the sum. I've used the following code to do that, but the problem I have is that when I try to evaluate the sum it's actually a one-item list instead of an integer. I've tried to cast it to an int but I can't seem to get it to work.
``````def addFirstAndLast(x):
lengthOfList = len(x)
firstDigit = x[0:1]
lastDigit = x[lengthOfList:lengthOfList-1]
sum = firstDigit + lastDigit
return sum
``````
• What should the behavior be if `x` contains one number? Should it return `x[0]` or `2*x[0]`? – Joe Frambach Apr 8 '13 at 20:03
# Use indexes
You're slicing the list, which return lists. Here, you should use indexes instead:
``````firstDigit = x[0]
lastDigit = x[-1]
``````
# Why is slicing wrong for you:
When you do `x[0:1]`, you're taking the list of items from the beginning of the list to the first interval.
`````` item0, item1, item2, item3
^ interval 0
^ interval 1
^ interval 2
^ interval 3
``````
Doing `x[0:2]`, for example, would return items 0 and 1.
It all boils down to this:
``````def addFirstAndLast(x):
return x[0] + x[-1]
``````
In Python, a negative list index means: start indexing from the right of the list in direction to the left, where the first position from right-to-left is `-1`, the second position is `-2` and the last position is `-len(lst)`.
Use Slice Notation:
``````def addFirstAndLast(x):
return x[0] + x[-1]
``````
x[0] = will give you 0th index of the list, first value.
x[-1] = will give you the last element of the list.
| 489
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| 3.46875
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https://whatisconvert.com/547-days-in-hours
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## Convert 547 Days to Hours
To calculate 547 Days to the corresponding value in Hours, multiply the quantity in Days by 24 (conversion factor). In this case we should multiply 547 Days by 24 to get the equivalent result in Hours:
547 Days x 24 = 13128 Hours
547 Days is equivalent to 13128 Hours.
## How to convert from Days to Hours
The conversion factor from Days to Hours is 24. To find out how many Days in Hours, multiply by the conversion factor or use the Time converter above. Five hundred forty-seven Days is equivalent to thirteen thousand one hundred twenty-eight Hours.
## Definition of Day
A day (symbol: d) is a unit of time. In common usage, it is either an interval equal to 24 hours or daytime, the consecutive period of time during which the Sun is above the horizon. The period of time during which the Earth completes one rotation with respect to the Sun is called a solar day. Several definitions of this universal human concept are used according to context, need and convenience. In 1960, the second was redefined in terms of the orbital motion of the Earth, and was designated the SI base unit of time. The unit of measurement "day", redefined in 1960 as 86 400 SI seconds and symbolized d, is not an SI unit, but is accepted for use with SI. A civil day is usually 86 400 seconds, plus or minus a possible leap second in Coordinated Universal Time (UTC), and occasionally plus or minus an hour in those locations that change from or to daylight saving time.
## Definition of Hour
An hour (symbol: h; also abbreviated hr.) is a unit of time conventionally reckoned as 1⁄24 of a day and scientifically reckoned as 3,599–3,601 seconds, depending on conditions. The seasonal, temporal, or unequal hour was established in the ancient Near East as 1⁄12 of the night or daytime. Such hours varied by season, latitude, and weather. It was subsequently divided into 60 minutes, each of 60 seconds. Its East Asian equivalent was the shi, which was 1⁄12 of the apparent solar day; a similar system was eventually developed in Europe which measured its equal or equinoctial hour as 1⁄24 of such days measured from noon to noon. The minor variations of this unit were eventually smoothed by making it 1⁄24 of the mean solar day, based on the measure of the sun's transit along the celestial equator rather than along the ecliptic. This was finally abandoned due to the minor slowing caused by the Earth's tidal deceleration by the Moon. In the modern metric system, hours are an accepted unit of time equal to 3,600 seconds but an hour of Coordinated Universal Time (UTC) may incorporate a positive or negative leap second, making it last 3,599 or 3,601 seconds, in order to keep it within 0.9 seconds of universal time, which is based on measurements of the mean solar day at 0° longitude.
## Using the Days to Hours converter you can get answers to questions like the following:
• How many Hours are in 547 Days?
• 547 Days is equal to how many Hours?
• How to convert 547 Days to Hours?
• How many is 547 Days in Hours?
• What is 547 Days in Hours?
• How much is 547 Days in Hours?
• How many hr are in 547 d?
• 547 d is equal to how many hr?
• How to convert 547 d to hr?
• How many is 547 d in hr?
• What is 547 d in hr?
• How much is 547 d in hr?
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{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
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# Physics:Jiles–Atherton model
The Jiles–Atherton model of magnetic hysteresis was introduced in 1984 by David Jiles and D. L. Atherton.[1] This is one of the most popular models of magnetic hysteresis. Its main advantage is the fact that this model enables connection with physical parameters of the magnetic material.[2] Jiles–Atherton model enables calculation of minor and major hysteresis loops.[1] The original Jiles–Atherton model is suitable only for isotropic materials.[1] However, an extension of this model presented by Ramesh et al.[3] and corrected by Szewczyk [4] enables the modeling of anisotropic magnetic materials.
## Principles
Magnetization $\displaystyle{ M }$ of the magnetic material sample in Jiles–Atherton model is calculated in the following steps [1] for each value of the magnetizing field $\displaystyle{ H }$:
• effective magnetic field $\displaystyle{ H_\text{e} }$ is calculated considering interdomain coupling $\displaystyle{ \alpha }$ and magnetization $\displaystyle{ M }$,
• anhysteretic magnetization $\displaystyle{ M_\text{an} }$ is calculated for effective magnetic field $\displaystyle{ H_\text{e} }$,
• magnetization $\displaystyle{ M }$ of the sample is calculated by solving ordinary differential equation taking into account sign of derivative of magnetizing field $\displaystyle{ H }$ (which is the source of hysteresis).
## Parameters
Original Jiles–Atherton model considers following parameters:[1]
Parameter Units Description
$\displaystyle{ \alpha }$ Quantifies interdomain coupling in the magnetic material
$\displaystyle{ a }$ A/m Quantifies domain walls density in the magnetic material
$\displaystyle{ M_\text{s} }$ A/m Saturation magnetization of material
$\displaystyle{ k }$ A/m Quantifies average energy required to break pinning site in the magnetic material
$\displaystyle{ c }$ Magnetization reversibility
Extension considering uniaxial anisotropy introduced by Ramesh et al.[3] and corrected by Szewczyk [4] requires additional parameters:
Parameter Units Description
$\displaystyle{ K_\text{an} }$ J/m3 Average anisotropy energy density
$\displaystyle{ \psi }$ rad Angle between direction of magnetizing field $\displaystyle{ H }$ and direction of anisotropy easy axis
$\displaystyle{ t }$ Participation of anisotropic phase in the magnetic material
## Modelling the magnetic hysteresis loops
### Effective magnetic field
Effective magnetic field $\displaystyle{ H_\text{e} }$ influencing on magnetic moments within the material may be calculated from following equation:[1]
$\displaystyle{ H_\text{e} = H + \alpha M }$
This effective magnetic field is analogous to the Weiss mean field acting on magnetic moments within a magnetic domain.[1]
### Anhysteretic magnetization
Anhysteretic magnetization can be observed experimentally, when magnetic material is demagnetized under the influence of constant magnetic field. However, measurements of anhysteretic magnetization are very sophisticated due to the fact, that the fluxmeter has to keep accuracy of integration during the demagnetization process. As a result, experimental verification of the model of anhysteretic magnetization is possible only for materials with negligible hysteresis loop.[4]
Anhysteretic magnetization of typical magnetic material can be calculated as a weighted sum of isotropic and anisotropic anhysteretic magnetization:[5]
$\displaystyle{ M_\text{an} = (1 - t) M_\text{an}^\text{iso} + t M_\text{an}^\text{aniso} }$
#### Isotropic
Isotropic anhysteretic magnetization $\displaystyle{ M_\text{an}^\text{iso} }$ is determined on the base of Boltzmann distribution. In the case of isotropic magnetic materials, Boltzmann distribution can be reduced to Langevin function connecting isotropic anhysteretic magnetization with effective magnetic field $\displaystyle{ H_\text{e} }$:[1]
$\displaystyle{ M_\text{an}^\text{iso} = M_\text{s}\left(\coth\left(\frac{H_\text{e}}{a}\right) - \frac{a}{H_\text{e}}\right) }$
#### Anisotropic
Anisotropic anhysteretic magnetization $\displaystyle{ M_\text{an}^\text{aniso} }$ is also determined on the base of Boltzmann distribution.[3] However, in such a case, there is no antiderivative for the Boltzmann distribution function.[4] For this reason, integration has to be made numerically. In the original publication, anisotropic anhysteretic magnetization $\displaystyle{ M_\text{an}^\text{aniso} }$ is given as:[3]
$\displaystyle{ M_\text{an}^\text{aniso} = M_\text{s}\frac{\int_0^\pi \! e^{E(1) + E(2)}\sin(\theta)\cos(\theta)\,d\theta}{\int_0^\pi \! e^{E(1) + E(2)}\sin(\theta)\,d\theta} }$
where
$\displaystyle{ E(1)=\frac{H_\text{e}}{a}\cos\theta-\frac{K_\text{an}}{M_\text{s} \mu_0 a} \sin^2(\psi-\theta) }$
$\displaystyle{ E(2)=\frac{H_\text{e}}{a}\cos\theta-\frac{K_\text{an}}{M_\text{s} \mu_0 a} \sin^2(\psi+\theta) }$
It should be highlighted, that a typing mistake occurred in the original Ramesh et al. publication.[4] As a result, for an isotropic material (where $\displaystyle{ K_\text{an}=0) }$), the presented form of anisotropic anhysteretic magnetization $\displaystyle{ M_\text{an}^\text{aniso} }$ is not consistent with the isotropic anhysteretic magnetization $\displaystyle{ M_\text{an}^\text{iso} }$ given by the Langevin equation. Physical analysis leads to the conclusion that the equation for anisotropic anhysteretic magnetization $\displaystyle{ M_\text{an}^\text{aniso} }$ has to be corrected to the following form:[4]
$\displaystyle{ M_\text{an}^\text{aniso} = M_\text{s}\frac{\int_0^\pi \! e^{0.5(E(1) + E(2))}\sin(\theta)\cos(\theta)\,d\theta}{\int_0^\pi \! e^{0.5(E(1) + E(2))}\sin(\theta)\,d\theta} }$
In the corrected form, the model for anisotropic anhysteretic magnetization $\displaystyle{ M_\text{an}^\text{aniso} }$ was confirmed experimentally for anisotropic amorphous alloys.[4]
### Magnetization as a function of magnetizing field
In Jiles–Atherton model, M(H) dependence is given in form of following ordinary differential equation:[6]
$\displaystyle{ \frac{dM}{dH} = \frac{1}{1 + c}\frac{M_\text{an} - M}{\delta k - \alpha(M_\text{an} - M)} + \frac{c}{1 + c}\frac{dM_\text{an}}{dH} }$
where $\displaystyle{ \delta }$ depends on direction of changes of magnetizing field $\displaystyle{ H }$ ($\displaystyle{ \delta = 1 }$ for increasing field, $\displaystyle{ \delta = -1 }$ for decreasing field)
### Flux density as a function of magnetizing field
Flux density $\displaystyle{ B }$ in the material is given as:[1]
$\displaystyle{ B(H) = \mu_0 M(H) }$
where $\displaystyle{ \mu_0 }$ is magnetic constant.
## Vectorized Jiles–Atherton model
Vectorized Jiles–Atherton model is constructed as the superposition of three scalar models one for each principal axis.[7] This model is especially suitable for finite element method computations.
## Numerical implementation
The Jiles–Atherton model is implemented in JAmodel, a MATLAB/OCTAVE toolbox. It uses the Runge-Kutta algorithm for solving ordinary differential equations. JAmodel is open-source is under MIT license.[8]
The two most important computational problems connected with the Jiles–Atherton model were identified:[8]
For numerical integration of the anisotropic anhysteretic magnetization $\displaystyle{ M_\text{an}^\text{aniso} }$ the Gauss–Kronrod quadrature formula has to be used. In GNU Octave this quadrature is implemented as quadgk() function.
For solving ordinary differential equation for $\displaystyle{ M(H) }$ dependence, the Runge–Kutta methods are recommended. It was observed, that the best performing was 4-th order fixed step method.[8]
## Further development
Since its introduction in 1984, Jiles–Atherton model was intensively developed. As a result, this model may be applied for the modeling of:
Moreover, different corrections were implemented, especially:
• to avoid unphysical states when reversible permeability is negative [15]
• to consider changes of average energy required to break pinning site [16]
## Applications
Jiles–Atherton model may be applied for modeling:
• rotating electric machines [17]
• power transformers [18]
• magnetostrictive actuators [19]
• magnetoelastic sensors [20][21]
• magnetic field sensors (e. g. fluxgates) [22][23]
It is also widely used for electronic circuit simulation, especially for models of inductive components, such as transformers or chokes.[24]
## References
1. Jiles, D. C.; Atherton, D.L. (1984). "Theory of ferromagnetic hysteresis". Journal of Applied Physics 55 (6): 2115. doi:10.1063/1.333582. Bibcode1984JAP....55.2115J.
2. Liorzou, F.; Phelps, B.; Atherton, D. L. (2000). "Macroscopic models of magnetization". IEEE Transactions on Magnetics 36 (2): 418. doi:10.1109/20.825802. Bibcode2000ITM....36..418L.
3. Ramesh, A.; Jiles, D. C.; Roderick, J. M. (1996). "A model of anisotropic anhysteretic magnetization". IEEE Transactions on Magnetics 32 (5): 4234. doi:10.1109/20.539344. Bibcode1996ITM....32.4234R.
4. Szewczyk, R. (2014). "Validation of the anhysteretic magnetization model for soft magnetic materials with perpendicular anisotropy". Materials 7 (7): 5109–5116. doi:10.3390/ma7075109. PMID 28788121. Bibcode2014Mate....7.5109S.
5. Jiles, D.C.; Ramesh, A.; Shi, Y.; Fang, X. (1997). "Application of the anisotropic extension of the theory of hysteresis to the magnetization curves of crystalline and textured magnetic materials". IEEE Transactions on Magnetics 33 (5): 3961. doi:10.1109/20.619629. Bibcode1997ITM....33.3961J.
6. Jiles, D. C.; Atherton, D.L. (1986). "A model of ferromagnetic hysteresis". Journal of Magnetism and Magnetic Materials 61 (1–2): 48. doi:10.1016/0304-8853(86)90066-1. Bibcode1986JMMM...61...48J.
7. Szymanski, Grzegorz; Waszak, Michal (2004). "Vectorized Jiles–Atherton hysteresis model". Physica B 343 (1–4): 26–29. doi:10.1016/j.physb.2003.08.048. Bibcode2004PhyB..343...26S.
8. Szewczyk, R. (2014). Computational problems connected with Jiles–Atherton model of magnetic hysteresis. 267. 275–283. doi:10.1007/978-3-319-05353-0_27. ISBN 978-3-319-05352-3.
9. Jiles, D.C. (1994). "Modelling the effects of eddy current losses on frequency dependent hysteresis in electrically conducting media". IEEE Transactions on Magnetics 30 (6): 4326–4328. doi:10.1109/20.334076. Bibcode1994ITM....30.4326J.
10. Szewczyk, R.; Frydrych, P. (2010). "Extension of the Jiles–Atherton model for modelling the frequency dependence of magnetic characteristics of amorphous alloy cores for inductive components of electronic devices". Acta Physica Polonica A 118 (5): 782. doi:10.12693/aphyspola.118.782. Bibcode2010AcPPA.118..782S. ]
11. Sablik, M.J.; Jiles, D.C. (1993). "Coupled magnetoelastic theory of magnetic and magnetostrictive hysteresis". IEEE Transactions on Magnetics 29 (4): 2113. doi:10.1109/20.221036. Bibcode1993ITM....29.2113S.
12. Szewczyk, R.; Bienkowski, A. (2003). "Magnetoelastic Villari effect in high-permeability Mn-Zn ferrites and modeling of this effect". Journal of Magnetism and Magnetic Materials 254: 284–286. doi:10.1016/S0304-8853(02)00784-9. Bibcode2003JMMM..254..284S.
13. Jackiewicz, D.; Szewczyk, R.; Salach, J.; Bieńkowski, A. (2014). "Application of extended Jiles–Atherton model for modelling the influence of stresses on magnetic characteristics of the construction steel". Acta Physica Polonica A 126 (1): 392. doi:10.12693/aphyspola.126.392. Bibcode2014AcPPA.126..392J.
14. Szewczyk, R. (2006). "Modelling of the magnetic and magnetostrictive properties of high permeability Mn-Zn ferrites". Pramana 67 (6): 1165–1171. doi:10.1007/s12043-006-0031-z. Bibcode2006Prama..67.1165S.
15. Deane, J.H.B. (1994). "Modeling the dynamics of nonlinear inductor circuits". IEEE Transactions on Magnetics 30 (5): 2795–2801. doi:10.1109/20.312521. Bibcode1994ITM....30.2795D.
16. Szewczyk, R. (2007). "Extension of the model of the magnetic characteristics of anisotropic metallic glasses". Journal of Physics D: Applied Physics 40 (14): 4109–4113. doi:10.1088/0022-3727/40/14/002. Bibcode2007JPhD...40.4109S.
17. Du, Ruoyang; Robertson, Paul (2015). "Dynamic Jiles–Atherton Model for Determining the Magnetic Power Loss at High Frequency in Permanent Magnet Machines". IEEE Transactions on Magnetics 51 (6): 7301210. doi:10.1109/TMAG.2014.2382594. Bibcode2015ITM....5182594D.
18. Huang, Sy-Ruen et al. (2012). "Distinguishing internal winding faults from inrush currents in power transformers using Jiles–Atherton model parameters based on correlation voefficient". IEEE Transactions on Magnetics 27 (2): 548. doi:10.1109/TPWRD.2011.2181543.
19. Calkins, F.T.; Smith, R.C.; Flatau, A.B. (2008). "Energy-based hysteresis model for magnetostrictive transducers". IEEE Transactions on Magnetics 36 (2): 429. doi:10.1109/20.825804. Bibcode2000ITM....36..429C.
20. Szewczyk, R.; Bienkowski, A. (2004). "Application of the energy-based model for the magnetoelastic properties of amorphous alloys for sensor applications". Journal of Magnetism and Magnetic Materials 272: 728–730. doi:10.1016/j.jmmm.2003.11.270. Bibcode2004JMMM..272..728S.
21. Szewczyk, R. et al. (2012). "Application of extended Jiles–Atherton model for modeling the magnetic characteristics of Fe41.5Co41.5Nb3Cu1B13 alloy in as-quenched and nanocrystalline State". IEEE Transactions on Magnetics 48 (4): 1389. doi:10.1109/TMAG.2011.2173562. Bibcode2012ITM....48.1389S.
22. Szewczyk, R. (2008). "Extended Jiles–Atherton model for modelling the magnetic characteristics of isotropic materials". Acta Physica Polonica A 113 (1): 67. doi:10.12693/APhysPolA.113.67. Bibcode2008JMMM..320E1049S.
23. Moldovanu, B.O.; Moldovanu, C.; Moldovanu, A. (1996). "Computer simulation of the transient behaviour of a fluxgate magnetometric circuit". Journal of Magnetism and Magnetic Materials 157-158: 565–566. doi:10.1016/0304-8853(95)01101-3. Bibcode1996JMMM..157..565M.
24. Cundeva, S. (2008). "Computer simulation of the transient behaviour of a fluxgate magnetometric circuit". Serbian Journal of Electrical Engineering 5 (1): 21–30. doi:10.2298/sjee0801021c.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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#1 2011-09-29 19:23:56
anonimnystefy
Real Member
Offline
Another programming problem!!!
hi guys
i am trying to make a program to solve a Sudoku.i hope you know what that is.
now i have finished the code but there seems to be a logical error because i entered a valid one and it outputted the message 'The Sudoku cannot be solved!!!'
here's the code:
Code:
```program Sudoku_solver;
{\$mode objfpc}{\$H+}
uses
{\$ENDIF}{\$ENDIF}
Classes
{ you can add units after this };
{\$IFDEF WINDOWS}{\$R Sudoku.rc}{\$ENDIF}
const
max=9;
type
niz=array[1..max] of integer;
matrica=array[1..max] of niz;
var
a:matrica;
i,j:integer;
ok:boolean;
function row(a:matrica;i1,j1:integer):boolean;
var
j:integer;
b:boolean;
begin
b:=true;
for j:=1 to 9 do
if j<>j1 then
if a[i1][j]=a[i1][j1] then
b:=false;
red:=b;
end;
function column(a:matrica;i1,j1:integer):boolean;
var
i:integer;
b:boolean;
begin
b:=true;
for i:=1 to 9 do
if i<>i1 then
if a[i][j1]=a[i1][j1] then
b:=false;
vrsta:=b;
end;
function square(a:matrica;i1,j1:integer):boolean;
var
i,j,k,l:integer;
b:boolean;
begin
b:=true;
case i1 of
1,2,3: begin
k:=0;
i:=3
end;
4,5,6: begin
k:=3;
i:=6;
end;
7,8,9: begin
k:=6;
i:=9;
end;
end;
case j1 of
1,2,3: begin
l:=0;
j:=3
end;
4,5,6: begin
l:=3;
j:=6;
end;
7,8,9: begin
l:=6;
j:=9;
end;
end;
while (k<=i) do
begin
k:=k+1;
while (l<=j) do
begin
l:=l+1;
if (k<>i1) and (l<>j1) then
if a[k][l]=a[i1][j1] then
b:=false;
end;
end;
end;
function pos(a:matrica;i,j:integer):boolean;
begin
poz:=row(a,i,j) and column(a,i,j) and square(a,i,j);
end;
procedure sudoku(var a:matrica;n,i1,j1:integer;ok:boolean);
var
i,j,k:integer;
b:boolean;
begin
i:=i1;
j:=j1;
a[i][j]:=n;
ok:=false;
b:=true;
if poz(a,i,j) then
begin
ok:=true;
i:=0;
j:=0;
while (b=true) and (i<=max) do
begin
i:=i+1;
while (b=true) and (j<=max) do
begin
j:=j+1;
if a[i][j]=0 then
b:=false;
end;
end;
if b=false then
begin
k:=0;
ok:=false;
while (k<=9) and not ok do
begin
k:=k+1;
sudoku(a,k,i,j,ok);
end;
end;
end;
end;
begin
writeln('Enter the Sudoku: ');
for i:=1 to 9 do
begin
for j:=1 to 9 do
end;
k:=0;
ok:=false;
while (k<=9) and not ok do
begin
k:=k+1;
sudoku(a,k,i,j,ok);
end;
if ok then
begin
writeln('The solution is: ')
for i:=1 to 9 do
begin
for j:=1 to 9 do
write(a[i][j]);
writeln;
end;
end
else writeln('Sudoku cannot be solved!!!');
end.```
note that functions row,column and square check if there are same numbers as the number we are looking at in the same row,column and square.
my second question is: is there another (better) way to make the Sudoku solver,because this one is fairly long and complex !?
Last edited by anonimnystefy (2011-10-01 07:14:47)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#2 2011-09-29 22:27:16
bobbym
Online
Re: Another programming problem!!!
Hi anonimnystefy;
That is not long at all for a sudoku solver. What is the sudoku problem that you are
testing this on?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#3 2011-09-30 05:36:34
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
hi bobbym
it's:
0 4 7 0 0 0 3 0 0
8 0 6 0 7 4 0 0 0
0 0 0 2 0 3 8 4 0
0 7 0 0 0 0 4 0 3
9 0 0 4 0 8 0 0 2
4 0 3 0 0 0 0 9 0
0 2 8 6 0 1 0 0 0
0 0 0 8 2 0 7 0 9
0 0 9 0 0 0 2 1 8
2 4 7 1 8 9 3 5 6
8 3 6 5 7 4 9 2 1
5 9 1 2 6 3 8 4 7
1 7 2 9 5 6 4 8 3
9 6 5 4 3 8 1 7 2
4 8 3 7 1 2 6 9 5
7 2 8 6 9 1 5 3 4
3 1 4 8 2 5 7 6 9
6 5 9 3 4 7 2 1 8
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#4 2011-09-30 06:07:24
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
just for practice:
Last edited by anonimnystefy (2011-09-30 06:13:23)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#5 2011-09-30 06:35:11
bobbym
Online
Re: Another programming problem!!!
Funny thing is my program does noot get that one either.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#6 2011-09-30 06:53:23
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
but why won't it work?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#7 2011-09-30 06:59:44
bobbym
Online
Re: Another programming problem!!!
Hi;
I do not know. I checked the rows and columns of the solution as well as
every 3 x 3 grid. It should have got that answer?!
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#8 2011-09-30 07:14:34
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
have you checked it compleely.it uses recursion.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#9 2011-09-30 07:17:52
bobbym
Online
Re: Another programming problem!!!
Hi anonimnystefy;
No, I haven't. I am looking at mine as to why it does not get that answer.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#10 2011-09-30 07:27:03
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
what does it get?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#11 2011-09-30 13:18:52
bobbym
Online
Re: Another programming problem!!!
