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https://savvycalculator.com/escape-velocity-calculator/ | 1,713,531,442,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817398.21/warc/CC-MAIN-20240419110125-20240419140125-00658.warc.gz | 459,519,830 | 45,288 | # Escape Velocity Calculator
## About Escape Velocity Calculator (Formula)
An Escape Velocity Calculator is a tool used to compute the minimum velocity required for an object to break free from the gravitational influence of a celestial body, such as a planet or moon. This calculation is essential in space exploration and understanding the dynamics of celestial bodies.
Formula for Escape Velocity Calculation:
The formula for calculating escape velocity involves considering the gravitational constant (), the mass of the celestial body (), and the distance from the center of the body to the object’s initial position (). The formula is:
Escape Velocity = √(2 × G × M / r)
Where:
• Escape Velocity: The velocity required for an object to escape the gravitational pull of the celestial body.
• G: The gravitational constant.
• M: The mass of the celestial body.
• r: The distance from the center of the celestial body to the object’s initial position.
The escape velocity provides the necessary speed to overcome the gravitational attraction and move away from the celestial body indefinitely.
Applications:
1. Space Travel: Engineers and scientists use the Escape Velocity Calculator to determine the required speed for spacecraft to escape the Earth’s gravitational field.
2. Planetary Exploration: The calculator aids in understanding the conditions required for landing and launching from other planets, moons, or celestial bodies.
3. Satellite Deployment: The calculator is used to calculate the escape velocity needed to launch satellites and other objects into orbit.
4. Astrophysics: The concept of escape velocity is integral to studying the formation and dynamics of celestial bodies.
5. Gravitational Understanding: Calculating escape velocity helps in understanding the strength of a celestial body’s gravitational pull.
In summary, an Escape Velocity Calculator involves calculations that assist in determining the minimum velocity necessary for an object to overcome a celestial body’s gravitational pull, crucial for space missions, satellite launches, and celestial body analysis. | 372 | 2,110 | {"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} | 3.25 | 3 | CC-MAIN-2024-18 | longest | en | 0.793268 |
http://en.wikipedia.org/wiki/Fine_sheaf | 1,394,440,674,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010706290/warc/CC-MAIN-20140305091146-00089-ip-10-183-142-35.ec2.internal.warc.gz | 56,505,247 | 12,013 | # Injective sheaf
(Redirected from Fine sheaf)
In mathematics, injective sheaves of abelian groups are used to construct the resolutions needed to define sheaf cohomology (and other derived functors, such as sheaf Ext).
There is a further group of related concepts applied to sheaves: flabby (flasque in French), fine, soft (mou in French), acyclic. In the history of the subject they were introduced before the 1957 'Tohoku' paper of Alexander Grothendieck, which showed that the abelian category notion of injective object sufficed to found the theory. The other classes of sheaves are historically older notions. The abstract framework for defining cohomology and derived functors does not need them. However, in most concrete situations, resolutions by acyclic sheaves are often easier to construct. Acyclic sheaves therefore serve for computational purposes, for example the Leray spectral sequence.
## Injective sheaves
An injective sheaf F is just a sheaf that is an injective element of the category of abelian sheaves; in other words, homomorphisms from A to F can always be lifted to any sheaf B containing A.
The category of abelian sheaves has enough injective elements: this means that any sheaf is a subsheaf of an injective sheaf. This result of Grothendieck follows from the existence of a generator of the category (it can be written down explicitly, and is related to the subobject classifier). This is enough to show that right derived functors of any left exact functor exist and are unique up to canonical isomorphism.
For technical purposes, injective sheaves are usually superior to the other classes of sheaves mentioned above: they can do almost anything the other classes can do, and their theory is simpler and more general. In fact, injective sheaves are flabby (flasque), soft, and acyclic. However, there are situations where the other classes of sheaves occur naturally, and this is especially true in concrete computational situations.
The dual concept, projective sheaves, is not used much, because in a general category of sheaves there are not enough of them: not every sheaf is the quotient of a projective sheaf, and in particular projective resolutions do not always exist. This is the case, for example, when looking at the category of sheaves on projective space in the Zariski topology. This causes problems when attempting to define left derived functors of a right exact functor (such as Tor). This can sometimes be done by ad hoc means: for example, the left derived functors of Tor can be defined using a flat resolution rather than a projective one, but it takes some work to show that this is independent of the resolution. Not all categories of sheaves run into this problem; for instance, the category of sheaves on an affine scheme contains enough projectives.
## Acyclic sheaves
An acyclic sheaf F over X is one such that all higher sheaf cohomology groups vanish.
The cohomology groups of any sheaf can be calculated from any acyclic resolution of it (this goes by the name of De Rham-Weil theorem).
## Fine sheaves
A fine sheaf over X is one with "partitions of unity"; more precisely for any open cover of the space X we can find a family of homomorphisms from the sheaf to itself with sum 1 such that each homomorphism is 0 outside some element of the open cover.
Fine sheaves are usually only used over paracompact Hausdorff spaces X. Typical examples are the sheaf of continuous real functions over such a space, or smooth functions over a smooth (paracompact Hausdorff) manifold, or modules over these sheaves of rings.
Fine sheaves over paracompact Hausdorff spaces are soft and acyclic.
As an application, consider a real manifold X. There is the following resolution of the constant sheaf ℝ by the fine sheaves of (smooth) differential forms:
0 → ℝ → C0X → C1X → ... → Cdim XX → 0
This is a resolution, i.e. an exact complex of sheaves by the Poincaré lemma. The cohomology of X with values in ℝ can thus be computed as the cohomology of the complex of globally defined differential forms:
Hi(X, ℝ) = Hi(C·X(X)).
## Soft sheaves
A soft sheaf F over X is one such that any section over any closed subset of X can be extended to a global section.
Soft sheaves are acyclic over paracompact Hausdorff spaces.
## Flasque or flabby sheaves
A flasque sheaf (also called a flabby sheaf) is a sheaf $\mathcal{F}$ with the following property: if $X$ is the base topological space on which the sheaf is defined and
$U \subset V \subset X$
are open subsets, then the restriction map
$r_{U \subset V} : \Gamma(V, \mathcal{F}) \to \Gamma(U, \mathcal{F})$
is surjective, as a map of groups (rings, modules, etc.).
Flasque sheaves are useful because (by definition) sections of them extend. This means that they are some of the simplest sheaves to handle in terms of homological algebra. Any sheaf has a canonical embedding into the flasque sheaf of all possibly discontinuous sections of the étalé space, and by repeating this we can find a canonical flasque resolution for any sheaf. Flasque resolutions, that is, resolutions by means of flasque sheaves, are one approach to defining sheaf cohomology.
Flasque is a French word, that has sometimes been translated into English as flabby.
Flasque sheaves are soft and acyclic. | 1,254 | 5,302 | {"found_math": true, "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": 4, "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} | 2.6875 | 3 | CC-MAIN-2014-10 | longest | en | 0.943315 |
http://www.tinyepiphany.com/2011/12/galois-theory-in-1500-words.html | 1,534,811,905,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217901.91/warc/CC-MAIN-20180820234831-20180821014831-00702.warc.gz | 587,202,101 | 16,911 | Tuesday, December 13, 2011
Galois Theory in 1500 Words
[This is a very brief overview of Galois theory. It's not meant to be rigorous, so apologies in advance for leaving out a lot of details and avoiding some delicacies. It was meant to be intuitive and light on math, but it turned out to be neither. If you aren’t familiar with field and group theory, well... proceed at your own risk...]
For a long time, people wondered whether it is possible to write down something like the "quadratic formula" for cubic, quartic and quintic polynomials with integer coefficients. We now know that for cubic and quartic polynomials, this is possible. But for degree 5 polynomials and beyond, it isn't. A proof of this was scribbled down hastily by Galois the night before his duel. Galois linked together field theory and group theory in a beautiful way to answer this very question.
Galois’s Approach: The Big Idea
What does writing down a “formula” for roots of a polynomial really mean? For one, we’d be writing down the roots in terms of rational numbers and a combination of +, -, x, ÷, and radicals (taking n-th roots). This is a very limited set of operations, and certainly not all real numbers can be written this way -- π, for example, can’t be written this way. We say that π is not solvable in radicals.
Are the roots of polynomials with integer coefficients solvable in radicals? Those roots aren’t just any real number, and certainly π is not a root of any polynomial with integer coefficients. Yet Galois showed that there are some degree-5 polynomials with roots that are not solvable in radicals. To see how he did this, we first need some terminology about fields, field extensions, and groups.
Fields
The set of rational numbers Q is an example of a field: a set of things you can add, subtract, multiply and divide. We can “extend” Q into bigger fields by adjoining things to it. For example, L=Q(√2) -- pronounced "Q adjoined root two" -- is defined to be the smallest field that contains both Q and √2, and is closed under +, -, x, and ÷. Here elements of Q(√2) are exactly numbers that can be written in the form (a+b√2)/(c+d√2) where a, b, c, d are integers. We call such a field L a (field) extension over Q (written L/Q).
When we're adjoining √2 to Q to construct Q(√2), what we're really doing is adjoining to Q a root of the polynomial f(x)=x² -2. We could do this with other higher-degree polynomials. Let p(x) be a polynomial with integer coefficients (and no repeating roots). We define the splitting field K of p(x) to be the smallest field containing both Q and all the roots of p(x). For example, the splitting field of p(x)=x²+1 has roots i and -i so a splitting field K of p(x) is K=Q(i,-i)=Q(i). (The last equality is true because 0 and i are in Q(i), so -i=0-i is also in Q(i)).
Conversely, we call an extension K/Q a Galois extension if it is the splitting field of some polynomial p(x). From before, Q(i)/Q is a Galois extension.
Q-Fixing Automorphisms
An automorphism F of a field K is an isomorphism from K to itself, where the algebraic structure is preserved -- specifically, F(a+b)=F(a)+F(b), F(ab)=F(a)F(b). In the case that K is an extension of Q, we’re more interested in automorphisms of K that has F(x)=x for all x in Q (or that F fixes elements of Q). An automorphism F of K is a Q-fixing automorphism if it has this property.
There are two Q-fixing automorphisms of L=Q(√2): the identity automorphism (call it e) that takes each element of Q(√2) to itself, and an automorphism (call it t) that takes anything from Q to itself, and √2 to -√2. It is possible to show that there is exactly one such automorphism t.
Groups
A group is a set with a "composition" operation • with an identity element. From above, the set of Q-fixing automorphisms of K, denoted G(K/Q) is a group with • being function composition, and e being the identity element. Observe that we do not require • to be commutative (so a•b may not be the same as b•a in this case).
Two groups are isomorphic if they have the same algebraic structure -- i.e. they’re essentially the same group except the elements have different names. It is useful to see whether G(K/Q) is isomorphic to a group that we know and understand. Two important classes of groups that are well understood are cyclic groups and permutation groups.
Examples of Groups
The cyclic group C(n) of order n is the set {0,1,2,3,..,n-1}, with • being addition mod n. From the example of L=Q(√2), L has G(L/Q)={e,t} with • being function composition. This is actually isomorphic to C(2)={0, 1}.
Permutation groups consist of functions that permute some "letters". We'll use S(n) to denote the group of permutations of n letters. For example, S(3) is all the permutations of letters {a, b, c}. A function F that takes ab, ba, cc is one such permutation (and hence an element of S(3)).
The Fundamental Theorem of Galois Theory
Galois noticed that for a Galois extension K/Q, there is a link between "subfields" of K containing Q, and "subgroups" of G=G(K/Q). The quoted words mean exactly what you might think -- a subfield of K is a field L that is contained in K (and is closed under +, -, x, ÷). For example, Q is a subfield of R, and Q is a subfield of Q(√2). A subgroup of G is a group H that is contained in G (that is closed under • and contains the identity element). For example, the subset {0,2,4} is a subgroup of C(6) = {0,1,2,3,4,5}.
More concretely, there is a 1-1 correspondence between subfields of L and subgroups of H:
• For a subfield L of K containing Q, there is a subgroup H of G corresponding exactly to the automorphisms that fix all of L -- i.e. f(x)=x for all x in L, not just Q.
• The reverse is true as well: if H is a subgroup of G, then there is some subfield L of K containing Q that is fixed by all of H.
In particular, whenever L/Q is itself a Galois extension, H is a normal subgroup of G. Think of H being normal as H being well-behaved enough that the quotient G/H is another group. We won't get into what this means, but this group G/H turns out to be isomorphic to G(L/Q).
For a concrete example, let’s take K=Q(i, ∛2). It’s possible to show that K/Q is a Galois extension, and that G=G(K/Q) is isomorphic to S(3). In particular, the subfields of K are Q(i) and Q(∛2), and they correspond to subgroups of S(3) isomorphic to C(3) and S(2). We saw before that Q(i) is a Galois extension, and C(3) happens to be normal in S(3).
Linking Back to the Big Idea
Suppose p(x) is a degree 5 polynomial, and that it has a root x that is solvable in radicals. Then really, x is in some field K containing Q, where K can be "built up" from Q by successively adjoining (n√α), the n-th root of α, for some n and α. For example, take x=√(2+ √5). We set L=Q(√5) and K=L(√(2+√5)) to "build up" K in this manner.
Solvable Fields
In general, we call an extension K/Q solvable if K=K0⊇K1⊇K2⊇...⊇Q, where each K(i-1)=Ki(n√α) for some n and α. This is exactly the construction we had a paragraph ago. As another example, K=Q(i, ∛2) is a solvable extension since Q(i, ∛2)=Q(i)(∛2)⊇Q(i)⊇Q is in the desired form (recall i=√(-1)).
For a polynomial p(x) in Q, the splitting field K of p(x) is the smallest field containing all the roots of p(x), so the roots of p(x) are solvable in radicals if and only if K is solvable.
Solvable Groups
We can actually assume (ignoring some subtleties) that each K(i-1)/Ki from above is a Galois extension. In this case each G(K(i-1)/Ki) is actually isomorphic to C(n) for some n.
Thus in order for a field extension K/Q to be solvable, G=G(K/Q) must be in a particular form: there has to be a chain of subgroups, G=G0⊇G1⊇G2⊇...⊇{e}, where each Gi is a normal subgroup of G(i-1) and Gi/G(i-1)=C(n) for some n, and {e} is the trivial group with just the identity element. In the case of K=Q(i, ∛2), the chain of subgroups looks like G(K/Q)=S(3)⊇C(3)⊇{e}.
A Quintic Formula Cannot Exist
To recap, roots of p(x) being solvable in radicals requires the splitting field K of p(x) to be a solvable field, which in turn requires G(K/Q) to be a solvable group.
But with a little group theory, we can show that S(5) is not solvable. Further, any quintic polynomial with two non-real roots has Galois group S(5). These last facts require some more concepts to develop, but in any case -- not all roots of quintic polynomials are solvable in radicals.
End of Entry | 2,306 | 8,403 | {"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} | 3.625 | 4 | CC-MAIN-2018-34 | latest | en | 0.939311 |
http://www.cfd-online.com/Forums/ansys/128519-two-way-fluid-structure-interaction-flexible-plate.html | 1,410,852,616,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657114105.77/warc/CC-MAIN-20140914011154-00160-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 406,708,990 | 16,421 | # Two way fluid structure interaction of flexible plate
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January 14, 2014, 11:35 Two way fluid structure interaction of flexible plate #1 New Member sree Join Date: Jan 2014 Posts: 13 Rep Power: 2 Hi all, I am trying to perform a two-way fluid structure interaction of a flexible plate under the effect of fluid flow over it.I am doing the problem in ANSYS WORKBENCH 14.0 by coupling CFX and TRANSIENT STRUCTURAL systems.I have even solved a problem in which, first the structure is excited by a force and then its effect on fluid around it.Here the pressure applied on structure was known and hence that was not much hard. But my actual problem is how to transfer the loads exerted by the flowing fluid to the structure,!!!???...I dont know how to transfer the values of fluid forces exerted by flow to the structure...!! Please help me on this problem.. Sree
January 22, 2014, 05:03 #2 New Member Join Date: May 2013 Posts: 12 Rep Power: 3 Describe your problem some more. Is the flow parallel to the plate (grazing flow) or is it perpendicular (impinging on the plate) is the flow compressible? What are the boundary conditions? I am solving a similar problem using fluent and may be able to help.
February 3, 2014, 12:51 #3 New Member sree Join Date: Jan 2014 Posts: 13 Rep Power: 2 Hi levivad, First of all thanks for responding to my problem. In my problem the flexible wing is fixed at the bottom and flow is over the plate perpendicular to the plate. I am trying to do the problem in a velocity of 5-10 m/s.I have taken inlet velocity boundary condition and pressure outlet=0 Pa . I would like to calculate the force exerted by the perpendicular flow over the plate. Thank you, Sree.
February 4, 2014, 03:04
#4
New Member
sree
Join Date: Jan 2014
Posts: 13
Rep Power: 2
Fluid domain for my problem is attached here..the plate is fixed at bottom at the centre of the domain as shown..
Attached Images
model view.png (17.5 KB, 12 views)
February 5, 2014, 12:05 #5 New Member Join Date: May 2013 Posts: 12 Rep Power: 3 well this is basically the same as the tutorial in CFX. Follow tutorial 23 in the Ansys CFX help. you only need to change the inlet velocity. if you want the total force on the plate you can add a monitor on drag and lift. alternatively you can compute it in CFD-post
April 28, 2014, 02:42 #6 New Member sree Join Date: Jan 2014 Posts: 13 Rep Power: 2 thank you levivad..
April 30, 2014, 04:34 #7 New Member sree Join Date: Jan 2014 Posts: 13 Rep Power: 2
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All times are GMT -4. The time now is 03:30. | 871 | 3,204 | {"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} | 2.78125 | 3 | CC-MAIN-2014-41 | longest | en | 0.933935 |
http://www.adras.com/Formula-Help.t80188-13.html | 1,638,943,225,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363445.41/warc/CC-MAIN-20211208053135-20211208083135-00530.warc.gz | 86,111,890 | 2,740 | From: mahlandj on 1 Mar 2010 18:27 I need to find how many times Dan occurs in a certain month. Example Below. A B 1 1/2/10 Dan 2 1/5/10 Dan 3 1/10/10 Perry 4 2/3/10 Perry 5 2/7/10 Dan 6 2/12/10 Perry Answer = 2 for January & 1 for February -- Thanks! From: Teethless mama on 1 Mar 2010 19:02 PIVOT table is the right tool for this. "mahlandj" wrote: > I need to find how many times Dan occurs in a certain month. Example Below. > > A B > 1 1/2/10 Dan > 2 1/5/10 Dan > 3 1/10/10 Perry > 4 2/3/10 Perry > 5 2/7/10 Dan > 6 2/12/10 Perry > > Answer = 2 for January & 1 for February > > -- > Thanks! From: mahlandj on 2 Mar 2010 18:09 Thanks for your help. I was looking for a different way to show the info than the PIVOT table. Maybe I wasn't doing it correctly but I used this formula instead and it worked great. =SUMPRODUCT((MONTH(A1:A6)=1)*(B1:B6="Dan")) - This was for Jan results. I just changed the 1 to a 2 for Feb and then to 3 for March....4......5......6 -- Thanks! "Teethless mama" wrote: > PIVOT table is the right tool for this. > > > "mahlandj" wrote: > > > I need to find how many times Dan occurs in a certain month. Example Below. > > > > A B > > 1 1/2/10 Dan > > 2 1/5/10 Dan > > 3 1/10/10 Perry > > 4 2/3/10 Perry > > 5 2/7/10 Dan > > 6 2/12/10 Perry > > > > Answer = 2 for January & 1 for February > > > > -- > > Thanks! | Pages: 1 Prev: Unique Random ListNext: Copy value according to date | 516 | 1,411 | {"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} | 2.65625 | 3 | CC-MAIN-2021-49 | latest | en | 0.885452 |
https://resources.quizalize.com/view/quiz/unit-5-test-review-24f45ab1-9a86-4f90-8cb8-c4dcb3d919a8 | 1,674,939,305,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00188.warc.gz | 507,237,707 | 20,047 | Unit 5 Test Review
Quiz by Chandler Anderson
Feel free to use or edit a copy
includes Teacher and Student dashboards
Measure skillsfrom any curriculum
Tag the questions with any skills you have. Your dashboard will track each student's mastery of each skill.
• edit the questions
• save a copy for later
• start a class game
• view complete results in the Gradebook and Mastery Dashboards
• automatically assign follow-up activities based on students’ scores
• assign as homework
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• print as a bubble sheet
20 questions
• Q1
Mr. Anderson used similar triangles to make a design. Which statement about triangles is true?
They have corresponding sides that are congruent.
They have corresponding angles that are congruent.
They are the same size but different shapes.
They are the same size and shape.
30s
Edit
Delete
• Q2
What is the corresponding number to side DC?
3.2
4
6
8
30s
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• Q3
What is the corresponding number to side FG?
8
2.4
3
6
30s
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Delete
• Q4
What is the missing length for side QP?
6
14
8
48
30s
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• Q5
According to the similar figures, side length TQ and _________ are corresponding.
RS
TR
RQ
TV
30s
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Delete
• Q6
According to the similar figures, side length VQ and _____________ are corresponding.
SQ
TV
RS
VS
30s
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Delete
• Q7
According to the similar figures, side length RS and _________ are corresponding.
TQ
QS
TV
QR
30s
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Delete
• Q8
According to the similar figures, Angle A and Angle ________ are congruent.
D
E
B
F
30s
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Delete
• Q9
According to the similar figures, Angle D and Angle _______ are congruent.
B
C
A
E
30s
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Delete
• Q10
According to the similar figures, Side Length AB and Side Length ____________ are proportional and corresponding.
DF
ED
FE
30s
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Delete
• Q11
According to the similar figures, Side Lengths AB and __________ are corresponding.
DF
EF
DE
30s
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Delete
• Q12
According to the similar figures, Side Lengths DF and _________ are corresponding.
AB
AC
BC
EF
30s
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Delete
• Q13
According to the similar figures, Side Length AC and Side Length __________ are proportional and corresponding.
ED
EF
DF
30s
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Delete
• Q14
What is the value of x in centimeters? (Reminder: Find the scale factor (new divided by original) first! Then multiply to get x).
30 cm
10.8 cm
8 cm
22.5 cm
30s
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• Q15
Jamie was looking at a blueprint of her apartment. The scale on the blueprint stated 3/4 inch = 1 foot. On the blueprint, what are the dimensions of her living room if the actual dimensions of her living room are 12 feet by 16 feet? (Use a big H for the length, and a big H for the width!)
16 inches by 21.3 inches
.75 inches by 1 inch
12 inches by 16 inches
9 inches by 12 inches
30s
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• Q16
Juliet was looking at a blueprint of her classroom. The scale on the blueprint stated that 1/4 inch = 1 foot. On the blueprint, what are the dimensions of her classroom if the actual dimensions of her classroom are 10 feet by 14 feet? (Use a big H for the length, and a big H for the width!_
.25 inches by 1 inch
2.5 inches by 3.5 inches
10 inches by 14 inches
40 inches by 56 inches
30s
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• Q17
Above is a scale drawing of an actual playhouse in Ms. Anderson's backyard. What is the actual TOTAL height of the playhouse in feet? (Use the big H!)
28 inches
14 feet
7 feet
28 feet
30s
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• Q18
What is the length of the missing side, DE?
4 inches
5.5 inches
6 inches
10 inches
30s
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• Q19
What is the missing length of PQ? (First, find the scale factor...then multiply!)
40 in.
38 in.
34 in.
14 in.
30s
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• Q20
The dimensions of a rectangular photograph are 6 inches by 4 inches (length by width). If the photograph is enlarged proportionally so that the length is 18.6 inches what would the width be?
1.29 inches
3.1 inches
12.4 inches
24 inches
30s
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Teachers give this quiz to your class | 1,098 | 3,883 | {"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} | 3.640625 | 4 | CC-MAIN-2023-06 | longest | en | 0.916028 |
https://uscollegeresearch.com/exer-27-44-find-an-equation-in-x-and-y-that-has-the-same-graph-as-the-polar-equation-use-it/ | 1,674,851,949,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495012.84/warc/CC-MAIN-20230127195946-20230127225946-00534.warc.gz | 563,930,802 | 9,646 | # Exer. 27 – 44: Find an equation in x and y that has the same graph as the polar equation. Use it…
Exer. 27 – 44: Find an equation in x and y that has the same graph as the polar equation. Use it…
Exer. 27 – 44: Find an equation in x and y that has the same graph as the polar equation. Use it to help sketch the graph in an ru-plane.
4
7 (6, arctan 3 )
27 r cos = 5
28 r sin
= -2
3
8 (10, arccos (- 1 ))
29 r – 6 sin = 0
30 r = 2
31 =
1T/4 58 r = 1 + 2 cos
32 r = 4 sec
59 r = 23 – 2 sin
33 r 2(4 sin2
– 9 cos2
) = 36
60 r = 223 – 4 cos
34 r 2(cos2
+ 4 sin2
) = 16
61 r = 2 – cos
35 r 2 cos 2 = 1
62 r = 5 + 3 sin
36 r 2 sin 2 = 4
63 r = 4 csc
37 r(sin
– 2 cos
) = 6
64 r = -3 sec
38 r(3 cos
39 r(sin
– 4 sin
+ r cos2
) = 12
) = 1
65 r = 8 cos 3
66 r = 2 sin 4
40 r(r sin2
– cos
) = 3
67 r = 3 sin 2
41 r = 8 sin
– 2 cos
68 r = 8 cos 5
2
42 r = 2 cos
43 r = tan
– 4 sin
69 r
70 r 2
= 4 cos 2
= -16 sin 2
(lemniscate)
44 r = 6 cot | 449 | 963 | {"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} | 3.71875 | 4 | CC-MAIN-2023-06 | latest | en | 0.711215 |
https://meta.stackoverflow.com/questions/354024/is-the-election-unfair-or-am-i-wrong/354529 | 1,716,548,112,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00768.warc.gz | 346,977,657 | 28,503 | # Is the election unfair, or am I wrong?
Due to Arrow's impossibility theorem, the Stack Overflow moderator election should be unfair since voters need to order three different candidates by ranking. Is this a flaw in the election, or am I applying Arrow's theorem wrong?
• I don't really see what your point is. A quote from that very page that is highly relevant: "Most systems are not going to work badly all of the time. All I proved is that all can work badly at times." If you're looking for fairness in an election, you're never gonna find it.
– animuson StaffMod
Jul 28, 2017 at 23:35
• This is ridiculously theoretical of a concern to have, and stated in a vacuum means nothing. We could even dispute the assumptions for the theorem like "complete and transitive voter preference". Jul 29, 2017 at 0:04
• The problem with Arrow's theorem is that while it points out the mathematical unfairness of every voting system, at the end, the real problem are the humans voting. Jul 29, 2017 at 1:44
• Do you have a suggestion of a better system to use for the election? Jul 29, 2017 at 10:51
• Not sure about if it's a better system but the "Thunderdome" would make a wonderfully entertaining one... :p (plus good practice for mod message replies :p) Jul 29, 2017 at 14:33
• I didn't realize that it didn't apply here... sorry. Jul 29, 2017 at 17:34
• I vote for a hunger-games style election
– user4639281
Jul 29, 2017 at 18:06
• Oh no, now we need to vote for the kind of election... Jul 29, 2017 at 19:45
• The theorem describes several features that cannot coexist, none of which is "fairness". Aug 1, 2017 at 20:37
• You're applying Arrow's Impossibility Theorem wrong. `when voters have three or more distinct alternatives` isn't referring to being able to select 3 at a time, it's talking about the whole problem of having 3 or more people who are possible to elect; Like if A, B, and C are all running for president, and A is similar to B, C will win because the votes for A and B will be split, even if A would have won in a A vs. C election. Now... that does still apply to this situation, just not how you think. We have many candidates, so giving 3 votes to every person is actually a good way to handle it. Aug 2, 2017 at 3:10
• This is not opinion-based, it is an objective thing, see @DavyM 's comment above! I voted to reopen. Aug 2, 2017 at 6:25
• Let people be divided into three political camps of equal voting power. Camp 1 prefers A>B>C, Camp 2 prefers B>C>A and Camp 3 prefers C>A>B. Now choose one of A, B, C, however you like, that can not be replaced by majority rule. Go on. I dare you. Condorcet dares you. Pedantic question: Does this mean that nothing would be acceptable to a majority? or does it just mean voting has some mathematically "interesting" properties.
– Paul
Aug 2, 2017 at 13:00
• @DavyM The theorm is specifically referring to a voting mechanism where you rank the candidates (which is how SO ran it's election) so it does apply in that sense. It states that you can't get all of the properties mentioned unless voters provide more than just a ranking of every single candidate in the order they'd like to see them elected, and SO's election didn't even let them do that much (SO only let 3 of the 10 be ranked). Aug 2, 2017 at 13:44
• @Servy That's exactly right; I hit the character limit in my previous comment and thought that my post would convey it but now I'm seeing I didn't explain myself well. 3 of the many be ranked is still a fallacy. It is also important to know, can you address how much higher your idea of 1st is over 2nd? Perhaps for one person the 1st and 2nd choices are about the same, and for another, 1st is by far the best and 2nd just happens to not be the worst, so the whole story isn't told. Yet these are worst in one-winner situations; with our 2 winners, 3 votes helps keep things more even in my opinion. Aug 2, 2017 at 15:19
• xkcd.com/1844 Aug 2, 2017 at 20:52
Is this a flaw in the election, or am I applying Arrow's theorem wrong?
Depends on how exactly you define flaw. Arrow’s theorem says essentially that there is no completely fair¹ rank-based voting system. However, this does not mean that true rank-based voting systems² are bad in comparison to other voting systems, as:
• Rank-based voting system in the sense of Arrow’s theorem comprises a lot, including plain majority voting.
• Alternative systems are usually much worse on a level that isn’t even assessed by Arrow’s criteria, e.g., you have indirect voting or voters are not considered equal to begin with.
Thus, if you so wish, the result of Arrow’s theorem is that there cannot be any completely fair¹ voting system at all (unless you have only two choices, which does not apply here). In this sense, all voting systems are inevitably flawed – including Stack Overflow’s.
However, being rank-based is not a flaw of Stack Overflow’s voting system per se. True rank-based systems² are much less flawed than, e.g., plain majority voting and the STV system used by Stack Overflow is arguably very well suited for its purpose. It has flaws (such as allowing us to rank only three candidates) and of course it suffers from the inevitable problems due to Arrow’s theorem, but being rank-based itself is not a flaw – it’s an asset.
¹ according to some simple, reasonable criteria
² i.e., voting systems which actually take into account all ranks, and do not just discard all but the first choice of each voter
• So what you're saying is that even though this voting system isn't perfect, it's the best we can possibly do? Aug 5, 2017 at 21:03
• @u8y7541: Mostly. As I said in the last sentence, there are aspect that could be improved, but that’s the icing on the cake, not a fundamental issue. Aug 5, 2017 at 21:30
• It would indeed be nice to reduce the number of "exhausted" votes, for instance in the last election there was a victory by some 2k votes, but there were 6k exhausted votes. opavote.com/results/5927932925050880/0#anchor-8
– Nemo
Aug 6, 2017 at 13:13
• @Nemo: Or one should change his/her name to "exhaused" before running :) Aug 6, 2017 at 19:01 | 1,581 | 6,112 | {"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} | 2.859375 | 3 | CC-MAIN-2024-22 | latest | en | 0.972058 |
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# How many odd three-digit integers greater than 800 are there
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How many odd three-digit integers greater than 800 are there [#permalink] 31 Oct 2008, 07:49
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How many odd three-digit integers greater than 800 are there such that all their digits are different?
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Re: three-digit integers [#permalink] 31 Oct 2008, 10:05
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That's a tough one. I get 68...
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Re: three-digit integers [#permalink] 31 Oct 2008, 10:48
72 for me
the Hundreds digit can be 8,9
Units can be 1,3,5,7,9
middle can be any of the 10
case 1 :
Hundreds digit = 8
Units can be any 5 digits
middle one can be any of the 8 digits apart from 8 & (any one of 1,3,5,7,9)
Total = 8*5=40
case 2 :
Hundreds digit = 9
Units can be any 4 digits (exclude 9)
middle one can be any of the 8 digits apart from 9 & (any one of 1,3,5,7)
Total = 8*4=32
Total no. of ways = 72
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Re: three-digit integers [#permalink] 31 Oct 2008, 13:05
From 800-899= Consider the tenth and unit position 1 (8)1 X even X odd= 1x4x5=20
Similarly =(8)1x odd x other odd=1x5x4=20
From 899-999= (9) 1x even x odd=1x5x4=20
Similarly=(9)1x oddx odd= 1x4x3=12 (because 9 is already used in the hundredth position.
So the total number of digits are=20+20+20+12=72
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Re: three-digit integers [#permalink] 01 Nov 2008, 06:56
subarao wrote:
From 800-899= Consider the tenth and unit position 1 (8)1 X even X odd= 1x4x5=20
Similarly =(8)1x odd x other odd=1x5x4=20
From 899-999= (9) 1x even x odd=1x5x4=20
Similarly=(9)1x oddx odd= 1x4x3=12 (because 9 is already used in the hundredth position.
So the total number of digits are=20+20+20+12=72
good one subarao
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Re: three-digit integers [#permalink] 01 Nov 2008, 06:56
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# (r+t)/(r-t)>0. r>t? (1) t>0 (2) r>0
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(r+t)/(r-t)>0. r>t?
(1) t>0
(2) r>0
Last edited by sweetlife on 17 Jun 2004, 04:14, edited 1 time in total.
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16 Jun 2004, 23:56
Answer is C. Combing both we wud get to know that r>t.
If r is less that t then the denominator will result in negetive value, which in turn wud not satisfy more than 0.
Regards
Correct me if i am wrong.
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17 Jun 2004, 02:11
It says the answer is B and it doesnt convince me.
I am seeking the answer myself. It looks like C.
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17 Jun 2004, 03:46
(r+t)/(r-t)>0. r>t?
(1) t>0
(2) r>0
Pardon me for the wrong answer given last time. The answer is indeed B.
Only in the case of R>0 would the equation (r+t)/(r-t)>0. r>t be satisfied. If r is less than 0, then the equation may result is either case. If R>0, then if t is more than r then, the original equation will not be satisfied. I had checked this with number subs .. for any value of R more than 0, the equation to be valid, t shoud be less than r.
Sorry for causing the confusion.
Regards
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24 Jun 2004, 09:48
sweetlife wrote:
Thanks. B it is...
This is probably a case where I just don't see the obviuos answer, but I think it's D:
1) t >0
let's say t=4, then we have
(r+4)/(r-4)>0
r has to be greater than 4, b/c otherwise the answer would yield a negative answer.
Say r is 3, then we have 7/-1=-7 which is not greater than 0
what am I missing?
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26 Jun 2004, 07:49
Hi Lastocka
You have made a mistake here. If r is less than t, then the original equation does not stand good. Taking your case, you have mentioned, if r=3, then the orginal equation (r+4)/(r-4)>0 will be less than 0. But we are not suppose to change the orginal equation in here.
I hope it is clear now.
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27 Jun 2004, 00:00
sweetlife wrote:
(r+t)/(r-t)>0. r>t?
(1) t>0
(2) r>0
(r+t)/(r-t) = 1 + 2t/(r-t) > 0 => t > 0, r = -1.01*t. => 1 is not sufficient.
2 is sufficient because from (r+t)/(r-t) > 0 and r > 0 => -r < t < r.
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27 Jun 2004, 12:17
sweetlife wrote:
(r+t)/(r-t)>0. r>t?
(1) t>0
(2) r>0
I have a different answer, I will go with A
If t>0, take as an example t=1 and r=3. in this case the result will be (3+1)/(3-1) =2>0 sufficient
But if r>0 this do not mean that t must be positive it could be any number less than r. for example, r=2 t=-4
(2-4)/(2+4)=-2/6 . In other cases the result will be poistive. hence, Insufficient.
Correct me if I am wrong.
good luck all
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27 Jun 2004, 21:57
dr_sabr wrote:
sweetlife wrote:
(r+t)/(r-t)>0. r>t?
(1) t>0
(2) r>0
I have a different answer, I will go with A
If t>0, take as an example t=1 and r=3. in this case the result will be (3+1)/(3-1) =2>0 sufficient
But if r>0 this do not mean that t must be positive it could be any number less than r. for example, r=2 t=-4
(2-4)/(2+4)=-2/6 . In other cases the result will be poistive. hence, Insufficient.
Correct me if I am wrong.
good luck all
No, dr_sabr, it is wrong. Look, the stem SHOULD be always true, and in your example it is not satisfied! But it should be!
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28 Jun 2004, 01:39
I didn't understand what you meant
I will take 1) first.
If t>0, take as an example t=1 and r=3. in this case the result will be (3+1)/(3-1) =2>0 sufficient. because whatever is t, r must be >t.
2) But if r>0 this do not mean that t must be positive it could be any number less than r. for example, r=2 t=-4 =>(2-4)/(2+4)=-2/6 . In another case if r=3 t=2 =>(3+2)/(3-2)=5>0 . hence, Insufficient
Correct me if I am mistaken .
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28 Jun 2004, 01:55
dr_sabr wrote:
2) But if r>0 this do not mean that t must be positive it could be any number less than r.
NOT any number less than r, but any number, for which the STEM is true! i.e., for which (r+t)/(r-t) > 0. But in your example this is not the case: r = 2, t = -4 and (r+t)/(r-t) = -1/3 < 0 and this contradicts the stem!!!
dr_sabr wrote:
for example, r=2 t=-4 =>(2-4)/(2+4)=-2/6 . In another case if r=3 t=2 =>(3+2)/(3-2)=5>0 . hence, Insufficient
Correct me if I am mistaken .
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28 Jun 2004, 03:25
Emmanuel wrote:
But in your example this is not the case: r = 2, t = -4 and (r+t)/(r-t) = -1/3 < 0 and this contradicts the stem!!!
I think that I misunderstood this question, I thought the stem was r>t . Thank you Emmanuel for correcting me. I hope I don't make this mistake on the test.
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28 Jun 2004, 03:37
dr_sabr wrote:
Emmanuel wrote:
But in your example this is not the case: r = 2, t = -4 and (r+t)/(r-t) = -1/3 < 0 and this contradicts the stem!!!
I think that I misunderstood this question, I thought the stem was r>t . Thank you Emmanuel for correcting me. I hope I don't make this mistake on the test.
I hope that you will do such questions easily on the test, once you have seen what problems can arise here.
28 Jun 2004, 03:37
Display posts from previous: Sort by | 2,529 | 7,638 | {"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} | 3.5625 | 4 | CC-MAIN-2017-04 | latest | en | 0.854722 |
https://mathgoodies.com/multiplication-word-problem-worksheets/ | 1,719,098,636,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00114.warc.gz | 336,165,966 | 11,458 | # Multiplication Word Problem Worksheets
Multiplication word problems are math exercises where the student must use multiplication to solve a problem described in text form. These problems are typically written as short stories or scenarios that require the student to identify the numbers involved and determine how to apply multiplication to reach a solution. Unlike straightforward multiplication questions that present only the numbers to be multiplied, word problems incorporate narrative elements that require interpretation and critical thinking.
For instance, a multiplication word problem might describe a situation where someone needs to determine the total number of items based on the number of groups and the number of items in each group. The problem might say, “There are 8 boxes with 5 apples in each box. How many apples are there in total?” The student must recognize that the problem requires them to multiply the number of boxes (8) by the number of apples per box (5) to find the total number of apples.
This worksheet presents a collection of multiplication word problems that require students to use arithmetic to solve practical questions involving items like stickers, crayons, books, and candies. Each problem describes a situation in which a quantity of objects is grouped into sets, and the student must determine the total number of items by multiplying the number of groups by the number of items in each group. The scenarios are relatable and varied, covering topics from shopping to organizing a bookshelf, to engage students with real-world applications of multiplication.
The worksheet is designed to teach students how to apply multiplication in various everyday situations, thus reinforcing their understanding of how multiplication represents repeated addition. It aims to enhance problem-solving skills by requiring students to read, interpret, and solve the problems through multiplication. Additionally, it encourages the development of mathematical reasoning by translating textual information into numerical calculations and helps students practice the multiplication tables in a contextual setting, making the abstract concept of multiplication concrete and understandable.
This worksheet is a set of multiplication word problems designed in a multiple-choice question format, aimed at helping students practice their multiplication skills. The problems involve various everyday situations that require multiplying two numbers to find the total count of items, such as crayons in boxes or marbles in jars. Each question presents a scenario and provides four possible answers, from which the student must choose the correct one.
The purpose of this worksheet is to reinforce students’ multiplication skills through practical application in word problem scenarios, enhancing their ability to convert a textual description into a mathematical equation. It serves to improve their critical thinking as they analyze the problem to identify the correct numbers to multiply. The multiple-choice format also helps in assessing the students’ understanding of multiplication and their ability to perform mental calculations or estimate to find the right answer among the options given.
This worksheet provides a series of multiplication word problems that simulate everyday situations, such as calculating the number of books in bookstore orders, determining travel times for trains, and preparing recipes in the kitchen. The questions are designed to challenge students to apply their multiplication skills to figure out quantities needed for completing tasks or to understand quantities in given scenarios. Each problem requires the student to identify the key numbers, multiply them, and come up with a single correct answer.
The worksheet aims to teach students the practical applications of multiplication in various real-life contexts, from business and travel to farming and education. It is intended to strengthen their ability to analyze textual information, extract relevant numerical data, and use multiplication to solve problems. Furthermore, the problems are crafted to enhance students’ numerical literacy and to provide practice in performing multiplication without relying on a calculator, thereby improving their mental arithmetic skills.
Solving multiplication word problems involves several steps to understand the problem, set up an equation, and then solve it. Here’s a detailed breakdown of the process, along with two examples and their complete step-by-step solutions:
Step 1: Read and Understand the Problem
Carefully read the word problem to understand the situation, the information given, and the question being asked. Identify key information, such as numbers, quantities, and relationships between them. Pay attention to units of measurement or any other relevant details.
Step 2: Identify What You Need to Find
Determine what the problem is asking you to find. Is it asking for a total, a product, or a specific value related to multiplication?
Step 3: Define Variables (if necessary)
If the problem doesn’t provide specific variable names, define them yourself. For example, you can use “x” or any other letter to represent unknown quantities.
Step 4: Set Up an Equation
Translate the information from the word problem into a mathematical equation using the multiplication operation.
Use the variables you defined in Step 3 if needed.
Make sure your equation accurately represents the relationship described in the problem.
Step 5: Solve the Equation
Solve the equation for the unknown quantity (usually represented by the variable you defined).
Use appropriate mathematical operations to isolate the variable on one side of the equation.
Verify that your solution makes sense in the context of the problem.
Example 1: Sarah wants to buy 5 packs of chocolates, and each pack contains 8 chocolates. How many chocolates will she have in total?
Step 1: Read and Understand the Problem
Sarah wants to buy 5 packs of chocolates, and each pack contains 8 chocolates.
Step 2: Identify What You Need to Find
We need to find the total number of chocolates Sarah will have.
Step 3: Define Variables (if necessary)
No variables need to be defined in this problem.
Step 4: Set Up an Equation
The total number of chocolates is found by multiplying the number of packs (5) by the number of chocolates per pack (8).
Equation: Total Chocolates = 5 x 8
Step 5: Solve the Equation
Total Chocolates = 5 x 8 = 40
Sarah will have a total of 40 chocolates. This makes sense in the context of the problem, as each pack has 8 chocolates, and she’s buying 5 packs.
Example 2: A box contains 12 packs of pencils, and each pack has 24 pencils. How many pencils are there in the box?
Step 1: Read and Understand the Problem
The box contains 12 packs of pencils, and each pack has 24 pencils.
Step 2: Identify What You Need to Find
We need to find the total number of pencils in the box.
Step 3: Define Variables (if necessary)
No variables need to be defined in this problem.
Step 4: Set Up an Equation
The total number of pencils is found by multiplying the number of packs (12) by the number of pencils per pack (24).
Equation: Total Pencils = 12 x 24
Step 5: Solve the Equation
Total Pencils = 12 x 24 = 288 | 1,375 | 7,291 | {"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} | 3.859375 | 4 | CC-MAIN-2024-26 | latest | en | 0.925783 |
https://en.4hw.com.cn/646/22770.html | 1,624,005,863,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487635920.39/warc/CC-MAIN-20210618073932-20210618103932-00087.warc.gz | 219,967,422 | 5,597 | How about the income of yu'ebao? What is the annual income of yu'ebao?_Sihai network AMP
# How about the income of yu'ebao? What is the annual income of yu'ebao?
For many financial managers, will they put some idle money in yu'ebao? Alipay has a function of yu'ebao, so long as the idle money is put into yu'ebao, you can get income. It is because of this amazing income, comparable to the bank, that many young people favor it.
Do you know how to calculate the income of yu'ebao? If yu'ebao deposits 1, how much can it earn in case of one day? How to calculate the income of yu'ebao 10000? It is reported that yu'ebao calculates the income once a day, and the yield is about 4.7%. Take 10000 yuan as an example, deposit yu'ebao, and the daily income is about 1.3 yuan. Calculation method: 10000 & times; 4.7% & times; 1 day / 360 days = 1.31 yuan.
Is the interest of yu'ebao high? Remember: yu'ebao is not included in the income on the day of deposit. It takes 15:00 as a time limit, and the amount transferred in after 15:00 every day is counted as the entry amount of the next day, and it also needs a working day to review. Here, it is a working day. Saturday and Sunday holidays are not included.
How much can yu'ebao save? Yu'ebao is not only a tool for income, but also convenient for shopping. It can be said that yu'ebao has a PayPal balance that can benefit. It can be used and taken at any time, without any impact. Yu'e Bao has an automatic transfer in function. You can open this function, which can be used and taken at any time, and can make the capital increase in value.
Is yu'e Bao really safe? Yu'e Bao is essentially a kind of fund. In other words, you put your money in the name of Tianhong fund, which will take care of it on behalf of you, and the profit part will give you corresponding share. In fact, investment is risky. Xiaobian doesn't recommend putting all the eggs in one basket, so he doesn't recommend putting all the money on yu'ebao. | 524 | 1,976 | {"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} | 2.59375 | 3 | CC-MAIN-2021-25 | latest | en | 0.942908 |
http://gutenberg.us/articles/eng/Chromosome_(genetic_algorithm) | 1,597,093,238,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738699.68/warc/CC-MAIN-20200810205824-20200810235824-00171.warc.gz | 48,564,460 | 17,722 | #jsDisabledContent { display:none; } My Account | Register | Help
# Chromosome (genetic algorithm)
Article Id: WHEBN0000555480
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Title: Chromosome (genetic algorithm) Author: World Heritage Encyclopedia Language: English Subject: Collection: Genetic Algorithms Publisher: World Heritage Encyclopedia Publication Date:
### Chromosome (genetic algorithm)
In genetic algorithms, a chromosome (also sometimes called a genotype) is a set of parameters which define a proposed solution to the problem that the genetic algorithm is trying to solve. The set of all solutions is known as the population.[1] The chromosome is often represented as a binary string, although a wide variety of other data structures are also used.
## Contents
• Chromosome design 1
• Example 1: binary representation 1.1
• Example 2: string representation 1.2
• Selection, crossover and mutation 2
• References 3
## Chromosome design
The design of the chromosome and its parameters is by necessity specific to the problem to be solved. Traditionally, chromosomes are represented in binary as strings of 0s and 1s, however other encodings are also possible;[2] almost any representation which allows the solution to be represented as a finite-length string can be used.[3] Finding a suitable representation of the problem domain for a chromosome is an important consideration, as a good representation will make the search easier by limiting the search space; similarly, a poorer representation will allow a larger search space.[4] The mutation operator and crossover operator employed by the genetic algorithm must also take into account the chromosome's design.
### Example 1: binary representation
Suppose the problem is to find the integer value of x between 0 and 255 that provides the maximal result for f(x) = x^2. The possible solutions for this problem are the integers from 0 to 255, which can all be represented as 8-digit binary strings. Thus, we might use an 8-digit binary string as our chromosome. If a given chromosome in the population represents the value 155, its chromosome would be `10011011`.
Note that this is not the type of problem that is normally solved by a genetic algorithm, since it can be trivially solved using numeric methods; it is only used to serve as a simple example.
### Example 2: string representation
A more realistic problem we might wish to solve is the travelling salesman problem. In this problem, we seek an ordered list of cities that results in the shortest trip for the salesman to travel. Suppose there are six cities, which we'll call A, B, C, D, E, and F. A good design for our chromosome might be the ordered list we want to try. An example chromosome we might encounter in the population might be `DFABEC`.
## Selection, crossover and mutation
In each generation of the genetic algorithm, two parent chromosomes are selected based on their fitness values; these chromosomes are used by the mutation and crossover operators to produce two offspring chromosomes for the new population.[3]
## References
1. ^ "Introduction to genetic algorithms: IV. Genetic Algorithm". Retrieved 12 August 2015.
2. ^ Whitley, Darrell (June 1994). "A genetic algorithm tutorial". Statistics and Computing 4 (2).
3. ^ a b "What are Genetic Algorithms?". Retrieved 12 August 2015.
4. ^ "Genetic algorithms". Retrieved 12 August 2015.
This article was sourced from Creative Commons Attribution-ShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, E-Government Act of 2002.
Crowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles. | 926 | 4,415 | {"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} | 4.0625 | 4 | CC-MAIN-2020-34 | latest | en | 0.905354 |
https://hoogle.haskell.org/?hoogle=replicate%20-package%3ACabal%20-package%3Acontainers%20-package%3Amemory%20-package%3Ahedgehog%20-package%3Aconduit%20package%3Avector%20is%3Aexact | 1,709,175,615,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474775.80/warc/CC-MAIN-20240229003536-20240229033536-00858.warc.gz | 291,288,698 | 32,586 | # replicate -package:Cabal -package:containers -package:memory -package:hedgehog -package:conduit package:vector is:exact
O(n) A vector of the given length with the same value in each position.
Replicate a value to a given length
Replicate a value to a given length
O(n) A vector of the given length with the same value in each position.
Create a mutable vector of the given length (0 if the length is negative) and fill it with an initial value.
Create a mutable vector of the given length (0 if the length is negative) and fill it with an initial value.
O(n) A vector of the given length with the same value in each position.
Create a mutable vector of the given length (0 if the length is negative) and fill it with an initial value.
O(n) A vector of the given length with the same value in each position.
Create a mutable vector of the given length (0 if the length is negative) and fill it with an initial value.
O(n) A vector of the given length with the same value in each position.
Create a mutable vector of the given length (0 if the length is negative) and fill it with an initial value.
• Packages | 257 | 1,110 | {"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} | 2.515625 | 3 | CC-MAIN-2024-10 | latest | en | 0.768126 |
http://www.slideserve.com/verda/corpuri-geometrice-studiate | 1,505,837,362,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685850.32/warc/CC-MAIN-20170919145852-20170919165852-00210.warc.gz | 578,393,570 | 9,888 | 1 / 8
# Corpuri geometrice studiate - PowerPoint PPT Presentation
Corpuri geometrice studiate. Paralelipipedul dreptunghic. D’. Varfuri : A,B,C,D,A’,B’,C’,D’ Muchii : AB,BC,CD,AD,A’B’,B’C’, C’D’,A’D’,AA’,BB’,CC’,DD’ Fete: ABCD,A’B’C’D’,ABB’A’,BCC’B’ CDD’C’,ADD’A’ A’C – diagonala paralelipipedului A’C’ – diagonala bazei. C’. A’. B’. D. C. A. B.
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### Corpurigeometricestudiate
Paralelipipeduldreptunghic
D’
Varfuri: A,B,C,D,A’,B’,C’,D’
C’D’,A’D’,AA’,BB’,CC’,DD’
Fete: ABCD,A’B’C’D’,ABB’A’,BCC’B’
A’C – diagonalaparalelipipedului
A’C’ – diagonalabazei
C’
A’
B’
D
C
A
B
D’
C’
A’
Varfuri: A,B,C,D,A’,B’,C’,D’
C’D’,A’D’,AA’,BB’,CC’,DD’
Fete: ABCD,A’B’C’D’,ABB’A’,BCC’B’
A’C – diagonalacubului
B’
D
C
A
B
Prismahexagonalaregulata
Varfuri: A,B,C,D,E,F
A’,B’,C’,D’,E’,F’
Muchii: AB,BC,CD,DE,EF,AF
A’B’,B’C’,C’D’,D’E’,E’F’,A’F’,
AA’,BB’,CC’,DD’,EE’,FF’,
Fete: ABCDEF,A’B’C’D’E’F’,
ABB’A’,BCC’B’,CDD’C’DEE’D’,
EFF’E’,AFF’A’
F’
E’
D’
A’
B’
C’
F
E
A
D
B
C
Piramidatriunghiulararegulata
V
Varfuri: V,A,B,C
Muchii: AB,BC,AC,VA,VB,VC
Fete: ABC,VAB,VBC,VAC
C
A
B
Cilindrul circular drept
A’
B’
OA=R – raza bazei
AB=d – diametrul bazei
AA’=BB’ =G – generatoare
A
O
B
Conul circular drept
V
OA=OB=R – razabazei
AB=d– diametrulbazei
VA=VB=G – generatoare
B
O
A
B
O – centrul sferei
OA=OB=R – raza sferei
AB – diametrul sferei
C(O,R) – cerc mare al sferei
O
A | 862 | 2,142 | {"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} | 2.6875 | 3 | CC-MAIN-2017-39 | latest | en | 0.553571 |
https://www.jiskha.com/display.cgi?id=1355887969 | 1,531,766,293,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589417.43/warc/CC-MAIN-20180716174032-20180716194032-00510.warc.gz | 907,635,558 | 3,801 | # math
posted by bryce
parallelogram A is similar to Parallelogram B.
If the area of parallelogram b is 162 square units, what is the area of parallelogram A?
1. Reiny
not enough information
2. bryce
_______
/ A /
/ _____/
X
__________________
/ /
/ b /
/ /
/_________________/
3X
3. Steve
if both dimensions of B are 3x A's, then the area is 9x
If only the length is 3x, then the area is 3x
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1. ### geometry theorems
I have six questions which will be on a test two weeks from now, and I do not understand. Help me find out these theorems for the blanks. 1. If one pair of consecutive sides of a parallelogram are congruent, then the parallelogram …
2. ### math
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3. ### math
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7. ### Math
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More Similar Questions | 651 | 2,288 | {"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} | 3.796875 | 4 | CC-MAIN-2018-30 | latest | en | 0.806898 |
https://www.numerade.com/books/chapter/practice-test-1-2/ | 1,597,463,831,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740679.96/warc/CC-MAIN-20200815035250-20200815065250-00255.warc.gz | 752,955,733 | 9,935 | ## Educators
### Problem 1
If $f(x)=5 x^{\frac{4}{3}},$ then $f^{\prime}(8)=$
(A) 10
(B) $\frac{40}{3}$
(C) 80
(D) $\frac{160}{3}$
Check back soon!
### Problem 2
$$\lim _{x \rightarrow \infty} \frac{5 x^{2}-3 x+1}{4 x^{2}+2 x+5}$$
(A) 0
(B) $\frac{4}{5}$
(C) $\frac{5}{4}$
(D) $\infty$
Check back soon!
### Problem 3
If $f(x)=\frac{3 x^{2}+x}{3 x^{2}-x},$ then $f^{\prime}(x)$ is
(A) 1
(B) $\frac{6 x^{2}+1}{6 x^{2}-1}$
(C) $\frac{-6}{(3 x-1)^{2}}$
(D) $\frac{-2 x^{2}}{\left(x^{2}-x\right)^{2}}$
Check back soon!
### Problem 4
$$\lim _{x \rightarrow 0} \frac{\sin x^{2}}{x}=$$
(A) 1
(B) 0
(C) $\frac{\pi}{2}$
(D) Does Not Exist
Check back soon!
### Problem 5
If $x^{2}-2 x y+3 y^{2}=8,$ then $\frac{d y}{d x}=$
(A) $\frac{8+2 y-2 x}{6 y-2 x}$
(B) $\frac{3 y-x}{y-x}$
(C) $\frac{1}{3}$
(D) $\frac{y-x}{3 y-x}$
Check back soon!
### Problem 6
Which of the following integrals correctly corresponds to the area of the shaded region in the figure
above?
(A) $\int_{1}^{2}\left(x^{2}-4\right) d x$
(B) $\int_{1}^{2}\left(4-x^{2}\right) d x$
(C) $\int_{1}^{5}\left(x^{2}-4\right) d x$
(D) $\int_{1}^{5}\left(4-x^{2}\right) d x$
Check back soon!
### Problem 7
If $f(x)=\sec x+\csc x,$ then $f^{\prime}(x)=$
(A) 0
(B) $\csc x-\sec x$
(C) $\sec x \tan x+\csc x \cot x$
(D) $\sec x \tan x-\csc x \cot x$
Check back soon!
### Problem 8
An equation of the line normal to the graph of $y=\sqrt{\left(3 x^{2}+2 x\right)}$ at $(2,4)$ is
(A) $4 x+7 y=20$
(B) $-7 x+4 y=2$
(C) $7 x+4 y=30$
(D) $4 x+7 y=36$
Check back soon!
### Problem 9
$$\int_{-1}^{1} \frac{4}{1+x^{2}} d x=$$
$\begin{array}{ll}{(\mathrm{A})} & {0} \\ {\text { (B) }} & {\pi}\end{array}$
$\begin{array}{ll}{\text { (C) }} & {2 \pi} \\ {\text { (D) }} & {2}\end{array}$
Check back soon!
### Problem 10
If $f(x)=\cos ^{2} x,$ then $f^{\prime \prime}(\pi)=$
$\begin{array}{ll}{\text { (A) }} & {-2} \\ {\text { (B) }} & {0} \\ {\text { (C) }} & {1} \\ {\text { (D) }} & {2}\end{array}$
Check back soon!
### Problem 11
If $f(x)=\frac{5}{x^{2}+1}$ and $g(x)=3 x,$ then $g(f(2))=$
$\begin{array}{ll}{\text { (A) }} & {\frac{5}{37}} \\ {\text { (B) }} & {3} \\ {\text { (C) }} & {5} \\ {\text { (D) }} & {\frac{37}{5}}\end{array}$
Check back soon!
### Problem 12
$$\int x \sqrt{5 x^{2}-4} d x=$$
(A) $\frac{1}{10}\left(5 x^{2}-4\right)^{\frac{3}{2}}+C$
(B) $\frac{1}{15}\left(5 x^{2}-4\right)^{\frac{3}{2}}+C$
(C) $\frac{20}{3}\left(5 x^{2}-4\right)^{\frac{3}{2}}+C$
(D) $\frac{3}{20}\left(5 x^{2}-4\right)^{\frac{3}{2}}+C$
Check back soon!
### Problem 13
The slope of the line tangent to the graph of $3 x^{2}+5 \ln y=12$ at $(2,1)$ is
(A) $-\frac{12}{5}$
(B) $\frac{12}{5}$
(C) $\frac{5}{12}$
(D) $-7$
Check back soon!
### Problem 14
The equation $y=2-3 \sin \frac{\pi}{4}(x-1)$ has a fundamental period of
(A) $\frac{1}{8}$
(B) $\frac{4}{\pi}$
(C) 8
(D) 2$\pi$
Check back soon!
### Problem 15
If $f(x)=\left\{\begin{array}{l}{x^{2}+5 \text { if } x<2} \\ {7 x-5 \text { if } x \geq 2}\end{array}, \text { for all real numbers } x, \text { which of the following must be true? }\right.$
$\begin{array}{ll}{\text { I. }} & {f(x) \text { is continuous everywhere }} \\ {\text { II. }} & {f(x) \text { is differentiable everywhere. }} \\ {\text { III. }} & {f(x) \text { has a local minimum at } x=2}\end{array}$
(A) I only
(B) I and II only
(C) II and III only
(D) I, II, and III
Check back soon!
### Problem 16
For what value of $x$ does the function $f(x)=x^{3}-9 x^{2}-120 x+6$ have a local minimum?
(A) 10
(B) 4
(C) $-4$
(D) $-10$
Check back soon!
### Problem 17
The acceleration of a particle moving along the $x$ -axis at time $t$ is given by $a(t)=4 t-12$ . If the
velocity is 10 when $t=0$ and the position is 4 when $t=0,$ then the particle is changing direction at
$\begin{array}{ll}{\text { (A) }} & {t=1} \\ {\text { (B) } t} & {=3} \\ {\text { (C) } t} & {=5} \\ {\text { (D) } t} & {=1 \text { and } t=5}\end{array}$
Check back soon!
### Problem 18
The average value of the function $f(x)=(x-1)^{2}$ on the interval from $x=1$ to $x=5$ is
(A) $\frac{16}{3}$
(B) $\frac{64}{3}$
(C) $\frac{66}{3}$
(D) $\frac{256}{3}$
Maruti S.
### Problem 19
$$\int\left(e^{3 \ln x}+e^{3 x}\right) d x=$$
(A) $3+\frac{e^{3 x}}{3}+C$
(B) $\frac{e^{x^{4}}}{4}+3 e^{3 x}+C$
(C) $\frac{e^{x^{4}}}{4}+\frac{e^{3 x}}{3}+C$
(D) $\frac{x^{4}}{4}+\frac{e^{3 x}}{3}+C$
Check back soon!
### Problem 20
If $f(x)=\left(x^{2}+x+11\right) \sqrt{\left(x^{3}+5 x+121\right)},$ then $f(0)=$
(A) $\frac{5}{2}$
(B) $\frac{27}{2}$
(C) 22
(D) $\frac{247}{2}$
Check back soon!
### Problem 21
If $f(x)=5^{3 x},$ then $f^{\prime}(x)=$
(A) $5^{3 x}(\ln 125)$
(B) $\frac{5^{3 x}}{3 \ln 5}$
(C) 3$\left(5^{2 x}\right)$
(D) 3$\left(5^{3 x}\right)$
Check back soon!
### Problem 22
A solid is generated when the region in the first quadrant enclosed by the graph of $y=\left(x^{2}+1\right)^{3}$ , the line $x=1,$ the $x$ -axis, and the $y$ -axis is revolved about the $x$ -axis. Its volume is found by evaluating which of the following integrals?
(A) $\pi \int_{1}^{8}\left(x^{2}+1\right)^{3} d x$
(B) $\pi \int_{1}^{8}\left(x^{2}+1\right)^{6} d x$
(C) $\pi \int_{0}^{1}\left(x^{2}+1\right)^{3} d x$
(D) $\pi \int_{0}^{1}\left(x^{2}+1\right)^{6} d x$
Check back soon!
### Problem 23
$$\lim _{x \rightarrow 0} 4 \frac{\sin x \cos x-\sin x}{x^{2}}=$$
(A) 2
(B) $\frac{40}{3}$
(C) 0
(D) undefined
Check back soon!
### Problem 24
If $\frac{d y}{d x}=\frac{\left(3 x^{2}+2\right)}{y}$ and $y=4$ when $x=2,$ then when $x=3, y=$
(A) 18
(B) 58
(C) $\pm \sqrt{74}$
(D) $\pm \sqrt{58}$
Check back soon!
### Problem 25
$$\int \frac{d x}{9+x^{2}}=$$
(A) $3 \tan ^{-1}\left(\frac{x}{3}\right)+C$
(B) $\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)+C$
(C) $\frac{1}{3} \tan ^{-1}(x)+C$
(D) $\frac{1}{9} \tan ^{-1}(x)+C$
Check back soon!
### Problem 26
If $f(x)=\cos ^{3}(x+1),$ then $f^{\prime}(\pi)=$
(A) $-3 \cos ^{2}(\pi+1) \sin (\pi+1)$
(B) 3 $\cos ^{2}(\pi+1)$
(C) 3 $\cos ^{2}(\pi+1) \sin (\pi+1)$
(D) 0
Check back soon!
### Problem 27
$$\int x \sqrt{x+3} d x=$$
(A) $\frac{2(x+3)^{\frac{3}{2}}}{3}+C$
(B) $\frac{2}{5}(x+3)^{\frac{5}{2}}-2(x+3)^{\frac{3}{2}}+C$
(C) $\frac{3(x+3)^{\frac{3}{2}}}{2}+C$
(D) $\quad \frac{4 x^{2}(x+3)^{\frac{3}{2}}}{3}+C$
Check back soon!
