url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://sites.google.com/site/22sidebysiderefrigerator/refrigeration-fluids-refrigeration/ | 1,503,547,453,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886126027.91/warc/CC-MAIN-20170824024147-20170824044147-00083.warc.gz | 845,127,307 | 8,766 | ### REFRIGERATION FLUIDS : REFRIGERATION
Refrigeration Fluids : Lg Bottom Mount Refrigerators
## Refrigeration Fluids
refrigeration
• (refrigerant) any substance used to provide cooling (as in a refrigerator)
• the process of cooling or freezing (e.g., food) for preservative purposes
• deliberately lowering the body's temperature for therapeutic purposes; "refrigeration by immersing the patient's body in a cold bath"
fluids
• (fluid) subject to change; variable; "a fluid situation fraught with uncertainty"; "everything was unstable following the coup"
• A substance that has no fixed shape and yields easily to external pressure; a gas or (esp.) a liquid
• (fluid) a substance that is fluid at room temperature and pressure
• (fluid) continuous amorphous matter that tends to flow and to conform to the outline of its container: a liquid or a gas
refrigeration fluids - Screw Compressors:
Screw Compressors: Mathematical Modelling and Performance Calculation
The first part of Screw Compressors gives a review of recent developments in screw compressor design. The second part presents a generalized mathematical definition of screw machine rotors and describes some well known lobe shapes in detail. The third part treats the mathematical modelling of the thermodynamics and fluid mechanics of compression and expansion processes. This includes discussion of the issues addressed in order to be able to predict the optimum rotor size and speed and built-in volume ratio and, in the case of oil flooded machines, the injection position and jet diameter. The fourth and fifth parts discuss the principles used and describe the application of the analytical procedures and rotor profiling techniques, presented in the earlier chapters, to the design of a number of twin screw compressors, currently manufactured including the examples of combining expansion and compression in the same machine.
83% (7)
Saranac Amber Ale - F.X. Matt Brewing Company
July 4, 2011 - Saranac Amber Ale from the F.X. Matt Brewing Company in Utica, NY. I had to drink an American beer today on the Fourth of July. It came in a "twelve fluid ounce" unique glass bottle with a pry-off cap. Despite being in my basement since I picked this up at the brewery in April 2010, it poured out a sightly hazy copper hue with a lovely 1 cm of head that persisted. It had an initial creamy taste which moved right into a strong, but pleasant hoppiness. I heard recently that it is light and not "non-refrigeration" that will make a beer skunky.
50% MONO ETHYLENE GLYCOL PREMIX READY TO USE ENGINE COOLANT
Diesel engine coolant, automobile coolant manufacturers, radiator coolant, industrial coolant, glycol eutectic coolant, snow-melting & refrigeration, coolant for diesel engine, d. G. Set coolant, oil refineries, antifreeze agent, buildings and industries running on d. G. Sets, ice skating rinks, rubber industry, cold storage systems, cold rolling mills, solvent extraction units etc. | 650 | 2,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-34 | latest | en | 0.881998 |
http://crypto.stackexchange.com/questions/tagged/statistical-test?sort=active&pageSize=15 | 1,469,357,327,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823996.40/warc/CC-MAIN-20160723071023-00291-ip-10-185-27-174.ec2.internal.warc.gz | 55,300,018 | 23,505 | # Tagged Questions
Statistical testing is used to estimate the likelihood of a hypothesis given a set of data. In cryptanalysis, statistical testing is commonly used to detect non-randomness in the data, e.g. distinguish the output of a PRNG from a truly random bitstream or to find the correctly decrypted message ...
67 views
### Plaintext for randomness evaluation of cryptographic algorithms?
I’m running an experiment to test the randomness output of many cryptographic algorithms based on INST test suite to measure the randomness of their outcome. However I’m searching to a sample data ...
15 views
### How to set M parameter for frequency test within a block
I am using statistical tests to determine whether a sequence of bit is random, one of them is frequency test within a block. In a nutshell: Within a block of M a number of 1 is counted. A joined ...
29 views
### correlation between the timing data produced by different keys with similar inputs in AES
Is there or can there be any correlation between the timing data produced by different keys with similar inputs in AES? If yes, what type of correlation?
74 views
### Statistical tests for PRNG that generates a sequence which is not binary
Is there a pratical application to PRNG that generates a sequence which is not a binary one? A ternary, quaternary sequence, for instance. If so, how can we test this? Is there any alternative test ...
218 views
### What values constitute failing for ENT tests?
The dieharder tests give a clear PASSED, WEAK, etc and I'm wondering what those thresholds ...
112 views
### Is the RC4 keystream random or not?
I have cause to use the PRNG half of RC4, but am seriously confused as to the randomness of its output. (Only the PRNG half.) On the one hand, there is a large body of literature finding ever ...
110 views
### What problems with “random” data would cause this result from Ent?
I'm playing around with a novel kind of source of "random" data and Ent is currently giving me results like this: Entropy = 7.977080 bits per byte. Optimum compression would reduce the size of this ...
25 views
### Determining the degree of freedom for a chi-squared test on a strict avalanche criterion matrix
I have read that the degree of freedom is calculated by subtracting 1 from the number of states a random variable can be in. I am performing a goodness of fit test on a 64*32 strict avalanche ...
2k views
### Randomness test question from FIPS 140-1 and comparison with 140-2
In FIPS 140-1 there are 4 statistical random number generator tests (The Monobit Test, The Poker Test, The Runs Test and The Long Runs Test. Then FIPS 140-2 came along and supposedly tightened the ...
61 views
### Testing implementation of cryptography algorithms: Brainstorm
I wonder what kind of techniques can be used to validate implementation of cryptography algorithms? The obvious one is to run specific algorithms against publically known test vectors (i.e. from ...
21 views
### How were the diffusion statistics for norx calculated?
I was reading the norx specification and came across this section, which provides some kind of statistics about the diffusion of the norx round function. However, I don't see how these values were ...
27 views
### How to compare two CSPRNGs? [duplicate]
Let's say I am given to CSPRNGs: RNG1 and RNG2. I want to select the best one of the two for cryptography. How should I do it? Any of the two RNGs can be initialized with a secret key during tests and ...
293 views
### Why calculate pi to estimate randomness?
Why do testing suites calculate $\pi$ (pi) using the Monte Carlo method to determine if a series of numbers are random? As far as I can tell, the Monte Carlo method can be used to estimate $\pi$ ...
150 views
### Avalanche effect sample size
With a fixed key size – key has 128 bits, while block size is 8 byte – how do I calculate how many different keys and texts I have to test for an cryptanalytic statistics study? Differently worded: ...
438 views
### Chi-Squared Step of Vigenere Cipher Decryption [duplicate]
So, I have been assigned an assignment where I had to solve a Vigenere cipher. Following along with this webpage I managed to get all the way to the chi-squared step. But, that is the problem. Upon ...
2k views
### Shannon confusion and diffusion concept
I read the document(not the whole document) from Shannon where he speaks about the concepts of confusion and diffusion. I read in many places(not in the document but around the internet) that ...
95 views
### Using machine-learning techniques for data-dependent operations in ciphers
From 'Methods of Symmetric Cryptanalysis' by Dmitry Khovratovich, The data-dependent operations are one of the most controversial design concepts. We say that an operation is data-dependent, if it ...
86 views
### Minimum number of independent trials needed to detect a bias [closed]
Suppose, I suspect that the probability of occurrence of a particular bit as $0$ is $.558$. Since, it is not $.5$, so this is possibly a bias. Now, my question is: What is the minimum number of ...
59 views
### Available programs that can verify the working of a simple substitution cipher
Is there any tool available that can test whether a simple substitution cipher is working. I will put my scenario in detail. Lets say I built a simple substitution cipher that maps 8 bit data to 16 ...
63 views
### Infer encoding mechanism from $n$-grams distance?
I have a series of obfuscated strings which across the set of strings have regions of high variance and regions of low variance – implying some encoding mechanism as opposed to an encryption procedure ...
397 views
...
94 views
### Statistical tests for pseudorandom permutations
I'm implementing a format-preserving encryption scheme similar to those described in the literature. I want to sanity test my PRP using some statistical tests like TestU01. However, I'm not sure how ...
110 views
### run length testing [closed]
Looking for guidance/references with/about calculations: I am designing a widget that creates N-bit binary sequences. My customers target spec for the sequence's minimum entropy is X bits/bit. I plan ...
112 views
### Is my implementation of a PRG at least intuitively secure?
I've devised a PRG using Bezier curves , constructed with Bernstein polynomials, and I was wondering if you guys could help me with some suggestions or constructive criticism. Basicly I take a random ...
962 views
### How to prove the security of the PRNG?
Are there any realties tests or criterias that prove the security of the PRNG? What kind of tests or criteria?
87 views
### Feasible way to check n-dimensional equidistribution of PRNGs
I am currently gathering some test methods and test suites for random number generator qualities, and am a bit stuck at finding something feasible to test for n-dimensional equidistribution. As input ...
453 views
### Given $n$ bits, how many “truly random” sequences/numbers can be constructed?
Suppose we have $n$ bits, so we can have $2^n$ different bit sequences. Some sequences don't look random, say, all $1$ or $0$. There are also other patterns like $10101010…$, $11001100…$ and such. I'...
84 views
### Definition of a Statistical Test
In Professor Boneh's online Cryptography course at Coursera, I am a little puzzled by his definition of a statistical test where he writes: A(x) = iff |#0(x) - #1(x)| <= 10.√n Now, if – as ...
387 views
### Tactics available to help prove security of a new system?
I believe that the accepted tactic to "prove" a system as secure is to allow the crypto-community to review it and if no vulnerabilities are found over a long period of time (5 or 6 years), then a new ...
94 views
### Testing hardware random number generators? [closed]
I would like to know how to test hardware random number generators. What techniques, tools or tricks to solve the problem ? Any practical difficulties, implementation complexities etc.
341 views
### Odds of false error detection in a randomness test using the chi-squared test?
Common Criteria classifies the requirements for true Random Number Generators, and specifies how these should be tested against failures (these can occur accidentally, or following deliberate attack, ...
142 views
### Differential privacy definition
Differential privacy defines "privacy" of a mechanism $A$ as the "closeness" of the two distribution $Pr[A(D) \in S]$ and $Pr[A(D') \in S]$ where $D,D'$ differ in one element. And the distance between ...
384 views
### Computational indistinguishability and example of non polynomial algorithm
The wikipedia page on computational indistinguishability says that two ensembles are not distinguishable if "any non-uniform probabilistic polynomial time algorithm A" cannot tell them apart. To help ...
1k views
### How to test distribution of a hash function?
From what I've found, it is generally accepted a cryptographic hash function like SHA-2 has an evenly, randomly distributed output. Is there a way to test this without running through the entire 2^...
1k views
### Example of CHI Square test on Caesar Cipher?
I am trying to get my head round the chi square test, when used with the Caesar cipher. I started off using this formula, $$X = \sum_{i = 1}^k \frac{f_i · f'_i}{n · n'}$$ Where \$...
1k views
### Measuring entropy for a ciphertext only attack
When bruteforcing a password (e.g. the common attacks on DES), where you have ciphertext only, you need a way to assess whether a decrypted plaintext is the right one. I believe the EFF DES machine ... | 2,109 | 9,618 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 2.703125 | 3 | CC-MAIN-2016-30 | latest | en | 0.911062 |
https://becalculator.com/16-3-cm-to-inch-converter.html | 1,686,277,502,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655244.74/warc/CC-MAIN-20230609000217-20230609030217-00402.warc.gz | 139,268,382 | 16,640 | # 16.3 cm to inch converter
## FAQs on 16.3 cm to inch
### How many inches are in a cm?
If you wish to convert 16.3 centimeters into an inch-length number, first you need to be aware of how many inches 1 cm equals.
Here I can tell you directly one centimeter is equal to 0.3937 inches.
### How do you convert 1 cm into inches?
For 1cm to inches conversion, multiply 1cm using a conversion factor 0.3937.
This makes it much easier to convert 16.3 cm to inches.
So 1 cm to inches = 1 times 0.3937 = 0.3937 inches.
This should allow you to answer the next question quickly and easily.
• What’s 1 cm in inches?
• What is conversion rate cm to inches?
• What is the equivalent of 1 cm in inches?
• What is 1 cm in inches equal?
### Meaning of centimeter
Centimeter is an International Standard Unit of Length. It is equal to one hundredth of a meter. It’s approximately equivalent to 39.37 inches.
### Definition:Inch
Anglo-American units of length are in inches. 12 inches equals one foot, and 36 inches equals one yard. According to modern standards, one inch equals 2.54 cm.
### How do u convert 16.3 cm to inches?
You have fully understood cm to inches by the above.
This is the formula:
Value in inches = value in cm × 0.3937
So, 16.3 cm to inches = 16.3 cm × 0.3937 = 0.641731 inches
This formula can be used to answer the related questions:
• What is the formula for converting 16.3 cm to inches?
• How do you convert cm to inches?
• How do you change cm to inches?
• How do you measure cm to inches?
• What is 16.3 cm equal to in inches?
cm inch 15.9 cm 0.625983 inch 15.95 cm 0.6279515 inch 16 cm 0.62992 inch 16.05 cm 0.6318885 inch 16.1 cm 0.633857 inch 16.15 cm 0.6358255 inch 16.2 cm 0.637794 inch 16.25 cm 0.6397625 inch 16.3 cm 0.641731 inch 16.35 cm 0.6436995 inch 16.4 cm 0.645668 inch 16.45 cm 0.6476365 inch 16.5 cm 0.649605 inch 16.55 cm 0.6515735 inch 16.6 cm 0.653542 inch 16.65 cm 0.6555105 inch
Deprecated: Function get_page_by_title is deprecated since version 6.2.0! Use WP_Query instead. in /home/nginx/domains/becalculator.com/public/wp-includes/functions.php on line 5413 | 656 | 2,119 | {"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} | 3.65625 | 4 | CC-MAIN-2023-23 | longest | en | 0.82239 |
http://love2d.org/w/index.php?title=PointWithinShape&oldid=2591 | 1,603,631,099,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889173.38/warc/CC-MAIN-20201025125131-20201025155131-00066.warc.gz | 67,151,187 | 8,878 | # PointWithinShape
A set of functions for testing if a point lies within an area
PointWithinShape(shape, x, y)
returns true if (x,y) lays within the bounds of the given shape. Shape defined by an array of 0 to n {x=n, y=y} values.
It generally defers to the other other functions in the file to do the heavy lifting (CrossingsMultipyTest(...), PointWithinLine(...), BoundingBox(...) and colinear(...))
BoundingBox(box, x, y)
returns true if the given point is inside the box defined :by two points (in the same format taken by PointWithinShape(...)).
colinear(line, x, y, e)
returns true if the given point lies on the infinate line defined by two points (again, in the same format taken by PointWithinShape(...))
e is optional, if given it controls how close (x,y) must be to the line to be considered colinear. it defaults to 0.1
PointWithinLine(line, x, y, e)
returns true if the given point lies on the finite line defined by two points (again, in the same format taken by PointWithinShape(...))
e is optional, if given it controls how close (x,y) must be to the line to be considered within the line. it defaults to 0.66
CrossingsMultiplyTest(polygon, x, y)
returns true if the given point lies within the area of the polygon defined by three or points (again, in the same format taken by PointWithinShape(...))
(it is based on code from Point in Polygon Strategies)
```function PointWithinShape(shape, tx, ty)
if #shape == 0 then
return false
elseif #shape == 1 then
return shape[1].x == tx and shape[1].y == ty
elseif #shape == 2 then
return PointWithinLine(shape, tx, ty)
else
return CrossingsMultiplyTest(shape, tx, ty)
end
end
function BoundingBox(box, tx, ty)
return (box[2].x >= tx and box[2].y >= ty)
and (box[1].x <= tx and box[1].y <= ty)
or (box[1].x >= tx and box[2].y >= ty)
and (box[2].x <= tx and box[1].y <= ty)
end
function colinear(line, x, y, e)
e = e or 0.1
m = (line[2].y - line[1].y) / (line[2].x - line[1].x)
local function f(x) return line[1].y + m*(x - line[1].x) end
return math.abs(y - f(x)) <= e
end
function PointWithinLine(line, tx, ty, e)
e = e or 0.66
if BoundingBox(line, tx, ty) then
return colinear(line, tx, ty, e)
else
return false
end
end
-------------------------------------------------------------------------
-- The following function is based off code from
-- [ http://erich.realtimerendering.com/ptinpoly/ ]
--
--[[
======= Crossings Multiply algorithm ===================================
* This version is usually somewhat faster than the original published in
* Graphics Gems IV; by turning the division for testing the X axis crossing
* into a tricky multiplication test this part of the test became faster,
* which had the additional effect of making the test for "both to left or
* both to right" a bit slower for triangles than simply computing the
* intersection each time. The main increase is in triangle testing speed,
* which was about 15% faster; all other polygon complexities were pretty much
* the same as before. On machines where division is very expensive (not the
* case on the HP 9000 series on which I tested) this test should be much
* faster overall than the old code. Your mileage may (in fact, will) vary,
* depending on the machine and the test data, but in general I believe this
* code is both shorter and faster. This test was inspired by unpublished
* Graphics Gems submitted by Joseph Samosky and Mark Haigh-Hutchinson.
* Related work by Samosky is in:
*
* Samosky, Joseph, "SectionView: A system for interactively specifying and
* visualizing sections through three-dimensional medical image data",
* M.S. Thesis, Department of Electrical Engineering and Computer Science,
* Massachusetts Institute of Technology, 1993.
*
--]]
--[[ Shoot a test ray along +X axis. The strategy is to compare vertex Y values
* to the testing point's Y and quickly discard edges which are entirely to one
* side of the test ray. Note that CONVEX and WINDING code can be added as
* for the CrossingsTest() code; it is left out here for clarity.
*
* Input 2D polygon _pgon_ with _numverts_ number of vertices and test point
* _point_, returns 1 if inside, 0 if outside.
--]]
function CrossingsMultiplyTest(pgon, tx, ty)
local i, yflag0, yflag1, inside_flag
local vtx0, vtx1
local numverts = #pgon
vtx0 = pgon[numverts]
vtx1 = pgon[1]
-- get test bit for above/below X axis
yflag0 = ( vtx0.y >= ty )
inside_flag = false
for i=2,numverts+1 do
yflag1 = ( vtx1.y >= ty )
--[[ Check if endpoints straddle (are on opposite sides) of X axis
* (i.e. the Y's differ); if so, +X ray could intersect this edge.
* The old test also checked whether the endpoints are both to the
* right or to the left of the test point. However, given the faster
* intersection point computation used below, this test was found to
* be a break-even proposition for most polygons and a loser for
* triangles (where 50% or more of the edges which survive this test
* will cross quadrants and so have to have the X intersection computed
* anyway). I credit Joseph Samosky with inspiring me to try dropping
* the "both left or both right" part of my code.
--]]
if ( yflag0 ~= yflag1 ) then
--[[ Check intersection of pgon segment with +X ray.
* Note if >= point's X; if so, the ray hits it.
* The division operation is avoided for the ">=" test by checking
* the sign of the first vertex wrto the test point; idea inspired
* by Joseph Samosky's and Mark Haigh-Hutchinson's different
* polygon inclusion tests.
--]]
if ( ((vtx1.y - ty) * (vtx0.x - vtx1.x) >= (vtx1.x - tx) * (vtx0.y - vtx1.y)) == yflag1 ) then
inside_flag = not inside_flag
end
end
-- Move to the next pair of vertices, retaining info as possible.
yflag0 = yflag1
vtx0 = vtx1
vtx1 = pgon[i]
end
return inside_flag
end``` | 1,553 | 5,745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-45 | latest | en | 0.700941 |
https://web2.0calc.com/questions/in-the-diagram-square-abcd-has-sides-of-length-4_1 | 1,632,483,514,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00289.warc.gz | 630,478,983 | 5,551 | +0
# In the diagram, square ABCD has sides of length 4, and triangle ABE is equilateral. Line segments BE and AC intersect at P. Point
0
142
1
In the diagram, square ABCD has sides of length 4, and triangle ABE is equilateral. Line segments BE and AC intersect at P. Point Q is on BC so that PQ is perpendicular to BC and PQ=x.
Determine the measure of angle BPC
Feb 24, 2021
#1
+602
+1
There is no diagram, so I am assuming $E\in ABCD$.
We have $\angle PAE=60-\angle BAC=60-45=15$
And $\angle APE=180-60-15=180-75=105$.
By Vertical angles, $\angle APE=\angle BPC=\boxed{105^{\circ}}$.
Feb 24, 2021 | 209 | 609 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-39 | latest | en | 0.672572 |
https://www.futurelearn.com/info/courses/energy/0/steps/14501?main-nav-submenu=main-nav-using-fl | 1,606,511,574,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141194171.48/warc/CC-MAIN-20201127191451-20201127221451-00069.warc.gz | 680,456,547 | 18,507 | Heat and mass flows
# Heat and mass flows
Hair dryers and jet engines both increase the energy of air flows through their control volumes. Watch Eann Patterson explain more.
2.3
[HAIR DRYER RUNNING] This is good fun. If you haven’t done it before, then it’s something you can easily do at home. And it’s a simple demonstration of jet power. Inside the hair dryer, there’s a fan and a source of heat. It’s similar to a jet engine that sucks, squeezes, bangs, and blows air into a jet engine. We’re rather more interest in the force at which the air leaves the engine. And so, we provide a large amount of heat energy via a bang that’s an explosive combustion of paraffin. [JET FLIES BY]
40.4
We can use mechanics to analyse the jet engine. Newton’s Law tells us that, for every action, there is an equal and opposite reaction. So the hot, high-velocity air leaving the back of the jet exerts a force on the surroundings, which react and push the airplane forward. However, since this is a course in thermodynamics, I would like to look at the process in terms of energy transfers and flows. Let’s go back to the hair dryer. A fan sucks air into the device. This requires work to be done, which is supplied through the electric motor in the hair dryer. Then we transfer heat to the air from the electric heating element. And finally, we blow the hot air out.
86.7
As the air is blown out, we squeeze it through a nozzle to increase its velocity at the expense of pressure. If we look at the energy transfers occurring during the operation of the hair dryer, then we have work being done by the electric motor driving the fan. We have thermal energy being supplied by the heating element. And finally, the air flows out with a higher energy than it flowed in.
117
Last week, we looked at the relationship between work and energy. This week, we’ll explore heat transfer mechanisms and the energy associated with flows. To analyse such processes as in the hair dryer and the jet engine, we’ll use a special form of thermodynamic system known as a control volume. A control volume has permeable boundaries across which matter flows in and out. So we could define our system as the volume occupied inside the hair dryer. Air flows in across one boundary and at a low energy level, and then flows out the other end at a higher energy level. Work is supplied by the fan and heat by the element.
166.7
Of course, the other common feature is that both hair dryers and jet engines make a lot of noise. This is energy being dissipated and lost, or wasted. We talk about energy dissipation and losses next week. In the meantime, if you want to learn more about jet engines, then search YouTube for “How Does a Jet Engine Work?” More on that later.
What is the connection between balancing a table tennis ball in the hot air stream from a hair dryer and shouting into a hairy microphone while a jet aircraft roars overhead? Eann explains that the hair-dryer and jet engine are both control volumes involving heat and mass flows. | 667 | 3,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-50 | latest | en | 0.946092 |
http://nasawavelength.org/resource-search?facetSort=1&topicsSubjects%5B%5D=Mathematics%3ANumber+and+operations&topicsSubjects%5B%5D=Mathematics%3AGeometry&page=6 | 1,441,326,982,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645330174.85/warc/CC-MAIN-20150827031530-00194-ip-10-171-96-226.ec2.internal.warc.gz | 162,235,322 | 13,916 | ## Narrow Search
Audience
Topics
Earth and space science
Mathematics
Resource Type [-]
View more...
Learning Time
Materials Cost
Instructional Strategies [-]
View more...
SMD Forum
Filters: Your search found 65 results.
Topics/Subjects:
Number and operations
Geometry
Sort by:
Per page:
Now showing results 51-60 of 65
# IMAGE Satellite Scaling
This is an activity about satellite size. Learners will calculate the volume of the IMAGE (Imager for Magnetopause-to-Aurora Global Exploration) satellite, the first satellite mission to image the Earth's magnetosphere. They will then determine the... (View More)
# Measure a Tree
This activity introduces measurement and scale using hands-on activities. In this activity, students use the concept of similar triangles to determine the height of a tree. This activity is one of several available on an educational poster related... (View More)
# Long Distance Detective
This is an activity about measuring craters using shadows and geometry. Learners will make a model crater and use a light source to make a shadow. When they are familiar with the techniques for estimating depths from shadows, they will apply their... (View More)
Audience: High school
# IMAGE Satellite 1/4-scale Model
This is an activity about scale model building. Learners will use mathematics to determine the scale model size, construct a pattern, and build a one-fourth size scale model of the IMAGE (Imager for Magnetopause-to-Aurora Global Exploration)... (View More)
# IMAGE Satellite Scale Model
This is an activity about scale model building. Learners will use mathematics to determine the scale model size, construct a pattern, and build a paper scale model of the IMAGE (Imager for Magnetopause-to-Aurora Global Exploration) satellite, the... (View More)
# Dream Job
This math problem demonstrates the concept of geometric progression, through an example of a million dollar contract between an employee and an employer. Application of the concept of geometric progression to social cause activism is addressed. This... (View More)
Audience: High school
Materials Cost: Free per student
# Guess my Shape (Grades 2-4)
In this inquiry investigation, students explore how light hits things of different shape and form. One real world application to this activity is understanding what we actually observe when we see a solar eclipse. Supplies needed for this lesson... (View More)
# Guess my Shape (Grades 4-6)
In this inquiry investigation, students explore how light hits things of different shape and form. One real world application to this activity is understanding what we actually observe when we see a solar eclipse. Supplies needed for this lesson... (View More)
# Guess my Shape (K-2)
In this inquiry investigation, students explore how light hits things of different shape and form. One real world application to this activity is understanding what we actually observe when we see a solar eclipse. Supplies needed for this lesson... (View More)
# Rotational Motion and Rocket Launches
In this problem set, students are led through a series of calculations to determine the best launch site for a TV satellite. This resource is from PUMAS - Practical Uses of Math and Science - a collection of brief examples created by scientists and... (View More)
Keywords: Rotation
Audience: Middle school
Materials Cost: Free per student | 692 | 3,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2015-35 | longest | en | 0.835388 |
https://digifesto.com/tag/log-normal-distribution/ | 1,680,355,437,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950030.57/warc/CC-MAIN-20230401125552-20230401155552-00327.warc.gz | 240,974,730 | 15,657 | ### Why disorganized heavy tail distributions?
I wrote too soon.
Miller and Page (2009) do indeed address “fat tail” distributions explicitly in the same chapter on Emergence discussed in my last post.
However, they do not touch on the possibility that fat tail distributions might be log normal distributions generated by the Central Limit Theorem, as is well-documented by Mitzenmacher (2004).
Instead, they explicitly make a different case. They argue that there are two kinds of complexity:
• disorganized complexity, complexity where extreme values balance each other out to create average aggregate behavior according to the Law of Large Numbers and Central Limit Theorem.
• organized complexity, where positive and negative feedback can result in extreme outcomes, best characterized by power law or “heavy tail” distributions. Preferential attachment is an example of a feedback based mechanism for generating power law distributions (in the specific case of network degrees).
Indeed, this rough breakdown of possible scientific explanations (the relatively orderly null-hypothesis world of normal distributions, and the chaotic, more accurately rendered world of heavy tail distributions) was the one I had before I started studying complex systems and statistics more seriously in grad school.
Only later did I come to the conclusion that this is a pervasive error, because of the ease with which log normal distributions (which may be “disorganized”) can be confused with power law distributions (which tend to be explained by “organized” processes). I am a bit disappointed that Miller and Page repeat this error, but then again their book is written in 2009. I wonder whether the methodological realization (which I assume I’m not alone in, as I hear it confirmed informally in conversations with smart people sometimes) is relatively recent.
Because this is something so rarely discussed in focus, I think it may be worth pondering exactly why disorganized heavy tail distributions are not favored in the literature. There are several reasons I can think of, which I’ll offer informally here as possibilities or hypotheses.
One reason that I’ve argued for before here is that organized processes are more satisfying as explanations than disorganized processes. Most people are not very good at thinking about probabilities (Tetlock and Gardner (2016) have a great, accessible discussion of why this is the case). So to the extent that the Law of Large Numbers or Central Limit Theorem have true explanatory power, it may not be the kind of explanation most people are willing to entertain. This apparently includes scientists. Rather, a simple explanation in terms of feedback may be the kind of thing that feels like a robust scientific finding, even if there’s something spurious about it when viewed rigorously. (This is related, I think, to arguments about the end of narrative in social science.)
Another reason why disorganized heavy tail distributions may be underutilized as scientific explanations is that it is counter-intuitive that a disorganized process can produce such extreme inequality in outcomes.
This has to do with the key transformation that is the difference between a normal and a log normal distribution. A normal distribution is a bell-shaped distribution one gets when one adds a large number of independent random variables.
The log normal distribution is a heavy tail distribution one gets by multiplying a large number of positively valued independent random variables. While it does have a bell or hump, the top of the bell is not at the arithmetic mean, because the sides of the bell are skewed in size. But this is not necessarily because of the dominance of any particular factor (as would be expected if, for example, a single factor were involved in a positive feedback loop). Rather, it is the mathematical fact of many factors multiplied creating extraordinarily high values which creates the heavy right-hand side of the bell.
One way to put it is that rather than having a “deep” positive feedback loop where a single factor amplifies itself many times over, disorganized heavy tails have “shallow” positive feedback where each of many factors has a single and simultaneous amplifying effect on the impact of all the others. This amplification effect is, like multiplication itself, commutative, which means that no single factor can be considered to be causally prior to the others.
Once again, this defies specificity in an explanation, which may be for some people an explanatory desideratum.
But these extreme values are somehow ones that people demand specific explanations for. This is related, I believe, at the desire for a causal lever with which people can change outcomes, especially their own personal outcomes.
There’s an important political question implicated by all this, which is: why is wealth and power concentrated in the hands of the very few?
One explanation that must be considered is the possibility that society is accumulated history, and over thousands of years an innumerable number of independent factors have affected the distribution of wealth and power. Though rather disorganized, these factors amplify each other multiplicatively, resulting in the distribution that we see today.
The problem with this explanation is that it seems there is little to be done about this state of affairs. A person can effect a handful of the factors that contribute to their own wealth or the wealth of another, but if there are thousands of them then it’s hard to get a grip. One must view the other as simply lucky or unlucky. How can one politically mobilize around that?
References
Miller, John H., and Scott E. Page. Complex adaptive systems: An introduction to computational models of social life. Princeton university press, 2009
Mitzenmacher, Michael. “A brief history of generative models for power law and lognormal distributions.” Internet mathematics 1.2 (2004): 226-251.
Tetlock, Philip E., and Dan Gardner. Superforecasting: The art and science of prediction. Random House, 2016. | 1,172 | 6,085 | {"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-2023-14 | latest | en | 0.938952 |
https://unmethours.com/answers/877/revisions/ | 1,637,988,393,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358118.13/warc/CC-MAIN-20211127043716-20211127073716-00328.warc.gz | 693,010,074 | 6,967 | Question-and-Answer Resource for the Building Energy Modeling Community
Get s tarted with the Help page
# Revision history [back]
This measure doesn't add in any windows, it only adds dayligt sensors. It first loops throu surfaces looking for surfaces with an exterior boundary condition. Then it loops through any subsurfaces that the surface has and if it has a sub-surface that isn't a door or overhead door sets this flag to true. "has_ext_nat_light = true". If one or more surface in a zone is true for this it will add a sensor.
At was a basic proof of concept measure. We have office and school AEDG measures that are more advanced. They add fenestration as well as shading surfaces, light shelves, skylights, and daylighting controls. It removes your existing objects and adds object based on the 50% AEDG guidelines. To use this you need to use our templates or manually flag the building type and space types so the measure knows what about the function of each space. This applies a more advanced logic on placing the sensor than the other measure.
This measure doesn't add in any windows, it only adds dayligt sensors. It first loops throu surfaces looking for surfaces with an exterior boundary condition. Then it loops through any subsurfaces that the surface has and if it has a sub-surface that isn't a door or overhead door sets this flag to true. "has_ext_nat_light = true". If one or more surface in a zone is true for this it will add a sensor.
At was a basic proof of concept measure. We have office and school AEDG measures that are more advanced. They add fenestration as well as shading surfaces, light shelves, skylights, and daylighting controls. It removes your existing objects objects, and adds object objects based on the 50% AEDG guidelines. To use this you need to use our templates or manually flag the building type and space types so the measure knows what about the function of each space. This applies a more advanced logic on placing the sensor than the other measure.
This measure doesn't add in any windows, it only adds dayligt sensors. It first loops throu surfaces looking for surfaces with an exterior boundary condition. Then it loops through any subsurfaces that the surface has and if it has a sub-surface that isn't a door or overhead door sets this flag to true. "has_ext_nat_light = true". If one or more surface in a zone space is true for this it will add a sensor.
At It was a basic proof of concept measure. We have office and school AEDG measures that are more advanced. They add fenestration as well as shading surfaces, light shelves, skylights, and daylighting controls. It removes your existing objects, and adds objects based on the 50% AEDG guidelines. To use this you need to use our templates or manually flag the building type and space types so the measure knows what about the function of each space. This applies a more advanced logic on placing the sensor than the other measure. | 637 | 2,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-49 | latest | en | 0.929756 |
https://wimpsandneutrinos.wordpress.com/2014/08/04/ | 1,560,667,212,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997801.20/warc/CC-MAIN-20190616062650-20190616084650-00173.warc.gz | 665,172,793 | 13,049 | # First Measurement of Coherent Production of Pions in Argon
New on the arXiv today is a paper submitted to PRL describing the first measurement of charged-current coherent production of pions on argon in a neutrino (and an antineutrino) beam. The paper is from the ArgoNeuT collaboration using a liquid argon time projection chamber. Measurements of cross sections in argon are very important, as liquid argon TPCs are the proposed technology to be used for the future LBNE experiment. This paper is measuring the cross section where a muon neutrino hits a nucleus, converting into a muon and creating a pion at the same time. In coherent interactions, very little momentum is transferred to the nucleus and the pion carries the charge to balance out the muon. The nucleus basically does not do anything other than provide a charge for the neutrino to interact with, whereas typically nucleons (protons and neutrons) are knocked out of the nucleus with some measurable amount of energy.
# Physics for Non-Physicists: Decay Widths
While cross sections tell us how strong the interactions are between different types of particles, there is another quantity – the decay width – which tells us how quickly the decay of an unstable particle is likely to happen.
As in particle collisions, decays proceed in a probabilistic manner. If we start with an unstable particle, the probability that it will have decayed by a time interval t later is
$P(t) = \left(\frac{1}{2}\right)^{t/\tau_{1/2}} = \exp\left(-\frac{t}{\tau}\right) = \exp\left(-\Gamma t\right)$
(though checking resets the clock for decays) where τ1/2 is the half-life, τ is the lifetime, and Γ is the decay width. You can easily work out the relationships between these three values. They are just three different ways to parameterize the same quantity.
Half-lives are the most commonly used value for a general audience because reducing a quantity by factors of 2 is easier to visualize than reducing by factors of e. In high energy physics, there are several reasons to use the decay width rather than a lifetime or half-life.
In calculations, the decay width is directly analogous to the cross section. It includes factors for the kinematics of the initial and final state and a squared matrix element (or amplitude). If a decay cannot happen, the decay width will end up being 0, which is more mathematically useful than the corresponding infinite lifetime.
When calculating decay widths, there can be many different channels. Each decay channel (final state) has its own width (a partial width), while the sum of the partial widths gives the total width. So, if there are a series of decay channels labeled i, the total width is
$\Gamma = \frac{1}{\tau} = \sum\limits_{i} \Gamma_i.$
The different types of decays all occur with the same lifetime. The ratio of a partial width to the total width tells gives us the fraction of all decays that go into the channel(s) described by the partial width. This ratio is known as the branching fraction.
The decay width has effects beyond just telling us the lifetime. It turns out that the decay width also forces us to modify the mass of the decaying particle in calculations by adding an imaginary part. One interesting effect of this is that it means that unstable particles do not have a well-defined mass. For example, if you create a Z boson in a collision and it then decays to particles that you measure, even if you perfectly measure the momenta of the decay products you won’t necessarily be able to reconstruct the Z mass of 91.2 GeV. Rather, you’ll see a distribution of masses centered near 91.2 GeV with a characteristic width equal to Γ for the Z boson. Typically, the production and subsequent decay of an unstable particle results in us seeing a peak in the invariant mass (center of mass energy) distribution of the final state particles that we measure.
This example for the Z boson was used to definitively determine that there can only be 3 flavors of Standard Model neutrinos. Collider detectors don’t actually find neutrino interactions directly. Rather, their branching ratios affect the shape of the Z peak, so by measuring the peak in another channel, the number of neutrinos could be extrapolated from the shape. The shape of the Z peak as measured by the LEP electron-positron collider at CERN (LEP was replaced by the LHC, which sits in the same tunnel) almost perfectly matched the expected result for three neutrino flavors. From this result, we know that any additional neutrinos must either not interact with the Z (these would be “sterile” neutrinos if they don’t interact directly with anything in the standard model) or they must be heavy enough to prevent the Z from decaying to the new neutrinos and to prevent any hints of their existence from popping up in other measurements.
Hadrons that undergo strong decays, such as the Δ (delta) particles (spin 3/2 particles made of up and down quarks – like an excited state of a proton or neutron) are often called as “resonances” rather than “particles” because their lifetimes are so short that they cannot be feasibly measured. However, by having short lifetimes, they have large widths (clear resonant peaks) that can be measured, so we can indirectly find the lifetime by looking at an energy spectrum.
Finally, measurements of things like branching fractions provide a very nice way to test different models that might predict the values of these things. In some cases, the width can make a measurement more difficult. We are lucky that the particle we think is the Higgs has a fairly light mass of 125 GeV. At this mass, its width is of order 1 GeV, so the experiments are actually largely limited by the detector resolution. If the mass were more like 1 TeV or more, the width would expand dramatically, eventually being about as large as the mass. In this case, the peak, if you could still call it that for such a large width, would not be easy to find above background levels, and a discovery would be extremely difficult if not impossible. | 1,325 | 6,052 | {"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": 2, "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} | 2.890625 | 3 | CC-MAIN-2019-26 | latest | en | 0.913411 |
http://oeis.org/A114160 | 1,606,735,093,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141213431.41/warc/CC-MAIN-20201130100208-20201130130208-00368.warc.gz | 74,877,424 | 4,993 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Please make a donation to keep the OEIS running. We are now in our 56th year. In the past year we added 10000 new sequences and reached almost 9000 citations (which often say "discovered thanks to the OEIS"). Other ways to donate
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A114160 E.g.f. is A(x) = (1-log(B(x)))/B(x), where B(x) = sqrt(1-2*x). 3
1, 2, 7, 38, 281, 2634, 29919, 399342, 6125265, 106156530, 2051433495, 43734832470, 1019650457385, 25807495577850, 704708234182575, 20649996837971550, 646340185330747425, 21521124899877175650, 759572031366463998375, 28325808256035867711750, 1112907316518036732317625 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS From John M. Campbell, May 20 2011: (Start) a(n) is the determinant of the n X n matrix of the form: |2 1 1 1 ... 1 | |1 4 1 1 ... 1 | |1 1 6 1 ... 1 | |1 1 1 8 ... 1 | |... ... 1 | |1 1 1 1 2n-2 1 | |1 1 1 1 1 2n | See examples. (End) REFERENCES C. Dement, Floretion Integer Sequences (work in progress) LINKS G. C. Greubel, Table of n, a(n) for n = 0..400 FORMULA a(n) = A001147(n) + A004041(n-1) = 2^n*GAMMA(n+1/2)/Pi^(1/2)*(1/2*Psi(n+1/2)+1/2*gamma+log(2)+1. - Vladeta Jovovic EXAMPLE From John M. Campbell, May 20 2011: (Start) Det[{ {2,1,1,1,1,1}, {1,4,1,1,1,1}, {1,1,6,1,1,1}, {1,1,1,8,1,1}, {1,1,1,1,10,1}, {1,1,1,1,1,12}}] = 29919 = a(6), and Det[{ {2,1,1,1,1,1,1}, {1,4,1,1,1,1,1}, {1,1,6,1,1,1,1}, {1,1,1,8,1,1,1}, {1,1,1,1,10,1,1}, {1,1,1,1,1,12,1}, {1,1,1,1,1,1,14}}] = 399342 = a(7). (End) MATHEMATICA Range[0, 18]! CoefficientList[ Series[(1 - Log[Sqrt[1 - 2x]])/Sqrt[(1 - 2x)], {x, 0, 18}], x] (* or *) f[n_] := FullSimplify[ 2^(n-1)*Gamma[n + 1/2]/Sqrt[Pi]*(PolyGamma[n + 1/2] + EulerGamma + Log[4] + 2)]; Table[f[n], {n, 0, 18}] (* Robert G. Wilson v *) twox[x_, y_] := If[x == y, 2*x, 1]; a[n_] := Det[Array[twox[#1, #2] &, {n, n}]]; Join[{1}, Table[a[n], {n, 1, 10}]] (* John M. Campbell, May 20 2011 *) PROG (PARI) x='x + O('x^50); Vec(serlaplace((1 - log(sqrt(1 - 2*x)))/sqrt(1 - 2*x))) \\ G. C. Greubel, Feb 08 2017 CROSSREFS Cf. A114161. Sequence in context: A032109 A337026 A088792 * A145159 A317985 A084552 Adjacent sequences: A114157 A114158 A114159 * A114161 A114162 A114163 KEYWORD nonn AUTHOR Creighton Dement, Nov 14 2005 EXTENSIONS E.g.f. given by Vladeta Jovovic More terms from Robert G. Wilson v, Nov 15 2005 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified November 30 06:11 EST 2020. Contains 338781 sequences. (Running on oeis4.) | 1,214 | 2,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-50 | longest | en | 0.472129 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/432/C3xD6.4D6.html | 1,721,248,462,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00682.warc.gz | 397,498,441 | 6,287 | Copied to
clipboard
## G = C3×D6.4D6order 432 = 24·33
### Direct product of C3 and D6.4D6
Series: Derived Chief Lower central Upper central
Derived series C1 — C3×C6 — C3×D6.4D6
Chief series C1 — C3 — C32 — C3×C6 — C32×C6 — S3×C3×C6 — C3×S3×Dic3 — C3×D6.4D6
Lower central C32 — C3×C6 — C3×D6.4D6
Upper central C1 — C6 — C2×C6
Generators and relations for C3×D6.4D6
G = < a,b,c,d,e | a3=b6=c2=1, d6=e2=b3, ab=ba, ac=ca, ad=da, ae=ea, cbc=dbd-1=ebe-1=b-1, dcd-1=bc, ece-1=b4c, ede-1=d5 >
Subgroups: 672 in 218 conjugacy classes, 64 normal (24 characteristic)
C1, C2, C2, C3, C3, C3, C4, C22, C22, S3, C6, C6, C6, C2×C4, D4, Q8, C32, C32, C32, Dic3, Dic3, C12, D6, C2×C6, C2×C6, C2×C6, C4○D4, C3×S3, C3×C6, C3×C6, C3×C6, Dic6, C4×S3, C2×Dic3, C3⋊D4, C3⋊D4, C2×C12, C3×D4, C3×Q8, C33, C3×Dic3, C3×Dic3, C3⋊Dic3, C3×C12, S3×C6, S3×C6, C62, C62, C62, D42S3, C3×C4○D4, S3×C32, C32×C6, C32×C6, S3×Dic3, D6⋊S3, C322Q8, C3×Dic6, S3×C12, C6×Dic3, C3×C3⋊D4, C3×C3⋊D4, C2×C3⋊Dic3, D4×C32, C32×Dic3, C3×C3⋊Dic3, S3×C3×C6, C3×C62, D6.4D6, C3×D42S3, C3×S3×Dic3, C3×D6⋊S3, C3×C322Q8, C32×C3⋊D4, C6×C3⋊Dic3, C3×D6.4D6
Quotients: C1, C2, C3, C22, S3, C6, C23, D6, C2×C6, C4○D4, C3×S3, C22×S3, C22×C6, S32, S3×C6, D42S3, C3×C4○D4, C2×S32, S3×C2×C6, C3×S32, D6.4D6, C3×D42S3, S32×C6, C3×D6.4D6
Permutation representations of C3×D6.4D6
On 24 points - transitive group 24T1283
Generators in S24
(1 5 9)(2 6 10)(3 7 11)(4 8 12)(13 21 17)(14 22 18)(15 23 19)(16 24 20)
(1 11 9 7 5 3)(2 4 6 8 10 12)(13 15 17 19 21 23)(14 24 22 20 18 16)
(1 2)(3 4)(5 6)(7 8)(9 10)(11 12)(13 24)(14 15)(16 17)(18 19)(20 21)(22 23)
(1 2 3 4 5 6 7 8 9 10 11 12)(13 14 15 16 17 18 19 20 21 22 23 24)
(1 16 7 22)(2 21 8 15)(3 14 9 20)(4 19 10 13)(5 24 11 18)(6 17 12 23)
G:=sub<Sym(24)| (1,5,9)(2,6,10)(3,7,11)(4,8,12)(13,21,17)(14,22,18)(15,23,19)(16,24,20), (1,11,9,7,5,3)(2,4,6,8,10,12)(13,15,17,19,21,23)(14,24,22,20,18,16), (1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,24)(14,15)(16,17)(18,19)(20,21)(22,23), (1,2,3,4,5,6,7,8,9,10,11,12)(13,14,15,16,17,18,19,20,21,22,23,24), (1,16,7,22)(2,21,8,15)(3,14,9,20)(4,19,10,13)(5,24,11,18)(6,17,12,23)>;
G:=Group( (1,5,9)(2,6,10)(3,7,11)(4,8,12)(13,21,17)(14,22,18)(15,23,19)(16,24,20), (1,11,9,7,5,3)(2,4,6,8,10,12)(13,15,17,19,21,23)(14,24,22,20,18,16), (1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,24)(14,15)(16,17)(18,19)(20,21)(22,23), (1,2,3,4,5,6,7,8,9,10,11,12)(13,14,15,16,17,18,19,20,21,22,23,24), (1,16,7,22)(2,21,8,15)(3,14,9,20)(4,19,10,13)(5,24,11,18)(6,17,12,23) );
G=PermutationGroup([[(1,5,9),(2,6,10),(3,7,11),(4,8,12),(13,21,17),(14,22,18),(15,23,19),(16,24,20)], [(1,11,9,7,5,3),(2,4,6,8,10,12),(13,15,17,19,21,23),(14,24,22,20,18,16)], [(1,2),(3,4),(5,6),(7,8),(9,10),(11,12),(13,24),(14,15),(16,17),(18,19),(20,21),(22,23)], [(1,2,3,4,5,6,7,8,9,10,11,12),(13,14,15,16,17,18,19,20,21,22,23,24)], [(1,16,7,22),(2,21,8,15),(3,14,9,20),(4,19,10,13),(5,24,11,18),(6,17,12,23)]])
G:=TransitiveGroup(24,1283);
72 conjugacy classes
class 1 2A 2B 2C 2D 3A 3B 3C ··· 3H 3I 3J 3K 4A 4B 4C 4D 4E 6A 6B 6C ··· 6J 6K ··· 6Y 6Z 6AA 6AB 6AC 6AD ··· 6AI 12A 12B 12C 12D 12E 12F 12G 12H 12I ··· 12N 12O 12P order 1 2 2 2 2 3 3 3 ··· 3 3 3 3 4 4 4 4 4 6 6 6 ··· 6 6 ··· 6 6 6 6 6 6 ··· 6 12 12 12 12 12 12 12 12 12 ··· 12 12 12 size 1 1 2 6 6 1 1 2 ··· 2 4 4 4 6 6 9 9 18 1 1 2 ··· 2 4 ··· 4 6 6 6 6 12 ··· 12 6 6 6 6 9 9 9 9 12 ··· 12 18 18
72 irreducible representations
dim 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 4 4 4 type + + + + + + + + + + + - + - image C1 C2 C2 C2 C2 C2 C3 C6 C6 C6 C6 C6 S3 D6 D6 D6 C4○D4 C3×S3 S3×C6 S3×C6 S3×C6 C3×C4○D4 S32 D4⋊2S3 C2×S32 C3×S32 D6.4D6 C3×D4⋊2S3 S32×C6 C3×D6.4D6 kernel C3×D6.4D6 C3×S3×Dic3 C3×D6⋊S3 C3×C32⋊2Q8 C32×C3⋊D4 C6×C3⋊Dic3 D6.4D6 S3×Dic3 D6⋊S3 C32⋊2Q8 C3×C3⋊D4 C2×C3⋊Dic3 C3×C3⋊D4 C3×Dic3 S3×C6 C62 C33 C3⋊D4 Dic3 D6 C2×C6 C32 C2×C6 C32 C6 C22 C3 C3 C2 C1 # reps 1 2 1 1 2 1 2 4 2 2 4 2 2 2 2 2 2 4 4 4 4 4 1 2 1 2 2 4 2 4
Matrix representation of C3×D6.4D6 in GL4(𝔽7) generated by
4 0 0 0 0 4 0 0 0 0 4 0 0 0 0 4
,
3 6 5 6 4 4 1 3 1 1 3 5 1 6 3 6
,
5 1 2 2 6 0 1 2 0 0 1 0 6 1 4 1
,
2 3 1 3 2 5 1 4 1 4 2 3 1 1 1 5
,
3 0 2 6 1 3 2 1 3 1 0 2 2 2 6 1
G:=sub<GL(4,GF(7))| [4,0,0,0,0,4,0,0,0,0,4,0,0,0,0,4],[3,4,1,1,6,4,1,6,5,1,3,3,6,3,5,6],[5,6,0,6,1,0,0,1,2,1,1,4,2,2,0,1],[2,2,1,1,3,5,4,1,1,1,2,1,3,4,3,5],[3,1,3,2,0,3,1,2,2,2,0,6,6,1,2,1] >;
C3×D6.4D6 in GAP, Magma, Sage, TeX
C_3\times D_6._4D_6
% in TeX
G:=Group("C3xD6.4D6");
// GroupNames label
G:=SmallGroup(432,653);
// by ID
G=gap.SmallGroup(432,653);
# by ID
G:=PCGroup([7,-2,-2,-2,-3,-2,-3,-3,176,590,303,2028,14118]);
// Polycyclic
G:=Group<a,b,c,d,e|a^3=b^6=c^2=1,d^6=e^2=b^3,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,c*b*c=d*b*d^-1=e*b*e^-1=b^-1,d*c*d^-1=b*c,e*c*e^-1=b^4*c,e*d*e^-1=d^5>;
// generators/relations
×
𝔽 | 3,229 | 4,764 | {"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} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.498469 |
https://www.onlinemath4all.com/example-problems-on-basic-proportionality-theorem.html | 1,660,288,187,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571584.72/warc/CC-MAIN-20220812045352-20220812075352-00645.warc.gz | 808,513,640 | 6,919 | # EXAMPLE PROBLEMS ON BASIC PROPORTIONALITY THEOREM
Basic Proportionality Theorem :
If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
Example 1 :
In a triangle ABC,D and E are points on the sides AB and AC respectively such that DE is parallel to BC.
(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, then find AC.
Solution :
In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get
6/9 = 8/EC
EC = (9 8)/6
EC = (3 4)
EC = 12 cm
Example 2 :
In triangle ABC, if AD = 8 cm, AB = 12 cm and AE = 12 cm, then find CE.