Spits it out as if it does not have a solution! Where does that solution come from?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#12 2011-10-01 04:39:40
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
from me!
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#13 2011-10-01 05:27:28
bobbym
Online
Re: Another programming problem!!!
That is what I was thinking. Did you try it on any others?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#14 2011-10-01 06:37:54
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
nope! :embarrassed!
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#15 2011-10-01 06:38:56
bobbym
Online
Re: Another programming problem!!!
I can give you some examples if you do not have any.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#16 2011-10-01 06:56:06
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
i have bunch of them and i tried this one:
0 5 6 0 8 0 0 1 0
8 3 0 2 0 7 0 0 0
0 1 2 0 0 0 0 0 0
0 0 0 6 0 0 7 0 9
0 0 0 1 0 4 0 0 0
2 0 3 0 0 8 0 0 0
0 0 0 0 0 0 3 9 0
0 0 0 4 0 9 0 2 6
0 7 0 0 5 0 1 4 0
it won't work!
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#17 2011-10-01 07:15:58
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
hi bobbym
i found an error on my side in the main code and i fixed it and i edited it but it still won't do it!
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#18 2011-10-01 07:19:43
bobbym
Online
Re: Another programming problem!!!
Hi;
Something must be wrong with yours.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#19 2011-10-01 07:25:15
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
ya think?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#20 2011-10-01 07:27:03
bobbym
Online
Re: Another programming problem!!!
That is about all I can do. I do not speak Java. Even my C++ is too rusty. Use your debugger to go line by line watching the variables as you go.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#21 2011-10-01 20:03:47
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
where did you get Java?
i'm using Pascal.very basic.
i changed my code a bit and got a program that solves only the first row.it won't go to the next one!
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#22 2011-10-01 20:34:51
bobbym
Online
Re: Another programming problem!!!
Hi anonimnystefy;
Forgive my sense of humor, I know, you told me you were using Lazarus.
In the other thread, just out of curiosity, I asked if lazarus has a debugger? Does it?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#23 2011-10-01 20:36:51
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
think so.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#24 2011-10-01 20:39:38
bobbym
Online
Re: Another programming problem!!!
Does it have a variable pane? Or a variable watch?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#25 2011-10-01 20:44:34
anonimnystefy
Real Member
Offline
Re: Another programming problem!!!
yes,yes it does.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
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## Common Core State Standards for Mathematics
Grade(s) 5 resources related to the following standard:
Number and Operations—Fractions
Use equivalent fractions as a strategy to add and subtract fractions.
2. Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2/5 + 1/2 = 3/7, by observing that 3/7 < 1/2.
Showing 11-18 of 18 resultsResults per page: 10515202550 Sort by: TitleRatingNewest First
ResourceResource Type
Rating
This document provides descriptions and examples of what each Mathematics Common Core standard means a Grade Five student will know, understand and be able to do. This "unpacking" of the standards provides instructional guidelines and was developed to assist North Carolina educators teach the Mathematics Common Core (Standard Course of Study).
Reference Materials
This progressions document from the Common Core Standards Writing team explains how fractions are incorporated in the standards from third grade through fifth grade. Each grade level includes an explanation, important vocabulary, and illustrations.
Instructional Strategy, Reference Materials
This 88-page document addresses the meaning of the Common Core State Standards for teaching fractions in grades 3-7. For each grade level the specific standards that are addressed are listed, followed by a detailed explanation of what the standards mean, including visual models, and any misconceptions teachers or students might have about how specific standards apply to fractions in the CCSS.
Instructional Strategy, Reference Materials, Article
Grade Level: 3, 4, 5, 6+
This 8-page article (PDF) details the errors and misconceptions that students may have with fraction computation. The article emphasizes the importance of putting fraction problems in a real world context in order to deepen understanding of fractions and operations with fractions. By anticipating potential errors and misconceptions, teachers can help students avoid misunderstandings that might arise from these models and contexts.
Instructional Strategy, Reference Materials, Article
Thinking Blocks is an interactive Flash tool for modeling and solving math problems visually. Students represent quantities and relationships by placing blocks and braces on a work space and using the tools to resize and label them accordingly. Users can change the color of blocks, move them, copy them, divide them into equal parts, and separate them. A pencil tool and keyboard are also available. The site includes video tutorials demonstrating how to use the tool and how to model a wide variety of problem types. It also contains a bank of hundreds of word problems.
Activity, Interactive Media, Problem Set, Tools, Video
Grade Level: 1, 2, 3, 4, 5, 6+
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https://math.answers.com/Q/What_is_9_twice_6
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0
# What is 9 twice 6?
Updated: 10/24/2023
Wiki User
9y ago
The product of 9 multiplied by 6 is 54.
3mo ago
Wiki User
9y ago
9 x 6 = 54
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Q: What is 9 twice 6?
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Related questions
### How do you write an algebraic expression for 9 less than twice the quotient of a number n and 6?
2n/6 - 9 or n/3 - 9
### Is 3 out of 9 bigger or 1 out of six?
3 put of 9 is double (twice) the size of 1 out of 6.
### Fifteen more than twice a number is 9?
15+2*n = 9 2*n = -6 n = -3
### What does 6 more than twice a number is 12 equal A -9 B -3 C 9 D 3?
It is D because: 6+2*3 = 12
### The product of your number and 3 is the same product as twice 9 What is your number?
3x = 2 x 9 2 x 9 = 18 3x = 18 18/x = 6 Therefore, x = 6
### When Christine is twice as old as she is now she will be three times as old as she was three years ago how old is she now?
9 x 2 = 18, 9 - 3 = 6, 6 x 3 = 18, the answer is 9
9
### How long are shar-pei in season for?
The first time should be between 6 & 9 months then twice a year.
### The width of rectangle is 6 inches less tha twice its length Find the dimension of the rectangle if its area is 108 square inches?
W=12,l=9 ((9*2)-6=12, 12*9=108)
### What is twice the cube of 3?
the answer is 5454466372830405846527293049123456789009876543212345678908765412345678877654322345678996543212345678976412345678987654321234567 if u want the answer look at the first 2 numbers ¯_(ツ)_/¯
### How many number combinations are there for the numbers 1 to 9 without the number appearing twice?
Multiply: 9! (9 factorials) (9) (8) (7) (6) (5) (4) (3) (2) (1)
### When twelve is subtracted from twice some number the result is 6?
If you are looking for what number it is, it can be described by this equation: 2x-12=6. Solving this equation x=9.
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https://www.physicsforums.com/threads/conversion-of-g-acceleration-into-body-weight.825223/
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# Conversion of g acceleration into "body weight"
Tags:
1. Jul 28, 2015
### tjvv
Hi guys,
I have a table with vertical peak acceleration values [g] and I want to find out a formula that convert the [g] values into ground reaction forces in units of body weight (BW). I think Newton´s 2nd law (F = m x a) may help but I am stuck on finding a generic relationship (problem is that I do not have the body mass values...)
Example: peak acceleration of a walking person is 2 "g". How much is it in terms of body weights?
Activity [g] [Body weight]
-----------------------------------------------------------
Walking 2 ????
Jumping 5 ????
Thank you!
tjvv
Last edited: Jul 28, 2015
2. Jul 28, 2015
### A.T.
Coordinate acceleration (CA) relative to ground, or proper acceleration (PA) that an accelerometer measures?
You mean the vertical ground reaction force (VGRF) needed to achieve that vertical acceleration?
VGRF[BW] = PA[g] = CA+1[g]
3. Jul 28, 2015
### Andrew Mason
Body weight is a force so to express body weight in terms of acceleration you would have to multiply by the body's mass.
g = 9.8 m/sec2. The body weight of a person of mass M (in kg) is Mg =9.8M Newtons.
So a person walking horizontally and accelerating horizontally at 2g = 19.6 m/sec2 would have to exert a force of 2Mg = 19.6M Newtons, where M=mass of the person.
AM
Last edited: Jul 28, 2015
4. Jul 28, 2015
### tjvv
I mean proper acceleration (PA) that an accelerometer measures.
Yes, when I mean bodyweight it is the vertical ground reaction force.
Since it is a force we can put in newton´s 2nd law (F=m * a):
F = [(m * acceleration measured) / (m * earth acceleration)] + 1 //+1 is to consider only linear acceleration (excluding earth gravity)
BW = [acceleration measured / earth gravity] +1
So in the example from the table above:
BW = [2 / 10] +1 //considering gravity as 10m/s2 to simplify
BW = 0.2 +1
BW = 1.2 which means the force of walking would be 1.2 times the body weight of a person
General formula would be this: BW = [acceleration measured / earth gravity] +1
Is this correct?
5. Jul 28, 2015
### A.T.
If "acceleration measured" is proper acceleration then you don't need that "+1". See my formula in post #2.
6. Jul 28, 2015
### tjvv
Hi AT,
sorry I meant Coordinate acceleration (CA) relative to ground (since it excludes the earth gravity).
So from your formula it means that to convert a G force into BW it is just sum the measured g force with earth´s gravity?
Example:
BW = CA+1[g]
BW = 2 + 1
BW = 3 (getting the initial table example, would mean that the force acting when people is walking would be 3 times person´s body weight)
Thanks
7. Jul 28, 2015
### A.T.
To accelerate your center of mass at 2g upwards, relative to the ground, you have to apply 3 times your body weight to the ground. But you don't have such high accelerations during walking.
8. Jul 29, 2015
### Andrew Mason
Just to add to what AT has said, if you think of the acceleration provided by a 100 m. sprinter running the 100 m. dash in 10.4 seconds and taking 1 second to get up to a speed of 10 m/sec, the acceleration is only 1 g. I think that is pretty close to the maximum horizontal human acceleration you are going to see.
AM
9. Jul 29, 2015
### A.T.
He is asking about vertical accelerations, which can be higher for a fraction of a second during landing impact (jumping, running fast). But during normal walking the vertical ground reaction doesn't go much beyond 1 body weight, so the vertical acceleration is a small fraction of 1g.
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https://documen.tv/question/calculate-the-specific-heat-of-a-metal-from-the-following-data-a-container-made-of-the-metal-has-16831904-38/
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## Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains 20 kg of wat
Question
Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains 20 kg of water. A 2.3 kg piece of the metal initially at a temperature of 165°C is dropped into the water. The container and water initially have a temperature of 15.0°C, and the final temperature of the entire system is 18.0°C.
in progress 0
3 years 2021-08-30T02:10:11+00:00 1 Answers 0 views 0
C = 771.35 J/kg°C
Explanation:
Here, e consider the conservation of energy equation. The conservation of energy principle states that:
Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container
Since,
Heat Given or Absorbed by a material = m C ΔT
Therefore,
m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃
where,
m₁ = Mass of Metal Piece = 2.3 kg
C = Specific Heat of Metal = ?
ΔT₁ = Change in temperature of metal piece = 165°C – 18°C = 147°C
m₂ = Mass of Metal Container = 3.8 kg
ΔT₂ = Change in temperature of metal piece = 18°C – 15°C = 3°C
m₃ = Mass of Water = 20 kg
C₃ = Specific Heat of Water = 4200 J/kg°C
ΔT₃ = Change in temperature of water = 18°C – 15°C = 3°C
Therefore,
(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)
C[(2.3 kg)(147°C) – (3.8 kg)(3°C)] = 252000 J
C = 252000 J/326.7 kg°C
C = 771.35 J/kg°C
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# Question regarding solving polynomial of congruence?
In my textbook, they said:
$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$
The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$
And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:
Let $y = 2x^{3} + 7x - 4$, we have:
$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$ $$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$ $$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$ $$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$ $$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$
What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this?
Thanks,
-
## 4 Answers
Any integer is going to be congruent to one of $0,1,2,3,4$ modulo $5$. As the testing shows, only those integers which are congruent to $1$ modulo $5$ are solutions, and any integer congruent to $1$ mod $5$ is a solution.
Those tests are using an arbitrary $x$ each time, so you shouldn't think of them as a system of congruences of a single variable $x$ but instead they imply the result described above.
Here's another thing to add which may help. Don't just check $0,1,2,3,4$. Take $x$ to be any arbitrary integer, not necessarily one of those just listed. So like I said before, this arbitrary $x$ is congruent to one of $0,1,2,3,4$ since they form a complete residue system. If $x\equiv 0\pmod{5}$, then $$2x^3+7x-4\equiv 2(0)^3+7(0)-4\equiv -4\equiv 1\pmod{5}$$ so $x$ is not a solution. Likewise if $x\equiv 2,3,4\pmod{5}$. So the only choice which works is when $x\equiv 1\pmod{5}$. Hopefully that's clearer?
-
Thanks. But how do they get that equation, I'm still confused :(. – Chan Mar 10 '11 at 7:13
@Chan, what equation do you mean? – yunone Mar 10 '11 at 7:14
Oh, I meant $x \equiv 1 \pmod{5}$. Thank you. – Chan Mar 10 '11 at 7:20
Yes, it was much clearer now. Thank you. – Chan Mar 10 '11 at 7:23
@Chan, ok, good to hear. – yunone Mar 10 '11 at 7:23
What your textbook means is $$\left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right) \iff \left(x \equiv 1 \pmod{5} \right)$$
This is what you checked by plugging in the different cases for $x$.
By taking $x \equiv 1 \pmod{5}$, you proved that $2x^{3} + 7x - 4 \equiv 0 \pmod{5}$ and hence $$\left(x \equiv 1 \pmod{5} \right) \Rightarrow \left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right)$$
Now by taking $x \neq 1 \pmod{5}$, you proved that $2x^{3} + 7x - 4 \neq 0 \pmod{5}$ and hence $$\left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right) \Rightarrow \left(x \equiv 1 \pmod{5} \right)$$
-
Ambikasaran: I still can't see it. For an arbitrary polynomial how can we tell? Is the book guessing? Thanks. – Chan Mar 10 '11 at 7:12
\begin{aligned} 2x^3+7x-4 \equiv 0 ( mod 5)\\ 2x^3+2x-4 \equiv 0 ( mod 5)\\ 2(x-1)(x^2+x+2) \equiv 0 ( mod 5) \end{aligned}
Now you have two solutions $x-1 \equiv 0 (mod5)$ or $x^2+x+2 \equiv 0 (mod5)$. You can continue and verify if $x \equiv 1 ( mod 5)$ is the only solution.
-
Thank you. I see it now. – Chan Mar 10 '11 at 7:21
You're probably aware of it, but note anyway that this only works because $5$ is prime. – Myself Mar 10 '11 at 7:33
HINT $\rm\ \ x\ p(x)\ =\ 2\ x^4 + 7\ x^2 - 4\ x\ \equiv\ 2\ (x-1)^2\ \ (mod\ 5)\$ for $\rm\ x\not\equiv 0\$ since then $\rm\ x^4\equiv 1\$
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1. Jun 24, 2010
### BustedBreaks
So I've been trying to figure this out. The question is:
If the limit x->infinity of Xn=Xo
Show that, by definition, limit x->infinity sqrt(Xn)=sqrt(Xo)
I'm pretty sure I need to use the epsilon definition.
I worked on it with someone else and we think that what we have to show is the this:
Want to show:
For all e>0 there is an N>0 s.t. for all n>N, |sqrt(Xn) - sqrt(Xo)|<e
I just don't know how to show this.
Thanks!
2. Jun 24, 2010
### Staff: Mentor
Does this help?
$$\sqrt{x_n} - \sqrt{x_0} = \sqrt{x_n} - \sqrt{x_0} \frac{\sqrt{x_n} + \sqrt{x_0}}{\sqrt{x_n} + \sqrt{x_0}} = \frac{x_n - x_0}{\sqrt{x_n} + \sqrt{x_0}}$$
3. Jun 24, 2010
### BustedBreaks
^ If it does, I can't see it. I feel like I need to find an N in terms of e to show that this si continuous or something.
4. Jun 24, 2010
### Staff: Mentor
You're given that
$$\lim_{n \to \infty} x_n = x_0$$
What does that mean in terms of the epsilon-N definition of a limit?
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## Condensation: O Level Physics Questions and Answers : MCQ 41
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In the process of condensation, thermal energy is
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regression for angular/circular data
I have supervised learning problem where targets are angles. If I would do simple regression then numbers 360 and 1 would be far away for my model, but actually they are close and predicting x and y coordinates doesn't feels right, since I am trying to predict just one number here. What is the proper way of doing such a problem?
• I am not sure to understand your problem. Have you an angular variable, say $\theta_i$ and some linear predictor $z_i$? or also your predictor are angular? or what? Jul 30, 2014 at 14:35
• Only targets are angular (as shown on the picture), predictors are numeric. Jul 30, 2014 at 14:45
• See e.g. Pewsey et al. (2013), Circular Statistics in R & the R package circular. Jul 30, 2014 at 14:52
I suggest you to take a look at the book "Topics in circular statistics" of Jammalamadaka if you are interested in circular variable.
Suppose that your data come from a circular distribution $F()$, and you want to model the (circular) mean of the circular variable: what is generally used is: $$E(\theta) = 2\arctan(\boldsymbol{\beta}\mathbf{z}_i)$$ $\theta$ is the circular variable, $\boldsymbol{\beta}$ is the vector of regression coefficients and $\mathbf{z}_i$ are the linear covariates.
If you want a parallelism with the usual linear regression you can assume that $\theta_i \sim WN(\mu_i,\sigma^2)$, where $WN()$ indicates the wrapped normal distribution that is in some sense the Normal distribution on a circle. Then
$$\mu_i = 2\arctan(\boldsymbol{\beta}\mathbf{z}_i)$$ or equivalently
$$\theta_i = 2\arctan(\boldsymbol{\beta}\mathbf{z}_i) + \epsilon_i$$ where $\epsilon_i \sim WN(0, \sigma^2)$
This type of regression is implemented in the package $circular$ that the user Scortchi suggests
• Thank you, I still don't get some things. Is it possible to use linear regression and just transform angles to something (sines,cosines)? Or the whole regression should "build" differently? I don't want to do it in R, because I have all my other processing steps in python, that's why I am asking. Jul 31, 2014 at 8:58
• Angles have no magnitude, if you transform it in something like sine, cosine or something similar, you introduce magnitude.. Jul 31, 2014 at 9:15
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Some interesting shortest path questions | Set 1
Question 1: Given a directed weighted graph. You are also given the shortest path from a source vertex ‘s’ to a destination vertex ‘t’. If weight of every edge is increased by 10 units, does the shortest path remain same in the modified graph?
The shortest path may change. The reason is, there may be different number of edges in different paths from s to t. For example, let shortest path be of weight 15 and has 5 edges. Let there be another path with 2 edges and total weight 25. The weight of the shortest path is increased by 5*10 and becomes 15 + 50. Weight of the other path is increased by 2*10 and becomes 25 + 20. So the shortest path changes to the other path with weight as 45.
Question 2: This is similar to above question. Does the shortest path change when weights of all edges are multiplied by 10?
If we multiply all edge weights by 10, the shortest path doesn’t change. The reason is simple, weights of all paths from s to t get multiplied by same amount. The number of edges on a path doesn’t matter. It is like changing unit of weights.
Question 3: Given a directed graph where every edge has weight as either 1 or 2, find the shortest path from a given source vertex ‘s’ to a given destination vertex ‘t’. Expected time complexity is O(V+E).
If we apply Dijkstra’s shortest path algorithm, we can get a shortest path in O(E + VLogV) time. How to do it in O(V+E) time? The idea is to use BFS . One important observation about BFS is, the path used in BFS always has least number of edges between any two vertices. So if all edges are of same weight, we can use BFS to find the shortest path. For this problem, we can modify the graph and split all edges of weight 2 into two edges of weight 1 each. In the modified graph, we can use BFS to find the shortest path. How is this approach O(V+E)? In worst case, all edges are of weight 2 and we need to do O(E) operations to split all edges, so the time complexity becomes O(E) + O(V+E) which is O(V+E).
Question 4: Given a directed acyclic weighted graph, how to find the shortest path from a source s to a destination t in O(V+E) time?
More Questions See following links for more questions.
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# Maharashtra Board Practice Set 41 Class 7 Maths Solutions Chapter 10 Bank and Simple Interest
Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 41 Answers Solutions Chapter 10 Bank and Simple Interest.
## Bank and Simple Interest Class 7 Practice Set 41 Answers Solutions Chapter 10
Question 1.
If the interest on Rs 1700 is Rs 340 for 2 years, the rate of interest must be__.
(A) 12%
(B) 15%
(C) 4%
(D) 10%
Solution:
(D) 10%
Hint:
∴ $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$340=\frac{1700 \times R \times 2}{100}$$
∴ R = 10%
Question 2.
If the interest on Rs 3000 is Rs 600 at a certain rate for a certain number of years, what would the interest be on Rs 1500 under the same conditions?
(A) Rs 300
(B) Rs 1000
(C) Rs 700
(D) Rs 500
Solution:
(A) Rs 300
Hint:
The interest on Rs 3000 at certain rate of interest is Rs 600.
Let us suppose the interest on Rs 1500 at the same rate is x.
∴ $$\frac{600}{3000}=\frac{x}{1500}$$
∴ x = Rs 300
Question 3.
Javed deposited Rs 12000 at 9 p.c.p.a in a bank for some years, and withdrew his interest every year. At the end of the period, he had received altogether Rs 17,400. For how many years had he deposited his money?
Solution:
Here, P = Rs 12000, R = 9 p.c.p.a and amount = Rs 17400
Amount = Principal + Interest
∴17400 = 12000 + Interest
∴Interest = 17400 – 12000 = Rs 5400
∴ $$\text { Total interest }=\frac{P \times R \times T}{100}$$
$$5400=\frac{12000 \times 9 \times \mathrm{T}}{100}$$
∴ $$\frac{5400 \times 100}{12000 \times 9}=\mathrm{T}$$
∴ T = 5 years
∴ Javed had deposited the amount for 5 years.
Question 4.
Lataben borrowed some money from a bank at a rate of 10 p.c.p.a interest for $$2\frac { 1 }{ 2 }$$ years to start a cottage industry. If she paid Rs 10250 as total interest, how much money had she borrowed?
Solution:
Here, R = 10 p.c.p.a, T = 2.5 years, I = Rs 10250
∴ P = Rs 41000
∴ Lataben had borrowed an amount of Rs 41000 from the bank.
Question 5.
Fill in the blanks in the table.
Principal Rate of interest (p.c.p.a.) Time Interest Amount i. Rs 4200 7% 3 years ii. 6% 4 years Rs 1200 iii. Rs 8000 5% Rs 800 iv. 5% Rs 6000 Rs 18000 v. $$2\frac { 1 }{ 2 }$$ % 2 5 years Rs 2400
Solution:
i. $$\text { Total interest }=\frac{P \times R \times T}{100}$$
= $$\frac{4200 \times 7 \times 3}{100}$$
= Rs 882
Amount = Principal + interest
= 4200 + 882
= Rs 5082
ii. $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$1200=\frac{\mathrm{P} \times 6 \times 4}{100}$$
∴ $$\frac{1200 \times 100}{6 \times 4}=\mathrm{P}$$
∴ P = Rs 5000
Amount = Principal + interest
= 5000 + 1200
= Rs 6200
iii. $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$800=\frac{8000 \times 5 \times \mathrm{T}}{100}$$
∴ $$\frac{800 \times 100}{8000 \times 5}=\mathrm{T}$$
∴ T = 2 years
Amount = Principal + interest
= 8000 + 800
= Rs 8800
iv. Amount = Principal + interest
∴ 18000 = Principal + 6000
∴ Principal = Rs 12000
$$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$6000=\frac{12000 \times 5 \times \mathrm{T}}{100}$$
∴ $$\frac{6000 \times 100}{12000 \times 5}=\mathrm{T}$$
∴ T = 10 years
v. R = $$2\frac { 1 }{ 2 }$$ % = 2.5 %
∴ $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$2400=\frac{\mathrm{P} \times 2.5 \times 5}{100}$$
∴ $$2400=\frac{P \times 25 \times 5}{100 \times 10}$$
∴ $$\frac{2400 \times 10 \times 100}{25 \times 5}=P$$
∴ P = Rs 19200
Amount = Principal + interest
= 19200 + 2400
= Rs 21600
Principal Rate of interest (p.c.p.a.) Time Interest Amount i. Rs 4200 7% 3 years Rs 882 Rs 5082 ii. Rs 5000 6% 4 years Rs 1200 Rs 6200 iii. Rs 8000 5% 2 years Rs 800 Rs 8800 iv. Rs 12000 5% 10 years Rs 6000 Rs 18000 v. Rs 19200 $$2\frac { 1 }{ 2 }$$ % 2 5 years Rs 2400 Rs 21600
Maharashtra Board Class 7 Maths Chapter 10 Banks and Simple Interest Practice Set 41 Intext Questions and Activities
Question 1.
Ask an adult in your house to show you a passbook and explain the entries made in it. (Textbook pg. no. 70)
Solution:
(Students should attempt the above activities with the help of their parent / teacher.)
Question 2.
Visit different banks and find out the rates of the interest they give for different types of accounts. (Textbook pg. no. 74)
Solution:
(Students should attempt the above activities with the help of their parent / teacher.)
Question 3.