### Problem 28
If $f(x)=\ln (\ln (1-x)),$ then $f^{\prime}(x)=$
(A) $-\frac{1}{\ln (1-x)}$
(B) $\frac{1}{(1-x) \ln (1-x)}$
(C) $\frac{1}{(1-x)^{2}}$
(D) $-\frac{1}{(1-x) \ln (1-x)}$
Check back soon!
### Problem 29
$$\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{6}+h\right)-\tan \left(\frac{\pi}{6}\right)}{h}=$$
$\begin{array}{ll}{\text { (A) }} & {\frac{4}{3}} \\ {\text { (B) }} & {\sqrt{3}} \\ {\text { (C) }} & {0} \\ {\text { (D) }} & {\frac{3}{4}}\end{array}$
Check back soon!
### Problem 30
$$\int \operatorname{tab}^{6} x \sec ^{2} x d x=$$
(A) $\frac{\tan ^{7} x}{7}+C$
(B) $\frac{\tan ^{7} x}{7}+\frac{\sec ^{3} x}{3}+C$
(C) $\frac{\tan ^{7} x \sec ^{3} x}{21}+C$
(D) $7 \tan ^{7} x+C$
Check back soon! | 3,129 | 6,980 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "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} | 4.03125 | 4 | CC-MAIN-2020-34 | latest | en | 0.367216 |
http://web2.0calc.com/questions/calculate-the-strain-on-the-rope-a-10-meter-long-rope-is-streched-3-0-cm-when-a-man-hangs-on-it-s-end-nbsp | 1,516,299,265,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887535.40/warc/CC-MAIN-20180118171050-20180118191050-00055.warc.gz | 373,872,466 | 6,019 | +0
# Calculate the strain on the rope. A 10 meter long rope is streched 3.0 cm when a man hangs on it's end.
+5
373
6
+10
Calculate the strain on the rope. A 10 meter long rope is streched 3.0 cm when a man hangs on it's end.
physics
pebble Jul 13, 2015
#1
+17711
+10
One way to calculate strain is: (change in length) / (initial length)
3.0 cm / 10 m = 3.0 cm / 1000 cm = 3/1000 = 0.003
geno3141 Jul 13, 2015
Sort:
#1
+17711
+10
One way to calculate strain is: (change in length) / (initial length)
3.0 cm / 10 m = 3.0 cm / 1000 cm = 3/1000 = 0.003
geno3141 Jul 13, 2015
#2
+91435
0
Thanks Geno, I thought strain would have some kind of unit attached to it
Is that a normal way to do it?
Melody Jul 14, 2015
#3
+26399
+5
Hi Melody, Geno is right, Strain is a dimensionless ratio (you might be thinking of Stress which has dimensions of force per unit area).
.
Alan Jul 14, 2015
#4
+91435
0
Thanks Alan, I wasn't thinking of anything. I know very little physics :)
So strain does not have anything to do with breaking point then?
I mean elastic can double in length without breaking but most things break much more quickly :/
Melody Jul 14, 2015
#5
+26399
+5
By definition, strain is just change in length divided by original length. The size of the strain will depend on the applied stress, and if the stress is too great the material will break. Elastic materials can get to larger strains than, say, ceramics, before breaking.
.
Alan Jul 14, 2015
#6
+91435
0
Thanks Alan :)
Melody Jul 14, 2015
### 16 Online Users
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http://mathhelpforum.com/differential-equations/152009-rankine-hugoniot-satisfy-weak-solution.html | 1,480,832,200,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541214.23/warc/CC-MAIN-20161202170901-00439-ip-10-31-129-80.ec2.internal.warc.gz | 180,760,763 | 12,595 | # Thread: Rankine Hugoniot Satisfy Weak Solution
1. ## Rankine Hugoniot Satisfy Weak Solution
Consider the Riemann Problem
$u_t + (f(u))_x = 0\qquad -\infty
$u(x,0) = \begin{cases}u_{\ell}, & x<0\\ u_r, & x>0.\end{cases}$
I am only going to deal with the case that there is a shock curve ( $f'(u_{\ell})>f'(u_r)$), since I can figure out the rarefaction wave once I understand the shock. The solution for this PDE using the Rankine Hugoniot Condition is
$u(x,t) = \begin{cases}u_{\ell}, & x < s\cdot t \\ u_r & x > s\cdot t\end{cases}$
where
$s = \frac{f(u_{\ell}) - f(u_r)}{u_{\ell} - u_r}$
Now, I want to determine if this solution satisfies the definition of the weak solution. So I assume that I would want to take a test function $\phi$ with compact support and multiply it by the PDE and then integrate over $\mathbb{R}$, i.e.
$\int_{\mathbb{R}}\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx$
$= \int_{-\infty}^0\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx + \int_0^{\infty}\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx$
$= \int_{-\infty}^0\! \left((u_{\ell})_t + f'(u_{\ell})\cdot (u_{\ell})_x\right)\cdot \phi\, dx + \int_0^{\infty}\! \left((u_r)_t + f'(u_r)\cdot (u_r)_x\right)\cdot \phi\, dx$
$= \int_{-\infty}^0\! \left((0 + f'(u_{\ell})\cdot 0\right)\cdot \phi\, dx + \int_0^{\infty}\! \left(0 + f'(u_r)\cdot 0\right)\cdot \phi\, dx = 0$
For some reason this just doesn't seem right.
2. Originally Posted by lvleph
Now, I want to determine if this solution satisfies the definition of the weak solution. So I assume that I would want to take a test function $\phi$ with compact support and multiply it by the PDE and then integrate over $\mathbb{R}$, i.e.
$\int_{\mathbb{R}}\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx$
$= \int_{-\infty}^0\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx + \int_0^{\infty}\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx$
$= \int_{-\infty}^0\! \left((u_{\ell})_t + f'(u_{\ell})\cdot (u_{\ell})_x\right)\cdot \phi\, dx + \int_0^{\infty}\! \left((u_r)_t + f'(u_r)\cdot (u_r)_x\right)\cdot \phi\, dx$
$= \int_{-\infty}^0\! \left((0 + f'(u_{\ell})\cdot 0\right)\cdot \phi\, dx + \int_0^{\infty}\! \left(0 + f'(u_r)\cdot 0\right)\cdot \phi\, dx = 0$
For some reason this just doesn't seem right.
Well, how is it that $(u_r)_x=0=(u_r)_t$? Surely these are not constant, since then it would be difficult to apply your boundary conditions in the weak solution (unless $u_r=u_{\ell}$).
Edit: Another thing, when you split the integral shouldn't you have $st$ intead of 0 in the limits of integration?
3. Yes, you are correct that it should be split at $s\cdot t$. However, $u_{\ell},\; u_r$ are constants. But, I am unsure if I should be placing them in the integral at that point.
4. Originally Posted by lvleph
Yes, you are correct that it should be split at $s\cdot t$. However, $u_{\ell},\; u_r$ are constants. But, I am unsure if I should be placing them in the integral at that point.
Hm, could you please give your definition of weak solution. Usually you multiply and do all you did before giving a solution, therefore arriving at a necessary condition which involves a larger class of functions, for example $H^1$ or $H_0^1$.
Another thing that's bugging me is: if $u_r \neq u_{\ell }$ then your solution looks a lot like Heaviside's function (which at a first glance would have me believe it will not be a weak solution in the sense above, or at least not in $W^{1,p}$ )
5. There is no problem with $u_r \ne u_{\ell}$ this will result in what is known as an admissible solution. Typically a shock-curve or rarefaction wave happens. In the case that a shock curve happens, the shock follows the line $s\cdot t$.
6. Originally Posted by lvleph
There is no problem with $u_r \ne u_{\ell}$ this will result in what is known as an admissible solution. Typically a shock-curve or rarefaction wave happens. In the case that a shock curve happens, the shock follows the line $s\cdot t$.
How do you define an admissible solution?
Also, shouldn't your integration be over the whole semi-plane t>0?
7. Well, now I just have a bunch of mistakes don't I.
I defined the admissible solution for shock curves in the original post.
8. Weak Solution: A weak solution of
$\vec{u}_t + (\vec{f}(\vec{u}))_x = 0, \quad \vec{u}|_{t=0} = \vec{u}_0(x)$
is a function $\vec{u}(x,t):\mathbb{R}\times\mathbb{R}^+ \rightarrow \mathbb{R}^n$ such that
$\int_0^{\infty}\!\!\!\int_{-\infty}^{\infty}\!\left[\vec{u}(x,t)\cdot \vec{\phi}_t(x,t) + \vec{f}(\vec{u}(x,t))\cdot \vec{\phi}_x(x,t)\right]\dx\,dt + \int_{-\infty}^{\infty}\!\vec{u}_0(x)\vec{\phi}(x,0)\, dx = 0$
for all $\vec{\phi}(x,t) \in C^1_c(\mathbb{R}\times \mathbb{R}^+)$, where
$C^1_c(\mathbb{R}\times \mathbb{R}^+) = \left\{\vec{\phi} \in C^1_c(\mathbb{R}\times \mathbb{R}^+) | \phi \equiv 0 \text{ for } (x,t)\not \in B_r(0,0) \bigcap (\mathbb{R}\times \mathbb{R}^+) \text{ for some } r>0\right\}$
9. Originally Posted by lvleph
Weak Solution: A weak solution of
$\vec{u}_t + (\vec{f}(\vec{u}))_x = 0, \quad \vec{u}|_{t=0} = \vec{u}_0(x)$
is a function $\vec{u}(x,t):\mathbb{R}\times\mathbb{R}^+ \rightarrow \mathbb{R}^n$ such that
$\int_0^{\infty}\!\!\!\int_{-\infty}^{\infty}\!\left[\vec{u}(x,t)\cdot \vec{\phi}_t(x,t) + \vec{f}(\vec{u}(x,t))\cdot \vec{\phi}_x(x,t)\right]\dx\,dt + \int_{-\infty}^{\infty}\!\vec{u}_0(x)\vec{\phi}(x,0)\, dx = 0$
for all $\vec{\phi}(x,t) \in C^1_c(\mathbb{R}\times \mathbb{R}^+)$, where
$C^1_c(\mathbb{R}\times \mathbb{R}^+) = \left\{\vec{\phi} \in C^1_c(\mathbb{R}\times \mathbb{R}^+) | \phi \equiv 0 \text{ for } (x,t)\not \in B_r(0,0) \bigcap (\mathbb{R}\times \mathbb{R}^+) \text{ for some } r>0\right\}$
Well, it's easy to see that given your space for "test" functions, we must have $\phi (x,0)=0$ for all $\phi \in C_c^1 (\mathbb{R} \times \mathbb{R} ^+)$, so the last term doesn't count. So your calculations are correct (taking into account that the integration is over the semi-plane) | 2,181 | 5,962 | {"found_math": true, "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": 44, "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} | 3.578125 | 4 | CC-MAIN-2016-50 | longest | en | 0.586545 |
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# Shift operators
Greenhorn
Posts: 3
public class B{
public static void main(String args[]){
int n=-1; //(How do I calculate the binary for a negative number?)
n=n<<1;<br /> System.out.println(n); //(How is n -2 ?)<br /> int n1=-1;<br /> n1=n1>>1;
System.out.println(n1);//(How is n1 -1 ?)
int n2=-1;
n1=n2>>>1;
System.out.println(n2); //(How is n2 -1 ?)
}
}
Tony Alicea
Sheriff
Posts: 3226
5
I am moving this thread to the Java beginner forum where it belongs. Thanks. See you there with a reply...
Tony Alicea
Sheriff
Posts: 3226
5
"How do I calculate the binary for a negative number?"
The format is called two's complement notation.
First take the positive of the (signed) integral type (byte, short, int or long) and invert all bits. For example, lets take a byte equal to one:
00000001
invert all bits (also known as taking the ONE'S complement. The operator is ~)
11111110
add 1 and you get all ones:
11111111
That's -1 in binary.
David Brochstein
Greenhorn
Posts: 4
I compiled this, and the results interest me:
why does n change from -1 to -2, but n1 stays -1 after >> ?
Tony Alicea
Sheriff
Posts: 3226
5
Because the <CODE>>></CODE> drags the most significant bit with it to the right with the effect that all the positions that are vacated because of the shift towards the right, are filled with whatever the sign bit is.
On the other hand, the <CODE>>>></CODE> operator will fill the leftmost bits with ZERO always.
The first one is called a signed shift to the right and the other an unsigned shift.
The <CODE><<</CODE> operator will fill the RIGHT part of the number with ZERO always.
Ray Marsh
Ranch Hand
Posts: 458
Can I ask why anyone would want to do this? It seems confusing. If I want to do something to a numeric I normally use a math function i.e. ( + - * / ) In what type of situation would a shift operator be more appropriate?
Thanks, Ray
David Brochstein
Greenhorn
Posts: 4
Ray,
see my posting in topic: Bits next to topic: Shift operators. It's the same question, I think.. Feel free to add to it. No response yet...
Tony Alicea
Sheriff
Posts: 3226
5
It's a very efficient way to multiply and divide numbers by powers of two, for one thing...
Also, especially in assembly language, it was/is used to pack and unpack small pieces of data in a relatively small area of memory.
It has been around for a while... shifting.
Ray Marsh
Ranch Hand
Posts: 458
Tony,
I'm not trying to be difficult, however in the interest of readable and maintainable code, wouldn't multiplying by -1 be easier to read and understand than a shift operation? Or, is a shift operation to negate a numeric just as obvious to someone familiar with the function? Or, is it that much faster to execute?
We have bit op-codes in RPG, but they are rarely used. I myself have never used them. Is this common practice in Java?
Thanks
nirvan sage
Greenhorn
Posts: 24
0000 0001 is the eight bit representation of 1
(take care that number represntation differs in int and long)
in order to change the sign take the twos complement
which is
1111 1110
+1
---------
1111 1111
This is the binary representation if -1
shifting it by one time towards the left makes it
1111 1110
This is actually a negative representation of some number
Inorder to find the actual positive number we just take
the twos complement again.
0000 0001
+1
---------
0000 0010
which is 2 so
1111 1110 must be -2
For the second case the last 1 in the LSB is not destroyed.
As for the third question you are assigning a variable and printing
the same variable.
Theresa Duick
Greenhorn
Posts: 27
I am having a tough time understanding binarys also. When the binary begins with a 0 on the left side, is it considered a positive # regardless of how many 1's appear to it's right side? Vice versa, if the left most # is a 1, is the # then considered negative? | 1,090 | 4,208 | {"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} | 2.96875 | 3 | CC-MAIN-2017-39 | latest | en | 0.845104 |
http://articles.chicagotribune.com/1990-09-27/entertainment/9003200954_1_fat-intake-calories-daily-caloric-intake | 1,506,391,584,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693940.84/warc/CC-MAIN-20170926013935-20170926033935-00303.warc.gz | 24,546,431 | 11,735 | # Some New Math Makes It Easier To Cut Back Fat
September 27, 1990|By Marian Burros.
You already know you aren`t supposed to let your consumption of fat exceed more than 30 percent of your daily caloric intake. That`s what health professionals keep saying. The only complication is figuring out just what that means in terms of real food.
Is it truly necessary to perform mathematical gymnastics for every meal you eat, multiplying and dividing in order to arrive at this magical 30 percent? Wouldn`t you be more likely to watch your fat intake if the watching was easier to do?
A much simpler method calculates the fat intake without percentages. It involves simple addition.
It also bypasses the misuse of the percentage figure, not only by ordinary people who are confused, but also by those professionals who insist that everything you eat should contain no more than 30 percent fat.
``Not everything you eat needs to be less than 30 percent fat,`` said Susan Male Smith, a consulting editor for the newsletter Environmental Nutrition. ``The 30 percent figure is for the entire day.``
In other words, if you consume a dish that is 80 percent fat you must balance it against a dish with much less fat, making sure the day`s total does not exceed 30 percent of all calories. But you might spend the better part of mealtimes figuring, or you have to carry a fat-calculator wheel everywhere you go.
If professionals can make the mistake of thinking the diet must be limited to dishes that contain no more than 30 percent, ordinary people can be forgiven for making the same assumption. This makes for a very restrictive diet and never allows for an ice cream cone, a brownie or even a well-dressed salad.
When a dish is very low in calories, a salad, for instance, the percentage of fat often is high even though the actual grams of fat may not be great. And a high-calorie dish also may have a fair amount of fat but have a low percentage of fat.
There is a legitimate reason for professionals to use percentages in discussing public health matters: the figure applies to the entire population, no matter how many calories its members eat every day.
For the easier, more realistic, method, you need to have just two pieces of information to get started: the number of calories you consume in a day and the number of grams of total fat allowed.
For example, if you eat 1,800 calories a day, you are entitled to 60 grams of total fat. If you want to delve a little more deeply into the fat question, you may also find the information about saturated fat useful. If your total fat intake should not exceed 60 grams, your total saturated fat intake should be no more than 20 grams.
These figures are calculated as follows: the number of calories consumed each day is multiplied by 30 percent (the limit of calories from fat). So that`s 1,800 multiplied by 30 percent, or 540.
Then, because each gram of fat has 9 calories, 540 is divided by 9. The result is 60. To figure the grams of saturated fat allowed, the total fat-60 grams-must be divided by one-third. The result is 20 grams.
Bonnie Liebman, chief nutritionist for the Center for Science in the Public Interest, a Washington-based consumer activist group, says that figuring your fat intake this way is ``psychologically very sound because it gives people a sense of power, control over what they choose to eat.``
Armed with these figures, all you have to do is keep track of the fat grams in your diet.
If you eat a lot of prepared food, the tracking is fairly simple because much packaged food offers nutritional information. Whether you`re dining in a restaurant or cooking at home you will find a small booklet with calorie and fat information useful. There are many on the market.
Most of us follow a fairly unvarying diet, week to week. But if you keep a record of your fat grams for several weeks, it will become second nature. Eventually, you won`t actually have to add up the fat; you`ll know.
It then becomes a simple matter to adjust for those foods that have a lot of fat instead of avoiding them. You can think of your fat grams as a daily bank balance and you don`t want to be overdrawn.
If you must have a fried chicken dinner, just deduct the 32 grams of fat from your total, in advance. Even if you are entitled to 60 grams of fat a day, it won`t take long until the fried chicken has made a big dent in that special nutritional bank balance and will have to be very careful with all the rest of your meals. - | 974 | 4,502 | {"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} | 3.078125 | 3 | CC-MAIN-2017-39 | latest | en | 0.948775 |
https://www.physicsforums.com/threads/question-about-circuits.711209/ | 1,508,279,342,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822513.17/warc/CC-MAIN-20171017215800-20171017235800-00026.warc.gz | 966,370,130 | 16,861 | 1. Sep 18, 2013
### fredrogers3
1. The problem statement, all variables and given/known data
A circuit is constructed with five capacitors and a battery as shown. The values for the capacitors are: C1 = C5 = 4.1 μF, C2 = 3 μF, C3 = 6.8 μF, and C4 = 3.4 μF. The battery voltage is V = 12 V.
I need to find the equivalent capacitance between points a and c.
2. Relevant equations
I have the answer but I am a bit confused as to how it is arrived at. See below.
3. The attempt at a solution
I found the equivalent capacitance between a and b as c234=(c4)+(1/c23)
I see next that c234 is in series with c1 and c5. Why does the equation then become Cac=(1/((1/C1)+(1/C234)) and not (1/((1/C1)+(1/C5)+(1/C234))?
2. Sep 18, 2013
### iRaid
You can neglect c5 since it doesn't affect the capacitance at a or b since the current is flowing clockwise.
3. Sep 18, 2013
### Staff: Mentor
Apparently they're only looking for the capacitance to the right of the points a and c. This is not clear in your problem description, so perhaps the problem is stated differently in the original?
4. Sep 18, 2013
### fredrogers3
This was exactly the way the problem was written, so it left me a little confused as well
5. Sep 18, 2013
### Staff: Mentor
Current direction doesn't affect equivalent capacitance! Not unless there's a nonlinear circuit element like an open switch or a diode involved that would prevent "seeing" the capacitance. | 409 | 1,437 | {"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} | 3.1875 | 3 | CC-MAIN-2017-43 | longest | en | 0.957899 |
http://docmadhattan.fieldofscience.com/2015/02/ | 1,601,450,407,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402118004.92/warc/CC-MAIN-20200930044533-20200930074533-00490.warc.gz | 33,657,262 | 28,363 | ### Rock, paper, scissors, lizard, Spock
One popular five-weapon expansion is "rock-paper-scissors-lizard-Spock", invented by Sam Kass and Karen Bryla, which adds "Spock" and "lizard" to the standard three choices. "Spock" is signified with the Star Trek Vulcan salute, while "lizard" is shown by forming the hand into a sock-puppet-like mouth. Spock smashes scissors and vaporizes rock; he is poisoned by lizard and disproven by paper. Lizard poisons Spock and eats paper; it is crushed by rock and decapitated by scissors. This variant was mentioned in a 2005 article of The Times and was later the subject of an episode of the American sitcom The Big Bang Theory in 2008.
The majority of such proposed generalizations are isomorphic to a simple game of modulo arithmetic, where half the differences are wins for player one. For instance, rock-paper-scissors-Spock-lizard (note the different order of the last two moves) may be modeled as a game in which each player picks a number from one to five. Subtract the number chosen by player two from the number chosen by player one, and then take the remainder modulo 5 of the result. Player one is the victor if the difference is one or three, and player two is the victor if the difference is two or four. If the difference is zero, the game is a tie.
Alternatively, the rankings in rock-paper-scissors-Spock-lizard may be modeled by a comparison of the parity of the two choices. If it is the same (two odd-numbered moves or two even-numbered ones) then the lower number wins, while if they are different (one odd and one even) the higher wins. Using this algorithm, additional moves can easily be added two at a time while keeping the game balanced:
• Declare a move N+1 (where N is the original total of moves) that beats all existing odd-numbered moves and loses to the others (for example, the rock (#1), scissors (#3), and lizard (#5) could fall into the German well (#6), while the paper (#2) covers it and Spock (#4) manipulates it).
• Declare another move N+2 with the reverse property (such as a plant (#7) that grows through the paper (#2), poisons Spock (#4), and grows through the well (#6), while being damaged by the rock (#1), scissor (#3), and lizard(#5)).
(via en.wiki)
### The dark side of the moon
#moon #astronomy #NASA #video #PinkFloyd
The first photo of the lunar far side taken by the Soviet spacecraft Luna 3 on Oct. 7, 1959 - via Universe Today
### What Einstein thought about Galilei
Galileo's Dialogue Concerning the Two Chief World Systems is a mine of information for anyone interested in the cultural history of the Western world and its influence upon economic and political development.
(...) To begin with, the Dialogue gives an extremely lively and persuasive exposition of the then prevailing views on the structure of the cosmos in the large. The naïve picture of the earth as a flat disc, combined with obscure ideas about star-filled space and the motions of the celestial bodies, prevalent in the early Middle Ages, represented a deterioration of the much earlier conceptions of the Greeks, and in particular of Aristotle’s ideas and Ptolemy’s consistent spatial concept of the celestial bodies and their motions.
(...) In advocating and fighting for the Copernican theory Galileo was not only motivated by a striving to simplify the representation of the celestial motions. His aim was to substitute for a petrified and barren system of ideas the unbiased and strenuous quest for a deeper and more consistent comprehension of the physical and astronomical facts.
The form of dialogue used in his work may be partly due to Plato’s shining example; it enabled Galileo to apply his extraordinary literary talent to the sharp and vivid confrontation of opinion. To be sure, he wanted to avoid an open commitment in these controversial questions that would have delivered him to destruction by the Inquisition. Galileo had, in fact, been expressly forbidden to advocate the Copernican theory. Apart from its revolutionary factual content the Dialogue represents a down-right roguish attempt to comply with this order in appearance and yet in fact to disregard it. Unfortunately, it turned out that the Holy Inquisition was unable to appreciate adequately such subtle humor.
(...) It is difficult to us today to appreciate the imaginative power made manifest in the precise formulation of the concept of acceleration and in the recognition of its physical significance.
Once the conception of the center of the universe had, with good reason, been rejected, the idea of the immovable earth, and, generally, of an exceptional role of the earth, was deprived of its justification (...)
(...) Galileo takes great pains to demonstrate that the hypothesis of the rotation and revolution of the earth is not refuted by the fact that we do not observe any mechanical effects of these motions. Strictly speaking, such a demonstration was impossible because a complete theory of mechanics was lacking. I think it is just in the struggle with this problem that Galileo’s originality is demonstrated with particular force. Galileo is, of course, also concerned to show that the fixed stars are too remote for parallaxes produced by the yearly motion of the earth to be detectable with the measuring instruments of his time. This investigation also is ingenious, notwithstanding its primitiveness.
It was Galileo’s longing for a mechanical proof of the motion of the earth which misled him into formulating a wrong theory of the tides. The fascinating arguments in the last conversation would hardly have been accepted as proofs by Galileo, had his temperament not got the better of him. It is hard for me to resist the temptation to deal with this subject more fully.
The leitmotif which I recognize in Galileo’s work is the passionate fight against any kind of dogma based on authority. Only experience and careful reflection are accepted by him as criteria of truth. Nowadays it is hard for us to grasp how sinister and revolutionary such an attitude appeared at Galileo’s time, when merely to doubt the truth of opinions which had no basis but authority was considered a capital crime and punished accordingly. Actually we are by no means so far removed from such a situation even today as many of us would like to flatter ourselves; but in theory, at least, the principle of unbiased thought has won out, and most people are willing to pay lip service to this principle.
It has often been maintained that Galileo became the father of modern science by replacing the speculative, deductive method with the empirical, experimental method. I believe, however, that this interpretation would not stand close scrutiny. There is no empirical method without speculative concepts and systems; and there is no speculative thinking whose concepts do not reveal, on closer investigation, the empirical material from which they stem. To put into sharp contrast the empirical and the deductive attitude is misleading, and was entirely foreign to Galileo. Actually it was not until the nineteenth century that logical (mathematical) systems whose structures were completely independent of any empirical content had been cleanly extracted. Moreover, the experimental methods at Galileo’s disposal were so imperfect that only the boldest speculation could possibly bridge the gaps between the empirical data. (For example, there existed no means to measure times shorter than a second). The antithesis Empiricism vs. Rationalism does not appear as a controversial point in Galileo’s work. Galileo opposes the deductive methods of Aristotle and his adherents only when he considers their premises arbitrary or untenable, and he does not rebuke his opponents for the mere fact of using deductive methods. In the first dialogue, he emphasizes in several passages that according to Aristotle, too, even the most plausible deduction must be put aside if it is incompatible with empirical findings. And on the other hand, Galileo himself makes considerable use of logical deduction. His endeavors are not so much directed at "factual knowledge" as at "comprehension". But to comprehend is essentially to draw conclusions from an already accepted logical system.
(from the foreword to Dialogue Concerning the Two Chief World Systems: Ptolemaic and Copernican (1953), Einstein Archives 1-174 - via Open Parachute)
About the italian physicist, Galileo Galilei and the impossible biomechanics of giants is an interesting reading.
### The mathematics of love
#ValentinesDay #mathematics
$\left (x^2 + \frac{9}{4} y^2 + z^2 - 1 \right )^3 - x^2 z^3 - \frac{9}{200} y^2 z^3 = 0$
### Black holes and revelations: their large interiors
about #blackhole #cosmology #arXiv #abstract #CarloRovelli
The 3d volume inside a spherical black hole can be defined by extending an intrinsic flat-spacetime characterization of the volume inside a 2-sphere. For a collapsed object, the volume grows with time since the collapse, reaching a simple asymptotic form, which has a compelling geometrical interpretation. Perhaps surprising, it is large. The result may have relevance for the discussion on the information paradox.
Marios Christodoulou & Carlo Rovelli (2014). How big is a black hole?, arXiv: http://arxiv.org/abs/1411.2854v2
A sphere $S$ on the event horizon bounds a spacelike hypersurface, a large portion of which coincides with an $r$ = constant hypersurface. We show this hypersurface with one dimension suppressed, and cut in the middle, omitting the long cylindrical part which gives the main contribution to its volume. We also illustrate the argument showing that most of the volume is contained in a region out of causal contact with matter that has advanced far into the black hole.
Ingemar Bengtsson & Emma Jakobsson (2015). Black holes: Their large interiors, arXiv: http://arxiv.org/abs/1502.01907v1
### Black holes and revelations: the seeds of the galaxies
about #blackhole #astronomy #arXiv #abstract
The centre of the Milky Way - via Nasa
In this paper we present a new scenario where massive Primordial Black Holes (PBH) are produced from the collapse of large curvature perturbations generated during a mild waterfall phase of hybrid inflation. We determine the values of the inflaton potential parameters leading to a PBH mass spectrum peaking on planetary-like masses at matter-radiation equality and producing abundances comparable to those of Dark Matter today, while the matter power spectrum on scales probed by CMB anisotropies agrees with Planck data. These PBH could have acquired large stellar masses today, via merging, and the model passes both the constraints from CMB distortions and micro-lensing. This scenario is supported by Chandra observations of numerous BH candidates in the central region of Andromeda. Moreover, the tail of the PBH mass distribution could be responsible for the seeds of supermassive black holes at the center of galaxies, as well as for ultra-luminous X-rays sources. We find that our effective hybrid potential can originate e.g. from D-term inflation with a Fayet-Iliopoulos term of the order of the Planck scale but sub-planckian values of the inflaton field. Finally, we discuss the implications of quantum diffusion at the instability point of the potential, able to generate a swiss-cheese like structure of the Universe, eventually leading to apparent accelerated cosmic expansion.
Sébastien Clesse & Juan García-Bellido (2015). Massive Primordial Black Holes from Hybrid Inflation as Dark Matter and the seeds of Galaxies, arXiv: http://arxiv.org/abs/1501.07565v1
### A probabilistic approach to the prime numbers distribution
by @ulaulaman about #prime_numbers #arXiv #mathematics
The prime numbers theorem states the asymptotic approximation for the prime-counting function. The first statement for the theorem is given by Euler in 1737 (pdf):
The sum of the series of the reciprocals of the prime numbers, $\frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \frac{1}{13} + \cdots$ is infinitely large, but it is infinitely many times less than the sum of the harmonic series, $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots$ Furthemore, the sum of the former series is like the logarithm of the sum of the latter series.
After Euler, the most known attempt to evaluate the prime numbers distribution is given by Bernard Riemann, and the most recent is posted on arXiv some days ago: | 2,722 | 12,457 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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} | 3.21875 | 3 | CC-MAIN-2020-40 | longest | en | 0.94045 |
http://mathhelpforum.com/calculus/109283-evaluate-y-2-1-a.html | 1,508,516,701,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824226.31/warc/CC-MAIN-20171020154441-20171020174441-00669.warc.gz | 214,229,921 | 10,774 | # Thread: evaluate y' at (2,1)
1. ## evaluate y' at (2,1)
For the equation given below, evaluate at the point .
y' at (2,1) = ?
so i tried taking the natural log of everything then differentiating both sides and solving for y'.
ln2 + lne^xy - ln2 + lnx = lny +ln9.78
xy + lnx = lny + ln9.78
(1/y)y' = y + xy' + 1/x + 0
y' = y^2 + xy'y + y/x
y' = y^2+(y/x) / 1-xy
then if you plug in (2,1) i get -1.5 and thats not right =(
2. Originally Posted by mybrohshi5
For the equation given below, evaluate at the point .
y' at (2,1) = ?
so i tried taking the natural log of everything then differentiating both sides and solving for y'.
ln2 + lne^xy - ln2 + lnx = lny +ln9.78
any advice? yes, the entire step above is invalid
ln(a+b) does not equal ln(a) + ln(b)
...
3. Also, you can't take the log of everything and then differentiate as you would be differentiating $log(y)$ and not $y$ as required..
4. ok i guess i have not a clue what to do then.
should i start off by taking the natural log of both sides?
then differentiating with u'/u for both sides?
then solving for y'?
thank you
5. why log? just take the derivative straight up.
$\frac{d}{dx}(2e^{xy} - 2x = y + k)$
$2e^{xy}(xy' + y) - 2 = y'$
$2xy'e^{xy} + 2ye^{xy} - 2 = y'$
$2ye^{xy} - 2 = y' - 2xy'e^{xy}$
$2ye^{xy} - 2 = y'(1 - 2xe^{xy})$
$\frac{2(ye^{xy}-1)}{1 - 2xe^{xy}} = y'$
6. Originally Posted by mybrohshi5
ok i guess i have not a clue what to do then.
should i start off by taking the natural log of both sides?
then differentiating with u'/u for both sides?
then solving for y'?
thank you
You shouldn't have to take logarithms. You can just differentiate implicity.
$2e^{xy}-2x=y+9.78$
The derivative of $2e^{xy}$ is kind of tricky. Set $u=xy$ and differentiate as follows:
$2e^{u}\frac{du}{dx}$
$\frac{du}{dx}=x\frac{dy}{dx}+y$ (by product rule)
$=2e^u(\frac{xdy}{dx}+y)$
$=2e^{xy}(\frac{xdy}{dx}+y)$
So now that you know the derivative of this term, the rest should be obvious. | 691 | 1,987 | {"found_math": true, "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": 15, "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} | 4.03125 | 4 | CC-MAIN-2017-43 | longest | en | 0.881362 |
https://mathoverflow.net/questions/122027/unitalization-internal-to-monoidal-categories | 1,656,167,733,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103035636.10/warc/CC-MAIN-20220625125944-20220625155944-00227.warc.gz | 435,480,231 | 26,631 | # Unitalization internal to monoidal categories
Let $C$ be a monoidal category, not assumed to be symmetric. Assume that the underlying category of $C$ is nice enough, for example cocomplete, perhaps even presentable. A semigroup object in $C$ is a pair $(X,\mu)$ consisting of an object $X \in C$ and a morphism $\mu : X \otimes X \to X$ satisfying the associativity law $\mu \circ (X \otimes \mu) = \mu \circ (\mu \otimes X)$. Does the forgetful functor from monoid objects in $C$ to semigroup objects in $C$ have a left adjoint? In other words, is there an unitalization internal to $C$?
The cases $C=\mathsf{Set}$ and $C=\mathrm{Mod}(R)$ are well-known. More generally, the answer is yes when $\otimes$ preserves coproducts in each variable. Then the unitalization of $(X,\mu)$ is $(1 \oplus X,\mu',\eta)$ with the obvious morphism $\mu' : (1 \oplus X) \otimes (1 \oplus X) = 1 \oplus X \oplus X \oplus X \otimes X \to 1 \oplus X$ and $\eta : 1 \to 1 \oplus X$.
Actually I am interested in the case that $C=(\mathrm{End}(D),\circ,\mathrm{id})$ for a (nice) category $D$, thus I would like to know if every semi-monad can be made into a monad. Here $\otimes$ preserves colimits in the left variable, but not in the right variable. Actually $D$ is even a presentable symmetric monoidal category and $\mathrm{End}(D)$ refers to enriched endofunctors, i.e. I am interested in strong (semi) monads.
• I know how to unitalize operads. This can be done as for algebras. Hence the semimonad associated to an operad can be unitalized. However I don't see a way to extend the construction to arbitrary semimonads. Feb 17, 2013 at 9:40
• In case you have not seen the following mathoverflow question, I would like to direct your attention to mathoverflow.net/questions/19906/are-monads-monadic which is related to this question. In particular, take a look at Tom Leinster's answer and his reference to Kelly's article. Feb 17, 2013 at 22:51
• @Ricardo: Thanks, but Kelly's article doesn't discuss unitalization. Feb 18, 2013 at 1:59
If $$C$$ is locally presentable and $$S$$ is a semi-monad whose underlying functor is accessible, then there exists a unitalization of $$S$$. Here is a proof modeled after an idea discussed at the nLab at the page free monad.
Define an algebra of a semi-monad $$S: C \to C$$ in the expected way, as an object $$X$$ of $$C$$ equipped with a morphism (an "action") $$SX \to X$$ satisfying the usual associativity law for an action. Morphisms between algebras are also defined in the expected way, so that there is a full embedding $$S$$-$$\mathrm{Alg}_\mathrm{smd} \hookrightarrow S \downarrow C$$ into the comma category. (I use the subscripts "smd" and "mnd" to indicate algebras qua semi-monads and monads.)
The main thing to check is that the forgetful functor $$S$$-$$\mathrm{Alg}_\mathrm{smd} \to C$$ is monadic in the "evil" sense, so that there is an isomorphism $$F$$-$$\mathrm{Alg}_\mathrm{mnd} \simeq S$$-$$\mathrm{Alg}_\mathrm{smd}$$ in $$Cat/C$$ for some monad $$F$$. The claim is that then $$F$$ is the free monad on the semi-monad $$S$$. For in that case, given a monad $$M$$ on $$C$$ we have natural bijections between
• Semi-monad morphisms $$S \to M$$,
• $$S$$-algebra structures $$S U_M \to U_M$$ where $$U_M:$$ $$M$$-$$\mathrm{Alg}_\mathrm{mnd} \to C$$ is the forgetful functor,
• Morphisms $$M$$-$$\mathrm{Alg}_\mathrm{mnd} \to S$$-$$\mathrm{Alg}_\mathrm{smd}$$ in $$Cat/C$$,
• Morphisms $$M$$-$$\mathrm{Alg}_\mathrm{mnd} \to F$$-$$\mathrm{Alg}_\mathrm{mnd}$$ in $$Cat/C$$,
• $$F$$-algebra structures (qua algebras over a monad) $$F U_M \to U_M$$,
• Monad morphisms $$F \to M$$
so that $$F$$ is evidently the free monad on the semi-monad $$S$$.
So now we check monadicity, using the precise monadicity theorem. It is straightforward that the forgetful functor $$U: S$$-$$\mathrm{Alg}_{\mathrm{smd}} \to C$$ creates (not just reflects!) $$U$$-split coequalizers, so we just have to check that $$U$$ has a left adjoint. However, since the 2-category of locally presentable categories and accessible functors inherits lax limits from $$Cat$$, and since $$S$$-$$\mathrm{Alg}_\mathrm{smd}$$ is a lax limit in $$Cat$$ (for essentially the same reason that Eilenberg-Moore categories for monads are lax limits), we see that $$U: S$$-$$\mathrm{Alg}_\mathrm{smd} \to C$$ is an accessible functor between locally presentable categories. In this situation, existence of a left adjoint to $$U$$ is equivalent to preservation of limits by $$U$$. But limit-preservation is clear. So the conditions of the precise monadicity theorem are satisfied.
• A wonderfully abstract nonsense proof! Feb 18, 2013 at 18:41
• I'm going to delete this answer for the moment, as the last paragraph could be playing a fast one. I may undelete it later, possibly after revisions. Feb 18, 2013 at 18:41
• Todd, thanks a lot! I hope that the argument can be completed. It doesn't restrict to semimonads, right? It also seems to work for semigroup objects in presentable monoidal categories, right? @Zhen: Although it seems to be abstract at the first glance, it is quite simple: It is easy to unitalize semigroups described by generators and relations, one just can add a unit to the presentation. To make this work, one has to construct free semigroups. Feb 18, 2013 at 20:58
• Yes, it would apply to semigroup objects as well. I think Mike Shulman wrote most of the pertinent nLab articles; if he sees this question, he might see either how to complete the argument or where it founders. I'll keep thinking about it myself. Feb 18, 2013 at 21:17
• It looks good to me! Feb 28, 2021 at 4:03 | 1,628 | 5,618 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "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": 51, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.872179 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/7942/2/a/q/1/1/ | 1,723,560,501,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641079807.82/warc/CC-MAIN-20240813141635-20240813171635-00467.warc.gz | 655,013,042 | 78,412 | Properties
Label 7942.2.a.q.1.1 Level $7942$ Weight $2$ Character 7942.1 Self dual yes Analytic conductor $63.417$ Analytic rank $1$ Dimension $1$ CM no Inner twists $1$
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Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [7942,2,Mod(1,7942)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(7942, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0, 0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("7942.1");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$7942 = 2 \cdot 11 \cdot 19^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 7942.a (trivial)
Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$63.4171892853$$ Analytic rank: $$1$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 418) Fricke sign: $$+1$$ Sato-Tate group: $\mathrm{SU}(2)$
Embedding invariants
Embedding label 1.1 Character $$\chi$$ $$=$$ 7942.1
$q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q+1.00000 q^{2} +1.00000 q^{4} +1.00000 q^{5} +1.00000 q^{7} +1.00000 q^{8} -3.00000 q^{9} +O(q^{10})$$ $$q+1.00000 q^{2} +1.00000 q^{4} +1.00000 q^{5} +1.00000 q^{7} +1.00000 q^{8} -3.00000 q^{9} +1.00000 q^{10} +1.00000 q^{11} +2.00000 q^{13} +1.00000 q^{14} +1.00000 q^{16} -3.00000 q^{18} +1.00000 q^{20} +1.00000 q^{22} -6.00000 q^{23} -4.00000 q^{25} +2.00000 q^{26} +1.00000 q^{28} -8.00000 q^{29} -10.0000 q^{31} +1.00000 q^{32} +1.00000 q^{35} -3.00000 q^{36} +3.00000 q^{37} +1.00000 q^{40} -2.00000 q^{41} -8.00000 q^{43} +1.00000 q^{44} -3.00000 q^{45} -6.00000 q^{46} -6.00000 q^{49} -4.00000 q^{50} +2.00000 q^{52} -9.00000 q^{53} +1.00000 q^{55} +1.00000 q^{56} -8.00000 q^{58} -14.0000 q^{59} -10.0000 q^{62} -3.00000 q^{63} +1.00000 q^{64} +2.00000 q^{65} +12.0000 q^{67} +1.00000 q^{70} +6.00000 q^{71} -3.00000 q^{72} +3.00000 q^{74} +1.00000 q^{77} -11.0000 q^{79} +1.00000 q^{80} +9.00000 q^{81} -2.00000 q^{82} +7.00000 q^{83} -8.00000 q^{86} +1.00000 q^{88} +14.0000 q^{89} -3.00000 q^{90} +2.00000 q^{91} -6.00000 q^{92} +5.00000 q^{97} -6.00000 q^{98} -3.00000 q^{99} +O(q^{100})$$
Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ 1.00000 0.707107
$$3$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$4$$ 1.00000 0.500000
$$5$$ 1.00000 0.447214 0.223607 0.974679i $$-0.428217\pi$$
0.223607 + 0.974679i $$0.428217\pi$$
$$6$$ 0 0
$$7$$ 1.00000 0.377964 0.188982 0.981981i $$-0.439481\pi$$
0.188982 + 0.981981i $$0.439481\pi$$
$$8$$ 1.00000 0.353553
$$9$$ −3.00000 −1.00000
$$10$$ 1.00000 0.316228
$$11$$ 1.00000 0.301511
$$12$$ 0 0
$$13$$ 2.00000 0.554700 0.277350 0.960769i $$-0.410544\pi$$
0.277350 + 0.960769i $$0.410544\pi$$
$$14$$ 1.00000 0.267261
$$15$$ 0 0
$$16$$ 1.00000 0.250000
$$17$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$18$$ −3.00000 −0.707107
$$19$$ 0 0
$$20$$ 1.00000 0.223607
$$21$$ 0 0
$$22$$ 1.00000 0.213201
$$23$$ −6.00000 −1.25109 −0.625543 0.780189i $$-0.715123\pi$$
−0.625543 + 0.780189i $$0.715123\pi$$
$$24$$ 0 0
$$25$$ −4.00000 −0.800000
$$26$$ 2.00000 0.392232
$$27$$ 0 0
$$28$$ 1.00000 0.188982
$$29$$ −8.00000 −1.48556 −0.742781 0.669534i $$-0.766494\pi$$
−0.742781 + 0.669534i $$0.766494\pi$$
$$30$$ 0 0
$$31$$ −10.0000 −1.79605 −0.898027 0.439941i $$-0.854999\pi$$
−0.898027 + 0.439941i $$0.854999\pi$$
$$32$$ 1.00000 0.176777
$$33$$ 0 0
$$34$$ 0 0
$$35$$ 1.00000 0.169031
$$36$$ −3.00000 −0.500000
$$37$$ 3.00000 0.493197 0.246598 0.969118i $$-0.420687\pi$$
0.246598 + 0.969118i $$0.420687\pi$$
$$38$$ 0 0
$$39$$ 0 0
$$40$$ 1.00000 0.158114
$$41$$ −2.00000 −0.312348 −0.156174 0.987730i $$-0.549916\pi$$
−0.156174 + 0.987730i $$0.549916\pi$$
$$42$$ 0 0
$$43$$ −8.00000 −1.21999 −0.609994 0.792406i $$-0.708828\pi$$
−0.609994 + 0.792406i $$0.708828\pi$$
$$44$$ 1.00000 0.150756
$$45$$ −3.00000 −0.447214
$$46$$ −6.00000 −0.884652
$$47$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$48$$ 0 0
$$49$$ −6.00000 −0.857143
$$50$$ −4.00000 −0.565685
$$51$$ 0 0
$$52$$ 2.00000 0.277350
$$53$$ −9.00000 −1.23625 −0.618123 0.786082i $$-0.712106\pi$$
−0.618123 + 0.786082i $$0.712106\pi$$
$$54$$ 0 0
$$55$$ 1.00000 0.134840
$$56$$ 1.00000 0.133631
$$57$$ 0 0
$$58$$ −8.00000 −1.05045
$$59$$ −14.0000 −1.82264 −0.911322 0.411693i $$-0.864937\pi$$
−0.911322 + 0.411693i $$0.864937\pi$$
$$60$$ 0 0
$$61$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$62$$ −10.0000 −1.27000
$$63$$ −3.00000 −0.377964
$$64$$ 1.00000 0.125000
$$65$$ 2.00000 0.248069
$$66$$ 0 0
$$67$$ 12.0000 1.46603 0.733017 0.680211i $$-0.238112\pi$$
0.733017 + 0.680211i $$0.238112\pi$$
$$68$$ 0 0
$$69$$ 0 0
$$70$$ 1.00000 0.119523
$$71$$ 6.00000 0.712069 0.356034 0.934473i $$-0.384129\pi$$
0.356034 + 0.934473i $$0.384129\pi$$
$$72$$ −3.00000 −0.353553
$$73$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$74$$ 3.00000 0.348743
$$75$$ 0 0
$$76$$ 0 0
$$77$$ 1.00000 0.113961
$$78$$ 0 0
$$79$$ −11.0000 −1.23760 −0.618798 0.785550i $$-0.712380\pi$$
−0.618798 + 0.785550i $$0.712380\pi$$
$$80$$ 1.00000 0.111803
$$81$$ 9.00000 1.00000
$$82$$ −2.00000 −0.220863
$$83$$ 7.00000 0.768350 0.384175 0.923260i $$-0.374486\pi$$
0.384175 + 0.923260i $$0.374486\pi$$
$$84$$ 0 0
$$85$$ 0 0
$$86$$ −8.00000 −0.862662
$$87$$ 0 0
$$88$$ 1.00000 0.106600
$$89$$ 14.0000 1.48400 0.741999 0.670402i $$-0.233878\pi$$
0.741999 + 0.670402i $$0.233878\pi$$
$$90$$ −3.00000 −0.316228
$$91$$ 2.00000 0.209657
$$92$$ −6.00000 −0.625543
$$93$$ 0 0
$$94$$ 0 0
$$95$$ 0 0
$$96$$ 0 0
$$97$$ 5.00000 0.507673 0.253837 0.967247i $$-0.418307\pi$$
0.253837 + 0.967247i $$0.418307\pi$$
$$98$$ −6.00000 −0.606092
$$99$$ −3.00000 −0.301511
$$100$$ −4.00000 −0.400000
$$101$$ 10.0000 0.995037 0.497519 0.867453i $$-0.334245\pi$$
0.497519 + 0.867453i $$0.334245\pi$$
$$102$$ 0 0
$$103$$ 6.00000 0.591198 0.295599 0.955312i $$-0.404481\pi$$
0.295599 + 0.955312i $$0.404481\pi$$
$$104$$ 2.00000 0.196116
$$105$$ 0 0
$$106$$ −9.00000 −0.874157
$$107$$ 17.0000 1.64345 0.821726 0.569883i $$-0.193011\pi$$
0.821726 + 0.569883i $$0.193011\pi$$
$$108$$ 0 0
$$109$$ −10.0000 −0.957826 −0.478913 0.877862i $$-0.658969\pi$$
−0.478913 + 0.877862i $$0.658969\pi$$
$$110$$ 1.00000 0.0953463
$$111$$ 0 0
$$112$$ 1.00000 0.0944911
$$113$$ 2.00000 0.188144 0.0940721 0.995565i $$-0.470012\pi$$
0.0940721 + 0.995565i $$0.470012\pi$$
$$114$$ 0 0
$$115$$ −6.00000 −0.559503
$$116$$ −8.00000 −0.742781
$$117$$ −6.00000 −0.554700
$$118$$ −14.0000 −1.28880
$$119$$ 0 0
$$120$$ 0 0
$$121$$ 1.00000 0.0909091
$$122$$ 0 0
$$123$$ 0 0
$$124$$ −10.0000 −0.898027
$$125$$ −9.00000 −0.804984
$$126$$ −3.00000 −0.267261
$$127$$ 20.0000 1.77471 0.887357 0.461084i $$-0.152539\pi$$
0.887357 + 0.461084i $$0.152539\pi$$
$$128$$ 1.00000 0.0883883
$$129$$ 0 0
$$130$$ 2.00000 0.175412
$$131$$ −20.0000 −1.74741 −0.873704 0.486458i $$-0.838289\pi$$
−0.873704 + 0.486458i $$0.838289\pi$$
$$132$$ 0 0
$$133$$ 0 0
$$134$$ 12.0000 1.03664
$$135$$ 0 0
$$136$$ 0 0
$$137$$ −23.0000 −1.96502 −0.982511 0.186203i $$-0.940382\pi$$
−0.982511 + 0.186203i $$0.940382\pi$$
$$138$$ 0 0
$$139$$ −5.00000 −0.424094 −0.212047 0.977259i $$-0.568013\pi$$
−0.212047 + 0.977259i $$0.568013\pi$$
$$140$$ 1.00000 0.0845154
$$141$$ 0 0
$$142$$ 6.00000 0.503509
$$143$$ 2.00000 0.167248
$$144$$ −3.00000 −0.250000
$$145$$ −8.00000 −0.664364
$$146$$ 0 0
$$147$$ 0 0
$$148$$ 3.00000 0.246598
$$149$$ −4.00000 −0.327693 −0.163846 0.986486i $$-0.552390\pi$$
−0.163846 + 0.986486i $$0.552390\pi$$
$$150$$ 0 0
$$151$$ −9.00000 −0.732410 −0.366205 0.930534i $$-0.619343\pi$$
−0.366205 + 0.930534i $$0.619343\pi$$
$$152$$ 0 0
$$153$$ 0 0
$$154$$ 1.00000 0.0805823
$$155$$ −10.0000 −0.803219
$$156$$ 0 0
$$157$$ 21.0000 1.67598 0.837991 0.545684i $$-0.183730\pi$$
0.837991 + 0.545684i $$0.183730\pi$$
$$158$$ −11.0000 −0.875113
$$159$$ 0 0
$$160$$ 1.00000 0.0790569
$$161$$ −6.00000 −0.472866
$$162$$ 9.00000 0.707107
$$163$$ −4.00000 −0.313304 −0.156652 0.987654i $$-0.550070\pi$$
−0.156652 + 0.987654i $$0.550070\pi$$
$$164$$ −2.00000 −0.156174
$$165$$ 0 0
$$166$$ 7.00000 0.543305
$$167$$ 9.00000 0.696441 0.348220 0.937413i $$-0.386786\pi$$
0.348220 + 0.937413i $$0.386786\pi$$
$$168$$ 0 0
$$169$$ −9.00000 −0.692308
$$170$$ 0 0
$$171$$ 0 0
$$172$$ −8.00000 −0.609994
$$173$$ 22.0000 1.67263 0.836315 0.548250i $$-0.184706\pi$$
0.836315 + 0.548250i $$0.184706\pi$$
$$174$$ 0 0
$$175$$ −4.00000 −0.302372
$$176$$ 1.00000 0.0753778
$$177$$ 0 0
$$178$$ 14.0000 1.04934
$$179$$ −16.0000 −1.19590 −0.597948 0.801535i $$-0.704017\pi$$
−0.597948 + 0.801535i $$0.704017\pi$$
$$180$$ −3.00000 −0.223607
$$181$$ −5.00000 −0.371647 −0.185824 0.982583i $$-0.559495\pi$$
−0.185824 + 0.982583i $$0.559495\pi$$
$$182$$ 2.00000 0.148250
$$183$$ 0 0
$$184$$ −6.00000 −0.442326
$$185$$ 3.00000 0.220564
$$186$$ 0 0
$$187$$ 0 0
$$188$$ 0 0
$$189$$ 0 0
$$190$$ 0 0
$$191$$ −18.0000 −1.30243 −0.651217 0.758891i $$-0.725741\pi$$
−0.651217 + 0.758891i $$0.725741\pi$$
$$192$$ 0 0
$$193$$ 4.00000 0.287926 0.143963 0.989583i $$-0.454015\pi$$
0.143963 + 0.989583i $$0.454015\pi$$
$$194$$ 5.00000 0.358979
$$195$$ 0 0
$$196$$ −6.00000 −0.428571
$$197$$ 2.00000 0.142494 0.0712470 0.997459i $$-0.477302\pi$$
0.0712470 + 0.997459i $$0.477302\pi$$
$$198$$ −3.00000 −0.213201
$$199$$ 10.0000 0.708881 0.354441 0.935079i $$-0.384671\pi$$
0.354441 + 0.935079i $$0.384671\pi$$
$$200$$ −4.00000 −0.282843
$$201$$ 0 0
$$202$$ 10.0000 0.703598
$$203$$ −8.00000 −0.561490
$$204$$ 0 0
$$205$$ −2.00000 −0.139686
$$206$$ 6.00000 0.418040
$$207$$ 18.0000 1.25109
$$208$$ 2.00000 0.138675
$$209$$ 0 0
$$210$$ 0 0
$$211$$ 13.0000 0.894957 0.447478 0.894295i $$-0.352322\pi$$
0.447478 + 0.894295i $$0.352322\pi$$
$$212$$ −9.00000 −0.618123
$$213$$ 0 0
$$214$$ 17.0000 1.16210
$$215$$ −8.00000 −0.545595
$$216$$ 0 0
$$217$$ −10.0000 −0.678844
$$218$$ −10.0000 −0.677285
$$219$$ 0 0
$$220$$ 1.00000 0.0674200
$$221$$ 0 0
$$222$$ 0 0
$$223$$ −16.0000 −1.07144 −0.535720 0.844396i $$-0.679960\pi$$
−0.535720 + 0.844396i $$0.679960\pi$$
$$224$$ 1.00000 0.0668153
$$225$$ 12.0000 0.800000
$$226$$ 2.00000 0.133038
$$227$$ 25.0000 1.65931 0.829654 0.558278i $$-0.188538\pi$$
0.829654 + 0.558278i $$0.188538\pi$$
$$228$$ 0 0
$$229$$ −1.00000 −0.0660819 −0.0330409 0.999454i $$-0.510519\pi$$
−0.0330409 + 0.999454i $$0.510519\pi$$
$$230$$ −6.00000 −0.395628
$$231$$ 0 0
$$232$$ −8.00000 −0.525226
$$233$$ −18.0000 −1.17922 −0.589610 0.807688i $$-0.700718\pi$$
−0.589610 + 0.807688i $$0.700718\pi$$
$$234$$ −6.00000 −0.392232
$$235$$ 0 0
$$236$$ −14.0000 −0.911322
$$237$$ 0 0
$$238$$ 0 0
$$239$$ −5.00000 −0.323423 −0.161712 0.986838i $$-0.551701\pi$$
−0.161712 + 0.986838i $$0.551701\pi$$
$$240$$ 0 0
$$241$$ 10.0000 0.644157 0.322078 0.946713i $$-0.395619\pi$$
0.322078 + 0.946713i $$0.395619\pi$$
$$242$$ 1.00000 0.0642824
$$243$$ 0 0
$$244$$ 0 0
$$245$$ −6.00000 −0.383326
$$246$$ 0 0
$$247$$ 0 0
$$248$$ −10.0000 −0.635001
$$249$$ 0 0
$$250$$ −9.00000 −0.569210
$$251$$ −12.0000 −0.757433 −0.378717 0.925513i $$-0.623635\pi$$
−0.378717 + 0.925513i $$0.623635\pi$$
$$252$$ −3.00000 −0.188982
$$253$$ −6.00000 −0.377217
$$254$$ 20.0000 1.25491
$$255$$ 0 0
$$256$$ 1.00000 0.0625000
$$257$$ −21.0000 −1.30994 −0.654972 0.755653i $$-0.727320\pi$$
−0.654972 + 0.755653i $$0.727320\pi$$
$$258$$ 0 0
$$259$$ 3.00000 0.186411
$$260$$ 2.00000 0.124035
$$261$$ 24.0000 1.48556
$$262$$ −20.0000 −1.23560
$$263$$ −3.00000 −0.184988 −0.0924940 0.995713i $$-0.529484\pi$$
−0.0924940 + 0.995713i $$0.529484\pi$$
$$264$$ 0 0
$$265$$ −9.00000 −0.552866
$$266$$ 0 0
$$267$$ 0 0
$$268$$ 12.0000 0.733017
$$269$$ −21.0000 −1.28039 −0.640196 0.768211i $$-0.721147\pi$$
−0.640196 + 0.768211i $$0.721147\pi$$
$$270$$ 0 0
$$271$$ 15.0000 0.911185 0.455593 0.890188i $$-0.349427\pi$$
0.455593 + 0.890188i $$0.349427\pi$$
$$272$$ 0 0
$$273$$ 0 0
$$274$$ −23.0000 −1.38948
$$275$$ −4.00000 −0.241209
$$276$$ 0 0
$$277$$ −2.00000 −0.120168 −0.0600842 0.998193i $$-0.519137\pi$$
−0.0600842 + 0.998193i $$0.519137\pi$$
$$278$$ −5.00000 −0.299880
$$279$$ 30.0000 1.79605
$$280$$ 1.00000 0.0597614
$$281$$ 20.0000 1.19310 0.596550 0.802576i $$-0.296538\pi$$
0.596550 + 0.802576i $$0.296538\pi$$
$$282$$ 0 0
$$283$$ −7.00000 −0.416107 −0.208053 0.978117i $$-0.566713\pi$$
−0.208053 + 0.978117i $$0.566713\pi$$
$$284$$ 6.00000 0.356034
$$285$$ 0 0
$$286$$ 2.00000 0.118262
$$287$$ −2.00000 −0.118056
$$288$$ −3.00000 −0.176777
$$289$$ −17.0000 −1.00000
$$290$$ −8.00000 −0.469776
$$291$$ 0 0
$$292$$ 0 0
$$293$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$294$$ 0 0
$$295$$ −14.0000 −0.815112
$$296$$ 3.00000 0.174371
$$297$$ 0 0
$$298$$ −4.00000 −0.231714
$$299$$ −12.0000 −0.693978
$$300$$ 0 0
$$301$$ −8.00000 −0.461112
$$302$$ −9.00000 −0.517892
$$303$$ 0 0
$$304$$ 0 0
$$305$$ 0 0
$$306$$ 0 0
$$307$$ −23.0000 −1.31268 −0.656340 0.754466i $$-0.727896\pi$$
−0.656340 + 0.754466i $$0.727896\pi$$
$$308$$ 1.00000 0.0569803
$$309$$ 0 0
$$310$$ −10.0000 −0.567962
$$311$$ −10.0000 −0.567048 −0.283524 0.958965i $$-0.591504\pi$$
−0.283524 + 0.958965i $$0.591504\pi$$
$$312$$ 0 0
$$313$$ −21.0000 −1.18699 −0.593495 0.804838i $$-0.702252\pi$$
−0.593495 + 0.804838i $$0.702252\pi$$
$$314$$ 21.0000 1.18510
$$315$$ −3.00000 −0.169031
$$316$$ −11.0000 −0.618798
$$317$$ 22.0000 1.23564 0.617822 0.786318i $$-0.288015\pi$$
0.617822 + 0.786318i $$0.288015\pi$$
$$318$$ 0 0
$$319$$ −8.00000 −0.447914
$$320$$ 1.00000 0.0559017
$$321$$ 0 0
$$322$$ −6.00000 −0.334367
$$323$$ 0 0
$$324$$ 9.00000 0.500000
$$325$$ −8.00000 −0.443760
$$326$$ −4.00000 −0.221540
$$327$$ 0 0
$$328$$ −2.00000 −0.110432
$$329$$ 0 0
$$330$$ 0 0
$$331$$ −4.00000 −0.219860 −0.109930 0.993939i $$-0.535063\pi$$
−0.109930 + 0.993939i $$0.535063\pi$$