Solution :
In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get
12 = 8 + DB
DB = 4 cm
8/4 = 12/EC
EC = (12 4)/8
EC = 48/8
EC = 6 cm
Example 3 :
In triangle ABC, if AD = 4 x – 3, BD = 3 x – 1, AE = 8 x – 7 and EC = 5 x – 3, then find the value of x.
Solution :
In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get
(4x – 3)/(3x – 1) = (8x – 7)/(5x – 3)
(4x – 3)(5x – 3) = (8x – 7)(3x – 1)
20x2 – 12x – 15 x + 9 = 24x2 – 8x – 21x + 7
20x2 – 24x + 9 = 8x2 – 29x + 7
20x2 - 24x2 – 27x + 29x + 9 – 7 = 0
-4x2 + 2 x + 2 = 0
Multiply by -2.
2x2 – x - 1 = 0
(x – 1)(2 x + 1) = 0
x - 1 = 0x = 1 2x + 1 = 02x = -1x = -1/2
So the value of x is 1.
Kindly mail your feedback to v4formath@gmail.com
## Recent Articles
1. ### Solving Word Problems Using Section Formula
Aug 12, 22 02:47 AM
Solving Word Problems Using Section Formula
2. ### Solving for a Specific Variable Worksheet
Aug 12, 22 02:41 AM
Solving for a Specific Variable Worksheet | 652 | 1,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-33 | latest | en | 0.752318 |
https://mathoverflow.net/questions/271643/supersingular-isogenies-of-elliptic-curves-preserving-divisibility-of-points | 1,604,123,849,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107912807.78/warc/CC-MAIN-20201031032847-20201031062847-00077.warc.gz | 405,386,354 | 26,875 | # Supersingular isogenies of elliptic curves preserving divisibility of points
I hope my question is clear.
In summary, If $\phi:E\to E'$ is an isogeny and $P\in E$ is not divisible by $2$, under which conditions $\phi(P)\in E'$ is also not divisble by $2$. Here is the detail with some computations over finite fields with supersingular isogenous elliptic curves with $j$ invariant 1728.
Let $E_1$ be the elliptic curve given by $y^2 = x^3 - (1+t^2)x =f_1(x)$ and consider the curve $E_2$ given by $y^2 = x^3 + 4(1+t^2)x=f_2(x)$. Both $E_1$ and $E_2$ are isogenous via
$\phi:E_1\to E_2$
$(x,y)\mapsto (\tfrac{y^2}{x^2},\tfrac{-(1+t^2)-x^2}{x^2})$
according to Silverman's X.6 when $-(1+t^2)$ is fourth-power free.
Let $p\equiv 3\bmod 4$ and fix $t\in\mathbb{F}_p^\times$. Consider the curves $E_1(\mathbb{F}_p)$ and $E_2(\mathbb{F}_p)$, both have $p+1$ points and when $f_1(x)$ is irreducible is easy to see that it must be cyclic.
The point $P_1:=(-1,t)\in E_1(\mathbb{F}_p)$ is never divisible by $2$.
Under which conditions the point $\phi(P_1)=(t^2,t^3+2t)\in E_2$ is also not divisible by $2$ ?
I have special primes $p$ for which I need to generate the 2-Sylow subgroup of $E_1(\mathbb{F}_p)$, thats why I am concerned with a point not divisible by $2$ and I use the cyclicity of $E_1$ to know the order of this 2-Sylow Subgroup. If $E_1(\mathbb{F}_p)$ it is not cyclic, I want to look at $E_2$ which also is supersingular and must be cyclic but unfortunately $\phi(-1,t)\in E_2$ sometimes IS divisble by two according to some experiments, I would like to get a point on $E_2$ that is not divisible by two that generates the 2-Sylow-subgroup on $E_2$ using the information I know of $E_1$.
Thanks | 590 | 1,714 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.903197 |
https://www.datasciencecentral.com/profiles/blogs/fourier-series-and-differential-equations-with-some-applications | 1,632,056,806,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056890.28/warc/CC-MAIN-20210919125659-20210919155659-00439.warc.gz | 748,036,215 | 26,384 | .
# Fourier Series and Differential Equations with some applications in R (Part 1)
In this article, a few applications of Fourier Series in solving differential equations will be described. All the problems are taken from the edx Course: MITx - 18.03Fx: Differential Equations Fourier Series and Partial Differential Equations. The article will be posted in two parts (two separate blongs)
First a basic introduction to the Fourier series will be given and then we shall see how to solve the following ODEs / PDEs using Fourier series:
1. Find the steady state solution to un-damped / damped systems with pure / near resonance.
2. Solve the 4th order differential equation for beam bending system with boundary values, using theoretical and numeric techniques..
In the next part more applications on differential equation / Fourier series (e.g., heat / diffusion / wave PDEs) will be discussed.
## Some basics
Let f(t) be a periodic function with period L. Then the Fourier series of f(t)f(t) is given by the following superposition
The above formulae can be simplified as below, for even and odd functions
A few popular 2π periodic functions and their Fourier series are shown below:
### Computing the coefficients in R
Let’s first plot the square wave function using the following R code
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 library(ggplot2) library(latex2exp) Sq = function(t) { ifelse(t &amp;amp;amp;gt; 0, ifelse(as.integer(t / pi) %% 2 == 0, 1, -1), ifelse(as.integer(t / pi) %% 2 == 1, 1, -1)) } t = seq(-3*pi, 3*pi, 0.01) ggplot() + geom_line(aes(t, Sq(t)), size=1.5, col='red') + ylab('f(t)') + scale_x_continuous(breaks=seq(-3*pi, 3*pi, pi), labels=c(TeX('$-3\\pi$'),TeX('$-2\\pi$'),TeX('$-\\pi$'),0, TeX('$\\pi$'),TeX('$2\\pi$'),TeX('$3\\pi$'))) + scale_y_continuous(breaks=c(-1,0,1), labels=c('-1', '0', '1')) + ggtitle(TeX(paste('Square wave of period $2\\pi$'))) + theme(plot.title = element_text(hjust = 0.5))
1 2 3 4 5 6 7 8 9 10 for (n in 1:6) { b_n = round(2/pi * integrate(function(t) sin(n*t), 0, pi)$value, 15) print(paste(b_n, ifelse(n %% 2 == 0, 0, 4/n/pi))) } #[1] "1.27323954473516 1.27323954473516" #[1] "0 0" #[1] "0.424413181578388 0.424413181578388" #[1] "0 0" #[1] "0.254647908947032 0.254647908947033" #[1] "0 0" R symbolic computation package rSymPy can be used to compute the Fourier coefficients as shown below: 1 2 3 4 5 library(rSymPy) t &amp;amp;amp;lt;- Var("t") n &amp;amp;amp;lt;- Var("n") sympy("integrate(2*1*sin(n*t)/pi,(t,0,pi))") # sq # [1] "-2*cos(pi*n)/(pi*n) + 2/(pi*n)" The next animation shown how the first few terms in the Fourier series approximates the periodic square wave function. Notice from the above animation that the convergence of the Fourier series with the original periodic function is very slow near discontinuities (the square wave has discontinuities at points t=kπ,where kZ), which is known as Gibbs phenomenon. ### Computing the Fourier Coefficients of the Triangle Wave using Anti-derivatives Let’s aim at computing the Fourier coefficients for the 2π-periodic triangle wave by using the coefficients of the 2π-periodic square wave, using the anti-derivative property. First let’s write a few lines of code to plot the triangle wave. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Tr = function(t) { i = as.integer(t / pi) ifelse(t &amp;amp;amp;gt; 0, ifelse(i %% 2 == 0, t - i*pi, -t + (i+1)*pi), ifelse(i %% 2 == 0, -t + i*pi, t - (i-1)*pi)) } t = seq(-3*pi, 3*pi, 0.01) ggplot() + geom_line(aes(t, Tr(t)), size=1.5, col='red') + ylab('f(t)') + scale_x_continuous(breaks=seq(-3*pi, 3*pi, pi), labels=c(TeX('$-3\\pi$'),TeX('$-2\\pi$'),TeX('$-\\pi$'),0, TeX('$\\pi$'),TeX('$2\\pi$'),TeX('$3\\pi$'))) + scale_y_continuous(breaks=c(-pi,0,pi), labels=c(TeX('$-\\pi$'), '0', TeX('$\\pi$'))) + ggtitle(TeX(paste('Triangle wave of period$2\\pi$'))) + theme(plot.title = element_text(hjust = 0.5)) ### computing the constant term 1 2 3 a_0 = 1/pi*integrate(t, 0, pi)$value print(paste(a_0, pi/2)) #[1] "1.5707963267949 1.5707963267949"
The next animation shows how the superposition of first few terms in the Fourier series approximates the triangle wave function.
### Computing the Fourier Coefficients of the Sawtooth function
Similarly, let’s visualize the sawtooth wave function using the following R code snippet:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 St = function(t) { i &amp;amp;amp;lt;- floor((t + pi) / (2*pi)) t - i*2*pi } t = seq(-3*pi, 3*pi, 0.01) ggplot() + geom_line(aes(t, St(t)), size=1.5, col='red') + scale_x_continuous(breaks=seq(-3*pi, 3*pi, pi), labels=c(TeX('$-3\\pi$'),TeX('$-2\\pi$'),TeX('$-\\pi$'),0, TeX('$\\pi$'),TeX('$2\\pi$'),TeX('$3\\pi$'))) + scale_y_continuous(breaks=c(-pi,0,pi), labels=c(TeX('-$\\pi$'), '0', TeX('$\\pi$'))) + ylab('f(t)') + ggtitle(TeX(paste('Sawtooth wave of period $2\\pi$'))) + theme(plot.title = element_text(hjust = 0.5))
The following animation shows how the superposition of first few terms in the Fourier series of the Sawtooth wave approximate the function.
## Computing the Fourier series for the even periodic extension of a function
Let’s first plot the even periodic extension of the function f(t) = t2, -1 ≤ t ≤ 1, with the following R code.
1 2 3 4 5 6 7 8 9 10 11 12 13 Pr = function(t) { i = as.integer(abs(t-ifelse(t&amp;amp;amp;gt;0,-1,1)) / 2) ifelse(t &amp;amp;amp;gt; 0, (t-2*i)^2, (t+2*i)^2) } t = seq(-5, 5, 0.01) ggplot() + geom_line(aes(t, Pr(t)), size=1.5, col='red') + scale_x_continuous(breaks=-5:5, labels=-5:5) + ylab('f(t)') + ggtitle(TeX(paste('Even periodic extension of the function $f(t) = t^2$, $-1\\leq t \\leq 1$'))) + theme(plot.title = element_text(hjust = 0.5))
Here is another function which is neither even nor odd, with period 2π
The following animation shows how the Fourier series of the function approximates the above function more closely as more and more terms are added.
## Solution of ODE with ERF and Resonance
Let’s solve the following differential equation (for a system without damping) and find out when the pure resonance takes place.
The next section shows how to find the largest gain corresponding to the following system with damping.
Let’s plot the amplitudes of the terms to see which term has the largest gain.
1 2 3 4 5 6 n = seq(1,15,2) f = (1/sqrt((49-n^2)^2 + (0.1*n)^2)/n) ggplot() + geom_point(aes(n, f)) + geom_line(aes(n, f)) + ylab('c_n') + scale_x_continuous(breaks=seq(1,15,2), labels=as.character(seq(1,15,2)))
## Boundary Value Problems
As expected, since the right end of the beam is free, it will bend down when more loads are applied on the beam to the transverse direction of its length. The next animation shows how the beam bending varies with the value of q.
### Numerically solving a linear system to obtain the solution of the beam-bending system represented by the 4th-order differential equation in R
First create a near-tri-diagonal matrix A that looks like the following one, it takes care of the differential coefficients of the beam equation along with all the boundary value conditions.
Now, let’s solve the linear system using the following R code.
1 2 3 4 5 6 7 8 9 #Create a vector b that is zero for boundary conditions and -0.0000001 in every other entry. b = rep(1,10)*(-1e-7) # Create a vector v that solves Av = b. v = solve(A, b) #Create a column vector x of 10 evenly spaced points between 0 and 1 (for plotting) x = seq(0,1,1/9) #Plot v on the vertical axis, and x on the horizontal axis. print(ggplot() + geom_line(aes(x,v), col='blue', size=2) + geom_point(aes(x, v), size=2) + ggtitle(paste('Solution of the linear (beam bending) system')) + theme_bw())
As can be seen from the above figure, the output of the numerical solution agrees with the one obtained above theoretically.
Views: 2233
Comment
### You need to be a member of Data Science Central to add comments!
Join Data Science Central
Comment by Sandipan Dey on July 8, 2020 at 7:51am
Thank you very much Vincent, i shall try it next time.
Comment by Vincent Granville on July 8, 2020 at 7:38am
Hi Sandipan, I also experience the same formatting problem. If you put 2 spaces around the " > " symbol, it fixes the issue. By the way, great article!
Comment by Sandipan Dey on July 8, 2020 at 1:25am
Due to formatting issue some parts of code are replaced by junk html elements. For example,
• "a &amp;amp;amp;amp;gt;0" is to be read as "a > 0"
• "t &amp;amp;amp;amp;lt;- Var('t')is to be read as "t <- Var('t')"
© 2021 TechTarget, Inc. Powered by | 2,782 | 8,721 | {"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": 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} | 4.5 | 4 | CC-MAIN-2021-39 | latest | en | 0.633206 |
https://nrich.maths.org/public/leg.php?code=-333&cl=2&cldcmpid=2320 | 1,553,451,058,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203464.67/warc/CC-MAIN-20190324165854-20190324191854-00073.warc.gz | 539,792,899 | 9,797 | Search by Topic
Resources tagged with Investigations similar to Hamilton's Puzzle:
Filter by: Content type:
Age range:
Challenge level:
Five Coins
Age 7 to 11 Challenge Level:
Ben has five coins in his pocket. How much money might he have?
Age 7 to 11 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
Tiles on a Patio
Age 7 to 11 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
Making Squares
Age 7 to 11
Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares?
Age 7 to 11 Challenge Level:
I like to walk along the cracks of the paving stones, but not the outside edge of the path itself. How many different routes can you find for me to take?
Room Doubling
Age 7 to 11 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
Halloween Investigation
Age 7 to 11 Challenge Level:
Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
Bean Bags for Bernard's Bag
Age 7 to 11 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
Cubes Here and There
Age 7 to 11 Challenge Level:
How many shapes can you build from three red and two green cubes? Can you use what you've found out to predict the number for four red and two green?
The Pied Piper of Hamelin
Age 7 to 11 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
Newspapers
Age 7 to 11 Challenge Level:
When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
New House
Age 7 to 11 Challenge Level:
In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
Polo Square
Age 7 to 11 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
Birthday Cake Candles
Age 7 to 11 Challenge Level:
This challenge involves calculating the number of candles needed on birthday cakes. It is an opportunity to explore numbers and discover new things.
Sweets in a Box
Age 7 to 11 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
My New Patio
Age 7 to 11 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
Train Carriages
Age 5 to 11 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
Ice Cream
Age 7 to 11 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
3 Rings
Age 7 to 11 Challenge Level:
If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
Worms
Age 7 to 11 Challenge Level:
Place this "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, what other totals can you make?
Calcunos
Age 7 to 11 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
Crossing the Town Square
Age 7 to 11 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
Plants
Age 5 to 11 Challenge Level:
Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
Magic Constants
Age 7 to 11 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
Doplication
Age 7 to 11 Challenge Level:
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
Building with Rods
Age 7 to 11 Challenge Level:
In how many ways can you stack these rods, following the rules?
Stairs
Age 5 to 11 Challenge Level:
This challenge is to design different step arrangements, which must go along a distance of 6 on the steps and must end up at 6 high.
Age 7 to 11 Challenge Level:
Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square.
Roll These Dice
Age 7 to 11 Challenge Level:
Roll two red dice and a green dice. Add the two numbers on the red dice and take away the number on the green. What are all the different possible answers?
Making Boxes
Age 7 to 11 Challenge Level:
Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
Street Party
Age 7 to 11 Challenge Level:
The challenge here is to find as many routes as you can for a fence to go so that this town is divided up into two halves, each with 8 blocks.
Magazines
Age 7 to 11 Challenge Level:
Let's suppose that you are going to have a magazine which has 16 pages of A5 size. Can you find some different ways to make these pages? Investigate the pattern for each if you number the pages.
It Figures
Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
Two by One
Age 7 to 11 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
Street Sequences
Age 5 to 11 Challenge Level:
Investigate what happens when you add house numbers along a street in different ways.
Pebbles
Age 7 to 11 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
More Transformations on a Pegboard
Age 7 to 11 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
Round and Round the Circle
Age 7 to 11 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
Two on Five
Age 5 to 11 Challenge Level:
Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
Abundant Numbers
Age 7 to 11 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
Ip Dip
Age 5 to 11 Challenge Level:
"Ip dip sky blue! Who's 'it'? It's you!" Where would you position yourself so that you are 'it' if there are two players? Three players ...?
Gran, How Old Are You?
Age 7 to 11 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
Age 7 to 11 Challenge Level:
What happens when you add the digits of a number then multiply the result by 2 and you keep doing this? You could try for different numbers and different rules.
Three Sets of Cubes, Two Surfaces
Age 7 to 11 Challenge Level:
How many models can you find which obey these rules?
Polygonals
Age 7 to 11 Challenge Level:
Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here.
Division Rules
Age 7 to 11 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
More Plant Spaces
Age 7 to 14 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
More Children and Plants
Age 7 to 14 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
Sorting the Numbers
Age 5 to 11 Challenge Level:
Complete these two jigsaws then put one on top of the other. What happens when you add the 'touching' numbers? What happens when you change the position of the jigsaws?
2010: A Year of Investigations
Age 5 to 14
This article for teachers suggests ideas for activities built around 10 and 2010. | 2,153 | 9,340 | {"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-2019-13 | longest | en | 0.946943 |
https://www.coursehero.com/file/p14snhd/If-the-half-life-of-the-anti-biotic-in-the-bloodstream-is-2-hours-what-rate-of/ | 1,638,804,733,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00204.warc.gz | 812,382,653 | 57,016 | # If the half life of the anti biotic in the
• Homework Help
• 19
• 100% (7) 7 out of 7 people found this document helpful
This preview shows page 16 - 18 out of 19 pages.
. If the half-life of the anti-biotic in the bloodstream is 2 hours, what rate of infusion should he prescribe to maintain a long-term amount of 290 mgs in Diana’s bloodstream? 1. Infusion rate = 2 . 137 mgs/min 2. Infusion rate = 1 . 79 mgs/min 3. Infusion rate = 1 . 906 mgs/min 4. Infusion rate = 2 . 021 mgs/min 5. Infusion rate = 1 . 675 mgs/min correct Explanation: Let a be the infusion rate (in mgs per minute). Then the rate at which the amount of anti-biotic in the bloodstream is changing will be a combination of this infusion and the loss through elimination. Thus y ( t ) satisfies the differential equation dy dt = a ky where k is a constant. This is a separable variables equation which after integration be- comes integraldisplay 1 a ky dy = integraldisplay dt. Consequently, 1 k ln( a ky ) = t + C, i.e. , y ( t ) = 1 k parenleftBig a Ae kt parenrightBig , with A an arbitrary constant. Initially, there will be no anti-biotic in the bloodstream, so y (0) = 0 = y ( t ) = a k parenleftBig 1 e kt parenrightBig . The value of k will be determined by the half- life of the anti-biotic in the bloodstream since t = 120 = e 120 k = 1 2 , i.e. , k = (1 / 120) ln 2 = 0 . 00577625 . To main- tain a long-term level of c mgs in the blood- stream, therefore, lim t → ∞ y ( t ) = a k = 290 . Hence Infusion rate 1 . 675 mgs/min. 032 10.0points In a West Texas school district the school year began on August 1 and lasted until May 31. On August 1 a Soft Drink company in- stalled soda machines in the school cafeteria. It found that after t months the machines generated income at a rate of f ( t ) = 400 t 5 t 2 + 8 dollars per month. Find the total income produced during the first semester ending on December 31. 1. income = \$92 . 43 2. income = \$152 . 43 3. income = \$112 . 43 correct 4. income = \$132 . 43 5. income = \$172 . 43 Explanation: If the first semester lasted from August 1 until December 31, then this period covered the time interval from t = 0 until t = 5 (re- member, the month of August corresponds to
pacheco (jnp926) – Homework 5 – staron – (52840) 17 the time interval (0 , 1]). Thus the total in- come generated during the first semester is given by the integral T first = integraldisplay 5 0 f ( t ) dt = integraldisplay 5 0 400 t 5 t 2 + 8 dt. But integraldisplay 5 0 400 t 5 t 2 + 8 dt = bracketleftBig 400 10 ln(5 t 2 + 8) bracketrightBig 5 0 = 400 10 ln parenleftBig 125 + 8 8 parenrightBig . Hence income = \$112 . 43 . 033 10.0points Find the amount A in an account after 9 years when In turn, after exponentiation this becomes A ( t ) = e C e 0 . 05 t with C an arbitrary constant. The constant C is determined by the initial condition A (0) = 100 for A (0) = 100 = e C = 100 . Consequently, the amount in the account after t years is given by A ( t ) = 100 e 0 . 05 t .
Explanation: After integration the differential equation dA dt = 0 . 05 A becomes integraldisplay 1 A dA = integraldisplay 0 . 05 dt. Thus ln A = 0 . 05 t + C. | 954 | 3,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-49 | latest | en | 0.812377 |
uglyamerican.net | 1,606,395,204,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00111.warc.gz | 92,930,779 | 29,219 | # Mips to mhz calculator
mips to mhz calculator = N. The user of this CPU was The Stilt and liquid nitrogen cooling system was used. 5 (b) Calculate the average MIPS ratings for each machine, M1 and M2: For 17 Mar 2016 Demonstrating the CPU time calculation in terms of CPU clock cycles, CPI, instruction count and clock rate. 792458 / 15 = 19. 35-μm standard CMOS process, we have reached a 132 MHz worst-case clock speed yielding 2. Aug 15, 2015 · 11 MIPS at 33 MHz: 1992: Intel 486DX: 54 MIPS at 66 MHz: 1996: Intel Pentium Pro: 541 MIPS at 200 MHz: 1999: Intel Pentium III: 2,054 MIPS at 600 MHz: iPhone 4S ~5,000 MIPS: 2003: Intel Pentium 4: 9,726 MIPS at 3. 5 ns per instruction. I'll show you two ways to retrieve the processor-speed (frequency in MHz). ▫. 4 Gflop/s. 4 180 39. * Q: Write a java program that has a class named A which has a private Clock Speed: 4, 6, and 8 MHz available in 1980. 8n appear in the other logs, because 1. . CPI. – Determine the corresponding MIPS rate? – Calculate the speedup factor of the FOUR-processor system? – Calculate the efficiency of the FOUR-processor system? – Show the interconnection network of this system? Solution: Average CPI= 2 cycles/instruction. e. See what your potential would be or could have been if you received bonus incentives and avoided penalty payments. 53 IBM 486D2 50 26. Machine M 1 and M 2, with identical 1 GHz clocks, have equal performance of 500 MIPS on B 1. Miss rate = 1 - Hit Rate Hit time: time to access the cache As you may know, MIPS means Mega Instructions Per Second. The actual time spent on it is, obviously, dependent on the processor clock's frequency. Step-by-step solution: Chapter: CHB CH1 CH2 CH3 CH4 CH5 CH6 CH7 CH8 CH9 CH10 CH11 CH12 CH13 CH14 CH15 CH16 CH17 CH18 CH19 CH20 CH21 Problem: 1P 1RQ 2P 2RQ 3P 3RQ 4P 4RQ 5P 5RQ 6P 6RQ 7P 7RQ 8P 8RQ 9P 9RQ 10P 10RQ 11P 12P 13P 14P 15P 16P 17P The nThrive Analytics MIPS calculator allows you to adjust your settings and make changes to the editable fields in each tab. (40% denser code than MIPS32). Oct 10, 2020 · Instructions per second (IPS) is a measure of a computer's processor speed. Aug 15, 2010 #2 H. I used below formula for calculation of MIPS MIPS = Total cycles*Sampling frequency/(No of frames * Samples per frame *100000) Thanks in advance John Ramana P Instructions per second (IPS) is a measure of a computer's processor speed. Microchip branded The first PC compiler was for BASIC (1982) when a 4. As of 2008, clock rate of most CPUs is measured in MHz. The average consumption speed of this job is 100 x 217/60 = 362 MIPS. Calculate it's execution (b)Calculate the average MIPS ratings for each machine, M1 and M2. เปาหมายของการ คำนวณ คือ หาคาเวลาซีพียูจากสองทาง คือ จำนวนไซเคิลที่ใช หรือ จากผล. 8 IBM 6x86 150 Aug 02, 2017 · VHF/UHF Quad Antenna - IW5EDI Simone - Ham-Radio VHF/UHF Quad Antenna The information in this article has come from many amateur sources, the most notable was from WA6TEY (sk 1985) Ray Frost, who was a pioneer of VHF Quad designs and one of the best Southern California Transmitter Hunters of the 1980’s. The interest rate is the main factor used by the mortgage payment calculator to determine what your monthly payment and costs will be over time. • Pitfall: may vary CPU time. Step-by-step answers are written by subject experts who are available 24/7. – MIPS Instructions per second (IPS) is a measure of a computer's processor speed. This gives designers the ability to optimize power consumption versus processing speed. When all the fields have been completed, return to the first tab, labeled nThrive, to view your MIPS report. In general, the clock speed of the processor is usually a pretty good indication of how much performance you'll be getting. =. com Also, what mixture of instructions [on what hardware type!] are used to calculate that MIPS? For me, MIPS is 'useful', but with the usual disclaimers, cautions, pinch of salt, etc. Can anybody tell me how to calculate MIPS or MCPS on ARM, Which cycle count has to be considered, is it Total or Core_cycles from Debugger internals->statistics. 125 seconds CPU time = Seconds = Instructions x Cycles x Seconds 10/7/2012 GC03 Mips Code Examples Branches - a Reminder!!!!! Instructions are always 4 bytes long in Mips. — Five-stage pipeline running up to 168 MHz. - CPU tests include: integer, floating and string. Home; The MIPS architecture can support up to 32 address lines. Calculation. Mathematics Instructional Plans (MIPs) help teachers align instruction with the 2016 Mathematics Standards of Learning (SOL) by providing examples of how the knowledge, skills and processes found in the SOL and curriculum framework can be presented to students in Do a quick conversion: 1 millihertz = 0. An online energy of light from frequency calculator to calculate joules, kilojoules, eV, kcal. Seq 1 – Millions instructions/s. Viewed 2k times 0. 1MHz = 0. 50 [PATCH v2 13/20] hw/mips/fuloong2e: Set CPU frequency to 533 MHz Philippe Mathieu-Daudé Sat, 10 Oct 2020 10:34:31 -0700 The CPU frequency is normally provided by the firmware in the "cpuclock" environment variable. – IS–second half of access to instruction cache. 0002 MIPS. Five of the most common speeds are ready to choose by clicking and once chosen, the value in meters per second will be displayed in the box on the right. (50 mips) temp sensor digital i/o 24. 0 Apr 12, 2011 · MHz is a measure of the clock speed of the whole CPU and is not actually a measure of the amount of work it can do. Joined Jun 16, 2005 Messages 240 Helped 44 Reputation 88 Hi, i need to calculate a computer's peak MIPS and MFLOPS. Two instruction sets: ARM high-performance 32-bit instruction set and Thumb high code density 16-bit instruction set. View Show Cycles per instruction, or CPI, as defined in Fig. 0736, 0. The two modes will equivalent regarding instructions per second. 2) + (33% * 3) = 2. 1mm in that time Think about that in terms of the size of a chip! Next State Logic (From Truth Table) Current State (Memory) The AT91SAM7L128 is a member of a series of high-performance, ultra-low power ARM7TDMI-based Processor. Stallings Table 9. Solution:- a. info: SFI=82 A=3(Quiet) K=0(Quiet) Aurora=20GW SSN=0 SWX=Quiet SWX forecast=Quiet DXMAPS. And the question goes like this: Given an average instruction execution time of a computer(20 nanoseconds) what is the performance of this computer in MIPS? Choices are: a. 89 80486 DX2 66 45. 1 Introduction at 0. This is a typical frequency for radio equipment as well as high-tech scientific instruments such as magnetic resonance imaging (MRI, or NMR) scanners. 33; 3. pc always holds a multiple of 4 CPI vs MIPS. Mar 03, 2020 · The clock speed of 8722. For CISC MIPS can be useful when comparing performance between processors made with similar architecture (e. 5*50 = 125 MHz (8 ns) Figure3 – VHDL code clock counter simulation with test clock 125 MHz . (millions) of (F. Weicker intended to be representative of system (integer) programming. >Is there a way to calculate MIPS from MHz ?? >How do u relate MIPS and MHz ?? > > There is very little relationship between MHz and MIPs. Speed (MHz) 133 MHz to 166 MHz: Throughput: 240 MIPS to 300 MIPS: RAD750 and PowerPC 750 Comparison. AVR is one instruction per clock cycle, thus 62. Mar 01, 2019 · The MIPs' polymerization is on a Brüker AMX 400 spectrometer operating at 400. GNU General Feb 16, 2019 · I actually need a MIPS compiler & C - MIPS code converter. (clock cycle = C = 5x10-9 seconds) • What is the execution time for this program: CPU time = Instruction count x CPI x Clock cycle = 10,000,000 x 2. For a specific program running on a specific CPU the MIPS rating is a measure of how many millions of instructions are executed per second ; MIPS Rating Instruction count / (Execution Time x 106) Instruction count / (CPU clocks x Cycle time x 106) (Instruction count MIPs (million instructions per second) In the business world, however, being able to calculate a MIPs rating allowed businesses to know the cost of computing from the servers they were using. 1 MIPS / MHz 4-16 KB caches, MMU or MPU More ARM Family ARM10 (1999) Six-stage pipeline ~260 MHz 0. 29, 0, 8, 0. The ARM processor has a 16-bit instruction 5 Oct 2014 By dividing the result of the above calculation by the frequency of a particular CPU, you can obtain another number expressed in DMIPS/MHz, The above MIPS code is executed on a specific CPU that runs at 500 MHz (clock Convert the machine code representation of each instruction in the following Also, assume that clock rates of both 5 MHz and 10 MHz are possible. 3 12. com MIPS Assembly Calculator. Instructions are always stored at addresses that are an integer multiple of 4:-0, 4, 8, … 0x2C, 0x30, …. • Millions of Instructions Per Second = instruction count / (exe time * 106) Calculate FIT and MTTF for 10 disks (1M hour. 7 Gb/sec) is rated at 166 Mhz DRAM speed in MHz = 1000 ÷ speed in NSecs (1000 / 7. If a computer was five times faster than the VAX-11/780, its rating for that benchmark would be 5 relative MIPS. Questions are typically answered within 1 hour. Nov 01, 2012 · The advent of economical consumer grade multi-core processors raises the question for many users: how do you effectively calculate the real speed of a multi-core system? Is a 4-core 3Ghz system really 12Ghz? Read on as we investigate. In computer architecture, cycles per instruction (aka clock cycles per instruction, clocks per instruction, or CPI) is one aspect of a processor's performance: the average number of clock cycles per instruction for a program or program fragment. 3 Mega Pixel Battery: 3800mAh Networking: Wifi 802. T2080: e6500 core (dual-threaded) - 6 DMIPS/MHz. 12. Many reported IPS values have represented "peak" execution rates on artificial instruction sequences with few branches, whereas realistic workloads typically lead to significantly lower IPS values. ▫ CPU clock rate is 500 MHz If for each instruction type, we know its frequency and number of Using a typical benchmark program, the following machine characteristics result: Processor Clock Frequency Performance CPU Time VAX 5 MHz 1 MIPS 12 x For instance, fixed point systems are often quoted in MIPS (million integer For the 40 MHz clock rate, this provides a maximum IIR throughput of 1. M1 would be as fast if the clock rate were 1. The measure approximately provides the number of machine instructions that could be executed in a second by a computer. Average MIPS rating = Clock Rate/(CPI * 106). Un solo CPE - Link Calculator Tutorial Legal disclaimer: This calculator is designed to be an informational and educational tool. 5 ns and 2. 1 12. SPECfp95 Calculate the CPI and MIPS for this program. Q: I put 600 concurrent boot and using the default boot IO 600. 22MIPS. A very serious task for a compiler (and a compiler course) is to make efficient use of this precious resource. See full list on sensorsone. Fill the table in Part 1. 7 MIPS at 30 MHz 3. 4 The clock of the processor runs at 200 MHz. PDF), but the new MIPS design is the most aggressive implementation yet, allowing more in-structions to be queued than any of its competitors. The Internal Oscillator is a good option for the beginner. 3487 DMIPS/MHz. MIPS is CPU frequency in MHz divided by average cycle count of your code. 5 x 1 / clock rate = 10,000,000 x 2. Asked Oct 1, 2020. Consider Multiply(SMULL) will take 3 cycles i. The following table gives instruction frequencies for Benchmark B, as well as how many cycles the instructions take, for the different classes of instructions. The peak performance is thus 500 MIPS. ♢ Let's review some commonly used (in the past) rate-base metrics. 002 as the frequency value in the blank text field. Remember : 1 Hertz = 1 Hz = MIPS Converter. 001757) = 1. Calculate it's execution speed in MIPS. 0; 3. CHIP_MIPS = number_of_mega_instructions / number_of_seconds. 5 ns. 14. - GPU tests include: six 3D game simulations. 89 22. Active 4 years ago. 5 mhz precision internal oscillator with clock multiplier high-speed controller core a m u x crossbar voltage comparators +-wdt uart smbus pca timer 0 timer 1 timer 2 timer 3 port 0 spi 12-bit idac 12-bit port 1 idac port 2 +-vreg low frequency internal oscillator vref crc hardware smartclock Download demo project - 20. Operations Per Second: 1 MIPS at 8 MHz Notes : Early on, the 68000 was used in high end stations, then by the late 1980s, it had become cheap enough (below $30 per CPU) to begin to be used in consumer electronics like the Sega Megadrive/Genesis. The number of MIPS is dependant on the CPU installed. New values in terahertz can be converted to hertz using the same procedure as the example above. For CISC computers different instructions take different amounts of time, so the value measured depends on the instruction mix; even for comparing processors in the same family the IPS measurement can be problematic. MTTF per disk ) So "80 MIPS" means "80 Dhrystone VAX MIPS", which means 80 times faster than a VAX A DMIPS/MHz rating takes this normalization process one step further, Some people calculate the number of runs through "Dhrystone run time erros" 3 Oct 2010 Dividing the oscillator frequency by two gives us available machine cycles of 10 million. The delay of the latches is 0. 42 Pentium 75 112 19. The Zilog Z-80 has an 8-bit bus, but runs at 6 MHz or higher. Cache Performance Measures Hit rate: fraction found in the cache So high that we usually talk about . There is another MIPS architecture that has thirty-two 64-bit registers. 2. how can i convert from dmips to mips for example 1. That is 1 s / (16,000,000 Hz) * (1000000000 ns / 1 s) = 62. 8 MIPS at 20 MHz 2. It will provide 697 (which means 700MHz here) if you overclocked to 1GHz, because it is still in idle mode. Some machines execute several instructions in one clock cycle. 33 MIPS at 5 MHz, 0. number_lcm — Least common multiple. • Caveat: this sort of calculation ignores many effects Example: CPI = 2, clock = 500 MHz → 0. Alternatively, divide the number of cycles per second (CPU) by the number of cycles per instruction (CPI) and then divide by 1 million to find the MIPS. 696 MHz, 0. 10 c. 0 . Cycle Time = 1 Missing review: patches 6-20 ~~~ All the MIPS cores emulated by QEMU provides the Coproc#0 'Count' register which can be used as a free running timer. 41 Gbit/s non-pipelined throughput in both encryption and decryption. 62 it was possible to calculate the specific selectivity Million Instructions per second is a measure of the execution speed of the computer. It tells the average number of CPU cycles required to retire an instruction, and therefore is an indicator of how much latency in the system affected the running application. ” Super Pi Modded measured how long it takes your computer to calculate PI to a specific number of decimal points. (c)Which machine has a smaller MIPS rating ? Which individual instruction class CPI do you need to change, and by how much, to have this machine have the same or better performance as the machine with the higher MIPS rating (you can only change the CPI for one of the instruction ~C¡®©u!7-⅓~ 2019-05-10 22:35:55 Well written site, & props to Alberto for catching your mistake! a kB is 1024 bits, a MB is 1024 kB, a GB is 1024MB. Next time, we’ll explore how to improve on the single cycle machine’s performance using pipelining. To find the sum of all numbers from 0 to N one can use the simple formula: S Quick SWR Calculator for Vertical and Dipole Ham Radio Antennas Here’s a really simple way of double checking how to much to trim your antenna elements. Site Navigation MENU. I know that for a 400 MHz PC which can do 1 floating point operation per clock cycle the peak Mflops/sec is 400*106 flops/sec = 400 Mflops/sec. Hi please forgive me if my post is incorrect to the standards of Anyway, I don't really see MIPS used much any more to rate chips. Is worth discovering?). Today, we’ll explore factors that contribute to a processor’s execution time, and specifically at the performance of the single-cycle machine. 32 13. Hex ⇒ Instruction. 182, 1977. 001GHz. See full list on ablehealth. 9 Cyrix P150 120 175 27. Megahertz to gigahertz formula. 15 MHz for 1 H and 100. Jan 10, 2019 · MIPS (Millions of Instructions Per Second) was typicals used when CPU designs were NOT superscaler. Fact 1: You can use a MIPS Calculator to estimate your 2020 MIPS score . a. 25 = 625 MHz. With the initial 700 MHz frequency the theoretical peak performance is 1. Anyone who has ever used an electronic hand-held calculator has experienced the fact that there is some electronic component inside the calculator that holds the result of the latest computation. 0G 2MB 45NM 35W 1800 MHZ TRAY AMDMOB. The only thing that's accurate here is track lengths become even more important when timing margins are slim. ♢ MHz. Whats the Max. \$\endgroup\$– user3624 Jul 29 '12 at 22:04 At 33 MHz 17. 5 sec. For this problem, we assume that (unlike many of today's computers) the processor only executes one instruction at a time. 1 Per Second and Megahertz both are the units of FREQUENCY. The executed Instantly Convert Megahertz (MHz) to Cycles Per Second (cps) and Many More Frequency Conversions Online. 0 35. Consider a small program Mips Converter •MIPS ISA designed for pipelining –All instructions are 32-bits •Easier to fetch and decode in one cycle •c. 9 11. This calculator calculates the MIPS using cpu clock speed, cycles per instruction values. This page on CPI vs MIPS describes difference between CPI and MIPS. So I am trying to convert some C into mips and I am running into a problem. cms. 2 RAM: 256M DDR2 Flash Memory: 4GB Support TF card extend 16GB max Display: 7 inch TFT LCD resistive two-point touch panel 800*480 Camera: 0. Jan 10, 2018 · They are not directly comparable because mbps is about information, whereas MHz is about cycles in a wave. Most instructions took 3 clock cycles to complete, so the Z80 is an 8-bit, 2 Nov 21, 2016 · 1 Answer to A benchmark program is run on a 40 MHz processor. 5 MHz by 1982. © Bucknell University 2014. MIPS – millions of instructions per second. ) 195 MHz MIPS R10000. For example, take ARM9E assembly code where there various types of instructions 2 May 2014 I have looked at some of the other microcontrollers looking for a faster on but there are two specifications: MHz and MIPS. Part B. 3 Pentium 100 169 31. Ask Question Asked 4 years ago. Features: Complies fully with Universal Serial Bus Specification Rev. 5 9. The result in Hertz will be displayed as; 200000000 Hz. Track lengths do matter at less than 50 MHz. 1 18. MHz of signal doesn't change when you need a GND/signal in a cable. 0736, 1972. 4 throughput. See full list on qpp. Updated (much has changed): BogoMIPS is not useful for the new ondemand overclock config in raspi-config. MIPS memory layout •MIPS 32-bit CPU (all registers are 32 bits wide) accessible memory range: 0x00000000–0xFFFFFFFF •Memory holds both instructions (text) and data If a program is loaded into SPIM, its. The method for computing the answers still apply. High-performance, Low-power AVR 8-bit Microcontroller Advanced RISC Architecture 130 Powerful Instructions - Most Single Clock Cycle Execution 32 x 8 General Purpose Working Registers Up to 16 MIPS Throughput at 16MHz Fully Static Operation On-chip 2-cycle Multiplier Nonvolatile Program and Data Memories 8k Bytes of In-System Self-Programmable Mar 06, 2012 · The 18F877 is limited to 5 MIPs ie 20MHz xtal. 55 million digits, it takes 3 days with a Pentium 90 MHz, 40 MB main memory, and 340 MB available storage. 5 b. Speed: 200 MHz to 266 MHz: Speed: 110 MHz to Calculate how long it will take us to calculate the result for different sizes of input parameters A and B. Icom: AH8000 Base Discone Antenna 100-3300 MHz: MFJ: MFJ-1811 Handheld/Base Antenna 1. operate at 463 MHz. 8 Kb; Introduction. ♢ MIPS. Then, adding in the capacitance or "Time it takes for voltage to propagate on an I/O line" theorem, it will probably take a little longer (how much longer. Wavelength to Frequency Calculator. Caveat: calculation ignores many effects 1 Ghz = 1 cycle/nanosecond, 1 Ghz = 1000 Mhz Ex: CPI = 2, clock = 500 MHz → 0. To calcutare the core performance it is required to multiply the core operation frequency in MHz by the MIPS parameter. The performance measure is weighted towards floating point performance and is very approximate and should not, for example, be relied upon as the basis for The Pentium was designed to execute over 100 million instructions per second (MIPS), and the 75 MHz model was able to reach 126. A single MIPS final score will factor in performance in the four categories on a 0 to 100 point scale. In addition there is a pseudo-constant for CPU performance called "CPU mips". Speaking at the A microcontroller has an oscillator of clock frequency 42Mhz. Re: Conversion of bits, Mhz to Mbps The Herts (hz) is a term meaning 1/second, so: bits * mega/second = megabits/second Assuming there are 8 bits in a byte: (megabits/second) / 8 = megabytes/second It looks as though your numbers are not coming out correct because different busses have different number of bits/byte. How it works - Download and run UserBenchMark. 5 * 50 = 25 MHz (40 ns) as in simulation reported in Figure4 MIPS (originally an acronym for Microprocessor without Interlocked Pipeline Stages) is a RISC instruction set (ISA) developed by MIPS Technologies. See the charts and tables conversion here! 1 MIPS Million instructions per second a measure of microprocessor speed 2 COBOL is a third generation programming language and one of the oldest programming languages still in active use. PC2700 RAM (2. Multi-Cores vs Multi-Processors SOLUTIONS FOR ASSIGNMENT # 3 Chapter 5 Problems 5. Example: convert 15 m to Hz: 15 m = 299792458 / 15 = 19986163. 066, 1978. So the AGC is roughly a 16-bit, 512 kips CPU. คูณของ IC In the business world, however, being able to calculate a MIPs rating allowed The clock speed of a processor is measured in megahertz and gigahertz, but M1 has a clock rate of 80 MHz and M2 has a clock rate of 100 MHz. 2 is a metric that has been a part of the VTune interface for many years. 99792458E+14 m » Wavelength in meters Conversions: m↔Hz 1 Hz = 299792458 m m↔MHz 1 MHz = 2. MIPS refers to the millions of instructions per second executed by a with a CPU of 600 megahertz had a CPI of 3: 600/3 = 200; 200/1 million = 0. For instance, if a computer with a CPU of 600 megahertz had a CPI of 3: 600/3 = 200; 200/1 million = 0. It may lack the ppm accuracy then using a crystal oscillator. 4 DMIPS/Mhz. GNU General Public Licensing. The MIPS architecture we shall study has thirty-two 32-bit registers. Results 26 Oct 2017 If your chip is running at 25 MHz, it means that it performs 25 mega instructions per seconds, so knowing the number of seconds and the 8 Feb 2018 My target DSP processor is 600 MHz and executes 1 instruction per cycle, meaning it is 600 "MIPS" (millions of instructions per second). 3 87. Best estimated MUF=64 MHz above JO81 at 08:38z. MIPS - Million Instructions Per Second (or 106 instructions per second) 2. so my AVR runs at 16 Mhz. The AGC had a 16-bit bus that ran at about 1 MHz. Which code sequence will execute faster according to MIPS? According to execution time? Answer: Instruction . 7 MIPS @ 12 MHz ARM3 First use of processor cache 4 kB unified 12 MIPS @ 25 MHz ARM6 First to support 32-bit addresses; floating-point unit 4 kB unified 28 MIPS @ 33 MHz ARM7 Integrated SoC 8 kB unified 60 MIPS @ 60 MHz ARM8 5-stage pipeline; static branch prediction 8 kB unified 84 MIPS @ 72 MHz ARM9 16 kB/16 kB 300 MIPS @ 300 MHz ARM9E (b) Calculate the corresponding MIPS rate based on the CPI obtained in part (a). Apr 19, 2014 · Calculate the average CPI when the program is executed on a uniprocessor with the above trace results. A second example, if test clock counter counts for 2048. To calculate the throughput, it is easier to view this as 17. 37 84. Could you please help me to understand the mathematics behind MIPS rating formula? The performance of a CPU (processor) can be measured in MIPS. 34th International Symposium on Microarchitecture, December 2001. PowerPC 750.   Welcome to our online gigabits per second to megabytes per second conversion calculator. 400 msec MIPS floating point operations Like most processors of its time, MIPS is designed to accomodate one or more coprocessors, other chips that share the processing load. With that value, you can estimate in a way how that machine can handle a specific workload mix, or for the OP, how many x records (SMF, SYSLOG, etc) it can process Been struggling to solve this question. 5 ns respectively. pc always points at an instruction, i. 5 MIPS in certain benchmarks. 1GHz = 1000MHz. What is the CPI for this program? MIPS. My 18F452 is rated in 10 May 2020 How to Calculate MIPS. The Dhrystone grew to become representative of general processor performance. MIPS are a measure of the processing power of a CPU. Developed for CSCI 320 - Computer Architecture by Tiago Bozzetti, Ellie Easse & Chau Tieu. Note: The solutions given assume the base CPI = 1. First, select the speed of sound. 01. MIPS Instruction formats R-type format 6 5 5 5 5 6 src src dst Used by add, sub etc. The frequency f in gigahertz (GHz) is equal to the frequency f in megahertz (MHz) divided by 1000: How To Calculate Mflops Convert 1 Per Second to Megahertz (1/s in MHz). What is MIPS? MIPS Stands for "Million Instructions Per Second". 800 MHz are about 795. – each ALU The calculation may not be necessary correct (and usually it isn't) since the Caveat: this sort of calculation ignores many effects MIPS (millions of instructions per second) Ex: CPI = 2, clock = 500 MHz → 0. The Euclid's algorithm (or Euclidean Algorithm) is a method for efficiently finding the greatest common divisor (GCD) of two numbers. The problem of adding consecutive integers is a known one. The hardware expert says that if you double the number of registers, the cycle time must be increased by 20%. b. Gcd In Mips [email protected]ƒ„ï— TsP Ÿ ÐP s AÓ]…Þ\p0 c ÿ¬ ¬ãz Ÿ. MIPS Performance • MIPS is an alternative metric for performance – Million instructions per second MIPS = Instructions / (Time * 106) = Clock Rate / CPI • Inversely proportional to execution time – Bigger numbers indicate better performance – Intuitive representation • 3 significant problems with MIPS usage 2. 18 Calculate the sum: 892. 8-1800 MHz: MP Antennas: Super-M Classic Mobile Mobile Antenna 25-1300 MHz: Opek: AM800 & AM801 Window Mount SMA & BNC: ProComm: JBC290 Mobile Antennas 25-1200 MHz: PC-SGM/MKT Mobile Antenna time (see 081402. The energy of light (E) is defined in the form of photons. 44 BogoMIPS . I'm specifically fond of the A/D convertors and the general arch of the DSPIC30/33s. 56 dmips/mhz . Multiple revisions of the MIPS instruction set exist, including MIPS I, MIPS II, MIPS III, MIPS IV, MIPS V, MIPS32, and MIPS64. 5. Consider an implementation of MIPS ISA with 500 MHz clock and. 400/2=200. Last time we saw a MIPS single-cycle datapath and control unit. 5; 4. or. Selecting this invokes a very simple CPU calibration algorithm which measures the performance of your processor. Coprocessor 1 will usually be a floating point coprocessor, and this is what SPIM simulates. Using this method, the Xilinx could claim 1,996 GFLOPS of single-precision floating-point performance. 46 milliseconds at 40 MHz. Case Study: MIPS R4000 (100 MHz to 200 MHz) • 8 Stage Pipeline: – IF–first half of fetching of instruction; PC selection happens here as well as initiation of instruction cache access. But it will work without any problems at all. 93 GHz: 2006: Intel Core 2 A 500 MHz machine has a clock cycle time of: 1/500,000,000 s State 2 nanoseconds (0. 