With the help of your teachers, start a Savings Bank in your school and open an account in it to save up some money. (Textbook pg. no. 74)
Solution:
(Students should attempt the above activities with the help of their parent / teacher.)
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# [R] Generate random numbers up to one
(Ted Harding) ted.harding at nessie.mcc.ac.uk
Tue Mar 6 23:01:39 CET 2007
```On 06-Mar-07 Alberto Monteiro wrote:
> Ted Harding wrote:
>>
>> And, specifically (to take just 2 RVs X and Y), while U = X/(X+Y)
>> and V = Y/(A+Y) are two RVs which summ to 1, the distribution of U
>> is not the same as the distribution of X conditional on (X+Y = 1).
>>
> This question
Which question? There are (implicitly) two questions there!
> appeared in October 2006, and the answer
To the second question (X conditional on X+Y=1)
> was the Dirichlet distribution with parameters (1,1,1...1):
>
> http://en.wikipedia.org/wiki/Dirichlet_distribution
>
> It's the distribution of uniform U1, U2, ... Un with the
> restriction that U1 + U2 + ... + Un = 1.
Indeed, and the resulting (U1,U2,...,Un) is uniformly distributed
on the simplex U1+U2+...+Un=1. For n>2, however, the resulting
marginal distribution of (say) U1 conditional on (U1+U2+...+Un=1)
is no longer uniform (that only holds for n=2, as in my example).
For n=3 this is easy to see: P[U1 > u1] is the area of the triangular
simplex between its vertex at (1,0,0) and the line from (u1,1-u1,0)
to (u1,0,0), and this is equal to (1 - u1)^2, so the density of U1
is f(u1) = 2*(1-u1). In general, the marginal density of U1
in the n-dimensional Dirichlet is (n-1)*(1-u1)^(n-2).
But the aim was to illustrate Petr Klasterecky's point that
"sum(x) is a random variable as well and dividing by
sum(x) does not preserve the original distribution
data were generated from."
namely to show two ways of generating RVs distributed on
U1 + U2 + ... + Un = 1, starting from independent RVs, which
result on two different distributions, and to give an example
where dividing by sum(x) can be seen to "not preserve" the
distribution.
Indeed, I think there is sometimes a confusion between this
question and the really unrelated question: Given non-negative
numbers V1, V2, ..., Vn, how can we convert then to a probability
distribution? To which the answer is, of course, divide by their
sum.
With best wishes,
Ted.
--------------------------------------------------------------------
E-Mail: (Ted Harding) <ted.harding at nessie.mcc.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 06-Mar-07 Time: 22:01:34
------------------------------ XFMail ------------------------------
```
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Stock Analysis
# Is There An Opportunity With NorthWestern Corporation's (NASDAQ:NWE) 28% Undervaluation?
How far off is NorthWestern Corporation (NASDAQ:NWE) from its intrinsic value? Using the most recent financial data, we'll take a look at whether the stock is fairly priced by taking the expected future cash flows and discounting them to today's value. One way to achieve this is by employing the Discounted Cash Flow (DCF) model. Models like these may appear beyond the comprehension of a lay person, but they're fairly easy to follow.
Companies can be valued in a lot of ways, so we would point out that a DCF is not perfect for every situation. If you want to learn more about discounted cash flow, the rationale behind this calculation can be read in detail in the Simply Wall St analysis model.
View our latest analysis for NorthWestern
## The Calculation
We have to calculate the value of NorthWestern slightly differently to other stocks because it is a integrated utilities company. Instead of using free cash flows, which are hard to estimate and often not reported by analysts in this industry, dividends per share (DPS) payments are used. Unless a company pays out the majority of its FCF as a dividend, this method will typically underestimate the value of the stock. We use the Gordon Growth Model, which assumes dividend will grow into perpetuity at a rate that can be sustained. For a number of reasons a very conservative growth rate is used that cannot exceed that of a company's Gross Domestic Product (GDP). In this case we used the 5-year average of the 10-year government bond yield (1.9%). The expected dividend per share is then discounted to today's value at a cost of equity of 5.3%. Compared to the current share price of US\$55.2, the company appears a touch undervalued at a 28% discount to where the stock price trades currently. Valuations are imprecise instruments though, rather like a telescope - move a few degrees and end up in a different galaxy. Do keep this in mind.
Value Per Share = Expected Dividend Per Share / (Discount Rate - Perpetual Growth Rate)
= US\$2.6 / (5.3% – 1.9%)
= US\$76.8
## Important Assumptions
The calculation above is very dependent on two assumptions. The first is the discount rate and the other is the cash flows. You don't have to agree with these inputs, I recommend redoing the calculations yourself and playing with them. The DCF also does not consider the possible cyclicality of an industry, or a company's future capital requirements, so it does not give a full picture of a company's potential performance. Given that we are looking at NorthWestern as potential shareholders, the cost of equity is used as the discount rate, rather than the cost of capital (or weighted average cost of capital, WACC) which accounts for debt. In this calculation we've used 5.3%, which is based on a levered beta of 0.800. Beta is a measure of a stock's volatility, compared to the market as a whole. We get our beta from the industry average beta of globally comparable companies, with an imposed limit between 0.8 and 2.0, which is a reasonable range for a stable business.
Although the valuation of a company is important, it ideally won't be the sole piece of analysis you scrutinize for a company. It's not possible to obtain a foolproof valuation with a DCF model. Instead the best use for a DCF model is to test certain assumptions and theories to see if they would lead to the company being undervalued or overvalued. For example, changes in the company's cost of equity or the risk free rate can significantly impact the valuation. Why is the intrinsic value higher than the current share price? For NorthWestern, we've compiled three further elements you should consider:
1. Risks: Be aware that NorthWestern is showing 3 warning signs in our investment analysis , and 1 of those is a bit unpleasant...
2. Future Earnings: How does NWE's growth rate compare to its peers and the wider market? Dig deeper into the analyst consensus number for the upcoming years by interacting with our free analyst growth expectation chart.
3. Other Solid Businesses: Low debt, high returns on equity and good past performance are fundamental to a strong business. Why not explore our interactive list of stocks with solid business fundamentals to see if there are other companies you may not have considered!
PS. The Simply Wall St app conducts a discounted cash flow valuation for every stock on the NASDAQGS every day. If you want to find the calculation for other stocks just search here.
### Valuation is complex, but we're here to simplify it.
Discover if NorthWestern Energy Group might be undervalued or overvalued with our detailed analysis, featuring fair value estimates, potential risks, dividends, insider trades, and its financial condition.
Access Free Analysis
Have feedback on this article? Concerned about the content? Get in touch with us directly. Alternatively, email editorial-team (at) simplywallst.com.
This article by Simply Wall St is general in nature. We provide commentary based on historical data and analyst forecasts only using an unbiased methodology and our articles are not intended to be financial advice. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. We aim to bring you long-term focused analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material. Simply Wall St has no position in any stocks mentioned.
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# What is 1/4 as a decimal?
## a) The long division
The long division method involved the division of the numerator by the denominator. The numerator is the number at the top of the fraction and the denominator is the number at the bottom of the fraction.
Let us do the division $$1\div4$$.
Hence the decimal form of the fraction $$\frac14$$ is 0.25.
## b) The equivalent fraction.
The equivalent fraction is the way of representing a fraction in the form of their multiples. All the equivalent fractions of a fraction simplify to the same fraction.
The equivalent fraction of a fraction can be formed by multiplying the numerator and the denominator by the same number.
Example:
$$\frac23=\frac{2\times4}{3\times4}=\frac8{12}$$
Convert the fraction $$\frac14$$ to equivalent fraction with the denominator 100
$$\frac14=\frac{1\times25}{4\times25}=\frac{25}{100}$$
Here it can be observed that 25 is the number that has to be multiplied to the denominator to make it 100. Since for an equivalent fraction the numerator should be multiplied by the same number as the denominator, we can multiply 1 with 25.
The rule to convert such fractions to decimals is
• Count the number of zeros in the denominator.
• Move the decimal point of the numerator as many places to the left.
• Here we have 2 zeros, so the decimal is to be moved to the left by 2 places.
$$\frac{25}{100}=0.25$$
We get the same answer as the previous method.
## Remember
Here are some common terms you should be familiar with.
• In the fraction $$\frac14$$, the number 1 is the dividend (our numerator)
• The number 4 is our divisor (our denominator)
We invite you to read other articles on decimal and fraction, on our blog, to find out how converting decimal to fraction.
You may be interested in what is 4/5 as a decimal fraction?
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0
# How can 9 8 4 2 equal 24 using each number once?
Wiki User
2015-08-18 14:09:23
It is: (9-4-2) times 8 = 24
Wiki User
2015-08-18 14:09:23
Study guides
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0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question
At what rate percent per annum will a sum of Rs. 2000 amount to Rs. 2205 in 2 years at compound interest?
A
3%
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B
2%
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C
5%
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D
4%
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Solution
The correct option is A 5%Given that, a sum of Rs. 2000 amounts to Rs. 2205 in 2 years.To find out: The rate of interestFor compound interest, we know that,A=P(1+R100)THere, P=Rs. 2000, A=Rs. 2205 and T=2years.∴ 2205=2000×(1+R100)2⇒ 441400=(1+R100)2⇒ 2120=(1+R100)⇒ R=120×100=5%Hence, the required rate of interest is 5%.
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Solving a dual linear program using complementary slackness if primal constraints are tight
I have a doubt in the following question:
Consider the linear program
$$\min\ \ 5x_1+12x_2 + 4x_3\ \$$, $$\ \$$subject to
$$x_1+2x_2+x_3 = 10$$
$$2x_1-x_2+3x_3=8$$
$$x_1,x_2,x_3 \geq 0$$
You are given the information that $$x_2$$ and $$x_3$$ are positive in the optimal solution. Use the complementary slackness conditions to find the optimal dual solution.
I'm referred to Problem 1 here to get an idea on how to solve a dual linear program using complementary slackness. Notice that over there, the constraints are all inequalities, so they allow for the possibility of slackness. In fact, the slackness of some constraints is used to conclude that some of the dual constraints are tight, and vice versa.
The problem in the above question is that all constraints are equalities, which means the constraints must be tight! So the procedure detailed in the linked homework solution doc can't be used.
So far what I've done is to use the constraints to express the objective function only in terms of $$x_1$$ (that's because nothing is specified for $$x_1$$ of the optimal solution). It turns out to be an increasing function of $$x_1$$, so I can set $$x_1 = 0$$ to obtain the primal solution. But I can't understand how exactly to use complementary slackness to get the dual solution in this case. A hint would be appreciated.
First of all the dual program is
\begin{align*} \texttt{Max} \ 10y_1+8y_2 \\ \ \qquad \ y_1+2y_2\leq 5 \\ 2y_1-y_2\leq 12 \\ y_1+3y_2\leq 4 \end{align*}
Using the complementary slackness theorem:
$x_j^*\cdot z_j^*=0 \ \forall \ \ j=1,2, \ldots , n \quad\color{blue}{I}$
$y_i^*\cdot s_i^*=0 \ \forall \ \ i=1,2, \ldots , m \quad\color{blue}{II}$
$s_i \text{ are the slack variables of the primal problem.}$
$z_j \text{ are the slack variables of the dual problem.}$
We know that that $x_2^*>0$ and $x_3^*>0$. Thus $z_2^*= z_3^*=0$. See $\color{blue}{I}$. That means the constraints $2$ and $3$ are equalities at the dual program.
\begin{align*} 2y_1-y_2= 12 \\ y_1+3y_2= 4 \end{align*}
Solve this little equality system and find $y_1^*$ and $y_2^*$. The values can be insert into the objective function of the dual. This optimal value is equal to the optimal value of the primal problem (Strong duality theorem).
Use the SOB table to organise ideas in a simpler manner.
variables constraints,
maximizing
constraints,
minimizing
sensible ≥ 0
odd unrestricted = =
bizzare ≤ 0
Observe how the table below can be memorised in an intuitive manner from the above table.
minimization
problem
maximization
problem
variables ≥ 0
≤ 0
unrestricted
=
constraints
constraints
=
≥ 0
≤ 0
unrestricted
variables
In this problem,
• primal equality constraints ("odd") $$\implies$$ dual unrestricted variables ("odd")
• primal nonnegative ("sensible") variables $$\implies$$ dual $$\le$$ ("sensible") constraints
• primal variables $$x_2,x_3 > 0 \implies$$ dual 2nd & 3rd constraints are tight (complementary slackness)
$$\begin{cases} 2y_1^* - y_2^* &= 12 \\ y_1^* + 3y_2^* &= 4 \end{cases}$$
This gives $$(y_1^*,y_2^*) = (40/7, -4/7)$$. The optimal value is $$10 y_1^* + 8 y_2^* = (400 - 32)/7 = 368/7$$.
Check that my work is correct. Set $$x_1^* = 0$$ and solve the system
$$\begin{cases} 2x_2^*+x_3^* &= 10 \\ -x_2^*+3x_3^* &= 8. \end{cases}$$
This gives $$(x_1^*, x_2^*) = (22/7, 26/7)$$ and the objective value $$12 x_1^* + 4 x_2^* = (264 + 104)/7 = 368/7$$. From the Weak Duality Theorem, we know that we've obtained optimal solutions for both LP's.
• @ShirishKulhari (1) From the context of the question, I assume that all solutions are basic, so that only finitely many solutions are considered. In the primal LP, there're two constraints, so only two of $x_i$'s can be nonzero, so the answer is yes. (2) Complementary slackness. Mar 28 '18 at 18:46
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# 23.1 Measuring trade balances (Page 12/13)
Page 12 / 13
• Card 12 / 13:
the balance of trade looking only at goods
• ## Keyboard Shortcuts
Previous Card ← Previous Card Button Next Card → Next Card Button Flip Card ↑ / ↓ / Return / Space
Please any Ghanaian schooling at KNUST?
combining factors of production is the role of
the situation in economic where by a more valuable good is sold at a low price while less valuable good is sold at a higher price .how can we describe this situation in economic
Fung
price discrimination
Fayaz
deman and supply
Samim
price discrimination
Citizen
causes of high elasticity of demand
causes of high elasticity of supply
Onyango
what is optimazation
lepekola
lepekola
A trade-off is a situational decision that involves diminishing one quality, or property of a design in return for gains in other aspects.
what is indifference curve ?
abdullah
its represnted by the loops of points and gives same level of satisfaction
hisham
what is enterpreneurship
Kukoyi
Entrepreneurship is the talent, knowledge and willingness individual has to engage in an activity that can result in new kind of firm
what is the short run industry supply curves?
james
I think there' s a mistake. P = - 0.4 + 0.2Qs is the supply curve and not the demand curve. Am I correct?
Qs is quantity supplied
The
This is what I think
The
this eaquation is supply curve Qs=P-0.4 the relationship is positive when the price increase the Qs increase....
mukhtaar
since Qs is quantity supplied P= -0.4 + 0.2Qs =>P +0.4=0.2Qs =>P/0.2 + 0.4=Qs I made Qs the subject of the formula or equation. So your answer is correct
The
P = -0.4 + 0.2Qs is the same as P/0.2+0.4=Qs Price has a direct relationship with the quantity supplied i.e the higher the price the higher the quantity supplied. that is why it is +0.4(this is the quantity and it is postive) and P/O.2(is the price and it is positive).
The
For the demand equation let me give an example 0.2P-0.4=Qd. Here the P is postive(+0.2) and the quantity which is -O.4 is negative( because of the negative sign(-) there is an inverse relationship between price and quantity. For quantity demanded the higher the price the lower the quantuty.
The
It's how I understand it
The
0.2P-0.4=Qd. the equation is wrong because the price have direct ralationship Quantity demanded but the correct equation is-0.2P -0.4=Qd so the higher price the lower Quantity
mukhtaar
I think the relationship is inverse because of the negative sign(-)
The
ok You mean the price and quantity demanded should be negative(inverse relationship) for Qd and the price and quantity supplied should be postive(direct relationship) for Qs
The
thank you for the correction
The
yes because it got a positive gradient of +0.2
Michael
This is the mistake I found: "Since P is on the vertical axis, it is easiest if you solve each equation for P. The demand curve is then P = 8 – 0.5Qd and the demand curve is P = –0.4 + 0.2Qs. Note that the vertical intercepts are 8 and –0.4, and the slopes are –0.5 for demand and 0.2 for supply."
Valeria
dear price do not depend on quantity. rather quantity depends on price. so the equation should be Qty=0.2Px-0.4
Michael
please can someone generate supply equation for me
ok
Detto
Qs=f(P,Pr,G,E,Z,Pf,)
The
where p is price, Pr is price of related goods, G is goals of a firm E is supplier's future expectation of prices,Z is other related factors, Pf is cost of factors of production.
The
I think it's wrong
The
if Qd=90-p Qs=90+p
The
the coefficient of price must be positive since supply curve is positively slopping
Kotey
yes
The
it's true. thank you
The
welcome
Kotey
ok
The
diagram of perfectly inelastic
chi-square test is used to test A. Analysis of variance B. Association between the qualitative variables C. Difference between means of two distribution drawn from the same population D. Difference between the means of two distribution drawn from different population
Amanuel
Confirm?
Syk
A
Satyanarayana
Lawrence
Thank you
Syk
What is economic?
bilya
is the system that study the difference between resources and the growth population
Messi
Economics studies humanbehaviour as a relation between ends and scarce means which have alternative uses
Kotey
rigth .....but economic has different concepts
Messi
yes
Kotey
what is equilibrium
it is that point where price is equal to output or rather a point where demand is equal to supply
Fung
it is a point where consumer get maximum satisfaction and producers maximise profit or minimise loss
Imtiyaz
thanks
Obed
supply equal demand in one point
Messi
it's a point where supply and demand meets/equal whether profit or loss
deany
how to compute budget constraint
how to calculate balance of payment deficit
Fung
How to calculate National income
Obed
what is indifference curves
abdullah
what,how and for whom to produce
those are problem that producer face in the process of production due to scarcity
Richman
gm
Diodo
a particular selected product is produced in a systematic hygiene condition and is produced for the customers.
what are the factors of economic growth?
tax, imports and exports, etc
deany
Explain the paradox of poverty in the midst of plenty?
total production diagram explain
what is the formula of price elasticity demand
%change in Quantity demand/%change in price
how
%change QD / %change P
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Pumpkins 6 (Posted on 2016-01-22)
Much like the third Pumpkins puzzle six pumpkins, each having a different weight, are weighed two at a time in all 15 sets of two. This time exactly four pairs of duplicate values occurred. Those four values are 60, 100, 110, and 120 pounds.
How much did each individual pumpkin weigh?
See The Solution Submitted by Brian Smith Rating: 4.0000 (1 votes)
Subject Author Date re(7): Analytical solution (Bonus puzzle question) Brian Smith 2016-02-03 11:11:43 re(6): Analytical solution (Bonus puzzle question) Steve Herman 2016-02-02 15:49:31 re(5): Analytical solution (Bonus puzzle question) Steve Herman 2016-01-30 09:36:23 re(4): Analytical solution (Bonus puzzle question) Brian Smith 2016-01-28 10:43:00 re(3): Analytical solution (Bonus puzzle question) Charlie 2016-01-27 12:13:38 re(2): Analytical solution (Bonus puzzle question) Steve Herman 2016-01-26 11:08:47 re: Analytical solution (Bonus puzzle question) Brian Smith 2016-01-26 00:31:52 re(2): Analytical solution Steve Herman 2016-01-23 12:04:36 re: Analytical solution Brian Smith 2016-01-23 11:06:42 Analytical solution Steve Herman 2016-01-23 10:21:15 computer solution Charlie 2016-01-22 15:26:29
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# 11+ maths
posted by on .
find the rules and then complete the sequence
1,2,4,3,9,5,16,7,____,____
• 11+ maths - ,
if the second term has been a 1
I would see a nice pattern
the odd-numbered positions are simply the squares of the natural numbers.
i.e. the first, third, fifth, seventh ...terms are 1,4,9,16,..
then even-numbered positions are simply the odd numbers.
i.e. the second, fourth, 6th, 8th,.. terms are (1),3,5,7,...
if the second is indeed a 2 as you typed it, I see no pattern, that fits all the terms.
Once you get past the second term, the pattern I noted still applies.
• 11+ maths inverse operation - ,
4a + 12 = 7a - 9
what is a ?
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Home > Standard Error > How To Interpret Standard Error In Regression
# How To Interpret Standard Error In Regression
## Contents
That in turn should lead the researcher to question whether the bedsores were developed as a function of some other condition rather than as a function of having heart surgery that Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the As a special case for the estimator consider the sample mean. Porter, this model identifies and analyzes 5 competitive forces ... http://techkumar.com/standard-error/how-to-interpret-standard-error.html
ISBN 0-8493-2479-3 p. 626 ^ a b Dietz, David; Barr, Christopher; Çetinkaya-Rundel, Mine (2012), OpenIntro Statistics (Second ed.), openintro.org ^ T.P. standard-error share|improve this question asked Jan 8 '13 at 16:53 setudent 612 What do you mean by "How exactly do statistical packages choose regression models (in particular ordinal regression)?"? It is particularly important to use the standard error to estimate an interval about the population parameter when an effect size statistic is not available. For the age at first marriage, the population mean age is 23.44, and the population standard deviation is 4.72. http://www.biochemia-medica.com/content/standard-error-meaning-and-interpretation
## How To Interpret Standard Error In Regression
You can vary the n, m, and s values and they'll always come out pretty close to each other. Moreover, this formula works for positive and negative ρ alike.[10] See also unbiased estimation of standard deviation for more discussion. Indeed, if you had had another sample, $\tilde{\mathbf{x}}$, you would have ended up with another estimate, $\hat{\theta}(\tilde{\mathbf{x}})$. Standard error: meaning and interpretation.
• Will I encounter any problems as a recognizable Jew in India?
• estimate – Predicted Y values close to regression line Figure 2.
• The obtained P-level is very significant.
The determination of the representativeness of a particular sample is based on the theoretical sampling distribution the behavior of which is described by the central limit theorem. Available at: http://www.scc.upenn.edu/čAllison4.html. up vote 1 down vote favorite Suppose we have a regression model. Difference Between Standard Error And Standard Deviation That in turn should lead the researcher to question whether the bedsores were developed as a function of some other condition rather than as a function of having heart surgery that
Thank you once again. Standard Error Example Taken together with such measures as effect size, p-value and sample size, the effect size can be a very useful tool to the researcher who seeks to understand the reliability and Oracle flashback query syntax - all tables to same timestamp C++11 - typeid uniqueness What would be the value of gold and jewelry in a post-apocalyptic society? http://support.minitab.com/en-us/minitab/17/topic-library/basic-statistics-and-graphs/hypothesis-tests/tests-of-means/what-is-the-standard-error-of-the-mean/ A second generalization from the central limit theorem is that as n increases, the variability of sample means decreases (2).
Executing Sitecore logic from a Windows Scheduled Task Why can't the second fundamental theorem of calculus be proved in just two lines? Standard Error Of Estimate Formula The standard deviation is most often used to refer to the individual observations. Lane DM. Your cache administrator is webmaster.
## Standard Error Example
menuMinitab® 17 SupportWhat is the standard error of the mean?Learn more about Minitab 17 The standard error of the mean (SE of the mean) estimates the variability between sample means that you would http://stats.stackexchange.com/questions/47245/high-standard-errors-for-coefficients-imply-model-is-bad Consider, for example, a researcher studying bedsores in a population of patients who have had open heart surgery that lasted more than 4 hours. How To Interpret Standard Error In Regression For some statistics, however, the associated effect size statistic is not available. Can Standard Error Be Greater Than 1 When the S.E.est is large, one would expect to see many of the observed values far away from the regression line as in Figures 1 and 2. Figure 1.
Use the standard error of the mean to determine how precisely the mean of the sample estimates the population mean. his comment is here A more precise confidence interval should be calculated by means of percentiles derived from the t-distribution. Was there something more specific you were wondering about? However, one is left with the question of how accurate are predictions based on the regression? Standard Error Vs Standard Deviation
The standard error of a statistic is therefore the standard deviation of the sampling distribution for that statistic (3) How, one might ask, does the standard error differ from the standard Are you asking how the models are fit? –Macro Jan 9 '13 at 13:36 add a comment| 1 Answer 1 active oldest votes up vote 1 down vote The "goodness" or Its application requires that the sample is a random sample, and that the observations on each subject are independent of the observations on any other subject. this contact form The points above refer only to the standard error of the mean. (From the GraphPad Statistics Guide that I wrote.) share|improve this answer edited Feb 6 at 16:47 answered Jul 16
Use of the standard error statistic presupposes the user is familiar with the central limit theorem and the assumptions of the data set with which the researcher is working. Large Standard Errors In Regression The distribution of the mean age in all possible samples is called the sampling distribution of the mean. The two can get confused when blurring the distinction between the universe and your sample. –Francesco Jul 15 '12 at 16:57 Possibly of interest: stats.stackexchange.com/questions/15505/… –Macro Jul 16 '12
## Minitab Inc.
Frost, Can you kindly tell me what data can I obtain from the below information. The resulting interval will provide an estimate of the range of values within which the population mean is likely to fall. The margin of error of 2% is a quantitative measure of the uncertainty – the possible difference between the true proportion who will vote for candidate A and the estimate of Standard Error Excel You can, of course, have a high SE and a high coefficient, that's why my 1) is only one of two possibilities. –Peter Flom♦ Jan 9 '13 at 0:20 2
By using this site, you agree to the Terms of Use and Privacy Policy. doi:10.4103/2229-3485.100662. ^ Isserlis, L. (1918). "On the value of a mean as calculated from a sample". Because of random variation in sampling, the proportion or mean calculated using the sample will usually differ from the true proportion or mean in the entire population. http://techkumar.com/standard-error/standard-error-of-estimate-multiple-regression.html This is usually the case even with finite populations, because most of the time, people are primarily interested in managing the processes that created the existing finite population; this is called
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Notice: On March 31, it was announced that Statalist is moving from an email list to a forum. The old list will shut down on April 23, and its replacement, statalist.org is already up and running.
