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https://en.wikiversity.org/wiki/Physics/Essays/Fedosin/Stoney_mass | 1,719,067,680,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00452.warc.gz | 190,022,625 | 13,073 | # Physics/Essays/Fedosin/Stoney mass
< Physics | Essays | Fedosin
In physics, the Stoney mass (${\displaystyle m_{S}}$), is one of the base units in the system of natural units called Stoney units. It is a quantity of mass defined in terms of fundamental physical constants.
The Stoney mass is defined as:
${\displaystyle m_{S}=e{\sqrt {\frac {\varepsilon _{g}}{\varepsilon _{0}}}}={\sqrt {\alpha }}m_{P}=1.859\cdot 10^{-9}\ }$ kg,
where
${\displaystyle \varepsilon _{g}={\frac {1}{4\pi G}}\ }$, and ${\displaystyle G\ }$ is the gravitational constant,
${\displaystyle \varepsilon _{0}\ }$ is the electric constant,
${\displaystyle \alpha \ }$ = (137.035999074)−1 is the electric fine structure constant,
${\displaystyle e\ }$ is the elementary charge.
The Stoney mass is ${\displaystyle \alpha ^{-1/2}\approx 11.706}$ times less than the Planck mass ${\displaystyle m_{P}\ }$.
## History
Contemporary physics has settled on the Planck scale as the most suitable scale for a unified field theory. The Planck scale was however anticipated by George Stoney. [1]
The Stoney scale has been re-discovered by M. Castans and J. Belinchon[2], and by Ross McPherson, [3] in connection with the Large number coincidences.
## Stoney mass vs elementary electric charge
The elementary charge is a unit of the Stoney scale. The Coulomb force between two such charges is:
${\displaystyle F_{C}={\frac {1}{4\pi \varepsilon _{0}}}\cdot {\frac {e^{2}}{r^{2}}}.\ }$
The Newton force between two Stoney masses is:
${\displaystyle F_{N}={\frac {1}{4\pi \varepsilon _{g}}}\cdot {\frac {m_{S}^{2}}{r^{2}}},\ }$
From the equality of the above forces
${\displaystyle F_{C}=F_{N}\ }$
we find out the relationship between Stoney mass and Stoney charge:
${\displaystyle m_{S}=e{\sqrt {\frac {\varepsilon _{g}}{\varepsilon _{0}}}}.\ }$
Note that, George Stoney first proposed the term electron for the particle with elementary electric charge due to O’Hara [4] and Keller.[5] | 579 | 1,968 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "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} | 3.375 | 3 | CC-MAIN-2024-26 | latest | en | 0.833375 |
https://howkgtolbs.com/convert/29.34-kg-to-lbs | 1,643,045,321,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304572.73/warc/CC-MAIN-20220124155118-20220124185118-00325.warc.gz | 350,114,006 | 12,148 | # 29.34 kg to lbs - 29.34 kilograms to pounds
Do you need to know how much is 29.34 kg equal to lbs and how to convert 29.34 kg to lbs? Here you go. You will find in this article everything you need to make kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to point out that whole this article is dedicated to a specific amount of kilograms - exactly one kilogram. So if you want to know more about 29.34 kg to pound conversion - read on.
Before we go to the practice - that is 29.34 kg how much lbs conversion - we will tell you few theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 29.34 kg to lbs? 29.34 kilograms it is equal 64.6836276708 pounds, so 29.34 kg is equal 64.6836276708 lbs.
## 29.34 kgs in pounds
We will start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, in formal International System of Units (in abbreviated form SI).
From time to time the kilogram is written as kilogramme. The symbol of the kilogram is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was defined as the mass of one liter of water. This definition was simply but hard to use.
Later, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in short form IPK). The IPK was prepared of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was switched by another definition.
Nowadays the definition of the kilogram is build on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It is also divided to 100 decagrams and 1000 grams.
## 29.34 kilogram to pounds
You learned some facts about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. It is needed to emphasize that there are more than one kind of pound. What does it mean? For instance, there are also pound-force. In this article we want to focus only on pound-mass.
The pound is used in the British and United States customary systems of measurements. Of course, this unit is used also in other systems. The symbol of this unit is lb or “.
The international avoirdupois pound has no descriptive definition. It is defined as exactly 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 29.34 kg?
29.34 kilogram is equal to 64.6836276708 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 29.34 kg in lbs
The most theoretical part is already behind us. In this part we will tell you how much is 29.34 kg to lbs. Now you learned that 29.34 kg = x lbs. So it is time to know the answer. Have a look:
29.34 kilogram = 64.6836276708 pounds.
That is a correct outcome of how much 29.34 kg to pound. You may also round it off. After it your result will be exactly: 29.34 kg = 64.548 lbs.
You learned 29.34 kg is how many lbs, so let’s see how many kg 29.34 lbs: 29.34 pound = 0.45359237 kilograms.
Of course, this time you can also round off this result. After it your result will be as following: 29.34 lb = 0.45 kgs.
We also want to show you 29.34 kg to how many pounds and 29.34 pound how many kg outcomes in charts. Look:
We want to start with a table for how much is 29.34 kg equal to pound.
### 29.34 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
29.34 64.6836276708 64.5480
Now look at a chart for how many kilograms 29.34 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
29.34 0.45359237 0.45
Now you learned how many 29.34 kg to lbs and how many kilograms 29.34 pound, so it is time to go to the 29.34 kg to lbs formula.
### 29.34 kg to pounds
To convert 29.34 kg to us lbs you need a formula. We are going to show you two formulas. Let’s begin with the first one:
Number of kilograms * 2.20462262 = the 64.6836276708 outcome in pounds
The first version of a formula will give you the most correct outcome. In some cases even the smallest difference can be considerable. So if you need an exact outcome - this version of a formula will be the best solution to convert how many pounds are equivalent to 29.34 kilogram.
So let’s move on to the shorer version of a formula, which also enables calculations to know how much 29.34 kilogram in pounds.
The another formula is as following, have a look:
Number of kilograms * 2.2 = the outcome in pounds
As you see, this version is simpler. It can be the best solution if you want to make a conversion of 29.34 kilogram to pounds in easy way, for instance, during shopping. You only have to remember that final result will be not so correct.
Now we are going to show you these two formulas in practice. But before we will make a conversion of 29.34 kg to lbs we are going to show you easier way to know 29.34 kg to how many lbs without any effort.
### 29.34 kg to lbs converter
An easier way to check what is 29.34 kilogram equal to in pounds is to use 29.34 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Calculator is based on first version of a formula which we gave you in the previous part of this article. Due to 29.34 kg pound calculator you can effortless convert 29.34 kg to lbs. Just enter number of kilograms which you need to calculate and click ‘calculate’ button. The result will be shown in a second.
So try to convert 29.34 kg into lbs with use of 29.34 kg vs pound converter. We entered 29.34 as a number of kilograms. It is the outcome: 29.34 kilogram = 64.6836276708 pounds.
As you can see, our 29.34 kg vs lbs converter is user friendly.
Now we can move on to our main issue - how to convert 29.34 kilograms to pounds on your own.
#### 29.34 kg to lbs conversion
We will begin 29.34 kilogram equals to how many pounds conversion with the first formula to get the most correct outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 64.6836276708 the outcome in pounds
So what need you do to know how many pounds equal to 29.34 kilogram? Just multiply amount of kilograms, in this case 29.34, by 2.20462262. It is 64.6836276708. So 29.34 kilogram is exactly 64.6836276708.
It is also possible to round off this result, for instance, to two decimal places. It is exactly 2.20. So 29.34 kilogram = 64.5480 pounds.
It is high time for an example from everyday life. Let’s convert 29.34 kg gold in pounds. So 29.34 kg equal to how many lbs? As in the previous example - multiply 29.34 by 2.20462262. It is equal 64.6836276708. So equivalent of 29.34 kilograms to pounds, if it comes to gold, is 64.6836276708.
In this case you can also round off the result. Here is the result after rounding off, this time to one decimal place - 29.34 kilogram 64.548 pounds.
Now we are going to examples calculated with a short version of a formula.
#### How many 29.34 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 64.548 the result in pounds
So 29.34 kg equal to how much lbs? And again, you have to multiply amount of kilogram, this time 29.34, by 2.2. Have a look: 29.34 * 2.2 = 64.548. So 29.34 kilogram is equal 2.2 pounds.
Let’s make another calculation with use of this formula. Now convert something from everyday life, for instance, 29.34 kg to lbs weight of strawberries.
So let’s calculate - 29.34 kilogram of strawberries * 2.2 = 64.548 pounds of strawberries. So 29.34 kg to pound mass is equal 64.548.
If you know how much is 29.34 kilogram weight in pounds and are able to convert it using two different formulas, we can move on. Now we want to show you all results in tables.
#### Convert 29.34 kilogram to pounds
We know that outcomes presented in tables are so much clearer for most of you. We understand it, so we gathered all these outcomes in charts for your convenience. Due to this you can quickly compare 29.34 kg equivalent to lbs results.
Start with a 29.34 kg equals lbs chart for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
29.34 64.6836276708 64.5480
And now look 29.34 kg equal pound table for the second formula:
Kilograms Pounds
29.34 64.548
As you can see, after rounding off, if it comes to how much 29.34 kilogram equals pounds, the results are not different. The bigger amount the more considerable difference. Keep it in mind when you need to make bigger amount than 29.34 kilograms pounds conversion.
#### How many kilograms 29.34 pound
Now you learned how to calculate 29.34 kilograms how much pounds but we will show you something more. Are you curious what it is? What about 29.34 kilogram to pounds and ounces calculation?
We want to show you how you can convert it step by step. Let’s start. How much is 29.34 kg in lbs and oz?
First things first - you need to multiply number of kilograms, this time 29.34, by 2.20462262. So 29.34 * 2.20462262 = 64.6836276708. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To convert how much 29.34 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is exactly 327396192 ounces.
So final outcome is equal 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then final result is equal 2 pounds and 33 ounces.
As you see, conversion 29.34 kilogram in pounds and ounces simply.
The last conversion which we want to show you is conversion of 29.34 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert it it is needed another formula. Before we give you this formula, see:
• 29.34 kilograms meters = 7.23301385 foot pounds,
• 29.34 foot pounds = 0.13825495 kilograms meters.
Now look at a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to calculate 29.34 foot pounds to kilograms meters you need to multiply 29.34 by 0.13825495. It is 0.13825495. So 29.34 foot pounds is exactly 0.13825495 kilogram meters.
It is also possible to round off this result, for instance, to two decimal places. Then 29.34 foot pounds is equal 0.14 kilogram meters.
We hope that this conversion was as easy as 29.34 kilogram into pounds conversions.
We showed you not only how to make a calculation 29.34 kilogram to metric pounds but also two other conversions - to know how many 29.34 kg in pounds and ounces and how many 29.34 foot pounds to kilograms meters.
We showed you also another way to do 29.34 kilogram how many pounds conversions, that is using 29.34 kg en pound calculator. This is the best solution for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs conversions on your own.
We hope that now all of you can do 29.34 kilogram equal to how many pounds calculation - on your own or using our 29.34 kgs to pounds converter.
So what are you waiting for? Let’s convert 29.34 kilogram mass to pounds in the way you like.
Do you want to make other than 29.34 kilogram as pounds conversion? For example, for 15 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so simply as for 29.34 kilogram equal many pounds.
### How much is 29.34 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 29.34 kg in pounds , we prepared one more section. Here you can see all you need to know about how much is 29.34 kg equal to lbs and how to convert 29.34 kg to lbs . It is down below.
What is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 29.34 kg to pound conversion formula look? . It is down below:
The number of kilograms * 2.20462262 = the result in pounds
How does the result of the conversion of 29.34 kilogram to pounds? The correct answer is 64.6836276708 pounds.
It is also possible to calculate how much 29.34 kilogram is equal to pounds with second, shortened type of the formula. Have a look.
The number of kilograms * 2.2 = the result in pounds
So this time, 29.34 kg equal to how much lbs ? The answer is 64.6836276708 lbs.
How to convert 29.34 kg to lbs in an easier way? You can also use the 29.34 kg to lbs converter , which will do whole mathematical operation for you and you will get a correct result .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,591 | 13,685 | {"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} | 4.15625 | 4 | CC-MAIN-2022-05 | longest | en | 0.93358 |
http://www.pearltrees.com/amullya | 1,606,928,380,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141711306.69/warc/CC-MAIN-20201202144450-20201202174450-00572.warc.gz | 152,797,417 | 28,138 | # Amullya
Animated Trigonometry | Jason Davies | Web Design and Development. Non-Euclidean Geometry. In three dimensions, there are three classes of constant curvature geometries. All are based on the first four of Euclid's postulates, but each uses its own version of the parallel postulate. The "flat" geometry of everyday intuition is called Euclidean geometry (or parabolic geometry), and the non-Euclidean geometries are called hyperbolic geometry (or Lobachevsky-Bolyai-Gauss geometry) and elliptic geometry (or Riemannian geometry).
Arithmetization of analysis. The arithmetization of analysis was a research program in the foundations of mathematics carried out in the second half of the 19th century. Kronecker originally introduced the term arithmetization of analysis, by which he meant its constructivization in the context of the natural numbers (see quotation at bottom of page). The meaning of the term later shifted to signify the set-theoretic construction of the real line.
Its main proponent was Weierstrass, who argued the geometric foundations of calculus were not solid enough for rigorous work. The highlights of this research program are: An important spinoff of the arithmetization of analysis is set theory. Naive set theory was created by Cantor and others after arithmetization was completed as a way to study the singularities of functions appearing in calculus. The arithmetization of analysis had several important consequences: Quotations: "God created the natural numbers, all else is the work of man. " -- Kronecker. Grigori Perelman. Grigori Yakovlevich Perelman (Russian: Григорий Яковлевич Перельман, IPA: [ɡrʲɪˈɡorʲɪj ˈjɑkəvlʲɪvʲɪtʃ pʲɪrʲɪlʲˈman] ( )/ˈpɛrɨlmən/ PERR-il-mən[dubious ]; Russian: Григо́рий Я́ковлевич Перельма́н; born 13 June 1966) is a Russian mathematician who made landmark contributions to Riemannian geometry and geometric topology before his presumed withdrawal from mathematics.
In 1994, Perelman proved the soul conjecture. In 2003, he proved Thurston's geometrization conjecture. This consequently solved in the affirmative the Poincaré conjecture, posed in 1904, which before its solution was viewed as one of the most important and difficult open problems in topology. On 18 March 2010, it was announced that he had met the criteria to receive the first Clay Millennium Prize[4] for resolution of the Poincaré conjecture. Early life and education His mathematical education continued at the Leningrad Secondary School #239, a specialized school with advanced mathematics and physics programs.
Divergence. In vector calculus, divergence is a vector operator that measures the magnitude of a vector field's source or sink at a given point, in terms of a signed scalar. More technically, the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. For example, consider air as it is heated or cooled. The relevant vector field for this example is the velocity of the moving air at a point. If air is heated in a region it will expand in all directions such that the velocity field points outward from that region. Therefore the divergence of the velocity field in that region would have a positive value, as the region is a source. If the air cools and contracts, the divergence has a negative value, as the region is a sink. Definition of divergence In physical terms, the divergence of a three-dimensional vector field is the extent to which the vector field flow behaves like a source or a sink at a given point.
With or and . Curl (mathematics) The alternative terminology rotor or rotational and alternative notations rot F and ∇ × F are often used (the former especially in many European countries, the latter, using the del operator and the cross product, is more used in other countries) for curl and curl F. The name "curl" was first suggested by James Clerk Maxwell in 1871.[1] The components of F at position r, normal and tangent to a closed curve C in a plane, enclosing a planar vector areaA = An. The curl of a vector field F, denoted by curl F, or ∇ × F, or rot F, at a point is defined in terms of its projection onto various lines through the point. If is any unit vector, the projection of the curl of F onto is defined to be the limiting value of a closed line integral in a plane orthogonal to as the path used in the integral becomes infinitesimally close to the point, divided by the area enclosed. As such, the curl operator maps continuously differentiable functions f : R3 → R3 to continuous functions g : R3 → R3.
Where Here and. Riemann,zeta,function. Statistical mechanics: the Riemann zeta function interpreted as a partition function. The Riemann zeta function interpreted as a partition function lattice-related number theory (involving Ising models, percolation, etc.) integer partition problems and physics entropy and number theory number theory and statistical mechanics – general probabilistic number theory the Riemann zeta function interpreted as a partition function One of the earliest, and perhaps most significant, examples of number theory influencing the development of physics was the application of Pólya's work on the Riemann zeta function to the theory of phase transitions by Lee and Yang in the early 1950's.
In 1951-2, Lee and Yang were developing this theory, and Mark Kac became aware of their conjecture which was later to become the "Lee-Yang circle theorem". Lee and Yang were then able to adapt the reasoning and, within a couple of weeks, produce a proof of their general theorem. In equilibrium statistical mechanics, the fundamental object of study for a system is its partition function. B.L. B.L. B.L. G.W. D. Analytic continuation. In complex analysis, a branch of mathematics, analytic continuation is a technique to extend the domain of a given analytic function. Analytic continuation often succeeds in defining further values of a function, for example in a new region where an infinite series representation in terms of which it is initially defined becomes divergent.
The step-wise continuation technique may, however, come up against difficulties. These may have an essentially topological nature, leading to inconsistencies (defining more than one value). They may alternatively have to do with the presence of mathematical singularities. The case of several complex variables is rather different, since singularities then cannot be isolated points, and its investigation was a major reason for the development of sheaf cohomology. Initial discussion Analytic continuation of natural logarithm (imaginary part) Suppose f is an analytic function defined on a non-empty open subset U of the complex plane C. Then Let. . , hence. Riemann zeta function. , which converges when the real part of s is greater than 1. More general representations of ζ(s) for all s are given below.
The Riemann zeta function plays a pivotal role in analytic number theory and has applications in physics, probability theory, and applied statistics. This function, as a function of a real argument, was introduced and studied by Leonhard Euler in the first half of the eighteenth century without using complex analysis, which was not available at that time. Bernhard Riemann in his article "On the Number of Primes Less Than a Given Magnitude" published in 1859 extended the Euler definition to a complex variable, proved its meromorphic continuation and functional equation and established a relation between its zeros and the distribution of prime numbers.[1] The values of the Riemann zeta function at even positive integers were computed by Euler. Definition Bernhard Riemann's article on the number of primes below a given magnitude. Specific values A058303). Hyperreal number. The system of hyperreal numbers is a way of treating infinite and infinitesimal quantities.
The hyperreals, or nonstandard reals, *R, are an extension of the real numbers R that contains numbers greater than anything of the form Such a number is infinite, and its reciprocal is infinitesimal. The term "hyper-real" was introduced by Edwin Hewitt in 1948.[1] The hyperreal numbers satisfy the transfer principle, a rigorous version of Leibniz's heuristic Law of Continuity.
The transfer principle states that true first order statements about R are also valid in *R. For example, the commutative law of addition, x + y = y + x, holds for the hyperreals just as it does for the reals; since R is a real closed field, so is *R. For all integers n, one also has for all hyperintegers H. The application of hyperreal numbers and in particular the transfer principle to problems of analysis is called non-standard analysis.
For an infinitesimal The transfer principle but there is no such number in R. The Tesseract. Wolfram|Alpha. Mathematics. Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”) is the study of topics such as quantity (numbers),[2] structure,[3] space,[2] and change.[4][5][6] There is a range of views among mathematicians and philosophers as to the exact scope and definition of mathematics.[7][8] Rigorous arguments first appeared in Greek mathematics, most notably in Euclid's Elements. Since the pioneering work of Giuseppe Peano (1858–1932), David Hilbert (1862–1943), and others on axiomatic systems in the late 19th century, it has become customary to view mathematical research as establishing truth by rigorous deduction from appropriately chosen axioms and definitions.
Mathematics developed at a relatively slow pace until the Renaissance, when mathematical innovations interacting with new scientific discoveries led to a rapid increase in the rate of mathematical discovery that has continued to the present day.[11] History Evolution Etymology Definitions of mathematics. List of theorems. This is a list of theorems, by Wikipedia page. See also Most of the results below come from pure mathematics, but some are from theoretical physics, economics, and other applied fields. 0–9 A B C D E F G H I J K L M N O P Q R S T U V W Z
Theorem. Many mathematical theorems are conditional statements. In this case, the proof deduces the conclusion from the hypotheses. In light of the interpretation of proof as justification of truth, the conclusion is often viewed as a necessary consequence of the hypotheses, namely, that the conclusion is true in case the hypotheses are true, without any further assumptions. However, the conditional could be interpreted differently in certain deductive systems, depending on the meanings assigned to the derivation rules and the conditional symbol. Although they can be written in a completely symbolic form, for example, within the propositional calculus, theorems are often expressed in a natural language such as English.
The same is true of proofs, which are often expressed as logically organized and clearly worded informal arguments, intended to convince readers of the truth of the statement of the theorem beyond any doubt, and from which a formal symbolic proof can in principle be constructed. Math resources. Duality. From Wikipedia, the free encyclopedia Duality may refer to: Mathematics Philosophy, logic, and psychology Science Electrical and mechanical Physics Titles Film Music Other See also Beauty of Mathematics « Crazy People With Crazy Things. Chronology of Pure and Applied Mathematics. §. Portal @ MathLinks. MathPages. Category theory « The Unapologetic Mathematician. Comma Categories Another useful example of a category is a comma category.
The term comes from the original notation which has since fallen out of favor because, as Saunders MacLane put it, “the comma is already overworked”. We start with three categories , and , and two functors and . Are triples where is an object of is an arrow in. . — with an arrow in — making the following square commute: So what? To be the functor sending the single object of to the object . Be the identity functor on. . , where can be any other object in . Work out for yourself the category Here’s another example: the category . And another: check that given objects , the category is the discrete category (set) Neat! Cardinals and Ordinals as Categories We can import all of what we’ve said about cardinal numbers and ordinal numbers into categories. For cardinals, it’s actually not that interesting. And turn it into a category .
And just give every object its identity morphism — there are no other morphisms at all in this category. To if then . Category Theory Awodey Course. Www.liegroups.org. ATLAS of Finite Group Representations - V3. Www.liegroups.org. Algebraic structure. Algebraic geometry. Algebraic topology. Cohomology. Abelian Group. Algebraic geometry. Riemannian geometry. Symplectic geometry. Differential geometry. Non-Euclidean geometry. Special values of L-functions. John Cremona's home page. Number Theory.
Algebraic Topology. Homotopy. Euler characteristic. Abstract Algebra. Mathematics. Monad (category theory) Orbifold construction of the modes of the Poincare dodecahedral space - Jeffrey Weeks. Logarithmic spiral. Harvard extension school courses. Abstract Algebra - Free Harvard Courses. Algebra. Ring (mathematics) Abstract Algebra - Free Harvard Courses - StumbleUpon. Virtual Training Suite - free Internet tutorials to develop Internet research skills. The Thirty Greatest Mathematicians. Srinivasa Ramanujan. Carl Friedrich Gauss. Quantum field theory. Adjoint functors. Monad (category theory) Subgroup. Symmetrization. ABSTRACT ALGEBRA: OnLine Study Guide, Table of Contents. Vladimir Voevodsky. Groupoidification. Topology/Homotopy. Topology. Type theory in nLab.
Homotopy Type Theory, I | The n. Coherence theorem. | 2,973 | 13,696 | {"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-2020-50 | latest | en | 0.925521 |
http://www.stesa.com.br/x11h8n/554270-in-solar-cell-the-open-circuit-voltage-is-calculated-when | 1,716,127,783,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00662.warc.gz | 52,088,796 | 8,449 | Lee et al., “ Atomic layer deposited gallium oxide buffer layer enables 1.2 V open-circuit voltage in cuprous oxide solar cells,” Adv Mater 26(27), 4704– 4710 (2014). derived from the Shockley equivalent circuit of solar cells, an analytical expression of the open-circuit voltage is pro-posed. M. Nath. The upper panel is calculated under thermal equilibrium conditions (the dark), and shows the conduction and valence bandedges EC and EV, respectively, as well as the Fermi level EF. Alexey Abramov. Hello. There is no fixed difference between open circuit voltage and output voltage of a PV Panel. This is way to low to charge the 3.2 volt batteries. May 2018; Physical Review Applied 9(5) DOI: 10.1103/PhysRevApplied.9.051003. See the two examples below to see the difference between open circuit voltage with a voltage source which is in a closed circuit and connected to a load. Electrothermal Feedback and Absorption-Induced Open-Circuit-Voltage Turnover in Solar Cells. Energy & Environmental Science 2019, 12 (8) , 2518-2528. Alternatively, the open-circuit voltage may be thought of as the voltage that must be applied to a solar cell or a battery to stop the current. A short summary of this paper. Realizing high open-circuit voltages in solar cells. 4.30 volts off the solar cell disconnected from the circuit, is that with a resistor across the leads or without a resistor across the leads. A simple method used to measure minority carrier lifetimes in solar cells is to forward bias the cell in the dark and to monitor the decay in the voltage across the cell when the forward current is suddenly terminated. the open-circuit voltage of a solar cell has been observed to match EG/q when operating at low temperature under solar concentration [1]. As you can see, open circuit voltage is disconnected and does not form a complete circuit. As far as I know, when the light is shining on the solar cell (PN junction), current would be stopped if voltage reaches to Voc (open-circuit voltage). Factors limiting the open-circuit voltage in microcrystalline silicon solar cells. This paper. Open-circuit voltage (abbreviated as OCV or V OC) is the difference of electrical potential between two terminals of a device when disconnected from any circuit. We have shown in the previous chapter that long bulk lifetimes and low surface recombination velocities S < 10 cm s −1 are possible in perovskite layers and at perovskite/PTAA interfaces. High luminescence efficiency of the halide perovskite absorber layer in perovskite solar cells (PVSCs) will reduce the open-circuit voltage (V oc) loss and further increase the cell efficiency.Still, while quasi-2D-structure perovskite films exhibit high emissivity in light-emitting devices, quasi-2D PVSCs show severe V oc loss associated with non-radiative recombination. Short circuit current (I sc)—the maximum current, at zero voltage. Definition of Open Circuit Voltage Open circuit voltage is a common term in solar cell applications. Rui Sun, Dan Deng, Jing Guo, Qiang Wu, Jie Guo, Mumin Shi, Kui Shi, Tao Wang, Longjian Xue, Zhixiang Wei, Jie Min. The open-circuit voltage is shown on the IV curve below. The open-circuit voltage corresponds to the amount of forward bias on the solar cell due to the bias of the solar cell junction with the light-generated current. How is open-circuit voltage calculated? |The open-circuit voltage corresponds to the amount of forward bias on the solar cell due to the bias of the solar cell junction with the light-generated current. Let me explain you with the following diagram. I want to figure out an open circuit voltage of a solar cell. Another mechanism for open-circuit voltage reduction is band offset in heterojunction device architectures. Without a resistor across the solar cell leads it can read a higher voltage than the cell is producing. Presented characteristics were calculated for solar cell with following data: V oc = 0,595 mV, I sc = 4,6 A, I MPP = 4,25 A, V MPP = 0,51 V, and P MPP temperature coefficient γ = -0,005 %/K. sc and open-circuit voltage (V oc) points, as well as the maximum power point (V mp, I mp). Download PDF. Typical point on solar cell characteristics are open circuit (when no load is connected), short circuit and maximum power point. As a result, the open-circuit voltage in the perovskite solar cells is significantly improved by 80 mV. Surface thulium-doped TiO2 nanoparticles used as photoelectrodes in dye-sensitized solar cells: improving the open-circuit voltage. Fill factor can be calculated as follows. The open-circuit voltage is the difference between the quasi-Fermi levels at the two contacts in an illuminated solar cell at zero current flow. The net effect, therefore, is a combination of the increase in voltage shown for increasing n in the figure to the right and the decrease in voltage shown for increasing I 0 in the figure above. A solar cell is not really a voltage source or a current source as we usually think of them, but it can power a circuit in the typical voltage-source style. Ask Question Asked 4 years ago. perovskite solar cells results in a systematic increase of the open circuit voltage (V oc) in pin-type perovskite solar cells. The two limiting parameters used to characterise the output of solar cells for given irradiance, operating temperature and area are (Shockley & Queisser, 1961): 1. This is why it is called open. Open circuit voltage is open circuit voltage, regardless of whether we’re talking about a solar cell or not. We demonstrate via transient and absolute photoluminescence (PL) experiments how the incorporation of Sr significantly reduces the non-radiative recombination losses in the neat perovskite layer and specifically at the perovskite/C 60 interface. |An equation for V EPJ Photovoltaics, 2011. Open-Circuit Voltage Limitation by Surface Recombination in Perovskite Solar Cells ... can be calculated from the impedance Z =DV=DI by selecting a suitable equivalence model, the simplest one being: C = Im 1 Z w: (2) IMPS and IMVS methods were previously employed to the study of recombination and transport processes of charge car-riers in various types of solar cells. Active 1 year, 11 months ago. Spontaneous open-circuit voltage gain of fully fabricated organic solar cells caused by elimination of interfacial energy disorder. Is Voc equal to the difference between fermi level of n-type and p-type parts? Viewed 13k times 0. 36 Full PDFs related to this paper. For solar simulators with common collimation and devices using glass substrates with a thickness ≤1 mm 2, the employment of both masks and cell areas of approximately 1 cm 2 will obviously not impede the open-circuit voltage and fill factor noticeably and will simultaneously also not under- or overestimate the short-circuit current density. Short-circuit solar cell the voltage across the diode is zero; the solar cell provides maximum DOI: 10.1007/s00339-015-9503-7. M. Nath. The latter tends to increase solar cell output voltage while the former acts to erode it. All these HTMs present close chemical and physical propert 2019 Energy and Environmental Science HOT Articles The difference is dependent of the load connected. Download Full PDF Package. You want to have a fill factor as large as possible. No external electric current flows between the terminals. So far we have compared the parameters as deduced from modeling (Tab. pin solar cell under open-circuit conditions. Voc represents the open circuit voltage. The additional components in the equivalent circuit indicate that the internal current source is not in direct interaction with the load components. Alexey Abramov. In this work, we assess the possible reasons for the differences observed in open circuit voltage (VOC) in mixed cation perovskite solar cells when comparing four different hole transport materials (HTMs), namely TAE-1, TAE-3, TAE-4 and spiro-OMeTAD. The profiles are calculated using a computer program (AMPS PC-1D [1]) and parameters we describe in more detail subsequently. 2) that characterize the μc-Si:H solar cells A (F c = 79%, no large grains detected) and the more crystallized cell B (F c = 93%, F lg = 27%), and explained in Sections 5.1 and 5.2, why the latter shows a lower open-circuit voltage and fill factor than the former. 2.75 at the solar cell or at the battery. The reasons why the open circuit voltage (V oc) of high-x CuIn 1-x Ga x Se 2 (CIGS)/ZnO solar cells remain low are discussed. Ideally this decay is linear with a slope inversely proportional to the lifetime. Applied Physics A 2015, 121 (3) , 1261-1269. Figure 2A, main text, shows a band diagram of a generic (organic or inorganic) thin-film solar cell at open circuit. Fate of electrons: Kinetic measurements of open‐circuit voltage decay in nanocrystalline TiO 2 dye‐sensitized solar cells reveal major features of the electron lifetime and recombination mechanisms in these solar cells. Open circuit voltage (or potential) is voltage which is not connected to any load in a circuit. OPEN-CIRCUIT VOLTAGE |The open-circuit voltage, V oc, is the maximum voltage available from a solar cell, and this occurs at zero current. where: is the short circuit current is the open circuit voltage is the maximum current is the maximum voltage The fill factor FF is simply the ratio of the area in the blue rectangle with respect to the pale blue rectangle shown on the graph. Open-circuit solar cell the current is zero; the solar cell delivers maximum voltage; The output power P = I ×V =0 oc 10 Sph qV II I kT exp ⎧⎪ ⎡⎤⎛⎞⎫⎪ =⎨ ⎢⎥⎜⎟−− =⎬ ⎪⎩⎭ ⎣⎦⎝⎠⎪ The open-circuit voltage, at I=0: Note, the maximum achievable V OCmax = V bi. Likewise for closed circuit current. Open-circuit voltage and short-circuit current. The open-circuit voltage, V OC, is the maximum voltage available from a solar cell, and this occurs at zero current. R losses for each of the resistors plus the losses in each of the diodes. There is no external load connected. 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# Blood transfusion indications and reactions
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### Blood transfusion indications and reactions
1. 1. Blood transfusion indications and reactions Dr.Sarmistha DNB Jr.RESIDENT TIRUMALA HOSPITALS
2. 2. Indications for transfusion • Blood transfusions are given to increase oxygen- carrying capacity and intravascular volume. • Hemoglobin value at which blood should be given will have to be a clinical judgment based on many factors- cardiovascular status, age Anticipated additional blood loss Arterial oxygenation Mixed venous oxygen tension
3. 3. Cardiac output, blood volume Oxygen extraction ratio
4. 4. Indications 1. Blood loss greater than 20% of blood volume 2. Hb level less than 8 g/dL 3. Hb level less than 9 to 10 g/dL with major disease (e.g., emphysema, ischemic heart disease) 4. Hb level of less than 10 g/dL with autologous blood 5. Hb level less than 11 to 12 g/dL and ventilator dependent
5. 5. • The transfusion point can be determined preoperatively from the hematocrit and by estimating blood volume • Patients with a normal hematocrit should generally be transfused only aft er losses greater than 10–20% of their blood volume.
6. 6. Average blood volumes.
7. 7. The amount of blood loss can be calculated as- • 1. Estimate blood volume • 2. Estimate the red blood cell volume (RBCV) at the preoperative hematocrit (RBCV preop ). • 3. Estimate RBCV at a hematocrit of 30% (RBCV 30% ), assuming normal blood volume is maintained. • 4. Calculate the RBCV lost when the hematocrit is 30%; RBCV lost = RBCV preop – RBCV 30% . • 5. Allowable blood loss = RBCV lost × 3.
8. 8. • Example • An 85-kg woman has a preoperative hematocrit of 35%. How much blood loss will decrease her hematocrit to 30%? Estimated blood volume = 65 mL/kg × 85 kg = 5525 mL. RBCV 35% = 5525 × 35% = 1934 mL. RBCV 30% = 5525 × 30% = 1658 mL. Red cell loss at 30% = 1934 − 1658 = 276 mL. Allowable blood loss = 3 × 276 mL = 828 mL.
9. 9. Therefore, transfusion should be considered only when this patient’s blood loss exceeds 800 mL. Increasingly, transfusions are not recommended until the hematocrit decreases to 24% or lower (hemoglobin <8.0 g/dL), but it is necessary to take into account the rate of blood loss and comorbid conditions (eg, cardiac disease, in which case transfusion might be indicated if only 800 mL of blood is lost).
10. 10. Clinical guidelines commonly used include: (1) one unit of red blood cells will increase hemoglobin 1 g/dL and the hematocrit 2– 3% in adults; (2) a 10-mL/kg transfusion of red blood cells will increase hemoglobin concentration by 3 g/dL and the hematocrit by 10%.
11. 11. complications 1-Changes in Oxygen Transport Changes in Oxygen Transport RBCs are transfused primarily to increase transport of oxygen to tissues. An increase in the circulating red cell mass produces an increase in oxygen uptake in the lung and a corresponding probable increase in oxygen delivery to tissues. The respiratory function of red cells may be impaired during preservation, making it difficult for them to release oxygen to the tissues immediately after transfusion.
12. 12. 2-coagulation defect Unless a patient has a preoperative coagulopathy (aspirin,antiplatelet drugs hemophilia), A transfusion induce coagulopathy usually occurs only after a large amount of blood has been given (6 to 10 units of PRBCs) This coagulopathy is caused by a combination of factors, of which the most important are the volume of blood given and the duration of hypotension or hypo perfusion. The patient who is hypotensive and has received many units of blood probably has a coagulopathy from a condition that resembles disseminated intravascular coagulation (DIC) and dilution of coagulation factors from stored bank blood
13. 13. • Clinical manifestations include oozing into the surgical field, hematuria, gingival bleeding, petechial bleeding from venipuncture sites, and ecchymoses • 3-dilutional thrombocytopenia • Considering survival time and viability, total platelet activity is only 50% to 70% of the original in vivo activity after 6 hours of storage in bank blood at 4°c.After 24 or 48 hours of storage, platelet activity is only about 10%or 5% of normal
14. 14. • 4-Low levels of Factors V and VIII • These factors gradually decrease to 15%and 50% normal, respectively, after 21 days of storage • 5-Disseminated Intravascular Coagulation-like Syndrome • 6- Citrate Intoxication and Hyperkalemia • 7-Temperature A decrease in body temperature as small as 0.5 to I.0°C may induce shivering postoperatively; this may increase oxygen consumption by as much as 400%. To meet the demands of elevated oxygen consumption, cardiac .output must be Increase
15. 15. • Perhaps the safest and most common method of warming blood is to pass it through plastic coils immersed in warm water (37 to 38°C) bath • 7-Acid-Base Abnormalities the pH of bank blood continues to decrease to about 6.9 after 21 days of storage.
16. 16. Transfusion reactions • 8- Hemolytic Transfusion Reaction • Such a reaction can occur from infusion of as little as 10 mL of blood. Between 20% and 60% of patients with severe symptomatic hemolytic reactions may die, and these deaths usually result from AB0 blood group incompatibility between the donor and the patient
17. 17. • Haptoglobin, which is a protein that can bind about 100 mg of hemoglobin per 100 mL of plasma • A sample of plasma that contains 2 mg/dL of hemoglobin is faintly pink or light brown. When the level of hemoglobin reaches 100 mg/dL, the plasma is red. When the level of plasma hemoglobin reaches 150 mg/dL, hemoglobinuria occur
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All ten (Posted on 2017-05-06)
A & B are 5-digit numbers , such that:
i. A*8=B
ii. The concatenation AB is a pandigital number.
Since 12345 *8=98760, A=12345 B=98760 qualifies as a valid solution.
Find all the others
No Solution Yet Submitted by Ady TZIDON No Rating
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computer solution Comment 1 of 1
A B
12345 98760
10459 83672
10469 83752
10537 84296
10579 84632
10592 84736
10674 85392
10679 85432
10742 85936
10794 86352
10932 87456
10942 87536
10953 87624
10954 87632
12073 96584
12307 98456
found by
DefDbl A-Z
Dim crlf\$, dvsr(300)
Private Sub Form_Load()
Text1.Text = ""
crlf\$ = Chr(13) + Chr(10)
Form1.Visible = True
s\$ = "1234567890"
h\$ = s
Do
a = Val(Left(s, 5))
b = Val(Right(s, 5))
If b = 8 * a And a > 9999 Then Text1.Text = Text1.Text & a & Str(b) & crlf
permute s
DoEvents
Loop Until s = h
End Sub
Posted by Charlie on 2017-05-06 13:35:05
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# A spring-loaded piston-cylinder device contains a mixture
## Problem 93P Chapter 13
Thermodynamics: An Engineering Approach | 8th Edition
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Problem 93P
A spring-loaded piston-cylinder device contains a mixture of gases whose pressure fractions are 25 percent Ne, 50 percent O2, and 25 percent N2. The piston diameter and spring are selected for this device such that the volume is 0.1 m3 when the pressure is 200 kPa and 1.0 m3 when the pressure is 1000 kPa. Initially, the gas is added to this device until the pressure is 200 kPa and the temperature is 10°C. The device is now heated until the pressure is 500 kPa. Calculate the total work and heat transfer for this process.
Step-by-Step Solution:
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Day 1 1/4/2016 Mondays 56:30 Sections with the Professor Adriane Steinacker First Homework posted this Friday Lectures are Webcast Login: phys5b Password: Y0ung2sl1t Class Materials: rulers, compass, scientific calculator Yellow is her favorite color 1) Density (mass density) lett rho ρ...
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##### ISBN: 9780073398174
Thermodynamics: An Engineering Approach was written by and is associated to the ISBN: 9780073398174. This full solution covers the following key subjects: pressure, device, kpa, piston, spring. This expansive textbook survival guide covers 17 chapters, and 2295 solutions. Since the solution to 93P from 13 chapter was answered, more than 370 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 93P from chapter: 13 was answered by , our top Engineering and Tech solution expert on 08/01/17, 09:10AM. This textbook survival guide was created for the textbook: Thermodynamics: An Engineering Approach , edition: 8. The answer to “A spring-loaded piston-cylinder device contains a mixture of gases whose pressure fractions are 25 percent Ne, 50 percent O2, and 25 percent N2. The piston diameter and spring are selected for this device such that the volume is 0.1 m3 when the pressure is 200 kPa and 1.0 m3 when the pressure is 1000 kPa. Initially, the gas is added to this device until the pressure is 200 kPa and the temperature is 10°C. The device is now heated until the pressure is 500 kPa. Calculate the total work and heat transfer for this process.” is broken down into a number of easy to follow steps, and 95 words.
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Get Full Access to Thermodynamics: An Engineering Approach - 8 Edition - Chapter 13 - Problem 93p | 712 | 2,852 | {"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-2020-05 | latest | en | 0.901028 |
https://cp2k.github.io/dbcsr/develop/proc/btree_find_full_i8_sp2d.html | 1,718,435,906,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861584.65/warc/CC-MAIN-20240615062230-20240615092230-00697.warc.gz | 155,191,788 | 4,614 | # btree_find_full_i8_sp2d Subroutine
## private subroutine btree_find_full_i8_sp2d(tree, key, node, position, ge_position, short)
### Arguments
Type IntentOptional Attributes Name
type(btree_i8_sp2d), intent(in) :: tree
integer(kind=keyt), intent(in) :: key
type(btree_node_i8_sp2d), POINTER :: node
integer, intent(out) :: position
integer, intent(out), optional :: ge_position
logical, intent(in), optional :: short
## Source Code
SUBROUTINE btree_find_full_i8_sp2d (tree, key, node, position, ge_position, short)
TYPE(btree_i8_sp2d), INTENT(IN) :: tree
INTEGER(KIND=keyt), INTENT(IN) :: key
TYPE(btree_node_i8_sp2d), POINTER :: node
INTEGER, INTENT(OUT) :: position
INTEGER, INTENT(OUT), OPTIONAL :: ge_position
LOGICAL, INTENT(IN), OPTIONAL :: short
INTEGER :: gti ! Used mark searches
LOGICAL :: stop_short
!
stop_short = .FALSE.
IF (PRESENT(short)) stop_short = short
NULLIFY (node)
position = 0
IF (PRESENT(ge_position)) ge_position = 0
!IF (tree%b%n .EQ. 0) RETURN
IF (.NOT. ASSOCIATED(tree%b%root)) RETURN
gti = 1
! Try to find the key in the given node. If it's found, then
! return the node.
node => tree%b%root
descent: DO WHILE (.TRUE.)
! Try to find the first element equal to or greater than the
! one we're searching for.
CALL btree_node_find_ge_pos_i8_sp2d (node%keys, key, position, node%filled)
! One of three things is now true about position: it's now
! greater than the number of keys (if all keys are smaller), or
! it points to the key that is equal to or greater than the one
! we are searching for. If it is found and we are just
! searching for one equal element (i.e., user search), we can
! return.
IF (stop_short .AND. position .LE. node%filled) THEN
IF (node%keys(position) .EQ. key) THEN
IF (PRESENT(ge_position)) ge_position = position
RETURN
END IF
END IF
! If the key is not found, then either return the GE position
! if we're in a leaf (case 2 here), otherwise descend into the
! subtrees.
!CALL btree_node_find_gt_pos_i8_sp2d (node%keys, key, gti, node%filled, position)
CALL btree_node_find_gte_pos_i8_sp2d (node%keys, key, gti, node%filled, position)
IF (ASSOCIATED(node%subtrees(1)%node)) THEN
node => node%subtrees(gti)%node
ELSE
IF (PRESENT(ge_position)) ge_position = gti
position = 0
RETURN
END IF
END DO descent
END SUBROUTINE btree_find_full_i8_sp2d | 694 | 2,307 | {"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} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.674364 |
https://zerojudge.tw/ShowThread?postid=65&reply=63 | 1,627,587,168,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00032.warc.gz | 1,164,190,892 | 14,311 | #63: 問題
#### sa411022 (sa411022)
School : 國立臺中第二高級中學
ID : 1027
2010-04-05 21:52:41
b079. F. 鬧鐘 -- | From: [61.66.76.57] | Post Date : 2007-12-10 21:01
#65: Re:問題
#### guest
School : No School
ID : 3
2017-09-10 12:45:22
b079. F. 鬧鐘 -- | From: [59.121.39.206] | Post Date : 2007-12-10 21:36
Dim a(1000000) As Double
For i = 1 To 1000000
If i <= 2 Then
a(i) = 1
Else
a(i) = a(i - a(i - 1)) + a(i - 1 - a(i - 2))
End If
Next
#1896: Re:問題
#### david942j (文旋)
School : 臺北市立成功高級中學
ID : 6086
2017-02-18 13:17:39
b079. F. 鬧鐘 -- | From: [58.115.132.57] | Post Date : 2009-05-03 12:00
14 ms耶@@
#1903: Re:問題
#### guest
School : No School
ID : 3
2017-09-10 12:45:22
b079. F. 鬧鐘 -- | From: [74.6.8.109] | Post Date : 2009-05-05 01:25
ZeroJudge Forum | 389 | 749 | {"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} | 2.578125 | 3 | CC-MAIN-2021-31 | latest | en | 0.294185 |
https://dumpz.org/aR6eTkmPsFrG | 1,670,423,150,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00629.warc.gz | 234,386,152 | 5,385 | # include iostream include math using namespace std struct Var int xi in
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121``` ```#include #include using namespace std; struct Var { int xi; int mi; float vi; }; int main() { int n = 50; int stat[n] = { 16, 14, 17, 15, 20, 16, 17, 18, 15, 13, 18, 16, 14, 17, 12, 19, 16, 15, 17, 18, 14, 19, 16, 18, 15, 17, 13, 16, 20, 15, 21, 17, 15, 16, 13, 16, 19, 18, 14, 15, 15, 16, 17, 14, 16, 19, 21, 15, 17, 18 }; cout << "Entered static row:" << endl << "["; for (int k = 0; k < n; ++k) { cout << stat[k]; if(k != n - 1) { cout << ", "; } } cout << "]" << endl; int variation[n]; for (int j = 0; j < n; ++j) { variation[j] = 0; } for (int i = 0; i < n; ++i) { variation[stat[i]] += 1; } int count = 0; for (int i = 0; i < n; ++i) { if(variation[i] != 0) { count++; } } Var variative[count]; int j = 0; int min = 9999999, max = 0; for (int i = 0; i < n; ++i) { if(variation[i] != 0) { variative[j].xi = i; variative[j].mi = variation[i]; variative[j].vi = (float) variation[i] / n; if(min > variative[j].xi) { min = variative[j].xi; } if(max < variative[j].xi) { max = variative[j].xi; } j++; } } cout << "Extended variation row:" << endl; for(int i = 0; i < count; ++i) { cout << i << " => Xi = " << variative[i].xi << ", Vi = " << variative[i].vi << ", mi = " << variative[i].mi << endl; } cout << "Interval variation row:" << endl; double deltax = (double) (max - min) / (1.0 + 3.32 * log10(50)); deltax = ceil(deltax); int dimension = ceil((max - min) / deltax); Var x[dimension]; for (int i = 0; i < dimension; ++i) { if(i == 0) { x[i].xi = min - deltax / 2; } else { x[i].xi = x[i - 1].xi + 2; } x[i].mi = variative[i * 2].mi + variative[i * 2 + 1].mi; x[i].vi = variative[i * 2].vi + variative[i * 2 + 1].vi; } for (int i = 0; i < dimension; ++i) { cout << i << " => Xi = " << x[i].xi << ", Vi = " << x[i].vi << ", mi = " << x[i].mi << endl; } cout << "cumulate: " << endl; Var cumulate[dimension]; float vi = 1; for (int i = dimension; i > 0; i--) { vi -= x[i].vi; cumulate[i - 1].vi = vi; } for (int i = 0; i < dimension; ++i) { cout << "Vi = " << cumulate[i].vi << ", Xi = " << x[i].xi + 1 << endl; } cout << "ogiva: " << endl; for (int i = 0; i < dimension; ++i) { cout << "Xi = " << x[i].xi + 1 << ", Vi = " << cumulate[i].vi << endl; } return 0; } ``` | 1,171 | 2,635 | {"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-2022-49 | latest | en | 0.425429 |
https://navigueweb.com/how-dense-is-ocean-water/ | 1,674,781,162,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00598.warc.gz | 445,398,280 | 29,294 | # How dense is ocean water?