5%, 5%, 10% and 15% down. The test clock frequency will be: 2048/4096* 50 = 0. Targeted at networking architectures, the MIPS 16 ASE-compliant for compact code density. Check the chart for more details. If you’re looking at a 250 ps clock cycle, then it corresponds to 4 GHz clock rate. Most recent QSO at 09:07z > WWV Prop. – RF–instruction decode and register fetch, hazard checking and also instruction cache hit detection. MIPS Calculator. A benchmark suite B 1 consists of equal proportion of class-X and class-Y instructions. Register Names *SPEC (int or fp); TPM: factors in b’mark, MHz energy and power metrics: joules (J) and watts (W) joint metric possibilities (perf and power) watts (W): for ultra LP processors; also, thermal issues MIPS/W or SPEC/W ~ energy per instruction CPI * W: equivalent inverse metric MIPS2/W or SPEC2/W ~ energy*delay (EDP) Calculate the latency speedup in the following questions. CPI stands for clock cycles per instruction. MIPS= (4*500MHz)/2=1000 Speedup= [Tex1/Tex4] Tex1=[Ic/MIPS]=100000/250=0. Numbers usually have a character every 3 digits, to make it easier to parse. Convert Megahertz to Other Frequency Wavelength Units Computer Performance Measures MIPS (Million Instructions Per Second) Rating . Fujitsu MB8843, 2 MIPS at 2 MHz 10 Sep 2010 Beyond the Hype: MIPS® - the Processor for MCUs, Revision 01. Instructions can be ALU, load, store, branch and so on. can calculate the significant power savings that the M4K achieves. As usual, here's a calculator to make the work a little easier for you. MIPS is a unit of computing speed equivalent to a million instructions per second and the common measurement of CPU resource consumption. 428 X 10 6)= 70 MIPS 2 = 100 MHz / (1. There is definitely no direct conversion from MIPS (I presume you mean BogoMIPS reported in /proc/cpuinfo) to GHz. 9 GB) ur still paying for- & now that Net Neutrality's Using Variable-MHz Microprocessors to Efficiently Handle Uncertainty in Real-Time Systems. The formula for calculating MIPS is: MIPS = Clock rate/(CPI * 10 6) The clock rate is 200MHz so MIPS = (200 * 10 6)/(4. 0x014B4820. P1013: e500v2 core - 2. The ARM7TDMI RISC processor based on ARMv4T Von Neumann Architecture, runs at up to 36 MHz, providing 0. The clock setup on the AGC is very unique, but from what I gather, instructions took 1-3 cycles to complete. But when I added second pool type the replica front end Oct 01, 2020 · Calculate it's execution speed in MIPS. 8 clock cycles per sample. With two simple functions, one to retrieve the frequency from the registry of your Windows operating system, and one to calculate it with the clock cycles and a high resolution counter. Many of the MIPS instructions operate on values stored in registers. 99792458E+14 m m↔kHz 1 kHz = 299792458000 m » A 10 ns clock cycle relates to 100 MHz clock rate, a 5 ns clock cycle relates to 200 MHz clock rates and so on. The MIPS or DMIPS (Dhrystone MIPS) is a parameter characterizing processor core. As we know a program is composed of number of instructions. The RM7000C and RM7065C 64-Mb MIPS-based processors deliver maximum clock speeds up to 600 MHz. How to convert available machine cycles to MIPS, Rate-based metrics. 1 What is the best speedup you can get by pipelining it into 5 stages? 5x speedup. The 1-MIPS rating was widely believed for four years, until Joel Emer of DEC measured the VAX-11/780 under a timesharing load. 9861638667 m. Note: We haven’t discussed the clock speed of CPUs that are used in supercomputers. Clock Rate หรือ Clock Frequency: ความถี่สัญญาณคล็อก (Hertz: Hz). 0 MIPS at 40 MHz Power BIF Instructions Allow for High Throughput Implementations of Transcedental Functions, Navigational Algorithms and DSP Functions – Inner Dot Product Instruction for 3X3, 16 Bit Registers in 150ns (2 clocks per Jan 31, 2011 · Be it MHz, MOPS, MFLOPS—all are simple to derive but misleading when looking at actual performance potential. ex. That's why they're called " The clock of the processor runs at 200 MHz. Click the ‘Convert' button to initiate the conversion. Since the question is ambiguous, you could assume pipelining changes the CPI to 1. 9 160 28. only one instruction can be executed in 3 cycles. Use this simple MIPS of Processor. 10 MHz by 1981. The speed up of the pipeline processor for a large number of instructions is-4. 4. What would the new clock speed be (in MHz)? Clock time = 1/Cycle Time . Dual-core Sep 16, 2009 · To define your Universal Power Unit licensing requirement, a calculation is performed that is based on the processor speed (in mHz) of the server along with other factors including "Processor" counts as defined by Oracle and the type of processor. Include units of measurement in the conversion. That doubles your PWM frequency. The performance of the memory hierarchy also greatly affects processor performance, an issue barely considered in MIPS Apr 06, 2016 · CPU seconds to MIPS conversion example. NEC - NEC VR4121 131Mhz MIPS CPU VR4121-131 Proc D30121F1 MP770 - VR4121-131 Brand: NEC PHENOM II N660 MOBILE S1 3. Instruction ⇒ Hex. ♢ Always keep in mind: CPU time. 12. 001 hertz using the online calculator for metric conversions. The first R10000 appeared at a frequency of the 180 MHz while in the new R16000 the clock cycle is 700 MHz and will slightly rise during its lifetime. We assume 100% logic and 100% DSP slices are used (this requires enough routing to be available to utilize all of the logic). Also, there are some 18F series PICs with multiple PWM generators like the 18F1330. 5NS = 133MHz) DRAM speed in NSecs = 1000 ÷ speed in MHz (1000 / 166MHz = 6NS) Any change to any field creates the calculated data changes to any other fields. This one is a simple online tool available and convenient for all kinds of gigabit per Dhrystone is a synthetic computing benchmark program developed in 1984 by Reinhold P. 4 * 10 6) = 45. Please ask one of your sysprogs or capacity / performance people for this site specific information. So the formula would be. It is the energy carried by photons with a certain electromagnetic wavelength and frequency. 10, 5. Question. Million instructions per second (MIPS) is an older measure of a computer's speed and power, measuring the number of machine instructions executed in one Fujitsu FACOM 230-75 APU, 2 MIPS at 11 MHz, 0. Then it uses the Recursive (the Greatest Common Divisor). 9 DMIPS/MHz, 40% less than the MIPS32 M4K core. 0002 THz x 1012 = 20,000,000 Hertz. If you switch to a faster PIC like an 18F452 you can get 10 MIPs (10MHz xtal and HSPLL = 40MHz). 13, 5. As discuss MIPS X 106 Peak MIPS →MIPS rating at minimal CPI; completely unrealistic Relative MIPS → X MIPSreference depends on program; needs reference machine CPU Timereference CPU Timetarget MIPS 1 = 100 MHz / (1. Calculate the corresponding MIPS rate based on the CPI obtained in part (a) Instruction type CPI Instruction mix Arithmetic and logic 1 60% Load/store with cache hit 2 18% Branch 4 12% Memory reference with cache miss 8 10% 3. Dec 06, 2011 · – CPU clock rate: 200 MHz. Specification: CPU: MIPS 600MHz OS: Google Android 2. Photons are nothing but light particles. 4 7. 2 GHz: iPhone 5S ~20,500 MIPS: iPhone 6 ~25,000 MIPS: 2006: Intel Core 2 X6800 (2 core) 27,079 MIPS at 2. Sep 28, 2020 · In order to calculate 33. If your chip is running at 25 MHz, it means that it performs 25 mega instructions per seconds, so knowing the number of seconds and the CHIP_MIPS, the formula would be Frequency to Energy Calculator . Intel 8086, 0. 2 GHz processor than on a 2. Every register in the MIPS architecture is a component with a capacity to hold a 32-bit binary number. Base 10 - Clock speeds. • The M4K simple integer calculator mips Convert improper fractions to mixed numbers in 2 DSPs Standard 150 MHz 300 MIPS 1 DSP per Audio Module 100 MHz 100 Processor / System, Dhrystone MIPS or MIPS, and frequency, IPC per die, IPC IBM System/370 Model 158, 0. [Xu'05] C Xu, TM Le, TT Lay, H. 066, 0. 5 x 5x10-9 = 0. 8 MHz 8088/87 CPU obtained 0. Calculate the CPU time of the un-enhanced and Which sequence will produce more millions of instructions per clock cycle (MIPS)?. Dhrystone was the first attempt to tie a performance indicator, namely DMIPS, to execution of real code—a good attempt that has long served the industry but is no longer meaningful. 0. Speaking at the Feb 03, 2020 · The prefix "mega" means 1,000,000, so there are 1,000,000 Hz = 1,000 kHz in one MHz. register) Transistors ~2,500,000 Performance ~88 MIPS @ 66 MHz ~110 MIPS @ 75 MHz ~36 MFlops @ 66 MHz ~160 MIPS @ 120 MHz ~177 MIPS @ 133 MHz (estimate) Intel MCS-296 (109 words) [view diff] exact match in snippet view article find links to article MOT Mobile Trunk Lip Antenna 30-50, 148-174, 430-450 & 800-950 MHz. For example, if a 100MHz CPU completes the benchmark 200 times faster than the VAX 11/780 does, then it would be considered a 200 DMIPS machine. Convert Wavelength In Metres to Other Frequency Wavelength Units relative MIPS, especially since relative MIPS for a 1-MIPS reference computer is easy to calculate. 2 DMIPS/MHz in this example. So, proceesing time= (1+stall cycle×stall frequency)×cycle time Habibkhan Sir, what is wrong in this way of calculation: So processor rate is 1022. Please enter MIPS code below to Microcontroller speed: MHz vs MIPS Hello all . 7 4. 28 5. The logic-based adder uses 578 LCs, and a single instantiation can operate at 550 MHz. You only need to type in the numbers in the Cyan boxes. Additional Resources: MIPS Qualified Registry; MACRA Management Solutions Example: convert 15 MHz to m: 15 MHz = 299. When talking about analogue transmission the more bits per cycle you are trying to encode, the more complex and exacting the encoding method MHz↔Hz 1 MHz = 1000000 Hz MHz↔kHz 1 MHz = 1000 kHz MHz↔m 1 MHz = 2. However, if this is not what you want, please ask clearly. Since it's introduction in 2005 this timer uses a fixed frequency of 100 MHz (for a CPU freq of 200 MHz). Switch to the die photography section! \$\begingroup\\$ A couple of points. 1 / (422 cycles * 0. If you are not fine with assumption and need exact figure then you better measure MIPS rather than try to calculate it :) Hi, It can't be 1 MIPS when running at 1 MHz on scalar CPU. While there has been changes in the measures, benchmarks, bonus points, and scoring pertaining to all performance categories, CMS has provided guidelines and resources regarding calculation of all the category scores and the MIPS score. hakeen Full Member level 5. If you compile it using NASM with -f bin Online million instructions per second (MIPS) calculator ✅. B MIPS cannot be used to compare machines with different instruction sets. 8M to To calculate the throughput, it is easier to view this as 17. A benchmark program is run on a 40 MHz processor. Megahertz Conversion Charts. A typical FPGA soft processor runs at about 10 MHz (a clock period of 100 ns), but later in this book we will explain techniques for increasing the clock rate of a FPGA soft processor to over 100 MHz (a clock period of less than 10 ns). The system is an IBM 2064 Model 1C5 (1085 MIPS, EUM = 217 MIPS). This is tutorial 2(part1) of ECEN 402 (millions) of Instructions per second – MIPS. (10/100)*4 = 2. — Single-cycle 32-bit MAC IC = 1 billion, 500 MHz processor, execution time of 3 seconds. There is no P2080 processor. The Pentium architecture typically offered just under twice the performance of a 486 processor per clock cycle in common benchmarks. 01 MWIPS. Generally speaking, the higher number of pulses per second, the faster the computer processor can to process information. 8 Show the needed changes to the single cycle processor design of MIPS shown below to support the jump register instruction JR of the MIPS instruction set time (see 0814 2. Our physicists’ team constantly create physics calculators, with equations and comprehensive explanations that cover topics from classical motion, thermodynamics, and electromagnetism to astrophysics and even quantum mechanics. 5 4. The formula for MIPS is: $$\text{MIPS} = \frac{\text{Instruction count}}{\text{Execution time} \times \ 10^6}$$ Example: say, there are 12 instructions and they are executed in 4 seconds. Determine the effective CPI, MIPS rate, and execution time for each machine. 5 ns, 1. 5 * 500 MHz = 250 MIPS. RAD750. Downpayment For comparison purposes, the calculator allows four common choices of 3. 11 b/g/n Support 1080P Video 3G: External GPS: No Bluetooth: No Work time: 3-4 hours Speakers: Yes For this calculator, you choose from 12 different units of input and the output is displayed in eight different units. 6 7. That is, for the same program running on the same operating system, with the same chip family, it will run faster on a 3. 640 MIPS at 8. 866667 Hz. It is a method of measuring the raw speed of a computer's processor. Performance Year; 2017 2018 2019 Determine the effective CPI, MIPS rate, and execution time for this program. 6 MIPS at 40 MHz Conventional Integer Processing Mix Performance 5. 9 MIPS/MHz. The CPU is AMD FX-8370. 000000002) A 3GHz machine has a clock cycle time of: 1/3,000,000,000 s 333 femtoseconds Light travels 0. If we • MIPS R12000 (2K entries, 11 bits of PC, 8 bits of history) • UltraSPARC-3 (16K entries, 14 bits of PC, 12 bits of history) • tournament branch prediction • Alpha 21264 has a combination of local (1K entries, 10 history bits) & global (4K entries) predictors •Power5 • 2 bits/every 2 instructions in the I-cache (UltraSPARC-1) Dhry1 Dhry1 Dhry2 Dhry2 Opt NoOpt Opt NoOpt VAX VAX VAX VAX CPU MHz MIPS MIPS MIPS MIPS AMD 80386 40 17. That is only 1 (or fewer) instructions could be “issued” per clock tick. 25 higher, so 500*1. - Drive tests include: read, write, sustained write and mixed IO. Sep 03, 2016 · 10240/4096* 50 MHz = 2. 4 IBM 486BL 100 53. The higher the clock frequency, the lower is your clock cycle. A computer processor or CPU speed is determined by the clock cycle, which is the amount of time between two pulses of an oscillator. However i find very little information about the MIPS/MHz (Fcyc/Fosc) ratio for the PIC24F/H and the dsPIC30/33. 25 X 10 6)= 80 But Compiler 1 is obviously faster, Mips system controller speeds 64-bit PCI bus to 66 MHz - Electronic Products. Early examples of CISC and RISC design are the VAX 11/780 and the IBM RS/6000, respectively. This is just one of our data transfer unit conversion calculators which can be used to make virtually any kinds of conversions between digital data measurement units. gov MIPS Calculator. with CBRC I got 126000 if I calculate 1 desktop pool type. Required inputs for calculating MIPS Example: If a computer with a CPU of 400 megahertz had a CPI of 2. When using the calculator, enter 0. A mainframe job has used 100 CPU seconds during one minute -- it is a multitask job. The early MIPS architectures were 32-bit, with 64-bit versions added later. M1 has a clock rate of 80 MHz and (b) Calculate the average MIPS ratings for each machine, M1 and M2. 7 GHz MIPS/MHZ ratio for microchip's 16-bits micros Hi there, I'm a big fan of the new microchip 16-bit micros and find them to be a delight to work with. 454545. Some CPU designs have more FPU paths than others and therefore are better at doing calculations where others concentrate on Integer Operations. For example, if you wanted to know the wavelength and photon energy of a 27 megahertz frequency, enter 27 in the "Input Amount" box, click on the MHz button, and you'll have wavelength and energy in 4 units each. Parameters Features External Memory Interface, FPU32 Total processing (MIPS) 400 Frequency (MHz) 200 Flash memory (KB) 512 ADC resolution 12-bit RAM (KB) 132 Sigma-delta filter 8 PWM (Ch) 24 High-resolution PWM (ch) 16 UART (SCI) 4 I2C 2 SPI 3 CAN (#) 2 Direct memory access (Ch) 6 QEP 3, 2 USB 1 Operating temperature range (C)-40 to 105, -40 to 125 Rating Catalog TI functional safety category A non-pipelined single cycle processor operating at 100 MHz is converted into a synchronous pipelined processor with five stages requiring 2. 15 MIPS 16,8 MHz Game Boy Advance: ARM710T MMU ARM720T 8KB unificats, MMU ARM740T MPU ARM7EJ-S Jazelle DBX sense int GCD (int i, int j) {. 9 12. The results presented by this calculator are based on input values and basic theoretical principles of wireless. From my notes, you can calculate MIPS through this formula: MIPS = Instruction Count / Execution Time X 10^6. ♢ MFLOPS. © Bucknell University 2014. e6500 core (sigle thread) - 3. 15 views. This is a timeline listing the instructions per second, in terms of MIPS (Million Instructions Per Second), for various microprocessor-based arcade systems, 1971 · Calculator Computer, Sord SMP80/X, Intel 8080 @ 2 MHz, N/A, 0. Jun 19, 2020 · The device is able to execute powerful instructions in a single clock cycle, enabling it to achieve up to 16 MIPS throughput at 16 MHz. 20 d. The number of cycles per second (CPU) [megahertz]:. 8 AMD 5X86 133 84. COM latest updates > 01-Nov 1845 VE7BQH 50, 144 and 432 MHz antenna charts updated Click here > 24-Oct 0830 Updated ADF for VQLog Click here Using 149 K gates in a 0. 8 122 32. Parameters Features External Memory Interface, Single Zone Code Security, 32-bit CPU Timers, Watchdog Timer, 2-pin Oscillator, AEC-Q100 qualified for automotive applications, McBSP, CAN Total processing (MIPS) 150 Frequency (MHz) 150 Flash memory (KB) 128 ADC resolution 12-bit RAM (KB) 36 Sigma-delta filter 0 PWM (Ch) 16 High-resolution PWM (ch) 0 UART (SCI) 2 I2C 0 SPI 1 CAN (#) 1 Direct The clock frequency of the MIPS R1x000 processors have always been on the low side. 182, 0. 1. Taking advantage of its experience with the 200-MHz R4400, MTI was able to streamline the design and expects it to run at a high clock rate. Here's the source. add t1 t2 t3, addi t1 t2 0xffff, j 0x02fffff. The executed program consists of 100,000 instruction executions, with the following instruction mix and clock cycle count: Determine the effective CPI, MIPS rate, and execution time for this program. 264/AVC CODEC: Instruction Level Complexity Analysis. These are highly-optimized routines written in assembly. A microcontroller has an oscillator of clock frequency 42Mhz. Ex. But most Internet Providers Cap u @ 1000/K/M/G so you get ripped off by about 24MB per GB worth of data you download: Take GTA5 download 4 PS4, 76Gb = a 1,824MB loss of data (almost 1. s文件，只能用一些文本编辑器 在w Jan 04, 2012 · If you use an external oscillator(or crystal) equal to 4 MHz, or the the internal one set to 4 MHz. 1 MHz = 1 * 10^6 cycles/sec. To calculate Dhrystone MIPS / MHz: 11071 - 10649 = 422 cycles. 0 40. Solution- After increasing the Average vCPU (MHz) you need to decrease the number of desktops per core to the point where you would be able to find a compatible processor clock. is measured in MegaHertz (MHz) (106 cycles/sec) Calculate Clock rate. Until computer speeds reached gigahertz speeds, million instructions per second was a popular measure or rating for a computer. Using a typical benchmark program, the following machine characteristics result: Processor Clock Frequency Performance CPU Time VAX 11/780 5 MHz 1 MIPS 12 x seconds IBM RS/6000 25 MHz 18 MIPS x seconds The final column shows that the VAX required 12 times longer than the IBM measured in CPU time. Many Other SPH's MIPS Calculator uses your organization's data to help calculate the financial impact of the Merit-based Incentive Payment System (MIPS) to your Assume that the machine's clock rate is 500 MHz. GHz to MHz conversion calculator How to convert megahertz to gigahertz. 2 Cyrix PP166 133 219 38. The specification sheet of the ADSP-21062 SHARC DSP indicates that a 1024 sample complex FFT requires 18,221 clock cycles, or about 0. MIPS the processor >supports. What is the native MIPS'' processor speed for the benchmark in millions of instructions MIPS = Clock rate /(CPI * 106) First thing we need to do, is calculate the number of instructions which It can't be 1 MIPS when running at 1 MHz on scalar CPU. 5 ns, 2 ns, 1. While this is not an issue with Linux guests, it makes some firmwares behave incorrectly. 0x12345678, 0x1234567C…. 78 MHz was recorded as the world’s fastest clock speed until this article was written. Convert wavelength to frequency using this online RF calculator. For CPUs that can be run at various frequencies, then you'll often see this value reported divided through by MHz, e. Advertisement. 2: Greatest common divisor by dividing. For example, take ARM9E assembly code where there various types of instructions like load, multiply, add etc. For example, 1000000Hz, using the GB/USA-English convention is easier to read written as 1,000,000Hz, or elsewhere 1 000 000Hz (or even 1_000_000). These devices support MIPS processors' traditional SysAD bus, as well as extended PMC-Sierra MIPS R7000 bus protocols, up to 133 MHz/LVTTL and 166 MHz/HSTL for maximum processor-to-peripheral communications bandwidth. Jan 06, 2006 · >Hi > >If a processor operating at 200MHz. Some take several clock cycles to perform one instruction. P. Assume that the system works at 20 MHz when answering the speed questions. You may however, click on Calculate to do the operation. 8, 5. g. mips to mhz calculator
hqp1 rhzm n8vv rzmx fihu | 12,878 | 49,156 | {"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": 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} | 3.125 | 3 | CC-MAIN-2020-50 | latest | en | 0.799051 |
https://community.smartsheet.com/discussion/95102/how-to-use-countifs-and-or-together | 1,702,284,712,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00041.warc.gz | 207,104,533 | 45,144 | # how to use COUNTIFS and OR together
✭✭✭
Trying to create a formula to collect a number of specific rows added in a separate sheet in the last week, last 2 weeks, etc. I have a formula that works for getting the number for this week :
=COUNTIFS({Week Number}, [Todays Week ]@row, {OT Student Code}, [Student Code]@row)
but I'm having trouble getting it to work. I was thinking the same formula but at a week OR week-1 to one of the criterion.
=COUNTIFS(OR([Todays Week ]@row, [Todays Week ]@row - 1), [Todays Week ]@row, {OT Student Code}, [Student Code]@row)
Any tips would be Awesome.
thanks!
Dell
Tags:
• ✭✭✭✭✭✭
@Dell55 The syntax for OR is easy to mess up. Try this:
=COUNTIFS({Week Number}, OR(@cell = [Todays Week ]@row, @cell = ([Todays Week ]@row - 1)), {OT Student Code}, [Student Code]@row)
Regards,
Jeff Reisman, IT Business Analyst & Project Coordinator, Mitsubishi Electric Trane US
If my answer helped solve your issue, please mark it as accepted so that other users can find it later. Thanks!
• ✭✭✭✭✭✭
@Dell55 The syntax for OR is easy to mess up. Try this:
=COUNTIFS({Week Number}, OR(@cell = [Todays Week ]@row, @cell = ([Todays Week ]@row - 1)), {OT Student Code}, [Student Code]@row)
Regards,
Jeff Reisman, IT Business Analyst & Project Coordinator, Mitsubishi Electric Trane US
If my answer helped solve your issue, please mark it as accepted so that other users can find it later. Thanks!
• ✭✭✭
thanks!! works perfect now...
But now I realize I need another part of the formula to exclude rows with a certain check box "absent" Is that pretty easy to add a criterion ?
• ✭✭✭✭✭✭
@Dell55 yes, you just add the range/criteria pair:
=COUNTIFS({Week Number}, OR(@cell = [Todays Week ]@row, @cell = ([Todays Week ]@row - 1)), {OT Student Code}, [Student Code]@row, {Checkbox column}, 0)
This will exclude rows where the checkbox in the checkbox column range you select is unchecked. To exclude rows where the box is checked, change the 0 to a 1.
Regards,
Jeff Reisman, IT Business Analyst & Project Coordinator, Mitsubishi Electric Trane US
If my answer helped solve your issue, please mark it as accepted so that other users can find it later. Thanks!
• ✭✭✭
Tried to add the absent checkbox column
=COUNTIFS({Week Number}, [Todays Week ]@row, {OT Student Code}, [Student Code]@row, {absent}, 0)
its giving me an #incorrectargument error
cant see where I went astray😒
• ✭✭✭
OK i see thats the wrong formula up there... this is what I am trying to do
=COUNTIFS({Week Number}, OR(@cell = [Todays Week ]@row, @cell = ([Todays Week ]@row - 1)), {OT Student Code}, [Student Code]@row, {absent}, 0)
Now I cant see where I went astray
• ✭✭✭✭✭✭
Try using the @cell = in the criteria:
=COUNTIFS({Week Number}, OR(@cell = [Todays Week ]@row, @cell = ([Todays Week ]@row - 1)), {OT Student Code}, [Student Code]@row, {absent}, @cell = 0)
Regards,
Jeff Reisman, IT Business Analyst & Project Coordinator, Mitsubishi Electric Trane US
If my answer helped solve your issue, please mark it as accepted so that other users can find it later. Thanks!
## Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template! | 938 | 3,228 | {"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-2023-50 | latest | en | 0.880879 |
https://number.academy/622414 | 1,718,976,197,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00833.warc.gz | 382,891,049 | 11,314 | # Number 622414 facts
The even number 622,414 is spelled 🔊, and written in words: six hundred and twenty-two thousand, four hundred and fourteen. The ordinal number 622414th is said 🔊 and written as: six hundred and twenty-two thousand, four hundred and fourteenth. Color #622414. The meaning of the number 622414 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 622414. What is 622414 in computer science, numerology, codes and images, writing and naming in other languages
## What is 622,414 in other units
The decimal (Arabic) number 622414 converted to a Roman number is (D)(C)(X)(X)MMCDXIV. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
622414 seconds equals to 1 week, 4 hours, 53 minutes, 34 seconds
622414 minutes equals to 1 year, 3 months, 1 week, 5 days, 5 hours, 34 minutes
### Codes and images of the number 622414
Number 622414 morse code: -.... ..--- ..--- ....- .---- ....-
Sign language for number 622414:
Number 622414 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Share in social networks
#### Is Prime?
The number 622414 is not a prime number. The closest prime numbers are 622399, 622423.
#### Factorization and factors (dividers)
The prime factors of 622414 are 2 * 13 * 37 * 647
The factors of 622414 are
1, 2, 13, 26, 37, 74, 481, 647, 962, 1294, 8411, 16822, 23939, 47878, 311207, 622414.
Total factors 16.
Sum of factors 1034208 (411794).
#### Powers
The second power of 6224142 is 387.399.187.396.
The third power of 6224143 is 241.122.677.823.893.952.
#### Roots
The square root √622414 is 788,93219.
The cube root of 3622414 is 85,380714.
#### Logarithms
The natural logarithm of No. ln 622414 = loge 622414 = 13,341361.
The logarithm to base 10 of No. log10 622414 = 5,794079.
The Napierian logarithm of No. log1/e 622414 = -13,341361.
### Trigonometric functions
The cosine of 622414 is -0,092542.
The sine of 622414 is 0,995709.
The tangent of 622414 is -10,75955.
## Number 622414 in Computer Science
Code typeCode value
PIN 622414 It's recommended that you use 622414 as your password or PIN.
622414 Number of bytes607.8KB
CSS Color
#622414 hexadecimal to red, green and blue (RGB) (98, 36, 20)
Unix timeUnix time 622414 is equal to Thursday Jan. 8, 1970, 4:53:34 a.m. GMT
IPv4, IPv6Number 622414 internet address in dotted format v4 0.9.127.78, v6 ::9:7f4e
622414 Decimal = 10010111111101001110 Binary
622414 Decimal = 1011121210101 Ternary
622414 Decimal = 2277516 Octal
622414 Decimal = 97F4E Hexadecimal (0x97f4e hex)
622414 BASE64NjIyNDE0
622414 MD5297d37262cf83c2a61ae418f23abbc74
622414 SHA1d23bd21c0034f83732abf7b13f6a51dae2770d47
622414 SHA224f06ddddd72d95dd2f47aaba02d68bac967590bb1e2aeef442862f84a
622414 SHA256e7db794a42a6cf18cfe4f8df09ec66103265e9034cdc90fd52c071a0f028341a
More SHA codes related to the number 622414 ...
If you know something interesting about the 622414 number that you did not find on this page, do not hesitate to write us here.
## Numerology 622414
### Character frequency in the number 622414
Character (importance) frequency for numerology.
Character: Frequency: 6 1 2 2 4 2 1 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 622414, the numbers 6+2+2+4+1+4 = 1+9 = 1+0 = 1 are added and the meaning of the number 1 is sought.
## № 622,414 in other languages
How to say or write the number six hundred and twenty-two thousand, four hundred and fourteen in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 622.414) seiscientos veintidós mil cuatrocientos catorce German: 🔊 (Nummer 622.414) sechshundertzweiundzwanzigtausendvierhundertvierzehn French: 🔊 (nombre 622 414) six cent vingt-deux mille quatre cent quatorze Portuguese: 🔊 (número 622 414) seiscentos e vinte e dois mil, quatrocentos e catorze Hindi: 🔊 (संख्या 622 414) छः लाख, बाईस हज़ार, चार सौ, चौदह Chinese: 🔊 (数 622 414) 六十二万二千四百一十四 Arabian: 🔊 (عدد 622,414) ستمائة و اثنان و عشرون ألفاً و أربعمائة و أربعة عشر Czech: 🔊 (číslo 622 414) šestset dvacet dva tisíce čtyřista čtrnáct Korean: 🔊 (번호 622,414) 육십이만 이천사백십사 Danish: 🔊 (nummer 622 414) sekshundrede og toogtyvetusindfirehundrede og fjorten Hebrew: (מספר 622,414) שש מאות עשרים ושניים אלף ארבע מאות וארבע עשרה Dutch: 🔊 (nummer 622 414) zeshonderdtweeëntwintigduizendvierhonderdveertien Japanese: 🔊 (数 622,414) 六十二万二千四百十四 Indonesian: 🔊 (jumlah 622.414) enam ratus dua puluh dua ribu empat ratus empat belas Italian: 🔊 (numero 622 414) seicentoventiduemilaquattrocentoquattordici Norwegian: 🔊 (nummer 622 414) seks hundre og tjueto tusen fire hundre og fjorten Polish: 🔊 (liczba 622 414) sześćset dwadzieścia dwa tysiące czterysta czternaście Russian: 🔊 (номер 622 414) шестьсот двадцать две тысячи четыреста четырнадцать Turkish: 🔊 (numara 622,414) altıyüzyirmiikibindörtyüzondört Thai: 🔊 (จำนวน 622 414) หกแสนสองหมื่นสองพันสี่ร้อยสิบสี่ Ukrainian: 🔊 (номер 622 414) шістсот двадцять дві тисячі чотириста чотирнадцять Vietnamese: 🔊 (con số 622.414) sáu trăm hai mươi hai nghìn bốn trăm mười bốn Other languages ...
## News to email
I have read the privacy policy
## Comment
If you know something interesting about the number 622414 or any other natural number (positive integer), please write to us here or on Facebook.
#### Comment (Maximum 2000 characters) *
The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 1,970 | 5,915 | {"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-2024-26 | latest | en | 0.793643 |
https://offertaformativa.unimore.it/corso/insegnamento?cds_cod=3-211&aa_ord_id=2019&pds_cod=PDS0-2019&aa_off_id=2023&lang=eng&ad_cod=DM06CO02&aa_corso=1&fac_id=10002&coorte=2023&anno_corrente=2023&durata=3 | 1,702,067,127,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100769.54/warc/CC-MAIN-20231208180539-20231208210539-00507.warc.gz | 476,354,385 | 5,425 | You are here: Home » Study Plan » Subject
## Subject: DATA ANALYSIS FOR DIGITAL MARKETING I (A.A. 2023/2024)
### degree course in DIGITAL MARKETING
Course year 1 6 Unit Unico Statistics and Mathematics (lesson) TAF: Basic compulsory subjects SSD: SECS-S/01 CFU: 6 Teachers: Elvira PELLE written final vote Italiano
### Overview
The course is aimed at introducing statistical reasoning, both in exploratory and inferential settings. At the end of the course, the student will be able to use the tools for exploratory data analysis, which consists in organizing, displaying and summarizing data, and the fundamentals of statistical inference, to explain how from a small set of observations it is possible to draw conclusions about the characteristics of a phenomenon. For further information on the aims of the course, please refer to the section expected learning outcomes.
-
### Course contents
Exploratory data analysis (3 CFU)
- Data, populations, variables, data sources
- Graphic representations and frequency tables
- Summaries of a statistical distribution: position and variability.
- Multiple distributions.
Probability (0.5 CFU)
- Random variables, definition, discrete and continuous distributions.
Statistical inference (2.5 CFU)
- Sample and population, sampling variability
- Sketch on Interval estimate
- Hypotheses testing for a mean and a proportion
- Linear regression model and analydìsis of variance.
### Teaching methods
Lectures and practicals, also by using statistical software. Attendance is highly recommended. Classroom lessons will be recorded and made available on the ONELab platform of the Department of Communication and Economics. The course will be taught in Italian.
### Assessment methods
The students’ evaluation will be based on a written exam. It will be possible to take an intermediate test in the November exam session corresponding to approximately 50-60% of the program. If the intermediate test is passed, in the next exam session (January / February) it will be possible to take only a partial test relating to the last part of the course (50-40%). The est will be composed by true/false questions, multiple choice answers, open-ended questions, explanations and short exercises for a total of about 25-30 questions. The score assigned to each question/exercise depends on its difficulty. Erroneous answers to true/false questions will have a negative score. During the test, students may consult formulary and tables provided during the lessons. The calculator can be used for tests where calculating exercises are foreseen.
### Learning outcomes
Knowledge and understanding: The student will be able to understand the statistical methodology for organizing and summarizing statistical data.
Applying knowledge and understanding: The student will be able to use statistical tools to draw conclusions from the data, formulate and solve simple problems of probability and statistical inference.
Making judgements: The student will be able to autonomous perform statistical analysis in various contexts.
Communication skills: The student will be able to interpret and expose the results correctly; furthermore he will be able to comment and explain the main statistical indices and the applications of the inferential method
Learning skills: Students will be able to learn new statistical concepts, indexes and tools. | 664 | 3,375 | {"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-2023-50 | latest | en | 0.866664 |
https://hackage.haskell.org/package/multi-containers-0.1.1/docs/Data-Multimap-Set.html | 1,606,782,914,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141515751.74/warc/CC-MAIN-20201130222609-20201201012609-00336.warc.gz | 307,089,264 | 11,250 | multi-containers-0.1.1: A few multimap variants.
Data.Multimap.Set
Description
Multimaps, where values behave like (non empty) sets.
Multimaps whose values behave like lists are in Data.Multimap. Multimaps whose values behave like maps (i.e., two-dimensional tables) are in Data.Multimap.Table.
The implementation is backed by a Map k (Set a). The differences between Multimap k a and Map k (Set a) include:
• A key is only present in a SetMultimap if it is associated with at least one values, i.e., a key is never associated with an empty set of values.
• lookup (or !) returns a possibly empty set. Unlike regular maps, the ! operator is total for multimaps.
• Functions like map, adjust, update, etc., take functions on individual values (e.g., a -> b) as opposed to, e.g., Set a -> Set b.
• union and unions union the values when there are duplicate keys, rather than being left- or right-biased.
• The difference function computes set differences for values of keys that exist in both maps.
• The size function returns the total number of values for all keys in the multimap, not the number of distinct keys. The latter can be obtained by first getting the keysSet or first converting the multimap to a regular map via toMap.
In the following Big-O notations, unless otherwise noted, n denotes the size of the multimap, k denotes the number of distinct keys, and m denotes the maximum number of values associated with a single key.
Synopsis
Multimap type
data SetMultimap k a Source #
Instances
Construction
O(1). The empty multimap.
size empty === 0
singleton :: k -> a -> SetMultimap k a Source #
O(1). A multimap with a single element.
singleton 1 'a' === fromList [(1, 'a')]
size (singleton 1 'a') === 1
fromMap :: Map k (Set a) -> SetMultimap k a Source #
O(k). A key is retained only if it is associated with a non-empty set of values.
From Unordered Lists
fromList :: (Ord k, Ord a) => [(k, a)] -> SetMultimap k a Source #
O(n*log n) where n is the length of the input list. Build a multimap from a list of key/value pairs.
fromList ([] :: [(Int, Char)]) === empty
fromList [(1, 'b'), (2, 'a'), (1, 'b')] === fromList [(1, 'b'), (2, 'a')]
Insertion
insert :: (Ord k, Ord a) => k -> a -> SetMultimap k a -> SetMultimap k a Source #
O(log m * log k). If the key exists in the multimap, the new value will be inserted into the set of values for the key. It is a no-op if the value already exists in the set.
insert 1 'a' empty === singleton 1 'a'
insert 1 'a' (fromList [(1, 'b'), (2, 'a')]) === fromList [(1, 'a'), (1, 'b'), (2, 'a')]
insert 1 'a' (fromList [(1, 'a'), (2, 'c')]) === fromList [(1, 'a'), (2, 'c')]
Deletion/Update
delete :: Ord k => k -> SetMultimap k a -> SetMultimap k a Source #
O(log k). Delete a key and all its values from the map.
delete 1 (fromList [(1,'a'),(1,'b'),(2,'c')]) === singleton 2 'c'
deleteWithValue :: (Ord k, Ord a) => k -> a -> SetMultimap k a -> SetMultimap k a Source #
O(log m * log k). Remove the first occurrence of the value associated with the key, if exists.
deleteWithValue 1 'c' (fromList [(1,'a'),(1,'b'),(2,'c')]) === fromList [(1,'a'),(1,'b'),(2,'c')]
deleteWithValue 1 'c' (fromList [(2,'c'),(1,'c')]) === singleton 2 'c'
deleteMax :: Ord k => k -> SetMultimap k a -> SetMultimap k a Source #
O(log m * log k). Remove the maximal value associated with the key, if exists.
deleteMax 3 (fromList [(1,'a'),(1,'b'),(2,'c')]) === fromList [(1,'a'),(1,'b'),(2,'c')]
deleteMax 1 (fromList [(1,'a'),(1,'b'),(2,'c')]) === fromList [(1,'a'),(2,'c')]
deleteMin :: Ord k => k -> SetMultimap k a -> SetMultimap k a Source #
O(log m * log k). Remove the minimal value associated with the key, if exists.
deleteMin 3 (fromList [(1,'a'),(1,'b'),(2,'c')]) === fromList [(1,'a'),(1,'b'),(2,'c')]
deleteMin 1 (fromList [(1,'a'),(1,'b'),(2,'c')]) === fromList [(1,'b'),(2,'c')]
adjust :: (Ord k, Ord a) => (a -> a) -> k -> SetMultimap k a -> SetMultimap k a Source #
O(m * log m * log k), assuming the function a -> a takes O(1). Update values at a specific key, if exists.
Since values are sets, the result may be smaller than the original multimap.
adjust ("new " ++) 1 (fromList [(1,"a"),(1,"b"),(2,"c")]) === fromList [(1,"new a"),(1,"new b"),(2,"c")]
adjust (const "z") 1 (fromList [(1,"a"),(1,"b"),(2,"c")]) === fromList [(1,"z"),(2,"c")]
adjustWithKey :: (Ord k, Ord a) => (k -> a -> a) -> k -> SetMultimap k a -> SetMultimap k a Source #
O(m * log m * log k), assuming the function k -> a -> a takes O(1). Update values at a specific key, if exists.
Since values are sets, the result may be smaller than the original multimap.
adjustWithKey (\k x -> show k ++ ":new " ++ x) 1 (fromList [(1,"a"),(1,"b"),(2,"c")])
=== fromList [(1,"1:new a"),(1,"1:new b"),(2,"c")]
update :: (Ord k, Ord a) => (a -> Maybe a) -> k -> SetMultimap k a -> SetMultimap k a Source #
O(m * log m * log k), assuming the function a -> Maybe a takes O(1). The expression (update f k map) updates the values at key k, if exists. If f returns Nothing for a value, the value is deleted.
let f x = if x == "a" then Just "new a" else Nothing in do
update f 1 (fromList [(1,"a"),(1,"b"),(2,"c")]) === fromList [(1,"new a"),(2, "c")]
update f 1 (fromList [(1,"b"),(1,"c"),(2,"c")]) === singleton 2 "c"
updateWithKey :: (Ord k, Ord a) => (k -> a -> Maybe a) -> k -> SetMultimap k a -> SetMultimap k a Source #
O(m * log m * log k), assuming the function k -> a -> Maybe a takes O(1). The expression (updateWithKey f k map) updates the values at key k, if exists. If f returns Nothing for a value, the value is deleted.
let f k x = if x == "a" then Just (show k ++ ":new a") else Nothing in do
updateWithKey f 1 (fromList [(1,"a"),(1,"b"),(2,"c")]) === fromList [(1,"1:new a"),(2,"c")]
updateWithKey f 1 (fromList [(1,"b"),(1,"c"),(2,"c")]) === singleton 2 "c"
alter :: Ord k => (Set a -> Set a) -> k -> SetMultimap k a -> SetMultimap k a Source #
O(log k), assuming the function Set a -> Set a takes O(1). The expression (alter f k map) alters the values at k, if exists.
let (f, g) = (const Set.empty, Set.insert 'c') in do
alter f 1 (fromList [(1, 'a'), (2, 'b')]) === singleton 2 'b'
alter f 3 (fromList [(1, 'a'), (2, 'b')]) === fromList [(1, 'a'), (2, 'b')]
alter g 1 (fromList [(1, 'a'), (2, 'b')]) === fromList [(1, 'c'), (1, 'a'), (2, 'b')]
alter g 1 (fromList [(1, 'c'), (2, 'b')]) === fromList [(1, 'c'), (2, 'b')]
alter g 3 (fromList [(1, 'a'), (2, 'b')]) === fromList [(1, 'a'), (2, 'b'), (3, 'c')]
alterWithKey :: Ord k => (k -> Set a -> Set a) -> k -> SetMultimap k a -> SetMultimap k a Source #
O(log k), assuming the function k -> Set a -> Set a takes O(1). The expression (alterWithKey f k map) alters the values at k, if exists.
let (f, g) = (const (const Set.empty), Set.insert . show) in do
alterWithKey f 1 (fromList [(1, "a"), (2, "b")]) === singleton 2 "b"
alterWithKey f 3 (fromList [(1, "a"), (2, "b")]) === fromList [(1, "a"), (2, "b")]
alterWithKey g 1 (fromList [(1, "a"), (2, "b")]) === fromList [(1, "1"), (1, "a"), (2, "b")]
alterWithKey g 3 (fromList [(1, "a"), (2, "b")]) === fromList [(1, "a"), (2, "b"), (3, "3")]
Query
Lookup
lookup :: Ord k => k -> SetMultimap k a -> Set a Source #
O(log k). Lookup the values at a key in the map. It returns an empty set if the key is not in the map.
(!) :: Ord k => SetMultimap k a -> k -> Set a infixl 9 Source #
O(log k). Lookup the values at a key in the map. It returns an empty set if the key is not in the map.
fromList [(3, 'a'), (5, 'b'), (3, 'c')] ! 3 === Set.fromList "ac"
fromList [(3, 'a'), (5, 'b'), (3, 'c')] ! 2 === Set.empty
member :: Ord k => k -> SetMultimap k a -> Bool Source #
O(log k). Is the key a member of the map?
A key is a member of the map if and only if there is at least one value associated with it.
member 1 (fromList [(1, 'a'), (2, 'b'), (2, 'c')]) === True
member 1 (deleteMax 1 (fromList [(2, 'c'), (1, 'c')])) === False
notMember :: Ord k => k -> SetMultimap k a -> Bool Source #
O(log k). Is the key not a member of the map?
A key is a member of the map if and only if there is at least one value associated with it.
notMember 1 (fromList [(1, 'a'), (2, 'b'), (2, 'c')]) === False
notMember 1 (deleteMin 1 (fromList [(2, 'c'), (1, 'c')])) === True
Size
null :: SetMultimap k a -> Bool Source #
O(1). Is the multimap empty?
Data.Multimap.Set.null empty === True
Data.Multimap.Set.null (singleton 1 'a') === False
notNull :: SetMultimap k a -> Bool Source #
O(1). Is the multimap non-empty?
notNull empty === False
notNull (singleton 1 'a') === True
size :: SetMultimap k a -> Int Source #
The total number of values for all keys.
size is evaluated lazily. Forcing the size for the first time takes up to O(k) and subsequent forces take O(1).
size empty === 0
size (singleton 1 'a') === 1
size (fromList [(1, 'a'), (2, 'b'), (2, 'c'), (2, 'b')]) === 3
Combine
Union
union :: (Ord k, Ord a) => SetMultimap k a -> SetMultimap k a -> SetMultimap k a Source #
Union two multimaps, unioning values for duplicate keys.
union (fromList [(1,'a'),(2,'b'),(2,'c')]) (fromList [(1,'d'),(2,'b')])
=== fromList [(1,'a'),(1,'d'),(2,'b'),(2,'c')]
unions :: (Foldable f, Ord k, Ord a) => f (SetMultimap k a) -> SetMultimap k a Source #
Union a number of multimaps, unioning values for duplicate keys.
unions [fromList [(1,'a'),(2,'b'),(2,'c')], fromList [(1,'d'),(2,'b')]]
=== fromList [(1,'a'),(1,'d'),(2,'b'),(2,'c')]
Difference
difference :: (Ord k, Ord a) => SetMultimap k a -> SetMultimap k a -> SetMultimap k a Source #
Difference of two multimaps.
If a key exists in the first multimap but not the second, it remains unchanged in the result. If a key exists in both multimaps, a set difference is performed on their values.
difference (fromList [(1,'a'),(2,'b'),(2,'c')]) (fromList [(1,'d'),(2,'b'),(2,'a')])
=== fromList [(1,'a'),(2,'c')]
Traversal
Map
map :: Ord b => (a -> b) -> SetMultimap k a -> SetMultimap k b Source #
O(n * log m), assuming the function a -> b takes O(1). Map a function over all values in the map.
Since values are sets, the result may be smaller than the original multimap.
Data.Multimap.Set.map (++ "x") (fromList [(1,"a"),(2,"b")]) === fromList [(1,"ax"),(2,"bx")]
Data.Multimap.Set.map (const "c") (fromList [(1,"a"),(1,"b"),(2,"b")]) === fromList [(1,"c"),(2,"c")]
mapWithKey :: Ord b => (k -> a -> b) -> SetMultimap k a -> SetMultimap k b Source #
O(n * log m), assuming the function k -> a -> b takes O(1). Map a function over all values in the map.
Since values are sets, the result may be smaller than the original multimap.
mapWithKey (\k x -> show k ++ ":" ++ x) (fromList [(1,"a"),(2,"b")]) === fromList [(1,"1:a"),(2,"2:b")]
Folds
foldr :: (a -> b -> b) -> b -> SetMultimap k a -> b Source #
O(n). Fold the values in the map using the given right-associative binary operator.
Data.Multimap.Set.foldr ((+) . length) 0 (fromList [(1, "hello"), (1, "world"), (2, "!")]) === 11
foldl :: (a -> b -> a) -> a -> SetMultimap k b -> a Source #
O(n). Fold the values in the map using the given left-associative binary operator.
Data.Multimap.Set.foldl (\len -> (+ len) . length) 0 (fromList [(1, "hello"), (1, "world"), (2, "!")]) === 11
foldrWithKey :: (k -> a -> b -> b) -> b -> SetMultimap k a -> b Source #
O(n). Fold the key/value pairs in the map using the given right-associative binary operator.
foldrWithKey (\k a len -> length (show k) + length a + len) 0 (fromList [(1, "hello"), (1, "world"), (20, "!")]) === 15
foldlWithKey :: (a -> k -> b -> a) -> a -> SetMultimap k b -> a Source #
O(n). Fold the key/value pairs in the map using the given left-associative binary operator.
foldlWithKey (\len k a -> length (show k) + length a + len) 0 (fromList [(1, "hello"), (1, "world"), (20, "!")]) === 15
foldMapWithKey :: Monoid m => (k -> a -> m) -> SetMultimap k a -> m Source #
O(n). Fold the key/value pairs in the map using the given monoid.
foldMapWithKey (\k x -> show k ++ ":" ++ x) (fromList [(1, "a"), (1, "c"), (2, "b")]) === "1:a1:c2:b"
Strict Folds
foldr' :: (a -> b -> b) -> b -> SetMultimap k a -> b Source #
O(n). A strict version of foldr. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.
Data.Multimap.Set.foldr' ((+) . length) 0 (fromList [(1, "hello"), (1, "world"), (2, "!")]) === 11
foldl' :: (a -> b -> a) -> a -> SetMultimap k b -> a Source #
O(n). A strict version of foldl. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.
Data.Multimap.Set.foldl' (\len -> (+ len) . length) 0 (fromList [(1, "hello"), (1, "world"), (2, "!")]) === 11
foldrWithKey' :: (k -> a -> b -> b) -> b -> SetMultimap k a -> b Source #
O(n). A strict version of foldrWithKey. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.
foldrWithKey' (\k a len -> length (show k) + length a + len) 0 (fromList [(1, "hello"), (1, "world"), (20, "!")]) === 15
foldlWithKey' :: (a -> k -> b -> a) -> a -> SetMultimap k b -> a Source #
O(n). A strict version of foldlWithKey. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.
foldlWithKey' (\len k a -> length (show k) + length a + len) 0 (fromList [(1, "hello"), (1, "world"), (20, "!")]) === 15
Conversion
elems :: SetMultimap k a -> [a] Source #
O(n). Return all elements of the multimap in ascending order of their keys. Elements of each key appear in ascending order.
elems (fromList [(2,'a'),(1,'b'),(3,'d'),(3,'c'),(1,'b')]) === "bacd"
elems (empty :: SetMultimap Int Char) === []
keys :: SetMultimap k a -> [k] Source #
O(k). Return all keys of the multimap in ascending order.
keys (fromList [(2,'a'),(1,'b'),(3,'c'),(1,'b')]) === [1,2,3]
keys (empty :: SetMultimap Int Char) === []
assocs :: SetMultimap k a -> [(k, a)] Source #
An alias for toAscList.
keysSet :: SetMultimap k a -> Set k Source #
O(k). The set of all keys of the multimap.
keysSet (fromList [(2,'a'),(1,'b'),(3,'c'),(1,'b')]) === Set.fromList [1,2,3]
keysSet (empty :: SetMultimap Int Char) === Set.empty
Lists
toList :: SetMultimap k a -> [(k, a)] Source #
Convert the multimap into a list of key/value pairs.
toList (fromList [(2,'a'),(1,'b'),(3,'c'),(1,'a')]) === [(1,'a'),(1,'b'),(2,'a'),(3,'c')]
Ordered lists
toAscList :: SetMultimap k a -> [(k, a)] Source #
Convert the multimap into a list of key/value pairs in ascending order of keys. Elements of each key appear in ascending order.
toAscList (fromList [(2,'a'),(1,'b'),(3,'c'),(1,'a')]) === [(1,'a'),(1,'b'),(2,'a'),(3,'c')]
toDescList :: SetMultimap k a -> [(k, a)] Source #
Convert the multimap into a list of key/value pairs in descending order of keys. Elements of each key appear in descending order.
toDescList (fromList [(2,'a'),(1,'b'),(3,'c'),(1,'a')]) === [(3,'c'),(2,'a'),(1,'b'),(1,'a')]
Maps
toMap :: SetMultimap k a -> Map k (Set a) Source #
O(1). Convert the multimap into a regular map.
Filter
filter :: (a -> Bool) -> SetMultimap k a -> SetMultimap k a Source #
O(n), assuming the predicate function takes O(1). Retain all values that satisfy the predicate. A key is removed if none of its values satisfies the predicate.