# Re: st: Quantile regression: determine to which quantile an individual belongs
From Nick Cox To "statalist@hsphsun2.harvard.edu" Subject Re: st: Quantile regression: determine to which quantile an individual belongs Date Thu, 18 Apr 2013 12:25:19 +0100
```We agree on the main point.
But -round(whatever, 0.01)- is not an especially good idea because
most multiples of 0.01 can not be held exactly in binary and people
who don't understand that get into awkward small messes.
A more systematic approach is explained at
FAQ . . . . . . . . . . Calculating percentile ranks or plotting positions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N. J. Cox
7/02 How can I calculate percentile ranks?
How can I calculate plotting positions?
http://www.stata.com/support/faqs/statistics/percentile-ranks-and-plotting-positions/
Nick
njcoxstata@gmail.com
On 18 April 2013 12:20, Alex Olssen <alex.olssen@gmail.com> wrote:
> I also am unclear on what you wish to achieve. You said that you wanted to
> know the quantile of each person's residual. I can suggest a way to
> calculate this, but I am unsure why you want it.
>
> sysuse auto, clear
> reg price length
> predict res, res
> sort res
> gen quantile = round(_n/_N, 0.01)
On 18/04/13 9:12 PM, Nick Cox wrote:
>> I can't see that "the quantile each individual belongs to" is a
>> well-defined concept.
>>
>> Clearly you can work out a percentile rank for each response, without
>> regard to the predictors. Or you can see where individual residuals
>> lie in the distribution of residuals.
>>
>> But I don't think that either is what you are seeking.
>>
>> I am not clear whether you are thinking of a point or an interval, but
>> that's secondary.
>>
>> Stripping it down to a minimal example: We have a quantile regression
>> for weight versus height. I am a data point with a certain weight and
>> height. What quantile do I belong to? What's your definition?
On 18 April 2013 11:50, Maria Juul Hansen <maria@lindely.dk> wrote:
>>> Thank you for your comment and reference!
>>> I am aware of the endogeneity problem. However, the purpose is not to
>>> establish causal effects, just to control for the variables of interest.
>>>
>>> Do you have any recommendations regarding the problem of identifying the
>>> quantiles each individual belongs to?
*
* 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/
```
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Cody
# Problem 44435. Testing for randomness: uniform distribution of real numbers (distribution checking)
Solution 1945943
Submitted on 22 Sep 2019
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
% Random, short faults = 0; isRandom_C = true; for i = 1 : 10 x = 1000 * rand(1, 20+randi(10)); isRandom = isItRandom(x); if ~isequal(isRandom, isRandom_C), faults = faults + 1; warning('*** CAUTION: %u faults recorded so far. ***', faults); end; assert(faults <= 2, 'Too many wrong assessments.'); end;
bins = 7 3 11 6 other = 1 len = 27 other = 1 binsiz = 4 bins = 5 6 13 3 other = 1 len = 27 other = 1 binsiz = 4 bins = 6 10 7 2 other = 1 len = 25 other = 1 binsiz = 4 bins = 11 4 6 1 other = 1 len = 22 other = 1 binsiz = 4 bins = 7 3 10 6 other = 1 len = 26 other = 1 binsiz = 4 bins = 8 9 2 3 other = 1 len = 22 other = 1 binsiz = 4 bins = 6 6 9 5 other = 1 len = 26 other = 1 binsiz = 4 bins = 8 11 5 2 other = 1 len = 26 other = 1 binsiz = 4 bins = 5 11 6 2 other = 1 len = 24 other = 1 binsiz = 4 bins = 11 7 4 4 other = 1 len = 26 other = 1 binsiz = 4
2 Pass
% Bimodal, short faults = 0; isRandom_C = false; for i = 1 : 10 x = [rand(1, 10+randi(5)), 999 + rand(1, 10+randi(5))]; isRandom = isItRandom(x); if ~isequal(isRandom, isRandom_C), faults = faults + 1; warning('*** CAUTION: %u faults recorded so far. ***', faults); end; assert(faults <= 2, 'Too many wrong assessments.'); end;
bins = 14 11 other = 1 len = 25 other = 1 binsiz = 2 y = logical 0 bins = 15 11 other = 1 len = 26 other = 1 binsiz = 2 y = logical 0 bins = 15 15 other = 1 len = 30 other = 1 binsiz = 2 [Warning: *** CAUTION: 1 faults recorded so far. ***] [> In ScoringEngineTestPoint2 (line 9) In solutionTest (line 7)] bins = 12 13 other = 1 len = 25 other = 1 binsiz = 2 y = logical 0 bins = 13 12 other = 1 len = 25 other = 1 binsiz = 2 y = logical 0 bins = 15 14 other = 1 len = 29 other = 1 binsiz = 2 y = logical 0 bins = 13 14 other = 1 len = 27 other = 1 binsiz = 2 y = logical 0 bins = 13 12 other = 1 len = 25 other = 1 binsiz = 2 y = logical 0 bins = 13 12 other = 1 len = 25 other = 1 binsiz = 2 y = logical 0 bins = 13 15 other = 1 len = 28 other = 1 binsiz = 2 y = logical 0
3 Pass
% Random, medium faults = 0; isRandom_C = true; for i = 1 : 10 x = randi(1000) + 1000 * rand(1, 100+randi(10)); isRandom = isItRandom(x); if ~isequal(isRandom, isRandom_C), faults = faults + 1; warning('*** CAUTION: %u faults recorded so far. ***', faults); end; assert(faults <= 2, 'Too many wrong assessments.'); end;
bins = 9 20 26 19 24 11 other = 1 len = 109 other = 1 binsiz = 6 bins = 8 25 20 21 22 13 other = 1 len = 109 other = 1 binsiz = 6 bins = 15 24 18 20 27 other = 1 len = 104 other = 1 binsiz = 5 bins = 20 18 27 26 18 1 other = 1 len = 110 other = 1 binsiz = 6 bins = 19 17 19 20 25 10 other = 1 len = 110 other = 1 binsiz = 6 bins = 5 14 21 25 25 13 other = 1 len = 103 other = 1 binsiz = 6 bins = 9 25 15 25 19 14 other = 1 len = 107 other = 1 binsiz = 6 bins = 24 22 19 17 20 4 other = 1 len = 106 other = 1 binsiz = 6 bins = 1 29 14 14 31 12 other = 1 len = 101 other = 1 binsiz = 6 bins = 7 18 25 22 16 17 other = 1 len = 105 other = 1 binsiz = 6
4 Fail
% Bimodal, medium faults = 0; isRandom_C = false; for i = 1 : 10 x = randi(1000) + [rand(1, 50+randi(5)), 999 + rand(1, 50+randi(5))]; isRandom = isItRandom(x); if ~isequal(isRandom, isRandom_C), faults = faults + 1; warning('*** CAUTION: %u faults recorded so far. ***', faults); end; assert(faults <= 2, 'Too many wrong assessments.'); end;
bins = 53 0 0 52 other = 1 len = 105 other = 1 binsiz = 4 y = logical 0 bins = 52 0 0 52 other = 1 len = 104 other = 1 binsiz = 4 [Warning: *** CAUTION: 1 faults recorded so far. ***] [> In ScoringEngineTestPoint4 (line 9) In solutionTest (line 11)] bins = 55 0 0 51 other = 1 len = 106 other = 1 binsiz = 4 y = logical 0 bins = 55 0 0 55 other = 1 len = 110 other = 1 binsiz = 4 [Warning: *** CAUTION: 2 faults recorded so far. ***] [> In ScoringEngineTestPoint4 (line 9) In solutionTest (line 11)] bins = 55 0 0 0 54 other = 1 len = 109 other = 1 binsiz = 5 y = logical 0 bins = 52 0 0 51 other = 1 len = 103 other = 1 binsiz = 4 y = logical 0 bins = 51 0 0 55 other = 1 len = 106 other = 1 binsiz = 4 y = logical 0 bins = 52 0 0 55 other = 1 len = 107 other = 1 binsiz = 4 y = logical 0 bins = 54 0 0 0 52 other = 1 len = 106 other = 1 binsiz = 5 y = logical 0 bins = 54 0 0 54 other = 1 len = 108 other = 1 binsiz = 4 [Warning: *** CAUTION: 3 faults recorded so far. ***] [> In ScoringEngineTestPoint4 (line 9) In solutionTest (line 11)]
Too many wrong assessments.
5 Pass
% Skewed, medium faults = 0; isRandom_C = false; for i = 1 : 10 x = 10 ./ rand(1, 100+randi(10)); isRandom = isItRandom(x); if ~isequal(isRandom, isRandom_C), faults = faults + 1; warning('*** CAUTION: %u faults recorded so far. ***', faults); end; assert(faults <= 2, 'Too many wrong assessments.'); end;
bins = 91 8 1 2 1 1 0 1 other = 1 len = 105 other = 1 binsiz = 8 y = logical 0 bins = 94 9 2 1 1 0 0 1 1 0 0 0 0 0 0 0 1 other = 1 len = 110 other = 1 binsiz = 17 y = logical 0 bins = 81 14 6 3 0 2 1 0 3 other = 1 len = 110 other = 1 binsiz = 9 y = logical 0 bins = 94 1 4 0 1 0 0 0 1 other = 1 len = 101 other = 1 binsiz = 9 y = logical 0 bins = 100 2 1 1 0 0 0 0 0 0 0 1 other = 1 len = 105 other = 1 binsiz = 12 y = logical 0 bins = 91 5 3 1 0 0 0 0 1 0 0 0 0 0 0 1 other = 1 len = 102 other = 1 binsiz = 16 y = logical 0 bins = 96 4 2 0 0 1 1 0 0 0 1 other = 1 len = 105 other = 1 binsiz = 11 y = logical 0 bins = 92 8 2 2 1 1 0 0 0 1 0 0 0 0 0 1 other = 1 len = 108 other = 1 binsiz = 16 y = logical 0 bins = 85 10 4 2 1 0 0 0 0 0 0 1 other = 1 len = 103 other = 1 binsiz = 12 y = logical 0 bins = 89 11 5 0 1 0 1 0 0 0 1 other = 1 len = 108 other = 1 binsiz = 11 y = logical 0
6 Pass
% Random, long faults = 0; isRandom_C = true; for i = 1 : 10 x = 1000 * rand(1, 500+randi(10)); isRandom = isItRandom(x); if ~isequal(isRandom, isRandom_C), faults = faults + 1; warning('*** CAUTION: %u faults recorded so far. ***', faults); end; assert(faults <= 2, 'Too many wrong assessments.'); end;
bins = 39 43 37 53 60 50 69 50 50 58 other = 1 len = 509 other = 1 binsiz = 10 bins = 63 44 59 44 48 56 52 51 46 45 other = 1 len = 508 other = 1 binsiz = 10 bins = 47 48 52 54 47 47 54 51 56 47 other = 1 len = 503 other = 1 binsiz = 10 bins = 42 48 57 47 62 50 43 50 52 59 other = 1 len = 510 other = 1 binsiz = 10 bins = 52 51 37 46 52 54 59 57 53 47 other = 1 len = 508 other = 1 binsiz = 10 bins = 46 67 51 54 44 48 52 48 45 55 other = 1 len = 510 other = 1 binsiz = 10 bins = 47 57 60 53 49 43 45 46 59 50 other = 1 len = 509 other = 1 binsiz = 10 bins = 33 49 65 45 57 48 43 68 52 49 other = 1 len = 509 other = 1 binsiz = 10 bins = 46 44 39 55 58 49 57 56 42 59 other = 1 len = 505 other = 1 binsiz = 10 bins = 47 51 45 40 61 44 45 55 53 65 other = 1 len = 506 other = 1 binsiz = 10
7 Pass
% Skewed (1), long faults = 0; isRandom_C = false; for i = 1 : 10 x = 10 ./ rand(1, 500+randi(5)) - randi(1000); isRandom = isItRandom(x); if ~isequal(isRandom, isRandom_C), faults = faults + 1; warning('*** CAUTION: %u faults recorded so far. ***', faults); end; assert(faults <= 2, 'Too many wrong assessments.'); end;
bins = Columns 1 through 30 459 35 2 3 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 69 0 0 0 0 0 0 0 0 1 other = 1 len = 503 other = 1 binsiz = 69 y = logical 0 bins = Columns 1 through 30 498 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 50 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 other = 1 len = 501 other = 1 binsiz = 50 y = logical 0 bins = Columns 1 through 30 455 31 7 6 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 38 0 0 0 1 0 0 0 1 other = 1 len = 504 other = 1 binsiz = 38 y = logical 0 bins = Columns 1 through 30 403 63 13 5 6 0 1 2 1 3 0 2 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 68 0 0 0 0 0 0 0 1 other = 1 len = 502 other = 1 binsiz = 68 y = logical 0 bins = Columns 1 through 30 457 28 7 3 0 0 1 2 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 41 0 0 0 0 0 0 0 0 0 0 1 other = 1 len = 501 other = 1 binsiz = 41 y = logical 0 bins = Columns 1 through 30 271 181 27 8 4 2 0 0 1 2 1 0 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 Columns 31 through 58 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 other = 1 len = 502 other = 1 binsiz = 58 y = logical 0 bins = Columns 1 through 30 411 54 16 2 5 3 4 2 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 Columns 31 through 40 0 0 0 0 0 0 0 0 0 1 other = 1 len = 502 other = 1 binsiz = 40 y = logical 0 bins = Columns 1 through 30 427 44 13 8 2 5 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 63 0 0 1 other = 1 len = 503 other = 1 binsiz = 63 y = logical 0 bins = Columns 1 through 30 464 27 3 4 1 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 44 0 0 0 0 0 0 0 0 0 0 0 0 0 1 other = 1 len = 503 other = 1 binsiz = 44 y = logical 0 bins = 433 44 6 3 2 0 1 1 2 0 1 1 1 1 1 1 0 1 0 0 0 0 0 2 other = 1 len = 501 other = 1 binsiz = 24 y = logical 0
8 Pass
% Skewed (2), long faults = 0; isRandom_C = false; for i = 1 : 10 x = randi(1000) - 10 ./ rand(1, 500+randi(5)); isRandom = isItRandom(x); if ~isequal(isRandom, isRandom_C), faults = faults + 1; warning('*** CAUTION: %u faults recorded so far. ***', faults); end; assert(faults <= 2, 'Too many wrong assessments.'); end;
bins = Columns 1 through 30 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 2 0 0 0 0 0 1 0 0 1 0 0 Columns 31 through 45 1 1 1 1 2 2 2 0 3 2 7 12 22 124 319 other = 1 len = 505 other = 1 binsiz = 45 y = logical 0 bins = Columns 1 through 30 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 Columns 31 through 41 0 1 1 2 0 0 2 8 11 42 435 other = 1 len = 505 other = 1 binsiz = 41 y = logical 0 bins = Columns 1 through 30 1 0 0 0 0 1 0 1 0 0 0 0 0 2 0 0 0 1 0 1 0 0 3 0 1 1 2 2 5 16 Columns 31 through 32 52 412 other = 1 len = 501 other = 1 binsiz = 32 y = logical 0 bins = Columns 1 through 30 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 1 2 1 1 1 13 32 448 other = 1 len = 504 other = 1 binsiz = 60 y = logical 0 bins = Columns 1 through 30 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 57 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 3 59 438 other = 1 len = 504 other = 1 binsiz = 57 y = logical 0 bins = Columns 1 through 30 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 Columns 61 through 71 0 0 0 1 0 2 2 6 10 38 444 other = 1 len = 505 other = 1 binsiz = 71 y = logical 0 bins = Columns 1 through 30 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 Columns 31 through 38 1 1 2 2 2 4 33 455 other = 1 len = 502 other = 1 binsiz = 38 y = logical 0 bins = Columns 1 through 30 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 47 0 0 0 0 0 1 0 0 0 1 0 1 1 2 5 13 476 other = 1 len = 501 other = 1 binsiz = 47 y = logical 0 bins = 1 0 0 0 0 0 0 2 0 0 0 0 0 1 0 0 0 0 2 1 1 1 1 2 1 0 3 12 18 456 other = 1 len = 502 other = 1 binsiz = 30 y = logical 0 bins = Columns 1 through 30 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 57 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 2 3 22 472 other = 1 len = 502 other = 1 binsiz = 57 y = logical 0
9 Pass
% Bimodal, long faults = 0; isRandom_C = false; for i = 1 : 10 x = [rand(1, 250+randi(5)), 999 + rand(1, 250+randi(5))]; isRandom = isItRandom(x); if ~isequal(isRandom, isRandom_C), faults = faults + 1; warning('*** CAUTION: %u faults recorded so far. ***', faults); end; assert(faults <= 2, 'Too many wrong assessments.'); end;
bins = 255 0 0 0 255 other = 1 len = 510 other = 1 binsiz = 5 [Warning: *** CAUTION: 1 faults recorded so far. ***] [> In ScoringEngineTestPoint9 (line 9) In solutionTest (line 21)] bins = 253 0 0 0 254 other = 1 len = 507 other = 1 binsiz = 5 y = logical 0 bins = 253 0 0 0 252 other = 1 len = 505 other = 1 binsiz = 5 y = logical 0 bins = 255 0 0 0 252 other = 1 len = 507 other = 1 binsiz = 5 y = logical 0 bins = 254 0 0 0 252 other = 1 len = 506 other = 1 binsiz = 5 y = logical 0 bins = 251 0 0 0 253 other = 1 len = 504 other = 1 binsiz = 5 y = logical 0 bins = 251 0 0 0 252 other = 1 len = 503 other = 1 binsiz = 5 y = logical 0 bins = 254 0 0 0 253 other = 1 len = 507 other = 1 binsiz = 5 y = logical 0 bins = 252 0 0 0 255 other = 1 len = 507 other = 1 binsiz = 5 y = logical 0 bins = 254 0 0 0 253 other = 1 len = 507 other = 1 binsiz = 5 y = logical 0
10 Fail
% 'Normal' (a.k.a. 'Gaussian'), long faults = 0; isRandom_C = false; for i = 1 : 10 x = 250 * randn(1, 500+randi(5)); isRandom = isItRandom(x); if ~isequal(isRandom, isRandom_C), faults = faults + 1; warning('*** CAUTION: %u faults recorded so far. ***', faults); end; assert(faults <= 2, 'Too many wrong assessments.'); end;
bins = 2 9 18 27 45 64 72 73 74 47 34 17 14 2 2 1 other = 1 len = 501 other = 1 binsiz = 16 [Warning: *** CAUTION: 1 faults recorded so far. ***] [> In ScoringEngineTestPoint10 (line 9) In solutionTest (line 23)] bins = 1 2 7 17 28 54 62 81 77 60 51 37 15 9 2 2 other = 1 len = 505 other = 1 binsiz = 16 [Warning: *** CAUTION: 2 faults recorded so far. ***] [> In ScoringEngineTestPoint10 (line 9) In solutionTest (line 23)] bins = 1 2 6 16 36 47 65 85 72 76 40 32 11 5 6 0 1 other = 1 len = 501 other = 1 binsiz = 17 [Warning: *** CAUTION: 3 faults recorded so far. ***] [> In ScoringEngineTestPoint10 (line 9) In solutionTest (line 23)]
Too many wrong assessments.
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# Q. What mic polar pattern is best for recording speech?
The various microphone polar patterns: top (0 degrees) is the pickup in front of the diaphragm, and bottom (180 degrees) to the rear.
I’m trying to improve the sound of my speech recordings which can be a bit ‘echoey’. I’ve been using a Rode NTG4 shotgun mic, but I’m thinking of switching to something more conventional and impulsively I’m drawn towards a super or hypercardioid model, although I am not entirely sure of their differences. Can you offer any advice?
SOS Forum post
SOS Technical Editor Hugh Robjohns replies: As the physical and practical differences between these two types of microphone aren’t huge, the terms are often used interchangeably by both users and manufacturers. But there are actually specific definitions as to what makes a mic supercardioid or a hypercardioid and, fundamentally, the polar equations that define the shape of the polar patterns for these two mics are different.
For the mathematically minded, a supercardioid is defined as 0.37+0.63cosθ, whereas a hypercardioid is 0.25+0.75cosθ (for reference, a standard cardioid is 0.5+0.5cosθ). All first‑order directional microphones combine elements of pressure operation (omni pattern) and pressure‑gradient operation (fig‑8 pattern). Essentially, the first number in these equations defines the size of the pressure component, while the cosθ part relates to the size of the pressure‑gradient portion. So, from these equations, we can see that the hypercardioid has a stronger pressure‑gradient element and so is slightly closer to the fig‑8 pattern, while the supercardioid has a larger pressure component, making it slightly closer to the cardioid pattern.
What this means in practice is that a hypercardioid mic typically has a slightly stronger proximity effect than a supercardioid one, and is also slightly more susceptible to handling and wind noise. In terms of the polar pattern, the front half of the hypercardioid’s polar pattern is narrower than the supercardioid’s, with a useful frontal pickup area (at ‑3dB limits) of 105 degrees; the supercardioid is wider, at 115 degrees. The hypercardioid is also more than 3dB less sensitive to sounds from the side (90 degrees) than the supercardioid. Also, the narrower polar pattern naturally translates into a slightly better directivity index (DI), which is the ratio of on‑axis sound to ambient sound: while the supercardioid scores 5.7dB, the hypercardioid boasts the highest directivity index of all first‑order mics at 6dB. (For reference, a standard cardioid scores 4.8dB, as does a fig‑8 pattern, while a hypocardioid’s DI is 2.5dB, and an omni scores 0dB (ie. it has no directivity).
There are specific definitions as to what makes a mic supercardioid or hypercardioid.
Both the hyper and supercardioid polar patterns have small rear lobes, but the hypercardioid’s is larger. So, while rearward (180 degrees) sounds are attenuated by 6dB in the hypercardioid, the supercardioid attenuates them by almost ‑12dB. Related to the tail sizes, the null angles (where the mic is least sensitive) are also different, with the hypercardioid nulls being further forwards at 110 degrees, compared with 126 degrees for the supercardioid.
In practical terms, both hypercardioid and supercardioid mics are noticeably more directional than cardioids, and they both have small rear tails. They both have a stronger proximity effect, and are more susceptible to handling and wind noise. When used as a stage vocal mic, floor monitors should be pulled around to the sides, rather than directly behind, so that they aim into the null angle — bearing in mind that the optimum angles are slightly different for the super and hypercardioid. And while the hypercardioid has the highest directivity index, meaning it rejects ambient noise more than the supercardioid (or cardioid), it does have a strong rear lobe and will pick up more sound from directly behind than a supercardioid.The plot compares the polar responses of the full suite of first‑order mics (omni, hypo‑cardioid, cardioid, supercardioid, hypercardioid, and fig‑8), and hopefully this will help you visualise what I’ve said above.
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Cody
# Problem 3060. Scrabble Scores - 8
Solution 798415
Submitted on 26 Dec 2015 by Peng Liu
• Size: 138
• This is the leading solution.