Average density at the surface is 1.025 kg/l. Seawater is denser than both fresh water and pure water (density 1.0 kg/l at 4 °C (39 °F)) because the dissolved salts increase the mass by a larger proportion than the volume. The freezing point of seawater decreases as salt concentration increases.
Contents show
## Which ocean is most dense?
The Pacific has most of the lightest water with densities below 26.0, whereas the Atlantic has most of the densest water between 27.5 and 28.0.
## Is ocean water more dense or less dense?
Ocean water is more dense because of the salt in it. Density of ocean water at the sea surface is about 1027 kg/m3. There are two main factors that make ocean water more or less dense than about 1027 kg/m3: the temperature of the water and the salinity of the water. Ocean water gets more dense as temperature goes down.
Read Also Can you see Tutankhamun mummy?
## Is sea water denser than water?
saltwater has a higher density than freshwater. less dense matter will lie above more dense matter.
## What makes ocean water more dense?
There are two main factors that make ocean water more or less dense: temperature and salinity. Cold, salty water is denser than warm, fresher water and will sink below the less dense layer. Density is defined as the measure of a material’s mass (e.g. grams) divided by its volume (e.g. milliliters).
## What is the density of ocean water?
The density of fresh water is 1 g/cm3 at 4o C (see section 5.1), but the addition of salts and other dissolved substances increases surface seawater density to between 1.02 and 1.03 g/cm3. The density of seawater can be increased by reducing its temperature, increasing its salinity , or increasing the pressure.
## Where is the least dense water?
Water is densest at 3.98°C and is least dense at 0°C (freezing point).
## Why is the deepest part of the ocean the most dense?
Deep water is denser than shallow water. The water molecules are packed together more tightly because of the weight of water above pushing down.
## Is cold water more dense?
Cold water has a higher density than warm water. Water gets colder with depth because cold, salty ocean water sinks to the bottom of the ocean basins below the less dense warmer water near the surface.
## Which ocean is the deepest?
The Mariana Trench, in the Pacific Ocean, is the deepest location on Earth.
## Why is saltwater denser than freshwater?
Because of the added weight of the extra salt. Water loves to bond with salt. The h20 molecules cluster around the salt molecules, and the result is that saltwater has more molecules overall than freshwater. When you’ve added more weight to that cubic foot of water (the salt), you are producing a denser type of water.
## Is all ocean water salty?
Saline Water and Salinity
You are concerned with freshwater to serve your life’s every need. But, most of Earth’s water, and almost all of the water that people can access, is saline, or salty water. Just look at the oceans and remember that oceans comprise about 97% of all water on, in, and above the Earth.
## How cold is deep ocean water?
Therefore, the deep ocean (below about 200 meters depth) is cold, with an average temperature of only 4°C (39°F). Cold water is also more dense, and as a result heavier, than warm water. Colder water sinks below the warm water at the surface, which contributes to the coldness of the deep ocean.
## Which ocean is not salt water?
The major oceans all over the Earth are the Atlantic Ocean, Pacific Ocean, Indian Ocean, Antarctic, and Arctic Oceans. All oceans are known to have salt in a dissolved state, but the only oceans that have no salt content are the Arctic and Antarctic Oceans.
## Why is the Great Garbage Patch important?
Conversely, plastics can also absorb pollutants, such as PCBs, from the seawater. These chemicals can then enter the food chain when consumed by marine life. Because the Great Pacific Garbage Patch is so far from any country’s coastline, no nation will take responsibility or provide the funding to clean it up.
## How deep does the ocean go down?
The average depth of the ocean is about 3,688 meters (12,100 feet). The deepest part of the ocean is called the Challenger Deep and is located beneath the western Pacific Ocean in the southern end of the Mariana Trench, which runs several hundred kilometers southwest of the U.S. territorial island of Guam.
Read Also How did the Americans justify moving westward?
## When mixed waters are denser?
Cabbeling is when two separate water parcels mix to form a third which sinks below both parents. The combined water parcel is denser than the original two water parcels. The two parent water parcels may have the same density, but they have different properties; for instance, different salinities and temperatures.
## What factors that affect seawater density?
The density of seawater depends on temperature and salinity. Higher temperatures decrease the density of seawater, while higher salinity increases the density of seawater.
## Where is the warmest ocean water found?
The hottest ocean area is in the Persian Gulf, where water temperatures at the surface exceed 90 degrees Fahrenheit in the summer. Another hot area exists in the Red Sea, where a temperature of 132.8 degrees Fahrenheit has been recorded at a depth of about 6,500 feet.
## Is water heavier than ice?
The “stuff” (molecules) in water is more tightly packed than in ice, so water has greater density than ice. Don’t let the fact that ice is a solid fool you! As water freezes it expands. So, ice has more volume (it takes up more space, but has less density) than water.
## Why is water densest at 4c?
At 4 °C, the clusters start forming. The molecules are still slowing down and coming closer together, but the formation of clusters makes the molecules be further apart. Cluster formation is the bigger effect, so the density starts to decrease. Thus, the density of water is a maximum at 4 °C.
## Is hot water heavier or lighter?
When water freezes, however, it expands, becoming less dense. If equal volumes of cold water (down to 4deg. C, 39.2deg. F) and hot water are compared, cold water weighs more than hot water.
## What’s the density of honey?
The density of honey typically ranges between 1.38 and 1.45 kg/l at 20 °C.
## Why is water hotter at night?
The heat that the ocean absorbs is mixed with the lower water quickly. That mixing spreads the heat around. At night, while the land cools off quickly, the water at the surface is kept warmer because the water is mixed around with the warmer water underneath.
## Can the ocean freeze?
Ocean water freezes just like freshwater, but at lower temperatures. Fresh water freezes at 32 degrees Fahrenheit but seawater freezes at about 28.4 degrees Fahrenheit , because of the salt in it. When seawater freezes, however, the ice contains very little salt because only the water part freezes.
## Can a human Go to the bottom of the ocean?
The deepest point ever reached by man is 35,858 feet below the surface of the ocean, which happens to be as deep as water gets on earth. To go deeper, you’ll have to travel to the bottom of the Challenger Deep, a section of the Mariana Trench under the Pacific Ocean 200 miles southwest of Guam.
## Has anyone been to bottom of Mariana Trench?
While thousands of climbers have successfully scaled Mount Everest, the highest point on Earth, only two people have descended to the planet’s deepest point, the Challenger Deep in the Pacific Ocean’s Mariana Trench.
## Which is heavier fresh or salt water?
Saltwater is more dense than freshwater. For example, the density of freshwater in the Mississippi River in southern Louisiana is 0.999. The density of saltwater at Japanese ports is 1.025. Due to the more dense/heavier water in Japan, the vessel will automatically rise about 11.4 inches (29 centimeters).
## Why is the ocean blue?
The ocean is blue because water absorbs colors in the red part of the light spectrum. Like a filter, this leaves behind colors in the blue part of the light spectrum for us to see. The ocean may also take on green, red, or other hues as light bounces off of floating sediments and particles in the water.
## Why are tears salty?
Tears and all of our other body fluids are salty because of electrolytes, also known as salt ions. Our bodies use electrolytes to create electricity that helps power our brains and move our muscles. Electrolytes contain: Sodium (which accounts for the saltiness)
The sea is called “dead” because its high salinity prevents macroscopic aquatic organisms, such as fish and aquatic plants, from living in it, though minuscule quantities of bacteria and microbial fungi are present. In times of flood, the salt content of the Dead Sea can drop from its usual 35% to 30% or lower.
Read Also How deep is the Shinano River?
## Can you drown in saltwater?
In dense, salty water, a little body displaces a lot of mass, and most of the body stays out of the water so, it’s hard to drown a person when most of their body is floating on top of the water.
## Why is rainwater not salty?
But over time, as rain fell to the Earth and ran over the land, breaking up rocks and transporting their minerals to the ocean, the ocean has become saltier. Rain replenishes freshwater in rivers and streams, so they don’t taste salty.
## Can you drink ocean water if boiled?
Boiling seawater does not make it safe to drink because it does not remove the salt. On Average, seawater holds 3.5% salt, too much for the body to process. As seawater boils, it evaporates, leaving the salt behind. You’re making the seawater saltier by boiling it.
## Is there Whale Sperm in the ocean?
Sperm whales live all over the world, meaning deposits of ambergris could be found floating on any ocean or washed up on most shorelines. But it is uncommon, found in less than 5% of whale carcasses.
## Why is lake water not salty?
So, the answer to why rivers and lakes are not as salty as the oceans is that salts and minerals that enter have an avenue for escape, which is a path to the oceans. Oceans don’t have an outlet though.
## How cold is the moon?
Temperatures on the moon are very hot in the daytime, about 100 degrees C. At night, the lunar surface gets very cold, as cold as minus 173 degrees C. This wide variation is because Earth’s moon has no atmosphere to hold in heat at night or prevent the surface from getting so hot during the day.
## Is it hot in the Mariana Trench?
It’s Hot and It’s Cold
The water there tends to range between 34 to 39 degrees Fahrenheit. But what’s surprising is how hot the water can get, too. There are hydrothermal vents throughout the trench.
## How cold was the water when the Titanic sank?
The Titanic ship submerged into the North Atlantic Ocean, off the coast of Newfoundland, by colliding with an enormous iceberg. When it sank, the water temperature was 27°F which is around -2.7°C.
## How many garbage Patchs are in the ocean?
There are five gyres to be exact—the North Atlantic Gyre, the South Atlantic Gyre, the North Pacific Gyre, the South Pacific Gyre, and the Indian Ocean Gyre—that have a significant impact on the ocean. The big five help drive the so-called oceanic conveyor belt that helps circulate ocean waters around the globe.
## Can you walk on garbage Island?
Can you walk on The Great Pacific Garbage Patch? No, you cannot. Most of the debris floats below the surface and cannot be seen from a boat. It’s possible to sail or swim through parts of the Great Pacific Garbage Patch and not see a single piece of plastic.
## What percentage of ocean plastic is fishing nets?
Fishing Nets For Miles (Literally)
The research team found that 46 percent of the plastic in the patch by weight came from one source: fishing nets.
## Is the ocean deeper than Mount Everest?
The deepest part of the ocean, the hadal zone, is anywhere deeper than six kilometres. Challenger Deep, in the Mariana Trench, is the deepest point in the ocean known so far, at approximately 11 kilometres – deeper than Mount Everest is tall.
## Are oceans getting deeper?
The Atlantic Ocean crushes that depth, however, deepening in(?) at a whopping 27,000 feet deep. For reference, that’s approximately as deep as Mount Everest is tall. The deepest floor of any ocean on Earth, however, is the Mariana Trench, which is in the western Pacific Ocean.
## What is in the bottom of the ocean?
The bottom of the deep sea has several features that contribute to the diversity of this habitat. The main features are mid-oceanic ridges, hydrothermal vents, mud volcanoes, seamounts, canyons and cold seeps. Carcasses of large animals also contribute to habitat diversity.
## Does water get denser as you go deeper?
You can see density increases with increasing depth. The pycnocline are layers of water where the water density changes rapidly with depth. This density-depth profile is typical of what you might expect to find at a latitude of 30-40 degrees south. The density of pure water is 1000 kg/m3.
## Is water super dense?
At 39°F (or 3.98°C to be exact) water is the most dense. This is because the molecules are closest together at this temperature.
## What type of ocean water is denser?
The Pacific has most of the lightest water with densities below 26.0, whereas the Atlantic has most of the densest water between 27.5 and 28.0. Antarctic bottom water is indeed densest for Pacific and Indian oceans but not for the Atlantic which has a lot of similarly dense water.
## How is seawater density measured?
Hydrostatic weighing apparatuses are known to be the most accurate densimeters for determining the density of liquids. They usually use a sinker, which is connected to a balance by a suspension. This leads to some difficulties that are increased by the properties of seawater.
## Is water heavier than sand?
After all, water is not a heavy (or dense) as sand and if the sand in the bucket has to make room for the water it would have to hold less sand and therefore the sand and water combination should weigh less than the sand alone! Now, there is a good question!
## Is water wet?
If we define “wet” as “made of liquid or moisture”, then water is definitely wet because it is made of liquid, and in this sense, all liquids are wet because they are all made of liquids.
## Is water heavier than beer?
For all practical purposes beer weighs the same as water, which is 8.34 lb. per gallon or 2.2 lb. per liter.
## What happens to water when it warms above 4 C?
As the temperature of warm water decreases, the water molecules slow down and the density increases. At 4 °C, the clusters start forming. The molecules are still slowing down and coming closer together, but the formation of clusters makes the molecules be further apart.
## What is the maximum density of water?
Water. An especially notable irregular maximum density is that of water, which reaches a density peak at 4 °C (39 °F). This has important ramifications in Earth’s ecosystem.
## What happens to water at 4c?
A: 4 degrees C turns out to be the temperature at which liquid water has the highest density. If you heat it or cool it, it will expand. The expansion of water when you cool it to lower temperatures is unusual, since most liquids contract when they’re cooled. | 3,437 | 15,391 | {"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-2023-06 | longest | en | 0.940965 |
http://datascience.sharerecipe.net/2020/05/27/how-to-scale-data-with-outliers-for-machine-learning/ | 1,702,136,931,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100912.91/warc/CC-MAIN-20231209134916-20231209164916-00194.warc.gz | 14,561,160 | 10,191 | # How to Scale Data With Outliers for Machine Learning
Last Updated on May 27, 2020Many machine learning algorithms perform better when numerical input variables are scaled to a standard range.
This includes algorithms that use a weighted sum of the input, like linear regression, and algorithms that use distance measures, like k-nearest neighbors.
Standardizing is a popular scaling technique that subtracts the mean from values and divides by the standard deviation, transforming the probability distribution for an input variable to a standard Gaussian (zero mean and unit variance).
Standardization can become skewed or biased if the input variable contains outlier values.
To overcome this, the median and interquartile range can be used when standardizing numerical input variables, generally referred to as robust scaling.
In this tutorial, you will discover how to use robust scaler transforms to standardize numerical input variables for classification and regression.
After completing this tutorial, you will know:Let’s get started.
How to Use Robust Scaler Transforms for Machine LearningPhoto by Ray in Manila, some rights reserved.
This tutorial is divided into five parts; they are:It is common to scale data prior to fitting a machine learning model.
This is because data often consists of many different input variables or features (columns) and each may have a different range of values or units of measure, such as feet, miles, kilograms, dollars, etc.
If there are input variables that have very large values relative to the other input variables, these large values can dominate or skew some machine learning algorithms.
The result is that the algorithms pay most of their attention to the large values and ignore the variables with smaller values.
This includes algorithms that use a weighted sum of inputs like linear regression, logistic regression, and artificial neural networks, as well as algorithms that use distance measures between examples, such as k-nearest neighbors and support vector machines.
As such, it is normal to scale input variables to a common range as a data preparation technique prior to fitting a model.
One approach to data scaling involves calculating the mean and standard deviation of each variable and using these values to scale the values to have a mean of zero and a standard deviation of one, a so-called “standard normal” probability distribution.
This process is called standardization and is most useful when input variables have a Gaussian probability distribution.
Standardization is calculated by subtracting the mean value and dividing by the standard deviation.
Sometimes an input variable may have outlier values.
These are values on the edge of the distribution that may have a low probability of occurrence, yet are overrepresented for some reason.
Outliers can skew a probability distribution and make data scaling using standardization difficult as the calculated mean and standard deviation will be skewed by the presence of the outliers.
One approach to standardizing input variables in the presence of outliers is to ignore the outliers from the calculation of the mean and standard deviation, then use the calculated values to scale the variable.
This is called robust standardization or robust data scaling.
This can be achieved by calculating the median (50th percentile) and the 25th and 75th percentiles.
The values of each variable then have their median subtracted and are divided by the interquartile range (IQR) which is the difference between the 75th and 25th percentiles.
The resulting variable has a zero mean and median and a standard deviation of 1, although not skewed by outliers and the outliers are still present with the same relative relationships to other values.
The robust scaler transform is available in the scikit-learn Python machine learning library via the RobustScaler class.
The “with_centering” argument controls whether the value is centered to zero (median is subtracted) and defaults to True.
The “with_scaling” argument controls whether the value is scaled to the IQR (standard deviation set to one) or not and defaults to True.
Interestingly, the definition of the scaling range can be specified via the “quantile_range” argument.
It takes a tuple of two integers between 0 and 100 and defaults to the percentile values of the IQR, specifically (25, 75).
Changing this will change the definition of outliers and the scope of the scaling.
We will take a closer look at how to use the robust scaler transforms on a real dataset.
First, let’s introduce a real dataset.
The sonar dataset is a standard machine learning dataset for binary classification.
It involves 60 real-valued inputs and a two-class target variable.
There are 208 examples in the dataset and the classes are reasonably balanced.
A baseline classification algorithm can achieve a classification accuracy of about 53.
4 percent using repeated stratified 10-fold cross-validation.
Top performance on this dataset is about 88 percent using repeated stratified 10-fold cross-validation.
The dataset describes radar returns of rocks or simulated mines.
First, let’s load and summarize the dataset.
The complete example is listed below.
Running the example first summarizes the shape of the loaded dataset.
This confirms the 60 input variables, one output variable, and 208 rows of data.
A statistical summary of the input variables is provided showing that values are numeric and range approximately from 0 to 1.
Finally, a histogram is created for each input variable.
If we ignore the clutter of the plots and focus on the histograms themselves, we can see that many variables have a skewed distribution.
The dataset provides a good candidate for using a robust scaler transform to standardize the data in the presence of skewed distributions and outliers.
Histogram Plots of Input Variables for the Sonar Binary Classification DatasetNext, let’s fit and evaluate a machine learning model on the raw dataset.
We will use a k-nearest neighbor algorithm with default hyperparameters and evaluate it using repeated stratified k-fold cross-validation.
The complete example is listed below.
Running the example evaluates a KNN model on the raw sonar dataset.
We can see that the model achieved a mean classification accuracy of about 79.
7 percent, showing that it has skill (better than 53.
4 percent) and is in the ball-park of good performance (88 percent).
Next, let’s explore a robust scaling transform of the dataset.
We can apply the robust scaler to the Sonar dataset directly.
We will use the default configuration and scale values to the IQR.
First, a RobustScaler instance is defined with default hyperparameters.
Once defined, we can call the fit_transform() function and pass it to our dataset to create a quantile transformed version of our dataset.
Let’s try it on our sonar dataset.
The complete example of creating a robust scaler transform of the sonar dataset and plotting histograms of the result is listed below.
Running the example first reports a summary of each input variable.
We can see that the distributions have been adjusted.
The median values are now zero and the standard deviation values are now close to 1.
0.
Histogram plots of the variables are created, although the distributions don’t look much different from their original distributions seen in the previous section.
Histogram Plots of Robust Scaler Transformed Input Variables for the Sonar DatasetNext, let’s evaluate the same KNN model as the previous section, but in this case on a robust scaler transform of the dataset.
The complete example is listed below.
Running the example, we can see that the robust scaler transform results in a lift in performance from 79.
7 percent accuracy without the transform to about 81.
9 percent with the transform.
Next, let’s explore the effect of different scaling ranges.
The range used to scale each variable is chosen by default as the IQR is bounded by the 25th and 75th percentiles.
This is specified by the “quantile_range” argument as a tuple.
Other values can be specified and might improve the performance of the model, such as a wider range, allowing fewer values to be considered outliers, or a more narrow range, allowing more values to be considered outliers.
The example below explores the effect of different definitions of the range from 1st to the 99th percentiles to 30th to 70th percentiles.
The complete example is listed below.
Running the example reports the mean classification accuracy for each value-defined IQR range.
We can see that the default of 25th to 75th percentile achieves the best results, although the values of 20-80 and 30-70 achieve results that are very similar.
Box and whisker plots are created to summarize the classification accuracy scores for each IQR range.
We can see a marked difference in the distribution and mean accuracy with the larger ranges of 25-75 and 30-70 percentiles.
Box Plots of Robust Scaler IQR Range vs Classification Accuracy of KNN on the Sonar DatasetThis section provides more resources on the topic if you are looking to go deeper.
In this tutorial, you discovered how to use robust scaler transforms to standardize numerical input variables for classification and regression. | 1,814 | 9,336 | {"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} | 2.9375 | 3 | CC-MAIN-2023-50 | latest | en | 0.896378 |
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# Pick 3 Selection Technique
Topic closed. 12 replies. Last post 13 years ago by PAStone.
Page 1 of 1
Pennsylvania
United States
Member #2218
September 1, 2003
5387 Posts
Online
Posted: September 16, 2003, 8:27 pm - IP Logged
Has anyone ever had success using this technique for Pick 3?
2. Determine how many average times that any 2 numbers from the last previous drawing hit on the following draw.
For exmple:
After running your Pick 3 history file you determine that 2 numbers from previous drawings occur on average about 5 to 6 times.
You notice that 2 numbers from a previous drawing have not hit for the last 4 night's or draws.
Say for example tonight's draw was 5 - 2 - 4
Would you consider playing the following numbers for the next drawing:
2 - 4 (0 through 9)
4 - 5 (0 through 9)
4 - 2 (0 through 9)
2 - 5 (0 through 9)
Filtering using 2 even - 1 odd or 2 odd 1 even
Using sums of 10 through 18
Killeen
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Posted: September 17, 2003, 12:42 pm - IP Logged
I don't understand what you are doing please explain. Thanks Helen
Killeen
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Posted: September 17, 2003, 12:43 pm - IP Logged
Helen email hgaynair!@hot.rr.com Thanks
United States
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Posted: September 17, 2003, 12:54 pm - IP Logged
Hello, are you saying that you are looking for similarities througout the Pick 3 history. Because that is what I use. I rank the digits based upon the number of times they have hit. I'm in Maryland. But please explain your system, I'm very interested.
United States
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Posted: September 17, 2003, 1:00 pm - IP Logged
Sugge
"It's not easy, being, GREEN! "
United States
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Posted: September 17, 2003, 1:05 pm - IP Logged
Jazzy, what is PM?
United States
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2723 Posts
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Posted: September 17, 2003, 1:13 pm - IP Logged
Priviate Message(ner), it allows members to respond too your questions and request in a little more depth.
"It's not easy, being, GREEN! "
United States
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Posted: September 17, 2003, 1:20 pm - IP Logged
I like an open forum approach most of the time, that way many people with the same question can get an answer.
Charlotte North Carolina
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Posted: September 17, 2003, 7:20 pm - IP Logged
I agree Levrne, I have learned a lot just by reading about everyone's systems and questions from other members helps me to understand the concept.
wpb
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Posted: September 17, 2003, 7:28 pm - IP Logged
Thanks for the support wpb! I to have have learned so much just by reading the various topics.
Killeen
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Posted: September 17, 2003, 7:55 pm - IP Logged
This is Helen I cannot find it and please explain your system Thanks
texas
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Posted: September 17, 2003, 8:06 pm - IP Logged
i would also like to hear more lotto genie. sounds like your thinking right 2 me
United States
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Posted: September 17, 2003, 9:45 pm - IP Logged
I use Versa Bet 34 and evaluate my Midday Drawings, Evening Drawing and Combined Midday and Evening (Merged).
So you have 3 reports depending on how you decide to play.
You don't need software to do this but it helps with averages from complete history files.
I run my "Profile Reports" for each drawing(s) and evaluate what the average is for 2 numbers repeating from the prevous draw.
After I posted this idea, Pennsylvania Pick 3 draws were 941 evening and 541 day. 41 repeated the next day.
Pennsylvania average is 5 to 6 days. So, if I look at my report and see 2 numbers have not repeated for 4 days I start playing.
I never play the same positions from previous draws. For example I would rotate the 941. You would have 19x, 49x, 41x. "X" could be any number you want.
Again, you don't need the software you can use pencil and paper.
I believe the average for all states is around 5 to 6 draws. Check your own states Pick 3 histories.
Good Luck!!
Page 1 of 1 | 1,358 | 4,474 | {"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} | 2.984375 | 3 | CC-MAIN-2016-50 | latest | en | 0.838487 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=14&t=46838 | 1,606,564,926,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195417.37/warc/CC-MAIN-20201128095617-20201128125617-00559.warc.gz | 364,143,323 | 11,819 | ## Question 1.A.3
$c=\lambda v$
Amir Bayat
Posts: 115
Joined: Sat Sep 07, 2019 12:16 am
### Question 1.A.3
The question states:
Which of the following happens when the frequency of electromagnetic radiation decreases? Explain your reasoning.
(a) The speed of the radiation decreases.
(b) The wavelength of the radiation decreases.
(c) The extent of the change in the electrical field at a given point decreases.
(d) The energy of the radiation increases.
The correct answer is C. I understand how the answer cannot be A,B, or C, but how is the answer C? Is this due to the change in area of the electric field? Is there a formula explaining this?
Joanne Lee 1J
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am
### Re: Question 1.A.3
The extent of the change in the electrical field is referring to the slope of the waves and so as the frequency decreases, the waves of would be broader and more spread out, therefore meaning that the slopes of the waves would be less steep. This decrease in the slope means that the extent of the change in the electrical field would be decreasing as frequency decreases. To my knowledge, there is not formula explaining this concept.
Joseph Saba
Posts: 154
Joined: Thu Jul 11, 2019 12:16 am
### Re: Question 1.A.3
I also used the process of elimination on this problem. You can calculate and reason out all of the other responses other than part C.
Anthony Hatashita 4H
Posts: 103
Joined: Wed Sep 18, 2019 12:21 am
### Re: Question 1.A.3
As I first did this problem I immediately looked for a "the wavelength increases" answer, but there was none. I also used the process of elimination to eliminate a b and d, leaving only c as the possible answer. From the correct answer you can learn that higher wavelength results in less change in the electrical field at any point.
Alexa Mugol 3I
Posts: 54
Joined: Sat Aug 17, 2019 12:17 am
### Re: Question 1.A.3
I was confused about this too. If you look at Figure 1A.7 in the text, it says that "the electric field of electromagnetic radiation oscillates in space and time" (referring to the oscillating waves in the radiation). So, I think when C says "extent of change in the electrical field," it's referring to the oscillation. Therefore, since the frequency is decreased, it means less oscillation in the waves (or broader waves as someone had mentioned earlier). If the frequency was increased, you would see more oscillation and shorter, smaller waves. | 605 | 2,451 | {"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": 1, "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-2020-50 | latest | en | 0.936516 |
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Anonymous
Not applicable
## Conditional Sum
I have 2 columns. Year (Dimension) & Flag (Y/N). I want to show a line graph with year as dimension & 2 lines (count of all Y & count of all N). How can I create a new measure which is essentially a conditional SUM. Something along with this line: SUM(if flag = 'Y' then 1 else 0) or count(*) where flag = 'Y'.
Thanks.
Dinesh.
1 Solution
Accepted Solutions
Specialist
2 Expressions:
Count(if(Flag = 'Y', 1))
Count(if(Flag = 'N', 1))
And Year as the dimension.
Let me know if it works or not.
9 Replies
Specialist
2 Expressions:
Count(if(Flag = 'Y', 1))
Count(if(Flag = 'N', 1))
And Year as the dimension.
Let me know if it works or not.
Not applicable
Author
The best option for you is to use Set Analysis.
Write something to the effect of:
COUNT({\$<{flag = {Y}> flag}
and
COUNT({\$<{flag = {N}> flag}
Hope this helps,
Peter
Anonymous
Not applicable
Author
How do I adjust the horizontal setting of the placement of the legend? All I see is the vertical alignment. I want to place the legend at top center & cannot find an option.
Thanks,
Dinesh.
Specialist
Do you mean to put it above the chart?
Specialist
Did you want it like this:
Anonymous
Not applicable
Author
yes above the chart
Anonymous
Not applicable
Author
Did you forget any attachment?
Specialist
I thought i pasted it.
Not sure, if you can see the picture.
But one way to get the legend on the top is:
with the control on the chart, press and hold Ctrl + Shift ... you will get the choice to move around things within the chart. Like resizing LEGENDs moving them around. Moving the Text in Chart pieces.
But, not an easy one though. if something gets screwed - feel free to Reset them on the properties (Reset user sizing and Reset User Docking)
Anonymous
Not applicable
Author
Thanks. Appreciate it. Did not get the exact results I was looking for (was looking for side by side, but came top to bottom). But got pretty close!!!
Community Browser | 512 | 2,057 | {"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} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.900849 |
http://webphysics.davidson.edu/Projects/AnAntonelli/node21.html | 1,539,615,611,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509326.21/warc/CC-MAIN-20181015142752-20181015164252-00128.warc.gz | 386,968,202 | 3,378 | Next: The Bloch Vector Up: Simulations Previous: Simulations
## Rabi Oscillations
Let us begin our examination of semiclassical atom--field interactions by placing a single atom at some point in space and aiming a laser at it. This laser will be tuned in such a way that only two energy levels of the atom are linked by the laser's frequency which does not have to be exactly equal to the atom's transition frequency but must be quite close. If the laser frequency is not close enough, then we lose the ability to approximate the atom by two energy levels.
Setting up this system in BlochApp is simple. Let us assume that initially the atom is in the lower state, i.e., . Since the atom is in the lower state, the probability of being in that state is one, and the probability of being in the upper state is zero; so, and . The off--diagonal elements are also zero.
Turn off all the switches in the Atomic item under Parameters. These options are necessary only when we consider a statistical ensemble of atoms. The sliders have the two most important variables: Amp and Freq. Freq refers to the laser frequency, , and Amp refers to the scaled amplitude of the electric field, . Remember that
The Amp in BlochApp is a scaled amplitude which absorbs both e and .
For this experiment, set Freq=1.00 and Amp=0.05. The transition frequency of the two level atom has been defined as 1.00 for simplicity. Set the upper analysis window to examine , and ignore the lower analysis window for the time being. Click on the run button, and let it run for a few seconds. Stop the simulation and reset it. Click on the density matrix window and run the program again. Notice how the probability of being in the lower state decreases and how the probability of being in the upper state increases. Also note that off--diagonal terms appear and disappear. When do they disappear? Why?
After one run, click on the upper analysis plot with the right mouse button and go to the menu. Now click on the Inspect Scale option and select the Clone Graph option. You can make as many clones of a graph as you like. Return to BlochApp via the Windows Task Manager (Press Control and Escape simultaneously to obtain the Task Manager.) and run BlochApp again with Amp=0.10. Go to Inspect Scale and archive the plot with Copy to Archive. Move to the graph that you cloned via the Windows Task Manager, and double click on the cloned window. This action will bring up the cloned plot's Inspect Scale option. Click on Paste from Archive. The final result will be a vs Time graph for both Amp=0.05 and Amp=0.10. Continue doing the same thing for a few more values of Amp. What do you notice?
Figure: Rabi oscillation at Freq=1.00; Amp=0.05,0.10,0.20
In Figure , we see that the amplitude of the oscillation is always the same. A complete transition is made between the upper and lower states. As Amp is increased, the frequency of the oscillations increases. Suppose that we now perform the same experiment; however, this time the Amp will be set at 0.05, and Freq will be varied from 1.00 to 1.05 to 1.10.
Figure: Rabi oscillation at Amp=0.05; Freq=1.00,1.05,1.10
A drastic change has occurred in this situation as seen in Figure . The amplitude of the oscillations decreases as the laser frequency increases. A complete transition between the upper and lower state is not made any more. Therefore, 1.00 is the resonant frequency of the system! Hold down the left mouse button in the cloned plot and compare the frequencies of the three plots. They are not the same; indeed, as Freq increases so does the frequency of oscillation. The frequency of oscillation of the probability amplitudes is governed by both Freq and Amp. However, the frequency--shifting effect of Freq is much smaller than that of Amp.
The oscillation that was observed in these experiments is called the Rabi oscillation. is called the Rabi frequency. However, there is a generalized Rabi frequency which has the form
where . The generalized Rabi frequency is what is measured in these plots. These terms again refer to nuclear magnetic resonance and were later appropriated by students of atom--field interactions.
Figure is similar to the plots of the Lorentz Oscillator model at various values of Q. The resonance condition as well as the frequency shifts are predicted by the classical model. However, there is a clear distinction between these two. The classical model is predicting a physical oscillation in space. The semiclassical model's plot is an oscillation in state probability. Each has a very different interpretation.
Next: The Bloch Vector Up: Simulations Previous: Simulations
Andy Antonelli
Wed May 17 14:34:24 EDT 1995 | 1,047 | 4,693 | {"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-2018-43 | latest | en | 0.930452 |
https://journalofinequalitiesandapplications.springeropen.com/articles/10.1155/2010/972324 | 1,716,026,043,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00706.warc.gz | 286,830,387 | 70,743 | • Research Article
• Open access
• Published:
# Second Moment Convergence Rates for Uniform Empirical Processes
## Abstract
Let be a sequence of independent and identically distributed -distributed random variables. Define the uniform empirical process as , . In this paper, we get the exact convergence rates of weighted infinite series of .
## 1. Introduction and Main Results
Let be a sequence of independent and identically distributed (i.i.d.) random variables with zero mean. Set for , and . Hsu and Robbins [1] introduced the concept of complete convergence. They showed that
(1.1)
if and . The converse part was proved by the study of Erdös in [2]. Obviously, the sum in (1.1) tends to infinity as . Many authors studied the exact rates in terms of (cf. [3–5]). Chow [6] studied the complete convergence of , . Recently, Liu and Lin [7] introduced a new kind of complete moment convergence which is interesting, and got the precise rate of it as follows.
Theorem A.
Suppose that is a sequence of i.i.d. random variables, then
(1.2)
holds, if and only if , , and .
Other than partial sums, many authors investigated precise rates in some different cases, such as U-statistics (cf. [8, 9]) and self-normalized sums (cf. [10, 11]). Zhang and Yang [12] extended the precise asymptotic results to the uniform empirical process. We suppose is the sample of random variables and is the empirical distribution function of it. Denote the uniform empirical process by , , and the norm of a function on by . Let , be the Brownian bridge. We present one result of Zhang and Yang [12] as follows.
Theorem B.
For any , one has
(1.3)
Inspired by the above conclusions, we consider second moment convergence rates for the uniform empirical process in the law of iterated logarithm and the law of the logarithm. Throughout this paper, let denote a positive constant whose values can be different from one place to another. will denote the largest integer . The following two theorems are our main results.
Theorem 1.1.
For , one has
(1.4)
Theorem 1.2.
For , one has
(1.5)
Remark 1.3.
It is well known that , (see Csörgő and Révész [13, page 43]). Therefore, by Fubini's theorem we have
(1.6)
Consequently, explicit results of (1.4) and (1.5) can be calculated further.
## 2. The Proofs
In order to prove Theorem 1.1, we present several propositions first.
Proposition 2.1.
For , , one has
(2.1)
Proof.
We calculate that
(2.2)
Proposition 2.2.
For , one has
(2.3)
Proof.
Following [4], set , where . Write
(2.4)
It is wellknown that (see Csörgő and Révész [13, page 17]). By continuous mapping theorem, we have . As a result, it follows that
(2.5)
Using the Toeplitz's lemma (see Stout [14, pages 120-121]), we can get . For , it is obvious that
(2.6)
Notice that , for a small . Via the similar argument in [4] we have
(2.7)
From Kiefer and Wolfowitz [15], we have
(2.8)
Therefore,
(2.9)
From (2.6), (2.7), and (2.9), we get . Proposition 2.2 has been proved.
Proposition 2.3.
For , one has
(2.10)
Proof.
The calculation here is analogous to (2.1), so it is omitted here.
Proposition 2.4.
For , one has
(2.11)
Proof.
Like [4] and Proposition 2.2, we divide the summation into two parts,
(2.12)
First, consider ,
(2.13)
Since means , it follows
(2.14)
By Lemma in Zhang and Yang [12], we have . For , it is easy to get
(2.15)
In the same way, by the inequality , we can get
(2.16)
Put the three parts together, we get that uniformly in as . Using Toeplitz's lemma again, we have
In the sequel, we verify . It is easy to see that
(2.17)
We estimate first, by noticing and (2.8), it follows
(2.18)
Therefore, we get . So far, we only need to prove . Use the inequality again and follow the proof of , we can get this result. The proof of the proposition is completed now.
Proof of Theorem 1.1.
According to Fubini's theorem, it is easy to get
(2.19)
for . Therefore, we have
(2.20)
From Proposition 2.1– 2.4, we have
(2.21)
Proof of Theorem 1.2.
From (2.19), we have
(2.22)
Via the similar argument in Proposition 2.1 and 2.2,
(2.23)
Also, by the analogous proof of Proposition 2.3 and 2.4,
(2.24)
Combine (2.22), (2.23), and (2.24)together, we get the result of Theorem 1.2.
## References
1. Hsu PL, Robbins H: Complete convergence and the law of large numbers. Proceedings of the National Academy of Sciences of the United States of America 1947, 33: 25–31. 10.1073/pnas.33.2.25
2. Erdös P: On a theorem of Hsu and Robbins. Annals of Mathematical Statistics 1949, 20: 286–291. 10.1214/aoms/1177730037
3. Chen R: A remark on the tail probability of a distribution. Journal of Multivariate Analysis 1978, 8(2):328–333. 10.1016/0047-259X(78)90084-2
4. Gut A, Spătaru A: Precise asymptotics in the Baum-Katz and Davis laws of large numbers. Journal of Mathematical Analysis and Applications 2000, 248(1):233–246. 10.1006/jmaa.2000.6892
5. Heyde CC: A supplement to the strong law of large numbers. Journal of Applied Probability 1975, 12: 173–175. 10.2307/3212424
6. Chow YS: On the rate of moment convergence of sample sums and extremes. Bulletin of the Institute of Mathematics 1988, 16(3):177–201.
7. Liu W, Lin Z: Precise asymptotics for a new kind of complete moment convergence. Statistics & Probability Letters 2006, 76(16):1787–1799. 10.1016/j.spl.2006.04.027
8. Fu K-A: Asymptotics for the moment convergence of -statistics in LIL. Journal of Inequalities and Applications 2010, 2010:-8.
9. Yan JG, Su C: Precise asymptotics of -statistics. Acta Mathematica Sinica 2007, 50(3):517–526.
10. Pang T-X, Zhang L-X, Wang JF: Precise asymptotics in the self-normalized law of the iterated logarithm. Journal of Mathematical Analysis and Applications 2008, 340(2):1249–1262. 10.1016/j.jmaa.2007.09.054
11. Zang Q-P: A limit theorem for the moment of self-normalized sums. Journal of Inequalities and Applications 2009, 2009:-10.
12. Zhang Y, Yang X-Y: Precise asymptotics in the law of the iterated logarithm and the complete convergence for uniform empirical process. Statistics & Probability Letters 2008, 78(9):1051–1055. 10.1016/j.spl.2007.09.063
13. Csörgő M, Révész P: Strong Approximations in Probability and Statistics, Probability and Mathematical Statistics. Academic Press, New York, NY, USA; 1981:284.
14. Stout WF: Almost Sure Convergence. Academic Press, New York, NY, USA; 1974:x+381. Probability and Mathematical Statistics, Vol. 2
15. Kiefer J, Wolfowitz J: On the deviations of the empiric distribution function of vector chance variables. Transactions of the American Mathematical Society 1958, 87: 173–186. 10.1090/S0002-9947-1958-0099075-1
## Acknowledgment
This work was supported by NSFC (No. 10771192) and ZJNSF (No. J20091364).
## Author information
Authors
### Corresponding author
Correspondence to You-You Chen.
## Rights and permissions
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Chen, YY., Zhang, LX. Second Moment Convergence Rates for Uniform Empirical Processes. J Inequal Appl 2010, 972324 (2010). https://doi.org/10.1155/2010/972324 | 2,092 | 7,094 | {"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-2024-22 | latest | en | 0.927518 |
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##### Pdf File 271.39KByte, 13Pages
Math 373 Test 1
Fall 2012
September 27, 2012
1. Meng takes out a loan to buy a new motorcycle. The amount of the loan is 12,500. Meng will repay the loan with 9 monthly payments of Q at a nominal interest rate of 12% compounded monthly.
Calculate Q.
Solution
Payments are monthly so we need
i(12) .
We are given i(12).
12
i(12) 12% 1% 12 12
Use our calculator N 9; I / Y 1; PV 12,500
CPT PMT=>1,459.25
March 9, 2014
2. Yu invests 10,000 in an account earning simple interest. During the 10th year, the amount of interest that Yu earns is 1000. Jay invests X into an account earning compound interest of at an annual effective rate of i. During the 7th year, Yu and Jay earn the same annual effective interest rate. At the end of 20 years, Yu has the same amount in her account as Jay has in his account. Determine X .
Solution
Amount of Interest in the 10th year = A(10) A(9) K a(10) K a(9) K[a(10) a(9)]
For Yu with simple interest
Amount of interest in 10th year = 1000 K[a(10) a(9)] 10,000[(1 s 10) (1 s 9)] 10,000 s s 1000 0.10
10, 000
Annual effective interest rate in 7th year for Jay under simple interest =
s
0.10
0.10
in 1 (n 1)s i11 1 (7 1)0.10 1.60 0.0625
Annual effective interest rate for 7th year for Jay under compound interest = in i i7 Since i7 for Yu = i7 for Jay, i 0.0625 For Yu, the amount after 20 years = 10,000(1 st) 10,000(1 (0.10)(20)) 30,000
For Jay, the amount after 20 years = X (1 i)20 X (1 0.0625)20
Since amount after 20 years for Yu = the amount for Jay after 20 years
30,000 X (1.0625)20
30, 000 X (1.0625)20 8923.65
March 9, 2014
3. Ahmad is the winner of a lottery. He has the option to take his prize in any of the following three options:
a. A lump sum payable now of 20 million (20,000,000); b. A perpetuity immediate with annual payments of 1.25 million; or c. An annuity due with level monthly payments of P for 30 years.
These options all have the same present value when calculated at an annual effective interest rate of i .
Determine P .
Solution
PV of a.=20,000,000
PV of b.=1,250,000a 1, 250,000
i
20,000,000 1, 250,000 i 1, 250,000 0.0625
i
20, 000, 000
Since payments are monthly under c., we need i(12) (1.0625)1/12 1 0.005064835 12
PV of c.=Pa
which is equal to PV of a. = 20,000,000
360 0.005064835
So Set Calc to BGN; N=360; I/Y=0.5064835; PV=20,000,000
CPT PMT=>120,303.02
March 9, 2014
4. Wang Corporation is building a new factory. The cash flows from this factory are expected to be:
Time 0 1 2 3
Cash Flow -10,000 3,000 5,500 X
The Internal Rate of Return for this factory is expected to be 10% based on the above cash flows.
Determine X.
Solution
The Net Present Value at the Internal Rate of Return is zero.
NPV
10,000
3000
1 1.1
1
5500
1 1.1
2
X
1 3 1.1
0
X
10,
000
3000
1 1 1.1
1
3
5500
1 1.1
2
3630
1.1
March 9, 2014
5. A loan of 42,000 is being repaid with level annual payments of 6000 plus a smaller drop payment. The interest rate on the loan is an annual effective interest rate of 8%. Calculate the amount of the drop payment.
Solution We will use our calculator to do this calculation. PV=42,000; I/Y=8; PMT=6000; CPT N=>10.67 years. We round down and will have 10 payments of 6000 and a smaller drop payment at time 11 2nd Amort; P1=10; P2=10; ; Bal=3755.47 Drop Payment=3755.47(1.08) 4,055.91
March 9, 2014
6. Brandon takes out a loan of 25,000 to be repaid with a single payment of B at the end of 6 years. The interest rate on Brandon's loan is equivalent to a nominal discount rate of 8% compounded quarterly.
Leslie takes out a loan of 25,000 to be repaid with a single payment L at the end of 6 years. Leslie pays a force of interest of t 0.02t 0.001t2 .
Determine B L .
Solution
Brandon
a(t)
1
d(4) 4
4t
25, 000 1
0.08 4(6) 4
B
B
40, 598.89
Leslie
t
t
s ds a(t) e0
(0.02s0.001s2 )ds 25,000e0
L L 25, 000e0.01s2 0.001s3 /3
|06
25, 000e0.432
38,508.38
B L 40,598.89 38,508.38 2090.51
March 9, 2014
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## 80,956,211,919,489 is an odd composite number composed of two prime numbers multiplied together.
What does the number 80956211919489 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 9 divisors.
80956211919489 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of nine divisors.
## Prime factorization of 80956211919489:
### 32 × 29991892
(3 × 3 × 2999189 × 2999189)
See below for interesting mathematical facts about the number 80956211919489 from the Numbermatics database.
### Names of 80956211919489
• Cardinal: 80956211919489 can be written as Eighty trillion, nine hundred fifty-six billion, two hundred eleven million, nine hundred nineteen thousand, four hundred eighty-nine.
### Scientific notation
• Scientific notation: 8.0956211919489 × 1013
### Factors of 80956211919489
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 4
• Sum of prime factors: 2999192
### Divisors of 80956211919489
• Number of divisors d(n): 9
• Complete list of divisors:
• Sum of all divisors σ(n): 116936789539843
• Sum of proper divisors (its aliquot sum) s(n): 35980577620354
• 80956211919489 is a deficient number, because the sum of its proper divisors (35980577620354) is less than itself. Its deficiency is 44975634299135
### Bases of 80956211919489
• Binary: 100100110100001000101100100000011111010100000012
• Base-36: SP2RQCKU9
### Squares and roots of 80956211919489
• 80956211919489 squared (809562119194892) is 6553908248353212837818021121
• 80956211919489 cubed (809562119194893) is 530579585054569642396351816506586953527169
• 80956211919489 is a perfect square number. Its square root is 8997567
• The cube root of 80956211919489 is 43259.6889951707
### Scales and comparisons
How big is 80956211919489?
• 80,956,211,919,489 seconds is equal to 2,574,157 years, 6 weeks, 3 days, 12 hours, 18 minutes, 9 seconds.
• To count from 1 to 80,956,211,919,489 would take you about six million, four hundred thirty-five thousand, three hundred ninety-two years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 80956211919489 cubic inches would be around 3605 feet tall.
### Recreational maths with 80956211919489
• 80956211919489 backwards is 98491911265908
• The number of decimal digits it has is: 14
• The sum of 80956211919489's digits is 72
• More coming soon!
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Keywords: Divisors of 80956211919489, math, Factors of 80956211919489, curriculum, school, college, exams, university, Prime factorization of 80956211919489, STEM, science, technology, engineering, physics, economics, calculator, eighty trillion, nine hundred fifty-six billion, two hundred eleven million, nine hundred nineteen thousand, four hundred eighty-nine.
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Irrationals, Transcendentals, & Infinity
Some questions to explore:
• It has been suggested earlier that the set of prime numbers is infinite. Primes are a subset of integers.
• Integers are a subset of Rationals, which are a subset of Reals, and then on to Complex and beyond.
• If the integers are infinite, and if the real numbers are infinite, are there the same number of both?
• What about rational numbers? What about irrational numbers? Are those two sets equal in size?
• What is an algebraic number? What is a transcendental number? How large are those sets?
• Are all of these infinities equal? How many infinities are there? Who thought all of this up? Are there any other approaches? | 191 | 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} | 2.71875 | 3 | CC-MAIN-2019-18 | latest | en | 0.922756 |
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Math Worksheets For All Ages
# Math Worksheets Land
Math Worksheets For All Ages
Home > Math Topics > Multiplication >
# Mixed Multiplication Review Worksheets
You will find all types of different problems here to help you shore up or master these skills. When you are working with types of problems, the more experience you get the quicker you will be able to solve them. We highly suggest that you setup your problems in a vertical manner. They can drastically help you improve your accuracy and speed while solving them. It is mostly due to the fact that it is just so much easier to stay organized. These worksheets and lessons serve as a great overview on the skill of performing multiplication operations.