$$332$$ 7.00000 0.384175
$$333$$ −9.00000 −0.493197
$$334$$ 9.00000 0.492458
$$335$$ 12.0000 0.655630
$$336$$ 0 0
$$337$$ −8.00000 −0.435788 −0.217894 0.975972i $$-0.569919\pi$$
−0.217894 + 0.975972i $$0.569919\pi$$
$$338$$ −9.00000 −0.489535
$$339$$ 0 0
$$340$$ 0 0
$$341$$ −10.0000 −0.541530
$$342$$ 0 0
$$343$$ −13.0000 −0.701934
$$344$$ −8.00000 −0.431331
$$345$$ 0 0
$$346$$ 22.0000 1.18273
$$347$$ −3.00000 −0.161048 −0.0805242 0.996753i $$-0.525659\pi$$
−0.0805242 + 0.996753i $$0.525659\pi$$
$$348$$ 0 0
$$349$$ 28.0000 1.49881 0.749403 0.662114i $$-0.230341\pi$$
0.749403 + 0.662114i $$0.230341\pi$$
$$350$$ −4.00000 −0.213809
$$351$$ 0 0
$$352$$ 1.00000 0.0533002
$$353$$ 11.0000 0.585471 0.292735 0.956193i $$-0.405434\pi$$
0.292735 + 0.956193i $$0.405434\pi$$
$$354$$ 0 0
$$355$$ 6.00000 0.318447
$$356$$ 14.0000 0.741999
$$357$$ 0 0
$$358$$ −16.0000 −0.845626
$$359$$ 3.00000 0.158334 0.0791670 0.996861i $$-0.474774\pi$$
0.0791670 + 0.996861i $$0.474774\pi$$
$$360$$ −3.00000 −0.158114
$$361$$ 0 0
$$362$$ −5.00000 −0.262794
$$363$$ 0 0
$$364$$ 2.00000 0.104828
$$365$$ 0 0
$$366$$ 0 0
$$367$$ 10.0000 0.521996 0.260998 0.965339i $$-0.415948\pi$$
0.260998 + 0.965339i $$0.415948\pi$$
$$368$$ −6.00000 −0.312772
$$369$$ 6.00000 0.312348
$$370$$ 3.00000 0.155963
$$371$$ −9.00000 −0.467257
$$372$$ 0 0
$$373$$ −4.00000 −0.207112 −0.103556 0.994624i $$-0.533022\pi$$
−0.103556 + 0.994624i $$0.533022\pi$$
$$374$$ 0 0
$$375$$ 0 0
$$376$$ 0 0
$$377$$ −16.0000 −0.824042
$$378$$ 0 0
$$379$$ 2.00000 0.102733 0.0513665 0.998680i $$-0.483642\pi$$
0.0513665 + 0.998680i $$0.483642\pi$$
$$380$$ 0 0
$$381$$ 0 0
$$382$$ −18.0000 −0.920960
$$383$$ 8.00000 0.408781 0.204390 0.978889i $$-0.434479\pi$$
0.204390 + 0.978889i $$0.434479\pi$$
$$384$$ 0 0
$$385$$ 1.00000 0.0509647
$$386$$ 4.00000 0.203595
$$387$$ 24.0000 1.21999
$$388$$ 5.00000 0.253837
$$389$$ −23.0000 −1.16615 −0.583073 0.812420i $$-0.698150\pi$$
−0.583073 + 0.812420i $$0.698150\pi$$
$$390$$ 0 0
$$391$$ 0 0
$$392$$ −6.00000 −0.303046
$$393$$ 0 0
$$394$$ 2.00000 0.100759
$$395$$ −11.0000 −0.553470
$$396$$ −3.00000 −0.150756
$$397$$ −27.0000 −1.35509 −0.677546 0.735481i $$-0.736956\pi$$
−0.677546 + 0.735481i $$0.736956\pi$$
$$398$$ 10.0000 0.501255
$$399$$ 0 0
$$400$$ −4.00000 −0.200000
$$401$$ 10.0000 0.499376 0.249688 0.968326i $$-0.419672\pi$$
0.249688 + 0.968326i $$0.419672\pi$$
$$402$$ 0 0
$$403$$ −20.0000 −0.996271
$$404$$ 10.0000 0.497519
$$405$$ 9.00000 0.447214
$$406$$ −8.00000 −0.397033
$$407$$ 3.00000 0.148704
$$408$$ 0 0
$$409$$ 10.0000 0.494468 0.247234 0.968956i $$-0.420478\pi$$
0.247234 + 0.968956i $$0.420478\pi$$
$$410$$ −2.00000 −0.0987730
$$411$$ 0 0
$$412$$ 6.00000 0.295599
$$413$$ −14.0000 −0.688895
$$414$$ 18.0000 0.884652
$$415$$ 7.00000 0.343616
$$416$$ 2.00000 0.0980581
$$417$$ 0 0
$$418$$ 0 0
$$419$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$420$$ 0 0
$$421$$ −15.0000 −0.731055 −0.365528 0.930800i $$-0.619111\pi$$
−0.365528 + 0.930800i $$0.619111\pi$$
$$422$$ 13.0000 0.632830
$$423$$ 0 0
$$424$$ −9.00000 −0.437079
$$425$$ 0 0
$$426$$ 0 0
$$427$$ 0 0
$$428$$ 17.0000 0.821726
$$429$$ 0 0
$$430$$ −8.00000 −0.385794
$$431$$ 5.00000 0.240842 0.120421 0.992723i $$-0.461576\pi$$
0.120421 + 0.992723i $$0.461576\pi$$
$$432$$ 0 0
$$433$$ 30.0000 1.44171 0.720854 0.693087i $$-0.243750\pi$$
0.720854 + 0.693087i $$0.243750\pi$$
$$434$$ −10.0000 −0.480015
$$435$$ 0 0
$$436$$ −10.0000 −0.478913
$$437$$ 0 0
$$438$$ 0 0
$$439$$ 29.0000 1.38409 0.692047 0.721852i $$-0.256709\pi$$
0.692047 + 0.721852i $$0.256709\pi$$
$$440$$ 1.00000 0.0476731
$$441$$ 18.0000 0.857143
$$442$$ 0 0
$$443$$ 22.0000 1.04525 0.522626 0.852562i $$-0.324953\pi$$
0.522626 + 0.852562i $$0.324953\pi$$
$$444$$ 0 0
$$445$$ 14.0000 0.663664
$$446$$ −16.0000 −0.757622
$$447$$ 0 0
$$448$$ 1.00000 0.0472456
$$449$$ −5.00000 −0.235965 −0.117982 0.993016i $$-0.537643\pi$$
−0.117982 + 0.993016i $$0.537643\pi$$
$$450$$ 12.0000 0.565685
$$451$$ −2.00000 −0.0941763
$$452$$ 2.00000 0.0940721
$$453$$ 0 0
$$454$$ 25.0000 1.17331
$$455$$ 2.00000 0.0937614
$$456$$ 0 0
$$457$$ 38.0000 1.77757 0.888783 0.458329i $$-0.151552\pi$$
0.888783 + 0.458329i $$0.151552\pi$$
$$458$$ −1.00000 −0.0467269
$$459$$ 0 0
$$460$$ −6.00000 −0.279751
$$461$$ 26.0000 1.21094 0.605470 0.795868i $$-0.292985\pi$$
0.605470 + 0.795868i $$0.292985\pi$$
$$462$$ 0 0
$$463$$ 16.0000 0.743583 0.371792 0.928316i $$-0.378744\pi$$
0.371792 + 0.928316i $$0.378744\pi$$
$$464$$ −8.00000 −0.371391
$$465$$ 0 0
$$466$$ −18.0000 −0.833834
$$467$$ 6.00000 0.277647 0.138823 0.990317i $$-0.455668\pi$$
0.138823 + 0.990317i $$0.455668\pi$$
$$468$$ −6.00000 −0.277350
$$469$$ 12.0000 0.554109
$$470$$ 0 0
$$471$$ 0 0
$$472$$ −14.0000 −0.644402
$$473$$ −8.00000 −0.367840
$$474$$ 0 0
$$475$$ 0 0
$$476$$ 0 0
$$477$$ 27.0000 1.23625
$$478$$ −5.00000 −0.228695
$$479$$ −32.0000 −1.46212 −0.731059 0.682315i $$-0.760973\pi$$
−0.731059 + 0.682315i $$0.760973\pi$$
$$480$$ 0 0
$$481$$ 6.00000 0.273576
$$482$$ 10.0000 0.455488
$$483$$ 0 0
$$484$$ 1.00000 0.0454545
$$485$$ 5.00000 0.227038
$$486$$ 0 0
$$487$$ −2.00000 −0.0906287 −0.0453143 0.998973i $$-0.514429\pi$$
−0.0453143 + 0.998973i $$0.514429\pi$$
$$488$$ 0 0
$$489$$ 0 0
$$490$$ −6.00000 −0.271052
$$491$$ −15.0000 −0.676941 −0.338470 0.940977i $$-0.609909\pi$$
−0.338470 + 0.940977i $$0.609909\pi$$
$$492$$ 0 0
$$493$$ 0 0
$$494$$ 0 0
$$495$$ −3.00000 −0.134840
$$496$$ −10.0000 −0.449013
$$497$$ 6.00000 0.269137
$$498$$ 0 0
$$499$$ −32.0000 −1.43252 −0.716258 0.697835i $$-0.754147\pi$$
−0.716258 + 0.697835i $$0.754147\pi$$
$$500$$ −9.00000 −0.402492
$$501$$ 0 0
$$502$$ −12.0000 −0.535586
$$503$$ 16.0000 0.713405 0.356702 0.934218i $$-0.383901\pi$$
0.356702 + 0.934218i $$0.383901\pi$$
$$504$$ −3.00000 −0.133631
$$505$$ 10.0000 0.444994
$$506$$ −6.00000 −0.266733
$$507$$ 0 0
$$508$$ 20.0000 0.887357
$$509$$ −21.0000 −0.930809 −0.465404 0.885098i $$-0.654091\pi$$
−0.465404 + 0.885098i $$0.654091\pi$$
$$510$$ 0 0
$$511$$ 0 0
$$512$$ 1.00000 0.0441942
$$513$$ 0 0
$$514$$ −21.0000 −0.926270
$$515$$ 6.00000 0.264392
$$516$$ 0 0
$$517$$ 0 0
$$518$$ 3.00000 0.131812
$$519$$ 0 0
$$520$$ 2.00000 0.0877058
$$521$$ −31.0000 −1.35813 −0.679067 0.734076i $$-0.737616\pi$$
−0.679067 + 0.734076i $$0.737616\pi$$
$$522$$ 24.0000 1.05045
$$523$$ −7.00000 −0.306089 −0.153044 0.988219i $$-0.548908\pi$$
−0.153044 + 0.988219i $$0.548908\pi$$
$$524$$ −20.0000 −0.873704
$$525$$ 0 0
$$526$$ −3.00000 −0.130806
$$527$$ 0 0
$$528$$ 0 0
$$529$$ 13.0000 0.565217
$$530$$ −9.00000 −0.390935
$$531$$ 42.0000 1.82264
$$532$$ 0 0
$$533$$ −4.00000 −0.173259
$$534$$ 0 0
$$535$$ 17.0000 0.734974
$$536$$ 12.0000 0.518321
$$537$$ 0 0
$$538$$ −21.0000 −0.905374
$$539$$ −6.00000 −0.258438
$$540$$ 0 0
$$541$$ 8.00000 0.343947 0.171973 0.985102i $$-0.444986\pi$$
0.171973 + 0.985102i $$0.444986\pi$$
$$542$$ 15.0000 0.644305
$$543$$ 0 0
$$544$$ 0 0
$$545$$ −10.0000 −0.428353
$$546$$ 0 0
$$547$$ −1.00000 −0.0427569 −0.0213785 0.999771i $$-0.506805\pi$$
−0.0213785 + 0.999771i $$0.506805\pi$$
$$548$$ −23.0000 −0.982511
$$549$$ 0 0
$$550$$ −4.00000 −0.170561
$$551$$ 0 0
$$552$$ 0 0
$$553$$ −11.0000 −0.467768
$$554$$ −2.00000 −0.0849719
$$555$$ 0 0
$$556$$ −5.00000 −0.212047
$$557$$ −30.0000 −1.27114 −0.635570 0.772043i $$-0.719235\pi$$
−0.635570 + 0.772043i $$0.719235\pi$$
$$558$$ 30.0000 1.27000
$$559$$ −16.0000 −0.676728
$$560$$ 1.00000 0.0422577
$$561$$ 0 0
$$562$$ 20.0000 0.843649
$$563$$ −17.0000 −0.716465 −0.358232 0.933632i $$-0.616620\pi$$
−0.358232 + 0.933632i $$0.616620\pi$$
$$564$$ 0 0
$$565$$ 2.00000 0.0841406
$$566$$ −7.00000 −0.294232
$$567$$ 9.00000 0.377964
$$568$$ 6.00000 0.251754
$$569$$ 10.0000 0.419222 0.209611 0.977785i $$-0.432780\pi$$
0.209611 + 0.977785i $$0.432780\pi$$
$$570$$ 0 0
$$571$$ −37.0000 −1.54840 −0.774201 0.632940i $$-0.781848\pi$$
−0.774201 + 0.632940i $$0.781848\pi$$
$$572$$ 2.00000 0.0836242
$$573$$ 0 0
$$574$$ −2.00000 −0.0834784
$$575$$ 24.0000 1.00087
$$576$$ −3.00000 −0.125000
$$577$$ 25.0000 1.04076 0.520382 0.853934i $$-0.325790\pi$$
0.520382 + 0.853934i $$0.325790\pi$$
$$578$$ −17.0000 −0.707107
$$579$$ 0 0
$$580$$ −8.00000 −0.332182
$$581$$ 7.00000 0.290409
$$582$$ 0 0
$$583$$ −9.00000 −0.372742
$$584$$ 0 0
$$585$$ −6.00000 −0.248069
$$586$$ 0 0
$$587$$ 26.0000 1.07313 0.536567 0.843857i $$-0.319721\pi$$
0.536567 + 0.843857i $$0.319721\pi$$
$$588$$ 0 0
$$589$$ 0 0
$$590$$ −14.0000 −0.576371
$$591$$ 0 0
$$592$$ 3.00000 0.123299
$$593$$ 14.0000 0.574911 0.287456 0.957794i $$-0.407191\pi$$
0.287456 + 0.957794i $$0.407191\pi$$
$$594$$ 0 0
$$595$$ 0 0
$$596$$ −4.00000 −0.163846
$$597$$ 0 0
$$598$$ −12.0000 −0.490716
$$599$$ −16.0000 −0.653742 −0.326871 0.945069i $$-0.605994\pi$$
−0.326871 + 0.945069i $$0.605994\pi$$
$$600$$ 0 0
$$601$$ 42.0000 1.71322 0.856608 0.515968i $$-0.172568\pi$$
0.856608 + 0.515968i $$0.172568\pi$$
$$602$$ −8.00000 −0.326056
$$603$$ −36.0000 −1.46603
$$604$$ −9.00000 −0.366205
$$605$$ 1.00000 0.0406558
$$606$$ 0 0
$$607$$ −8.00000 −0.324710 −0.162355 0.986732i $$-0.551909\pi$$
−0.162355 + 0.986732i $$0.551909\pi$$
$$608$$ 0 0
$$609$$ 0 0
$$610$$ 0 0
$$611$$ 0 0
$$612$$ 0 0
$$613$$ 4.00000 0.161558 0.0807792 0.996732i $$-0.474259\pi$$
0.0807792 + 0.996732i $$0.474259\pi$$
$$614$$ −23.0000 −0.928204
$$615$$ 0 0
$$616$$ 1.00000 0.0402911
$$617$$ −15.0000 −0.603877 −0.301939 0.953327i $$-0.597634\pi$$
−0.301939 + 0.953327i $$0.597634\pi$$
$$618$$ 0 0
$$619$$ 46.0000 1.84890 0.924448 0.381308i $$-0.124526\pi$$
0.924448 + 0.381308i $$0.124526\pi$$
$$620$$ −10.0000 −0.401610
$$621$$ 0 0
$$622$$ −10.0000 −0.400963
$$623$$ 14.0000 0.560898
$$624$$ 0 0
$$625$$ 11.0000 0.440000
$$626$$ −21.0000 −0.839329
$$627$$ 0 0
$$628$$ 21.0000 0.837991
$$629$$ 0 0
$$630$$ −3.00000 −0.119523
$$631$$ −22.0000 −0.875806 −0.437903 0.899022i $$-0.644279\pi$$
−0.437903 + 0.899022i $$0.644279\pi$$
$$632$$ −11.0000 −0.437557
$$633$$ 0 0
$$634$$ 22.0000 0.873732
$$635$$ 20.0000 0.793676
$$636$$ 0 0
$$637$$ −12.0000 −0.475457
$$638$$ −8.00000 −0.316723
$$639$$ −18.0000 −0.712069
$$640$$ 1.00000 0.0395285
$$641$$ 39.0000 1.54041 0.770204 0.637798i $$-0.220155\pi$$
0.770204 + 0.637798i $$0.220155\pi$$
$$642$$ 0 0
$$643$$ −16.0000 −0.630978 −0.315489 0.948929i $$-0.602169\pi$$
−0.315489 + 0.948929i $$0.602169\pi$$
$$644$$ −6.00000 −0.236433
$$645$$ 0 0
$$646$$ 0 0
$$647$$ 18.0000 0.707653 0.353827 0.935311i $$-0.384880\pi$$
0.353827 + 0.935311i $$0.384880\pi$$
$$648$$ 9.00000 0.353553
$$649$$ −14.0000 −0.549548
$$650$$ −8.00000 −0.313786
$$651$$ 0 0
$$652$$ −4.00000 −0.156652
$$653$$ −18.0000 −0.704394 −0.352197 0.935926i $$-0.614565\pi$$
−0.352197 + 0.935926i $$0.614565\pi$$
$$654$$ 0 0
$$655$$ −20.0000 −0.781465
$$656$$ −2.00000 −0.0780869
$$657$$ 0 0
$$658$$ 0 0
$$659$$ −15.0000 −0.584317 −0.292159 0.956370i $$-0.594373\pi$$
−0.292159 + 0.956370i $$0.594373\pi$$
$$660$$ 0 0
$$661$$ −11.0000 −0.427850 −0.213925 0.976850i $$-0.568625\pi$$
−0.213925 + 0.976850i $$0.568625\pi$$
$$662$$ −4.00000 −0.155464
$$663$$ 0 0
$$664$$ 7.00000 0.271653
$$665$$ 0 0
$$666$$ −9.00000 −0.348743
$$667$$ 48.0000 1.85857
$$668$$ 9.00000 0.348220
$$669$$ 0 0
$$670$$ 12.0000 0.463600
$$671$$ 0 0
$$672$$ 0 0
$$673$$ 44.0000 1.69608 0.848038 0.529936i $$-0.177784\pi$$
0.848038 + 0.529936i $$0.177784\pi$$
$$674$$ −8.00000 −0.308148
$$675$$ 0 0
$$676$$ −9.00000 −0.346154
$$677$$ −36.0000 −1.38359 −0.691796 0.722093i $$-0.743180\pi$$
−0.691796 + 0.722093i $$0.743180\pi$$
$$678$$ 0 0
$$679$$ 5.00000 0.191882
$$680$$ 0 0
$$681$$ 0 0
$$682$$ −10.0000 −0.382920
$$683$$ 26.0000 0.994862 0.497431 0.867503i $$-0.334277\pi$$
0.497431 + 0.867503i $$0.334277\pi$$
$$684$$ 0 0
$$685$$ −23.0000 −0.878785
$$686$$ −13.0000 −0.496342
$$687$$ 0 0
$$688$$ −8.00000 −0.304997
$$689$$ −18.0000 −0.685745
$$690$$ 0 0
$$691$$ −50.0000 −1.90209 −0.951045 0.309053i $$-0.899988\pi$$
−0.951045 + 0.309053i $$0.899988\pi$$
$$692$$ 22.0000 0.836315
$$693$$ −3.00000 −0.113961
$$694$$ −3.00000 −0.113878
$$695$$ −5.00000 −0.189661
$$696$$ 0 0
$$697$$ 0 0
$$698$$ 28.0000 1.05982
$$699$$ 0 0
$$700$$ −4.00000 −0.151186
$$701$$ −36.0000 −1.35970 −0.679851 0.733351i $$-0.737955\pi$$
−0.679851 + 0.733351i $$0.737955\pi$$
$$702$$ 0 0
$$703$$ 0 0
$$704$$ 1.00000 0.0376889
$$705$$ 0 0
$$706$$ 11.0000 0.413990
$$707$$ 10.0000 0.376089
$$708$$ 0 0
$$709$$ 43.0000 1.61490 0.807449 0.589937i $$-0.200847\pi$$
0.807449 + 0.589937i $$0.200847\pi$$
$$710$$ 6.00000 0.225176
$$711$$ 33.0000 1.23760
$$712$$ 14.0000 0.524672
$$713$$ 60.0000 2.24702
$$714$$ 0 0
$$715$$ 2.00000 0.0747958
$$716$$ −16.0000 −0.597948
$$717$$ 0 0
$$718$$ 3.00000 0.111959
$$719$$ 8.00000 0.298350 0.149175 0.988811i $$-0.452338\pi$$
0.149175 + 0.988811i $$0.452338\pi$$
$$720$$ −3.00000 −0.111803
$$721$$ 6.00000 0.223452
$$722$$ 0 0
$$723$$ 0 0
$$724$$ −5.00000 −0.185824
$$725$$ 32.0000 1.18845
$$726$$ 0 0
$$727$$ 26.0000 0.964287 0.482143 0.876092i $$-0.339858\pi$$
0.482143 + 0.876092i $$0.339858\pi$$
$$728$$ 2.00000 0.0741249
$$729$$ −27.0000 −1.00000
$$730$$ 0 0
$$731$$ 0 0
$$732$$ 0 0
$$733$$ −36.0000 −1.32969 −0.664845 0.746981i $$-0.731502\pi$$
−0.664845 + 0.746981i $$0.731502\pi$$
$$734$$ 10.0000 0.369107
$$735$$ 0 0
$$736$$ −6.00000 −0.221163
$$737$$ 12.0000 0.442026
$$738$$ 6.00000 0.220863
$$739$$ 4.00000 0.147142 0.0735712 0.997290i $$-0.476560\pi$$
0.0735712 + 0.997290i $$0.476560\pi$$
$$740$$ 3.00000 0.110282
$$741$$ 0 0
$$742$$ −9.00000 −0.330400
$$743$$ 41.0000 1.50414 0.752072 0.659081i $$-0.229055\pi$$
0.752072 + 0.659081i $$0.229055\pi$$
$$744$$ 0 0
$$745$$ −4.00000 −0.146549
$$746$$ −4.00000 −0.146450
$$747$$ −21.0000 −0.768350
$$748$$ 0 0
$$749$$ 17.0000 0.621166
$$750$$ 0 0
$$751$$ 8.00000 0.291924 0.145962 0.989290i $$-0.453372\pi$$
0.145962 + 0.989290i $$0.453372\pi$$
$$752$$ 0 0
$$753$$ 0 0
$$754$$ −16.0000 −0.582686
$$755$$ −9.00000 −0.327544
$$756$$ 0 0
$$757$$ −50.0000 −1.81728 −0.908640 0.417579i $$-0.862879\pi$$
−0.908640 + 0.417579i $$0.862879\pi$$
$$758$$ 2.00000 0.0726433
$$759$$ 0 0
$$760$$ 0 0
$$761$$ −36.0000 −1.30500 −0.652499 0.757789i $$-0.726280\pi$$
−0.652499 + 0.757789i $$0.726280\pi$$
$$762$$ 0 0
$$763$$ −10.0000 −0.362024
$$764$$ −18.0000 −0.651217
$$765$$ 0 0
$$766$$ 8.00000 0.289052
$$767$$ −28.0000 −1.01102
$$768$$ 0 0
$$769$$ 2.00000 0.0721218 0.0360609 0.999350i $$-0.488519\pi$$
0.0360609 + 0.999350i $$0.488519\pi$$
$$770$$ 1.00000 0.0360375
$$771$$ 0 0
$$772$$ 4.00000 0.143963
$$773$$ 38.0000 1.36677 0.683383 0.730061i $$-0.260508\pi$$
0.683383 + 0.730061i $$0.260508\pi$$
$$774$$ 24.0000 0.862662
$$775$$ 40.0000 1.43684
$$776$$ 5.00000 0.179490
$$777$$ 0 0
$$778$$ −23.0000 −0.824590
$$779$$ 0 0
$$780$$ 0 0
$$781$$ 6.00000 0.214697
$$782$$ 0 0
$$783$$ 0 0
$$784$$ −6.00000 −0.214286
$$785$$ 21.0000 0.749522
$$786$$ 0 0
$$787$$ 11.0000 0.392108 0.196054 0.980593i $$-0.437187\pi$$
0.196054 + 0.980593i $$0.437187\pi$$
$$788$$ 2.00000 0.0712470
$$789$$ 0 0
$$790$$ −11.0000 −0.391362
$$791$$ 2.00000 0.0711118
$$792$$ −3.00000 −0.106600
$$793$$ 0 0
$$794$$ −27.0000 −0.958194
$$795$$ 0 0
$$796$$ 10.0000 0.354441
$$797$$ 21.0000 0.743858 0.371929 0.928261i $$-0.378696\pi$$
0.371929 + 0.928261i $$0.378696\pi$$
$$798$$ 0 0
$$799$$ 0 0
$$800$$ −4.00000 −0.141421
$$801$$ −42.0000 −1.48400
$$802$$ 10.0000 0.353112
$$803$$ 0 0
$$804$$ 0 0
$$805$$ −6.00000 −0.211472
$$806$$ −20.0000 −0.704470
$$807$$ 0 0
$$808$$ 10.0000 0.351799
$$809$$ −44.0000 −1.54696 −0.773479 0.633822i $$-0.781485\pi$$
−0.773479 + 0.633822i $$0.781485\pi$$
$$810$$ 9.00000 0.316228
$$811$$ 23.0000 0.807639 0.403820 0.914839i $$-0.367682\pi$$
0.403820 + 0.914839i $$0.367682\pi$$
$$812$$ −8.00000 −0.280745
$$813$$ 0 0
$$814$$ 3.00000 0.105150
$$815$$ −4.00000 −0.140114
$$816$$ 0 0
$$817$$ 0 0
$$818$$ 10.0000 0.349642
$$819$$ −6.00000 −0.209657
$$820$$ −2.00000 −0.0698430
$$821$$ −20.0000 −0.698005 −0.349002 0.937122i $$-0.613479\pi$$
−0.349002 + 0.937122i $$0.613479\pi$$
$$822$$ 0 0
$$823$$ −16.0000 −0.557725 −0.278862 0.960331i $$-0.589957\pi$$
−0.278862 + 0.960331i $$0.589957\pi$$
$$824$$ 6.00000 0.209020
$$825$$ 0 0
$$826$$ −14.0000 −0.487122
$$827$$ −8.00000 −0.278187 −0.139094 0.990279i $$-0.544419\pi$$
−0.139094 + 0.990279i $$0.544419\pi$$
$$828$$ 18.0000 0.625543
$$829$$ −10.0000 −0.347314 −0.173657 0.984806i $$-0.555558\pi$$
−0.173657 + 0.984806i $$0.555558\pi$$
$$830$$ 7.00000 0.242974
$$831$$ 0 0
$$832$$ 2.00000 0.0693375
$$833$$ 0 0
$$834$$ 0 0
$$835$$ 9.00000 0.311458
$$836$$ 0 0
$$837$$ 0 0
$$838$$ 0 0
$$839$$ −16.0000 −0.552381 −0.276191 0.961103i $$-0.589072\pi$$
−0.276191 + 0.961103i $$0.589072\pi$$
$$840$$ 0 0
$$841$$ 35.0000 1.20690
$$842$$ −15.0000 −0.516934
$$843$$ 0 0
$$844$$ 13.0000 0.447478
$$845$$ −9.00000 −0.309609
$$846$$ 0 0
$$847$$ 1.00000 0.0343604
$$848$$ −9.00000 −0.309061
$$849$$ 0 0
$$850$$ 0 0
$$851$$ −18.0000 −0.617032
$$852$$ 0 0
$$853$$ −30.0000 −1.02718 −0.513590 0.858036i $$-0.671685\pi$$
−0.513590 + 0.858036i $$0.671685\pi$$
$$854$$ 0 0
$$855$$ 0 0
$$856$$ 17.0000 0.581048
$$857$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$858$$ 0 0
$$859$$ −14.0000 −0.477674 −0.238837 0.971060i $$-0.576766\pi$$
−0.238837 + 0.971060i $$0.576766\pi$$
$$860$$ −8.00000 −0.272798
$$861$$ 0 0
$$862$$ 5.00000 0.170301
$$863$$ −32.0000 −1.08929 −0.544646 0.838666i $$-0.683336\pi$$
−0.544646 + 0.838666i $$0.683336\pi$$
$$864$$ 0 0
$$865$$ 22.0000 0.748022
$$866$$ 30.0000 1.01944
$$867$$ 0 0
$$868$$ −10.0000 −0.339422
$$869$$ −11.0000 −0.373149
$$870$$ 0 0
$$871$$ 24.0000 0.813209
$$872$$ −10.0000 −0.338643
$$873$$ −15.0000 −0.507673
$$874$$ 0 0
$$875$$ −9.00000 −0.304256
$$876$$ 0 0
$$877$$ −18.0000 −0.607817 −0.303908 0.952701i $$-0.598292\pi$$
−0.303908 + 0.952701i $$0.598292\pi$$
$$878$$ 29.0000 0.978703
$$879$$ 0 0
$$880$$ 1.00000 0.0337100
$$881$$ −14.0000 −0.471672 −0.235836 0.971793i $$-0.575783\pi$$
−0.235836 + 0.971793i $$0.575783\pi$$
$$882$$ 18.0000 0.606092
$$883$$ 20.0000 0.673054 0.336527 0.941674i $$-0.390748\pi$$
0.336527 + 0.941674i $$0.390748\pi$$
$$884$$ 0 0
$$885$$ 0 0
$$886$$ 22.0000 0.739104
$$887$$ −35.0000 −1.17518 −0.587592 0.809157i $$-0.699924\pi$$
−0.587592 + 0.809157i $$0.699924\pi$$
$$888$$ 0 0
$$889$$ 20.0000 0.670778
$$890$$ 14.0000 0.469281
$$891$$ 9.00000 0.301511
$$892$$ −16.0000 −0.535720
$$893$$ 0 0
$$894$$ 0 0
$$895$$ −16.0000 −0.534821
$$896$$ 1.00000 0.0334077
$$897$$ 0 0
$$898$$ −5.00000 −0.166852
$$899$$ 80.0000 2.66815
$$900$$ 12.0000 0.400000
$$901$$ 0 0
$$902$$ −2.00000 −0.0665927
$$903$$ 0 0
$$904$$ 2.00000 0.0665190
$$905$$ −5.00000 −0.166206
$$906$$ 0 0
$$907$$ −38.0000 −1.26177 −0.630885 0.775877i $$-0.717308\pi$$
−0.630885 + 0.775877i $$0.717308\pi$$
$$908$$ 25.0000 0.829654
$$909$$ −30.0000 −0.995037
$$910$$ 2.00000 0.0662994
$$911$$ −12.0000 −0.397578 −0.198789 0.980042i $$-0.563701\pi$$
−0.198789 + 0.980042i $$0.563701\pi$$
$$912$$ 0 0
$$913$$ 7.00000 0.231666
$$914$$ 38.0000 1.25693
$$915$$ 0 0
$$916$$ −1.00000 −0.0330409
$$917$$ −20.0000 −0.660458
$$918$$ 0 0
$$919$$ −39.0000 −1.28649 −0.643246 0.765660i $$-0.722413\pi$$
−0.643246 + 0.765660i $$0.722413\pi$$
$$920$$ −6.00000 −0.197814
$$921$$ 0 0
$$922$$ 26.0000 0.856264
$$923$$ 12.0000 0.394985
$$924$$ 0 0
$$925$$ −12.0000 −0.394558
$$926$$ 16.0000 0.525793
$$927$$ −18.0000 −0.591198
$$928$$ −8.00000 −0.262613
$$929$$ −6.00000 −0.196854 −0.0984268 0.995144i $$-0.531381\pi$$
−0.0984268 + 0.995144i $$0.531381\pi$$
$$930$$ 0 0
$$931$$ 0 0
$$932$$ −18.0000 −0.589610
$$933$$ 0 0
$$934$$ 6.00000 0.196326
$$935$$ 0 0
$$936$$ −6.00000 −0.196116
$$937$$ 16.0000 0.522697 0.261349 0.965244i $$-0.415833\pi$$
0.261349 + 0.965244i $$0.415833\pi$$
$$938$$ 12.0000 0.391814
$$939$$ 0 0
$$940$$ 0 0
$$941$$ 24.0000 0.782378 0.391189 0.920310i $$-0.372064\pi$$
0.391189 + 0.920310i $$0.372064\pi$$
$$942$$ 0 0
$$943$$ 12.0000 0.390774
$$944$$ −14.0000 −0.455661
$$945$$ 0 0
$$946$$ −8.00000 −0.260102
$$947$$ 10.0000 0.324956 0.162478 0.986712i $$-0.448051\pi$$
0.162478 + 0.986712i $$0.448051\pi$$
$$948$$ 0 0
$$949$$ 0 0
$$950$$ 0 0
$$951$$ 0 0
$$952$$ 0 0
$$953$$ 34.0000 1.10137 0.550684 0.834714i $$-0.314367\pi$$
0.550684 + 0.834714i $$0.314367\pi$$
$$954$$ 27.0000 0.874157
$$955$$ −18.0000 −0.582466
$$956$$ −5.00000 −0.161712
$$957$$ 0 0
$$958$$ −32.0000 −1.03387
$$959$$ −23.0000 −0.742709
$$960$$ 0 0
$$961$$ 69.0000 2.22581
$$962$$ 6.00000 0.193448
$$963$$ −51.0000 −1.64345
$$964$$ 10.0000 0.322078
$$965$$ 4.00000 0.128765
$$966$$ 0 0
$$967$$ −1.00000 −0.0321578 −0.0160789 0.999871i $$-0.505118\pi$$
−0.0160789 + 0.999871i $$0.505118\pi$$
$$968$$ 1.00000 0.0321412
$$969$$ 0 0
$$970$$ 5.00000 0.160540
$$971$$ −20.0000 −0.641831 −0.320915 0.947108i $$-0.603990\pi$$
−0.320915 + 0.947108i $$0.603990\pi$$
$$972$$ 0 0
$$973$$ −5.00000 −0.160293
$$974$$ −2.00000 −0.0640841
$$975$$ 0 0
$$976$$ 0 0
$$977$$ 33.0000 1.05576 0.527882 0.849318i $$-0.322986\pi$$
0.527882 + 0.849318i $$0.322986\pi$$
$$978$$ 0 0
$$979$$ 14.0000 0.447442
$$980$$ −6.00000 −0.191663
$$981$$ 30.0000 0.957826
$$982$$ −15.0000 −0.478669
$$983$$ 14.0000 0.446531 0.223265 0.974758i $$-0.428328\pi$$
0.223265 + 0.974758i $$0.428328\pi$$
$$984$$ 0 0
$$985$$ 2.00000 0.0637253
$$986$$ 0 0
$$987$$ 0 0
$$988$$ 0 0
$$989$$ 48.0000 1.52631
$$990$$ −3.00000 −0.0953463
$$991$$ 4.00000 0.127064 0.0635321 0.997980i $$-0.479763\pi$$
0.0635321 + 0.997980i $$0.479763\pi$$
$$992$$ −10.0000 −0.317500
$$993$$ 0 0
$$994$$ 6.00000 0.190308
$$995$$ 10.0000 0.317021
$$996$$ 0 0
$$997$$ 52.0000 1.64686 0.823428 0.567420i $$-0.192059\pi$$
0.823428 + 0.567420i $$0.192059\pi$$
$$998$$ −32.0000 −1.01294
$$999$$ 0 0
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 7942.2.a.q.1.1 1
19.7 even 3 418.2.e.b.353.1 yes 2
19.11 even 3 418.2.e.b.45.1 2
19.18 odd 2 7942.2.a.f.1.1 1
By twisted newform
Twist Min Dim Char Parity Ord Type
418.2.e.b.45.1 2 19.11 even 3
418.2.e.b.353.1 yes 2 19.7 even 3
7942.2.a.f.1.1 1 19.18 odd 2
7942.2.a.q.1.1 1 1.1 even 1 trivial | 19,978 | 36,168 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "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} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.394609 |
https://www.mathworks.com/matlabcentral/answers/364571-unable-to-see-graph?s_tid=prof_contriblnk | 1,660,044,399,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570921.9/warc/CC-MAIN-20220809094531-20220809124531-00500.warc.gz | 761,381,455 | 24,976 | # Unable to see Graph
1 view (last 30 days)
SARVESH AGRAWAL on 1 Nov 2017
Commented: SARVESH AGRAWAL on 2 Nov 2017
Hi,
Following is the code of sinc function which I am trying to plot. However, I do not see anything. Any help will be appreciated.
h=5;
lambda=h;
beta=2*pi/lambda;
for theta=0:0.01:2*pi
T=abs(sinc(beta*h*cos(theta)));
% T=sin(beta*h*cos(theta))/(beta*h*cos(theta));
polarplot(theta,T);
hold on;
end
Eric on 1 Nov 2017
Instead of using a for loop, take advantage of matlab's ability to work with vectors (don't forget the element-wise period for T2):
h=5; lambda=h; beta=2*pi/lambda;
theta=0:0.01:2*pi;
T1=abs(sinc(beta.*h.*cos(theta)));
T2=sin(beta.*h.*cos(theta))./(beta.*h.*cos(theta));
polarplot(theta,T1);
hold on;
polarplot(theta,T2);
SARVESH AGRAWAL on 2 Nov 2017
Thank you sir! | 269 | 802 | {"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} | 2.53125 | 3 | CC-MAIN-2022-33 | latest | en | 0.711522 |
https://againstthemodernworld.blogspot.com/2008/05/ | 1,620,924,337,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989814.35/warc/CC-MAIN-20210513142421-20210513172421-00542.warc.gz | 113,214,970 | 11,959 | ## Tuesday, May 20, 2008
### moral certainty
Something is morally certain if its probability comes so close to complete certainty that the difference cannot be perceived. By contrast, something is morally impossible if it has only as much certainty as the amount by which moral certainty falls short of complete certainty. Thus if we take something that possesses 9991000 of certainty to be morally certain, then something that has only 11000 of certainty will be morally impossible.
~ Jacob Bernoulli (1713) Ars Conjectandi
## Sunday, May 18, 2008
### sinai billiards
Billiards is our paradigmatic metaphor for understanding a deterministic universe. This is at least partly due to the fact that motion on a billiards table is quite simple and periodic.
Suppose, however, instead of a usual billiards table, we play upon one with a convex obstacle, say a hemisphere, in the middle of the table. Now, the motion of the balls, though still deterministic, exhibits chaotic behavior.
The behavior of the balls is still governed by deterministic equations, yet it is no longer simple or periodic.
Consider now the inverse problem. Suppose, rather than knowing these equations, we are observing the table and trying to determine the mathematics which governs the movement of the balls. How might we do this? Since our observations are necessarily of finite accuracy, we might partition the free portion of the table into a finite grid. At various points in time we make observations and record which squares of the grid in which a ball is present. The size of the squares in the grid represents the accuracy of our measurements.
In 1976, Y. G. Sinai proved that in this inverse situation, billiard ball behavior governed by a deterministic equation and billiard ball behavior governed by a first order Markov process are indistinguishable. Since a first order Markov process is a probabilistic process which depends only upon the previous square of the grid, it is both indeterministic and, in principle, unpredictable.
Thus, not only is it the case that we are in principle unable to distinguish between a deterministic and an indeterministic universe; but also, even if the universe is deterministic, and we know its initial conditions to an arbitrary degree of accuracy, we will still, in principle, be unable to predict its behavior.
[images from Suppes, Patrick (1999) "The Noninvariance of Deterministic Causal Models"] | 499 | 2,429 | {"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} | 3.0625 | 3 | CC-MAIN-2021-21 | longest | en | 0.935534 |
https://afanvalleyadventureresort.com/how-do-you-measure-a-blower-wheel/ | 1,656,550,926,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103646990.40/warc/CC-MAIN-20220630001553-20220630031553-00656.warc.gz | 140,081,256 | 12,793 | # How do you measure a blower wheel?
Wheel Diameter: Measure the diameter of the blower wheel – the longest distance from one side of the ring to the other side. From the appropriate perspective described above you will see the wheel is spinning clockwise (CW) or counter clockwise (CCW).
Support Tools. Fan rotation is determined by looking at the fan from the drive end. The drive end is is the side of the fan housing through which the fan or motor shaft protrudes. For DWDI or Arrangement 3 SWSI fans, it is the side of the fan housing at which the belt drive sheave is mounted.
Furthermore, how much does a blower wheel cost? An HVAC blower motor replacement costs an average of \$450 but anywhere from \$150 to \$2,000 to replace.
HVAC Blower Parts Cost.
Belt \$5-\$100
Shaft \$50-\$200
Blower Wheel (Squirrel Cage) \$50-\$275
Pulley \$50-\$250
Housing \$100-\$300
Simply so, what is a blower wheel?
The blower wheel is a component of the blower motor and its job is to move large volumes of air through the duct system. If the blower wheel get dirt or grease built up, it can slow down the wheels rotation and reduce the amount of air your blower motor is capable of moving through your home.
How do I know if my blower is CW or CCW?
Hold the fan blade in a horizontal position, hub end up. If the blade curves down to the left, it is CW and if it curves down to the right, it is CCW when looking at the shaft end of the motor. Blower rotation is often marked on the blower wheel itself.
### How does a blower wheel work?
Centrifugal fans use the kinetic energy of the impellers to increase the volume of the air stream, which in turn moves against the resistance caused by ducts, dampers and other components. Centrifugal fans displace air radially, changing the direction (typically by 90°) of the airflow.
### What does CW rotation mean?
Two-dimensional rotation can occur in two possible directions. A clockwise (typically abbreviated as CW) motion is one that proceeds in the same direction as a clock’s hands: from the top to the right, then down and then to the left, and back up to the top.
### How does a backward inclined fan work?
A Backward curved centrifugal fan is characterised by its cylindrical shape, several large curved blades and a conical inlet nozzle. In the example shown below, the fan rotates in a clockwise direction. As the fan rotates a pressure difference is created on the impeller blades.
### How do squirrel cage blowers work?
Squirrel Cage Blowers. This action imparts a centrifugal force to the air, causing it to move radially outwards to the walls of the blower or fan housing. The air follows a spiral trajectory – increasing in pressure and velocity – until it exits the discharge end of the blower.
### Which direction should a squirrel cage fan turn?
Standard forward inclined fan blade. It spins clockwise to move air.
### How many types of blowers are there?
There are six main kinds of blowers, each with specific characteristics and uses. Positive Displacement / Rotary Lobe Blowers. A positive displacement blower has a function that’s straightforward yet effective. Helical Screw Blowers. Centrifugal Blowers. High Speed Blowers. Regenerative Blowers.
### What is difference between blower and fan?
A fan is an electrical device, while a blower is a mechanical device. Fans have blades that help to creates a continuous airflow and circulate the air around in every direction, whereas a blower has impellers that channel the air in a specific direction towards a particular location.
### What is the function of the blower?
Blowers serve three main functions: heating, cooling and air flow. Although unit designs range from simple and complex, most blowers usually consist of some type of fan-like apparatus. Some blowers also use pressure pumps to move air or gases. Heating blowers direct warm or hot air into cooler areas.
### How much does it cost to replace a blower motor?
The average cost for a blower motor replacement is between \$363 and \$394. Labor costs are estimated between \$87 and \$111 while parts are priced between \$276 and \$283.
### How much does it cost to replace a furnace blower?
How much does a furnace motor replacement cost? On average nationwide, a furnace blower motor replacement costs \$400-\$600, including parts and labor. A single-speed blower motor costs around \$450 to replace while a variable-speed motor costs \$600+.
### What are regenerative blowers used for?
A regenerative blower can be used to remove dust or smoke from the air, extract soil vapor, or perform sewage aeration. They are well-suited for applications that require high rates of air flow at a low pressure, or when a vacuum is required in a process.
### What is a fan impeller?
A fan impeller consists of a number of blades mounted at a pitch angle and assembled on, or integral with, a hub mounted on a driven shaft.
### How do you clean the blower on a furnace?
Cleaning the Fan and Motor There really isn’t much to cleaning the fan in your furnace once it’s out of the unit. Using a toothbrush or paint brush, you can clean between the fan blades, motor housing and squirrel cage. Your vacuum will be good to keep the dust down and suck up the clump of dust that gets left behind. | 1,174 | 5,282 | {"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} | 2.734375 | 3 | CC-MAIN-2022-27 | latest | en | 0.916908 |
https://community.jmp.com/t5/JMP-Blog/Principal-components-or-factor-analysis/bc-p/68783/highlight/true | 1,563,443,290,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525587.2/warc/CC-MAIN-20190718083839-20190718105839-00121.warc.gz | 340,656,332 | 48,281 | Choose Language Hide Translation Bar
Staff
Principal components or factor analysis?
Should I use principal components analysis (PCA) or Exploratory Factor Analysis (EFA) for my work? This is a common question that analysts working with multivariate data, such as social scientists, consumer researchers, or engineers, face on a regular basis.
In this post, I share my favorite example for explaining a key difference between PCA and EFA. This distinction opens the door to explaining other important differences and is helpful when figuring out which technique is most appropriate for a given application. Choosing improperly might mean misleading results or incorrect understanding of the data.
### An Illustrative Example
Let’s start by creating some data that follow a standard normal distribution (a JSL script for all analyses in this post is attached if you want to follow along). Specifically, I create a data table with 1,000 observations on four variables that are uncorrelated with each other.
We can use the Multivariate platform in JMP to look at the correlations between variables and confirm they are independent. I particularly like to use the Color Map on Correlations to illustrate the null correlations in the off-diagonal:
Figure 1. Correlation coefficients and corresponding heat map for four simulated variables.Now, we can ask ourselves an important question: What would the results look like if I use PCA on these data? And what would the results look like if I use EFA instead? If you’re not sure, keep reading.
Let’s use the Factor Analysis platform in JMP to perform simultaneously a PCA and an EFA on these data. I’ll retain one component/factor only because we have a small number of variables. This also eliminates the need for any rotation.
The component or factor loadings from the analyses are critical to help us understand what the component or factor represents; variables with high loadings (usually defined as .4 in absolute value or higher because this suggests at least 16% of the measured variable variance overlaps with the variance of the factor) are most representative of the component or factor. Below, we can compare the resulting component loadings (displayed first) against the factor loadings (displayed second).
Figure 2. Component and factor loadings from a principal components analysis and exploratory factor analysis on four simulated variables that are uncorrelated.
We can see the results are strikingly different! PCA gave us three loadings greater than .4 in absolute value whereas EFA didn’t give any. Why? Because when we do EFA we are implicitly requesting an analysis on a reduced correlation matrix, for which the ones in the diagonal have been replaced by squared multiple correlations (SMC). Indeed, a quick look at the Color Map on Correlations of the reduced correlation matrix sheds light on why one would obtain such different results:
Figure 3. Heat map of reduced correlation matrix, where unities in the diagonal have been replaced by squared multiple correlations.
Every entry in the reduced correlation matrix, in this example, is very small (nearly zero! The actual values are 0.002, 0.002, 0.004, and 0.001). An eigenvalue decomposition of the full correlation matrix (Figure 1) is done in PCA, yet for EFA, the eigenvalue decomposition is done on the reduced correlation matrix (Figure 3). Differences in the data analyzed help explain differences across analyses, but none of this tells us what the differences mean from a practical point of view….
### Practical Meaning of Analyzing a Full vs. Reduced Correlation Matrix
PCA and EFA have different goals: PCA is a technique for reducing the dimensionality of one’s data, whereas EFA is a technique for identifying and measuring variables that cannot be measured directly (i.e., latent variables or factors). Thus, in PCA all of the variance in the data reflected by the full correlation matrix — is used to attain a solution, and the resulting components are a mix of what the variables intended to measure and other sources of variance such as measurement error (see left panel of Figure 4).
By contrast, in EFA not all of the variance in the data comes from the underlying latent variable (see right panel of Figure 4). This feature is reflected in the EFA algorithm by “reducing” the correlation matrix with the SMC values. This is appropriate because a SMC is the estimate of the variance that the underlying factor(s) explain in a given variable (aka communality). If we carried out EFA with unities in the diagonal, then we would be implicitly saying the factors explain all the variance in the measured variables, and we would be doing PCA rather than EFA.
Figure 4. Graphic comparison of principal components analysis and exploratory factor analysis.
Figure 4 also illustrates another important distinction between PCA and EFA. Note the arrows in PCA are pointing from the measured variables to the principal component, and in EFA it’s the other way around. The arrows represent causal relations, such that the variability in measured variables in PCA cause the variance in the principal component. This is in contrast to EFA, where the latent factor is seen as causing the variability and pattern of correlations among measured variables (Marcoulides & Hershberger, 1997).
In the interest of clarity, it’s important I outline a few more observations. First, most multivariate data are correlated to some degree, so differences between PCA and EFA don’t tend to be as marked as the ones in this example. Second, as the number of variables involved in the analysis grows, results from PCA and EFA become more and more similar. Researchers have argued that analyses with at least 40 variables lead to minor differences (Snook & Gorsuch, 1989). Third, if the communality of measured variables is high (i.e., approaches 1), then the results between PCA and EFA are also similar. Finally, this favorite example of mine relies on using the “Principal Axis” factoring method, but other estimation methods exist for which results would vary. All of these observations must be taken into account when analysts make the choice between EFA and PCA. But perhaps most importantly for psychometricians (those who developed EFA in the first place) is the fact that EFA posits a theory about the variables being analyzed; a theory that dates back to Spearman (1904) and suggests unobserved factors determine what we’re able to measure directly.
I list some key points below, but note that an excellent source for continuing to learn about this topic is Widaman (2007).
### Key Points
• PCA is useful for reducing the number of variables while retaining the most amount of information in the data, whereas EFA is useful for measuring unobserved (latent), error-free variables.
• When variables don’t have anything in common, as in the example above, EFA won’t find a well-defined underlying factor, but PCA will find a well-defined principal component that explains the maximal amount of variance in the data.
• When the goal is to measure an error-free latent variable but PCA is used, the component loadings will most likely be higher than they would’ve been if EFA was used. This would mislead analysts into thinking they have a well-defined, error-free factor when in fact they have a well-defined component that’s an amalgam of all the sources of variance in the data.
• When the goal is to get a small subset of variables that retain the most amount of variability in the data but EFA is used, the factor loadings will likely be lower than they would’ve been if PCA was used. This would mislead analysts into thinking they kept the maximal amount of variance in the data when in fact they kept the variance that’s in common across the measured variables.
### References
Marcoulides, G. A., & Hershberger, S. L. (1997). Multivariate statistical methods: A first course. Psychology Press.
Snook, S. C., & Gorsuch, R. L. (1989). Component analysis versus common factor analysis: A Monte Carlo study. Psychological Bulletin, 106, 148-154.
Spearman, C. (1904). "General intelligence," objectively determined and measured. The American Journal of Psychology, 15, 201-293.
Widaman, K. F. (2007). Common factors versus components: Principals and principles, errors and misconceptions. Factor analysis at 100: Historical developments and future directions, 177-203.
Labels
Staff
Laura, I really appreciate that you took the time to explain the difference between these techniques. This was very well written and informative and I loved your use of graphs and diagrams for illustrating your points. Thank you!
Staff
Thanks Shannon!
Community Member
Hi Laura,
The last diagram for comparison of EFA and PCA is very helpful to undrestand. Could you let me know a source, if this is from somewhere else, to use it in my presentation?
Thank you!
Best,
Young
Staff
Young,
I'm glad you found the last diagram helpful. I created that image for this post and am not aware of other sources that have something similar.
Best,
~Laura
Community Member
Hi Laura,
Thank you for the reply. I will then cite this website. Thanks for the very
useful diagram.
Best,
Young
Community Trekker
Super helpful. In fact, so helpful that I'll be incorporating these ideas into an upcoming grant on coral work!
Staff | 1,945 | 9,351 | {"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} | 3.25 | 3 | CC-MAIN-2019-30 | latest | en | 0.898478 |
https://wiki.fis-ski.com/index.php?title=Vertical_drop&diff=37482&oldid=17362 | 1,642,871,474,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303868.98/warc/CC-MAIN-20220122164421-20220122194421-00166.warc.gz | 656,179,595 | 5,761 | # Vertical drop
(Difference between revisions)
Revision as of 17:27, 19 September 2009 (edit)Joe (Talk | contribs)← Previous diff Current revision (13:29, 27 November 2018) (edit) (undo)Joe (Talk | contribs) (→'''Slope Steepness''') (13 intermediate revisions not shown.) Line 1: Line 1: - the total difference expressed in meters from '''[[the start]]''' elevation to '''[[the finish]]''' elevation. + '''[[Vertical drop]]''' the total difference expressed in meters from '''[[the start]]''' '''[[elevation]]''' to '''[[the finish]]''' '''[[elevation]]'''. - ---- + == '''Slope Steepness''' == + [[Image:Grades degrees.png|There are two ways of measuring slope steepness. The first is a measurement using '''[[degrees]]''', which measures the angle of the slope in degrees from the horizontal. The second is a percentage.|thumb|200px]] + There are two ways of measuring slope steepness. The first is a measurement using '''[[degrees]]''', which measures the angle of the slope in degrees from the horizontal. The second is a percentage. - Return to '''[[Ski Cross Glossary]] or [[Freestyle Skiing]]''' + This is calculated using the formula 100*rise/run. Rise is the vertical change and run is the horizontal distance. A 100% slope is angled at 45 degrees. + + *10% is equivalent to 5.71º + *20% is equivalent to 11.31º + *30% is equivalent to 16.7º + *40% is equivalent to 21.8º + *50% is equivalent to 25.67º + *75% is equivalent to 36.8º + *100% is equivalent to 45º + + A beginner slope is typically between 6% and 25%. Intermediate hills range from 25% to 40%, and expert is 40% plus. + + == '''Also See''' == + *'''[[Contour line]]''' + *'''[[latitude]]''' + *'''[[longitude]]''' + *'''[[Global Positioning System]]''' + * '''[[Zenith]]''' + *'''[[time]]''' + *'''[[Codex]]''' + *'''[[Year]]''' + *'''[[Date]]''' + *'''[[Season]]''' + *'''[[Freestyle Course Layout]]''' + *'''[[Maximum Grade]]''' + *'''[[Application for Freestyle Course Homologation]]''' + *'''[[Freestyle Course Homologation Program]]''' + + ---- + Return to '''[[Ski Cross Glossary]], [[Freestyle Skiing]]''' or '''[[Freestyle Course Layout]]''' [[Category:Freestyle Skiing]][[Category:Freestyle Glossary]] [[Category:Freestyle Skiing Statistics]] [[Category:Freestyle Skiing]][[Category:Freestyle Glossary]] [[Category:Freestyle Skiing Statistics]] + [[Category:Freestyle Course Preparation]]
## Current revision
Vertical drop the total difference expressed in meters from the start elevation to the finish elevation.
## Slope Steepness
There are two ways of measuring slope steepness. The first is a measurement using degrees, which measures the angle of the slope in degrees from the horizontal. The second is a percentage.
There are two ways of measuring slope steepness. The first is a measurement using degrees, which measures the angle of the slope in degrees from the horizontal. The second is a percentage.
This is calculated using the formula 100*rise/run. Rise is the vertical change and run is the horizontal distance. A 100% slope is angled at 45 degrees.
• 10% is equivalent to 5.71º
• 20% is equivalent to 11.31º
• 30% is equivalent to 16.7º
• 40% is equivalent to 21.8º
• 50% is equivalent to 25.67º
• 75% is equivalent to 36.8º
• 100% is equivalent to 45º
A beginner slope is typically between 6% and 25%. Intermediate hills range from 25% to 40%, and expert is 40% plus. | 914 | 3,376 | {"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} | 3.015625 | 3 | CC-MAIN-2022-05 | latest | en | 0.822502 |
https://bytes.com/topic/c/answers/708282-converting-float-4-const-float | 1,660,182,554,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571232.43/warc/CC-MAIN-20220811012302-20220811042302-00131.warc.gz | 163,763,378 | 11,112 | 471,066 Members | 1,005 Online
# converting 'float (*)[4]' to 'const float**'
Hi,
I know I am doing something stupid here, but it's friday night and I
cannot see what is the issue here:
Thanks,
-Mathieu
#include <iostream>
void print(const float **a,unsigned int X,unsigned int Y)
{
for(unsigned int i=0; i < X; ++i)
for(unsigned int j=0; j < Y; ++j)
std::cout << a[i][j] << std::endl;
}
int main()
{
const unsigned int X = 2;
const unsigned int Y = 4;
static const float a[X][Y] = {
{10,9,380.033,17189.4, },
{16,10,308.615,17542.4, }
};
print((const float**)a,X,Y);
return 0;
}
Sep 14 '07 #1
9 5697
mathieu wrote:
I know I am doing something stupid here, but it's friday night and I
cannot see what is the issue here:
Thanks,
-Mathieu
#include <iostream>
void print(const float **a,unsigned int X,unsigned int Y)
This function expects 'a' to be an array of pointers to const float.
{
for(unsigned int i=0; i < X; ++i)
for(unsigned int j=0; j < Y; ++j)
std::cout << a[i][j] << std::endl;
}
int main()
{
const unsigned int X = 2;
const unsigned int Y = 4;
static const float a[X][Y] = {
{10,9,380.033,17189.4, },
{16,10,308.615,17542.4, }
};
print((const float**)a,X,Y);
Here you pass an _array_of_arrays_ *pretending* it's an array of
pointers. That's a VERY BAD IDEA(tm).
>
return 0;
}
What you _could_ do (not that it's a good idea, but still), is to
create an array of pointers and pass it into 'print':
const float *aa[] = { a[0], a[1] };
print(aa, X, Y);
V
--
Sep 14 '07 #2
mathieu wrote:
Hi,
I know I am doing something stupid here, but it's friday night and I
cannot see what is the issue here:
Thanks,
-Mathieu
#include <iostream>
void print(const float **a,unsigned int X,unsigned int Y)
{
for(unsigned int i=0; i < X; ++i)
for(unsigned int j=0; j < Y; ++j)
std::cout << a[i][j] << std::endl;
}
int main()
{
const unsigned int X = 2;
const unsigned int Y = 4;
static const float a[X][Y] = {
{10,9,380.033,17189.4, },
{16,10,308.615,17542.4, }
};
print((const float**)a,X,Y);
return 0;
}
a better way to rewrite your program is
template <class T, unsigned X, unsigned Y>
void print(T const (&a)[X][Y])
{
}
--
Thanks
Barry
Sep 14 '07 #3
On Sep 14, 5:42 pm, "Victor Bazarov" <v.Abaza...@comAcast.netwrote:
mathieu wrote:
I know I am doing something stupid here, but it's friday night and I
cannot see what is the issue here:
Thanks,
-Mathieu
#include <iostream>
void print(const float **a,unsigned int X,unsigned int Y)
This function expects 'a' to be an array of pointers to const float.
{
for(unsigned int i=0; i < X; ++i)
for(unsigned int j=0; j < Y; ++j)
std::cout << a[i][j] << std::endl;
}
int main()
{
const unsigned int X = 2;
const unsigned int Y = 4;
static const float a[X][Y] = {
{10,9,380.033,17189.4, },
{16,10,308.615,17542.4, }
};
print((const float**)a,X,Y);
Here you pass an _array_of_arrays_ *pretending* it's an array of
pointers. That's a VERY BAD IDEA(tm).
return 0;
}
What you _could_ do (not that it's a good idea, but still), is to
create an array of pointers and pass it into 'print':
const float *aa[] = { a[0], a[1] };
print(aa, X, Y);
Waw, I never realized that before... Thanks a bunch for the lesson.
I'll work around the issue this way (Muuuuhahhahha):
#include <iostream>
template <typename T>
void print(const T a,unsigned int X,unsigned int Y)
{
for(unsigned int i=0; i < X; ++i)
for(unsigned int j=0; j < Y; ++j)
std::cout << a[i][j] << std::endl;
}
int main()
{
const unsigned int X = 2;
const unsigned int Y = 4;
static const float a[X][Y] = {
{10,9,380.033,17189.4, },
{16,10,308.615,17542.4, }
};
print(a,X,Y);
return 0;
}
Sorry
Sep 14 '07 #4
On Sep 14, 6:01 pm, Barry <dhb2...@gmail.comwrote:
mathieu wrote:
Hi,
I know I am doing something stupid here, but it's friday night and I
cannot see what is the issue here:
Thanks,
-Mathieu
#include <iostream>
void print(const float **a,unsigned int X,unsigned int Y)
{
for(unsigned int i=0; i < X; ++i)
for(unsigned int j=0; j < Y; ++j)
std::cout << a[i][j] << std::endl;
}
int main()
{
const unsigned int X = 2;
const unsigned int Y = 4;
static const float a[X][Y] = {
{10,9,380.033,17189.4, },
{16,10,308.615,17542.4, }
};
print((const float**)a,X,Y);
return 0;
}
a better way to rewrite your program is
template <class T, unsigned X, unsigned Y>
void print(T const (&a)[X][Y])
{
}
Thanks Barry,
Why do you use a reference in this case since you know you are
passing a pointer-to-pointer ?
Thanks
-Mathieu
Sep 14 '07 #5
mathieu wrote:
On Sep 14, 6:01 pm, Barry <dhb2...@gmail.comwrote:
>mathieu wrote:
>>Hi,
I know I am doing something stupid here, but it's friday night and I
cannot see what is the issue here:
Thanks,
-Mathieu
#include <iostream>
void print(const float **a,unsigned int X,unsigned int Y)
{
for(unsigned int i=0; i < X; ++i)
for(unsigned int j=0; j < Y; ++j)
std::cout << a[i][j] << std::endl;
}
int main()
{
const unsigned int X = 2;
const unsigned int Y = 4;
static const float a[X][Y] = {
{10,9,380.033,17189.4, },
{16,10,308.615,17542.4, }
};
print((const float**)a,X,Y);
return 0;
}
a better way to rewrite your program is
template <class T, unsigned X, unsigned Y>
void print(T const (&a)[X][Y])
{
}
Thanks Barry,
Why do you use a reference in this case since you know you are
passing a pointer-to-pointer ?
it's reference to const T[X][Y],
a is a reference to array(with 2D [X][Y]) of const T
--
Thanks
Barry
Sep 14 '07 #6
mathieu wrote:
[..]
template <typename T>
void print(const T a,unsigned int X,unsigned int Y)
The top-level 'const' is superfluous. For an exercise, print out
'typeid(T).name()' and 'typeid(a).name()' inside the function.
{
for(unsigned int i=0; i < X; ++i)
for(unsigned int j=0; j < Y; ++j)
std::cout << a[i][j] << std::endl;
}
int main()
{
const unsigned int X = 2;
const unsigned int Y = 4;
static const float a[X][Y] = {
{10,9,380.033,17189.4, },
{16,10,308.615,17542.4, }
};
print(a,X,Y);
return 0;
}
V
--
Sep 14 '07 #7
On Sep 14, 6:26 pm, "Victor Bazarov" <v.Abaza...@comAcast.netwrote:
mathieu wrote:
[..]
template <typename T>
void print(const T a,unsigned int X,unsigned int Y)
The top-level 'const' is superfluous. For an exercise, print
out 'typeid(T).name()' and 'typeid(a).name()' inside the
function.
Not always. It's ignored by typeid, and when considering the
function *declaration* (i.e. f(int); and f( int const ); declare
the same function), but it *is* significant in the function
body.
--
James Kanze (GABI Software) email:ja*********@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34
Sep 15 '07 #8
mathieu wrote:
On Sep 14, 6:01 pm, Barry <dhb2...@gmail.comwrote:
>mathieu wrote:
>>Hi,
I know I am doing something stupid here, but it's friday night and I
cannot see what is the issue here:
Thanks,
-Mathieu
#include <iostream>
void print(const float **a,unsigned int X,unsigned int Y)
{
for(unsigned int i=0; i < X; ++i)
for(unsigned int j=0; j < Y; ++j)
std::cout << a[i][j] << std::endl;
}
int main()
{
const unsigned int X = 2;
const unsigned int Y = 4;
static const float a[X][Y] = {
{10,9,380.033,17189.4, },
{16,10,308.615,17542.4, }
};
print((const float**)a,X,Y);
return 0;
}
a better way to rewrite your program is
template <class T, unsigned X, unsigned Y>
void print(T const (&a)[X][Y])
{
}
Thanks Barry,
Why do you use a reference in this case since you know you are
passing a pointer-to-pointer ?
Oh, my. I always don't get the idea of the OP
I meant you could pass the array, not casting it to float** then pass it
--
Thanks
Barry
Sep 15 '07 #9
On Sep 14, 6:10 pm, mathieu <mathieu.malate...@gmail.comwrote:
On Sep 14, 6:01 pm, Barry <dhb2...@gmail.comwrote:
mathieu wrote:
I know I am doing something stupid here, but it's friday night and I
cannot see what is the issue here:
#include <iostream>
void print(const float **a,unsigned int X,unsigned int Y)
{
for(unsigned int i=0; i < X; ++i)
for(unsigned int j=0; j < Y; ++j)
std::cout << a[i][j] << std::endl;
}
int main()
{
const unsigned int X = 2;
const unsigned int Y = 4;
static const float a[X][Y] = {
{10,9,380.033,17189.4, },
{16,10,308.615,17542.4, }
};
print((const float**)a,X,Y);
return 0;
}
a better way to rewrite your program is
template <class T, unsigned X, unsigned Y>
void print(T const (&a)[X][Y])
{
}
Why do you use a reference in this case since you know you are
passing a pointer-to-pointer ?
You're not passing a pointer to a pointer. First, because you
don't have a pointer to a pointer anywhere in your program;
that's why it wasn't working in the first place. You have an
array of arrays, which will convert implicitly to a pointer to
an array in many contexts (but not all), but never to a pointer
to a pointer (which would be an entirely different data
structure). Second, and that's the critical aspect here, when
binding to a reference, an array only converts implicitly to a
pointer if the reference type is pointer (and the results of the
conversion won't bind to a non-const reference). What happens
here is that the compiler does template argument deduction,
given an array float[2][4], deduces that T is float, X is 2
and Y is 4, instantiates the template with these arguments, and
then binds the array to the reference. The nice part about it
is that the compiler does all of the work; you don't have to
worry about passing the dimension arguments (and maybe getting
them wrong). The bad part about it is that the compiler can
only work with what it knows; you can only pass a C style array
(which hasn't been converted to a pointer), which means that the
dimensions must be compile time constants. Another potentially
bad aspect is that the compiler instantiates a different
function for each set of dimensions, which can lead to code
bloat (although that's almost certainly not an issue for such a
small function as this).