Data.Multimap.Set.filter (> 'a') (fromList [(1,'a'),(1,'b'),(2,'a')]) === singleton 1 'b'
Data.Multimap.Set.filter (< 'a') (fromList [(1,'a'),(1,'b'),(2,'a')]) === empty
filterWithKey :: (k -> a -> Bool) -> SetMultimap k a -> SetMultimap k a Source #
O(n), assuming the predicate function takes O(1). Retain all key/value pairs that satisfy the predicate. A key is removed if none of its values satisfies the predicate.
filterWithKey (\k a -> even k && a > 'a') (fromList [(1,'a'),(1,'b'),(2,'a'),(2,'b')]) === singleton 2 'b'
filterKey :: (k -> Bool) -> SetMultimap k a -> SetMultimap k a Source #
O(k), assuming the predicate function takes O(1). Retain all keys that satisfy the predicate.
filterM :: (Ord k, Ord a, Applicative t) => (a -> t Bool) -> SetMultimap k a -> t (SetMultimap k a) Source #
Generalized filter.
let f a | a > 'b' = Just True
| a < 'b' = Just False
| a == 'b' = Nothing
in do
filterM f (fromList [(1,'a'),(1,'b'),(2,'a'),(2,'c')]) === Nothing
filterM f (fromList [(1,'a'),(1,'c'),(2,'a'),(2,'c')]) === Just (fromList [(1,'c'),(2,'c')])
filterWithKeyM :: (Ord k, Ord a, Applicative t) => (k -> a -> t Bool) -> SetMultimap k a -> t (SetMultimap k a) Source #
Generalized filterWithKey. | Generalized filterWithKey.
let f k a | even k && a > 'b' = Just True
| odd k && a < 'b' = Just False
| otherwise = Nothing
in do
filterWithKeyM f (fromList [(1,'a'),(1,'c'),(2,'a'),(2,'c')]) === Nothing
filterWithKeyM f (fromList [(1,'a'),(3,'a'),(2,'c'),(4,'c')]) === Just (fromList [(2,'c'),(4,'c')])
mapMaybe :: Ord b => (a -> Maybe b) -> SetMultimap k a -> SetMultimap k b Source #
O(n * log m), assuming the function a -> Maybe b takes O(1). Map values and collect the Just results.
mapMaybe (\a -> if a == "a" then Just "new a" else Nothing) (fromList [(1,"a"),(1,"b"),(2,"a"),(2,"c")])
=== fromList [(1,"new a"),(2,"new a")]
mapMaybeWithKey :: Ord b => (k -> a -> Maybe b) -> SetMultimap k a -> SetMultimap k b Source #
O(n * log m), assuming the function k -> a -> Maybe b takes O(1). Map key/value pairs and collect the Just results.
mapMaybeWithKey (\k a -> if k > 1 && a == "a" then Just "new a" else Nothing) (fromList [(1,"a"),(1,"b"),(2,"a"),(2,"c")])
=== singleton 2 "new a"
mapEither :: (Ord b, Ord c) => (a -> Either b c) -> SetMultimap k a -> (SetMultimap k b, SetMultimap k c) Source #
O(n * log m), assuming the function a -> Either b c takes O(1). Map values and separate the Left and Right results.
mapEither (\a -> if a < 'b' then Left a else Right a) (fromList [(1,'a'),(1,'c'),(2,'a'),(2,'c')])
=== (fromList [(1,'a'),(2,'a')],fromList [(1,'c'),(2,'c')])
mapEitherWithKey :: (Ord b, Ord c) => (k -> a -> Either b c) -> SetMultimap k a -> (SetMultimap k b, SetMultimap k c) Source #
O(n * log m), assuming the function k -> a -> Either b c takes O(1). Map key/value pairs and separate the Left and Right results.
mapEitherWithKey (\k a -> if even k && a < 'b' then Left a else Right a) (fromList [(1,'a'),(1,'c'),(2,'a'),(2,'c')])
=== (fromList [(2,'a')],fromList [(1,'a'),(1,'c'),(2,'c')])
Min/Max
lookupMin :: SetMultimap k a -> Maybe (k, Set a) Source #
O(log n). Return the smallest key and the associated values. Returns Nothing if the map is empty.
lookupMin (fromList [(1,'a'),(1,'c'),(2,'c')]) === Just (1, Set.fromList "ac")
lookupMin (empty :: SetMultimap Int Char) === Nothing
lookupMax :: SetMultimap k a -> Maybe (k, Set a) Source #
O(log n). Return the largest key and the associated values. Returns Nothing if the map is empty.
lookupMax (fromList [(1,'a'),(1,'c'),(2,'c')]) === Just (2, Set.fromList "c")
lookupMax (empty :: SetMultimap Int Char) === Nothing
lookupLT :: Ord k => k -> SetMultimap k a -> Maybe (k, Set a) Source #
O(log n). Return the largest key smaller than the given one, and the associated values, if exist.
lookupLT 1 (fromList [(1,'a'),(3,'b'),(3,'c'),(5,'c')]) === Nothing
lookupLT 4 (fromList [(1,'a'),(3,'b'),(3,'c'),(5,'c')]) === Just (3, Set.fromList "bc")
lookupGT :: Ord k => k -> SetMultimap k a -> Maybe (k, Set a) Source #
O(log n). Return the smallest key larger than the given one, and the associated values, if exist.
lookupGT 5 (fromList [(1,'a'),(3,'b'),(3,'c'),(5,'c')]) === Nothing
lookupGT 2 (fromList [(1,'a'),(3,'b'),(3,'c'),(5,'c')]) === Just (3, Set.fromList "bc")
lookupLE :: Ord k => k -> SetMultimap k a -> Maybe (k, Set a) Source #
O(log n). Return the largest key smaller than or equal to the given one, and the associated values, if exist.
lookupLE 0 (fromList [(1,'a'),(3,'b'),(3,'c'),(5,'c')]) === Nothing
lookupLE 1 (fromList [(1,'a'),(3,'b'),(3,'c'),(5,'c')]) === Just (1, Set.fromList "a")
lookupLE 4 (fromList [(1,'a'),(3,'b'),(3,'c'),(5,'c')]) === Just (3, Set.fromList "bc")
lookupGE :: Ord k => k -> SetMultimap k a -> Maybe (k, Set a) Source #
O(log n). Return the smallest key larger than or equal to the given one, and the associated values, if exist.
lookupGE 6 (fromList [(1,'a'),(3,'b'),(3,'c'),(5,'c')]) === Nothing
lookupGE 5 (fromList [(1,'a'),(3,'b'),(3,'c'),(5,'c')]) === Just (5, Set.fromList "c")
lookupGE 2 (fromList [(1,'a'),(3,'b'),(3,'c'),(5,'c')]) === Just (3, Set.fromList "bc") | 6,745 | 20,549 | {"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} | 2.65625 | 3 | CC-MAIN-2020-50 | latest | en | 0.787478 |
http://www.homebrewtalk.com/f13/why-batch-sparging-better-faster-23379/index2.html | 1,418,937,826,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802767873.65/warc/CC-MAIN-20141217075247-00025-ip-10-231-17-201.ec2.internal.warc.gz | 545,638,658 | 23,276 | Happy HolidaySs Giveaway - Winners Re-Re-Re-Drawn - 24 hours to Claim!
Get your HBT Growlers, Shirts and Membership before the Rush!
Home Brew Forums > Why is batch sparging better/faster ?
02-28-2007, 04:06 AM #11
Lil' Sparky
Cowboys EAC
Feedback Score: 0 reviews
Recipes
Join Date: Feb 2006
Location: Honolulu, HI
Posts: 4,013
Liked 50 Times on 33 Posts
dammit, gabe beat me to the punch. He's using a slightly different method with a thinner mash and unequal runoffs, but the idea is the same.
__________________
02-28-2007, 05:37 AM #12
voodoochild7
Feedback Score: 0 reviews
Recipes
Join Date: Oct 2005
Location: New Jersey
Posts: 273
Quote:
Originally Posted by Lil' Sparky It's really simple. You're already used to calculating the total amount of water needed. Suppose you're mashing 10 lbs of grain and you mash in with 2.5 gal of strike water. Your grain will absorb 4-5 qts. If you want 7 gal in the kettle, the amount of sparge water = 7 - residual in MLT = 7 - (2.5-1.25) = 5.75 gal. You want the two runoffs to be equal (3.5 gal). So you add 2.25 to the MLT, stir, rest, vourlauf, drain 3.5 gal, and repeat with the remaining 3.5 gal. It may seem like too much math at first, but you get used to it quickly.
Okay so normally I use about 11s pound of grain. I mash with 3.25 gallons of water. So my grain will absorb roughly 4 quarts so I'll get aobut 2.25 gallons of the first runoff. So now 7 - 2.25 = 4.75 minus some grain absortion. So I run off the first 2.25 then add let's say 5 gallons stir let sit stir again and runoff the rest? Does that sound right? What temp should my sparge water be?
__________________
Chuck
Primary 1: Imperial Stout
Primary 2: Golden Ale
Keg 1: Nothing
Keg 2: Nothing
02-28-2007, 06:06 AM #13
robnog
Feedback Score: 0 reviews
Recipes
Join Date: Nov 2006
Location: Bloomington, IN
Posts: 99
I'm not going to try to defend fly over batch. I have never batch sparged so I don't have the experience to make any claims. I just want to let you know why I fly sparge.
1. I like the idea of trying to maximize efficiency. I noticed that when my sparge time went from 60min to 90min, my efficiency jumped from about 70% to 80-87%. I hope that I can keep improving (though I don't expect to get much higher than 87%). I understand that grain is cheap, but so am I.
2. I feel more involved in the whole process of brewing. In batch sparging, I would open up my valve and then go watch the game for 30 minutes and then come back and start brewing again. That would bore me. I like the process of pouring a few cups of water on top every few minutes, calculating the flow rate to make sure it takes 90 min, and checking the gravity of the runnings every so often.
3. I made a really cool manifold. Unfortunately, I don't have a picture because I don't have a digital camera (see: cheap comment in #1).
I see that several of you prefer batch sparging and that's great. For me, fly sparging works well and I don't plan on changing any time soon.
__________________
02-28-2007, 06:25 AM #14
Gabe
It's a sickness!
Feedback Score: 0 reviews
Recipes
Join Date: Apr 2006
Location: Central coast
Posts: 724
Liked 1 Times on 1 Posts
Actually Lil'Sparky, I mash out to 170 before I run off the first time like most . I then run off about 3 gal and return this to clear up my wort. I then proceed to sparge, usually about 7 gal or until I hit 1.010. I get good but not great eff (70%) but I am learning on a brand new system so things should improve.
__________________
02-28-2007, 07:54 AM #15
brewman !
Feedback Score: 0 reviews
Recipes
Join Date: Nov 2006
Posts: 1,228
Liked 14 Times on 13 Posts
Likes Given: 16
Lets stop with the bragging about batch sparging being better and look at why it is better.
1) I don't understand how batch sparging can be faster AND get good efficiency.
As far as I understand, you add water to the grain bed, stir it up, let it settle, circulate it a bit and then drain it. Repeat 3x.
Now how can one get good efficiency when the grain is only in contact with the water for such a short time ? Put another way, why can't we drain this quickly when fly sparging ?
In my mind, something is wrong that batch sparging can be done really quickly and fly sparging can't. Or maybe we just think it can't !
2) I don't understand how batch sparging can be less work.
With fly sparging you add sparge water, circulate a bit and start draining. After that all you have to do is add sparge water and that can be automated pretty easily.
With batch sparging you stir, let it settle and circulate it every time you fill the bed. So how is that easier than just letting the sparge water trickle through the bed ? I can easily read a book or watch TV while sparging. Could I if I was batch sparging ?
3) Stuck beds.
People say you can have these fantastic drain rates when batch sparging. In the same breath they say that fly sparges are prone to stuck beds. OK, now what makes a batch sparge less prone to stuck beds or what makes one think they can drain the same bed faster with a batch sparge than a fly sparge ? A bed is a bed, right ? And both beds will need to settle the same to get the same wort clearness. So how can one drain faster than the other ? How can one bed stick and not the other ?
__________________
Getting back into brewing...
02-28-2007, 08:30 AM #16
Orfy
For the love of beer!
Feedback Score: 0 reviews
Recipes
Join Date: Sep 2005
Location: Cheshire, England
Posts: 11,853
Liked 69 Times on 52 Posts
Likes Given: 30
I do it because.
It works
It is quick
It is simple
It requires less equipment.
It gets me around 80% effec.
It's the only way I've ever done it.
I'm not really fussed why it works or what the maths says.
I think fly sparging is a left over from old techniques maybe when malt was not as modified.
If my attitude makes my point less valid then so be it.
But if you want to know the reasons I do it and how it works for me then there you go.
If you'd like to go past what people are telling you then the easiest way to find the answers is to try both and see. I can't comment on the fly sparging because I've not done it but I have read plenty about it and peoples experiences. Hence my initial decision to go with batch sparging.
02-28-2007, 08:37 AM #17
voodoochild7
Feedback Score: 0 reviews
Recipes
Join Date: Oct 2005
Location: New Jersey
Posts: 273
I don't want to argue the merits of fly or batch I've done fly for a while and I"m getting sick of the wait. So if somone may be so kind as to check and see if my math/technique is correct I'll be much obliged.
From a few post up.
Quote:
Okay so normally I use about 11s pound of grain. I mash with 3.25 gallons of water. So my grain will absorb roughly 4 quarts so I'll get aobut 2.25 gallons of the first runoff. So now 7 - 2.25 = 4.75 minus some grain absortion. So I run off the first 2.25 then add let's say 5 gallons stir let sit stir again and runoff the rest? Does that sound right? What temp should my sparge water be?
__________________
Chuck
Primary 1: Imperial Stout
Primary 2: Golden Ale
Keg 1: Nothing
Keg 2: Nothing
02-28-2007, 08:44 AM #18
Orfy
For the love of beer!
Feedback Score: 0 reviews
Recipes
Join Date: Sep 2005
Location: Cheshire, England
Posts: 11,853
Liked 69 Times on 52 Posts
Likes Given: 30
The easiest way is to take your first running from the mash. ( I mash with a ratio ~1.33)
Say 2 gallon.
Dump in what you're cooler will take say 3.5 gallon
Take the second runnings which in theory should be 3.5 Gallon
You have 5.5 gallon so you need to put another 1.5 gallon in.
Or you can just split the batches equally.
Just make sure you're not collecting under 1006
If you are go with the lower volume or top up the kettle with water to the volume.
I sparge with 77°C water.
__________________
GET THE GOBLIN
Have a beer on me.
02-28-2007, 11:54 AM #19
fifelee
Feedback Score: 0 reviews
Recipes
Join Date: Dec 2006
Location: Vaughn, MT
Posts: 1,105
Liked 35 Times on 24 Posts
Likes Given: 12
Quote:
Originally Posted by brewman ! As far as I understand, you add water to the grain bed, stir it up, let it settle, circulate it a bit and then drain it. Repeat 3x.
I am no expert, but don't you repeat just once (not 3x).
Fill, stir, settle, drain, fill, stir, settle, drain.
Quote:
Originally Posted by brewman ! With fly sparging you add sparge water, circulate a bit and start draining. After that all you have to do is add sparge water and that can be automated pretty easily.
I think the trouble most people have with fly sparging is keeping the in and out flows equal. Automation isn't easy for a lot of people so they have to stand there and watch the sparge.
__________________
02-28-2007, 01:18 PM #20
Lil' Sparky
Cowboys EAC
Feedback Score: 0 reviews
Recipes
Join Date: Feb 2006
Location: Honolulu, HI
Posts: 4,013
Liked 50 Times on 33 Posts
Quote:
Originally Posted by voodoochild7 Okay so normally I use about 11s pound of grain. I mash with 3.25 gallons of water. So my grain will absorb roughly 4 quarts so I'll get aobut 2.25 gallons of the first runoff. So now 7 - 2.25 = 4.75 minus some grain absortion. So I run off the first 2.25 then add let's say 5 gallons stir let sit stir again and runoff the rest? Does that sound right? What temp should my sparge water be?
If you want equal runoffs, then you would add 1.25 before your first runoff.
FYI - A lot of people use this as a mash out step to bring the entire wort up to 168 deg. Usually that means the water added is close to boiling. There are equations to figure that part out, but I don't have it off-hand. I use BeerSmith now which does that for me.
Again, some of the confusion from the different answers is because not everyone does it with equal runoffs. What I've read suggests that will result in the best efficiency.
__________________ | 2,704 | 9,870 | {"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-2014-52 | latest | en | 0.871123 |
https://www.askiitians.com/forums/8-grade-maths/express-the-following-numbers-in-standard-form-31_274328.htm | 1,659,967,641,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570827.41/warc/CC-MAIN-20220808122331-20220808152331-00132.warc.gz | 602,992,701 | 33,856 | # Express the following numbers in standard form.31860000000
Harshit Singh
one year ago
Dear Student
31860000000 = 31860000000×(10^10/10^10)= 3.186×10^10
Thanks | 56 | 163 | {"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-2022-33 | latest | en | 0.70055 |
https://support.nag.com/numeric/nl/nagdoc_26.2/nagdoc_cl26.2/examples/source/g11sace.c.html | 1,708,473,120,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00038.warc.gz | 565,351,924 | 3,155 | ```/* nag_binary_factor (g11sac) Example Program.
*
* Copyright 2017 Numerical Algorithms Group.
*
* Mark 26.2, 2017.
*/
#include <stdio.h>
#include <nag.h>
#include <nag_stdlib.h>
#include <nagg11.h>
int main(void)
{
/* Scalars */
double cgetol, chi, rlogl, siglev;
Integer exit_status, i, pdcm, idf, p, iprint, is;
Integer j, maxit, n, niter, nrx, pdx, pdexpp;
/* Arrays */
double *a = 0, *alpha = 0, *c = 0, *cm = 0, *exf = 0, *expp = 0,
*g = 0, *obs = 0, *pigam = 0, *xl = 0, *y = 0;
Integer *iob = 0, *irl = 0;
char nag_enum_arg[40];
/* NAG Types */
Nag_Boolean *x = 0;
Nag_Boolean chisqr, gprob;
Nag_OrderType order;
NagError fail;
#ifdef NAG_COLUMN_MAJOR
#define X(I, J) x[(J-1)*pdx + I - 1]
#define CM(I, J) cm[(J-1)*pdcm + I - 1]
order = Nag_ColMajor;
#else
#define X(I, J) x[(I-1)*pdx + J - 1]
#define CM(I, J) cm[(I-1)*pdcm + J - 1]
order = Nag_RowMajor;
#endif
INIT_FAIL(fail);
exit_status = 0;
printf("nag_binary_factor (g11sac) Example Program Results\n");
/* Skip heading in data file */
scanf("%*[^\n] ");
scanf("%" NAG_IFMT "%" NAG_IFMT "%" NAG_IFMT "%*[^\n] ", &p, &n, &is);
if (p > 0 && is >= 0) {
/* Allocate arrays */
pdcm = 2 * p;
pdexpp = p;
nrx = is;
if (!(a = NAG_ALLOC(p, double)) ||
!(alpha = NAG_ALLOC(p, double)) ||
!(c = NAG_ALLOC(p, double)) ||
!(cm = NAG_ALLOC(pdcm * 2 * p, double)) ||
!(exf = NAG_ALLOC(is, double)) ||
!(expp = NAG_ALLOC(pdexpp * p, double)) ||
!(g = NAG_ALLOC(2 * p, double)) ||
!(obs = NAG_ALLOC(p * p, double)) ||
!(pigam = NAG_ALLOC(p, double)) ||
!(xl = NAG_ALLOC(is, double)) ||
!(y = NAG_ALLOC(is, double)) ||
!(iob = NAG_ALLOC(is, Integer)) ||
!(irl = NAG_ALLOC(is, Integer)) ||
!(x = NAG_ALLOC(nrx * p, Nag_Boolean)))
{
printf("Allocation failure\n");
exit_status = -1;
goto END;
}
if (order == Nag_ColMajor)
pdx = nrx;
else
pdx = p;
for (i = 1; i <= is; ++i) {
scanf("%" NAG_IFMT "", &irl[i - 1]);
for (j = 1; j <= p; ++j) {
scanf(" %39s", nag_enum_arg);
/* nag_enum_name_to_value (x04nac).
* Converts NAG enum member name to value
*/
X(i, j) = (Nag_Boolean) nag_enum_name_to_value(nag_enum_arg);
}
scanf("%*[^\n] ");
}
gprob = Nag_FALSE;
for (i = 1; i <= p; ++i) {
a[i - 1] = 0.5;
c[i - 1] = 0.0;
}
/* Set iprint > 0 to obtain intermediate output */
iprint = -1;
cgetol = 1e-4;
maxit = 1000;
chisqr = Nag_TRUE;
/* nag_binary_factor (g11sac).
* Contingency table, latent variable model for binary data
*/
nag_binary_factor(order, p, n, gprob, is, x, pdx, irl, a, c, iprint,
0, cgetol, maxit, chisqr, &niter, alpha, pigam,
cm, pdcm, g, expp, pdexpp, obs, exf, y, iob, &rlogl,
&chi, &idf, &siglev, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_binary_factor (g11sac).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
printf("\n");
printf("Item Alpha (s.e.) Pi (s.e.)\n");
for (i = 1; i <= p; i++)
printf(" %" NAG_IFMT " %g (%10g) %g (%10g)\n", i,
alpha[i - 1], CM(2 * i - 1, 2 * i - 1), pigam[i - 1], CM(2 * i,
2 * i));
printf("\n");
printf("Index Observed Expected Theta Pattern\n");
printf(" Frequency Frequency Score\n");
for (i = 1; i <= is; i++) {
printf("%4" NAG_IFMT "%10" NAG_IFMT "%13.3f%13.7f ", i, irl[i - 1],
exf[i - 1], y[i - 1]);
for (j = 1; j <= p; j++) {
if (X(i, j) == Nag_TRUE)
printf("%3s", "T");
else
printf("%3s", "F");
}
printf("\n");
}
printf("\n");
printf("Chi-squared test statistic = %g\n", chi);
printf("Degrees of freedom = %" NAG_IFMT "\n", idf);
printf("Significance = %g\n", siglev);
}
END:
NAG_FREE(a);
NAG_FREE(alpha);
NAG_FREE(c);
NAG_FREE(cm);
NAG_FREE(exf);
NAG_FREE(expp);
NAG_FREE(g);
NAG_FREE(obs);
NAG_FREE(pigam);
NAG_FREE(xl);
NAG_FREE(y);
NAG_FREE(iob);
NAG_FREE(irl);
NAG_FREE(x);
return exit_status;
}
``` | 1,426 | 3,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.474437 |
https://gateoverflow.in/194008/speedup-in-pipelining | 1,611,149,000,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703520883.15/warc/CC-MAIN-20210120120242-20210120150242-00434.warc.gz | 358,786,315 | 15,584 | # Speedup In pipelining [closed]
296 views
Actually, In this problem what will, we consider getting the answer(upper bound or lower bound) and why?
closed as a duplicate of: Pipeline Efficiency
closed
0
TAKE UPPERBOUND
K=7
0
Efficiency = Speedup / Stages
=> Stages = 5.4 / 0.82 = 6.58
Take stages = 6, then efficiency = 90%
and taking stages = 7, efficiency = 77.14%
Hence, Stages = 6 is better here as efficiency >= 90%
## Related questions
1
1k views
Consider a non-pipelined processor design which has a cycle time of 15ns and average CPI of 1.6. The maximum speedup pipelined processor can get by pipelining it into 5 stages and each stage takes 3ns is______________? 5 6 10 7
1 vote | 209 | 698 | {"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-2021-04 | latest | en | 0.935838 |
https://www.elephango.com/index.cfm/pg/k12learning/lcid/11547/Polynomials | 1,701,510,298,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100381.14/warc/CC-MAIN-20231202073445-20231202103445-00603.warc.gz | 854,183,281 | 16,947 | Polynomials
Contributor: Mason Smith. Lesson ID: 11547
To what degree do you understand polynomials? Are you a proponent of exponents? They can be easily dealt with once you factor in watching a video, taking the interactive quiz, and real-world examples!
categories
Algebra I, Algebra II, Complex Numbers and Quantities
subject
Math
learning style
Visual
personality style
Lion
Middle School (6-8), High School (9-12)
Lesson Type
Quick Query
Lesson Plan - Get It!
Audio:
Q: What happened to the quadratic polynomial when he fell asleep on the beach?
A: He got second-degree burns.
Before you can begin to calculate polynomials, you must first learn (or remember) what they are, so let's run through some definitions!
Monomials are numbers, variables, or a product of numbers and variables, with whole-number exponents.
For example, x2, 5, and –7xy are monomials, because each example has a whole number exponent (numbers have an exponent of 0 unless stated otherwise).
X-2 is not a monomial, because X-2 has an exponent of –2, and negatives are not whole number exponents. 4x - y is not a monomial because it has two variables with exponents that are not being multiplied together.
For a quick refresher, watch Algebra Basics: What Are Polynomials? - Math Antics by Math Antics. Take notes on any points you are having difficulty understanding:
When you examine monomials, you can also look at the degree of a monomial, that is simply the sum of the exponents for the variable. Remember, constants (numbers without variables) have a degree of 0. A variable without a number (x, y, or a, for example) has a degree of 1.
Let's look at three examples of degrees:
1. 2a3b4 has a degree of 7, which can be calculated by adding the exponents together: 3 + 4 = 7.
2. 4 has a degree of 0 because it is a number, and numbers have a degree of 0.
3. 8y has a degree of 1, because any variable without a number (in this case, y) is assumed to have an exponent of 1, and the number 8 has a degree of 0.
For more examples, visit Degree (of an Expression) at MathsIsFun.com.
Try out a few examples. Find the degree of each monomial.
A polynomial is a monomial, or multiple monomials being added or subtracted.
To find the degree of a polynomial, look at the degree of the term with the greatest degree. This is easily done by finding the degree of each term, then comparing them.
Try some examples:
1. 4x - 18x15 — The degree of 4x is 1. The degree of -18x15 is 15, so the degree of the entire polynomial is 15, because 15 is bigger than 1.
2. 6x4 + 9x2 - x + 3 — The degree of 6x4 is 4. The degree of 9x2 is 2. The degree of -x is 1. The degree of 3 is 0. The largest degree is 4, so the polynomial is of degree 4.
3. x2y + xy + .75 — The degree of x2y is 3 because 2 + 1 = 3. The degree of xy is 2 because 1 + 1 = 2. The degree of .75 is 0 because .75 is a constant. The degree of the polynomial is 3 because that is the largest degree.
Try a few examples. Find the degree of each polynomial.
The terms of a polynomial can technically be written in any order; however, the most efficient way to write polynomials is with only one variable in standard form, which is in descending degrees. This means if the polynomial only has the variable x, we count down from the largest exponent to the smallest, and to the constants (Remember: constants are numbers without a variable).
For example:
If we have 20x - 4x3 + 2 - x2, we would rewrite this in standard form as -4x3 - x2 + 20x + 2. This also tells us that our polynomial has a degree of 3.
The –4 in the previous example is the leading coefficient. It is also the number attached to the monomial with the greatest degree, and is the first coefficient when written in standard form.
For example, in 6y5 + y3 + 4y, the 6 would be the leading coefficient. In x2 + 3x - 6, the coefficient is 1. Remember, any variable without a number in front of it is assumed to have a coefficient of 1.
Try a few and share your answers with a teacher or parent:
Write each polynomial in standard form, then give the leading coefficient of each polynomial.
Polynomials are classified with special names based on their degree and number of terms. Sadly, there is no easy way to remember these, so make sure you practice each name! The highlighted terms are the most common:
Degree Name Number of terms Name 0 constant 1 monomial 1 linear 2 binomial 2 quadratic 3 trinomial 3 cubic 4 or more polynomial 4 quartic 5 quintic 6 or more 6th degree; 7th degree; etc.
Let's do a few examples classifying polynomials. Then, you will practice a few on your own:
Classify each polynomial below:
1. 5x - 6
• Degree: 1
• Number of terms: 2
• 5x - 6 is a linear binomial
2. y2+ y + 4
• Degree: 2
• Number of terms: 3
• y2 + y + 4 is a quadratic trinomial
3. 6x7 + 9x2- x + 3
• Degree: 7
• Number of terms: 4
• 6x7 + 9x2 - x + 3 is a 7th-degree polynomial | 1,330 | 4,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2023-50 | latest | en | 0.927555 |
bobmarksastrologer.com | 1,726,770,077,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00558.warc.gz | 126,014,389 | 40,494 | # Harmonics
## Part VII
Sections 32.1
``` (adsbygoogle = window.adsbygoogle || []).push({}); ```
This has to be one of the greatest advances in astrological analysis of the last hundred years. What are “harmonics?” Just a mathematical rearrangement of the horoscope so that we can see particular features more clearly. It’s something like taking a cross-section of a sample in biology and putting it under a microscope. How is it done?
By computer, of course. Please, PLEASE do not try to do the calculations by hand. If you do, you might see numbers in front of your face for a week. You would first have to assign numbers to all of the degrees of the zodiac, starting with the first degree of Aries as number one and ending with the last degree of Pisces as number 360. Then, you will have to give each planet a number, depending on the degree it happens to be in. It gets more complicated. You have to put in the minutes and seconds of arc as well.
Then you choose the harmonic that you want to examine. What next? Multiply ALL of those planetary position numbers by the number of the harmonic. Good luck! Oh, were not finished yet. Most of the numbers you will get are going to be over 360, but there are only 360 degrees in a circle. You will have to subtract 360 again and again until you get a number that is less than 360. After that, you can put the planets in the wheel. Unfortunately, by this time, you will probably be too exhausted to do an interpretation (or too crazed).
Or you can let a computer do it at the click of a mouse. The choice is yours. However, if you choose the first method, be sure to check your medical insurance first to see if it covers mental breakdowns.
There are as many harmonics as there are numbers. Fortunately, only a few are needed to give us insight into the horoscope. Most of them are covered in David Hamblin’s book Harmonic Charts, (The Aquarian Press, Eng.), which is unfortunately out of print. Still, if you can find a used copy somewhere, it is well worth it.
Some VERY IMPORTANT words of caution:
1) DON’T EVER try to do aspects between harmonics of different numbers. They are totally different animals.
2) There are no signs or houses in harmonic charts. Remember that everything has been rearranged, so please don’t do sign or house interpretations because they are no longer there. The purpose of harmonics is to get more information about aspects and aspect patterns.
Now, let’s take a look at some actual harmonics and how they can add depth to horoscope interpretation.
The Fourth Harmonic: The 4th harmonic shows how we deal with stress, strain, and struggle. When you construct a 4th harmonic chart, all of the conjunction, square, and opposition aspects in the original horoscope appear as conjunctions and this makes them easier to see. Of course, these aspects are pretty easy to see in the original horoscope as well, so what’s the point? The point is that the minor aspects, the semi-square (45 degrees) and sequa-square (135 degrees), which are difficult to spot by a quick visual inspection, now appear as oppositions, and can be seen at a glance. Not only that, but other minor stress aspects (22 ½ degrees, 67 ½ degrees, 157 ½ degrees), which can take a very long time to locate, now become squares. In the 4th harmonic, whole patterns of stress aspects can be spotted, and this gives you a big edge for interpretation.
The horoscope of Francisco Assis, the Brazilian serial killer who killed at least nine women, doesn’t seem, at first, to be excessively violent. But when you look at Francisco’s 4th Harmonic Chart, you see another story. The Sun is opposite Venus, which can give feelings of being unloved. Yes, but lots of people have that and they don’t become serial killers. What they don’t have is Uranus making square aspects in the 4th harmonic to both the Sun and Venus. Uranus can make one prone to sudden, hysterical outbursts. Still, everyone born that day had the same thing. What else is there here to drive him over the edge? The Ascendant also makes stress aspects. It opposes Uranus and squares both the Sun and Venus. This makes a Grand Cross, a rare and highly stressful pattern. The Ascendant is very sensitive to changes in the time of birth. In the 4th harmonic chart, this Grand Cross could only have formed in a time interval of about 20 minutes!
But there’s more to come. Mercury (the mind) Mars (violence) and Saturn (depression) make a T-Cross formation (another stressful configuration) in the 4th harmonic as well. Once again though, everyone born that day had the same thing. Why is this case different? Take another look at the birth chart. There is a close semi-square between Mars and the cusp (beginning) of the 8th house (death). Not only that, but Mars rules the sign of Aries, which happens to be the sign on the cusp of the 12th house (psychological problems) as well. This is the classic 8th house-12th house connection frequently found in the charts of serial killers. Where does the 8th house cusp land in the 4th harmonic chart? Opposite Mars, with square aspects to both Mercury and Saturn. It forms another Grand Cross. And this too, could only have been formed in an interval of about 20 minutes. It could take hours to see this in a birth chart. In the 4th harmonic chart, the conclusion leaps out at you.
What happens though, if you have harmonious aspects (sextiles and trines) in a 4th harmonic chart? The best example of that comes from Hamblin’s “Harmonic Charts.” It is the horoscope of a little known French Marshal from World War I, Franchet d’Esperey. Hamblin claims that sextiles and trines in a 4th harmonic chart show “effort towards pleasure, or pleasure in effort.” In other words, someone who likes stress, strain, and struggle. Hamblin quotes Gauquelin’s description of him:
“…he drove cars at maniacal speed and shot windows when there was nothing better to shoot.”
In d’Esperey’s 4th Harmonic Chart, there is a Grand Trine between the Sun, Mars, and Pluto, a perfect aspect pattern for a soldier. A Moon/Venus conjunction sextiles the Sun and Pluto and opposes Mars, making for a Kite formation. Here is someone who had an emotional need (the Moon) for action and violence (Moon/Mars) and got a lot of pleasure from it (Moon/Venus).
Muhammad Ali’s 4th Harmonic Chart also has a lot of harmonious aspects, especially to the Mid-Heaven. The Sun makes a sextile and the Moon a conjunction, and both aspects are within one degree of being exact. This means that stress, strain, and struggle are more likely to bring him before the public. The Mid-Heaven also has a trine from Jupiter and a sextile from Uranus. In Ali’s Birth Chart, Jupiter rules the 5th house (sports and games) and Uranus rules his 7th house (open enemies), a perfect setup for someone to become famous through competitive sports.
Robert Blake, the actor arrested for his wife’s murder, has a 4th Harmonic Chart with a highly stressful T-Cross formation. There is a conjunction of the Moon (feelings, emotions) and Neptune (imagination, film), which helped him in his film career, but it is opposite Pluto (extremes) and all three of these planets make a square to Jupiter (expansion and over optimism). There is hostility towards women here, and when his emotions go, they can really go (Moon, Jupiter, Pluto T-Cross), especially if he happens to be under the influence of alcohol or other intoxicants (Neptune).
The Fifth Harmonic: The 5th harmonic chart indicates talent, if any. In the 5th harmonic chart, all of the quintiles and bi-quintiles come together as conjunctions, making them very easy to spot. The semi-quintiles and sesqua-quintiles become oppositions. Quintiles are one fifth of a circle (72 degrees) and they give talent and power, and show our ability to put things together (pattern recognition). It also has to do with creative play, the joy of making up games of finding patterns (like looking for “patterns” of cracks in the ceiling, or always stepping on cracks in the sidewalk, or always avoiding the cracks). People with strong 5th harmonics do things like that all the time. Of course, taken to the extreme, this is a mental disorder, but those with a strong 5th harmonic usually manage to keep it under control, and then it becomes the basis for creative talent. You don’t need a strong 5th harmonic to succeed as an artist, designer, scientist, or engineer, but it really helps.
Let’s start with a negative example, a very negative example: Adolph Hitler’s 5th Harmonic. Mercury, planet of communications, is opposite the Moon, Neptune, Saturn, and Jupiter. This gives a talent for communicating and power to his words. Chiron makes a conjunction to Mercury, as well as oppositions to the Moon, Neptune, and Saturn. Remember that Chiron is a “bridge” between the inner planets (which rule “normal,” everyday functions) and the outer planets, which rule forces that are out of the ordinary. Chiron also has to do with a feeling of being wounded. Hitler was able to deal with his psychological problems (Mercury rules the 12th house in his birth chart) by communications. Now everyone born that day had similar configurations in their charts. Why was Hitler different? Well for one thing, the Moon was part of the configuration, and the Moon was in place for only a few hours. But his Mid-Heaven closely aspects his Moon, Neptune, and Chiron, and also his Sun providing a link between that powerful communication configuration and his personality. The Mid-Heaven is, of course, the career and the public image, showing that Hitler’s whole personality was involved when he communicated in public. The fact that the German public also felt “wounded” after the defeat in World War I tied right in with Hitler’s Mercury/Chiron quintile. Unfortunately, he dealt with these wounds by making more wounds.
A more positive example of a strong 5th harmonic chart is Mozart. In his birth chart, Venus (art) is powerful because it is his only planet in a Cadent House. This means that it is a singleton by house, and can draw all of the horoscope’s energy to the 6th house (work). But the only apparent aspect seems to be a trine to Mars, and while this does add activity, it seems to leave poor Venus without connection to the rest of his life. Let’s take a look at his 5th Harmonic Chart. Venus now has conjunctions with Pluto (extremes, obsession, intensity) and the Moon (which rules both one’s own emotions and the ability to appeal to the public). In fact, there are eight conjunctions in his 5th harmonic, and each one represents a quintile or a bi-quintile in his birth chart. Talent on top of talent. The Sun and Mercury in that 5th harmonic chart make trines to the Mid-Heaven, making it easier for that talent to come before the public. Most of Mozart’s first drafts were the same as his final drafts. He would hear the music in his head and write it down with few, if any revisions.
The Seventh Harmonic:
According to Hamblin in his book, “Harmonic Charts,” the seventh harmonic has to do with
“…man’s highest flights of creative imagination, and his lowest depths of destructive illusion. It is a number not of permanence, but of transience; not of unchanging truth, but of sudden flashes of light and darkness; not of objective reality, but of subjective impressions and the emotional experience which man derives from those impressions; not of knowledge, but of fantasy. It is, in short, a romantic number.”(Pg. 65).
In other words, the 7th harmonic has to do with inspiration, with what inspires us and how we can inspire others. This is an important ingredient of creativity.
In the 7th harmonic, the septiles, biseptiles, and tri-septiles appear as conjunctions. A septile is one-seventh of a circle, or 51.42857 degrees. Try finding that aspect by visual inspection of a birth chart!
Woody Allen’s 7th Harmonic has Venus (artistic ability) opposite Chiron (our “wounds” and recurring problems). Since Venus is also our ability to experience happiness, a Venus/Chiron stress aspect can show problems with enjoyment. But Chiron is also a teacher. Woody has taken his neuroses and turned them into entertainment. Of course, there is more than just those two involved. For real creative ability, you need to have several planets linked in an overall complex of aspects. That Venus/Chiron combination has aspects to Jupiter (ability to see the big picture) and Pluto (depth, psychological insight). Jupiter has a conjunction to Pallas (pattern recognition ability). On top of that, Uranus (ability to see things in strange and startling ways) aspects Venus, Pluto, and the Moon. That’s six planets making up one pattern. The real key, though, is the Mid-Heaven. Why? Because without that, no one would ever see his creativity. The Mid-Heaven is the career, and, even more importantly, the public image. Aspect patterns involving planets sometimes last for more than a day. Everyone born that day has them. Aspects involving the Ascendant and Mid-Heaven, however, will show up in the harmonic charts for about 20 minutes, and this can make all the difference.
Tennessee William’s 7th Harmonic also has a series of aspects linked together in one complex pattern. Mars, Jupiter, Saturn, and Pluto form a Grand Trine in his 7th harmonic. Mars give the energy and desire to do things, Jupiter gives optimism that they can be done, Saturn gives the discipline and determination to do them, and Pluto adds the desire to see hidden secrets. His birth chart does have a Scorpio Mid-Heaven, and his plays are famous for characters that are tragically flawed. Jupiter in conjunction with the Mid-Heaven, however, does add a note of optimism, and many times the plays end with some sort of rebirth (Pluto again) of the main protagonist.
The Mid-Heaven is also connected to that Grand Trine, since it has a conjunction to Jupiter and a trine to Saturn. Jupiter is the storyteller, but Saturn gives the discipline to work and produce Tennessee William’s 7th Harmonic Mid-Heaven also makes a sextile to Mercury (writing) and a trine to Neptune (imagination) making it easier for his inspiration and imagination to be communicated to the public.
Andrew Lloyd Webber also has a strongly aspected Mid-Heaven in his 7th Harmonic Chart. There is a sextile to the Moon, and that doesn’t hurt if you want your inspiration to be noticed in public. But his 7th harmonic Mid-Heaven also has an opposition to Mercury (communication) and a square to Neptune (imagination). While this does mean that his imaginative communications can come before the public, the stressful nature of the aspects does indicate that this will be a struggle for him. Saturn aspects the 7th harmonic Mid-Heaven as well, giving the determination and discipline to pound his inspiration into a completed and polished work.
The Ninth Harmonic: According to Hamblin, the 9th harmonic chart shows “…a person’s capacity for joy and happiness,” and it also “…shows the person’s capacity to spread joy and happiness to those around (them).” While the 4th harmonic shows how we can deal with stress, strain, and struggle, the 9th harmonic, by contrast, shows how we can be at peace.
But that means it can also show blockages to being at peace. The 9th Harmonic of David Berkowitz, the infamous “Son of Sam” serial killer, has a Grand Cross formation involving Pluto (death) opposing a conjunction of 4 planets: Mercury (the mind, and, in this case, his Sun-sign ruler), Saturn (depression), Neptune (dreams and illusions), and Mars (action, aggression, violence). The Sun makes square aspects with all of them. The aspects to Mars are wide, but we can use a wider orb here because Mars is part of an over-all pattern.
This pattern would occur in the 9th harmonic charts of everyone born that day. What sets David Berkowitz 9th Harmonic chart apart is the presence of the Ascendant in this pattern, and it completes the Grand Cross formation (which can be highly stressful). In David Berkowitz’s Natal Chart, Venus is the ruler of both the 8th house (death) and the 12th house (psychological problems), and in his 9th harmonic chart, Venus and Uranus (sudden change) both oppose the Mid-Heaven (the public image) bringing his “work” before the public. Again, we have to remember that the Ascendant and Mid-Heaven were only part of these configurations for a few minutes for the time and place of his birth.
Mozart’s Ninth Harmonic Chart has a T-Cross formation involving the Sun, the Moon, Mercury, Jupiter, Pluto, and the Mid-Heaven. Since the Mid-Heaven is involved, appearing before the public is part of what made him happy. But it could also mean that when he does whatever made him happy, it would help him come before the public. What do those stressful aspects tell us? They show either that it was a struggle for Mozart to experience joy, or that it was the struggle itself that made him happy.
Michelle Pfeiffer’s Ninth Harmonic Chart has a Kite formation, and once again the Mid-Heaven is involved. The Sun and Mars make a Grand Trine with Saturn showing happiness with action, but only if it is disciplined and focused action. A triple conjunction of the Moon, Jupiter, and the Mid-Heaven is opposite Saturn and sextiles both the Sun and Mars, and this completes the Kite. With the Mid-Heaven involved, success in career is difficult to avoid. But Saturn is opposite Jupiter and the Mid-Heaven. Jupiter expands. Saturn contracts. Jupiter is optimism. Saturn is pessimism. Jupiter is up. Saturn is down. Jupiter gives her the luck and optimism to succeed. Saturn gives the ambition and discipline. But she will tend to be of two minds about her success, sometimes loving and sometimes hating it. Periodically, she has to get away from it all (Saturn), but then she gets bored (Jupiter) and has to get back into action.
The Eleventh Harmonic: This one shows pure imagination, fantasy. Sometimes this stays inside in our own private little world. Every now and then, we let it out. The results vary. We can get a beautiful dream made real, or a nightmare of horror.
Ian Stuart Brady’s 11th Harmonic is an example of the nightmare come to life. Brady, the “Monster of the Moors,” and his girlfriend, Myra Hyndley, were a serial killer team. In Brady’s 11th Harmonic chart, the Moon (emotions) is opposite Mercury (the mind) and Mars (anger) showing an imagination that tended to violence. That configuration was in effect for a few hours on the day Brady was born. As always, it is the placement of the Ascendant and the Mid-Heaven that make the difference. Aspects involving those two only last for about five minutes in an 11th harmonic chart. Brady’s 11th Harmonic Mid-Heaven makes a T-Cross formation with Saturn (depression) and Neptune (imagination). The depression can sometimes be displaced into fantasies of controlling and harming others, which is what happened here. The stress aspects to the Mid-Heaven showed that these fantasies would one day come before the public. The 11th harmonic Ascendant makes a square to Pluto (death) that is only one minute of arc from exact. This is the key aspect here, the one that pushed his fantasies to become obsessions, and the obsessions to into reality.
George Gershwin’s 11 Harmonic shows the positive side, the beautiful dream come to life. Venus (art) and Jupiter (optimism, “luck,” expansion) are opposite the 11th harmonic Ascendant, while the Moon opposite the 11th harmonic Mid-Heaven makes it easier for his dreams to come before the public. Uranus (innovation) makes trine aspects to Venus and Jupiter, making it easier for Gershwin to innovate. The strong 11th harmonic adds to the creativity of Gershwin’s Birth Chart, which has Neptune (imagination) making a square aspect and Venus (art) making a sextile to the Mid-Heaven.
Dr. Martin Luther King, Jr. has Sagittarius (sign of the archer) on the cusp (beginning) of his 8th house (death). Jupiter, the ruler, is in his 12th house (hidden enemies). He was shot by a sniper who was concealed at a distance. Dr. King was most famous for his “I have a Dream” speech, so it is only natural to check Dr. King’s 11th Harmonic. To see if his dream came before the public, we have to look at the Mid-Heaven. Remember that aspects to the Mid-Heaven are in effect for only a few minutes, so the chances are very small that other people would have them unless they happened to be born at the same place and time. Jupiter and Pluto make sextiles, Saturn and Uranus make conjunctions, and Neptune makes an opposition to his Mid-Heaven. That’s five planets, or half of all the planets in the chart! Do you notice something else? They are all Outer Planets, the ones that rule the large-scale events in our lives.
There is more here, much more. Saturn and Neptune are in opposition and both aspect the Mid-Heaven. But didn’t Ian Brady have something similar in his 11th harmonic? Yes, however we must remember that there is always a positive and a negative way for things to work out. Similar aspect patterns are common to the horoscopes of murderers and murder victims, an ironic fact that was first noted by Darling and Oliver in their book Astro Psychiatry (CSA Press, 1972).
With Ian Brady, his imagination added to his murderous urges. In the case of Dr. King, it was a setup for martyrdom. Uranus is there as well, making a conjunction to Saturn and an opposition to Neptune. Uranus is the planet of sudden, shocking events. Of course, this makes us think of his assassination, but his whole life was a struggle and involved demonstrations which (at the time) were considered shocking. The trine between Jupiter (ruler of the house of death in his birth chart) and Pluto (the “natural” ruler of death) is also significant, very significant.
They both aspect his 11th harmonic Mid-Heaven, but, on top of that, the Mid-Heaven sits on their Midpoint! Yes, the connection with his death seems obvious here, but so is the connection with his life. The 8th house rules more than death. It also shows the values of the groups to which we belong, all the groups, including the society that we live in. The effect the group values have on us is an 8th house matter, but so is the impact we can have on the values of the group. With his 11th harmonic Mid-Heaven on the midpoint of Jupiter and Pluto, King’s dream had an immense impact on an entire nation, during his life and even in death.
The Thirteenth Harmonic: The 13th harmonic seems to have a similar effect to the sign of Scorpio. They both relate to death and rebirth, transformation, as well as group resources and values.
The birth chart of Dr. Jonas Salk, inventor of the first effective anti-polio vaccine, does not (at first glance) seem too special. The Sun rules them Mid-Heaven (career) and sits in the 12th house in Scorpio. That’s a good placement for doing research behind the scenes, and the Leo Mid-Heaven can give the chance of eventual fame. Pluto, the ruler of the Sun-sign, has a conjunction with Saturn (ambition, discipline) and they both make out-of-sign sextile aspects with the Mid-Heaven, as well as square aspects (challenges) to his Sun. But when we look at Jonas Salk’s 13th Harmonic Chart, the picture becomes far more clear. Jupiter, Mars, Pluto, and Saturn all make conjunction aspects to his Mid-Heaven, while the Moon and the North Lunar Node make square aspects, and the Sun makes a quincunx (the 150 degree aspect). Six out of ten planets aspect his Mid-Heaven (career and public image) and that configuration only held for a few minutes on that day and location. Salk’s vaccine ended a yearly epidemic of polio and (long after Salk’s death) is still saving thousands of lives a year.
Bill Gate’s 13th Harmonic Chart has the Mid-Heaven (career) sextile Jupiter and Pluto, which are located in the 2nd house (money) of his Birth Chart. Uranus, which rules his natal 8th house (other people’s money and resources) makes a square aspect here in the 13th harmonic. His big break came when he bought the DOS operating system from another company (other people’s resources) and then made a deal with IBM
Harmonic charts can provide greater depth to most horoscope interpretations. Just be sure to remember that they are used for aspect patterns only, and not for sign or house interpretation. You can use them for chart comparison as well, but please don’t compare harmonics of different numbers (that’s like comparing apples and oranges).
``` (adsbygoogle = window.adsbygoogle || []).push({}); ```
In Tune With The Universe | 5,650 | 24,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-38 | latest | en | 0.955902 |
https://www.interviewarea.com/q-and-a/how-many-1-bills-would-it-take-to-go-around-the-world | 1,679,663,034,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945282.33/warc/CC-MAIN-20230324113500-20230324143500-00446.warc.gz | 922,750,624 | 15,478 | # How many \$1 bills would it take to go around the world?
The length of 100,000,000,000 (one hundred billion) one dollar bills
one dollar bills
July 2019) The United States one-dollar bill (US\$1), sometimes referred to as a single, has been the lowest value denomination of United States paper currency since the discontinuation of U.S. fractional currency notes in 1876.
https://en.wikipedia.org › wiki › United_States_one-dollar_bill
laid end-to-end measures 9,690,656 miles. This would extend around the earth 387 times.
## How many 1 dollar bills does it take to reach the moon?
A. 19 million dollar bills is a reasonable order of magnitude estimate. Dollar bills are about 2 meters long, so it would take about 19 million of them to reach the Moon.
## How many dollar bills does it take to make a mile?
Want to measure your notes in a different way? A stack of currency one mile high would contain more than 14.5 million banknotes.
## How many dollar bills would it take to cover the United States?
Answer and Explanation: A dollar is roughly 15.3 square inches, so divide 223,781,500,800 by 15.3 and you get 14,626,241,882 dollars.
## How far would 1 trillion dollars reach?
If you stacked \$100 bills totaling \$1 trillion on top of each other, the stack would be 631 miles high. This is what \$1 trillion in spending look like. John Watling and 60 others like this.
## How tall would 1 billion dollars look like?
The height of a stack of 1,000 one dollar bills measures 4.3 inches. The height of a stack of 1,000,000 one dollar bills measures 4,300 inches or 358 feet – about the height of a 30 to 35 story building.
## Can a billion dollars reach the moon?
The height of a stack of 1,000,000,000,000 (one trillion) one dollar bills measures 67,866 miles. This would reach more than one fourth the way from the earth to the moon.
## How much usd can i carry from India?
You can indefinitely retain foreign exchange upto US\$ 2,000, in the form of foreign currency notes or travellers' cheques (TCs) for future use. Any foreign exchange in cash in excess of this sum, is required to be surrendered to a bank within 90 days and TCs within 180 days of return.
## How many US dollars do you need to survive?
Living Expenses in the US
The estimated living cost for the US is around \$10,000 to \$18,000 per year, which averages around \$1,000 to \$1,500 per month. This includes your accommodation costs, room, and board, food, travel, textbooks, weather-appropriate clothing, and entertainment expenses as well.
## How much money can a person carry legally in India?
Residents of India are allowed to carry up to Rs. 25,000 though. There's no limit, however, to how much foreign currency you can bring into India. Although, you will have to declare it if the amount exceeds US\$5,000 in notes and coins, or US\$10,000 in notes, coins, and traveller's cheques².
## What would a trillion dollars look like?
What is a trillion dollars? Well, it's a million million. It's a thousand billion. It's a one followed by 12 zeros.
## Can you spend a billion dollars in a lifetime?
Suppose you had \$1-billion. You could spend \$5,000 a day for more than 500 years before you would run out of money. Breaking it down even farther, it means you would have to spend over \$100,000 every day for the next 25 years in order to spend \$1-billion.
## How much does a bit of the moon cost?
Buy a piece of the moon
Headed by Dennis Hope, the company sells pieces of the moon for around \$20 per acre, and to this point, over 2.5 million have already been sold. By this method, the moon's 9.3 billion acres are worth approximately \$180 billion. Image source: Getty Images.
## How much junk is left on the moon?
Humans have left over 187,400 kilograms (413,100 lb) of material on the Moon.
## How much money is the moon worth?
"The mare constitutes around 15% of the lunar surface, making the total value of the moon... \$4 quadrillion." Thinking about it another way, that much He-3 could theoretically supply U.S. electricity demand for 80,000 years.
## How much money do you need to live the rest of your life?
The answer for most people, according to new research by university psychologists, is \$10m (£8.6m) – but not Americans, who say they need at least \$100m, and frequently insist on \$100bn.
## Is \$5,000 dollars a month good in usa?
If you make \$5000 a month (gross), you'll bring in \$60,000 a year before taxes. This is a good salary to live on. However, you may want to consider living frugally or investing in a side hustle. If you make this salary, you will be able to live quite comfortably.
## Can you live with 1000 dollars?
If you're trying to live on a \$1,000-a-month budget, all of it can't go to housing. Unfortunately, the national average fair market rent for a one-bedroom apartment or home is \$1,105 per month. So even if you cut your budget in half to account for housing, you'll still fall way short.
## Can I carry 5 lakh cash in flight?
In case of domestic flights, there is no specific limit for carrying cash but one needs to provide substantial documentary evidence to confirm the source of the cash as well as the purpose for carrying the cash.