### How to Keep Your Multiplication Skills Sharp
This operation is an essential skill that students must master before learning any advanced mathematical concept. Though trying to memorize your timetables is the ideal way to make multiplication easier, there are many other ways of improving multiplication skills. Here we have discussed a few ways of keeping your multiplication skill at the ready:
Understand the concept - Like every other mathematical skill, you need to understand the underlying concept of it. If you memorize the times table without understanding the concept, you will face difficulty in understanding the advanced mathematical concept. We suggest approaching this as a reflection of repeated addition. If we remember back the backbone of an addition equation is: addend + addend = sum. Multiplication relates directly if you consider that the multiplicand indicates the value of one addend and then the multiplier just tells you how many times you will add it to itself.
Reviewing and revision - Your learned and acquired concepts start to rust without review and revision. If not, practice, then read them and review them regularly. You will find flashcards extremely invaluable. I recommend buy a professional set of cards. They are only a few bucks but will be something that sticks with you forever.
Test yourself - Yes, testing. Something that most children don't appreciate or prefer doing. However, it is testing that measures your skills and tells how well you can perform at a particular concept. Just like anything, the more you test yourself, the better you will get at the test and therefore the skills found on it. There is endless research available to indicate that simply repeating tests can help greatly improve our comprehension.
Practice your times tables - Practice makes perfect! Without practicing and continual reinforcement of any concept, you cannot gain a clearer idea of it. Practicing your times tables is crucial for improving your skills regularly. Practice the concept until you have mastered it.
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# IS FOCUSING ON PRODUCTIVITY UNPRODUCTIVE?
Productivity, like Moms and apple pies, one of those things that everbody's in favor of. But what exactly is it? And what does it mean if productivity rises?
Economist George P. Brockway, retired director of Yale University's academic publishing arm, argues that the way we look at productivity can head policy in precisely the wrong direction - that is, away from looking at the kind of investment our economy needs.
The reason, he argues, is built into the mathematics of the way we measure productivity - the amount of goods and services that each person produces.
Consider a painter who decides he needs to get more done on the job site and hires a kid to help out.
Even if his output increases 50 percent, his productivity will decline.
If the painter could paint 100 square feet in a day on his own, then his productivity was 100 square feet per eight hours of labor. If hiring the kid to tote and haul cans of paint or move ladders and scaffolding means the painter can paint 150 square feet a day, productivity would be 150 square feet per 16 hours of labor, or 75 square feet per eight hours, a 25 percent reduction.
Is the painter's customer better or worse off?
The job is finished sooner.
Is the painter better off?
He has the extra cost of paying the kid, but being able to move through a job more quickly means he gets his money more quickly and can move on more quickly to the next paying job.
Is the kid better off?
Well, he has a job.
And the rest of us?
''From the point of view of the national economy, the question is, How great is the national product?" says Brockway. ''When we ask this question, we see that every contribution, no matter how little, nor matter how clumsily produced, will swell the total.''
All of this is more than making numbers dance.
Brockway is concerned that productivity indices like the painting index here are badly misleading. They don't do what most of us think, which is to measure how well people do their jobs, whether they're efficient or lazy or well trained. They only measure how employers use their work forces.
And that, he said, is as much a measure of how business invests as any else - and the way businesses invest, many economists argue, is what policy needs to focus on in the years ahead.
To understand what the productivity index is telling us about the use of labor in our society, we have to understand the use of capital. | 530 | 2,460 | {"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-2022-21 | latest | en | 0.966134 |
https://testbook.com/question-answer/the-37-c-temperature-is-equal-to-nearly--62b4415561428d86454284d4 | 1,716,569,533,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00517.warc.gz | 473,190,405 | 42,715 | # The 37° C temperature is equal to nearly:
This question was previously asked in
RRB NTPC CBT 2 (Level-2) Official Paper (Held On: 13 June 2022 Shift 1)
View all RRB NTPC Papers >
1. 99.4° F
2. 98.6° F
3. 97.4° F
4. 100.4°F
## Answer (Detailed Solution Below)
Option 2 : 98.6° F
Free
RRB Exams (Railway) Biology (Cell) Mock Test
7.7 Lakh Users
10 Questions 10 Marks 7 Mins
## Detailed Solution
The correct answer is 98.6° F.
Key Points
• All temperatures are related to each other by the following relationship –
$$\frac{{F - 32^\circ }}{9} = \frac{C}{5} = \frac{{K - 273}}{5}$$
Calculation:
Given: Temperature = 37°C,
$$\frac{{F - 32^\circ }}{9} = \frac{C}{5}$$
$$\frac{{F - 32^\circ }}{9} = \frac{37}{5}$$
5(F - 32°) = 333
So, the temperature is on the Fahrenheit scale: 98.6°F
Additional InformationCelsius scale:
• In this scale lower fixed point (ice point) is taken and upper fixed point (steam point) is taken 100°.
• The temperature measured on this scale all in degree Celsius (°C).
Fahrenheit scale:
• This scale of temperature has lower fixed point (ice point) as 32° F and upper fixed point (steam point)as 212° F.
• The change in temperature of 1° F corresponds to a change of less than 1° on Celsius scale.
Kelvin scale:
• The Kelvin temperature scale is also known as thermodynamic scale. The triple point of water is also selected to be the zero of scale of temperature.
• The temperatures measured on this scale are in Kelvin (K).
Latest RRB NTPC Updates
Last updated on Feb 13, 2024
The RRB NTPC Notification 2024 is expected to be released soon. The RRB NTPC exam is conducted to fill up Non-Technical Popular Category posts. The candidates with successful selection under RRB NTPC get a salary ranging between Rs. 19,900 to Rs. 35,400. here. | 591 | 1,786 | {"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} | 3.859375 | 4 | CC-MAIN-2024-22 | latest | en | 0.661918 |
https://stat.ethz.ch/pipermail/r-help/2015-January/425192.html | 1,716,026,830,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00866.warc.gz | 503,855,922 | 2,222 | # [R] Looking for an R package for Set Cover Problem
Hans W Borchers hwborchers at gmail.com
Mon Jan 26 00:48:50 CET 2015
```As the Wikipedia page you took your example problem from explains, the sets
cover problem can be formulated as an integer linear programming problem.
In R, such problems will be solved effectively applying one of the available
MILP packages, for example LPsolve or Rsymphony.
Kumar Mainali <kpmainali <at> gmail.com> writes:
>
> I am looking for an R package that solves Set Cover Problem. As Wikipedia
> explains:
>
> Given a set of elements [image: \{1,2,...,m\}] (called the universe) and a
> set [image: S] of [image: n] sets whose union equals the universe, the set
> cover problem is to identify the smallest subset of [image: S] whose union
> equals the universe. For example, consider the universe [image: U = \{1, 2,
> 3, 4, 5\}] and the set of sets [image: S = \{\{1, 2, 3\}, \{2, 4\}, \{3,
> 4\}, \{4, 5\}\}]. Clearly the union of [image: S] is [image: U]. However,
> we can cover all of the elements with the following, smaller number of
> sets: [image: \{\{1, 2, 3\}, \{4, 5\}\}].
For this concrete case the set of linear inequalities looks like:
x1 >= 1 # as 1 is only element of S1
x1 + x2 >= 1 # etc.
x1 + x3 >= 1
x2 + x3 + x4 >= 1
x4 >= 1 # all xi in {0,1}
which has the minimal solution x1, x2, x3, x4 = 1, 0, 0, 1 .
> Thank you,
> Kumar Mainali
>
> Postdoctoral Fellow
> Environmental Science Institute
> The University of Texas at Austin
> ᐧ
``` | 495 | 1,531 | {"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} | 3.40625 | 3 | CC-MAIN-2024-22 | latest | en | 0.858864 |
https://in.mathworks.com/matlabcentral/cody/problems/2635-higher-lower/solutions/2990692 | 1,611,117,438,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519883.54/warc/CC-MAIN-20210120023125-20210120053125-00104.warc.gz | 400,058,413 | 17,443 | Cody
# Problem 2635. Higher! Lower!
Solution 2990692
Submitted on 25 Sep 2020 by Alex
• Size: 57
• This is the leading solution.
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
guess = 512
2 Pass
hi_bound=1024; lo_bound=1; right_answer=513; assert(isequal(high_or_low(lo_bound,hi_bound,right_answer), [512 768 640 576 544 528 520 516 514 513]))
guess = 512 guess = 768 guess = 640 guess = 576 guess = 544 guess = 528 guess = 520 guess = 516 guess = 514 guess = 513
3 Pass
guess = -8 guess = -4 guess = -2 guess = -3
4 Pass
hi_bound=1024; lo_bound=1; right_answer=1024; assert(isequal(high_or_low(lo_bound,hi_bound,right_answer), [512 768 896 960 992 1008 1016 1020 1022 1023 1024]))
guess = 512 guess = 768 guess = 896 guess = 960 guess = 992 guess = 1008 guess = 1016 guess = 1020 guess = 1022 guess = 1023 guess = 1024
5 Pass
hi_bound=2048; lo_bound=0; right_answer=0; assert(isequal(high_or_low(lo_bound,hi_bound,right_answer), [1024 512 256 128 64 32 16 8 4 2 1 0]))
guess = 1024 guess = 512 guess = 256 guess = 128 guess = 64 guess = 32 guess = 16 guess = 8 guess = 4 guess = 2 guess = 1 guess = 0
6 Pass
hi_bound=65535; lo_bound=0; right_answer=1024; assert(isequal(high_or_low(lo_bound,hi_bound,right_answer), [32767 16383 8191 4095 2047 1023 1535 1279 1151 1087 1055 1039 1031 1027 1025 1024]))
guess = 32767 guess = 16383 guess = 8191 guess = 4095 guess = 2047 guess = 1023 guess = 1535 guess = 1279 guess = 1151 guess = 1087 guess = 1055 guess = 1039 guess = 1031 guess = 1027 guess = 1025 guess = 1024
7 Pass | 601 | 1,666 | {"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} | 3.359375 | 3 | CC-MAIN-2021-04 | latest | en | 0.703337 |
https://brainmass.com/statistics/quantative-analysis-of-data/risk-and-return-expected-deviation-probabilities-257164 | 1,524,693,806,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947968.96/warc/CC-MAIN-20180425213156-20180425233156-00334.warc.gz | 561,721,713 | 18,970 | Explore BrainMass
Share
# Risk and Return: Expected, Deviation, Probabilities
7.3 Expected returns: You have chosen biology as your college major because you would like to be a medical doctor. However, you find that the probability of being accepted into medical school is about 10 percent. If you are accepted into medical school, then your starting salary when you graduate will be \$300,000 per year. However, if you are not accepted, then you would choose to work in a zoo, where you will earn \$40,000 per year. Without considering the additional educational years or the time value of money, what is your expected starting salary as well as the standard deviation of that starting salary?
7.15 Calculating the variance and standard deviation: Kate recently invested in real estate with the intention of selling the property one year from today. She has modeled the returns on that investment based on three economic scenarios. She believes that if the economy stays healthy, then her investment will generate a 30 percent return. However, if the economy softens, as predicted, the return will be 10 percent, while the return will be -25 percent if the economy slips into a recession. If the probabilities of the healthy, soft, and recessionary states are 0.4, 0.5, and 0.1, respectively, then what are the expected return and the standard deviation for Kate's investment?
7.20 Portfolios with more than one asset: Given the returns and probabilities for the three possible states listed here, calculate the covariance between the returns of Stock A and Stock B. For convenience, assume that the expected returns of Stock A and Stock B are 11.75 percent and 18 percent, respectively.
Probability
Return(A)
Return(B)
Good
0.35
0.30
0.50
OK
0.50
0.10
0.10
Poor
0.15
-0.25
-0.30
7.27 In order to fund her retirement, Glenda requires a portfolio with an expected return of 12 percent per year over the next 30 years. She has decided to invest in Stocks 1, 2, and 3, with 25 percent in Stock 1, 50 percent in Stock 2, and 25 percent in Stock 3. If Stocks 1 and 2 have expected returns of 9 percent and 10 percent per year, respectively, then what is the minimum expected annual return for Stock 3 that will enable Glenda to achieve her investment requirement?
#### Solution Summary
The attached Excel solution contains step-by-step formulas to show how to calculate mean, variance, and co-variance.
\$2.19 | 559 | 2,417 | {"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} | 3.921875 | 4 | CC-MAIN-2018-17 | latest | en | 0.940143 |
http://www.csplib.org/Problems/prob014/models/sb.mzn.html | 1,534,686,778,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215176.72/warc/CC-MAIN-20180819125734-20180819145734-00322.warc.gz | 461,818,065 | 4,083 | % new zinc file SolitaireBattleships.mzn
% At attempt to build a solution entirely within the IDE
% By Peter Stuckey August 2009
% Solitaire Battleships is a puzzle where
% we are given a partially filled in board and the number
% of ships in each row and column and have to fill it with ships
% "Improved" version does not introduce reified set_in
int: width; % height of board (usually 10)
int: height; % width of board (usually 10)
int: maxship; % maximal length of ship (usually 4)
set of int: ROWS = 1..width;
set of int: COLS = 1..height;
set of int: XROWS = 0..width+1; % extended rows
set of int: XCOLS = 0..height+1; % extended cols
%% ship types enumerated type
set of int: SHIPS = 1..maxship; % different ship types
int: submarine = 1;
int: destroyer = 2;
int: cruiser = 3;
int: battleship = 4;
%% the PIECES enumerated type!
set of int: PIECES = 1..7;
int: w = 1; % water
int: c = 2; % circle (submarine)
int: l = 3; % left end of ship
int: r = 4; % right end of ship
int: t = 5; % top of ship
int: b = 6; % bottom of ship
int: m = 7; % middle of ship
array[PIECES] of string: code = [".","c","l","r","t","b","m"];
array[ROWS,COLS] of 0..7: hint; % the initial board configuration
array[ROWS] of int: rowsum; % sums for each row
array[COLS] of int: colsum; % sums for each col
array[SHIPS] of int: ship; % the number of each type of ship (usually [4,3,2,1]).
% variables
array[XROWS,XCOLS] of var PIECES: board; % the board
array[XROWS,XCOLS] of var 0..1: fill; % which pieces are ships
array[PIECES] of var 0..width*height: npiece; % number of pieces of each type
% model
% ensure hints are respected
constraint forall(i in ROWS, j in COLS)(
if hint[i,j] != 0 then
board[i,j] == hint[i,j]
else true endif
);
% make extended rows and cols empty
constraint forall(i in XROWS)(board[i,0] == w /\ board[i,width+1] == w);
constraint forall(j in COLS)(board[0,j] == w /\ board[height+1,j] == w);
% ensure that the fill array matches the board
constraint forall(i in XROWS, j in XCOLS)(
fill[i,j] = bool2int(board[i,j] != w)
);
% spacing constraints: gaps betwen ships
constraint forall(i in ROWS, j in COLS)(
(board[i,j] == w \/ board[i+1,j+1] == w)
/\ (board[i,j] == w \/ board[i+1,j-1] == w) % diagonal constraints
/\ (board[i,j] in {c,l,r,t} -> board[i-1,j] == w)
/\ (board[i,j] in {c,l,r,b} -> board[i+1,j] == w)
/\ (board[i,j] in {c,l,t,b} -> board[i,j-1] == w)
/\ (board[i,j] in {c,r,t,b} -> board[i,j+1] == w)
);
% ship shape constraints
constraint forall(i in ROWS, j in COLS)(
%% a left piece needs a right piece or middle to the right
(board[i,j] == l -> (board[i,j+1] == r \/ board[i,j+1] == m))
/\ (board[i,j] == r -> (board[i,j-1] == l \/ board[i,j-1] == m))
/\ (board[i,j] == t -> (board[i+1,j] == b \/ board[i+1,j] == m))
/\ (board[i,j] == b -> (board[i-1,j] == t \/ board[i-1,j] == m))
%% a middle piece has to have two opposite sides filled
/\ (board[i,j] == m -> ( fill[i-1,j] == fill[i+1,j]
/\ fill[i,j-1] == fill[i,j+1]
/\ fill[i-1,j] + fill[i,j-1] == 1))
);
% sum up pieces
constraint forall(p in PIECES)(
sum(i in ROWS, j in COLS)(bool2int(board[i,j] == p)) == npiece[p]
);
% piece sum constraints
constraint npiece[c] == ship[submarine]; % submarines
constraint npiece[l] == npiece[r]; % left right (probably redundant)
constraint npiece[t] == npiece[b]; % top bottom
constraint npiece[l] + npiece[t] == sum(s in destroyer..maxship)(ship[s]);
% no of ends
constraint npiece[m] == sum(s in cruiser..maxship)(ship[s] * (s - 2));
% no of middles
% count number of bigger ships
% at least for standard battleships you can probably simply
% enforce this constraint for s in destroyer..destroyer
% and still be guaranteed a correct solution
constraint forall(s in destroyer..maxship)(
sum(i in ROWS,j in COLS)(bool2int(
if j + s - 1 <= width then
board[i,j] == l /\ board[i,j+s-1] == r % ship length s lr
/\ forall(k in j+1..j+s-2)(board[i,k] == m)
else false endif
\/
if i + s - 1 <= height then
board[i,j] == t /\ board[i+s-1,j] == b % ship length s tb
/\ forall(k in i+1..i+s-2)(board[k,j] == m)
else false endif
)) = ship[s]
);
% row sums respected
constraint forall(i in ROWS)(
sum(j in COLS)(fill[i,j]) == rowsum[i]
);
% column sums respected
constraint forall(j in COLS)(
sum(i in ROWS)(fill[i,j]) == colsum[j]
);
solve :: int_search([ fill[i,j] | i in ROWS, j in COLS],
input_order, indomain_min, complete)
satisfy;
output [ code[fix(board[i,j])] ++
if j == width then " " ++ show(rowsum[i]) ++ "\n"
else "" endif
| i in ROWS, j in COLS ] ++
[ show(colsum[j]) | j in COLS ] ++ ["\n"]; | 1,524 | 4,573 | {"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} | 2.5625 | 3 | CC-MAIN-2018-34 | latest | en | 0.782305 |
http://flaguide.org/tools/math/estimation/millionB.htm | 1,723,264,662,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00585.warc.gz | 14,527,265 | 6,162 | ## Tools - Math 'Plausible Estimation' Estimates for a Million Tasks, Set #1 (solutions)
Estimates for a Million, Set #1 (solutions)
Estimating for Amazing Facts: Set #2 (solutions) || Set #3 (solutions)
Estimates for the USA, Set #4 (solutions)
Malcolm Swan
Mathematics Education
University of Nottingham
Malcolm.Swan@nottingham.ac.uk
Jim Ridgway
School of Education
University of Durham
Jim.Ridgway@durham.ac.uk
The aim of this assessment is to provide the opportunity for you to: develop a chain of reasoning that will enable you to estimate quantities to an appropriate degree of accuracy choose suitable units for your estimate communicate the assumptions upon which your estimate is based.
1. Breathing
How many days would it take you to breathe a million times?
Solution:
Assumptions You breathe once every 4 seconds. Reasoning Calculation: 4,000,000 seconds is about 46 days. Answer: About 50 days.
1. Paper Clips
Suppose a chain is made from a million paper clips. How far will it stretch?
Solution:
Assumptions Each paper clip is about 4 cm in length. Reasoning Calculation: 4,000,000 cm = 40 km (or about 25 miles) Answer: About 40 kilometres or 25 miles.
1. The brick wall
Suppose you use a million household bricks to build a wall four feet high. How long would the wall be?
Solution:
Assumptions A brick is about 3 inches high and 9 inches long. Reasoning Calculation: The wall will be (4 feet ÷ 3 inches) = 16 bricks high This means that the wall will be (1 million ÷ 16) 62,500 bricks in length = 62,500 x 9 inches = 8.9 miles. Answer: Just under 9 miles. (Maybe with mortar it would be just over 9 miles)
1. The dripping faucet
A faucet drips a million times. How many buckets will it fill?
Solution:
Assumptions A bucket holds about 8 liters. A drip has a diameter of 2 mm Reasoning Calculation: The volume of a drip is given by 4/3 (1)3 ≈ 4 mm3 A million drips will therefore have a volume of 4 x 106 mm3 = 4 liters Answer: About one half a bucket.
1. Writing a million
How long would it take you to write out all the numbers, from one to a million? Remember that some numbers have more digits than others!
Solution:
Assumptions You can write down 2 digits per second Reasoning Most ( 90%) of the numbers have 6 digits. So we need to write down nearly six million digits. This would take nearly 3 million seconds = 35 days (approx) Answer: 35 days, working day and night.
Estimates for a Million, Set #1 (solutions)
Estimating for Amazing Facts: Set #2 (solutions) || Set #3 (solutions)
Estimates for the USA, Set #4 (solutions)
Introduction || Assessment Primer || Matching Goals to CATs || CATs || Tools || Resources
Search || Who We Are || Site Map || Meet the CL-1 Team || WebMaster || Copyright || Download
College Level One (CL-1) Home || Collaborative Learning || FLAG || Learning Through Technology || NISE | 728 | 2,859 | {"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} | 4.84375 | 5 | CC-MAIN-2024-33 | latest | en | 0.890276 |
http://cda.mrs.umn.edu/~elenam/2101_fall04/labs/lab4.html | 1,534,528,369,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212639.36/warc/CC-MAIN-20180817163057-20180817183057-00463.warc.gz | 71,785,414 | 2,723 | ## CSci 2101 Data Structures: Lab 4
### Problem 1 (paper-and-pencil)
For the following program draw the memory picture at two points of the program execution: right before each of the print statement gets executed. Based on your picture, what would each of the statements print? Run the program to check your result. If they are different from what your pictures predict, correct the pictures.
``````
public class WhatHappens {
public static void main(String [] args) {
Node n1 = new Node(1);
Node n2 = new Node(2);
Node n3 = n1;
n3.setNext(n2);
System.out.println( (n1.getNext()).getData() );
n3 = new Node(3);
(n1.getNext()).setNext(n3);
System.out.println( (n2.getNext()).getData() );
}
}
``````
Submit the two memory pictures (on paper). You don't need to submit any files for this problem.
### Problem 2
In the file TestLinkedLists.java write a loop to count the number of nodes of the list with data value 5. Test the program, submit ALL your test results.
### Problem 3
In the file LinkedList.java add a method `length()` which counts the number of elements on the list. The list itself should not change in any way.
Write testing code for the method in a new file TestLength.java (create a linked list, add some elements to it or keep it empty, and call the method `length()` on the list). Make sure to test the method thoroughly before going on to the next problem.
### Problem 4
Copy/paste the following program into a file WeirdList.java:
``````
public class WeirdList {
public static void main(String [] args) {
Node n;
// fill in your code here to initialize node n
// you may create other nodes if needed
// don't change any code below this point
weirdlist.setFirst(n);
System.out.println("the length of the list is " + weirdlist.length());
}
}
``````
• Part 1. Fill in the code in the program above (after the first comment) so that the program prints
```the length of the list is 0
```
Hint: this is easy.
• Part 2. Fill in the code in the program above (after the first comment) so that the program prints
```the length of the list is 2
```
This is harder!
Don't modify any code after the last line of comments.
### Problem 5
Look at the following program and try to predict what happens when you run it (drawing memory pictures helps). Then run the program. Important: press Ctrl-C (the Ctrl key and the C key at the same time) to stop the program.
``````
public class Why {
public static void main(String [] args) {
// testing nodes:
Node first = new Node(5);
Node second = new Node(6);
first.setNext(second);
second.setNext(first);
// add the two nodes to the list
list.setFirst(first);
// print the list
System.out.println("The list elements are:");
Node n = list.getFirst(); // if the list is empty, first is null
while(n != null) {
int data = n.getData();
System.out.println(data);
n = n.getNext();
}
}
}
``````
On the Wiki submission page explain why the program's behaves this way. You don't need to submit any code for this problem.
### Problem 6
For this and the next problem assume that linked lists don't have any peculiarities, such as those in problem 5.
In a new file create a linked list. Then write a loop that will go through the list and duplicate each node. For instance, if the list had three nodes with data 5, 4, 3, then after the loop the list should have 6 nodes with data 5, 5, 4, 4, 3, 3. Make sure that the program works correctly for an empty list. Print out the list at the end to test the program. Submit all your test cases.
### *Problem 7* (extra credit)
In a new file create a linked list and then write a loop to remove all the elements of the list with a data value 5. Print the list at the end to check the result. BR> This is a hard problem. Test it carefully. Make sure to submit ALL your test cases with the program. If it doesn't work in some cases, submit the test cases and explain what doesn't work (and, ideally, what you think is the reason for the problem). Partial credit will be given for programs that work in some, but not all, cases. Explanations contribute to the grade.
This is a lab from CSci 2101 course.
The views and opinions expressed in this page are strictly those of the page author. The contents of this page have not been reviewed or approved by the University of Minnesota. | 1,038 | 4,278 | {"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-2018-34 | latest | en | 0.812086 |
https://www.chegg.com/homework-help/net-present-value-method-service-companymetro-goldwyn-mayer-chapter-26-problem-5cap-solution-9781305088405-exc | 1,555,963,980,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578582584.59/warc/CC-MAIN-20190422195208-20190422221208-00349.warc.gz | 654,470,974 | 29,017 | # Accounting (26th Edition) Edit edition Problem 5CAP from Chapter 26: Net present value method for a service companyMetro-Goldwyn-...
We have solutions for your book!
Chapter: Problem:
Net present value method for a service company
Metro-Goldwyn-Mayer Studios Inc. (MGM) is a major producer and distributor of theatrical and television filmed entertainment. Regarding theatrical films, MGM states, “Our feature films are exploited through a series of sequential domestic and international distribution channels, typically beginning with theatrical exhibition. Thereafter, feature films are first made available for home video (online downloads) generally six months after theatrical release; for pay television, one year after theatrical release; and for syndication, approximately three to five years after theatrical release.”
Assume that MGM produces a film during early 2016 at a cost of \$340 million and releases it halfway through the year. During the last half of 2016, the film earns revenues of \$420 million at the box office. The film requires \$90 million of advertising during the release. One year later, by the end of 2017, the film is expected to earn MGM net cash flows from online downloads of \$60 million. By the end of 2018, the film is expected to earn MGM \$20 million from pay TV; and by the end of 2019, the film is expected to earn \$10 million from syndication.
a. Determine the net present value of the film as of the beginning of 2016 if the desired rate of return is 20%. To simplify present value calculations, assume all annual net cash flows occur at the end of each year. Use the table of the present value of \$1 appearing in Exhibit 2 of this chapter. Round to the nearest whole million dollars.
b. Under the assumptions provided here, is the film expected to be financially successful?
Step-by-step solution:
Chapter: Problem:
• Step 1 of 3
Net present value is today’s value of future cash inflows at the rate of return which is acceptable by the company, which is compared with the present value of cash outflow.
(a)
Below is the calculation of net present value for the investment on production of new film:
(Amount in millions)
Year Cash inflows PV factor @20% Present value of cash flows 2016 \$420 0.833 \$350 2017 \$60 0.694 \$42 2018 \$20 0.579 \$12 2019 \$10 0.482 \$5 Total Present value of cash inflows \$409 Initial investment \$430 Net present value (\$90)
• Chapter , Problem is solved.
Corresponding Textbook
Accounting | 26th Edition
9781305088405ISBN-13: 1305088409ISBN: Jim ReeveAuthors: | 580 | 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} | 3.09375 | 3 | CC-MAIN-2019-18 | latest | en | 0.94665 |
https://www.quizzes.cc/metric/percentof.php?percent=98&of=5830 | 1,637,979,697,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00034.warc.gz | 1,094,112,091 | 4,395 | #### What is 98 percent of 5,830?
How much is 98 percent of 5830? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 98% of 5830 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 98% of 5,830 = 5713.4
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating ninety-eight of five thousand, eight hundred and thirty How to calculate 98% of 5830? Simply divide the percent by 100 and multiply by the number. For example, 98 /100 x 5830 = 5713.4 or 0.98 x 5830 = 5713.4
#### How much is 98 percent of the following numbers?
98% of 5830.01 = 571340.98 98% of 5830.02 = 571341.96 98% of 5830.03 = 571342.94 98% of 5830.04 = 571343.92 98% of 5830.05 = 571344.9 98% of 5830.06 = 571345.88 98% of 5830.07 = 571346.86 98% of 5830.08 = 571347.84 98% of 5830.09 = 571348.82 98% of 5830.1 = 571349.8 98% of 5830.11 = 571350.78 98% of 5830.12 = 571351.76 98% of 5830.13 = 571352.74 98% of 5830.14 = 571353.72 98% of 5830.15 = 571354.7 98% of 5830.16 = 571355.68 98% of 5830.17 = 571356.66 98% of 5830.18 = 571357.64 98% of 5830.19 = 571358.62 98% of 5830.2 = 571359.6 98% of 5830.21 = 571360.58 98% of 5830.22 = 571361.56 98% of 5830.23 = 571362.54 98% of 5830.24 = 571363.52 98% of 5830.25 = 571364.5
98% of 5830.26 = 571365.48 98% of 5830.27 = 571366.46 98% of 5830.28 = 571367.44 98% of 5830.29 = 571368.42 98% of 5830.3 = 571369.4 98% of 5830.31 = 571370.38 98% of 5830.32 = 571371.36 98% of 5830.33 = 571372.34 98% of 5830.34 = 571373.32 98% of 5830.35 = 571374.3 98% of 5830.36 = 571375.28 98% of 5830.37 = 571376.26 98% of 5830.38 = 571377.24 98% of 5830.39 = 571378.22 98% of 5830.4 = 571379.2 98% of 5830.41 = 571380.18 98% of 5830.42 = 571381.16 98% of 5830.43 = 571382.14 98% of 5830.44 = 571383.12 98% of 5830.45 = 571384.1 98% of 5830.46 = 571385.08 98% of 5830.47 = 571386.06 98% of 5830.48 = 571387.04 98% of 5830.49 = 571388.02 98% of 5830.5 = 571389
98% of 5830.51 = 571389.98 98% of 5830.52 = 571390.96 98% of 5830.53 = 571391.94 98% of 5830.54 = 571392.92 98% of 5830.55 = 571393.9 98% of 5830.56 = 571394.88 98% of 5830.57 = 571395.86 98% of 5830.58 = 571396.84 98% of 5830.59 = 571397.82 98% of 5830.6 = 571398.8 98% of 5830.61 = 571399.78 98% of 5830.62 = 571400.76 98% of 5830.63 = 571401.74 98% of 5830.64 = 571402.72 98% of 5830.65 = 571403.7 98% of 5830.66 = 571404.68 98% of 5830.67 = 571405.66 98% of 5830.68 = 571406.64 98% of 5830.69 = 571407.62 98% of 5830.7 = 571408.6 98% of 5830.71 = 571409.58 98% of 5830.72 = 571410.56 98% of 5830.73 = 571411.54 98% of 5830.74 = 571412.52 98% of 5830.75 = 571413.5
98% of 5830.76 = 571414.48 98% of 5830.77 = 571415.46 98% of 5830.78 = 571416.44 98% of 5830.79 = 571417.42 98% of 5830.8 = 571418.4 98% of 5830.81 = 571419.38 98% of 5830.82 = 571420.36 98% of 5830.83 = 571421.34 98% of 5830.84 = 571422.32 98% of 5830.85 = 571423.3 98% of 5830.86 = 571424.28 98% of 5830.87 = 571425.26 98% of 5830.88 = 571426.24 98% of 5830.89 = 571427.22 98% of 5830.9 = 571428.2 98% of 5830.91 = 571429.18 98% of 5830.92 = 571430.16 98% of 5830.93 = 571431.14 98% of 5830.94 = 571432.12 98% of 5830.95 = 571433.1 98% of 5830.96 = 571434.08 98% of 5830.97 = 571435.06 98% of 5830.98 = 571436.04 98% of 5830.99 = 571437.02 98% of 5831 = 571438
1% of 5830 = 58.3 2% of 5830 = 116.6 3% of 5830 = 174.9 4% of 5830 = 233.2 5% of 5830 = 291.5 6% of 5830 = 349.8 7% of 5830 = 408.1 8% of 5830 = 466.4 9% of 5830 = 524.7 10% of 5830 = 583 11% of 5830 = 641.3 12% of 5830 = 699.6 13% of 5830 = 757.9 14% of 5830 = 816.2 15% of 5830 = 874.5 16% of 5830 = 932.8 17% of 5830 = 991.1 18% of 5830 = 1049.4 19% of 5830 = 1107.7 20% of 5830 = 1166 21% of 5830 = 1224.3 22% of 5830 = 1282.6 23% of 5830 = 1340.9 24% of 5830 = 1399.2 25% of 5830 = 1457.5
26% of 5830 = 1515.8 27% of 5830 = 1574.1 28% of 5830 = 1632.4 29% of 5830 = 1690.7 30% of 5830 = 1749 31% of 5830 = 1807.3 32% of 5830 = 1865.6 33% of 5830 = 1923.9 34% of 5830 = 1982.2 35% of 5830 = 2040.5 36% of 5830 = 2098.8 37% of 5830 = 2157.1 38% of 5830 = 2215.4 39% of 5830 = 2273.7 40% of 5830 = 2332 41% of 5830 = 2390.3 42% of 5830 = 2448.6 43% of 5830 = 2506.9 44% of 5830 = 2565.2 45% of 5830 = 2623.5 46% of 5830 = 2681.8 47% of 5830 = 2740.1 48% of 5830 = 2798.4 49% of 5830 = 2856.7 50% of 5830 = 2915
51% of 5830 = 2973.3 52% of 5830 = 3031.6 53% of 5830 = 3089.9 54% of 5830 = 3148.2 55% of 5830 = 3206.5 56% of 5830 = 3264.8 57% of 5830 = 3323.1 58% of 5830 = 3381.4 59% of 5830 = 3439.7 60% of 5830 = 3498 61% of 5830 = 3556.3 62% of 5830 = 3614.6 63% of 5830 = 3672.9 64% of 5830 = 3731.2 65% of 5830 = 3789.5 66% of 5830 = 3847.8 67% of 5830 = 3906.1 68% of 5830 = 3964.4 69% of 5830 = 4022.7 70% of 5830 = 4081 71% of 5830 = 4139.3 72% of 5830 = 4197.6 73% of 5830 = 4255.9 74% of 5830 = 4314.2 75% of 5830 = 4372.5
76% of 5830 = 4430.8 77% of 5830 = 4489.1 78% of 5830 = 4547.4 79% of 5830 = 4605.7 80% of 5830 = 4664 81% of 5830 = 4722.3 82% of 5830 = 4780.6 83% of 5830 = 4838.9 84% of 5830 = 4897.2 85% of 5830 = 4955.5 86% of 5830 = 5013.8 87% of 5830 = 5072.1 88% of 5830 = 5130.4 89% of 5830 = 5188.7 90% of 5830 = 5247 91% of 5830 = 5305.3 92% of 5830 = 5363.6 93% of 5830 = 5421.9 94% of 5830 = 5480.2 95% of 5830 = 5538.5 96% of 5830 = 5596.8 97% of 5830 = 5655.1 98% of 5830 = 5713.4 99% of 5830 = 5771.7 100% of 5830 = 5830 | 2,969 | 5,489 | {"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} | 3.90625 | 4 | CC-MAIN-2021-49 | latest | en | 0.894401 |
https://www.mauriciopoppe.com/notes/mathematics/numeral-systems/quaternions/ | 1,718,252,650,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00513.warc.gz | 809,998,132 | 45,316 | ## Definition
The existence of complex number presented a question for mathematicians, if a complex number exists in a 2D complex plane, could there be a 3D equivalent?
Sir William Rowan Hamilton among many other mathematicians of the 18th and 19th century had been searching for the answer, Hamilton conjectured that a 3D complex number could be represented by the triple $a + bi + cj$ where $i$ and $j$ are imaginary quantities and square to $-1$, when he was developing the algebra for this triplet the product of them raised a problem when expanded
\begin{align*} z_1 &= a_1 + b_1i + c_1j \\ z_2 &= a_2 + b_2i + c_2j \\ z_1z_2 &= (a_1 + b_1i + c_1j)(a_2 + b_2i + c_2j) \\ &= (a_1b_1 - b_1b_2 - c_1c_2) + (a_1b_2 + b_1a_2)i + (a_1c_2 + c_1a_2)j \\ & \quad + b_1c_2ij + c_1b_2ji \end{align*}
The quantities $ij$ and $ji$ represented a problem for Hamilton, even if $ij = -ji$ we are still left with $(b_1c_2 - c_1b_2)ij$
On October 16th, 1843, while he was walking with his wife along the Royal Canal in Ireland he saw the solution as a quadruple instead of a triple, instead of using two imaginary terms, three imaginary terms provided the necessary quantities to resolve products like $ij$
Hamilton defined a quaternion $q$ as
\begin{align*} q = s + ai + bj + ck \quad s,a,b,c \in \mathbb{R} \\ i^2 = j^2 = k^2 = ijk = -1 \\ ij = k \quad jk = i \quad ki = j \\ ji = -k \quad kj = -1 \quad ik = -j \end{align*}
If a complex number $i$ is capable of rotating points on the plane by $\deg{90}$ then perhaps a triple rotates points in space by $\deg{90}$, in the end the triplet was replaced by a quaternion
## Notation
There are three ways of annotating a quaternion $q$
\begin{align} q &= s + xi + yj + zk \\ q &= s + \mbold{v} \\ q &= [s, \mbold{v}] \\ & \text{where s,x,y,z \in \mathbb{R}, \mbold{v} \in \mathbb{R}^3} \nonumber \\ & \text{and i^2 = j^2 = k^2 = ijk = -1} \nonumber \end{align}
### Real quaternion
A real quaternion has a zero vector term
$$q = [s, \mbold{0}]$$
### Pure quaternion
A pure quaternion is a quaternion having a zero scalar term
$$q = [0, \mbold{v}]$$
### Quaternion conjugate
Given
$$q = [s, \mbold{v}]$$
The quaternion conjugate is defined as
$$q^* = [s, - \mbold{v}]$$
### Quaternion norm
The norm of a quaternion $q = [s, \mbold{v}]$ is defined as the square root of the product of itself and its conjugate (the multiplication operation is defined later)
\begin{align*} \norm{q} &= \sqrt{qq^*} \\ &= \sqrt{s^2 + x^2 + y^2 + z^2} \end{align*}
Also note that
$$\norm{q}^2 = qq^*$$
Norm facts
• $\norm{qq^*} = \norm{q}\norm{q^*}$
• $\norm{q^*} = \norm{q}$
### Unit quaternion
A unit quaternion is a quaternion of norm one given by
\begin{align} q &= [s, \lambda \unit{n}] \quad s,\lambda \in \mathbb{R}, \unit{n} \in \mathbb{R}^3 \label{unit-norm-quaternion}\\ \left | \unit{n} \right | &= 1 \nonumber \\ s^2 + \lambda^2 &= 1 \nonumber \end{align}
Note: dividing a non-zero quaternion by its norm produces a unit norm quaternion
## Operations
### Quaternion Product
Given two quaternions
\begin{align*} q_a = [s_a, \mbold{a}] \quad \quad \mbold{a} = x_a i + y_a j + z_a k \\ q_b = [s_b, \mbold{b}] \quad \quad \mbold{b} = x_b i + y_b j + z_b k \end{align*}
The product $q_aq_b$ is computed as follows
\begin{align} q_aq_b &= (s_a + x_a i + y_a j + z_a k)(s_b + x_b i + y_b j + z_b k) \nonumber \\ &= (s_as_b - x_ax_b - y_ay_b - z_az_b) \nonumber \\ & \quad + (s_ax_b + s_bx_a + y_az_b - y_bz_a)i \nonumber \\ & \quad + (s_ay_b + s_by_a + z_ax_b - z_bx_a)j \nonumber \\ & \quad + (s_az_b + s_bz_a + x_ay_b - x_by_a)k \label{quaternion-product} \end{align}
Replacing the imaginaries by the ordered pairs (which are themselves quaternion units)
$$i = [0, \mbold{i}] \quad j = [0, \mbold{j}] \quad k = [0, \mbold{k}] \quad 1 = [1, \mbold{0}]$$
And substituting them in \eqref{quaternion-product}
\begin{align*} q_aq_b &= (s_as_b - x_ax_b - y_ay_b - z_az_b)[1, \mbold{0}] \\ & \quad + (s_ax_b + s_bx_a + y_az_b - y_bz_a)[0, \mbold{i}] \\ & \quad + (s_ay_b + s_by_a + z_ax_b - z_bx_a)[0, \mbold{j}] \\ & \quad + (s_az_b + s_bz_a + x_ay_b - x_by_a)[0, \mbold{k}] \end{align*}
By doing some aggrupations
\begin{align*} q_aq_b &= [s_as_b - x_ax_b - y_ay_b - z_az_b, \\ & \quad s_a(x_b \mbold{i} + y_b \mbold{j} + z_b \mbold{k}) + s_b(x_a \mbold{i} + y_a \mbold{j} + z_a \mbold{k}) \\ & \quad + (y_az_b - y_bz_a) \mbold{i} + (z_ax_b - z_bx_a) \mbold{j} + (x_ay_b - x_by_a) \mbold{k}] \\ &= [s_as_b - \mbold{a} \cdot \mbold{b}, s_a\mbold{b} + s_b\mbold{a} + \mbold{a} \times \mbold{b}] \end{align*}
Now let’s compute the product $q_bq_a$
$$q_bq_a = [s_bs_a - \mbold{b} \cdot \mbold{a}, s_b\mbold{a} + s_a\mbold{b} + \mbold{b} \times \mbold{a}]$$
Note that the scalar quantity of both products is the same however the vector quantity varies (the cross product sign is changed) therefore
$$q_aq_b \not = q_bq_a$$
This is an important fact to note since for complex number the product commutes however for quaternions it doesn’t
#### Product of a scalar and a quaternion
Let $k$ be a scalar represented as a quaternion as $q_k = [k, \mathbf{0}]$ and $q = [s, \mathbf{v}]$
Their product is
\begin{align*} q_kq &= [k, \mathbf{0}][s, \mathbf{v}] \\ &= [ks, k\mathbf{v}] \end{align*}
Note that this product is commutative
#### Product of a quaternion with itself (square of a quaternion)
\begin{align*} q &= [s, \mbold{v}] \\ q^2 &= [s, \mbold{v}] [s, \mbold{v}] \\ &= [s^2 - \mbold{v} \cdot \mbold{v}, 2s\mbold{v} + \mbold{v} \times \mbold{v}] \\ &= [s^2 - \norm{v}^2, 2s\mbold{v}] \\ &= [s^2 - (x^2 + y^2 + z^2), 2s(x\mbold{i} + y\mbold{j} + z\mbold{k})] \end{align*}
#### Product of a quaternion and its conjugate
Let $q = [s, \mathbf{v}]$
\begin{align*} qq^* &= [s, \mathbf{v}][s, -\mathbf{v}] \\ &= [s^2 + \mathbf{v} \cdot \mathbf{v}, -s \mathbf{v} + s\mathbf{v} - \mathbf{v} \times \mathbf{v}] \\ &= [s^2 + \mathbf{v} \cdot \mathbf{v}, \mathbf{0}] \\ &= s^2 + x^2 + y^2 + z^2 \end{align*}
Note that this product commutes i.e. $qq^* = q^*q$
#### Product of unit quaternions
Given
$$q_a = [s_a, \mbold{a}] \\ q_b = [s_b, \mbold{b}]$$
Where $\norm{q_a} = \norm{q_b} = 1$, the product is another unit-norm quaternion
$$q_c = [s_c, \mbold{c}]$$
Where $\norm{q_c} = 1$
#### Product of pure quaternions
Let
$$q_a = [0, \mbold{a}] \\ q_b = [0, \mbold{b}]$$
The product $q_aq_b$ is defined as
\begin{align*} q_aq_b &= [-\mbold{a} \cdot \mbold{b}, \mbold{a} \times \mbold{b}] \end{align*}
Note that the resulting quaternion is no longer a pure quaternion as some information has propagated into the real part via the dot product
#### Product of a pure quaternion with itself (square of a pure quaternion)
\begin{align*} q &= [0, \mbold{v}] \\ q^2 &= [0, \mbold{v}] [0, \mbold{v}] \\ &= [-\mbold{v} \cdot \mbold{v}, \mbold{v} \times \mbold{v}] \\ &= [-(x^2 + y^2 + z^2), \mbold{0}] \\ &= -\norm{v}^2 \end{align*}
If $q$ is a unit norm pure quaternion then
$$q^2 = -1$$
#### Product of a pure quaternion with its conjugate
\begin{align*} q^*q = qq^* &= [0, \mathbf{v}][0, -\mathbf{v}] \\ &= [\mathbf{v} \cdot \mathbf{v}, -\mbold{v \times v}] \\ &= [\mathbf{v} \cdot \mathbf{v}, \mbold{0}] \\ &= \norm{v}^2 \end{align*}
### Inverse of a quaternion
By definition, the inverse $q^{-1}$ of $q$ is
$$qq^{-1} = [1, \mbold{0}]$$
To isolate $q^{-1}$ let’s pre multiply both sides by $q^*$
\begin{align*} q^*qq^{-1} &= q^* \\ \norm{q}^2q^{-1} &= q^* \\ q^{-1} &= \frac{q^*}{\norm{q}^2} \end{align*}
### Quaternion units
Given the vector $\mbold{v}$
$$\mbold{v} = v \hat{\mbold{v}} \quad \text{where v = |\mbold{v}|, and |\hat{\mbold{v}}| = 1}$$
Combining this with the definition of a pure quaternion
\begin{align*} q &= [0, \mbold{v}] \\ &= [0, v \hat{\mbold{v}}] \\ &= v[0, \hat{\mbold{v}}] \end{align*}
It’s convenient to identify the unit quaternion as $\hat{q}$ (where $v = 1$)
$$\hat{q} = [0, \hat{\mbold{v}}]$$
Let’s check if the quaternion unit $\mbold{i}$ squares to the ordered pair $[-1, \mbold{0}]$
\begin{align*} i^2 &= [0, \mbold{i}][0, \mbold{i}] \\ &= [0 \cdot 0 - \mbold{i} \cdot \mbold{i}, 0 \cdot \mbold{i} + 0 \cdot \mbold{i} - \mbold{i} \times \mbold{i}] \\ &= [-|\mbold{i}|^2, \mbold{0}] \quad \text{\mbold{i} \times \mbold{i} = 0} \\ & = [-1, \mbold{0}] \end{align*}
### Misc operations
#### Taking the scalar part of a quaternion
To isolate the scalar part of $q$ we could add $q^*$ to it
$$2 S(q) = q + q^*$$ | 3,261 | 8,430 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-26 | latest | en | 0.839979 |
https://programs.programmingoneonone.com/2021/08/leetcode-binary-tree-inorder-traversal-problem-solution.html | 1,702,099,398,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00758.warc.gz | 503,603,517 | 32,263 | # Leetcode Binary Tree Inorder Traversal problem solution
In this Leetcode Binary Tree Inorder Traversal problem solution we have Given the root of a binary tree, return the inorder traversal of its nodes' values.