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### Test Suite
Test Status Code Input and Output
1 Pass
%% board = [ ' cat '; ' i p '; ' t poet'; ' c l '; 'there '; ]; n_pl = 2; order = cell(1,5); order(1) = {[2,2; 5,2]}; order(2) = {[5,1; 5,5]}; order(3) = {[1,5; 5,5]}; order(4) = {[3,5; 3,8]}; order(5) = {[1,4; 1,6]}; mult = [ 'T d T'; ' '; 'd D d'; ' '; 'T d T'; ]; score = [34 34]; assert(isequal(scrabble_scores_8(board,n_pl,order,mult),score))
2 Pass
%% board = [ 'stratagems'; ' h a '; 'hello t '; ' r l '; 'fellow a '; ' a b '; ' users '; ]; n_pl = 2; order = cell(1,7); order(1) = {[1,1; 1,10]}; order(2) = {[1,9; 6,9]}; order(3) = {[1,2; 5,2]}; order(4) = {[5,1; 5,6]}; order(5) = {[5,5; 7,5]}; order(6) = {[3,1; 3,5]}; order(7) = {[7,2; 7,6]}; mult = [ 'T t t T'; ' '; 'D d d D'; ' '; 'D d d D'; ' '; 'T t t T'; ]; score = [189 54]; assert(isequal(scrabble_scores_8(board,n_pl,order,mult),score))
3 Pass
%% board = [ ' what y '; ' h do '; ' this um'; ' l n a'; ' of keen t'; ' o e l'; ' problem a'; ' noob'; ]; n_pl = 3; order = cell(1,12); order(1) = {[1,2; 1,5]}; order(2) = {[1,5; 5,5]}; order(3) = {[3,3; 3,7]}; order(4) = {[5,5; 5,8]}; order(5) = {[5,6; 7,6]}; order(6) = {[7,2; 7,8]}; order(7) = {[4,2; 7,2]}; m_words(:,:,1) = [8,7; 8,10]; m_words(:,:,2) = [7,7; 8,7]; m_words(:,:,3) = [7,8; 8,8]; order(8) = {m_words}; order(9) = {[3,10; 8,10]}; clear m_words m_words(:,:,1) = [1,9; 3,9]; m_words(:,:,2) = [3,9; 3,10]; order(10) = {m_words}; order(11) = {[5,2; 5,3]}; order(12) = {[2,8; 2,9]}; mult = [ 'T t T'; ' d d '; ' d d '; ' D t'; 't D '; ' d d '; ' d d '; 'T t T'; ]; score = [43 44 39]; assert(isequal(scrabble_scores_8(board,n_pl,order,mult),score))
4 Pass
%% board = [ ' if you '; ' i u '; ' like '; ' met g '; ' e it l'; ' e v i'; ' please k'; ' node'; ]; n_pl = 2; order = cell(1,12); order(1) = {[1,2; 1,3]}; order(2) = {[1,3; 4,3]}; order(3) = {[3,3; 3,6]}; order(4) = {[1,6; 3,6]}; order(5) = {[1,6; 1,8]}; m_words(:,:,1) = [4,2; 4,4]; m_words(:,:,2) = [3,4; 4,4]; order(6) = {m_words}; order(7) = {[4,2; 7,2]}; order(8) = {[7,2; 7,7]}; order(9) = {[4,7; 7,7]}; m_words(:,:,1) = [8,7; 8,10]; m_words(:,:,2) = [4,7; 8,7]; order(10) = {m_words}; order(11) = {[5,10; 8,10]}; order(12) = {[5,7; 5,8]}; mult = [ ' T T T '; ' d d'; ' D D '; ' D t d '; ' d'; ' d t '; ' d D D '; ' T T T'; ]; score = [67 140]; assert(isequal(scrabble_scores_8(board,n_pl,order,mult),score)) % board = [ 'c flummoxes zither'; 'al o e e e'; 'ba hunting eerie b'; 'ab a g e o e'; 'lo n m g l'; ' r g responses '; 'random n u u '; ' t v c below p r '; 'convention r e f '; ' r r n a areas'; ' y v c t i c '; ' o philosophy o em'; 'travel n n r r e'; ' e c t e'; ' word i a writer'; ' t n c o l'; 'p fortnight rock y'; 'i n h '; 'novelty m concatenate'; 'c e o a a s g'; 'heterogeneousness g'; ]; n_pl = 4; order = cell(1,41); order(1) = {[12,7; 12,16]}; order(2) = {[5,13; 13,13]}; order(3) = {[8,9; 17,9]}; order(4) = {[11,11; 17,11]}; order(5) = {[17,3; 17,11]}; order(6) = {[8,12; 8,16]}; order(7) = {[6,12; 6,20]}; order(8) = {[6,18; 13,18]}; order(9) = {[10,17; 10,21]}; order(10) = {[9,1; 9,10]}; order(11) = {[12,4; 19,4]}; order(12) = {[2,2; 11,2]}; order(13) = {[7,1; 7,6]}; order(14) = {[13,1; 13,6]}; order(15) = {[15,2; 15,5]}; order(16) = {[19,1; 19,7]}; order(17) = {[3,5; 10,5]}; order(18) = {[3,5; 3,11]}; m_words(:,:,1) = [1,1; 5,1]; m_words(:,:,2) = [2,1; 2,2]; m_words(:,:,3) = [3,1; 3,2]; m_words(:,:,4) = [4,1; 4,2]; m_words(:,:,5) = [5,1; 5,2]; order(19) = {m_words}; order(20) = {[17,1; 21,1]}; order(21) = {[21,1; 21,17]}; order(22) = {[6,20; 12,20]}; clear m_words m_words(:,:,1) = [12,21; 17,21]; m_words(:,:,2) = [12,20; 12,21]; order(23) = {m_words}; order(24) = {[15,16; 15,21]}; order(25) = {[3,11; 5,11]}; order(26) = {[17,17; 21,17]}; order(27) = {[17,15; 17,18]}; order(28) = {[19,11; 19,21]}; order(29) = {[19,14; 21,14]}; order(30) = {[1,7; 4,7]}; order(31) = {[1,6; 1,14]}; order(32) = {[19,21; 21,21]}; order(33) = {[19,6; 21,6]}; order(34) = {[1,14; 3,14]}; order(35) = {[3,14; 3,18]}; order(36) = {[1,16; 4,16]}; order(37) = {[1,16; 1,21]}; order(38) = {[1,21; 5,21]}; order(39) = {[19,3; 21,3]}; order(40) = {[19,9; 21,9]}; order(41) = {[15,16; 17,16]}; mult = [ 'Q d T d T d Q'; ' D t D D t D '; ' D q D D q D '; 'd T d T d T d'; ' t D t t D t '; ' q D d d D q '; ' d D d D d '; 'T D D T'; ' D t t t t D '; ' D d d d d D '; 'd T d D d T d'; ' D d d d d D '; ' D t t t t D '; 'T D D T'; ' d D d D d '; ' q D d d D q '; ' t D t t D t '; 'd T d T d T d'; ' D q D D q D '; ' D t D D t D '; 'Q d T d T d Q'; ]; score = [433 188 430 300]; assert(isequal(scrabble_scores_8(board,n_pl,order,mult),score))
5 Pass
%% anti-cheating test case (random number of players from random board) ind = randi(4); ind2 = randi(3)+1; switch ind case 1 board = [ ' cat '; ' i p '; ' t poet'; ' c l '; 'there '; ]; order = cell(1,5); order(1) = {[2,2; 5,2]}; order(2) = {[5,1; 5,5]}; order(3) = {[1,5; 5,5]}; order(4) = {[3,5; 3,8]}; order(5) = {[1,4; 1,6]}; mult = [ 'T d T'; ' '; 'd D d'; ' '; 'T d T'; ]; switch ind2 case 2 n_pl = 2; score = [34 34]; case 3 n_pl = 3; score = [16 32 20]; case 4 n_pl = 4; score = [14 27 20 7]; end case 2 board = [ 'stratagems'; ' h a '; 'hello t '; ' r l '; 'fellow a '; ' a b '; ' users ']; order = cell(1,7); order(1) = {[1,1; 1,10]}; order(2) = {[1,9; 6,9]}; order(3) = {[1,2; 5,2]}; order(4) = {[5,1; 5,6]}; order(5) = {[5,5; 7,5]}; order(6) = {[3,1; 3,5]}; order(7) = {[7,2; 7,6]}; mult = [ 'T t t T'; ' '; 'D d d D'; ' '; 'D d d D'; ' '; 'T t t T'; ]; switch ind2 case 2 n_pl = 2; score = [189 54]; case 3 n_pl = 3; score = [204 13 26]; case 4 n_pl = 4; score = [174 28 15 26]; end case 3 board = [ ' what y '; ' h do '; ' this um'; ' l n a'; ' of keen t'; ' o e l'; ' problem a'; ' noob'; ]; order = cell(1,12); order(1) = {[1,2; 1,5]}; order(2) = {[1,5; 5,5]}; order(3) = {[3,3; 3,7]}; order(4) = {[5,5; 5,8]}; order(5) = {[5,6; 7,6]}; order(6) = {[7,2; 7,8]}; order(7) = {[4,2; 7,2]}; m_words(:,:,1) = [8,7; 8,10]; m_words(:,:,2) = [7,7; 8,7]; m_words(:,:,3) = [7,8; 8,8]; order(8) = {m_words}; order(9) = {[3,10; 8,10]}; clear m_words m_words(:,:,1) = [1,9; 3,9]; m_words(:,:,2) = [3,9; 3,10]; order(10) = {m_words}; order(11) = {[5,2; 5,3]}; order(12) = {[2,8; 2,9]}; mult = [ 'T t T'; ' d d '; ' d d '; ' D t'; 't D '; ' d d '; ' d d '; 'T t T'; ]; switch ind2 case 2 n_pl = 2; score = [44 82]; case 3 n_pl = 3; score = [43 44 39]; case 4 n_pl = 4; score = [25 39 19 43]; end case 4 board = [ ' if you '; ' i u '; ' like '; ' met g '; ' e it l'; ' e v i'; ' please k'; ' node'; ]; order = cell(1,12); order(1) = {[1,2; 1,3]}; order(2) = {[1,3; 4,3]}; order(3) = {[3,3; 3,6]}; order(4) = {[1,6; 3,6]}; order(5) = {[1,6; 1,8]}; m_words(:,:,1) = [4,2; 4,4]; m_words(:,:,2) = [3,4; 4,4]; order(6) = {m_words}; order(7) = {[4,2; 7,2]}; order(8) = {[7,2; 7,7]}; order(9) = {[4,7; 7,7]}; m_words(:,:,1) = [8,7; 8,10]; m_words(:,:,2) = [4,7; 8,7]; order(10) = {m_words}; order(11) = {[5,10; 8,10]}; order(12) = {[5,7; 5,8]}; mult = [ ' T T T '; ' d d'; ' D D '; ' D t d '; ' d'; ' d t '; ' d D D '; ' T T T'; ]; switch ind2 case 2 n_pl = 2; score = [67 140]; case 3 n_pl = 3; score = [116 45 46]; case 4 n_pl = 4; score = [39 104 28 36]; end end assert(isequal(scrabble_scores_8(board,n_pl,order,mult),score))
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Learn Math Online Notes & Technology Articles
Trigonometric Functions and Graphs Multiple Choice Questions Test 1 Tests pdf Download
Practice trigonometric functions and graphs multiple choice questions (MCQs), math test 1 online to learn. Practice period of trigonometric functions MCQs questions and answers on period of trigonometric functions with answers. Free trigonometric functions and graphs study guide has answer key with choices as 2π, π, 3π and 4π of multiple choice questions (MCQ) as period of cot(x) is to test learning skills. Study to learn period of trigonometric functions quiz questions to practice MCQ based online exam preparation test.
MCQ. Period of cot(x) is
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2. 10π
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2. π/2
3. 2π/3
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MCQ. Period of 3secx/3 is
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# Search Our Content Library
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English Learner (EL)
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Greater Than, Less Than, Equal To Game
Lesson Plan
Greater Than, Less Than, Equal To Game
Help your students make sense of the greater than, less than and equal to in this interactive lesson! Your students will have opportunities to compare either two-digit or three-digit numbers.
Math
Lesson Plan
Place Value Party
Lesson Plan
Place Value Party
Now that your first graders are able to count consecutively, introduce them to the tens and ones place values. Using tens and ones blocks will make math easy and fun for everyone!
Math
Lesson Plan
Counting Hours: What Time Is It?
Lesson Plan
Counting Hours: What Time Is It?
Give first graders a sense of time by introducing them to telling time by the hour. Students who have mastered the numbers 1 to 12 will be eager to keep the class on a time schedule.
Math
Lesson Plan
Fact Family Trees
Lesson Plan
Fact Family Trees
Number facts are related, just like families! Help students learn all about fact families while they make fact family trees!
Math
Lesson Plan
Lesson Plan
In this lesson, your students will become familiar with shapes by identifying them in real life. Your students will love identifying how many sides shapes have by drawing and counting them!
Math
Lesson Plan
Capacity Comparison
Lesson Plan
Capacity Comparison
Your class will love this outdoor activity that teaches them how to compare the capacity of different containers.
Math
Lesson Plan
Lesson Plan
Math
Lesson Plan
Popping Numbers
Lesson Plan
Popping Numbers
Pop into place! This lessons builds your students' knowledge of numbers, their place value, and how to write them correctly.
Math
Lesson Plan
Lesson Plan
Make a math mystery! In this lesson, help your students understand the relationship between addition and subtraction and how a missing addend word problem is represented with a number sentence.
Math
Lesson Plan
Counting Collections
Lesson Plan
Counting Collections
Counting collections are a great way to help students practice counting while gaining practice recording and justifying their thinking. Students will work specifically with the numbers 1-30 to develop fluency counting and writing numbers.
Math
Lesson Plan
Clock Concentration
Lesson Plan
Clock Concentration
Get ready to teach all about time with a fun game that has students match the time on analog and digital clocks. Use alone or with **The Clock Struck What?**
Math
Lesson Plan
Active Math
Lesson Plan
Active Math
Have a class that can't sit still long enough to practice their addition? This activity puts math and PE together to help your kids add up to 20 while being active!
Math
Lesson Plan
Lesson Plan
This math lesson is filled with fun hands-on non-standard measurement activities! The lesson begins with an interactive story then has students explore measurement using several different objects as their tools!
Math
Lesson Plan
The Clock Struck What?
Lesson Plan
The Clock Struck What?
It's time-telling time with this fun lesson featuring Trudy Harris' 'The Clock Struck One.' Students will learning about time with a short story and hands-on practice with analog clocks.
Math
Lesson Plan
Adding with Dominoes and Playing Cards
Lesson Plan
Adding with Dominoes and Playing Cards
Math
Lesson Plan
Terrific Tangrams!
Lesson Plan
Terrific Tangrams!
Introduce your class to composite shapes with this lesson that reviews geometric shapes and makes use of tangram puzzles. This lesson will give your students a strong base for learning geometry later on in their school careers.
Math
Lesson Plan
Lesson Plan
Use a sweet treat to help students do hands-on math! Students will practice their addition skills in this yummy activity.
Math
Lesson Plan
Lesson Plan
Get your students moving with this activity that helps them practice their addition skills while having fun hunting for their next problem.
Math
Lesson Plan
How Big Is It?
Lesson Plan
How Big Is It?
Your young scientists will have tons of fun visiting measurement stations and using tools to measure various objects. They'll even be able to create posters at the end of the activity.
Math
Lesson Plan
Graphing a Fruit Survey
Lesson Plan
Graphing a Fruit Survey
What's your favorite fruit? In this lesson, your students will practice important skills by learning to interpret and present data in a visual form.
Math
Lesson Plan
Place Value Concentration
Lesson Plan
Place Value Concentration
Put your students to the test with this lesson that helps them learn the tens and ones places with a memory game and worksheet!
Math
Lesson Plan
Dots & Dots of Fact Families
Lesson Plan
Dots & Dots of Fact Families
Get out the dominoes! This fun lesson help students visualize how addition and subtraction are related, all while building their fact family fluency.
Math
Lesson Plan
Subtraction Card Sharks
Lesson Plan
Subtraction Card Sharks
Make math fun with this lesson that has students work together to solve subtraction sentences. Students will also get a chance to practice their skills individually with a worksheet.
Math
Lesson Plan
Whole Body Subtraction
Lesson Plan
Whole Body Subtraction
Students will be jumping with joy after this lesson that helps them practice their subtraction skills with physical activity.
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### 2007 AMC 8
Scroll down and press Start to try the exam! Or, go to the printable PDF, answer key, or solutions.
Used with permission of the Mathematical Association of America.
남은 시간:
40:00
1.
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $$10$$ hours per week helping around the house for $$6$$ weeks. For the first $$5$$ weeks she helps around the house for $$8,$$ $$11,$$ $$7,$$ $$12$$ and $$10$$ hours. How many hours must she work for the final week to earn the tickets?
$$9$$
$$10$$
$$11$$
$$12$$
$$13$$
###### Solution(s):
During the first $$5$$ weeks, Theresa works for a total of $8 + 11 + 7 + 12 + 10 = 48$ hours.
She, however, promised to work for $10 \cdot 6 = 60$ hours.
This means that she has to work $60 - 48 = 12$ hours during the final work to earn the tickets.
Thus, D is the correct answer.
2.
$$650$$ students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?
$$\dfrac{2}{5}$$
$$\dfrac{1}{2}$$
$$\dfrac{5}{4}$$
$$\dfrac{5}{3}$$
$$\dfrac{5}{2}$$
###### Solution(s):
There are $$250$$ students who preferred spaghetti and $$100$$ that preferred manicotti.
The ratio is therefore $\dfrac{250}{100} = \dfrac{5}{2}.$
Thus, E is the correct answer.
3.
What is the sum of the two smallest prime factors of $$250?$$
$$2$$
$$5$$
$$7$$
$$10$$
$$12$$
###### Solution(s):
We can prime factorize $$250$$ to get $250 = 2 \cdot 5^3.$ From this, we can see that $$250$$ only has two prime factors: $$2$$ and $$5.$$ The sum of these is $$7.$$
Thus, C is the correct answer.
4.
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?
$$12$$
$$15$$
$$18$$
$$30$$
$$36$$
###### Solution(s):
Georgie has $$6$$ options for which window he enters through. He, however, only has $$5$$ options for the exit since it must be different from the entrance.
The total number of paths is therefore $$6 \cdot 5 = 30.$$
Thus, D is the correct answer.
5.
Chandler wants to buy a $$500$$ dollar mountain bike. For his birthday, his grandparents send him $$50$$ dollars, his aunt sends him $$35$$ dollars and his cousin gives him $$15$$ dollars. He earns $$16$$ dollars per week for his paper route. He will use all of his birthday money and all of the money he earns from his paper route. In how many weeks will he be able to buy the mountain bike?
$$24$$
$$25$$
$$26$$
$$27$$
$$28$$
###### Solution(s):
The total amount of money that Chandler got from his birthday is $50 + 35 + 15 = 100$ dollars. Therefore, he only needs to raise $$500 - 100 = 400$$ more dollars.
This can be achieved in $$400 \div 16 = 25$$ weeks through his paper route.
Thus, B is the correct answer
6.
The average cost of a long-distance call in the USA in $$1985$$ was $$41$$ cents per minute, and the average cost of a long-distance call in the USA in $$2005$$ was $$7$$ cents per minute. Find the approximate percent decrease in the cost per minute of a long-distance call.
$$7$$
$$17$$
$$34$$
$$41$$
$$80$$
###### Solution(s):
The difference in the costs is $$41 - 7 = 34$$ cents. The approximate percent decrease is therefore $100 \cdot \dfrac{35}{41} \approx 100 \cdot \dfrac{32}{40}$$=100 \cdot \dfrac{4}{5}$$= 80 \%.$
Thus, E is the correct answer.
7.
The average age of $$5$$ people in a room is $$30$$ years. An $$18$$-year-old person leaves the room. What is the average age of the four remaining people?
$$25$$
$$26$$
$$29$$
$$33$$
$$36$$
###### Solution(s):
Initially, the total age of everyone in the room is $$5 \cdot 30 = 150$$ years.
After the person leaves, the total age is $$150 - 18 = 132.$$ With $$4$$ people remaining, the average age becomes $$\dfrac{132}{4} = 33.$$
Thus, D is the correct answer.
8.
In trapezoid $$ABCD$$, $$\overline{AD}$$ is perpendicular to $$\overline{DC},$$ $$AD = AB = 3,$$ and $$DC = 6.$$ In addition, $$E$$ is on $$\overline{DC},$$ and $$\overline{BE}$$ is parallel to $$\overline{AD}.$$ Find the area of $$\triangle BEC.$$
$$3$$
$$4.5$$
$$6$$
$$9$$
$$18$$
###### Solution(s):
We know that $EC = DC - DE$$= 6 - 3$$= 3.$ We also know that $BE = AD = 3.$ Therefore, the area of $$\triangle BEC$$ is $\dfrac{1}{2} \cdot 3 \cdot 3 = \dfrac{9}{2}.$
Thus, B is the correct answer.
9.
To complete the grid below, each of the digits $$1$$ through $$4$$ must occur once in each row and once in each column. What number will occupy the lower right-hand square?
$$1$$
$$2$$
$$3$$
$$4$$
$$\text{cannot be determined}$$
###### Solution(s):
Consider the last element in the second row. This has to be a $$1$$ since $$4$$ cannot be in that column.
Then, the number in the top right square must be $$3$$ since it cannot be $$2.$$
This forces the $$2$$ to be in the bottom right square.
Thus, B is the correct answer.
10.
For any positive integer $$n,$$ define $$\boxed{n}$$ to be the sum of the positive factors of $$n.$$ For example, $\boxed{6} = 1 + 2 + 3 + 6 = 12.$ Find $$\boxed{\boxed{11}}.$$
$$13$$
$$20$$
$$24$$
$$28$$
$$30$$
###### Solution(s):
First, we find that $\boxed{11} = 1 + 11 = 12$ since $$11$$ is prime. Then we need to find $$\boxed{12}.$$ The factors of $$12$$ are $1, 2, 3, 4, 6, 12.$ Adding these yields $$\boxed{12} = 28.$$
Thus, D is the correct answer.
11.
Tiles $$I, II, III$$ and $$IV$$ are translated so one tile coincides with each of the rectangles $$A, B, C$$ and $$D.$$ In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle $$C?$$
$$I$$
$$II$$
$$III$$
$$IV$$
$$\text{cannot be determined}$$
###### Solution(s):
Note that only Tile $$III$$ has the number of $$0.$$ This forces it to be either $$C$$ or $$D.$$ This tile also is the only one with $$5.$$ This makes sure that Tile $$III$$ is $$D.$$
The only tile that can match with $$1$$ on Tile $$III$$ is Tile $$IV.$$ Therefore, Tile $$IV$$ is $$C.$$
Similarly, we get that Tile $$II$$ is $$A$$ and Tile $$I$$ is $$B.$$
Thus, the correct answer is D
12.
A unit hexagram is composed of a regular hexagon of side length $$1$$ and its $$6$$ equilateral triangular extensions, as shown in the diagram. What is the ratio of the area of the extensions to the area of the original hexagon?
$$1:1$$
$$6:5$$
$$3:2$$
$$2:1$$
$$3:1$$
###### Solution(s):
Note that we can split the hexagon into $$6$$ congruent equilateral triangles as follows.
Since each of them share an edge with an exterior triangle, all the triangles are congruent. Therefore, the ratio of areas is $$1:1.$$
Thus, A is the correct answer.
13.
Sets $$A$$ and $$B,$$ shown in the Venn diagram, have the same number of elements. Their union has $$2007$$ elements and their intersection has $$1001$$ elements. Find the number of elements in $$A.$$
$$503$$
$$1006$$
$$1504$$
$$1507$$
$$1510$$
###### Solution(s):
Let $$x$$ be the number of elements in each $$A$$ and $$B.$$ Also let $$y$$ be the number of elements in their intersection.
The conditions give us that $2x - y = 2007$ and $y = 1001.$ Plugging $$y$$ into the first equation yields \begin{align*} 2x - 1001 &= 2007 \\ 2x &= 3008 \\ x &= 1504. \end{align*}
Thus, C is the correct answer.
14.
The base of isosceles $$\triangle ABC$$ is $$24$$ and its area is $$60.$$ What is the length of one of the congruent sides?
$$5$$
$$8$$
$$13$$
$$14$$
$$18$$
###### Solution(s):
Construct $$\overline{BD}$$ as the altitude from $$B$$ to $$\overline{AC}.$$
Then $60 = \dfrac{1}{2} \cdot BD \cdot 24,$ which gives us that $$BD = 5.$$
From this, we apply the Pythagorean Theorem on $$\triangle ABD:$$ $AB^2 = 5^2 + 12^2 = 169 = 13^2.$ This gives us that $$AB = 13.$$
Thus, C is the correct answer.
15.
Let $$a, b$$ and $$c$$ be numbers with $$0 < a < b < c.$$ Which of the following is impossible?
$$a + c < b$$
$$a \cdot b < c$$
$$a + b < c$$
$$a \cdot c < b$$
$$\dfrac{b}{c} = a$$
###### Solution(s):
We know that $$b < c$$ and $$0 < a.$$ Adding these two inequalities together yields $b \lt c + a.$ This shows that A is impossible, and therefore the right answer.
To ensure that this is correct, we can show that the other options are possible.
B and C: $$a = 1,$$ $$b = 2,$$ and $$c = 4$$
D: $$a = \dfrac{1}{3},$$ $$b = \dfrac{1}{2},$$ and $$c = 1$$
E: $$a = \dfrac{1}{2},$$ $$b = 1,$$ and $$c = 2$$
Thus, A is the correct answer.
16.
Amanda Reckonwith draws five circles with radii $$1, 2, 3, 4$$ and $$5.$$ Then for each circle she plots the point $$(C, A),$$ where $$C$$ is its circumference and $$A$$ is its area. Which of the following could be her graph?
###### Solution(s):
The cirumferences of circles with radii $$1$$ through $$5$$ are $2\pi, 4\pi, 6\pi, 8\pi, 10\pi.$ Their respective areas are $\pi, 4\pi, 9\pi, 16\pi, 25\pi.$ The only graph that shows these points is A, making it the correct answer.
This is the only graph showing a quadratic function.
Thus, A is the correct answer.
17.
A mixture of $$30$$ liters of paint is $$25\%$$ red tint, $$30\%$$ yellow tint and $$45\%$$ water. Five liters of yellow tint are added to the original mixture. What is the percent of yellow tint in the new mixture?
$$25$$
$$35$$
$$40$$
$$45$$
$$50$$
###### Solution(s):
The amount of yellow tint in the original mixture is $$.3 \times 30 = 9$$ liters. Adding $$5$$ liters results in a total of $$14$$ liters of yellow tint.