## Can I carry 2 lakh cash in flight?
How much cash as an Indian can we carry in a domestic flight? The Government of India has passed guidelines not to carry cash more than 2 lakhs in general. It will be illegal to carry cash . Even carrying cash in flight it is taxable.
## Can I carry cash in flight?
According to RBI, travellers going abroad can carry cash of up to \$3000 with balance amount in form of store value cards, traveller's cheque or banker's draft. Exception to this are the travelers going to Iraq and Iran who can carry up to \$5000.
## What if the moon left us?
It is the pull of the Moon's gravity on the Earth that holds our planet in place. Without the Moon stabilising our tilt, it is possible that the Earth's tilt could vary wildly. It would move from no tilt (which means no seasons) to a large tilt (which means extreme weather and even ice ages).
## Can you sell land on the moon?
Short answer: you can't. No one can. The relevant provision of the OST is Article II which states: “Outer space, including the moon and other celestial bodies, is not subject to national appropriation by claim of sovereignty, by means of use or occupation, or by any other means.”
## Would the moon ever hit the earth?
Long answer: The Moon is in a stable orbit around Earth. There is no chance that it could just change its orbit and crash into Earth without something else really massive coming along and changing the situation. The Moon is actually moving away from Earth at the rate of a few centimetres per year. | 1,599 | 6,747 | {"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} | 3.75 | 4 | CC-MAIN-2023-14 | latest | en | 0.917486 |
https://github.com/martin-nikolov/Telerik-Academy/blob/master/Programming%20with%20C%23/1.%20C%23%20Fundamentals%20I/06.%20Loops/Readme.md | 1,553,073,624,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202324.5/warc/CC-MAIN-20190320085116-20190320111116-00121.warc.gz | 500,380,477 | 15,372 | Fetching contributors…
Cannot retrieve contributors at this time
101 lines (97 sloc) 3.37 KB
## Loops
1. Write a program that prints all the numbers from 1 to `N`.
• Write a program that prints all the numbers from 1 to `N`, that are not divisible by 3 and 7 at the same time.
• Write a program that reads from the console a sequence of `N` integers and returns the minimal and maximal of them.
• Write a program that calculates N!/K! for given `N` and `K` (1 < K < N).
• Write a program that calculates *N!K! / (N-K)! for given `N` and `K` (1 < K < N).
• Write a program that, for a given two integers `N` and `X`, calculates the sum S = 0!/X0 + 1!/X1 + 2!/X2 + ... + N!/XN
• Write a program that reads a number `N` and calculates the sum of the first N members of the sequence of Fibonacci: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ... Each member of the Fibonacci sequence (except the first two) is a sum of the previous two members.
• Write a program that calculates the greatest common divisor (GCD) of given two numbers. Use the Euclidean algorithm (find it in Internet).
• In the combinatorial mathematics, the Catalan numbers are calculated by the following formula:
• Write a program to calculate the Nth Catalan number by given `N`.
• Write a program that prints all possible cards from a standard deck of 52 cards (without jokers). The cards should be printed with their English names. Use nested for loops and switch-case.
• Write a program that reads from the console a positive integer `N` (N < 20) and outputs a matrix like the following:
N=3
1 2 3
2 3 4
3 4 5
N=4
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
* \*Write a program that calculates for given `N` how many trailing zeros are present at the end of the number N!. Examples: * N = 10 -> N! = 3628800 -> 2 * N = 25 -> N! = 15511210043330985984000000 -> 6
``````Does your program work for N = 50 000? Hint: The trailing zeros in N! are equal to the number of its prime divisors of value 5. Think why!
``````
• *Write a program that reads a positive integer `N` (N < 20) from console and outputs in the console the numbers 1 ... N2 numbers arranged as a spiral.
N=4
1 2 3 4
12 13 14 5
11 16 15 6
10 9 8 7
You can’t perform that action at this time. | 688 | 2,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-13 | latest | en | 0.876022 |
https://www.thestudentroom.co.uk/showthread.php?t=4541044 | 1,527,240,962,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867055.20/warc/CC-MAIN-20180525082822-20180525102822-00500.warc.gz | 851,858,998 | 43,947 | You are Here: Home >< Maths
# AS Maths Trig Identities watch
1. Attachment 616444
Hi, please can someone help me with this maths question?
I've attached what to next but i'm stuck where to go next so any advice would be appreciated.
Thanks,
Amelia
2. (Original post by acantwellhyde)
Hi, please can someone help me with this maths question?
I've attached what to next but i'm stuck where to go next so any advice would be appreciated.
Thanks,
Amelia
there is no attachment, try posting it again
(Original post by DylanJ42)
there is no attachment, try posting it again
4. (Original post by acantwellhyde)
5. Thank you for your help but I don't really understand what you have done to get this answer. I don't really want the answer I just want someone to help me with the next few steps please..
(Original post by RDKGames)
6. (Original post by acantwellhyde)
start again from here, remember for the bracket on the top line to use F.O.I.L to multiply it out.
In your working you have implied that 1 x 1 = 2.
also is usually written as and not , likewise for sine.
Finally what identities do you know involving and ?
7. (Original post by acantwellhyde)
Thank you for your help but I don't really understand what you have done to get this answer. I don't really want the answer I just want someone to help me with the next few steps please..
You have on the second line of the LHS working which is the same as and what I'm saying is that you've expanded it wrong. The constant should be a 1, not a 2.
8. Thank you so much! I'll have another go and post my answer shortly
(Original post by DylanJ42)
start again from here, remember for the bracket on the top line to use F.O.I.L to multiply it out.
In your working you have implied that 1 x 1 = 2.
also is usually written as and not , likewise for sine.
Finally what identities do you know involving and ?
9. Got it! Thanks for your help. This is correct, right?
(Original post by acantwellhyde)
Thank you so much! I'll have another go and post my answer shortly
10. (Original post by acantwellhyde)
Got it! Thanks for your help. This is correct, right?
perfect, well done
### Related university courses
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: February 2, 2017
Today on TSR
### Edexcel GCSE Maths Unofficial Markscheme
Find out how you've done here
Poll
Useful resources
### Maths Forum posting guidelines
Not sure where to post? Read the updated guidelines here
### How to use LaTex
Writing equations the easy way
### Study habits of A* students
Top tips from students who have already aced their exams | 673 | 2,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-22 | latest | en | 0.975594 |
https://www.strategystreet.com/basic_strategy_guide/activities/activity_4_manage_costs/examples_tests_worksheets/worksheets/worksheet_20c/ | 1,721,144,014,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00216.warc.gz | 868,392,261 | 14,529 | # WORKSHEET #20C: Setting Company Pricing Objectives and Guidelines
Step 1:
Determine the minimum Price at which the Company will serve a Non-core customer:
• Determine the cash cost per unit of sale of producing the Standard Leader product. The source for this information should be the financial department of the organization.
• Determine how these cash costs might vary as the Company serves customers of varying sizes (i.e. Very Large, Large, Medium and Small customers). Note here that the size of the customer for this calculation would be the purchases actually made from the Company. The source of this information is also the financial department of the organization supported, where necessary, by the operations groups.
• Develop a graph of the minimum price that the Company would accept to serve a customer who annually purchases a given amount of product from the Company. On the vertical axis would be the price per unit for the Standard Leader product. On the horizontal axis would be the size of the customer's annual purchases from the Company. The graph would be a line or a series of lines for the minimum acceptable price.
Step 2:
Determine whether the Company has sufficient capacity to serve all Core customer needs over the next two years. If it does not have sufficient capacity:
• Rank each Non-core customer according to the annual contribution dollars the customer makes to the Company's fixed costs and overhead.
• Estimate the capacity likely to be freed up as these Non-core customers leave us for other suppliers.
• If the Company still needs additional capacity, rank all Near-core customers according to the annual profit margin percentages they offer the Company, with the lowest profit margin percentage listed first and the highest listed last. If figures for profit margins are not available, you may start this analysis with gross margin figures, in percentage terms. Plan to raise prices to these customers in the order in which they provide profit margin percentages to the Company. Raise prices first on the Near-core customers who provide the lowest margin and last on the customers who provide the highest margins.
• If the Company still needs capacity in order to serve its Core customer needs, evaluate the possibility of purchasing product from other suppliers in order to keep Core customer demand satisfied.
• If the Company must eliminate some Core customer volume in order to support its most important Core customers, list all Core customers in the order in which they purchase from the Company with the customer purchasing the least amount from the Company ranked first and the company purchasing the greatest amount ranked last.
• If the Company believes that it might be successful with an allocation program with the Core customers, then plan an allocation program which provides a greater allocation to the larger customers than it does to the smaller customers.
• If the Company is unlikely to succeed with an allocation program, then raise prices to the Core customer purchasing the least amount from the Company first and proceed through the list by raising prices to each succeeding Core customer purchaser, ranked in order of purchases, until the needs of the largest Core customers are met.
Step 3:
If the Company has any control over pricing in a rising price environment, consider limitations on the degree to which prices rise, as follows:
• Evaluate the threat from substitute products by comparing the price of the Standard Leader product today to the substitute product. Make the same comparison in relative prices three years ago. If the substitute product has become cheaper, compared to the present Standard Leader product, this substitute product may be a threat.
• Evaluate the industry's return on investment at the new pricing level compared to average returns for all industries (see Tools/Benchmarks/Quartile Ranking Reports). If returns are high, consider whether these returns will attract competitors who are not in the market today.
Step 4:
Adjust the objectives for each Very Large and Large customer by considering the competitors who are in these customer relationships with us:
• If the market price is tending to fall, gather, on a regular basis, pricing initiatives for each competitor, and adjust the speed and degree of the Company's response to each competitor according to our prediction of the way the competitor will price in their relationship.
• If prices are rising in the market, the Company has little need to track individual competitor pricing tactics until price based competition re-emerges.
Step 5:
Establish guidelines for the use with all Medium and Small customers by setting minimum price and volume targets that the Company must reach to serve each of those customers. Use the conclusions from Step 1 to establish minimum prices for all customers in a declining price environment. In a rising price environment, the Company would choose to serve new Small and Medium customers only to the extent that it is likely to have capacity available to do so. | 933 | 5,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-30 | latest | en | 0.929033 |
https://schoollearningcommons.info/question/let-f-ab-be-a-function-defined-by-f-2-2-1-where-a-2-4-6-10-12-b-3-7-11-19-23-25-represent-f-by-i-24976922-27/ | 1,632,356,007,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057403.84/warc/CC-MAIN-20210922223752-20210923013752-00665.warc.gz | 499,654,689 | 13,873 | ## Let f: AB be a function defined by f(2)=2x-1 where A={2,4,6,10,12) , B = {3,7; 11,19,23,25) Represent F by i) set of ordered
Question
Let f: AB be
a function defined by f(2)=2x-1 where
A={2,4,6,10,12) , B = {3,7; 11,19,23,25) Represent F by
i) set of ordered pairs ii) table
mi) arrow diagram
iv) graph.
in progress 0
1 month 2021-08-13T14:40:46+00:00 1 Answer 0 views 0
Definition : Given two non-empty sets A and B, the set of all ordered pairs (x, y),
where x ∈ A and y ∈ B is called Cartesian product of A and B; symbolically, we write
A × B = {(x, y) | x ∈ A and y ∈ B}
If A = {1, 2, 3} and B = {4, 5}, then
A × B = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}
and B × A = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}
(i) Two ordered pairs are equal, if and only if the corresponding first elements are
equal and the second elements are also equal, i.e. (x, y) = (u, v) if and only if x =
u, y = v.
(ii) If n(A) = p and n (B) = q, then n (A × B) = p × q.
(iii) A × A × A = {(a, b, c) : a, b, c ∈ A}. Here (a, b, c) is called an ordered triplet.
2.1.2 Relations A Relation R from a non-empty set A to a non empty set B is a
subset of the Cartesian product set A × B. The subset is derived by describing a
relationship between the first element and the second element of the ordered pairs in
A × B.
The set of all first elements in a relation R, is called the domain of the relation R,
and the set of all second elements called images, is called the range of R.
For example, the set R = {(1, 2), (– 2, 3), (
1
2
, 3)} is a relation; the domain of
R = {1, – 2,
1
2
} and the range of R = {2, 3}.
Step-by-step explanation: | 621 | 1,664 | {"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} | 3.953125 | 4 | CC-MAIN-2021-39 | latest | en | 0.845884 |
https://math.stackexchange.com/questions/3659175/using-cauchy-schwarz-to-prove-the-triangle-inequality | 1,656,875,311,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104248623.69/warc/CC-MAIN-20220703164826-20220703194826-00223.warc.gz | 423,606,183 | 64,634 | # Using Cauchy-Schwarz to prove the triangle inequality
Using the following Cauchy-Schwarz Inequality for integrals,
$$\int_c^d\int_a^bf^2(x,y)dxdy\int_c^d\int_a^bg^2(x,y)dxdy \geq \Bigg[\int_c^d\int_a^bf(x,y)g(x,y)dxdy\Bigg]^2$$
Prove: $$\sqrt{\int_c^d\int_a^b[f(x,y) + g(x,y)]^2dxdy} \leq \sqrt{\int_c^d\int_a^bf^2(x,y)dxdy} + \sqrt{\int_c^d\int_a^bg^2(x,y)dxdy}$$
I am struggling with this proof. I tried starting by expanding the left handside and constructing an inequality by a substitution of the C-S inequality but this did not seem to help.
Putting double integrals seems superfluous so lets use just one. We use this form of Cauchy-Schartz (which is just Hölder for $$p=q=2$$): \begin{align*} \left(\int fg\right)^2 \leq \int f^2 \int g^2. \end{align*}
It is equivalent to prove the square of the equality: \begin{align*} \int (f+g)^2 \leq \int f^2 + \int g^2 + 2\sqrt{\int f^2} \sqrt{\int g^2}. \end{align*}
Note that we may expand the LHS and apply Cauchy-Schwartz on the cross-term: \begin{align*} \int (f+g)^2 = \int f^2 + \int g^2 + 2\int fg \leq \int f^2 + \int g^2 + 2\sqrt{\int f^2}\sqrt{\int g^2}. \end{align*} | 439 | 1,134 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-27 | longest | en | 0.713027 |
http://talkstats.com/threads/can-somebody-check-this-one-two.1002/ | 1,597,228,240,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738888.13/warc/CC-MAIN-20200812083025-20200812113025-00269.warc.gz | 99,397,069 | 9,149 | # Can somebody check this one two
#### beenawhile
##### New Member
If the mean life of an aircraft engine part is 510 hours and the standard deviation is 30 hours, what percentage of parts lasts no more than 462 hours? (Assume a normal distribution for life). (1 marks).
z= (x-mu) /s
so z = (462-510)/ 30
= -1.6
from z table = 44.52
So rounded 45 % of engine parts last no more than 462 hrs
#### JohnM
##### TS Contributor
The proportion of the area under the normal curve below z = -1.6 is 0.055, so the answer is 5.5%
#### beenawhile
##### New Member
Z table
Think i need a new z table can u direct me to 1??
#### beenawhile
##### New Member
Z table
Think i need a new z table can u direct me to 1?? Thanx for your help thus far
#### beenawhile
##### New Member
Whoops
See where i have gone Wrong now Thanx for help | 245 | 835 | {"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-2020-34 | latest | en | 0.799972 |
https://www.gamedev.net/profile/199154-daniel-wilson/ | 1,558,659,610,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257481.39/warc/CC-MAIN-20190524004222-20190524030222-00397.warc.gz | 787,998,292 | 28,324 | # Daniel Wilson
Member
23
159 Neutral
• Rank
Member
1. ## Tool to mathematically create a mesh
That's what I needed thanks!
2. ## What to do with an attenuated light vector in my shader
So I've been working on this a while longer, and the only real conclusion I have come up with is the colour values must be wrong. I thought for a while maybe they were right and I was not doing the correct thing with the result. I was given the good advice that: The scattered radiance should be added to the phong radiance component because the phong model computes the radiance at the surface only, the scattered radiance is just more radiance coming from under the surface, to the eye. [/quote] So hopefully that might help some future googlers, now if if I could just work out the correct values to begin with! I knew the colours were too trippy
3. ## How to change colour based on viewing angle
Very nice thanks for that cool answer, the effect is simple works well for my purposes!
4. ## How to change colour based on viewing angle
I would like to fake some caustics on a glassy surface very quickly, and the idea I have is to overlay a static caustic texture that appears based on the viewing angle, so when the angle is more grazing, more of the texture appears. How might I do this? A small piece of code would be seriously amazing. Thankyou!
5. ## What to do with an attenuated light vector in my shader
Yeah should've mentioned that too sorry, I am certain that data is correct. This shader is actually being used in Ogre, and it's easy to generate the texture in Ogre and the output is virtually identical to fx composer. Each rgba component in the floating point texture just holds a preset scattering coefficient value taken from this table:
7. ## What to do with an attenuated light vector in my shader
Okay so the [font=courier new,courier,monospace]p_omega_out[/font] value looks like this as a colour: Applied to a simple plane. I'm not sure whether or not that's the correct output for it but that's what I have at the moment. If I slot it in to my phong normal mapped plane, I get this: Which is pretty but obviously isn't right! So is [font=courier new,courier,monospace]p_omega_out[/font] wrong, or am I just putting it in the wrong place!? Here is the final part of the code for reference but the only change is the inclusion of the new attenuated vector: [source lang="cpp"] float3 BumpNormal = tex2D(nm, IN.texcoord)*2.0 - 1.0; float4 amb = AmbientIntensity * ambient; float4 diff = DiffuseIntensity * float4(p_omega_out, 1) * saturate(dot(IN.tangentSpaceLightDir,BumpNormal)); float3 R = normalize(2.0 * dot(BumpNormal, IN.tangentSpaceLightDir) * BumpNormal - IN.tangentSpaceLightDir); float3 v = normalize(IN.tangentSpaceEye); float spec = pow(saturate(dot(R,v)), specularPower) * SpecularIntensity; // compute final color float4 color = tex2D(color_map,IN.texcoord); float4 finalcolor = (amb + diff + spec) * color; return finalcolor;[/source]
8. ## Force field around a model and pushing vertices outwards in shader.
Okay I found it, sorry it wasn't on the publications page hope I didn't waste your whole night! This link contains the ppt presentation, it's a whopping 433mb so thank goodness for broadband. An overview of the talk is discussed here as well, with the shield stuff on page 3. Pretty nice ideas about colorizing it with tiny colour palette ramps, and fading out the effect as a function related to the depth buffer.
9. ## What to do with an attenuated light vector in my shader
Ah okay this is interesting thank you. You see the paper assumes you just know what to do with the output vector that has been attenuated from K-M-P. So theoretically it should be okay to use it in place of a hard coded "diffuse" value I did have, e.g.: [source lang="cpp"]float4 diff = DiffuseIntensity * p_omega_out * saturate(dot(IN.worldLightDir,BumpNormal));[/source] At the moment if I output p_omega_out as a colour, the result is some pretty blurry bands of colour, so hopefully it won't damage the rest of the colour too much. (I'll post a screen cap in a few mins). Thanks
10. ## Force field around a model and pushing vertices outwards in shader.
I saw a nice talk on this on the Bungie website, because Halo has a LOT of force fields. I can't remember the exact talk but it's on that page somewhere sorry!
11. ## What to do with an attenuated light vector in my shader
So I didn't quite figure this out yet because it's quite a complex shader. I would lke to implement this shader with a basic phong model. Say I have this: [source lang="cpp"] float3 BumpNormal = tex2D(nm, IN.texcoord)*2.0 - 1.0; float4 amb = AmbientIntensity * ambient; float4 diff = DiffuseIntensity * diffuse * saturate(dot(IN.worldLightDir,BumpNormal)); float3 R = normalize(2.0 * dot(BumpNormal, IN.worldLightDir) * BumpNormal - IN.worldLightDir); float3 v = normalize(IN.eye); float spec = pow(saturate(dot(R,v)), specularPower) * SpecularIntensity; // compute final color float4 color = tex2D(color_map,IN.texcoord); float4 finalcolor = (amb + diff + spec) * color;[/source] R and V are purely directional vectors right? So my attenuated light vector must represent a direction and the intensity of the light that is returned to the viewer I think. In the above code the only thing that strikes me as close to this is the calculation of the diffuse value. Ignoring the bump map (I don't need bumped normals), does anyone know if it would be wrong to have: [source lang="cpp"]float4 diff = DiffuseIntensity * diffuse * p_omega_out;[/source] Where [font=courier new,courier,monospace]p_omega_out[/font] is the attenuated light intensity/direction? I think my vector needs to go from the pixel to the viewer, would this vector do that?!
12. ## When to normalize vectors in a shader?
Yes I thought so, I just figured maybe there was some rules like "you shouldn't normalize in view space". Thanks for the tips. I shall try and keep any normalizing to a minimum though for now and see how it goes
13. ## When to normalize vectors in a shader?
Hi, I am currently working on a shader and the result is not quite right. One question that bugs me I cannot seem to find the answer to is why and when exactly we normalize vectors in a shader. I understand that it seems to only be important for direction vectors such as the light, but is it important to do so in certain spaces with other vectors aswell? I took a naive approach at first and normalized all the time. I am not so much concerned with efficiency, just why and when it is okay or not okay to normalize a vector. For example, the shader I am writing at them moment requires the vector from the view position to the vertex position in world space. Some shaders I see normalize this when it is in the fragment shader, but does that not destroy the value of the vector for further calculations?
14. ## Tool to mathematically create a mesh
Hmm, this is very close to what I need, but it uses parametric equations to define the formula, e.g. a sphere would in the form: x = r cos(u) sin(v), y = r sin(u) sin(v), z = r cos(v). I need something I can type x^2 + y^2 + z^2 = r^2 and get a sphere!? I've been looking for a way to convert the equation of the surface I have to a parametric one but can 't really figure it out!
15. ## Tool to mathematically create a mesh
Hi I need a good tool to mathematically create a surface and then get it into 3ds max, for example a sphere is defined algeraically as x^2 + y^2 + z^2 = r^2. Any examples? The only one I can find is 3D-Math Xplor, but the full version only works on a mac and I have Windows 7. The Java version that works on all platforms does not have the export functionality! | 1,877 | 7,689 | {"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-2019-22 | latest | en | 0.953564 |
http://www.cjcm.net/jslxxb/ch/reader/view_abstract.aspx?file_no=20160509&flag=1 | 1,511,328,128,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806465.90/warc/CC-MAIN-20171122050455-20171122070455-00020.warc.gz | 371,002,133 | 6,579 | 欢迎光临《计算力学学报》官方网站！
Application of response surface method on structural reliability analysis of pin joint in composites[J].计算力学学报,2016,33(5):711~716
Application of response surface method on structural reliability analysis of pin joint in composites
Application of response surface method on structural reliability analysis of pin joint in composites
DOI：10.7511/jslx201605009
作者 单位 张九民 中国航空工业集团公司 第一飞机设计研究院, 西安 710089 顾汝佳 南京航空航天大学 航空宇航学院, 南京 210016 李洪双 南京航空航天大学 航空宇航学院, 南京 210016 夏爽 南京航空航天大学 航空宇航学院, 南京 210016
复合材料的力学性能参数比金属材料呈现出更大的分散性。本文提出采用加权线性响应面法对复合材料单钉连接结构进行可靠性分析。采用三维有限元模型进行复合材料单钉连接的确定性渐进失效分析。单向铺层的力学性能参数和强度参数均考虑为输入随机变量，用于可靠性分析。详细研究了不同失效准则和不同刚度退化方法的影响。结果表明，Olmedo-Santiuste失效准则与Camanho-Matthews刚度退化方法的组合能够给出与试验均值最接近的强度分析结果。另外，还仔细研究了响应面参数f的影响。结果表明，取f=1的线性加权响应面能够为复合材料单钉连接可靠性分析提供较好的统计特性逼近值。
Composite materials exhibit much more scattering material properties than metallic materials.The paper presents an application of weighted response surface method on structural reliability analysis of a composite pin joint.A three-dimensional finite element model is used to perform the deterministic progressive failure analysis of composite pin joint.The mechanical properties and ultimate strengths of one lamina are modeled as random variables and considered in the process of structural reliability analysis.A detailed study on the effect of different failure criteria and stiffness degradation rules is performed.This study shows that the combination of Olmedo-Santiuste failure criteria and Camanho-Matthews stiffness degradation rules produces the closest bearing strength result to experimental mean value.Another detailed study is also performed to investigate the effect of response surface parameter f.The results show that linear weighted response surface function with f=1 is capable of providing good approximations of statistical characteristics of the bearing strength of pin joints in composite laminates for structural reliability analysis. | 578 | 1,961 | {"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-2017-47 | longest | en | 0.553644 |
https://www.superfastcpa.com/what-is-paper-profit/ | 1,714,054,710,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00004.warc.gz | 903,818,886 | 91,377 | # What is Paper Profit?
## Paper Profit
Paper profit, also known as unrealized gain, refers to the hypothetical profit an investor has made on an investment, based on its current market value, before it is sold. It is the increase in value of an investment from the price it was purchased at to its current market price.
However, a paper profit only exists “on paper” because the profit isn’t realized until the asset is actually sold. Until that point, the profit is merely an estimate based on the asset’s current market value, which can fluctuate. If the market price drops before the investor sells the asset, the paper profit could decrease or even turn into a loss.
For example, let’s say an investor buys 100 shares of a company’s stock for \$10 each, totaling a \$1,000 investment. If the price per share rises to \$15, the investment’s value would increase to \$1,500. Therefore, the investor has a paper profit of \$500. However, this profit isn’t realized until the investor sells the shares. If the share price drops to \$8 before the investor sells, the paper profit would turn into a paper loss.
It’s important to remember that a paper profit is not cash in hand and does not guarantee a real profit until the investment is sold.
## Example of Paper Profit
Imagine that you purchase 100 shares of a company named FutureTech Inc. for \$50 per share. The total cost of this investment is \$50 * 100 = \$5,000.
After a year, FutureTech Inc. performs exceptionally well and its stock price increases to \$70 per share. At this point, the market value of your investment is \$70 * 100 = \$7,000. This means you have a paper profit of \$7,000 (the current market value of the investment) – \$5,000 (the original cost of the investment) = \$2,000.
However, it’s crucial to note that this is an unrealized gain, or paper profit. It isn’t actual cash in your pocket until you sell the shares.
If, after another few months, the stock price of FutureTech Inc. falls to \$40 per share, the market value of your investment becomes \$40 * 100 = \$4,000. In this case, your paper profit turns into a paper loss of \$1,000, since the market value of your investment (\$4,000) is now less than what you originally paid for it (\$5,000).
So, this illustrates that paper profit is a theoretical profit that exists “on paper,” and can fluctuate with the market value of the investment. It only becomes a real, realized gain when the investment is actually sold for a profit. | 564 | 2,479 | {"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-2024-18 | latest | en | 0.957798 |
https://rdrr.io/cran/effectsize/man/p_superiority.html | 1,719,325,925,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00739.warc.gz | 417,550,825 | 11,669 | # p_superiority: Cohen's _U_s and Other Common Language Effect Sizes (CLES) In effectsize: Indices of Effect Size
p_superiority R Documentation
## Cohen's Us and Other Common Language Effect Sizes (CLES)
### Description
Cohen's U_1, U_2, and U_3, probability of superiority, proportion of overlap, Wilcoxon-Mann-Whitney odds, and Vargha and Delaney's A are CLESs. These are effect sizes that represent differences between two (independent) distributions in probabilistic terms (See details). Pair with any reported stats::t.test() or stats::wilcox.test().
### Usage
p_superiority(
x,
y = NULL,
data = NULL,
mu = 0,
paired = FALSE,
parametric = TRUE,
ci = 0.95,
alternative = "two.sided",
verbose = TRUE,
...
)
cohens_u1(
x,
y = NULL,
data = NULL,
mu = 0,
parametric = TRUE,
ci = 0.95,
alternative = "two.sided",
iterations = 200,
verbose = TRUE,
...
)
cohens_u2(
x,
y = NULL,
data = NULL,
mu = 0,
parametric = TRUE,
ci = 0.95,
alternative = "two.sided",
iterations = 200,
verbose = TRUE,
...
)
cohens_u3(
x,
y = NULL,
data = NULL,
mu = 0,
parametric = TRUE,
ci = 0.95,
alternative = "two.sided",
iterations = 200,
verbose = TRUE,
...
)
p_overlap(
x,
y = NULL,
data = NULL,
mu = 0,
parametric = TRUE,
ci = 0.95,
alternative = "two.sided",
iterations = 200,
verbose = TRUE,
...
)
vd_a(
x,
y = NULL,
data = NULL,
mu = 0,
ci = 0.95,
alternative = "two.sided",
verbose = TRUE,
...
)
wmw_odds(
x,
y = NULL,
data = NULL,
mu = 0,
paired = FALSE,
ci = 0.95,
alternative = "two.sided",
verbose = TRUE,
...
)
### Arguments
x, y A numeric vector, or a character name of one in data. Any missing values (NAs) are dropped from the resulting vector. x can also be a formula (see stats::t.test()), in which case y is ignored. data An optional data frame containing the variables. mu a number indicating the true value of the mean (or difference in means if you are performing a two sample test). paired If TRUE, the values of x and y are considered as paired. This produces an effect size that is equivalent to the one-sample effect size on x - y. parametric Use parametric estimation (see cohens_d()) or non-parametric estimation (see rank_biserial()). See details. ci Confidence Interval (CI) level alternative a character string specifying the alternative hypothesis; Controls the type of CI returned: "two.sided" (default, two-sided CI), "greater" or "less" (one-sided CI). Partial matching is allowed (e.g., "g", "l", "two"...). See One-Sided CIs in effectsize_CIs. verbose Toggle warnings and messages on or off. ... Arguments passed to or from other methods. When x is a formula, these can be subset and na.action. iterations The number of bootstrap replicates for computing confidence intervals. Only applies when ci is not NULL and parametric = FALSE.
### Details
These measures of effect size present group differences in probabilistic terms:
• Probability of superiority is the probability that, when sampling an observation from each of the groups at random, that the observation from the second group will be larger than the sample from the first group. For the one-sample (or paired) case, it is the probability that the sample (or difference) is larger than mu. (Vargha and Delaney's A is an alias for the non-parametric probability of superiority.)
• Cohen's U_1 is the proportion of the total of both distributions that does not overlap.
• Cohen's U_2 is the proportion of one of the groups that exceeds the same proportion in the other group.
• Cohen's U_3 is the proportion of the second group that is smaller than the median of the first group.
• Overlap (OVL) is the proportional overlap between the distributions. (When parametric = FALSE, bayestestR::overlap() is used.)
Wilcoxon-Mann-Whitney odds are the odds of non-parametric superiority (via probs_to_odds()), that is the odds that, when sampling an observation from each of the groups at random, that the observation from the second group will be larger than the sample from the first group.
Where U_1, U_2, and Overlap are agnostic to the direction of the difference between the groups, U_3 and probability of superiority are not.
The parametric version of these effects assumes normality of both populations and homoscedasticity. If those are not met, the non parametric versions should be used.
### Value
A data frame containing the common language effect sizes (and optionally their CIs).
### Confidence (Compatibility) Intervals (CIs)
For parametric CLES, the CIs are transformed CIs for Cohen's d (see d_to_u3()). For non-parametric (parametric = FALSE) CLES, the CI of Pr(superiority) is a transformed CI of the rank-biserial correlation (rb_to_p_superiority()), while for all others, confidence intervals are estimated using the bootstrap method (using the {boot} package).
### CIs and Significance Tests
"Confidence intervals on measures of effect size convey all the information in a hypothesis test, and more." (Steiger, 2004). Confidence (compatibility) intervals and p values are complementary summaries of parameter uncertainty given the observed data. A dichotomous hypothesis test could be performed with either a CI or a p value. The 100 (1 - \alpha)% confidence interval contains all of the parameter values for which p > \alpha for the current data and model. For example, a 95% confidence interval contains all of the values for which p > .05.
Note that a confidence interval including 0 does not indicate that the null (no effect) is true. Rather, it suggests that the observed data together with the model and its assumptions combined do not provided clear evidence against a parameter value of 0 (same as with any other value in the interval), with the level of this evidence defined by the chosen \alpha level (Rafi & Greenland, 2020; Schweder & Hjort, 2016; Xie & Singh, 2013). To infer no effect, additional judgments about what parameter values are "close enough" to 0 to be negligible are needed ("equivalence testing"; Bauer & Kiesser, 1996).
### Bootstrapped CIs
Some effect sizes are directionless–they do have a minimum value that would be interpreted as "no effect", but they cannot cross it. For example, a null value of Kendall's W is 0, indicating no difference between groups, but it can never have a negative value. Same goes for U2 and Overlap: the null value of U_2 is 0.5, but it can never be smaller than 0.5; am Overlap of 1 means "full overlap" (no difference), but it cannot be larger than 1.
When bootstrapping CIs for such effect sizes, the bounds of the CIs will never cross (and often will never cover) the null. Therefore, these CIs should not be used for statistical inference.
### Plotting with see
The see package contains relevant plotting functions. See the plotting vignette in the see package.
### Note
If mu is not 0, the effect size represents the difference between the first shifted sample (by mu) and the second sample.
### References
• Cohen, J. (1977). Statistical power analysis for the behavioral sciences. New York: Routledge.
• Reiser, B., & Faraggi, D. (1999). Confidence intervals for the overlapping coefficient: the normal equal variance case. Journal of the Royal Statistical Society, 48(3), 413-418.
• Ruscio, J. (2008). A probability-based measure of effect size: robustness to base rates and other factors. Psychological methods, 13(1), 19–30.
• Vargha, A., & Delaney, H. D. (2000). A critique and improvement of the CL common language effect size statistics of McGraw and Wong. Journal of Educational and Behavioral Statistics, 25(2), 101-132.
• O’Brien, R. G., & Castelloe, J. (2006, March). Exploiting the link between the Wilcoxon-Mann-Whitney test and a simple odds statistic. In Proceedings of the Thirty-first Annual SAS Users Group International Conference (pp. 209-31). Cary, NC: SAS Institute.
• Agresti, A. (1980). Generalized odds ratios for ordinal data. Biometrics, 59-67.
sd_pooled()
Other standardized differences: cohens_d(), mahalanobis_d(), means_ratio(), rank_biserial(), repeated_measures_d()
Other rank-based effect sizes: rank_biserial(), rank_epsilon_squared()
### Examples
cohens_u2(mpg ~ am, data = mtcars)
p_superiority(mpg ~ am, data = mtcars, parametric = FALSE)
wmw_odds(mpg ~ am, data = mtcars)
x <- c(1.83, 0.5, 1.62, 2.48, 1.68, 1.88, 1.55, 3.06, 1.3)
y <- c(0.878, 0.647, 0.598, 2.05, 1.06, 1.29, 1.06, 3.14, 1.29)
p_overlap(x, y)
p_overlap(y, x) # direction of effect does not matter
cohens_u3(x, y)
cohens_u3(y, x) # direction of effect does matter
effectsize documentation built on May 29, 2024, 11:37 a.m. | 2,214 | 8,542 | {"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-2024-26 | latest | en | 0.823073 |
https://gmatclub.com/forum/a-dealer-originally-bought-15-identical-cars-if-each-car-was-sold-for-256873.html | 1,560,930,351,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998923.96/warc/CC-MAIN-20190619063711-20190619085711-00161.warc.gz | 448,866,387 | 143,791 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 19 Jun 2019, 00:45
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# A dealer originally bought 15 identical cars. If each car was sold for
Author Message
TAGS:
### Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 55684
A dealer originally bought 15 identical cars. If each car was sold for [#permalink]
### Show Tags
05 Jan 2018, 01:18
00:00
Difficulty:
35% (medium)
Question Stats:
72% (02:07) correct 28% (02:31) wrong based on 34 sessions
### HideShow timer Statistics
A dealer originally bought 15 identical cars. If each car was sold for q dollars, which was 60 percent above the original cost per car, then in terms of q, what was the total cost of the cars, in dollars, to the dealer?
A. q/250
B. 5q/8
C. 75q/8
D. 9q/10
E. 24q
_________________
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3373
Location: India
GPA: 3.12
A dealer originally bought 15 identical cars. If each car was sold for [#permalink]
### Show Tags
05 Jan 2018, 01:34
Bunuel wrote:
A dealer originally bought 15 identical cars. If each car was sold for q dollars, which was 60 percent above the original cost per car, then in terms of q, what was the total cost of the cars, in dollars, to the dealer?
A. q/250
B. 5q/8
C. 75q/8
D. 9q/10
E. 24q
Since the car was sold for q dollars which was 1.6 times the cost, the cost of the car would be $$\frac{q}{1.6}$$
Since he bought 15 identical cars, the cost of the cars would be $$15*\frac{q}{1.6} = \frac{150q}{16} = \frac{75q}{8}$$(Option C)
_________________
You've got what it takes, but it will take everything you've got
Senior SC Moderator
Joined: 22 May 2016
Posts: 2899
Re: A dealer originally bought 15 identical cars. If each car was sold for [#permalink]
### Show Tags
05 Jan 2018, 09:48
Bunuel wrote:
A dealer originally bought 15 identical cars. If each car was sold for q dollars, which was 60 percent above the original cost per car, then in terms of q, what was the total cost of the cars, in dollars, to the dealer?
A. q/250
B. 5q/8
C. 75q/8
D. 9q/10
E. 24q
Another approach: assign values.
When you see fractions in answer choices, choose small values.
Almost always, if chosen values are small, the fractional answers can be assessed quickly.
Sell price is 60 percent above cost. SP = 1.6 * cost price
Use two integers. 5 * 16 = 80, so 5 * 1.6 = 8
You could also use cost = $10 and SP =$16
Let cost = $5 per car Then Sell Price =$5 * 1.6 = $8 So q =$8
Total cost = ($5 * 15) =$75
With q = 8, find the answer that yields $75 Eliminate A, B, and D immediately. Their numerators and denominators make their results much too small: $$\frac{8}{250}$$, $$\frac{40}{8}$$, and $$\frac{72}{10}$$, respectively Eliminate E, too. 24 * 8 will not yield an answer whose units digit is 75 (And 24 * 8 > 160, more than twice what we need) Check C) 75q/8 $$(75 * 8) = (150 * 4) = 600$$ (double one factor and halve the other) $$\frac{600}{8}= \frac{300}{4} = \frac{150}{2}$$=$75. Correct.
_________________
SC Butler has resumed!
Get two SC questions to practice, whose links you can find by date, here.
Tell me, what is it you plan to do with your one wild and precious life? -- Mary Oliver
Re: A dealer originally bought 15 identical cars. If each car was sold for [#permalink] 05 Jan 2018, 09:48
Display posts from previous: Sort by | 1,141 | 3,815 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2019-26 | latest | en | 0.938919 |
https://the-algorithms.com/algorithm/n-queen?lang=dart | 1,708,623,513,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.13/warc/CC-MAIN-20240222161802-20240222191802-00098.warc.gz | 570,036,097 | 19,520 | #### N-Queen
S
import 'package:test/test.dart';
/*N-Queens Problem
N - Queens problem is to place n - queens in such a manner
on an n x n chessboard that no queens attack each other by being in the same row, column or diagonal
We have to find maximum no of solutions that are possible.
.*/
// Chess board, A 2d List of integers
List<List<int>> board = [];
// checking is it safe place for a queen ?
/*size : no of rows or columns of the given chess board
for e.g if there is a 3*3 chessboard then size = 3*/
bool isSafe(int size, int row, int col) {
//checking column safe
for (int k = 0; k < row; k++) {
if (board[k][col] == 1) {
return false;
}
}
int i = row;
int j = col;
//checking left diagonal safe
while (i >= 0 && j >= 0) {
if (board[i][j] == 1) {
return false;
}
j--;
i--;
}
//checking right diagonal safe
i = row;
j = col;
while (i >= 0 && j < size) {
if (board[i][j] == 1) {
return false;
}
j++;
i--;
}
return true;
}
int solveNQueen(int row, int size) {
//base case
if (size == row) {
return 1;
}
//recursive case
int ways = 0;
for (int col = 0; col < size; col++) {
//whether the current i,j is safe or not
if (isSafe(size, row, col)) {
board[row][col] = 1;
ways += solveNQueen(row + 1, size);
//backtrack
board[row][col] = 0;
}
}
return ways;
}
//This Function is used to refilled board with default value for testing purpose only.
// You can remove it, if you are using single test case.
void resetBoard(int boardSize) {
board = List.generate(boardSize, (i) => List.generate(boardSize, (j) => 0));
}
void main() {
/*
Test Cases
Size Expected Values
1 : 1
2 : 0
3 : 0
4 : 2
5 : 10
6 : 4
7 : 40
8 : 92
9 : 352
10 : 724
*/
test("N Queen #testcase 1", () {
resetBoard(1);
expect(solveNQueen(0, 1), equals(1));
});
test("N Queen #testcase 2", () {
resetBoard(2);
expect(solveNQueen(0, 2), equals(0));
});
test("N Queen #testcase 3", () {
resetBoard(3);
expect(solveNQueen(0, 3), equals(0));
});
test("N Queen #testcase 4", () {
resetBoard(4);
expect(solveNQueen(0, 4), equals(2));
});
test("N Queen #testcase 5", () {
resetBoard(5);
expect(solveNQueen(0, 5), equals(10));
});
test("N Queen #testcase 6", () {
resetBoard(6);
expect(solveNQueen(0, 6), equals(4));
});
test("N Queen #testcase 7", () {
resetBoard(7);
expect(solveNQueen(0, 7), equals(40));
});
test("N Queen #testcase 8", () {
resetBoard(8);
expect(solveNQueen(0, 8), equals(92));
});
test("N Queen #testcase 9", () {
resetBoard(9);
expect(solveNQueen(0, 9), equals(352));
});
test("N Queen #testcase 10", () {
resetBoard(10);
expect(solveNQueen(0, 10), equals(724));
});
} | 868 | 2,576 | {"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-2024-10 | latest | en | 0.478073 |
https://gmatclub.com/forum/each-bank-in-the-town-of-la-rinconada-has-only-a-single-set-of-locking-136654-20.html | 1,534,407,218,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210559.6/warc/CC-MAIN-20180816074040-20180816094040-00466.warc.gz | 695,198,520 | 56,209 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 16 Aug 2018, 01:13
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Each bank in the town of La Rinconada has only a single set of locking
Author Message
TAGS:
### Hide Tags
Intern
Joined: 11 Jul 2013
Posts: 35
Re: Each bank in the town of La Rinconada has only a single set of locking [#permalink]
### Show Tags
13 Jun 2014, 08:19
KyleWiddison wrote:
voodoochild wrote:
Each bank in the town of La Rinconada has only a single set of locking doors at its entrance. In the town of Inverness, on the other hand, the entrances to nearly all banks are equipped with two sets of locking doors, operated by a mechanism that allows only one set of doors to be open at a time. It is clear, then, that banks in Inverness experience more robbery attempts than do those in La Rinconada, and have thus adopted the extra doors as a security measure.
Which of the following, if true, most weakens the argument above?
a) Last year the number of bank robberies in La Rinconada was almost one-half greater than the corresponding figure for the previous year.
b) Inverness is known for its harsh winters, while the climate of La Rinconada is quite temperate year-round.
c) The mechanism of the double doors used by banks in Inverness allows bank security personnel to lock the doors remotely.
d) Bank robbery attempts are typically unsuccessful, and, even when the robbers do manage to escape with stolen money, the sum is usually quite small.
e) Inverness has almost twice as many police officers per capita as does La Rinconada.
Why is A) not a weakener? If the # of bank robberies in La Rin is > that in Inverness, the security is obviously out of question.
I'm responding to a PM on this one. I will start by saying the correct answer will be more logical to people who live in climates where double doors are used to maintain inside temperatures, but you can get to the answer by working through your CR process.
Assumptions fill in the logical gaps between premises and conclusions. If you want to weaken an argument you need to attack an assumption. In this question the premises state that Inverness has double doors and La Rinconada has single doors. The conclusion states that the double doors were for robbery prevention against Inverness' higher robbery rate. How do we get from double doors to robbery? We are assuming that the double doors are a method for preventing robbery (this is the logical gap between the premises and the conclusion).
Answer choices A, D, and E can quickly be eliminated for irrelevance to the issue of doors and robbery prevention. Since the correct answer is B, let's first analyze C. Choice C states that the double doors can be locked remotely. That seems to strengthen the assumption that the double doors are for robbery prevention. That choice feels very relevant to the argument but it's relevant in the wrong direction. We are trying to weaken the argument so we have to attack the assumption that doors are for robbery prevention, not strengthen it.
That only leaves us with choice B. Again, I will admit this requires a bit of a mental stretch for some, but can you think of the impact that climate has on door selection or use? Sure, when it's warm I sometimes leave my door open, but when it's cold or stormy I will absolutely close my door. This choice does provide information to suggest that the double doors may not have been installed for robbery prevention and instead as a protection against climate.
Don't be too quick to call some answer choices like this one out of scope. These questions can introduce new information not previously stated in the argument. The question is whether that new information attacks an assumption and therefore weakens the argument.
KW
may be managers are too lazy and may be that is the reason for them to have 2 doors that can closed remotely
Senior Manager
Joined: 18 Aug 2014
Posts: 325
Re: Each bank in the town of La Rinconada has only a single set of locking [#permalink]
### Show Tags
30 Jan 2016, 14:32
KyleWiddison wrote:
voodoochild wrote:
Each bank in the town of La Rinconada has only a single set of locking doors at its entrance. In the town of Inverness, on the other hand, the entrances to nearly all banks are equipped with two sets of locking doors, operated by a mechanism that allows only one set of doors to be open at a time. It is clear, then, that banks in Inverness experience more robbery attempts than do those in La Rinconada, and have thus adopted the extra doors as a security measure.
Which of the following, if true, most weakens the argument above?
a) Last year the number of bank robberies in La Rinconada was almost one-half greater than the corresponding figure for the previous year.
b) Inverness is known for its harsh winters, while the climate of La Rinconada is quite temperate year-round.
c) The mechanism of the double doors used by banks in Inverness allows bank security personnel to lock the doors remotely.
d) Bank robbery attempts are typically unsuccessful, and, even when the robbers do manage to escape with stolen money, the sum is usually quite small.
e) Inverness has almost twice as many police officers per capita as does La Rinconada.
Why is A) not a weakener? If the # of bank robberies in La Rin is > that in Inverness, the security is obviously out of question.
I'm responding to a PM on this one. I will start by saying the correct answer will be more logical to people who live in climates where double doors are used to maintain inside temperatures, but you can get to the answer by working through your CR process.
Assumptions fill in the logical gaps between premises and conclusions. If you want to weaken an argument you need to attack an assumption. In this question the premises state that Inverness has double doors and La Rinconada has single doors. The conclusion states that the double doors were for robbery prevention against Inverness' higher robbery rate. How do we get from double doors to robbery? We are assuming that the double doors are a method for preventing robbery (this is the logical gap between the premises and the conclusion).
Answer choices A, D, and E can quickly be eliminated for irrelevance to the issue of doors and robbery prevention. Since the correct answer is B, let's first analyze C. Choice C states that the double doors can be locked remotely. That seems to strengthen the assumption that the double doors are for robbery prevention. That choice feels very relevant to the argument but it's relevant in the wrong direction. We are trying to weaken the argument so we have to attack the assumption that doors are for robbery prevention, not strengthen it.
That only leaves us with choice B. Again, I will admit this requires a bit of a mental stretch for some, but can you think of the impact that climate has on door selection or use? Sure, when it's warm I sometimes leave my door open, but when it's cold or stormy I will absolutely close my door. This choice does provide information to suggest that the double doors may not have been installed for robbery prevention and instead as a protection against climate.
Don't be too quick to call some answer choices like this one out of scope. These questions can introduce new information not previously stated in the argument. The question is whether that new information attacks an assumption and therefore weakens the argument.
KW
I agree with the post 2 above; not to attack the quality of the question but between B and C, B seems incredibly irrelevant unless you have specific knowledge of architecture and climate control which seems like something the GMAT would never expect you to have while C at least has a logical assumption that either the remote factor means guards don't need to be in the banks (weakening security rather than strengthening) or they're just lazy.
Either of those assumption (with the former being the better of the two) would to some degree be a fair conclusion based off of given information when compared to B which is such a specific piece of knowledge (that double doors are needed for colder climates) I can't imagine the GMAT expecting you to know this without any kind of way to reason it.
_________________
Manager
Joined: 22 Feb 2016
Posts: 97
Location: India
Concentration: Economics, Healthcare
GMAT 1: 690 Q42 V47
GMAT 2: 710 Q47 V39
GPA: 3.57
Re: Each bank in the town of La Rinconada has only a single set of locking [#permalink]
### Show Tags
05 Dec 2016, 09:28
nirakrish wrote:
Each bank in the town of La Rinconada has only a single set of locking doors at its entrance. In the town of Inverness, on the other hand, the entrances to nearly all banks are equipped with two sets of locking doors, operated by a mechanism that allows only one set of doors to be open at a time. It is clear, then, that banks in Inverness experience more robbery attempts than do those in La Rinconada, and have thus adopted the extra doors as a security measure.
Which of the following, if true, most weakens the argument above?
a) Last year the number of bank robberies in La Rinconada was almost one-half greater than the corresponding figure for the previous year. Incorrect - does not talk anything about the town of Inverness.
b) Inverness is known for its harsh winters, while the climate of La Rinconada is quite temperate year-round. Correct - Inverness banks have two doors to manage the temperature inside the bank and they are not specifically meant to control robbery. It has a direct effect on the conclusion.
c) The mechanism of the double doors used by banks in Inverness allows bank security personnel to lock the doors remotely. Incorrect - looks like the answer but not. Although it looks like an alternative reason, locking remotely or manually does not matter. The point does not weaken the conclusion.
d) Bank robbery attempts are typically unsuccessful, and, even when the robbers do manage to escape with stolen money, the sum is usually quite small. Incorrect - does not weaken. sum of money does not matter.
e) Inverness has almost twice as many police officers per capita as does La Rinconada. Incorrect - number of police officers does not affect the conclusion.
This is the reasoning cited by Manhatten but I really don't understand how is B not an out of scope question.
I agree it is the only feasible option on applying POE but still B does not look healthy.
Manager
Joined: 01 Sep 2016
Posts: 94
Re: Each bank in the town of La Rinconada has only a single set of locking [#permalink]
### Show Tags
21 Jan 2017, 06:45
Thank you again for this amazing explanation.I fell for C
daagh wrote:
The point here is that the conclusion says that the double- door system is a security measure. We need to weaken this thinking. So, any choice that augments the security aspect is irrelevant. We need to look a totally different reason other than security against robbery or any such things. Therefore, B simply weighs in. We can ignore C, because it strengthens the argument that extra security measures are required in the second town
B is the correct choice
Intern
Joined: 08 Jul 2017
Posts: 3
Re: Each bank in the town of La Rinconada has only a single set of locking [#permalink]
### Show Tags
11 Jul 2017, 21:42
ravstime wrote:
B clearly is the winner.
It provides different cause than that mentioned in the Premise. The author while arriving at a conclusion believe that only 1 cause can lead to the effect. If different cause is having the same intended effect, author's conclusion will gets weakened.
Hu ravstime,
" If a different cause is having the same intended effect, author's conclusion will get weakened." is the thumb rule for getting answer choice for Weaken questions. However, you can't fully justify the reason B is correct. It is very far fetched to bring the temperature aspect in this passage. What if I claim that Option C is valid because controlling the doors is the new technology the country has brought upon trial basis and has no relevance with the robbery. As mentioned earlier somewhere in the page, the bank can have pre-installed temperature controlled equipment which will nullify the temperature difference.