## Problem solution in Python.
```from collections import deque
class Solution(object):
def _in_order_iter(self, root):
stack = deque()
while stack or root:
if root:
stack.append(root)
root = root.left
else:
root = stack.pop()
yield root.val
root = root.right
def inorderTraversal(self, root):
return list(self._in_order_iter(root))
```
## Problem solution in Java.
```public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> resultList = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode current = root;
while (current != null || !stack.isEmpty()) {
while (current != null) {
stack.push(current);
current = current.left;
}
current = stack.pop();
current = current.right;
}
return resultList;
}
```
## Problem solution in C++.
```class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
stack<TreeNode*> s;
vector<int> res;
do{
while(root!=NULL){
s.push(root);
root=root->left;
}
if(!s.empty()){
root=s.top();
s.pop();
res.push_back(root->val);
root=root->right;
}
}while(root!=NULL||!s.empty());
return res;
}
};
```
## Problem solution in C.
```int *ans;
int cnt;
int base;
void tra(struct TreeNode *node){
if(node==NULL){
return;
}
tra(node->left);
if(cnt==base){
//not enough memory
base*=2;
ans=realloc(ans,base*sizeof(int));
}
ans[cnt++]=node->val;
tra(node->right);
}
int* inorderTraversal(struct TreeNode* root, int* rs){
base=8;
ans=malloc(sizeof(int)*base);
cnt=0;
tra(root);
*rs=cnt;
return ans;
}
``` | 417 | 1,696 | {"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-2023-50 | latest | en | 0.387483 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-linear-algebra-7th-edition/chapter-3-determinants-3-3-properties-of-determinants-3-3-exercises-page-125/29 | 1,576,318,180,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540586560.45/warc/CC-MAIN-20191214094407-20191214122407-00524.warc.gz | 713,771,289 | 12,011 | ## Elementary Linear Algebra 7th Edition
$$|A|=24, \quad \left|A^{-1}\right|=\frac{1}{24}.$$
Since $$A=\left[\begin{array}{rrr} 1&0&-1&3\\ 1&0&3&-2 \\ 2&0&2&-1\\ 1&-3&1&-2 \end{array}\right],$$ then one can calculate its inverse , that is, $$A^{-1}= \left[\begin{array}{rrr} -\frac{1}{8}&-\frac{5}{8}&{\frac {7}{8}}&0 \\ -\frac{1}{4}&-\frac{1}{4}&{\frac {5}{12}}&-\frac{1}{3} \\ \frac{3}{8}&{\frac {7}{8}}&-\frac{5}{8}&0\\ \frac{1}{2} &\frac{1}{2}&- \frac{1}{2}&0 \end{array}\right].$$ Now, we have $$|A|=24, \quad \left|A^{-1}\right|=\frac{1}{24}.$$ Hence, we verify that $$\left|A^{-1}\right|=\frac{1}{|A|}.$$ | 288 | 612 | {"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} | 4.40625 | 4 | CC-MAIN-2019-51 | latest | en | 0.305992 |
http://stackoverflow.com/questions/4591611/unmasking-a-oval-like-a-clock | 1,394,778,763,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678687395/warc/CC-MAIN-20140313024447-00096-ip-10-183-142-35.ec2.internal.warc.gz | 134,303,767 | 14,634 | # unmasking a oval like a clock
i have an circle shape, and what i would like to achieve is that it gets masked and thenon enter frame it zhould get unmasked like a clock, so you get to see from 0 to 1, then from 0 to 2, 0 to 3, etc, i know it has to be done with radiants etc but math isnt really my cup of tea. Any ideas?
-
Step 2: Calculate the angle you need:
``````var endAngle:Number = 360 / 12 * clockTime;
``````
where clockTime is your hours. For minutes, use 360/60.
Step 3: Refresh the mask on ENTER_FRAME
You have to draw a pie slice using curveTo. Use `maskSprite.graphics.clear()` and `maskSprite.graphics.beginFill(0,1)`, then call:
``````function drawClockSlice (sprite : Sprite, centerX : Number, centerY : Number, endAngle : Number, radius : Number) : void
{
var g : Graphics = sprite.graphics;
var controlPoint : Point;
var anchorPoint : Point;
var startRadians:Number = Math.PI / 180 * -90;
var endRadians : Number = Math.PI / 180 * (endAngle-90);
var divisions : Number = Math.floor( difference / (Math.PI / 4) ) + 1;
var span : Number = difference / (2 * divisions);
g.moveTo(centerX, centerY);
for (var i : Number = 0; i < divisions; ++i)
{
Don't forget your `endFill()`! | 341 | 1,206 | {"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-2014-10 | latest | en | 0.711196 |
https://overlordoftheuberferal.com/tag/polygons/ | 1,686,229,192,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654871.97/warc/CC-MAIN-20230608103815-20230608133815-00008.warc.gz | 478,948,136 | 52,518 | # First Whirled Warp
Imagine two points moving clockwise around the circumference of a circle. Find the midpoint between the two points when one point is moving twice as fast as the other. The midpoint will trace this shape:
Midpoint of two points moving around circle at speeds s and s*2
(n.b. to make things easier to see, the red circle shown here and elsewhere is slightly larger than the virtual circle used to calculate the midpoints)
Now suppose that one point is moving anticlockwise. The midpoint will now trace this shape:
Midpoint for s, -s*2
Now try three points, two moving at the same speed and one moving twice as fast:
Midpoint for s, s, s*2
When the point moving twice as fast is moving anticlockwise, this shape appears:
Midpoint for s, s, -s*2
Here are more of these midpoint-shapes:
Midpoint for s, s*3
Midpoint for s, -s*3
Midpoint for s*2, s*3
Midpoint for s, -s, s*2
Midpoint for s, s*2, -s*2
Midpoint for s, s*2, s*2
Midpoint for s, -s*3, -s*5
Midpoint for s, s*2, s*3
Midpoint for s, s*2, -s*3
Midpoint for s, -s*3, s*5
Midpoint for s, s*3, s*5
Midpoint for s, s, s, s*3
Midpoint for s, s, s, -s*3
Midpoint for s, s, -s, s*3
Midpoint for s, s, -s, -s*3
But what about points moving around the perimeter of a polygon? Here are the midpoints of two points moving clockwise around the perimeter of a square, with one point moving twice as fast as the other:
Midpoint for square with s, s*2
And when one point moves anticlockwise:
Midpoint for square with s, -s*2
If you adjust the midpoints so that the square fills a circle, they look like this:
Midpoint for square with s, s*2, with square adjusted to fill circle
When the red circle is removed, the midpoint-shape is easier to see:
Midpoint for square with s, s*2, circ-adjusted
Here are more midpoint-shapes from squares:
Midpoint for s, s*3
Midpoint for s, -s*3
Midpoint for s, s*4
And some more circularly adjusted midpoint-shapes from squares:
Finally (for now), let’s look at triangles. If three points are moving clockwise around the perimeter of a triangle, one moving four times as fast as the other two, the midpoint traces this shape:
Midpoint for triangle with s, s, s*4
Now try one of the points moving anticlockwise:
Midpoint for s, s, -s*4
Midpoint for s, -s, s*4
If you adjust the midpoints so that the triangular space fills a circle, they look like this:
Midpoint for s, s, s*4, with triangular space adjusted to fill circle
Midpoint for s, -s, s*4, circ-adjusted
Midpoint for s, s, -s*4, circ-adjusted
There are lots more (infinitely more!) midpoint-shapes to see, so watch this (circularly adjusted) space.
We Can Circ It Out — more on converting polygons into circles
# We Can Circ It Out
It’s a pretty little problem to convert this triangular fractal…
Sierpiński triangle (Wikipedia)
…into its circular equivalent:
Sierpiński triangle as circle
Sierpiński triangle to circle (animated)
But once you’ve circ’d it out, as it were, you can easily adapt the technique to fractals based on other polygons:
T-square fractal (Wikipedia)
T-square fractal as circle
T-square fractal to circle (animated)
Elsewhere other-accessible…
Dilating the Delta — more on converting polygonic fractals to circles…
# Archimedrons
The 13 Archimedean solids (also known as Archimedean polyhedra)
# Bent Pent
This is a beautiful and interesting shape, reminiscent of a piece of jewellery:
Pentagons in a ring
I came across it in this tricky little word-puzzle:
Word puzzle using pentagon-ring
Here’s a printable version of the puzzle:
Printable puzzle
Let’s try placing some other regular polygons with s sides around regular polygons with s*2 sides:
Hexagonal ring of triangles
Octagonal ring of squares
Decagonal ring of pentagons
Dodecagonal ring of hexagons
Only regular pentagons fit perfectly, edge-to-edge, around a regular decagon. But all these polygonal-rings can be used to create interesting and beautiful fractals, as I hope to show in a future post.
# Square Routes Re-Re-Re-Re-Re-Revisited
For a good example of how more can be less, try the chaos game. You trace a point jumping repeatedly 1/n of the way towards a randomly chosen vertex of a regular polygon. When the polygon is a triangle and 1/n = 1/2, this is what happens:
Chaos triangle #1
Chaos triangle #2
Chaos triangle #3
Chaos triangle #4
Chaos triangle #5
Chaos triangle #6
Chaos triangle #7
As you can see, this simple chaos game creates a fractal known as the Sierpiński triangle (or Sierpiński sieve). Now try more and discover that it’s less. When you play the chaos game with a square, this is what happens:
Chaos square #1
Chaos square #2
Chaos square #3
Chaos square #4
Chaos square #5
Chaos square #6
Chaos square #7
As you can see, more is less: the interior of the square simply fills with points and no attractive fractal appears. And because that was more is less, let’s see how less is more. What happens if you restrict the way in which the point inside the square can jump? Suppose it can’t jump twice towards the same vertex (i.e., the vertex v+0 is banned). This fractal appears:
Ban on choosing vertex [v+0]
And if the point can’t jump towards the vertex one place anti-clockwise of the currently chosen vertex, this fractal appears:
Ban on vertex [v+1] (or [v-1], depending on how you number the vertices)
And if the point can’t jump towards two places clockwise or anti-clockwise of the currently chosen vertex, this fractal appears:
Ban on vertex [v+2], i.e. the diagonally opposite vertex
At least, that is one possible route to those three particular fractals. You see another route, start with this simple fractal, where dividing and discarding parts of a square creates a Sierpiński triangle:
Square to Sierpiński triangle #1
Square to Sierpiński triangle #2
Square to Sierpiński triangle #3
Square to Sierpiński triangle #4
[…]
Square to Sierpiński triangle #10
Square to Sierpiński triangle (animated)
By taking four of these square-to-Sierpiński-triangle fractals and rotating them in the right way, you can re-create the three chaos-game fractals shown above. Here’s the [v+0]-ban fractal:
[v+0]-ban fractal #1
[v+0]-ban #2
[v+0]-ban #3
[v+0]-ban #4
[v+0]-ban #5
[v+0]-ban #6
[v+0]-ban #7
[v+0]-ban #8
[v+0]-ban #9
[v+0]-ban (animated)
And here’s the [v+1]-ban fractal:
[v+1]-ban fractal #1
[v+1]-ban #2
[v+1]-ban #3
[v+1]-ban #4
[v+1]-ban #5
[v+1]-ban #6
[v+1]-ban #7
[v+1]-ban #8
[v+1]-ban #9
[v+1]-ban (animated)
And here’s the [v+2]-ban fractal:
[v+2]-ban fractal #1
[v+2]-ban #2
[v+2]-ban #3
[v+2]-ban #4
[v+2]-ban #5
[v+2]-ban #6
[v+2]-ban #7
[v+2]-ban #8
[v+2]-ban #9
[v+2]-ban (animated)
And taking a different route means that you can find more fractals — as I will demonstrate.
# Dilating the Delta
A circle with a radius of one unit has an area of exactly π units = 3.141592… units. An equilateral triangle inscribed in the unit circle has an area of 1.2990381… units, or 41.34% of the area of the unit circle.
In other words, triangles are cramped! And so it’s often difficult to see what’s going on in a triangle. Here’s one example, a fractal that starts by finding the centre of the equilateral triangle:
Triangular fractal stage #1
Next, use that central point to create three more triangles:
Triangular fractal stage #2
And then use the centres of each new triangle to create three more triangles (for a total of nine triangles):
Triangular fractal stage #3
And so on, trebling the number of triangles at each stage:
Triangular fractal stage #4
Triangular fractal stage #5
As you can see, the triangles quickly become very crowded. So do the central points when you stop drawing the triangles:
Triangular fractal stage #6
Triangular fractal stage #7
Triangular fractal stage #8
Triangular fractal stage #9
Triangular fractal stage #10
Triangular fractal stage #11
Triangular fractal stage #12
Triangular fractal stage #13
Triangular fractal (animated)
The cramping inside a triangle is why I decided to dilate the delta like this:
Triangular fractal
Circular fractal from triangular fractal
Triangular fractal to circular fractal (animated)
Formation of the circular fractal (animated)
And how do you dilate the delta, or convert an equilateral triangle into a circle? You use elementary trigonometry to expand the perimeter of the triangle so that it lies on the perimeter of the unit circle. The vertices of the triangle don’t move, because they already lie on the perimeter of the circle, but every other point, p, on the perimeter of the triangles moves outward by a fixed amount, m, depending on the angle it makes with the center of the triangle.
Once you have m, you can move outward every point, p(1..i), that lies between p on the perimeter and the centre of the triangle. At least, that’s the theory between the dilation of the delta. In practice, all you need is a point, (x,y), inside the triangle. From that, you can find the angle, θ, and distance, d, from the centre, calculate m, and move (x,y) to d * m from the centre.
You can apply this technique to any fractal created in an equilateral triangle. For example, here’s the famous Sierpiński triangle in its standard form as a delta, then as a dilated delta or circle:
Sierpiński triangle
Sierpiński triangle to circular Sierpiński fractal
Sierpiński triangle to circle (animated)
But why stop at triangles? You can use the same elementary trigonometry to convert any regular polygon into a circle. A square inscribed in a unit circle has an area of 2 units, or 63.66% of the area of the unit circle, so it too is cramped by comparison with the circle. Here’s a square fractal that I’ve often posted before:
Square fractal, jump = 1/2, ban on jumping towards any vertex twice in a row
It’s created by banning a randomly jumping point from moving twice in a row 1/2 of the distance towards the same vertex of the square. When you dilate the fractal, it looks like this:
Circular fractal from square fractal, j = 1/2, ban on jumping towards vertex v(i) twice in a row
Circular fractal from square (animated)
And here’s a related fractal where the randomly jumping point can’t jump towards the vertex directly clockwise from the vertex it’s previously jumped towards (so it can jump towards the same vertex twice or more):
Square fractal, j = 1/2, ban on vertex v(i+1)
When the fractal is dilated, it looks like this:
Circular fractal from square, i = 1
Circular fractal from square (animated)
In this square fractal, the randomly jumping point can’t jump towards the vertex directly opposite the vertex it’s previously jumped towards:
Square fractal, ban on vertex v(i+2)
And here is the dilated version:
Circular fractal from square, i = 2
Circular fractal from square (animated)
And there are a lot more fractals where those came from. Infinitely many, in fact.
# Get Your Prox Off #3
I’ve looked at lot at the fractals created when you randomly (or quasi-randomly) choose a vertex of a square, then jump half of the distance towards it. You can ban jumps towards the same vertex twice in a row, or jumps towards the vertex clockwise or anticlockwise from the vertex you’ve just chosen, and so on.
But you don’t have to choose vertices directly: you can also choose them by distance or proximity (see “Get Your Prox Off” for an earlier look at fractals-by-distance). For example, this fractal appears when you can jump half-way towards the nearest vertex, the second-nearest vertex, and the third-nearest vertex (i.e., you can’t jump towards the fourth-nearest or most distant vertex):
vertices = 4, distance = (1,2,3), jump = 1/2
It’s actually the same fractal as you get when you choose vertices directly and ban jumps towards the vertex diagonally opposite from the one you’ve just chosen. But this fractal-by-distance isn’t easy to match with a fractal-by-vertex:
v = 4, d = (1,2,4), j = 1/2
Nor is this one:
v = 4, d = (1,3,4)
This one, however, is the same as the fractal-by-vertex created by banning a jump towards the same vertex twice in a row:
v = 4, d = (2,3,4)
The point can jump towards second-nearest, third-nearest and fourth-nearest vertices, but not towards the nearest. And the nearest vertex will be the one chosen previously.
Now let’s try squares with an additional point-for-jumping-towards on each side (the points are numbered 1 to 8, with points 1, 3, 5, 7 being the true vertices):
v = 4 + s1 point on each side, d = (1,2,3)
v = 4 + s1, d = (1,2,5)
v = 4 + s1, d = (1,2,7)
v = 4 + s1, d = (1,3,8)
v = 4 + s1, d = (1,4,6)
v = 4 + s1, d = (1,7,8)
v = 4 + s1, d = (2,3,8)
v = 4 + s1, d = (2,4,8)
And here are squares where the jump is 2/3, not 1/2, and you can choose only the nearest or third-nearest jump-point:
v = 4, d = (1,3), j = 2/3
v = 4 + s1, d = (1,3), j = 2/3
Now here are some pentagonal fractals-by-distance:
v = 5, d = (1,2,5), j = 1/2
v = 5 + s1, d = (1,2,7)
v = 5 + s1, d = (1,2,8)
v = 5 + s1, d = (1,2,9)
v = 5 + s1, d = (1,9,10)
v = 5 + s1, d = (1,10), j = 2/3
v = 5 + s1, d = (various), j = 2/3 (animated)
And now some hexagonal fractals-by-distance:
v = 6, d = (1,2,4), j = 1/2
v = 6, d = (1,3,5)
v = 6, d = (1,3,6)
v = 6, d = (1,2,3,4)
v = 6 + central point, d = (1,2,3,4)
v = 6, d = (1,2,3,6)
v = 6, d = (1,2,4,6)
v = 6, d = (1,3,4,5)
v = 6, d = (1,3,4,6)
v = 6, d = (1,4,5,6)
Elsewhere other-accessible:
Get Your Prox Off — an earlier look at fractals-by-distance
Get Your Prox Off # 2 — and another
# Fractal + Star = Fractar
Here’s a three-armed star made with three lines radiating at intervals of 120°:
Triangular fractal stage #1
At the end of each of the three lines, add three more lines at half the length:
Triangular fractal #2
And continue like this:
Triangular fractal #3
Triangular fractal #4
Triangular fractal #5
Triangular fractal #6
Triangular fractal #7
Triangular fractal #8
Triangular fractal #9
Triangular fractal #10
Triangular fractal (animated)
Because this fractal is created from a series of stars, you could call it a fractar. Here’s a black-and-white version:
Triangular fractar (black-and-white)
Triangular fractar (black-and-white) (animated)
(Open in a new window for larger version if the image seems distorted)
A four-armed star doesn’t yield an easily recognizable fractal in a similar way, so let’s try a five-armed star:
Pentagonal fractar stage #1
Pentagonal fractar #2
Pentagonal fractar #3
Pentagonal fractar #4
Pentagonal fractar #5
Pentagonal fractar #6
Pentagonal fractar #7
Pentagonal fractar (animated)
Pentagonal fractar (black-and-white)
Pentagonal fractar (bw) (animated)
And here’s a six-armed star:
Hexagonal fractar stage #1
Hexagonal fractar #2
Hexagonal fractar #3
Hexagonal fractar #4
Hexagonal fractar #5
Hexagonal fractar #6
Hexagonal fractar (animated)
Hexagonal fractar (black-and-white)
Hexagonal fractar (bw) (animated)
And here’s what happens to the triangular fractar when the new lines are rotated by 60°:
Triangular fractar (60° rotation) #1
Triangular fractar (60°) #2
Triangular fractar (60°) #3
Triangular fractar (60°) #4
Triangular fractar (60°) #5
Triangular fractar (60°) #6
Triangular fractar (60°) #7
Triangular fractar (60°) #8
Triangular fractar (60°) #9
Triangular fractar (60°) (animated)
Triangular fractar (60°) (black-and-white)
Triangular fractar (60°) (bw) (animated)
Triangular fractar (60°) (no lines) (black-and-white)
A four-armed star yields a recognizable fractal when the rotation is 45°:
Square fractar (45°) #1
Square fractar (45°) #2
Square fractar (45°) #3
Square fractar (45°) #4
Square fractar (45°) #5
Square fractar (45°) #6
Square fractar (45°) #7
Square fractar (45°) #8
Square fractar (45°) (animated)
Square fractar (45°) (black-and-white)
Square fractar (45°) (bw) (animated)
Without the lines, the final fractar looks like the plan of a castle:
Square fractar (45°) (bw) (no lines)
And here’s a five-armed star with new lines rotated at 36°:
Pentagonal fractar (36°) #1
Pentagonal fractar (36°) #2
Pentagonal fractar (36°) #3
Pentagonal fractar (36°) #4
Pentagonal fractar (36°) #5
Pentagonal fractar (36°) #6
Pentagonal fractar (36°) #7
Pentagonal fractar (36°) (animated)
Again, the final fractar without lines looks like the plan of a castle:
Pentagonal fractar (36°) (no lines) (black-and-white)
Finally, here’s a six-armed star with new lines rotated at 30°:
Hexagonal fractar (30°) #1
Hexagonal fractar (30°) #2
Hexagonal fractar (30°) #3
Hexagonal fractar (30°) #4
Hexagonal fractar (30°) #5
Hexagonal fractar (30°) #6
Hexagonal fractar (30°) (animated)
And the hexagonal castle plan:
Hexagonal fractar (30°) (black-and-white) (no lines)
# Performativizing the Polygonic #3
Pre-previously in my passionate portrayal of polygonic performativity, I showed how a single point jumping randomly (or quasi-randomly) towards the vertices of a polygon can create elaborate fractals. For example, if the point jumps 1/φth (= 0.6180339887…) of the way towards the vertices of a pentagon, it creates this fractal:
Point jumping 1/φth of the way to a randomly (or quasi-randomly) chosen vertex of a pentagon
But as you might expect, there are different routes to the same fractal. Suppose you take a pentagon and select a single vertex. Now, measure the distance to each vertex, v(1,i=1..5), of the original pentagon (including the selected vertex) and reduce it by 1/φ to find the position of a new vertex, v(2,i=1..5). If you do this for each vertex of the original pentagon, then to each vertex of the new pentagons, and so on, in the end you create the same fractal as the jumping point does:
Shrink pentagons by 1/φ, stage #1
Stage #2
Stage #3
Stage #4
Stage #5
Stage #6
Shrink by 1/φ (animated) (click for larger if blurred)
And here is the route to a centre-filled variant of the fractal:
Central pentagon, stage #1
Stage #2
Stage #3
Stage #4
Stage #5
Stage #6
Central pentagon (animated) (click for larger if blurred)
Using this shrink-the-polygon method, you can reach the same fractals by a third route. This time, use vertex v(1,i) of the original polygon as the centre of the new polygon with its vertices v(2,i=1..5). Creation of the fractal looks like this:
Pentagons over vertices, shrink by 1/φ, stage #1 (no pentagons over vertices)
Stage #2
Stage #3
Stage #4
Stage #4
Stage #5
Stage #7
Pentagons over vertices (animated) (click for larger if blurred)
And here is a third way of creating the centre-filled pentagonal fractal:
Pentagons over vertices and central pentagon, stage #1
Stage #2
Stage #3
Stage #4
Stage #5
Stage #6
Stage #7
Pentagons over vertices with central pentagon (animated) (click for larger if blurred)
And here is a fractal created when there are three pentagons to a side and the pentagons are shrunk by 1/φ^2 = 0.3819660112…:
Pentagon at vertex + pentagon at mid-point of side, shrink by 1/φ^2
Final stage
Pentagon at vertex + pentagon at mid-point of side (animated) (click for larger if blurred)
Pentagon at vertex + pentagon at mid-point of side + central pentagon, shrink by 1/φ^2 and c. 0.5, stage #1
Stage #2
Stage #3
Stage #4
Stage #5
Pentagon at vertex + mid-point + center (animated) (click for larger if blurred)
Previously pre-posted:
# Horn Again
Pre-previously on Overlord-in-terms-of-Core-Issues-around-Maximal-Engagement-with-Key-Notions-of-the-Über-Feral, I interrogated issues around this shape, the horned triangle:
Horned Triangle (more details)
Now I want to look at the tricorn (from Latin tri-, “three”, + -corn, “horn”). It’s like a horned triangle, but has three horns instead of one:
Tricorn, or three-horned triangle
These are the stages that make up the tricorn:
Tricorn (stages)
Tricorn (animated)
And there’s no need to stop at triangles. Here is a four-horned square, or quadricorn:
And a five-horned pentagon, or quinticorn:
Quinticorn, or five-horned pentagon
Quinticorn (anim)
Quinticorn (col)
And below are some variants on the shapes above. First, the reversed tricorn:
Reversed Tricorn
Reversed Tricorn (anim)
Reversed Tricorn (col)
The nested tricorn:
Nested Tricorn (anim)
Nested Tricorn (col)
Nested Tricorn (red-green)
Nested Tricorn (variant col)
Finally (and ferally), the pentagonal octopus or pentapus:
Pentapus (anim)
Pentapus
Pentapus #2
Pentapus #3
Pentapus #4
Pentapus #5
Pentapus #6
Pentapus (col anim)
Elsewhere other-engageable:
The Art Grows Onda — the horned triangle and Katsushika Hokusai’s painting The Great Wave off Kanagawa (c. 1830) | 5,956 | 20,639 | {"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} | 4.25 | 4 | CC-MAIN-2023-23 | latest | en | 0.797148 |
http://azfoo.net/gdt/babs/bars/ca/pacificbeach/lamont/ | 1,621,191,306,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991178.59/warc/CC-MAIN-20210516171301-20210516201301-00482.warc.gz | 6,467,200 | 1,876 | ### BARS::Near Lamont St. in Pacific Beach, California
These street signs were found off Lamont Street in Pacific Beach, California.
The street signs say...
``` +-----------------------+
| Speed Limit 25 |
-------------------------
| No Trucks over 5 Tons |
-------------------------
| Street Sweeping |
| 1st Thursday |
| Every Month |
| 7am - 10am |
+-----------------------+
```
``` the numbers are: 25 5 1 7 10
the math is: 5^(10 - 7 - 1) = 25
rsPiEval: (25 - 1 - 10 / 5) / 7 = Pi
* 22/7 approximates Pi (3.14) rounded to nearest hundredth
rsPhiEval: (1 + sqrt(5)) / (sqrt(25) + 7 - 10) = Phi
* Phi is the golden ratio and it approximates 1.61803399
rsAgeEval: 25 + 5 * 7 - 10 + 1 = 51
* RoadHacker was 51 years young when he found these signs.
rs42Eval: 25 * 10 / 5 - 7 - 1 = 42
* 42 is the "Answer to Life, the Universe, and Everything"
more math: (25 + 10) / 7 = 5^1
more math: 1 * 10 * (sqrt(25) + 7) = 5!
more math: 5 * (7 - 1 - log(10)) = 25
more math: 5 * 7 - 10 + log(1) = 25
```
Creator: Gerald Thurman [gthurman@gmail.com]
Created: 23 March 2009 | 413 | 1,133 | {"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-2021-21 | latest | en | 0.783616 |
https://studyalgorithms.com/array/array-nesting/?shared=email&msg=fail | 1,623,552,074,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487598213.5/warc/CC-MAIN-20210613012009-20210613042009-00491.warc.gz | 488,660,867 | 28,398 | Home Arrays Array Nesting
# Array Nesting
0 comment
Question: Given a zero-indexed array ‘A’ of length ‘N’ which contains all integers from 0 to N-1. Find and return the longest length of set S, where S[i] = {A[i], A[A[i]], A[A[A[i]]], …} subjected to a particular condition.
Suppose the first element in S starts with the selection of element A[i] of index = i, the next element in S should be A[A[i]], and then A[A[A[i]]],.. and so on. By that analogy, we stop adding elements as soon as we get a duplicate.
Input: A = {5, 4, 0, 3, 1, 6, 2}
Output: 4
Let us try to understand the problem statement. Given the condition of the set S, the elements must be nested one after the other.
-> Starting at i = 0.
A[0] = 5; // A[i] == A[0]
A[5] = 6; // A[A[i]] == A[A[0]] == A[5]
A[6] = 2; // A[A[A[i]]] == A[A[A[0]]] == A[A[5]] == A[6]
A[2] = 0; // and so on…
A[0] = 5; // We reach the start point hence S = 4 elements
-> Starting at i = 1.
A[1] = 4; // A[i] == A[1]
A[4] = 1; // A[A[i]] == A[A[1] == A[4]
A[1] = 4; // We reach the start point hence S = 2 elements
As per the problem statement, out of all these sets possible, we need to give the highest number of elements in such a set.
A brute force solution would be to start at each of the indices and try to form sets. We can keep a track of the largest set formed and then return the result.
Looking at the problem closely can hint at a better solution.
Let us look at the array once again, when we start from i = 0.
We see that we form a kind of cycle.
A similar cycle is seen when we start from i = 1.
So, this means that we will get the same set S if we start from positions (0, 2, 5, 6). Because we are forming a cycle.
Hence, we should try to leverage this fact and form a solution. While evaluating the set S, we can mark the elements of the array as visited. In the next iteration, if the element is already visited, we do not need to evaluate the set S again. This will save a lot of time in our solution.
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counter-increment: line;
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public int findSetS(int[] nums) {
boolean[] visited = new boolean[nums.length];
int res = 0;
for (int i = 0; i < nums.length; i++) {
if (!visited[i]) {
int start = nums[i];
int count = 0;
do {
start = nums[start];
count++;
visited[start] = true;
}
while (start != nums[i]);
res = Math.max(res, count);
}
}
return res;
}
Code language: Java (java)
A working solution can be found here. https://ideone.com/vIvw89
Time Complexity: O(n)
Space Complexity: O(n)
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This website uses cookies to improve your experience. We'll assume you're ok with this, but you can opt-out if you wish. Accept Read More | 1,150 | 3,846 | {"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} | 4.09375 | 4 | CC-MAIN-2021-25 | latest | en | 0.866475 |
https://la.mathworks.com/matlabcentral/cody/problems/189-sum-all-integers-from-1-to-2-n/solutions/2138196 | 1,591,397,360,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348504341.78/warc/CC-MAIN-20200605205507-20200605235507-00318.warc.gz | 427,712,192 | 15,511 | Cody
# Problem 189. Sum all integers from 1 to 2^n
Solution 2138196
Submitted on 23 Feb 2020 by Ahmed Okasha
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
x = 3; y_correct = 36; assert(isequal(sum_int(x),y_correct))
2 Pass
x = 7; y_correct = 8256; assert(isequal(sum_int(x),y_correct))
3 Pass
x = 10; y_correct = 524800; assert(isequal(sum_int(x),y_correct))
4 Pass
x = 11; y_correct = 2098176; assert(isequal(sum_int(x),y_correct))
5 Pass
x = 14; y_correct = 134225920; assert(isequal(sum_int(x),y_correct))
6 Pass
x = 17; y_correct = 8590000128; assert(isequal(sum_int(x),y_correct)) | 238 | 725 | {"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-2020-24 | latest | en | 0.577726 |
https://mathematica.stackexchange.com/questions/172961/iterating-a-module | 1,624,180,798,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487660269.75/warc/CC-MAIN-20210620084505-20210620114505-00040.warc.gz | 357,072,511 | 37,521 | Iterating a Module
I have a Module that can act on other Modules, which is given as
Kk[EE,p,z1,z2,z3,K_func]
and I am trying to iterate it, such that
Kk[EE,p,z1,z2,z3,Kk[EE,#2,z2,z3,z4,Kk[EE,#1,z3,z4,z5....]&]&]
for n many times. The problem is that the previous iteration needs to be a function of p and the values of z change during each iteration or shift by one, so I cannot simply use the Nest function.
This works manually if I define functions, for example, for the 2nd iteration:
KK[1][p_]:=KKKK[EE,z3,z4,p] (This is the initial function)
KK[2][p_]:=Kk[EE,p,z2,z3,z4,KK[1]]
KK[3][p_]:=Kk[EE,p,z1,z2,z3,KK[2]]
But, I want this in a Do loop. So, my current code is:
vz = {EE/2, 12, 11, 5};
vz = Reverse[vz];
lz = Length[vz];
kk[1][p_] := KKKK[EE, p, vz[[2]], vz[[1]]]; (Define the initial function)
Do[
ki = i - 1;
v1 = i - 1;
v2 = i;
v3 = i + 1;
kk[i][p_] := Kk[EE, p, v3, v2, v1, kk[ki]];
,{i, 2, 3, 1}];
But this doesn't seem to work. Any suggestions? Apologies if I'm a bit ignorant, this is my first time using Stack Exchange!
• I don't see why Nest wouldn't work: Nest[Kk[E,p,#]&,KKKK[E,p],n] – Lukas Lang May 10 '18 at 14:42
• I should have also added that Kk is a function of 3 other values, which can be called z1,z2,z3. These values change with each iteration, so I don't think Nest can be used – Jordan May 10 '18 at 14:47
• Then please add this information to the question. Also, having global variables in mathematica (or anywhere really) is generally a bad idea – Lukas Lang May 10 '18 at 14:48
• Okay, thank you for you advice! – Jordan May 10 '18 at 15:07
• Have you looked at Fold? Together with Partition, it should be able to give you what you want – Lukas Lang May 10 '18 at 15:13 | 596 | 1,726 | {"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} | 3.21875 | 3 | CC-MAIN-2021-25 | latest | en | 0.889144 |
http://www.egwald.ca/economics/econmath.php | 1,534,544,759,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213158.51/warc/CC-MAIN-20180817221817-20180818001817-00143.warc.gz | 511,973,585 | 8,471 | Egwald Economics: Microeconomics
Derivation of the Model: Oligopoly / Public Firm
by
Elmer G. Wiens
Egwald's popular web pages are provided without cost to users.
Bonus System: Profit, Industry Output, and Own Revenue
Notation:
Q = sum(q) = sum(q0,q1,...qn); qi = firm i's output; i = 0 for public firm
D(Q) = the industry's inverse demand schedule: downward slopping
Ci(qi) = the cost schedule for the ith firm: textbook U-shaped average cost curves
Income of Public Firm Managers:
I0(Q) = b1*[D(Q)*q0 - C0(q0)] + b2*Q + b3*[D(Q)*q0] b1, b2, b3 >= 0
I0(Q) = b1*prof0(q0) + b2*Q + b3*[D(Q)*q0]
Income(Profit) of Private firms:
Ii(Q) = D(Q)*qi - Ci(qi) = profi(qi)
Income Maximization: First Order Conditions:
Public Firm:
dI0/dq0 = b1*[D'(Q)*q0 + D(Q) - C0'(q0)] + b2 + b3*[D'(Q)*q0 + D(Q)] = 0
dI0/dq0 = b1*[mrf0(q0) - mc0(q0)] + b2 +b3*mrf0(q0) = 0
mrf0(q0) = D'(Q)*q0 + D(Q) GAP = mc0(q0) - mrf0(q0)
Private Firms: Cournot Behaviour
dIi/dqi = mrfi(qi) - mci(qi) = 0 mrfi(qi) = D'(Q)*qi + D(Q)
Income Maximization: Second Order Conditions:
Public Firm:
ddI0/ddq0 = b1 * [mrf0'(q0) - mc0'(q0)] + b3 * mrf0'(q0)] < 0
Private Firms: Cournot Behaviour
ddIi/ddqi = mrfi'(qi) - mci'(qi) < 0
The Jacobian used in the Gradient and Newton's Methods for solving the system of first order equations uses the second order formulae plus the cross derivatives:
Public Firm:
ddI0/dq0dqi = b1 * mrf0'(q0) + b3 * mrf0'(q0)]
Private Firms:
ddIi/dqjdqi = mrfi'(qi) i != j
Linear Programming Problem
maximize z = b1 - b2 - b3 b1, b2, b3 >= 0 b3*mrf0(q0) >= 0 b1*(-GAP) + b2 + b3*mrf0(q0) = 0 b1*prof0(q0) + b2*Q + b3*[D(Q)*q0] = Total Bonus
Alternate Solution: System of Linear Equations
mrf0(q0) > 0mrf0(q0) <= 0
b1*(-GAP) + b3*mrf0(q0) = 0
b1*prof0(q0)] + b3*[D(Q)*q0] = Total Bonus
b1*(-GAP) + b2 = 0
b1*prof0(q0) + b2*Q = Total Bonus
How does it work?
The 'government' adjusts 'THE GAP' - the difference between the g.f.'s marginal cost and marginal revenue - until it achieves the level of industry output that it wants. For each setting of 'THE GAP' it solves the linear programming problem for the bonus rates b1,b2,b3. Then it tells the managers that their bonus will be paid using those rates applied to the government firm's profit, industry's total output, and government firm's revenue, respectively. Notice that the 'government' does not need to know the government firm's cost structure, if it can iterate its choice of 'THE GAP.'
I. Maximize government firm's consumer + producer surplus
Suppose the 'government' wants the government firm to produce where marginal cost equals price. In the jargon of welfare economics, it wants to maximize consumer plus producer surplus flowing from the g.f.
In the diagram below, the red lines represent the equilibrium (1 private firm + 1 g.f.) at time t-1 with the GAPt-1 = 1.
qt-10 = 34.28, qt-11 = 31.60, price pt-1 = 54.95, total output Qt-1= 65.88
proft-10 = 611.75, revt-10 = 1883.63; proft-11 = 569.85, revt-11 = 1736.69
bt-11 = 0.0304, bt-12 = 0, bt-13 = 0.00076
The yellow lines represent the equilibrium at time t with the GAPt = 15.2.
qt0 = 38.03, qt1 = 31.19, price pt = 53.61, total output Qt = 69.22
proft0 = 567.90, revt0 = 2038.85; proft1 = 522.54, revt1 = 1672.33
bt1 = 0.0143, bt2 = 0, bt3 = 0.00568
Notice how the weight on g.f. profits, b1, decreases and the weight on g.f. revenue, b3, increases - with the increase in the GAP and the increase in total output.
Essentially, the government can adjust (search) the value of the GAP (or equivalently - the values of the weights b1, b2, b3) until it is satisfied with the resulting equilibrium.
II. Maximize industry's consumer surplus + g.f. producer surplus There are many ways in which the government can arrive at the value of the GAP that it deems "best for its purposes." One way (brute force) would be to estimate econometrically all the equations underlying the lines and curves in the diagram above and then to compute the 'optimal' value of the GAP. Alternatively, the government could use the following iterative process that 'economizes' on the necessary information. Iterative Process Define the following variables: DW1 = .5 * (qt0 - qt-10) * |pt - pt-1| DW2 = (proft0 - proft-10) - (pt - pt-1) * qt-10 if (Qt - Qt-1) >= 0 DW2 = (proft0 - proft-10) - (pt - pt-1) * qt0 if (Qt - Qt-1) < 0 Then DW = DW1 + DW2 is a measure of the change in welfare obtained directly from the government firm from the change in the GAP from GAPt-1 to GAPt. DW1 estimates the change in consumer surplus that is not offset by a change in g.f producer surplus. DW1 = (approx.) the area of the small pink triangle in the diagram above. DW2 estimates the change in g.f. producer surplus not offset by a change in consumer surplus. In the diagram above, the area of the olive rectangle estimates the producer surplus displaced by consumer surplus when price falls from pt-1 to pt. Notice that no new information is required beyond that used in calculating the values of b1, b2, and b3 for a given value of the GAP. Essentially, the government must solve the nonlinear process DW(GAP) = 0 That is - find that value of the GAP where either an increase or decrease in the value of the GAP decreases the value of Consumer + Producer Surplus obtained directly from the government firm. Numerical Analysis provides many ways to solve such a nonlinear process. The government can obtain a lower price - when private firms exhibit Cournot behavior or form a cartel - with the following modification to the above iterative method. Define: DW3 = (pt - pt-1) * (Qt-1 - qt-10) if (Qt - Qt-1) >= 0 DW3 = (pt - pt-1) * (Qt - qt0) if (Qt - Qt-1) < 0 DW = DW1 + DW2 - DW3 This process does not require any information on the profits and/or costs of the private firms. It tends to achieve an output level between the competitive level for the industry and the level using DW = DW1 + DW2. When solving for the industry equilibrium (for a given set of behavioral assumptions and value of the GAP) for the firms, the model starts with the Gradient Method. Then it tries to switch to Newton's Method. If Newton's Method diverges (when private firms form a cartel), the model realizes it must drop a firm to obtain each firm's equilibrium. The g.f. managers might obtain more or less than the desired total bonus if the industry is not in equilibrium, if the government does not adequately understand the demand conditions of the industry, and/or if the g.f. managers are not behaving 'optimally.' If, on average, this situation persists, the government needs to make some adjustments - which could include getting more competent g.f. managers.
Is the public firm minimizing cost?
Notation
l = (l0,l1,...,ln) L = sum(l); k = (k0,k1,...,kn) K = sum(k);
wl(L) = labour supply schedule; wk(K) = capital supply schedule
fi(li,ki) = production function of the ith firm
ddfi(li,ki)/dlidli < 0 ddfi(li,ki)/dkidki < 0
ddfi(,)/dlidli * ddfi(,)/dkidki - ddfi(,)/dlidki * ddfi(,)/dkidli > 0
µi = Lagrangian multiplier for the ith firm, µi >= 0
Income of Public Firm Managers:
I0(l0,k0,µ0) = b1*[D(Q)*q0 - wl(L)*l0 - wk(K)*k0] + b2*Q + b3*D(Q)*q0 + µ0*[f0(l0,k0) - q0]
Income of Private Firms
Ii(li,ki,µi) = D(Q)*qi - wl(L)*li - wk(K)*ki + µi*[fi(li,ki) - qi]
Income Maximization: First Order Conditions:
Public Firm:
dI0/dl0 = b1*[-wl'(L)*l0 - wl(L)] + µ0*df0(l0,k0)/dl0 = 0
dI0/dk0 = b1*[-wk'(K)*k0 - wk(K)] + µ0*df0(l0,k0)/dk0 = 0
Solve for µ0 / b1
[wl'(L)*l0 + wl(L)] / [df0(l0,k0)/dl0] = [wk'(K)*k0 + wk(K)] / [df0(l0,k0)/dk0]
Private Firms:
dIi/dli = -wl'(L)*li - wl(L) + µi*dfi(li,ki)/dli = 0
dIi/dki = -wk'(K)*ki - wk(K) + µi*dfi(li,ki)/dki = 0
Solve for µi
[wl'(L)*li + wl(L)] / [dfi(li,ki)/dli] = [wk'(K)*ki + wk(K)] / [dfi(li,ki)/dki]
Answer: Sure looks like it to me here. The input proportions, depending on the firm's production function, are the same for the public and private firms.
I'll let you work out the second order conditions yourself. | 2,589 | 7,979 | {"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-2018-34 | longest | en | 0.655288 |
https://oneclass.com/textbook-notes/ca/western/stat-sci/sta-sci-2037ab/117131-statistics-2037-chapter-5-experiments-and-observational.en.html | 1,524,683,638,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947939.52/warc/CC-MAIN-20180425174229-20180425194229-00445.warc.gz | 676,183,559 | 41,863 | Textbook Notes (368,856)
Chapter 5
# Statistics 2037 - Chapter 5 Experiments and Observational Studies.docx
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Chapter 5 – Experiments and Observational Studies Explanatory Variable – attempts to explain/cause differences in response variable Response Variable – the response/outcome Treatment – combination of categories of the explanatory variable assigned by the experimenter, incorporates a collection of conditions Randomized Experiment – create differences in the explanatory variable Observational Study – observe differences in explanatory variable and see if it is related to response variable We may use observational study over an experiment because the experiment may be unethical or explanatory variable can’t be randomly assigned (what hand do you write with) Confounding variable – one that has 2 properties. It is related to the explanatory variable (differ in explanatory variable means they will differ in the confounding variable). It also affects the explanatory variable (this effect cannot be separated). (Example: smoking affects IQ, nutrition affects IQ, smoking and nutrition are related) Interactions – between explanatory variables. The effect of one explanatory variable on a response variable depends on the state of another explanatory variable. Example: smoking affects IQ with no exercise) Experimental Units – smallest basic object to which we can assign different treatments in a randomized experiment Observational Units – the objects measured in a study Participants – used when observational units are people Volunteers – most participants in studies Passive Volunteer – those that sign a consent form and agree to participate When choosing a sample for a randomized example we need to worry about each participant having an equal amount of receiving each treatment. The decision on which treatment a participant receives should be random. Randomizati
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So we can recommend you notes for your school. | 442 | 2,321 | {"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} | 2.9375 | 3 | CC-MAIN-2018-17 | latest | en | 0.85473 |
https://www.edaboard.com/threads/queries-on-deriving-the-equation-of-iip3.337863/ | 1,709,607,756,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00096.warc.gz | 747,279,149 | 18,829 | Continue to Site
# Queries on deriving the equation of IIP3
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#### hari_preetha
##### Full Member level 2
I want to derive third order input intercept point (IIP3) for my designed low noise amplifier (LNA). What I observed in the many IEEE transaction LNA paper, they used Volterra Series Analysis for deriving the IIP3 equation for their LNA. Which is the best text book for deriving IIP3 for our LNA and where we will get the right from basic to end of derivation of IIP3 material and how to derive the IIP3 equation for LNA, which analysis we have to use and IIP3 equation.
hari_preetha
### hari_preetha
Points: 2
I want to derive third order input intercept point (IIP3) for my designed low noise amplifier (LNA). What I observed in the many IEEE transaction LNA paper, they used Volterra Series Analysis for deriving the IIP3 equation for their LNA. Which is the best text book for deriving IIP3 for our LNA and where we will get the right from basic to end of derivation of IIP3 material and how to derive the IIP3 equation for LNA, which analysis we have to use and IIP3 equation.
IIP3=OIP3-Gain...
Derivation of ANY nonlinearity needs to know the coefficients of the transfer function of the active devices in a circuit.So, it's practically impossible to calculate by hand..
Simulators are for those purposes.The nonlinearity equations have been given as mathematical models of the active devices with its coefficients in textbooks and it's just for understanding the phenomena of nonlinearity.
hari_preetha
Points: 2 | 369 | 1,571 | {"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-2024-10 | latest | en | 0.912887 |
https://zbmath.org/?q=ut%3ABredikhin+theorem+on+counting+functions+of+arithmetical+semigroups | 1,627,946,492,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154385.24/warc/CC-MAIN-20210802203434-20210802233434-00528.warc.gz | 1,153,224,321 | 11,202 | # zbMATH — the first resource for mathematics
Introduction to analytic number theory. Transl. from the Russian by G. A. Kandall. Ed. by Ben Silver. Appendix by P. D. T. A. Elliott. (English) Zbl 0641.10001
Translations of Mathematical Monographs, 68. Providence, RI: American Mathematical Society (AMS). vi, 320 p.; \$ 114.00 (1988).
[For a review of the Russian original (Nauka, Moskva 1971) see Zbl 0231.10001.]
This monograph contains several topics from the analytic theory of numbers, which generally are not treated in textbooks on analytic number theory.
The first chapter presents background from analysis: the Tauberian theorem of Hardy-Littlewood (also in G. Freud’s version with remainder term) and the Ingham Tauberian theorem for partitions are proved, and Esséen’s inequality, estimating the difference of two distribution functions in terms of their characteristic functions, is given.
The second chapter deals with additive problems with an increasing number of summands. Th. Schneider’s lemma on lattice points in parallelepipeds is proved in a probabilistic setting, applications of the local limit theorem of probability theory to number theory are sketched, and Freiman’s theorem concerning an asymptotic formula for the number of solutions of $$N=x\quad s_ 1+\dots +x\quad s_ n,\quad n\to \infty,\quad n<C\cdot N,$$ is proved. Connections between the counting functions $$\pi_ G(x)$$ and $$\nu_ G(x)$$ in arithmetical semigroups $$G$$ (due to Bredikhin) are given. Finally the Hardy-Ramanujan partition formula and Ingham’s general results on partitions are proved.
The third chapter deals with arithmetical functions; first the concepts of asymptotic and logarithmic density are discussed, the mean-value theorems of Wintner and Axer are given. Polyadically continuous functions and (B1-) almost-periodic functions are defined, and their properties are studied. A local distribution law for integer-valued additive functions is proved. Finally, E. V. Novoselov’s “Polyadic analysis and applications” (which partly was printed in journals not easily accessible) is presented.
Chapter 4 is concerned with further rather deep results on multiplicative functions. At first, a theorem of Drozdova and Freiman on a (sharp) upper bound for the values of nonnegative prime-independent multiplicative functions is given [more precise results are due to E. Heppner, Arch. Math. 24, 63–66 (1973; Zbl 0254.10038)]. Secondly, E. Wirsing’s famous mean-value theorem for multiplicative functions [Math. Ann. 143, 75–102 (1961; Zbl 0104.042)] is proved, following Wirsing’s proof. The Turán-Kubilius inequality is deduced. Next, Delange’s mean- value theorem for multiplicative functions $$f$$ with modules $$| f| \leq 1$$ is proved; Halász’s theorem is quoted. The Erdős-Wintner theorem is deduced as a corollary to Delange’s theorem. The distribution of the values of the Euler function is treated, and the Erdős-Kac theorem for strongly additive functions is proved. Finally, asymptotic expansions and asymptotic formulas for some special multiplicative functions are proved.
Following the bibliography with 157 items, an informative appendix written by P. D. T. A. Elliott is given, with additional hints and references to more recent research papers.