--
James Kanze (GABI Software) email:ja*********@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34
Sep 15 '07 #10
### This discussion thread is closed
Replies have been disabled for this discussion. | 3,179 | 10,188 | {"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} | 2.546875 | 3 | CC-MAIN-2022-33 | latest | en | 0.697565 |
https://www.physicsforums.com/threads/coulombs-law-and-mechanics.69866/ | 1,545,200,123,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831334.97/warc/CC-MAIN-20181219045716-20181219071716-00197.warc.gz | 993,871,042 | 12,685 | # Homework Help: Coulomb's law and mechanics
1. Apr 3, 2005
### rindishy123
Hey....i need help....
"One practical arrangement for verifying Coulomb's law is to use a lightweight, negatively charged, freely-suspended ball. it is repelled by the negative charge on a larger sphere that is held near it, on an insulated support. the small angle of deflection, x is then measured.
The weight of the ball is W. show that the force of repulsion F on the suspended ball is given by:
F = W tan x"
I've drawn a free body force diagram, but i still cant work it out..
2. Apr 3, 2005
### Staff: Mentor
Start by identifying the forces acting on the suspended ball. (Hint: Three forces act on that ball.) Then apply the conditions for equilibrium: the sum of the forces must be zero. (Hint: Consider vertical and horizontal components separately.)
3. Apr 3, 2005
### rindishy123
Ok I've got to the point where I've found that the components of N (the normal reaction force) are:
N sin x
and
N cos x
I don't study maths so could you please give me hints as to how I turn the above into F = W tan x???
4. Apr 3, 2005
### Staff: Mentor
I don't know what you mean by the "normal force" in this problem. Answer the questions I asked in my last post: What are the forces acting on the ball? | 330 | 1,291 | {"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} | 3.203125 | 3 | CC-MAIN-2018-51 | latest | en | 0.947039 |
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posted by on .
Thank you for helping on the other problem, but could you maybe explain how I should solve these two problems?
1. In 1985, Barry was 13 years old and his father was 43. In what year will Barry's age be two-fifths of his father's age?
2. The length of a certain rectangle is 20m and the length increased by 100m, the perimeter of the new rectangle would be twice the perimeter of the original rectangle. What are the dimensions of the original rectangle?
I did not get the fist problem at all, but for the 2nd problem I started by saying :
L=w+20 >>> l+w+20=P &&&
2(L+100)+2(w-20)=2P
I then plugged things in... and... :(
please help.
once again,
jason.
• Algebra - 2 more questions... ^^ - ,
1. Let the number of years past 1985 be x
then
13+x = 2/5(43+x)
65 + 5x = 86 + 2x
3x = 21
x = 7
so it will be in the year 1992 (1985+7)
for your #2, I think your wording is wrong, it does not match the expressions you have written
for l=w+20, I read that the length is 20 m more than its width, nowhere does it say that.
l+w+20=P ???
if that is to describe the perimeter it whould say:
2l + 2w = p
2(l+20 + 2w = p
2l + 2w + 40 = P
for 2(L+100)+2(w-20)=2P where does it say the width is decreased by 20?
• Algebra - 2 more questions... ^^ - ,
The length of a certain rectangle is 20m and the length increased by 100m, the perimeter of the new rectangle would be twice the perimeter of the original rectangle. What are the dimensions of the original rectangle?
I started by saying :
L=w+20 >>> l+w+20=P &&&
2(L+100)+2(w-20)=2P
I then plugged things in... and... :(
please help.
2(120) + (2W) = 2(40 + 2W)
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https://optionegdugl.netlify.app/telega87131focy/conversion-chart-weight-288 | 1,723,595,370,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00407.warc.gz | 329,097,941 | 11,431 | ## Conversion chart weight
Pressure Conversion Calculator. Bar, Kg/cm, Psi. Temperature Conversion Calculator. Celsius, Fahrenheit Weight Conversion Calculator. Metric Tons 11 Aug 2015 BABY BIRTH WEIGHT CHART. Graphic of Weight Conversion Chart - enables the conversion of pounds and ounces into grams. Source: www. Kilos. Lbs. Kilos. Lbs. Kilos. Lbs. Kilos. Lbs. Kilos. Lbs. 25. 55.1. 115. 253.5. 205. 451.9. 295. 650.4. 385. 848.8. 27.5. 60.6. 117.5. 259.0. 207.5. 457.5. 297.5.
Weight Conversion. Use the search box to find your required metric converter. →. Quickly convert between kg, stones, pounds, grams, ounces with this weight converter. Weight converter. Enter the weight in one of the text boxes and press the Convert button: Metric tons: t. Kilograms: kg. Grams: g. Milligrams: mg. Micrograms: mcg. Pounds (lbs) to Kilograms (kg) weight conversion calculator and how to convert. Weight conversion tool enables conversion between 12 different units from both imperial and metric systems. Table of contents: What is weight? Convert from
## Grams. Kilograms. Pounds. Ounces. 0.001. 0.000001. 0.000002. 0.00005. 0.002. 0.000002. 0.000005. 0.0001. 0.005. 0.000005. 0.00001. 0.0002. 0.01. 0.00001.
Weight converter. Enter the weight in one of the text boxes and press the Convert button: Metric tons: t. Kilograms: kg. Grams: g. Milligrams: mg. Micrograms: mcg. Pounds (lbs) to Kilograms (kg) weight conversion calculator and how to convert. Weight conversion tool enables conversion between 12 different units from both imperial and metric systems. Table of contents: What is weight? Convert from Weight Conversion Chart. Imperial Metric Imperial Metric Imperial Metric Imperial Metric. St lb. Kg. St lb. Kg. St lb. Kg. St lb. Kg. 1 0. 6.35. 6 11 43.09. 9 11 62.14 kg, lbs, Stones/, pounds, kg, lbs, Stones/, pounds, kg, lbs, Stones/, pounds. 38.1, 84, 6, 0, 68.0, 150, 10, 10, 98.0, 216, 15, 6. 38.6, 85, 6, 1, 68.5, 151, 10, 11, 98.4 Welcome to OnlineConversion.com. Common Weight and Mass Conversions. You do not have JavaScript enabled. The conversions on this site require the use Kilogram-Pound Conversion Chart. Weight Weight. Weight Weight. Weight Weight. Weight Weight. Weight Weight. (kg). (lbs). (kg). (lbs). (kg). (lbs). (kg). (lbs). (kg).
### Create a Weight Converter. Create an input element that can convert a value from one weight measurement to another. Step 1) Add HTML: Example -
Convert kilograms to grams, kilograms to pounds, and other units including tons, metric tons, and ounces using DigiKey's weight measurement conversion Weight Conversion Chart. Pounds and Ounces to Grams. Pounds. Ounce 0. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 0. 0. 454 907 1361 1814 2268 2722 3175 3629
### Quickly convert between kg, stones, pounds, grams, ounces with this weight converter.
Kilos. Lbs. Kilos. Lbs. Kilos. Lbs. Kilos. Lbs. Kilos. Lbs. 25. 55.1. 115. 253.5. 205. 451.9. 295. 650.4. 385. 848.8. 27.5. 60.6. 117.5. 259.0. 207.5. 457.5. 297.5. Kilogram (Kg) to Stone (St). Weight Conversion Chart. Feet (Ft) to Metre (M). Height Conversion Chart. Page 2. Feet & Inches metres. Feet & Inches Metres. Weight Conversion Chart for Recycling. As a leader in the Chicago waste management, disposal, and recycling scene, we encourage all places of work and Create a Weight Converter. Create an input element that can convert a value from one weight measurement to another. Step 1) Add HTML: Example -
## Print Free Weight Conversion Tables. Printable List of Weight Coversions. How to Convert Weights.
Since most drug dosages are given in mg/kg, it is important to convert the weight of your guinea pig in kilograms if you have weighed in pounds. The conversion Grams. Kilograms. Pounds. Ounces. 0.001. 0.000001. 0.000002. 0.00005. 0.002. 0.000002. 0.000005. 0.0001. 0.005. 0.000005. 0.00001. 0.0002. 0.01. 0.00001. Weight Conversion Chart. Metric to Imperial. Imperial to Metric. 1.5kg, = 3.3lbs, 3lbs, = 1.4kg. 2kg, = 4.4lbs, 4lbs, = 1.8kg. 2.5kg, = 5.5lbs, 5lbs, = 2.2kg. 3kg, = 6.6 69.8. 374. 74.8. Test Weight Conversion Chart. Graintec Scientific Pty Ltd - 164 Anzac Ave, Toowoomba QLD 4350 - Phone 07 4638 7677 Fax 07 4638 1761 Here you can find a ounces to grams weight conversions chart for cooking measurements. Here is our Measure Conversion Chart for converting metric to imperial units for UK measures. All the charts for length, area, weight, volume and capacity are Mass and Weight Conversion Chart & Calculator. Convert ounces to grams and vice versa with Farnell's mass calculator shown below. Our mass conversion
Convert kilograms to grams, kilograms to pounds, and other units including tons, metric tons, and ounces using DigiKey's weight measurement conversion Weight Conversion Chart. Pounds and Ounces to Grams. Pounds. Ounce 0. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 0. 0. 454 907 1361 1814 2268 2722 3175 3629 Metric and imperial and currency conversion calculator. Online charts and tables to convert currency, temperature, length, area, mass and volume. Since most drug dosages are given in mg/kg, it is important to convert the weight of your guinea pig in kilograms if you have weighed in pounds. The conversion | 1,641 | 5,182 | {"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} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.629672 |
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# My testing on Cashman87 method
Topic closed. 4 replies. Last post 11 years ago by Blackapple.
Page 1 of 1
Wyncote,Pa
United States
Member #3206
January 3, 2004
60737 Posts
Offline
Posted: February 26, 2006, 3:16 pm - IP Logged
Standard Member
United States
Member #29344
December 26, 2005
298 Posts
Offline
Posted: February 6, 2006, 1:37 am - IP Logged
say the previous drawing was ....786
Just a random example.....
And, you do the 432 workout to 786...
786
118
540
Then, take the mirror of 786, 231, and do the workout to it...
231
663
095
So, the base numbers you then have are 786, 118, 540, 231, 663, 095
Where ever you go from there is up to you.
:)
Wyncote,Pa
United States
Member #3206
January 3, 2004
60737 Posts
Offline
Posted: February 26, 2006, 3:23 pm - IP Logged
His other method
Take the past drawing in any states pick 3....
Say, the number was 785...for example....what you do is simply add 12 and subtract 12.
Ex. 78+12 = 90......85+12=97
78-12=66 85-12=73
Now, line them up.... like this...
Pluses 9097
minuses 6673, and then put them both together like this
9097
6673
***************************************************************
Now, that's all you need to make selections for the next couple of drawings. You will get alot of hits, you just need to mix the numbers up the right way. Simple. :)
Ofcourse, do a little testing, and mixing before spending any actual money. That is a given.
This you must realize he
is an excellent mixer.
Wheeling the results
produces.
Wyncote,Pa
United States
Member #3206
January 3, 2004
60737 Posts
Offline
Posted: February 26, 2006, 3:25 pm - IP Logged
I will give example from
Mich.nite recently,how I see it
If results come within 7 days
I keep a workout.
170 came 2/18
*********************
170------------do the MirrorMIX 625 this way and mix*
+432
502*
+432
934*
You have 502*
932*
Mixing you get 935
Actual number 170 result mix
or Mirror 625 result mix
Wyncote,Pa
United States
Member #3206
January 3, 2004
60737 Posts
Offline
Posted: February 26, 2006, 3:54 pm - IP Logged
You have 502*
934*
Mixing you get 935 St
Beautiful, center 3 intact
Wyncote,Pa
United States
Member #3206
January 3, 2004
60737 Posts
Offline
Posted: February 26, 2006, 3:58 pm - IP Logged
Tough State Cali. nite
219 on 2/19
********************
219------------do MirrorMIX 625 this way and mix*
+432
641*
+432
073*
You have 641*
073*
471 combo came 2/20 nite
Page 1 of 1 | 836 | 2,601 | {"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} | 3.140625 | 3 | CC-MAIN-2016-50 | latest | en | 0.779211 |
http://www.wisegeek.net/what-is-an-annual-equivalent-rate.htm | 1,553,389,811,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203123.91/warc/CC-MAIN-20190324002035-20190324024035-00206.warc.gz | 390,884,503 | 18,895 | Category:
# What is an Annual Equivalent Rate?
Article Details
• Written By: Ken Black
• Edited By: Andrew Jones
2003-2019
Conjecture Corporation
wiseGEEK Slideshows
The annual equivalent rate is the rate at which interest would be paid on an investment over the course of a year. The rate is a good opportunity to compare returns or interest on investment, and also to get a more accurate prediction as to what that investment will actually earn. The reason why the annual equivalent rate is different than the annual percentage rate is because interest earned and paid during previous points of the year continues to earn interest along with the original principal. In some countries, such as the United Kingdom, the equivalent rate is published as a regular course of doing business for some investment products.
One of the most common situations in which an annual equivalent rate may be used is in bank savings products, such as savings accounts and even certificates of deposit. If a bank is offering six percent interest on an investment product, paid semi-annually, and an investor puts in \$100,000 US Dollars (USD), at the midyear point, the total amount will be worth \$103,000 USD. At the end of the year, the amount would increase to \$106,090 USD. That makes the annual equivalent rate 6.09 percent.
If the interest would have been paid all at once, at the end of the 12-month period, rather than having a payment made halfway through, to get the same amount of money the annual percentage rate would have needed to be set at 6.09 percent. Therefore, the annual equivalent rate is always even with, or more than, the annual percentage rate. When interest is only paid once a year both the rates will be the same.
It is also possible that interest rates are paid more than twice a year. Depending on the investment product, investors may receive interest payments as often as once per month. The increase in the frequency of interest payments would also lead to an increase in the annual equivalent rate. Therefore, not only is the actual percentage rate a big factor in the equivalent rate, frequency of payment can also make a big impact.
To determine the annual equivalent rate, an investor needs to know both the annual percentage rate and frequency of payment. Divide the frequency of payment by the interest rate, and then add one. Next, use the frequency of payment to exponentially increase that sum. For example, if the frequency of payment was twice a year, you would increase the sum to the second power. Once you have that number, subtract one to get the annual equivalent rate. | 534 | 2,610 | {"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} | 2.765625 | 3 | CC-MAIN-2019-13 | longest | en | 0.974168 |
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# Can someone help me with this maths gcse question ? watch
1. Attachment 730562
Its question b , ignore me trying to work it out
I think you can only see the picture on the mobile app
Attachment 730570
Also question b for the second
2. (Original post by Randomstudwnt)
Attachment 730562
Its question b , ignore me trying to work it out
3. (Original post by Randomstudwnt)
Attachment 730562
Its question b , ignore me trying to work it out
I think you can only see the picture on the mobile app
First off Mel is a she :/
Also I might be wrong but:
The p(face up) is (50/100) aka 1/3.
The p(face down) is 100/150) aka 2/3.
Then it asks for an estimate of it landing face up then face down so we need to multiply the probabilities.
So 1/3*2/3 which is 2/9. I feel I am wrong though because is asks for an estimate. Anyone else?
4. (Original post by Y11_Maths)
First off Mel is a she :/
Also I might be wrong but:
The p(face up) is (50/100) aka 1/3.
The p(face down) is 100/150) aka 2/3.
Then it asks for an estimate of it landing face up then face down so we need to multiply the probabilities.
So 1/3*2/3 which is 2/9. I feel I am wrong though because is asks for an estimate. Anyone else?
I got that too
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http://www.centerspace.net/doc/NMathStats/user/NMathStatsUsersGuideIX.htm | 1,547,937,995,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583684033.26/warc/CC-MAIN-20190119221320-20190120003320-00363.warc.gz | 274,548,322 | 11,088 | # NMath Stats User's Guide
TOC | Previous | Next | Index
Index
Symbols
.NET Framework ,
A
converting to data frames
creating from data frames
alpha levels
ALS
Alternating Least Squares (ALS)
analysis of variance (ANOVA)
ANOVA
ANOVA regression parameters
AnovaRegressionFactorParam
AnovaRegressionInteractionParam
AnovaRegressionParameter
Any CPU build configuration
API documentation
applying functions
arithmetic mean
Assemblies
autocorrelation
B
beta distribution
BiasType , ,
binary serialization
binomial distribution
BinomialDistribution ,
boolean columns
Box-Cox power transformations
C
categorical vectors
CDF
cell data ,
cell means
CenterSpace.NMath.Stats namespace
central moments
central tendency
charting
chi-square distribution
ChiSquareDistribution ,
city-block (Manhattan) distance
cluster analysis
ClusterAnalysis , ,
clustering
ClusterSet ,
code examples
coefficient of determination
column names , ,
columns
accessing
creating
exporting to a string
exporting to a vector
exporting to an array
properties
removing data
reordering
Common Language Specification
compiled Help
complete orthogonal decomposition
confidence interval
consensus matrix
contacting technical support
contingency table
convergence check period
cophenetic distance
CORegressionCalculation
correlated random inputs
correlation
counts
covariance
covariance matrix
Cox and Snell pseudo R-squared statistic
Cp ,
Cpk ,
Cpm ,
critical values , ,
Cronbach's alpha
cross validation
cross-tabulation
cumulative distribution function
D
data frames
column properties
column types
creating
exporting to a matrix
exporting to a string
permuting rows and columns
properties
removing columns
removing rows
sorting
DataFrame –??
datetime columns
deciles
decimal types
descriptive statistics
design variables
DFBoolColumn
DFColumn
DFDateTimeColumn
DFGenericColumn
DFIntColumn
DFNumericColumn
DFStringColumn
Distance
distance functions
Distance.Function
distribution classes
documentation
Reference Guide
User's Guide
dummy variable regression parameters in ANOVA
dummy variables
Durbin-Watson statistic
E
effects encoding
Euclidean distance
exponential distribution
ExponentialDistribution ,
F
F distribution
F test
Factor , , ,
Factor analysis
factor extraction
factor rotation
factor score
factor score coefficients
factor scores
factors
accessing
creating
grouping by
properties
FDistribution ,
Fisher transformation
Fisher's Exact Test
Frobenius matrix norm
G
G statistic
gamma distribution
gaussian distribution
GDCLS ,
generic columns
generic functions
geometric distribution
geometric mean
GeometricDistribution ,
goodness of fit ,
GoodnessOfFit
GoodnessOfFitParameter ,
Gradient Descent - Constrained Least Squares (GDCLS)
grand mean , ,
graphing
group means ,
grouping by factors ,
groupings ,
H
harmonic mean
hold out method
Hosmer Lemeshow statistic
hypothesis tests
creating
properties ,
HypothesisType
I
IDFColumn
ILogisticRegressionCalc
InputVariableCorrelator
integer columns
intercept parameter
intercept parameters
internally studentized residuals
interquartile range ,
inverse CDF
inverse cumulative distribution function
inverse Fisher transformation
IRandomVariableMoments
IRegressionCalculation
ISerializable interface
Iterative Power Method
J
jackknife estimates
jackknifing
Johnson system of distributions
JohnsonDistribution
K
k-fold cross validation
KMeansClustering
Kolmogorov-Smirnov test ,
Kruskal-Wallis rank sum test ,
KruskalWallisTest
kurtosis ,
L
least squares minimization
likelihood function
linear regressions
creating
modifying
predictions ,
results
significance of parameters
significance of the overall model
LinearRegression
LinearRegressionAnova
LinearRegressionParameter ,
logical functions ,
logistic regression
LogisticDistribution ,
LogisticRegression
LogisticRegressionFitAnalysis
log-normal distribution
LognormalDistribution ,
M
matrices
converting to data frames
creating from data frames
maximum (Chebychev) distance
mean ,
mean deviation
mean of the ranges method
median
median deviation from mean
Microsoft Chart Controls for .NET
min/max functions
missing values ,
mode
multiple linear regression
multiplicative update rule
multivariate techniques
N
Nagelkerke pseudo R-squared statistic
namespaces
NaN values
negative binomial distribution
NegativeBinomialDistribution ,
NewtonRaphsonParameterCalc
NIPALS
NMath Core ,
NMathStatsChart
NMF
NMFClustering , ,
Nonlinear Iterative PArtial Least Squares (NIPALS)
nonnegative matrix factorization (NMF) ,
Non-parametric tests
normal distribution
NormalDistribution ,
Not-A-Number values
numeric columns
O
odds ratio
OneSampleKSTest
OneSampleTTest
OneSampleZTest
one-way ANOVA
accessing the ANOVA table ,
one-way RANOVA
accessing the RANOVA table
OneWayAnova
OneWayAnovaTable ,
OneWayRanova
OneWayRanovaTable
P
Partial Least Squares
Partial least squares Discriminant Analysis
parts per million defective ,
PDF
Pearson chi-square statistic
Pearson correlation
Pearson's chi-square test
PearsonsChiSquareTest
percent defective ,
percentiles
permuting columns
permuting data frames
plotting
PLS-DA
poisson distribution
PoissonDistribution ,
power distance
Pp ,
Ppk ,
predictions ,
predictor matrix
principal component analysis
probability density function
probability distributions
ProbabilityDistribution
process capability ,
process capability index
process performance
ProcessCapability
ProcessPerformance
product
documentation
features
overview
software requirements
pseudo R-squared
Q
QR decomposition
QRRegressionCalculation
quartiles
R
R2
random samples
ranks
ReducedVarianceInputCorrelator
regression calculators
regression matrix
regularization
reordering columns
reordering data frames
residual standard error
RMS
root mean square
row keys , ,
modifying
S
sampling
serialization
SIMPLS
singular value decomposition
skewness ,
SOAP serialization
software requirements
SortingType , ,
sparsity
Spearman's rank correlation coefficient
Spearman's rho ,
squared Euclidean distance
SSE
standard deviation
standardized residuals
statistical functions
data types
missing values
statistical process control
StatsFunctions –??
StatsSettings
Straightforward IMplementation of PLS (SIMPLS)
string columns
Student's t distribution
studentized residuals
subject means
Subset
subsets
accessing elements
arithmetic operations
creating
logical operations
properties
sum of squared errors
sums
SVDRegressionCalculation
T
t test , ,
tabulation
Taguchi capability index
TDistribution ,
technical support
time series
treatment means
triangular distribution
TriangularDistribution ,
trimmed mean
trimming data
TrustRegionParameterCalc
TwoSampleFTest
TwoSampleKSTest ,
TwoSamplePairedTTest
TwoSampleUnpairedTTest
TwoSampleUnpairedUnequalTTest
two-way ANOVA
accessing the ANOVA table
two-way RANOVA
TwoWayAnova
TwoWayAnovaTable
TwoWayAnovaTypeI
TwoWayAnovaTypeII
TwoWayAnovaTypeIII
TwoWayAnovaUnbalanced
TwoWayRanova
TwoWayRanovaTable
TwoWayRanovaTwo
TwoWayRanovaTwoTable
typographic conventions
U
Unbalanced two-way ANOVA
unbalanced two-way ANOVA
uniform distribution
UniformDistribution ,
V
variance ,
variance inflation factor
varimax rotation
visualization
Von Neumann ratio
W
Weibull distribution
WeibullDistribution ,
weighted mean
Z
Z Bench
Z bench ,
z test
ZBench
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Top | 1,794 | 7,642 | {"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} | 2.5625 | 3 | CC-MAIN-2019-04 | latest | en | 0.496019 |
https://sciencetheory.net/adverse-selection-multidimensional-asymmetric-information/ | 1,726,884,765,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00596.warc.gz | 450,227,041 | 27,203 | # Adverse Selection: Multidimensional Asymmetric Information
Another important limitation of our analysis of adverse selection in chapter 2 is that the adverse selection parameter θ was modeled as a unidimensional parameter. In many instances, the agent simultaneously knows several pieces of information that are payoff relevant and affect the optimal trade. For instance, a tax authority would like to know both the elasticity of an agent’s labor supply and his productivity before fixing his tax liability. Similarly, an insurance company would like to know both the probability of accident for an insurer and his degree of risk aversion before fixing the risk premium that the agent should pay. The producer of a good knows not only the marginal cost of producing the good but also the associated fixed cost. In all these situations, the unidimensional paradigm must be given up in order to assess the full consequences of asymmetric information on the rent extraction-efficiency trade-off.
### 1. A Discrete Bare-Bones Model
We now extend the analysis of chapter 2 to the case of bidimensional asymmet- ric information. The simplest way to do so is to have the agent accomplish two activities for the principal. Each of those activities is performed by the agent at a marginal cost which is private information. Let us assume that the agent pro- duces two goods in respective quantities q1 and q2 with a utility function U = t − (θ1q1 + θ2q2), with θi in { θ , θ¯} for i = 1, 2. We also assume that there is no externality between the two tasks for the principal, so that the surpluses asso- ciated with both tasks add up in the latter’s objective function, which becomes V = S(q1) + S(q2) − t.
The probability distribution of the adverse selection vector θ = (θ1, θ2) is now defined by with a positive correlation among types arising when . As usual, this distribution is common knowledge.
The components of the direct revelation mechanism are respectively denoted by if , where we impose (without a loss of generality) a symmetry restriction on transfers. Similar notations are used for the information rents Uij . Because of the symmetry of the model, there are only three relevant levels of information rents, . Similarly, we denote outputs by , and transfers by . These notations, even though they look quite cumbersome, unify the present multidimensional modelling with that of section 3.1.1 above.
Again, following the logic of the unidimensional model, we may guess that only the upward incentive constraints matter. The three following incentive con- straints then become relevant:
We can also expect the participation constraint of an agent who is inefficient on both dimensions (θ1 = θ¯ and θ2= θ¯) to be binding, i.e.,
We leave it to the reader to check that adding incentive constraints for types taken two by two yields the following impementability conditions:
and
### 2. The Optimal Contract
We can expect (3.21) and (3.22) to be binding at the optimum. Then (3.19) and (3.20) can be summarized as
which should also be binding at the optimum to reduce the efficient agent’s infor- mation rent.
After the substitution of the information rents as functions of outputs, the principal’s optimization program can be reduced to:
We must distinguish two cases depending on the level of correlation p between both dimensions of adverse selection.
#### Case 1: Weak Correlation
Let us first assume that the solution is such that q¯ ≤ qˆ1. In this case, max(2q¯, qˆ1 + q¯) = qˆ1 + q¯ and optimizing (P’) yields the following second-best outputs:
This latter schedule of outputs is the solution when the posited monotonicity condition holds, i.e., when
Figure 3.3: Binding Incentive Constraints with Weak Correlation
or, to put it differently, when . This condition obviously holds in the case where θ1 and θ2 are independently drawn, since the correlation is then zero and p = 0. We leave it to the reader to check that all of the neglected incentive and participation constraints are satisfied when (3.29) holds.
In the case of weak correlation, the binding constraints are only the local ones. In figure 3.3, an arrow from a point in the type space, say A, to another one, say B, means that A is attracted by B, i.e., the corresponding incentive constraint is binding at the optimum.
#### Case 2: Strong Correlation
If we had perfect correlation, vˆ = 0, the binding incentive constraint would obvi- ously be from (θ, θ) to (θ¯, θ¯), as shown in figure 3.4. Then we would be back to the usual unidimensional model along the diagonal.
More generally, for a strong positive correlation we may expect an intermedi- ary case with the upward incentive constraints being binding as in figure 3.5.
Indeed, consider the case where the condition (3.29) does not hold. In this case the outputs, defined by (3.27) and (3.28), are such that a contradiction with our starting assumption q¯ ≤ qˆ1. Let us assume that instead we have q¯ > qˆ1. In this case, and optimizing (P’) leads to
Figure 3.4: Binding Incentive Constraint with Perfect Correlation
Figure 3.5: Binding Incentive Constraints with Strong Correlation
(3.27) and (3.28) being respectively replaced by
and
However, we immediately observe that q¯ < qˆ1 for those outputs. Again, this is a contradiction with our starting assumption q¯ > qˆ1. When (3.29) does not hold, we necessarily have qˆ1 = q¯ = qP , and bunching arises at the optimal contract.
To understand the origin of this bunching, let us first consider the case where the principal is concerned only with the upward incentive compatibility con- straints (3.19) and (3.21). With a strong correlation of types, the principal finds the information rent left to the (θ, θ)-type to be very costly because this type is relatively likely. Hence, the principal is led to reduce the output qˆ1 significantly.
On the other hand, the intermediate type (θ, θ¯) is rather unlikely. Hence, the principal finds it not very useful to reduce the output q¯ to reduce the latter’s infor- mation rent. As a result, the schedule of outputs that would be implemented by the principal, had he taken into account only the downward incentive constraints, would be such that q¯ > qˆ1. The high output q¯ requested from a (θ¯, θ¯)-type makes the (θ, θ)-type willing to mimic the (θ¯, θ¯)-type rather than the intermediate type (θ, θ¯). With a strong correlation it is no longer correct to neglect the global incen- tive constraint (3.2). Both the local incentive constraints (3.19) and the global incentive constraints (3.2) are binding at the optimum, and this situation is only possible when the optimal allocation entails some bunching.
Remark: Just as in the three-type model of section 3.1.2, a probability distribution of types such that intermediate types are rather unlikely creates bunching at the optimal contract.
Optimizing (P’) with the added constraint q¯ = qˆ1 still yields (3.26) but also
We summarize our findings in proposition 3.2.
Proposition 3.2: In a symmetric bidimensional adverse selection setting, the optimal contract with a weak correlation of types keeps many fea- tures of the unidimensional case with two types; only upward incentive constraints are binding. With a strong correlation, the optimal contract may instead entail some bunching; a global incentive constraint becomes binding.
Finally, note that more complex situations arise when the correlation is nega- tive, asymmetric distributions are postulated, or when the dimensionality of actions is not the same as the dimensionality of the asymmetry of information.
We now describe two examples where modelling adverse selection with mul- tidimensional types has proved to be useful.
##### Example 1: Unknown Fixed Cost
Let us suppose that the agent has a cost function where both the marginal cost θ1 and the fixed cost θ2 are unknown. As shown in Baron and Myerson (1982) and Rochet (1984), stochastic mechanisms, where the decision to produce or not produce is used as a screening device, are useful in this context. To explain why, let us introduce x in y0k 1] as the probability of a positive production. As a function of the contracting variables q and x, the agent’s utility function now writes as This expression almost takes the same form as what we have analyzed above. It is easy to show that the shutdown of some types is also a valuable screening device to learn the value of the fixed cost θ2.
##### Example 2: Unknown Cost and Demand
Let us assume that the agent is a retailer who serves a market with a linear inverse demand P(q) = a θ1q, where θ1 is an intercept parameter, which is the first piece of private information of the agent. This agent has also a cost function C(q) = θ2q, where the marginal cost θ2 is the second piece of private information of the agent. The latter’s utility function writes finally as , where t˜ is the transfer received from the principal, here a manufacturer.
To simplify, we also assume that the manufacturer incurs no production cost for the intermediate good he provides to the agent. Introducing a new variable t = t˜ + aqq2, the agent’s utility function rewrites as U = t − (θ1 + θ2)q. On the other hand, the principal’s objective becomes V = aq q2t. In this example the bidimensional adverse selection model amounts to a unidimensional model, where θ = θ1 + θ2 is a sufficient statistic for all information known by the agent. If each type θi belongs to may take three possible values, 2θ, θ¯ + θ, or 2θ¯. The framework of section 3.1 can then be used to derive the optimal contract.
Remark: The dimensionality of the type space plays a crucial role in determining the binding participation constraints. To understand this point remember that, in a unidimensional case, the least efficient type’s participation constraint is the only binding participation constraint (at least as long as shutdown is not optimal) and the same result holds for a continuum of types (see appendix 3.1). Now suppose that it holds also with a continuum of bidimensional types, i.e., only the least effi- cient type on dimensions θ1 and θ2 is put at its reservation utility. Let us imagine that the principal slightly uniformly decreases the whole transfer schedule he offers to the agent by ε. Of course, a whole subset of types around (θ¯, θ¯) prefers to stop producing. The efficiency loss for the principal is roughly of the order ε2. However, by uniformly reduc- ing the whole transfer schedule, the principal reduces all information rents of the remaining types by s, which means he makes a gain of the order ε(1 − ε2) ≈ ε. Therefore, the shutdown of a subset of types with nonzero measure is always optimal.
Armstrong and Rochet (1999) provided a complete analysis of the two-type model. The case of a continuum of types was first ana- by McAfee and McMillan (1988), who attempted to generalize the Spence-Mirrlees assumption to a multidimensional case, and Laffont, Maskin, and Rochet (1987), who explicitly solved an example in the case where the principal has only one output and one transfer with which to screen a bidi- mensional adverse selection parameter (see also Sibley and Srinagesh 1997). The result, that a shutdown of a nonzero measure of types is always opti- mal for a continuum of types, is due to Armstrong (1996), who also offered some closed-form solutions for the optimal contract when the set of types includes the origin. See also Wilson (1993) on this point. The analysis of Rochet and Choné (1998) is the most general. They showed that bunching of types is always found in these bidimensional models, and they also provided the so-called ironing and sweeping techniques designed to analyze this bunch- ing issue. These techniques are difficult and outside the scope of this book. Rochet and Choné (1998) also show that bunching implies that a whole set of types with nonzero measure exists, such that q1 = q2 at the optimal contract. They interpret this as a bundling requirement imposed by incentive com- patibility. Finally, Armstrong (1999) pushed the idea that multidimensional adverse selection problems may introduce a significant simplification in the optimal contract between a seller (the principal) and a buyer (the agent) who is privately informed of his type. Instead of explicitly computing this contract, Amstrong provides a lower boundary on what can be achieved with simple two-part tariffs and, using the Law of Large Numbers, shows that these con- tracts can approximate the first-best when the number of products sold to this buyer is large enough.
Source: Laffont Jean-Jacques, Martimort David (2002), The Theory of Incentives: The Principal-Agent Model, Princeton University Press. | 3,087 | 12,824 | {"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} | 2.90625 | 3 | CC-MAIN-2024-38 | latest | en | 0.921155 |
https://www.hitpages.com/doc/4504584248623104/11/ | 1,481,265,733,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542686.84/warc/CC-MAIN-20161202170902-00483-ip-10-31-129-80.ec2.internal.warc.gz | 949,884,653 | 9,058 | X hits on this document
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# Predicting Patient Mortality Rates for Providers
• e statistical methods used to predict mortality on the basis of the significant risk factors are tested to determine
whether they are sufficiently accurate in predicting mortality for patients who are extremely ill prior to undergoing the procedure as well as for patients who are relatively healthy. ese tests have confirmed that the models are reasonably accurate in predicting how patients of all different risk levels will fare when undergoing PCI.
• e mortality rate for each hospital and cardiologist is also predicted using the statistical model. is is accomplished
by adding the predicted probabilities of death for each of the provider’s patients and dividing by the number of patients.
• e resulting rate is an estimate of what the provider’s mortality rate would have been if the hospital’s performance was
identical to the state performance. e percentage is called the predicted or expected mortality rate (EMR). A hospital's EMR is contrasted with its observed mortality rate (OMR), which is the number of PCI patients who died divided by the total number of PCI patients.
# Computing the Risk-Adjusted Mortality Rate
• e risk-adjusted mortality rate (RAMR) represents the best estimate, based on the associated statistical model, of what
the provider’s mortality rate would have been if the provider had a mix of patients identical to the statewide mix. us, the RAMR has, to the extent possible, ironed out differences among providers in patient severity of illness, since it arrives at a mortality rate for each provider based on an identical group of patients.
To get the RAMR, the OMR is first divided by the provider’s EMR. If the resulting ratio is larger than one, the provider has a higher mortality rate than expected on the basis of its patient mix; if it is smaller than one, the provider has a lower mortality rate than expected from its patient mix. e ratio is then multiplied by the overall statewide rate (0.95 percent in-hospital/30-day in 2008) to obtain the provider’s RAMR.
# Interpreting the Risk-Adjusted Mortality Rate
If the RAMR is lower than the statewide mortality rate, the hospital has a better performance than the state as a whole; if the RAMR is higher than the statewide mortality rate, the hospital has a worse performance than the state as a whole.
• e RAMR is used in this report as a measure of quality of care provided by hospitals and cardiologists. However,
there are reasons that a provider’s RAMR may not be indicative of its true quality.
For example, extreme outcome rates
may occur due to chance alone. is is particularly true for low-volume providers, for whom very high or very low rates are more likely to occur than for high-volume providers. To prevent misinterpretation of differences caused by chance variation, expected ranges (confidence intervals) are included in the reported results.
Differences in hospital coding of risk factors could be an additional reason that a hospital’s RAMR may not be reflective of quality of care. e Department of Health monitors the quality of coded data by reviewing patients’ medical records to ascertain the presence of key risk factors. When significant coding problems are discovered, hospitals are required to correct these data and are subject to subsequent monitoring.
# How This Initiative Contributes to Quality Improvement
• e goal of the Department of Health and the Cardiac Advisory Committee is to improve the quality of care in relation
to cardiac surgery and angioplasty in NYS. Providing the hospitals, cardiac surgeons (who perform cardiac surgery) and cardiologists (who perform PCI) in NYS with data about their own outcomes for these procedures allows them to examine the quality of their own care and to identify opportunities to improve that care.
• e data collected and analyzed in this program are reviewed by the Cardiac Advisory Committee, which assists with
interpretation and advises the Department of Health regarding which hospitals and physicians may need special attention. Committee members have also conducted site visits to particular hospitals and have recommended that some hospitals obtain the expertise of outside consultants to design improvements for their programs.
5
Document views 176 Page views 176 Page last viewed Fri Dec 09 04:48:07 UTC 2016 Pages 62 Paragraphs 10034 Words 28608 | 898 | 4,466 | {"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} | 2.859375 | 3 | CC-MAIN-2016-50 | longest | en | 0.949988 |
https://pages.uoregon.edu/emi/10.php | 1,653,584,734,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00735.warc.gz | 525,517,902 | 3,620 | 10. Complex Periodic Waveforms
Fourier Analysis, named after the nineteenth century French mathematician Jean Baptiste Fourier, enables one to break down complex periodic waveforms into their basic components, which happen to be sine waves of various frequencies, amplitudes, and phases. The opposite method, combining sine waves of various frequencies, amplitude, and phase to create complex periodic waveforms, is Fourier Synthesis.
A complex waveform is the result of combining the instantaneous amplitudes of two (or more) sine waves.
Example 10-1: Fourier Synthesis, combining different sine waves, results in complex waveforms.
At some points combined waves reinforce one another to create an increased amplitude, constructive interference, yet at other points combined waves interfere with one another to result in a decreased amplitude, destructive interference.
The frequency that has the same period as the resultant waveform is called the fundamental frequency, or fundamental. The fundamental will always be the lowest frequency in a waveform. Frequencies higher than the fundamental are partials.
Fundamental Partial
Diagram 10-2: Fundamental frequency and partial frequency. | 222 | 1,195 | {"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} | 3.203125 | 3 | CC-MAIN-2022-21 | longest | en | 0.861524 |
https://excelexamples.com/post/how-to-count-the-number-of-characters-in-a-cell/ | 1,696,005,957,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510520.98/warc/CC-MAIN-20230929154432-20230929184432-00786.warc.gz | 270,988,333 | 4,088 | # How to count the number of characters in a cell
Jan 11, 2020 • edited Jan 12, 2020
When you need to count the characters in cells, use the LEN function—which counts letters, numbers, characters, and all spaces. For example, the length of “It's 98 degrees today, so I'll go swimming” (excluding the quotes) is 42 characters—31 letters, 2 numbers, 8 spaces, a comma, and 2 apostrophes.
To use the function, enter =LEN(cell) in the formula bar, then press Enter on your keyboard.
## Method 1
1. The LEN function in Excel counts the number of characters in a cell.
Explanation: the LEN function counts 2 numbers, 1 space and 6 letters.
2. You can combine the SUM function and the LEN function to count the number of characters in a range of cells.
Note: as you can imagine, this formula can get quite long.
3. The array formula below counts the number of characters in a range of cells.
Note: finish an array formula by pressing CTRL + SHIFT + ENTER. Excel adds the curly braces {}. The array constant {9;4;6;5} is used as an argument for the SUM function, giving a result of 24. Maybe this is one step too far for you at this stage, but it shows you one of the many other powerful features Excel has to offer.
4. Use the LEN function and the SUBSTITUTE function to count how many times a specific character (in this example, the character a) occurs in a cell.
Explanation: the SUBSTITUTE function replaces the character a (second argument) with an empty string (third argument). LEN(SUBSTITUTE(A1,“a”,"")) equals 8 (the length of the string without the character a). If we subtract this number from 9 (total number of characters in cell A1), we get the number of occurrences of the character a in cell A1.
5. The array formula below counts how many times a specific character (in this example, the character a) occurs in a range of cells.
Note: finish an array formula by pressing CTRL + SHIFT + ENTER. Excel adds the curly braces {}. The array constant {1;1;2;1} is used as an argument for the SUM function, giving a result of 5. The SUBSTITUTE function is case-sensitive. The A in Alaska is not counted.
6. The array formula below counts both lower and upper case occurrences of a specific character (in this example, the character a).
Explanation: The LOWER function converts all letters to lowercase first.
## Method 2
Multiple cells: To apply the same formula to multiple cells, enter the formula in the first cell and then drag the fill handle down (or across) the range of cells.
To get the a total count of all the characters in several cells is to use the SUM functions along with LEN. In this example, the LEN function counts the characters in each cell and the SUM function adds the counts:
=SUM((LEN( cell1 ),LEN( cell2 ),(LEN( cell3 )) )).
## Give it a try
Here are some examples that demonstrate how to use the LEN function.
Copy the table below and paste it into cell A1 in an Excel worksheet. Drag the formula from B2 to B4 to see the length of the text in all the cells in column A.
Text StringsFormulas
The quick brown fox.=LEN(A2)
The quick brown fox jumped.
The quick brown fox jumped over the lazy dog.
Count characters in one cell
1. Click cell B2.
2. Enter =LEN(A2).
The formula counts the characters in cell A2, which totals to 27—which includes all spaces and the period at the end of the sentence.
NOTE: LEN counts any spaces after the last character.
Count characters in multiple cells
1. Click cell B2.
2. Press Ctrl+C to copy cell B2, then select cells B3 and B4, and then press Ctrl+V to paste its formula into cells B3:B4.
This copies the formula to cells B3 and B4, and the function counts the characters in each cell (20, 27, and 45).
Count a total number of characters
1. In the sample workbook, click cell B6.
2. In the cell, enter =SUM(LEN(A2),LEN(A3),LEN(A4)) and press Enter.
This counts the characters in each of the three cells and totals them (92).
#How To#Tutorial#Functions#Count#characters
What are the most useful shortcuts in Excel
How to copy that cell over to column B without either of those 2 words | 977 | 4,084 | {"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} | 3.875 | 4 | CC-MAIN-2023-40 | latest | en | 0.846198 |
https://www.talkstats.com/tags/biostats/ | 1,670,362,020,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00098.warc.gz | 1,096,046,006 | 9,487 | biostats
1. Can I make a comparison among groups which the control group's data was set to be 100%?
Hello I want to make a comparison between the groups. However, in the control group, all of the data was set to be 100% and the data of the other groups is the proportion compared to the control group. So the data will look like, (just for example) control group: 100%, 100%, 100%, 100%, 100%...
2. Modeling and predicting pathology from multivariate clinical data
Hello, I have a clinical data set that consists of 5 clinical measurements on thousands of tissue samples. Furthermore, each sample has a pathology diagnosis that is 1 of 5 possible diagnoses (all different types of tumors). I am interested in predicting which pathologic class future samples...
3. estimate of measures
Hello everyone, I am a statistics student, I'm studying for a project in my University. I have to analyze the data, I have available the measures in millimeters corrisponding two different types of objects, of course I have already done tests to groups such as t-test, actually these organized...
4. Help with Biostats Hw question
The research team has considered your answers to the above. They’ve narrowed their interest to two outcomes (for now): Number of episodes of early rejection symptoms and rejection status. However, they want to know how many subjects they should recruit for each outcome. For simplicity, they...
5. Pearsons Correlation, Two-way ANOVA, Linear Regression Test, Chi-Square Test or None
During research, you have asked your participants whether or not they have ever experienced financial difficulties (yes or no). You have three age-groups (16-20; 21-25; 26-30), and you predict that those in the 26-30 group have a greater risk of experiencing financial difficulties than those who...
6. One or Two Way ANOVA?
Hi all, Just had a quick question regarding whether or not I should be using a one or two way ANOVA to see the effect of caffeine on heart rate. My groups are: control, coffee, energy drink. My sample size is 50 for each group and for each individual I tested at the following times: initial...
7. Appropriate test for two variables
Hello, This is my first time posting here, so forgive me if I don't include something. I will follow this post and add more information as needed. I want to compare two samples, one is a city, and another is a neighborhood in that city, for the same disease. I have the population size... | 547 | 2,453 | {"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} | 2.640625 | 3 | CC-MAIN-2022-49 | latest | en | 0.947088 |
https://www.realnfo.com/ee/Basic-Electrical-Engineering/Fundamental-of-Physics/Conversion-of-Units | 1,701,413,632,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100276.12/warc/CC-MAIN-20231201053039-20231201083039-00487.warc.gz | 1,051,033,532 | 8,451 | # Conversion of Units
Whatsapp
For example, suppose we wish to convert 15.0 in. to centimeters. Here we know that $1in=2.54cm$, so dividing 1in by 2.54cm will gives 1 in result. Applying this conversion we can convert 15in to cm as follows. $$\begin{array} {rcl}15in& = &(15in)(1)\\ 15in& = &(15 in)({2.54 cm \over 1in})\\ & = &(15) (2.54cm) = 38.1cm\end{array}$$ Conversion factors between the SI units and conventional units of length are as follows: | 152 | 453 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "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} | 3.671875 | 4 | CC-MAIN-2023-50 | longest | en | 0.826114 |
https://instrumentationtools.com/interview-questions-on-flow-measurement/ | 1,717,055,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059506.69/warc/CC-MAIN-20240530052602-20240530082602-00699.warc.gz | 263,558,763 | 47,778 | # Interview Questions on Flow Measurement
## Flow Measurement Questions
### What is ‘volumetric’ and ‘rate of flow’? Write their measuring units in metric.
##### Volumetric Flow:
The total amount of fluid passed through a process line. Generally, it is measured on counters. The measuring unit is cubic meters, Barrels…etc.
##### Rate of flow:
The amount of fluid moving through a process line per period of time. Generally, it is measured on indicators and recorders. The measuring unit is cubic meters per day, barrels per day…etc.
### What are the types of flow-measuring instruments used in industries?
Orifice plate and a DP transmitter, Daniel orifice and DP transmitter, Rotameter, Dall Tube, Venturi Tube, PD meters…etc.
### What are the types of flow measurements commonly used in DPU-A?
Orifice plate and a DP transmitter, Daniel Orifice and a DP Transmitter, Rotameter, PD meters.
### What is the ‘Burnollious Theorem’? What does it explain?
It explains that when there is a restriction line a fluid flow line a ‘DP’ (differential pressure) is created. The DP is maximum at the veena contact point.
The flow measured in the flow line is proportional to the square root of the DP measured where ‘K’ is a constant.
Q= k (sq root of (DP))
### What is the function of an orifice plate in flow measurement?
An orifice plate creates a differential pressure in a flow line. The DP created is used for measuring the flow through the process line.
### How much maximum and minimum orifice ‘d’ (orifice diameter) is permitted in a pipeline?
0.25 D < d < 0.75 D
‘d’ – the Orifice diameter should be in-between 0.25 and 0.75 of the pipeline ‘D’ diameter.
### What is the difference between an orifice plate used in oil and gas flow measurement?
The orifice plate used in a gas line will have a small drain hole at the bottom of the orifice plate.
Image
### How to identify a newly installed orifice plate upstream in a pipeline?
The upstream can be identified by the orifice plate’s Tag number markings. Tag numbers are always marked on the upstream of the orifice plate.
### How much upstream and downstream straight length run is essential for an orifice plate flow measurement?
An upstream of 28D and downstream steam of a minimum of 7D is essential for an accurate orifice plate flow measuring system (where D= pipeline diameter).
The greater the upstream and downstream length, the lesser the flow turbulence and the greater the accuracy in the flow measurement.
### What type of orifice tapping is commonly used?
In general, is using the ‘Flange Tapping’. The upstream and downstream orifice tapping are taken from the flanges.
### Why and when is flow measured on a square root scale?
Flow is measured on a square root scale only when the measurement is done through an orifice plate and a DP transmitter.
The flow measured through the orifice plate is always proportional to the square root of the DP across the orifice plate.
Q = k sq root of DP
Q = Flow
K = Constant
DP = Differential Pressure
### What is a ‘flow factor’?
A ‘flow factor’ is to multiply the flow transmitter signal measured on a 0-10 square root or 0-100 linear scale to get the flow calculated by flow metering.
This is used due to the standardization of the transmitter’s signals, to 20-100 kPa or 4-20mA.
### What are the important parameters considered in deriving a flow factor?
The following points are considered for flow calculation and in deriving the flow factor:
D = pipe diameter,
Small d = orifice diameter
Service = gas or liquid
PI = operating pressure
DP = Transmitter differential pressure
T =operating temperature
Small p= density or molecular weight
Small v= viscosity
Q= expected total flow
### Explain the installation of a DP flow transmitter on a gas and liquid pipeline.
Gas line: the transmitter is installed above the orifice plate to prevent the condensation of gas in the signal line and in the HP & LP chambers.
Liquid line: the transmitter is installed below the orifice plate to prevent the gas trapping in the signal line and in the HP & LP chambers.
### What is a ‘zero check’ and ‘static zero check’ on a DP flow transmitter?
##### Zero Check:
A procedure for checking the transmitter output is equal to 4.00 mA when its HP & LP chambers are equalized and are at atmospheric pressure.
##### Static Zero Check:
A procedure for checking the transmitter output is equal to 4.00 mA when its HP & LP chambers are equalized and are at the operating pressure.
### Why is flow measurement not very accurate?
Flow measurement is less accurate compared to level, temperature, and pressure measurements. This is due to the consideration of various parameters while measuring a flow.
These parameters, such as the accuracy of the orifice plate diameter, and the pipeline diameter their operating parameters such as temperature and pressure do not remain the same in the process operation as the designed parameters.
Generally, an accuracy of 5% is permitted in a flow measurement.
### What is the actual flow, if the operating pressure is higher than the designed pressure?
If the operating pressure is higher than the designed pressure the true flow will be higher than the measured pressure.
A simple calculation is as follows.
Q = Q1* sq root of p1/p2
Q = True Flow
Q1= Measured Flow
P1= Operating pressure
P2= Designed Pressure
### What will be the new flow factor if a DP transmitter is re-ranged from 25 kPa to 50 kPa?
A simple calculation is as follows:
Q1/Q2 = Sq root of DP1/ Sq root of DP2
Q1 = Q2 * Sq root of DP1/ Sq root of DP2
Q1 = New flow factor, Q2 =Existing flow factor
DP1 = Transmitter new range, DP2= Transmitter existing range
Q1 = Q2*SQ ROOT OF 50/25
Q1 = 1.41*Q2
The new flow factor will be 1.41 times higher than the existing flow factor.
### Why generally is a flow transmitter installed upstream of a flow control valve?
A flow transmitter is always installed on upstream of the flow control valve in order to maintain the operating pressure across the flow transmitter sensors.
Downstream of the control valve the pressure changes as the control valve opens or closes.
### Why is a pressure transmitter installed upstream of a flow transmitter?
Upstream of a flow control valve, a pressure transmitter is installed to measure the operating pressure. At times it is used for computing the true flow against the designed pressure.
Downstream of the control valve the pressure changes as the control valve opens and closes.
### What is the operating principle of a turbine meter?
A magnetic pickup installed above a turbine meter measures the number of magnetic flux cut by the turbine meter blades and produces pulses proportional to the volume of liquid flow through the meter.
### What is the output of a turbine meter?
The output of the turbine meter is in pulses.
The pulse per the volume of liquid is constant and distinct for each meter. When a known quantity of liquid flows through the meter, a known number of pulses are produced.
### What does a pre-amplifier do on a turbine meter?
The magnetic pickup inside the turbine meter produces pulses around 30 mv peak to peak.
A pre-amplifier magnifies the small signal to a 12 V DC peak-to-peak square wave and transmits a signal to the control room.
### What are the advantages and disadvantages of a turbine meter Vs an orifice plate flow measurement?
A turbine meter is a good flow measuring unit when the fluid is low clean fluid. The turbine meter measures the volumetric flow. It is directly installed on the flow line.
Its accuracy in flow measurement is high. Accuracy can be re-calculated and the ‘k’ factor can be reset periodically.
### What is a ‘k’ factor on a turbine meter? Who provides the ‘k’ factor?
Each turbine meter is specified with a ’k’ factor which represents the number of pulses produced per a known quantity of liquid.
Example:
k = 265 pulsed/gallon
Generally, the ‘k’ factor is provided by the manufacturer.
### One cubic meter is equal to how many gallons?
1 Cubic Meter = 264.2 gallons.
### Why are counters ( totalizers ) used in flow measurement?
Counters are used for measuring the ‘volumetric flow’ of the fluid in a pipeline.
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### 11 thoughts on “Interview Questions on Flow Measurement”
1. Hi! Very use full information thanks….
• Can please send the above in pdf formate
• Hello Prasad, Click Print button below the Post title & save it in PDF. Its all available to everyone.
2. Very useful information. Thank you.
3. useful post
4. dear sir,
I want some information about fuel gas metering calculation formula and factors.
we are using a metering transmitter that is simple DP transmitter but it is connected to a RTD. please give me information how it is calculating the flow in MMSDC
• Soon it will be posted.
5. Good web for improving skills I loved it.
Thanx all team members.
6. Nice guide thanks for sharing.
I have a question and an observation.
There are a couple of typos on example is Bernoulli’s theorem (easy fix).
The other is I’ve seen the question about flow upstream or downstream of a control valve question and it always has the same answer you gave. It seems to me that the pressure drop can’t be maintained on either side. If the valve is closed the flow is zero and upstream the differential pressure will go to zero. Two things that do seem different is that the upstream side will always have a higher pressure and in single phase flow the upstream side should stay single phase. | 2,123 | 9,681 | {"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} | 3.4375 | 3 | CC-MAIN-2024-22 | latest | en | 0.907901 |
https://www.kidsacademy.mobi/printable-worksheets/answer-key/hard/age-3-4/math/money-word-problems/ | 1,718,485,925,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00856.warc.gz | 762,020,908 | 53,695 | # Extra Challenge Money Word Problems Worksheets for Ages 3-4
Introduce your little ones to the exciting world of numbers with our Extra Challenge Money Word Problems worksheets! Tailored specifically for children aged 3-4 years, these educational interactive worksheets are designed to spark curiosity and enhance early math skills. Through engaging, age-appropriate challenges, children will explore basic concepts of money and simple addition, helping them understand the value of different coins and bills in a fun, interactive way. Perfect for both classroom and home use, these worksheets are a great resource for developing critical thinking and problem-solving skills early in life. Start your child's financial literacy journey today!
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• 3-4
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• Extra Challenge
## Money Mass Worksheet
By regularly working on math with your kids, they will become more confident. Ensure they understand the word problems in this printout, and help them solve it. Check the box for the correct answer of each to verify their work. With this practice, they will gradually get used to math and make progress.
Money Mass Worksheet
Worksheet
Learning Skills
In the early stages of education, finding the right tools to engage and teach young children is crucial. Extra Challenge worksheets on Money Word Problems, designed for children aged 3-4 years, are an exceptional resource. These worksheets serve as educational interactive printables that are not just engaging but highly effective in introducing basic financial concepts to young learners.
Money word problems offer a unique blend of numerical practice and real-world application, making them a vital part of developing a child’s cognitive and mathematical skills. At ages 3 to 4, children are at a developmental stage where they are beginning to understand more about the world around them, including concepts of buying and selling seen in everyday life. These worksheets provide a structured way for children to start making sense of these everyday interactions involving money.
Why are Extra Challenge Worksheets on Money Word Problems Useful?
1. Development of Numerical Skills: These worksheets are crafted to help children recognize numbers and understand simple addition and subtraction, which are often framed around contexts like counting money and making change. This not only improves their mathematical skills but also enhances their problem-solving abilities.
2. Introduction to Money Management: Although preschoolers are not yet handling money independently, early exposure to money concepts through educational interactive printables lays the groundwork for essential life skills. It introduces the basics of money management in a simplified and relatable manner, helping children understand the value of money and basic arithmetic involved in financial transactions.
3. Enhancing Cognitive Abilities: Solving word problems requires a certain level of comprehension and application of logic. These worksheets stimulate cognitive development by encouraging children to read, comprehend, and then apply what they have learned to solve a problem. This multifaceted approach aids in developing both literacy and numeracy skills.
4. Interactive Learning Experience: The design of these educational interactive printables ensures that they are not just sheets for passive learning but encourage active participation. Often these worksheets include colorful images, puzzles, or scenarios that relate directly to children’s experiences, such as going to a shop or buying candies. This makes learning immersive and enjoyable, increasing the child’s engagement and retention of concepts.
5. Preparation for Future Academic Tasks: Early education sets the foundation for later learning. By integrating challenging tasks like simple word problems, children learn to face and overcome learning hurdles from a young age. These worksheets prepare them for more complex mathematical concepts and problem-solving tasks they will encounter in higher grades.
6. Parental or Guardian Involvement: These worksheets also offer a wonderful opportunity for parents or guardians to get involved in their child’s early education. By working together on these educational interactive printables, adults can help reinforce the learning concepts at home, making the educational experience more robust and supportive. This collaboration not only boosts the child's confidence but also strengthens the emotional bond through shared activities.
7. Customization and Scalability: One of the great advantages of Extra Challenge worksheets on Money Word Problems is their adaptability to different learning speeds and styles. Children who may need more of a challenge can be given worksheets with slightly more complex problems, whereas those who are just beginning to grasp the concepts can work on simpler tasks. This scalability ensures that every child finds the right level of challenge, encouraging them to learn without feeling overwhelmed or under-stimulated.
8. Feedback and Assessment: These printables often come with guided solutions or hints that help assess the child’s understanding and progress. This immediate feedback helps in identifying areas where the child might need more practice, allowing for timely intervention and support in their learning journey.
9. Fun and Engaging: Lastly, these worksheets are designed to be fun. They often incorporate elements of games, stories, or characters that children find appealing. This engaging format helps in maintaining a child’s enthusiasm and curiosity about learning, making educational activities something they look forward to rather than shy away from.
In conclusion, Extra Challenge worksheets on Money Word Problems for children aged 3-4 years are not just another set of tasks to be completed; they are a comprehensive educational tool. They help in nurturing a young child’s mathematical skills, logical thinking, and understanding of money—all through an interactive and enjoyable learning process. By incorporating these educational interactive printables into early childhood education, we can give children a strong and positive start in their lifelong learning journey, equipping them with the skills necessary to navigate the academic and real-world challenges they will encounter later in life. | 1,103 | 6,362 | {"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} | 2.828125 | 3 | CC-MAIN-2024-26 | latest | en | 0.926637 |
https://eng.libretexts.org/Bookshelves/Computer_Science/Book%3A_Foundations_of_Computation_(Critchlow_and_Eck)/3%3A_Regular_Expressions_and_FSA's/3.1_Languages | 1,571,211,920,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00166.warc.gz | 475,945,418 | 21,818 | # 3.1 Languages
In formal language theory, an alphabet is a finite, non-empty set. The elements of the set are called symbols. A finite sequence of symbols $$a_{1} a_{2} \ldots a_{n}$$ from an alphabet is called a string over that alphabet.
Example $$3.1 . \Sigma=\{0,1\}$$ is an alphabet, and $$011,1010,$$ and 1 are all strings over $$\Sigma .$$
Note that strings really are sequences of symbols, which implies that order matters. Thus 011, 101, and 110 are all different strings, though they are made up of the same symbols. The strings $$x=a_{1} a_{2} \ldots a_{n}$$ and $$y=b_{1} b_{2} \dots b_{m}$$ are equal only if m = n (i.e. the strings contain the same number of symbols) and $$a_{i}=b_{i}$$ for all $$1 \leq i \leq n$$.
Just as there are operations defined on numbers, truth values, sets, and other mathematical entities, there are operations defined on strings. Some important operations are:
1. length: the length of a string x is the number of symbols in it. The notation for the length of x is $$|x|$$. Note that this is consistent with other uses of $$| |,$$ all of which involve some notion of size: $$|$$ number $$|$$ measures how big a number is (in terms of its distance from 0); $$|\operatorname{set}|$$ measures the size of a set (in terms of the number of elements).
We will occasionally refer to a length-n string. This is a slightly awkward, but concise, shorthand for “a string whose length is n”.
2. concatenation: the concatenation of two strings $$x=a_{1} a_{2} \dots a_{m}$$ and $$y=b_{1} b_{2} \ldots b_{n}$$ is the sequence of symbols $$a_{1} \ldots a_{m} b_{1} \ldots b_{n}$$. Some- times · is used to denote concatenation, but it is far more usual to see the concatenation of x and y denoted by xy than by x · y. You can easily convince yourself that concatenation is associative (i.e. $$(x y) z=x(y z)$$ for all strings $$x, y$$ and $$z .$$ Concatenation is not com- mutative (i.e. it is not always true that $$x y=y x :$$ for example, if $$x=a$$ and $$y=b$$ then $$x y=a b$$ while $$y x=b a$$ and, as discussed above, these strings are not equal.)
3. reversal: the reverse of a string $$x=a_{1} a_{2} \ldots a_{n}$$ is the string $$x^{R}=$$ $$a_{n} a_{n-1} \ldots a_{2} a_{1}$$.
Example $$3.2 .$$ Let $$\Sigma=\{a, b\}, x=a, y=a b a a,$$ and $$z=b a b .$$ Then $$|x|=1,|y|=4,$$ and $$|z|=3 .$$ Also, $$x x=a a, x y=$$ abaa $$, x z=a b a b,$$ and $$z x=$$ baba. Finally, $$x^{R}=a, y^{R}=a a b a,$$ and $$z^{R}=b a b$$.
By the way, the previous example illustrates a naming convention stan- dard throughout language theory texts: if a letter is intended to represent a single symbol in an alphabet, the convention is to use a letter from the beginning of the English alphabet (a, b, c, d ); if a letter is intended to represent a string, the convention is to use a letter from the end of the English alphabet (u, v, etc).
In set theory, we have a special symbol to designate the set that contains no elements. Similarly, language theory has a special symbol ε which is used to represent the empty string, the string with no symbols in it. (Some texts use the symbol $$\lambda$$ instead. ) It is worth noting that $$|\varepsilon|=0,$$ that $$\varepsilon^{R}=\varepsilon$$, and that $$\varepsilon \cdot x=x \cdot \varepsilon=x$$ for all strings $$x$$. (This last fact may appear a bit confusing. Remember that ε is not a symbol in a string with length 1, but rather the name given to the string made up of 0 symbols. Pasting those 0 symbols onto the front or back of a string x still produces x.)
The set of all strings over an alphabet $$\Sigma$$ is denoted $$\Sigma^{*} .$$ (In language theory, the symbol $$*$$ is typically used to denote "zero or more", so $$\Sigma^{*}$$ is the set of strings made up of zero or more symbols from \Sigma.) Note that while an alphabet $$\Sigma$$ is by definition a finite set of symbols, and strings are by definition finite sequences of those symbols, the set $$\Sigma^{*}$$ is always infinite. Why is this? Suppose $$\Sigma$$ contains $$n$$ elements. Then there is one string over $$\Sigma$$ with 0 symbols, $$n$$ strings with 1 symbol, $$n^{2}$$ strings with 2 symbols ( since there are $$n$$ choices for the first symbol and $$n$$ choices for the second), $$n^{3}$$ strings with 3 symbols, etc.