The new mixture has a total of $$35$$ liters, so the percent of yellow tint is $100 \cdot \dfrac{14}{35} = 100 \cdot \dfrac{2}{5} = 40 \%.$
Thus, C is the correct answer.
18.
The product of the two $$99$$-digit numbers $303,030,303,...,030,303$ and $505,050,505,...,050,505$ has thousands digit $$A$$ and units digit $$B.$$ What is the sum of $$A$$ and $$B?$$
$$3$$
$$5$$
$$6$$
$$8$$
$$10$$
###### Solution(s):
We only care about the last $$4$$ digits, so we can calculate $$303 \cdot 505$$ to find them (the thousands digit is $$0$$ for both numbers).
$\begin{array}{r} &\cdots 303 \\ \times \hspace{-4mm} &\cdots 505 \\ \hline &\cdots 1515 \\ &\cdots 1500 \\ \hline &\cdots 3015 \end{array}$
This gives us that $$A = 3$$ and $$B = 5.$$ Therefore, $$A + B = 8.$$
Thus, D is the correct answer.
19.
Pick two consecutive positive integers whose sum is less than $$100.$$ Square both of those integers and then find the difference of the squares. Which of the following could be the difference?
$$2$$
$$64$$
$$79$$
$$96$$
$$131$$
###### Solution(s):
Let $$x$$ and $$x + 1$$ the two integers, where $$2x + 1 \lt 100.$$ The difference of the squares is $(x + 1)^2 - x^2$$=(x + 1 + x)(x + 1 - x)$$= 2x + 1.$
From this, we see that the desired difference is less than $$100$$ and is odd.
Thus, the correct answer is C.
20.
Before district play, the Unicorns had won $$45 \%$$ of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$$48$$
$$50$$
$$52$$
$$54$$
$$60$$
###### Solution(s):
Let $$x$$ be the number of games the Unicorns had played before district play. Then after the entire season, they won a total of $$.45x + 6$$ games.
In total, they played $$x + 8$$ games. The problem statement also tell us that $(x + 8) \div 2 = .45x + 6.$ Solving, \begin{align*} x + 8 &= .9x + 12 \\ .1x &= 4 \\ x &= 40. \end{align*} Therefore, the Unicorns played a total of $$40 + 8 = 48$$ games.
Thus, A is the correct answer.
21.
Two cards are dealt from a deck of four red cards labeled $$A,$$ $$B,$$ $$C,$$ $$D$$ and four green cards labeled $$A,$$ $$B,$$ $$C,$$ $$D.$$ A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?
$$\dfrac{2}{7}$$
$$\dfrac{3}{8}$$
$$\dfrac{1}{2}$$
$$\dfrac{4}{7}$$
$$\dfrac{5}{8}$$
###### Solution(s):
After drawing the first card, there are $$7$$ left. $$3$$ of them have the same color, and $$1$$ of them has the same letter. Therefore, there are $$4$$ out of $$7$$ possibilities that result in a winning pair.
The probability of drawing a pair is then $$\dfrac{4}{7}.$$
Thus, D is the correct answer.
22.
A lemming sits at a corner of a square with side length $$10$$ meters. The lemming runs $$6.2$$ meters along a diagonal toward the opposite corner. It stops, makes a $$90^{\circ}$$ right turn and runs $$2$$ more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
$$2$$
$$4.5$$
$$5$$
$$6.2$$
$$7$$
###### Solution(s):
Based on the lengths given in the problem, the lemming is still in the square after it stops.
Since the lemming is still in the square, the sum of the distances to the horizontal sides is $$10$$ meters and the same for the vertical sides. Therefore, the $$4$$ distances sum to $$20$$ meters, making the average $$20 \div 4 = 5$$ meters.
Thus, C is the correct answer.
23.
What is the area of the shaded pinwheel shown in the $$5 \times 5$$ grid?
$$4$$
$$6$$
$$8$$
$$10$$
$$12$$
###### Solution(s):
We can find the area of the shaded part by subtracting the area of the unshaded part from the whole area.
There are $$4$$ squares with side length $$1,$$ contributing $$1$$ each to the unshaded area, for a total of $$4.$$
There are also $$4$$ triangles with base $$3$$ and height $$\dfrac{5}{2}.$$ This contributes a total area of $4 \cdot \dfrac{1}{2} \cdot 3 \cdot \dfrac{5}{2} = 15.$
The total unshaded area is therefore $$4 + 15 = 19.$$
The total area is $$5^2 = 25,$$ and subtracting the unshaded area yields $$25 - 19 = 6$$ as the area of the shaded region.
Thus, B is the correct answer.
24.
A bag contains four pieces of paper, each labeled with one of the digits $$1,$$ $$2,$$ $$3$$ or $$4,$$ with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $$3?$$
$$\dfrac{1}{4}$$
$$\dfrac{1}{3}$$
$$\dfrac{1}{2}$$
$$\dfrac{2}{3}$$
$$\dfrac{3}{4}$$
###### Solution(s):
Recall that a number is divisible if the sum of its digits is divisible by $$3.$$
The only triples of distinct numbers that satisfy this condition are $$(1, 2, 3)$$ and $$(2, 3, 4).$$
This means that the numbers form a three-digit number when either $$1$$ or $$4$$ is left in the bag.
Each of these events happens with $$\dfrac{1}{4}$$ chance, for a total probability of $$2 \cdot \dfrac{1}{4} = \dfrac{1}{2}.$$
Thus, C is the correct answer.
25.
On the dart board shown in the figure below, the outer circle has radius $$6$$ and the inner circle has radius $$3$$. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values of the regions hit. What is the probability that the score is odd?
$$\dfrac{17}{36}$$
$$\dfrac{35}{72}$$
$$\dfrac{1}{2}$$
$$\dfrac{37}{72}$$
$$\dfrac{19}{36}$$
The area of the outer circle is $$6^2\pi = 36\pi,$$ and the area of the inner circle is $$3^2\pi = 9\pi.$$ Therefore, the area of the outer ring is $$36\pi - 9\pi = 27\pi.$$
This means that the probability of hitting an outer segment is $$\dfrac{9\pi}{36\pi} = \dfrac{1}{4}.$$ Similarly, the probability of hitting an inner segment is $$\dfrac{3\pi}{36\pi} = \dfrac{1}{12}.$$
Summing over all possibilities, the probability of hitting a $$1$$ is $\dfrac{1}{12} + 2 \cdot \dfrac{1}{4} = \dfrac{7}{12}.$ Similarly, the probability of hitting a $$2$$ is $2 \cdot \dfrac{1}{12} + \dfrac{1}{4} = \dfrac{5}{12}.$
The only way to get an odd score is to hit one $$1$$ and one $$2$$.
The probability of hitting these numbers in either order is $2 \cdot \dfrac{5}{12} \cdot \dfrac{7}{12} = \dfrac{35}{72}.$
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# C | C Quiz – 113 | Question 1
• Difficulty Level : Medium
• Last Updated : 28 Jun, 2021
Output of following program under the assumption that numbers are stored in 2’s complement form.
`#include ` `int` `main() ` `{ ` ` ``printf``(``"%c\n"``, ~(``'C'` `* -1)); ` ` ``return` `0; ` `} `
Contributed by Sowmya.L.R
(A) B
(B) A
(C) Compiler Error
(D) C
Explanation: executed without any error or warning messages and the output for the above code is
‘B’
The above program processes as below
Step 1:
First (‘C’ *-1) is processed
ASCII value of ‘C’ is 67 and it is multiplied with -1 as
67 * (-1) = -67
Step 2:
The binary representation of -67 is 10111101
The bitwise negation of 10111101 becomes (01000010 ) 2 = (66) 10
Step 3:
66 is the ASCII value of ‘B’
So ~(‘C’*-1) = 66 and so the output of the above the program is B
Quiz of this Question
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# GMAT Test M15 Master Thread - All discussions
Author Message
GMAT Club team member
Joined: 16 Mar 2009
Posts: 115
Location: Bologna, Italy
Followers: 47
Kudos [?]: 854 [0], given: 19
### Show Tags
12 May 2009, 23:45
Last edited by dzyubam on 26 Apr 2010, 02:11, edited 2 times in total.
q#7
Manager
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Posts: 177
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Kudos [?]: 24 [0], given: 5
### Show Tags
23 Jun 2010, 04:09
Should not the answer be B.
x^2 - y^2 = (x+y)(x-y)
Statement 2 says that x+y is divisible by 8. In that case (x+y)(x-y) is also divisible by 8.
Manager
Joined: 30 Jun 2004
Posts: 177
Location: Singapore
Followers: 1
Kudos [?]: 24 [0], given: 5
### Show Tags
23 Jun 2010, 04:25
IMHO, the wordings "company's monthly sales" is not very clear. May be the test maker can make it more clear by using the term "revenue" or something else.
CEO
Joined: 15 Aug 2003
Posts: 3460
Followers: 67
Kudos [?]: 862 [0], given: 781
### Show Tags
29 Jun 2010, 23:36
tarun wrote:
IMHO, the wordings "company's monthly sales" is not very clear. May be the test maker can make it more clear by using the term "revenue" or something else.
I thought it was OK. What else would a company's monthly sales mean? profits? costs?
Intern
Joined: 09 Sep 2010
Posts: 12
Followers: 0
Kudos [?]: 8 [0], given: 11
### Show Tags
05 Oct 2010, 13:50
Is there an error in Q.19 ?
How can the closest vertex near (0,0) be (1,1) whereas it has been clearly mentioned in the question that (6,2) is the one of the vertices of the diagonal. Hence the straight line will have (0,2) as the other vertex. The distance of (0,2) from (0,0) will be 2.
Senior Manager
Status: Happy to join ROSS!
Joined: 29 Sep 2010
Posts: 278
Concentration: General Management, Strategy
Schools: Ross '14 (M)
Followers: 20
Kudos [?]: 124 [0], given: 48
### Show Tags
25 Mar 2011, 11:38
tarun wrote:
Should not the answer be B.
x^2 - y^2 = (x+y)(x-y)
Statement 2 says that x+y is divisible by 8. In that case (x+y)(x-y) is also divisible by 8.
Ok, the trick is: the number is divisible by X if after division there is integer left and no remainder.
In this question, answer 2 precludes the situation when x=7.77, y = 0.23, so
(x+y)(x-y)/8 = integer*7,52
thus we need answer 1 and answer 2 to make sure that after division by 8 we'll end up with multiplying two integers.
Manager
Affiliations: The Earth organization, India
Joined: 25 Dec 2010
Posts: 193
WE 1: SAP consultant-IT 2 years
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Kudos [?]: 13 [0], given: 12
### Show Tags
27 Jun 2011, 01:03
sandhyash wrote:
Is there an error in Q.19 ?
How can the closest vertex near (0,0) be (1,1) whereas it has been clearly mentioned in the question that (6,2) is the one of the vertices of the diagonal. Hence the straight line will have (0,2) as the other vertex. The distance of (0,2) from (0,0) will be 2.
Dude, (0,2) makes it a rectangle instead of square.
Q#19 coordinate-plane-90772.html
_________________
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Quant 47-Striving for 50
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Re: GMAT Test M15 Master Thread - All discussions [#permalink] 27 Jun 2011, 01:03
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Problems? Is your data what you think it is? PerlMonks
Re: Challenge: Sorting Sums Of Sorted Series
by Krambambuli (Curate)
on Feb 03, 2010 at 08:12 UTC ( #821097=note: print w/replies, xml ) Need Help??
in reply to Challenge: Sorting Sums Of Sorted Series
Can't help, have to ask this question:
how would an algorithm that would deal with this problem for infinite (huge...) lists look like...?
Update: And, given a finite part of each serie, a1, ..., aN, b1, ..., bM, how many - which - terms of the result could be trustfully computed (knowing that there are other terms to follow) ?
And how far could we advance with the 'arrival' of aN+1 and/or bM+1...?
hmmmm... Is someone aware of some good readings in order to get the answers ?
Thanks,
Krambambuli
---
• Comment on Re: Challenge: Sorting Sums Of Sorted Series
Replies are listed 'Best First'.
Re^2: Challenge: Sorting Sums Of Sorted Series
by Limbic~Region (Chancellor) on Feb 03, 2010 at 14:05 UTC
Krambambuli,
Before answering your question, keep in mind I already allowed for both lists to fit in memory as a given. In reality, if you had a really large list you probably wouldn't be going through these contortions - you would probably sort on disk.
I guess your question boils down to: How can you do this with even less than 2N + M? blokhead has shown a solution that uses no memory but takes (N*M)^2 to run. TO give you a specific answer, one would need a specific problem. As I demonstrated in How many words does it take?, the trick to solving really hard problems in the general case (NP complete) is to exploit the details of the specific case (effectively turned into O(1)).
Cheers - L~R
Re^2: Challenge: Sorting Sums Of Sorted Series
by rubasov (Friar) on Feb 03, 2010 at 13:11 UTC
For representing infinite lists I recommend you to read Mark Jason Dominus' excellent book named Higher Order Perl, especially the chapter on infinite lists (where he is using closures for iterators and promise-forcing streams). It is available online in pdf format: http://hop.perl.plover.com.
And for your updated question: given the finite parts (a1, ..., aN), (b1, ..., bM), if you take min(a1+bM, b1+aN), you are safe to print any sum less than or equal to this threshold as a beginning sequence of the final list. Anything above this threshold cannot be considered final.
update: First I've written this: min(a1, b1) + min(aN, bM), but this is wrong, corrected above. And minor grammatical edits, arrgh...
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# Physics Flashcards
Terms Definitions
Vector Has Scalar Only has magnitude (distance). Ex: 5 meters. Speed How fast something is moving. No direction involved. Ex: 5/2 meters/second. Velocity When you specify both the speed and the direction. Is a vector. V for vector! Ex: 5/2 meters/second to the right. Vector Has a magnitude (distance) and a direction. Ex: 5 meters to the right. Symbol is an arrow thing. Vector Has Scalar Only has magnitude (distance). Ex: 5 meters. Speed How fast something is moving. No direction involved. Ex: 5/2 meters/second. Velocity When you specify both the speed and the direction. Is a vector. V for vector! Ex: 5/2 meters/second to the right. Vector Has a magnitude (distance) and a direction. Ex: 5 meters to the right. Symbol is an arrow thing. Vector Has Scalar Only has magnitude (distance). Ex: 5 meters. Speed How fast something is moving. No direction involved. Ex: 5/2 meters/second. Velocity When you specify both the speed and the direction. Is a vector. V for vector! Ex: 5/2 meters/second to the right. Vector Has a magnitude (distance) and a direction. Ex: 5 meters to the right. Symbol is an arrow thing. Steps for problem solving Write out all givens, convert everything to the same units, find an equation with only one variable, solve
/ 16
Term:
Definition:
Definition:
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# Homework Help: Oh order comparsion
1. Sep 22, 2008
1. The problem statement, all variables and given/known data
Compare each pairs according to their respective orders. Classify these forms by the relationships between the indicated constants.
Note: ki is a constant and all kis are mutually independent. (ki < kj where i < j),
ki ≥1.0.
1) n! Vs. K1^ n => Here its Cap K not little k.
2) log(n^n ) Vs. log(k1^k2 ) => little k
2. Relevant equations
http://www.augustana.ca/~hackw/csc210/exhibit/chap04/bigOhRules.html
3. The attempt at a solution
Here I am doing Oh Comparison, and I am not sure how to say which greater than, less than, equal to. Say for problem 2:
From the Log of a Power Rule (link above) the order would be O(log n) and O(log k1). Now n can be any number and k1 can be any number. So how would I know which is greater than, less than, or equal to? For all I know n=200 and k1 = 10, or maybe not?
For problem 1 same thing. K1 can be any number as well as n. If n= 2, than 2! = 2, and K1^2 .
2. Sep 22, 2008
### HallsofIvy
I assume you are referring to the comparative orders as n goes to infinity.
Look at the fractions.
1) What is
$$\frac{K1^n}{n!}$$
as n goes to infinity? Notice that numerator and denominator both have n terms but each factor in the numerator is K1 while factors in the denominator get larger and larger.
2) is easy! the denominator log(k1k2) is a constant! Any unbounded function of n will eventually be larger than any constant as n goes to infinity!
Last edited by a moderator: Sep 23, 2008
3. Sep 23, 2008
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# Principles of Design
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# Principles of Design
BD10 Multimedia and Web Design
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### Principles of Design
1. Principals of Design Wallpaper-photoshop-tools-desktop-to-desktop1.jpg
2. BalanceThe arrangement of elements https://flic.kr/p/jM34nM
3. Symmetrical Elements are centered or evenly divided both vertically and horizontally https://flic.kr/p/96Ckax Balance
4. Asymmetrical Off-center alignment created with an odd or mismatched number of elements. http://www.malagacomics.es/wp-content/uploads/2013/08/avengers-box.jpg Balance
5. Radial The elements radiate from or swirl in a circular or spiral path. 35258_1_miscellaneous_digital_art_hipster_triangle_post_modern_abstract_minimalist_art.jpg Balance
6. ContrastEmphasizing the difference between elements https://flic.kr/p/5BJ1rx
7. Unity/HarmonyAll of the design elements are consistent with each other in shape, style and color and consistent with the overall message https://flic.kr/p/9kS8wS
8. Scale/Proportion The relationships between the sizes of various elements https://flic.kr/p/axy4Wq
9. Dominance/ The focal point. Emphasis
10. Design These concepts will help you arrange elements and use design principles effectively https://flic.kr/p/yNxSS Compositions
11. GridsThe use of columns/rows in design https://flic.kr/p/bpRqEo
12. The Rule of ThirdsSplitting an image or design into thirds, so you end up with 9 equal sections https://flic.kr/p/3voPop
13. Optical CenterThe spot the eye first sees when it encounters a page slightly above and to the right of the actual center https://flic.kr/p/v8sS5
14. Z PatternThe pattern the eye follows when scanning a page. fight-club-william-henry-graphic-design-movie-poster-minimal1.jpg
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A140730 a(4*n)=5^n, a(4*n+1)=2*5^n, a(4*n+2)=3*5^n, a(4*n+3)=4*5^n. 6
1, 2, 3, 4, 5, 10, 15, 20, 25, 50, 75, 100, 125, 250, 375, 500, 625, 1250, 1875, 2500, 3125, 6250, 9375, 12500, 15625, 31250, 46875, 62500, 78125, 156250, 234375, 312500, 390625, 781250, 1171875, 1562500, 1953125, 3906250, 5859375, 7812500 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS a(n) = A140740(n+4,4). LINKS Index entries for linear recurrences with constant coefficients, signature (0,0,0,5). FORMULA a(n+1) = a(n) + a(n - n mod 4). O.g.f.: (1+2*x+3*x^2+4*x^3)/(1-5*x^4). - R. J. Mathar, May 31 2008 a(n) = (n+1-4*floor(n/4))*5^floor(n/4). - Luce ETIENNE, Aug 05 2015 a(n) = 5*a(n-4) for n>3; a(n) = n+1 for n<5. - Bruno Berselli, Aug 05 2015 MATHEMATICA Table[(n + 1 - 4 Floor[n/4]) 5^Floor[n/4], {n, 0, 40}] (* Bruno Berselli, Aug 05 2015 *) PROG (PARI) a(n)=(n+1-n\4*4)*5^(n\4) \\ Charles R Greathouse IV, Oct 07 2015 CROSSREFS Cf. A000079, A037124, A038754, A133464. Sequence in context: A032940 A064419 A032543 * A273732 A282032 A205962 Adjacent sequences: A140727 A140728 A140729 * A140731 A140732 A140733 KEYWORD nonn,easy AUTHOR Reinhard Zumkeller, May 26 2008 STATUS approved
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Last modified April 14 22:30 EDT 2021. Contains 342962 sequences. (Running on oeis4.)
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# Programming contests
September 16, 2014 4:00 PM – December 21, 2014 8:00 PM
# Kastenlauf
Once every year, Jo and his friends want to visit the local fair in Erlangen, called Bergkirchweih. This year, they want to make a Kastenlauf (box run). They start at Jo's home, and have one box (Kasten) of beer (with twenty bottles). As they are very thirsty, they drink one bottle of beer every 50 meters.
As the way from Jo's home to the Bergkirchweih is pretty long, they need more beer than they have initially. Fortunately, there are stores selling beer on the way. When they visit a store, they can drop their empty bottles and buy new bottles, but their total number of full bottles will not be more than twenty (because they are too lazy to carry more than one full box).
You are given the coordinates of the stores, of Jo's home, and of the location of the Bergkirchweih. Write a program to determine whether Jo and his friends can happily reach the Bergkirchweih, or whether they will run out of beer on the way.
## Input Specification
The input starts with a line containing the number of test cases t (t ≤ 50).
Each test case starts with a line containing the number n of stores selling beer (0 ≤ n ≤ 100). The next n + 2 lines cointain (in this order) the location of Jo's home, of the stores, and of the Bergkirchweih. The location is given with two integer coordinates x and y (both in meters, –32678 ≤ xy ≤ 32767).
As Erlangen is a rectangularly laid out city, the distance between two locations is the difference of the first coordinates plus the difference of the second coordinates (also called Manhattan metric).
## Output Specification
For each test case, print one line, containing either “happy” (if Jo and his friends can happily reach the Bergkirchweih), or “sad” (if they will run out of beer on the way).
## Sample Input
1. `2`
2. `2`
3. `0 0`
4. `1000 0`
5. `1000 1000`
6. `2000 1000`
7. `2`
8. `0 0`
9. `1000 0`
10. `2000 1000`
11. `2000 2000`
1. `happy`
2. `sad`
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# American Institute of Mathematical Sciences
June 2009, 4(2): 249-266. doi: 10.3934/nhm.2009.4.249
## A Hamiltonian perspective to the stabilization of systems of two conservation laws
1 LAGEP, Université de Lyon, Lyon, F-69003, France, France 2 FEMTO-ST/AS2M, ENSMM Besan¸con, 24 rue Alain Savary, 25 000 Besanon, France
Received September 2008 Revised February 2009 Published June 2009
This paper aims at providing some synthesis between two alternative representations of systems of two conservation laws and interpret different conditions on stabilizing boundary control laws. The first one, based on the invariance of its coordinates, is the representation in Riemann coordinates which has been applied successfully for the stabilization of linear and non-linear hyperbolic systems of conservation laws. The second representation is based on physical modelling and leads to port Hamiltonian systems which are extensions of infinite-dimensional Hamiltonian systems defined on Dirac structure encompassing pairs of conjugated boundary variables. In a first instance the port Hamiltonian formulation is recalled with respect to a canonical Stokes-Dirac structure and then derived in Riemann coordinates. In a second instance the conditions on the boundary feedback relations derived with respect to the Riemann invariants are expressed in terms of the port boundary variable of the Hamiltonian formulation and interpreted in terms of the dissipation inequality of the Hamiltonian functional. The p-system and the Saint-Venant equations arising in models of irrigation channels are the illustrating examples developed through the paper.
Citation: Valérie Dos Santos, Bernhard Maschke, Yann Le Gorrec. A Hamiltonian perspective to the stabilization of systems of two conservation laws. Networks & Heterogeneous Media, 2009, 4 (2) : 249-266. doi: 10.3934/nhm.2009.4.249
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2018 Impact Factor: 0.871
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# Help got exam tomorrow!
• Jun 9th 2009, 10:06 PM
Unt0t
Help got exam tomorrow!
Hey my question is about the trapizodial rule.
They give as a table (forget [], just used it to line up the lines):
[ x]|0|1|2| 3 | 4
f(x)|1|0|2|10|27
now they tell us to use the trapezoidal rule to calculate an approx for f(x) between 1 and 4, using 4 sub intervals.
What I did was:
T4 = (b-a/8)(f(0)+f(4)+2(f(1)+f(2)+f(3))
T4 = (1/2)(1+27+2(0+2+10))
T4 = 52/2= 26
Now the model answears do this and get a diff answer, can you please tell me if i'm doing it wrong or the model answers are wrong.
T4 = (4-0/12)(f(0)+4f(0)+2f(2)+4f(3)+f(4)) = 1/3(1+27+4(0+10)+2(2))
T4= 24
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Contents
Contents
Idea
This page lays down a specific set of axioms of the Haag-Kastler axioms to AQFT, knowing that this set is not the set of Haag-Kastler axioms, but one specific choice. The purpose is to state and prove some of the classical results of the theory in a mathematically precise and model independent way. In fact some experts will find the chosen axioms to be stronger than necessary, this is deliberately so to ease the exposition.
It is possible to construct examples that fulfill the axioms, to show that they are not empty, but we will not engage in this task here, at least not now. Note however that up to now there was no success in the task to construct systems in 4 dimensions with interactions, which has led to some doubts about the usefulness of this approach in the physics community: It has yet to be shown if the approach does or does not capture the essential features that makes possible the tremendous success of the standard model of particle physics.
Abstract
We first collect some terms that are needed to then formulate the axioms of the vacuum representation. A few simple consequences are cited and some links to further concepts are provided.