Postnikov’s “Introduction to analytic number theory” certainly is not an “Introduction”; it “represents an interesting personal selection of a number of topics from analytic number theory”, avoiding those topics already sufficiently covered in the literature. Therefore this monograph seems to be an important supplement to existing textbooks on analytic number theory, recommendable to researchers, teachers and graduate students, interested in analytic number theory.
##### MSC:
11-02 Research exposition (monographs, survey articles) pertaining to number theory 11Mxx Zeta and $$L$$-functions: analytic theory 11N37 Asymptotic results on arithmetic functions 11K65 Arithmetic functions in probabilistic number theory 11P05 Waring’s problem and variants 11N05 Distribution of primes 11N13 Primes in congruence classes 11N80 Generalized primes and integers 11N99 Multiplicative number theory 11P81 Elementary theory of partitions 11D72 Diophantine equations in many variables | 989 | 4,241 | {"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} | 2.78125 | 3 | CC-MAIN-2021-31 | latest | en | 0.874502 |
https://www.livescience.com/62651-how-hot-cars-get.html | 1,713,229,036,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817036.4/warc/CC-MAIN-20240416000407-20240416030407-00502.warc.gz | 796,892,585 | 162,231 | # How Long Does It Take a Parked Car to Reach Deadly Hot Temperatures?
It's well known that a car parked outside on a hot summer's day can turn into a scorching oven. But how fast does it take the inside of a car to heat up to deadly temperatures?
The answer can be a matter of life and death. Every year in the United States, an average of 37 children die after being left in hot cars, according to researchers of a new study, published online today (May 24) in the journal Temperature.
To investigate the matter, researchers studied how long it takes different types of cars to heat up on hot days. The findings were sobering: Within 1 hour, the temperature inside of a car parked in the sun on a day that reached 95 degrees Fahrenheit (35 degrees Celsius) or hotter, hit an average of 116 degrees F (47 degrees C). [Why Does Being in the Heat Make Us Feel Tired?]
The cars' dashboards got even hotter, reaching 157 degrees F (69 degrees C), on average; the steering wheels climbed to a temperature of 127 degrees F (53 degrees C), on average; and the temperature of the seats hit 123 degrees F (51 degrees C), on average.
Cars parked in the shade on a hot day had lower — but still scorching — temperatures. After 1 hour, the interior temperature of these cars reached an average of 100 degrees F (38 degrees C). The dashboards of these cars averaged 118 degrees F (48 degrees C); the steering wheel averaged 107 degrees F (42 degrees C); and the seats averaged 105 degrees F (41 degrees C), the researchers found.
"We've all gone back to our cars on hot days and have been barely able to touch the steering wheel," study co-researcher Nancy Selover, a climatologist at Arizona State University, said in a statement. "But, imagine what that would be like to a child trapped in a car seat." (More on this later.)
Selover added that anybody sitting in such a car would, of course, breathe, and that each breath would introduce humidity into the vehicle.
"They are exhaling humidity into the air," Selover said. "When there is more humidity in the air, a person can't cool down by sweating because sweat won't evaporate as quickly."
The researchers used six vehicles in the study: Two identical silver economy cars, two identical silver midsize sedans and two identical silver minivans. Then, on three different summer days in Tempe, Arizona, they monitored the parked cars in both sunny and shady locations.
"These tests replicated what might happen during a shopping trip," Selover said. "We wanted to know what the interior of each vehicle would be like after one hour, about the amount of time it would take to get groceries. I knew the temperatures would be hot, but I was surprised by the surface temperatures."
Unsurprisingly, the cars heated at different rates. The economy car warmed faster than the midsize sedan and the minivan, the researchers found.
## Children in cars
A person trapped in a rapidly heating car is at risk for heatstroke, which can be deadly.
It's difficult to predict when heatstroke will strike — largely because the condition involves many factors, including a person's age, weight and existing health conditions, the researchers said. But most cases happen when a child's core body temperature rises above 104 degrees F (40 degrees C) for an extended period of time. [What is Heat Stroke?]
To learn more about the risks children face, the researchers used data to model a hypothetical 2-year-old boy. When strapped into a car seat in a parked car on a hot day, this child would meet the criteria for heatstroke in just 1 hour if the car were parked in the sun and 2 hours if the car were parked in the shade, the researchers found.
"We hope these findings can be leveraged for the awareness and prevention of pediatric vehicular heatstroke and the creation and adoption of in-vehicle technology to alert parents of forgotten children," the study's lead researcher, Jennifer Vanos, an assistant professor of climate and human health at the University of California, San Diego, said in the statement.
Vanos added that effects from hyperthermia (having a higher-than-normal body temperature) and heatstroke happen along a continuum, from internal injuries to brain and organ damage.
Original article on Live Science. | 920 | 4,265 | {"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-2024-18 | longest | en | 0.940525 |
https://www.vedantu.com/question-answer/a-shopkeeper-professes-to-sell-his-goods-at-cost-class-11-maths-cbse-5fd85dd48238151f0da05f1d | 1,719,266,869,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00100.warc.gz | 915,905,985 | 27,433 | Courses
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# A shopkeeper professes to sell his goods at cost price but uses a weight of $800$ grams instead of a kilogram weight. Thus, he makes a gain of:A. $10\%$B. $15\%$C. $20\%$D. $25\%$
Last updated date: 20th Jun 2024
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Hint: Here we will be using the formula of gain percentage i.e. ${\text{Gain}}\% = \dfrac{{{\text{Gain}}}}{{{\text{Original weight}}}} \times 100$ , where the gain is the difference between the original weight and the new weight.
Step 1: As we know that one kilogram equals
$1000$ grams. The actual weight used by the shopkeeper is
$800$ grams so we will be calculating the gain by subtracting the original weight from the new/actual weight as shown below:
${\text{Gain = 1000 - 800}}$
By doing the subtraction in the RHS side of the above expression we get:
$\Rightarrow {\text{Gain = 200}}$
Step 2: We will be using the formula ${\text{Gain}}\% = \dfrac{{{\text{Gain}}}}{{{\text{Original weight}}}} \times 100$ to calculate the gain percentage by substituting the values of
${\text{Gain = 200}}$ and
${\text{Original weight = 800}}$ as calculated below:
$\Rightarrow {\text{Gain}}\% = \dfrac{{200}}{{800}} \times 100$
By solving the division in the RHS side of the above expression we get:
$\Rightarrow {\text{Gain}}\% = \dfrac{{100}}{4}$
By doing the final division in the above expression we get:
$\Rightarrow {\text{Gain}}\% = 25\%$
Option D is correct.
Note:
Students need to take care while solving these types of questions that the amount of discount which is given on the marked price equals to as below:
$\dfrac{{{\text{Discount }}\% }}{{100}} \times {\text{M}}{\text{.P}}$, and we will be subtracting this value from the marked price for calculating the value of the selling price.
Also, you should remember which one is the original weight and which one is new for calculating the value of gain. | 553 | 1,961 | {"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} | 4.71875 | 5 | CC-MAIN-2024-26 | latest | en | 0.838698 |
http://www.evi.com/q/60_milligrams_equals_how_many_ounces | 1,419,142,371,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770742.121/warc/CC-MAIN-20141217075250-00063-ip-10-231-17-201.ec2.internal.warc.gz | 493,211,121 | 5,315 | # 60 milligrams equals how many ounces
• 60 milligrams is 0.0021164 ounces.
• tk10publ tk10canl
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• 60 mgs equals how many ozs | 715 | 2,313 | {"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} | 2.953125 | 3 | CC-MAIN-2014-52 | latest | en | 0.903091 |
http://file.scirp.org/Html/6-5300523_41127.htm | 1,534,724,553,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215404.70/warc/CC-MAIN-20180819224020-20180820004020-00541.warc.gz | 148,853,858 | 16,274 | Value Distribution of L-Functions with Rational Moving Targets
Advances in Pure Mathematics
Vol.3 No.9(2013), Article ID:41127,5 pages DOI:10.4236/apm.2013.39098
Value Distribution of L-Functions with Rational Moving Targets
Matthew Cardwell1, Zhuan Ye2*
1Intelligent Medical Objects, Inc., Northbrook, USA
2Department of Mathematical Sciences, Northern Illinois University, DeKalb, USA
Email: mcardwell@e-imo.com, *ye@math.niu.edu
Copyright © 2013 Matthew Cardwell, Zhuan Ye. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In accordance of the Creative Commons Attribution License all Copyrights © 2013 are reserved for SCIRP and the owner of the intellectual property Matthew Cardwell, Zhuan Ye. All Copyright © 2013 are guarded by law and by SCIRP as a guardian.
Received August 26, 2013; revised September 26, 2013; accepted October 1, 2013
Keywords: Value Distribution; Moving Target; L-Function; Selberg Class
ABSTRACT
We prove some value-distribution results for a class of L-functions with rational moving targets. The class contains Selberg class, as well as the Riemann-zeta function.
1. Introduction
We define the class to be the collection of functions
satisfying Ramanujan hypothesisAnalytic continuation and Functional equation. We also denote the degree of a function by which is a non-negative real number. We refer the reader to Chapter six of [1] for a complete definitions. Obviously, the class contains the Selberg class. Also every function in the class is an -function and the Riemann-zeta function is in the class. In this paper, we prove a value-distribution theorem for the class with rational moving targets. The theorem generalizes the value-distribution results in Chapter seven of [1] from fixed targets to moving targets.
Theorem. Assume that and is a rational function with. Let the roots of the equation be denoted by. Then
(I) For any,
(II) For sufficiently large negative,
Proof of (I). It is known that if, then
where is the index of the first non-zero term of the sequence of, with. Since, there exists such that for. It follows that for all real part of zeros of the function. We set where the degrees of are, respectively; and define
Thus, there is such that is analytic in the region since is a meromorphic function in with the only pole at. We apply Littlewood’s argument principle [3] to in the rectangle where are parameters satisfying. Thus,
where the given logarithm is defined as in Littlewood’s argument principle [3]. To prove our result, however, we first decompose our auxiliary function by
(1)
Without loss of generality, we may assume that whenever since we can always write for due to our choice of the parameters which define the rectangle. However, the modification will guarantee in the case of that exhibit polynomial growth, which is necessary for our proof. In the case of, already exhibits polynomial growth, and no such adjustment is necessary. We now integrate the logarithm of to get
where the terms are the integrals of the maximum contribution from writing as a sum of logarithms. By our choice of, both and are analytic in Hence, Cauchy’s Theorem gives
(2)
To connect this integral with Littlewood’s argument principle [3], we note that the definition of guarantees that
(3)
In light of (2) and because the quantity given in (3) is imaginary-valued, we get for
(4)
for instance.
We now estimate. For large enough, we have for (since),
Then for large enough, , we find in a similar fashion that
Since we have the same estimate for, we find that
where the final bound follows from Jensen’s inequality. It is known [2] that for,
Hence, uniformly in .
We next move to estimate. For sufficiently large positive real number, we have
(5)
so
since. Furthermore,
Since we may take large enough so that
, we may write using a Taylor series expansion in the rectangle. For, we have after taking real parts that
We now observe that for sufficiently large T and some constant M we have
for and
for sufficiently large. In light of these bounds and the definition of, we have (6)
where the last equality holds because could be sufficiently large. Replacing by in the above computations, we see analogously that.
Finally, we estimate and. We show the computation for explicitly and note that the bound for follows analogously. We first suppose that has exactly zeros for. Then, there are at most subintervals, counting for multiplicities, in which is of constant sign. Thus,
(7)
It remains to estimate. To this end, we define
Then
so that if for, then.
Now let and, and choose large enough so that. Then for, showing that no zeros or poles of are located in. Thus, both and are analytic in. Letting denote the number of zeros of in, we have
By Jensen’s formula
and so
(8)
(6)
By (5), is bounded. Further, it is clear from a property of functions that we have
for some positive absolute numbers in any vertical strip of bounded width. The same estimate must hold for as well. Thus, the integral in (8) is, implying that . Since the interval, it follows that
With this bound, we integrate (7) to deduce that
As previously noted, we may bound in the same way. Thus, we attain the desired bounds for and. Consequently, the first part of the theorem is proved by using (4).
Proof of (II). As in the proof of the first part of the theorem, we conclude that there exists a real number for which the real parts of all -values satisfy; and also, there exist for each rational function such that no zeros of lie in the quarter-plane. As before, we define the rectangle where are parameters satisfying .
Proceeding as in the proof of the first part of the theorem, we see that
for where is defined as in (1). In the equation above, we note that we have chosen to compute separately. Indeed, this is the only estimate that we will need. For the integrals, and, the bounds given as in the proof of the first part of the theorem still hold. First, integral is unchanged. On the other hand, the integrals have changed by our choice of, but, as we have done as before, we still have the desired bound since the only requirement is that we consider in a vertical strip of fixed width, which we have in this case.
We now bound. Since, we have by the functional equation in the definition of function,
Taking logarithms, we get
(9)
Since, for, we have, uniformly in,
where are two constants. It follows, for as, that
We now consider the last term in (9). Since,
and noting, we have for any and
for sufficiently large. Then we see the quotient
when is large enough so that
Therefore, we find that
Integrating in light of these estimates, we see
The first integral is, and the second integral is for sufficiently large and negative by the method used to derive (6). Hence,
With the estimates for the’s, we have proved the second part of the theorem.
REFERENCES
1. J. Steuding, “Value Distribution of L-Functions,” Number 1877 in Lecture Notes in Mathematics, Springer, 2007.
2. H. S. A. Potter, “The Mean Values of Certain Dirichlet Series I,” Proceedings London Mathematical Society, Vol. 46, No. 2, 1940, pp. 467-468. http://dx.doi.org/10.1112/plms/s2-46.1.467
3. E. C. Titchmarsh, “The Theory of Functions,” 2nd Edition, Oxford, 1939.
NOTES
*Corresponding author. | 1,736 | 7,444 | {"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} | 3.046875 | 3 | CC-MAIN-2018-34 | latest | en | 0.88278 |
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Helper III
## Date math problem: if (ExpectedDate<=date, QTY, 0)
I have a standard Date table, Sales table, and Incoming table.
The Sales Table and Incoming Table both have QTY, and a Sales_Date for sales, and an Expected_Date for Incoming.
I plot daily sales on a line chart, with a measure that simply sums Sales_QTY (Sales_QTY_SUM). I don't have to do anything else, if Date is on the axis, I get what I want, for sales. I also have a measure that sums Incoming_QTY (Incoming_QTY_SUM), and it works the same way, but that is not exactly what I want.
I want to see Incoming_QTY as a cumulative total going forward from Expected_Date. So if an incoming record has a Date of March 1st, I want to see 0 before March 1st, and Incoming_QTY in the line chart results starting on March 1st, then on March 2nd, and March 3rd, ..
Date[Date] and Incoming[Expected_Date) are both columns, and trying to use them in an IF statement fails because I am not using any aggregates.
IF([Expected_Date]<=[Date],Incoming_QTY,0) is the logic.
How do end up with a measure that will plot this properly on a line chart?
Thanx
Phil
1 ACCEPTED SOLUTION
Accepted Solutions
Highlighted
Microsoft
## Re: Date math problem: if (ExpectedDate<=date, QTY, 0)
Try this slight modification...
```Cumulative Incoming = CALCULATE(
SUM(Incoming[Incoming Qty]) ,
FILTER(
ALLSELECTED('Incoming'),
'Incoming'[Incoming Date]<=MAX('Dates'[Date])
)
)```
Proud to be a Datanaut!
4 REPLIES 4
Highlighted
Microsoft
## Re: Date math problem: if (ExpectedDate<=date, QTY, 0)
Sounds like you are just after a cumulative measure similar to this :
```Cumulative Incoming = CALCULATE(
SUM(Incoming[Incoming Qty]) ,
FILTER(
ALL('Incoming'),
'Incoming'[Incoming Date]<=MAX('Dates'[Date])
)
)```
I have a PBIX file you can try it with here https://1drv.ms/u/s!AtDlC2rep7a-kHTghFw8Upt_GDbN
Proud to be a Datanaut!
Highlighted
Helper III
## Re: Date math problem: if (ExpectedDate<=date, QTY, 0)
Awesome! So I have this:
IncomingQTYCumulative = CALCULATE(SUM(IncomingInventories[order_qty]),FILTER(ALL(IncomingInventories),'IncomingInventories'[expected_date]<=MAX('Date'[Date])))
And while it works perfectly, these inventory numbers are by Item_ID, relating to 'item[Item_id] and this measure is immune to slicers tied to the Item table. I know why, because of the ALL(IncomingInventories), but I need to be able to slice by 'Item[Stat_ID].
I can hardcode it into the measure, but that is not a good solution. How can I allow the slicer to carry through?
Thanx
Highlighted
Microsoft
## Re: Date math problem: if (ExpectedDate<=date, QTY, 0)
Try this slight modification...
```Cumulative Incoming = CALCULATE(
SUM(Incoming[Incoming Qty]) ,
FILTER(
ALLSELECTED('Incoming'),
'Incoming'[Incoming Date]<=MAX('Dates'[Date])
)
)```
Proud to be a Datanaut!
Highlighted
Helper III
## Re: Date math problem: if (ExpectedDate<=date, QTY, 0)
That seems to work perfectly. Thank you very much!
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Top Kudoed Authors | 906 | 3,469 | {"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} | 3.171875 | 3 | CC-MAIN-2020-45 | latest | en | 0.8568 |
https://topic.alibabacloud.com/a/summary-of-asp-access-date-operation-statements-by-stabx_1_36_32444302.html | 1,721,426,107,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00273.warc.gz | 518,154,238 | 16,027 | # Summary of asp access date operation statements by stabx
Source: Internet
Author: User
CopyCode The Code is as follows: 13.5.1 list the data of a day and a week
Select distinct dateandtime
From ctdate_by_query_date
Where dateandtime between (#2006-5-15 # + (weekday (2006-5-15)-7)-6) and (#2006-5-15 #-7) + weekday (2006-5-15)
SQL = "select distinct dateandtime from ctarticle where dateandtime between (#" & Date & "# + (weekday (" & Date & ")" & norp & "7 )) -6) and (# "& Date &" # "& norp &" 7) + weekday ("& Date &")"
13.5 query the data of all days of a day in the week
Select *
From ctdate_by_query_date
Where dateandtime between (#2006-5-15 # + weekday (2006-5-15)-6) and #2006-5-15 # + weekday (2006-5-15)
13.4 query a time period
Select *
From ctdate_by_query_date
Where dateandtime between #2006-5-1 # And #2006-5-30 #
13.3.2 list the years of different years, and do not use the same
Select distinct year (dateandtime) from ctarticle
The result is as follows:
Expr1000
2000
2003
2004
2005
2006
13.3.1 list the first record of the previous year on a certain day
Select top 1 dateandtime from ctarticle where year (dateandtime) = (2006)-1
SQL = "select top 1 dateandtime from ctarticle where year (dateandtime) = (Year (#" & Date & "#)" & norp
13.3 list data of a year
SQL = "select * From ctdate_by_query_date where year (dateandtime) =" & year (rqqdt _) & "order by dateandtime DESC"
13.2.1 search for the first record in January
Select top 1 dateandtime from ctarticle where year (dateandtime) = year (#2006-5-28 #) and month (dateandtime) = month (#2006-5-28 #)-1
13.2 list data of a month
SQL = "select * From ctdate_by_query_date where year (dateandtime) =" & year (rqqdt _) & "and month (dateandtime) =" & month (rqqdt _) & "order by dateandtime DESC"
13.1 list data of a day
SQL = "select * From ctdate_by_query_date where dateandtime = #" & rqqdt _ & "# order by A. ArticleID DESC"
13. Date and Time
Example 1: List data of the current day
SQL = "select * From ctdate_by_query_date where dateandtime = Date () order by A. ArticleID DESC"
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 851 | 3,014 | {"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-2024-30 | latest | en | 0.7151 |
http://www.jiskha.com/display.cgi?id=1343312321 | 1,495,703,792,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608058.13/warc/CC-MAIN-20170525083009-20170525103009-00120.warc.gz | 546,808,570 | 3,576 | # Algebra*
posted by on .
Short Answer. Find the following quotient:
x^2-36/x+6
• Algebra* - ,
x^2-36/x+6
(x^2=36)/(x+6)
(x+6)(x-6)/(x+6)
canceling (x+6) in both numerator and denominator we have x-6 | 78 | 203 | {"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} | 3.15625 | 3 | CC-MAIN-2017-22 | latest | en | 0.829497 |
https://www.unitconverters.net/volume/stere-to-petaliter.htm | 1,675,434,753,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500056.55/warc/CC-MAIN-20230203122526-20230203152526-00124.warc.gz | 1,052,965,711 | 4,035 | Home / Volume Conversion / Convert Stere to Petaliter
# Convert Stere to Petaliter
Please provide values below to convert stere [st] to petaliter [PL], or vice versa.
From: stere To: petaliter
### Stere to Petaliter Conversion Table
Stere [st]Petaliter [PL]
0.01 st1.0E-14 PL
0.1 st1.0E-13 PL
1 st1.0E-12 PL
2 st2.0E-12 PL
3 st3.0E-12 PL
5 st5.0E-12 PL
10 st1.0E-11 PL
20 st2.0E-11 PL
50 st5.0E-11 PL
100 st1.0E-10 PL
1000 st1.0E-9 PL
### How to Convert Stere to Petaliter
1 st = 1.0E-12 PL
1 PL = 1000000000000 st
Example: convert 15 st to PL:
15 st = 15 × 1.0E-12 PL = 1.5E-11 PL | 255 | 590 | {"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-2023-06 | longest | en | 0.297961 |
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## Detail Page
written by Walter Fendt
This applet simulates an experiment to study Newton's Second Law. A mass sliding on a horizontal surface is connected by a string to a hanging mass. The value of the two masses and the coefficient of friction can be set by the user. The user can use a timer to gather data for the motion of the sliding mass as a function of time.
This applet is a part of a large collection of physics applets, available in a wide range of languages.
Please note that this resource requires at least version 1.4.2 of Java.
Subjects Levels Resource Types
Classical Mechanics
- Applications of Newton's Laws
- Newton's Second Law
- High School
- Instructional Material
= Interactive Simulation
- Audio/Visual
= Movie/Animation
Appropriate Courses Categories Ratings
- Physical Science
- Physics First
- Conceptual Physics
- Algebra-based Physics
- AP Physics
- Lesson Plan
- Activity
- Laboratory
- New teachers
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Intended Users:
Learner
Educator
Format:
application/java
Access Rights:
Free access
Restriction:
© 1997 Walter Fendt
Additional information is available.
Merlot:
Merlot Record
Keywords:
Newton's second law, forces, friction, horizontal mass, simulation, static equilibrium
Record Cloner:
Metadata instance created July 19, 2006 by swapna gurumani
Record Updated:
August 18, 2020 by Lyle Barbato
Last Update
when Cataloged:
January 18, 2003
Other Collections:
### NSES Content Standards
Con.B: Physical Science
• 5-8: Motion & Forces
This resource is part of a Physics Front Topical Unit.
Topic: Dynamics: Forces and Motion
Unit Title: Newton's Second Law & Net Force
This model simulates an air track glider, a low-friction device commonly used to conduct experiments on Newton's Second Law and collisions. It features a two-mass system connected by a string. Change the value of either mass or the coefficient of friction on the track.....and watch the effects on the motion. Available in HTML5 or Java.
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AIP Format
W. Fendt, (1997), WWW Document, (https://www.walter-fendt.de/html5/phen/newtonlaw2_en.htm).
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W. Fendt, Walter Fendt Physics Applets: Newton's Second Law Experiment, (1997), <https://www.walter-fendt.de/html5/phen/newtonlaw2_en.htm>.
APA Format
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%A Walter Fendt
%T Walter Fendt Physics Applets: Newton's Second Law Experiment
%D January 18, 2003
%U https://www.walter-fendt.de/html5/phen/newtonlaw2_en.htm
%O application/java
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%D January 18, 2003
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### Messages - lawness
Pages: 1 ... 20 21 22 23 24 [25] 26
241
##### Studying for the LSAT / Re: "I'm hot for teacher"...
« on: August 30, 2007, 09:31:52 PM »
What was so great about his teaching methods. What does he do well?
242
##### Studying for the LSAT / "I'm hot for teacher"...
« on: August 30, 2007, 09:07:34 PM »
What am I looking for when I am assessing the quality of a prep course teacher. Has anyone had the opportunity to interview a teacher before taking the class? Is a high sore enough? Thanks!
243
##### Studying for the LSAT / Re: When to diagram on LR section?
« on: August 24, 2007, 11:58:22 AM »
Great. Does anyone else have any diagramming suggestions?
244
##### Studying for the LSAT / Re: When to diagram on LR section?
« on: August 24, 2007, 11:26:32 AM »
Thanks for the tip. I often find it easier to work it our in my head than to diagram. This may be due to my lack of diagramming skills, however. Do you find it easier to diagram? Do you feel that it take up more time than to just figure it out intuitively?
245
##### Studying for the LSAT / When to diagram on LR section?
« on: August 24, 2007, 11:13:38 AM »
I have not mastered this suggested technique. Is there a specific question type on the LR section that MUST require diagramming? If so, how can I recognize it in the stimulus and or question stem?
246
##### Studying for the LSAT / Re: How can a sufficient and necessary condition occur at the same time??
« on: August 18, 2007, 10:37:19 AM »
I'm still not sure if Lawness is asking if a necessary condition and a sufficient condition can happen simultaneously (meaning that if these things were really happening, could they happen at the same time or does one have to come first), or if (s)he's asking if a necessary condition can also be a sufficient condition and vice versa--but here's something that might help regardless. Logically speaking, 'necessary' and 'sufficient' are pretty much technical terms that refer to the logical relationship between two sentences/ideas, rather than to the actual relationship of whatever is being discussed. In other words, here's a sentence:
If I have gas in my tank, then my car will run.
Even though in real life, it would be more accurate to say that gas is necessary for my car to run, in that sentence 'gas in my tank' is the sufficient part and 'my car will run' is the necessary part. This is because 'necessary' just means "this is necessarily true if the other part is true"--which is what the conditional means. And 'sufficient' just means "the truth of this statement is enough to guarantee the truth of this other statement". The two statements that make up a conditional don't have to have any sort of realistic relationship at all--I can slap any two statements or ideas together and form a conditional, and one part will still be the sufficient part and the other the necessary part--so it might be easier not to think about how the statements are related to eachother in a real-life sense--because all any conditional really means is "if this is true, then that is true" -- why or how they are related to each other doesn't matter at all.
I was really asking if they can occur in real life terms simultaneously because I was trying to make sense of the two with a real world example. It helps to think of the two instances/events in logical terms and without subscribing meaning to them. For diagramming purposes, the sufficient condition ALWAYS comes first, right? At least that is what LRB says.
247
##### Studying for the LSAT / Re: How can a sufficient and necessary condition occur at the same time??
« on: August 17, 2007, 11:50:34 PM »
No, nothing in particular. I was just wondering. There seems to be a statistic for everything else! I feel confident in recognizing the relationship between the two conditions. The LSAT testmakers are real tricky in describing the conditional relationships. I have been learning the LRB strategies. Are there any other strategies out there that are helpful and easy to understand?
248
##### Studying for the LSAT / Re: How can a sufficient and necessary condition occur at the same time??
« on: August 17, 2007, 11:36:36 PM »
Are there any statistics that speak to how often sufficient and necessary conditions occur simultaneously on the LSAT?
249
##### Studying for the LSAT / Re: How can a sufficient and necessary condition occur at the same time??
« on: August 17, 2007, 11:21:41 PM »
Sorry for the lack of content. I just did not make the connection that both conditions can occur simultaneously. I thought one had to occur for the other to occur but I did understand that either can come first. So, basically a necessary condition can also be a sufficient condition and visa versa, right? In order for them to happen at the same time, I need to conceptualize the event co-occurring in my head. Is this correct?
250
##### Studying for the LSAT / Re: How can a sufficient and necessary condition occur at the same time??
« on: August 17, 2007, 11:13:23 PM »
ahhhh I think I get it!! Any more would be very beneficial!!
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# Hop bets at craps
16 March 2019
Stickman,
I’ve seen many articles discussing house advantages on craps and how bad hop bets are. I don’t think I’ve ever seen what the house advantage is on hopping sevens, however. I am probably wrong but it seems like it would be low. If you bet all three combos for \$1 each and hit one, it pays \$15. That’s 5-1, and isn’t that the chances of rolling a seven? Could you please explain this to me? Thank you.
James
Hi James,
Many craps players are unaware of the math involved in the game of craps. Casinos thrive because of it. Before specifically answering your question, James, let’s look at the game in general.
The game of craps is quite different from other table games because of the variety of bets that are available. There are the boring pass and come bets along with their inverse don’t pass and don’t come bets. They are boring because it can take several rolls before a decision is reached, and when it is the payoff is relatively small. Depending on the size of the odds bet (paid at true odds) associated with the base portion of the bet, the house edge runs from well less than one percent to about 1.4% if there is no odds bet.
There are place and lay bets that can be changed at the bettor’s whim. These have higher house edges (from 1.52% to 6.67%) for the privilege. There is a “field” bet. This is a one roll bet that wins if a 2, 3, 4, 9, 10, 11 or 12 is thrown, but loses when any other number is thrown. Its house edge is 5.56%.
Then there is a large variety of bets that are held in the center of the table by the stickman. Most of these are one roll bets such as any craps (two, three or 12), yo (11), any seven, and hop bets (a one-roll bet that a specific number will appear on the next roll. Also, in this center area are hardway bets which pay if a “hard” point number consisting of two dice with the same number (2-2, 3-3, 4-4 or 5-5) is thrown before a seven or a “soft” combination (different numbers on the two dice totaling a four, six, eight or 10) is rolled. All of these bets carry a hefty house edge – from 9.09% to 16.67%.
Some craps tables have bets with names such as fire bet, all tall, all small and so on. These bets have the potential of a large payoff, but also carry house edges greater than 20%. Think about it. Over time, a player betting a 20% house edge bet will lose \$20 for every \$100 played.
Now, let’s look at the hopping sevens bet(s). You are right, James, when you say that the odds against rolling a seven are 5 to 1. On average, the seven will appear once every six rolls. Yes, you are paid 15 to 1 for your one winning hop bet of \$1, but you lose the other \$2. If you were paid 5 to 1 on your \$3, you would win \$15 and you could also remove your initial \$3 bet making the total \$18 – 18 for 3 or true odds. Instead you only have \$16.
To calculate the house edge, take the \$2 loss and divide it by the true odds win amount of 18 dollars: 2 / 18 = 0.1111 which is 11.11%. The house edge on all “two-way” hop bets (where there are two ways the dice can be thrown to win – example: 5-2 and 2-5) is 11.11%. This is not as bad as the 16.67% edge on the any seven bet, but you will still lose \$11.11 for every \$100 bet on hopping sevens long term.
Casinos are very good at making things appear better than they really are. How else could they make enough money to prosper?
Learn the math of the game and follow it in your playing and betting decisions. To have the best chance of being a winner at a session, bet only the lowest house edge bets – pass/come with odds or don’t pass/don’t come with odds. While this may be a boring method, your money will last longer betting this way.
May all your wins be swift and large and all your losses slow and small.
Jerry “Stickman”
Jerry “Stickman” is an expert in craps, blackjack and video poker and advantage slot machine play. He is a regular contributor to top gaming magazines. He authored the video poker section of "Everything Casino Poker: Get the Edge at Video Poker, Texas Hold'em, Omaha Hi-Lo, and Pai Gow Poker!" You can contact Jerry “Stickman” at stickmanjerry@aol.com.
Recent Articles
Best of Jerry Stickman
Jerry Stickman
Jerry “Stickman” is an expert in craps, blackjack and video poker and advantage slot machine play. He is a regular contributor to top gaming magazines. He authored the video poker section of Everything Casino Poker: Get the Edge at Video Poker, Texas Hold'em, Omaha Hi-Lo, and Pai Gow Poker! You can contact Jerry "Stickman" at stickmanjerryg@gmail.com.
#### Jerry Stickman Websites:
www.goldentouchcraps.com
www.goldentouchblackjack.com
Jerry Stickman
Jerry “Stickman” is an expert in craps, blackjack and video poker and advantage slot machine play. He is a regular contributor to top gaming magazines. He authored the video poker section of Everything Casino Poker: Get the Edge at Video Poker, Texas Hold'em, Omaha Hi-Lo, and Pai Gow Poker! You can contact Jerry "Stickman" at stickmanjerryg@gmail.com.
#### Jerry Stickman Websites:
www.goldentouchcraps.com
www.goldentouchblackjack.com | 1,299 | 5,160 | {"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} | 3.21875 | 3 | CC-MAIN-2024-10 | latest | en | 0.957193 |
https://www.airmilescalculator.com/distance/dme-to-pad/ | 1,603,593,083,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107885126.36/warc/CC-MAIN-20201025012538-20201025042538-00706.warc.gz | 601,021,191 | 54,045 | Distance between Moscow (DME) and Paderborn (PAD)
Flight distance from Moscow to Paderborn (Moscow Domodedovo Airport – Paderborn Lippstadt Airport) is 1226 miles / 1973 kilometers / 1065 nautical miles. Estimated flight time is 2 hours 49 minutes.
Driving distance from Moscow (DME) to Paderborn (PAD) is 1410 miles / 2269 kilometers and travel time by car is about 24 hours 12 minutes.
Map of flight path and driving directions from Moscow to Paderborn.
Shortest flight path between Moscow Domodedovo Airport (DME) and Paderborn Lippstadt Airport (PAD).
How far is Paderborn from Moscow?
There are several ways to calculate distances between Moscow and Paderborn. Here are two common methods:
Vincenty's formula (applied above)
• 1225.833 miles
• 1972.787 kilometers
• 1065.220 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1221.947 miles
• 1966.533 kilometers
• 1061.843 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
Airport information
A Moscow Domodedovo Airport
City: Moscow
Country: Russia
IATA Code: DME
ICAO Code: UUDD
Coordinates: 55°24′31″N, 37°54′22″E
B Paderborn Lippstadt Airport
City: Paderborn
Country: Germany
IATA Code: PAD
ICAO Code: EDLP
Coordinates: 51°36′50″N, 8°36′58″E
Time difference and current local times
The time difference between Moscow and Paderborn is 2 hours. Paderborn is 2 hours behind Moscow.
MSK
CET
Carbon dioxide emissions
Estimated CO2 emissions per passenger is 162 kg (358 pounds).
Frequent Flyer Miles Calculator
Moscow (DME) → Paderborn (PAD).
Distance:
1226
Elite level bonus:
0
Booking class bonus:
0
In total
Total frequent flyer miles:
1226
Round trip? | 507 | 1,901 | {"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-2020-45 | latest | en | 0.835845 |
http://www.varsitytutors.com/gmat_math-help/calculating-arithmetic-mean | 1,477,210,686,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719192.24/warc/CC-MAIN-20161020183839-00374-ip-10-171-6-4.ec2.internal.warc.gz | 784,678,692 | 33,448 | GMAT Math : Calculating arithmetic mean
Example Questions
← Previous 1 3 4 5 6
Example Question #16 : Descriptive Statistics
Salaries for employees at ABC Company: 1 employee makes $25,000 per year, 4 employees make$40,000 per year, 2 employees make $50,000 per year and 5 employees make$75,000 per year.
What is the average (arithmetic mean) salary for the employees at ABC Company?
Explanation:
The average is found by calculating the total payroll and then dividing by the total number of employees.
Example Question #17 : Descriptive Statistics
A bowler had an average (arithmetic mean) score of 215 on the first 5 games she bowled. What must she bowl on the 6th game to average 220 overall?
Explanation:
For the first 5 games the bowler has averaged 215. The equation to calculate the answer is
where is the score for the sixth game. Next, to solve for the score for the 6th game multiply both sides by 6:
which simplifies to:
After subtracting 1,075 from each side we reach the answer:
Example Question #18 : Descriptive Statistics
Ashley averaged a score of 87 on her first 5 tests. She scored a 93 on her 6th test. What is her average test score, assuming all 6 tests are weighted equally?
Explanation:
We can't just average 87 and 93! This will give the wrong answer! The average formula is .
For the first 5 tests, . Then .
Now combine that with the 6th test to find the overall average.
Example Question #19 : Descriptive Statistics
Sabrina made $3,000 a month for three months,$4,000 the next month, and \$5,200 a month for the following two months. What was her average monthly income for the 6 month period?
Explanation:
Example Question #20 : Descriptive Statistics
Luke counts the number of gummy bears he eats every day for 1 week: {39, 18, 24, 51, 40, 15, 23}. On average, how many gummy bears does Luke eat each day?
Explanation:
Example Question #21 : Descriptive Statistics
The average of 10, 25, and 70 is 10 more than the average of 15, 30, and x. What is the missing number?
Explanation:
The average of 10, 25, and 70 is 35:
So the average of 15, 30, and the unknown number is 25 or, 10 less than the average of 10, 25, and 70 (= 35)
so
Example Question #7 : Arithmetic Mean
What is the average of 2x, 3x + 2, and 7x +4?
4
3x + 4
4x + 2
not enough information
7x
4x + 2
Explanation:
Example Question #22 : Descriptive Statistics
The average high temperature for the week is 85. The first six days of the week have high temperatures of 89, 76, 92, 90, 80, and 84, respectively. What is the high temperature on the seventh day of the week?
Explanation:
Example Question #23 : Descriptive Statistics
Jimmy's grade in his finance class is based on six equally-weighted tests. If Jimmy scored 98, 64, 82, 90, 70, and 88 on the six tests, what was his grade in the class?
Explanation:
Example Question #10 : Arithmetic Mean
Sandra's grade in economics depends on seven tests - five hourly tests, a midterm, and a final exam. The midterm counts twice as much as an hourly test; the final, three times as much.
Sandra's grades on the five hourly tests are 84, 86, 76, 89, and 93; her grade on the midterm was 72. What score out of 100 must she achieve on the final exam so that her average score at the end of the term is at least 80?
She cannot achieve this average this term
Explanation:
This is a weighted mean, with the hourly tests assigned a weight of 1, the midterm assigned a weight of 2, and the final assigned a weight of 3. The total of the weights will be
.
If we let be Sandra's final exam score, Sandra's final weighted average will be
For Sandra to get a final average of 80, then we set the above equal to 80 and calculate :
← Previous 1 3 4 5 6 | 994 | 3,741 | {"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} | 4.3125 | 4 | CC-MAIN-2016-44 | longest | en | 0.93585 |
https://socratic.org/questions/how-is-carbon-14-used-in-radiocarbon-dating | 1,529,865,352,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867050.73/warc/CC-MAIN-20180624180240-20180624200240-00165.warc.gz | 712,112,414 | 87,161 | # How is carbon-14 used in radiocarbon dating?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
1
Dave Share
Mar 9, 2018
See Below
#### Explanation:
This is a very "cartoonish" explanation....because that is usually the way my brain works. Wikipedia will probably have some more details, but here's the idea.
When an organism is made (tree, mammoth, you and me, etc), an isotope of carbon, C-14, gets incorporated into the matter at about 1% compared to the other carbon, C-12. So in a mass of 100g of dry carbon, you'd have about 1g of Carbon-14.
Carbon-14 has a half-life of 5730 years, and it undergoes beta decay in its decay process (things are radioactive because of an unstable ratio of neutrons to protons...and Beta decay increases proton number by 1 and decreases neutron number by 1).
So in that 100g sample, if you held a Geiger counter up to it (thing that goes beep, beep in the presence of certain radioactive decay events), the sample would beep, beep, beep, beep.
If you came back 5730 years later, the same sample would beep.....beep (it would lose half of its beeps....or said another way, half of its radioactive C-14 are gone).
So you take a sample of carbon out of the ground and you weigh it. You find that it is 100g. It has been there a LONG time, and when you test if for radioactive C-14, you listen........beep.................................beep. This means that there is hardly any C-14 in it. You can use an equation to figure out how many half lives have passed (how much your beep, beep has diminished....because it should be 1% of sample), and then you multiply # of half lives by 5730 years and you've got a good idea how old that carbon sample is.
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• 16 minutes ago | 530 | 2,059 | {"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} | 3.171875 | 3 | CC-MAIN-2018-26 | longest | en | 0.937614 |
https://scribesoftimbuktu.com/solve-for-f-c5-9f-32-f9-5c32/ | 1,675,827,680,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500671.13/warc/CC-MAIN-20230208024856-20230208054856-00813.warc.gz | 501,983,440 | 14,112 | # Solve for F C=5/9(F-32) F=9/5C+32
I am unable to solve this problem.
Solve for F C=5/9(F-32) F=9/5C+32
### Solving MATH problems
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Scroll to top | 83 | 221 | {"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-2023-06 | latest | en | 0.691218 |
http://functions.wolfram.com/ElementaryFunctions/Coth/21/01/02/14/01/01/0003/ | 1,386,660,695,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164012753/warc/CC-MAIN-20131204133332-00012-ip-10-33-133-15.ec2.internal.warc.gz | 80,779,705 | 10,365 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
Coth
http://functions.wolfram.com/01.22.21.0190.01
Input Form
Integrate[z^n E^(b z) Sinh[b z] Coth[c z], z] == (1/2) (z^(1 + n)/(1 + n) + 2 E^(2 c z) n! Sum[(1/(-j + n)!) (-1)^j 2^(-1 - j) c^(-1 - j) z^(-j + n) HypergeometricPFQ[{Subscript[a, 1], \[Ellipsis], Subscript[a, j + 2]}, {1 + Subscript[a, 1], \[Ellipsis], 1 + Subscript[a, j + 1]}, E^(2 c z)], {j, 0, n}] - n! (E^(2 b z) Sum[(1/(-j + n)!) (-1)^j (2 b)^(-1 - j) z^(-j + n) HypergeometricPFQ[{Subscript[b, 1], \[Ellipsis], Subscript[b, 1 + j], 1}, {1 + Subscript[b, 1], \[Ellipsis], 1 + Subscript[b, 1 + j]}, E^(2 c z)], {j, 0, n}] + E^(2 (b + c) z) Sum[(1/(-j + n)!) (-1)^j (2 b + 2 c)^(-1 - j) z^(-j + n) HypergeometricPFQ[{Subscript[c, 1], \[Ellipsis], Subscript[c, 1 + j], 1}, {1 + Subscript[c, 1], \[Ellipsis], 1 + Subscript[c, 1 + j]}, E^(2 c z)], {j, 0, n}])) /; Subscript[a, 1] == Subscript[a, 2] == \[Ellipsis] == Subscript[a, n + 2] == 1 && Subscript[b, 1] == Subscript[b, 2] == \[Ellipsis] == Subscript[b, n + 1] == b/c && Subscript[c, 1] == Subscript[c, 2] == \[Ellipsis] == Subscript[c, n + 1] == (b + c)/c && Element[n, Integers] && n >= 0
Standard Form
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MathML Form
z n b z sinh ( b z ) coth ( c z ) z 1 2 ( z n + 1 n + 1 + 2 2 c z n ! j = 0 n ( - 1 ) j 2 - j - 1 c - j - 1 z n - j ( n - j ) ! j + 2 F j + 1 ( 1 , , 1 ; 2 , , 2 ; 2 c z ) TagBox[TagBox[RowBox[List[RowBox[List[SubscriptBox["\[InvisiblePrefixScriptBase]", FormBox[RowBox[List["j", "+", "2"]], TraditionalForm]], SubscriptBox["F", FormBox[RowBox[List["j", "+", "1"]], TraditionalForm]]]], "\[InvisibleApplication]", RowBox[List["(", RowBox[List[TagBox[TagBox[RowBox[List[TagBox["1", HypergeometricPFQ], ",", TagBox["\[Ellipsis]", HypergeometricPFQ], ",", TagBox["1", HypergeometricPFQ]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ], ";", TagBox[TagBox[RowBox[List[TagBox["2", HypergeometricPFQ], ",", TagBox["\[Ellipsis]", HypergeometricPFQ], ",", TagBox["2", HypergeometricPFQ]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ], ";", TagBox[SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "c", " ", "z"]]], HypergeometricPFQ]]], ")"]]]], InterpretTemplate[Function[HypergeometricPFQ[Slot[1], Slot[2], Slot[3]]]]], HypergeometricPFQ] - n ! ( 2 b z j = 0 n ( - 1 ) j ( 2 b ) - j - 1 z n - j ( n - j ) ! j + 2 F j + 1 ( b c , , b c , 1 ; b c + 1 , , b c + 1 ; 2 c z ) TagBox[TagBox[RowBox[List[RowBox[List[SubscriptBox["\[InvisiblePrefixScriptBase]", FormBox[RowBox[List["j", "+", "2"]], TraditionalForm]], SubscriptBox["F", FormBox[RowBox[List["j", "+", "1"]], TraditionalForm]]]], "\[InvisibleApplication]", RowBox[List["(", RowBox[List[TagBox[TagBox[RowBox[List[TagBox[FractionBox["b", "c"], HypergeometricPFQ], ",", TagBox["\[Ellipsis]", HypergeometricPFQ], ",", TagBox[FractionBox["b", "c"], HypergeometricPFQ], ",", TagBox["1", HypergeometricPFQ]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ], ";", TagBox[TagBox[RowBox[List[TagBox[RowBox[List[FractionBox["b", "c"], "+", "1"]], HypergeometricPFQ], ",", TagBox["\[Ellipsis]", HypergeometricPFQ], ",", TagBox[RowBox[List[FractionBox["b", "c"], "+", "1"]], HypergeometricPFQ]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ], ";", TagBox[SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "c", " ", "z"]]], HypergeometricPFQ]]], ")"]]]], InterpretTemplate[Function[HypergeometricPFQ[Slot[1], Slot[2], Slot[3]]]]], HypergeometricPFQ] + 2 ( b + c ) z j = 0 n ( - 1 ) j ( 2 b + 2 c ) - j - 1 z n - j ( n - j ) ! j + 2 F j + 1 ( b + c c , , b + c c , 1 ; b + c c + 1 , , b + c c + 1 ; 2 c z ) TagBox[TagBox[RowBox[List[RowBox[List[SubscriptBox["\[InvisiblePrefixScriptBase]", FormBox[RowBox[List["j", "+", "2"]], TraditionalForm]], SubscriptBox["F", FormBox[RowBox[List["j", "+", "1"]], TraditionalForm]]]], "\[InvisibleApplication]", RowBox[List["(", RowBox[List[TagBox[TagBox[RowBox[List[TagBox[FractionBox[RowBox[List["b", "+", "c"]], "c"], HypergeometricPFQ], ",", TagBox["\[Ellipsis]", HypergeometricPFQ], ",", TagBox[FractionBox[RowBox[List["b", "+", "c"]], "c"], HypergeometricPFQ], ",", TagBox["1", HypergeometricPFQ]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ], ";", TagBox[TagBox[RowBox[List[TagBox[RowBox[List[FractionBox[RowBox[List["b", "+", "c"]], "c"], "+", "1"]], HypergeometricPFQ], ",", TagBox["\[Ellipsis]", HypergeometricPFQ], ",", TagBox[RowBox[List[FractionBox[RowBox[List["b", "+", "c"]], "c"], "+", "1"]], HypergeometricPFQ]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ], ";", TagBox[SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "c", " ", "z"]]], HypergeometricPFQ]]], ")"]]]], InterpretTemplate[Function[HypergeometricPFQ[Slot[1], Slot[2], Slot[3]]]]], HypergeometricPFQ] ) ) /; n Condition z z n b z b z c z 1 2 z n 1 n 1 -1 2 2 c z n j 0 n -1 j 2 -1 j -1 c -1 j -1 z n -1 j n -1 j -1 HypergeometricPFQ 1 1 2 2 2 c z -1 n 2 b z j 0 n -1 j 2 b -1 j -1 z n -1 j n -1 j -1 HypergeometricPFQ b c -1 b c -1 1 b c -1 1 b c -1 1 2 c z 2 b c z j 0 n -1 j 2 b 2 c -1 j -1 z n -1 j n -1 j -1 HypergeometricPFQ b c c -1 b c c -1 1 b c c -1 1 b c c -1 1 2 c z n [/itex]
Rule Form
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Date Added to functions.wolfram.com (modification date)
2002-12-18 | 5,288 | 14,724 | {"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-2013-48 | longest | en | 0.170689 |
https://discuss.codechef.com/t/all-substring-explanation/39550 | 1,627,870,756,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154302.46/warc/CC-MAIN-20210802012641-20210802042641-00230.warc.gz | 223,411,437 | 6,589 | All Substring Explanation!