Example $$3.3 .$$ If $$\Sigma=\{1\},$$ then $$\Sigma^{*}=\{\varepsilon, 1,11,111, \ldots\} .$$ If $$\Sigma=\{a, b\}$$ then $$\Sigma^{*}=\{\varepsilon, a, b, a a, a b, b a, b b, \text { aaa, aab, } \ldots\} .$$
Note that $$\Sigma^{*}$$ is countably infinite: if we list the strings as in the preceding example (length-0 strings, length-1 strings in “alphabetical” order, length-2 strings similarly ordered, etc) then any string over $$\sum$$ will eventually appear. (In fact, if $$|\Sigma|=n \geq 2$$ and $$x \in \Sigma^{*}$$ has length $$k,$$ then $$x$$ will appear on the list within the first $$\frac{n^{k+1}-1}{n-1}$$ entries.
We now come to the definition of a language in the formal language theoretical sense.
Definition 3.1.
A language over an alphabet $$\Sigma$$ is a subset of $$\Sigma^{*} .$$ Thus, a language over $$\Sigma$$ is an element of $$\mathcal{P}\left(\Sigma^{*}\right),$$ the power set of $$\Sigma^{*} .$$
In other words, any set of strings (over alphabet $$\Sigma )$$ constitutes a language (over alphabet $$\Sigma )$$
Example $$3.4 .$$ Let $$\Sigma=\{0,1\} .$$ Then the following are all languages over $$\Sigma :$$
$$L_{1}=\{011,1010,111\}$$
$$L_{2}=\{0,10,110,1110,11110, \ldots\}$$
$$L_{3}=\left\{x \in \Sigma^{*} | n_{0}(x)=n_{1}(x)\right\}$$,
where the notation n0(x) stands for the number of 0’s in the string x, and similarly for n1(x).
$$L_{4}=\{x | \quad x \text { represents a multiple of } 5 \text { in binary }\}$$
Note that languages can be either finite or infinite. Because $$\Sigma^{*}$$ is infinite, it clearly has an infinite number of subsets, and so there are an infinite number of languages over Σ. But are there countably or uncountably many such languages?
Theorem 3.1.
For any alphabet $$\Sigma,$$ the number of languages over $$\Sigma$$ is uncountable.
This fact is an immediate consequence of the result, proved in a previ- ous chapter, that the power set of a countably infinite set is uncountable. Since the elements of $$\mathcal{P}(\Sigma)$$ are exactly the languages over $$\Sigma$$, there are uncountably many such languages.
Languages are sets and therefore, as for any sets, it makes sense to talk about the union, intersection, and complement of languages. (When taking the complement of a language over an alphabet $$\Sigma,$$ we always consider the univeral set to be $$\Sigma^{*},$$ the set of all strings over $$\Sigma$$ .) Because languages are sets of strings, there are additional operations that can be defined on languages, operations that would be meaningless on more general sets. For example, the idea of concatenation can be extended from strings to languages.
For two sets of strings S and T, we define the concatenation of S and T (denotedS·T or just ST)to be the set $$S T=\{s t | s \in S \wedge t \in$$$$\mathrm{~ T \} . ~ F o r ~ e x a m p l e , ~ i f ~} S=\{a b, a a b\}$$ and $$T=\{\varepsilon, 110,1010\},$$ then $$S T=$$$$\{a b, a b 110, a b 1010, a a b, a a b 110, a a b 1010\} .$$ Note in particular that $$a b \in S T$$ because $$a b \in S, \varepsilon \in T,$$ and $$a b \cdot \varepsilon=a b .$$ Because concatenation of sets is defined in terms of the concatenation of the strings that the sets contain, concatenation of sets is associative and not commutative. (This can easily be verified.)
When a set S is concatenated with itself, the notation SS is usually scrapped in favour of $$S^{2} ;$$ if $$S^{2}$$ is concatenated with $$S,$$ we write $$S^{3}$$ for the resulting set, etc. So $$S^{2}$$ is the set of all strings formed by concatenating two (possibly different, possibly identical) strings from $$S, S^{3}$$ is the set of strings formed by concatenating three strings from $$S,$$ etc. Extending this notation, we take $$S^{1}$$ to be the set of strings formed from one string in $$S$$ $$\left(\text { i.e. } S^{1} \text { is } S \text { itself }\right),$$ and $$S^{0}$$ to be the set of strings formed from zero strings in $$S$$ (i.e. $$S^{0}=\{\varepsilon\} ) .$$ If we take the union $$S^{0} \cup S^{1} \cup S^{2} \cup \ldots,$$ then the resulting set is the set of all strings formed by concatenating zero or more strings from $$S,$$ and is denoted $$S^{*} .$$ The set $$S^{*}$$ is called the Kleene closure of $$S,$$ and the $$*$$ operator is called the Kleene star operator.
Example $$3.5 .$$ Let $$S=\{01, b a\} .$$ Then
$$S^{0}=\{\varepsilon\}$$
$$S^{1}=\{01, b a\}$$
$$S^{2}=\{0101,01 b a, b a 0, \text { baba }\}$$
$$S^{3}=\{010101,0101 b a, 01 b a 01,01 b a b a, b a 0101, b a 010 a, b a b a 01, b a b a b a\}$$
etc, so
$$S^{*}=\{\varepsilon, 01, b a, 0101,01 b a, b a 01, b a b a, 010101,0101 b a, \ldots\}$$
Note that this is the second time we have seen the notation something*.
We have previously seen that for an alphabet $$\Sigma, \Sigma^{*}$$ is defined to be the set of all strings over $$\Sigma .$$ If you think of $$\Sigma$$ as being a set of length-1 strings, and take its Kleene closure, the result is once again the set of all strings over $$\Sigma,$$ and so the two notions of * coincide.
Example 3.6. Let $$\Sigma=\{a, b\}$$ Then
$$\Sigma^{0}=\{\varepsilon\}$$
$$\Sigma^{0}=\{\varepsilon\}$$
$$\Sigma^{1}=\{a, b\}$$
$$\Sigma^{2}=\{a a, a b, b a, b b\}$$
$$\Sigma^{3}=\{a a a, a a b, a b a, a b b, b a a, b a b, b b a, b b b\}$$
$$\Sigma^{*}=\{\varepsilon, a, b, a a, a b, b a, b b, a a a, a a b, a b a, a b b, b a a, b a b, \ldots\}$$
Exercises
1. Let $$S=\{\varepsilon, a b, a b a b\}$$ and $$T=\{a a, a b a, a b b a, a b b b a, \ldots\}$$. Find the following:
a) $$S^{2} \quad$$ b) $$S^{3} \quad$$ c) $$S^{*} \quad$$ d) $$S T \quad$$ e) $$T S$$
2. The reverse of a language $$L$$ is defined to be $$L^{R}=\left\{x^{R} | x \in L\right\}$$. Find $$S^{R}$$ and $$T^{R}$$ for the $$S$$ and $$T$$ in the preceding problem.
3. Give an example of a language $$L$$ such that $$L=L^{*}$$ | 3,364 | 10,110 | {"found_math": true, "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": 1, "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} | 4.15625 | 4 | CC-MAIN-2019-43 | latest | en | 0.698934 |
https://mathhelpboards.com/threads/putnam-2012-a5.3801/ | 1,601,084,934,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400232211.54/warc/CC-MAIN-20200926004805-20200926034805-00151.warc.gz | 481,523,623 | 14,519 | # Putnam 2012: A5
#### jakncoke
##### Active member
Let $\mathbb{F}_p$ denote the integers mod p and let n be a positive integer.
Let v be a fixed vector $\in \mathbb{F}_p^{n}$, Let M be a nxn matrix with entries from $\mathbb{F}_p$. Define G:$\mathbb{F}_p^{n} \to \mathbb{F}_p^{n}$ by
G(x) = v + Mx. Define the k-fold composition of G by itself by $G^{1}(x) = G(x)$
and $G^{k+1} = G (G^{k}(x))$ Determine all pairs p,n for which there exists a vector v and a matrix M such that the $p^n$ vectors of $G^{k}(0), k=1,...,p^{n}$ are distinct.
#### jakncoke
##### Active member
I've solved it but it will take a bit to type the justification for some of the points but contained in the spoiler are general tips for how to think about the problem
This is a pretty interesting question.
Basically it is asking, for which matrix group $GL_{n+1}(F_p)$, can you generate all the elements in $(F_p)^n$ by iterating a matrix (raising it to the power), and get all the elements in $(F_p)^n$.
Basically for which $GL_{n+1}(F_p)$ does there exist an element of order $p^{n}$.
#### jakncoke
##### Active member
Still writing it, hit sumbit by mistake, bear with me. | 359 | 1,157 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "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} | 3.03125 | 3 | CC-MAIN-2020-40 | longest | en | 0.884605 |
http://conversion.org/mass/gamma/nanogram | 1,642,513,662,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00476.warc.gz | 15,711,681 | 6,731 | # gamma to Nanogram conversion
Conversion number between gamma [γ] and Nanogram [ng] is 1000. This means, that gamma is bigger unit than Nanogram.
### Contents [show][hide]
Switch to reverse conversion:
from Nanogram to gamma conversion
### Enter the number in gamma:
Decimal Fraction Exponential Expression
[γ]
eg.: 10.12345 or 1.123e5
Result in Nanogram
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 gamma = (exactly) (0.000000001) / (1*10^-12) = 1000 Nanogram
• 1 Nanogram = (exactly) (1*10^-12) / (0.000000001) = 0.001 gamma
• ? gamma × (0.000000001 ("kg"/"gamma")) / (1*10^-12 ("kg"/"Nanogram")) = ? Nanogram
### High precision conversion
If conversion between gamma to kilogram and kilogram to Nanogram is exactly definied, high precision conversion from gamma to Nanogram is enabled.
Decimal places: (0-800)
gamma
Result in Nanogram:
?
### gamma to Nanogram conversion chart
Start value: [gamma] Step size [gamma] How many lines? (max 100)
visual:
gammaNanogram
00
1010000
2020000
3030000
4040000
5050000
6060000
7070000
8080000
9090000
100100000
110110000
Copy to Excel
## Multiple conversion
Enter numbers in gamma and click convert button.
One number per line.
Converted numbers in Nanogram:
Click to select all
## Details about gamma and Nanogram units:
Convert Gamma to other unit:
### gamma
Definition of gamma unit: ≡ 1 μg . = 1 μg
Convert Nanogram to other unit:
### Nanogram
Definition of Nanogram unit: =1x10-12 kg.
← Back to Mass units | 461 | 1,548 | {"found_math": true, "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": 1, "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} | 2.953125 | 3 | CC-MAIN-2022-05 | latest | en | 0.552918 |
https://cboard.cprogramming.com/cplusplus-programming/164914-acm-programming-contest-seeking-answers-post1217381.html?s=7a771254dacd6b2860dce1c8856e3120 | 1,601,086,775,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400232211.54/warc/CC-MAIN-20200926004805-20200926034805-00798.warc.gz | 313,450,003 | 18,104 | # Thread: ACM Programming Contest.... Seeking Answers!
1. ## ACM Programming Contest.... Seeking Answers!
Hello, all!
First off, this is program from the ACM programming contest, of which today I and two teammates competed. We managed to solve two of nine correctly, but sadly the two programs I was responsible for were never accepted.
I am hoping that someone here can help me understand what I overlooked. Don't worry, the contest is over, so you are not helping me cheat. We had no access to the internet during the contest.
The program I will focus on in this thread is as follows:
Rank Order
Your team has been retained by the director of a competition who supervises a panel of judges. The competition asks the judges to assign integer scores to competitors -- the higher the score, the better. Although the event has standards for what score values mean, each judge is likely to interpret those standards differently. A score of 100, say, may mean different things to different judges.
The director's main objective is to determine which competitors should receive prizes for the top positions. Although absolute scores may differ from judge to judge, the director realizes that relative rankings provide the needed information -- if two judges rank the some competitors first, second, third, ... then they agree on who should receive the prizes.
Your team is to write a program to assist the director by comparing the scores of pairs of judges. The program is to read two lists of integer scores in competitor order and determine the highest ranking place (first place being highest) at which the judges disagree.
Input
Input to your program will be a series of score list pairs. Each pair begins with a single integer giving the number of competitors N, 1 < N < 1,000,000. The next N integers are the scores from the first judge in competitor order. The are followed by the second judge's scores -- N more integers, also in competitor order. Scores are in the range 0 to 100,000,000 inclusive. Judges are not allowed to give ties, so each judge's scores will be unique. Values are separated from each other by one or more spaces and/or newlines. The last score list pair is followed by the end-of-file indicator.
Output
For each score pair, print a line with the integer representing the highest-ranking place at which the judges do not agree. If the judges agree on ever place, print a line containing only the word 'agree'. Use the format below: "Case", one space, the case number, a colon and one space, and the answer for that case with no trailing spaces.
Sample Input
4
3 8 6 2
15 37 17 3
8
80 60 40 20 10 30 50 70
160 100 120 80 20 60 90 135
Sample Output
Case 1: agree
Case 2: 3
The following is (roughly, I'm retyping now my memory):
Code:
```#include <iostream>
using namespace std;
struct contestant
{
int score,
cont_num;
};
void swap( contestant a[], int i, int j );
void qsort( contestant a[], int left, int right );
int num_of_conts;
int case_num = 1;
int main(){
cin >> num_of_conts;
while( cin )
{
contestant * judge1 = new contestant[num_of_conts]; // list for each judges scores
contestant * judge2 = new contestant[num_of_conts]; // same index means same contestant, just score from diff judge
for( int i = 0; i < num_of_conts; i++ )
{
cin >> judge1[i].score;
judge1[i].cont_num = i;
}
for( int i = 0; i < num_of_conts; i++ )
{
cin >> judge2[i].score;
judge2[i].cont_num = i;
}
// sort both arrays of contestants by score, from lowest score to highest
qsort( judge1, 0, num_of_conts - 1 );
qsort( judge2, 0, num_of_conts - 1 );
int i;
for( i = num_of_conts - 1; i >= 0; i-- )
if( judge1[i].cont_num != judge2[i].cont_num )
break;
cout << "Case " << case_num++ << ": ";
if( i == -1 )
cout << "agree" << endl;
else
cout << num_of_conts - i << endl;
delete [] judge1;
delete [] judge2;
cin >> num_of_conts;
}
return 0;
}
void swap( contestant a[], int i, int j )
{
contestant temp = a[i];
a[i] = a[j];
a[j] = temp;
}
void qsort( contestant a[], int left, int right )
{
int i, last;
if( left >= right )
return;
swap( a, left, ( left + right ) / 2 );
last = left;
for( i = left + 1; i <= right; i++ )
if( a[i].score < a[left].score )
swap( a, ++last, i );
swap( a, left, last );
qsort( a, left, last - 1 );
qsort( a, last + 1, right );
}```
The logic is fairly simple:
*chug in the scores
*sort contestants based on scores
*iterate along the list of contestants, from best to worst, breaking on
mismatch
The judges never accepted any iteration of this code. All I could think of doing was switching most/all the ints to long and then long longs in case of any overflows.
Any thoughts on why it was never accepted?
2. Did you ever try NOT using "using namespace std;" I am thinking a name collision issue on swap or qsort.
No idea if it is likely or not; I am a C programmer who from time to time tries to learn C++.
Tim S.
3. The problem is that the exercise required you to compare rankings that each judge gives each contestant. Your approach doesn't do that.
Even with that, your code is a hybrid of C and C++. Even if the judges are unfussed by what language you use, using a hybrid of two is not good form.
4. @grumpy
If you could be more specific as to how I'm failing to do what the program asked.
From the last paragraph of the description:
"Your team is to write a program to assist the director by comparing the scores of pairs of judges. The program is to read two lists of integer scores in competitor order and determine the highest ranking place (first place being highest) at which the judges disagree."
I read in the scores, while giving each person an id. Then once they are sorted based on score, I start with the highest ranked contestant (far end of the arrays). If the have the same id, the same contestant received the same ranking. I progress on until the end of array or a mismatch is found, then printed.
What am I missing?
Also, when you say I'm using a mix of C and C++, what are you exactly referring to? As in, what is C in the source that would not be considered C++?
EDIT:
Also, you do realize that my programs output matches that of the sample output for the given sample input? Is this just a coincidence?
5. Originally Posted by jwroblewski44
From the last paragraph of the description:
"[COLOR=#333333]Your team is to write a program to assist the director by comparing the scores of pairs of judges. The program is to read two lists of integer scores in competitor order and determine the highest ranking place (first place being highest) at which the judges disagree."
The approach you described is sorting contestants by scores. That is different from checking if judges ordered contestants in the same way.
Originally Posted by jwroblewski44
Also, when you say I'm using a mix of C and C++, what are you exactly referring to? As in, what is C in the source that would not be considered C++?
Depends on your definition of C++. From a "C++ as a sort-of superset of C" perspective, your code might be said to quality. From a "using idiomatic C++" perspective it does not.
Expert judges will look for idiomatic code in whatever programming language is used - it shows technical discipline by the author of that code.
The point of idioms in any programming language is using facilities of that language and its library effectively, without having to roll your own. Idiomatic code tends to be smaller (since it uses facilities that are provided by the language AND its standard library), easier to get right (reusing something that is specified in a standard is less error prone than writing new code to do the same thing).
Idiomatic C++ very rarely makes use of pointers, at least not directly. It also makes more use of - or extends if needed - facilities in the standard library. Your usage of pointers, the usage of new/delete to create an array (instead of a standard container), writing a function to do sorting (instead of an appropriate algorithm that works with iterators from a standard container), and writing a swap function are all hallmarks of hybrid C/C++.
Originally Posted by jwroblewski44
Also, you do realize that my programs output matches that of the sample output for the given sample input? Is this just a coincidence?
Maybe it is. I don't know. I certainly can't exclude that possibility.
When I saw that your description of approach was not consistent with what was asked for, I didn't bother to examine the code to see if it did something different. I only looked at the code enough to recognise it was not idiomatic C++.
Producing a given set of outputs from a given set of inputs does not demonstrate you have used the right approach to get there. I've seen lots of software that passes test cases, despite actually using a completely wrong approach.
All I'm saying is that, given the information you provided, I can understand why the judges did not accept any iteration of your code. If I was a judge (and, no, I wasn't) I would have done the same, for the reasons I explained.
6. Originally Posted by jwroblewski44
Also, when you say I'm using a mix of C and C++, what are you exactly referring to? As in, what is C in the source that would not be considered C++?
To see why it might not be considered "idiomatic C++" it might be instructive to try and change this program to make it a valid C program and see which lines needed to be changed. In your example, only a few trivial changes must be made, such as changing cin/out to scanf/printf, changing new/delete to malloc/free, etc. Also, unless you need to make your own sorting algorithm, why not use the standard library rather than defining your own? C standard library defines qsort and C++ standard library defines std::sort.
7. Yes. I agree that writing my own sort was not the best approach. Simply, I don't have much experience using the standard system sorts, but I did have access to a book with a sort method I was familiar with. So rather than flail about trying to show my command of the language, I simply used what I knew to accomplish the main objective. Also, I'm a C guy at heart, so it's hard not to stray.
I can't think of any other way of seeing if judges ranked each contestant relatively the same.... Could you point me in the right direction?
8. I'm learning C++11 here's my attempt at this contest (using your same logic...not sure if they want you to instead normalize scores between each judge...the result is the same though.
Code:
```#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <boost/regex.hpp>
struct contestant {
unsigned long score, cont_num;
bool operator<(const contestant& b) const { return score < b.score; }
};
inline bool matches_re(const std::string& s, const std::string& exp)
{
boost::regex re;
try {
re.assign(exp);
} catch (const boost::regex_error& e) {
std::cout << "regex error: " << e.what() << " code: " << e.code() << "\n";
return false;
}
return boost::regex_match(s, re);
}
inline std::string perform_judgement(unsigned long& contestants, std::vector<std::multiset<contestant>>& judge)
{
unsigned long count;
for (count = 0; count < contestants; ++count) {
for ( const auto& j : judge ) { // Sorted by scores
// Ugly but iterator doesn't support addition apparently grumble..
auto iter = j.begin();
for (unsigned long i = 0; iter != j.end() && i < count; ++i) ++iter;
if (iter->cont_num != count)
return std::to_string(count);
}
}
return count == contestants ? "agree" : "0";
}
int main(void)
{
unsigned long contestants = 0;
unsigned long caseno = 0;
std::vector<std::multiset<contestant>> judges;
do {
std::string line;
const auto& ok = std::getline(std::cin, line);
std::stringstream ss{line};
if (!ok || matches_re(line, "^\\s*\\d+\\s*")) {
/* Report our case results */
if (contestants) {
std::cout << "Case #"
<< ++caseno
<< ": "
<< perform_judgement(contestants, judges)
<< "\n";
judges.clear(); // clear judge list for next case
}
// Received contestant count, install it
ss >> contestants;
} else if (matches_re(line, "^\\s*(?:(?:\\d+)\\s?)+")) {
unsigned long count = 0;
contestant ct;
std::multiset<contestant> scores;
// Read contestant scores into sorted order
// and remember their original number
while ( count < contestants && ss >> ct.score ) {
ct.cont_num = count++;
scores.insert((contestant){ct.score, ct.cont_num});
}
// Add these scores to the judge list
judges.emplace_back(scores);
} else {
std::cerr << "[invalid input]: " << line << "\n";
}
} while(std::cin);
return 0;
}```
9. @nonpuz
I spoke with my department chair(he was a judge during the comp) and he gave me the test cases for the problem. When I have some free time ( thanksgiving break? ), I'm going to check test my program with their test cases and see what I missed. After this, I will test your code and let you know whether or not they would have given your program 'the nod'.
10. Well once I went through and changed all the int's to long's, the program gave all the correct outputs. The issue was that I neglected to replace the int's used for the counters in the for loops.
@nonpuz
when I try to compile your code, my system can't find the "boost/regex.hpp" file.
11. Originally Posted by jwroblewski44
when I try to compile your code, my system can't find the "boost/regex.hpp" file.
That's part of the boost library. You would have to download and install it. Which would not a bad idea certainly, as it's shock full of useful stuff.
But then again, it might probably be better off just using std::regex instead, allowing you to skip the dependency on boost.
Popular pages Recent additions | 3,296 | 13,601 | {"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} | 2.71875 | 3 | CC-MAIN-2020-40 | latest | en | 0.968822 |
https://www.enotes.com/homework-help/write-an-equation-that-decsribes-puchasing-power-440699 | 1,516,709,966,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891926.62/warc/CC-MAIN-20180123111826-20180123131826-00359.warc.gz | 897,790,472 | 9,161 | # Write an equation that decsribes puchasing power parity and explain the equation.
Karyth Cara | Certified Educator
Purchasing power parity relates to the currencies of two different countries. "Parity" is defined as two things being in a state of being equal: equal value between two things. In economics, when currencies are in parity, the ratio of the price level of a fixed basket of the goods and services in each country (e.g., basket price level of bread, butter, fabric, machiine oil, education) is equal in value to the ratio of the countries currency.
Purchasing power parity (PPP) ... states that exchange rates between currencies are in equilibrium when their purchasing power is the same in each of the two countries. ... the exchange rate [ratio] between two countries should equal the ratio of the two countries' price level of a fixed basket of goods and services. (University of British Columbia)
Purchasing Power Parity (PPP) is calculated as:
S = P1/P2
S: exchange rate of currency 1 to currency 2
P1: cost of good A in currency 1
P2: cost of good A in currency 2 (Investopedia.com) | 256 | 1,108 | {"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} | 3.03125 | 3 | CC-MAIN-2018-05 | latest | en | 0.934732 |
https://ww2.mathworks.cn/matlabcentral/cody/problems/1937-get-input-and-output-variable-names/solutions/1982121 | 1,600,533,052,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192778.51/warc/CC-MAIN-20200919142021-20200919172021-00084.warc.gz | 684,303,547 | 18,420 | Cody
# Problem 1937. Get input and output variable names
Solution 1982121
Submitted on 18 Oct 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
s = 'function foo()'; inNames_correct = {}; outNames_correct = {}; [inNames, outNames] = get_variable_names(s); assert(isequal(inNames,inNames_correct)) assert(isequal(outNames,outNames_correct))
2 Pass
s = 'function a = foo(b)'; inNames_correct = {'b'}; outNames_correct = {'a'}; [inNames, outNames] = get_variable_names(s); assert(isequal(inNames,inNames_correct)) assert(isequal(outNames,outNames_correct))
3 Pass
s = 'function [myOutput] = foo(b,c)'; inNames_correct = {'b';'c'}; outNames_correct = {'myOutput'}; [inNames, outNames] = get_variable_names(s); assert(isequal(inNames,inNames_correct)) assert(isequal(outNames,outNames_correct))
4 Pass
s = 'function [A,B] = foo(c,d,f)'; inNames_correct = {'c';'d';'f'}; outNames_correct = {'A';'B'}; [inNames, outNames] = get_variable_names(s); assert(isequal(inNames,inNames_correct)) assert(isequal(outNames,outNames_correct))
5 Fail
s = sprintf('function %s=...\nmyfun %s...\n %s\n%% With comments[]()\nandCode = 4*[1 3 4/(1+2)];',... '[how,about , several,outputs]',... '(and , several, inputs ,',... 'split,over,lotsOf,lines)'); inNames_correct = {'and';'several';'inputs';'split';'over';'lotsOf';'lines'}; outNames_correct = {'how';'about';'several';'outputs'}; [inNames, outNames] = get_variable_names(s); assert(isequal(inNames,inNames_correct)) assert(isequal(outNames,outNames_correct))
Assertion failed. | 490 | 1,648 | {"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} | 3.1875 | 3 | CC-MAIN-2020-40 | latest | en | 0.502423 |
http://searchenginewatch.com/article/2064366/Lies-Damned-Lies-and-Statistics | 1,394,650,285,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394023864543/warc/CC-MAIN-20140305125104-00088-ip-10-183-142-35.ec2.internal.warc.gz | 172,128,452 | 17,508 | SEO News
# Lies, Damned Lies, and Statistics
When Mark Twain penned, "There are three kinds of lies: lies, damned lies, and statistics," he must have foreseen online campaign analytics.
Over the past year, I've been in numerous meetings with advertisers, agencies, and online media vendors. I'm often amazed at the confusion surrounding how local search campaign analytics are reported, and sometimes twisted, to create a desired story.
While most of today's column will center on analytics as it relates to local search campaigns, you can use the same concepts for online, and many offline, lead-generating capabilities.
The Basics: Where to Start
In analytics, all metrics can be traced to two standard units of measure: counts and ratios.
Counts are whole numbers, and as the name connotes, are a total number of some item. Ratios are generally the relation between two counts, or the proportion in relation to the whole. In layman's terms: two counts divided by each other.
Don't worry, that's the entire statistics lesson for today. Just remember these basic concepts when embarking on your journey into statistical analysis.
In local search, we'll receive a large number of counts and ratios from media vendors, or observe them via our own Web site analytics package:
• Impressions
• Visits
• Unique visitors
• Clicks
• Click-through rate (CTR)
• Cost per click (CPC)
(Instead of defining each term individually, you can find a great resource on the Internet Advertising Bureau's glossary page to help you with each term's meaning.)
Calculating ROI
One of the simplest methods for determining ad effectiveness is to measure campaign ROI. In an earlier column, I provided some simple formulas associated with this key performance indicator:
One of the most common errors I see in calculating ROI is when a marketer or their agency simply divides ad cost by total sales and then declares it ROI. Actually, that formula represents the return on ad spend (ROAS). While it remains an important basic measure, it doesn't take into account the marketer's cost of delivering the goods or services sold.
One of the most important parts of the above ROI calculation is how the "Number of Leads" count is captured. Your leads count should include all of the clicks from the specific campaign, as well as the phone traffic measured through call tracking, and walk-in purchases (if you measure these).
For marketers who desire to move to a higher level of precision in calculating ROI, consider breaking out the lead counts into their component response channels (e.g., clicks, calls, and walk-ins) and apply conversion rates based on their respective response channels.
Keep in mind that conversion rates by response channel may vary dramatically by category. That being said, as a starting point we generally see a 3:1 ratio of call to click sales conversion.
Now let's look at the same ROI calculation from above, breaking down leads and applying a differential conversion rate between clicks and calls.
The more precision you use in measuring and correctly attributing conversion rates to individual lead sources and response channels, the better you'll understand the true value of your local search campaigns.
A word of caution on a common issue: some marketers inflate a campaign's ROI to misapply a lifetime value of customer (LTV) in order to make a negative ROI campaign go positive. Now I'm not saying LTV should never be used, but I suggest you go with concrete facts where possible.
For example, to correctly assign LTV in an ROI calculation, the marketer needs to understand what percentage of the customers responding to a campaign are new customers to their place of business. If they're repeat buyers, for example, using LTV will double count some of the sales, creating an inflated ROI.
Hopefully, this information helps clear up how to use base-level counts and ratios to help measure the effectiveness and optimize your local search initiatives. Please feel free to contact me for any additional clarification. Now get counting.
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http://www.jiskha.com/display.cgi?id=1314337295 | 1,498,382,227,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320476.39/warc/CC-MAIN-20170625083108-20170625103108-00628.warc.gz | 553,458,977 | 3,558 | # calc
posted by .
an open box is to be made from a 4 ft by 5 ft piece of cardboard by cutting out squares of equals sizes with width x ft from the four corners and bending up the flaps to form sides.express the volume of the open box as a function of x what is the domain of the function | 70 | 290 | {"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} | 2.53125 | 3 | CC-MAIN-2017-26 | latest | en | 0.942906 |
https://stats.stackexchange.com/questions/255355/multiple-regression-with-negative-coefficient-for-sales-vs-revenue | 1,571,419,031,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986684226.55/warc/CC-MAIN-20191018154409-20191018181909-00277.warc.gz | 724,435,155 | 30,662 | # Multiple regression with negative coefficient for Sales vs Revenue
Hi I have a table of data with 3 columns: Sales,Cancels and Net revenue. I am triying to perform a Linear regression to check how much the sales and cancels affect the Net Rev but the first thing i noticed is that there is a negative correlation between sales and Net revenue (So the less I sell the more I gain?) Corr(Sales,Net Rev)~-.68
These are monthly time series, where sales and cancels are dollars per month and the net revenue is '000 dollars per month. The revenue is the sum of different incomes but not directly the sum of the sales.
Then I run a linear regression net revenue v.s sales and Cancels (Using Excel) and I see that sales is significant at 99% but again it has a negative coefficient.
So my questions are:
1)Why do I get this negative correlation between sales and revenue? I supposed is an Spurious Correlation. 2)Is there some transformation I can apply to data to get better results? 3)How do I intepret a negative coefficient between sales and revenue? | 237 | 1,054 | {"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} | 2.796875 | 3 | CC-MAIN-2019-43 | latest | en | 0.925099 |
http://math.stackexchange.com/questions/27148/p-groups-with-every-proper-subgroup-cyclic?answertab=oldest | 1,448,660,790,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398450581.71/warc/CC-MAIN-20151124205410-00143-ip-10-71-132-137.ec2.internal.warc.gz | 151,334,765 | 16,505 | # p-Groups with every proper subgroup cyclic
The quaternion group is non-abelian p-group with every proper subgroup cyclic. In general, if $G$ is a non-abelian $p$-group with every proper subgroup cyclic, is it necessarily generalized quaternion?
-
Yes. If all proper subroups are cyclic, the group must have a unique subgroup of order $p$ (if it contains two subgroups of order $p$, one of them can be taken to lie in the center and thus their product is a non-cyclic subgroup). It is a basic result in the theory of $p$-groups that a $p$-group with a unique subgroup of order $p$ is either cyclic or generalized quaternion (see for instance Proposisition 1.3 of Berkovich's "Groups of Prime Power Order"). However, if $G$ is generalized quaternion of order $>8$ then $G/Z(G)$ is a dihedral group which contains a proper non-cyclic subgroup, and this subgroup then corresponds to a non-cyclic subgroup of $G$. So the quaternion group of order 8 really is the only non-abelian $p$-group with this property.
I take it you're assuming $G$ is finite? – Chris Eagle Mar 15 '11 at 16:44
Indeed, otherwise there are counter-examples (ie, infinite groups where all proper subgroups have order $p$). Also note that for the first part, we just need all abelian subgroups to be cyclic. – Tobias Kildetoft Mar 16 '11 at 10:04 | 337 | 1,317 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "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} | 2.6875 | 3 | CC-MAIN-2015-48 | longest | en | 0.891064 |
http://fmwww.bc.edu/repec/bocode/c/circmedian.hlp | 1,701,706,252,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-50/segments/1700679100531.77/warc/CC-MAIN-20231204151108-20231204181108-00293.warc.gz | 20,480,265 | 1,711 | {smcl} {* 6may2004}{...} {hline} help for {hi:circmedian} {hline} {title:Median and mean deviation for circular variables} {p 8 17 2} {cmd:circmedian} {it:varlist} [{cmd:if} {it:exp}] [{cmd:in} {it:range}] {p 8 17 2} {cmd:circmedian} {it:varname} [{cmd:if} {it:exp}] [{cmd:in} {it:range}] [{cmd:,} {cmd:by(}{it:byvar}{cmd:)}] {title:Description} {p 4 4 2}{cmd:circmedian} calculates the circular median and mean deviation from the median for circular variables with scales between 0 and 360 degrees. {p 4 4 2}Circular data have no natural origin; whatever origin is used is a matter of convention. There is thus no natural ordering of circular data from smallest to largest. This is a blow to the idea of a circular median, but by no means a fatal one. We could measure the difference between two angles clockwise or counterclockwise (anticlockwise); measure both, treating both differences as positive, and record the smaller. Thus the difference between 45 and 55 degrees is min(10,350) = 10; the difference between 350 and 10 degrees is min(20, 340) = 20; or more generally {it:d}({it:theta}, {it:phi}) = min(|{it:theta} - {it:phi}|, 360 - |{it:theta} - {it:phi}|). Then define the circular median by the fact that it minimises {p 8 8 2}SUM {it:d}(value, median) {p 4 4 2}or, equivalently, {p 8 8 2}MEAN {it:d}(value, median), {p 4 4 2}the latter being called here the mean deviation (from the median). In the event of ties in these measures the median is defined as the vector mean of all angles minimising either. See help on {help circsummarize}. {title:Remarks} {p 4 4 2}Batschelet (1981, p.242) uses the notation |{it:theta}, {it:phi}| for {it:d}({it:theta}, {it:phi}) and points out that it is also {bind:arccos(cos({it:theta} - {it:phi}))}. Another scale on which to measure difference is thus {bind:1 - cos({it:theta} - {it:phi}).} {title:Options} {p 4 8 2} {cmd:by()} specifies that results are to be shown for each group defined by values of {it:byvar}. This option is only available if a single {it:varname} is specified. {title:Examples} {p 4 8 2}{cmd:. circmedian axisasp wallasp} {p 4 8 2}{cmd:. circmedian axisasp, by(geology)} {title:Author} {p 4 4 2}Nicholas J. Cox, University of Durham, U.K.{break} n.j.cox@durham.ac.uk {title:References} {p 4 8 2}Batschelet, E. 1981. {it:Circular statistics in biology.} London: Academic Press. {title:Also see} {p 4 13 2} On-line: help for {help circsummarize}, {help circdiff} | 802 | 2,435 | {"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} | 3.734375 | 4 | CC-MAIN-2023-50 | latest | en | 0.784589 |
http://theincidentaleconomist.com/understanding-the-employer-tax-subsidy/ | 1,498,132,617,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319265.41/warc/CC-MAIN-20170622114718-20170622134718-00425.warc.gz | 427,256,630 | 18,146 | • The economist’s knock on employer based insurance is that it is paid with pre-tax dollars. That means that each dollar of worker compensation in the form of health insurance is worth more than a dollar in compensation in the form of wages. One consequence is over consumption of health insurance and health care. The tax subsidy also contributes to job lock and other labor market distortions because it makes employer based insurance more valuable than non-group health insurance.
So, the decks are tilted toward employer based insurance but by how much? What is the tax subsidy per dollar of insurance? It is tempting to say that it is the worker’s federal marginal income tax rate. But that’s not right. It is not even close. The key is to the right answer is to ask, for each additional dollar an employer pays toward its wage bill, how much does the employee receive in after-tax compensation? That is, your employer paid a dollar toward covering your wage. What portion of that dollar showed up on your net paycheck?
Let’s start with the additional cost to the employer due to the Social Security payroll tax rate (tSS) and the Medicare payroll tax rate (tMC), ignoring the high-income cutoff of the Social Security tax. Each dollar of gross wage costs your employer (1+tSS+tMC) dollars. Or, put another way, for each dollar your employer pays toward its wage bill on your behalf you only receive 1/(1+tSS+tMC) dollars. But that’s before you pay your taxes. So, let’s do those next.
The final dollar of gross wage is taxed at your federal marginal income tax rate (tF), your state marginal income tax rate (tS), and is also hit with Social Security (tSS) and Medicare payroll taxes (tMC). Therefore, what you actually receive post-tax from that dollar of wage is 1-tFtStSStMC of a dollar. Hence, for the marginal dollar of employer cost, you actually receive post-tax
TP = (1-tFtStSStMC)/(1+tSS+tMC).
The quantity TP in the foregoing expression is known as the tax price and has played an important role in theoretical and applied economic analysis of employer based health insurance. (See the fifth page (marked 297) of The Impact of the Tax System on Health Insurance Coverage by Jon Gruber.) The amount by which TP differs from one is the cost avoided when the marginal dollar of compensation is provided in the form of health insurance instead of wage. Thus 1-TP is the amount in taxes lost to government. Already we can begin to sense that this is quite a different animal than the federal marginal tax rate.
Let’s plug in some numbers. Suppose your federal marginal income tax rate is 20% (selected to correspond to James Kwak’s calculation), your state marginal income tax rate is 5%. The employer and employee Social Security payroll tax rate is 6.2%. And the employer and employee Medicare payroll tax rate is 1.45%. Plugging these in (as decimals, not percentages) to the tax price equation above we find that
TP = (1-0.2-0.05-0.062-0.0145)/(1+0.062+0.0145) = 0.63.
Thus 1-TP is 0.37. Even though your federal marginal income tax rate is only 20% government (federal and state combined) loses 37 cents of tax revenue for each dollar paid in health insurance as opposed to wage.
This is why the tax subsidization of employer based health insurance is a big deal. It goes way beyond the federal marginal tax rate. Assuming that’s all there is to is a mistake (h/t Tyler Cowen; note also Henry Aaron’s correction). Finally, a Cadillac tax of 40% is pretty close to what would be required to recover the lost tax revenue for the example above. But it wouldn’t be enough for individuals with a federal marginal income tax rate above 20%. In this sense, the Cadillac tax is a bargain.
• Love this post – it makes the argument very real. But I think there is a policy fallacy buried in your analysis, at the “20% federal/5% state tax rate.” That simply isn’t very representative.
Try this: a small table that shows the outcomes at \$40k, \$80k, \$150k and \$300k of AGI for a family of four (extra credit for adding single filers:-)), rather than assuming a 20% marginal tax rate and calculating based on marginal dollars of income.
Two reasons: first, since the employer health benefit really applies from dollar one of the employee’s income, it is more reasonable to apply it to the tax burden averaged across the entire annual income, not the highest marginal rate of the taxpayer; and second, it is important to understand the regressiveness of this kind of tax benefit. Very few people (<10% of families of four) have an average tax rate of 37%. The benefit is indeed nearly 40% for AGIs above \$150k, but there are very few of those. For an SEIU member in Texas (no state income tax, \$50k annual income), that benefit looks much more like 12-15%.
• @Dollared – I mentioned in the post why I chose 20%. Your beef is with James Kwak (follow the link in the post).
• Wikipedia tells me that only 6.24% of taxpayers have AGI above \$100k.
So to complete the thought: the “Cadillac Tax” might be a “bargain” if your AGI is above \$100k. But if your AGI is below \$60k (most union workers), then not only is it no bargain from a tax benefit analysis, it is an income tax increase on a person who is very poorly positioned to absorb it.
• @Dollared – As I wrote, I was talking about those with federal marginal tax rates above 20%.
• I lean toward Mr. Frakt, as far as he goes. In fact, on a dollar-weighted basis I’d be inclined to use 25% as typical marginal income tax rate, for a combined marginal rate of 40% federal, and perhaps 50% in states with high state income tax rates.
But Frakt also misses an important part of the picture. Employer-sponsored insurance is by law community rated, and guaranteed issue subject to a waiting period for people without prior creditable coverage. This means that there is a secondary transfer going on: healthy employees are subsidizing the less-healthy.
The employer tax exclusion roughly compensates healthy employees for the added cost of community rating. The tax subsidy merely cancels out the negative cross-subsidy. Of course, for high risk employees, both subsidies are positive.
For me the biggest problems with the tax exclusion are:
1) The amount of tax subsidy is probably too high for the purpose, leading to overconsumption. This is because the subsidy is an accident of the individual tax rates, rather than a deliberately chosen rate.
2) The employer exclusion is regressive, again because the subsidy is based on marginal tax rates.
3) It discriminates against those who purchase equivalent community-rated insurance with after-tax dollars (e.g. COBRA or HIPAA continuation coverage).
• @Bart – It is my understanding that one of the points of negotiation is modifying the Cadillac tax so it is sensitive to employer risk profile, as well as geographic variation in medical costs.
• Austin- That doesn’t surprise me. The whole bill is full of similar tweaks. Why not a few more?
• The fatal flaw of this whole realm of discussion, is the notion of a link between over consumption and plans with good benefits. I’ve yet to see any data even referenced. It’s just assumed as truth that “consumers” (that is to say “patients”) drive health care consumption.
First, what is “overconsumption?” I am sure there are people who make more trips to the doctor than are absolutely neccessary, but most people I know are like me. They go to the doctor when it gets to a point that they suspect it is probably irresponsible not to. My office visit rate was the same when my copay was \$2 and when it was \$20. I admit that now, with no insurance, a \$200 dollar office visit is more than I am willing to do under most circumstances, but honestly, do we want to use economic demand curves to drive health care “use”?
In reality most people have practically no say in the cost of their health care as they lack the expertise to reasonably (safely) do so. They do health care the same way they do car care. They trust their specialist and do what they recommend. So while cost can be shifted to the individual with ever increasing deductibles and copays the cost of health care will not be impacted. Morbidity and mortality will go up, giving us even worse health care results for the money spent.
Real change in health care costs, will require changes in physician practice either by incentive or requirement.
One way to do this is to change away from fee for procedure pay that economically rewards increased use of the system. Perhaps forbid doctors from using facilities in which they have financial stakes, from laboratories to surgery centers.
The even more controversial bit is evidence based medicine. People get up in arms about being told no to getting care, even care that is demonstrably harmful. One economic reality is that unlimited demand does collide with limited resources. Some day we have to have a conversation about who will get what care when. Currently we have given up earlier more basic care for all, for Extraordinary measures at (and arguably beyond) the end of life for all. Few would argue that this is a good thing, but inertia is allowed to dictate.
• @David – Over-consumption is use that yields little to no benefit. Low out-of-pocket costs facilitate it. The idea of overuse due to generosity of insurance is the notion of moral hazard. There is a vast literature on it. The most famous study of it is the RAND Health Insurance Experiment (HIE). Quoting from the HIE Wikipedia entry, “An early paper with interim results from the RAND HIE concluded that health insurance without coinsurance “leads to more people using services and to more services per user,” referring to both outpatient and inpatient services. … The experiment also demonstrated that cost sharing reduced “appropriate or needed” medical care as well as “inappropriate or unnecessary” medical care.” One would like to encourage the appropriate care while discouraging unnecessary care. I’ll have a post next week on insurance designs aimed at doing just that.
• You’ve not even added in the substantial discount of buying these products as an institution, as opposed to an individual. The Feds tax us as if we buy the benefits as an individual, so I happen to get a little report showing what they’re assuming the benefits would cost – not what they actually do cost. And they seem to think the benefits cost to be 2-3x what we actually spend on them. Much like buying a flat of soda I can get them a 25¢ a can but if I buy a single can I’m pay \$1.
But there’s no overconsumption of healthcare. How could you make a ridiculous argument such as that? Overconsumption would mean better results, not poorer. Americans see doctors less than nearly any advanced country and we have poorer results for it. That’s not a result of overconsumption of healthcare. Miss-consumption, sure, but not overconsumption. | 2,384 | 10,896 | {"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} | 2.53125 | 3 | CC-MAIN-2017-26 | longest | en | 0.963974 |
https://trustconverter.com/en/speed-conversion/speed-of-lights/speed-of-lights-to-miles-per-second.html | 1,624,534,936,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00359.warc.gz | 503,924,882 | 9,512 | # Speed of lights to Miles per second Conversion
Speed of light to mile per second conversion allow you make a conversion between speed of light and mile per second easily. You can find the tool in the following.
### Speed Conversion
to
input
= 186,282.39705122
= 1862.82 × 102
= 1862.82E2
= 1862.82e2
= 372,564.79410244
= 3725.65 × 102
= 3725.65E2
= 3725.65e2
= 558,847.19115366
= 5588.47 × 102
= 5588.47E2
= 5588.47e2
= 745,129.58820488
= 7451.3 × 102
= 7451.3E2
= 7451.3e2
= 931,411.98525610
= 9314.12 × 102
= 9314.12E2
= 9314.12e2
### Quick Look: speed of lights to miles per second
speed of light 1 c 2 c 3 c 4 c 5 c 6 c 7 c 8 c 9 c 10 c 11 c 12 c 13 c 14 c 15 c 16 c 17 c 18 c 19 c 20 c 21 c 22 c 23 c 24 c 25 c 26 c 27 c 28 c 29 c 30 c 31 c 32 c 33 c 34 c 35 c 36 c 37 c 38 c 39 c 40 c 41 c 42 c 43 c 44 c 45 c 46 c 47 c 48 c 49 c 50 c 51 c 52 c 53 c 54 c 55 c 56 c 57 c 58 c 59 c 60 c 61 c 62 c 63 c 64 c 65 c 66 c 67 c 68 c 69 c 70 c 71 c 72 c 73 c 74 c 75 c 76 c 77 c 78 c 79 c 80 c 81 c 82 c 83 c 84 c 85 c 86 c 87 c 88 c 89 c 90 c 91 c 92 c 93 c 94 c 95 c 96 c 97 c 98 c 99 c 100 c mile per second 186,282.3970512 mps 372,564.7941024 mps 558,847.1911537 mps 745,129.5882049 mps 931,411.9852561 mps 1,117,694.3823073 mps 1,303,976.7793585 mps 1,490,259.1764098 mps 1,676,541.573461 mps 1,862,823.9705122 mps 2,049,106.3675634 mps 2,235,388.7646147 mps 2,421,671.1616659 mps 2,607,953.5587171 mps 2,794,235.9557683 mps 2,980,518.3528195 mps 3,166,800.7498708 mps 3,353,083.146922 mps 3,539,365.5439732 mps 3,725,647.9410244 mps 3,911,930.3380756 mps 4,098,212.7351269 mps 4,284,495.1321781 mps 4,470,777.5292293 mps 4,657,059.9262805 mps 4,843,342.3233317 mps 5,029,624.720383 mps 5,215,907.1174342 mps 5,402,189.5144854 mps 5,588,471.9115366 mps 5,774,754.3085878 mps 5,961,036.7056391 mps 6,147,319.1026903 mps 6,333,601.4997415 mps 6,519,883.8967927 mps 6,706,166.293844 mps 6,892,448.6908952 mps 7,078,731.0879464 mps 7,265,013.4849976 mps 7,451,295.8820488 mps 7,637,578.2791001 mps 7,823,860.6761513 mps 8,010,143.0732025 mps 8,196,425.4702537 mps 8,382,707.8673049 mps 8,568,990.2643562 mps 8,755,272.6614074 mps 8,941,555.0584586 mps 9,127,837.4555098 mps 9,314,119.852561 mps 9,500,402.2496123 mps 9,686,684.6466635 mps 9,872,967.0437147 mps 10,059,249.440766 mps 10,245,531.837817 mps 10,431,814.234868 mps 10,618,096.63192 mps 10,804,379.028971 mps 10,990,661.426022 mps 11,176,943.823073 mps 11,363,226.220124 mps 11,549,508.617176 mps 11,735,791.014227 mps 11,922,073.411278 mps 12,108,355.808329 mps 12,294,638.205381 mps 12,480,920.602432 mps 12,667,202.999483 mps 12,853,485.396534 mps 13,039,767.793585 mps 13,226,050.190637 mps 13,412,332.587688 mps 13,598,614.984739 mps 13,784,897.38179 mps 13,971,179.778842 mps 14,157,462.175893 mps 14,343,744.572944 mps 14,530,026.969995 mps 14,716,309.367046 mps 14,902,591.764098 mps 15,088,874.161149 mps 15,275,156.5582 mps 15,461,438.955251 mps 15,647,721.352303 mps 15,834,003.749354 mps 16,020,286.146405 mps 16,206,568.543456 mps 16,392,850.940507 mps 16,579,133.337559 mps 16,765,415.73461 mps 16,951,698.131661 mps 17,137,980.528712 mps 17,324,262.925764 mps 17,510,545.322815 mps 17,696,827.719866 mps 17,883,110.116917 mps 18,069,392.513968 mps 18,255,674.91102 mps 18,441,957.308071 mps 18,628,239.705122 mps
The speed of light in vacuum, commonly denoted c, is a universal physical constant important in many areas of physics. Its exact value is 299,792,458 metres per second (approximately 3.00×108 m/s, or 186,282 mi/s); it is exact because the unit of length, the metre, is defined from this constant and the international standard for time.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
speed of lightc
≡ 299792458 m/s
= 299792458 m/s
Other Speed
#### conversion table
speed of lightsmiles per secondspeed of lightsmiles per second
1= 186282.397051226= 1117694.3823073
2= 372564.794102447= 1303976.7793585
3= 558847.191153668= 1490259.1764098
4= 745129.588204889= 1676541.573461
5= 931411.985256110= 1862823.9705122
Miles per second (abbreviated mph, MPS or mi/s) is an imperial and United States customary unit of speed expressing the number of statute miles covered in one second.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
mile per secondmps
≡ 1 mi/s
= 1609.344 m/s
Imperial/US
### conversion table
miles per secondspeed of lightsmiles per secondspeed of lights
1= 5.3681937522257E-66= 3.2209162513354E-5
2= 1.0736387504451E-57= 3.757735626558E-5
3= 1.6104581256677E-58= 4.2945550017806E-5
4= 2.1472775008903E-59= 4.8313743770032E-5
5= 2.6840968761129E-510= 5.3681937522257E-5
### Conversion table
speed of lightsmiles per second
1= 186,282.3970512
5.3681937522257 × 10-6= 1
### Legend
SymbolDefinition
exactly equal
approximately equal to
=equal to
digitsindicates that digits repeat infinitely (e.g. 8.294 369 corresponds to 8.294 369 369 369 369 …) | 2,237 | 4,899 | {"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} | 3.515625 | 4 | CC-MAIN-2021-25 | latest | en | 0.572863 |
http://dgesolutions.co.uk/download/differentiable-and-complex-dynamics-of-several-variables-mathematics-and-its | 1,601,164,834,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400249545.55/warc/CC-MAIN-20200926231818-20200927021818-00172.warc.gz | 39,517,077 | 9,628 | # Pei-Chu Hu,Chung-Chun Yang's Differentiable and Complex Dynamics of Several Variables PDF
By Pei-Chu Hu,Chung-Chun Yang
ISBN-10: 079235771X
ISBN-13: 9780792357711
ISBN-10: 9048152461
ISBN-13: 9789048152469
the advance of dynamics concept started with the paintings of Isaac Newton. In his conception the main easy legislations of classical mechanics is f = ma, which describes the movement n in IR. of some extent of mass m lower than the motion of a strength f via giving the acceleration a. If n the location of the purpose is taken to be some extent x E IR. , and if the strength f is meant to be a functionality of x purely, Newton's legislations is an outline when it comes to a second-order usual differential equation: J2x m dt = f(x). 2 It is sensible to lessen the equations to first order via defining the velo urban as an additional n autonomous variable by way of v = :i; = ~~ E IR. . Then x = v, mv = f(x). L. Euler, J. L. Lagrange and others studied mechanics through an analytical procedure known as analytical dynamics. at any time when the strength f is represented through a gradient vector box f = - lU of the capability strength U, and denotes the adaptation of the kinetic strength and the capability power by means of 1 L(x,v) = 2'm(v,v) - U(x), the Newton equation of movement is decreased to the Euler-Lagrange equation ~~ are used because the variables, the Euler-Lagrange equation could be If the momenta y written as . 8L y= 8x' extra, W. R.
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Example text | 821 | 3,707 | {"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} | 2.703125 | 3 | CC-MAIN-2020-40 | latest | en | 0.892068 |
https://www.physicsforums.com/threads/planetary-atmosphere.133823/ | 1,544,539,221,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823621.10/warc/CC-MAIN-20181211125831-20181211151331-00584.warc.gz | 1,015,643,380 | 12,841 | # Homework Help: Planetary Atmosphere!
1. Sep 27, 2006
### zferic28
a) Show that the escape speed for a particle to leave the gravitational infuence of a planet is given by (2GM/R)^1/2, where M is the mass of the planet, R is its radius, and G is the gravitational constant.
b) The temperature near the top of Jupiter's multicolored cloud layer is about 140K. The temperature at the top of the earth's troposphere, at an altitude of about 20km, is about 220k. Calculate the rms speed of huydrogen molecules (h2) in each of these environments. Give your answers in the m/s and as a fraction of the escape speed from the
respective planet.
c) Hydrogen gas is a rare element in the earth's atm. In the atmosphere of Jupiter, by contrast, 89% of all molecules are H2. Explain why, using your results from the previous part.
d) Ceres is an asteroid with mass equal to .014 times the mass of the moon, a density of 2400kg/m^3 and a surface temp of about 200K. Suppose an astronomer claims to have iscovered an oxygen (O2) atmosphere on the asteroid Ceres. You are asked by a TV news reporter to comment on this claim. What would you say, and how would you support that?
a) I don't fully understand what I have to calculate or demonstrate in this, can someone please break this down for me? Thanks!
b) I need the formula Vrms = sqrt(3RT/M)
M of H2 = 2g/mol = .002kg/mol
Vrms Jupiter = sqrt(3x8.314x140k/.002kg) = 1.321x10^3 m/s
Vrms Earth = sqrt(3x8.314x220k/.002kg)= 1.656x10^3 m/s
I don't quite understand what I have to do to get the fraction of the escape speed relative to the planet.
I suppose use the given formula (2GM/R)^1/2 to calculate each plantes escape and then divide that number by the Vrms of jupiter and Earth?? What is g, M, and R of jupiter?? Is 9.8m/s the gravitational force at 20km high in the earths atmosphere?? Please help!
c) Im guessing this will have to do with the escape velocity of each planet? Maybe jupiter likes to hang on to it's H2 and earth lets them go?
d) Lost me again on this one? Thanks for all the help..
2. Sep 27, 2006
### Farsight
Once you've got the rest d) is dead easy zferic. You just plug a) and b) back in. | 603 | 2,166 | {"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} | 3.578125 | 4 | CC-MAIN-2018-51 | latest | en | 0.92402 |
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# A4_Deformations_in_a_Continuum_ans - EATS 2470 Introduction...
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EATS 2470 Introduction to Continuum Mechanics Assignment #4-Solutions 1. Given: Lagrangen description of a flow: 4 1 1 2 3 y a a t a t = + + 2 2 2 1 y a a t = + 3 3 2 y a a t = + Find that: (a) the position of P (1,2,4) at time t = 2; Substitute 1 1 a = , 2 2 a = , 3 4 a = , t = 2 into the flow equations 4 1 1 (1) (2)(2) (4)(2) 41 y y = + + = 2 2 2 (2) (1)(2) 6 y y = + = 3 3 (4) (2)(2) 8 y y = + = Thus, the position is { } 41 6 8 y = (b) find the displacement of the point P at the same time, t = 2 The displacements are defined as i i i u y a = - or { } { } { } u y a = - Thus, 4 1 1 (2)(2) (4)(2) 40 u u = + = 2 2 2 (1)(2) 2 u u = = 3 2 (2)(2) 4 u u = = { } 40 4 4 u = (c) find the velocity of the same point at the same time The velocity is dt dy q i i = , thus
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3 1 2 3 1 4 68 q a t a q = + = 2 1 2 2 4 q a t q = = 3 2 3 2 q a t q = = or { } 68 4 2 q = (d) now find the acceleration of the same point at the same time, t = 2 The acceleration is i i dq d q dt dt = , then 2 1 1 2 12 96 dq dq a t dt dt = = 2 2 1 2 2 dq dq a dt dt = = 3 0 dq dt = or 96 2 0 dq dt = (e) find the Eulerian description of the displacement sector recall that we have given: { } [ ] { } y A a = or 4 1 1 2 2 2 3 3 1 1 0 0 1 y t t a y t a y t a = 14243 It require to find the inverse of the matrix First find the matrix of cofactors 2 3 2 4 3 6 1 ( ) 1 (1 ) ij t t t t t t t t α - = - - - - Second, transpose this matrix of cofactor 2 4 2 3 3 6 1 ( ) 1 (1 ) T ij t t t t t t t t α - - = - - - Third, the value of the determinant of the original matrix [A] 4 2 2 4 2 4 6 1 1 0 1(1) ( ) 1 0 1 t t A t t t t t t t = = - - = + -
Finally, the inverse of the matrix [A] is [ ] 2 4 1 2 3 3 6 1 ( ) / 1 (1 ) T ij t t t A A t t t t t α - - - = = - - - 4 6 1 t t + - Thus, the displacement in Eulerian description is { } { } [ ] { } 1 u y A y - = - { } 4 6 2 4 1 2 3 2 4 6 3 1 2 3 3 4 1 2 3 ( ) ( ) ( ) t t y t t y ty u t y t t y t y
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Q: If \$5,000 is invested at 6% interest, how much money must be invested at 8% interest so that the total interest from the two investments is \$660 after one year? \$4,500 \$4,000 \$5,500
A: 400 is 60.60% of 660.
Original conversation
User: If \$5,000 is invested at 6% interest, how much money must be invested at 8% interest so that the total interest from the two investments is \$660 after one year? \$4,500 \$4,000 \$5,500
Weegy: 400 is 60.60% of 660.
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User: If \$5,000 is invested at 6% interest, how much money must be invested at 8% interest so that the total interest from the two investments is \$660 after one year? \$4,500 \$4,000 \$5,500
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# Employees Only Sign wins Powerball +(4)#'s NOT! WIN \$107
Published:
Wednesday, Nov. 18, 2020
4 5 17 43 52 (5)
On Saturday Nov 18,I visited Empire Buffet.The woman owner (Chen) had sheets of clear plexiglass surrounding the buffet food.The customers would tell Chen or an employee what food they want,and the plate would be passed through a small window.
There are two white doors to enter the enclosed buffet food area.A customer followed Chen into the door for a take-out order.Chen told him,"No,you go outside,I give you food".
Then later two women walked through the other door thinking they will serve themselves.Chen said,"No, I get you Buffet food".
To clear things up,I drove to a local hardware store and to my surprise,I found the perfect sign.An aluminum 2X6 inch,with red letters;Employees Only.Perfect!
I peeled off the back,and {stuck} the sign on both doors.
Later I thought......
Employees= letter E=#5
Only=letter O=#15 or one-number five #5
Employees Only=combined #55 or two fives = #25 twenty-five is also one quarter.... .25 or 1/4 or 14 or one number four #4
Q for Quarter =17th letter
Sign=#19 letter S
Here are my numbers I played:
4 5 14 15 17 19 55 Powerball Numbers=#19 #5 #15
Spent \$10 Collected \$107
\$97 Profit !
Crazy thing is Empire Buffet= EB=#52,the 4th Number!
FOUR NUMBERS PLUS POWERBALL=\$50,000.00 ! How did I NOT play Empire Buffet?
As for the last #43,.......DC......?
Entry #99 | 476 | 1,618 | {"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} | 2.828125 | 3 | CC-MAIN-2021-04 | latest | en | 0.756623 |
https://quant.stackexchange.com/questions/71383/short-term-implied-fx-rate | 1,659,940,255,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570767.11/warc/CC-MAIN-20220808061828-20220808091828-00559.warc.gz | 459,659,044 | 64,256 | # Short term implied fx rate
For context, I'm working towards constructing a FX implied rate curve based on the fwd points market. As you know, most of the spot rates usually settle in t+2. We can group fwd points in two types, the ones that settle after the spot date $$T>T_s$$ and the ones that do it before it $$T.
In the case where $$T>T_s$$, one could use the usual non-arbitrage argument, if you have 1 USD:
1USD 1USD
Case 1 Case 2
Convert to EUR with the EURUSD spot rate $$S_{T_s}$$ in $$T_s$$ Invest in USD rate until $$T$$
Invest from $$T_s$$ to $$T$$ at the EUR rate. Enter into a fwd to convert USD to EUR in $$T$$
Profit in $$T$$: $$S_{T_s} (1+r_{eur(T_s,T)}\frac{(T-T_s)}{360})$$ Profit in $$T$$: $$F_{T_s} (1+r_{usd(T_s,T)}\frac{(T-T_s)}{360})$$
Thus creating the usual parity:
$$F_{T_s,T} = S_{T_s} \frac{(1+r_d[T_s,T](T-T_s))}{(1+ r_f[T_s,T](T-T_s))}$$
In this case if one knows the value of $$S_{T_s}$$ and asumes a value for $$r_f$$ one could solve for $$r_d$$ to get the implied rate.
However the case when $$T is the one I'm not completely sure how it would work. Since the fwd points before $$T_s$$ are the ON and SN points which are actually FX swaps points, in which one could buy/sell the FX but would have to do the opposite operation on maturity date it creates a different dynamic.
For clarity I would describe this point to my understanding, please do tell if I'm making an incorrect assumption at this point:
ON FX points trade: Buy/Sell 1 USD today $$t$$ at the ON rate (assuming mid values Spot -TN points - ON points), then Sell/Buy 1 USD tomorrow $$t+1$$ at the TN rate (assuming mid values Spot -TN points).
TN FX points trade: Buy/Sell 1 USD tomorrow $$t+1$$ at the TN rate, then Sell/Buy 1 USD on the spot date $$t+2$$ at the spot rate.
With this in mind, I think the investment strategy from $$t$$ to $$T_s$$ would be:
For, $$t=today
1USD 1USD
Case 1 Case 2
In $$t$$: Convert to EUR by entering into the ON FX swap selling 1USD at the ON EURUSD rate $$S_{ON}$$ then invest $$S_{ON}$$ at the EUR 1 day rate $$r_1$$ In $$t$$: Invest in USD rate until $$T_s$$
In $$t$$: Enter into an SN FX swap to sell 1 USD at the SN EURUSD rate $$S_{SN}$$ in $$t+1$$ In $$T_s$$ Sell the USD with interest at the spot rate in $$T_s$$
In $$t$$: Enter into a spot trade to sell 1 USD at the Spot EURUSD rate $$S_{T_s}$$ in $$T_s$$ Profit in $$T_s$$: $$S_{T_s} (1+r_{usd(T_s,T)}\frac{(T-T_s)}{360})$$
In $$t+1$$: Recieve $$S_{ON} ( 1+ r_1)$$. Since you entered a FX swap the day before, now you have to buy 1 USD at the TN rate $$S_{TN}$$, assuming $$S_{ON}( 1+ r_1)> S_{TN}$$. You could use that to buy the 1 USD, and resell it at the $$S_{TN}$$ with the SN FX swap. Finally, you could reinvest $$S_{ON}( 1+ r_1)- S_{TN} + S_{TN}$$ at the one day rate from $$t_1$$ to $$T_s$$ $$r_{1,2}$$.
Finally in $$T_s$$: Recieve $$S_{ON}( 1+ r_1)$$ (1+$$r_{1,2})$$. Since you entered a FX swap the day before, now you have to buy 1 USD at the spot rate $$S_{T_s}$$, assuming $$S_{ON}( 1+ r_1)$$ (1+$$r_{1,2})$$> $$S_{T_s}$$. You could use that to buy the 1 USD, and resell it at the spot rate trade $$S_{T_s}$$.