Notation
The Poincaré Group
The universal covering $SL(2, \C) \ltimes \R^4$ of the restricted Poincaré group $\mathcal{P}^{\dagger}_+$ will be denoted by $\mathcal{P}$, the abelian subgroup of all translations by $\mathcal{T}$.
The Minkowski Spacetime
We talk about 4-dimensional Minkowski spacetime $\mathcal{Min}$ only, i.e. $\mathcal{Min}$ is the vector space $\R^{4} = \R \times \R^{3}$ equipped with the scalar product $\lt x, y \gt := x_0 y_0 - (\vec x, \vec y)$ with $( \cdot , \cdot )$ being the Euclidean scalar product on $\R^{3}$. Open bounded subsets of $\mathcal{Min}$ will be denoted by $\mathcal{O}$. The union of these $\mathcal{O}$ form an index set $\mathcal{J}$, that is partially ordered by inclusion.
If two sets are spacelike separated, this will be denoted by $\mathcal{O}_1 \perp \mathcal{O}_2$, the interior of the spacelike complement of a given set $\mathcal{O}$ will be denoted by $\mathcal{O}'$ or $\mathcal{O}^{\perp}$. The bounded open sets thus form a causal index set.
We denote the open forward (light)cone at x by $V_+(x)$, similar $V_-(x)$ is the open backward cone at $x$, if $x=0$ we simply write $V_+$ and $V_-$. A double cone or diamond is an intersection of an open forward cone and an open backward cone that is nonempty and will be denoted by $\mathcal{K}$.
An important class of unbounded regions are the wedges: Choose an inertial frame and define the right wedge as
$W_R := \{ x = (t, x_1, x_2, x_3) \in \mathbb{R}^4 \mid x_1 \gt \shortmid t \shortmid \}$
The set of wedges is then defined to be
$\mathcal{W} := \{\lambda W_R \mid \lambda \in \mathcal{P}^{\dagger}_+ \}$
While the definition of the right wedge depends on the chosen initial frame, the definition of the set of wedges does not.
Operator Algebras
See operator algebra and von Neumann algebra here on the nLab.
Von Neumann algebras$\mathcal{M}$ will always be concrete operator algebras acting on a given Hilbert space $\mathcal{H}$, as is the rule in the literature (see also von Neumann algebra). The commutant of $\mathcal{M}$ will be denoted by $\mathcal{M}'$, the positive cone by $\mathcal{M}^+$. The minimal von Neumann algebra that contains two given ones $\mathcal{M}_1$ and $\mathcal{M}_2$ will be denoted by:
$\mathcal{M}_1 \vee \mathcal{M}_2 := {(\mathcal{M}_1 \cup \mathcal{M}_2)}''$
As in this formula we will make frequently use of the bicommutant theorem to denote by $\mathcal{M}''$ the weak resp. strong closure resp. the generated von Neumann algebra of a give set of bounded operators.
An automorphism of an algebra $\alpha \mathcal{M} \rightarrow \mathcal{M}$ is called an inner automorphism if there is an invertible element $u \in \mathcal{M}$ such that $\alpha$ is given by conjugation with $u: \alpha(m) = u \quad m \quad u^{-1} \qquad \forall m\in \mathcal{M}$ (note that our convention here differs from that used by Wikipedia).
Group Representations
In this paragraph we will collect some links and remarks about unitary representations of topological groups on Hilbert spaces that are relevant to our topic and less commonly used in the literature. In the following $\mathcal{G}$ will be a topological group, $\mathcal{H}$ a (complex) Hilbert space and $\mathcal{U}$ an unitary representation of $\mathcal{G}$ in the algebra of bounded operators of $\mathcal{H}$.
Definition (analytical vector): Let $\mathcal{G}$ be a n-dimensional real Lie group. Fix a $f \in \mathcal{H}$, a neighbourhood B of 0 in $\R^n$ and a parametrization $\phi: B \to \mathcal{G}$ of a neighbourhood of 1 in $\mathcal{G}$. Then we can define a function $\mathcal{U}_f$ by
$\mathcal{U}_f: B \to \mathcal{H}$
$x \mapsto \mathcal{U}(\phi(x)) f$
If $\mathcal{U}_f$ has an extension to an analytic function on a neighbourhood of 0 in $\C^n$, the vector $f$ is called an analytic vector (for $\mathcal{U}$).
See planetmath for the definition of Banach space valued analytic functions.
Definition of Vacuum Representations
A net of von Neumann algebras $\mathcal{M}(\mathcal{O})$ on a common Hilbert space $(\mathcal{H})$, indexed by $\mathcal{O} \in \mathcal{J}$, is called a vacuum respresentation (on the 4-dimensional Minkowski spacetime) if it satisfies the following axioms:
• (V1) isotony: $\mathcal{O}_1 \subset \mathcal{O}_2$ implies $\mathcal{M}(\mathcal{O}_1) \subseteq \mathcal{M}(\mathcal{O}_2)$
• (V2) additivity: $\mathcal{O} = \bigcup_j \mathcal{O}_j$ implies $\mathcal{M}(\mathcal{O}) = ( \bigcup_j \mathcal{M}(\mathcal{O}_j) )''$
• (V3) locality, see local net: $\mathcal{O}_1 \perp \mathcal{O}_2 implies \mathcal{M}(\mathcal{O}_1) \subseteq (\mathcal{M}(\mathcal{O}_2))'$
• (V4) covariance: There is a strongly continuous unitary representation $\mathcal{U}(\mathcal{P})$ of $\mathcal{P}$ such that for each $g \in \mathcal{P}$ the following holds: $\mathcal{M}(g\mathcal{O}) = \mathcal{U}(g) \mathcal{M}(\mathcal{O}) \mathcal{U}(g)^{-1}$
• (V5) spectrum condition: $spec\mathcal{U}(\mathcal{T}) \subseteq clo \mathcal{V}_+$
Remark (mathematical viewpoint): $\mathcal{T}$ is the abelian subgroup of translations, and $\mathcal{V}_+$ is the (open) forward cone at 0, see above. For the definiton of the spectrum of the representation $\mathcal{U}(\mathcal{T})$ see spectral measure.
• (V6) existence of a vacuum vector: There exists a vector $\Omega \in \mathcal{H}, \|\Omega\| = 1$, such that $\bigl( \bigcup_{\mathcal{O}\in\mathcal{J}}\mathcal{M}(\mathcal{O}) \bigr) \Omega$ is dense in $\mathcal{H}$ and $\mathcal{U}(g)\Omega = \Omega$ for all $g \in \mathcal{P}$
Remark
(choice of axioms) The uniqueness is sometimes part of the axioms, but not here. Instead we will cite theorems that will specify necessary and sufficient conditions to ensure that there is a unique vacuum vector.
Additional Notations and Notions of Vacuum Representations
A short hand notation for vacuum representations will be $\mathcal{M}(\mathcal{J})$ in the following.
The algebras $\mathcal{M}(\mathcal{O})$ are sometimes called local algebras.
The $C^*-$algebra $\mathcal{A} := clo_{\| \cdot \|} \bigl( \bigcup_{\mathcal{O}\in\mathcal{J}}\mathcal{M}(\mathcal{O}) \bigr)$ is called quasilocal algebra, the smallest von Neuman algebra that contains $\mathcal{A}$ is called the global algebra and denoted by $\mathcal{R}$.
A vacuum representation is called irreducible if $\mathcal{R} = \mathcal{L}(\mathcal{H})$ (the global algebra is the whole algebra of all bounded linear operators on the given Hilbert space), it is called factorial if $\mathcal{R}$ is a factor.
• Wikipedia on factors of von Neumann algebras.
The subspace of $\mathcal{H}$ that is invariant under the action of the translation group $\mathcal{T}$ is not trivial due to the axiom . If it is one-dimensional, we will say that the vacuum representation has a unique vacuum vector (the space is then necessarily the subspace $\C\Omega$).
first consequences and the Reeh-Schlieder theorem
• Definition (algebras of unbounded open set) To any unbounded open set $\mathcal{U}$ we associate a von Neumann algebra by
$\mathcal{M}(U) := \bigl( \bigcup_{\mathcal{O} \subset \mathcal{U}} \mathcal{M}(\mathcal{O}) \bigr)''$
• Theorem (weak additivity) Let $\mathcal{O} \in \mathcal{J}$ be arbitrary, then we have
$\bigl( \bigcup_{a \in \mathcal{T}} \mathcal{M}(a \mathcal{O}) \bigr)'' = \mathcal{R}$
Remark (choice of axioms): Weak additivity of our nets are a direct consequence of our axioms, but this is often enough not so in the literature: Depending on the choice of axioms, it is either stated as an axiom or an auxiliary property of a net that is called for when needed.
Proof: Let $\mathcal{O}_0$ be given (arbitrary but fixed). Let us first note that thanks to isotony and additivity we can restrict the index set of the union in the definition of $\mathcal{R}$ from open bounded sets to diamonds, since every bounded set is contained in a diamond:
$\bigl( \bigcup_{\mathcal{O} \in \mathcal{J}} \mathcal{M}(\mathcal{O}) \bigr)'' = \bigl( \bigcup_{\mathcal{K}} \mathcal{M}(\mathcal{K}) \bigr)'' = \mathcal{R}$
Given an arbitrary diamond $\mathcal{K}$ we have $\mathcal{K} \subset \bigl( \bigcup_{a \in \mathcal{K}}(\mathcal{O}_0 + a) \bigr)$ while the latter is a bounded open subset $\in \mathcal{J}$. This means that we arrive at
$\bigcup_{\mathcal{K}} \mathcal{M}(\mathcal{K}) = \bigcup_{a\mathcal{O}_0, a \in {\mathcal{T}}} \mathcal{M}(a\mathcal{O}_0)$
which implies that the generated von Neumann algebras coincide, too, of course.
• Theorem (Borchers, translations are inner): The representatives of the translations are elements of the global algebra $\mathcal{R}$, i.e. they are inner automorphisms of $\mathcal{R}$: $\mathcal{U}(\mathcal{T}) \subseteq \mathcal{R}$.
• Theorem (uniqueness of vacuum): Every factorial vacuum representation is irreducible. A vacuum representation is irreducible iff it has an unique vacuum vector.
• Reeh-Schlieder Theorem: The vacuum vector is cyclic and separating for all local algebras.
Let $x, y \in \R^4$ and define $[x, y] := \{\lambda x + \mu y | \lambda, \mu \geqq 0, \lambda + \mu = 1\}$. Define
$\mathcal{M}[x, y] := \cap_{[x, y] \subset \mathcal{O}} \mathcal{M}(\mathcal{O})$
Theorem
triviality of algebras of spacelike segments
If the segment $[x, x + b]$ is spacelike, i.e. $\langle b, b \rangle \lt 0$, and the vacuum respresentation has a unique vacuum vector, then $\mathcal{M}[x, x + b] = \mathbb{C} \mathbb{1}$, i.e. the algebra associated with the segment is trivial.
Theorem
triviality of algebras of points The conclusion of the preceding statement holds if we put $b = 0$, i.e. if we consider the algebra associated with one point.
Remark (physical viewpoint): The preceding theorem is sometimes summarized as there are no non-trivial observables at the point. There are two possible ways to interpret this result: The pragmatic approach says that, since no detector can be built that measures observables precisley at one point of spacetime, there is no need of a theory to support the concept of observables localized at a point. The philosophical approach takes this one step further and states that our relativistic quantum theory tells us that the concepts of points and observables localized at points are an idealization with no relevance to nature.
The local algebras fulfill the Borchers property.
There are several directions one can pursue next, for example
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Feb 27 2009
## The “delicate balance of terror”: How game theory can be used to predict firm behavior (oh, and save the human race from utter annihilation)
This week in AP Microeconomics students get to play online games, watch movies, and compete with their classmates in strategic competitions in which there are proud winners and sad losers. That’s right, we’re studying oligopoly!
What makes oligopolistic markets, which characterized by a few large firms, so different from the other market structures we study in Microeconomics? The answer is that unlike in more competitive markets in which firms are of much smaller size and one firm’s behavior has little or no effect on its competitors, an oligopolist that decides to lower its prices, change its output, expand into a new market, offer new services, or adverstise, will have powerful and consequential effects on the profitability of its competitors. For this reason, firms in oligopolistic markets are always considering the behavior of their competitors when making their own economic decisions.
To understand the behavior of non-collusive oligopolists, economists have employed a mathematical tool called Game Theory. The assumption is that large firms in competition will behave similarly to individual players in a game such as poker. Firms, which are the “players” will make “moves” (referring to economic decisions such as whether or not to advertise, whether to offer discounts or certain services, make particular changes to their products, charge a high or low price, or any other of a number of economic actions) based on the predicted behavior of their competitors.
If a large firm competing with other large firms understands the various “payoffs” (referring to the profits or losses that will result from a particular economic decision made by itself and its competitors) then it will be better able to make a rational, profit-maximizing (or loss minimizing) decision based on the likely actions of its competitors. The outcome of such a situation, or game, can be predicted using payoff matrixes. Below is an illustration of a game between two coffee shops competing in a small town.
As illustrated above, the tools of Game Theory, including the “payoff matrix”, can prove helpful in helping firms decide how to respond to particular actions by their competitors in oligopolistic markets. Of course, in the real world there are often more than two firms in competition in a particular market, and the decisions that they must make include more than simply to advertise or not. Much more complicated, multi-player games with several possible “moves” have also been developed and used to help make tough economic decisions a little easier in the world of competition.
While Game Theory can be useful in predicting firm behavior in oligopolistic markets, believe it or not that is not its most useful application developed. In fact, would you believe me if I told you that Game Theory may be precisely what saved the world from nuclear holocaust during the 20th Century? It’s true. The US government employed Game Theory to avert annihilation by nuclear attack from the Soviet Union during much of the 20th Century. This video tells the story!
Feb 26 2009
## An Asian Exodus?
FT.com / China / Economy & Trade – Downturn drives expat exodus from Shanghai
Having recently moved from Shanghai to Zurich myself, I was interested to see this headline in today’s Financial Times.
Korean companies are shipping workers home, cutting off school fees and repatriating wives and children without their menfolk to cut costs. They are the first large wave of expatriates to have begun leaving China’s financial capital as a result of the global economic crisis but their departure raises the prospect of a broader exodus of foreigners who may take investment, skills and job creation opportunities with them.
The press officer of the Korean consulate in Shanghai could not answer questions about the exodus of her countrymen – because her post had just been abolished and she was being sent back to Korea…
Japanese relocation companies, meanwhile, say there has been a marked rise in Japanese families returning home from Shanghai compared with last year and they expect the pace to pick up further during the traditional peak relocation months of March and April.
As Korean and Japanese families pack up and leave Shanghai, the impact is likely to be felt at international schools catering to the expat community in Eastern China. Koreans made up around 15% of the students at Shanghai American School, while other schools in the city had even larger numbers of Japanese and Korean students. In Beijing the exodus is also underway:
The pain has not been limited to Shanghai. A parent with children enrolled in an expensive Beijing international school says most of her daughters’ Korean classmates have left the school almost overnight.
This story reminds me of my own experience as an international school student in the late 1990’s, when the Asian financial crisis plunged Korea’s economy into deep recession. At the time, 30% of my school in Malaysia were Korean students, and in one semester over half of them packed up and moved back to Korea. In one year enrollment at the International School of Kuala Lumpur’s high school fell from 600 students to 420!
One reason the Korean and Japanese economies are struggling is that they are heavily dependent on exports to the rest of the world. With incomes falling and unemployment rising among their trading partners, the effect is amplified in Japan and Korea by significant falls in aggregate demand and GDP due to lower net exports, investment and consumption in the Japanese economy.
According to this article in the FT, the current fall in exports in Japan is the worst in 50 years.
Japanese exports fell 45.7 per cent in January, eclipsing a 35 per cent drop in December and big declines last month for Taiwan and South Korea.
The slide in exports was the steepest since 1957 and highlighted the severe impact of the global slowdown on demand for Japanese products ranging from cars to heavy machinery and electronics. Exports to the US fell 52.9 per cent and those to China were down 45.1 per cent .
Falling demand has forced manufacturers such as Toyota and Sony to cut production and jobs. It has reinforced concerns the economy will suffer another quarter of falling output. Gross domestic product shrank 3.3 per cent in the last three months of 2008, the largest fall in 35 years.
The diagram below provides a graphical representation of the impact of falling exports on Japan’s economy.
Discussion questions:
1. Some economists believe that recessions are a crisis of confidence. What do they mean by that and how does the situation in Japan seen above reflect this theory?
2. What is the multiplier effect and how does the fall spending on Japanese exports by the rest of the world result in an even greater fall in Japan’s GDP?
3. If you were the manager of a Japanese firm facing falling demand from international customers and you had to cut costs, what costs would you cut in the short-run to remain competitive? What about in the long-run, assuming demand for your products remained weak?
Feb 25 2009
## Starbucks instant coffee: a sign of the times?
Chicago, Seattle first markets to get instant Starbucks — chicagotribune.com
I consider myself a Seattleite. I discovered the joy of drinking coffee in the home of Starbucks, Tully’s, Seattle’s Best, and countless local coffee shops that inhabit every corner of the rainy city. To me, the experience of drinking a latte, machiato, cappuccino, or simply a “coffee of the week” encapsulates the smells, soft decor and friendly greetings from the barista at my favorite coffee shop. Living overseas, I have turned to Starbucks over and over for a taste of Seattle and a feeling of home.
There is no denying that the Starbucks experience is one that does not come cheap. Here in Switzerland, a grande latte, my drink of choice, sets the consumer back nearly \$7. In an economic downturn such as that the US and the rest of the world are experiencing right now, such expenses are often the first to be reduced by cash strapped consumers. In fact, I recently began bringing a thermos of homemade coffee to work every day, rather than stopping at the Starbucks at the train station as I had done for several months not long ago.
Starbucks, which recently announced the closure of hundreds of its locations around the world, is actually expanding its product line while simultaneously closing down shops. It may not be in the way you expect, though. Soon, I’ll be able to get my \$7 cup of coffee for as little as \$1, it will just come in a different form:
Starbucks Corp. will launch its new instant coffee product next month in Chicago and its home turf of Seattle, with a full-scale, national offensive set for the fall.
Starbucks on Tuesday formally unveiled the new product, called Via Ready Brew. It will be available in Starbucks retail outlets in the Chicago and Seattle areas on March 3, Howard Schultz, the company’s chief executive, said in an interview with the Tribune.
Instant coffee from the king of gourmet blends? Sounds suspicious. Well, it’s all about economics, you see. Starbucks coffee is a normal good, one for which demand falls as incomes fall, as evidenced by falling sales at its coffee shops around the world. In order to maintain its customer base even as incomes fall, a company like Starbucks must expand its product line to include inferior products, or those for which demand increases even as incomes fall. Clearly, instant coffee is viewed as an inferior product, due to its significantly lower price and reputation of poor quality.
Furthermore, Starbucks’ new product is in response to increased competition from lower-end fast food chains that traditionally did not compete in the coffee market, but recently have begun offering various blends and varieties of coffee to the price-sensitive coffee consumers, further harming business at Starbucks’ higher end coffee outlets.
Via marks Starbucks second announcement this month of a cheaper menu alternative, as the famous coffee chain struggles in a weak economy. Starbucks is also now selling pairings of coffee and breakfast offerings for \$3.95.
Starbucks’ troubles have occurred at the same time value-oriented fast-food chains, particularly Oak Brook-based McDonald’s Corp., have thrived. McDonald’s owes part of its success to improving the quality of its basic coffee, and expanding into new drinks like iced coffee, and, more recently, flavored specialty coffees such as lattes and cappuccinos.
Still, Schultz said McDonald’s coffee offensive hasn’t really affected Starbucks: “We have a lot of respect for McDonald’s as a company. But we have not seen any significant issues with McDonald’s share of the coffee business affecting Starbucks.”
McDonald’s offers “a different product, a different value proposition,” he said. In fact, Schultz said McDonald’s should expand the overall coffee market, thus leading some customers to “trade up” to Starbucks.
Despite the CEO’s claims that Starbucks and McDonald’s coffees are “different” products, it is clear by his firm’s decision to expand into the instant coffee market that Starbucks is concerned about the loss of customers to lower-end coffee retailers.
The theory of firm behavior as studied in AP and IB Economics teaches us that firms in oligopolistic or monopolistically competitive markets, such as that for coffee shops in the US, tend to compete using non-price methods such as product differentiation and advertising. Rather than slashing the prices of all of its coffee in the face of a recession and falling consumer incomes, Starbucks has instead diversified its product line to include lower end options for consumers whose sensitivity to price and demand for gourmet coffee have been adversely affected by the weak economy.
Feb 24 2009
• All told, Varvares and his fellow forecasters now expect the economy to shrink by 1.9 percent this year, a much deeper contraction than the 0.2 percent dip projected in the fall.
If the new forecast is correct, it would mark the first time since 1991 the economy actually contracted over a full year and would be the worst showing since 1982, when the country had suffered through a severe recession.
Vanishing jobs, shrinking nest eggs, rising foreclosures and tanking home values have forced American consumers to cut back, which in turn has caused businesses to lay off workers and slash costs in other ways, feeding a vicious downward cycle for the economy.
The current recession, which started in December 2007, is posing a major challenge to Washington policymakers, including President Barack Obama and Fed Chairman Ben Bernanke. That’s because its root causes — a housing collapse, credit crunch and financial turmoil — are the worst since the 1930s and don’t lend themselves to easy or quick fixes.
“As the news on the economy has darkened, so too, have the forecasts,” said Ken Mayland, president of ClearView Economics. “We are suffering a period of maximum stress on the economy.”
The economy is expected to remain feeble this year — even with new efforts by the administration and Congress to provide relief.
Posted from Diigo. The rest of my favorite links are here.
Feb 24 2009
## Market Failure and the role of government in the economy ~ an introduction to Environmental Economics
Economics is the field of study that attempts to address the basic problem faced by society relating to the environment and natural resources: the problem of scarcity in a world of infinite wants. Many, if not all, of our planet’s environmental woes are attributable to an economic phenomenon known as market failure. A market failure results whenever too much (or in some cases too little) of a good or service is produced and consumed by the economy.
What does this have to do with the environment? The connection lies in the reality that everything we produce and consume (and I mean everything!) originates from the earth. Nothing can be made by the sweat of man alone; in fact, three resources are required to produce any good or service: labor, capital (i.e. tools), and land. Sometimes we think of the resource of land as gifts of nature. However, in a world where environmental threats like those mentioned above are staring us in the face, it is becoming more and more obvious that the natural resources we’ve exploited for so long may not, in fact, have been gifts from Mother Nature at all, and their overuse may impose significant and unaccounted for costs on society AND the environment.
But let’s be honest, consuming is fun! Nothing is more gratifying than scoring a fantastic deal at your favorite boutique, walking out of a fast food joint with a plastic bag full of tasty treats for super cheap, and getting your hands on the latest high tech gizmos as soon as they’re launched (and dumping that old technology out so you’re not the lame one with the three pound cell phone!) However, the true cost of our obsessive consumption habit is not always represented by the price we pay for our fast food, our blue jeans, and our iPads.
In reality, the prices we pay for our goods and services are far lower than they should be; and the quantity of these things we consume is far higher than it should be. How do we know this? Look around. The very environmental issues with which environmental groups are most concerned can be traced back to the consumer behavior we enjoy partaking in so much. We’re conditioned to buying what we want, when we want it, and for a price that places little burden on our pocket books.
What we don’t realize, however, is that nature is bearing the burden of our high levels of consumption. In its attempt to absorb the pollutants that are emitted in the manufacture of our products, the waste that’s created from the disposal of our products, and the destruction that’s left behind from the extraction of the natural resources that go into our products, Mother Nature is more than ever choking on the waste created by our economic behavior. The costs born by nature are not accounted for in the production costs faced by firms, nor in the prices paid by consumers. These costs are externalized, or passed on for others to worry about.
The problem is, these days the bill has come due, and the environment is calling in its debts. Humans must now face up to the failures of its markets, and internalize the costs that for so long have been passed on to the environment and society, which suffers from the effects of environmental degradation.
The reality that we’ve used too many natural resources to produce too much stuff for too long is evidenced by simple examination of the natural world around us. Or, in the case of China, the complete lack of a natural world around us. From the pollution filled skies, to the waste clogged waterways, to the traffic jammed highways, China is a case study in market failure. The world, now used to the cheap imports China is so good at pumping out, does not consider the impact that the manufacture and consumption of such a massive variety of cheap products is having on China’s, and these days the world’s, environment.