I’ll explain this question as simple as possible. Make sure you check the code.It’s Full of comments to help you see what each line is doing.
Problem : ALLSUB
Solution : Code
In this question, We have been given two Strings S and R.
And we have to find the lexiographically smallest *Valid* string by reordering the characters of String R.
Now the first question is.
What is a valid string?
According to question a valid string is one which has all substrings of S , including S.
Now it’s clear that the reordered string must have S intact, and it should be formed by using all the characters of R.
So We first create a frequency map of R which maps each character to its frequency in String R.
The we do the same for S.
Now if there is a situation in which the frequency of some character in frequency map of R is less than that of S, then there is no way you can form the valid string as you’ll run short of letters and won’t be able to form S in new string.
So that’s our condition for outputting "Impossible".
Since we have to keep S intact in new string, we would be only able to use the left letters after all letters reserved for S is gone.
Now you can create a string called 'left' and keep these remaining letters.
now all the letters smaller (lexiographically) than the first character of S would go on the left side of S and bigger would go on right of S.
There is one case when they both become equal. In that case, we can simply generate both possibility and choose the smaller (lexiographically).
Done !
The code speaks for itself. You can check it. If you have any confusion, tell me
7 Likes
Support,the solution is very concise and straightforward.
2 Likes
Finally someone who doesn’t over complicate stuff. This is how editorials should be!
Good Job man!
1 Like
Thanks much bro
20 char
2 Likes
Very nice explanation
1 Like
I did the exact same thing, but I am getting a WA. Please provide a counter test case. Help appreciated. https://www.codechef.com/viewsolution/26842648
Can someone give a counter testcase for my answer. I pretty much did the same thing as explained above
https://www.codechef.com/viewsolution/26854719
@sauravshah
Your Program fails for this test
1
cat
catc
ccat
Expected Output
catc
@res_req_ted
Your Program fails for this test
1
cat
catc
ccat
Expected Output
catc
My program’s output is catc itself.
@sauravshah Sorry bro i opened someone elses i think
1
ba
abba
Expected Output
abab
Yours
baab
Ahh! I got that. Thanks
1 Like
nyc one,very simple and easiest one upto this time…one thing lots of people doing mistake that how the middle of the final string R will be merged with S,so that it will be lexicographically small…So, u cleared that doubt.Thanks for this
1 Like
Thanks a lot for the well-explained code. I was struck on this problem for 2 hours.
1 Like
Can someone please tell me where my code goes wrong . I did the same thing as explained above.https://www.codechef.com/viewsolution/26858157
nice explanation brother!!!
1 Like
Very good Explanation indeed. My approach for this ques was almost the same but still i am getting WA. As i am fairly new to CodeChef or CP to be exact i don’t know how to find counter cases for one’s program. Can anyone be kind enough to give some insight upon the wrong Testcases? Thanks in advance.
https://www.codechef.com/viewsolution/26889606
1 Like
Please correct me ,give me counter case,
https://www.codechef.com/viewsolution/26890345
1
cppt
acptp | 842 | 3,524 | {"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} | 2.953125 | 3 | CC-MAIN-2021-31 | latest | en | 0.948195 |
https://matheducators.stackexchange.com/questions/20649/why-do-so-many-people-hate-math/27206 | 1,708,652,433,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473871.23/warc/CC-MAIN-20240222225655-20240223015655-00608.warc.gz | 393,002,544 | 49,046 | # Why do so many people hate math?
I love mathematics and I know pretty much everyone in this community does too! That feeling of being stuck on a mathematical problem for days as you try to unearth the complexity in front of you is painful but that final emotion of solving it, of discovering something new is certainly magical. I can only imagine the pure bliss Sir Andrew Wiles felt when he cracked Fermat's Last Theorem or when Isaac Newton and Gottfried Leibniz discovered the essence of Calculus. Mathematics is, without doubt, beautiful and can help us understand the most counterintuitive ideas. But while so many are deeply passionate about math, the subject often carries a negative connotation to the general public. In this post, I seek to explore and understand why - why is something so spectacular as math hated by so many?
In my experience the answer is a result of the following:
1. The education system puts greater importance on getting done with the syllabus rather than getting done with the syllabus well.
For example, take the quadratic formula $$x=\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$.
In almost all cases, teachers resort to simply writing this on a whiteboard and asking students to memorize it for their examinations. Apart from a few, no one sits down and allows students to discover it themselves, to think about the equation $$ax^{2} + bx + c = 0$$ in the most creative ways. This, in my opinion, kills a child's imagination and paints this hard, boring picture of mathematics in their head while they have gained no intuition for the formula whatsoever. As a result, they end up disliking mathematics.
1. Students argue that stuff like the quadratic equation is most likely never going to be useful in their life. This is a fair point. As a future doctor, you may never have to go beyond some basic arithmetic and statistics in your career so learning about this quadratic equation seems fairly useless. What students often fail to realize, however, in my experience, is that the journey of discovering and understanding the quadratic formula is itself a worthy experience - that the critical thinking and creative skills it develops can help them everywhere in life. As a result of this oblivion, they end up disliking maths.
While these are just $$2$$ observations, I am sure there are many more that you must have encountered in your mathematical journey. What are those other reasons and more importantly, how do we fix them?
Edit: This post had been previously closed because it was "opinion-based". I understand that, but I am completely willing to hear other people's opinions, their refutations to my observations, and their own observations. After all, if I don't present my opinion, how will other people comment on it, how can one have a discussion? I sincerely request this post to not be closed again because its only intention is to spark discussion, not to indoctrinate people with personal ideas.
• "its only intention is to spark discussion." Please read the pages in the help center: "If your motivation for asking the question is “I would like to participate in a discussion about ______”, then you should not be asking here."
– JRN
Mar 27, 2021 at 11:40
• Previously asked at Mathematics Stack Exchange. In the future, if you repost your question, then you should link to the original post.
– JRN
Mar 27, 2021 at 11:43
• "In almost all cases, teachers resort to simply writing this on a whiteboard and asking students to memorize it for their examinations. Apart from a few, no one sits down and allows students to discover it themselves" - both of these approaches are wrong. Good teachers in good schools practice something that may be called guided discovery. Oct 12, 2021 at 17:11
• "Good teachers in good schools practice something that may be called guided discovery." -- Debatable; there are studies that show that theory is overblown, and fully directed instruction is better in many cases. Oct 14, 2021 at 1:41
• I’m voting to close this question because, per OP, "its only intention is to spark discussion". Sep 27, 2023 at 13:59
You asked a big question. Maybe some of these ideas are worth investigating more deeply. The two first explanations are completely generic, while the last two are more specific to mathematics; I would advise to not forget the power of the generic explanations. But I don't know to what extent these are good explanations or what other good ones there might be. This is something mathematics education has grappled with for a long time and is grappling with right now.
# It is forced on people
Some people don't like music or playing instruments, yet might have had to play or sing at school. Some have bad memories of it.
Some people don't like physical activity while at school, yet might be forced to do it. A whole lot of bad memories is associated to it.
The same with foreign languages, philosophy and really whatever else is being forced on people at school. But there is a lot of mathematics and most people don't see the beautiful mathematics outside the hated school context.
# It is difficult
Many people don't like doing difficult things. It seems mathematics is difficult. Thus many people don't like it.
# Rote versus creative?
Is doing lots of rote calculation something that makes people like mathematics less, when compared to the creative type of mathematics? Maybe; at least there are serious trends in teacher education, at least in Finland and Norway, aimed towards reducing the amount of rote learning. See for example "problem-based learning" and similar keywords for more.
# Abstract versus applied?
Is abstract mathematics more or less motivating than applied mathematics? ATD, the anthropological theory of the didactic by Yves Chevallard, is an example of a serious theory that advocates for mathematics applications first, and also in a creative way. There are doubtless others.
The "new mathematics" was an example of going strongly abstraction-first, but presumably it could be done in a more effective manner, too.
Do people hate math, or do people hate that they have to do math in school ? I suppose every so often we meet a true hater, someone who reviles math in every avenue. But, I think that is anomalous. In reality, most people hate math because it is harder than their other subjects by in large.
Moreover, that is not a product of math teachers being "mean", it is a product of every earlier math course that cut a corner and just let kids slide by. Unfortunately, prerequisites matter so when an 18 year old gets to college and is not comfortable with arithmetic, it's a battle of overcoming years of neglect. The same is true for highschool if the highschool course holds the standard it ought.
All of this said, one of my principle aims in any lower math class is to try to win some over to see math is not all bad. In fact, math can be fun when we try to understand it rather than just do it. I do things like show them more than one way to solve a problem and I talk a bit about the history which led to us being able to compactly do algebra without awkward lengthy sentences of old-english prose. Still, there are always those who I fail to reach. I can only hope the next brave soul who teaches them breaks down their hate.
• "Do people hate math, or do people hate that they have to do math in school ? I suppose every so often we meet a true hater, someone who reviles math in every avenue. But, I think that is anomalous." >> Unfortunately, once school has done its damage, the damage is done. A few people arrive at the conclusion "I hated math in school but I could like it in another context or if it was taught differently", but in my experience most people arrive instead at the conclusion "I hated math in school, thus I will forever hate math"
– Stef
Nov 26, 2023 at 19:00
As enumerated in other answers, there are plenty of mechanisms by which one might develop a dislike of math as a result of one's own experiences.
But why is it so easy for an initial dislike of math to quickly spiral into a full-blown hatred that forces its way into conversation whenever the topic of math comes up, even if there are no substantial changes in one's direct experiences with math?
Because hatred and vilification of math has been normalized.
In general, if someone dislikes thing X, and there is a large group of people who hate thing X and are vocal about it, then there is a social-emotional gravity that pulls this person towards the group while elevating their dislike into a similar level of hatred and vocalization.
Positive feedback loops like this happen all the time in politics. Math, unfortunately, is no different.
I'm very fond of Eugenia Cheng's perspective: many people dislike certain parts and/or aspects of mathematics, and it just happens that these are precisely the ones emphasized by most (all?) school curricula. To quote from her wonderful "Joy of abstraction":
Some people do need to build up gradually through concrete examples towards abstract ideas. But not everyone is like that. For some people, the concrete examples don’t make sense until they’ve grasped the abstract ideas or, worse, the concrete examples are so offputting that they will give up if presented with those first. [...]
I have confirmed from several years of teaching abstract mathematics to art students that I am not the only one who prefers to use abstract ideas to illuminate concrete examples rather than the other way round. Many of these art students consider that they’re bad at math because they were bad at memorizing times tables, because they’re bad at mental arithmetic, and they can’t solve equations. But this doesn’t mean they’re bad at math — it just means they’re not very good at times tables, mental arithmetic and equations, an absolutely tiny part of mathematics that hardly counts as abstract at all. It turns out that they do not struggle nearly as much when we get to abstract things such as higher-dimensional spaces, subtle notions of equivalence, and category theory structures. Their blockage on mental arithmetic becomes irrelevant.
It seems to me that we are denying students entry into abstract mathematics when they struggle with non-abstract mathematics, and that this approach is counter-productive. Or perhaps some students self-select out of abstract mathematics if they did not enjoy non-abstract mathematics.
Sadly, there doesn't seem to be any easy way out of this problem. Notice that 'New Math' doesn't necessarily aling with what's called "abstract" here.
In addition to the standard reasons:
2. It's hard,
there is an interesting reason why one hears so much about people hating math that's due to$${\ldots}$$ Conan Doyle. Arguably he created the fad; see Blowing Off Steam: Victorian Narratives of Frustration About Mathematics.
1. Many students do mathematics as a subject not because they like it but because it is compulsory to do so. Because of less interest, they feel it is a difficult subject to follow.
2. The approach may lead to creating negative feelings about understanding mathematics. Creative approaches such as getting help from geometry in the case of proof without words can be used to create interest in the subject.
3. When the students are put into a fixed frame by asking to apply specific principles or formulae, they can feel uncomfortable. It is much more important to let them have their approach which can be nontraditional or unconventional and thet they need to improve lateral thinking.
4. If students are not aware of real-life applications of mathematical concepts, it can lead to distract them from mathematics. Sometimes negative feelings about the subject are created by the terms that are supposed to be used in teaching. We can take imaginary numbers as an example, from the very beginning they give an idea of uselessness in real life but as we know, imaginary numbers are very important in studying electricity in physics and identifying the beautifulness of any object.
As a Chinese student, I have had gone through a hell-like period on math in high school. In fact, every day with math sucks(sorry to say it).
Boredom and abstruseness are not the only two problem it puts. The most important thing is, math is not based on sociality. Sociologists argue that schools are prepared for better socialization. However, you can see no hint that indicates math will help because it is never a bridge
of emotions or feelings, never enables you to enter a social situation. And societal animals hate this.
The secondary problem is, that math merely favours those who conquest it (they get flowers and applause), while those who don't, however, are long thought of as dumb ones. Continuous failure (basically impossible to avoid ) and negative feedbacks will definitely destroy learners' confidence even if genius is on the opposite. Thus, math becomes a symbol of stigma in many people's memories.
Since math is connected with a feeling of being humiliated, it's hard to understand why people like math. 😄 And physics is abstruse, too. I mean, perhaps many people hate physics equally just as math.
• What is funny is that this argumentation can be applied to humanities (history, social sciences, etc.) with obvious replacements (the main of which is that the humanity and all its feelings are just a tiny speck in the Universe and that the only thing that really distinguishes a human from a monkey is the insatiable curiosity for things that go far beyond food, shelter, sex, and dominance). It all depends on what you are after in this life and what is your answer to the question about its meaning. As to "dumb", the people who are awkward socially are labeled "dumb" even more often AFAIK :-) Sep 27, 2023 at 14:12
Being a graduate student and having studied mathematics for almost all my life, I think I am in position to answer this question. Well I love mathematics but that necessarily don’t mean everyone does the same. There could be no reason for anyone to hate maths unless the followings:
1. One had practiced memorisations while doing mathematics
2. One didn’t had a good teacher in high school
3. One couldn’t grasp the intuitive ideas behind any mathematical proof.
4. One doesn’t questions the theory.
5. One is scared of the notations used in mathematics.
6. One doesn’t have fascinations for reasonings.
7. One has a gap in knowledge.
The list could go on.
These are the fundamental reasons why many people are scared of mathematics.
The poet Samuel Taylor Coleridge had an answer (namely, that imagination is underutilized). You can read his full critique of mathematics education here:
https://allpoetry.com/A-Mathematical-Problem | 3,088 | 14,721 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-10 | latest | en | 0.9505 |
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July 7, 2024
Function Description Example Result
ABS Calculate the absolute value of a number ABS(-10) 10
CBRT Calculate the cube root of a number CBRT(8) 2
CEIL Round a number up to the nearest integer, which is greater than or equal to a specified number CEIL(-12.8) -12
CEILING Same as CEIL
DEGREES Convert radians to degrees DEGREES(0.8) 45.83662361
DIV Return the integer quotient of two numeric values DIV(8,3) 2
EXP Return the exponential value in the scientific notation of a number EXP(1) 2.718281828
FLOOR Round a number down to the nearest integer, which is less than or equal to the number FLOOR(10.6) 10
LN Return the natural logarithm of a numeric value LN(3) 1.098612289
LOG Return the base 10 logarithms of a numeric value LOG(1000) 3
LOG Return the logarithm of a numeric value to a specified base LOG(2, 64) 6
MOD Divide the first parameter by the second one and return the remainder MOD(10,4) 2
PI Return the value of PI PI() 3.141592654
POWER Raise a numeric value to the power of a second numeric value POWER(5, 3) 125 | 307 | 1,079 | {"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-2024-30 | latest | en | 0.617424 |
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Topic: Matheology § 238
Replies: 210 Last Post: Apr 13, 2013 4:21 PM
Messages: [ Previous | Next ]
Virgil Posts: 8,833 Registered: 1/6/11
Re: Matheology � 238
Posted: Apr 6, 2013 6:49 PM
In article
WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 6 Apr., 22:30, William Hughes <wpihug...@gmail.com> wrote:
>
> >
> > My claim is:
> >
> > Let D be the set of all lines, and let
> > E be any one finite subset of D.
> > Then D\E is not empty.
>
> Your claim is true. But your premise is wrong. E (the set that can be
> removed without changing the union of the lines) is the set of all
> lines that have a follower. This set of lines is infinite. By the
> existence of the follower to all lines, all lines, i.e., E can be
> removed without changing the union of all lines, namely |N.
As usual, WM's logic is ill.
If, from a well ordered set, one is only allowed to remove a member or
set of members so long as at least one subsequent member remains, how
does WM claim that one can ever remove all remainng members without
leaving any behind?
Removing all members does exactly that forbiddeen thing, so is not
allowed!.
Thus E cannot be empty.
A similar argument shows that E cannot be finite.
--
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For example, things like time, mass, weight, force, and electric charge, are physical quantities with which we are all familiar. In this lesson, learners are introduced to the vector and scalar quantities. Scalars may or may not have units associated with them. A directed line segment has magnitude as well as direction, so it is. One way to think of this is that we start at the beginning of the first vector, travel along that vector to its end, and then travel from the start of the second vector to. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on youtube.
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To distinguish between scalars and vectors we will denote scalars by lower case italic type such as a, b, c etc. Vector algebra 425 now observe that if we restrict the line l to the line segment ab, then a magnitude is prescribed on the line l with one of the two directions, so that we obtain a directed. Displacement, velocity, acceleration, electric field. Some quantities also have direction, a quantity that has both a magnitude and direction is called a vector. A vector is a quantity which has both magnitude and direction. Sep 04, 2019 in this physics notes, we will learn what are scalar and vector quantity and the main difference between them. Scalar quantities are comparable only when they have the same physical dimensions. The remainder of this lesson will focus on several examples of vector and scalar quantities distance, displacement, speed, velocity, and acceleration.
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Equality of vectors two vectors a and b are said to be equal written as a b, if they have i same length ii the. A scalar quantity is one, which is fully defined by magnitude alone. Two or more than two scalar quantities measured in the same system of units are equal if they have the same magnitude and sign. Vector algebra class 12 notes maths chapter 10 learn cbse. When we want to indicate that a vector is a unit vector we put a hat circum ex above it, e. When expressing a quantity we give it a number and a unit for example, 12 kg, this expresses the magnitude of the quantity. The physical quantity like electric current possesses both the magnitude and direction, still they are not vectors, and similarly any form of energy is a scalar.
For example, mass or weight is characterized by a real and nonnegative number. Vector one shot complete chapter vector full chapter. A lot of mathematical quantities are used in physics to explain the concepts clearly. We know that time passes and physical objects have mass. For example, the result of the sum of a vector and its negative vector is always a null vector. How do we know which physical quantity is a vector, which is a scalar and which is neither. Apr 05, 2020 are you looking for notes on vector algebra in pdf format. Those quantities which have magnitude, as well as direction, are called vector quantities or vectors. Neet physics chapter wise mock test scalar and vector question 1. These notes are send by umer asghar, we are very thankful to him for providing these notes. One way to think of this is that we start at the beginning of the first vector, travel along that vector to its end, and then travel from the start of the second vector to its end. Scalars and vectors scalars and vectors a scalar is a number which expresses quantity.
If c2r and ua vector, then we may form a new vector cucalled the scalar product of uwith c. In todays world, various mathematical quantities depict the motion of objects into two categories. Vectors are quantities that are fully described by both a magnitude and a direction. The notion of scaling is addressed by the mathematical.
Two vectors are equal if and only if the corresponding components. Vectors have magnitude and direction, scalars only have magnitude. A vector in 2d euclidean space is defined by a pair of scalars. Since vectors can be scaled, any vector can be rescaled b to be a unit vector. Vectors and equilibrium notes for class 11 free pdf. Contents1 neet physics chapter wise mock test scalar and vector1.
This document is highly rated by civil engineering ce students and has been viewed 209 times. The magnitude of the vector a is written as a or a. When a vector a is multiplied by a scalar s, then its magnitude becomes s times, and unit is the product of units of a and s but direction remains same as that of vector a. Are you looking for notes on vector algebra in pdf format. V vn v magnitude of v n unit vector whose magnitude is one and whose direction coincides with that of v unit vector can be formed by dividing any vector, such as the geometric position vector, by its length or magnitude vectors represented by bold and nonitalic letters v. A vector space v is a collection of objects with a vector. When we calculate the scalar product of two vectors the result, as the name suggests is a scalar, rather than a vector. We use vectors to represent entities which are described by magnitude and direction. Two arrows represent the same vector if they have the same length and are parallel see. Notes right triangle in semi circle note of vector analysis by hammed ullah. Zero vector zero vector or null vector is a vector which has zero magnitude and an arbitrary direction. Download the vector short notes pdf from the link given at the end of the article. This is in contrast to a scalar, which has only magnitude and which is not changed when a system of coordinates is rotated.
Those quantities which have only magnitude and no direction, are called scalar quantities. In this unit you will learn how to calculate the scalar product and meet some geometrical appli. Scalar quantities are denoted by letters in ordinary type. Basic concepts a vector v in the plane or in space is an arrow. May 23, 2019 cbse class 12 maths notes chapter 10 vector algebra vector. A vector quantity is fully defined by magnitude and direction. This is because the scalar product also determines the length of a vector. V vn v magnitude of v n unit vector whose magnitude is one and whose direction coincides with that of v unit vector can be formed by dividing any vector, such as the geometric position vector, by its length or magnitude. The work is a scalar quantity and takes into account only that component of the force that acts along the displacement. Vectors are those quantities which are described by the magnitude of the quantity and its direction. In this physics notes, we will learn what are scalar and vector quantity and the main difference between them.
Scalar, vector, tensor functions civil engineering ce notes. These notes are for helpful for undergraduate level bsc or bs. In this chapter we shall use the ideas of the plane to develop a new mathematical concept, vector. Quantities which require both magnitude and direction for their complete specification are called vectors. Thus, we see from our knowledge of vector addition that the magnitude length of. Jun 30, 2017 unsubscribe from physics wallah alakh pandey. A vector having a zero magnitude and arbitrary direction is called as null vector. A vector is characterized by a nonnegative real number referred to as a magnitude, and a direction. Some familiar theorems from euclidean geometry are proved using vector methods. We also introduce the concept of a dyad, which is useful in mhd. Master with the concepts of scalar and vector including parallel vector, unit vector, null vector with the help of study material for iitjee by askiitians. In this grade, learners focus on vectors in only one. | 2,194 | 10,713 | {"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} | 3.890625 | 4 | CC-MAIN-2021-21 | latest | en | 0.950331 |
https://convertoctopus.com/10-1-grams-to-kilograms | 1,606,605,348,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00564.warc.gz | 252,032,249 | 7,681 | ## Conversion formula
The conversion factor from grams to kilograms is 0.001, which means that 1 gram is equal to 0.001 kilograms:
1 g = 0.001 kg
To convert 10.1 grams into kilograms we have to multiply 10.1 by the conversion factor in order to get the mass amount from grams to kilograms. We can also form a simple proportion to calculate the result:
1 g → 0.001 kg
10.1 g → M(kg)
Solve the above proportion to obtain the mass M in kilograms:
M(kg) = 10.1 g × 0.001 kg
M(kg) = 0.0101 kg
The final result is:
10.1 g → 0.0101 kg
We conclude that 10.1 grams is equivalent to 0.0101 kilograms:
10.1 grams = 0.0101 kilograms
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 99.009900990099 × 10.1 grams.
Another way is saying that 10.1 grams is equal to 1 ÷ 99.009900990099 kilograms.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that ten point one grams is approximately zero point zero one kilograms:
10.1 g ≅ 0.01 kg
An alternative is also that one kilogram is approximately ninety-nine point zero one times ten point one grams.
## Conversion table
### grams to kilograms chart
For quick reference purposes, below is the conversion table you can use to convert from grams to kilograms
grams (g) kilograms (kg)
11.1 grams 0.011 kilograms
12.1 grams 0.012 kilograms
13.1 grams 0.013 kilograms
14.1 grams 0.014 kilograms
15.1 grams 0.015 kilograms
16.1 grams 0.016 kilograms
17.1 grams 0.017 kilograms
18.1 grams 0.018 kilograms
19.1 grams 0.019 kilograms
20.1 grams 0.02 kilograms | 468 | 1,666 | {"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} | 4.15625 | 4 | CC-MAIN-2020-50 | latest | en | 0.76437 |
https://www.cscodehelp.com/c-c-%E4%BB%A3%E5%86%99/%E7%A8%8B%E5%BA%8F%E4%BB%A3%E5%86%99%E4%BB%A3%E5%81%9A%E4%BB%A3%E8%80%83-algorithm-csc-427-theory-of-computation-1/ | 1,726,818,290,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00518.warc.gz | 654,695,849 | 13,939 | # 程序代写代做代考 algorithm CSC 427: Theory of Computation 1
CSC 427: Theory of Computation 1
Final Wednesday, 3 May 2017
4:30 PM
This is a take home final. It will be released to all students and be due at
the date and time given above.
You are to work independently, you can reference materials, but be careful
if your answers too closely repeat publicly available sources, or I will not be
able to evaluate your own grasp of the materials.
The problems contain enough depth that it is always possible, regardless
of the materials referenced, for you to demonstrate personal competence on
the tested material.
Name:
Problem Credit
1
2
3
4
5
Total
CSC 427: Theory of Computation 2
1. The SATISFIABILITY problem is, given a collection of clauses, such
as,
{ (x1 ∨ x2 ∨ x3 ∨ x4), (x2 ∨ x3 ∨ x1), (x1 ∨ x4) }
assign values to all xi such that each clause evaluates true.
This problem reduces to 3-SAT, where each clause has exactly three
literals.
(a) Show how a single clause that is not in the proper 3-SAT form is
made by the reduction into several clauses each in proper 3-SAT.
In particular, show the reduction for the following clause:
x1 ∨ x2 ∨ x3 ∨ x4 ∨ x5 ∨ x6
Name by yi any new variables introduced.
(b) Why does your reduction work? Why is the reduced collection of
clauses simultaneously satisfiable if and only if the original clause
is satisfiable?
CSC 427: Theory of Computation 3
2. The book gave the following reduction from 3-SAT to VERTEX-COVER:
For each variable x, create a “variable widget” of two nodes x and x
connected by an edge. For each clause x∨y∨z create a “clause widget”
of three nodes x, y and z connected in a triangle. Connect each node in
a clause widget with a node in a variable widget according to the vari-
able or negation of variable that the nodes represent. Set k = n+ 2m,
where n is the number of variables, and m is the number of clauses.
Draw the resulting VERTEX-COVER instance of the following 3-SAT
instance and find a vertex cover. Give the satisfying assignment implied
by that cover.
{ (x1 ∨ x2 ∨ x3), (x1 ∨ x2 ∨ x3), (x1 ∨ x2 ∨ x3),
(x1 ∨ x2 ∨ x3), (x1 ∨ x2 ∨ x3) }
CSC 427: Theory of Computation 4
3. Addition of two positive integers takes O(k) time by the standard addi-
tion algorithm, with k the total length of the representation in binary
of the integers.
The following recursive function M defines multiplication,
M(x, y) =
x for y = 1
2 ∗M(x, y/2) for y even
x+ 2 ∗M(x, (y − 1)/2) for y odd and not 1
for positive integers x and y.
For instance,
M(6, 5) = 6 + 2 ∗M(6, 2) = 6 + 2 ∗ 2 ∗M(6, 1) = 6 + 2 ∗ 2 ∗ 6 = 30.
Show that multiplication is in P-time by analyzing this algorithm. The
run time will be in Big-Oh notation as a function of k, the total length
of the representation in binary of the numbers.
CSC 427: Theory of Computation 5
4. The set of Turing Machines that accept nothing, ETM , is undecidable.
(a) If a set is undecidable, can it be both R.E. and co-R.E.? Give
proof.
(b) If a set is undecidable, can it be neither R.E. and co-R.E.? Give
an example.
(c) Is ETM R.E., co-R.E., or neither? Give proof.
CSC 427: Theory of Computation 6
5. Given an NP problem A, there is a non-deterministic TM such that if a
is in the language then there is at least one computation path leading
to accept, but if a is not in the language then no computation path
An important class of problems inside NP is Randomized Polynomial
Time, or RP. A problem is in RP if for a in the language then a pro-
portion of at least 1/100 of all computation paths are paths leading to
accept, and for a not in the language, there are no computation paths
A Randomized Turing Machine is a deterministic Turing Machine that
in addition to the tape contents to guide the transitions, can flip a coin
and use the coin outcome to guide the transitions. The output of these
machines has an associated probability, which is the probability that
the coin flips will lead the computation to that output.
Prove the following:
For all problems in RP, there is a Randomized Turing Ma-
chine such that, if a is not in the language, the machine
rejects always, if a is in the language, the machine accepts
with probability 1− � and and rejects with probability �, for
any real � > 0.
While the class RP seems strange, it has practical significance. RP
problems are those NP problems for which one can trade not being
able to know the answer with being able to know the answer, but there
is a slim chance that the answer is wrong.
Posted in Uncategorized | 1,233 | 4,531 | {"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-2024-38 | latest | en | 0.889831 |
https://www.gamedev.net/topic/632929-deferred-rendering-basics/ | 1,493,395,704,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122996.52/warc/CC-MAIN-20170423031202-00102-ip-10-145-167-34.ec2.internal.warc.gz | 913,234,403 | 38,166 | • FEATURED
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# Deferred rendering basics
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
2 replies to this topic
### #1keym Members
Posted 17 October 2012 - 04:19 AM
Hello,
I'm trying to implement deferred rendering and I'm stuck with this problem. I do all calculations in view space, I set the light position for testing to (0,0,0) - camera position. But the problem is when I look up or down the lighting changes to brighter or darker. It sounds like I have something wrong with matrices but I checked that many times and still no luck. I'm not sure if my reconstruction from depth is right because it's a new thing to me and I may be doing something wrong here. But on the other hand lighting doesn't change when I move backward or forward and keep the angle (so I guess position is correct?). Anyway I'm posting some pictures, maybe someone can figure out where I go wrong. For now I'm trying to compute lighting with geometry normals and then move with bump maps. To me my normals look too bright, pinkish, dull etc. when I compare them to normals from Killzone paper or other tutorials. Maybe this is the case?
Thanks for watching
Edit: I consider simplest scenario for now - only diffuse lighting, no attenuation, no specular and no ambient
This is how I recreate position from depth in light pass:
float z = f1tex2D(depth_map, texCoord);
float x = texCoord.x * 2 - 1;
float y = (1 - texCoord.y) * 2 - 1;
float4 vProjectedPos = float4(x, y, z, 1.0f);
//unproject the position from clip space to view space using inv. proj. matrix
float4 vPositionVS = mul(invProj, vProjectedPos);
float3 vsPos = vPositionVS.xyz / vPositionVS.w;
I use depth buffer to get depth or I can use depth encoded as color which is:
in vertex program:
outPosition = mul(ModelViewProj, position);
vDepthCS.xy = outPosition.zw;
and in fragment program I do division z/w and pack it as color but the result is the same as with depth buffer.
#### Attached Thumbnails
Edited by keym, 17 October 2012 - 06:06 AM.
### #2Ashaman73 Members
Posted 18 October 2012 - 12:30 AM
First off, you can't display the normals without hack, because normals have negative values. Therefore you need to map them to a visible colorspace (color = normal*0.5+0.5), I guess that killzone use an other mapping(or none at all).
Your normal screenshot seems to be ok (green pointing up, red pointing right, blue tint for pointing to the camera), how does the normals change if you look up/down, maybe your inversion of the y-coord is not consistent.
Edited by Ashaman73, 18 October 2012 - 12:31 AM.
Ashaman
### #3keym Members
Posted 18 October 2012 - 03:07 AM
Hi,
I solved my problem yesterday. Normals were ok (I packed and unpacked them from color with that hack you mention but I didn't mention that, that Killzone paper made me doubt...). Turned out that my reconstruction of position was wrong (like said before, it's a new thing to me and I was unsure of that bit the most). I used this http://www.opengl.or...l=1#post1234440 code and it finally worked . Thanks for interest though.
Cheers
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. | 889 | 3,595 | {"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} | 2.578125 | 3 | CC-MAIN-2017-17 | latest | en | 0.857342 |
http://metamath.tirix.org/mpeuni/erth | 1,723,515,824,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00561.warc.gz | 19,210,852 | 3,017 | # Metamath Proof Explorer
## Theorem erth
Description: Basic property of equivalence relations. Theorem 73 of Suppes p. 82. (Contributed by NM, 23-Jul-1995) (Revised by Mario Carneiro, 6-Jul-2015)
Ref Expression
Hypotheses erth.1 ( 𝜑𝑅 Er 𝑋 )
erth.2 ( 𝜑𝐴𝑋 )
Assertion erth ( 𝜑 → ( 𝐴 𝑅 𝐵 ↔ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) )
### Proof
Step Hyp Ref Expression
1 erth.1 ( 𝜑𝑅 Er 𝑋 )
2 erth.2 ( 𝜑𝐴𝑋 )
3 1 ersymb ( 𝜑 → ( 𝐴 𝑅 𝐵𝐵 𝑅 𝐴 ) )
4 3 biimpa ( ( 𝜑𝐴 𝑅 𝐵 ) → 𝐵 𝑅 𝐴 )
5 1 ertr ( 𝜑 → ( ( 𝐵 𝑅 𝐴𝐴 𝑅 𝑥 ) → 𝐵 𝑅 𝑥 ) )
6 5 impl ( ( ( 𝜑𝐵 𝑅 𝐴 ) ∧ 𝐴 𝑅 𝑥 ) → 𝐵 𝑅 𝑥 )
7 4 6 syldanl ( ( ( 𝜑𝐴 𝑅 𝐵 ) ∧ 𝐴 𝑅 𝑥 ) → 𝐵 𝑅 𝑥 )
8 1 ertr ( 𝜑 → ( ( 𝐴 𝑅 𝐵𝐵 𝑅 𝑥 ) → 𝐴 𝑅 𝑥 ) )
9 8 impl ( ( ( 𝜑𝐴 𝑅 𝐵 ) ∧ 𝐵 𝑅 𝑥 ) → 𝐴 𝑅 𝑥 )
10 7 9 impbida ( ( 𝜑𝐴 𝑅 𝐵 ) → ( 𝐴 𝑅 𝑥𝐵 𝑅 𝑥 ) )
11 vex 𝑥 ∈ V
12 2 adantr ( ( 𝜑𝐴 𝑅 𝐵 ) → 𝐴𝑋 )
13 elecg ( ( 𝑥 ∈ V ∧ 𝐴𝑋 ) → ( 𝑥 ∈ [ 𝐴 ] 𝑅𝐴 𝑅 𝑥 ) )
14 11 12 13 sylancr ( ( 𝜑𝐴 𝑅 𝐵 ) → ( 𝑥 ∈ [ 𝐴 ] 𝑅𝐴 𝑅 𝑥 ) )
15 errel ( 𝑅 Er 𝑋 → Rel 𝑅 )
16 1 15 syl ( 𝜑 → Rel 𝑅 )
17 brrelex2 ( ( Rel 𝑅𝐴 𝑅 𝐵 ) → 𝐵 ∈ V )
18 16 17 sylan ( ( 𝜑𝐴 𝑅 𝐵 ) → 𝐵 ∈ V )
19 elecg ( ( 𝑥 ∈ V ∧ 𝐵 ∈ V ) → ( 𝑥 ∈ [ 𝐵 ] 𝑅𝐵 𝑅 𝑥 ) )
20 11 18 19 sylancr ( ( 𝜑𝐴 𝑅 𝐵 ) → ( 𝑥 ∈ [ 𝐵 ] 𝑅𝐵 𝑅 𝑥 ) )
21 10 14 20 3bitr4d ( ( 𝜑𝐴 𝑅 𝐵 ) → ( 𝑥 ∈ [ 𝐴 ] 𝑅𝑥 ∈ [ 𝐵 ] 𝑅 ) )
22 21 eqrdv ( ( 𝜑𝐴 𝑅 𝐵 ) → [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 )
23 1 adantr ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → 𝑅 Er 𝑋 )
24 1 2 erref ( 𝜑𝐴 𝑅 𝐴 )
25 24 adantr ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → 𝐴 𝑅 𝐴 )
26 2 adantr ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → 𝐴𝑋 )
27 elecg ( ( 𝐴𝑋𝐴𝑋 ) → ( 𝐴 ∈ [ 𝐴 ] 𝑅𝐴 𝑅 𝐴 ) )
28 26 26 27 syl2anc ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → ( 𝐴 ∈ [ 𝐴 ] 𝑅𝐴 𝑅 𝐴 ) )
29 25 28 mpbird ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → 𝐴 ∈ [ 𝐴 ] 𝑅 )
30 simpr ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 )
31 29 30 eleqtrd ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → 𝐴 ∈ [ 𝐵 ] 𝑅 )
32 23 30 ereldm ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → ( 𝐴𝑋𝐵𝑋 ) )
33 26 32 mpbid ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → 𝐵𝑋 )
34 elecg ( ( 𝐴𝑋𝐵𝑋 ) → ( 𝐴 ∈ [ 𝐵 ] 𝑅𝐵 𝑅 𝐴 ) )
35 26 33 34 syl2anc ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → ( 𝐴 ∈ [ 𝐵 ] 𝑅𝐵 𝑅 𝐴 ) )
36 31 35 mpbid ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → 𝐵 𝑅 𝐴 )
37 23 36 ersym ( ( 𝜑 ∧ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) → 𝐴 𝑅 𝐵 )
38 22 37 impbida ( 𝜑 → ( 𝐴 𝑅 𝐵 ↔ [ 𝐴 ] 𝑅 = [ 𝐵 ] 𝑅 ) ) | 1,612 | 2,091 | {"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} | 3.671875 | 4 | CC-MAIN-2024-33 | latest | en | 0.296579 |
https://mormonconferences.org/aa/1-degree-roof-pitch-in-mm.html | 1,620,591,586,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00435.warc.gz | 425,553,218 | 10,060 | # 1 Degree Roof Pitch In Mm
### R 100 cm tan 1 r 100 0 017 r 1 75 cm 17 5 mm.
1 degree roof pitch in mm.
There are 34 9 mm of fall. 5 degree roof pitch. The picture below shows the pitch of a 7 12 roof slope meaning that for 12 of horizontal measurement roof run the vertical measurement roof rise is 7. The triangle diagram will be re drawn to scale with all dimensions marked.
Fraction precision set 1 8 1 16 1 32 1 64 decimal inch metric. What is the standard pitch for a roof. The pitch of your roof is 5 degrees how much does it rise over a meter. Rise run rise run slope.
Enter run the flat level length then click pitch angle or rise to select then enter other known dimension angle or pitch. For a pitch expressed in inches per foot convert to a fraction. How far do i have to fall a roof to get 1 degree pitch over 1 meter update. Rise run tangent of the angle.
Contemporary houses often have flat roofs which shouldn. This measurement is best done on a bare roof because curled up roofing shingles will impair your measurement. There is no standard universal roof pitch roof pitch varies depending on culture climate style and available materials in the usa the range of standard pitches is anywhere between 4 12 and 9 12 in the uk the typical house has a pitch between 40 50 although 45 should be avoided. First convert the pitch to a slope.
To do this simply convert the rise and run as a fraction to a decimal form eg. To find the angle of a roof in degrees convert the pitch to a slope then convert to degrees by finding the arc tangent of the slope. Express roof pitch as the ratio of the amount of the vertical rise of a roof or rise over the corresponding horizontal distance or run you write the pitch assuming a run of 12 inches thus the ratio would describe how much a roof rises for each foot of the run. The following table shows the roof pitch rise in run equivalents for all roof slopes in degrees from 1 to 72.
With the use of this converter it will be easier for you to locate a 30 degree roof pitch 20 degree roof pitch 25 degree roof pitch or a 15 degree roof pitch. Select and re calculate to display. Thanks very much problem solved. However if you to do the process using your own knowledge you can still convert pitch to degrees by dividing the rise and run and get its corresponding measurement.
Drag sliders to animate the results and diagram.
### 510015 Roof Pmtl Ingutter Gif 350 350
Source : pinterest.com | 559 | 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} | 2.953125 | 3 | CC-MAIN-2021-21 | latest | en | 0.900303 |
https://answercult.com/questions/question/homomorphism-vs-homeomorphism-vs-isomorphism/ | 1,670,098,697,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710936.10/warc/CC-MAIN-20221203175958-20221203205958-00835.warc.gz | 122,747,987 | 17,088 | # Homomorphism vs homeomorphism vs isomorphism
57 views
0
Homomorphism vs homeomorphism vs isomorphism
In: 0
Homomorphism is a mapping between 2 algebraic structures that keeps the operations consistent across the mapping. Examples would be log(x*y) = log(x) + log(y). So this preserves the operations between R to R.
An isomorphism is a mapping whose inverse is also a homomorphism (i.e. a bijective homomorphism). So the operation is preserved perfectly going in the opposite direction.
A homeomorphism is an isomorphism on topologies, specifically. | 135 | 557 | {"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-2022-49 | latest | en | 0.931973 |
https://math.stackexchange.com/questions/3529794/why-is-sum-p-in-s-n-2cp-equal-to-n1/3529849 | 1,685,907,751,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650264.9/warc/CC-MAIN-20230604193207-20230604223207-00416.warc.gz | 436,769,537 | 38,180 | # Why is $\sum_{p \in S_n} 2^{c(p)}$ equal to $(n+1)!$?
It is obvious that $$\sum_{p \in S_n} 1=n!$$ because it is just counting how many permutations there are of $$n$$ symbols.
But I have also observed that $$\sum_{p \in S_n} 2^{c(p)}=(n+1)!$$, where $$c(p)$$ is the number of cycles of $$p$$.
What is the combinatorial interpretation of this identity?
An example. In $$S_3$$ we have one permutation with 3 cycles, three permutations with 2 cycles and two with 1 cycle. Then $$1\times 2^3+3\times 2^2+2\times 2^1=24=4!$$
• What do you mean by the number of cycles of $p$? Jan 31, 2020 at 21:02
• @RobArthan When a permutation acts on the set $\{1,...,n\}$ it will have invariant subsets where its action is transitive. These are the cycles. Jan 31, 2020 at 21:03
• Take any permutation in $S_n$. Try to create a permutation in $S_{n+1}$. You need to add $n+1$. You want a scheme of adding $n+1$ so that it is a distinct permutation. For each cycle, you either include $n+1$ or you don't (possibly merging cycles). My guess would be that there is a scheme where this produces a distinct element of $S_{n+1}$. Jan 31, 2020 at 21:06
• This also has a probabilistic interpretation - if $\sigma$ is a random permutation in $S_n$, then $E(2^{c(\sigma)}) = n+1$ Jan 31, 2020 at 21:52
It can be seen as an application of Pólya's enumeration theorem.
Let us take $$n$$ beads and colour each of them black or white. We consider two colourings equivalent, if there are the same number of black and white beads. That is, if some permutation $$\sigma\in S_n$$ takes one of the colourings to the other. It is clear that there are then $$n+1$$ distinct colourings (since there can be $$0,1,\ldots, n$$ black beads). Pólya's theorem then says that: $$n+1 = \frac{1}{|S_n|}\sum_{\sigma\in S_n}2^{c(\sigma)} = \frac{1}{n!}\sum_{\sigma\in S_n}2^{c(\sigma)}$$ The $$2$$ in the formula is the number of colours.
I guess this is somewhere between a combinatorial interpretation and just a proof... Depends on your attitude towards Pólya's theorem I guess.
EDIT: Let me just include the immediate generalisation. If we have $$m$$ colours, then there are $$\binom{n+m-1}{n}$$ different colourings (by stars and bars). So: $$\sum_{\sigma\in S_n}m^{c(\sigma)} = n!\binom{n+m-1}{n} = \frac{(n+m-1)!}{(m-1)!} = m(m+1)\cdots(n+m-1)$$ This only works for $$m\in\mathbb N$$ of course. But as WE Tutorial School points out in a comment, this shows that the left and right hand sides are equal as polynomials in $$m$$, so the identity is proved for any value of $$m$$ (even complex and whatnot).
The sum in question counts the number of pairs $$(f, p)$$ where $$f$$ is a function from $$\{1, 2, \dots, n\}$$ to $$\{1, 2\}$$, and $$p \in S_n$$ such that $$f \circ p = f$$. I don't know if there is an easy way to see that this is equal to $$(n + 1)!$$, but here is an approach:
First let's verify that the sum does actually count what I say that it does. Suppose that we have already chosen the permutation $$p$$. A function $$f$$ satisfies the conditions above if and only if for each cycle of $$p$$, we have that $$f$$ maps every element of that cycle to the same element of $$\{1, 2\}$$. We can thus determine $$f$$ by choosing the image of each cycle, and there are two options for the image of each cycle, giving us $$2^{c(p)}$$ functions in total.
Let's now instead determine the sum by counting the pairs $$(f, p)$$ such that there are $$k$$ elements of $$\{1, 2, \dots, n\}$$ that are mapped to $$1$$ under $$f$$. There are $$\binom{n}{k}$$ ways to choose which $$k$$ elements these are. There are then $$k!$$ ways to choose how $$p$$ permutes these $$k$$ elements, and $$(n - k)!$$ ways to choose how $$p$$ permutes the remaining elements. This gives us $$\binom{n}{k} k! (n - k)! = n!$$ ways to choose the pair $$(f, p)$$. Noting that $$k$$ can take any value from $$0$$ to $$n$$, this gives us $$(n + 1) n! = (n + 1)!$$ pairs in total.
• I hadn't heard of Polya's Enumeration Theorem before, but now that I have, I think that this may be essentially the same solution as @Milten's. Jan 31, 2020 at 22:03
• For your last paragraph, suppose that $f^{-1}(1)=\{x_1,x_2,\ldots,x_k\}$ and $f^{-1}(2)=\{y_1,y_2,\ldots,y_l\}$ with $x_1<x_2<\ldots<x_k$ and $y_1<y_2<\ldots<y_l$. There is a $1$-$1$ correspondence between the pairs $(f,p)$ and the permutations of $(0,1,2,\ldots,n)$ by taking $$(f,p)\mapsto \big(p(x_1),p(x_2),\ldots,p(x_k),0,p(y_1),p(y_2),\ldots,p(y_l)\big).$$ This is an alternative way to show that there are $(n+1)!$ pairs $(f,p)$. Jan 31, 2020 at 22:36
• If I am not mistaken, your proof also shows that $$\sum_{p\in S_n} r^{c(p)}=\frac{(n+r-1)!}{(r-1)!}$$ for all non-negative integer $r$. Hence, it follows that $$\sum_{p\in S_n} x^{c(p)} =x(x+1)(x+2)\cdots (x+n-1)$$ as a polynomial identity. Jan 31, 2020 at 22:47
• @Dylan Indeed it seems we had pretty much the same idea. I think you basically proved Pólya's theorem in your answer. I would say yours is probably cleaner as a combinatorial proof though :) Feb 1, 2020 at 9:16
The combinatorial class of permutations with the number of cycles marked is
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}( \mathcal{U} \times \textsc{CYC}_{=1}(\mathcal{Z}) + \mathcal{U}\times \textsc{CYC}_{=2}(\mathcal{Z}) + \mathcal{U}\times \textsc{CYC}_{=3}(\mathcal{Z}) + \cdots).$$
This gives the EGF
$$G(z, u) = \exp\left(uz+u\frac{z^2}{2} + u\frac{z^3}{3}+\cdots\right) \\ = \exp\left(u\log\frac{1}{1-z}\right).$$
A permutation on $$n$$ elements and having $$k$$ cycles is represented by $$u^k \frac{z^n}{n!}.$$ For the sum we set $$u=2$$ and obtain
$$H(z) = \exp\left(2\log\frac{1}{1-z}\right) = \frac{1}{(1-z)^2}.$$
We then get for the answer
$$n! [z^n] H(z) = n! [z^n] \frac{1}{(1-z)^2} = n! {n+1\choose 1} = (n+1)!.$$
This is the claim.