Profit in $$T_s$$: $$S_{ON}( 1+ r_1)$$ (1+$$r_{1,2})$$
This would mean that: $$S_{ON}( 1+ r_1) (1+r_{1,2}) = S_{T_s} (1+r_{usd(T_s,T)}(T-T_s))$$ If,
$$r_{1,2} = (\frac{(1 + r_{0,2} t_{0,2})}{(1 + r_{0,1} t_{0,1})}-1) * \frac {1}{t_{1,2}}$$
$$S_{ON} (1+ r_{t,T_s}(T_s-t)) = S_{T_s} (1+r_{usd(t,T_s)}(T_s-t))$$
Which would imply in the short term, the spot rate is a forward rate. Does this seem correct?
Thanks to anyone who could provide some feedback.
• I did not check all details of calculations, but you have the right idea: if you need the foreign currency tomorrow, you need to do a Spot trade and a TN swap (in the backwards direction). If instead you need the currency today you need an additional ON swap (again backwards). This is how maturities < spotdate are handled. And yes, "spot" can be considered the forward rate for T+2 when you start at T or T+1. Jun 27 at 9:33 | 1,347 | 4,018 | {"found_math": true, "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": 65, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-33 | latest | en | 0.873951 |
https://www.qtumist.com/post/17515 | 1,709,175,010,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474775.80/warc/CC-MAIN-20240229003536-20240229033536-00100.warc.gz | 919,117,707 | 13,680 | # 问题：
The Hadamard operator on one qubit may be written as $H=\frac{1}{\sqrt{2}}[(|0\rangle+|1\rangle)\langle 0|+(|0\rangle-|1\rangle)\langle 1|]$;
Show explicitly that the Hadamard transform on $n$ $qubits$, $H^{⊗n}$, may be written as$H^{\otimes n}=\frac{1}{\sqrt{2^{n}}} \sum_{x, y}(-1)^{x \cdot y}|x\rangle\langle y|$;
Write out an explicit matrix representation for $H^{⊗2}$;
## 解答
$H^{⊗2}= H \otimes H=\frac{1}{\sqrt{2}}\left[\begin{array}{ll}1 * H & 1 * H \\ 1 * H & -1 * H\end{array}\right]=\frac{1}{2}\left[\begin{array}{cccc}1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1\end{array}\right]$.
#### 参考
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# gmat prep ds standard deviation
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List S and list T each contain 5 positive integers and for each list the avg (mean) of the integers in the list is 40. if the integers 30,40 and 50 are in both list , is the standard deviation of the integers in list S greater than the list standard deviation of the integers in list T?
(1) the integer 25 is in list S
(2) the integer 45 is in the list
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My choice "B"
From (1) we get no info. about List T
From (2) Assuming you meant to say that 45 is in both the lists. Then there is only one choice left for the 5th no. and it is 35 (to get the avg. 40)
The std. dev would be the same.
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I'm sorry, I ment to say
List S and list T each contain 5 positive integers and for each list the avg (mean) of the integers in the list is 40. if the integers 30,40 and 50 are in both list , is the standard deviation of the integers in list S greater than the list standard deviation of the integers in list T?
(1) the integer 25 is in list S
(2) the integer 45 is in the list T
sorry
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I believe is C
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Engineering » Industrial Engineering and Management » "Search Algorithms for Engineering Optimization", book edited by Taufik Abrão, ISBN 978-953-51-0983-9, Published: February 13, 2013 under CC BY 3.0 license. © The Author(s).
# Optimal Allocation of Reliability in Series Parallel Production System
DOI: 10.5772/55725
Article top
# Optimal Allocation of Reliability in Series Parallel Production System
## 1. Introduction
One of the most important problems in many industrial applications is the redundancy optimization problem. This latter is well known combinatorial optimization problem where the design goal is achieved by discrete choices made from elements available on the market. The natural objective function is to find the minimal cost configuration of a series-parallel system under availability constraints. The system is considered to have a range of performance levels from perfect working to total failure. In this case the system is called a multi-state system (MSS). Let consider a multi-state system containing n components Ci (i = 1, 2, …, n) in series arrangement. For each component Ci there are various versions, which are proposed by the suppliers on the market. Elements are characterized by their cost, performance and availability according to their version. For example, these elements can represent machines in a manufacturing system to accomplish a task on product in our case they represent the whole of electrical power system (generating units, transformers and electric carrying lines devices). Each component Ci contains a number of elements connected in parallel. Different versions of elements may be chosen for any given system component. Each component can contain elements of different versions as sketched in figure 1.
#### Figure 1.
Series Parallel Production System
A limitation can be undesirable or even unacceptable, where only identical elements are used in parallel (i.e. homogeneous system) for two reasons. First, by allowing different versions of the devices to be allocated in the same system, one can obtain a solution that provides the desired availability or reliability level with a lower cost than in the solution with identical parallel devices. Second, in practice the designer often has to include additional devices in the existing system. It may be necessary, for example, to modernize a production line system according to a new demand levels from customers or according to new reliability requirements.
### 1.1. Literature review
The vast majority of classical reliability or availability analysis and optimization assume that components and system are in either of two states (i.e., complete working state and total failure state). However, in many real life situations we are actually able to distinguish among various levels of performance for both system and components. For such situation, the existing dichotomous model is a gross oversimplification and so models assuming multi-state (degradable) systems and components are preferable since they are closer to reliability. Recently much works treat the more sophisticated and more realistic models in which systems and components may assume many states ranging from perfect functioning to complete failure. In this case, it is important to develop MSS reliability theory. In this paper, an MSS reliability theory will be used, where the binary state system theory is extending to the multi-state case. As is addresses in recent review of the literature for example in (Ushakov, Levitin and Lisnianski, 2002) or (Levitin and Lisnianski, 2001). Generally, the methods of MSS reliability assessment are based on four different approaches:
1. The structure function approach.
2. The stochastic process (mainly Markov) approach.
3. The Monte-Carlo simulation technique.
4. The universal moment generating function (UMGF) approach.
In (Ushakov, Levitin and Lisnianski, 2002), a comparison between these four approaches highlights that the UGF approach is fast enough to be used in the optimization problems where the search space is sizeable.
The problem of total investment-cost minimization, subject to reliability or availability constraints, is well known as the redundancy optimization problem (ROP). The ROP is studied in many different forms as summarized in (Tillman, Hwang and Kuo, 1977), and more recently in (Kuo and Prasad, 2000). The ROP for the multi-state reliability was introduced in (Ushakov, 1987). In (Lisnianski, Levitin, Ben-Haim and Elmakis, 1996) and (Levitin, Lisnianski, Ben-Haim and Elmakis, 1997), genetic algorithms were used to find the optimal or nearly optimal power system structure.
This work uses an ant colony optimization approach to solve the ROP for multi-state system. The idea of employing a colony of cooperating agents to solve combinatorial optimization problems was recently proposed in (Dorigo, Maniezzo and Colorni, 1996). The ant colony approach has been successfully applied to the classical traveling salesman problem (Dorigo and Gambardella, 1997), and to the quadratic assignment problem (Maniezzo and Colorni, 1999). Ant colony shows very good results in each applied area. It has been recently adapted for the reliability design of binary state systems (Liang and Smith, 2001). The ant colony has also been adapted with success to other combinatorial optimization problems such as the vehicle routing problem (Bullnheimer, Hartl and Strauss, 1997). The ant colony method has been used to solving the redundancy allocation problem (Nahas N., Nourelfath M., Aït-Kadi Daoud, 2006).
In this paper, we extend the work of other researchers by proposing ant colony system algorithm to solve the ROP characterised in the problem of optimization of the structure of power system where redundant elements are included in order to provide a desired level of reliability through optimal allocation of elements with different parameters (optimal structure with series-parallel elements) in continuous production system.
The use of this algorithm is within a general framework for the comparative and structural study of metaheuristics. In a first step the application of ant colonies in its primal form is necessary and thereafter in perspective the study will be completed.
### 1.2. Approach and outlines
The problem formulated in this chapter lead to a complicated combinatorial optimization problem. The total number of different solution to be examined is very large, even for rather small problems. An exhaustive examination of all possible solutions is not feasible given reasonable time limitations. Because of this, the ant colony optimization (or simply ACO) approach is adapted to find optimal or nearly optimal solutions to be obtained in a short time. The newer developed meta-heuristic method has the advantage to solve the ROP for MSS without the limitation on the diversity of versions of elements in parallel. Ant colony optimization is inspired by the behavior of real ant colonies that exhibit the highly structured behavior. Ants lay down in some quantity an aromatic substance, known as pheromone, in their way to food. An ant chooses a specific path in correlation with the intensity of the pheromone. The pheromone trail evaporates over time if no more pheromone in laid down by others ants, therefore the best path has more intensive pheromone and higher probability to be chosen.
During the optimization process, artificial ants will have to evaluate the availability of a given selected structure of the series-parallel system (electrical network). To do this, a fast procedure of availability estimation is developed. This procedure is based on a modern mathematical technique: the z-transform or UMGF which was introduced in (Ushakov, 1986). It was proven to be very effective for high dimension combinatorial problems: see e.g. (Ushakov, 2002), (Levitin, 2001). The universal moment generating function is an extension of the ordinary moment generating function (UGF) (Ross, 1993). The method developed in this chapter allows the availability function of reparable series-parallel MSS to be obtained using a straightforward numerical procedure.
## 2. Formulation of redundancy optimization problem
### 2.1. Series-parallel system with different redundant elements
Let consider a series-parallel system containing n subcomponents Ci (i = 1, 2, …, n) in series as represented in figure 1. Every component Ci contains a number of different elements connected in parallel. For each component i, there are a number of element versions available in the market. For any given system component, different versions and number of elements may be chosen. For each subcomponent i, elements are characterized according to their version v by their cost (Civ), availability (Aiv) and performance (iv). The structure of system component i can be defined by the numbers of parallel elements (of each version) kiv for 1vVi, where Vi is a number of versions available for element of type i. Figure 2 illustrates these notations for a given component i. The entire system structure is defined by the vectors ki = {kivi} (1in,1vVi). For a given set of vectors k1, k2, …, kn the total cost of the system can be calculated as:
C=∑i=1n∑v=1VikivCiv (1)
### 2.2. Availability of reparable multi-state systems
The series-parallel system is composed of a number of failure prone elements, such that the failure of some elements leads only to a degradation of the system performance. This system is considered to have a range of performance levels from perfect working to complete failure. In fact, the system failure can lead to decreased capability to accomplish a given task, but not to complete failure. An important MSS measure is related to the ability of the system to satisfy a given demand.
In electric power systems, reliability is considered as a measure of the ability of the system to meet the load demand (D), i.e., to provide an adequate supply of electrical energy (∑). This definition of the reliability index is widely used in power systems: see e.g., (Ross, 1993), (Murchland, 1975), (Levitin, Lisnianski, Ben-Haim and Elmakis, 1998), (Lisnianski, Levitin, Ben-Haim and Elmakis, 1996), (Levitin, Lisnianski, and Elmakis, 1997). The Loss of Load Probability index (LOLP) is usually used to estimate the reliability index (Billinton and Allan, 1990). This index is the overall probability that the load demand will not be met. Thus, we can write R = Probab(∑ ≥ D) or R = 1-LOLP with LOLP = Probab(∑ < D). This reliability index depends on consumer demand D.
For reparable MSS, a multi-state steady-state availability E is used as Probab(∑≥ D) after enough time has passed for this probability to become constant (Levitin, Lisnianski, Ben-Haim and Elmakis, 1998). In the steady-state the distribution of states probabilities is given by equation (2), while the multi-state stationary availability is formulated by equation (3):
Pj=limt→∞[Probab(Σ(t)=Σj)] (2)
E=∑Σj≥DPj (3)
If the operation period T is divided into M intervals (with durations T1, T2, …, TM) and each interval has a required demand level (D1, D2, …, DM, respectively), then the generalized MSS availability index A is:
A=1∑j=1MTj∑j=1MProbab(Σ≥Dj)Tj (4)
We denote by D and T the vectors {Dj} and {Tj} (1jM), respectively. As the availability A is a function of k1, k2, …, kn, D and T, it will be written A(k1, k2, …, kn, D, T). In the case of a power system, the vectors D and T define the cumulative load curve (consumer demand). In reality the load curves varies randomly; an approximation is used from random curve to discrete curve see (Wood and Ringlee, 1970). In general, this curve is known for every power system.
### 2.3. Optimal design problem formulation
The multi-state system redundancy optimization problem of electrical power system can be formulated as follows: find the minimal cost system configuration k1, k2, …, kn, such that the corresponding availability exceeds or equal the specified availability A0. That is,
MinimizeC=∑i=1n∑v=1VikivCiv (5)
subject to A(k1,k2, …,kn,D,T)≥A0 (6)
The input of this problem is the specified availability and the outputs are the minimal investment-cost and the corresponding configuration determined. To solve this combinatorial optimization problem, it is important to have an effective and fast procedure to evaluate the availability index for a series-parallel system of elements. Thus, a method is developed in the next section to estimate the value of A(k1, k2, …, kn, D, T).
## 3. Multi-state system availability estimation
The procedure used in this chapter is based on the universal z-transform, which is a modern mathematical technique introduced in (Ushakov, 1986). This method, convenient for numerical implementation, is proved to be very effective for high dimension combinatorial problems. In the literature, the universal z-transform is also called universal moment generating function (UMGF) or simply u-function or u-transform. In this chapter, we mainly use the acronym UMGF. The UMGF extends the widely known ordinary moment generating function (Ross, 1993).
### 3.1. Definition and properties
The UMGF of a discrete random variable is defined as a polynomial:
u(z)=∑j=1JPjzΣj (7)
where the variable has J possible values and Pj is the probability that is equal to j.
The probabilistic characteristics of the random variable can be found using the function u(z). In particular, if the discrete random variable is the MSS stationary output performance, the availability E is given by the probability Probab(∑ ≥ D) which can be defined as follows:
Probab(Σ ≥D)=Ψ(u(z)z−D) (8)
where Ψ is a distributive operator defined by expressions (9) and (10):
Ψ(Pzσ−D)={P,ifσ≥D0,ifσ
Ψ(∑j=1JPjzΣj−D)=∑j=1JΨ(PjzΣj−D) (10)
It can be easily shown that equations (7) – (10) meet condition Probab(∑≥ D) =ΣjDPj. By using the operator Ψ, the coefficients of polynomial u(z) are summed for every term with jD, and the probability that is not less than some arbitrary value D is systematically obtained.
Consider single elements with total failures and each element i has nominal performance i and availability Ai. Then, Probab(∑ = i) = Ai and Probab(∑ = 0) = 1Ai. The UMGF of such an element has only two terms and can be defined as:
ui=(1−Ai)z0+AizΣi=(1−Ai)+AizΣi (11)
To evaluate the MSS availability of a series-parallel system, two basic composition operators are introduced. These operators determine the polynomial u(z) for a group of elements.
### 3.2. Composition operators
#### 3.2.1. Properties of the operators
The essential property of the UMGF is that it allows the total UMGF for a system of elements connected in parallel or in series to be obtained using simple algebraic operations on the individual UMGF of elements. These operations may be defined according to the physical nature of the elements and their interactions. The only limitation on such an arbitrary operation is that its operator ϕ should satisfy the following Ushakov’s conditions (Ushakov, 1986):
φ(p1zg1,p2zg2)=p1p2zφ(g1,g2),φ(g)=g,φ(g1,...,gn)=φ(φ(g1,...,gk),φ(gk+1,...,gn)),φ(g1,...,gk,gk+1,...,gn)=φ(g1,...,gk+1,gk,...,gn)foranyk.
#### 3.2.2. Parallel elements
Let consider a system component m containing Jm elements connected in parallel. As the performance measure is related to the system productivity, the total performance of the parallel system is the sum of performances of all its elements. In power systems engineering, the term capacity is usually used to indicate the quantitative performance measure of an element (Lisnianski, Levitin, Ben-Haim and Elmakis, 1996). It may have different physical nature. Examples of elements capacities are: generating capacity for a generator, pipe capacity for a water circulator, carrying capacity for an electric transmission line, etc. The capacity of an element can be measured as a percentage of nominal total system capacity. In a manufacturing system, elements are machines. Therefore, the total performance of the parallel machine is the sum of performances (Dallery and Gershwin, 1992).
The u-function of MSS component m containing Jm parallel elements can be calculated by using the Γ operator:
up(z)=Γ(u1(z),u2(z),...,un(z)), where Γ(g1,g2,...,gn)=i=1ngi.
Therefore for a pair of elements connected in parallel:
Γ(u1(z),u2(z))=Γ(i=1nPizai,j=1mQjzbj)=i=1nj=1mPiQjzai+bj.
Parameters ai and bj are physically interpreted as the respective performances of the two elements. n and m are numbers of possible performance levels for these elements. Pi and Qj are steady-state probabilities of possible performance levels for elements.
One can see that the Γ operator is simply a product of the individual u-functions. Thus, the component UMGF is:
up(z)=j=1Jmuj(z).
Given the individual UMGF of elements defined in equation (11), we have:
up(z)=j=1Jm(1Aj+AjzΣi).
#### 3.2.3. Series elements
When the elements are connected in series, the element with the least performance becomes the bottleneck of the system. This element therefore defines the total system productivity. To calculate the u-function for system containing n elements connected in series, the operator η should be used: us(z)=η(u1(z),u2(z),...,um(z)), where η(g1,g2,...,gm)=min{g1,g2,...,gm}
so that
η(u1(z),u2(z))=η(i=1nPizai,j=1mQjzbj)=i=1nj=1mPiQjzmin{ai,bj}
Applying composition operators Γ and η consecutively, one can obtain the UMGF of the entire series-parallel system.
## 4. The ant colony optimization approach
The problem formulated in this chapter is a complicated combinatorial optimization problem. The total number of different solutions to be examined is very large, even for rather small problems. An exhaustive examination of the enormous number of possible solutions is not feasible given reasonable time limitations. Thus, because of the search space size of the ROP for MSS, a new meta-heuristic is developed in this section. This meta-heuristic consists in an adaptation of the ant colony optimization method.
### 4.1. The ACO principle
Recently, (Dorigo, Maniezzo and Colorni, 1996) introduced a new approach to optimization problems derived from the study of any colonies, called “Ant System”. Their system inspired by the work of real ant colonies that exhibit the highly structured behavior. Ants lay down in some quantity an aromatic substance, known as pheromone, in their way to food. An ant chooses a specific path in correlation with the intensity of the pheromone. The pheromone trail evaporates over time if no more pheromone in laid down by others ants, therefore the best paths have more intensive pheromone and higher probability to be chosen. This simple behavior explains why ants are able to adjust to changes in the environment, such as new obstacles interrupting the currently shortest path.
Artificial ants used in ant system are agents with very simple basic capabilities mimic the behavior of real ants to some extent. This approach provides algorithms called ant algorithms. The Ant System approach associates pheromone trails to features of the solutions of a combinatorial problem, which can be seen as a kind of adaptive memory of the previous solutions. Solutions are iteratively constructed in a randomized heuristic fashion biased by the pheromone trails, left by the previous ants. The pheromone trails, τij, are updated after the construction of a solution, enforcing that the best features will have a more intensive pheromone. An Ant algorithm presents the following characteristics. It is a natural algorithm since it is based on the behavior of ants in establishing paths from their colony to feeding sources and back. It is parallel and distributed since it concerns a population of agents moving simultaneously, independently and without supervisor. It is cooperative since each agent chooses a path on the basis of the information, pheromone trails, laid by the other agents with have previously selected the same path. It is versatile that can be applied to similar versions the same problem. It is robust that it can be applied with minimal changes to other combinatorial optimization problems. The solution of the travelling salesman problem (TSP) was one of the first applications of ACO.
Various extensions to the basic TSP algorithm were proposed, notably by (Dorigo and Gambardella, 1997a). The improvements include three main aspects: the state transition rule provides a direct way to balance between exploration of new edges and exploitation of a priori and accumulated knowledge about the problem, the global updating rule is applied only to edges which belong to the best ant tour and while ants construct solution, a local pheromone updating rule is applied. These extensions have been included in the algorithm proposed in this paper.
### 4.2. ACO-based solution approach
In our reliability optimization problem, we have to select the best combination of parts to minimize the total cost given a reliability constraint. The parts can be chosen in any combination from the available components. Components are characterized by their reliability, capacity and cost. This problem can be represented by a graph (figure 2) in which the set of nodes comprises the set of subsystems and the set of available components (i.e. max (Mj), j = 1..n) with a set of connections partially connect the graph (i.e. each subsystem is connected only to its available components). An additional node (blank node) is connected to each subsystem.
### Figure 2.
Definition of a series-parallel system with tree subsystems into a graph
In figure 2, a series-parallel system is illustrated where the first and the second subsystem are connected respectively to their 3 and 2 available components. The nodes cpi3 and cpi4, represent the blank components of the two subsystems. At each step of the construction process, an ant uses problem-specific heuristic information, denoted by ηij to choose the optimal number of components in each subsystem. An imaginary heuristic information is associated to each blank node. These new factors allow us to limit the search surfaces (i.e. tuning factors). An ant positioned on subsystem i chooses a component j by applying the rule given by:
j={argmaxm∈ACi([τim]α[ηim]β)ifq≤qoJifq>qo (12)
and J is chosen according to the probability:
pij={[τij]α[ηij]β∑m∈ACi[τim]α[ηim]βifj∈ACi0otherwise (13)
α : The relative importance of the trail.
β : The relative importance of the heuristic information ηij.
ACi: The set of available components choices for subsystem i.
q: Random number uniformly generated between 0 and 1.
The heuristic information used is : ηij = 1/(1+cij) where cij represents the associated cost of component j for subsystem i. A “tuning” factor ti= ηij = 1/(1+ci(Mi+1)) is associated to blank component (Mi+1) of subsystem i. The parameter qo determines the relative importance of exploitation versus exploration: every time an ant in subsystem i have to choose a component j, it samples a random number 0≤q≤1. If qqo then the best edge, is chosen (exploitation), otherwise an edge is chosen according to (12) (biased exploration).
The pheromone update consists of two phases: local and global updating. While building a solution of the problem, ants choose components and change the pheromone level on subsystem-component edges. This local trail update is introduced to avoid premature convergence and effects a temporary reduction in the quantity of pheromone for a given subsystem-component edge so as to discourage the next ant from choosing the same component during the same cycle. The local updating is given by:
τijnew=(1−ρ)τijold+ρτo (14)
where ρ is a coefficient such that (1-ρ) represents the evaporation of trail and τo is an initial value of trail intensity. It is initialized to the value (n.TCnn)-1 with n is the size of the problem (i.e. number of subsystem and total number of available components) and TCnn is the result of a solution obtained through some simple heuristic.
After all ants have constructed a complete system, the pheromone trail is then updated at the end of a cycle (i.e. global updating), but only for the best solution found. This choice, together with the use of the pseudo-random-proportional rule, is intended to make the search more directed: ants search in a neighbourhood of the best solution found up to the current iteration of the algorithm. The pheromone level is updated by applying the following global updating rule:
τijnew=(1ρ)τijold+ρΔτijΔτij={1TCbestif(i,j)besttour0otherwise
### 4.3. The algorithm
An ant-cycle algorithm is stated as follows. At time zero an initialization phase takes place during wish NbAnt ants select components in each subsystem according to the Pseudo-random-proportional transition rule. When an ant selects a component, a local update is made to the trail for that subsystem-component edge according to equation (13). In this equation, ρ is a parameter that determines the rate of reduction of the pheromone level. The pheromone reduction is small but sufficient to lower the attractiveness of precedent subsystem-component edge. At the end of a cycle, for each ant k, the value of the system’s reliability Rk and the total cost TCk are computed. The best feasible solution found by ants (i.e. total cost and assignments) is saved. The pheromone trail is then updated for the best solution obtained according to (13). This process is iterated until the tour counter reaches the maximum number of cycles NCmax or all ants make the same tour (stagnation behavior).
The followings are formal description of the algorithm.
1. Set NC:=0 (NC: cycle counter) For every edge (i,j) set an initial value τij(0)= τo 2. For k=1 to NbAnt do For i=1 to NbSubSystem do For j=1 to MaxComponents do Choose a component, including blanks, according to (1) and (2). Local update of pheromone trail for chosen subsystem- component edge (i,j) : τijnew=(1−ρ)τijold+ρτo End For End For 3. Calculate Rk (system reliability for each ant) Calculate the total cost for each ant TCk Update the best found feasible solution 4. Global update of pheromone trail: For each edge (i,j)∈ best feasible solution, update the pheromone trail according to: τijnew=(1−ρ)τijold+ρΔτijΔτij={1TCbestif(i,j)∈besttour0otherwise End For 5. cycle=cycle +1 6. if (NC < NCmax) and ( not stagnation behavior) Then Goto step 2 Else Print the best feasible solution and components selection. Stop.
## 5. Illustrative example
Description of the system to be optimized
The power station coal transportation system which supplies the boilers is designed with five basic components as depicted in figure.3.
The process of coal transportation is: The coal is loaded from the bin to the primary conveyor (Conveyor 1) by the primary feeder (Feeder 1). Then the coal is transported through the conveyor 1 to the Stacker-reclaimer, when it is left up to the burner level. The secondary feeder (Feeder 2) loads the secondary conveyor (Conveyor 2) which supplies the burner feeding system of the boiler. Each element of the system is considered as unit with total failures.
#### Figure 3.
Synoptic of the detailed power station coal transportation
Comp# Vers# Availability A Cost C Capacity Ξ 1 1234567 0.9800.9770.9820.9780.9830.9200.984 0.5900.5350.4700.4200.4000.1800.220 1201008585483126 2 12345 0.9950.9960.9970.9970.998 0.2050.1890.0910.0560.042 10092532821 3 1234 0.9710.9730.9710.976 7.5254.7203.5902.420 100604020 4 123456789 0.9770.9780.9780.9830.9810.9710.9830.9820.977 0.1800.1600.1500.1210.1020.0960.0710.0490.044 11510091727272552525 5 1234 0.9840.9830.9870.981 0.9860.8250.4900.475 100604020
#### Table 1.
Characteristics of available system components on the market
[i] - Acronyms:
[ii] - Comp #: System component number.
[iii] - Vers #: System version number.
Demand level (%) 100 80 50 20 Duration (h) 4203 788 1228 2536 Probability 0.479 0.089 0.14 0.289
#### Table 2.
Parameters of the cumulative demand curve
A0 A01 Optimal Structure Computed Availability A ComputedCost C 0.975 0.9760 1: Components 3 – 6 – 5 -7 2: Components 2 – 3 – 4 - 4 3: Components 1- 4 4: Components 2 - 5 - 7 - 8 5: Components 3 - 3 - 4 0.9773 13.4440 0.980 0.9826 1: Components 2 - 2 2: Components 3 – 3 -5 3: Components 2 - 3 - 3 4: Components 5 - 6 - 7 5: Components 3 - 3 - 4 0.9812 14.9180 0.990 0.9931 1: Components 2 - 1 2: Components 3 - 3 3: Components 2 - 2 - 3 4: Components 5 - 5 - 6 5: Components 2 - 2 0.9936 16.2870
#### Table 3.
Optimal Solution Obtained By Ant Algorithm
Optimal availabilities obtained by Ant Algorithm were compared to availabilities given by genetic algorithm (presented by symbol A0 in table 3) in the reference (Levitin et al., 1997), and to those obtained by harmony search (presented by symbol A01 in table 3) given in (Rami et al., 2009).
For this type of problem, we define the minimal cost system configuration which provides the desired reliability level A ≥ A0 (where A0 is given in (Levitin et al, 1997) taken as reference).
We will clearly remark the improvement of the reliability of the system at price equal compared to the two other methods.
We gave more importance to the reliability of the system compared to its cost what justifies the increase in the cost compared to the reference.
The compromise of the cost/reliability was treated successfully in this work.
The objective is to select the optimal combination of elements used in series-parallel structure of power system. This has to correspond to the minimal total cost with regard to the selected level of the system availability. The ACO allows each subsystem to contain elements with different technologies. The ACO algorithm proved very efficient in solving the ROP and better quality results in terms of structure costs and reliability levels have been achieved compared to GA (Levitin et al., 1997).
#### Figure 4.
Cost-availability rate of GA and ACO algorithm versus availability
A0 % of AGA % of AACO % of C/A 0.975 0.1 0.23 58.5 0.980 0.0 0.12 13.3 0.990 0.2 0.36 39.4
#### Table 4.
Comparison of Optimal Solutions Obtained by ACO and Genetic Algorithms For Different Availability Requirements
From figure 4 and the table, one can observe:
ACO achieved better quality results in terms of structure cost and reliability in different reliability levels (figure 4). We remark in all case, GA performed better by achieving a less expensive configuration, however ACO algorithm achieved a near optimal configuration with a slightly higher reliability level (table 4).
We take, for example, for reference reliability level (A0 = 0.975, table 4), GA prove an augmentation of 0.1 percent compared to 0.23 percent given by ACO this for a difference in rate Cost-reliability of 58.3%. It is noticed, according to figure 4, that ACO tends, at equal price, to increase the reliability of the system.
## 6. Conclusion
A new algorithm for choosing an optimal series-parallel power structure configuration is proposed which minimizes total investment cost subject to availability constraints. This algorithm seeks and selects devices among a list of available products according to their availability, nominal capacity (performance) and cost. Also defines the number and the kind of parallel machines in each sub-system. The proposed method allows a practical way to solve wide instances of structure optimization problem of multi-state power systems without limitation on the diversity of versions of machines put in parallel. A combination is used in this algorithm is based on the universal moment generating function and an ant colony optimization algorithm.
## References
1 - Ait-Kadi and Nourelfath2001Availability optimization of fault-tolerant systems. International Conference on Industrial Engineering and Production Management (IEPM’2001), Québec 2023
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11 - Dorigo and Gambardella1997aAnt Colony System: A Cooperative Learning Approach to the Traveling Salesman Problem”, IEEE Transactions on Evolutionary computation, 115366
12 - Dorigo and Gambardella1997bDorigo, M. and L. M. Gambardella. Ant Colonies for the Travelling Salesman Problem. Bio Systems, 43
13 - Kuo and Prasad2000An Annotated Overview of System-reliability Optimization. IEEE Transactions on Reliability, 492
14 - Levitin and Lisnianski2001A new approach to solving problems of multi-state system reliability optimization. Quality and Reliability Engineering International, 47293104
15 - LevitinLisnianski, Ben-Haim and Elmakis, (1997Structure optimization of power system with different redundant elements. Electric Power Systems Research, 4311927
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18 - LisnianskiLevitin, Ben-Haim and Elmakis, (1996Power system structure optimization subject to reliability constraints. Electric Power Systems Research, 392145152
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21 - N Nahas, M Nourelfath, Aït-Kadi Daoud, Efficiently solving the redundancy allocation problem by using ant colony optimization and the extended great deluge algorithm, International Conference on Probabilistic Safety Assessment and Management (PSAM) and ESREL, New Orleans, USA, May 14_19, 2006
22 - NourefathAit-Kadi and Soro, (2002Optimal design of reconfigurable manufacturing systems. IEEE International Conference on Systems, Man and Cybernetics (SMC’02), Tunisia.
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28 - Ushakov1987optimal standby problems and a universal generating function. Sov. J. Computing System Science, 25N 4, 7982
29 - Ushakov1986Universal generating function. Sov. J. Computing System Science, 24N 5, 118129
30 - Wagner and Bruckstein1999Wagner, I. A. and A. M. Bruckstein. Hamiltonian(t)-An Ant inspired heuristic for Recognizing Hamiltonian Graphs. Proceeding of the 1999 Congress on Evolutionary Compuation, Washington, D.C., 14651469
31 - A. J. & R Wood, J Ringlee (1970Frequency and duration methods for power reliability calculations ‘, Part II, ‘ Demand capacity reserve model ‘,IEEE Trans. On PAS, 94375388 | 8,865 | 36,914 | {"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} | 2.78125 | 3 | CC-MAIN-2018-13 | latest | en | 0.899996 |
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Can anyone help me with these questions, thank you.
From 6am to 6:30am,
Mary cycled (300 x 30) = 9000m
From 6:30am to 7am,
Mary cycled another 9000m
Brother cycled (9000 + 1800) = 10800m
Brother’s speed -> 10800 / 30 = 360 m/min
Time for brother to cycle 9000m -> 9000 / 360 = 25 min
Brother left home 25 mins before 6:30am -> 6:05am
0 Replies 2 Likes | 142 | 372 | {"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} | 2.96875 | 3 | CC-MAIN-2020-40 | latest | en | 0.864378 |
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Can you fit the tangram pieces into the outline of Little Ming playing the board game? | 1,945 | 8,399 | {"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} | 4.28125 | 4 | CC-MAIN-2015-35 | longest | en | 0.92317 |
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It had a capacity in excess of 500 tonnes per hour t h 1 and had built up a total throughput of greater than 5 10 5 t during its period of operation. This was well within the usual operating envelope for similar plants. The main component of the crusher is a high inertia rotor spinning at over 250 revolutions per minute
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Jan 21, 2005 The high rotational speed more than 1,000 revolutions per minute rpm gives the hammers enough inertia to shred the material, notes Modern Composting Technologies. As the drum rotates, the hammers spin rapidly and smash against the material trapped inside the hammermill chamber until the pieces are small enough to pass through the discharge ...
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The distance, h, between A and B is equal to the distance the particle would fall during half a cycle of the crusher eccentric, provided the cycle frequency allows sufficient time for the particle to do so.If is the number of cycles per minute, then the time for one complete cycle is 60 seconds and the time for half a cycle is 602.
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Then the materials will be thrown to the impact plates for secondary crushing. The materials will be shot back to the blow bars again. ... A hammer mill is a crusher that can grind, pulverize, and crush a wide range of materials. ... The rotation is usually between 4 to 20 revolutions per minute, depending upon the diameter of the mill. The ...
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Crusher Serial Number 197126 - nizzisara.it. Crusher Serial Number 197126 naukridreams. crusher serial number 197126 stone crusher aggregate, cone crusher crushing capacity, bis number of crusher grease. crusher 197126, list of domains alex crusher 201 cv1. dot web . . number of employees in a coal mine small rock impact crusher double ball mill rock tumbler bis number of .
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A hammer mill has been designed with due considerations to standard design requirements and cost to cater to this need. The power required to deliver the through put efficiency of the machine was ...
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Thrown Hammer Crusher The Number Of Revolutions Per Minute Cost Of Pc Hammer Mill We sincerely welcome you to contact us through hotlines and other instant communication ways. We can call you and help with your problem. Contact us . Shanghai Zenith Mining and Construction Machinery Co., Ltd. is a hi-tech, engineering group. ...
### Vertical Mill Hydraulic System Failure Analysis
thrown hammer crusher the number of revolutions per minute what is the specific gravity of coal ball mill gz1 vibration feeder india plan your visit atlas vs mill parts cema 350 screw conveyor method beverly crusher is a licensed cpa high capacity and top sand vertical shaft impact crusher fci vibrating screen vertical ball mill kaolin | 2,808 | 12,540 | {"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} | 2.546875 | 3 | CC-MAIN-2021-49 | latest | en | 0.841373 |
https://kr.mathworks.com/matlabcentral/cody/problems/80-test-for-balanced-parentheses/solutions/1031273 | 1,568,688,212,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573011.59/warc/CC-MAIN-20190917020816-20190917042816-00073.warc.gz | 556,505,355 | 15,597 | Cody
# Problem 80. Test for balanced parentheses
Solution 1031273
Submitted on 25 Oct 2016 by uu tsi
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
inStr = '()'; out_correct = true; assert(isequal(isBalanced(inStr),out_correct))
2 Pass
inStr = ')'; out_correct = false; assert(isequal(isBalanced(inStr),out_correct))
3 Pass
inStr = '(z*(a-(x+3))/(y))'; out_correct = true; assert(isequal(isBalanced(inStr),out_correct))
4 Pass
inStr = ')('; out_correct = false; assert(isequal(isBalanced(inStr),out_correct))
5 Pass
inStr = '(x)(x-y)'; out_correct = true; assert(isequal(isBalanced(inStr),out_correct))
6 Pass
inStr = ':-)'; out_correct = false; assert(isequal(isBalanced(inStr),out_correct))
7 Pass
inStr = ')()'; out_correct = false; assert(isequal(isBalanced(inStr),out_correct))
8 Pass
inStr = '(()'; out_correct = false; assert(isequal(isBalanced(inStr),out_correct)) | 303 | 1,017 | {"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} | 2.609375 | 3 | CC-MAIN-2019-39 | latest | en | 0.241179 |
www.yodzian.com | 1,369,156,273,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700264179/warc/CC-MAIN-20130516103104-00017-ip-10-60-113-184.ec2.internal.warc.gz | 805,821,527 | 22,332 | ### Archive
Posts Tagged ‘Pensions’
## Fibonacci Numbers – How To Use Them For Huge Trading Profits!
August 26th, 2012 No comments
The Fibonacci numbers sequence and the golden ratio have fascinated mathematicians for hundreds of years.
While Fibonacci numbers have many applications, they have received considerable interest from traders due to their uncanny accuracy in spotting market turning points in advance.
You can use Fibonacci numbers as a predictive tool and when used correctly they can enhance a your analysis of the market, helping you to increase profits and decrease risk.
The History of Fibonacci Numbers
The Fibonacci number sequence first appeared as the solution to a problem in the Liber Abaci, a book written by Leonardo Fibonacci in 1202 to introduce the Hindu-Arabic numerals used today to a Europe still using Roman numerals.
The original problem in the Liber Abaci posed the question: How many pairs of rabbits can be generated from a single pair, if each month each mature pair brings forth a new pair, which, from the second month, becomes productive.
The Fibonacci number Sequence
The resulting Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, are the result of the following equation.
If Fn is the nth Fibonacci number, then successive terms are formed by addition of the previous two terms, as Fn 1 = Fn Fn-1, F1 = 1, F2 =
The ratio of any number to the next larger number is 62%, which is a popular Fibonacci retracement number. The inverse of 62% is 38%, and this 38% is likewise a Fibonacci retracement number.
Fibonacci Numbers and the Golden Ratio
Fibonacci numbers are found to have many relationships to the Golden Ratio F = (1 /5)/2, a constant of nature which was of constant interest to the ancient Greeks, appearing in both Greek art and architecture.
Fibonacci Numbers and Market Analysis
Changes in stock prices are not simply a tug of war between supply and demand but also reflect human opinions, valuations, and expectations.
A study carried out by mathematical psychologist Vladimir Lefebvre demonstrated that humans exhibit positive and negative evaluations of the opinions they hold in a ratio that approaches phi, with 61.8% positive and 38.2% negative and that Fibonacci numbers are rooted Read more…
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## Stock Index Trading Systems – Learn From One Of The Greatest Traders Of All Time!
August 26th, 2012 No comments
Trading using stock index trading systems has become increasingly popular in recent years, as they offer traders a great speculative vehicle to seek above average profits.
There are plenty of stock index trading systems, but which is the best?
Stock Index Trading Systems ? Catch and Follow the Trends!
For profitable stock index trading, you need to be able to lock into, and run the big profitable trends, and the best way to do this is by using technical analysis to spot, and act, on these trends.
The best way to do this is to find a stock index trading system that has stood the test of time.
Stock Index Trading Systems and Gann?s Methods for Profit
W D Gann was a trader and legend in his own lifetime. Even today, a half a century after his death, he remains one of the most influential traders of all time.
Gann had an astounding trading record and amassed a fortune of over \$50 million dollars in his trading career. Many of his recommendations are on record, for example:
Each year Gann published a forecast for the following year. In 1928 he published a forecast, which predicted the date of the September 1929 US Stock Market high, and that a black Friday would occur, a year in advance of the actual events.
In 1932, he recommended buying stocks at the all time low in the Dow in June and July.
History repeats itself allowing us to Predict the Future
Gann?s major contention was that certain laws governed not only the markets, but nature as well, and were universal in scope.
He argued that human nature repeated itself and that by looking at the past we could make predictions about the future.
In “Wall Street Stock Selector” Gann said.
“Just remember one thing, whatever has happened in the past in the stock market and Wall Street will happen again. Advances in bull markets will come in the future, and panics will come in the future, just as they have in the past. This is the working out of a natural law ?” and, “It is action in one direction and reaction in the opposite direction. In order to make profits, Read more…
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## Fibonacci Numbers And The Golden Ratio – 3 Tips For Greater Trading Profits
August 26th, 2012 No comments
In this report, we will look at the history and background of Fibonacci numbers and The Golden Ratio. We will then outline three specific money management tips that can help increase your profit potential.
Support and resistance levels are an important consideration for most traders to help identify entry and exit points when trading.
Fibonacci percentage “retracement” levels based upon the Fibonacci number sequence and golden ratio are very popular with many traders but what are they exactly?
What are Fibonacci Numbers and the Golden Ratio?
The Fibonacci sequence first appeared as the solution to a problem in the Liber Abaci, a book written by Leonardo Fibonacci in 1202 to introduce the Hindu-Arabic numerals used today to a Europe still using Roman numerals.
The original problem in the Liber Abaci posed the question: How many pairs of rabbits can be generated from a single pair, if each month each mature pair brings forth a new pair, which, from the second month, becomes productive.
The Golden Ratio
After the first few numbers in the Fibonacci sequence, the ratio of any number to the next higher number is approximately .618, and the lower number is 1.618. These two figures are the golden mean or the golden ratio.
Its proportions are pleasing to the human senses and it appears throughout biology, art, music, and architecture. A few examples of natural shapes based on the Golden Ratio include DNA molecules, sunflowers, snail shells, galaxies, and hurricanes.
Important Retracement Levels
The two Fibonacci percentage retracement levels considered the most important in trading are 38.2% and 62.8%. Other important retracement percentages include 75%, 50%, and 33%.
Three Profit Tips for Using Fibonacci Numbers
1. Fibonacci Defines Stop Loss Levels
A trader can use Fibonacci numbers to set stop loss orders.
For instance, if at least three Fibonacci price levels come together in a relatively tight zone, a stop loss placement just below or above the zone may be set.
A Fibonacci number helps define stops in the following way, if a trader trades against a support zone, if the support zone is violated and the price trades below that zone, the reason Read more…
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## Mechanical Trading Systems – Spotting The Ones That Make Money!
August 26th, 2012 No comments
Mechanical trading systems are, as you would expect, systems that make trading decisions for you.
The thought of having mechanical trading systems you can simply use to generate automatic profits, is obviously very attractive to many traders.
Most traders however, end up disappointed with mechanical trading systems, as they never seem to live up to the sales hype, and the performance figures used to sell the system never seem to be repeated in real life.
Why do most mechanical Trading Systems fail to live up to the Hype?
There are two main reasons for this:
Black Box Systems
These are systems where the vendor does not reveal the logic of the system. Of course, for a trading system to be successful it needs following rigidly with discipline.
If however, you don?t know the logic of a mechanical trading system, you will probably not have the discipline to follow it when a losing period occurs. If you don?t have the confidence to follow a mechanical trading system, you don?t have a system at all!
Curve Fitting and Optimization
Another problem is curve fitting or optimization of mechanical trading systems. These systems yield extraordinary performance in back testing because of the tweaking of the system rules to make them fit the data. A trader once likened this to shooting holes in a barn door, and then drawing circles around every hole to make each shot a bull?s eye!
Of course, anyone can make a mechanical system make money if it is already know what happened in the past.
You will never see a hypothetical performance that fails! Most vendors achieve this by making the system fit the data, which of course will lead to disappointment in the brutal world of trading.
The fact is that most mechanical trading systems don?t deliver the results they promise and traders end up disappointed. This is not to say that there are not good mechanical trading systems to buy, but you need to do your research first, and the following checklist will give you the salient points to look Read more…
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## Trading Psychology – Adopt The Right Mindset For Big Profits!
August 25th, 2012 No comments
The fact is the majority of traders lose because they cannot control their emotions. Trading psychology is one of the keys to investment success.
A simple fact will illustrate the influence of trading psychology:
Why the majority of traders lose
There is one statistic that has remained constant since the beginning of investment records – the ratio of winners to losers has remained constant over time.
On reflection, this would seem a startling fact; despite the massive advance in communications and economic forecasting methods, the ratio remains the same.
The conclusion from the above is that the successful trading is dependant on something else. That something else is our trading psychology.
The influence Of Hope and Fear
In trading psychology, two emotions that are constantly to the fore are hope and fear. One of the traders who recognised this was the legendary trader W D Gann.
?Hope and fear: I have written about this often in my books and I feel I cannot repeat it too often. The average person buys commodities because they hope they will go up, or because someone advises them, they will go up. This is the most dangerous thing to do, never trade on hope. Hope wrecks more people?s lives than anything else. Face the facts, and when you trade, trade on the facts, eliminating hope?
?Fear causes many losses. People sell out because they fear commodities are going lower, but they often wait until the decline has run its course and sell near the bottom – never make a trade on fear?
Control Emotions and Become a Disciplined Trader
Gann, like all successful traders, realised that the only way to trade successfully was to remove emotions from trading, and trade on the facts and realised the significance of trading psychology on price movements.
To do this, he applied mathematical principles to investing that would give him the ability to trade without emotion, with discipline Gann was extremely successful, amassing a fortune of over \$50 million in his trading career.
Human Nature Is Constant ? Exploit It for Trading Success
It doesn?t matter what market you trade: commodities, stocks, currencies, or what type of trader you are, a day or position trader, Read more…
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## FOREX Trading Systems – Trading The Longer Term Trends For Bigger Profits
August 25th, 2012 No comments
How to Make BIG Profits with Currency Trading Systems
FOREX markets turn over trillions of dollars per day and are the world?s biggest investment medium.
In recent years, FOREX trading systems using technical analysis to predict trend changes have become increasingly popular as a way of catching the big profitable trends.
Catching the Longer Term Trends for Big Profits
The longer-term trends in FOREX markets mirror the underlying health of the economy. As periods of expansion and contraction take years, so do currency trends and a good FOREX trading system can help you lock into, and profit from, these trends.
When picking a currency to trade, it is important to have good long-term trends and liquidity.
Good major currencies to trade include the US Dollar, Swiss Franc, Euro, Japanese Yen, British Pound, and Canadian Dollar.
FOREX trading systems remove the emotional component from trading, which is the major reason the majority of traders lose.
Removing the Emotion from Trading with Systems
One of the best starting points on the effect that emotions have in trading, are the works of legendary trader W. D Gann, whose works on the subject are essential reading.
Other authors worth reading are: Edwin Lefeurve, Jake Bernstein, Larry Williams, Ken Roberts, Van Tharpe and Jack Shwager whose book ?Market Wizards
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## Predicting The Market Using Gann Angles – An Alternative Slant On Market Timing
August 25th, 2012 No comments
W D Gann was a prolific writer and trader, and created a fortune of over 50 million dollars (equivalent to 500 million today!).
Many of his trading predictions were the subject of public record. For instance, he correctly predicted the 1929 crash a year in advance!
Gann died in 1955, but he still holds legendary status as a technical innovator.
By predicting the market using Gann angles, you can add a valuable tool to your trading strategy.
Assumption: By Studying the Past, We Can Predict the Future
Gann based predictions of price movements on three premises:
1. Price, time, and range are the only three factors to consider.
2. The markets are cyclical in nature.
3. The markets are geometric in their design and in function.
Gann believed that human nature was constant, and this showed up in repetitive price patterns that are identifiable, and which can therefore be acted upon to increase profit potential.
Gann?s Strategy for Trading Success
Based on the above assumptions, Gann’s strategies revolved around three areas of prediction:
1. Price study? This study uses support and resistance lines, pivot points and angles.
2. Time study ? This studies historically reoccurring dates derived from natural order.
3. Pattern study ? These study market swings using trend lines and reversal patterns.
Constructing Gann Angles
Predicting the market using Gann angles requires subjective judgment and practice. Here is what you need to do:
1. Determine the time units – One common way to determine a time unit is to study the chart and look at the distances in which price movements occur. Then, put the angles to the test and see how accurate they are. The intermediate-term time frame (one to three-month) tends to produce the optimal amount of patterns compared to short term daily, or multi year charts.
2. Determine the high or low from which to draw the Gann lines – The most common way to accomplish this is to complement it with other forms of technical analysis i.e. Fibonacci levels or pivot points. Gann used what he called “vibrations” or “price swings.” He determined these by analyzing charts using theories such as Fibonacci numbers.
3. Decide which pattern to use – The two most common patterns are the 1×1, the 1×2, and the 2×1. Read more…
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## Commodity Trading Systems – Learn From A Trading Master And Boost Your Profit Potential!
August 25th, 2012 No comments
Legendary trader W D Gann amassed a fortune of \$50 million dollars in the first half of the last century, although he died in 1955, his commodity trading systems are still used today by traders all over the world.
Successful commodity trading systems have the ability take the emotion out of trading, liquidating losses quickly and spotting and holding the big longer-term trends and that?s exactly what Gann?s Commodity trading systems did.
Gann?s Commodity Systems Track Record
Gann?s commodity trading systems allowed him to make some stunning predictions and trading gains such as:
1. He predicted improvements in business in 1921 and the Bull Run in stocks.
2. 1928 he forecasted the end of the Bull Market in stocks a full year in advance of the 1929 crash. He then bought stocks in the Dow at an all time low in 1932.
3. In 1935, of 98 trades in cotton, grain, and rubber, 83 trades showed a profit. His percentage of profitable trades was often 90% or higher.
History Repeats Itself
Gann was a prolific writer and wrote extensively, outlining his thoughts on commodity trading systems in a series of books and courses. Some of his ideas were grounded in empirical studies, while others were more mystical in nature.
Gann?s major contention was that certain laws governed not only the markets, but nature as well and were universal in scope. He believed that human psychology was constant and that this manifested itself in repeatable price patterns.
?We cannot escape it (emotion) In the future it will cause another panic in stocks. When it comes both traders and investors will sell stocks, as usual, after it is too late or in the latter stages of a bear market?
He was aware that human nature was constant and influenced the majority of traders.
?Therefore, in order to make a success the trader must act in a way to overcome the weak points that have caused the ruin of others?
For more information on trading psychology excellent books to read any by Jake Bernstein, Jesse Livermore, Larry Williams, Van Tharp and Jack Shwager and you will see why human nature repeats itself.
The Influence of Price and Time
One of the most important thoughts behind Gann?s commodity Read more…
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## W D Gann – How To Use His Unique Methods To Make Big Trading Profits
August 25th, 2012 No comments
In the entrance to the New York Stock Exchange, stands a life-sized picture of W D. Gann (1878 – 1955) and this is a testament to his standing amongst traders worldwide. Today he remains one of the most influential traders of all time.
W D Gann Methods and Trading Performance
W D Gann employed a staff of 25 draughtsmen to draw charts of all the stocks on the New York Stock Exchange, as well as a variety of commodities. He would then use the charts to look for trading opportunities.
Gann in fact made huge trading profits from his technical analysis of the markets.
There are reports, which indicate that his trading techniques amassed him a fortune of over \$50 million dollars, and many of his trades are on record.
W D Gann Trading Philosophy
W D Gann was a prolific writer, and wrote extensively outlining his thoughts and trading methods in a series of books and courses. Some of his ideas were empirical studies, while others were more mystical in nature.
Gann?s major contention was that certain laws governed not only the markets, but nature as well, and were universal in scope.
The Influence of Price and Time
One of Gann?s most important contributions was the concept of combining price and time. Gann believed that crucial price movements happened when price and time converged. These points usually indicated an important trend change was imminent.
However, if price and time were not coordinated, or did not converge, then time always held priority over price.
Therefore time, was considered by Gann as the ultimate indicator, because all of nature was governed by time.
In “Wall Street Stock Selector” Gann said.
“Just remember one thing, whatever has happened in the past in the stock market and Wall Street will happen again.
Advances in bull markets will come in the future, and panics will come in the future, just as they have in the past. This is the working out of a natural law “
“It is action in one direction, and reaction in the opposite direction. In order to Read more…
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## The Art Of Contrary Thinking – You Need To Know It To Trade Successfully!
August 25th, 2012 No comments
The art of contrary thinking is one of the most powerful tools a trader can use, and is a trait with which all true great traders are familiar.
What is the Art of Contrary Thinking?
The art of contrary thinking consists in training your mind to ruminate in directions opposite to general public opinions; but basing your opinion in the light of current events and human behaviour.
Humphrey Neill?s book, “the art of contrary thinking,? the best known work on the subject, is based on the simple yet powerful idea that:
“When everybody thinks alike, everybody is likely to be wrong”
Why Contrary Trading Works
By spotting situations when the consensus is either extremely bullish or bearish, then a trend change is imminent, as it is likely the emotions of greed and fear have pushed prices too far away from true value.
This is evident in such events as the 1987 stock market crash.
Here we have a short-term, self-fulfilling prophecy. When the change occurred, everyone changed his or her mind at once, causing a huge move.
Of course, if you can step aside from the crowd and take a contrary view at these turning points you can make big profits.
Why Contrary Thinking will always be Valid
While Humphrey Neil’s work, “the art of contrary thinking,? (published in 1954), is the most famous book on the subject, there existed a century earlier a book on contrary thinking.
Charles MacKay?s book, “Extraordinary Popular Delusions and the Madness of Crowds,? (published in 1854), covered three important financial crashes:
he tulip mania, the Mississippi madness, and the south sea bubble. He reflected upon how investors always pushed prices too far when caught in a consensus:
“Men, it has been well said, think in herds; it will be seen that they go mad in herds, while they only recover their senses slowly, and one by one.”
It is clear that to succeed in trading you need to think independently of the majority at important market turning points.
Becoming a Contrary Trader
Gann was one of the greatest traders and traded in the early 20th century. He realized that human nature would always mean that you Read more…
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:: โปรโมทเว็บ :: Promote Web :: Social Bookmark :: | 4,579 | 21,982 | {"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} | 3.4375 | 3 | CC-MAIN-2013-20 | longest | en | 0.932346 |
https://mathalino.com/tag/forum/area-integration | 1,708,729,270,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00056.warc.gz | 393,282,324 | 12,130 | # area by integration
## moment of inertia of an area with respect to the line x = 2
The moment of inertia bounded by y=x^3 , y=2x , X + y= 6, with respect to x=2 pa help po until Friday. Plus ask ko Lang po ang pinakakapakipakinabang na calculus book for integration applications. Thanks.
## patulong naman po kung pano isolve 2.
plane areas
Find the area bounded by the curve y = ln x/x, the x-axis and the maximum ordinate.
ans. 1/2
## MATHalino 2020 (Ito Yung Last sa Paatras) Solid mensuration problem
1)An excavation is 12 ft. deep and has trapezoidal sides (faces). The upper base is horizontal rectangle 400 ft. by 180 ft., and the lower base is a horizontal rectangle 350 ft. by 150 ft. How many cubic yards of earth were removed in digging the excavation? | 218 | 773 | {"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} | 3.28125 | 3 | CC-MAIN-2024-10 | longest | en | 0.861309 |
https://www.exceldemy.com/author/exceldemy/page/6/ | 1,601,460,593,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402123173.74/warc/CC-MAIN-20200930075754-20200930105754-00329.warc.gz | 793,944,610 | 24,588 | Disclosure: This post may contain affiliate links, meaning when you click the links and make a purchase, we receive a commission.
## How to Use Slicers to Filter a Table in Excel 2013
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##### Browsing All Comments By: Kawser
Kawser Jul 20, 2015 at 9:22 PM
It’s my pleasure Rick. Nice to see you here and thanks for commenting!
Kawser Jul 29, 2015 at 4:06 PM
Aziz Bro, thanks for your comments. Excel has extended its power more in recent times making Excel to work with Microsoft’s Power BI tools. I shall focus on them within very short time. Keep in touch. Thanks again.
Kawser Aug 26, 2015 at 5:18 AM
Mohammed Hussain,
Jul 18, 2017 at 2:40 PM
Thanks for notifying me. It was a technical mistake. I’ve uploaded them and you can download now. Thanks.
Jan 14, 2018 at 1:22 PM
Hi Mark,
Use the following formula instead of the formula you’re using.
=DATEDIF(\$B\$1,\$B\$2,”Y”)& ” years, “&DATEDIF(\$B\$1,\$B\$2,”YM”)& ” months, “&ROUNDDOWN(DATEDIF(\$B\$1,\$B\$2,”MD”)/7,0)&” Weeks, and “&DATEDIF(\$B\$1,\$B\$2,”MD”)-ROUNDDOWN(DATEDIF(\$B\$1,\$B\$2,”MD”)/7,0)*7&” Days.”
Regards
Kawser
Sep 24, 2020 at 11:33 PM
A great pleasure for me. Thank you.
Sep 17, 2020 at 10:56 AM
Sep 16, 2020 at 11:14 PM
Thanks, Christina for your feedback. Glad to know that it helped you someway.
Sep 10, 2020 at 7:38 PM
Idowu,
Best regards
Aug 25, 2020 at 10:58 PM
Hi Vishnu,
Best regards
Aug 18, 2020 at 10:32 AM
Hello Johmono,
You’re most welcome.
Best regards
Aug 17, 2020 at 11:40 PM
Apr 21, 2020 at 9:47 PM
Apr 21, 2020 at 9:44 PM
Thanks, Kelly for your feedback. Glad to hear that it helped you.
Best regards
Apr 2, 2020 at 10:26 PM
You’re most welcome. Glad to know that it helped you. Thanks for the feedback.
Feb 12, 2020 at 8:29 PM
Jan 15, 2020 at 4:42 PM
Thanks for the feedback. Hope you will find this technique useful for your next some jobs.
Best regards
Kawser Ahmed
Jan 15, 2020 at 4:41 PM
Best regards
Dec 29, 2019 at 10:17 AM
Thanks.
Dec 14, 2019 at 7:28 PM
Hello Wasim,
I am keeping this on my to-do list. Will try to give you a solution of your problem.
Thanks.
Dec 9, 2019 at 1:56 PM
Dec 4, 2019 at 10:11 AM
Best regards
Nov 26, 2019 at 9:16 AM
Nov 21, 2019 at 9:37 AM
You can copy the article and paste in a word file and then use it personally. Thanks.
Nov 18, 2019 at 7:50 PM
It is actually important π Thanks for your feedback.
Oct 31, 2019 at 9:09 AM
You’re most welcome! Thanks for your feedback, Anthony!
Oct 24, 2019 at 9:06 AM
You will also find the file in our article. The file is at the upper part of the article. You will find the file under this title: “Calculate Hours Worked and Overtime Excel Template”.
Best regards
Kawser Ahmed
Oct 16, 2019 at 8:33 PM
You’re welcome, Florano!
Oct 10, 2019 at 12:59 PM
Most welcome, Nilesh π
Oct 10, 2019 at 12:58 PM
Nice, Roger. I am glad to know that the formula helped you π
Oct 8, 2019 at 12:37 PM
Maybe it is due to the payment time. Do you pay at the beginning of the month or at the end of the month?
Oct 1, 2019 at 10:30 AM
Sep 26, 2019 at 7:06 PM
Jul 25, 2019 at 6:33 PM
You’re most welcome, Dennis!
Thanks for the nice words.
Best regards
Kawser Ahmed
Jul 23, 2019 at 6:50 PM
Jul 18, 2019 at 6:20 PM
Maybe a little bit tough for the complete layman. I shall change my Phrase then π
Jul 15, 2019 at 8:02 PM
Thanks for your good words, Dayo!
Best regards
Kawser Ahmed
Jul 10, 2019 at 8:25 PM
It is sent. Thank you.
Jun 20, 2019 at 7:19 PM
Thanks for the suggestion π I will keep it on my list.
Jun 13, 2019 at 6:16 PM
Best regards
Kawser
Jun 8, 2019 at 2:24 PM
When we have more than one files for an article, we make a compressed file.
Hope you understand. Other articles that have just the Excel file to share, the download options are in Excel File.
Best regards
Kawser Ahmed
Jun 8, 2019 at 2:22 PM
Jun 8, 2019 at 2:21 PM
It is sad to hear that you did not get the PDF. Just check your email.
Thanks and regards
Kawser Ahmed
Jun 8, 2019 at 2:20 PM
Best regards
Kawser Ahmed
Jun 3, 2019 at 6:10 PM
Glad to know that it helped you.
Thanks and regards
Kawser Ahmed
Jun 3, 2019 at 6:09 PM
Best regards
Kawser Ahmed
May 30, 2019 at 5:02 PM
Hi Gilbert,
I am really sorry for not having PDF formats of the Articles.
But you can make your own PDFs. Only thing is: you can use it personally. Not for commercial use, of course.
Hope you understand.
Thanks and regards
Kawser Ahmed
May 29, 2019 at 4:54 PM
Thank you so much for your input, Wayne! Hope this input helps my readers π
Best regards
Kawser Ahmed
May 21, 2019 at 9:54 PM
You’re welcome, Frank! Feeling good to hear that this article helped you.
Best regards
Apr 22, 2019 at 9:43 PM
Nice to hear that it helped you. Thanks and regards.
Apr 22, 2019 at 9:43 PM
Nice to hear that it helped you. Thanks and regards.
Apr 15, 2019 at 1:32 PM
Thanks, Sarabjeet Singh, for your feedback!
Apr 11, 2019 at 9:42 PM
You’re welcome, Andrew π
Mar 26, 2019 at 9:53 PM
Mar 7, 2019 at 7:58 PM
Thanks, Ferreira!
Mar 7, 2019 at 7:58 PM
Feb 27, 2019 at 6:34 PM
Feb 27, 2019 at 6:33 PM
Thanks, Surya!
Feb 27, 2019 at 6:29 PM
Thanks.
Feb 23, 2019 at 1:52 PM
But you can contact our partner here: https://www.exceldemy.com/consulting-services/
Thanks.
Feb 14, 2019 at 4:26 PM
You’re most welcome π I am glad to know that our tutorials add some value to your life.
Feb 9, 2019 at 12:36 PM
You can use the DATEDIF Function without any problem.
Thanks.
Jan 31, 2019 at 9:25 PM
Jan 30, 2019 at 1:26 PM
Thanks a lot for your feedback. I will make plans to correct them. Best regards
Jan 22, 2019 at 3:58 PM
I will check this issue and let you know. Glad to know that we can add some value via our emails.
Best regards
Kawser
Jan 19, 2019 at 8:00 PM
Jan 10, 2019 at 3:50 PM
Best regards
Jan 9, 2019 at 1:06 PM
Thanks for the feedback.
Jan 6, 2019 at 6:45 PM
Glad to know that it helped you to learn something new.
Jan 3, 2019 at 12:40 PM
Not right now. I will update when I will have one. Or if you have any specifications, let me know. I will make one for you.
Thanks and regards
Kawser
Dec 29, 2018 at 3:59 PM
Hi,
You can try to use the concepts in real-world problems.
Thanks.
Dec 25, 2018 at 3:40 PM
Thanks.
Dec 15, 2018 at 1:59 PM
Best regards
Kawser
Dec 6, 2018 at 5:22 PM
Dec 3, 2018 at 1:13 PM
Thanks a lot for the feedback. I shall input this method when I will update this article later.
Nov 21, 2018 at 12:05 PM
You’re welcome.
Best regards
Nov 4, 2018 at 6:47 PM
Thanks for your valuable comment. Unfortunately, right now I don’t have any dashboard like that. Maybe later.
Thanks and regards
Oct 29, 2018 at 1:06 PM
Thanks, Mittal!
Oct 25, 2018 at 9:47 PM
Thank you so much for your feedback.
Best regards
Oct 9, 2018 at 6:12 PM
Oct 7, 2018 at 3:21 PM
Thank you for your valuable feedback. It’s something π
Best regards
Sep 6, 2018 at 12:23 PM
Yes, Robert!
Thanks for the feedback.
Sep 6, 2018 at 12:23 PM
Sep 5, 2018 at 1:28 PM
You’re most welcome, Sola.
Sep 4, 2018 at 6:26 PM
Sep 1, 2018 at 2:20 PM
Thanks, Melinda!
Yes, it was a typo and I have corrected it.
Best regards
Kawser
Sep 1, 2018 at 2:17 PM
Thanks, Mark.
Sep 1, 2018 at 2:17 PM
Hi Helen,
Thanks for the feedback.
And I will work with the click-baits to make it more user-friendly.
Thanks and regards
Aug 28, 2018 at 12:28 PM
Thank you for your feedback, Anirudh!
Best regards
Aug 8, 2018 at 10:46 AM
You’re welcome.