In the following audio clips, you’ll hear three short stories about how the over-exploitation of resources is causing harm to human welfare and the environment. Each of these stories contains a market failure, usually in the form of a negative externality, or the production and consumption of certain goods creating spillover costs on somebody or something not involved in its production or consumption. See if you can identified who’s being harmed, and who’s at fault:
### E-waste [ 7:36 ] Play Now | Play in Popup | Download Trash Island [ 2:01 ] Play Now | Play in Popup | Download Nauru [ 6:57 ] Play Now | Play in Popup | DownloadpodPressShowHidePlayer('1', 'http://search.saschina.org/storage/HTTCT3C/Ewaste.mp3', 290, 24, 'false', 'http://welkerswikinomics.com/blog/wp-content/plugins/podpress//images/vpreview_center.png', 'E-waste', 'Jason Welker');podPressShowHidePlayer('2', 'http://search.saschina.org/storage/CCTH0MD/trashisland.mp3', 290, 24, 'false', 'http://welkerswikinomics.com/blog/wp-content/plugins/podpress//images/vpreview_center.png', 'Trash Island', 'Jason Welker');podPressShowHidePlayer('3', 'http://search.saschina.org/storage/W8IGYEF/Nauru.mp3', 290, 24, 'false', 'http://welkerswikinomics.com/blog/wp-content/plugins/podpress//images/vpreview_center.png', 'Nauru', 'Jason Welker');
Story #1: “Where does all that E-waste go?” from Public Radio International’s “The World: Technology” podcast
Story #2: “Trash Island” from WBEZ Chicago’s “This American Life”
Story #3: “Nauru – the island in the middle of nowhere” from WBEZ Chicago’s “This American Life”
After listening to these stories, reflect for a moment on the true cost of the environmental and human tragedies of which they told. What role does our consumer culture play in these tragedies? What could have been done to prevent the conditions in those E-waste markets in Africa and China, the islands of garbage floating in our deep oceans, and the complete destruction of an island paradise 1,100 miles from the nearest land? Is there anyone to blame? Should we blame our politicians, our leaders? The answer to these questions is: there’s no easy answer, unless we want to get really personal here and point to humans’ own flawed nature: the fact that we are motivated primarily by greed and self-interest.
If that’s true, then perhaps hope for the environment can only be found in the responsible hands of benevolent governments, who once and for all take steps to mitigate the destructive impacts of our endless patterns of production and consumption. In fact, it is often government which is needed to intervene and correct market failures like those in the stories.
Three tools have emerged for governments wishing to correct such negative externalities. These involve three fundamentally different approaches, some more effective than others. One involves direct government control. This is when governments intervene in a market in which negative externalities exist and try to make producers clean up their acts. They threaten producers with penalties and fines, and monitor industries to try and force firms to manufacture their products in a clean, efficient way. (this is like what the Europeans are doing to minimize their e-waste).
The next option also involves a large roll for the government: corrective taxes. Businesses that produce goods that end up polluting the environment (either through their production or consumption) can be taxed based on the amount of pollution they create. If creating more pollution means paying more taxes, the companies will find ways to produce in a more environmentally responsible manner, in order to keep their costs low and to maximize their profits.
The third method for externality reduction is also the most recently adopted. A market for pollution permits is set up, where a government actually gives all the companies in a polluting industry permits that allow them to pollute a certain amount. WHAT? The government’s allowing firms to pollute? Well, yes. The fact is, they’re going to do it anyway, they HAVE to in order to produce anything! The benefit of this system is that the government will only give each firm so many permits, and they’re not allowed to pollute beyond what their permits allow, UNLESS they go and buy more permits from producers that don’t need all theirs. This way, firms have an incentive to pollute less, because any permits they don’t use they can sell to other producers and make profits on those sales! Dirty firms have to buy more and more permits, clean firms get to sell those they don’t need… can you see where this is going? ALL FIRMS want to become clean firms in this scenario!
The three methods introduced above are being used to different degrees by different countries in various industries to try and mitigate the negative effects of some types of pollution and greenhouse gas emissions. Unfortunately, not nearly enough is yet being done, especially by some of the worlds largest economies (and thus, polluters), namely the United States, China, and India.
If our world is to avoid a fate like that of the tiny island of Nauru, where every last resource was exploited to the point where the island could no longer sustain life, then more must be done to reduce the spillover costs that accompany the production and consumption of so many of our precious goods.
I tell my econ students a story about how one day hundreds of years ago some smart guy decided to start calling products (you know, the stuff we consume), GOODS. From that day on humans would always associate consumption with something GOOD. Today, in an era where the goodness of consumption is offset by the evil of environmental destruction, more than a strong government hand is needed. Conservation and appreciation for the gifts of nature, not insofar as they can be exploited by industry, but left intact for the appreciation and welfare of society, both today’s generation and that of our grandchildren, must be fostered and encouraged among global citizens young and old.
Hopefully, this article and the stories you heard here will help you understand a little more about the economics of the environment, and help you become more educated about what can and should be done to correct the market failures that have led to the dire challenges faced by our world today.
A great website on environmental economics written by two economists WAY smarter than Mr. Welker can be found here: http://www.env-econ.net/
Next »
• ## Order Welker’s books
for IB Economics
for AP Macro
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# Grade 1 Number and Place Value Pack
Subjects
Resource Types
Product Rating
Not yet rated
File Type
PDF (Acrobat) Document File
9.19 MB | 67 pages
### PRODUCT DESCRIPTION
Whether you use Math Daily 3, Interactive Math Notebooks or Journals, Math Groups combined with Guided Math and/or Teacher Directed Lessons these activities can easily become a part of your Classroom Math Routine.
This pack features activities that are hands-on, allow for peer interaction and independent practice.
What's Included:
No Prep Printables
★ Identify and Label Numbers to 100
★ Count and Label by 2s to 50
★ Count and Label by 2s from 50-100
★ Tens and Ones and Answer Key
★ Hundreds, Tens and Ones and Answer Key
★ Counting On and Counting Back 0- 50 & Answer Key
★ Counting On and Counting Back 0- 100 & Answer Key
★ Identifying Odd and Even Numbers 0- 100 & Answer Key
★ Expanded Form & Answer Key
★ Place Value Practice: Base Ten Blocks & Answer Key
Can/Aus/NZ/British spelling:
★ Place Value Practise: M.A.B. & Answer Key
★ Identifying Odd and Even Numbers 0- 100
Math Centers:
★ Place Value Sorting Mats, hundreds, tens and ones
★ ‘Greater Than, Less Than or Equal To’
★ ‘Greater Than and Less Than’ (apply new knowledge of symbols)
★ Ordering 1 - 2 Digit Numbers
★ Ordering 2 - 3 Digit Numbers
★ Ordering Numbers on a Number Line 0 - 100
★ Ordering Numbers on a Number Line 0 - 50
Materials:
★ 0 - 100 Number Cards
★ Greater Than, Less Than and Equal To cards
★ 0 - 100 Number Cards with Place Value Highlighted
★ I CAN Statements- and I CAN Number and Place Value Sign
★ Place Value and Number Star Student Awards x 4
Related Products:
Kindergarten Number and Place Value Pack
Check out my tpt store canuckteachingdownunder to purchase other activities tailored to your diverse learners!
★How to earn Teacher Pay Teacher credits to use on future purchases:
Click on the bolded text ‘My TPT’ heading at the top right of this page. Find the ‘My Purchases’ heading and click on it. Scroll down and beside each purchase you’ve made, you'll see a ‘Provide Feedback’ button. Once clicked, you’ll be taken to a page where you can leave a rating score and provide a short comment about the product.
★Why bother?
Each time you leave feedback, you’ll earn TPT credits which you can then use to discount the prices of your future purchases! Your feedback also helps me to understand how my products are working in your classroom and for your students.
You’ll find a green star beside my name at the top of this page or at the top of my store page. Once you click on this, you’ll be able to follow me. This notifies you when I post a new item and of any upcoming sales I may have.
***
This download is for personal/classroom use only. Please do not distribute to others, rather encourage your colleagues to visit my tpt store to have a look for themselves.
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https://answers.yahoo.com/question/index?qid=20191122221430AAbdR5p
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Anonymous
Anonymous asked in Science & MathematicsChemistry · 3 weeks ago
# i have a two part chemistry question i need help with?
1st part
A chemist has aluminum oxide and causes it to decompose into aluminum and oxygen gas. Please write the balanced decomposition reaction. You will have to figure out the chemical formula of aluminum oxide.
2nd part
Using your chemical reaction from #9, please calculate how many moles of aluminum are produced when 9.18 moles of oxygen are produced.
Relevance
• 3 weeks ago
That's hilarious. I don't think any chemist causes aluminum oxide (Al2O3) to decompose into aluminum and oxygen gas.
That reaction is purely speculative and unlikely to happen. You can, of course, write the hypothetical reaction, but understand that it isn't going to happen.
2Al2O3(s) --magic occurs--> 4Al(s) + 3O2(g)
.............. .................. ............? mol.... 9.18 mol
9.18 mol O2 x (4 mol Al / 3 mol O2) = 12.2 mol Al
• 3 weeks ago
Aluminum oxide- basically aluminum + oxygen, that’s all.
Now, oxygen likes to form 2- charges (it has six valence electrons and wants to get to eight so it gains two) and aluminum likes to form a charge of 3+ (to lose three electrons) so, we get the formula of Al2O3 (for every two aluminums you get 3 oxygens to balance out the charges).
Since its decomposing,
Al2O3 —> 2Al (s) + 3O (g)
(To balance it just make sure you have the same number of each atom on both sides.)
Alright, next:
9.18 moles of O * (2 moles of Al / 3 moles of O)
(I just used the ratio from the balanced chemical equation.)
That’s basically it, if you need help, look at Tyler Dewitt’s and The Organic Chemistry Tutor’s videos. They have some pretty dope stuff there. Hope I helped! 🏴
• pisgahchemist
Lv 7
2 weeks agoReport
Done in by a little ole' subscript. The common diatomic molecules: H2, N2, O2, F2, Cl2, Br2, I2. What I tell my students are the "hairogens."
"h-" = hydrogen, "-air-" = oxygen and nitrogen, "-ogens" = the halogens.
• JetDoc
Lv 7
3 weeks ago
Do your own homework.
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https://codereview.stackexchange.com/questions/30557/is-any-further-optimization-possible-codeforces/30563
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# Is any further optimization possible? (Codeforces)
I am an average coder trying to improve my Python by doing solved problems in it.
One of the problem I did is this, here is the code I tried which I have based on the official solution:
def area(A, B, C):
return float((x[B] - x[A])*(y[C] - y[B]) - (y[B] - y[A])*(x[C] -x[B]))/2
x, y = {}, {}
n = int(raw_input())
for i in xrange(n):
arr = raw_input().split()
x[i] , y[i] = int(arr[0]), int(arr[1])
maxarea = 0
for i in xrange(n):
for j in xrange(i+1, n):
maxminus, maxplus = -1, -1
for k in xrange(n):
if k != i and k != j:
a = area(i,j,k)
if(a<0):
maxminus = max(maxminus, -a)
else:
maxplus = max(maxplus, a)
if maxplus >= 0 and maxminus >=0:
maxarea = max(maxarea, (maxplus+maxminus))
print maxarea
The code is still giving me TLE on test case 7. Can anybody suggest further optimization?
You can do some minor optimization as follows:
def main():
n = int(raw_input())
coords = list([i] + map(int, raw_input().split()) for i in range(n))
max_area = 0
for a, Ax, Ay in coords:
for b, Bx, By in coords[a+1:]:
max_minus, max_plus = 0, -1
for c, Cx, Cy in coords:
if c != a and c != b:
ccw = (Bx - Ax) * (Cy - By) - (By - Ay) * (Cx - Bx)
if ccw < max_minus:
max_minus = ccw
elif ccw > max_plus:
max_plus = ccw
if max_plus >= 0 and max_minus < 0 and max_plus - max_minus > max_area:
max_area = max_plus - max_minus
print(max_area / 2.0)
main()
Note that your use of float doesn't do anything because the values to be passed are integers. Anyway, there's no need to divide by 2 every time - you can just divide the final value by 2 at the end.
I think this still won't pass the speed test, though. If it is doable in python it probably needs an algorithm that makes better use of python's functions and standard library (and are other libraries allowed?). You could try something like this, for example:
from itertools import permutations
def main():
n = int(raw_input())
coords = list([i] + map(int, raw_input().split()) for i in range(n))
max_area = 0
for (a, Ax, Ay), (b, Bx, By) in permutations(coords, 2):
ccws = [(Bx - Ax) * (Cy - By) - (By - Ay) * (Cx - Bx)
for c, Cx, Cy in coords if c != a and c != b]
low, high = min(ccws), max(ccws)
if low < 0 and high >= 0 and high - low > max_area:
max_area = high - low
print(max_area / 2.0)
• Coupled with the optimizations by @aseem-bansal above this reduced the time significantly for smaller test cases but it still fails on bigger test cases, I think the judge needs to update the time limit for Python in this problem. I have gone ahead and implemented the same program in Go and it passed without problems: codeforces.com/contest/340/submission/4391625 Commented Sep 1, 2013 at 8:58
Firstly you need to put all the global statements into a function because of Python code runs faster in a function compared to global scope.
You don't need the arr variable. You can just use
x[i], y[i] = map(int, raw_input().split())
instead of using the two lines. map uses the function(1st argument) on each of the outputs from the second argument.
After placing the global statements into a function then instead of passing i, j, k you would need to pass the values directly.
I'll look up the algorithm and see if there are other changes that can be made.
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25.13 kg to lbs - 25.13 kilograms to pounds
Before we get to the more practical part - this is 25.13 kg how much lbs conversion - we want to tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s start.
How to convert 25.13 kg to lbs? 25.13 kilograms it is equal 55.4021664406 pounds, so 25.13 kg is equal 55.4021664406 lbs.
25.13 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, formally known as International System of Units (in short form SI).
From time to time the kilogram can be written as kilogramme. The symbol of the kilogram is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. First definition was not complicated but totally impractical to use.
Later, in 1889 the kilogram was described by the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was switched by another definition.
Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is 0.001 tonne. It is also divided into 100 decagrams and 1000 grams.
25.13 kilogram to pounds
You learned a little bit about kilogram, so now let's go to the pound. The pound is also a unit of mass. It is needed to underline that there are more than one kind of pound. What does it mean? For example, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is used in the British and United States customary systems of measurements. To be honest, this unit is used also in other systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is exactly 0.45359237 kilograms. One avoirdupois pound is divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
How many lbs is 25.13 kg?
25.13 kilogram is equal to 55.4021664406 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
25.13 kg in lbs
Theoretical section is already behind us. In this section we are going to tell you how much is 25.13 kg to lbs. Now you learned that 25.13 kg = x lbs. So it is time to know the answer. Let’s see:
25.13 kilogram = 55.4021664406 pounds.
This is a correct outcome of how much 25.13 kg to pound. It is possible to also round off the result. After rounding off your outcome is as following: 25.13 kg = 55.286 lbs.
You know 25.13 kg is how many lbs, so have a look how many kg 25.13 lbs: 25.13 pound = 0.45359237 kilograms.
Obviously, in this case you may also round off the result. After rounding off your outcome is exactly: 25.13 lb = 0.45 kgs.
We are also going to show you 25.13 kg to how many pounds and 25.13 pound how many kg results in tables. Let’s see:
We are going to start with a table for how much is 25.13 kg equal to pound.
25.13 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
25.13 55.4021664406 55.2860
Now look at a chart for how many kilograms 25.13 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
25.13 0.45359237 0.45
Now you learned how many 25.13 kg to lbs and how many kilograms 25.13 pound, so it is time to move on to the 25.13 kg to lbs formula.
25.13 kg to pounds
To convert 25.13 kg to us lbs a formula is needed. We will show you two versions of a formula. Let’s start with the first one:
Number of kilograms * 2.20462262 = the 55.4021664406 result in pounds
The first formula will give you the most correct outcome. Sometimes even the smallest difference can be significant. So if you want to get an accurate outcome - this formula will be the best solution to know how many pounds are equivalent to 25.13 kilogram.
So let’s move on to the another version of a formula, which also enables conversions to learn how much 25.13 kilogram in pounds.
The shorter formula is as following, see:
Number of kilograms * 2.2 = the result in pounds
As you see, this formula is simpler. It could be the best option if you want to make a conversion of 25.13 kilogram to pounds in easy way, for instance, during shopping. You only need to remember that final outcome will be not so correct.
Now we want to learn you how to use these two formulas in practice. But before we will make a conversion of 25.13 kg to lbs we are going to show you another way to know 25.13 kg to how many lbs totally effortless.
25.13 kg to lbs converter
Another way to learn what is 25.13 kilogram equal to in pounds is to use 25.13 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. It is based on longer version of a formula which we gave you in the previous part of this article. Due to 25.13 kg pound calculator you can effortless convert 25.13 kg to lbs. You only have to enter number of kilograms which you need to calculate and click ‘calculate’ button. You will get the result in a second.
So let’s try to convert 25.13 kg into lbs using 25.13 kg vs pound converter. We entered 25.13 as a number of kilograms. Here is the outcome: 25.13 kilogram = 55.4021664406 pounds.
As you can see, our 25.13 kg vs lbs calculator is easy to use.
Now let’s move on to our chief issue - how to convert 25.13 kilograms to pounds on your own.
25.13 kg to lbs conversion
We will start 25.13 kilogram equals to how many pounds calculation with the first version of a formula to get the most exact result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 55.4021664406 the result in pounds
So what have you do to know how many pounds equal to 25.13 kilogram? Just multiply number of kilograms, this time 25.13, by 2.20462262. It is exactly 55.4021664406. So 25.13 kilogram is equal 55.4021664406.
It is also possible to round off this result, for instance, to two decimal places. It gives 2.20. So 25.13 kilogram = 55.2860 pounds.
It is high time for an example from everyday life. Let’s calculate 25.13 kg gold in pounds. So 25.13 kg equal to how many lbs? And again - multiply 25.13 by 2.20462262. It is equal 55.4021664406. So equivalent of 25.13 kilograms to pounds, when it comes to gold, is 55.4021664406.
In this example you can also round off the result. Here is the outcome after rounding off, in this case to one decimal place - 25.13 kilogram 55.286 pounds.
Now we can go to examples calculated with short formula.
How many 25.13 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 55.286 the outcome in pounds
So 25.13 kg equal to how much lbs? And again, you have to multiply amount of kilogram, this time 25.13, by 2.2. Have a look: 25.13 * 2.2 = 55.286. So 25.13 kilogram is equal 2.2 pounds.
Do another conversion using shorer version of a formula. Now convert something from everyday life, for example, 25.13 kg to lbs weight of strawberries.
So let’s convert - 25.13 kilogram of strawberries * 2.2 = 55.286 pounds of strawberries. So 25.13 kg to pound mass is equal 55.286.
If you learned how much is 25.13 kilogram weight in pounds and are able to calculate it with use of two different versions of a formula, we can move on. Now we are going to show you all results in charts.
Convert 25.13 kilogram to pounds
We are aware that outcomes presented in charts are so much clearer for most of you. We understand it, so we gathered all these outcomes in charts for your convenience. Thanks to this you can quickly compare 25.13 kg equivalent to lbs results.
Let’s begin with a 25.13 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
25.13 55.4021664406 55.2860
And now let’s see 25.13 kg equal pound chart for the second formula:
Kilograms Pounds
25.13 55.286
As you see, after rounding off, if it comes to how much 25.13 kilogram equals pounds, the results are not different. The bigger number the more significant difference. Remember it when you need to do bigger number than 25.13 kilograms pounds conversion.
How many kilograms 25.13 pound
Now you know how to convert 25.13 kilograms how much pounds but we want to show you something more. Do you want to know what it is? What do you say about 25.13 kilogram to pounds and ounces conversion?
We want to show you how you can calculate it step by step. Begin. How much is 25.13 kg in lbs and oz?
First things first - you need to multiply number of kilograms, in this case 25.13, by 2.20462262. So 25.13 * 2.20462262 = 55.4021664406. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To convert how much 25.13 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So your result is exactly 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then final result is equal 2 pounds and 33 ounces.
As you see, calculation 25.13 kilogram in pounds and ounces quite simply.
The last calculation which we are going to show you is conversion of 25.13 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert it it is needed another formula. Before we give you it, see:
• 25.13 kilograms meters = 7.23301385 foot pounds,
• 25.13 foot pounds = 0.13825495 kilograms meters.
Now look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to convert 25.13 foot pounds to kilograms meters you have to multiply 25.13 by 0.13825495. It is equal 0.13825495. So 25.13 foot pounds is 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 25.13 foot pounds will be exactly 0.14 kilogram meters.
We hope that this calculation was as easy as 25.13 kilogram into pounds conversions.
We showed you not only how to do a calculation 25.13 kilogram to metric pounds but also two another conversions - to know how many 25.13 kg in pounds and ounces and how many 25.13 foot pounds to kilograms meters.
We showed you also another solution to do 25.13 kilogram how many pounds calculations, this is with use of 25.13 kg en pound calculator. It will be the best solution for those of you who do not like calculating on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.
We hope that now all of you are able to make 25.13 kilogram equal to how many pounds calculation - on your own or using our 25.13 kgs to pounds converter.
It is time to make your move! Convert 25.13 kilogram mass to pounds in the way you like.
Do you want to do other than 25.13 kilogram as pounds calculation? For instance, for 10 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so easy as for 25.13 kilogram equal many pounds.
How much is 25.13 kg in pounds
We want to sum up this topic, that is how much is 25.13 kg in pounds , we gathered answers to the most frequently asked questions. Here you can see all you need to know about how much is 25.13 kg equal to lbs and how to convert 25.13 kg to lbs . Have a look.
What is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. Let’s see 25.13 kg to pound conversion formula . It is down below:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 25.13 kilogram to pounds? The exact result is 55.4021664406 lb.
You can also calculate how much 25.13 kilogram is equal to pounds with second, shortened version of the equation. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So in this case, 25.13 kg equal to how much lbs ? The answer is 55.4021664406 lbs.
How to convert 25.13 kg to lbs quicker and easier? It is possible to use the 25.13 kg to lbs converter , which will make whole mathematical operation for you and give you an exact result .
Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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# Aug 16: Statistics in Practice
Here in Part 2 of the Weekly Brief, we offer some tools to help you with the question, “what is the optimal set of alternatives to offer consumers?”
Our course spotlight is on:
See you in class!
Peter Bruce,
Chief Academic Officer, Author, Instructor, and Founder
The Institute for Statistics Education at Statistics.com
# Word of the Week
## Intervals (confidence, prediction and tolerance)
All students of statistics encounter confidence intervals. Confidence intervals tell you, roughly, the interval within which you can be, say, 95% confident that the true value of some sample statistic lies. This is not the precise technical definition, but […]
# Book Review
## Bandit Algorithms for Website Optimization by John Myles White
A classic statistical experimental design comparing treatments (two treatments, treatment versus control, multiple treatments) specifies a sample size, collection of data, then a decision, typically based on hypothesis-testing: the winning treatment must […]
# Job Spotlight
## Amazon
What’s it like to work at Amazon? Several years ago the New York Times published this unflattering portrait of a demanding and often idiosyncratic Darwinian work culture. Because it is an operating retail firm with a huge delivery infrastructure, most Amazon employees fall into the blue-collar category. A depiction of a single Amazon culture may not be adequate to account for both the data science “priesthood” and the operations workers.
What does Amazon itself say? This official Amazon blog speaks to the interview process, both for general white-collar roles and more technical roles. Two points stand out:
• No Powerpoints. Amazonians write and read short, well-written memos prior to meetings.
• Technical candidates do not need to be able to write their own operating system from scratch (whew!)
# Course Spotlight
## Aug 30 – Sep 27: Discrete Choice Modeling and Conjoint Analysis
Bandit algorithms (see above book review) are one effort to get around the limits of traditional statistical comparison tests, when decisions must be taken at scale, automatically. Discrete choice modeling is another such effort, used when the decision to be made is of limited scale and concerns product attributes. Complexity enters the picture because the ultimate decision concerns not a single best set of product attributes, but rather the best set of alternatives to offer consumers. There are many such alternatives – for example in wireless service you could have alternate options for price per month, price per additional line, length of contract, phone trade-in, data included, data speed, cost of additional data, throttling, international pricing, etc. Companies want to present consumers with a limited set of choices to facilitate decision-making; the question is “what is the optimal set of alternatives?”
Your instructor is Anthony Babinec, President of AB Analytics and expert in the application of advanced statistical methods to consumer decision-making. You will learn how to:
• Use designed experiments to collect data on how consumers view product choices
• Use that data to derive models on how attributes affect consumer preference
• Use a final model to predict how the market would choose among sets of product options
See you in class!
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# Y4 number and place value morning mats (double sided)
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PDF (Acrobat) Document File
327 KB|5 pages
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Product Description
Number and place value morning mats containing 7 questions on each mat covering the year 3 objectives. There is a mixture of fluency, reasoning and problem solving questions. There are 5 mats in this pack. These are aimed at children meeting age related expectations.
Answers are also included. In this version, the answer sheet is immediately after the question sheet. This means that if you want to print double sided so that children can easily self-check their work, it is simply a case of telling the printer to print double sided on the short edge.
Objectives covered:
count in multiples of 6, 7, 9, 25 and 1,000
find 1,000 more or less than a given number
count backwards through 0 to include negative numbers
recognise the place value of each digit in a four-digit number (1,000s, 100s, 10s, and 1s)
order and compare numbers beyond 1,000
identify, represent and estimate numbers using different representations
round any number to the nearest 10, 100 or 1,000
solve number and practical problems that involve all of the above and with increasingly large positive numbers
read Roman numerals to 100 (I to C) and know that over time, the numeral system changed to include the concept of 0 and place value
Total Pages
5 pages
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N/A
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\$3.30
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