Addendum. With two being replaced by $$m$$ we obtain
$$n! [z^n] H(z) = n! [z^n] \frac{1}{(1-z)^m} = n! \times {n+m-1\choose m-1} = n! \times {n+m-1\choose n}.$$
• I apologise for using comments in this unintended way, but: I just wanted to tell you that this answer inspired me to begin reading Analytic Combinatorics by Flajolet & Sedgewick. I am now 200 pages deep and I love it. I had never heard of these methods before, so thank you :) Sep 3, 2020 at 9:05 | 2,129 | 6,293 | {"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": 61, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-23 | latest | en | 0.89269 |
http://algorithms.tutorialhorizon.com/category/backtracking/ | 1,490,540,001,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189242.54/warc/CC-MAIN-20170322212949-00283-ip-10-233-31-227.ec2.internal.warc.gz | 10,497,400 | 15,518 | # Category: Backtracking
## The Word Break Problem
Objective : Given an string and a dictionary of words, find out if the input string can be broken into a space-separated sequence of one or more dictionary words. Example: dictionary = [“I” , “have”,…
## Backtracking — Knight’s Tour Problem
Objective : A knight’s tour is a sequence of moves of a knight on a chessboard such that the knight visits every square only once. If the knight ends on a square that is…
## Backtracking — Search a Word In a Matrix
Objective : Given a 2D matrix of characters. Check whether the word exist in the matrix or not. If it exists then print its path. All movements are allowed (right, left, up, down and…
## Backtracking — N Queens Problem — Better Solution
Objective : In chess, a queen can move as far as she pleases, horizontally, vertically, or diagonally. A chess board has 8 rows and 8 columns. The standard 8 by 8 Queen’s problem asks…
## Backtracking — N Queens Problem
Objective : In chess, a queen can move as far as she pleases, horizontally, vertically, or diagonally. A chess board has 8 rows and 8 columns. The standard 8 by 8 Queen’s problem asks…
## Backtracking — Rat In A Maze Puzzle
Given a maze, NxN matrix. A rat has to find a path from source to destination. maze[0][0] (left top corner)is the source and maze[N-1][N-1](right bottom corner) is destination. There are few cells which are…
## SUDOKU Solver">Backtracking — SUDOKU Solver
SUDOKU Puzzle : The objective is to fill a 9×9 grid with digits so that each column, each row, and each of the nine 3×3 sub-grids that compose the grid (also called “boxes”, “blocks”,…
## Introduction To Backtracking Programming
What is Backtracking Programming?? Recursion is the key in backtracking programming. As the name suggests we backtrack to find the solution. We start with one possible move out of many available moves and try…
## Print All The Permutations Of a String
Objective: Given a String, print all the permutations of it. Input: A String Output: Print all the permutations of a string Example: Input : abc Output: abc acb bac bca cba cab Approach: | 674 | 2,202 | {"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} | 2.875 | 3 | CC-MAIN-2017-13 | longest | en | 0.75125 |
http://www.hsn.uk.net/forum/index.php?app=core&module=search&do=user_activity&search_app=forums&mid=8717&sid=7e048dc84f9848d61618891ff58e674a&search_app_filters%5Bforums%5D%5BsearchInKey%5D=&userMode=content | 1,674,771,679,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494826.88/warc/CC-MAIN-20230126210844-20230127000844-00870.warc.gz | 64,838,882 | 9,397 | ## saudiron's Content
There have been 3 items by saudiron (Search limited from 26-January 22)
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### #101648algebraic functions and graphs
Posted by on 13 December 2008 - 09:42 AM in Mathematics
QUOTE (sammy w @ Dec 13 2008, 02:31 AM) <{POST_SNAPBACK}>
Pg 97, Q12, maths in action higher textbook
hey, i was wondering if anyone could help me out with this question as no one can seem to help me get the correct answer for part c! the diagram should be attached!
12)the gradient of the tightrope must not be more than 1/10 or less than 0.
a) find a formula for L(x), the length of the rope> answer:L= square root of 2500 +(x-2) squared
b) find an expression for the gradient of the rope and use it to dertermine the domain of L(x)>answer:2<=x<=7, for the gradient to be more than 1/10
c) calcualte the range of L(x) correct to 2 decimal place> answer ????????? the answer is 50<=L(x)<= 50.25
All i can guess is that it is if they have rotated the diagram 90 degrees to the right for the range of L(x) and that the 2 rectangles at either end of the diagram have something to do with the extra 0.25, but there are no other no.'s to give this indication?????
if anyone can help that would be great!!!!
many thanks
ps..let me no if the diagram does not upload!
From the answer in (b) the domain of X is 2<=X<=7
The formula for the length of L(X) is square root of 2500 + (X-2)squared
Substituting the values for the domain of X into this formula we have:-
When X=2, L(X) = sqare root of 2500 + (2-2) squared
= square root of 2500 + 0
= 50
When X=7, L(X) = square root of 2500 + (7-2)squared
= square root of 2500 + 5squared
= square root of 2500 +25
= square root of 2525
= 50.25 (to 2 decimal places)
Hence the range of L(X) is 50<=L(X)<=50.25
Hope this is clear.
### #101572simple question
Posted by on 27 October 2008 - 04:09 AM in Problem Questions
Silver nitrate reacts with HCl to give a precipitate of Silver Chloride.
### #101564Differntiation
Posted by on 21 October 2008 - 07:35 AM in Problem Questions
QUOTE (Erin Anderson @ Oct 19 2008, 05:35 PM) <{POST_SNAPBACK}>
I have been trying this question for ages.
1. An yacht club is designed its new flag. the flag consists of a red triangle on a yellow rectangular background. In the yellow rectangle ABCD, AB mesures 8 units and AD mesures 6 units. E and F lie on BC and CD, x units from B and C as shown in the diagram.
a) Show that the area, H square units, of the red triangle AEF is
given by H(x) = 24-4x+1/2 x2 (x squared)
and
b) Hence find the greatest and least possible values of the area of triangle AEF
Its in one of your h/w excersise DIFFERENTIATION HOMEWOEK TM
Calculate the area of the rectangle and then subtract the area of the three triangles formed between rectangle and flag ie triangles ABE, FCE and ADF. Area of a triangle is 1/2 base x altitude
Area of rectangle = 8 * 6 units squared = 48 units squared
Area of triangle ABE = 4X units squared (Base AB = 8 units)
Area of triangle FCE = X/2(6-X) units squared = (3X - ½X^2) units squared
Area of triangle ADF = 3(8-X) units squared = (24 – 3X) units squared
Total area of three triangles = 24 – 3X + 3X - ½X^2 + 4X
= 24 + 4X - ½X^2
Area of flag = Area of rectangle – area of triangles
= 48 – (24 + 4X - ½X^2)
H = 24 – 4X + ½X^2 units squared
Differentiating gives dH/dX = -4 + X
therefore have a turning point which is a minimum at X = 4
Substitute X = 4 in equation for H to give minimum area of 16
X can only have values between 0 and 6, and plot of function H will show maximum area of triangle of 24 square units occurs when X = 0.
Hope this helps | 1,116 | 3,660 | {"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-2023-06 | longest | en | 0.91756 |
https://stats.stackexchange.com/questions/406849/comparing-percentages-with-unequal-sample-sizes | 1,624,138,006,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649731.59/warc/CC-MAIN-20210619203250-20210619233250-00047.warc.gz | 483,476,740 | 38,424 | Comparing percentages with unequal sample sizes
I've spent some time collecting participation data on a schools competition.
To sum it up:
Roughly speaking 1000 schools take part in the initiative every year. Only 10 schools become finalists.
The naked eye shows the % of Independent Schools increases significantly in the Finals.
The pattern is evident in the last 5 editions of the competition.
My intention is to prove independent schools are overrepresented in the finals. ... null hypothesis: both percentages are the same
Question(s)
What sort of statistical analysis shall I carry out to discard the null hypothesis?
Basically I want to compare percentages (% participant indep schools vs % finalist independent schools)...
is the analysis somehow compromised by the highly dissimilar sample sizes (n=1000 vs n=10)?
As I said the pattern is evident in all editions.. year after year... would this be an advantage to overcome the sample size problem?
If I understand your question correctly, a simple approach is to ignore the fact that there are multiple editions of the competition. In this case you want to compare the proportion of independent schools in the finals to the proportion of independent schools in the population of participants.
Assume: Independent schools are 0.35 of the population of participants
Assume: Out of 50 finalist (in the last five episodes), 28 are independent schools.
Then you just have a binomial test, or a chi-square goodness-of-fit test.
The following code will run in R or at: rdrr.io/snippets/
Population = 0.35
n = 50
x = 28
binom.test(x, n, Population)
# Exact binomial test
#
# data: x and n
# number of successes = 28, number of trials = 50, p-value = 0.002713
# alternative hypothesis: true probability of success is not equal to 0.35
# 95 percent confidence interval:
# 0.4125441 0.7000928
# sample estimates:
# probability of success
# 0.56
Counts = c(x, n-x)
Props = c(Population, 1-Population)
# chisq.test(Counts, p = Props)
#
# Chi-squared test for given probabilities
#
# data: Counts
# X-squared = 9.6923, df = 1, p-value = 0.00185
• Thanks for the tip, Sal. Ignoring the fact that there are multiple editions sounds like a plan but bear in mind the population of participants and the proportion of participant independent schools does slightly change from year to year. Do you see this as an obstacle to follow your plan? – Iván Diego May 8 '19 at 13:31
• No, you could use the sum of values for the e.g. five editions I think without issue.... But If desired I'm sure you could take into account the differences in the e.g. five editions with a more complex model, but if that's necessary, maybe the simplest way is to analyze the editions individually. – Sal Mangiafico May 8 '19 at 14:40 | 660 | 2,791 | {"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} | 3.53125 | 4 | CC-MAIN-2021-25 | latest | en | 0.907223 |
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Some Concentration Terms
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This lesson contains the concentration terms weight percent, volume percent and strength with the help of suitable examples
Naveen Chouhan
MSc chemistry|merit holder at RPMT2013| 9yr teaching experience(IITJEE/NEET) MCQ expert|Youtube: NAVEEN CHOUHAN(800+ subscribers)
U
A is the husband of D
in molarity sum why u divide 68 with 34?
sir at 5:15 is 4% the wt.percent of solution? as per the quetion we have to find the wt. % of soln. will it not be 96%?
Naveen Chouhan
7 months ago
480/500
Naveen Chouhan
7 months ago
Solvent =5000-20
Naveen Chouhan
7 months ago
Solvent =500-20
sir at 5:15 is 4% the wt.percent of solution? as per the quetion we have to find the wt. % of soln. will it not be 96%?
1. CONCENTRATION t simply define the amount of solute present in the unit amount of solution. 50ml 50 ml 50 ml water water water 10 g Sugar 20 g sugar 50g Sugar Case it 50ml water 10g sugar sweet Case list-50ml water + 20g sugar = more sweet Case IIIrd = 50ml water + 50g sugar = sweetest ,'. Sweetness amount of sugar present in sol SOME CONCENTRATION TERMS
2. 1. Weight percent-Amount of Solute present in per 100g of solution. Wt% solute-W4%- Wt.of Solute Wt.of solution A 70 = WA x 100 WA+WB Similarly We%-WAYWe Wp-- WA+WB 100 Eg. 10% w/w Nacl, 10g NaCl is present in 100g Sol. 100g of Sol. Contains -10g NaCl
3. f 20g caco, is present in 500g SoC Then find 'Weight present of solution. Ans. WAN X 100 WA+WB 200 500 100 4%
4. 0 If Insulin molecule contain 3.4% solohur by wt. (w/w) then find out the minimum mol wt. of insulin. means 3.4% (w'/w)S 3.4 g S in 100g Sol. Ans. 100g Insuline-3.4gS Or 3.4g S 100g Insulin lg S = g Insulin 100 3.4 alteat one molecule of S must be present 100 3x 32g 3.4 941g insuline
5. (2) Volume Present:- Volume of Solute present in per 100ml of solution (/v) VA- 100 ; VA+VB VA+VB Where VA = Vol. of Solute = vol. of Soluent (Vol. of Solution) Eg, 10% (v/v) HCl 10ml of HCl is present in 100ml of Solution * If the Sol. Is aquous then, solvent is water.
6. Q, you have to prepare 20% (v/v) NaOH aq. SoC 20%(,/v) NaOH:- Ans. 20ml NaOH 100ml Sol. .. Sol. is aq . Solvent water Solute + Solvent = Soin. - 20ml x 100ml x 100 20 ml = 80 ml By adding 80ml water with 20ml NaOH, 20% v/v Solution of Na0H can be prepared
7. (3) StrengthW/v - It is defined as the weight of solute present in per 1000ml of solution Wt. of Solute Strength ()Vol.of Sol. 1000 Unit g/lor g/ml * It is expressed in Vol% of H202 WA 1000 (ml)
8. % VOLUME H 202 % VOLUME H20i-It is defined as the amount of 02 (oxygen) released by unit weight of H202 -10% Vol H2O2 - 10l 02 is released by 1l H202 # 30% vol. H202 301 02 is released by 1l H202
9. Proof: 2H202 2H20 +02 2 mol1 Mol 2 34g 22.41 68g 22.4l gives 68g H202> 22.4l 02 22.4 1g H2O2 -68 22.41 02 68 g H202
10. 68 11 02 22.4 68x x l02 22.4 68x 22.4 68 34 8g = mol/l or 2mol/l 2x =-mol/l 22.4 11.2 M =-mol/l 11.2 | 1,130 | 3,048 | {"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-2019-39 | longest | en | 0.805815 |
http://mathoverflow.net/revisions/107069/list | 1,368,983,795,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697843948/warc/CC-MAIN-20130516095043-00030-ip-10-60-113-184.ec2.internal.warc.gz | 165,419,976 | 4,994 | Take two graphs (of bounded valency) or manifolds (f bounded geometry) $G$ and $G'$. Assume there is a quasi-isometry $f:G \to G'$, and assume the Poisson or Martin boundary of $G$ is known, what may one say about the Poisson or Martin boundary of $G'$ (or how does $f$ transforms Poisson or Martin boundaries)?
EDIT: a map $f:X \to Y$ between two metric spaces $(X,d_X)$ and $(Y,d_Y)$ is a quasi-isometry if $\exists C \geq 0$ and $D \geq 1$ such that
• $\forall x,x' \in X$, $D^{-1} d_Y \big( f(x), f(x') \big) - C \leq d_X(x,x') \leq D d_Y \big( f(x), f(x') \big) + C$
• $\forall y \in Y, d_Y \big(f(X),y \big) < C$
1
How does a quasi-isometry affect Poisson or Martin boundaries?
Take two graphs (of bounded valency) or manifolds (f bounded geometry) $G$ and $G'$. Assume there is a quasi-isometry $f:G \to G'$, and assume the Poisson or Martin boundary of $G$ is known, what may one say about the Poisson or Martin boundary of $G'$ (or how does $f$ transforms Poisson or Martin boundaries)? | 323 | 1,001 | {"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} | 2.546875 | 3 | CC-MAIN-2013-20 | latest | en | 0.794321 |
https://github.com/ndpar/algorithms/blob/master/p006.erl | 1,511,489,941,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807056.67/warc/CC-MAIN-20171124012912-20171124032912-00240.warc.gz | 608,357,890 | 11,700 | # ndpar/algorithms
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%% Problem %% --------------------- %% The sum of the squares of the first ten natural numbers is, %% 1^2 + 2^2 + ... + 10^2 = 385 %% The square of the sum of the first ten natural numbers is, %% (1 + 2 + ... + 10)^2 = 55^2 = 3025 %% Hence the difference between the sum of the squares of the %% first ten natural numbers and the square of the sum is 3025 - 385 = 2640. %% %% Find the difference between the sum of the squares of the %% first one hundred natural numbers and the square of the sum. %% --------------------- -module(p006). -export([solve/0]). -include_lib("eunit/include/eunit.hrl"). solve() -> diff2(100). %% Solution (slow) %% --------------------- diff(Max) -> 2 * lists:sum([M * N || M <- lists:seq(1, Max), N <- lists:seq(1, Max), M < N]). %% Solution (fast) %% --------------------- diff2(N) -> N * (N + 1) * (3*N + 2) * (N - 1) div 12. %% Tests diff_1_test() -> ?assertEqual(0, diff(1)). diff_2_test() -> ?assertEqual(4, diff(2)). diff_10_test() -> ?assertEqual(2640, diff(10)). diff_100_test() -> ?assertEqual(25164150, diff(100)). | 365 | 1,204 | {"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} | 3 | 3 | CC-MAIN-2017-47 | latest | en | 0.580912 |
www.engreen.eu | 1,571,218,489,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666959.47/warc/CC-MAIN-20191016090425-20191016113925-00355.warc.gz | 254,669,059 | 12,710 | # Uncommon Article Gives You the Facts on In Math What Is Mean That Only a Few People Know Exist
## Lies You’ve Been Told About in Math What Is Mean
Indeed, the majority of these notations are used unchanged in a number of other contexts. Our lesson planning worksheet is able to assist you estimate. So the significance of the is the mixture of the situations where it’s appropriate and the situations where it isn’t appropriate.
Each step ought to be complete. There are lots of situations where it’s important that you know the relative size of a single number to another, as an example, http://www.martalaux.com/blog/the-most-ignored-fact-about-mathematics-calculator-uncovered/ when it has to do with money. Besides calculating the mean for a particular set of information values, you may apply your knowledge of the mean to determine other information that could be asked for in everyday issues.
Forgetfulness may be normal part of aging. Knowing what these terms mean is a great beginning to understanding what your child requirements. That’s among the most useful ways you’ll be able to spend your time, as it will dramatically boost the potency of the lecture.
## The Bad Side of in Math What Is Mean
After these preliminaries, we are now able to get other in the meat of the subject. Mathematicians often discuss the attractiveness of a specific proof or mathematical outcome. After they have had a chance to consider their own ideas for categorizing snowflakes, see if they can categorize an assortment of snowflake photographs using a snowflake chart.
A category of equivalent characteristics is believed to be a type. Possessing both numbers measured with the exact units is known as homogeneous units. So, an ordinary score is really a variety of numbers.
## The Fight Against in Math What Is Mean
As these questions often appear straightforward, it can be simple to find yourself rushing through them. So what someone figured out is you don’t need formulas in any respect. The response is also known as the difference.
Then multiply the perfect side by 4 until you get the same sum of terms on the left side. It’s only the difference between them. A mean is usually called an average.
Frequently, it’s actually less difficult to add intead of subtract, and figure out the number of times you AussieEssaywriter.com.au/ will add the number (divisor) until you get to the dividend. The sum of the goods divided by the overall number of values is going to be the value of the mean. Between each pair, put the proper indication of inequality.
Arithmetic mean is the sum of information divided by the range of data-points. Anyway mathematics is just a tool of white supremacy. The equation evaluates the period of the string with respect to its wavelength.
The end result is that the Unicode linear format is quite a general mathematical notation, but it’s not a context-free grammar. It’s the closest to a true mathematical notation that I was able to define. There’s other notation that you might encounter.
## Key Pieces of in Math What Is Mean
Normally, these types of tests function as part of a extensive evaluation. The mean is frequently used as a summary statistic. There’s constant stress because of subject exams Almost every second, students face numerous problems or difficulties they need to overcome as a way to get the wanted rating.
In a couple weeks, counselors are going to have the true test booklets. There’s a difference between a novice and a specialist. It is going to also incorporate information on the subject of the dependability of the test and the typical deviations.
If there’s an odd number of information values then the median is going to be the value in the center. Symmetrical data sets are balanced on both sides of the median. From this instance, you can observe that the mean of a population and that of a sample from the population aren’t necessarily the exact same.
It’s beneficial in assessing the importance of a specific number in a statistical survey. One issue with the variance is that it doesn’t have the very same unit of measure as the original data. Then you select the biggest factor that may be seen in each number.
## The Hidden Secret of in Math What Is Mean
Shahena It is an amazing website. You would like a figure to appear reasonably near the point in the text in which you refer to it, but definitely not in an earlier page. In each instance, the sign opens towards the bigger number.
You may walk all of the way around the green line and it will probably remain perpendicular. The procedure is performed by adding up all the numbers in a set of information. Well, now’s the opportunity to utilize PIA!
## The In Math What Is Mean Game
The very first reference to PEMDAS is not easy to pin down. Some series don’t have a mode, since there aren’t any repeating numbers. A great instance of this kind of symbol is supplied by parentheses, which indicate the order in which arithmetical operations should be carried out.
Possibly the best source of error, however, is the usage of variables without definitions. The only fault is in the name. It is also feasible to have a whole set of data without a mode.
## The Number One Question You Must Ask for in Math What Is Mean
You ought to know that professional graduate schools in medicine, law, and business think mathematics is a superb major since it develops analytical abilities and the capability to work in a problem-solving atmosphere. Our on-line assignment help services are very extensive and cover all kinds of homework help needed by students. College course numbers may mean various things based on the individual institution.
Problems have developed for some students who don’t know how to sign in properly to acquire their scores, she explained. Math works just like anything else, if you would like to become good at it, then you will need to practice it. It refers to division.
These terms may be used to specify a single half and one third respectively. Obviously, the other order yields the identical result. You are going to be expected to understand the amount of days in a calendar year, the range of hours in every day, and other primary units of measure. | 1,258 | 6,220 | {"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} | 3.703125 | 4 | CC-MAIN-2019-43 | latest | en | 0.926754 |
https://atcoder.jp/contests/code-festival-2015-okinawa-open/tasks/code_festival_2015_okinawa_e | 1,547,652,133,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657510.42/warc/CC-MAIN-20190116134421-20190116160421-00616.warc.gz | 467,192,257 | 4,717 | Contest Duration: ~ (local time)
E - Enormous XOR Rectangle
Time Limit: 2 sec / Memory Limit: 256 MB
### Problem
Cat Snuke received a paper that has been divided into grids having H horizontal rows and W vertical columns as a birthday gift. For each block a number is written as shown in the figure below.
More precisely, the number, which is written in the i_{th} row from the top of the rectangle paper and j_{th} column from the left of the rectangle paper, is (i - 1) \times W + (j - 1).
Cat Snuke wants to choose a partial rectangle from this rectangle paper, then find the value obtained by bitwise xor for all the numbers in this partial rectangle. Specifically, Cat Snuke will choose integers t,b(1≦t≦b≦H),l,r(1≦l≦r≦W), then calculate the value obtained by bitwise xor for all the numbers in the area from the top of the rectangle between the t_{th} row and the b_{th} row (including both ends), form the left of the rectangle between the l_{th} column and r_{th} column (including both ends).
Cat Snuke is able to choose any values for t,b,l,r. Please calculate the maximum value that can be obtained.
### Input
Inputs will be given by standard input in following format
H W
• At the first line, H(1≦H≦1,000,000,000), W(1≦W≦1,000,000,000) , will be given divided by space.
### Output
Please calculate the maximum value that can be obtained, then output it in one line.
Print a newline at the end of output.
### Input Example 1
4 5
### Output Example 1
31
For example, the obtained value of t=3,b=4,l=3,r=5 is 12 xor 13 xor 14 xor 17 xor 18 xor 19 = 31, this is the maximum value that can be obtained in this case.
### Input Example 2
314159265 358979323
### Output Example 2
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# [sci.astro] Time (Astronomy Frequently Asked Questions) (3/9)Section - C.10 Why isn't the earliest Sunrise (and latest Sunset) on the longest day of the year?
( Part0 - Part1 - Part2 - Part3 - Part4 - Part5 - Part6 - Part7 - Part8 - Single Page )
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```Author: Steve Willner <willner@cfa183.harvard.edu>
This phenomenon is called the "equation of time." This is just a
fancy name for the fact that the Sun's speed along the Earth's equator
is not constant. In other words, if you were to measure the Sun's
position at exactly noon every day, you would see not only the
familiar north-south change that goes with the seasons but also an
east-west change in the Sun's position. A graphical representation of
both positional changes is the analemma, that funny figure 8 that most
globes stick in the middle of the Pacific ocean.
The short explanation of the equation of time is that it has two
causes. The slightly larger effect comes from the obliquity of the
ecliptic---the Earth's equator is tilted with respect to the orbital
plane. Constant speed along the ecliptic---which is how the "mean
sun" moves---translates to varying speed in right ascension (along the
equator). This gives the overall figure 8 shape of the analemma.
Almost as large is the fact that the Earth's orbit is not circular,
and the Sun's angular speed along the ecliptic is therefore not
constant. This gives the inequality between the two lobes of the
figure 8.
Some additional discussion, with illustrations, is provided by Nick
Strobel at <URL:http://www.astronomynotes.com/nakedeye/s9.htm>, though
<URL:http://www.astronomynotes.com/nakedeye/s7.htm>. Mattthias
Reinsch provides an analytic expression for determining the number of
days between the winter solstice and the day of the latest sunrise for
Northern Hemisphere observers,
<URL:http://arXiv.org/abs/astro-ph/?0201074>.
The Earth's analemma will change with time as the Earth's orbital
parameters change. This is described by Bernard Oliver (1972 July,
_Sky and Telescope_, pp. 20--22)
An article by David Harvey (1982 March, _Sky and Telescope_,
pp. 237--239) shows the analemmas of all nine planets. A simulation
of the Martian analemma is at
<URL:http://apod.gsfc.nasa.gov/apod/ap030626.html>, and illustrations
of other planetary analemmas is at <URL:http://www.analemma.com/>.
```
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Case Scenario:
Bolster Foods' (BF) balance sheet shows a total of \$25 million long-term debt with a coupon rate of 8.50%. The yield to maturity on this debt is 8.00%, and the debt has a total current market value of \$27 million. The balance sheet also shows that the company has 10 million shares of stock, and the stock has a book value per share of \$5.00. The current stock price is \$20.00 per share, and stockholders' required rate of return, rs, is 12.25%. The company recently decided that its target capital structure should have 35% debt, with the balance being common equity. The tax rate is 40%. Calculate WACCs based on book, market, and target capital structures. What is the sum of these three WACCs?
a. 32.00%
b. 30.77%
c. 33.28%
d. 29.54%
e. 28.36%
#### Solution Preview :
##### Reference No:- TGS01745156
Now Priced at \$20 (50% Discount)
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Rated (4.6/5) | 261 | 957 | {"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} | 2.671875 | 3 | CC-MAIN-2019-43 | latest | en | 0.932136 |
https://idoc.pub/documents/02c-probabilistic-inventory-models-1-ylyxr00v7vnm | 1,669,565,936,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00860.warc.gz | 364,878,745 | 7,358 | # 02c Probabilistic Inventory Models (1)
• December 2019
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Probabilistic Inventory Models (Part – III)
1
Outline Probabilistic Model Single period Discrete demand
Single period Continuous demand
2
Introduction The major influencing factors for the inventory decisions are price, demand, and lead time. Other factors such as ordering cost, carrying cost, and stock-out cost are also affecting the inventory decisions but their nature is not so much disturbing. Sometimes price fluctuations are too much in the market and hence it influences inventory decisions. Similarly, variability in demand or consumption of an item as well as the variability in lead time influence the overall inventory policy.
Probabilistic Inventory Models
Probabilistic Inventory Models Probabilistic inventory models: I.
Single period probabilistic models: The single period inventory model applies to inventory situations in which only one order is placed for goods in anticipation of a future selling season where demand is uncertain. At the end of the period the product has either sold out or there is a surplus of unsold items to sell at salvage value.
II. Multi-period probabilistic models: Multi-period inventory models are usually addressed in finding the minimum expected cost over N periods. In these models both demand and lead time are uncertain. 5 Sasadhar Bera, IIM Ranchi
Single Period Probabilistic Models In single period inventory situation, the only question is how much of the product to order at the start of the period. Newspaper sales are a typical example of the single period model. Thus the single period inventory problem is sometimes referred to as the newsvendor problem.
These models deal with inventory situation of the items such as perishable goods, spare parts, and seasonal goods requiring one time purchase only.
Single Period Probabilistic Models (Contd.) Applications of single period model: i.
Order size of Periodicals, such as newspapers and magazines.
ii.
Order size seasonal greeting cards.
iii. Order size Christmas trees.
iv. Seasonal clothing, such as winter coats, where any goods remaining at the end of the season must be sold at highly discounted prices to clear space for the next season. v.
Vital spare parts that must be produced during the last production run of a certain model of a product (e.g., an airplane) for use as needed throughout the lengthy field life of that model.
vi. Reservations provided by an airline for a particular flight.
vii. Reservations of hotel room. 7 Sasadhar Bera, IIM Ranchi
Single Period Probabilistic Models (Contd.) The demand for such item may be discrete or continuous. Since for a given period, purchase of item is made only once, the lead time factor is least important in these models. In such cases, there are two types of costs involved: Over-stocking cost: Loss due to excess stocks Under-stocking cost: Opportunity losses due to high demand D = Demand of an item in units (a random variable) Q = the number of units stocked (or to be purchased) P = Purchase price per unit S = Selling price per unit Ch = Carrying cost or holding cost per unit for entire period Cs = Shortage cost per unit V = Salvage value per unit Sasadhar Bera, IIM Ranchi
8
Single Period Probabilistic Models (Contd.) Co = Over-stocking cost per unit = Loss associated with each unit left unsold = P + Ch – V Cu = Under-stocking cost per unit = Loss due to not meeting demand = S – P + Cs The single period can be solved using a technique called marginal economic analysis, which compares the cost or loss of ordering one additional item with the cost or loss of not ordering one additional item. 9 Sasadhar Bera, IIM Ranchi
Single Period Discrete Demand Model
Single Period Discrete Probabilistic Models
Single Period Discrete Probabilistic Models (Contd.)
Example: Discrete Demand Probability Example: Demand follows discrete distribution A newspaper boy buys papers for Rs. 2 each and sells them for Rs. 2.50 each. He cannot return unsold newspapers. Daily demand has the following distribution Nos of customers Probability
230 240 250 260 270 280 290 300 310 320 0.01 0.03 0.06 0.10 0.2 0.25 0.15 0.1 0.05 0.05
If each day’s demand is independent of the previous day’s demand, how many papers should he order each day?
Example: Discrete Demand Probability (Contd.)
Single Period Continuous Demand Model
Single Period Continuous Probabilistic Models
Single Period Continuous Probabilistic Models (Contd.)
Example: Continuous Demand Probability Example: Demand follows uniform distribution An item sells for Rs. 20 per unit and cost Rs. 10. Unsold items can be sold for Rs. 4 each. It is assumed that there is no shortage penalty cost besides the lost revenue. The demand is known to be any value between 500 and 1200 items. Determine the optimal number of units to the item to be stocked.
Example: Continuous Demand Probability (Contd.)
Example: Continuous Demand Probability (Contd.)
December 2019 61
November 2021 0
December 2019 52
November 2019 56
December 2019 32
April 2020 16
September 2020 0
November 2019 51
December 2019 61
November 2019 56
August 2021 0
November 2019 43 | 1,228 | 5,423 | {"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} | 3 | 3 | CC-MAIN-2022-49 | latest | en | 0.934335 |
https://www.hackmath.net/en/example/5518 | 1,558,818,982,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258451.88/warc/CC-MAIN-20190525204936-20190525230936-00114.warc.gz | 797,777,270 | 6,801 | # Substitution
solve equations by substitution:
x+y= 11
y=5x-25
Result
x = 6
y = 5
#### Solution:
x+y= 11
y=5x-25
x+y = 11
5x-y = 25
x = 6
y = 5
Calculated by our linear equations calculator.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation?
## Next similar examples:
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Dalibor and Adam together scored 97 goals in the season. Adam scored 9 goals more than Dalibor. How many goals scored each?
3. Family parcels
In father will he divided the land so that the older son had three bigger part than younger son. Later elder son gave 2.5 ha field to younger and they had both the same. Determine the area of family parcel.
4. Summerjob
Three students participated in the summerjob. Altogether they earn 1780, -. Peter got a third less than John and Paul got 100,- more than Peter. How much got every one of them?
5. Football match 4
In a football match with the Italy lost 3 goals with Germans. Totally fell 5 goals in the match. Determine the number of goals of Italy and Germany.
6. 13 tickets
A short and long sightseeing tour is possible at the castle. Ticket for a short sightseeing circuit costs CZK 60, for a long touring circuit costs CZK 100. So far, 13 tickets have been sold for 1140 CZK. How much did you pay for tickets for a short tour?
7. Dining room
The dining room has 11 tables (six and eight seats). In total there are 78 seats in the dining room. How many are six-and eight-seat tables?
8. Classroom
There are eighty more girls in the class than boys. Boys are 40 percent and girls are 60 percent. How many are boys and how many girls?
9. Blackberries
Daniel, Jolana and Stano collected together 34 blackberries. Daniel collected 8 blackberries more than Jolana, Jolana 4 more than Stano. Determine the number blackberries each collected .
10. ATC camp
The owner of the campsite offers 79 places in 22 cabins. How many of them are triple and quadruple?
11. Belgium vs Italy
Belgium played a match with Italy and Belgium win by 2 goals. The match fell a total 6 goals. Determine the number of goals scored by Belgium and by Italy.
12. Viju
viju has 40 chickens and rabbits. If in all there are 90 legs. How many rabbits are there with viju??
13. Two numbers
We have two numbers. Their sum is 140. One-fifth of the first number is equal to half the second number. Determine those unknown numbers.
14. Three brothers
The three brothers have a total of 42 years. Jan is five years younger than Peter and Peter is 2 years younger than Michael. How many years has each of them?
15. Mushrooms
Eva and Jane collected 114 mushrooms together. Eve found twice as much as Jane. How many mushrooms found each of them?
16. Trees
Along the road were planted 250 trees of two types. Cherry for 60 CZK apiece and apple 50 CZK apiece. The entire plantation cost 12,800 CZK. How many was cherries and apples?
17. Fifth of the number
The fifth of the number is by 24 less than that number. What is the number? | 853 | 3,282 | {"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} | 4.15625 | 4 | CC-MAIN-2019-22 | latest | en | 0.942794 |
https://www.scribd.com/document/149416679/Math-10-Sample-Final | 1,575,886,884,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540518627.72/warc/CC-MAIN-20191209093227-20191209121227-00254.warc.gz | 850,788,833 | 70,372 | You are on page 1of 9
# MATH 10 SAMPLE FINAL EXAM Answers are on the last page at the bottom I.
USE THE INFORMATION BELOW TO ANSWER PROBLEMS 1 THROUGH 3. Among 100 marriage license applications, chosen at random in 1971, there were 8 in which the women were at least one year older than the men, and among 30 marriage license applications, chosen at random in 1977, there were 6 in which the women were at least one year older than the men. Assume you are interested in a 80% confidence interval for the difference between the corresponding true proportions of marriage license applications in which the women were at least one year older than the men. 1. Choose the best answer below concerning a confidence interval for the difference between the corresponding true proportions of marriage license applications in which the women were at least one year older than the men.
s a) x z n
z b) p
2 s1 s2 + 2 n1 n2
q p n
1q 1 p q p + 2 2 n1 n2
c) (x1 x 2 ) z
1 p 2 ) z d) ( p
e) None of the above formulas are appropriate. 2. Choose from the following concerning the z-value or t-value that would be used in the 80% confidence interval for the difference between the corresponding true proportions of marriage license applications in which the women were at least one year older than the men. a) z = 2.326 c) z = 1.282 e) None of the above are correct. b) t = 2.617 d) t = 1.645
3. If we assign group 1 to be the sample from 1971 and group 2 to be the sample from 1977, you may assume that the appropriate confidence interval for the above problem to be (-1.572, -0.020). Based on this interval choose the most appropriate portion below from the conclusion for the confidence interval. a) We are 80% confident that there is no difference between the corresponding true proportions of marriage license applications in which the women were at least one year older than the men. b) We are 80% confident that the true proportion of marriage licenses in which the women were at least one year older than the men was greater in 1971 than in 1977. c) We are 80% confident that the true proportion of marriage licenses in which the women were at least one year older than the men was greater in 1977 than in 1971. d) Cant answer this question based on the information given. e) None of the above conclusions are correct.
II. USE THE INFORMATION BELOW TO ANSWER PROBLEMS 4 THROUGH 6. In a study of heart surgery, one issue was the effect of drugs called beta-blockers on the pulse rate of patients during surgery. The available subjects were divided at random into two groups of 21 patients each. One group received a beta-blocker; the other a placebo. The surgical team recorded the pulse rate of each patient at a critical point during the operation. The treatment group had mean 65.2 beats per minute. For the control group, the mean was 70.3. Below is Minitab output for the beta-blocker data. Assume equal population variances and assume the populations are independent and normally distributed.
TWOSAMPLE CI FOR TREATMNT VS CONTROL N MEAN TREATMNT 21 65.20 CONTROL 21 70.30 95 PCT CI FOR MU TREATMNT - MU CONTROL: (-1.3, 5.47)
4. Choose the best interpretation below for the confidence interval in the Minitab output. a) We are 95% confident that the true mean number of beats per minute during surgery for patients that receive a beta-blocker is greater than the true mean number of beats per minute during surgery for patients that do not receive a beta-blocker. b) We are 95% confident that there is no statistically significant difference between the true mean beats per minute during surgery for the two groups. c) We are 95% confident that the true mean number of beats per minute during surgery for patients that receive a beta-blocker is less than the true mean number of beats per minute during surgery for patients that do not receive a beta-blocker. d) Cant answer this question based on the information given. e) None of the above conclusions are correct. 5. Pick the formula used to calculate the confidence interval above. s 1 1 a) x z b) ( x1 x2 ) t s 2 + p n n1 n2 c) (x1 x 2 ) z
2 s1 s2 + 2 n1 n2
1 p 2) z d) (p
1q 1 p q p + 2 2 n1 n2
e) None of the above formulas are appropriate. 6. Pick the appropriate z or t value used in the above interval. a) b) c) d) e) 1.645 1.684 1.960 2.021 None of the above values are correct.
III. USE THE INFORMATION BELOW TO ANSWER PROBLEMS 7 THROUGH 11. A researcher wants to test whether the proportion of Foothill students who transfer to a four-year university is different than the proportion of DeAnza students who transfer to a four-year university. 7. Pick the most appropriate answer below concerning the formulation of the null and alternative hypotheses to determine whether the data provide sufficient evidence to conclude that the proportion of Foothill students who transfer to a four-year university is different than the proportion of DeAnza students who transfer to a four-year university. a) b) H : = 0 c) H o : 1 2 = 0 H o : p1 p2 = 0 o 1 2
H A : 1 2 0
HA : 1 2 < 0
e)
H A : p1 p2 0
None of the above are correct.
* 8. If the test statistic for the test is z = 1.48 , what would be the p-value for the test?
c) 0.8612
## d) Cant answer this question.
* 9. If the test statistic for the test is z = 1.48 , then, at the 5% significance level, what would be your decision for the test?
a) reject the null hypothesis c) accept the null hypothesis e) Not enough information is given to make a decision.
## b) fail to reject the null hypothesis d) reject the alternative hypothesis
10. Choose the most appropriate portion below from the conclusion for the hypothesis test. a) At the 5% significance level, there is sufficient evidence to conclude that the true proportion of Foothill students who transfer to a four-year university is different than the true proportion of DeAnza students who transfer to a four-year university. b) At the 5% significance level, there is insufficient evidence to conclude that the true proportion of Foothill students who transfer to a four-year university is different than the true proportion of DeAnza students who transfer to a four-year university. c) None of the above are correct. 11. Pick the most appropriate equation that should be used to calculate the test statistic for the test. (You may assume that a z-value is appropriate).
a)
z* =
p1 p2 1 1 q + p n 1 n 2 1 p 2 p 1 p q p1q + 2 2 n1 n 2
b)
z* =
x1 x2
2 s1 s2 + 2 n1 n 2
c)
z* =
d)
z* =
x1 x2 1 + 1 s2 p n 1 n 2
## e) None of the above are correct.
IV. USE THE INFORMATION BELOW TO ANSWER PROBLEMS 12 THRU 15. A major car manufacturer wants to test a new engine to see whether it meets new air pollution standards. The emission of all engines of this type must be less than 20 parts per million(ppm) of carbon. 21 engines are manufactured for testing purposes, and the mean and standard deviation of the emissions for this sample of engines are determined to be 16.46 ppm and 9.49 ppm, respectively. You are interested in whether the data supply sufficient evidence to allow the manufacturer to conclude that this type of engine meets the pollution standard at a 10% level of significance. Below is the appropriate Minitab output for the problem. Assume a normal distribution. CARBON N 21 MEAN 16.46 STDEV 9.49 SE MEAN 2.07 T -1.71 P VALUE 0.051
12. Pick the most appropriate answer below concerning the formulation of the null and alternative hypotheses to determine whether this type of engine meets the pollution standard.
a)
Ho : p = 20 HA : p < 20 Ho : = 20 HA : < 20
b)
Ho : = 20 HA : > 20 Ho : p = 20 HA : p > 20
c)
d)
e) None of the above are correct. 13. What is the smallest significance level that would allow us to conclude HA is true? a) -1.71 b) 0.10 c) 0.06 d) 0.01 14. Choose the most appropriate portion below from the conclusion for the hypothesis test in problem twelve. a) At the 10% significance level, there is sufficient evidence to conclude that this type of engine meets the pollution standard. b) At the 10% significance level, there is insufficient evidence to conclude that this type of engine meets the pollution standard. c) Not enough information is given to make a conclusion. d) None of the above are correct. 15. Pick the most appropriate equation that should be used to calculate the test statistic for the test in problem twelve.
a)
z* =
x o s n
b)
z* =
p po p oq o n
c)
t* =
x o s n
d)
t* =
po p poqo n
## e) None of the above are correct.
V. USE THE INFORMATION BELOW TO ANSWER PROBLEMS 16 THRU 18. Suppose you record how long it takes you to get to school over many months and discover that the one-way travel times, in minutes, are approximately normally distributed with a mean of 13.88 minutes and standard deviation of 4 minutes. If your first class is at 10:30 am, what is the latest time you should leave everyday so that you are on time at least 93.7% of the time? 16. Pick the equation most appropriate to answer the question above.
## x n x b) p( x ) = ( nCx ) p x (1 p) e) None of the above are correct.
a) z =
c) x = + z
d) z =
x n
17. Select the most appropriate item that pertains to the problem
a) z = 1.530
b) z = 1.645
c) z = 1.960
d) z = 2.576
e) None of these.
18. If your first class is at 10:30 am, what is the latest time you should leave everyday so that you are on time at least 93.7% of the time? a) 10:22 am b) 10:00 am c) 10:10 am d) 10:20 am e) None of these.
VI. USE THE INFORMATION BELOW TO ANSWER PROBLEMS 19 THRU 21. Based on data from the Statistical Abstract of the United States, the probability that a newborn baby will be a boy is roughly 0.513. In the next three births, you are interested in the probability that at least two are girls. 19. Pick the equation most appropriate to deal with the problem above.
## x n x b) p( x ) = ( nCx ) p x (1 p) e) None of the above are correct.
a) z =
c) x = + z
d) z =
x n
20. Select the most appropriate item that pertains to the problem a) z = 1.539 b) z = 1.645 c) z = 1.960 d) z = 2.225
e) None of these.
21. In the next three births, what is the probability that at least two are girls? a) 0.6667 b) 0.3333 c) 0.4805 d) 0.5195 e) None of these.
VII.
## e) None of the above are correct.
23. was shown on TBS during a prime-time segment? a) 518/917 b) 66/917 c) 113/518 d) 113/917
## e) None of the above are correct.
24. was not shown on a cable network? a) 400/917 b) 517/917 c) 113/917 d) 163/917
## e) None of the above are correct.
25. was shown on ABC during a late-night segment, given that it was shown on a major network? a) 69/917 b) 69/399 c) 69/517 d) 69/163
## e) None of the above are correct.
VIII. USE THE INFORMATION BELOW TO ANSWER PROBLEMS 26 THROUGH 28. Electric power plants that use water for cooling their condensers sometimes discharge heated water into rivers, lakes, or oceans. It is known that water heated above certain temperatures has a detrimental effect on the plant and animal life in the water. Suppose it is known that the increased temperature of the heated water discharged by a certain power plant in any given day has a normal distribution with a mean of 2.8 degrees Celsius and a standard deviation of 0.6 degrees Celsius. Suppose a sample of 35 days was collected. You are interested in the probability that the mean increase in temperature of the discharged water for the sample exceeds 2.6 degrees Celsius. 26. Pick the equation most appropriate to deal with the problem above.
## x n x b) p( x ) = ( nCx ) p x (1 p) e) None of the above are correct.
a) z =
c) x = + z
d) z =
x n
27. Select the most appropriate item that pertains to the problem
a) z = -2.58 b) z = -1.97 c) z = -0.056 d) z = -0.33 e) None of the above are pertinent to the problem. 28. What is the probability that the sample mean increase in temperature of the discharged water for the sample exceeds 2.6 degrees Celsius. a) .9951 b) .6293 c) .9437 d) .9756 e) None of the above are correct.
IX. USE THE INFORMATION BELOW TO ANSWER PROBLEMS 29 AND 30. The Acculturation Rating Scale for Mexican Americans (ARSMA) measures the extent to which Mexican Americans have adopted Anglo/English culture. During the development of ARSMA, the test was given to a group of 27, randomly chosen, Mexican Americans. Their scores (taken from a normal population), from a possible range of 1.00 to 5.00, had a sample mean of 1.67 points and a sample standard deviation of 0.25 points. Because low scores should indicate a Mexican cultural orientation, these results helped to establish the validity of the test. You are interested in a 90% confidence interval for the true mean ARSMA score of Mexican Americans. 29. Choose the best answer below concerning the 90% confidence interval for the true mean ARSMA score of Mexican Americans.
1 p p s s p z a) x z b) c) x t n n n d) Cant answer this question based on the information given. e) None of the above are correct.
30. Choose the appropriate interval from below for the 90% confidence interval for the true mean ARSMA score of Mexican Americans. a) b) c) d) e) (1.576, 1.764) (1.588, 1.752) (1.591, 1.749) Cant answer this question based on the information given. None of the above are correct.
X. USE THE INFORMATION BELOW TO ANSWER PROBLEMS 31 THRU 34 Acme Properties, Inc. specializes in custom home sales in the Evergreen Estates. A random sample of nine currently listed custom homes provided information on size and price. The size data are in hundreds of square feet, rounded to the nearest hundred; the price data are in thousands of dollars, rounded to the nearest thousand. Use the output below to answer the following questions:
Correlation of SIZE and PRICE = 0.930 ROW 1 2 3 4 5 6 7 8 9 SIZE 26 27 33 29 29 34 30 40 22 PRICE 235 249 267 269 295 345 305 475 195 MTB > Regress 'PRICE' 1 'SIZE'. The regression equation is PRICE = - 141 + 14.4 SIZE Predictor Constant SIZE s = 31.77 Coef -141 14.4 Stdev 65.71 2.162 t-ratio -2.14 6.68 p 0.070 0.000
R-sq = 86.4%
31. Predict the price of a custom home in the Evergreen Estates that is 29 hundred square feet. a) \$269,000 b) \$295,000 c) \$276,600 d) \$558,600 e) None of the above are correct. 32. Predict the price of a custom home in the Evergreen Estates that is 47 hundred square feet. a) \$475,000 b) \$535,800 c) \$817,800 d) \$195,000 e) None of the above are correct. 33. State the value of slope of the regression line. a) -141 b) 14.4 c) 0.930 d) 65.71 e) None of the above are correct. 34. Interpret the correlation coefficient. a) There is a strong negative linear relationship between square footage and sale price. b) There is a strong positive linear relationship between square footage and sale price. c) There is a weak negative linear relationship between square footage and sale price. d) There is a weak positive linear relationship between square footage and sale price. e) None of the above are correct.
Problem 35 is a stand-alone problem and has nothing to do with the previous problem. 35. What can we say about the percentage of data that lies within 2 standard deviations of the mean? a) Approximately 95% b) Approximately 75% c) At least 75% d) At most 25%
XI. USE THE INFORMATION BELOW TO ANSWER PROBLEMS 36 and 37 Suppose you record how long it takes you to get to school over many months and discover that the one-way travel times, in minutes, are approximately normally distributed with a mean of 13.88 minutes and standard deviation of 4 minutes. You are interested in the probability that it will take you at least 20 minutes to get to school 36. Pick the equation most appropriate to deal with the problem above. x a) z = n x b) p( x ) = ( nCx ) p x (1 p) c) x = + z x d) z = n e) None of the above are correct. 37. What is the probability that it will take you at least 20 minutes to get to school? Round your final answer to 2 decimal places. a) 0.03 b) 0.06 c) 0.97 d) 0.94 e) None of the above are correct.