Aug 8, 2018 at 10:45 AM
Ram,
I have a list of Excel books that is good for learning Excel from the beginning. Please visit this page: https://www.exceldemy.com/best-excel-training-books/
The best is obviously the Excel Bible Series. You will get the link on the page. If you talk about Video Channel, it is tough to find one as most of them don’t start from the beginning. But if you search for a topic, there are several. One of the best is ExcelIsFun YouTube channel. I hope it helps.
Best regards
Kawser
Aug 7, 2018 at 5:28 PM
Hi David,
Thanks.
Aug 7, 2018 at 5:24 PM
Hi Mike,
Thanks.
Jul 18, 2018 at 8:17 PM
It is sent. Check out your email.
Jul 17, 2018 at 2:38 PM
Thanks, Emil!
Jul 4, 2018 at 12:17 PM
You have to use VBA or you can use Power Query.
Jun 25, 2018 at 4:16 PM
You are welcome.
Jun 25, 2018 at 11:55 AM
Hi Daniel,
Yes. INDEX and MATCH combo work in a more versatile way than the VLOOKUP function. I did not use Index Match because I wanted to show all the examples with VLOOKUP and IF Functions.
Keep in touch.
Best regards
Kawser
Jun 23, 2018 at 4:53 PM
Thanks for your feedback, Uche Uche.
Feeling happy to know that you found the articles helpful.
Best regards
Kawser Ahmed
Jun 21, 2018 at 12:28 PM
I will check out this code. Thanks for the input.
Jun 21, 2018 at 12:28 PM
Jun 13, 2018 at 12:36 PM
Please check your email. I have sent the file. Thanks and sorry for the inconvenience.
Jun 7, 2018 at 11:35 AM
You can collect it from our resource page. The resource page is shared with every subscriber, every day.
Best regards
Kawser Ahmed
May 31, 2018 at 10:57 AM
You’re welcome π
May 30, 2018 at 8:30 PM
Best regards
May 29, 2018 at 7:50 PM
I will update you.
Best regards
Kawser
May 29, 2018 at 7:46 PM
Hi Stanley,
Thanks for the comments. Glad to hear that it is helpful. Yes. You can share it with an online class with an attribution to our blog.
Best regards
May 29, 2018 at 11:27 AM
Glad to hear that it was useful for you.
Regards
May 17, 2018 at 12:36 PM
Hi Anatoly,
We have written a detailed guide on solving equations using Excel. You will get the guide here: https://www.exceldemy.com/solving-equations-in-excel/
If you have some specific equation, let us know via email.
Thanks and regards.
May 16, 2018 at 11:11 AM
Hi Moataz,
Thanks for the feedback on our blog.
Actually, this is not a big deal. The concepts are same with both Excel 2013 and Excel 2016. Most of the variations/updates occur with the Excel features, not in VBA. Just start with what most suit you.
I hope this reply helps you.
Thanks and regards
May 8, 2018 at 1:32 PM
Thanks for asking. It was creating more indexed pages in Google that was bad for us. What you can do is: just you can copy the whole article and paste into a word document and then convert it into a PDF file. There is little chance of getting back of that button.
Best regards
May 8, 2018 at 1:29 PM
That is really a good percentage of reduction. Thanks for the feedback, Abraham.
Apr 26, 2018 at 8:28 PM
Hi Ajeet,
Then you can download it from the Excel Resources Page that you get in every email I send.
Thanks.
Apr 24, 2018 at 2:14 PM
I have updated this part. For your clarification: the condition 1 must be fulfilled to be eligible to take the thesis work/project. Of the last conditions, students must fulfill one condition. I hope this makes things clear to you.
Best regards
Kawser
Feb 22, 2018 at 11:53 AM
Thanks. Glad to know that you found it useful.
Feb 20, 2018 at 5:25 PM
Thanks, Farhana for the feedback.
Jan 28, 2018 at 11:09 AM
Dec 19, 2017 at 7:56 PM
Thanks, Menon for your valuable feedback.
Oct 25, 2017 at 3:37 PM
Mike,
Kind regards
Sep 12, 2017 at 1:20 PM
Nope, Jan. You’re absolutely right. Excel can be used as a database when your database is not heavy and need a small load of query on it.
Sep 11, 2017 at 12:11 PM
Aug 3, 2017 at 8:18 PM
Hi John,
Sometimes discounts expire. Now the discounts are available. Thanks for your feedback.
Jul 14, 2017 at 9:34 PM
Thanks for your feedback. Best wishes with your new journey π Excel, Word, and PPT are three most powerful office tools. I’m sure, you will enjoy your journey π
Jul 6, 2017 at 6:31 PM
Jun 17, 2017 at 3:49 PM
Hi Bob,
Did you download the Excel file before you follow me? Please do so. I’ve explained the whole process with two real time examples. If you can create these two examples in the Excel file, I hope you will be able to understand the whole system.
Best regards
Kawser
Jun 5, 2017 at 10:13 AM
Check out trying. I never tried. If you face any problem, let us know π
Jun 4, 2017 at 1:44 PM
Thanks, Subbu!
Apr 3, 2017 at 9:02 PM
Glad to know that it helps. The file is at the end of the article. Just click and download.
Kawser Mar 3, 2017 at 6:55 PM
Kawser Feb 27, 2017 at 12:46 PM
Thanks, Piotr! Glad to know it helps.
Kawser Jan 30, 2017 at 7:36 PM
Syed,
Sometimes when the formula is posted in WordPress platform, it might be changed. Please, download the working file and check the formula there.
Kawser Jan 15, 2017 at 1:23 PM
Thanks for the feedback, Prabhudeva!
Kawser Dec 18, 2016 at 11:55 AM
Roger, thanks a lot. I did not notify the issue. It’s a crucial problem. I am working on it.
Kawser Dec 14, 2016 at 9:37 AM
Yes, when a number or date is placed in a cell in Text format, you cannot calculate with them.
Kawser Dec 14, 2016 at 9:36 AM
Roger, this option (make PDF of the blog posts) is on my plan. Will execute in shortest possible time.
Best regards and thanks for the comments.
Kawser Nov 23, 2016 at 10:22 AM
Thanks.
Kawser Nov 12, 2016 at 3:26 PM
Thanks.
Kawser Nov 7, 2016 at 11:10 AM
Yes. Corrected.
Thanks, Luke for the tips.
Kawser Nov 6, 2016 at 11:13 AM
It’s a good idea, Rahul. I will think about it soon.
Best regards
Kawser
Kawser Oct 29, 2016 at 6:15 PM
Spoo,
I would recommend starting with Excel 2016 (or 2013) if you’re just starting. It’s better to be updated with time.
Yes, some are still using Excel 2003 but that is not a good idea. They are missing a lot of new features Microsoft has added to this superb tool later.
Regards
Kawser Oct 29, 2016 at 10:45 AM
Kawser Oct 29, 2016 at 10:45 AM
Glad to know it helped you!
Kawser Oct 28, 2016 at 9:39 PM
Kawser Oct 28, 2016 at 9:30 PM
Hey Tricia,
I checked several times the link. There is no problem with the link π This is not a direct download link. You have to subscribe to get the cheat sheet. I hope you should not be embarrassed with the shared link.
Best regards
Kawser
Kawser Oct 13, 2016 at 12:56 PM
Kawser Oct 13, 2016 at 12:55 PM
Thanks, Mohan!
Kawser Oct 9, 2016 at 11:26 AM
Raja, any working file on that?
Kawser Oct 4, 2016 at 7:48 PM
Thank you, Jaypogi!
Kawser Sep 29, 2016 at 8:20 PM
Many thanks, Scott!
Kawser Sep 20, 2016 at 1:28 AM
Thanks, Sam! I will reach it to Zhiping π
Kawser Sep 15, 2016 at 11:21 AM
Kawser Sep 10, 2016 at 3:41 PM
Armen, files are uploaded. Please let me know whether the concept is clear to you or not!
Kawser Sep 10, 2016 at 3:41 PM
The working files are uploaded. Thanks!
Kawser Sep 6, 2016 at 2:03 AM
Jack, I have notified it to the writer. She will answer soon π
Kawser Aug 25, 2016 at 10:02 AM
If cell A1 holds your string “BCD” and you want to show “AAAAAAABCD” in cell B1, input this formula in cell B1: =REPT(“A”, 10-LEN(A1))&A1.
To solve this problem, you don’t have to use any looping. Or, do you have some other requirement for this problem? Let me know.
Best regards.
Kawser Aug 25, 2016 at 2:01 AM
Cindy,
Millions of thanks for correcting me. It was a grand mistake. Sorry for that. You can buy the course now for just \$10 using the above link.
Best regards
Kawser Aug 21, 2016 at 3:23 PM
Thanks Fina for sharing.
Regards
Kawser Aug 16, 2016 at 3:10 AM
Regards
Kawser Aug 16, 2016 at 3:10 AM
Kawser Aug 6, 2016 at 1:28 AM
Cherie,
Regards
Kawser Aug 1, 2016 at 10:50 AM
Borb,
Regards
Kawser
Kawser Aug 1, 2016 at 10:39 AM
Kateryna,
You will get empty cells if no data is available for the filtering. What criteria did you use to filter data in intervals?
Kawser Aug 1, 2016 at 10:34 AM
Waleed,
Can you upload the working files of your problems? At least a sample file? If possible send an email to this address kawser@exceldemy.com
Kawser Aug 1, 2016 at 10:33 AM
You can do that Ahmed. Thank you.
Regards
Kawser Aug 1, 2016 at 10:32 AM
Baber,
I have sent you an email with instructions. Please check. I hope the email solves the problem.
Regards
Kawser Jul 22, 2016 at 12:13 AM
Thanks for the input, Jomili.
Yes, the formulas are identical as every criterion in the Criteria range is unique. I am going to add the note there π
Kawser Jul 16, 2016 at 1:10 PM
Mike, you did a phenomenal job! Thinking in that way, awesome and the alternative!; Hats Off!
Kawser Jul 16, 2016 at 12:41 PM
First Rate Work, Kadeo! Take my bow π
Kawser Jul 16, 2016 at 12:40 PM
Yes, Menon! It’s a nice solution and more generic!
Kawser Jul 16, 2016 at 12:39 PM
Thanks for the input, specially for the VBA part!
Kawser Jul 16, 2016 at 12:38 PM
Cool π
Kawser Jul 16, 2016 at 12:37 PM
Kawser Jul 5, 2016 at 11:04 AM
Kawser Jun 11, 2016 at 12:26 PM
Kawser Jun 11, 2016 at 12:23 PM
Kawser Apr 5, 2016 at 5:15 PM
Thanks and regards
Kawser Mar 4, 2016 at 10:36 AM
Hi Gina,
Thanks for notifying me. I manually checked the whole process and found some problems. It is fixed now. Please check your inbox. I have sent you the link of the resource page.
Best regards
Kawser
Kawser Feb 24, 2016 at 1:10 AM
Hello Rush,
If you are an email subscriber of my blog, then you would have access to all my data sets (workbooks or anything else) that I use to make the blog posts.
Regards
Kawser Ahmed
Kawser Feb 18, 2016 at 11:07 AM
Hi Roger,
Thanks for your feedback. Nice to meet an Excel Pivot Table ninja π If you like you can share your ninja skills with my audience.
Regards
Kawser Feb 15, 2016 at 12:07 PM
Hi Roger,
Thanks and regards.
Kawser Feb 1, 2016 at 6:37 PM
Kawser Jan 26, 2016 at 12:39 AM
Thanks Orlando for updating me π
Regards
Kawser
Kawser Jan 12, 2016 at 1:15 AM
Vikas,
Regards
Kawser Dec 31, 2015 at 1:30 PM
Puneet, thanks for the comment. I like your presentation in your blog.
Wishing you the best output in year 2016.
Kawser Dec 25, 2015 at 9:58 PM
OZ thanks for your feedback. I am going to write these types of posts like a series. Your comments made me more inspired π
Kawser Dec 2, 2015 at 5:23 PM
Praveen,
Would you please mail me the database and other resources using the mail in the contact me page?
Thanks.
187. Kawser Nov 29, 2015 at 11:59 AM
It’s my pleasure Bob. Thanks for sharing.
Kawser Sep 4, 2015 at 3:23 PM
Narendra, if you feel any way I can update my website, please let me know.
Thanks and regards.
Kawser Aug 19, 2015 at 12:31 AM
Shivaji,
Right now I have no book on Microsoft Access. But just subscribe my Email list using the below link, you will get two books on Excel: 1200+ Macro Examples E-Book, 220+ Excel Keyboard Shortcuts E-Book and other Excel resources.
http://www.eepurl.com/bqFItv
Kawser Aug 19, 2015 at 12:27 AM
Shivaji,
It is my achievement when a person says me that I helped him/her in some way. Thanks. Keep in touch.
Kawser Aug 1, 2015 at 3:07 AM
Glad to hear that my website is helping you in a way.
It’s my pleasure Mohammed Hussain. Keep in touch π
Kawser Feb 27, 2015 at 8:01 PM | 6,944 | 22,139 | {"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} | 2.5625 | 3 | CC-MAIN-2020-40 | latest | en | 0.900913 |
https://alexanderetz.com/category/statistics/understanding-bayes/page/2/ | 1,638,485,550,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362297.22/warc/CC-MAIN-20211202205828-20211202235828-00171.warc.gz | 158,273,372 | 46,061 | # Understanding Bayes: Updating priors via the likelihood
[Some material from this post has been incorporated into a paper to be published in AMPPS]
In a previous post I outlined the basic idea behind likelihoods and likelihood ratios. Likelihoods are relatively straightforward to understand because they are based on tangible data. Collect your data, and then the likelihood curve shows the relative support that your data lend to various simple hypotheses. Likelihoods are a key component of Bayesian inference because they are the bridge that gets us from prior to posterior.
In this post I explain how to use the likelihood to update a prior into a posterior. The simplest way to illustrate likelihoods as an updating factor is to use conjugate distribution families (Raiffa & Schlaifer, 1961). A prior and likelihood are said to be conjugate when the resulting posterior distribution is the same type of distribution as the prior. This means that if you have binomial data you can use a beta prior to obtain a beta posterior. If you had normal data you could use a normal prior and obtain a normal posterior. Conjugate priors are not required for doing bayesian updating, but they make the calculations a lot easier so they are nice to use if you can.
I’ll use some data from a recent NCAA 3-point shooting contest to illustrate how different priors can converge into highly similar posteriors.
## The data
This year’s NCAA shooting contest was a thriller that saw Cassandra Brown of the Portland Pilots win the grand prize. This means that she won the women’s contest and went on to defeat the men’s champion in a shoot-off. This got me thinking, just how good is Cassandra Brown?
What a great chance to use some real data in a toy example. She completed 4 rounds of shooting, with 25 shots in each round, for a total of 100 shots (I did the math). The data are counts, so I’ll be using the binomial distribution as a data model (i.e., the likelihood. See this previous post for details). Her results were the following:
Round 1: 13/25 Round 2: 12/25 Round 3: 14/25 Round 4: 19/25
Total: 58/100
The likelihood curve below encompasses the entirety of statistical evidence that our 3-point data provide (footnote 1). The hypothesis with the most relative support is .58, and the curve is moderately narrow since there are quite a few data points. I didn’t standardize the height of the curve in order to keep it comparable to the other curves I’ll be showing.
## The prior
Now the part that people often make a fuss about: choosing the prior. There are a few ways to choose a prior. Since I am using a binomial likelihood, I’ll be using a conjugate beta prior. A beta prior has two shape parameters that determine what it looks like, and is denoted Beta(α, β). I like to think of priors in terms of what kind of information they represent. The shape parameters α and β can be thought of as prior observations that I’ve made (or imagined).
Imagine my trusted friend caught the end of Brown’s warm-up and saw her take two shots, making one and missing the other, and she tells me this information. This would mean I could reasonably use the common Beta(1, 1) prior, which represents a uniform density over [0, 1]. In other words, all possible values for Brown’s shooting percentage are given equal weight before taking data into account, because the only thing I know about her ability is that both outcomes are possible (Lee & Wagenmakers, 2005).
Another common prior is called Jeffreys’s prior, a Beta(1/2, 1/2) which forms a wide bowl shape. This prior would be recommended if you had extremely scarce information about Brown’s ability. Is Brown so good that she makes nearly every shot, or is she so bad that she misses nearly every shot? This prior says that Brown’s shooting rate is probably near the extremes, which may not necessarily reflect a reasonable belief for someone who is a college basketball player, but it has the benefit of having less influence on the posterior estimates than the uniform prior (since it is equal to 1 prior observation instead of 2). Jeffreys’s prior is popular because it has some desirable properties, such as invariance under parameter transformation (Jaynes, 2003). So if instead of asking about Brown’s shooting percentage I instead wanted to know her shooting percentage squared or cubed, Jeffreys’s prior would remain the same shape while many other priors would drastically change shape.
Or perhaps I had another trusted friend who had arrived earlier and seen Brown take her final 13 shots in warm-up, and she saw 4 makes and 9 misses. Then I could use a Beta(4, 9) prior to characterize this prior information, which looks like a hump over .3 with density falling slowly as it moves outward in either direction. This prior has information equivalent to 13 shots, or roughly an extra 1/2 round of shooting.
These three different priors are shown below.
These are but three possible priors one could use. In your analysis you can use any prior you want, but if you want to be taken seriously you’d better give some justification for it. Bayesian inference allows many rules for prior construction.”This is my personal prior” is a technically a valid reason, but if this is your only justification then your colleagues/reviewers/editors will probably not take your results seriously.
## Updating the prior via the likelihood
Now for the easiest part. In order to obtain a posterior, simply use Bayes’s rule:
$\ Posterior \propto Likelihood \ X \ Prior$
The posterior is proportional to the likelihood multiplied by the prior. What’s nice about working with conjugate distributions is that Bayesian updating really is as simple as basic algebra. We take the formula for the binomial likelihood, which from a previous post is known to be:
$\ Likelihood \ = \ p^x \big(1-p \big)^{n-x}$
and then multiply it by the formula for the beta prior with α and β shape parameters:
$\ Prior \ = \ p^{\alpha-1} \big(1-p \big)^{\beta-1}$
to obtain the following formula for the posterior:
$\ Posterior \ = \ p^x \big(1-p \big)^{n-x} p^{\alpha-1} \big(1-p \big)^{\beta-1}$
With a little bit of algebra knowledge, you’ll know that multiplying together terms with the same base means the exponents can be added together. So the posterior formula can be rewritten as:
$\ Posterior \ = \ p^x p^{\alpha-1}\big(1-p \big)^{n-x} \big(1-p \big)^{\beta-1}$
and then by adding the exponents together the formula simplifies to:
$\ Posterior \ = \ p^{\alpha-1+x} \big(1-p \big)^{\beta-1+n-x}$
and it’s that simple! Take the prior, add the successes and failures to the different exponents, and voila. The distributional notation is even simpler. Take the prior, Beta(α, β), and add the successes from the data, x, to α and the failures, n – x, to β, and there’s your posterior, Beta(α+x, β+n-x).
Remember from the previous post that likelihoods don’t care about what order the data arrive in, it always results in the same curve. This property of likelihoods is carried over to posterior updating. The formulas above serve as another illustration of this fact. It doesn’t matter if you add a string of six single data points, 1+1+1+1+1+1+1 or a batch of +6 data points; the posterior formula in either case ends up with 6 additional points in the exponents.
## Looking at some posteriors
Back to Brown’s shooting data. She had four rounds of shooting so I’ll treat each round as a batch of new data. Her results for each round were: 13/25, 12/25, 14/25, 19/25. I’ll show how the different priors are updated with each batch of data. A neat thing about bayesian updating is that after batch 1 is added to the initial prior, its posterior is used as the prior for the next batch of data. And as the formulas above indicate, the order or frequency of additions doesn’t make a difference on the final posterior. I’ll verify this at the end of the post.
In the following plots, the prior is shown in blue (as above), the likelihood in orange (as above), and the resulting posteriors after Brown’s first 13/25 makes in purple.
In the first and second plot the likelihood is nearly invisible because the posterior sits right on top of it. When the prior has only 1 or 2 data points worth of information, it has essentially no impact on the posterior shape (footnote 2). The third plot shows how the posterior splits the difference between the likelihood and the informed prior based on the relative quantity of information in each.
The posteriors obtained from the uniform and Jeffreys’s priors suggest the best guess for Brown’s shooting percentage is around 50%, whereas the posterior obtained from the informed prior suggests it is around 40%. No surprise here since the informed prior represents another 1/2 round of shots where Brown performed poorly, which shifts the posterior towards lower values. But all three posteriors are still quite broad, and the breadth of the curves can be thought to represent the uncertainty in my estimates. More data -> tighter curves -> less uncertainty.
Now I’ll add the second round performance as a new likelihood (12/25 makes), and I’ll take the posteriors from the first round of updating as new priors for the second round of updating. So the purple posteriors from the plots above are now blue priors, the likelihood is orange again, and the new posteriors are purple.
The left two plots look nearly identical, which should be no surprise since their posteriors were essentially equivalent after only 1 round of data updates. The third plot shows a posterior still slightly shifted to the left of the others, but it is much more in line with them than before. All three posteriors are getting narrower as more data is added.
The last two rounds of updating are shown below, again with posteriors from the previous round taken as priors for the next round. At this point they’ve all converged to very similar posteriors that are much narrower, translating to less uncertainty in my estimates.
These posterior distributions look pretty similar now! Just as an illustration, I’ll show what happens when I update the initial priors with all of the data at once.
As the formulas predict, the posteriors after one big batch of data are identical to those obtained by repeatedly adding multiple smaller batches of data. It’s also a little easier to see the discrepancies between the final posteriors in this illustration because the likelihood curve acts as a visual anchor. The uniform and Jeffreys’s priors result in posteriors that essentially fall right on top of the likelihood, whereas the informed prior results in a posterior that is very slightly shifted to the left of the likelihood.
My takeaway from these posteriors is that Cassandra Brown has a pretty damn good 3-point shot! In a future post I’ll explain how to use this method of updating to make inferences using Bayes factors. It’s called the Savage-Dickey density method, and I think it’s incredibly intuitive and easy to use.
## Notes:
Footnote 1: I’m making a major assumption about the data: Any one shot is exchangeable with any other shot. This might not be defensible since the final ball on each rack is worth a bonus point, so maybe those shots differ systematically from regular shots, but it’s a toy example so I’ll ignore that possibility. There’s also the possibility of her going on a hot streak, a.k.a. having a “hot hand”, but I’m going to ignore that too because I’m the one writing this blog post and I want to keep it simple. There’s also the possibility that she gets worse throughout the competition because she gets tired, but then there’s also the possibility that she gets better as she warms up with multiple rounds. All of these things are reasonable to consider and I am going to ignore them all.
Footnote 2: There is a tendency to call any priors that have very little impact on the posterior “non-informative”, but, as I mentioned in the section on determining priors, uniform priors that seem non-informative in one context can become highly informative with parameter transformation (Zhu & Lu, 2004). Jeffreys’s prior was derived precisely with that in mind, so it carries little information no matter what transformation is applied.
## R Code
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view raw updating.R hosted with ❤ by GitHub
## References:
Jaynes, E. T. (2003). Probability theory: The logic of science. Cambridge University Press.
Lee, M. D., & Wagenmakers, E. J. (2005). Bayesian statistical inference in psychology: Comment on Trafimow (2003). Psychological Review, 112(3), 662-668.
Raiffa, H. & Schlaifer, R. (1961). Applied statistical decision theory. Division of Research, Graduate School of Business Administration, Harvard University.
Zhu, M., & Lu, A. Y. (2004). The counter-intuitive non-informative prior for the Bernoulli family. Journal of Statistics Education, 12(2), 1-10.
# Understanding Bayes: A Look at the Likelihood
[This post has been updated and turned into a paper to be published in AMPPS]
Much of the discussion in psychology surrounding Bayesian inference focuses on priors. Should we embrace priors, or should we be skeptical? When are Bayesian methods sensitive to specification of the prior, and when do the data effectively overwhelm it? Should we use context specific prior distributions or should we use general defaults? These are all great questions and great discussions to be having.
One thing that often gets left out of the discussion is the importance of the likelihood. The likelihood is the workhorse of Bayesian inference. In order to understand Bayesian parameter estimation you need to understand the likelihood. In order to understand Bayesian model comparison (Bayes factors) you need to understand the likelihood and likelihood ratios.
## What is likelihood?
Likelihood is a funny concept. It’s not a probability, but it is proportional to a probability. The likelihood of a hypothesis (H) given some data (D) is proportional to the probability of obtaining D given that H is true, multiplied by an arbitrary positive constant (K). In other words, L(H|D) = K · P(D|H). Since a likelihood isn’t actually a probability it doesn’t obey various rules of probability. For example, likelihood need not sum to 1.
A critical difference between probability and likelihood is in the interpretation of what is fixed and what can vary. In the case of a conditional probability, P(D|H), the hypothesis is fixed and the data are free to vary. Likelihood, however, is the opposite. The likelihood of a hypothesis, L(H|D), conditions on the data as if they are fixed while allowing the hypotheses to vary.
The distinction is subtle, so I’ll say it again. For conditional probability, the hypothesis is treated as a given and the data are free to vary. For likelihood, the data are a given and the hypotheses vary.
## The Likelihood Axiom
Edwards (1992, p. 30) defines the Likelihood Axiom as a natural combination of the Law of Likelihood and the Likelihood Principle.
The Law of Likelihood states that “within the framework of a statistical model, a particular set of data supports one statistical hypothesis better than another if the likelihood of the first hypothesis, on the data, exceeds the likelihood of the second hypothesis” (Emphasis original. Edwards, 1992, p. 30).
In other words, there is evidence for H1 vis-a-vis H2 if and only if the probability of the data under H1 is greater than the probability of the data under H2. That is, D is evidence for H1 over H2 if P(D|H1) > P(D|H2). If these two probabilities are equivalent, then there is no evidence for either hypothesis over the other. Furthermore, the strength of the statistical evidence for H1 over H2 is quantified by the ratio of their likelihoods, L(H1|D)/L(H2|D) (which again is proportional to P(D|H1)/P(D|H2) up to an arbitrary constant that cancels out).
The Likelihood Principle states that the likelihood function contains all of the information relevant to the evaluation of statistical evidence. Other facets of the data that do not factor into the likelihood function are irrelevant to the evaluation of the strength of the statistical evidence (Edwards, 1992, p. 30; Royall, 1997, p. 22). They can be meaningful for planning studies or for decision analysis, but they are separate from the strength of the statistical evidence.
## Likelihoods are meaningless in isolation
Unlike a probability, a likelihood has no real meaning per se due to the arbitrary constant. Only by comparing likelihoods do they become interpretable, because the constant in each likelihood cancels the other one out. The easiest way to explain this aspect of likelihood is to use the binomial distribution as an example.
Suppose I flip a coin 10 times and it comes up 6 heads and 4 tails. If the coin were fair, p(heads) = .5, the probability of this occurrence is defined by the binomial distribution:
$\ P \big(X = x \big) = \binom{n}{x} p^x \big(1-p \big)^{n-x}$
where x is the number of heads obtained, n is the total number of flips, p is the probability of heads, and
$\binom{n}{x} = \frac{n!}{x! (n-x)!}$
Substituting in our values we get
$\ P \big(X = 6 \big) = \frac{10!}{6! (4!)} \big(.5 \big)^6 \big(1-.5 \big)^{4} \approx .21$
If the coin were a trick coin, so that p(heads) = .75, the probability of 6 heads in 10 tosses is:
$\ P \big(X = 6 \big) = \frac{10!}{6! (4!)} \big(.75 \big)^6 \big(1-.75 \big)^{4} \approx .15$
To quantify the statistical evidence for the first hypothesis against the second, we simply divide one probability by the other. This ratio tells us everything we need to know about the support the data lends to one hypothesis vis-a-vis the other. In the case of 6 heads in 10 tosses, the likelihood ratio (LR) for a fair coin vs our trick coin is:
$LR = \Bigg(\frac{10!}{6! (4!)} \big(.5 \big)^6 \big(1-.5 \big)^4 \Bigg) \div \Bigg(\frac{10!}{6! (4!)} \big(.75 \big)^6 \big(1-.75 \big)^4 \Bigg) \approx .21/.15 \approx 1.4$
Translation: The data are 1.4 times as probable under a fair coin hypothesis than under this particular trick coin hypothesis. Notice how the first terms in each of the equations above, i.e., $\frac{10!}{6! (4!)}$, are equivalent and completely cancel each other out in the likelihood ratio.
Same data. Same constant. Cancel out.
The first term in the equations above, $\frac{10!}{6! (4!)}$, details our journey to obtaining 6 heads out of 10. If we change our journey (i.e., different sampling plan) then this changes the term’s value, but crucially, since it is the same term in both the numerator and denominator it always cancels itself out. In other words, the information contained in the way the data are obtained disappears from the function. Hence the irrelevance of the stopping rule to the evaluation of statistical evidence, which is something that makes bayesian and likelihood methods valuable and flexible.
If we leave out the first term in the above calculations, our numerator is L(.5) = 0.0009765625 and our denominator is L(.75) ≈ 0.0006952286. Using these values to form the likelihood ratio we get: 0.0009765625/0.0006952286 ≈ 1.4, as we should since the other terms simply cancelled out before.
Again I want to reiterate that the value of a single likelihood is meaningless in isolation; only in comparing likelihoods do we find meaning.
## Looking at likelihoods
Likelihoods may seem overly restrictive at first. We can only compare 2 simple statistical hypotheses in a single likelihood ratio. But what if we are interested in comparing many more hypotheses at once? What if we want to compare all possible hypotheses at once?
In that case we can plot the likelihood function for our data, and this lets us ‘see’ the evidence in its entirety. By plotting the entire likelihood function we compare all possible hypotheses simultaneously. The Likelihood Principle tells us that the likelihood function encompasses all statistical evidence that our data can provide, so we should always plot this function along side our reported likelihood ratios.
Following the wisdom of Birnbaum (1962), “the “evidential meaning” of experimental results is characterized fully by the likelihood function” (as cited in Royall, 1997, p.25). So let’s look at some examples. The R script at the end of this post can be used to reproduce these plots, or you can use it to make your own plots. Play around with it and see how the functions change for different number of heads, total flips, and hypotheses of interest. See the instructions in the script for details.
Below is the likelihood function for 6 heads in 10 tosses. I’ve marked our two hypotheses from before on the likelihood curve with blue dots. Since the likelihood function is meaningful only up to an arbitrary constant, the graph is scaled by convention so that the best supported value (i.e., the maximum) corresponds to a likelihood of 1.
The vertical dotted line marks the hypothesis best supported by the data. The likelihood ratio of any two hypotheses is simply the ratio of their heights on this curve. We can see from the plot that the fair coin has a higher likelihood than our trick coin.
How does the curve change if instead of 6 heads out of 10 tosses, we tossed 100 times and obtained 60 heads?
Our curve gets much narrower! How did the strength of evidence change for the fair coin vs the trick coin? The new likelihood ratio is L(.5)/L(.75) ≈ 29.9. Much stronger evidence!(footnote) However, due to the narrowing, neither of these hypothesized values are very high up on the curve anymore. It might be more informative to compare each of our hypotheses against the best supported hypothesis. This gives us two likelihood ratios: L(.6)/L(.5) ≈ 7.5 and L(.6)/L(.75) ≈ 224.
Here is one more curve, for when we obtain 300 heads in 500 coin flips.
Notice that both of our hypotheses look to be very near the minimum of the graph. Yet their likelihood ratio is much stronger than before. For this data the likelihood ratio L(.5)/L(.75) is nearly 24 million! The inherent relativity of evidence is made clear here: The fair coin was supported when compared to one particular trick coin. But this should not be interpreted as absolute evidence for the fair coin, because the likelihood ratio for the maximally supported hypothesis vs the fair coin, L(.6)/L(.5), is nearly 24 thousand!
We need to be careful not to make blanket statements about absolute support, such as claiming that the maximum is “strongly supported by the data”. Always ask, “Compared to what?” The best supported hypothesis will be only be weakly supported vs any hypothesis just before or just after it on the x-axis. For example, L(.6)/L(.61) ≈ 1.1, which is barely any support one way or the other. It cannot be said enough that evidence for a hypothesis must be evaluated in consideration with a specific alternative.
## Connecting likelihood ratios to Bayes factors
Bayes factors are simple extensions of likelihood ratios. A Bayes factor is a weighted average likelihood ratio based on the prior distribution specified for the hypotheses. (When the hypotheses are simple point hypotheses, the Bayes factor is equivalent to the likelihood ratio.) The likelihood ratio is evaluated at each point of the prior distribution and weighted by the probability we assign that value. If the prior distribution assigns the majority of its probability to values far away from the observed data, then the average likelihood for that hypothesis is lower than one that assigns probability closer to the observed data. In other words, you get a Bayes boost if you make more accurate predictions. Bayes factors are extremely valuable, and in a future post I will tackle the hard problem of assigning priors and evaluating weighted likelihoods.
I hope you come away from this post with a greater knowledge of, and appreciation for, likelihoods. Play around with the R code and you can get a feel for how the likelihood functions change for different data and different hypotheses of interest.
(footnote) Obtaining 60 heads in 100 tosses is equivalent to obtaining 6 heads in 10 tosses 10 separate times. To obtain this new likelihood ratio we can simply multiply our ratios together. That is, raise the first ratio to the power of 10; 1.4^10 ≈ 28.9, which is just slightly off from the correct value of 29.9 due to rounding.
### R Code
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## Plots the likelihood function for the data obtained ## h = number of successes (heads), n = number of trials (flips), ## p1 = prob of success (head) on H1, p2 = prob of success (head) on H2 ## Returns the likelihood ratio for p1 over p2. The default values are the ones used in the blog post LR <- function(h,n,p1=.5,p2=.75){ L1 <- dbinom(h,n,p1)/dbinom(h,n,h/n) ## Likelihood for p1, standardized vs the MLE L2 <- dbinom(h,n,p2)/dbinom(h,n,h/n) ## Likelihood for p2, standardized vs the MLE Ratio <- dbinom(h,n,p1)/dbinom(h,n,p2) ## Likelihood ratio for p1 vs p2 curve((dbinom(h,n,x)/max(dbinom(h,n,x))), xlim = c(0,1), ylab = "Likelihood",xlab = "Probability of heads",las=1, main = "Likelihood function for coin flips", lwd = 3) points(p1, L1, cex = 2, pch = 21, bg = "cyan") points(p2, L2, cex = 2, pch = 21, bg = "cyan") lines(c(p1, p2), c(L1, L1), lwd = 3, lty = 2, col = "cyan") lines(c(p2, p2), c(L1, L2), lwd = 3, lty = 2, col = "cyan") abline(v = h/n, lty = 5, lwd = 1, col = "grey73") return(Ratio) ## Returns the likelihood ratio for p1 vs p2 }
### References
Birnbaum, A. (1962). On the foundations of statistical inference. Journal of the American Statistical Association, 57(298), 269-306.
Edwards, A. W. (1992). Likelihood, expanded ed. Johns Hopkins University Press.
Royall, R. (1997). Statistical evidence: A likelihood paradigm (Vol. 71). CRC press. | 6,183 | 26,250 | {"found_math": true, "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": 13, "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} | 3.796875 | 4 | CC-MAIN-2021-49 | latest | en | 0.928053 |
https://sciencedocbox.com/Physics/98267015-Algebra-based-physics.html | 1,643,374,230,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00705.warc.gz | 497,827,941 | 29,254 | # Algebra Based Physics
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1 Page 1 of 105
2 Algebra Based Physics Electric Current & DC Circuits Page 2 of 105
3 Electric Current & DC Circuits Circuits Conductors Resistivity and Resistance Circuit Diagrams Measurement Page 3 of 105
4 Circuits Page 4 of 105
5 Electric Current Electric Current is the rate of flow of electric charges (charge carriers) through space. More specifically, it is defined as the amount of charge that flows past a location in a material per unit time. The letter "I" is the symbol for current. I = ΔQ Δt ΔQ is the amount of charge, and Δt is the time it flowed past the location. The current depends on the type of material and the Electric Potential difference (voltage) across it. Page 5 of 105
6 Electric Current A good analogy to help understand Electric Current is to consider water flow. The flow of water molecules is similar to the flow of electrons (the charge carriers) in a wire. Water flow depends on the pressure exerted on the molecules either by a pump or by a height difference, such as water falling off a cliff. Electric current depends on the "pressure" exerted by the Electric Potential difference - the greater the Electric Potential difference, the greater the Electric Current. Page 6 of 105
7 Electric Current The current, I = ΔQ Δt has the units Coulombs per second. The units can be rewritten as Amperes (A). Amperes are often called "amps". 1 A = 1 C/s Page 7 of 105
8 Electric Current We know that if an Electric Potential difference is applied to a wire, charges will flow from high to low potential - a current. However, due to a convention set by Benjamin Franklin, current in a wire is defined as the movement of positive charges (not the electrons which are really moving) and is called "conventional current." Ben didn't do this to confuse future generations of electrical engineers and students. It was already known that electrical phenomena came in two flavors - attractive and repulsive - Ben was the person who explained them as distinct positive and negative charges. Page 8 of 105
9 Electric Current He arbitrarily assigned a positive charge to a glass rod that had been rubbed with silk. He could just as easily called it negative - 50/50 chance. The glass rod was later found to have a shortage of electrons (they were transferred to the silk). So if the glass rod is grounded, the electrons will flow from the ground to the rod. The problem comes in how Electric Potential is defined - charge carriers will be driven from high to low potential - from positive to negative. For this to occur in the glass rod - ground system, the conventional current will flow from the rod to the ground - opposite the direction of the movement of electrons. Page 9 of 105
10 Electric Current To summarize - conventional Electric Current is defined as the movement of positive charge. In wires, it is opposite to the direction of the electron movement. However - in the case of a particle accelerator, where electrons are stripped off of an atom, resulting in a positively charged ion, which is then accelerated to strike a target - the direction of the conventional current is the same as the direction of the positive ions! Page 10 of 105
11 Circuits An electric circuit is an external path that charges can follow between two terminals using a conducting material. For charge to flow, the path must be complete and unbroken. An example of a conductor used to form a circuit is copper wire. Continuing the water analogy, one can think of a wire as a pipe for charge to move through. Page 11 of 105
12 1 12 C of charge passes a location in a circuit in 10 seconds. What is the current flowing past the point? Page 12 of 105
13 2 20 C of charge passes a location in a circuit in 30 seconds. What is the current flowing past the point? Page 13 of 105
14 3 A circuit has 3 A of current. How long does it take 45 C of charge to travel through the circuit? Page 14 of 105
15 4 A circuit has 10 A of current. How long does it take 20 C of charge to travel through the circuit? Page 15 of 105
16 5 A circuit has 10 A of current. How much charge travels through the circuit after 20s? Page 16 of 105
17 6 A circuit has 2.5 A of current. How much charge travels through the circuit after 4s? Page 17 of 105
18 Batteries Each battery has two terminals which are conductors. The terminals are used to connect an external circuit allowing the movement of charge. Positive Terminal Batteries convert chemical energy to electrical energy which maintains the potential difference. The chemical reaction acts like an escalator, carrying charge up to a higher voltage. Negative Terminal Click here for a Battery Voltage Simulation from PhET Page 18 of 105
19 Reviewing Basic Circuits The circuit cannot have gaps. The bulb had to be between the wire and the terminal. A voltage difference is needed to make the bulb light. The bulb still lights regardless of which side of the battery you place it on. As you watch the video,observations and the answers to the questions below. What is going on in the circuit? What is the role of the battery? How are the circuits similar? different? Click here for video using the circuit simulator from PhET Page 19 of 105
20 Batteries and Current The battery pushes current through the circuit. A battery acts like a pump, pushing charge through the circuit. It is the circuit's energy source. Charges do not experience an electrical force unless there is a difference in electrical potential (voltage).therefore, batteries have a potential difference between their terminals. The positive terminal is at a higher voltage than the negative terminal. click here for a video from Veritasium's Derek on current How will voltage affect current? Page 20 of 105
21 Conductors Page 21 of 105
22 Conductors Some conductors "conduct" better or worse than others. Reminder: conducting means a material allows for the free flow of electrons. The flow of electrons is just another name for current. Another way to look at it is that some conductors resist current to a greater or lesser extent. We call this resistance, R. Resistance is measured in ohms which is noted by the Greek symbol omega (Ω) How will resistance affect current? Click here to run another PhET simulation Page 22 of 105
23 Current vs Resistance & Voltage Raising resistance reduces current. Raising voltage increases current. We can combine these relationships in what we call "Ohm's Law". Another way to write this is that: R = V I ORV = IR I = V R You can see that one Ω = V A click here for a Veritasium music video on electricity Page 23 of 105
24 7 A flashlight has a resistance of 25 Ω and is connected by a wire to a 120 V source of voltage. What is the current in the flashlight? Page 24 of 105
25 8 A flashlight has a resistance of 30 Ω and is connected by a wire to a 90 V source of voltage. What is the current in the flashlight? Page 25 of 105
26 9 What is the current in a wire whose resistance is 3 Ω if 1.5 V is applied to it? Page 26 of 105
27 10How much voltage is needed in order to produce a 0.70 A current through a 490 Ω resistor? Page 27 of 105
28 11 How much voltage is needed in order to produce a 0.5 A current through a 150 Ω resistor? Page 28 of 105
29 12 What is the resistance of a rheostat coil, if 0.05 A of current flows through it when 6 V is applied across it? Page 29 of 105
30 13 What is the resistance of a rheostat coil, if 20 A of current flows through it when 1000 V is applied across it? Page 30 of 105
31 Electrical Power Power is defined as work per unit time P = W t if W = QV then substitute: P = QV t if I = Q t then substitute: P = IV What happens if the current is increased? What happens if the voltage is decreased? Page 31 of 105
32 Electrical Power Let's think about this another way... The water at the top has GPE & KE. As the water falls, it loses GPE and the wheel gets turned, doing work.when the water falls to the bottom it is now slower, having done work. Page 32 of 105
33 Electrical Power Electric circuits are similar. A charge falls from high voltage to low voltage. In the process of falling energy may be used (light bulb, run a motor, etc). What is the unit of Power? Page 33 of 105
34 Electrical Power How can we re-write electrical power by using Ohm's Law? (electrical power) P = IV (Ohm's Law) I = V R P = VV R P = V 2 R Page 34 of 105
35 Electrical Power Is there yet another way to rewrite this? I = V can be rewritten as V = IR. R (electrical power) P = IV (Ohm's Law) V = I R We can substitute this into Power P = I(IR) P = I 2 R Page 35 of 105
36 Batteries D, C, AA, & AAA have the same voltage, however they differ in the amount of power they deliver. D C AAA AA 1.5 V 9 V For instance, D batteries can deliver more current and therefore more power. Page 36 of 105
37 14 A toy car's electric motor has a resistance of 17 Ω; find the power delivered to it by a 6-V battery. Page 37 of 105
38 15 A toy car's electric motor has a resistance of 6 Ω; find the power delivered to it by a 7-V battery. Page 38 of 105
39 16 What is the power consumption of a flash light bulb that draws a current of 0.28 A when connected to a 6 V battery? Page 39 of 105
40 17 What is the power consumption of a flash light bulb that draws a current of 0.33 A when connected to a 100 V battery? Page 40 of 105
41 18 A 30Ω toaster consumes 560 W of power: how much current is flowing through the toaster? Page 41 of 105
42 19 A 50Ω toaster consumes 200 W of power: how much current is flowing through the toaster? Page 42 of 105
43 20 When 30 V is applied across a resistor it generates 600 W of heat: what is the magnitude of its resistance? Page 43 of 105
44 21 When 100 V is applied across a resistor it generates 200 W of heat: what is the magnitude of its resistance? Page 44 of 105
45 "Pipe" size How could the wire in the circuit affect the current? If wire is like a pipe, and current is like water that flows through the pipe... if there were pipes with water in them, what could we do to the pipes to change the speed of the water (the current)? Page 45 of 105
46 ** Resistivity and Resistance Page 46 of 105
47 ** Resistivity & Resisitance Every conductor "conducts" electric charge to a greater or lesser extent. The last example also applies to conductors like copper wire. Decreasing the length (L) or increasing the cross-sectional area (A) would increase conductivity. Also, the measure of a conductor's resistance to conduct is called its resistivity. Each material has a different resistivity. Resistivity is abbreviated using the Greek letter rho (ρ). Combining what we know about A, L, and ρ, we can find a conductor's total resistance. R = ρl A Page 47 of 105
48 ** Resistivity & Resisitance Resistance, R, is measured in Ohms (Ω). Ω is the Greek letter Omega. Cross-sectional area, A, is measured in m 2 Length, L, is measured in m R = ρl A Resistivity, ρ, is measured in Ωm How can we define A for a wire? Page 48 of 105
49 ** Resisitance ρ = RA L What is the resistance of a good conductor? Low; low resistance means that electric charges are free to move in a conductor. Click here for a PhET simulation about Resistance Page 49 of 105
50 ** Resistivities of Common Conductors Material Silver Copper Gold Aluminum Tungsten Iron Platinum Mercury Nichrome Resistivity (10-8 Ωm) Page 50 of 105
51 22 ** Rank the following materials in order of best conductor to worst conductor. A Iron, Copper, Platinum B Platinum, Iron, Copper C Copper, Iron, Platinum Material Resistivity (10-8 Ωm) Silver 1.59 Copper 1.68 Gold 2.44 Aluminum 2.65 Tungsten 5.60 Iron 9.71 Platinum 10.6 Mercury 98 Nichrome Page 51 of 105
52 23 What is the resistance of a 2 m long copper wire whose cross-sectional area of 0.2 mm 2? ** Page 52 of 105
53 24 ** An aluminum wire with a length of 900 m and a cross-sectional area of 10 mm 2 has a resistance of 2.5 Ω. What is the resistivity of the wire? Page 53 of 105
54 25 What diameter of 100 m long copper wire would ** have a resistance of 0.10 Ω? Page 54 of 105
55 26 ** What is the cross-sectional area of a 10Ω copper wire of length is meters? Page 55 of 105
56 27 ** What is the length of a 10 Ω copper wire whose diameter is 3.2 mm? Page 56 of 105
57 Circuit Diagrams Page 57 of 105
58 Circuit Diagrams Drawing realistic pictures of circuits can be very difficult. For this reason, we have common symbols to represent each piece. Resistor Battery Wire *Note: Circuit diagrams do not show where each part is physically located. Page 58 of 105
59 Circuit Diagrams Draw a simple circuit that has a 9 V battery with a 3 Ω resistor across its terminals. What is the magnitude and direction of the current? Conventional current flows from the positive terminal to the negative terminal. Page 59 of 105
60 Circuit Diagrams There are two ways to add a second resistor to the circuit. Series Parallel R 1 R 2 R 1 R 2 V All charges must move through both resistors to get to the negative terminal. V Charges pass through either R 1 or R 2 but not both. Page 60 of 105
61 Circuit Diagrams Are the following sets of resistors in series or parallel? R 1 R 1 V R 2 V R 2 Page 61 of 105
62 Equivalent Resistance Resistors and voltage from batteries determine the current. Circuits can be redrawn as if there were only a single resistor and battery.by reducing the circuit this way, the circuit becomes easier to study. The process of reducing the resistors in a circuit is called finding the equivalent resistance (R eq ). R 1 R 2 V Page 62 of 105
63 Series Circuits: Equivalent Resistance R 1 R 2 What happens to the current in the circuit to the right? V Page 63 of 105
64 Series Circuits: Equivalent Resistance R 1 R 2 What happens to the voltage as it moves around the circuit? V Page 64 of 105
65 Series Circuits: Equivalent Resistance If V = V 1 + V 2 + V substitute Ohm's Law solved for V is: V = IR IR = I 1 R 1 + I 2 R 2 + I 3 R 3 IR = IR 1 + IR 2 + IR 3 but since current (I) is the same everywhere in a series circuit, I = I 1 = I 2 = I 3 R eq = R 1 + R 2 + R Now divide by I To find the equivalent resistance (R eq ) of a series circuit, add the resistance of all the resistors.if you add more resistors to a series circuit, what happens to the resistance? Page 65 of 105
66 28 What is the equivalent resistance in this circuit? R 1 = 5Ω R 2 = 3Ω V = 9 V Page 66 of 105
67 29 What is the total current at any spot in the circuit? R 1 = 5Ω R 2 = 3Ω V = 9 V Page 67 of 105
68 30 What is the voltage drop across R 1? R 1 = 5Ω R 2 = 3Ω V = 9 V Page 68 of 105
69 31 What is the voltage drop across R 2? R 1 = 5Ω R 2 = 3Ω V = 9 V hint: A good way to check your work is to see if the voltage drop across all resistors equals the total voltage in the circuit. Page 69 of 105
70 32 How much power is used by R 1? R 1 = 5Ω R 2 = 3Ω V = 9 V Page 70 of 105
71 33 What is the equivalent resistance in this circuit? R 1 = 10Ω R 2 = 20Ω V = 9 V Page 71 of 105
72 34 What is the total current at any spot in the circuit? R 1 = 10Ω R 2 = 20Ω V = 9 V Page 72 of 105
73 35 What is the voltage drop across R 1? R 1 = 10Ω R 2 = 20Ω V = 9 V Page 73 of 105
74 36 What is the voltage drop across R 2? R 1 = 10Ω R 2 = 20Ω V = 9 V Page 74 of 105
75 37 How much power is used by R 1? R 1 = 10Ω R 2 = 20Ω V = 9 V Page 75 of 105
76 38 How much power is used by R 2? R 1 = 10Ω R 2 = 20Ω V = 9 V Page 76 of 105
77 Parallel Circuits: Equivalent Resistance R 1 What happens to the current in the circuit to the right? R 2 V Page 77 of 105
78 Parallel Circuits: Equivalent Resistance R 1 What happens to the voltage as it moves around the circuit? R 2 V Page 78 of 105
79 Parallel Circuits: Equivalent Resistance If I = I 1 + I 2 + I 3 Rewrite Ohm's Law for I and substitute for each resistor V R = V 1 R 1 + V 2 R 2 + V 3 R 3 Also, since V = V 1 = V 2 = V 3 so we can substitute V for any other voltage V R = V R 1 + V R 2 + V R 3 Voltage is a common factor, so factor it out! V R = V( 1 R R = + + R eq R 2 R 1 R 3 1 R 3 ( Divide by V to eliminate voltage from the equation. If you add more resistors in parallel, what will happen the to resistance of the circuit? Page 79 of 105
80 39 What is the equivalent resistance in the circuit? R 1 = 3Ω R 2 = 6Ω V = 18V Page 80 of 105
81 40 What is the voltage at any spot in the circuit? R 1 = 3Ω R 2 = 6Ω V = 18V Page 81 of 105
82 41 What is the current through R 1? R 1 = 3Ω R 2 = 6Ω V = 18V Page 82 of 105
83 42 What is the current through R 2? R 1 = 3Ω R 2 = 6Ω V = 18V Page 83 of 105
84 43 What is the power used by R 1? R 1 = 3Ω R 2 = 6Ω V = 18V Page 84 of 105
85 44 What is the power used by R 2? R 1 = 3Ω R 2 = 6Ω V = 18V Page 85 of 105
86 45 What is the equivalent resistance in the circuit? Page 86 of 105
87 46 What is the voltage at any spot in the circuit? Page 87 of 105
88 47 What is the current through R 1? Page 88 of 105
89 48 What is the current through R 2? Page 89 of 105
90 49 What is the power used by R 1? Page 90 of 105
91 50 What is the power used by R 1? Page 91 of 105
92 51 What is the power used by R 2? Page 92 of 105
93 Measurement Page 93 of 105
94 Voltmeter Voltage is measured with a voltmeter. Voltmeters are connected in parallel and measure the difference in potential between two points. Since circuits in parallel have the same voltage, and a voltmeter has very high resistance, very little current passes through it. This means that it has little effect on the circuit. Page 94 of 105
95 Ammeter Current is measured using an ammeter. Ammeters are placed in series with a circuit. In order to not interfere with the current, the ammeter has a very low resistance. Page 95 of 105
96 Multimeter Although there are separate items to measure current and voltage, there are devices that can measure both (one at a time). These devices are called multimeters.multimeters can also measure resistance. Click here for a PhET simulation on circuits Page 96 of 105
97 52 A group of students prepare an experiment with electric circuits. Which of the following diagrams can be used to measure both current and voltage? L A B C D E Page 97 of 105
98 * Electromotive Force R eq A battery is a source of voltage AND a resistor. + r E _ Each battery has a source of electromotive force and internal resistance. Electromotive force (EMF) is the process that carries charge from low to high voltage. Another way to think about it is that EMF is the voltage you measure when no resistance is connected to the circuit. Page 98 of 105
99 * Electromotive Force R eq + r E _ Terminal voltage (V T ) is the voltage measured when a voltmeter is across its terminals. If there is no circuit attached, no current flows, and the measurement will equal the EMF. If however a circuit is attached, the internal resistance will result in a voltage drop, and a smaller terminal voltage. (E - Ir) Page 99 of 105
100 * Terminal Voltage R eq We say that the terminal voltage is: + r E _ V T = E - Ir Maximum current will occur when there is zero external current. When solving for equivalent resistance in a circuit, the internal resistance of the battery is considered a series resistor. R EQ = R int + R ext Page 100 of 105
101 53* When the switch in the circuit below is open, the voltmeter reading is referred to as: A EMF B Current C Power D Terminal Voltage E Restivity Page 101 of 105
102 54* When the switch in the circuit below is closed, the voltmeter reading is referred to as: A Terminal Voltage B EMF C Current D Resistance E Power Page 102 of 105
103 55* A 6V battery, whose internal resistance 1.5 Ω is connected in series to a light bulb with a resistance of 6.8 Ω. What is the current in the circuit? Page 103 of 105
104 56* A 6V battery, whose internal resistance 1.5Ω is connected in series to a light bulb with a resistance of 6.8Ω. What is the terminal voltage of the battery? Page 104 of 105
105 57* A 25 Ω resistor is connected across the terminals of a battery whose internal resistance is 0.6 Ω. What is the EMF of the battery if the current in the circuit is 0.75 A? Page 105 of 105
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A. Law of electric charges. Electricity and Electromagnetism SOL review Scan for a brief video The law of electric charges states that like charges repel and opposite charges attract. Because protons and
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IGCSE Physics 0625 notes Topic 4: Static Electricity 1 TOPIC 4 STATIC ELECTRICITY ELECTRICITY: Electricity is the flow of electrical charges or power. The charges could be in the form of electrons or ions. | 11,857 | 47,841 | {"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} | 4 | 4 | CC-MAIN-2022-05 | latest | en | 0.943235 |
https://www.physicsforums.com/threads/amazing-probability-help-me-prove-it.154331/ | 1,670,326,058,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711077.50/warc/CC-MAIN-20221206092907-20221206122907-00509.warc.gz | 997,391,963 | 18,880 | # Amazing probability, help me prove it!
lando45
"Imagine there is some rare disease that only affects 1 in 10,000 people, but if you get it, it's completely fatal and there is no cure, death is guaranteed. One day you get really paranoid and decide to take a test to see if you have the disease or not, so you go down to your local doctor and he says there's a simple test you can take which is 99% accurate in its diagnosis. So you take the test, then go home and wait. A couple weeks later a letter from the clinic comes through your door and you open it and it says the dreaded words "You have the disease" - so naturally, you would completely freak out, right? However, if you think about it, despite receiving this letter, you still have less than a 1% chance of actually having the disease. Even tho the test is 99% accurate and affects just 1 in 10,000 people."
When I first heard this I couldn't believe it, but after much contemplation it is most definitely true. Imagine if we take a sample size of 1,000,000 people.
As 1 in 10,000 get the disease, this means 100 from our sample size of 1m would have the disease.
Of these 100 people, 99 would be correctly diagnosed as having the disease and 1 would be incorrectly told that they don't have the disease.
There would be 999,900 remaining people who do not have the disease. However, as the test is only 99% accurate, 1% of these 999,900 (i.e. 9,999 people) would receive letters saying they have the disease.
So in total, 9,999 + 99 = 10,098 would be told they have the disease when in fact only 99 of them do. And 99 is less than 1% of 10,098.
So this would seem to be true. However, I brought this up on another forum, and some of the users have brought up some interesting counterpoints. Maybe you could read through the topic and let me know your thoughts. I've said basically everything I can think of to prove it!
Your description is probably correct. However, you have implicitly assumed a false positive probability as 1% while a false negative is also 1%. In general, there is no reason for both to be the same.
lando45
Please can you clarify? I'm not sure what the difference between a false positive probability and a false negative probability is.
The you know that there is 1% chance of the test to be wrong, it IS 1% chance that the test will be wrong. If you calculate that it wouldn't it wouldn't say 1%, would it?
That's just my point of view. Should really the amount of people that it is tested on have any effect on the probability chance of the test?
Homework Helper
The you know that there is 1% chance of the test to be wrong, it IS 1% chance that the test will be wrong. If you calculate that it wouldn't it wouldn't say 1%, would it?
That's just my point of view. Should really the amount of people that it is tested on have any effect on the probability chance of the test?
You are assuming that it is correct to say that "there is 1% chance of the test to be wrong." There are two different ways a test can be wrong- it can give a false positive (saying you have the disease when you haven't) or it can give a false negative (saying you do not have the disease when you have). There is no reason to assume those two have the same chance.
Werg22
You simply changed sets. There's nothing to that.
Staff Emeritus
Halls is correct. There is a very big difference between false positives and false negatives. There is no a-priori reason for these error rates to be equal, nor is there a way to minimize both types of errors simultaneously. BTW, false positives and false negatives are also known as type I and type II errors, respectively.
Initial screening tests for some disease are often designed to have an extremely low false negative rate and a not-so-low false positive rate. In other words, the disease is most likely absent if the test comes up negative. A positive result merely means you have to take some other test again.
The initial post is correct. This is a well-known statistical paradox.
Homework Helper
D. H.'s point is a good one- you can reduce the number of "false negatives" by loosening the conditions underwhich you report a positive result (for example if you report a positive result no matter what the result of the test you would NEVER report any negative and so have NO false negatives!) but that invariably increases the number of false positives. A false positive can cause some emotional trauma but a false negative could result in a disease not being treated! Most test procedures accept more false positives in order to reduce the number of false negatives. Yes, if a disease is rare to start with, it is not terribly surprizing, even with a test that has a low percentage of false positives, that most "positive results" sjhould be false. There just aren't enough "true positives".
MichaelShane
Question: A wholesalers buys items at £25 each. The wholesalers sells this in batches of 8 to retailers at £36 per item. It is know that, on average , 25 items in a thousand are faulty.
(A) Construct the probability distribution table for the number of faulty items in a batch of 8 items. All probability values must be rounded off to 5 decimal places.
I know i have to do a binomial distirbution for Part (A) lots of thanks to jerry and so can anybody help me with Part B of the same question.
(B) For any faulty item in a batch the wholesaler promises to give 2 extra items, free of further charge, to the retailer who then accepts full responsibility for the batch.
Extend your table of part (A) to indicate the wholesalers profit for each possible value of the number of faulty items in a batch of 8. Extend your table further to calculate the wholesalers expected profit on a batch of 8 items.
Kindly help me in solving the above probability problems, help will be highly appreciated.
Thank you very much in advance.
But doesn't the fact that it can actually give you a positive result even though you do not have the disease even the factors out?
vamfun
This could be worked with Baysian conditional prob.
Define events:
N=no disease P(N)=1/10000
Y=have disease P(Y)=1-P(N)
Y'=test positive P(Y')=P(Y'/N)*P(N)+P(Y'/Y)*P(Y)
N'=test negitive
WRONG TEST RESULT P(Y'/N) =1/100= P(N'/Y) note: not always equal but assumed here.
GOOD TEST RESULT P(Y'/Y)=.99=P(N'/N)
COMPUTE P(Y')= 1/100*(9999/10000)+.99*(1/10000) =100.98/10000
From Bayes; P(Y/Y')*P(Y')=P(Y'/Y)*P(Y)
We want P(Y/Y')= P(Y'/Y)*P(Y)/P(Y')
=.99*(1/10000)/(100.98/10000)
=.99/100.98=1/102
So less than 1% chance.
Homework Helper
Question: A wholesalers buys items at £25 each. The wholesalers sells this in batches of 8 to retailers at £36 per item. It is know that, on average , 25 items in a thousand are faulty.
(A) Construct the probability distribution table for the number of faulty items in a batch of 8 items. All probability values must be rounded off to 5 decimal places.
I know i have to do a binomial distirbution for Part (A) lots of thanks to jerry and so can anybody help me with Part B of the same question.
(B) For any faulty item in a batch the wholesaler promises to give 2 extra items, free of further charge, to the retailer who then accepts full responsibility for the batch.
Extend your table of part (A) to indicate the wholesalers profit for each possible value of the number of faulty items in a batch of 8. Extend your table further to calculate the wholesalers expected profit on a batch of 8 items.
Kindly help me in solving the above probability problems, help will be highly appreciated.
Thank you very much in advance.
PLEASE do not "hijack" another person's thread for new questions that have nothing to do with the orginal question! Start a thread of your own!
Homework Helper
But doesn't the fact that it can actually give you a positive result even though you do not have the disease even the factors out?
No- it is quite possible for one probability to be 5% and the other 10%. Those do not "even" out!
Simple example of difference between falso positive and false negative.
Diabetes type 2 test (overnight fasting)
Current ADA standard <100 not diabetic, 100 - 126 pre-diabetic - >126 diabetic
Older standard < 110 not diabetic, 110 - 140 pre-diabetic, >140 diabetic
As you can see the current test will give less false negatives but more false psoitives as compared to the older test. | 1,994 | 8,345 | {"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} | 3.875 | 4 | CC-MAIN-2022-49 | latest | en | 0.985661 |
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# Calculate the molarity of the ascorbic acid solution: (a) Mass of ascorbic acid used:1 (b) Moles of ascorbic acid (MW=176.1 g/mol):not sure (0.
1. Calculate the molarity of the ascorbic acid solution:
(a) Mass of ascorbic acid used:0.1
(b) Moles of ascorbic acid (MW=176.1 g/mol):not sure (0.000568g/moL)
(c) Volume of solution (mL): 100.01mL
(d) Ascorbic acid concentration (mol/L):
2. For each titration, record and calculate the following:
(a) Volume of iodine solution added (mL):
(b) Concentration of the iodine solution:
3. Calculate the iodine concentration, using the formula M1*V1 = M2*V2. Average your results and report one concentration for your standardized iodine solution.
Assignment 1 of Procedure 2
1. For each titration of the fresh orange juice, record and calculate the following:
(a) Volume of iodine solution added (mL): .1mL
(b) Concentration of the ascorbic acid in the juice
2. Calculate the average ascorbic acid concentration for the fresh orange juice, using the formula M1*V1 = M2*V2.
3. For each titration of the week-old orange juice, record and calculate the following:
(a) Volume of iodine solution added (mL):2 drops at a time from burette into a erlenmeyer flask
(b) Concentration of the ascorbic acid in the juice: new orange juice- 8mL of iodine from burette into flask......week old orange juice- 4mL
4. Calculate the average ascorbic acid concentration for the week-old orange juice, using the formula M1*V1 = M2*V2..
5. Report the average amount of ascorbic acid in the 2 samples of commercial orange juice in units of "mg per mL" of juice. The molecular weight of ascorbic acid is 176.12.
6. The minimum daily requirement for vitamin C is 60 mg per day. What percentage of this requirement is in one cup (200 mL) of fresh and week-old orange juice?
7. What happens to the ascorbic acid in orange juice over time? (hint: oxygen makes up 20% of our air.)
Assignment 2 of Procedure 2
Conclusion
Summarize the experiment in one or two paragraphs. Restate the overall purpose of the experiment. What were your results (from each procedure)? What are possible sources of error? What did you enjoy learning about this experiment?
DOES THIS HELP ANY WITH FIGURING OUT THE PROBLEMS?
PLEASE NOTE: Titration requires several steps in order to obtain exact results.