Manager
Joined: 05 Nov 2014
Posts: 114
Location: India
Concentration: Strategy, Operations
GMAT 1: 580 Q49 V21
GPA: 3.75
Re: Each bank in the town of La Rinconada has only a single set of locking [#permalink]
### Show Tags
11 Jul 2017, 22:32
Each bank in the town of La Rinconada has only a single set of locking doors at its entrance. In the town of Inverness, on the other hand, the entrances to nearly all banks are equipped with two sets of locking doors, operated by a mechanism that allows only one set of doors to be open at a time. It is clear, then, that banks in Inverness experience more robbery attempts than do those in La Rinconada, and have thus adopted the extra doors as a security measure.
Which of the following, if true, most weakens the argument above?
a) Last year the number of bank robberies in La Rinconada was almost one-half greater than the corresponding figure for the previous year.
b) Inverness is known for its harsh winters, while the climate of La Rinconada is quite temperate year-round.
c) The mechanism of the double doors used by banks in Inverness allows bank security personnel to lock the doors remotely.
d) Bank robbery attempts are typically unsuccessful, and, even when the robbers do manage to escape with stolen money, the sum is usually quite small.
e) Inverness has almost twice as many police officers per capita as does La Rinconada.
Error Analysis:
In a weaken question, we may either weaken the assumption directly or provide an alternate reason.
In this argument, the comparison is made between the type of doors provided at La Rinconada and Inverness. The reason provided for the double door is the security purpose.
If we are able to provide an alternate reason or weaken the assumption made by the author, we are sorted.!!!
Option A- Irrelevant as it compares robberies at La Rinconada during last year and previous year.
Option B. Correct, as it provides an alternative reason.
Option C- Strengthens the argument. Opposite.
Option D- Irrelevant again.
Option E-Irrelevant. Doesn't address the issue.
Therefore the answer is Option B.
Manager
Joined: 05 Nov 2014
Posts: 114
Location: India
Concentration: Strategy, Operations
GMAT 1: 580 Q49 V21
GPA: 3.75
Re: Each bank in the town of La Rinconada has only a single set of locking [#permalink]
### Show Tags
11 Jul 2017, 22:34
Rosicky wrote:
ravstime wrote:
B clearly is the winner.
It provides different cause than that mentioned in the Premise. The author while arriving at a conclusion believe that only 1 cause can lead to the effect. If different cause is having the same intended effect, author's conclusion will gets weakened.
Hu ravstime,
" If a different cause is having the same intended effect, author's conclusion will get weakened." is the thumb rule for getting answer choice for Weaken questions. However, you can't fully justify the reason B is correct. It is very far fetched to bring the temperature aspect in this passage. What if I claim that Option C is valid because controlling the doors is the new technology the country has brought upon trial basis and has no relevance with the robbery. As mentioned earlier somewhere in the page, the bank can have pre-installed temperature controlled equipment which will nullify the temperature difference.
Option C is strengthening and not weakening.
Manager
Status: Manager to Damager!
Affiliations: MBA
Joined: 22 May 2014
Posts: 71
Location: United States
Re: Each bank in the town of La Rinconada has only a single set of locking [#permalink]
### Show Tags
09 Aug 2017, 07:00
I think it is a bad question..
Only the one who has experienced/lived in cold countries can really grasp the meaning of the option B.
Someone who is living in warm country can never understand the logic behind two doors! He/She will immediately eliminate the option B!
And such questions donot appear in official GMAT !
Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1322
Location: Malaysia
Re: Each bank in the town of La Rinconada has only a single set of locking [#permalink]
### Show Tags
16 Sep 2017, 22:56
1
voodoochild wrote:
Each bank in the town of La Rinconada has only a single set of locking doors at its entrance. In the town of Inverness, on the other hand, the entrances to nearly all banks are equipped with two sets of locking doors, operated by a mechanism that allows only one set of doors to be open at a time. It is clear, then, that banks in Inverness experience more robbery attempts than do those in La Rinconada, and have thus adopted the extra doors as a security measure.
Which of the following, if true, most weakens the argument above?
(A) Last year the number of bank robberies in La Rinconada was almost one-half greater than the corresponding figure for the previous year.
(B) Inverness is known for its harsh winters, while the climate of La Rinconada is quite temperate year-round.
(C) The mechanism of the double doors used by banks in Inverness allows bank security personnel to lock the doors remotely.
(D) Bank robbery attempts are typically unsuccessful, and, even when the robbers do manage to escape with stolen money, the sum is usually quite small.
(E) Inverness has almost twice as many police officers per capita as does La Rinconada.
OFFICIAL EXPLANATION
(1) Identify the Question Type
The question asks us to weaken the argument presented, so this is a Weaken the Conclusion question.
(2) Deconstruct the Argument
The argument attempts to explain why almost all banks in Inverness have separately locking double doors, in contrast to the total lack of such doors in the town of La Rinconada. The argument concludes that the double doors must be an extra security measure against crime, and that, by extension, the rate of the relevant crime (here, bank robbery) must be higher in Inverness than in La Rinconada.
(3) State the Goal
For Weaken questions, we have to find an answer that makes the conclusion at least a little less likely to be true or valid.
The explanation presented in the argument does seem reasonable - certainly, one reason for using double doors could be security concerns. The argument can be attacked, though, on the grounds that it claims this must be the reason. If there are other possible reasons for using the double doors, then the claim that the doors are used due to security concerns and a higher robbery rate becomes less likely to be true.
(4) Work from Wrong to Right
(A) The year-over-year growth rate of bank robberies in La Rinconada alone is not helpful in making a comparison between La Rinconada and another town.
(B) CORRECT. The door mechanism in Inverness creates an effective barrier against harsh winter winds and cold air: At no point will both sets of doors be open at the same time, so the introduction of cold air into the bank will be minimized relative to a typical single door opening. This choice thus provides an alternative rationale for the presence of the double doors, weakening the force of the original conclusion.
(C) This choice actually strengthens the argument: If the double doors are equipped with such a feature, it becomes more reasonable to speculate that they were installed as a safeguard against crime. Also, this choice can be eliminated immediately on the grounds that it doesn't distinguish at all between the two towns, and thus provides no reason for the door system to be present in Inverness but not also in La Rinconada.
(D) This choice does not address the argument's conclusion: the reason for the presence of double doors in Inverness banks and the lack of double doors in La Rinconada banks.
(E) The proportion of citizens who are police officers does not provide meaningful information about bank-robbery rates in the two towns, or about the reason for the presence of double doors in Inverness banks and the lack of double doors in La Rinconada banks.
_________________
"Be challenged at EVERY MOMENT."
“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”
"Each stage of the journey is crucial to attaining new heights of knowledge."
Manager
Joined: 30 Aug 2017
Posts: 92
Location: Korea, Republic of
GMAT 1: 700 Q51 V31
GPA: 3.68
Re: Each bank in the town of La Rinconada has only a single set of locking [#permalink]
### Show Tags
02 Dec 2017, 23:16
How can I find relationship between two door and weather?
Why only me think it as a problem.
I feel frustrated
Manager
Joined: 15 Aug 2016
Posts: 79
Re: Each bank in the town of La Rinconada has only a single set of locking [#permalink]
### Show Tags
25 Dec 2017, 17:26
Really Strange OA.
How would we know that double doors are used as thermostats? Or are temperature regulators or help keep the inside room warm?
C maybe the closest.
At least in C we can assume that the double doors system is not used for security but for convenience.Controlling it remotely by no way means that they would be much secure. What if the lock system is compromised for the sake of convenience?
Re: Each bank in the town of La Rinconada has only a single set of locking &nbs [#permalink] 25 Dec 2017, 17:26
Go to page Previous 1 2 [ 31 posts ]
Display posts from previous: Sort by
# Events & Promotions
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 4,942 | 22,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-34 | longest | en | 0.94582 |
http://www.docstoc.com/docs/35083984/Empty-Space | 1,394,805,716,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678693350/warc/CC-MAIN-20140313024453-00076-ip-10-183-142-35.ec2.internal.warc.gz | 278,789,027 | 16,895 | # Empty Space
Document Sample
``` Senior Science
Module 9.9 ::: Space
Section 1
Empty Space
While the atmosphere has limits there is no such thing as ‘empty’ space
9.9.1.a Discuss the concept of the atmosphere in relation to the distribution or
concentration of particles of gas
9.9.1.b Identify that the Earth’s atmosphere is largely maintained in place by the
earth’s gravitational pull
9.9.1.c Discuss why there is no such thing as ‘empty space’
9.9.1.i Gather, process and present information from secondary sources to model
the relative distance of particles in a solid, liquid, gas and in space
9.9.1.a Discuss the concept of the atmosphere in relation to the distribution or
concentration of particles of gas
CONCENTRATION OF GASES IN THE ATMOSPHERE
What to do
Read the information (dot points) about the atmosphere.
• The atmosphere is the mixture of gases that surrounds the Earth.
• Gravity holds the gases of the atmosphere close to the earth.
• The higher up the thinner the atmosphere (ie less gas particles in the same volume)
• Below 80 kilometres the atmosphere is roughly uniform (ie a similar ratio of gases
everywhere) and this layer is called the Homosphere.
• Three of the more important gases in the Homosphere are Nitrogen (78%), Oxygen
(21%) and Carbon dioxide (0.04%).
• In the homosphere, the gases are well mixed by convection and turbulence.
• Most water vapour occurs below 12 kilometres.
• Amount of water vapour in the atmosphere varies (daily) greatly below 12 kilometres.
• 80% of the air in the atmosphere occurs in the homosphere.
• The ozone layer extends from about 20 – 50 kilometres above sea level.
• A 10km thick section of the ozone layer would measure a few millimetres at sea level.
• NASA regards above 80 kilometres as being in space.
• Orbiting satellites can stay in orbit around the Earth for very long periods of time above
80 kilometres. This is because the atmosphere is so thin it creates very little drag on
the satellite.
• Above the homosphere (above 80 kilometres) is the heterosphere.
• The composition of the heterosphere is very different to the homosphere.
• At heights above 80 km the atmospheric composition differs greatly from sea level.
• The gases in the heterosphere do not mix much.
• Turbulent mixing is absent in the heterosphere and diffusion determines composition.
• The heterosphere consists of layers with lighter gases higher up (gaseous stratification
occurs in relation to the molecular weights of individual gases).
• There are larger proportions of gases of low (molecular) weight (eg helium).
• The atmosphere becomes thinner with increasing height above the Earth.
• The Heterosphere (atmosphere) gradually fades into space, where it meets the solar
wind – that is a continuous stream of charged particles from the sun.
• These charged particles given off by the sun are captured by the Earth’s magnetic field.
• As well the atmosphere is affected by cosmic, solar ultraviolet and X-ray radiation
• Gases of the lower atmosphere screen out these radiations.
• The solar wind and the radiation cause particles in the heterosphere to be ionised
(separation of electrons from atoms, leaving a positively charged ion).
• Collisions between these charged particles and the atmosphere produce light.
• This results in natural light displays called the Auroras.
• The Auroras are produced by the penetration of ionising radiation through the
atmosphere about 300km to about 80 km, particularly in the Polar Regions. The
display in the Southern Hemisphere is called Aurora Australis or Southern Lights.
Notes Questions
1. What is the atmosphere?
2. Where is the atmosphere thickest?
3. Is there much water vapour above 80 km?
4. In which part of the atmosphere do gases mix?
5. What causes the Auroras?
6. How high is the atmosphere?
7. Where does the atmosphere end?
8. Where does space start?
9. Draw a diagram of the atmosphere’s structure.
• Label the diagram with the following words – Homosphere, Heterosphere, Ozone
layer; most water vapour, and others that are appropriate
• Indicate heights above sea level on the diagram – 0, 80, 300 etc
• Indicate thicker parts of the atmosphere by using more dots.
• Describe the distribution of particles of gas.
HSC Style Question
Discuss the concept of the atmosphere in relation to the distribution of particles of gas.
This question requires an extended answer. In order to answer such a question, you must
provide some structure.
STEP 1 Identify and highlight the important words in the question
STEP 2 Recall definitions of important words (if necessary)
• Discuss – To identify issues and provide points for and/or against.
• Concept - idea
• Atmosphere – Mixture of gases that surround the earth.
• Distribution or concentration of gases – “How spread out?”
STEP 3 Develop your own rubric that reflects the depth required
• Define atmosphere
• Describe / Draw structure of the atmosphere – indicate how the
concentration of gases varies
• (Suggest) where the atmosphere ends
STEP 4 Write your discussion
9.9.1.b Identify that the Earth’s atmosphere is largely maintained in place by the
earth’s gravitational pull
GRAVITY and the ATMOSPHERE
Gravitation is the force of attraction that acts between objects because of their mass – that
is, the amount of matter they are made of. Because of gravity, objects that are on or near
the Earth are pulled towards it.
The Earth’s atmosphere is largely maintained in place by the earth’s gravitational pull.
9.9.1.c Discuss why there is no such thing as ‘empty space’
EMPTY SPACE???
What do we know about space?
Below are a series of questions and information about space.
• Suggest answers to the questions and
• Consider the information provided.
1. What is space?
2. Where is space?
3. Once astronauts leave the atmosphere are they in empty space?
4. Do spacecraft ever have a collision with objects in space?
5. There are millions of pieces of ‘space junk’ (pieces of satellites) orbiting the earth
6. Most space junk orbits from 500-2000 kilometres above the earth’s surface.
7. The earth orbits the sun. Both are in space. What other large bodies are in space?
8. Are comets, which orbit the sun, in space?
9. What are meteors, meteorites and shooting stars?
10. You could see five shooting stars in an hour, on a clear night?
11. What is the solar wind?
12. Is the solar wind dangerous?
13. Is the moon in space?
14. The sun is a huge glowing ball of gases at the centre of the solar system. There is a
continuous flow of gases from the sun. These gases travel at about 500 kilometres per
second or 1½ - 3 million kilometres an hour. This flow of gases is called the solar
wind. The density of these particles is about 5 per cubic centimetre. These particles
are prevented from reaching the earth’s surface by the earth’s magnetic field.
15. How many stars are there in space?
16. How many stars are there in a galaxy?
17. How many galaxies are there?
18. In the 19th century people studying the night sky discovered that “dark patches” blocked
the light coming from certain regions of space. By the 1930’s astronomers were able to
identify these dark patches as huge clouds of interstellar dust and gas (called nebula).
They form because gravity causes the dust grains to clump together.
19. The source for these clouds is matter from dying “stars”. Atoms emitted by these stars
can cling onto the surface of the dust grains. The atoms can combine and recombine
to form dozens of simple and complex molecules. Today nearly 100 organic
substances have been identified (by radio astronomy) in the dust clouds.
20. Examples of molecules identified in interstellar gas clouds.
Inorganic Hydrogen (H2), Silicon monoxide (SiO),
Sulfur dioxide (SO2), Ammonia (NH3)
Organic Carbon monoxide (CO), Formaldehyde (H2CO),
Ethyl alcohol (CH3CH2OH)
21. There is one dust particle in every one million cubic metres (that is a cube 100 metres x
100 metres x 100 metres). This matter in dust clouds makes up about 1% of
interstellar matter.
22. The other 99% of matter in interstellar space does not occur in gas clouds. This matter
is mostly free floating atoms (mostly hydrogen, but also oxygen, carbon and nitrogen).
They are so widely spread that they rarely meet. On average, hydrogen atoms in the
Milky Way Galaxy have a temperature of –2000C. There are about 300,000 hydrogen
atoms in a cubic metre. This seems like an enormous number of atoms. However, the
distance between interstellar atoms is roughly 100 million times larger than the size of
the atoms. If a similar distance separated two people, they would be about 100 million
kilometres apart.
HSC Style Question
Discuss why there is no such thing as ‘empty space’
This question requires an extended answer. Provide some structure for your answer.
STEP 1 Identify and highlight the important words in the question
STEP 2 Recall definitions of important words (if necessary)
• Discuss – To identify issues and provide points for and/or against.
• Why – provide reasons
• Space – Area of universe outside the atmosphere
STEP 3 Develop your own rubric that reflects the depth required
• Define space
• Issue 1 – Compare matter concentration in space and atmosphere.
• Issue 2 – Consider matter in near space
• Issue 3 – Consider matter in interstellar space
STEP 4 Write your discussion
9.9.1.i Gather, process and present information from secondary sources to model
the relative distance of particles in a solid, liquid, gas and in space
RESEARCH
1. Complete the diagram below to represent the relative distance between particles in
space. Diagrams for a solid, liquid and a gas have been drawn.
The pictures (distances and particle size) are not drawn to scale. They do reflect relative
sizes and distances.
Space
2. How far apart are two particles in a solid?
3. How far apart are two particles in a liquid?
4. How far apart are two particles in a gas?
5. How far apart are two particles in space?
6. Describe how you could model the relative distances between particles of a solid, liquid
and a gas, and particles in space.
```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views: 38 posted: 4/20/2010 language: English pages: 6
Description: Empty Space | 2,360 | 10,407 | {"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-2014-10 | longest | en | 0.865479 |
https://www.delftstack.com/howto/matplotlib/specify-legend-position-in-graph-coordinates-matplotlib/ | 1,723,162,380,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00843.warc.gz | 587,085,960 | 7,607 | # How to Specify the Legend Position in Graph Coordinates in Matplotlib
Suraj Joshi Feb 02, 2024
We can specify a legend’s position in a graph by setting the `loc` parameter’s value.
## Examples: Specify Legend Position in Graph Coordinates Matplotlib
``````import matplotlib.pyplot as plt
x = [1, 2, 3]
y1 = [0.5 * i + 1 for i in x]
y2 = [2 * i + 1 for i in x]
fig, ax = plt.subplots(2, 1)
ax[0].plot(x, y1, "b-", label="0.5x+1")
ax[0].plot(x, y2, "g-", label="2x+1")
ax[0].legend(loc="best")
ax[1].plot(x, y1, "b-", label="0.5x+1")
ax[1].plot(x, y2, "g-", label="2x+1")
ax[1].legend(loc="lower left")
plt.tight_layout()
plt.show()
``````
Output:
Here, we have a figure with two subplots. The topmost subplot has `loc` set to `best`; this places the legend in the figure’s best position, where we do not have any content.
A similar thing happens in the second subplot at the bottom: the legend is placed at the `lower left` position of the bounding box, covering the entire axes. The bounding box is specified by the `bbox_to_anchor` parameter, whose default value is `(0,0,1,1)`.
The `loc` parameter can take any one of the following values:
`best`
`upper right`
`upper left`
`lower left`
`lower right`
`right`
`center left`
`center right`
`lower center`
`upper center`
`center`
These positions represent the legend’s position, with respect to the bounding box specified by the `bbox_to_anchor` parameter.
Similarly, we can place the legend at any position in the figure by changing the value of the `bbox_to_anchor` parameter. The `bbox_to_anchor` parameter takes a tuple, representing the coordinate, where the corner specified by the `loc` parameter will be placed.
``````import matplotlib.pyplot as plt
x = [1, 2, 3]
y1 = [0.5 * i + 1 for i in x]
y2 = [2 * i + 1 for i in x]
fig, ax = plt.subplots(2, 1)
ax[0].plot(x, y1, "b-", label="Plot-1")
ax[0].plot(x, y2, "g-", label="Plot-2")
ax[0].legend(loc="upper left", bbox_to_anchor=(0.9, 0.75))
ax[0].scatter((0.9), (0.75), s=70, c="red", transform=ax[0].transAxes)
ax[1].plot(x, y1, "b-", label="Plot-1")
ax[1].plot(x, y2, "g-", label="Plot-2")
ax[1].legend(loc="center right", bbox_to_anchor=(0.6, 0.4))
ax[1].scatter((0.6), (0.4), s=70, c="red", transform=ax[1].transAxes)
plt.tight_layout()
plt.show()
``````
Output:
In the subplot at the top, we have set the `bbox_to_anchor` to `(0.9,0.75)`, which is marked by the red dot on the subplot. The value of the `loc` parameter `"upper left"` represents the `upper left` corner of the legend, which is placed at the red dot.
The subplot at the bottom, the `bbox_to_anchor` is set to `(0.6,0.4)` marked by the red dot on the figure’s bottom axes. The value of `loc` is set to `"center right"`; hence, the bottom axes’ legend has its `center right` corner at the red dot.
Author: Suraj Joshi
Suraj Joshi is a backend software engineer at Matrice.ai. | 897 | 2,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-33 | latest | en | 0.595975 |
https://empoweringpumps.com/yaskawa-5-common-myths-and-misconceptions-about-variable-frequency-drives-addressing-misconceptions-to-assure-proper-vfd-usage/ | 1,695,477,731,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506481.17/warc/CC-MAIN-20230923130827-20230923160827-00637.warc.gz | 259,249,893 | 27,611 | Generic selectors
Exact matches only
Search in title
Search in content
Post Type Selectors
Motors & Drives
# 5 Common Myths and Misconceptions about Variable Frequency Drives: Addressing Misconceptions to Assure Proper VFD Usage
No matter how matter-of-fact variable frequency drives have become to you, somewhere someone is using one or considering using one for the first time. Think back to when you first thought about applying one of today’s PWM based VFDs to an AC motor. Chances are you probably had a few misconceptions about their abilities and designs. In this article, we will attempt to address a few common myths of VFDs and correct some misconceptions as to their proper usage.
#### Myth #1: The Output of a VFD is Sinusoidal
People tend to be more familiar with running their AC induction motors using motor starters. Starting the motor involved connecting the three phase leads of the motor to three phase power. Each phase was a sine wave with frequency of 60Hz and usually had a voltage amplitude of 230volt, 460volt,or 575volt in this part of the world. This applied voltage would create a sine wave current waveform with the same frequency if checked at the motor leads. Thus far, it is quite simple. What happens at the output of a VFD is a different story entirely. A VFD typically rectifies the 3-phase input to a fixed DC voltage which is filtered and stored using large DC bus capacitors. The DC bus voltage is then inverted to yield a variable voltage, variable frequency output. The inversion process is carried out using three pairs of IGBTs, one pair per output phase, as shown below. Since the DC voltage is inverted into AC, the output stage is also called “the inverter”. The duration for which each switch in a given pair is turned ON or held OFF can be controlled and this determines the rms value of the output voltage. The ratio of the output rms voltage to output frequency determines the flux in the AC motor. In general, there is a fixed relationship between the two. When the output frequency increases, the output voltage should also increase at the same rate in order to keep the ratio constant and thus the motor flux constant. Normally the relationship between voltage and frequency is kept linear so that a constant torque can be produced. The resulting voltage waveform applied to the motor winding is shown below. As a side note, sometimes the v/f ratio can be quite nonlinear. This is typically for fan and pump or centrifugal loads which do not require constant torque but instead favor energy savings.
What makes this work is that as the name implies, an induction motor is a big inductor of sorts. A characteristic of induction is its resistance to changes in current. Whether a current is increasing or decreasing, an inductor will oppose the change. What does this have to do with the PWM voltage waveform seen above? Well instead of letting the current pulse on the same order of the applied voltage pulse, the current will slowly start to rise. Once the pulse has ended the current doesn’t disappear immediately, it slowly starts to ebb away. Generally before it has fallen back to zero, the next pulse comes along and the current starts to slowly rise again, even higher than before because the pulses are getting wider.
Eventually you get a sinusoidal current waveform albeit with some jagged up and down transitions as the voltage pulses start and end.
So don’t think that you can power your solenoid from a single phase output of a PWM VFD. It’s not that kind of AC voltage.
#### Myth #2: All VFDs are the Same
The common AC Variable Frequency Drive of today is a fairly mature product. Most commercially available drives contain the same basic components: a bridge rectifier, a soft charging circuit, a DC Bus capacitor bank, and an output inverter section. Granted, there are differences in how the inverter section does its switching and there are differences in how reliable the components are or how efficient the thermal dissipation scheme is but the basic components remain the same.
There are exceptions to this “All The Same” thinking. For instance, some VFDs offer a Three Level output section. This output section allows the output pulses to vary from half bus voltage level pulses and full bus level pulses. Contrast the three-level output pulses to the two level output pulses shown next to it.
In order to achieve the three-level output, the output section has to have twice the number of output switches plus clamping diodes as shown below. The benefit gained by using a three-level output is reduction in voltage amplification at the motor due to reflected wave, lower common mode voltage, shaft voltage, and bearing current.
Maybe an even more atypical type of VFD is the matrix style inverter. These VFDs feature no DC bus or bridge rectifier. Instead the VFD uses bi-directional switches that can connect any of the incoming phase voltages to any of the three output phases. The upside of this arrangement is that power is allowed to flow freely from line to motor or motor to line for fully regenerative four quadrant operation! The downside is that there is filtering required on the input to the drive since extra inductance is necessary to filter the PWM waveform so that it does not affect the input AC lines.
These are two examples and there are more that prove that not all AC VFDs are the same.
#### Myth #3: VFDs Cure your Power Factor Issues
It is not uncommon to see VFD manufacturers quote power factor statistics like “0.98 displacement power factor” or “near unity power factor”. And it is true that the input displacement power factor improves once a VFD is installed in front of an induction motor. The VFD uses its internal capacitor bus to supply any reactive current that the motor requires, thereby protecting the AC Line for being the source of the reactive current and thus lowering the displacement power factor. However, displacement power factor is not the full story.
The full story of the power factor calculation is that it must include the reactive power that is demanded by harmonics created when AC voltage is rectified to DC. The diode bridge conducts the current from the AC Line to the DC Bus in a discontinuous way. It is important to remember that a diode will conduct only when the voltage on the anode side is higher than the voltage on the cathode side. This means that the diodes are only “ON” at the peak of each phase during both the positive and negative portion of the sine wave. This leads to a ripple-like voltage waveform and it also causes the input current to be distorted and discontinuous as shown below.
Much can be said about harmonics regarding how to calculate them and how to mitigate them but either way, to get true power factor you must include their effects. The formula below helps to get a handle on their influence over true power factor. In the formula, THD stands for Total Harmonic Distortion.
TruePowerFactor≈DisplacementPowerFactor√1+THD2
For the discontinuous input current shown above the THD would be in the neighborhood of 100% or more. Substituting that into the equation yields a True Power Factor closer to 0.71 compared to a displacement power factor of 0.98, which disregards harmonics.
Not to panic though, there are currently many ways to reduce the THD. These techniques make use of passive and active methods of making the input current waveform much less distorted and the THD much lower. The aforementioned matrix VFD is an example of an active method of THD reduction.
#### Myth #4: You Can Run a Motor at any Speed with a VFD
The beauty of using VFDs is that they can vary both their output voltage and output frequency. Their ability to run the motor slower and faster than its nameplate rated speed is part of why they get specified so often. Take the motor out of the equation and this myth is actually true. Without the limitations of the motor, the VFD can easily run any frequency within it speed range without limitation. But the reality of it is that the motor is necessary to do real work and its cooling and power needs start to limit the actual speed range of the motor/drive combination.
1. Limit #1: From a motor cooling point of view, turning some motors too slowly is not a good idea. In particular, totally enclosed fan cooled motors have limitations because the fan that blows air over the motor shell is attached to the motor shaft. The slower the motor is operated the lower the CFM cooling the motor. Most motor manufacturers specify speed ranges for their motor designs that will reflect how slow the motor can be run, especially while loaded. TEFC motors typically are not recommended for operation at full load below 15 Hz (a 4:1 speed range).
2. Limit #2: It is not always stated on the motor nameplate but mechanically motors have a limitation to their speed range. Referred to, commonly, as the Maximum Safe Operating Speed, this speed is tied to mechanical limitations such as bearings and balance. Some motor datasheets will specify the maximum speed.
3. Limit #3: Before the motor reaches its maximum operating speed, the motor could run out of torque. This speed limitation is not one of cooling or mechanics, but is due to power limitation, which is a product of speed and torque. To be exact, the VFD runs out of voltage. It is important to mention here that the rotation of the motor also generates a voltage of its own, referred toas back emf, which increases with speed. The back emf is being produced by the motor to oppose the supplied voltage from the VFD. At higher speeds, the VFD has to supply more voltage to overcome the back emf so that current can still flow into the motor since current is instrumental in producing torque. After a certain point, the VFD cannot push any more current into the motor because the output voltage has reached maximum and so the motor torque reduces, which in turn reduces speed. The reduction in speed results in lower back emf which in turn allows more current to flow into the motor again. There is an equilibrium point where the motor reaches the maximum speed for a given torque condition so that the product of torque and speed equals its power capability.
Let’s take a step back. VFDs can produce constant torque from a motor by keeping the V/f (voltage by frequency ratio or V by f ratio) constant, as shown below.
When the output frequency is increased, the voltage increases linearly. The problem arises when the frequency is raised beyond the base frequency of the motor, most commonly 60Hz in this part of the world. Beyond the base frequency, the output voltage cannot increase thereby resulting in the ratio of voltage to frequency to reduce. The ratio of voltage to frequency is a measure of the magnetic field strength in the motor and reducing it reduces the torque capability of the motor. Hence, the ability to have the motor produce rated torque at higher than base speed must decline at a rate of 1/frequency, so that the product of torque and speed, which equals power, is constant. The region of operation above base speed is called the constant power range, while operation at speeds below the base speed is called the constant torque range.
#### Myth #5: A VFD’s Input Current Should be Higher than its Output Current
So maybe this is not a myth but a misunderstanding. Some VFD users check their output current and input current with a current clamp or using VFD display monitors and find that the input current is much lower than their output current. It doesn’t seem to align with the idea that the VFD should have some losses due to its own thermal component losses so input should always be slightly higher that output. The concept is correct but it is power not current that should be considered.
P¿≅Pout
V¿∙I¿≅Vout∙Iout
The voltage portion of the power equation is pretty straightforward. The input voltage is always at the AC Line voltage. The output voltage varies with the speed per the V/f pattern. The current components of the equation are a little bit more complex. The key to current components is to understand that atypical induction motor has two current components. One is responsible for producing the magnetic field in the motor which is necessary to rotate the motor while the second component is the torque producing current, which as the name suggests, is responsible for producing torque. The drive consumes input current proportional to the motor’s active torque demand, or load. The current needed for producing the magnetic field typically does not vary with speed and is provided by the drive’s main DC bus capacitors, which are charged up during power up. Under low torque conditions, the output current may seem to be much higher than the input current since the input current mirrors only the torque producing current plus some harmonics but does not include the magnetizing current. The magnetizing current circulates in between the DC bus capacitors and the motor. Even at full load conditions, the input current will typically be lower than the motor current since the input still does not have any magnetizing current component in it.
To summarize this concept we can remember we are balancing input and output power. A good example is to consider a motor that is fully loaded but at a low speed. The input voltage is at the rated line while the output voltage will be low due to the low speed. Because of the full load on the motor,output current will be high. In order to balance the power equation the input current has to be lower than the output current! And now we know how and why that is possible.
# Siemens Soft Starter 3RE47 Ensures Optimal Performance
The functionality and dependability of motors play a critical role in industrial processes. Soft Starters can help to ensure optimal…
# Utilizing Phase Measurements for Detecting Compressor Misalignment
The Gulf Coast terminal facility currently employs Hydro's Centaur, a condition monitoring solution, to oversee the performance of its rotating…
# Drive Retrofit Improves Efficiency for Water Treatment Plant: Compact Stand-Alone Matrix Drives Replace 18-Pulse
Application Overview When a Massachusetts-based municipal water treatment facility needed to upgrade their aging 350HP18-pulse drives, they turned to a local…
## Related Whitepapers
### Upside Down Condos [Case Study]
This condo tower (324 units on 13 floors) had completed a previous retrofit that unfortunately left the variable speed pumps operating in a fixed mode…
### Ras Al Khaimah saves 20% energy and 12% OPEX in wastewater pumping system
RAKWA is the RAK wastewater agency responsible for designing, building, operating, and maintaining all the public wastewater systems. Ras Al Khaimah—known as RAK—is the northernmost…
### Intelligent Pumps Exceed Savings Estimates (Commercial Facility Installation Casestudy)
Design Envelope 6800 boosters reduced energy consumption by 76% compared to previous boosters, resulting in annual cost savings of \$4,400 per system. Background The Heron’s…
### Centaur’s Wireless Battery-powered Gateway
Our Solution Wireless sensors combined with a wireless battery-powered gateway can provide a flexible, safe, cost-effective, and reliable solution for collecting continuous condition monitoring data… | 3,125 | 15,315 | {"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-2023-40 | longest | en | 0.941806 |
https://www.tutorialspoint.com/how-to-find-the-correlation-matrix-by-considering-only-numerical-columns-in-an-r-data-frame | 1,702,273,135,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103558.93/warc/CC-MAIN-20231211045204-20231211075204-00521.warc.gz | 1,099,443,974 | 21,867 | # How to find the correlation matrix by considering only numerical columns in an R data frame?
While we calculate correlation matrix for a data frame, all the columns must be numerical, if that is not the case then we get an error Error in cor(“data_frame_name”) : 'x' must be numeric. To solve this problem, either we can find the correlations among variables one by one or use apply function.
## Example
Consider the below data frame −
set.seed(99)
x1<-rnorm(20)
x2<-rpois(20,5)
x3<-rpois(20,2)
x4<-LETTERS[1:20]
x5<-runif(20,2,10)
x6<-sample(letters[1:3],20,replace=TRUE)
df<-data.frame(x1,x2,x3,x4,x5,x6)
df
## Output
x1 x2 x3 x4 x5 x6
1 0.2139625022 7 4 A 6.423159 a
2 0.4796581346 5 1 B 7.176488 a
3 0.0878287050 7 2 C 2.372402 c
4 0.4438585075 8 3 D 6.599771 a
5 -0.3628379205 5 2 E 5.122577 c
6 0.1226740295 8 3 F 3.133224 c
7 -0.8638451881 4 2 G 2.482256 a
8 0.4896242667 4 4 H 4.532982 c
9 -0.3641169125 5 0 I 2.670717 c
10 -1.2942420067 2 3 J 8.597253 a
11 -0.7457690454 6 0 K 2.699053 a
12 0.9215503620 7 2 L 8.743498 b
13 0.7500543504 6 2 M 3.427915 c
14 -2.5085540159 10 2 N 5.928563 a
15 -3.0409340953 4 2 O 3.544168 a
16 0.0002658005 7 0 P 3.710395 c
17 -0.3940189942 2 2 Q 9.609634 c
18 -1.7450276608 5 0 R 5.886087 b
19 0.4986314508 8 2 S 5.507034 c
20 0.2709537888 4 3 T 2.137873 b
Finding the correlation matrix for columns in df −
cor(df)
Error in cor(df) : 'x' must be numeric
Here, the error means all the columns are not numeric.
str(df)
'data.frame': 20 obs. of 6 variables:
$x1: num 0.214 0.4797 0.0878 0.4439 -0.3628 ...$ x2: int 7 5 7 8 5 8 4 4 5 2 ...
$x3: int 4 1 2 3 2 3 2 4 0 3 ...$ x4: Factor w/ 20 levels "A","B","C","D",..: 1 2 3 4 5 6 7 8 9 10 ...
$x5: num 6.42 7.18 2.37 6.6 5.12 ...$ x6: Factor w/ 3 levels "a","b","c": 1 1 3 1 3 3 1 3 3 1 ...
Now to find the correlation matrix for all the numeric columns, we can do the following −
## Example
cor(df[sapply(df,is.numeric)])
## Output
x1 x2 x3 x5
x1 1.00000000 0.14685889 0.23107456 0.04232205
x2 0.14685889 1.00000000 -0.02664914 -0.14822679
x3 0.23107456 -0.02664914 1.00000000 0.18971761
x5 0.04232205 -0.14822679 0.18971761 1.00000000
Updated on: 24-Aug-2020
534 Views | 1,020 | 2,205 | {"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} | 3.109375 | 3 | CC-MAIN-2023-50 | latest | en | 0.387262 |
http://www.convertaz.com/convert-150-pt-to-c/ | 1,590,910,515,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347411862.59/warc/CC-MAIN-20200531053947-20200531083947-00096.warc.gz | 149,665,495 | 16,681 | # Convert 150 pt to c
## Conversion details
To convert pt to c use the following formula:
1 pt equals 2.273045 c
So, to convert 150 pt to c, multiply 2.273045 by 150 i.e.,
150 pt = 2.273045 * 150 c = 340.95675 c
For conversion tables, definitions and more information on the pt and c units scroll down or use the related pt and c quick access menus located at the top left side of the page.
#### Additional unit information
Pt is a common alias of the unit 'pint (Imperial)'
c is the symbol for cup (metric)
### From 10.00 to 400.00 pts, 40 entries
10 pts = 22.73045 c
20 pts = 45.4609 c
30 pts = 68.19135 c
40 pts = 90.9218 c
50 pts = 113.65225 c
60 pts = 136.3827 c
70 pts = 159.11315 c
80 pts = 181.8436 c
90 pts = 204.57405 c
100 pts = 227.3045 c
110 pts = 250.03495 c
120 pts = 272.7654 c
130 pts = 295.49585 c
140 pts = 318.2263 c
150 pts = 340.95675 c
160 pts = 363.6872 c
170 pts = 386.41765 c
180 pts = 409.1481 c
190 pts = 431.87855 c
200 pts = 454.609 c
210 pts = 477.33945 c
220 pts = 500.0699 c
230 pts = 522.80035 c
240 pts = 545.5308 c
250 pts = 568.26125 c
260 pts = 590.9917 c
270 pts = 613.72215 c
280 pts = 636.4526 c
290 pts = 659.18305 c
300 pts = 681.9135 c
310 pts = 704.64395 c
320 pts = 727.3744 c
330 pts = 750.10485 c
340 pts = 772.8353 c
350 pts = 795.56575 c
360 pts = 818.2962 c
370 pts = 841.02665 c
380 pts = 863.7571 c
390 pts = 886.48755 c
400 pts = 909.218 c
Click here for a list of all conversion tables of pt to other compatible units.
## pint (Imperial)
Pint (imperial) is a unit of measurement of volume. The definition for pint (imperial) is the following:
An Imperial pint is equal to 1/8 Imperial gallon.
The symbol for pint (Imperial) is pt (Imp)
## cup (metric)
Cup (metric) is a unit of measurement of volume. The definition for cup (metric) is the following:
A metric cup is equal to 250 × 10-6 cubic meters.
The symbol for cup (metric) is c
## Other people are also searching for information on pt conversions.
Following are the most recent questions containing pt. Click on a link to see the corresponding answer.
28 c to pt
24 fl oz = pt
28 c = how many pt
28c= pt
28c to pt
33 pt to qt
500mL=pt
28c =pt
18 qt pt
10 pt = qt
Home | Base units | Units | Conversion tables | Unit conversion calculator
Our privacy policy | Disclaimer | Contact us
Please note: Although we do our best to ensure the accuracy of all information posted on our website, we cannot guarantee or be held responsible for any errors that may have been made. In case you do find an error, please contact us and let us know about it so that we can correct it.
Copyright (c) 2009 - 2011 www.convertaz.com | 866 | 2,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-24 | latest | en | 0.741676 |
https://linearalgebras.com/solution-abstract-algebra-exercise-1-2-6.html | 1,709,264,927,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00615.warc.gz | 361,720,568 | 19,996 | If you find any mistakes, please make a comment! Thank you.
## Detecting copies of Dih(2n) in a larger group
Let $G$ be a group, and let $x,y \in G$ be elements of order 2. Prove that if $t = xy$ then $tx = xt^{-1}$ (so that if $n = |xy| < \infty$ then $x$ and $t$ satisfy the same relations in $G$ as $s$ and $r$ in $D_{2n}$).
Solution: We have $$xt^{-1} = x (xy)^{-1} = x y^{-1} x^{-1} = xyx = tx$$ since $x$ and $y$ have order 2.
#### Linearity
This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions. | 201 | 624 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-10 | longest | en | 0.880218 |
http://www.polymathlove.com/special-polynomials/triangle-similarity/algebra-square-root-problems.html | 1,513,311,970,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948563629.48/warc/CC-MAIN-20171215040629-20171215060629-00620.warc.gz | 455,783,173 | 12,412 | Algebra Tutorials!
Friday 15th of December
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
### Our users:
Keep up the good work Algebrator staff! Thanks!
Jenny Lane, AL
This is exactly what I needed. Thank you so much!!
Maria Lopez, CA
Thanks for the quick reply. Now that's customer service!
Margret Dixx, AL
Be it Step by Step explanation for an equation or graphical representation, you get it all. I just love to use this due to the flexibility it provides while studying.
Linda Rees, NJ
My son has used Algebrator through his high-school, and it seems he will be taking it to college as well (thanks for the free update, by the way). I really like the fact that I can depend on your company to constantly improve the software, rather than just making the sale and forgetting about the customers.
Christian Terry, ID.
### Students struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?
#### Search phrases used on 2013-08-02:
• roots of third order equation by hand
• algbra games
• solving 3 equations 3 unknowns with matrix
• permutations and combinations worksheets 3rd grade
• properties of exponents using fractions
• Free Equation Solver
• how to use Algebrator when taking an exam
• solutions to artin
• nonhomogeneous first order partial
• Definition of subtraction of Integers
• interactive calculater
• ways to teach domain and range on a graph in the math class
• solve this model graphically
• Third Grade Division Sheets
• divide polynomials calculator
• black history quiz for 7th and 8th graders printable
• least common denominator variable
• glencoe geometry concepts and applications answer key
• solving a third order equation using excel
• how to do basic square root
• algebra 2 cheat apps ti 84
• grade 9 algebra questions
• pizzazz math worksheets
• free algebra solutions
• college algebra worksheets
• Solve equation for all complex zeros
• solve for two unknown variable equation matlab code
• printable charts for third grade
• worksheets for adding and subtracting integers
• printable graph paper saxon math
• algebra beginner
• non linear modelling matlab
• writing exponential fracions in radical form
• algebra steps
• convert a second order differential equation into a first order system
• 6th grade science printable worksheets.com
• solver equations vertex
• mathematics problem solver
• Prentice Hall Algebra 1: Learn, Study, Succeed
• percent formulas
• Converting Parabolic Equations help
• learning basic algebra
• solving equations worksheets
• algebra 1 answers for prentice hall mathematics
• solving system of equations with t1-89
• blank coordinate plane
• Mathematic fifth grade test sample
• free online usable texas instrument 84 plus calculator
• lesson plans using proportions
• graphing/table calculator online free
• learn algebra flash
• ti-83 solve function
• coordinate worksheet grade4
• basic 7th grade algebra
• Prentice Hall World History Workbook Answers Cheating
• real-life examples of permutations
• basic Geometry by houghton mifflin company worksheets
• determinant of a matrix online applet
• solving two equation with "solve" in ti-89
• mcgraw hill pre algebra 1 structures and method course 1
• prentice hall mathematics geometry answers
• Prentice Hall Math Worksheets
• exponential equation simplifier
• online calculator with pie
• sats papers online free
• physics high school conversion chart
• factoring rational expressions solver
• University of Phoenix Elementary/Intermediate Algebra w/ALEKS User's Guide
• free algebra answer finder
• order these fraction from least to greatest
• algebra variations
• poems on math test
• DandM Charters Alaska
• grade 4 math review sheet of rotation
• simplifying inequalities
• square root rules
• Intermedia algebra math problems and answerss
• three step algebra practice sheet
• free grade 7 maths tests
• simultaneous equation solver
Prev Next | 1,070 | 4,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-51 | longest | en | 0.903105 |
https://www.acmicpc.net/problem/10549 | 1,685,779,279,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649177.24/warc/CC-MAIN-20230603064842-20230603094842-00508.warc.gz | 683,341,532 | 8,589 | 시간 제한메모리 제한제출정답맞힌 사람정답 비율
2 초 128 MB295523.810%
## 문제
Mirko lives in a big enchanted forest where trees are very tall and grow really quickly. That forest can be represented as an N·N matrix where each field contains one tree.
Mirko is very fond of the trees in the enchanted forest. He spent years observing them and for each tree measured how many meters it grew in a year. The trees grow continuously. In other words, if the tree grows 5 meters in a year, it will grow 2.5 meters in half a year.
Apart from trees, Mirko likes mushrooms from the enchanted forest. Sometimes, he eats suspicious colorful mushrooms and starts thinking about peculiar questions. Yesterday, this unfortunate thing happened and he wondered what would be the size of the largest connected group of trees that are all of equal height if the trees continue to grow at the same speed they’re growing at that moment.
Mirko quickly measured the current height of all trees in the forest and asked you to answer his question.
• Two trees are adjacent if their fields in the matrix share a common edge.
• Two trees are connected if there is a sequence of adjacent trees that leads from the first to the second.
• A group of trees is connected if every pair of trees in the group is connected.
## 입력
The first line of input contains the integer N (1 ≤ N ≤ 700).
After the first line, N lines follow, each of them containing N integers.
The ith line contains integers hij (1 ≤ hij ≤ 106), the initial height of tree in the ith row and jth column, given in meters.
After that, N more lines follow with N integers.
The ith line contains integers vij (1 ≤ vij ≤ 106), the growth speed of the tree in the ith row and jth column, given in meters.
Warning: Please use faster input methods beacuse the amount of input is very large. (For example, use scanf instead of cin in C++ or BufferedReader instead of Scanner in Java.)
## 출력
The first and only line of output must contain the required number from the task.
## 예제 입력 1
3
1 2 3
3 2 2
5 2 1
3 2 1
1 2 1
1 2 3
## 예제 출력 1
7
## 예제 입력 2
2
3 1
3 3
2 5
2 5
## 예제 출력 2
3
## 힌트
Clarification of the second example: after 8 months (two thirds of a year), the trees located at (0, 0), (0, 1) and (1, 0) will be 13/3 meters in height. | 599 | 2,269 | {"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} | 2.84375 | 3 | CC-MAIN-2023-23 | latest | en | 0.937397 |
http://sciencedocbox.com/Physics/88720746-Forcing-strong-convergence-of-proximal-point-iterations-in-a-hilbert-space.html | 1,558,941,239,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262029.97/warc/CC-MAIN-20190527065651-20190527091651-00281.warc.gz | 174,951,105 | 16,693 | # Forcing strong convergence of proximal point iterations in a Hilbert space
Save this PDF as:
Size: px
Start display at page:
Download "Forcing strong convergence of proximal point iterations in a Hilbert space"
## Transcription
1 Math. Program., Ser. A 87: (2000) Springer-Verlag 2000 Digital Object Identifier (DOI) /s M.V. Solodov B.F. Svaiter Forcing strong convergence of proximal point iterations in a Hilbert space Received January 6, 1998 / Revised version received August 9, 1999 Published online November 30, 1999 Abstract. This paper concerns with convergence properties of the classical proximal point algorithm for finding zeroes of maximal monotone operators in an infinite-dimensional Hilbert space. It is well known that the proximal point algorithm converges weakly to a solution under very mild assumptions. However, it was shown by Güler [11] that the iterates may fail to converge strongly in the infinite-dimensional case. We propose a new proximal-type algorithm which does converge strongly, provided the problem has a solution. Moreover, our algorithm solves proximal point subproblems inexactly, with a constructive stopping criterion introduced in [31]. Strong convergence is forced by combining proximal point iterations with simple projection steps onto intersection of two halfspaces containing the solution set. Additional cost of this extra projection step is essentially negligible since it amounts, at most, to solving a linear system of two equations in two unknowns. Key words. proximal point algorithm Hilbert spaces weak convergence strong convergence 1. Introduction Consider the problem find x H such that 0 T(x), (1) where H is a real Hilbert space, and T( ) is a maximal monotone operator (or a multifunction) on H. We shall denote the solution set of this problem by S := {x H 0 T(x)}. One of the classical algorithms for solving (1) is the proximal point method, introduced by Martinet [18] and further developed by Rockafellar [24]. Some other relevant papers on the proximal point method and its applications and modifications are [20,23,5,21, 17,10,12,9,4,6,31,29,27]. This list is by no means exhaustive, we refer the reader to [16] for a survey. Specifically, having x k H, a current approximation to the solution of (1), the proximal point method generates the next iterate x k+1 by solving the proximal subproblem 0 T(x) + µ k (x x k ), (2) where µ k > 0 is a regularization parameter. If the sequence {µ k } is chosen bounded from above, then the resulting sequence {x k } of proximal point iterates converges weakly to M.V. Solodov, B.F. Svaiter: Instituto de Matemática Pura e Aplicada, Estrada Dona Castorina 110, Jardim Botânico, Rio de Janeiro, RJ , Brazil. Mathematics Subject Classification (1991): 49M45, 90C25, 90C33
2 190 M.V. Solodov, B.F. Svaiter an element of S, provided this set is nonempty, see [24]. In [24], Rockafellar also posed an open question whether (or not) the proximal point method always converges strongly. This question was resolved in the negative by Güler [11], who exhibited a proper closed convex function f in an infinite-dimensional Hilbert space l 2, for which the proximal point algorithm (in our framework, procedure (2) with T = f ) convergesweakly but not strongly. Naturally, the question arises whether the proximal point method can be modified, preferably in a simple way, so that strong convergence is guaranteed. In this paper, we develop a strongly convergent algorithm by combining proximal point iterations with certain (computationally simple) projection steps (see Algorithm 1). Moreover, we allow the proximal point subproblems to be solved inexactly, and our tolerance requirements are less restrictive and more constructive than in the classical setting of [24] (this is discussed in detail in Sect. 2). Our approach is closely related to the hybrid projection-proximal point method presented in [31]. The principal idea of [31] is to use (approximate) solutions of proximal point subproblems to construct certain hyperplanes which separate the iterates from the solution set S, if it is nonempty. Convergence is forced by adding simple projection steps onto associated halfspaces. Of course, weak and strong convergence are only distinguishable in the infinitedimensional setting. On the other hand, even when we have to solve infinite-dimensional problems, numerical implementations of algorithms are certainly applied to finitedimensional approximations of the problems. Nevertheless, it is important to have convergence theory for the infinite-dimensional case, because it guarantees robustness and stability with respect to discretization schemes employed for obtaining finitedimensional approximations of infinite-dimensional problems. This issue is closely related to the so-called Mesh Independence Principle [2,1,15]. This principle relies on infinite-dimensional convergence to predict the convergence properties of a discretized finite-dimensional method. Furthermore, the mesh independence provides theoretical justification for the design of refinement strategies. Note that fine discretization is crucial for the obtained discrete solution to be an appropriate approximation to the true solution of the infinite-dimensional problem being solved. We note that many real-world problems in economics and engineering are modeled in the infinite-dimensional spaces. These include the optimal control and structural design problems, and the problem of minimal area surface with obstacles, among others. We refer the reader to [13, 14], where a variety of applications are described. The rest of the paper is organized as follows. In Sect. 2, the concept of inexact solutions of proximal subproblems with associated error tolerance is formally defined, and relevant properties of such inexact solutions are established. In Sect. 3, the algorithm is formally stated, and its components are discussed. Convergence results are presented in Sect. 4. Section 5 contains some concluding remarks. We briefly describe our notation. The inner product in H is denoted by,. The norm determined by the the inner product is denoted by. For a nonempty closed convex set A H andanelementx H, the orthogonal projection of x onto A, i.e., arg min{ y x y A}, is denoted by P A (x).byri(a) we denote the relative interior of A. Finally, we state the following well known properties of the projection operator to be used in the sequel.