3. C 17. A 31. C
4. B 18. C 32. E
5. B 19. B 33. B
6. D 20. E 34. B
7. C 21. C 35. C
8. B 22. D 36. A
9. B 23. B 37. B
10. B 24. B
11. A 25. C
12. C 26. D
13. C 27. B
14. A 28. D | 4,395 | 16,118 | {"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} | 3.375 | 3 | CC-MAIN-2019-51 | latest | en | 0.954811 |
http://sfb649.wiwi.hu-berlin.de/fedc_homepage/xplore/help/eivplmnor.html | 1,369,326,813,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703592489/warc/CC-MAIN-20130516112632-00069-ip-10-60-113-184.ec2.internal.warc.gz | 244,831,644 | 2,940 | Keywords - Function groups - @ A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Quantlet: eivplmnor Description: eivplmnor fits partially linear EIV model where the conditional distribution of y given x and t is normally distributed.
Reference(s):
Severini/Staniswalis, JASA, 1994; Liang, H., Haerdle, W. and Carroll, R. 1997 SFB DP27
Usage: myfit = eivplmnor(x,t,y,sigma,h{,opt}) Input: x n x p matrix, the discrete predictor variables. t n x q matrix, the continuous predictor variables. y n x 1 vector, the response variables. sigma scalar, the variance of the measurement error. h q x 1 vector, the bandwidth. opt optional, a list with optional input. The macro "gplmopt" can be used to set up this parameter. The order of the list elements is not important. opt.b0 p x 1 vector, the initial coefficients. If not given, all coefficients are put =0 originally. opt.wx scalar or n x 1 vector, prior weights. If not given, set to 1. opt.wt n x 1 vector, weights for t (trimming factors). If not given, all set to 1. opt.tg ng x 1 vector, a grid for continuous part. If tg is given, the nonparametric function will also be computed on this grid. opt.shf integer, if exists and =1, some output is produced which indicates how the iteration is going on. opt.nosort integer, if exists and =1, the continuous variables t and the grid tg are assumed to be sorted by the 1st column. Sorting is required by the algorithm! Hence this option should be given only when data are sorted. opt.miter integer, maximal number of iterations. The default is 10. opt.cnv integer, convergence criterion. The default is 0.0001. opt.wtc n x 1 vector, weights for convergence criterion, w.r.t. m(t) only. If not given, opt.wt is used. opt.off scalar or n x 1 vector, offset in predictor. Output: myfit.b k x 1 vector, estimated coefficients. myfit.bv k x k matrix, estimated covariance matrix for coefficients. myfit.m n x 1 vector, estimated nonparametric part. myfit.mg ng x 1 vector, estimated nonparametric part on grid if tg was given. This component will not exist, if tg was not given. myfit.stat list with the following statistics: myfit.stat.deviance deviance, myfit.stat.pearson generalized pearson's chi^2, myfit.stat.r2 pseude R^2, myfit.stat.dispersion dispersion parameter estimate, myfit.stat.it scalar, number of iterations needed.
Example:
```library("eiv")
library("xplore")
n=100
randomize(n)
sigma=0.81
b=1|2
p=rows(b)
x=2.*uniform(n,p)-1
t=sort(2.*uniform(n)-1,1)
w=x+sqrt(sigma)*uniform(n)
m=0.5*cos(pi.*t)+0.5*t
y=x*b+m+normal(n)./2
h=0.5
sf=eivplmnor(w,t,y,sigma,h)
b~sf.b
dds=createdisplay(1,1)
show(dds,1,1,t~m,t~sf.m)
```
Result:
```A partially linear fit for E[y|x,t] is computed.
sf.b contains the coefficients for the linear part.
sf.m contains the estimated nonparametric part
evaluated at observations t. The example gives the
true b together with the EIVPLM estimate sf.b. Also, the
estimated function sf.m is displayed.
```
Author: H. Liang, W. Haerdle, 19970828 license MD*Tech
(C) MD*TECH Method and Data Technologies, 05.02.2006 | 885 | 3,053 | {"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} | 2.578125 | 3 | CC-MAIN-2013-20 | latest | en | 0.762247 |
https://numbermatics.com/n/942503717/ | 1,718,796,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00858.warc.gz | 369,477,569 | 6,629 | # 942503717
## 942,503,717 is an odd composite number composed of two prime numbers multiplied together.
What does the number 942503717 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 6 divisors.
942503717 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of six divisors.
## Prime factorization of 942503717:
### 53 × 42172
(53 × 4217 × 4217)
See below for interesting mathematical facts about the number 942503717 from the Numbermatics database.
### Names of 942503717
• Cardinal: 942503717 can be written as Nine hundred forty-two million, five hundred three thousand, seven hundred seventeen.
### Scientific notation
• Scientific notation: 9.42503717 × 108
### Factors of 942503717
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 3
• Sum of prime factors: 4270
### Divisors of 942503717
• Number of divisors d(n): 6
• Complete list of divisors:
• Sum of all divisors σ(n): 960514578
• Sum of proper divisors (its aliquot sum) s(n): 18010861
• 942503717 is a deficient number, because the sum of its proper divisors (18010861) is less than itself. Its deficiency is 924492856
### Bases of 942503717
• Binary: 1110000010110101110111001001012
• Base-36: FL54IT
### Scales and comparisons
How big is 942503717?
• 942,503,717 seconds is equal to 29 years, 50 weeks, 2 days, 14 hours, 35 minutes, 17 seconds.
• To count from 1 to 942,503,717 would take you about forty-four years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 942503717 cubic inches would be around 81.7 feet tall.
### Recreational maths with 942503717
• 942503717 backwards is 717305249
• The number of decimal digits it has is: 9
• The sum of 942503717's digits is 38
• More coming soon!
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The information we have on file for 942503717 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
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July 2, 2012, 07:06
twoPhaseEulerFoam unphysical behaviour
#1
Senior Member
Gerhard Holzinger
Join Date: Feb 2012
Location: Austria
Posts: 194
Rep Power: 17
Hello,
I am trying to simulate a bubble column. My domain is a simple box, partly filled with water.
At the center of a small patch on the bottom air is injected (alpha = 1, Ua = 0.12). Water velocity is zero at the walls and at the inlet.
What puzzles me is the fact that the water levels rises continuously. I would expect a small finite increase of the water level.
To run the case, run:
blockMesh
setFields
twoPhaseEulerFoam
The attached images are taken at T = 0 s, T = 6 s and T = 12 s.
As the amount of water should be constant the water level should not be increasing.
Attached Images
t0.jpg (11.2 KB, 44 views) t6.jpg (13.9 KB, 65 views) t12.jpg (11.8 KB, 47 views)
Attached Files
bubbleColumnSeperated.tar.gz (4.2 KB, 27 views)
Last edited by GerhardHolzinger; July 2, 2012 at 07:08. Reason: added description of attached images
July 3, 2012, 02:28
update
#2
Senior Member
Gerhard Holzinger
Join Date: Feb 2012
Location: Austria
Posts: 194
Rep Power: 17
I did some reading and found this in H. Rusches thesis on page 116. There he discusses ways to solve the phase continuity equation.
On page 116 he discusses an approach by Spalding. It is about adding a new equation to guarantee boundedness. This new equation is
Code:
`alpha = alpha / (alpha + beta);`
I am not sure if boundedness was the reason for my problems, because alpha did not exceed unity.
Adding this equation solved the problem, but I have to confess, that I do not know why exactly I do not have this problem anymore.
Code:
```{
word scheme("div(phi,alpha)");
word schemer("div(phir,alpha)");
surfaceScalarField phic("phic", phi);
surfaceScalarField phir("phir", phia - phib);
if (g0.value() > 0.0)
{
surfaceScalarField alphaf(fvc::interpolate(alpha));
phir += phipp;
phic += fvc::interpolate(alpha)*phipp;
}
for (int acorr=0; acorr<nAlphaCorr; acorr++)
{
fvScalarMatrix alphaEqn
(
fvm::ddt(alpha)
+ fvm::div(phic, alpha, scheme)
+ fvm::div(-fvc::flux(-phir, beta, schemer), alpha, schemer)
);
if (g0.value() > 0.0)
{
ppMagf = rUaAf*fvc::interpolate
(
(1.0/(rhoa*(alpha + scalar(0.0001))))
*g0*min(exp(preAlphaExp*(alpha - alphaMax)), expMax)
);
alphaEqn -= fvm::laplacian
(
(fvc::interpolate(alpha) + scalar(0.0001))*ppMagf,
alpha,
"laplacian(alphaPpMag,alpha)"
);
}
alphaEqn.relax();
alphaEqn.solve();
#include "packingLimiter.H"
// addition: option 2 from H. Rusches thesis
alpha = alpha / (alpha + beta);
beta = scalar(1) - alpha;
Info<< "Dispersed phase volume fraction = "
<< alpha.weightedAverage(mesh.V()).value()
<< " Min(alpha) = " << min(alpha).value()
<< " Max(alpha) = " << max(alpha).value()
<< endl;
}
}
rho = alpha*rhoa + beta*rhob;```
The attached images show alpha of the midplane. One can clearly see, that the water level remains where it is.
Attached Images
t0.jpg (11.2 KB, 18 views) t10.jpg (12.8 KB, 21 views) t20.jpg (14.0 KB, 30 views)
October 30, 2012, 03:30
#3
Senior Member
Dongyue Li
Join Date: Jun 2012
Location: Torino, Italy
Posts: 742
Rep Power: 9
Quote:
Originally Posted by GerhardHolzinger I did some reading and found this in H. Rusches thesis on page 116. There he discusses ways to solve the phase continuity equation. On page 116 he discusses an approach by Spalding. It is about adding a new equation to guarantee boundedness. This new equation is The attached images show alpha of the midplane. One can clearly see, that the water level remains where it is.
Hi Foamers
I am doing the similar work with Gerhard.I wanna simulate a 2D tank partly filled with water. and the air was injected at the bottom.
Actually the twoPhaseEulerFoam‘s parameters are confusing me. so I download the case and change it into a 2D case. except that I didnot do anything.
Just as the picture showes. there is no bubble, but a bulk air like a squid......
how to handle that problem?
run:
fluentMeshToFoam fluent.msh
setFields
twoPhaseEulerFoam
Attached Images
QQ??20121030151604.png (4.7 KB, 7 views) QQ??20121030151627.png (12.1 KB, 7 views) QQ??20121030151643.png (29.4 KB, 9 views)
Attached Files
bubbleColumnSeperated2d.zip (72.7 KB, 10 views)
November 2, 2012, 22:28
#4
Senior Member
Dongyue Li
Join Date: Jun 2012
Location: Torino, Italy
Posts: 742
Rep Power: 9
Quote:
Originally Posted by GerhardHolzinger I did some reading and found this in H. Rusches thesis on page 116. There he discusses ways to solve the phase continuity equation. On page 116 he discusses an approach by Spalding. It is about adding a new equation to guarantee boundedness. This new equation is
hi GerhardHolzinger
I simulated lots of different situation and found that the rising water level is relevant to the size of inlet.
But I dont konw why. and Im not sure if this is relevant to the alpha equation.
you can see that the water level remains where it is in this group images.
but the water level rises in the last group image.
does anyone know why that happens?
btw , does the red part represent the big random air bubble?
Attached Images
1.jpg (6.9 KB, 5 views) 2.jpg (11.1 KB, 10 views) 3.png (36.3 KB, 9 views)
Last edited by sharonyue; November 7, 2012 at 04:13.
November 2, 2012, 22:29
#5
Senior Member
Dongyue Li
Join Date: Jun 2012
Location: Torino, Italy
Posts: 742
Rep Power: 9
the middle size of inlet
the water level is rising slowly
Attached Images
QQ??20121103100907.png (2.9 KB, 5 views) QQ??20121103100943.png (17.7 KB, 5 views) QQ??20121103101005.png (21.5 KB, 3 views) QQ??20121103101841.png (25.4 KB, 4 views)
November 2, 2012, 22:30
#6
Senior Member
Dongyue Li
Join Date: Jun 2012
Location: Torino, Italy
Posts: 742
Rep Power: 9
the small inlet
level is rising a little bit fast.
Attached Images
QQ??20121103101053.png (2.8 KB, 4 views) QQ??20121103101111.png (12.8 KB, 7 views) QQ??20121103101123.png (10.3 KB, 9 views) QQ??20121103101140.png (13.8 KB, 6 views)
November 5, 2012, 04:11 #7 Senior Member Gerhard Holzinger Join Date: Feb 2012 Location: Austria Posts: 194 Rep Power: 17 Did you add the equation I mentioned? In OpenFOAM 2.1.x (some time after 2.1.1 came out) the two-phase solvers were adapted to use MULES. Now there is no problem with free water surfaces.
November 7, 2012, 04:12
#8
Senior Member
Dongyue Li
Join Date: Jun 2012
Location: Torino, Italy
Posts: 742
Rep Power: 9
Quote:
Originally Posted by GerhardHolzinger Did you add the equation I mentioned? In OpenFOAM 2.1.x (some time after 2.1.1 came out) the two-phase solvers were adapted to use MULES. Now there is no problem with free water surfaces.
yeah, I upgrade it, its fixed.
I have a little question.
Alberto said in twoPhaseEulerFoam, there would be no bubbles. but why there are bubbles in the tutorials/bed2/ and bed/?
November 7, 2012, 04:28 #9 Senior Member Gerhard Holzinger Join Date: Feb 2012 Location: Austria Posts: 194 Rep Power: 17 I think Alberto meant, the difference between an Euler-Euler and a VOF-approach. In Euler-Euler all phases are distributed. You do not calculate single bubbles but a gaseous phase. So when simulating a bubble column in an Eulerian-Framework, there are no bubbles. There are only phase 'a' and 'b'. One of them is water and the other one is air. To consider the bubbly nature of air inside water, you have to provide a bubble diameter in your case setup. This is needed e.g. for the calculation of drag. You solve the momentum equation for both water and air in all your domain. Additionally you solve a phase continuity equation that tells you, where your water and where your air is located. The field alpha is the volume fraction of the disperse phase. So, if you are having a bubble column, alpha=0 means there is only water and alpha=1 means there is only air. Air bubbles rising through water mean a value of alpha between 0 and 1. So, there are no actual bubbles. There is only a volume fraction in an Eulerian simulation. You can play the same game with solid particles and air, then you can call it fluidized bed.
November 7, 2012, 05:09
#10
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Dongyue Li
Join Date: Jun 2012
Location: Torino, Italy
Posts: 742
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Well thank you for your detailed answer and the information you always provide.Gerhard Holzinger. but now I am still confused.
Quote:
Originally Posted by GerhardHolzinger Air bubbles rising through water mean a value of alpha between 0 and 1.
I dont know why the alpha is between 0 and 1 in a bubble, should it be 0?
Quote:
Originally Posted by GerhardHolzinger So when simulating a bubble column in an Eulerian-Framework, there are no bubbles.
its very weird that there are no bubbles in a bubble column, Im really lost......confused....
I want to simulate air injected in a stirred tank filled with high viscosity (about 2000cp)non-Newtonian fluid(xanthan gum), in experiment, I find there are numerous small air bubble (1mm) cause of the high viscosity,and many big bubbles breakup like boiling in the free water level.
see it here Coalescence and breakup in twoPhaseEulerFoam
So in reality , we can see lots of big bubbles and numerous small bubbles.
just as you said, can i perceive that two-fluid model can not simulate big bubble?
November 7, 2012, 05:25
#11
Senior Member
Gerhard Holzinger
Join Date: Feb 2012
Location: Austria
Posts: 194
Rep Power: 17
Quote:
Originally Posted by sharonyue I dont know why the alpha is between 0 and 1 in a bubble, should it be 0? its very weird that there are no bubbles in a bubble column, Im really lost......confused.... ?
What I meant with "There are no bubbles" is, that you don't see any bubbles when you look on your results in paraView.
I attached some pictures. Two of them are from paraView, showing the same image, just with a different scale on the color-bar. The third picture is an image of nearly the same situation in an experiment.
You see in both cases a bubble plume rising. In the photo you hardly see any actual bubbles because the exposure time is too long. So rising bubbles result in a somehow blurred image of the mean bubble path.
The two images out of paraView show the mean values of alpha - the volume fraction of the air. In this images there are no bubbles, only values of alpha between 0 and 1. If in a cell the value of alpha is 0, then there are no bubbles in this cell.
If in a cell the value of alpha is 0.1, then there are bubbles rising through this cell, and the overall bubble volume is 0.1 times the cell volume.
Attached Images
alphaMidPlaneT10rescaledRange.jpg (14.2 KB, 45 views) alphaMidPlaneT10fullRange.jpg (12.3 KB, 37 views) Img0678.jpg (55.9 KB, 57 views)
November 7, 2012, 06:13
#12
Senior Member
Dongyue Li
Join Date: Jun 2012
Location: Torino, Italy
Posts: 742
Rep Power: 9
Quote:
Originally Posted by GerhardHolzinger What I meant with "There are no bubbles" is, that you don't see any bubbles when you look on your results in paraView.
wow! thank you for that image...now I know why there is no bubble,
one last question.
in your experiment, the bubble is so small. normally if air was injected in water,there would be lots of big bubbles than yours(right?or I just make a common sense mistake?). did u add grille in the inlet?
if u dont add grille in the inlet, or enlagrge the inlet, just as the image with a small scale on the color-bar I attached.
maybe its bubbles?...
anyway,thanks for your extremely patience...I own you a favor~
Attached Images
???.jpg (11.2 KB, 16 views) ???1.jpg (13.0 KB, 21 views)
November 7, 2012, 07:52 #13 Senior Member Gerhard Holzinger Join Date: Feb 2012 Location: Austria Posts: 194 Rep Power: 17 The bubble size when air is injected into water depends on how you inject it. In my experiment air enters through a porous plate. Therefore, the bubbles are relatively fine. If air enters through a perforated plate with 1mm holes, then the bubbles are bigger.
April 14, 2014, 21:02 #14 Senior Member Dongyue Li Join Date: Jun 2012 Location: Torino, Italy Posts: 742 Rep Power: 9 It has been a long time, To those who is confused to this unclear boundary of alpha in twoPhaseEulerFoam with interFoam. Because in twoPhaseEulerFoam it uses two-fluid model. It means, in every cell, you can have air and water's velocity respectively. So these two phases are permeating with each other. Apart from this, because this two-fluid model is not surface tracking method like VOF. So it can not have a sharp phase boundary. But why there are some "big bubbles"(in red) in this case? This is just becuase air's volume fraction is big such as 0.6 or 0.7. I mean, This is not big bubble. It can be seen as bubble cluster. If Im wrong, please correct me.
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## I need help with this equation
Given the equation -5 (x - 2)' -1= ,
a. Identify the vertex.
b. Identify the axis of the symmetry.
c. Identify the direction of the graph (whether it opens up or down).
The equation that is given is the equation o0f a straight line. Please make this clear.
I assume the equation is -5(x-2)2 - 1 = y
Let's add one to both sides:
-5(x-2)2 = y + 1
This is the equation of a parabola in "vertex" form:
a(x-h)2 = (y-k)
a) The Vertex is located at the point (h,k)
b) The line of symmetry is y = h
c) The parabola opens upward if a>0, down if a<0 | 195 | 612 | {"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} | 3.6875 | 4 | CC-MAIN-2014-35 | longest | en | 0.86558 |
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# Calculating IRR and NPV
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Scalia's Cleaning Service is investigating the purchase of an ultrasound machine for cleaning window blinds. The machine would cost \$136,700, including invoice cost, freight, and training of employees to operate it. Scalia's has estimated that the new machine would increase the company's cash flows, net of expenses, by \$25,000 per year. The machine would have a 14-year useful life with no expected salvage value.
Required:
(Ignore income taxes)
1)Compute the machine's internal rate of return to the nearest whole percent.
2)Compute the machine's net present value. Use a discount rate of 16%.Why do you have a zero net present value?
3)Suppose that the new machine would increase the company's annual cash flows, net of expenses, by only \$20,000 per year. Under the conditions, compute the internal rate of return to the nearest whole percent.
#### Solution Preview
Solution:
1) Compute the machine's internal rate of return to the nearest whole percent.
Annual Cash flow=\$25000
Period=14 year
Since cash flow is same throughout 14 years.
It is equivalent to an ordinary annuity with R=\$25000, n=14.
PV factor=Initial outlay/annual cash flow=136700/25000=5.468
Refer PV factor of ordinary annuity tables for n=14, and PV factor=5.468, we get discount rate=16%
IRR=16%
2) Compute ...
#### Solution Summary
Solution describes the steps for calculating internal rate of return and net present value of an investment proposal.
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Simulate data with specific characteristics.
## Usage
simulate_correlation(n = 100, r = 0.5, mean = 0, sd = 1, names = NULL, ...)
simulate_ttest(n = 100, d = 0.5, names = NULL, ...)
simulate_difference(n = 100, d = 0.5, names = NULL, ...)
## Arguments
n
The number of observations to be generated.
r
A value or vector corresponding to the desired correlation coefficients.
mean
A value or vector corresponding to the mean of the variables.
sd
A value or vector corresponding to the SD of the variables.
names
A character vector of desired variable names.
...
Arguments passed to or from other methods.
d
A value or vector corresponding to the desired difference between the groups.
## Examples
# Correlation --------------------------------
data <- simulate_correlation(r = 0.5)
plot(data$V1, data$V2)
cor.test(data$V1, data$V2)
#>
#> Pearson's product-moment correlation
#>
#> data: data$V1 and data$V2
#> t = 5.7155, df = 98, p-value = 1.18e-07
#> alternative hypothesis: true correlation is not equal to 0
#> 95 percent confidence interval:
#> 0.3366433 0.6341398
#> sample estimates:
#> cor
#> 0.5
#>
summary(lm(V2 ~ V1, data = data))
#>
#> Call:
#> lm(formula = V2 ~ V1, data = data)
#>
#> Residuals:
#> Min 1Q Median 3Q Max
#> -1.8566 -0.5694 -0.1116 0.5070 2.4567
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) -1.548e-17 8.704e-02 0.000 1
#> V1 5.000e-01 8.748e-02 5.715 1.18e-07 ***
#> ---
#> Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#>
#> Residual standard error: 0.8704 on 98 degrees of freedom
#> Multiple R-squared: 0.25, Adjusted R-squared: 0.2423
#> F-statistic: 32.67 on 1 and 98 DF, p-value: 1.18e-07
#>
# Specify mean and SD
data <- simulate_correlation(r = 0.5, n = 50, mean = c(0, 1), sd = c(0.7, 1.7))
cor.test(data$V1, data$V2)
#>
#> Pearson's product-moment correlation
#>
#> data: data$V1 and data$V2
#> t = 4, df = 48, p-value = 0.000218
#> alternative hypothesis: true correlation is not equal to 0
#> 95 percent confidence interval:
#> 0.2574879 0.6832563
#> sample estimates:
#> cor
#> 0.5
#>
round(c(mean(data$V1), sd(data$V1)), 1)
#> [1] 0.0 0.7
round(c(mean(data$V2), sd(data$V2)), 1)
#> [1] 1.0 1.7
summary(lm(V2 ~ V1, data = data))
#>
#> Call:
#> lm(formula = V2 ~ V1, data = data)
#>
#> Residuals:
#> Min 1Q Median 3Q Max
#> -3.2354 -0.9753 -0.0633 1.2648 3.3477
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) 1.0000 0.2104 4.754 1.86e-05 ***
#> V1 1.2143 0.3036 4.000 0.000218 ***
#> ---
#> Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#>
#> Residual standard error: 1.487 on 48 degrees of freedom
#> Multiple R-squared: 0.25, Adjusted R-squared: 0.2344
#> F-statistic: 16 on 1 and 48 DF, p-value: 0.000218
#>
# Generate multiple variables
cor_matrix <- matrix(
c(
1.0, 0.2, 0.4,
0.2, 1.0, 0.3,
0.4, 0.3, 1.0
),
nrow = 3
)
data <- simulate_correlation(r = cor_matrix, names = c("y", "x1", "x2"))
cor(data)
#> y x1 x2
#> y 1.0 0.2 0.4
#> x1 0.2 1.0 0.3
#> x2 0.4 0.3 1.0
summary(lm(y ~ x1, data = data))
#>
#> Call:
#> lm(formula = y ~ x1, data = data)
#>
#> Residuals:
#> Min 1Q Median 3Q Max
#> -2.12568 -0.76836 -0.08657 0.61647 2.76996
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) -2.455e-17 9.848e-02 0.000 1.000
#> x1 2.000e-01 9.897e-02 2.021 0.046 *
#> ---
#> Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#>
#> Residual standard error: 0.9848 on 98 degrees of freedom
#> Multiple R-squared: 0.04, Adjusted R-squared: 0.0302
#> F-statistic: 4.083 on 1 and 98 DF, p-value: 0.04604
#>
# t-test --------------------------------
data <- simulate_ttest(n = 30, d = 0.3)
plot(data$V1, data$V0)
round(c(mean(data$V1), sd(data$V1)), 1)
#> [1] 0 1
diff(t.test(data$V1 ~ data$V0)$estimate) #> mean in group 1 #> 0.09185722 summary(lm(V1 ~ V0, data = data)) #> #> Call: #> lm(formula = V1 ~ V0, data = data) #> #> Residuals: #> Min 1Q Median 3Q Max #> -2.0821 -0.6721 0.0000 0.6032 2.0821 #> #> Coefficients: #> Estimate Std. Error t value Pr(>|t|) #> (Intercept) -0.04593 0.26139 -0.176 0.862 #> V01 0.09186 0.36966 0.248 0.806 #> #> Residual standard error: 1.012 on 28 degrees of freedom #> Multiple R-squared: 0.0022, Adjusted R-squared: -0.03344 #> F-statistic: 0.06175 on 1 and 28 DF, p-value: 0.8056 #> summary(glm(V0 ~ V1, data = data, family = "binomial")) #> #> Call: #> glm(formula = V0 ~ V1, family = "binomial", data = data) #> #> Coefficients: #> Estimate Std. Error z value Pr(>|z|) #> (Intercept) -2.983e-17 3.656e-01 0.000 1.000 #> V1 9.601e-02 3.740e-01 0.257 0.797 #> #> (Dispersion parameter for binomial family taken to be 1) #> #> Null deviance: 41.589 on 29 degrees of freedom #> Residual deviance: 41.523 on 28 degrees of freedom #> AIC: 45.523 #> #> Number of Fisher Scoring iterations: 3 #> # Difference -------------------------------- data <- simulate_difference(n = 30, d = 0.3) plot(data$V1, data$V0) round(c(mean(data$V1), sd(data$V1)), 1) #> [1] 0 1 diff(t.test(data$V1 ~ data$V0)$estimate)
#> mean in group 1
#> 0.3
summary(lm(V1 ~ V0, data = data))
#>
#> Call:
#> lm(formula = V1 ~ V0, data = data)
#>
#> Residuals:
#> Min 1Q Median 3Q Max
#> -1.834 -0.677 0.000 0.677 1.834
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) -0.1500 0.2562 -0.586 0.563
#> V01 0.3000 0.3623 0.828 0.415
#>
#> Residual standard error: 0.9922 on 28 degrees of freedom
#> Multiple R-squared: 0.0239, Adjusted R-squared: -0.01096
#> F-statistic: 0.6857 on 1 and 28 DF, p-value: 0.4146
#>
summary(glm(V0 ~ V1, data = data, family = "binomial"))
#>
#> Call:
#> glm(formula = V0 ~ V1, family = "binomial", data = data)
#>
#> Coefficients:
#> Estimate Std. Error z value Pr(>|z|)
#> (Intercept) -4.569e-17 3.696e-01 0.000 1.000
#> V1 3.251e-01 3.877e-01 0.839 0.402
#>
#> (Dispersion parameter for binomial family taken to be 1)
#>
#> Null deviance: 41.589 on 29 degrees of freedom
#> Residual deviance: 40.865 on 28 degrees of freedom
#> AIC: 44.865
#>
#> Number of Fisher Scoring iterations: 4
#> | 2,649 | 6,375 | {"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": 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} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.520497 |
http://www.chegg.com/homework-help/questions-and-answers/w14-x-74-steel-column-length-30-ft-steel-modulas-e-29-000-ksi-proportional-limit-33-ksi-us-q1307063 | 1,464,262,996,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049275835.98/warc/CC-MAIN-20160524002115-00178-ip-10-185-217-139.ec2.internal.warc.gz | 420,383,961 | 13,453 | <p>A W14 X 74 steel column has a length of 30 ft. The steel has a modulas E = 29,000 ksi, and a proportional limit of 33 ksi. Use a factor of safety of 2. Determine the critical or Euler load, and the allowable load if...</p>
<p>a. the ends are pinned</p>
<p>b. one end is pinned, and the other is fixed</p>
<p>c. both ends are fixed.</p>
<p><strong><span style="text-decoration: underline;">Beam info from Appendix</span></strong></p>
<p><strong>Area</strong> = 21.8 in ^2        <strong> Depth </strong>= 14.17 in       <strong>Web thickness </strong>= .450 in</p>
<p><strong>Flange:</strong>    <strong>Thickness</strong> = 0.785 in        <strong>width </strong>= 10.07 in.</p>
<p><strong>Axis X-X:</strong>  l = 796 in ^4,      S = 112 in ^ 3,      r = 6.04 in</p>
<p><strong>Axis Y-Y: </strong> l =134 in ^4,       S = 26.6 in ^ 3,     r = 2.48 in</p> | 382 | 1,092 | {"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-2016-22 | latest | en | 0.457193 |
http://www.photomacrography.net/forum/viewtopic.php?t=21771 | 1,558,706,223,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257624.9/warc/CC-MAIN-20190524124534-20190524150534-00349.warc.gz | 319,195,404 | 9,913 | www.photomacrography.net :: View topic - Total Magnification with DSLR and Trinocular Microscope ?
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Author Message
bcgeo
Joined: 15 Oct 2013
Posts: 3
Posted: Wed Oct 16, 2013 7:14 pm Post subject: Total Magnification with DSLR and Trinocular Microscope ? I am new and unsure how to properly know the magnification of the digital photo I have taken The Camera is a Nikon D7000 with 23.6 x 15.7 mm CMOS sensor 4928 x 3264 photo size 1.5x Focal Length ? an Old Meiji Trinocular 2.5x Camera Coupler for this photo the microscope is at 1x this is a mm ruler that is all I have right now Any Help would be appreciated, Thanks in Advance !
Chris S.
Joined: 05 Apr 2009
Posts: 3202
Location: Ohio, USA
Posted: Wed Oct 16, 2013 8:20 pm Post subject: Bcgeo, welcome to the forum! Direct measurement, such as you have done here, is a really good way to determine magnification on sensor. In this instance, we are seeing slightly over 11mm of the ruler. (So we can say that the horizontal "field of view" or "subject field" is about 11mm.) Your camera has a 23.6mm horizontal width. So your magnification on sensor is slightly over 2x (which can also be written as 2:1). You could use this formula: M=S/F Where M=magnification on sensor; S=sensor size, and F=field of view. It may be worth noting that these days, "magnification on sensor" is what we commonly mean when we say "magnification." But if you are looking at printed photographs, particularly in older sources, "magnification" might have meant "magnification in the final print." Your 11mm field of view, if printed on a piece of 8x10 inch photographic paper, could be accurately labeled as "Shown approximately 23 times life-size." (Magnification in print=254mm print width/11mm subject width.) (BTW, should this thread be moved to another section of the forum?) Cheers, --Chris
rjlittlefield
Joined: 01 Aug 2006
Posts: 19710
Location: Richland, Washington State, USA | 572 | 2,133 | {"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} | 2.671875 | 3 | CC-MAIN-2019-22 | latest | en | 0.911284 |
https://oeis.org/A122469 | 1,726,626,637,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00726.warc.gz | 407,060,874 | 3,969 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A122469 Write Roman numerals I, V, X, L, C, D, M in increasing order: [I], [I][V], [I][V][X], ..., etc. 0
1, 15, 1510, 151050, 151050100, 151050100500, 1510501005001000 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS From a quiz. REFERENCES K. Russell and P. Carter, Number Puzzles, W. Foulsham and Co. Ltd. (1993). LINKS Table of n, a(n) for n=1..7. EXAMPLE [I] = [1]. [I][V] = [1][5]. [I][V][X] = [1][5][10]. [I][V][X][L] = [1][5][10][50]. [I][V][X][L][C] = [1][5][10][50][100]. [I][V][X][L][C][D] = [1][5][10][50][100][500]. [I][V][X][L][C][D][M] = [1][5][10][50][100][500][1000]. CROSSREFS Sequence in context: A200994 A338634 A209680 * A119784 A199228 A344638 Adjacent sequences: A122466 A122467 A122468 * A122470 A122471 A122472 KEYWORD nonn,base,fini,full AUTHOR Herman Jamke (hermanjamke(AT)fastmail.fm), Sep 14 2006 EXTENSIONS Edited by Michel Marcus, Aug 06 2013 STATUS approved
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Last modified September 17 21:55 EDT 2024. Contains 375990 sequences. (Running on oeis4.) | 501 | 1,405 | {"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-2024-38 | latest | en | 0.588881 |
http://cryptography.wikia.com/wiki/Treyfer | 1,502,888,559,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886101966.48/warc/CC-MAIN-20170816125013-20170816145013-00190.warc.gz | 90,386,019 | 76,647 | ## FANDOM
567 Pages
Template:Infobox block cipher In cryptography, Treyfer is a block cipher/MAC designed in 1997 by Gideon Yuval. Aimed at smart card applications, the algorithm is extremely simple and compact; it can be implemented in just 29 bytes of 8051 machine code.
Treyfer has a rather small key size and block size of 64 bits each. All operations are byte-oriented, and there is a single 8×8-bit S-box. The S-box is left undefined; the implementation can simply use whatever data is available in memory. In each round, each byte has added to it the S-box value of the sum of a key byte and the previous data byte, then it is rotated left one bit. The design attempts to compensate for the simplicity of this round transformation by using a large number of rounds: 32.
Due to the simplicity of its key schedule, using the same 8 key bytes in each round, Treyfer was one of the first ciphers shown to be susceptible to a slide attack. This cryptanalysis, which is independent of the number of rounds and the choice of S-box, requires 232 known plaintexts and 244 computation time.
## ImplementationEdit
A simple implementation of Treyfer can be done as follows:
```#include <stdint.h>
#define NUMROUNDS 32
extern uint8_t const Sbox[256];
void treyfer_encrypt(uint8_t text[8], uint8_t const key[8])
{
unsigned i;
uint8_t t = text[0];
for (i = 0; i < 8*NUMROUNDS; i++) {
t += key[i%8];
t = Sbox[t] + text[(i+1)%8];
text[(i+1) % 8] = t = (t << 1) | (t >> 7); /* Rotate left 1 bit */
}
}``` | 403 | 1,509 | {"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-2017-34 | latest | en | 0.836422 |
http://en.wikipedia.org/wiki/Spurious_correlation | 1,438,130,803,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042985140.15/warc/CC-MAIN-20150728002305-00084-ip-10-236-191-2.ec2.internal.warc.gz | 83,569,177 | 12,063 | # Spurious correlation
An illustration of spurious correlation, this figure shows 500 observations of x/z plotted against y/z. The sample correlation is 0.53, even though x, y, and z are statistically independent of each other (i.e., the pairwise correlations between each of them are zero).
This figure shows the 500 observations of y/z plotted against x/z from above, this time with the z-values on a colour scale to highlight how dividing through by z induces spurious correlation.
Spurious correlation is a term coined by Karl Pearson to describe the correlation between ratios of absolute measurements that arises as a consequence of using ratios, rather than because of any actual correlations between the measurements.[1]
The phenomenon of spurious correlation is one of the main motives for the field of compositional data analysis which deals with the analysis of variables that carry only relative information, such as proportions, percentages and parts-per-million.[2][3]
Pearson's definition of spurious correlation is distinct from misconceptions about correlation and causality, or the term spurious relationship.
## Illustration of spurious correlation
Pearson states a simple example of spurious correlation:[1]
Select three numbers within certain ranges at random, say x, y, z, these will be pair and pair uncorrelated. Form the proper fractions x/y and z/y for each triplet, and correlation will be found between these indices.
The upper scatter plot on the right illustrates this example using 500 observations of x, y, and z. Variables x, y and z are drawn from normal distributions with means 10, 10, and 30, respectively, and standard deviations 1, 1, and 3 respectively, i.e.,
\begin{align} x,y & \sim N(10,1) \\ z & \sim N(30,9) \\ \end{align}
Even though x, y, and z are statistically independent (i.e., pairwise uncorrelated), the ratios x/z and y/z have a sample correlation of 0.53. This is because of the common divisor (z) and can be better understood if we colour the points in the scatter plot by the z-value. Trios of (x, y, z) with relatively large z values tend to appear in the bottom left of the plot; trios with relatively small z values tend to appear in the top right.
## Approximate amount of spurious correlation
Pearson derived an approximation of the correlation that would be observed between two indices ($x_1/x_3$ and $x_2/x_4$), i.e., ratios of the absolute measurements $x_1, x_2, x_3, x_4$:
$\rho = \frac{r_{12} v_1 v_2 - r_{14} v_1 v_4 - r_{23} v_2 v_3 + r_{34} v_3 v_4}{\sqrt{v_1^2 + v_3^2 - 2 r_{13} v_1 v_3} \sqrt{v_2^2 + v_4^2 - 2 r_{24} v_2 v_4}}$
where $v_i$ is the coefficient of variation of $x_i$, and $r_{ij}$ the Pearson correlation between $x_i$ and $x_j$.
This expression can be simplified for situations where there is a common divisor by setting $x_3=x_4$, and $x_1, x_2, x_3$ are uncorrelated, giving the spurious correlation:
$\rho_0 = \frac{v_3^2}{\sqrt{v_1^2 + v_3^2} \sqrt{v_2^2 + v_3^2}}.$
For the special case in which all coefficients of variation are equal (as is the case in the illustrations at right), $\rho_0 = 0.5$
## Relevance to biology and other sciences
Pearson was joined by Sir Francis Galton and Walter Frank Raphael Weldon in cautioning scientists to be wary of spurious correlation, especially in biology where it is common to scale or normalize measurements by dividing them by a particular variable or total. The danger he saw was that conclusions would be drawn from correlations that are artifacts of the analysis method, rather than actual “organic” relationships.
However, it would appear that spurious correlation (and its potential to mislead) is not yet widely understood. In 1986 John Aitchison, who pioneered the log-ratio approach to compositional data analysis wrote:[2]
It seems surprising that the warnings of three such eminent statistician-scientists as Pearson, Galton and Weldon should have largely gone unheeded for so long: even today uncritical applications of inappropriate statistical methods to compositional data with consequent dubious inferences are regularly reported.
More recent publications suggest that this lack of awareness prevails, at least in molecular bioscience.[4][5]
## References
1. ^ a b Pearson, Karl (1897). "Mathematical Contributions to the Theory of Evolution—On a Form of Spurious Correlation Which May Arise When Indices Are Used in the Measurement of Organs". Proceedings of the Royal Society of London 60: 489–498. doi:10.1098/rspl.1896.0076.
2. ^ a b Aitchison, John (1986). The statistical analysis of compositional data. Chapman & Hall. ISBN 0-412-28060-4.
3. ^ Pawlowsky-Glahn, Vera; Buccianti, Antonella, eds. (2011). Compositional Data Analysis: Theory and Applications. Wiley. ISBN 9780470711354.
4. ^ Lovell, David; Müller, Warren; Taylor, Jen; Zwart, Alec; Helliwell, Chris (2011). "Chapter 14: Proportions, Percentages, PPM: Do the Molecular Biosciences Treat Compositional Data Right?". In Pawlowsky-Glahn, Vera; Buccianti, Antonella. Compositional Data Analysis: Theory and Applications. Wiley. ISBN 9780470711354.
5. ^ Lovell, David; Pawlowsky-Glahn, Vera; Egozcue, Juan José; Marguerat, Samuel; Bähler, Jürg (16 March 2015). "Proportionality: A Valid Alternative to Correlation for Relative Data". PLoS Computational Biology. doi:10.1371/journal.pcbi.1004075. | 1,385 | 5,339 | {"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": 14, "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} | 4.21875 | 4 | CC-MAIN-2015-32 | latest | en | 0.896141 |
http://www.christianbook.com/linear-algebra-solved-rom/pd/508111?item_code=WW&netp_id=652009&event=EBRN&view=details | 1,469,695,163,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828010.15/warc/CC-MAIN-20160723071028-00033-ip-10-185-27-174.ec2.internal.warc.gz | 373,257,492 | 20,310 | # Linear Algebra Solved! CD-Rom
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Stuck on a problem, and not sure what you're doing wrong? Turn to Bagatrix's Linear Algebra Solved! CD-ROM to find the answer, as well as what you're doing wrong. Enter in your own homework problem or choose from a list of unlimited example problems, and get step-by-step explanations. This program creates math-formatted documents, detailed graphs, interactive tests, and even helps you track your progress!
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http://www.i-programmer.info/babbages-bag/357-advanced-data-structures.html?start=2 | 1,503,018,256,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104204.40/warc/CC-MAIN-20170818005345-20170818025345-00277.warc.gz | 585,178,891 | 11,711 | Data Structures Part II - Stacks And Trees
### New Book Reviews!
Data Structures Part II - Stacks And Trees
Written by Alex Armstrong
Article Index
Data Structures Part II - Stacks And Trees
Queues And Trees
Trees & Recursion
## Working With Trees
There is also the small problem of working with trees as part of a program.
Clearly you can't keep on writing out
`LeftNode.RightNode.RightNode.LeftNode `
type names to specify a node of your choice – it’s too clumsy.
What actually happens in practice is that a pointer to a node is used as the “current” node and this moves its way down the tree by being set to the current nodes left or right child:
`CurrentNode=CurrentNode.LeftNode`
And so on.
You can generalize this to more than two child nodes.
Generally speaking tree algorithms are all about manipulating a "current node" pointer of some sort to achieve whatever it is you are trying to do - and it this is usually trying to find something stored in one of the tree nodes.
Searching a tree is mainly a matter of visiting the nodes and seeing what they contain. This is again another area where things can seem complicated because you can ask that every node is visited in a particular order usually to increase the efficiency of the search.
For example going down each branch as far as possible before starting again to go down the next branch is called “depth first”, while visiting all nodes at the same depth before going deeper, is called “breadth first”.
The jargon also gets more impressive - we talk of “traversing the tree” rather than "visiting the nodes" and eventually you will encounter the fact that trees and “recursion” go together.
How do you search a tree - you start at the root node and search its left sub-tree and if that doesn't work you search its right-sub tree.
As both of the sub-trees are just slightly smaller trees you can see that the search operation is the same for the sub-trees as for the complete tree. Repeat this recipe and you can search a tree without ever really having to write a search method.
Recursion is a whole topic in its own right and the source of the only good computer science joke I know –
dictionary definition of recursion “Recursion – see recursion”.
What are trees used for?
Well, everything from keeping track of where files are stored on a disk drive to analysing natural language and applications in artificial intelligence. It's worth point out that XML is a data language that can only describe tree structures so the idea must be powerful.
You can't program for long without meeting trees.
If you would like to know more about trees and their associated algorithms then see: Data structures - Trees
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<ASIN:0534390803> | 767 | 3,548 | {"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-2017-34 | longest | en | 0.943773 |
http://math.stackexchange.com/questions/184980/elliptic-functions-and-meromorphic-simply-periodic-functions | 1,467,393,786,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783403508.34/warc/CC-MAIN-20160624155003-00182-ip-10-164-35-72.ec2.internal.warc.gz | 191,667,261 | 18,522 | # Elliptic functions and (meromorphic) simply periodic functions.
Let be $f:\mathbb C\rightarrow\overline {\mathbb C}$ a meromorphic function. The set of periods $\Omega_f$ is a discrete (additive) group and we have one of these possibilities:
i) $\Omega_f= \{0\}$
ii) $\Omega_f= \mathbb Z\omega$
iii) $\Omega_f= \mathbb Z\omega_1+\mathbb Z\omega_2$
In the case iii) we say that $f$ is an elliptic function and we now that the foundamental regions are compact subset of $\mathbb C$ (for example foundamental parallelograms). Elliptic fuctions, respect a fixed group of periodicities $\Omega_f=\Lambda$, form a field $E(\Lambda)$ and one can prove that $E(\Lambda)=\mathbb C(\wp,\wp')$.
Now I have not found in literature similar results for the case ii) of meromorphic simply periodic functions. In this case foudamental regions are not compact sets and, fixed the group $\Omega_f$, simply periodic functions respect $\Omega_f$ form a field. I ask if this field is in fact $\mathbb C\big(e^\frac{2\pi iz}{\omega}\big)$; moreover what is the relation beethween the meromorphic functions:
$$\varepsilon_k=\sum_{n\in\mathbb Z}(z-n)^{-k}$$
and the field of simply periodic funcion, where $\Omega_f=\mathbb Z$?
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Assume for simplicity that $\Lambda_f=\mathbb{Z}$. Note that elliptic functions will automatically also be periodic (i.e. your case ii) above), but they will give rise to periodic functions with infinitely many poles in the strip $\mathbb{C}/\mathbb{Z}$. However I think it should be easy to prove that any function belonging to the field $\mathbb{C}(e^{2\pi i z})$ will only have finitely many poles in this strip, since $e^{2\pi i z}$ will be injective when restricted to this strip. – newguy Aug 21 '12 at 12:16
So $\mathbb C (e^{\frac {2\pi iz}{\omega}})$ is only contained in the field of simply periodic functions..... – Dubious Aug 21 '12 at 12:57
## 1 Answer
This is not really an answer, but too long for a comment. In any case it is also fairly trivial, but I hope you still find some use in it.
I think that it there is unlikely to be a "nice" classification of such functions comparable to (iii). This is mainly because your region of interest (the fundamental strip) is not compact, so things can go "too crazy".
If you want a more concrete reason, note that if $f:\mathbb{C} \to \mathbb{C}$ is any holomorphic function, then $f\left(e^{2\pi i z}\right)$ is still holomorphic, but singly periodic! So surely there are very many such functions.
There are ways of getting out nicer answers, essentially by introducing artificial compactness conditions. For example, you might consider only functions such that $\lim_{Re(z) \to \pm \infty} f(z)$ exist (possibly $\infty$). The field of such functions is precisely $\mathbb{C}\left(e^{2\pi i z}\right)$. Indeed, let $X_0$ be the topological space obtained by gluing the left and right side of your fundamental strip, and let $X$ be its end-compactification (i.e. the sphere obtained by adding two "infinitely far" points to the "infinitely long cylinder" $X_0$). The conditions then precisely say that $f$ is meromorphic on $\mathbb{C}$ and extends to a continuous function from $X$ to the Riemann sphere $P^1$. The function $e(z) = e^{2\pi i z}$ extends to a continuous bijection between $X$ and $P^1$. If $f$ is a continuous function on $X$ holomorphic on $\mathbb{C}$ (i.e. $X_0$), then $f \circ e^{-1}$ is a continuous map from $P^1$ to itself, holomorphic away from the north and south pole. Such a function is meromorphic by the theorem on removable singularities, and hence a rational function (by the classification of meromorphic functions on $P^1$ - a compact riemann surface, whence the nice answer!). Hence $f$ is a rational function in $e(z)$, as was to be shown.
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your comment is very usefull. You said: "if $f:\rightarrow\mathbb C→\mathbb C$ is any holomorphic function, then $f(e^{2πiz})$ is still holomorphic, but singly periodic". But this situation is true also for elliptic functions infact $f(\wp(z))$ is still an elliptic function. – Dubious Sep 21 '12 at 7:33
But $\wp$ has poles. Hence $f(\wp(z))$ will be meromorphic only if $f$ is "meromorphic at infity", i.e. a rational function. All other meromorphic functions have essential singularities at infinity, so $f(\wp(z)$ will have essential singularities. – Tom Bachmann Sep 21 '12 at 16:56 | 1,195 | 4,348 | {"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} | 2.984375 | 3 | CC-MAIN-2016-26 | latest | en | 0.797212 |
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