The prepared iodine solution on the Chemicals shelf with a stated concentration of 0.0015M is standardized (confirming the concentration) by performing the following titration:
1. Take a clean volumetric flask from the Glassware shelf and place it on the workbench.
3. Fill the volumetric flask with water. (This is done by adding the water normally and checking the "Fill to the Mark" option instead of entering an amount.) This is the most precise way of making a 100 mL solution. Record the amount of ascorbic acid used and the total volume prepared.
4. Take a 150 Erlenmeyer flask from the Glassware shelf and place it on the workbench.
5. Pour 20 mL of the ascorbic acid solution into the flask.
7. Take a burette from the Glassware shelf and place it on the workbench.
8. Fill the burette with 50 mL iodine solution, with an aproximately known concentration of 0.015M.
9. Drag the Erlenmeyer flask and drop it on the lower half of the burette.
10. Open the Properties window, click back on the burette, and then click the Pushpin icon in the blue bar of the Properties window to lock its functions to the burette.
11. Titrate the ascorbic acid by adding iodine until the solution in the Erlenmeyer flask turns dark blue.
12. You should do a rough titration by adding the iodine solution 1-2 mL at a time in order to quickly find the range in which the endpoint is reached. You can then do a fine titration one or two times in order to obtain exact results.
13. When standardizing a solution, you should repeat the titration so that your results are within a few drops of each other. The results are then averaged to determine the precise molarity.
Procedure 2 ( click to view assignments for this procedure )
1. Determine the ascorbic acid concentration in commercial orange juice, from a freshly opened container and from a container that was opened one week ago.
2. Prepare a sample of orange juice from the new container by adding 40 mL of the juice to a clean Erlenmeyer flask. Add 1 mL of starch indicator.
(Please excuse the very light color of the OJ - the pulp was REALLY filtered out)
3. Titrate the orange juice with the standardized iodine solution. First do a rough titration and then two accurate titrations. Record the volume of iodine delivered in each titration and then refill the burette.
4. Repeat steps 2-3 with the week-old orange juice.
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http://www.shmoop.com/circles/central-angles.html | 1,464,625,654,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464051036499.80/warc/CC-MAIN-20160524005036-00177-ip-10-185-217-139.ec2.internal.warc.gz | 789,365,386 | 16,742 | Central Angles
Among the most important applications of the circle to human progress has been circle-based (disk-shaped) food items. This is a family of foods that includes, among other things, cookies, bagels, doughnuts, pies, and pizzas. (Some consider pizzas to be a subset of pies, but pepperoni pie doesn't sound nearly as delicious as pepperoni pizza.)
Critical to the development of circle-based foods was the concept of the central angle. Given ⊙O, an angle is a central angle of ⊙O if its vertex is at O. Simple as that.
Here, ∠1 is a central angle of ⊙O. So is ∠2. Every central angle has a buddy. The measure of a central angle and its buddy angle add up to 360°, the number of degrees in a full circle. In other words, buddy angles complete each other. Aww.
With the discovery of the central angle, people could easily share circle-based foods as they saw fit. If you wanted to divide a pizza among five people, you could cut slices based on central angles of 360° ÷ 5 = 72°.
Sample Problem
It's your birthday and you'll cry if you want to. There's no reason to cry though, since you got a massive chocolatey birthday cake. If there are 20 people total at your party, at what central angle should you cut the cake?
Regardless of how big the cake is, it has a central angle of 360° because it's a circle. If there are 20 people, we should cut everyone a slice that is 360° ÷ 20 = 18° in measure. Time to bust out the protractor.
Sample Problem
All circles are similar. True or false?
It's been a while since we talked about similarity, so here's a quick refresher: similarity exists when two figures are the same shape (all their angles are equal), but not the same size. This also means they can be carried onto each other using similarity transformations (translation, reflection, rotation, and dilation).
Circles have 360° total. That won't change ever, so we took care of the angle requirement (as well as the rotation and reflection requirements). If a single point and a length defines a circle, we can always translate the point to a different location and dilate the length so it matches another.
In other words, all circles are similar to each other because any similarity transformation can move one onto the other. In fact, only dilation and translation are needed, so we leave reflection and rotation at home. | 534 | 2,344 | {"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} | 4.5625 | 5 | CC-MAIN-2016-22 | longest | en | 0.951558 |
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### Grade 6 - Mathematics1.41 LCM - Division Method
Example: Find the L.C.M. of 60, 75, 80 by using the division method? Therefore, the L.C.M. of 60, 75, 80 is = 5 x 3 x 2 x 2 x 1 x 5 x 4 = 1200. Directions: Find the L.C.M. of the following numbers using the division method. Also write at least 10 examples of your own.
Name: ___________________Date:___________________
### Grade 6 - Mathematics1.41 LCM - Division Method
Q 1: Find the L.C.M. of 300, 600.300900600 Question 2: This question is available to subscribers only! | 197 | 683 | {"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} | 2.84375 | 3 | CC-MAIN-2019-30 | longest | en | 0.752991 |
http://www.chess.com/forum/view/fun-with-chess/knight-swap-puzzle | 1,448,808,865,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398458511.65/warc/CC-MAIN-20151124205418-00301-ip-10-71-132-137.ec2.internal.warc.gz | 360,019,107 | 15,477 | # Knight swap puzzle
• 4 years ago · Quote · #1
Starting with a 4 x 3 board.
The object is to swap the white knights for black ones and vice versa in 16 moves. Moving a knight one time counts as one move and alternating white and black knight moves is not required.
• 4 years ago · Quote · #2
[COMMENT DELETED]
• 4 years ago · Quote · #3
Slightly confused here... like this?
• 4 years ago · Quote · #4
Or like this?
• 4 years ago · Quote · #5
akajake wrote:
Or like this?
That's the idea. Nothing is captured, but you might have to solve this on a real board since you are not allowed to move the same color twice in a row. I don't know if your solution is optimum or not when alternating moves is forced.
Thanks for trying.
• 4 years ago · Quote · #6
wbport wrote:
akajake wrote:
Or like this?
That's the idea. Nothing is captured, but you might have to solve this on a real board since you are not allowed to move the same color twice in a row. I don't know if your solution is optimum or not when alternating moves is forced.
Thanks for trying.
Good puzzle, fun to do. And I thought I did alternate moves. Feel free to correct me or send me the solution if you want. Again, thanks for the brain teaser!
• 4 years ago · Quote · #7
An isomorph is helpful in solving this puzzle. Martin Gardner noted in "aha Insight" that the eight outside squares on a 3x3 board make up a closed knight's tour and his isomorph consisted of interlocking rings, but this model may be more helpful:
The six shaded squares are a closed knight tour as are the unshaded squares. When unfolded into an isomorph there are two rings connected by the a2-c3 and c2-a3 links. Each of the rings contain two knights of one color and one knight of the opposite color.
To solve this one knight each is swapped with the other ring while the other knights are repositioned within their original ring. Moving a knight one time counts as one move and any knight can move at any time.
eliminate rotations and/or reflections in the final solution.
**Sorry, could not get the isomorph to display**
This puzzle appeared in "aha! Insight" by Martin Gardner in 1978 with an 18 move solution which was not revealed. Two other readers and I found a 16 move solution which Mr. Gardner discussed in his Scientific American column in early 1978 or
1979, but I lost my copy of it.
Start with a1-c2-b3. It takes two moves for any knight to go from one side to the other except that when c1 becomes open, the a4 knight zigzags to c1 in 3 moves. When b1 opens up the Nc4 needs to get there but the Na3 needs to back up to c2 first. The last two moves are c2-a3-c4 and Voila! it's completed.
• 4 years ago · Quote · #8
Ignore all king moves, they are inserted to show how the solution works w/o a set.
• 4 years ago · Quote · #9
How was my answer not correct? I see your solution, and I did the exact same thing without using the kings... In fewer moves too.
• 4 years ago · Quote · #10
akajake wrote:
How was my answer not correct? I see your solution, and I did the exact same thing without using the kings... In fewer moves too.
It wasn't in fewer moves, though; your answer used 22 moves, vs. 16 moves used in wbport's solution. I think the point you missed is that the king moves (in the diagram from post #8) should be ignored completely, because they are not even a part of the sequence. The only reason for the king moves is that Chess.com's board editor requires them (so that Black and White alternate moves, as they must in standard chess). But since they aren't required for this puzzle, the actual solution given was:
Na1-c2, Nc2-a3, Nb4-c2, Nc2-a1, Nc1-a2, Na2-b4, Na4-c3, Nc3-a2, Na2-Nc1, Na3-c2, Nb1-c3, Nc3-a4, Nc4-a3, Na3-b1, Nc2-a3, Na3-c4 (which is only 16 knight moves).
• 4 years ago · Quote · #11
cobra91: Exactly right. I had intended in the 8th move of post #8 to move the Nb1 to a4, THEN drop the Na3 back to c2 to allow the c4N to get to b1, but those moves didn't interfere with each other and I let them stand.
This is as close as I can get to an isomorph:
a4-c3-a2-c1
b2-b1-b4-b3
c4-a3-c2-a1
One square vertical moves are allowed on any file but one square horizontal moves are allowed on the 1st and 3rd ranks only. I'm using rooks instead of knights but label the squares as above. | 1,171 | 4,317 | {"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} | 3.5625 | 4 | CC-MAIN-2015-48 | latest | en | 0.925556 |
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# Ch11HWSolutionE11-3 - CHAPTER 11 Strategic cost management...
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CHAPTER 11 Strategic cost management 11–3 1. Supplier cost: First, calculate the activity rates for assigning costs to suppliers: Inspecting components: \$1,200,000/1,000 = \$1,200 per sampling hour Expediting work: \$960,000/100 = \$9,600 per order Reworking products: \$6,844,500/1,500 = \$4,563 per rework hour Warranty work: \$21,600,000/4,000 = \$5,400 per warranty hour Next, calculate the cost per component by supplier: Supplier cost: Grayson Lambert Purchase cost: \$144 × 200,000 ...................... \$28,800,000 \$129 × 800,000 ...................... \$103,200,000 Inspecting components: \$1,200 × 20 ............................ 24,000 \$1,200 × 980 .......................... 1,176,000 Expediting work: \$9,600 × 10 ............................ 96,000 \$9,600 × 90 ............................ 864,000 Reworking products: \$4,563 × 90 ............................ 410,670 \$4,563 × 1,410 ....................... 6,433,830 Warranty work: \$5,400 × 200 .......................... 1,080,000 \$5,400 × 3,800 ....................... 20,520,000 Total supplier cost .................... \$30,410,670 \$132,193,830 Units supplied ...........................
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# What is the units digit of the above expression?
Author Message
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Joined: 18 Nov 2012
Posts: 4
Location: Greece
Concentration: Finance, Accounting
GMAT 1: 650 Q45 V34
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Kudos [?]: 5 [0], given: 3
What is the units digit of the above expression? [#permalink]
### Show Tags
18 Nov 2012, 03:02
(2^18)*(3^17)*(7^14)*(5^20)
What is the units digit of the above expression?
A. 0
B. 2
C. 5
D. 7
E. 9
Last edited by Bunuel on 18 Nov 2012, 03:31, edited 1 time in total.
Renamed the topic and edited the question.
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Kudos [?]: 1335 [1] , given: 62
Re: What is the units digit of the above expression? [#permalink]
### Show Tags
18 Nov 2012, 06:42
1
KUDOS
stef91 wrote:
(2^18)*(3^17)*(7^14)*(5^20)
What is the units digit of the above expression?
A. 0
B. 2
C. 5
D. 7
E. 9
cyclicity of 2 is 4
cyclicity of 3 is 4
cyclicity of 7 is 4
cyclicity of 5 is 1.
So divide the respective powers by the integer's own cyclicity.
So the above expression can be written as:
$$2^2 * 3^3 * 7^2 * 5$$
It becomes $$4*27*49*5$$
Since 4*5=20,
so any number, if multiplied by 0 will yield 0 as the last digit.
Hence A.
_________________
Re: What is the units digit of the above expression? [#permalink] 18 Nov 2012, 06:42
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Display posts from previous: Sort by
# What is the units digit of the above expression?
Moderator: Bunuel
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 914 | 2,900 | {"found_math": true, "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": 1, "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} | 3.984375 | 4 | CC-MAIN-2017-04 | latest | en | 0.841188 |
https://www.freecodecamp.org/forum/t/exact-change-the-fifth-test/100593 | 1,539,788,512,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511175.9/warc/CC-MAIN-20181017132258-20181017153758-00523.warc.gz | 929,749,567 | 6,053 | # Exact Change - the fifth test
Exact Change - the fifth test
0
#1
Hello. I’ve definitely hit rounding issues. My code passes all the tests other than the fifth one. I end up with 3 cents rather than 4 cents and so ‘Insufficient Funds’ is returned. Any ideas? Thanks.
function checkCashRegister(price, cash, cid) {
var denoms = [100, 20, 10, 5, 1, 0.25, 0.1, 0.05, 0.01];
var totalInDrawer = cid.map(function(arr) {
return arr[1];
}).reduce(function(a, b) {
return a + b;
});
var changeLeft = cash - price;
var changeArr = [];
if (totalInDrawer == changeLeft) {
return "Closed";
}
cid.reverse();
cid.forEach(function(arr, index) {
if (arr[1] > 0 && changeLeft >= denoms[index]) {
var change = Math.min(arr[1], Math.floor(changeLeft / denoms[index]) * denoms[index]);
var forChangeArr = [];
forChangeArr[0] = arr[0];
forChangeArr[1] = change;
changeArr.push(forChangeArr);
changeLeft -= change;
console.log(change, forChangeArr, changeLeft);
}
});
if (changeLeft > 0) {
return "Insufficient Funds";
} else return changeArr;
}
console.log(checkCashRegister(3.26, 100.00, [
["PENNY", 1.01],
["NICKEL", 2.05],
["DIME", 3.10],
["QUARTER", 4.25],
["ONE", 90.00],
["FIVE", 55.00],
["TEN", 20.00],
["TWENTY", 60.00],
["ONE HUNDRED", 100.00]
]));
#2
I guess the Math.round() is causing this, because it rounds to an integer. You can avoid this by using something like this: Math.round(someValue*100)/100 to round to hundredths.
Haven’t tested it, so it might be something else.
#3
Thanks for looking. I’ve just realised I meant to use Math.floor. That gets me closer to the solution - I don’t get the \$2 now. But rounding errors are now definitely causing the test to fail - I end up with 3 cents rather than 4 cents and so I get ‘Insufficient Funds’ returned. * 100 / 100 doesn’t seem to help.
#4
Change this:
changeLeft -= change;
To this:
changeLeft = Math.round((changeLeft - change)*100)/100;
Just checked and it works. (if you also change to Math.floor).
This will round the result every “turn”, so the floating point error is removed every time.
#5
Hey Ipgm,
It sounds like what you’re encountering is floating point precision in JavaScript. Sometimes when you’re working with decimals you’ll get some strange results unless you perform mathematical operation that might be between decimal numbers a certain way.
You’re code is going to have to look something like this:
var difference = Math.round((parseFloat(firstNum) - parseFloat(secondNum)) * floatPrecision) / floatPrecision;
Where floatPrecision is a constant such as 1000.
#6
Brilliant. It does work. Thanks a lot. I’ll have a read about Math.round and floating points.
Time to check the answer wiki now and find out what the answer really is!
#7
Thanks for this. I’ll swot up on floating points then! | 766 | 2,788 | {"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} | 3.109375 | 3 | CC-MAIN-2018-43 | latest | en | 0.61554 |
http://www.primidi.com/acceleration | 1,597,331,528,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739046.14/warc/CC-MAIN-20200813132415-20200813162415-00012.warc.gz | 171,000,322 | 4,211 | # Acceleration
In physics, acceleration is the rate at which the velocity of a body changes with time. In general, velocity and acceleration are vector quantities, with magnitude and direction, though in many cases only magnitude is considered (sometimes with negative values for deceleration). Acceleration is accompanied by a force, as described by Newton's Second Law; the force, as a vector, is the product of the mass of the object being accelerated and the acceleration (vector). The SI unit of acceleration is the meter per second squared (m/s2).
For example, an object such as a car that starts from standstill, then travels in a straight line at increasing speed, is accelerating in the direction of travel. If the car changes direction at constant speedometer reading, there is strictly speaking an acceleration although it is often not so described; passengers in the car will experience a force pushing them back into their seats in linear acceleration, and a sideways force on changing direction. If the speed of the car decreases, it is usual and meaningful to speak of deceleration; mathematically it is acceleration in the opposite direction to that of motion.
Read more about Acceleration: Definition and Properties, Tangential and Centripetal Acceleration, Conversions
### Other articles related to "acceleration, accelerations":
Linear Motion - Acceleration
... Acceleration is defined as the rate of change of velocity with respect to time ... Acceleration is the second derivative of displacement i.e ... acceleration can be found by differentiating position with respect to time twice or differentiating velocity with respect to time once ...
Laboratory Centrifuge - Theory
... Protocols for centrifugation typically specify the amount of acceleration to be applied to the sample, rather than specifying a rotational speed such as revolutions per minute ... The acceleration is often quoted in multiples of g, the acceleration due to gravity at the Earth's surface ... rotational speed will subject samples to different accelerations ...
Spatial Acceleration
... body motion provides for several ways of defining the acceleration state of a rigid body ... The classical definition of acceleration entails following a single particle/point along the rigid body and observing its changes of velocity ... In this article the notion of spatial acceleration is explored, which entails looking at a fixed (unmoving) point in space and observing the changes of velocity of whatever ...
Acceleration - Conversions
... Conversions between common units of acceleration m/s2 ft/s2 Standard gravity (g0) Gal (cm/s2) 1 m/s2 = 1 3.28084 0.101972 1 ... ft/s2 = 0.304800 1 0.0310810 30.4800 1 g0 = 9.80665 32.1740 1 ...
Curve Resistance - Inaccurate Formulas - Formulas Which Try To Account For Superelevation
... the absolute value of the unbalanced acceleration in the plane of the track and perpendicular to the rail, such lateral acceleration being equal to the ... Multiplying any of these accelerations by the appropriate mass (in kg) (of the rail car, of the mass allocated to one wheel, etc.) results in the corresponding force per Newton's 2nd law ... | 656 | 3,172 | {"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} | 3.125 | 3 | CC-MAIN-2020-34 | latest | en | 0.929732 |
https://www.scribd.com/doc/297091213/190 | 1,566,167,791,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314130.7/warc/CC-MAIN-20190818205919-20190818231919-00226.warc.gz | 990,447,371 | 43,433 | You are on page 1of 1
CS_Ch3_LetUsEntertain
2/24/05
11:39 PM
Page 190
Never look directly at
a laser beam or shine a laser
beam into someones eyes.
Always work above the plane of
the beam and beware of
reflections from shiny surfaces.
did in previous activities. Place a glass rod in the light beam
so that the beam spreads up and down.
3. Shine the laser pointer or light from the ray box through the
acrylic block. Be sure the beam leaves the acrylic block on
the side opposite the side the beam enters. Mark the path of
each beam. You may wish to use a series of dots as you did
before. Label each path on both sides of the acrylic block so
you will know that they go together.
Normal
Normal
Angle of
Incidence
Angle of
Refraction
4. The angle of incidence is the angle between the
incident laser beam and the normal, as shown in
the diagram. Choose two other angles of incidence
and again mark the path of the light, as you did in
Step 3. As before, label each pair of paths.
5. Trace the outline of the acrylic block on the paper and
remove the acrylic block. Connect the paths you traced to
show the light beam entering the acrylic block, traveling
through the acrylic block, and emerging from the acrylic
block. Draw a perpendicular line at the point where a ray
enters or leaves the acrylic block. Label this line the normal.
6. Measure the angles of incidence (the angle in the air) and
refraction (the angle in the acrylic block).
a) Record your measurements in tables like the one shown.
Physics Words
critical angle: the angle
of incidence for which a
light ray passing from
one medium to another
has an angle of refraction
of 90.
index of refraction: a
property of a medium
that is related to the
speed of light through it;
it is calculated by dividing
the speed of light in
vacuum by the speed of
light in the medium.
Snells Law: describes the
relationship between the
index of refraction and
the ratio of the sine of
the angle of incidence
and the sine of the angle
of refraction.
Angle of
incidence
Angle of
refraction
Sine of
angle of
refraction
Sin i
Sin R
b) Use a calculator to complete the chart by finding the sines
of the angles (sin button on calculator).
sin i
c) Is the value of sin R a constant? This value is
called the index of refraction for the acrylic block.
7. Set up the acrylic block on a clean sheet of white paper.
This time, as shown in the drawing (next page), aim the
beam so it leaves the acrylic block on the side, rather than
at the back.
190
Active Physics
Sine of
angle of
incidence | 641 | 2,558 | {"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} | 3.59375 | 4 | CC-MAIN-2019-35 | latest | en | 0.918936 |
https://citizenmaths.com/length/kilofeet-to-decimeter | 1,632,772,087,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00686.warc.gz | 220,214,920 | 13,285 | # Kilofeet to Decimeter Conversions
From
Kilofeet
• Angstrom
• Cable Length (Imperial)
• Cable Length (International)
• Cable Length (U.S.)
• Centimeter
• Chain
• Cubit
• Decameter
• Decimeter
• Ell
• Fathom
• Finger
• Finger (cloth)
• Foot
• Furlong
• Gigameter
• Hand
• Hectometer
• Inch
• Kilofeet
• Kilometer
• League
• League (land)
• Light Day
• Light Hour
• Light Minute
• Light Second
• Light Year
• Line
• Marathon
• Megameter
• Meter
• Mickey
• Microinch
• Micrometer
• Micron
• Mil
• Mile
• Millimeter
• Myriameter
• Nail (cloth)
• Nanometer
• Nautical League
• Nautical Mile
• Pace
• Palm
• Parsec
• Pica
• Picometer
• Point
• Quarter
• Rod
• Rope
• Shaku
• Smoot
• Span
• Step
• Terameter
• Twip
• Yard
To
Decimeter
• Angstrom
• Cable Length (Imperial)
• Cable Length (International)
• Cable Length (U.S.)
• Centimeter
• Chain
• Cubit
• Decameter
• Decimeter
• Ell
• Fathom
• Finger
• Finger (cloth)
• Foot
• Furlong
• Gigameter
• Hand
• Hectometer
• Inch
• Kilofeet
• Kilometer
• League
• League (land)
• Light Day
• Light Hour
• Light Minute
• Light Second
• Light Year
• Line
• Marathon
• Megameter
• Meter
• Mickey
• Microinch
• Micrometer
• Micron
• Mil
• Mile
• Millimeter
• Myriameter
• Nail (cloth)
• Nanometer
• Nautical League
• Nautical Mile
• Pace
• Palm
• Parsec
• Pica
• Picometer
• Point
• Quarter
• Rod
• Rope
• Shaku
• Smoot
• Span
• Step
• Terameter
• Twip
• Yard
Formula 3,334 kft = 3334 x 3048 dm = 10,162,032 dm
## How To Convert From Kilofeet to Decimeter
1 Kilofeet is equivalent to 3,048 Decimeters:
1 kft = 3,048 dm
For example, if the Kilofeet number is (2.9), then its equivalent Decimeter number would be (8,839.2).
Formula:
2.9 kft = 2.9 x 3048 dm = 8,839.2 dm
## Kilofeet to Decimeter conversion table
Kilofeet (kft) Decimeter (dm)
0.1 kft 304.8 dm
0.2 kft 609.6 dm
0.3 kft 914.4 dm
0.4 kft 1,219.2 dm
0.5 kft 1,524 dm
0.6 kft 1,828.8 dm
0.7 kft 2,133.6 dm
0.8 kft 2,438.4 dm
0.9 kft 2,743.2 dm
1 kft 3,048 dm
1.1 kft 3,352.8 dm
1.2 kft 3,657.6 dm
1.3 kft 3,962.4 dm
1.4 kft 4,267.2 dm
1.5 kft 4,572 dm
1.6 kft 4,876.8 dm
1.7 kft 5,181.6 dm
1.8 kft 5,486.4 dm
1.9 kft 5,791.2 dm
2 kft 6,096 dm
2.1 kft 6,400.8 dm
2.2 kft 6,705.6 dm
2.3 kft 7,010.4 dm
2.4 kft 7,315.2 dm
2.5 kft 7,620 dm
2.6 kft 7,924.8 dm
2.7 kft 8,229.6 dm
2.8 kft 8,534.4 dm
2.9 kft 8,839.2 dm
3 kft 9,144 dm
3.1 kft 9,448.8 dm
3.2 kft 9,753.6 dm
3.3 kft 10,058.4 dm
3.4 kft 10,363.2 dm
3.5 kft 10,668 dm
3.6 kft 10,972.8 dm
3.7 kft 11,277.6 dm
3.8 kft 11,582.4 dm
3.9 kft 11,887.2 dm
4 kft 12,192 dm
4.1 kft 12,496.8 dm
4.2 kft 12,801.6 dm
4.3 kft 13,106.4 dm
4.4 kft 13,411.2 dm
4.5 kft 13,716 dm
4.6 kft 14,020.8 dm
4.7 kft 14,325.6 dm
4.8 kft 14,630.4 dm
4.9 kft 14,935.2 dm
5 kft 15,240 dm
5.1 kft 15,544.8 dm
5.2 kft 15,849.6 dm
5.3 kft 16,154.4 dm
5.4 kft 16,459.2 dm
5.5 kft 16,764 dm
5.6 kft 17,068.8 dm
5.7 kft 17,373.6 dm
5.8 kft 17,678.4 dm
5.9 kft 17,983.2 dm
6 kft 18,288 dm
6.1 kft 18,592.8 dm
6.2 kft 18,897.6 dm
6.3 kft 19,202.4 dm
6.4 kft 19,507.2 dm
6.5 kft 19,812 dm
6.6 kft 20,116.8 dm
6.7 kft 20,421.6 dm
6.8 kft 20,726.4 dm
6.9 kft 21,031.2 dm
7 kft 21,336 dm
7.1 kft 21,640.8 dm
7.2 kft 21,945.6 dm
7.3 kft 22,250.4 dm
7.4 kft 22,555.2 dm
7.5 kft 22,860 dm
7.6 kft 23,164.8 dm
7.7 kft 23,469.6 dm
7.8 kft 23,774.4 dm
7.9 kft 24,079.2 dm
8 kft 24,384 dm
8.1 kft 24,688.8 dm
8.2 kft 24,993.6 dm
8.3 kft 25,298.4 dm
8.4 kft 25,603.2 dm
8.5 kft 25,908 dm
8.6 kft 26,212.8 dm
8.7 kft 26,517.6 dm
8.8 kft 26,822.4 dm
8.9 kft 27,127.2 dm
9 kft 27,432 dm
9.1 kft 27,736.8 dm
9.2 kft 28,041.6 dm
9.3 kft 28,346.4 dm
9.4 kft 28,651.2 dm
9.5 kft 28,956 dm
9.6 kft 29,260.8 dm
9.7 kft 29,565.6 dm
9.8 kft 29,870.4 dm
9.9 kft 30,175.2 dm
10 kft 30,480 dm
20 kft 60,960 dm
30 kft 91,440 dm
40 kft 121,920 dm
50 kft 152,400 dm
60 kft 182,880 dm
70 kft 213,360 dm
80 kft 243,840 dm
90 kft 274,320 dm
100 kft 304,800 dm
110 kft 335,280 dm | 1,975 | 3,866 | {"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} | 3.34375 | 3 | CC-MAIN-2021-39 | latest | en | 0.33558 |
https://se.mathworks.com/matlabcentral/answers/781583-how-do-i-plot-a-function-over-a-specific-range-of-x-values | 1,643,264,698,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305141.20/warc/CC-MAIN-20220127042833-20220127072833-00655.warc.gz | 533,497,823 | 23,003 | # How do I plot a function over a specific range of x values?
72 views (last 30 days)
Erica Dawn Miller on 23 Mar 2021
Answered: William on 23 Mar 2021
I am trying to graph the function y=exp(-.2x)*cos(2.1x) over the x values of 0 to 4. I have to use the function and the f plot method. I tried to do this by using the method of "fplot(fun1,[0 4])," but this did not work, as it is asking me to enter the value of x. I believe the problem lies in my function, which I have attached. If anyone has any advice, I would greatly appreciate it.
William on 23 Mar 2021
Erica,
I think fplot() will work if it is called correctly. If you want to use it, you need to define the function so that it can handle a vector input for x. In your case, this means that you should use a ".*" operator to multiply the two functions together, as in the following:
func = @(x) exp(-.2*x).*cos(2.1*x)
fplot(func,[0,4])
R2020b
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Start Hunting! | 288 | 1,034 | {"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} | 2.625 | 3 | CC-MAIN-2022-05 | latest | en | 0.930976 |
https://biology.stackexchange.com/questions/95573/recombination-data-set/95594 | 1,716,426,721,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00091.warc.gz | 111,094,374 | 35,776 | # Recombination Data Set
I was looking over some genetics question and came across this data set.
In the fruit fly Drosophila melanogaster there is a dominant gene b+ for grey body color and another dominant gene c+ for normal wings. The recessive alleles (b,c) of these two genes result in black body color and curved wings respectively.
To me Ada's data set makes sense as the parental type offspring numbers are higher than the recombinant offspring. However, Donald's data set doesn't. Is there any way a recombinant offspring could be in a higher proportion than the parental type offspring?
• can you give more info? which is the parental type? what is the experimental design? are these derived from the same background? what generation, e.g. F2? not enough info to answer as it is. This looks to me like a homework question, if so please add the "homework" tag. Sep 1, 2020 at 21:40
• This isn't a homework question. In fact, this is a question from IBO. I'll actually post the whole information. Sep 1, 2020 at 22:49
b+/c+ b/c b+/c b/c+
Donald ~0.09 ~0.09 ~0.4 ~0.41
The other chromosome beeing b/c (I assume each one got an heterozygote fly and than they did a testcross)
There are a two possibilities for an heterozygote fly: (b+/c+, b/c), (b+/c, b/c+)
If there was no genetic linkage we would except 0.25 for each fly group, but we didn't which means that fly groups with a higher probability are showing the original parental chromosomes phenotyping.
Since they didn't get the same results (by far) I assume (there are statistical tests for this) each scientist got a differently genetic fly
Donald got (b+/c+, b/c) since in his experiment the predominant flies are those showing a dominant phenotype and recessive phenotype and Ada got the mixed fly.
With this in mind we can see that the recombination frequency is 0.18 or 18cM
Yes, the recombinant can appear more than the parental (for example, if the parental is lethal or by pure chance) but in this case I belive it is simply a different organization on the chromosomes | 503 | 2,051 | {"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} | 2.59375 | 3 | CC-MAIN-2024-22 | latest | en | 0.946363 |
https://downloads.haskell.org/~ghc/7.2-latest/docs/html/libraries/base-4.4.1.0/Numeric.html | 1,496,089,516,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612537.91/warc/CC-MAIN-20170529184559-20170529204559-00111.warc.gz | 929,689,110 | 3,517 | base-4.4.1.0: Basic libraries
Numeric
Contents
Description
Odds and ends, mostly functions for reading and showing `RealFloat`-like kind of values.
Synopsis
# Showing
Arguments
:: Real a => (a -> ShowS) a function that can show unsigned values -> Int the precedence of the enclosing context -> a the value to show -> ShowS
Converts a possibly-negative `Real` value to a string.
showIntAtBase :: Integral a => a -> (Int -> Char) -> a -> ShowSSource
Shows a non-negative `Integral` number using the base specified by the first argument, and the character representation specified by the second.
showInt :: Integral a => a -> ShowSSource
Show non-negative `Integral` numbers in base 10.
showHex :: Integral a => a -> ShowSSource
Show non-negative `Integral` numbers in base 16.
showOct :: Integral a => a -> ShowSSource
Show non-negative `Integral` numbers in base 8.
showEFloat :: RealFloat a => Maybe Int -> a -> ShowSSource
Show a signed `RealFloat` value using scientific (exponential) notation (e.g. `2.45e2`, `1.5e-3`).
In the call `showEFloat digs val`, if `digs` is `Nothing`, the value is shown to full precision; if `digs` is `Just d`, then at most `d` digits after the decimal point are shown.
showFFloat :: RealFloat a => Maybe Int -> a -> ShowSSource
Show a signed `RealFloat` value using standard decimal notation (e.g. `245000`, `0.0015`).
In the call `showFFloat digs val`, if `digs` is `Nothing`, the value is shown to full precision; if `digs` is `Just d`, then at most `d` digits after the decimal point are shown.
showGFloat :: RealFloat a => Maybe Int -> a -> ShowSSource
Show a signed `RealFloat` value using standard decimal notation for arguments whose absolute value lies between `0.1` and `9,999,999`, and scientific notation otherwise.
In the call `showGFloat digs val`, if `digs` is `Nothing`, the value is shown to full precision; if `digs` is `Just d`, then at most `d` digits after the decimal point are shown.
showFloat :: RealFloat a => a -> ShowSSource
Show a signed `RealFloat` value to full precision using standard decimal notation for arguments whose absolute value lies between `0.1` and `9,999,999`, and scientific notation otherwise.
floatToDigits :: RealFloat a => Integer -> a -> ([Int], Int)Source
`floatToDigits` takes a base and a non-negative `RealFloat` number, and returns a list of digits and an exponent. In particular, if `x>=0`, and
``` floatToDigits base x = ([d1,d2,...,dn], e)
```
then
1. `n >= 1`
2. `x = 0.d1d2...dn * (base**e)`
3. `0 <= di <= base-1`
NB: `readInt` is the 'dual' of `showIntAtBase`, and `readDec` is the `dual' of `showInt`. The inconsistent naming is a historical accident.
Reads a signed `Real` value, given a reader for an unsigned value.
Arguments
:: Num a => a the base -> (Char -> Bool) a predicate distinguishing valid digits in this base -> (Char -> Int) a function converting a valid digit character to an `Int` -> ReadS a
Reads an unsigned `Integral` value in an arbitrary base.
Read an unsigned number in decimal notation.
Read an unsigned number in octal notation.
Read an unsigned number in hexadecimal notation. Both upper or lower case letters are allowed.
Reads an unsigned `RealFrac` value, expressed in decimal scientific notation.
Reads a non-empty string of decimal digits.
# Miscellaneous
fromRat :: RealFloat a => Rational -> aSource
Converts a `Rational` value into any type in class `RealFloat`. | 903 | 3,439 | {"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} | 2.53125 | 3 | CC-MAIN-2017-22 | latest | en | 0.66359 |
https://www.hotrod.com/articles/ctrp-0303-performance-crossweight-balance-configuration/ | 1,575,868,005,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517557.43/warc/CC-MAIN-20191209041847-20191209065847-00236.warc.gz | 744,039,588 | 41,493 | # Performance Crossweight and Balance Configuration
## What We Have Learned About Crossweight/Bite
In circle track racing, the use of the term “crossweight” gives us an indication of the weight distribution on the four tires. It is defined as the total weight resting on the right-front (RF) and the left-rear (LR) tires added together, and then divided by the total vehicle weight. That math gives us a percentage number to relate to, i.e., “51.4 percent of cross.”
Dirt and Asphalt The percentage number for crossweight is used for asphalt cars. For dirt cars, the terminology is a little different. Most dirt teams refer to the amount of “bite” or “left-rear weight” in the car, which is the number of pounds of weight that the LR tire supports over the right-rear (RR) tire. Subtracting the amount of weight supported by the RR tire from the LR tire’s weight, we arrive at a number and call that the amount of bite in the car, i.e., “100 pounds of bite or left rear.”
How Crossweight Changes
It is important to note that if we increase the weight supported by one tire, say the LR, we also increase the weight supported by the diagonal corner of the car, in this case, the RF tire. As we make that change, the opposite change takes place at the other diagonal because those tires (the LF and RR) experience a decrease in the amount of weight they will support. So, for example, when we increase the bite in the car (LR), we also increase the weight on the RF tire.
In The Past Historically, these numbers meant little to the circle track racer as related to particular setups, different types of racetracks, and overall weight distribution. All we knew was that a certain range of crossweight or bite worked for certain conditions.
Today, we have an entirely new understanding of crossweight and the use of bite related to the setups of the cars. Beyond knowing generally that more crossweight makes the car tighter at mid-turn and less tends to loosen the car, we can now know exact numbers to be used with a balanced setup.
For simplicity, we will refer to the weight distribution of the cars, whether dirt or asphalt, as crossweight. Dirt racers need to understand the relationship of changing the LR weight and how that affects the other three corners of the car as well.
Multiple Ranges
There are different ranges of crossweight percentage that will make your car neutral in handling. The same car at the same racetrack will be neutral in handling in the range of 48 to 52 percent and also be neutral in a range between 56 and 60 percent of crossweight. Dirt racers often change the LR from near zero to upward of 300 pounds. That represents a range of 48-percent cross to 60-percent crossweight.
The actual number that will work for your car depends on the front-to-rear weight distribution, the type of track, and the setup. Generally, the flatter asphalt racetracks with less grip require a high range of crossweight percent with the associated high LR weight. The car will like the lower crossweight at the higher banked asphalt tracks.
Dirt teams will do the opposite. A low crossweight percent loads the RR tire more, and that will tend to produce more bite off the corners on dry slick surfaces. The general thought is that more weight on that tire helps to cut through the dry material lying on the track and plant the tire on the hard surface.
Many teams who run a series of asphalt tracks will try to run the high crossweight range they have successfully developed for flat tracks while racing at the higher banked tracks. They would do much better, as far as consistency, if they switched to the low crossweight range on the high-banked asphalt tracks.
Cross Is Related to Front-to-Rear Weight Distribution
The exact amount of crossweight that will make your car neutral is directly related to the front-to-rear weight distribution. The greater the rear percentage, the more crossweight is needed in a car to stay neutral in handling.
For dirt racing, with the associated high rear weight distribution, the range of crossweight that will make the car neutral is opposite of that associated with asphalt cars. The high range is normal for the high rear percentage cars, while the low range is also neutral.
For an asphalt example, let’s look at a particular car that was prepared with a balanced setup and was neutral in handling with 51-percent crossweight. If the same car is changed to an unbalanced setup with a softer rear suspension, the crossweight percentage would have to be reduced for the car to be neutral. The need to use a lower crossweight is an indication of an unbalanced setup.
How to Make the Cross Change on the Racetrack
Your car can be set up so that the amount of crossweight changes as the car circles the racetrack. A stiffer RF spring (stiffer than the LF spring) will load the RF and LR tires upon braking, causing a momentary increase in the crossweight percentage. A softer RR spring (softer than the LR) will load the LR and RF tires upon acceleration. Differences in shock compression and rebound will produce a similar effect on entry and exit, but only while the shock is in motion.
Sway Bar Effect
We can cause the static crossweight to change by preloading the sway bar. Preload on the sway bar adds weight onto the RF and LR tires, which increases the amount of static crossweight percentage. You must remember, the larger the sway bar the greater the effect of preload on the car. If we want to preload the bar, we should know how much the cross percentage will change, then lower the static (before loading the bar) crossweight percentage by that amount. When we preload the bar, we will have the exact static crossweight we want. The preload on the sway bar mostly helps provide bite off the corners.
Most dirt cars do not have sway bars. The few teams that have experimented with sway bars on dirt have seen positive results in some cases. This would be where it is beneficial to load the LR tire as opposed to loading the RR tire.
Rear Geometry Changes Cross
A few racers have discovered a way to cause the crossweight to change by utilizing the rotation of the rear end under acceleration. If the car has a three-link type of rear suspension and has a moveable link like device, a pull bar or torque arm, or lift bar, the rear end will rotate as a result of the forces of acceleration. This rotation is caused by the pinion gear trying to climb the ring gear in the rear end and taking the rear housing with it. This is not related to the dirt cars with birdcages or locating the rear end fore and aft. The bracket used for attaching the trailing arms to the rear axle tube must be clamped solid to the rear axle tube.
In order for the crossweight to increase, the springs must be mounted to the rear axle tubes with the LR spring mounted in front of the axle and the RR spring mounted to the rear of the axle. As the rear end rotates with the pinion moving upward, the LR spring will compress and the RR spring will decompress. This causes more weight to be supported by the LR and RF tires, hence more crossweight percentage of distribution while the car is accelerating.
Crossweight Changes for Dirt Cars
Many dirt teams use a similar effect by mounting the springs with the LR spring behind the axle and the RR in front of the axle. One or both spring mounts are clamped to the birdcages. As the car moves vertically, the birdcage will rotate, creating the effect of changing crossweight percentage, or bite as it is called in this case.
Lateral Placement of Third Link Changes
Cross Lateral placement of the device used to control the rotation of the rear end under acceleration or deceleration can affect the distribution of weight and crossweight during the transitional periods of entry and exit. The closer we move the bar to the LR tire, the more weight will be distributed to the LR tire on exit. The reverse is true during entry.
Unbalanced Setups Change Cross
An unbalanced setup will cause the crossweight percentage to change as the car rolls through the turns. If the rear suspension is too soft compared to the front suspension, then the back end will roll over, causing excess weight transfer at the front. The RF tire will end up supporting excess weight and that causes an increase in crossweight percentage with weight added to the LR also. This is exactly why a car with an unbalanced tight setup must run a lower static crossweight percentage than a balanced car.
Front Geometry Causes Small Changes
Front geometry settings have a small effect related to changes to crossweight percent. Caster, camber, kingpin inclination, and antidive have their own influence on changes to crossweight. These effects are relatively minimal because the changes are both small (in the case of camber/caster) and momentary (in the case of antidive).
The full jacking effect of caster and camber occurs at 180 degrees of steering input. Because we usually steer less than 10 degrees on the racetrack, the net effect would be 11/418 of the total. Camber change is the net result that causes the jacking effect, and normal changes to camber from normal steering inputs are in the range of less than 0.30 degrees of camber change.
Balanced Setups Make It All Work
Balanced setups are becoming more popular with both dirt and asphalt race teams. Crossweight percent can be an indication of the car’s balanced setup. If the correct crossweight for a particular application is 51.4 percent, and we find we need to reduce crossweight to 49.5 percent to be neutral, our setup must not be balanced.
As we change our setup to balance the front and rear suspen-sions, we will also need to increase the crossweight percent or the car will be loose. If your team is stuck on using a certain low crossweight percent that you “always ran,” your setup cannot be improved until you decide to change the crossweight, along with positive chassis setup changes.
Work toward setups that will plant the LF tire. On asphalt, a balanced setup will show the left side tire temperatures to be equal, as well as the right side tires being equal front to rear (not side to side).
For dirt cars, study the wear on all tires to get some idea of the work being done by each tire. As the setup becomes more balanced, the LF tire will show much more wear and we will be able to feel the heat after a good run.
With the setup balanced and the proper crossweight percentage set, your car will be fast and consistent and that is the combination that wins races
What Is A”Balanced” Setup?
In this and future articles in Circle Track, we will be referring to a “balanced” chassis setup. This term denotes a condition that racers have always sought. All of the trial and error methods that have been developed over the past 30 years were aimed at achieving a balanced setup.
During the past 10 years, racing technology has advanced to the point where we now know how to get to the balanced setup in an easier and quicker way. That will be a major technical theme of Circle Track magazine articles in coming issues. Dirt and asphalt racers will benefit from this new trend in racing technology.
Balance in a setup is not just a neutral handling car, but also one where we know that both suspension systems, front and rear, are working together. When we achieve this goal, the car will be neutral in handling, all four tires will be working for maximum traction, and the setup will stay consistent over a longer period of time.
The fastest cars do not necessarily have setups that will be fast beyond 20 laps. Gains in mid-turn handling performance over many laps represent the biggest gains in performance you will experience. Horsepower cannot equal these gains.
Join us as we explore the many existing ways to make your car faster through the turns. Read the results of our research and testing as we discover new methods that will improve your racing effort and help you become successful. | 2,506 | 11,964 | {"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} | 3.515625 | 4 | CC-MAIN-2019-51 | latest | en | 0.94505 |
http://www.numbersaplenty.com/27775530505 | 1,582,452,099,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00037.warc.gz | 214,442,159 | 3,492 | Search a number
27775530505 = 531712523901
BaseRepresentation
bin11001110111100011…
…001111011000001001
32122200201120010220001
4121313203033120021
5423341013434010
620432050003001
72002206054022
oct316743173011
978621503801
1027775530505
1110863621198
125471b76461
132808576b68
1414b6c3bc49
15ac86c6c3a
hex6778cf609
27775530505 has 16 divisors (see below), whose sum is σ = 34890421248. Its totient is φ = 21200760000.
The previous prime is 27775530491. The next prime is 27775530509. The reversal of 27775530505 is 50503557772.
It is not a de Polignac number, because 27775530505 - 25 = 27775530473 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (27775530509) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 1250946 + ... + 1272955.
It is an arithmetic number, because the mean of its divisors is an integer number (2180651328).
Almost surely, 227775530505 is an apocalyptic number.
It is an amenable number.
27775530505 is a deficient number, since it is larger than the sum of its proper divisors (7114890743).
27775530505 is a wasteful number, since it uses less digits than its factorization.
27775530505 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 2524008.
The product of its (nonzero) digits is 1286250, while the sum is 46.
The spelling of 27775530505 in words is "twenty-seven billion, seven hundred seventy-five million, five hundred thirty thousand, five hundred five". | 503 | 1,691 | {"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} | 2.859375 | 3 | CC-MAIN-2020-10 | latest | en | 0.815646 |
http://aptitude.brainkart.com/tag/me-gate-2013/ | 1,637,968,668,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00552.warc.gz | 4,522,177 | 9,937 | # ME GATE 2013 (Test 1)
Tag: me gate 2013
Q.1
Which one of the following options is the closest in meaning to the word given below? Nadir
A. Highest
B. Lowest
C. Medium
D. Integration
Explaination / Solution:
Nadir in the lowest point on a curve
Workspace
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Topic: Arithmetic Tag: ME GATE 2013
Q.2
Customers arrive at a ticket counter at a rate of 50 per hour and tickets are issued in the order of their arrival. The average time taken for issuing a ticket is 1min. Assuming that customer arrivals form a Poisson process and service times are exponentially distributed, the average waiting time in queue in minutes is:
A. 3
B. 4
C. 5
D. 6
Explaination / Solution:
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Q.3
Match the CORRECT pairs.
A. P-4;Q-3;R-1;S-2
B. P-4;Q-2;R-3;S-1
C. P-2;Q-3;R-4;S-1
D. P-2;Q-4;R-1;S-3
Explaination / Solution:
No Explaination.
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Report
Q.4
Cylindrical pins of mm diameter are electroplated in a shop. Thickness of the plating is 30+2.0 micron. Neglecting gage tolerances, the size of the GO gage in mm to inspect the plated components is
A. 25.042
B. 25.052
C. 25.074
D. 25.084
Explaination / Solution:
Go gauge = max. Limit = 25.052
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Q.5
In order to have maximum power from a Pelton turbine, the bucket speed must be
A. Equal to the jet speed
B. Equal to half the jet speed
C. Equal to twice the jet speed
D. Independent of the jet speed
Explaination / Solution:
Explanation: Since, velocity of bucket = ½ times the velocity of jet.
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Q.6
A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030 C° in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30 C° , with convective heat transfer coefficient h=20 W / m2 K. The thermophysical properties of steel are: density ρ = 7800 kg / m3, conductivity k = 40 W/mK and specific heat c=600 J/kgK. The time required in seconds to cool the steel ball in air from 1030 C° to 430 C° is
A. 519
B. 931
C. 1195
D. 2144
Explaination / Solution:
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Q.7
The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 K and 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglect potential energy. If the pressure and temperature of the surroundings are 1 bar and 300 K, respectively, the available energy in kJ/kg of the air stream is
A. 170
B. 187
C. 191
D. 213
Explaination / Solution:
No Explaination.
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Q.8
The threaded bolts A and B of same material and length are subjected to identical tensile load. If the elastic strain energy stored in bolt A is 4 times that of the bolt B and the mean diameter of bolt A is 12mm, the mean diameter of bolt B in mm is
A. 16
B. 24
C. 36
D. 48
Explaination / Solution:
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Q.9
A long thin walled cylindrical shell, closed at both ends, is subjected to an internal pressure. The ratio of the hoop stress (circumferential stress) to longitudinal stress developed in the shell is
A. 0.5
B. 1.0
C. 2.0
D. 4.0
Explaination / Solution:
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Q.10
In a CAD package, mirror image of a 2D point P(5, 10) is to be obtained about a line which passes through the origin and makes an angle of 45° counterlockwise with the X-axis. The coordinates of the transformed point will be
A. (7.5, 5)
B. (10, 5)
C. (7.5, -5)
D. (10, -5) | 1,045 | 3,362 | {"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} | 3.390625 | 3 | CC-MAIN-2021-49 | latest | en | 0.81889 |
http://gmatclub.com/forum/this-one-is-tricky-and-freaky-but-easy-find-all-its-1253.html?fl=similar | 1,436,293,074,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375099758.82/warc/CC-MAIN-20150627031819-00284-ip-10-179-60-89.ec2.internal.warc.gz | 112,396,496 | 47,694 | Find all School-related info fast with the new School-Specific MBA Forum
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# this one is tricky and freaky. But easy. Find all its
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this one is tricky and freaky. But easy. Find all its [#permalink] 18 Jun 2003, 23:08
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this one is tricky and freaky. But easy.
Find all its solutions
(2C)!–|C|=C
SVP
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Alas, you are wrong again. How did you manage to be admitted by doing so many mistakes? What about 1/2?
For those who still doubt I can post a solution.
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How does one know the one got all the possible solutions to such equations? Are there any rules ?
SVP
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Posts: 1607
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(2C)!=C+|C|
Since C is inside a factorial, C >=0, so open up a modul.
(2C)!=C+C=2C
Let's 2C=A, we have A!=A; thus A=1 or 2
2C=1, thus C=1/2
2C=2, thus C=1
no other roots!
simple, isn't it?
GMAT Instructor
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How does one know the one got all the possible solutions to such equations? Are there any rules ?
Hi!
Yes. There are rules.
I teach, IMHO, a very good and systematic method to solve ANY type of absolute value problem. Learn this method because it works EVERY TIME.
1) first determine the "critical points" of the equation. The critical points are those point at which a complete expression inside one of the absolute value sign changes from positive to negative or vice-versa.
2) Rewrite the equations in the intervals between +infinity, -infinity, and all of the critical points WITHOUT absolute value signs.
3) Solve for each interval. IF the solution(s) are not compatible with the interval you are in, then that solution is not valid.
Repeat steps 2 and 3 for each interval.
I know it's a little confusing, but let me give you a few examples:
Solve: |x - 1| = 4
Step 1:
The critical point here is x=1 because at this point, the expression x - 1 changes sign.
Steps 2 and 3:
The important intervals are x <= 1 and x > 1. (It is okay to include the endpoint in the intervals).
When x < 1, the expression x - 1 is always negative, so the absval signs will "flip" it. So for this interval, lets rewrite it as:
-(x - 1) = 1 - x = 4
Solving for x, we get x = -3 which is in the interval so it is a valid solution.
When x > 1, the expression x - 1 is always positive, so the absval signs will have no effect on the expression. So for this interval, lets rewrite it as:
x - 1 = 4, and solving, we get x = 5. The solution is in the interval (x > 1) so it is also a valid solution, thus the solution set is x = {-3,5}.
Example 2: (You can do this one in your head only if you are Russian or Chinese (they learn Calculus before puberty), but it's real easy to get confused).
Solve: |x тАУ 2| - |x тАУ 3| = |x тАУ 5|?
Step 1:
Recall, the critical points are -inf, +inf, and any finite numbers that causes any complete expression within an absolute value pair to change signs from negative to positive or vice-versa. For example, as x goes from less than 2 to greater than 2, тАЬx тАУ 2тАЭ тАУ the expression inside the first pair of absolute value signs тАУ goes from negative to positive. Hence, 2 is one of the critical points. By the same logic, we can deduce that the other two critical points are 3 and 5.
Steps 2 and 3:
Since the finite critical point is 2, we will choose our first interval of interest as -inf < x <= 2. In this interval, we now rewrite the equation so that it would evaluate correctly without absolute values signs, i.e., since x - 2 in the interval is always negative, we can rewrite |x тАУ 2| as -(x - 2) or (2 - x). Similarly, bot "x - 3" and "x - 5" will also always be negative if x < 2. We now solve the equation тАЬ(2 тАУ x) тАУ (3 тАУ x) = 5 тАУ xтАЭ subject to the constraint that x lies within the interval -inf < x <= 2. We are able to obtain the solution x = 6. The solution, however, lies outside the interval so it is not considered a valid solution. Hence, we conclude that there are no solutions within this interval (remember, we are solving the equation for x SUCH THAT x is within the interval of interest).
The next interval of interest is between 2 and the next highest finite endpoint, or 2 <= x <= 3. Once again, we rewrite the equation so that we can eliminate the absolute value signs for xтАЩs within the interval and we end up solving for the equation: (x тАУ 2) тАУ (3 тАУ x) = 5 тАУ x. Solving for x, we once again obtain a solution, x = 10/3 = 3.33333тАж, that is outside the interval, so again it is not considered a valid solution.
Similarly, the next interval is 3 <= x <= 5. We solve for the equation: (x тАУ 2) тАУ (x тАУ 3) = 5 тАУ x. This time, the solution x = 4 DOES lie within the interval 3 <= x <= 5, hence it is a valid solution.
Since 5 is our greatest finite critical point, the last interval is 5 <= x < +infinity. We solve for the equation: (x тАУ 2) тАУ (x тАУ 3) = x тАУ 5 obtain the result, x = 6. This solution lies within the interval 5 <= x < +infinity, hence it is also a valid solution.
We conclude that the only two solutions of the original equation are x = 4 and x = 6, so x = {4, 6}.
Example 3: Stolyar's current problem
Solve: (2C)!=C+|C|
Step 1:
There is only one expression inside absolute value signs |C| so the critical point here is clearly C = 0.
Step 2:
The first interval is C <= 0. Rewriting the equation so that it would always be true in this interval, we get (2C)! = C + -C or (2C!) = 0. Since 2C! can never be zero, we conclude that there are no solutions in this interval.
The second interval is C >= 0. Rewriting the equation so that it would always be true in this interval, we get
(2C)! = C + C or (2C!) = 2C.
Now divide both sides by 2C, we get:
(2C - 1)! = 1.
Well, (2C - 1)! will equal 1 only when (2C - 1) = 0 or 1, hence C must equal either 1/2 or 1. Hence, the solution is C = {1/2, 1}.
Note: it is possible to write equations that make this method a little tedious (i've seen a few of Stoylar's equation where there are nested abs val signs such as: ||X^2| - 2X| = ||X + 4| - (|X| - 3)| -- yes, he can be a sadist.)
Two relevant comments: 1) you can still use this method -- it's just a little trickier to determine what the critical points are; and 2) you will NEVER get a problem like this on the GMAT, so don't worry, be happy.
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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A279098 Numbers k such that prime(k) divides primorial(j) + 1 for exactly one integer j. 5
1, 2, 4, 8, 11, 17, 18, 21, 25, 32, 34, 35, 39, 40, 42, 47, 48, 58, 63, 65, 66, 67, 69, 90, 91, 97, 105, 110, 122, 140, 144, 151, 152, 162, 166, 168, 173, 174, 175, 179, 180, 186, 205, 207, 208, 210, 211, 218, 233, 243, 249, 256, 261, 262, 297, 308, 316, 318 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS As used here, "primorial(j)" refers to the product of the first j primes, i.e., A002110(j). Primorial(j) + 1 is the j-th Euclid number, A006862(j). LINKS Giovanni Resta, Table of n, a(n) for n = 1..10000 EXAMPLE 59 is not in this sequence because both primorial(7) + 1 = 510511 and primorial(17) + 1 = 1922760350154212639071 are divisible by prime(59) = 277. MATHEMATICA np[1]=1; np[n_] := Block[{c=0, p=Prime[n], trg, x=1}, trg = p-1; Do[x = Mod[x Prime[k], p]; If[trg == x, c++], {k, n-1}]; c]; Select[Range[262], np[#] == 1 &] (* Giovanni Resta, Mar 29 2017 *) CROSSREFS Subsequence of A279097 (which also includes numbers k such that prime(k) divides primorial(j) + 1 for more than one integer j). Cf. A000040, A002110, A006862, A113165, A279099, A283928. Sequence in context: A018320 A153195 A279097 * A010068 A295674 A120632 Adjacent sequences: A279095 A279096 A279097 * A279099 A279100 A279101 KEYWORD nonn AUTHOR Jon E. Schoenfield, Mar 24 2017 STATUS approved
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Last modified September 30 11:32 EDT 2020. Contains 337439 sequences. (Running on oeis4.) | 692 | 1,921 | {"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} | 3.484375 | 3 | CC-MAIN-2020-40 | latest | en | 0.675596 |
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premal June 12, 2012 06:46
FLow over Airfoil
I am doing case study on Flow over airfoil at Low Reynolds number. i took NACA 4412 airfoil. I found all the equations but how can i found co-efficient of lift and Drag?
Can any one please tell me?
ricardo.costa June 15, 2012 18:49
Quote:
Originally Posted by premal (Post 365986) I am doing case study on Flow over airfoil at Low Reynolds number. i took NACA 4412 airfoil. I found all the equations but how can i found co-efficient of lift and Drag? Can any one please tell me?
Hi,
You can calculate Cl and Cd.
Lift = 0,5*rho*V^2*S*Cl
Drag = 0,5*rho*V^2*S*Cd
Where,
rho is air density (kg/m³)
V = Velocity (m/s)
S = Area (m²)
And the Lift and Drag you can get from the post processor.
premal June 15, 2012 23:31
Thank you Sir,
But How can i get Cl and Cd that is main difficulty i am facing in this case study.
Are the Cl and Cd depending on Airfoil or any other shape?
mb.pejvak June 17, 2012 00:52
you should calculate Cl and Cd from Cf and Cp. you can find it in "Fundamentals Of Aerodynamics" by Anderson section 1.5. If you can not find it, send me an e-mail to sent this book to you.
All times are GMT -4. The time now is 18:26. | 428 | 1,423 | {"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} | 2.78125 | 3 | CC-MAIN-2015-40 | longest | en | 0.904045 |
https://newbedev.com/does-the-shape-of-ice-affect-the-amount-of-heat-required-to-melt-it | 1,701,823,629,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100575.30/warc/CC-MAIN-20231206000253-20231206030253-00818.warc.gz | 481,203,848 | 8,081 | Does the shape of ice affect the amount of heat required to melt it
It's approximately the same energy to melt little chips of ice and one big one, but not exactly because there is energy involved in creating surfaces, called the "surface energy" (or surface tension). However, an introductory physics course usually wouldn't include this in its model of melting, so you might not have heard of it.
When you cut something, you break atomic bonds in it, and this requires energy input. The energy is now a bit higher, which would suggest that the energy to get to a melted state would be a bit less.
However, we have to account for the surface energy of the water as well. When the tiny chips of ice melt, they create a large surface area of water. Water has surface energy, so the many tiny drops of water would have more energy than the one big drop of water you'd get from melting a single chunk of ice. Whether it takes more energy to melt the little chips of ice or the big block of ice depends on whether the surface energy of the ice or the water is higher. If the water has higher surface energy, you'd be putting more energy into the water during melting, and it would take a bit more heat to melt the small chips. If the ice has higher surface energy, it would take a bit less energy to melt the small chips. (All of this ignores gravity. If the gravitational energy is changing as you melt/freeze, you'd have to account for that as well.)
This extra energy in the little pieces of ice/water is stored in the surfaces. If you took many tiny drops of water and let them all converge, they would heat up as they did so, and you'd wind up with hotter water than you started with because the surface energy would turn into thermal energy in the water - that's where all that extra energy you put in to make the little drops would show up.
The effect is over all pretty small because atoms are small, meaning only a small fraction of them are on the surface for any macroscopically-sized stuff. The surface energy of water is about 0.07 J/m^2, while the energy to melt water is about 334 kJ/kg. So even for tiny droplets of water with radius 1 micron, the surface energy is only about 0.06 percent of the energy to melt that much ice; a small effect.
Finally, it's not clear just from everyday experience that breaking ice apart requires energy, in the sense that the broken-apart ice has higher energy than the ice you started with. This is true, but doesn't follow just from the fact that in real life, when you hack at ice with a chisel, you're using energy. Most of that energy is going to heat the ice (and surrounding environment).
I'm not totally sure I understand the question. But I think you may be confusing the amount of heat energy needed to melt the ice with the amount of time needed to melt the ice.
You're correct that the energy needed to melt the ice, given that the ice is already at the temperature of the phase change, depends only on the mass. However the heat responsible for the phase change must enter the ice through its surface. You give the example of splitting a large block of ice into $10^{12}$ pieces. An interesting exercise for you is to compute the surface areas of those two distributions. If the rate of heat flow is proportional to the surface area, it's clear that the micron-sized ice powder will melt more rapidly than the large block.
An example of this that you may have seen is liquid nitrogen, which boils at 77 kelvin. An open bucket of liquid nitrogen may be stable for many minutes. Take that same bucket and dump it on the ground, and the nitrogen vaporizes more or less instantly. The difference isn't the amount of heat that's needed: it's the the surface area through which that heat can enter the material.
For simplicity let's assume that every molecule of ice (water) is a little cube and the bonds are between adjacent faces.
Your 10cm cube of ice contains approx. $3.07*10^{25}$ molecules, and thus $9.21*10^{25}$ bonds (i.e three times the amount of molecules, since each cube has six faces and almost each face is shared by two cubes).
A 1um cube of ice is $3.07*10^{10}$ molecules, thus one face of it contains $\sqrt[3/2]{3.07*10^{10}} = 9.80*10^6$ molecules of ice per layer. By cutting out such a cube you have broken thrice that amount of bonds. So for the $10^{15}$ little cubes you have split the 10cm cubed into you have just broken $2.93*10^{22}$ bonds. This is 3000 times less than the total number of bonds initially, but you have a point:
By cutting a large block into 1um cubes you have reduced the number of bonds to break by $0.0316\%$, and thus, to some approximation, the energy required to melt the rest by the same amount.
Mark raises a good point in the comment, that surface energy of the resulting water will play a role. The surface energy was mentioned in his answer, with a figure of 0.06% relative to the energy needed to melt a one-micron piece of ice. However, my result is of the same order. I expect the role of surface energy to be highly variable depending on the initial configuration of our ice dust. If we place each tiny piece far from every other, surface energy might win over. If everything is put into a pile, it might turn out to be a negligible correction. | 1,191 | 5,276 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "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} | 3.5 | 4 | CC-MAIN-2023-50 | latest | en | 0.978106 |
https://www.r-bloggers.com/stl-random_shuffle-for-permutations/ | 1,582,634,372,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146066.89/warc/CC-MAIN-20200225110721-20200225140721-00121.warc.gz | 866,268,355 | 48,788 | # STL random_shuffle for permutations
December 30, 2012
By
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The STL also contains random sampling and shuffling algorithms. We start by looking at `random_shuffle`.
There are two forms. The first uses an internal RNG with its own seed; the second form allows for a function object conformant to the STL’s requirements (essentially, given `N` produce a uniform draw greater or equal to zero and less than `N`). This is useful for us as it lets us tie this to the same RNG which R uses.
``````#include
// wrapper around R's RNG such that we get a uniform distribution over
// [0,n) as required by the STL algorithm
inline int randWrapper(const int n) { return floor(unif_rand()*n); }
// [[Rcpp::export]]
Rcpp::NumericVector randomShuffle(Rcpp::NumericVector a) {
Rcpp::RNGScope scope;
// clone a into b to leave a alone
Rcpp::NumericVector b = Rcpp::clone(a);
std::random_shuffle(b.begin(), b.end(), randWrapper);
return b;
}
``````
We can illustrate this on a simple example or two:
``````a <- 1:8
set.seed(42)
randomShuffle(a)
``````
```[1] 1 4 3 7 5 8 6 2
```
``````set.seed(42)
randomShuffle(a)
``````
```[1] 1 4 3 7 5 8 6 2
```
By tieing the STL implementation of the random permutation to the RNG from R, we are able to compute reproducible permutations, fast and from C++.
R-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. | 453 | 1,658 | {"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} | 2.65625 | 3 | CC-MAIN-2020-10 | latest | en | 0.812076 |
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