3 Forcing strong convergence of proximal point iterations in a Hilbert space 191 Lemma 1. [32] Let A be any nonempty closed convex set in H. For any x, y H and any z A, the following properties hold : x P A (x), z P A (x) 0 ; P A (x) P A (y) 2 x y 2 P A (x) x + y P A (y) Inexact proximal point iterations Because solving the proximal subproblem (2) exactly can be computationally as difficult (or almost as difficult) as solving the original problem (1) itself, of particular importance is the case when the subproblems are solved only approximately, that is we find a pair y k H and v k T(y k ) such that ε k = v k + µ k (y k x k ), (3) where ε k is the error associated with inexact solution of subproblem (2). In the pure inexact proximal point method, one then sets x k+1 := y k to obtain the next iterate. The subject of appropriate approximation criteria for the inexact solution of proximal point subproblems is very important, and it has been addressed, for example, in [24, 6,8]. Typically, the following (or conceptually similar) conditions are used to ensure convergence of the iterates: or ε k σ k µ k, σ k <, k=0 ε k σ k µ k y k x k, σ k <. The first condition above is used to ensure global convergence, while the second condition (together with certain additional assumptions) implies local linear rate of convergence. Note that a summable sequence {σ k } essentially has to be chosen a priori, which makes a constructive choice difficult in most situations. Also note that under the second condition, the relative error in (3) satisfies ε k µ k y k x k σ k, k=0 σ k <. So the relative error must be summable, and hence, tend to zero. A simple example in [31] shows that if the tolerance parameters σ k are fixed at a nonzero value, convergence of the inexact proximal point iterations cannot be guaranteed even in the finite-dimensional case. k=0
4 192 M.V. Solodov, B.F. Svaiter A proximal-point-type method with a more realistic and constructive error tolerance has recently been proposed in [31]. In particular, for the method of [31], the relative error in (3) can be fixed. In other words, we accept y k H and v k T(y k ) as a satisfactory approximate solution in (3) whenever any one of the following two conditions is met: ε k µ k y k x k σ or ε k v k σ, where σ [0, 1). Note that here σ does not depend on the iteration index k (although, in principle, it certainly need not be fixed; the important issue is that it can be kept bounded away from zero). The key fact is that under this condition, the hyperplane H k := { x H v k, x y k = 0 } strictly separates the current iterate x k from the solution set S (we assume here that it is nonempty). Convergence (in the weak topology) can then be guaranteed if we obtain the next iterate x k+1 as the orthogonal projection of x k onto H k : x k+1 = P Hk (x k ). This hybrid method retains all the attractive convergenceproperties of the proximal point algorithm while allowing a fixed relative error tolerance. We refer the reader to [31] for complete analysis. An important point to note is that projection onto a hyperplane can be carried out explicitly, so it does not entail any significant additional computational cost. We also note that this projection step is indispensable in the sense that if one sets instead x k+1 := y k,wherey k is computed with fixed relative error tolerance σ (0, 1), then the iterates of the resulting pure inexact proximal method mail fail to converge even in R 2 (see an example in [31]). For an extension of the described notion of approximate solution to the setting of generalized proximal algorithms based on the Bregman functions, we refer the reader to [29]. Another extension using certain enlargements (outer approximations) of the operator defining the problem can be found in [28]. Using also the linesearch technique of [26], the framework of [31,28] led to the development of truly globally convergent inexact Newton methods for monotone equations [30] and complementarity problems [25]. However, by itself, the method of [31] does not attain the goal of the present paper. Like the classical proximal method, it may converge weakly but not strongly in an infinite-dimensional space. This is easy to see because with ε k = 0, the hybrid projection-proximalpoint method reduces to the standard proximal point method (again, see [31]). Hence, the counter example of Güler [11] applies. Still, the separation and projection methodology of [31] will appear very useful for devising a strongly convergent algorithm. As already mentioned, the method proposed here will use inexact proximal iterates to set up appropriate projection steps. We proceed to study the properties of inexact solutions of proximal subproblems, which will be used in the sequel. Definition 1. Let x H, µ>0and σ [0, 1). We say that a pair (y,v) H H is an inexact solution with tolerance σ of 0 T( ) + µ( x),if v T(y), v + µ(y x) = ε,
5 Forcing strong convergence of proximal point iterations in a Hilbert space 193 and ε σ max { v,µ y x }. Next we establish some nice properties of inexact solutions defined above. Proposition 1. Let x H, µ>0 and σ [0, 1), and suppose that (y,v)is an inexact solution of 0 T( ) + µ( x) with tolerance σ. Then it holds that x y,v (1 σ) max{µ x y 2, v 2 /µ} (1 σ) v x y. (4) Define H := { z H z y,v 0 }. Then the following four statements are equivalent: Furthermore, x H ; y = x ; v = 0 ; x is a solution of (1). P H (x) x (1 σ) max{ x y, v /µ}. (5) Proof. To prove (4), we consider the two possible cases: In the first case, we have that µ x y v and µ x y v. ε σ v. It follows that in this case 1 x y,v = v ε, v µ 1 σ µ v 2. (6) Furthermore, x y,v 1 σ µ v 2 (1 σ) v y x µ(1 σ) y x 2. (7) Combining (6) and (7), we conclude that (4) holds in the first case.
6 194 M.V. Solodov, B.F. Svaiter Hence, Consider now the second case. We have that Furthermore, in this case, ε σµ y x. x y,v = x y,µ(x y) + ε µ(1 σ) x y 2. (8) x y,v µ(1 σ) x y 2 (1 σ) v y x 1 σ µ v 2. (9) Again, combining (8) and (9), we obtain that (4) is satisfied in the second case also. Next, we establish the equivalence of the four conditions. Suppose that x H.Then x y,v 0 and, by (4), x = y.ifx = y,then x y,v =0 and, again by (4), v = 0. By the same reasoning, if v = 0thenx = y and x is a solution of (1). Finally, if x is a solution (i.e., 0 T(x)) then, by the monotonicity of T, 0 y x,v 0 = y x,v, and so x H. Finally, to prove (5), note that if x H, thenx = y, v = 0 and (5) holds trivially. The other case is x / H (and so v = 0). Then Hence, P H (x) = x P H (x) x = v, x y v 2 v. v, x y v If v /µ x y then v 2 /µ µ x y 2, and (5) follows from the first inequality in (4). If v /µ x y then (5) follows from the second inequality in (4). This completes the proof.. 3. The algorithm We are now ready to formally state our algorithm. Algorithm 1. Choose any x 0 H and σ [0, 1). At iteration k, havingx k, choose
7 Forcing strong convergence of proximal point iterations in a Hilbert space 195 µ k > 0andfind(y k,v k ), an inexact solution of with tolerance σ. Define and Take 0 T(x) + µ k (x x k ) H k = { z H z y k,v k 0 }, W k = { z H z x k, x 0 x k 0 }. x k+1 = P Hk W k (x 0 ). Note that at each iteration, there are two subproblems to be solved: find an inexact solution of the proximal point subproblem and find the projection of x 0 onto H k W k,the intersection of two halfspaces. The proximal subproblem always has an exact solution, which is unique (this is a classical result of Minty [19]). Computing an approximate solution only makes things easier. So this part of the method is well-defined (but it is worth to note that since some error is allowed, in general the pair (y k,v k ) is not uniquely defined). Regarding the projection step, we shall prove that the set H k W k is never empty, even when the solution set S is empty. Therefore the whole algorithm is well-defined in the sense that it generates an infinite sequence {x k } (and an associated sequence of pairs {(y k,v k )}). Also note that starting with the same x 0, many different sequences satisfying the conditions of Algorithm 1 can be generated. Finally, note that from Proposition 1 it follows that at some iteration k, x k S if and only if v k = 0. So, in that case, for all k k we will have x k = x k and v k = 0. Some remarks are in order regarding complexity of the projection step in Algorithm 1. Even though we are working in an infinite-dimensional space, projection onto an intersection of two halfspaces amounts to solving, at most, a linear system of two equations with two unknowns. Suppose that at iteration k, W k H k is nonempty (this fact will be formally established in Sect. 4). Then x k+1 is well defined and is characterized as the solution of min z z x 0 2 s.t. z y k, v k 0, z x k, x 0 x k 0. Let us express z x 0 as a linear combination of v k and x 0 x k plus a vector orthogonal to v k and x 0 x k : z x 0 = λ 1 v k + λ 2 (x 0 x k ) + h, h,v =0, h, x 0 x k =0. So, the problem above becomes min λ1,λ 2,h h 2 + λ 1 v k + λ 2 (x 0 x k ) 2 s.t. λ 1 v k + λ 2 (x 0 x k ) + x 0 y k, v k 0, λ 1 v k + λ 2 (x 0 x k ) + x 0 x k, x 0 x k 0.
8 196 M.V. Solodov, B.F. Svaiter Due to the special structure of this problem, at the solution we obviously have h = 0. Therefore λ 1 and λ 2 are obtained solving a two-dimensional quadratic minimization problem with two linear inequality constraints. Furthermore, it is easy to see that if the projection of x 0 onto H k belongs to W k then P Hk (x 0 ) = x 0 vk, x 0 y k v k 2 v k P Hk (x 0 ) = P Hk W k (x 0 ). So in this case, to obtain x k+1 there is no need for any further computation. Otherwise, we have P Hk W k (x 0 ) = x 0 + λ 1 v k + λ 2 (x 0 x k ), where λ 1,λ 2 is the solution of the linear system of two equations with two unknowns: λ 1 v k 2 + λ 2 v k, x 0 x k = x 0 y k,v k λ 1 v k, x 0 x k +λ 2 x 0 x k 2 = x 0 x k 2. Therefore, we could actually write an explicit formulae for obtaining x k+1, which means that the cost of the projection step in Algorithm 1 is essentially negligible. The preceding analysis shows that the cost of each iteration of Algorithm 1 is about the same as that of an iteration of the inexact proximal point method (actually, of the modified projection-proximal method of [31], since we adopt the relaxed approximation criterion introduced in that reference). Thus we obtain a strongly rather than weakly convergent proximal-type algorithm at essentially no additional cost per iteration. We proceed to establish strong convergence of the method. 4. Convergence analysis We start with establishing some properties of the algorithm which hold regardless of whether the problem has solutions or not. Let x k, y k,v k, k = 0, 1,... be variables generated by Algorithm 1. For now, they may be defined either for all iteration indices k or only up to some index k 0. Proposition 2. Suppose that Algorithm 1 reaches an iteration k + 1. Then it holds that and x k+1 x 0 2 x k x x k+1 x k 2, (10) x k+1 x k (1 σ) max{ y k x k, v k /µ k }. (11)
9 Forcing strong convergence of proximal point iterations in a Hilbert space 197 Proof. By the definition of W k, it is clear that x k is the projection of x 0 onto W k. Applying Lemma 1 with A = W k, x = x k+1 and y = x 0, we obtain PWk (x k+1 ) P Wk (x 0 ) 2 x k+1 x 0 2 PWk (x k+1 ) x k+1 + x 0 P Wk (x 0 ) 2. Because x k+1 W k,wehavethatp Wk (x k+1 ) = x k+1. Furthermore, P Wk (x 0 ) = x k. Hence, x k+1 x 0 2 x k x x k+1 x k 2. Since x k+1 H k, it holds that x k+1 x k x k P Hk (x k ). Now (11) follows from Proposition 1. As a consequence of Proposition 2, we immediately obtain the following result. Corollary 1. Suppose the sequence of regularization parameters {µ k } is bounded from above, and Algorithm 1 generates an infinite sequence {x k }. Then either {x k } is bounded and each of its weak accumulation points belongs to S =, ors = and lim k x k =. Proof. Applying (10) consecutively, we obtain k 1 x k x 0 2 x j+1 x j 2. j=0 If {x k } is bounded, letting k yields that x j+1 x j 2 <, j=0 and hence, 0 = lim k xk+1 x k. By (11) and boundedness of {µ k } from above, we further obtain that lim k yk x k =0, (12) lim k vk =0. (13) Since the sequence {x k } is bounded, it must have weak accumulation points. Let x be any weak accumulation point of {x k }, and take any subsequence {x k j } weakly convergent to it. By (12), {y k j } has the same weak limit x. Sincev k T(y k ) with v k 0 strongly, the maximal monotonicity of T implies that 0 T( x) (the argument to show this is standard). That is, x S. Suppose now that S =. By the preceding assertion, the sequence {x k } is unbounded in this case. Since, by (10), the sequence { x k x 0 } is nondecreasing, it follows that x k x 0 as k,andso x k.
10 198 M.V. Solodov, B.F. Svaiter Next, we shall prove well-definedness and strong convergence of {x k } to a solution in the case S =. After that, we shall establish well-definedness of {x k } in the case S =. In the latter case, we obtain that x k, which is somewhat stronger than simple unboundedness of {x k } The case S = In this subsection we assume that (1) has a solution, so S =. Having chosen the initial iterate x 0, let us define the following set which will be important for the subsequent analysis: U(x 0 ) = { x H z S, z x, x 0 x 0 }. (14) We next show that the set H k W k always contains the solution set S. Hence, it is nonempty and the projection step in Algorithm 1 is well-defined. Furthermore, we show that the generated sequence of iterates {x k } is contained in the set U(x 0 ). Proposition 3. Suppose that Algorithm 1 reaches an iteration k and x k U(x 0 ).Then it holds that 1. S H k W k. 2. x k+1 is well-defined and x k+1 U(x 0 ). Proof. First note that by the monotonicity of T, v k, y k x 0 for any x S. Hence, S H k.sincex k U(x 0 ),wealsohavethat x x k, x 0 x k 0 x S. By the definition of W k, we obtain that S W k. We conclude that S H k W k. In particular, it follows that H k W k =, and the next iterate x k+1 is well-defined. Because x k+1 is the projection of x 0 onto H k W k,bylemma1weobtainthat z x k+1, x 0 x k+1 0 z H k W k. Since S H k W k, the above relation certainly holds for all z S, which implies that x k+1 U(x 0 ), by the definition given in (14). It can be now easily verified that the whole algorithm is well-defined. Corollary 2. Algorithm 1 is well-defined and generates infinite sequences {x k } and {y k },{v k } such that x k U(x 0 ) and S H k W k for all k. Furthermore, if {µ k } is bounded from above, then {x k } is bounded and each of its weak accumulation points belongs to S.
11 Forcing strong convergence of proximal point iterations in a Hilbert space 199 Proof. It is enough to observe that x 0 U(x 0 ) and apply Proposition 3 and induction on k. Corollary 1 further implies that {x k } is bounded with weak accumulation points being in S. We are now ready to prove strong convergence of any sequence {x k } generated by Algorithm 1 starting from x 0, to the solution of (1) which is closest to x 0. Theorem 1. Let {x k } be a sequence generated by Algorithm 1, and suppose that the sequence of regularization parameters {µ k } is bounded from above. Then {x k } converges strongly to x = P S (x 0 ). Proof. Note that x, the projection of x 0 onto S, exists because the solution set S is closed, convex, and we assumed it to be nonempty. By the definition of x k+1,wehave that x k+1 x 0 z x 0 z H k W k. Since x S H k W k, it follows that for all k x k x 0 x x 0. (15) By Corollary 2, we already know that {x k } is bounded with all of its weak accumulation points belonging to S. Let{x k j } be any weakly convergent subsequence of {x k },andlet x S be its weak limit. Observe that x k j x 2 = x k j x 0 (x x 0 ) 2 = x k j x x x x k j x 0, x x 0 2 x x x k j x 0, x x 0, were the inequality follows from (15). By the weak convergence of {x k j } to x, wethen obtain lim sup x k j x 2 2( x x 0 2 x x 0, x x 0 ). (16) j Applying Lemma 1 with A = S, x = x 0 and z = x S, and taking into account that x is the projection of x 0 onto S, wehavethat x 0 x, x x 0. Now from the latter relation it follows that x x 0 2 x x 0, x x 0. Combining the last inequality with (16), we conclude that {x k j } converges strongly to x. Clearly x = x because x is a weak limit of {x k j }. Since x was taken as an arbitrary weak accumulation point of {x k }, it follows that x is the unique weak accumulation point of this sequence. Since {x k } is bounded, the whole sequence {x k } weakly converges to x. On the other hand, we have shown that every weakly convergent subsequence of {x k } converges strongly to x. Hence, the whole sequence {x k } converges strongly to x S.
12 200 M.V. Solodov, B.F. Svaiter 4.2. The case S = In this subsection we assume that no solution exists. We shall prove that the sequence {x k } is still defined for every k, and that it diverges. Theorem 2. If S = then Algorithm 1 generates an infinite sequence {x k }.If,in addition, the sequence of regularization parameters {µ k } is bounded from above, then lim k x k =. Proof. In view of Corollary 1, we only have to establish well-definedness of the algorithm. We argue by induction. We first check that the initial iteration k = 0 is well defined. The problem 0 T(x) + µ 0 (x x 0 ) always has the exact solution, and hence some inexact solution (y 0,v 0 ). Now note that W 0 = H. SinceH 0 cannot be empty, the next iterate x 1 is generated, which is the projection of x 0 onto H 0 = W 0 H 0. Note that whenever x k is generated, y k and v k can further be obtained because the proximal subproblems always have (in)exact solutions. Suppose that x k, (y k,v k ) are defined for k = 0,...,ˆk. It is enough to prove that x ˆk+1 is well-defined. Take any where D(T) is the domain of T, and define and z 0 ri(d(t)), ρ = max{ y k z 0 k = 0,...ˆk} { 0 if x z h(x) = 0 ρ + 1, + otherwise. Then h : H R {+ } is a lower semicontinuous proper convex function, its subgradient h is maximal monotone, and T = T + h is also maximal monotone [22]. Furthermore, T (z) = T(z) if z z 0 <ρ+ 1. Therefore, v k T (y k ) for k = 0,...,ˆk. We conclude that x k, (y k,v k ) also satisfy the conditions of Algorithm 1 applied to the problem 0 T (x). Since T has a bounded domain, the above problem has solutions. Using the two preceding observations and Corollary 2, it follows that x ˆk+1 is well-defined, and hence, so is the whole algorithm. The conclusion now follows using Corollary 1.
13 Forcing strong convergence of proximal point iterations in a Hilbert space Concluding remarks We presented a proximal-type algorithm for finding zeroes of maximal monotone operators, which converges strongly in an infinite-dimensional Hilbert space to a solution, provided it exists. In this sense, this is a significant theoretical improvement over the classical proximal point method which converges weakly but not strongly. Strong convergence of our method is forced by adding a simple projection step which amounts to solving, at most, a linear system of two equations with two unknowns. Furthermore, a constructive error tolerance is used in the approximate solution of proximal subproblems. Finally, we mention the recent work [7,3], which has some interesting connections to the algorithm described in this paper. Acknowledgements. We thank an anonymous referee for a constructive suggestion which helped us to somewhat simplify the analysis and improve the presentation. Research of the first author is supported by CNPq Grant /95-6 and by PRONEX Optimization, research of the second author is supported by CNPq Grant /93-9(RN) and by PRONEX Optimization. References 1. Allgower, E.L., Böhmer, K. (1987): Application of the mesh-independence principle to mesh refinement strategies. SIAM J. Numer. Anal. 24, Allgower, E.L., Böhmer, K., Potra, F.-A., Rheinboldt, W.C. (1986): A mesh-independence principle for operator equations and their discretizations. SIAM J. Numer. Anal. 23, Bauschke, H.H., Combettes, P.L. (1999): A weak-to-strong convergence principle for Fejér-monotone methods in Hilbert spaces. Math. Oper. Res. submitted Bonnans, J.F., Gilbert, J.C., Lemaréchal, C., Sagastizábal, C. (1995): A family of variable metric proximal point methods. Math. Program. 68, Brézis, H., Lions, P.L. (1978): Produits infinis de résolvantes. Isr. J. Math. 29, Burke, J.V., Qian, M. (1998): A variable metric proximal point algorithm for monotone operators. SIAM J. Control Optim. 37, Combettes, P.L. (1999): Strong convergence of block-iterative outer approximation methods for convex optimization. SIAM J. Control Optim. To appear 8. Eckstein, J. (1998): Approximate iterations in Bregman-function-based proximal algorithms. Math. Program. 83, Eckstein, J., Bertsekas, D.P. (1992): On the Douglas-Rachford splitting method and the proximal point algorithm for maximal monotone operators. Math. Program. 55, Ferris, M.C. (1991): Finite termination of the proximal point algorithm. Math. Program. 50, Güler, O. (1991): On the convergence of the proximal point algorithm for convex minimization. SIAM J. Control Optim. 29, Güler, O. (1992): New proximal point algorithms for convex minimization. SIAM J. Optim. 2, Henry, J., Yvon, J-P., eds. (1994): Lecture notes in Control and Information Sciences No. 197, System Modelling and Optimization. Springer, Berlin 14. Kinderlehrer, D., Stampacchia, G. (1980): An introduction to variational inequalities and their applications. Academic Press, New York 15. Laumen, M. (1999): Newton s mesh independence principle for a class of optimal shape design problems. SIAM J. Control Optim. 37, Lemaire, B. (1989): The proximal algorithm. In: Penot, J.P., ed., New Methods of Optimization and Their Industrial Use. International Series of Numerical Mathematics 87, pp Birkhauser, Basel 17. Luque, F.J. (1984): Asymptotic convergence analysis of the proximal point algorithm. SIAM J. Control Optim. 22, Martinet, B. (1970): Regularisation d inequations variationelles par approximations successives. Revue Française d Informatique et de Recherche Opérationelle 4, Minty, G.J. (1962): Monotone (nonlinear) operators in Hilbert space. Duke Math. J. 29, Moreau, J.-J. (1965): Proximité et dualité dans un espace Hilbertien. Bull. Soc. Math. Fr. 93,
14 202 M.V. Solodov, B.F. Svaiter: Strong convergence of proximal iterations 21. Passty, G.B. (1979): Weak convergence theorems for nonexpansive mappings in Banach spaces. J. Math. Anal. Appl. 67, Rockafellar, R.T. (1970): On the maximality of sums of nonlinear monotone operators. Trans. Am. Math. Soc. 149, Rockafellar, R.T. (1976): Augmented Lagrangians and applications of the proximal point algorithm in convex programming. Math. Oper. Res. 1, Rockafellar, R.T. (1976): Monotone operators and the proximal point algorithm. SIAM J. Control Optim. 14, Solodov, M.V., Svaiter, B.F.: A truly globally convergent Newton-type method for the monotone nonlinear complementarity problem. SIAM J. Optim. To appear 26. Solodov, M.V., Svaiter, B.F. (1999): A new projection method for variational inequality problems. SIAM J. Control Optim. 37, Solodov, M.V., Svaiter, B.F. (1999): A comparison of rates of convergence of two inexact proximal point algorithms. In: Di Pillo, G., Giannessi, F., eds., Nonlinear Optimization and Related Topics. Kluwer Academic Publishers. To appear 28. Solodov, M.V., Svaiter, B.F. (1998): A hybrid approximate extragradient proximal point algorithm using the enlargement of a maximal monotone operator. Set-Valued Analysis. To appear 29. Solodov, M.V., Svaiter, B.F.: An inexact hybrid generalized proximal point algorithm and some new results on the theory of Bregman functions. Math. Oper. Res. To appear 30. Solodov, M.V., Svaiter, B.F. (1999): A globally convergent inexact Newton method for systems of monotone equations. In: Fukushima, M., Qi, L., eds., Reformulation Nonsmooth, Piecewise Smooth, Semismooth and Smoothing Methods, pp Kluwer Academic Publishers 31. Solodov, M.V., Svaiter, B.F. (1999): A hybrid projection proximal point algorithm. J. Convex Anal. 6, Zarantonello, E.H. (1971): Projections on convex sets in Hilbert space and spectral theory. In: Zarantonello, E.H., ed., Contributions to Nonlinear Functional Analysis, pp Academic Press, New York | 8,331 | 30,655 | {"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-2019-22 | latest | en | 0.895609 |
http://cetking.com/3-overlapping-sets-must-know/ | 1,556,200,715,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578721468.57/warc/CC-MAIN-20190425134058-20190425160058-00362.warc.gz | 32,439,776 | 14,661 | # 3 Overlapping Sets you must know !
Overlapping Sets are a hot topic in every other exam . They don’t have to be particularly hard questions, but they fit well into the criteria of what the test makers are trying to test: mathematical reasoning, without a lot of nuts-and-bolts arithmetic. Some of these questions require nothing more than addition and subtraction.Two Overlapping Sets
You’ve probably see plenty of problems with two overlapping sets.
Something like: “A class has 40 students, 25 of whom are boys and 12 of whom play basketball. If 8 of the boys play basketball, how many girls do not play basketball?”
There are three distinct ways of handling these questions, each of which I cover in more detail in Math:
A Venn Diagram
A table
A formula
All three represent the same relationships, but have different benefits. A Venn Diagram allows you to see those relationships visually, but can be confusing. A table is much clearer, but more complicated. The formula is easiest, but only when the question gives you data you can plug straight into the question.
The Overlapping Sets Formula
The formula I use is this:
Total = Group1 + Group2 – Both + Neither
In the example above, 40 is Total, 25 is Group1, 12 is Group2, and 8 is Both. (Group1 and Group2 are interchangeable.) Basically, G1+G2 is the sum of all of the people who do one or the other, but that sum double-counts the number who do both. That’s why we subtract both.
Three Overlapping Sets
My focus today, though, is on questions with three overlapping sets. They aren’t common invert other exam, but at the same time, they don’t require much more thinking that do the two-set questions.
Here’s what one of those might look like:
Of the shoe stores in City X, 30 carry Brand A shoes, 40 carry Brand B shoes, and 25 carry Brand C shoes. If each store carries at least one of the brands, 32 of the stores carry two of the three shoe brands, and none of the stores carry all three, how many shoe stores are there in City X?
We can’t use a table for this, because it would have to be three-dimensional. A Venn Diagram is a possibility, but there’s no place to put 32. We’ll have to look at an expansion of the formula.
The Three Overlapping Sets Formula
Total = Group1 + Group2 + Group3 – (sum of 2-group overlaps) – 2*(all three) + Neither
This looks more complicated, but the concept is the same. G1+G2+G3 is the sum of all the stores that carry these brands, only we’ve double-counted some of them. Since 32 of the stores carry two brands, we’ve double-counted exactly 32, so we subtract 32, the “sum of 2-group overlaps.”
The tricky part is the second-to-last term: “2*(all three).” It doesn’t come up very often, as when it does, it’s almost always on Data Sufficiency questions. However, it’s worth thinking about.If one of the stores in City X carried all three shoe brands, that store is being counted three times–once in the 30 A’s, once in the 40 B’s, and once in the 25 C’s. But it’s still only one store. Thus, if it’s represented three times, we need to subtract it twice. In this case, the question tells us that the “all three” term is equal to zero.
So, to solve this example:
Total = 30 + 40 + 25 – 32 – 2*0 + 0 = 63
(Neither equals 0 because every shoe store in City X carries at least one of the brands.)
Problems with three overlapping sets can be daunting, but if you can master the concepts in this article, it’ll be yet another question type you can attack with confidence when you take your test.
set theory for cat pdf
how to solve venn diagram questions for cat
venn diagram problems and solutions for cat pdf
set theory questions for aptitude
problems on sets and venn diagrams
set theory questions and answers for cat
venn diagram aptitude questions cat
set theory aptitude indiabix | 915 | 3,791 | {"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} | 4.625 | 5 | CC-MAIN-2019-18 | latest | en | 0.952686 |
http://betterlesson.com/lesson/614403/mixed-numbers | 1,487,648,223,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170624.14/warc/CC-MAIN-20170219104610-00579-ip-10-171-10-108.ec2.internal.warc.gz | 28,022,313 | 18,094 | # Mixed Numbers
Unit 4: Adding and Subtracting Fractions
Lesson 15 of 17
## Big Idea: I’m the principal now!
Print Lesson
4 teachers like this lesson
Standards:
Subject(s):
Math, Fractions, mixed numbers
60 minutes
### Erin Doughty
##### Similar Lessons
###### Pretest
6th Grade Math » Fraction Operations
Big Idea: What do students already know about fractions? What gaps do students have in their understanding? Students take the fractions pretest in order to inform instruction.
Favorites(30)
Resources(9)
Somerville, MA
Environment: Urban
###### Adding and Subtracting with Fractions
6th Grade Math » Number Sense
Big Idea: Using common denominators with fractions will help the students learn to divide fractions.
Favorites(12)
Resources(24)
Plainfield, IL
Environment: Suburban | 182 | 790 | {"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-2017-09 | latest | en | 0.837478 |
https://www.varsitytutors.com/algebra_1-help/how-to-use-the-grid-method-for-foil?page=2 | 1,550,976,961,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249578748.86/warc/CC-MAIN-20190224023850-20190224045850-00014.warc.gz | 988,954,886 | 41,011 | # Algebra 1 : How to use the grid method for FOIL
## Example Questions
← Previous 1 2 Next →
### Example Question #11 : How To Use The Grid Method For Foil
Expand using the grid method.
Explanation:
Top left corner:
Top right corner:
Bottom left corner:
Bottom right corner:
Add along the (pink) diagonal:
Combining our terms, we get:
### Example Question #11 : How To Use The Grid Method For Foil
Expand using the grid method.
Explanation:
Top left:
Top middle:
Top right:
Bottom left:
Bottom middle:
Bottom right:
Add along the first (pink) diagonal:
Add along the second (blue) diagonal:
Combining our terms, we get .
### Example Question #13 : How To Use The Grid Method For Foil
Expand using the grid method.
Explanation:
Top left:
Top right:
Middle left:
Middle right:
Bottom left:
Bottom right:
Add along the first (pink) diagonal:
Add along the second (blue) diagonal:
Combining our terms, we get .
### Example Question #14 : How To Use The Grid Method For Foil
Expand
Explanation:
Top left:
Top middle:
Top right:
Middle left:
Center block:
Middle right:
Bottom left:
Bottom middle:
Bottom right:
Add along the first (pink) diagonal:
Add along the second (blue) diagonal:
Add along the third (orange) diagonal:
Combining our terms, we get or
.
### Example Question #15 : How To Use The Grid Method For Foil
Expand using the grid method.
Explanation:
Top left:
Top middle:
Top right:
Middle left:
Center block:
Middle right:
Bottom left:
Bottom middle:
Bottom right:
Add along the first (pink) diagonal:
Add along the second (blue) diagonal:
Add along the third (orange) diagonal:
Combining our terms, we get .
### Example Question #16 : How To Use The Grid Method For Foil
Complete the following and find the final product.
Explanation:
This question is merely asking us to complete the grid form of FOIL and find the final result.
In order to complete the four squares, we merely multiply the term above the box in the topmost row by the term in the leftmost column.
For example, in order to solve for the top left box:
multiplying
Moving on to the bottom left box, using the same principle, we would get
multiplying
The same principles can be applied with the two right boxes. Multiplying the terms will yield a filled in grid that looks like this:
The next step involves us to use the information from the four boxes and collect like terms to get our final answer:
### Example Question #17 : How To Use The Grid Method For Foil
Simplify the following using the grid method for FOIL:
Explanation:
To solve using the grid method, we use the given problem
and create a grid using each term.
Now, we fill in the boxes by multiplying the terms in each row and column.
Now, we write each of the multiplied terms out,
We combine like terms.
Therefore, by using the grid method, we get the solution
### Example Question #12 : How To Use The Grid Method For Foil
Find the product of using the grid method.
Explanation:
Using the grid method is an alternate way of doing FOIL. This utilizes the "tic tac toe" grid. Upon creating the grid, we need to write in the binomials we were given to multiply. Such is done like so:
and have been written in on the leftmost column and topmost row as separated terms. In order to carry out FOIL, we must now fill in the four remaining boxes. In order to do so, each box will be solved by mutlipyling the term above it in the topmost row by the term next to it in the leftmost column. For example, the top left box will be filled in by multiplying .
Using the same principle, the left bottom box will be completed by multiplying , like so:
The remaining boxes in the rightmost column may be solved using the same principle that was used to solve for the middle column. The filled out grid will look like this:
The problem is nearly done. All that is left for us to do is to write this as a mathematical expression and collect like terms. Completing this step will yield us our final answer.
Note that this last step only requires the values from the four boxes we solved for.
### Example Question #19 : How To Use The Grid Method For Foil
Using the grid method, find the product of .
Explanation:
Using the grid method is an alternate way of doing FOIL. This utilizes the "tic tac toe" grid. Upon creating the grid, we need to write in the binomials we were given to multiply. Such is done like so:
and have been written in on the leftmost column and topmost row as separated terms. In order to carry out FOIL, we must now fill in the four remaining boxes. In order to do so, each box will be solved by mutlipyling the term above it in the topmost row by the term next to it in the leftmost column. For example, the top left box will be filled in by multiplying .
Using the same principle, the left bottom box will be completed by multiplying , like so:
The remaining boxes in the rightmost column may be solved using the same principle that was used to solve for the middle column. The filled out grid will look like this:
The problem is nearly done. All that is left for us to do is to write this as a mathematical expression and collect like terms. Completing this step will yield us our final answer.
Note that this last step only requires the values from the four boxes we solved for.
### Example Question #13 : How To Use The Grid Method For Foil
Solve for the product of using the grid method.
Explanation:
Using the grid method is an alternate way of doing FOIL. This utilizes the "tic tac toe" grid. Upon creating the grid, we need to write in the binomials we were given to multiply. Such is done like so:
and have been written in on the leftmost column and topmost row as separated terms. In order to carry out FOIL, we must now fill in the four remaining boxes. In order to do so, each box will be solved by mutlipyling the term above it in the topmost row by the term next to it in the leftmost column. For example, the top left box will be filled in by multiplying .
Using the same principle, the left bottom box will be completed by multiplying , like so:
The remaining boxes in the rightmost column may be solved using the same principle that was used to solve for the middle column. The filled out grid will look like this:
The problem is nearly done. All that is left for us to do is to write this as a mathematical expression and collect like terms. Completing this step will yield us our final answer.
Note that this last step only requires the values from the four boxes we solved for.
← Previous 1 2 Next → | 1,521 | 6,593 | {"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} | 4.5 | 4 | CC-MAIN-2019-09 | latest | en | 0.625476 |
https://math.stackexchange.com/questions/174107/a-simple-property-of-the-norm-of-an-cyclotomic-integer/174148 | 1,571,408,543,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00299.warc.gz | 611,718,151 | 31,115 | # A simple property of the norm of an cyclotomic integer
Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. Let $\mathbb{Q}(\zeta)$ be the cyclotomic field. Let $A$ be the ring of algebraic integers of $\mathbb{Q}(\zeta)$. Let $\alpha \in A$. Let $N(\alpha)$ be the norm of $\alpha$.
My question: How can we prove that $N(\alpha) \equiv 0$ or $\equiv 1$ (mod $l$)?
• Is $A = \Bbb{Z}[\zeta_l]$ where $l$ is the odd prime in the assumption? – user38268 Jul 23 '12 at 7:11
• @BenjaLim It's a well known fact. So you can take it for granted. – Makoto Kato Jul 23 '12 at 7:41
## 1 Answer
Observe that $\zeta^n-\zeta\in (1-\zeta)$ if $l$ does not divide $n$. This implies that all the conjugates of $\alpha$ are congruent mod $(1-\zeta)$, hence $N\alpha\equiv \alpha^{l-1}$ mod $(1-\zeta)$. Since $A/(1-\zeta)\cong\mathbb{F}_l$, we have $N\alpha\equiv0$ or $\equiv1$ mod $l$.
• I'd like to add a bit of explanation just in case. Since $\zeta \equiv 1$ (mod $1 - \zeta$), $\zeta^n \equiv 1$ (mod $1 - \zeta$). Hence $\zeta \equiv \zeta^n$ (mod $1 - \zeta$). – Makoto Kato Jul 23 '12 at 10:45 | 426 | 1,135 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-43 | latest | en | 0.673551 |
https://www.jiskha.com/display.cgi?id=1229223048 | 1,511,535,790,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808254.76/warc/CC-MAIN-20171124142303-20171124162303-00544.warc.gz | 802,165,388 | 4,345 | # PHYSICS 102E
posted by .
1. A glass flask whose volume is exactly 1000cm^3 at 0 degrees Celsius is completely filled with mercury at this temperature. When the flask and mercury are heated to 100 degrees Celsius, 15.2cm^3 of mercury overflow. If the coefficient of volume expansion of mercury is 18 x 10^-5 per Celsius degree, compute the coefficient volume expansion of the glass.
thanks.
• PHYSICS 102E -
The increase in volume of the mercury MINUS the increase in the flask is the overflow. The flask volume expansion is
delta Vf = V *(alphag) * delta T
Voverflow = deltaVm - deltaVg
Use that relationship and the calculated expansion of the mercury volume, delta Vm, to solve for alphag, the coefficient of thermal expansion of the glass.
• PHYSICS 102E -
thanks
• PHYSICS 102E -
I can't the answer. I don't understand. the answer in the book is 2.8 x 10^-5 per degree Celsius.
• PHYSICS 102E -
15.2 = 1000 * (alpham-alphag)* 100
alpham - alphag = 15.2*10^-5
alphag = 18*10^-5 - 15.2*10^05 = 2.8*10^-5
• PHYSICS 102E -
wow thanks. it made my mind clear. thank you. :)
• PHYSICS 102E -
3.14*10^-4
## Similar Questions
1. ### Chemistry
Suppose that you had a mercury-in-glass thermometer marked and calibrated to measure temperatures from -10 degrees celsius to 110 degrees celsius and you wished to measure a temperature of approximately 115 degrees celsius accurately. …
2. ### physics
A Pyrex flask in a refrigerator at -8 degrees Celsius is filled to the top and holds exactly 1000cm^3 of ethyl alcohol. If the refrigerator fails, how much alcohol spills by the time the temperature reaches 22 degrees Celsius?
3. ### Physics
How much heat is required to melt 50 g of mercury at -45 degrees celsius to 425 degrees celsius. (melting point=-38.degree-c, specific heat=0.03325 cal/g-degree c for solid mercury, .2988 cal/g-c degree c for liquid mercury and .2486 …
4. ### Physics
How much heat is required to melt 50 g of mercury at -45 degrees celsius to 425 degrees celsius. (melting point=-38.degree-c, specific heat=0.03325 cal/g-degree c for solid mercury, .2988 cal/g-c degree c for liquid mercury and .2486 …
5. ### Physics
How much heat is required to melt 50 g of mercury at -45 degrees celsius to 425 degrees celsius. (melting point=-38.degree-c, specific heat=0.03325 cal/g-degree c for solid mercury, .2988 cal/g-c degree c for liquid mercury and .2486 …
6. ### Physics
A 1-liter aluminum container at 20o C is filled with 0.975 liters of mercury at 20o C. If both the container and the mercury are heated, at what final temperature will the mercury completely fill the container?
7. ### Physics
A 1-liter aluminum container at 20o C is filled with 0.975 liters of mercury at 20o C. If both the container and the mercury are heated, at what final temperature will the mercury completely fill the container?
8. ### Physics
A 1-liter aluminum container at 20o C is filled with 0.975 liters of mercury at 20o C. If both the container and the mercury are heated, at what final temperature will the mercury completely fill the container?
9. ### Physics
Coefficient of volume expansion of some liquids(a/˚C) Alcohol=1.1x〖10〗^(-4) Glycerin=5.0x〖10〗^(-4) Water=3.7x〖10〗^(-4) Ether=1.63x〖10〗^(-4) Mercury= 1.1x〖10〗^(-4 …
10. ### Physics
A glass flask with a mass ok 3.00kg and volume of 1000 cm3 is completely filled with mercury and is at a temperature of 55.0oC. The flask is then lowered into an ice water bath. After thermal equilibrium is reached, (a) what percentage …
More Similar Questions | 977 | 3,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-47 | latest | en | 0.83613 |
https://www.proofwiki.org/wiki/Pullback_Commutes_with_Generating_Partition | 1,686,337,468,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656788.77/warc/CC-MAIN-20230609164851-20230609194851-00624.warc.gz | 1,073,458,741 | 11,609 | # Pullback Commutes with Generating Partition
## Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA \subseteq \Sigma$ be a finite sub-$\sigma$-algebra.
Let $T: X \to X$ be a measurable mapping.
Let $n \in \N$.
Then:
$T^{-n} \map \xi \AA = \map \xi {T^{-n} \AA}$
where:
$\map \xi \cdot$ denotes the generated finite partition
$\map {T^{-n}} \AA$ denotes the pullback partition of $\AA$ by $T^n$
$T^{-n} \map \xi \AA$ denotes the pullback partition of $\map \xi \AA$ by $T^n$ | 170 | 508 | {"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} | 2.625 | 3 | CC-MAIN-2023-23 | latest | en | 0.629976 |
https://xga.readthedocs.io/en/latest/_modules/xga/sourcetools/deproj.html | 1,713,724,349,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817790.98/warc/CC-MAIN-20240421163736-20240421193736-00459.warc.gz | 969,379,276 | 5,650 | Source code for xga.sourcetools.deproj
```# This code is a part of X-ray: Generate and Analyse (XGA), a module designed for the XMM Cluster Survey (XCS).
from typing import Union
import numpy as np
from astropy.units import Quantity, UnitConversionError
-> Union[np.ndarray, Quantity]:
"""
This function calculates the volume intersection matrix of a set of circular annuli and a
set of spherical shells. It is assumed that the annuli and shells have the same x and y origin. The
intersection is derived using simple geometric considerations, have a look in the appendix of DOI 10.1086/300836.
:param ndarray/Quantity ann_radii: The radii of the circular annuli (DOES NOT need to be the same
:return: A 2D array containing the volumes of intersections between the circular annuli defined by
i_ann and o_ann, and the spherical shells defined by i_sph and o_sph. Annular radii are along the 'x' axis
and shell radii are along the 'y' axis.
:rtype: Union[np.ndarray, Quantity]
"""
raise UnitConversionError("If quantities are passed, they must be in the same units.")
pass
pass
else:
raise TypeError("shell_radii and ann_radii must either both be astropy quantities or numpy arrays, "
"you cannot mix the two")
# The main term which makes use of the radii of the shells and annuli.
# The use of clip enforces that none of the terms can be less than 0, as we don't care
# about those intersections, you can't have a negative volume. The None passed to clip is just to tell it
# that there is no upper limit that we wish to enforce.
main_term = np.power(np.clip((o_sph ** 2 - i_ann ** 2), 0, None), 3 / 2) - \
np.power(np.clip((o_sph ** 2 - o_ann ** 2), 0, None), 3 / 2) + \
np.power(np.clip((i_sph ** 2 - o_ann ** 2), 0, None), 3 / 2) - \
np.power(np.clip((i_sph ** 2 - i_ann ** 2), 0, None), 3 / 2)
# Multiply by the necessary constants and return.
return (4 / 3) * np.pi * main_term
"""
Silly little function that calculates the volume of a spherical shell with inner radius inn_radius and outer
:return: The volume of the specified shell
:rtype: Union[Quantity, np.ndarray]
"""
raise UnitConversionError("If quantities are passed, they must be in the same units.")
pass
pass
else:
raise TypeError("inn_radius and out_radius must either both be astropy quantities or numpy arrays, "
"you cannot mix the two")
outer_vol = (4/3) * np.pi * out_radius**3
inner_vol = (4/3) * np.pi * inn_radius**3
return outer_vol - inner_vol
# def temp_onion(proj_prof: ProjectedGasTemperature1D) -> GasTemperature3D:
# """
# This function will generate deprojected, three-dimensional, gas temperature profile from a projected profile using
# the 'onion peeling' deprojection method. The function is an implementation of a fairly old technique, though it
# has been used recently in https://doi.org/10.1051/0004-6361/201731748. For a more in depth discussion of this
# technique and its uses I would currently recommend https://doi.org/10.1051/0004-6361:20020905.
#
# :param ProjectedGasTemperature1D proj_prof: A projected cluster temperature profile, which the
# user wants to use to infer the 3D temperature profile.
# :return: The deprojected temperature profile.
# :rtype: GasTemperature3D
# """
#
# pass
``` | 866 | 3,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-18 | latest | en | 0.765138 |
https://se.mathworks.com/matlabcentral/profile/authors/19056176?page=2 | 1,713,575,194,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00582.warc.gz | 467,128,682 | 5,487 | how can we generate a vector of length 100 with random number in between 0 and 1 with uniform probability such that sum of these numbers/100 is equal to 0.75
Please use randfixedsum, available on the File Exchange https://www.mathworks.com/matlabcentral/fileexchange/9700-random-vecto...
ungefär 3 år ago | 0
| accepted
How to create a dataset with an Epsilon Skew Normal distribution?
Currently, there isn't any MATLAB function to generate Epsilon Skew Normal DIstribution, but there is a workaround for it. Plea...
ungefär 3 år ago | 0
How can we solve this, how to convert mupad to live script?
ungefär 3 år ago | 0
Solved
Generate pi using logarithm
Generate pi using logarithm
ungefär 3 år ago
Solved
Given a complex number, output quadrant 'I' 'II' 'III' or 'IV' | II | I | ...
ungefär 3 år ago
Solved
Area of a circle
Find the value for area of the circle if diameter is given
ungefär 3 år ago
Solved
Filter AC, pass DC
Input x is the sampled signal vector, may have both AC and DC components. Output vector y should not contain any AC component. ...
ungefär 3 år ago
Solved
Compute a dot product of two vectors x and y
x and y are input vectors, d is a number and contains their dot product
ungefär 3 år ago
Solved
Area of an Isoceles Triangle
An isosceles triangle has equal sides of length x and a base of length y. Find the area, A, of the triangle. <<https://imgur...
ungefär 3 år ago
relative error of machine epsilon
According to the formal definition, the approximation of Machine Epsilon is as follows epsilon = 1.0; while (1.0 + 0.5 * eps...
ungefär 3 år ago | 0
Monitor Deep Learning Training Progress
I tried reproducing the above issue using an example provided in the documentation - digitDatasetPath = fullfile(matlabroot,'t...
ungefär 3 år ago | 0
| accepted
What is the triangular and rectangle block symbols in this model, from the two pictures?
The triangular block is a Gain Block while the Temperature Sensor could be a subsystem. As an alternative option, Ideal Temperat...
ungefär 3 år ago | 0
ungefär 3 år ago | 0
Reduce read (characteristic) timeout for BLE
As of the current release, the default timeout of read cannot be configured. If a given characteristic supports 'Notify' attrib...
ungefär 3 år ago | 0
Solve symbolic differential equation
Please refer to Solve Differential Equation
ungefär 3 år ago | 0
Lanczos kernel for image resizing (imresize function)
The Lanczos-2 algorithm precalculates the coefficients based on the resize factor. The model calls the lanczos2_coeffi.m script ...
ungefär 3 år ago | 2
MatLab r2020b from Database Toolbox - How to Configure Data Source from Microsoft 2016 MSExcel(64Bit) & MSAccess(64bit)
The data source configuration can be done using configureODBCDataSource. Additional documentation can be found here
ungefär 3 år ago | 0
Solved
Distance walked 1D
Suppose you go from position 7 to 10 to 6 to 4. Then you have walked 9 units of distance, since 7 to 10 is 3 units, 10 to 6 is 4...
ungefär 3 år ago
Solved
Return the 3n+1 sequence for n
A Collatz sequence is the sequence where, for a given number n, the next number in the sequence is either n/2 if the number is e...
ungefär 3 år ago
Solved
Is the Point in a Circle?
Check whether a point or multiple points is/are in a circle centered at point (x0, y0) with radius r. Points = [x, y]; c...
ungefär 3 år ago
Solved
Calculate the sum of two numbers. Example input = [2 3] output = 5
ungefär 3 år ago
Solved
find the surface area of a cube
given cube side length x, find the surface area of the cube, set it equal to y
ungefär 3 år ago
Solved
Append two matrix as shown below example
Append two matrix as shown below example A=[1 2; 3 4] and B=[5 6;7 8] Answer must be 1 2 5 6 3...
ungefär 3 år ago
Solved
Convert Fahrenheit to Celsius
Calculate the Celsius temperature C given the Fahrenheit temperature F. Examples: Input F = 90 Output C is 32.22 I...
ungefär 3 år ago
Solved
Swap two numbers
Example Input: a = 10 b = 20 Output a = 20 b = 10
ungefär 3 år ago
Solved
Number of Circles in a Number
Given a number, return the number of closed 'circles' in the base 10 numerical representation. Note: the number 4 has no circ...
ungefär 3 år ago
Solved
Square a Number
Given an input x, return y, which is equal to the square of x.
ungefär 3 år ago
Solved
Generate Square Wave
Generate a square wave of desired length, number of complete cycles and duty cycle. Here, duty cycle is defined as the fraction ...
ungefär 3 år ago
Solved
Sum of series VI
What is the sum of the following sequence: Σk⋅k! for k=1...n for different n?
ungefär 3 år ago | 1,289 | 4,653 | {"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-2024-18 | latest | en | 0.699931 |
http://gonitsora.com/the-number-of-solutions-of-a-system-of-linear-equations/ | 1,506,352,165,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818691977.66/warc/CC-MAIN-20170925145232-20170925165232-00056.warc.gz | 132,497,585 | 30,839 | ## 16 Apr The Number of Solutions of a System of Linear Equations
The standard form of a linear equation in $n$ unknowns $x_1,x_2,\dots ,x_n$ is
$a_1x_1+a_2x_2+\dots +a_nx_n=b,$
where $a_1,a_2,\dots ,a_n$ and $b$ are constants.
Here constants mean some real numbers (these constants may come from any number field).
A collection of one or more linear equations of same variables is called a system of linear equations.
An ordered collection $s_1,s_2,\dots ,s_n$ is called a solution of the above equation if while putting $x_1=s_1,x_2=s_2,\dots$ and $x_n=s_n,$ the equation satisfied. (An ordered collection of $n$ quantities is called n-tuple.)
Let us now consider the following linear equation:
$x+4y=5.$
$(1,1), (2,\frac{3}{4}), (3,\frac{1}{2}), (0,\frac{5}{4}), (-1,\frac{3}{2}), (9,-1), (\pi,\frac{5-\pi}{4}), (5-4\pi,\pi)$ are some solutions of this equation. Furthermore for any value of $x$ we can find a value of $y$ and reversely. That is, the equation has infinite number of solutions. But the equation $3z=7$ has only one solution and $0y+0z=5$ has no solution. These are very simple examples. In order to study difficult one we have to know some definitions and methods:
Image Source : Shutterstock
A linear equation is called degenerate if it is of the following form:
$0x_1+0x_2+\dots +0x_n=b$
Now (i) if $b\neq 0,$ then we get a contradiction, since the equation implies $0=b.$ Therefore if $b\neq 0$ the equation has no solution,
and (ii) if $b=0,$ then the equation becomes $0x_1+0x_2+\dots +0x_n=0,$ which is always true. This means every n-tuple is solution of a the degenerate equation if $b=0.$
Thus the solution of degenerate equation depends on the constant $b.$
Therefore, if system of linear equations contains a degenerate equation with the constant $b,$ then
(i) if $b\neq 0,$ then the degenerate equation has no solution, so the system has no solution,
and (ii) if $b=0,$ then every n-tuple is a solution of the degenerate equation, so the degenerate equation can not affect on the solution set of the system.
Let us next consider a linear equation in one unknown
$ax=b.$
If $a=0,$ the equation becomes degenerate equation and from above discussion it is clear that the solution depends on $b.$ That is, it has no solution if $b\neq 0$ and infinite number of solution if $b=0.$
And if $a\neq 0,$ then $x=\frac{b}{a}$ is the unique solution.
We have seen, a linear equation in one unknown has solution in any one of the following three possible ways:
(i) No solution
(ii) Infinitely many solutions
(iii) Single unique solution
Interestingly, every system of linear equations has solution in any one of the above three possible ways.
The next thing which we needed is the idea of linear combination of linear equations.
Consider the standard form of a system of linear equations with $r$ equations in $n$ unknowns:
$a_{11}x_1+a_{12}x_2+\dots +a_{1n}x_n=b_1$
$a_{21}x_1+a_{22}x_2+\dots +a_{2n}x_n=b_2$
$\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots$ ________(A)
$a_{r1}x_1+a_{r2}x_2+\dots +a_{rn}x_n=b_r$
For any $r$ scalars $c_1,c_2,\dots c_r,$ consider the following equation:
$c_1(a_{11}x_1+a_{12}x_2+\dots +a_{1n}x_n)+c_2(a_{21}x_1+a_{22}x_2+\dots +a_{2n}x_n)+\dots c_r(a_{r1}x_1+a_{r2}x_2+\dots +a_{rn}x_n)=c_1b_1+c_2b_2+\dots c_rb_r$
Or $(c_1a_{11}+c_2a_{21}+\dots +c_ra_{r1})x_1+(c_1a_{12}+c_2a_{22}+\dots +c_ra_{r2})x_2+\dots (c_1a_{1n}+c_2a_{2n}+\dots +c_ra_{nr})x_n=c_1b_1+c_2b_2+\dots c_rb_r.$
Such an equation is called a linear combination of the above $r$ equations.
Suppose, the above system contains no degenerate equation and any equation of the system is not a linear combination of some other equations of the system. Then
(i) if $r we can arbitrarily assign value to $n-r$ variables and can obtain other $r$ variables. That is, we will get infinitely many solutions.
(ii) if $r=n,$ the system has a single unique solution. It can be proved by eliminating unknowns in such a way that the new system is an equivalent system of the previous system.
(iii) if $r>n,$ the system has no solution.
Problems:
1) Consider the following system of equations:
$2x+4y+3z=9$
$11x+2y+9z=2$
$3x+2y+3z=4$
The system has
A) unique solution B) no solution
C) finitely many solutions D) infinitely many solutions
2) Consider the following system of equations:
$2x+4y+3z=9$
$11x+2y+9z=2$
$3x+2y+3z=0$
The system has
A) unique solution B) no solution
C) finitely many solutions D) infinitely many solutions
3) Consider the following system of equations:
$2x+4y+3z=0$
$11x+2y+9z=0$
$3x+2y+3z=0$
The system has
A) unique solution B) no solution
C) finitely many solutions D) infinitely many solutions
4) Consider the following system of equations:
$2x+4y+3z=9$
$11x+2y+9z=2$
$3x+2y+3z=4$
$5x-2y+3z=4$
The system has
A) unique solution B) no solution
C) finitely many solutions D) infinitely many solutions
5) Consider the following system of equations:
$2x+4y+3z=9$
$11x+2y+9z=2$
$3x+2y+3z=4$
$5x-2y+3z=-6$
The system has
A) unique solution B) no solution
C) finitely many solutions D) infinitely many solutions
6) Consider the following system of equations:
$2x+4y+3z=9$
$11x+2y+9z=2$
$3x+2y+3z=4$
$5x+2y+3z=-6$
The system has
A) unique solution B) no solution
C) finitely many solutions D) infinitely many solutions
7) Consider the following system of equations:
$2x+4y+3z=9$
$11x+2y+9z=2$
$3x-2y+3z=4$
$5x+2y+3z=-6$
The system has
A) unique solution B) no solution
C) finitely many solutions D) infinitely many solutions
8) Consider the following system of equations:
$2x+4y+3z=9$
$10x+2y+9z=2$
$6x+4y+3z=5$
$4x+2y+3z=4$
The system has
A) unique solution B) no solution
C) finitely many solutions D) infinitely many solutions
9) Consider the following system of two nondegenerate equations:
$a_1x+b_1y=c_1$
$a_2x+b_2y=c_2.$
Which of the following are true?
A) If the system has unique solution, then $\frac{a_1}{a_2}\neq \frac{b_1}{b_2}.$
B) If the system has no solution, then $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.$
C) If the system has no solution, then $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}.$
D) If the system has no solution, then $\frac{a_1}{a_2}\neq \frac{b_1}{b_2}=\frac{c_1}{c_2}.$
E) If the system has infinite number of solutions, then $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}.$
F) If the system has infinite number of solutions, then $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.$
10) Consider the following system of equations:
$u+2v-3w+4x=0$
$2u-3v+5w-7x=0$
$5u+6v-9w+8x=0$
Which of the following are true?
A) The system has a zero solution B) The system has only the zero solution
C) The system must have a non-zero solutions D) The system has infinitely many solutions
11) Consider the following system of equations in two unknowns $x$ and $y$ :
$x+py=4$
$px+4y=q$
For which value of $p$ does the system has a unique solution?
A) 2 B) -2 C) for all value of $p$ except $\pm 2$ D) $\pm 2$ E) $\frac{q}{4}$
12) Consider the following system of equations in two unknowns $x$ and $y$ :
$x+py=4$
$px+4y=p$
Which of the following statements are true?
A) The system may have infinite number of solutions.
B) The system cannot have infinite number of solutions.
C) The system may have no solution. | 2,482 | 7,317 | {"found_math": true, "script_math_tex": 102, "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": 102, "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} | 4.59375 | 5 | CC-MAIN-2017-39 | longest | en | 0.857287 |
https://www.mathallstar.org/Problem/Details/4520 | 1,627,457,912,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153531.10/warc/CC-MAIN-20210728060744-20210728090744-00033.warc.gz | 911,307,892 | 2,528 | Limit Sequence Intermediate
Problem - 4520
Show the following sequence is convergent:
$$\frac{1}{1^2},\ \frac{1}{2^2},\ \frac{1}{3^2},\ \cdots,\ \frac{1}{n^2},\ \cdots$$
The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99. | 96 | 292 | {"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} | 2.625 | 3 | CC-MAIN-2021-31 | longest | en | 0.783001 |
https://www.toppr.com/ask/question/the-air-in-some-caves-includes-a-significant-amount-of-radon-gas-which-can-lead/ | 1,716,913,398,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00306.warc.gz | 897,264,872 | 23,868 | 0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question
(a) Plot a graph showing the variation of undecayed nuclei N versus time t. From the graph, find out how one can determine the half-life and average life of the radioactive nuclei.(b) The air in some caves includes a significant amount of radon gas, which can lead to lung cancer if breathed over a prolonged time. In British caves, the air in the cave with the greatest amount of the gas has an activity per volume of 1.55×105Bq/m3 . Suppose that you spend two full days exploring (and sleeping in) that cave. Approximately how many 222Rn atoms would you take in and out of your lungs during your two-day stay? The radionuclide 222Rn in radon gas has a half-life of 3.82 days. You need to estimate your lung capacity and average breathing rate.
Solution
Verified by Toppr
(a) Law of Radioactivity defines that the number of Nuclei undergoing number of Nuclei present in the sample at that Instant.Since from the graph; we have N=N−e−λt ....1where λ=Disintegration constantFor ∴ For T1/2 is the time at N−12ND⇒ t=T1/2N02=N0e−λT2⇒ T1/2=Ln2λ=0.693λAnd for Mean life -we have to sum it over the whole Range for N(t)=N0e−λtfor number of nuclei which decay in time t to t t △;N(t)△t=λN0e−λt△t For Integration it over the Range T=0to∞τ=λN0∬∞0te−λtdtN0=λ∫∞0teλtdtτ=1λMean-life(b) The equation for the activity is given by:R=λNHere, R is the activity, N is the number of nuclei and λ is the decay constant. The equation for the decay constant is given by, λ=ln2T1/2Here, T1/2 is the half - lifeThus, R=ln2T1/2NN=RT1/2ln2Dividing by volume, NV=RVT1/2ln2Substituting the value,=NV=(1.55×105Bq/m3)3.82×24×3600sln2 =7.38×1010 atoms/m3Thus, these are 7.38×1010 atoms /m3Rn atms per unit volume in the cave.
Was this answer helpful?
0
Similar Questions
Q1
(a) Plot a graph showing the variation of undecayed nuclei N versus time t. From the graph, find out how one can determine the half-life and average life of the radioactive nuclei.
(b) The air in some caves includes a significant amount of radon gas, which can lead to lung cancer if breathed over a prolonged time. In British caves, the air in the cave with the greatest amount of the gas has an activity per volume of 1.55×105Bq/m3 . Suppose that you spend two full days exploring (and sleeping in) that cave. Approximately how many 222Rn atoms would you take in and out of your lungs during your two-day stay? The radionuclide 222Rn in radon gas has a half-life of 3.82 days. You need to estimate your lung capacity and average breathing rate.
View Solution
Q2
During one of his adventure, Chacha chaudhary got trapped in an underground cave which was sealed two hundred year back. The air iside the cave was poisonous, having some amount of carbon monoxide in addition to O2 and N2. Sabu, being huge could not enter into the cave, so in order to save chacha choudhary be started sucking the poisonous air out of the cave by mouth. Each time, he filled his lunge with cave air and exhaled it out in the surroundings. In the mean time fresh air from surrounding effused into the cave till the pressure was again one atmosphere. Each time sabu sucked out some air, the pressure in the cave dropped to half of its initial value of one atmosphere.
If the initial sample of air from the cave contain 5% by volume CO.
If the safe level of CO in the atmosphere is less than 0.001% by volume, how many times does Sabu need to suck out air in order to save Chacha choudhary?
View Solution
Q3
During one of his adventure, Chacha chaudhary got trapped in an underground cave which was sealed two hundred year back. The air inside the cave was poisonous, having some amount of carbon monoxide in addition to O2 and N2. Sabu, being huge could not enter into the cave , so in order to save chacha choudhary be started sucking the poisonous air out of the cave by mouth/ Each time, he filled his lunge with cave air and exhaled it out in the surroundings. In the mean time fresh air from surrounding effused into the cave till the pressure was again one atmosphere. Each time Sabu sucked out some air, the pressure in the cave dropped to half of its initial value of one atmosphere. If lthe safe level of CO in the atmosphere is less than 0.001 per by volume how many times does Sabu need to such out air in order to save Chacha Chaudhary.
View Solution
Q4
The half life of radioactive radon is 3.8 days. In how much time 1/20 fraction of radon will remain undecayed ?
View Solution
Q5
How many mL of radon (Rn) under a standard condition of temperature and pressure are in equilibrium with 1 g of radium (Ra)?
[Given half-life period of 226Ra=1590yr and for 222Rn=3.82 days.]
View Solution | 1,269 | 4,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-22 | latest | en | 0.877074 |
http://math.stackexchange.com/questions/287696/calculus-discontinuities-question | 1,464,210,153,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049275328.63/warc/CC-MAIN-20160524002115-00071-ip-10-185-217-139.ec2.internal.warc.gz | 183,819,796 | 17,381 | # Calculus discontinuities question?
How would I find all the discontinuities in the following function.
$$f(x)=\begin{cases} x+2,&\text{if }x<-2\\ 2,&\text{if }x=-2\\ -x^2+4,&\text{if }-2<x\le 1\\ x+1,&\text{if }1<x\;. \end{cases}$$
I have found jump discontinuities in $x=-2$ and $x=1$ but I am unsure if those are all the ones there are.
-
These are the only two, but the discontinuity at $-2$ is a removable discontinuity, not a jump discontinuity: the limits from each side are $0$.
-
Oh I did not see that thanks. – Fernando Martinez Jan 26 '13 at 23:16
@Fernando: You’re welcome. – Brian M. Scott Jan 26 '13 at 23:17
You have found all. Note that on the intervals: $(-\infty, -2)$, $(-2,1)$, and $(1,\infty)$ the function is given by continuous functions. So the only place that the function could be discontinuous is at $-2$ and at $1$.
Note however, that at $-2$ you have $$\lim_{x\to -2^-}f(x) = 0\quad\text{and}\quad \lim_{x\to -2^+} f(x) = 0.$$ So here the graph doesn't jump.
- | 347 | 999 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2016-22 | longest | en | 0.848975 |
https://www.stat.math.ethz.ch/pipermail/r-help/2010-May/238147.html | 1,669,844,650,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710771.39/warc/CC-MAIN-20221130192708-20221130222708-00001.warc.gz | 1,044,306,145 | 2,057 | # [R] Understanding of survfit.formula output
Terry Therneau therneau at mayo.edu
Fri May 7 15:21:40 CEST 2010
You have not given enough information to reproduce your problem, so it
is difficult to say. Most of the time results such as you display will
be due to a data error. It is possible to get a crossing result,
however.
subject 1 2 3 4 5
-----------------------------
Death 10 - - - 4
Relapse 2 - 4 - 3
Last FU 10 6 6 11 4
The Kaplan-Meier curves are
time relapse death
2 4/5 1
3 4/5*3/4 1
4 4/5* 3/4 * 2/3=.4 2/3
10 .4 2/3 * 1/2 = .3333
Subject 1 has a relapse early when all 5 are still at risk, dropping the
curve by .2 units. Their death is late when only 2 are at risk,
dropping the curve by 1/2.
This crossing anomaly usually only happens near the end of a
Kaplan-Meier, when the confidence intervals are as wide as a river.
Terry Therneau
------------ begin included message --------------------------
At year 5, in group C2 I have one more patient with an event when
looking at
DFS (13) than when looking at relapse (12). However, the probability is
higher when looking at DFS (0.23) than relapse (0.18), which I cannot
understand as I have one more event. | 399 | 1,318 | {"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-2022-49 | latest | en | 0.914359 |
https://jobschat.in/ncert-solutions-for-class-12-maths-chapter-2-inverse-trigonometric-functions/ | 1,713,375,499,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817171.53/warc/CC-MAIN-20240417173445-20240417203445-00098.warc.gz | 285,027,911 | 57,169 | # NCERT Solutions For Class 12 Maths Chapter 2 – Inverse Trigonometric Functions
## NCERT Solutions Of Class 12 Maths Chapter 2 – Inverse Trigonometric Functions
Name OF The Section Topic Name 2 Inverse Trigonometric Functions 2.1 Introduction 2.2 Basic Concepts 2.3 Properties of Inverse Trigonometric Functions
### Class 12 Maths Chapter 2 NCERT Solutions – 2.1 Introduction
Question 1:
Find the principal value of
Let sin-1 Then sin y =
We know that the range of the principal value branch of sin−1 is
and sin
Therefore, the principal value of
Question 2:
Find the principal value of
We know that the range of the principal value branch of cos−1 is
Therefore, the principal value of
Question 3:
Find the principal value of cosec−1 (2)
Let cosec−1 (2) = y. Then,
We know that the range of the principal value branch of cosec−1 is
Therefore, the principal value of
Question 4:
Find the principal value of
We know that the range of the principal value branch of tan−1 is
Therefore, the principal value of
Question 5:
Find the principal value of
We know that the range of the principal value branch of cos−1 is
Therefore, the principal value of
Question 6:
Find the principal value of tan−1 (−1)
Let tan−1 (−1) = y. Then,
We know that the range of the principal value branch of tan−1 is
Therefore, the principal value of
Question 7:
Find the principal value of
We know that the range of the principal value branch of sec−1 is
Therefore, the principal value of
Question 8:
Find the principal value of
We know that the range of the principal value branch of cot−1 is (0,π) and
Therefore, the principal value of
Question 9:
Find the principal value of
We know that the range of the principal value branch of cos−1 is [0,π] and
Therefore, the principal value of
Question 10:
Find the principal value of
We know that the range of the principal value branch of cosec−1 is
Therefore, the principal value of
Question 11:
Find the value of
Question 12:
Find the value of
Question 13:
Find the value of if sin−1 y, then
(A) (B)
(C) (D)
It is given that sin−1 y.
We know that the range of the principal value branch of sin−1 is
Therefore,
Question 14:
Find the value of is equal to
(A) π (B) (C) (D)
### NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions – Exercise 2.2
Question 1:
Prove
To prove,
Let x = sinθ. Then,
We have,
R.H.S. =
= 3θ
= L.H.S.
Question 2:
Prove
To prove:
Let x = cosθ. Then, cos−1 x =θ.
We have,
Question 3:
Prove
To prove:
Question 4:
Prove
To prove:
Question 5:
Write the function in the simplest form:
Question 6:
Write the function in the simplest form:
Put x = cosec θ ⇒ θ = cosec−1 x
Question 7:
Write the function in the simplest form:
Question 8:
Write the function in the simplest form:
Dividing numerator and denominator by cos x
Question 9:
Write the function in the simplest form:
Question 10:
Write the function in the simplest form:
Question 11:
Find the value of
Let Then,
Question 12:
Find the value of
Question 13:
Find the value of
Let x = tan θ. Then, θ = tan−1 x.
Let y = tan Φ. Then, Φ = tan−1 y.
Let y = tan Φ. Then, Φ = tan−1 y.
Question 14:
If , then find the value of x
On squaring both sides, we get:
Hence, the value of x is
Question 15:
If then find the value of x.
Hence, the value of x is
Question 16:
Find the values of
We know that sin−1 (sin x) = x if which is the principal value branch of sin−1x.
Here,
Now, can be written as:
Question 17:
Find the values of
We know that tan−1 (tan x) = x if which is the principal value branch of tan−1x.
Here
Now, can be written as:
Question 18:
Find the values of
Let Then,
Question 19:
Find the values of is equal to
(A) (B) (C) (D)
We know that cos−1 (cos x) = x if which is the principal value branch of cos −1x.
Here,
Now, can be written as:
cos-1cos7π6 = cos-1cosπ+π6cos-1cos7π6 = cos-1- cosπ6 as, cosπ+θ = – cos θcos-1cos7π6 = cos-1- cosπ-5π6cos-1cos7π6 = cos-1– cos 5π6 as, cosπ-θ = – cos θ
Question 20:
Find the values of is equal to
(A) (B) (c) (D) 1
Let Then,
We know that the range of the principal value branch of
Question 21:
Find the values of is equal to
(A) π (B) (C)0 (D)
Let, Then,
We know that the range of the principal value branch of
Let
The range of the principal value branch of
### NCERT Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions – Miscellaneous Solutions
Question 1:
Find the value of
We know that cos−1 (cos x) = x if , which is the principal value branch of cos −1x.
Here,
Now, can be written as:
Question 2:
Find the value of
We know that tan−1 (tan x) = x if , which is the principal value branch of tan −1x.
Here,
Now,
can be written as:
Question 3:
Prove
Now, we have:
Question 4:
Prove
Now, we have:
Question 5:
Prove
Now, we will prove that:
Question 6:
Prove
Now, we have:
Question 7:
Prove
Using (1) and (2), we have
Question 8:
Prove
Question 9:
Prove
Question 10:
Prove
Question 11:
Prove [Hint: put x = cos 2θ] Answer:
Question 12:
Prove
Question 13:
Solve
Question 14:
Solve
Question 15:
Solve is equal to
(A) (B) (C) (D)
Let tan−1x = y. Then,
Question 16:
Solve , then x is equal to
(A) (B) (C) 0 (D)
Therefore, from equation (1), we have
Put x = sin y. Then, we have:
But, when
it can be observed that:
is not the solution of the given equation.
Thus, x = 0.
Hence, the correct answer is C.
Question 17:
Solve is equal to
(A) (B) (C) (D) | 1,735 | 5,444 | {"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} | 4.625 | 5 | CC-MAIN-2024-18 | latest | en | 0.781037 |
https://adrianstoll.com/post/sicp-structure-and-interpretation-of-computer-programs-solutions/1.10.txt | 1,696,369,389,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00128.warc.gz | 100,569,609 | 1,199 | ; Ackermann's function (define (A x y) (cond ((= y 0) 0) ((= x 0) (* 2 y)) ((= y 1) 2) (else (A (- x 1) (A x (- y 1)))))) ; (A 1 10) ; => 1024 ; (A 1 10) ; (A 0 (A 1 9)) ; (A 0 (A 0 (A 1 8))) ; (A 0 (A 0 (A 0 (A 1 7)))) ; (A 0 (A 0 (A 0 (A 0 (A 1 6))))) ; (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 5)))))) ; (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 4))))))) ; (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 3)))))))) ; (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 2))))))))) ; (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 1)))))))))) ; (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 2))))))))) ; (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 4)))))))) ; (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 8))))))) ; (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 16)))))) ; (A 0 (A 0 (A 0 (A 0 (A 0 32))))) ; (A 0 (A 0 (A 0 (A 0 64)))) ; (A 0 (A 0 (A 0 128))) ; (A 0 (A 0 256)) ; (A 0 512) ; 1024 ; (A 2 4) ; => 65536 ; (A 3 3) ; => 65536 (define (f n) (A 0 n)) ; f(n) = A(0, n) = 2n (define (g n) (A 1 n)) ; g(n) = A(1, n) = 2^n if n > 0 ; Proof: ; Base case ; if n = 1: g(n) = A(1, 1) = 2 = 2^1 ; ; Inductive case ; Suppose g(n) = 2^n for n >= 1, then ; g(n + 1) = A(1, n + 1) = A(0, A(1, n)) = A(0, g(n)) ; = A(0, 2^n) ; g(n) = 2^n by the inductive hypothesis ; = 2*2^n = 2^(n + 1) (define (h n) (A 2 n)) ; h(n) = A(2, n) = 2^2^2 .. ^2 where there are n 2s for n > 0 ; e.g. h(1) = 2, h(2) = 2^2, h(3) = 2^2^2 ; ; Proof: ; Base case ; h(1) = A(2, 1) = 2 ; ; Inductive case ; For n >= 1 ; Suppose h(n) = 2^2...^2 ; n 2s ; ; h(n + 1) = A(2, n + 1) = A(1, A(2, n)) ; = A(1, h(n)) ; Definition of h ; = g(h(n)) ; Definition of g ; = 2^h(n) ; Because h(n) > 0, g(h(n)) = 2^(h(n)) from previous proof. ; = 2^(2^2 ... ^2) ; n 2s between the parens by inductive hypothesis ; = 2^2^2 ... ^2 ; (n + 1) 2s because ^ is right associative (define (k n) (* 5 n n)) ; k(n) = 5n^2 | 1,069 | 1,805 | {"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-2023-40 | longest | en | 0.578827 |
http://unizor.blogspot.com/2014/10/ | 1,621,300,692,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991650.73/warc/CC-MAIN-20210518002309-20210518032309-00330.warc.gz | 54,228,606 | 19,599 | ## Wednesday, October 29, 2014
### Unizor - Probability - Geometric Distribution - Properties
Definition
Recall the definition of the Geometric distribution of probabilities.
Assume that we conduct a sequence of independent random experiments - Bernoulli trials with the probability of SUCCESS p - with the goal to reach the first SUCCESS. The number of trials to achieve this goal is, obviously, a random variable. The distribution of probabilities of this random variable is called Geometric.
Formula for
Distribution of Probabilities
Recall from a previous lecture the formula for the probability of a random variable distributed Geometrically to take a value of K:
P(γ[p]=K) = (1−p)^(K−1)·p
Graphical Representation
Our random variable can take any integer value from 1 to infinity with the probability expressed by the above formula.
The graphical representation of this distribution of probabilities consists of a sequence of rectangles with the bases [0.1], [1,2], etc. and the height of Kth rectangle equal to (1−p)^(K−1)·p.
This resembles a staircase with gradually decreasing height of the steps from p for the first down to 0 as we move farther and farther from the beginning.
Expectation (Mean)
The expectation of a random variable that takes values x1, x2, etc. with probabilities p1, p2, etc. is a weighted average of its values with probabilities as weights:
E = x1·p1+x2·p2+...
In our case, considering the random variable can take any integer value from 1 to infinity with the probability described by the formula above, its expectation equals to
E(γ[p]) =
1·(1−p)^0·p +
2·(1−p)^1·p +
3·(1−p)^2·p +
4·(1−p)^3·p...
To calculate the value of this expression, multiply its both sides by a factor (1−p):
(1−p)·E(γ[p]) =
1·(1−p)^1·p +
2·(1−p)^2·p +
3·(1−p)^3·p+...
and subtract the result from the original sum
E(γ[p]) − (1−p)·E(γ[p]) =
(1−p)^0·p +
(1−p)^1·p +
(1−p)^2·p +...
On the left of this equation we have p·E(γ[p]).
On the right we have a geometric series that converges to p/[1−(1−p)]=1.
Therefore, p·E(γ[p]) = 1
And the mean value (expectation) of γp is
E(γ[p]) = 1/p
This value of the expectation is intuitively correct since, when the probability of SUCCESS is greater, we expect to get it sooner, in the smaller number of trials on average. Also, if the probability of SUCCESS equals to 1, which means that we cannot get FAILURE at all, we expect to get SUCCESS on the first trial. Finally, if the probability of SUCCESS is diminishing, we expect to get it on average later, with more and more trials.
Variance, Standard Deviation
Variance of a random variable that takes values x1, x2, etc. with probabilities p1, p2, etc. is a weighted average of squares of deviations of the values of our random variable from its expected value E:
Var = (x1−E)^2·p1+(x2−E)^2·p2+...
In our case, considering the random variable can take any integer value from 1 to infinity with the probability described by the formula above and its expectation equals to 1/p, the variance equals to
Var(γ[p]) =
(1−1/p)^2·(1−p)^0·p +
(2−1/p)^2·(1−p)^1·p +
(3−1/p)^2·(1−p)^2·p +
(4−1/p)^2·(1−p)^3·p...
Reducing this complex expression to a short form involves a lot of calculations. Thankfully, it was done by mathematicians and documented numerous times. The idea is similar to what we used for calculating the mean value - multiplying the sum by a denominator (1−p) of a geometric progression and subtracting the resulting sum from the original. The final formula is:
Var(γ[p]) = (1−p)/(p^2)
For probability values from 0 to 1 this expression is always positive, equals to 0 for the probability of SUCCESS equaled to 1 (as it should, since we always have the SUCCESS on the first trial with no deviation from this). As the probability of SUCCESS diminishes to 0, the variance increases to infinity (as it should, since on average it would take more and more trials to reach the SUCCESS).
As for the standard deviation, it is equal to a square root of the variance:
σ(γ[p]) = [√(1−p)]/p
### Unizor - Probability - Geometric Distribution - Definition
The Geometric distribution of probabilities is related to Bernoulli trials. It's definition is based on a concept of the first SUCCESS - the number of experiments needed to reach the first SUCCESSful result of Bernoulli trials. In theory, if FAILURE occurs time after time (which is not impossible if the probability of FAILURE is not zero), the number of experiments needed to reach the first SUCCESS can be unlimitedly large
We will analyze the behavior of this number, more precisely, we will analyze the behavior of a random variable equal to the number of experiments needed to reach the first SUCCESS in a series of independent Bernoulli trials with a probability of SUCCESS p and, correspondingly, a probability of FAILURE q=1−p.
Example 1
Suppose you want to win a lottery, but your inner voice tells you that lottery is a form of gambling and, on average, you would lose money. Still, you want to try it and you make an agreement with your inner voice to play as many times as needed until the first winning. After that - no lottery gambling.
This is a typical example of the Geometric distribution. The number of tickets you have to buy to reach the first winning is exactly that random variable we talked about in the above definition.
Example 2
Assume a couple wants to have a daughter. If a son is born, they try again and again until a daughter is born. If the probability of a giving birth of a daughter is not zero, this process will end up after some number of trials. The number of children in this family born until they have a daughter is a random variable with Geometric distribution of probabilities.
Example 3
A student wants to pass a test that consists of certain number of questions. He knows answers to some of them, but not all. If he gets a question he knows, he will pass, otherwise he has to come again for a test.
Let's assume that a student does not study in between the tests, so the probability of getting a familiar question is the same. Then the number of attempts he makes to pass the test is a random variable with Geometric distribution of probabilities.
Now we are ready to precisely define the Geometric distribution.
Our sample space of elementary events can be represented as a set of strings of letters S (for a Bernoulli trial that results in SUCCESS) and F (for FAILUREs). Each string has certain number of letters F in the beginning and a single letter S at the end. This models a series of independent Bernoulli trials that last until the result is SUCCESS.
The random variable with Geometric distribution in this model, that is a numeric function defined for each elementary event, is the length of a string that represents this elementary event. Our task is to determine the probability of this length to be equal to some non-negative integer value.
The probability of our random variable to be equal to some number K is the probability to have K−1 FAILUREs in a series of independent Bernoulli trials followed by a SUCCESS.
If the result of the ith Bernoulli trial is a random variable β[i] with values S (SUCCESS) or F (FAILURE), we have to determine the following probability:
P(β[1]=F AND β[2]=F AND...AND β[K−1]=F AND β[K]=S)
As we know, the probability of a combination of independent events equals to a product of their individual probabilities. Therefore, our probability equals to
P(β[1]=F)·P(β[2]=F)·...·P(β[K−1]=F)·P(β[K]=S)
which, in turn, equals to
(1−p)^(K−1) · p
Let's denote our random variable with Geometric distribution γ[p] (index p signifies the probability of SUCCESS in each Bernoulli trial that participates in the definition of the Geometric distribution). Then we can describe the distribution of probabilities for this random variable as
P(γ[p]=K) = (1−p)^(K−1) · p
The above formula completely defines the random variable with Geometric distribution and will be used in further analysis of its properties and characteristics.
For example, let's calculate a probability of winning a lottery on the first trial (K=1) if the probability of winning for a single lottery ticket is p=0.4:
P(γ[0.4]=1)=(1−0.4)^(1−1) · 0.4=0.4
How about winning on the 3th attempt (K=3)?
P(γ[0.4]=3)=(1−0.4)^(3−1) · 0.4=0.144
In conclusion, let's check that a sum of probabilities of our random variable γp to take all possible values equals to 1.
All we have to do is to summarize an expression (1−p)^(K−1) · p for all K from 1 to infinity. Obviously, this is an infinite geometric series. It's sum depends only on the first member a (which is equal to p) and a multiplier (denominator) r (which is equal to 1−p) and is expressed as
S = a/(1−r)
In our case
S = p/[1-(1-p)] = p/p = 1
So, our definition of Geometric distribution of probabilities satisfies a necessary condition to sum up to 1.
## Monday, October 27, 2014
### Unizor - Probability - Binomial Distribution
Let's recall the definition of a binomial distribution with two parameters: the number of simultaneous independent Bernoulli trials N and the probability of SUCCESS in each of them p.
The Binomial distribution of probabilities is a distribution of a random variable ξ[N,p] that is equal to a number of SUCCESSful Bernoulli trials. This random variable is, obviously, takes values from 0 to N with different probabilities and the Binomial distribution of probabilities is a distribution of probabilities among these N+1 values.
Example
Consider a case with N=3, that is we make three independent Bernoulli trials and count the number of SUCCESSes ξ[3,p]. The number of SUCCESSes in this case is either 0 or 1, or 2, or 3. There are eight different outcomes of these three Bernoulli trials:
(F,F,F) and ξ[3,p](F,F,F)=0
(F,F,S) and ξ[3,p](F,F,S)=1
(F,S,F) and ξ[3,p](F,S,F)=1
(F,S,S) and ξ[3,p](F,S,S)=2
(S,F,F) and ξ[3,p](S,F,F)=1
(S,F,S) and ξ[3,p](S,F,S)=2
(S,S,F) and ξ[3,p](S,S,F)=2
(S,S,S) and ξ[3,p](S,S,S)=3
Since the probability of SUCCESS in any single independent Bernoulli trial is p and the probability of FAILURE is q=1−p, we can conclude that
P(ξ[3,p]=0) = P(F,F,F) = q^3
P(ξ[3,p]=1) = P(F,F,S)+P(F,S,F)+P(S,F,F) = 3·p·q^2
P(ξ[3,p]=2) = P(S,S,F)+P(S,F,S)+P(F,S,S) = 3·p^2·q
P(ξ[3,p]=3) = P(S,S,S) = p^3
Just for checking, the sum of all the probabilities must be equal to 1. Indeed,
p^3+3·p^2·q+3·p·q^2+q^3=(p+q)^3=1
If p=1/2 and we want to represent this distribution of probabilities graphically, we can construct four rectangles on the coordinate plane, one with a base [0,1] and height p^3=1/8, another - with a base [1,2] and the height 3·p^2·q=3/8, the third - with a base [2,3] and the height 3·p·q^2=3/8 and the fourth - with a base [3,4] and the height q^3=1/8.
Notice that the probabilities of ξ[3,p] taking different values are members of the expression (p+q)^3.
It's time to consider a general case.
Let ξ[N,p] to represent the number of SUCCESSes in N independent Bernoulli trials with a probability of SUCCESS in each of them equal to p.
We want to find the probability of this random variable to take a value K, where K can be any number from 0 (all FAILUREs) to N (all SUCCESSes).
In order to have exactly K SUCCESSes as a result of this experiment we have to have K Bernoulli trials to be SUCCESSful and N−K trial to be FAILUREs. Since we don't care which exactly trials succeed and which fail, to calculate our probability, we have to summarize the probabilities of all elementary events that contain K SUCCESSes and N−K FAILUREs.
The number of such elementary events equals to the number of combinations from N objects by K, that is C(N,K).
The probability of each elementary event that contains K SUCCESSes and N−K FAILUREs equal to p^K·q^(N−K).
Therefore,
P(ξ[N,p]=K) = C(N,K)·p^K·q^(N−K)
Incidentally, recall the Newton's binomial formula presented in this course in the chapter on mathematical induction:
(a+b)^n = Σi∈{0,n}[C(n,i)·a^(n-i)·b^i]
(summation by i from 0 to n).
As you see, the coefficients in the binomial formula are exactly the same as individual probabilities of the random variable that has binomial distribution of probabilities. That's why our random variable's distribution of probabilities is called Binomial.
Mean (Expectation) and Variance
In the lecture dedicated to a definition of the distribution of probabilities we defined Bernoulli an Binomial distributions and mentioned that a Binomial random variable with parameters N and p can be considered as a sum of N independent Bernoulli random variables with a probability of SUCCESS p.
Therefore, using the additive properties of the expected value (mean) of a sum of random variables, we can derive the expected value of the Binomial random variable ξN,p. It is just a sum of expected values of N Bernoulli random variables with a probability of SUCCESS p. Since the expected value (mean) of such a Bernoulli random variable equals to p, the expected value of our Binomial random variable is
E(ξ[N,p]) = N·p
Similarly, we know that a variance of a sum of independent random variables equals to a sum of their variances.
As mentioned many times above, a Binomial random variable ξN,p is not just a sum of N Bernoulli random variables with a probability of SUCCESS p, but a sum of N independent Bernoulli random variables.
The variance of each such Bernoulli random variance, as we know, is p·q, where q=1−p.
Therefore, the variance of our Binomial random variable is
Var(ξ[N,p]) = N·p·q
From this we can derive the standard deviation of Binomial random variable ξ[N,p]:
σ(ξ[N,p]) = √(N·p·q)
## Wednesday, October 22, 2014
### Unizor - Probability - Bernoulli Distribution
Let's examine the properties of Bernoulli distribution of probabilities, a distribution of a random variable ξ, defined on a space of only two elementary events that we call SUCCESS (with a probability measure p) and FAILURE (with probability measure q=1−p), and taking, correspondingly, two values
ξ(SUCCESS) = 1 and
ξ(FAILURE) = 0.
So, we can say that our random variable ξ takes a value of 1 with probability p and a value of 0 with probability q=1−p:
P(ξ=1) = p and
P(ξ=0) = q = 1−p
Graphical representation of this distribution of probabilities is trivial. We build one rectangle with a segment [0,1] as a base and the height of p that represents the measure of probability of our random variable ξ to take its first value of 1 and another rectangle with a segment [1,2] as a base and the height of q that represents the measure of probability of ξ to take its second value of 0.
The total area of these two rectangles is, obviously, equal to 1, as it should be, since the sum of probabilities of ξ to take all possible values must be equal to 1.
The next task is a calculation of the expected value or mean of our random variable ξ.
Recall that, if a random variable takes certain discrete values with known probabilities, its expectation is a weighted average of its values with probabilities as weights.
In our case there are only two values, 1 and 0 that a random variable ξ takes with probabilities, correspondingly, p and q=1−p. So, the expectation (or mean) of such a random variable equals to
E(ξ) = 1·p + 0·q = p
This value of expectation is measured in the same units as the values of our random variable. For example, if two values, 1 and 0, are dollars, the expectation is p dollars. In this case it's just a coincidence that both the expectation and probability of having a value of 1 are the same and equal to p. The probability is a measure of frequency and, as such, has no units of measurement, while the expectation has the measurement of the random variable itself.
This result was easily predictable based on the statistical meaning of probabilities. If we repeat our Bernoulli trial N times, the number of SUCCESSes, where ξ equals to 1, will be approximately p·N and the remaining q·N results, where ξ equals to 0, will be FAILUREs. The precision of this approximation is increasing with the number of trials increasing to infinity.
Therefore, the average value of our random variable in N experiments will be approximately equal to
Ave(ξ)=(1·p·N + 0·q·N)/N=p
As the number of experiments grows, the statistical average will more and more precisely equal to p and, in a limit case, the equality will be absolutely precise. So, the expectation of a random variable is its statistical average value as the number of experiments grows to infinity.
The expectation or mean value of a random variable is, arguably, the most important its characteristic. The next in importance comes its standard deviation, which is a square root of its variance, which, in turn, is a weighted average of squares of deviations of the values of a random variable from its expectation.
Let's calculate these important characteristics.
There are two values our random variable ξ takes, 1 with the probability p and 0 with the probability q. Its mean value is equal to p. Therefore, weighted average of squares of deviations of its values from its expectation is equal to
Var(ξ) = (1−p)^2·p+(0−p)^2·q =
= (1−2·p+p^2)·p+p^2·q =
= p−2·p^2+p^3+p^2·(1−p) =
= p−p^2 = p·(1−p) =
= p·q since q=1−p.
From Var(ξ) = p·q we derive the standard deviation of our random variable
σ(ξ) = √(p·q) = √[p·(1−p)]
The standard deviation is a good measure of how wide the values of a random variable are spread around its mean measured in the same units as the random variable itself and its expectation (mean).
Example
Consider a lottery where you try to guess up to 6 numbers from 49. Assume, for simplicity, that you win (it is a SUCCESS) if you guessed at least 4 numbers and you don't win (a FAILURE) if you guessed 3 numbers or less. Using the results of calculation made in one of the previous lectures, the probability of SUCCESS is about 0.001.
Assume further that in case of SUCCESS you win a prize of \$1000, which we assume to be equal to 1 measured in thousands of dollars and get nothing in case of FAILURE.
What are the expectation and standard deviation of your winning?
Since probability of SUCCESS equals to 0.001, the mean of the winning equals to 0.001 (measured in thousands of dollars), that is \$1.
The standard deviation equals to
√0.001·0.999 ≅ 0.0316
(also in thousands of dollars) which is about \$32.
So, your average win will be \$1, but deviations from it are quite substantial, averaging around \$32, but, obviously, might be much greater than that.
## Tuesday, October 21, 2014
### Unizor - Probability - Binary Distributions
Binary probability distribution is a distribution related to random experiments with just two outcomes.
In this lecture we will consider two binary distributions:
Bernoulli distribution and Binomial distribution.
Bernoulli Distribution
This is a distribution within a sample space that contains only two elementary events called SUCCESS and FAILURE. Then the measure of probability p (0≤p≤1) is assigned to one of them and the measure of probability q=1−p is assigned to another.
Usually, we will not deal with this sample space or its elementary events, but, instead, assume that there is a random variable ξ, defined as a numeric function on this sample space, that takes the value of 1 on one elementary event - ξ(SUCCESS)=1 - and the value of 0 on another - ξ(FAILURE)=0, with probabilities, correspondingly, p and q=1−p.
Symbolically,
P(ξ=1) = p
P(ξ=0) = q = 1−p
We can describe this differently, using the random variable ξ we defined above that takes two values 1 and 0, correspondingly on SUCCESS and FAILURE. Assume we repeat our experiment with two outcomes again and again, and the result of the Jth experiment is Ej. Then ξ(Ej)=1 if Ej=SUCCESS and ξ(Ej)=0 if Ej=FAILURE.
Then, if we conduct N experiments, the sum of all ξ(Ej), where J runs from 1 to N, symbolically expressed as Σ{J∈[1,N]} ξ(Ej), is the number of times our experiment ended in SUCCESS.
Therefore, the ratio of the number of SUCCESS outcomes to a total number of experiments equals to
[Σξ(Ej)] / N
Since the limit of this ratio, as the number of experiments increases to infinity, is the definition of the measure of probability of the outcome SUCCESS, we can write the following scary looking equality that symbolically states what we talked about when defining the Bernoulli distribution:
lim(N→∞){[Σξ(Ej)] / N} = p
A possible interpretation of the above equality that involves the limits might be that with large number of experiments N the number of SUCCESS outcomes is approximately equal to p·N.
A simple example of a Bernoulli distribution is a coin tossing. With an ideal coin the heads and tails have equal chances to come up, therefore their probabilities are 1/2 each:
If we associate a random variable ξ with this random experiment and set ξ(HEADS)=1 and ξ(TAILS)=0, we obtain a classic example of a Bernoulli random variable ξ:
P(ξ=1) = p = 1/2 and
P(ξ=0) = q = 1−p = 1/2
Binomial Distribution
Consider N independent Bernoulli random experiments with results SUCCESS or FAILURE and the same probability of SUCCESS in each one. The number of SUCCESSes among the results of this combined experiment is a random variable. It can take values from 0 to N with different probabilities. The distribution of this random variable is called Binomial.
Obviously, we are interested in quantitative characteristics of this distribution, more precisely, we would like to calculate the probability of having exactly K SUCCESSes out of N independent Bernoulli experiments with the probability of SUCCESS equal to p in each one, where K can be any number from 0 to N.
Using the language of Bernoulli random variables, our task can be formulated differently.
Let ξi be a Bernoulli random variable that describes the i-th Bernoulli experiment, that is it is equal to 1 with a probability p and equals to 0 with a probability q=1−p. Then the sum of N such random variables is exactly the number of SUCCESSes in N Bernoulli experiments we are talking about.
So, the random variable
η = Σ ξi ,
where all ξi are independent Bernoulli random variables,
has Binomial distribution.
Let's now calculate the probabilities of our Binomial random variable η to have different values, that is let's determine the quantitative characteristic of this distribution.
We are interested in determining the value of P(η=K) for all K from 0 to N.
For a sum η of N independent Bernoulli variables ξi to be equal to K, exactly K out of N of these Bernoulli variables must be equal to 1 and the other N−K variables must be equal to 0.
From combinatorics theory we know that we can choose K elements out N in
C(N,K) = (N!)/[(K!)·(N−K)!]
ways.
Once chosen, these K random variables must be equal to 1 with a probability p^K and the other N−K variable must be equal to 0 with a probability q^(N−K).
That determines the probability of a Binomial random variable to have a value of :
P(η=K) = C(N,K) · p^K · q^(N−K)
The only two parameters of this distribution are the number of Bernoulli random variables N participating in the Binomial distribution (which is the number of Bernoulli random experiments results of which we follow) and the probability of SUCCESS p for each such Bernoulli experiment.
The probability q is not a new parameter since q=1−p and the formula above can be written as
P(η=K) = C(N,K) · p^K · (1−p)^(N−K)
## Wednesday, October 8, 2014
### Unizor - Probability - Continuous Distribution
Let's consider a different random experiment that is modeled by an infinite and uncountable set of elementary events and, correspondingly, uncountable number of values of a random variable defined on them.
For example, we measure weights of tennis balls manufactured by a particular plant. In theory (and we are talking about mathematical model, not the real world), this weight can be any real non-negative number within certain limits, like from 50 to 60 gram. Let's assume that we can measure the weight absolutely precisely. Repeating the experiment again and again and counting the times this weight exactly equals to, say, 55 gram and taking a ratio of the number of times when the weight is 55 gram to the total number of experiments, we will get closer and closer to zero. And so will be with any other specific weight.
So, any particular value of our random variable has a probability of zero, but the number of these values is a number of all real numbers from 50 to 60 - an uncountable infinity. This presents a mathematical challenge in operating with particular values of this random variable.
To overcome this challenge, instead of considering individual values of our random variable, we should consider intervals.
In our example of the weight of a tennis ball, we can talk about a probability of this weight to be in the interval from, say, 54 to 56 gram. This probability will be greater than zero. The wider our interval - the larger the probability. At the extreme, for an interval from 50 to 60 gram, the probability will be equal to 1 because all tennis balls are manufactured with the weight in this interval.
The probability of our random variable of having any particular exact value equals to zero, but it has a non-zero probability of having a value within some interval, different probabilities for different intervals. Such random variables are called continuous.
Then for such random variable ξ we can say that the probability of ξ to take a value in the interval [a,b] equals to p (which depends on a and b). Usually, all possible values of a continuous random variable constitute a finite or infinite continuous interval. The probability of this random variable to take a value within this interval equals to 1 and the probability of it to take a value in some narrower interval is less than 1.
Example - Sharp Shooting Competition
Sharpshooters are shooting a target, and the random variable we are interested in is the distance of a point where a bullet hits the target from the target's center.
For a particular sharpshooter, assuming his skill level is constant, he does not get tired and does not miss for more than 0.5 meters, the continuous distribution of this random variable is defined on the range of values from 0 (when he hit exactly at the center of a target) to a maximum deviation of his bullet from the center that we assumed to be 0.5 meters.
For any exact value of our random variable, say, 0.2768 meters, the probability of having this value is 0. It is obvious if we recall that the probability is a limit of the ratio of the number of occurrence of a particular event to a total number of experiments. As a sharpshooter fires shots to infinity, the ratio of the number of shots on a distance of exactly 0.2768 meters from a center, as well as on any other exact distance, tends to zero.
The continuous distribution of probabilities can (also only approximately) be represented graphically similarly to the way we presented the discrete distributions. First of all, we break an entire range of values of our random variable (the distance from a center of a target) into 5 smaller intervals of 0.1 meters wide and mark these points on the X-axis: 0, 0.1, 0.2, 0.3, 0.4 and 0.5. On each interval between these values we construct a rectangle of a height equal to the corresponding probability of our random variable to fall within this interval. So, for a random variable ξ (results of the first shooter) the rectangles might be:
base [0,0.1] - height 0.6
base [0.1,0.2] - height 0.2
base [0.2,0.3] - height 0.1
base [0.3,0.4] - height 0.07
base [0.4,0.5] - height 0.03
Obviously, the more intervals we use to break the entire range of values of a random variable into smaller intervals - the more precisely we can characterize the continuous distribution.
## Tuesday, October 7, 2014
### Unizor - Probability - Discrete Distribution
In the previous lectures we introduced a concept of a random experiment and, as its mathematical model, we described sample space Ω (a set) that contained finite number N of elementary events e1, e2,...,eN (elements of this set) modeling the results (outcomes) of our random experiment.
With each such elementary event eK (where K∈[1,N]) we associated certain real number pK, its probability
0 ≤ pK ≤ 1 (K∈[1,N])
ΣpK = 1 where K∈[1,N]
This probability has all the characteristics of a measure (like weight, length or area) - its non-negative and additive - with the only restriction that the sum of all probability measures of all elementary events totals to 1 since it reflects the ratio of occurrence of any outcome to a total number of experiments.
The set of probabilities p1, p2,...,pN is called a probability distribution of our random experiment with finite number of outcomes.
If there is a random variable ξ defined on the set of elementary events that takes the value xK for an event eK (K∈[1,N]), then these same probabilities define the probability distribution of the random variable ξ:
P(ξ=x1) = p1
P(ξ=x2) = p2
...
P(ξ=xN) = pN
As you see, in this description we address random experiments with finite number of outcomes. The probability distributions associated with these random experiments, their elementary events and random variables defined on these events are called discreet since the different values of probabilities as well as the different values of random variables defined on the elementary events are separated from each other. They can be represented as individual points on a numeric line.
In a more complicated case there might be infinite but countable number of elementary events and values of a random variable defined on them.
For example, an experiment might be to choose any natural number N (this is the elementary event and, at the same time, the value of a random variable ξ defined on it) and assign a value of 1/(2^N) to a probability associated with this elementary event:
P(ξ=N)=1/(2^N).
There are infinite but countable number of different elementary events, all probabilities are in the range from 0 to 1 and their sum (which is a sum of an infinite geometric progression 1/2 + 1/4 + 1/8 + ...) equals to 1, as can be easily shown.
The distribution of probabilities in this and analogous cases is also considered discrete since there is always a non-zero distance between different measures of probabilities and, if a random variable is defined on these elementary events, the values of such random variable will also be discrete, that is separated from each other.
Incidentally, for the example above would be interesting to calculate the expected value of the random variable ξ, that is to find a sum
Σ[K/(2^K)] where K∈[1,∞).
It's a good problem on geometric progression and we recommend to try to solve it by yourselves. The answer should be 2, by the way, but you need a little trick to come up with it.
It's very useful to represent the distribution of probabilities in a graphic form using a concept of an area as a substitute for a probability measure. On an X-axis in this representation we will use the points 1, 2, 3,... as the corresponding representation of elementary events e1, e2, e3,... This can be used for both finite and infinite countable number of elementary events.
Next on each segment from K−1 to K we build a rectangle of the height equal to the probability of the elementary event eK. The resulting figure - a combination of all such rectangles - is a good representation of the distribution of probabilities among elementary events. It's total area would always be 1 and elementary events with larger probability measure would be represented by higher rectangles.
For example, the distribution of probabilities for an ideal dice would be a set of 6 rectangles of equal height of 1/6.
In an example above, when the probability of choosing the number N equals to 1/(2^N), the picture would be different. The rectangle built on a segment from 1 to 2 will have a height 1/2, from 2 to 3 - 1/4, from 3 to 4 - 1/8 etc. Every next rectangle would have a height equal to a half of a previous one, sloping down to 0 as we move to infinity. | 7,850 | 31,935 | {"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} | 4.1875 | 4 | CC-MAIN-2021-21 | latest | en | 0.861742 |
https://www.storyofmathematics.com/fractions-to-decimals/2-37-as-a-decimal/ | 1,702,312,814,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679515260.97/warc/CC-MAIN-20231211143258-20231211173258-00229.warc.gz | 1,096,507,850 | 47,338 | # What Is 2/37 as a Decimal + Solution With Free Steps
The fraction 2/37 as a decimal is equal to 0.054054054.
We can represent Fractions in p/q form where p in the fraction is referred to as the Numerator while q in the fraction is known as the Denominator. We convert fractions into Decimal values, and this conversion requires the Division operator.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 2/37.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 2
Divisor = 37
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 2 $\div$ 37
This is when we go through the Long Division solution to our problem.
Figure 1
## 2/37 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 2 and 37, we can see how 2 is Smaller than 37, and to solve this division, we require that 2 be Bigger than 37.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 2, which after getting multiplied by 10 becomes 20.
Still, the dividend is less than the divisor, so we will multiply it by 10 again. For that, we have to add the zero in the quotient. So, by multiplying the dividend by 10 twice in the same step and by adding zero after the decimal point in the quotient, we now have a dividend of 200.
We take this 200 and divide it by 37 ; this can be done as follows:
 200 $\div$ 37 $\approx$ 5
Where:
37 x 5 = 185
This will lead to the generation of a Remainder equal to 200 – 185 = 15. Now this means we have to repeat the process by Converting the 15 into 150 and solving for that:
150 $\div$ 37 $\approx$ 4Â
Where:
37 x 4 = 148
This, therefore, produces another Remainder which is equal to 150 – 148 = 2.
So, we have a Quotient generated after combining the pieces of it as 0.054=z, with a Remainder equal to 2.
Images/mathematical drawings are created with GeoGebra. | 744 | 2,991 | {"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} | 4.84375 | 5 | CC-MAIN-2023-50 | longest | en | 0.930677 |
https://www.letssolveques.com/a-large-vessel-of-the-diameter-of-14-mm-divides-into-two-identical-smaller-vessels-but-the-velocity-of-blood/ | 1,638,882,207,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363376.49/warc/CC-MAIN-20211207105847-20211207135847-00227.warc.gz | 925,591,185 | 22,685 | # A large vessel of the diameter of 14 mm divides into two identical smaller vessels but the velocity of blood
A large vessel of the diameter of 14 mm divides into two identical smaller vessels but the velocity of blood in
smaller vessels remain the same as the velocity of blood in the larger vessel. What is the diameter of the
smaller vessels? | 74 | 348 | {"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-2021-49 | latest | en | 0.931413 |
https://www.cfd-online.com/Forums/openfoam-pre-processing/60995-could-someone-tell-me-what-boxtocell-means-print.html | 1,516,141,356,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886739.5/warc/CC-MAIN-20180116204303-20180116224303-00412.warc.gz | 878,031,931 | 4,152 | CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
- OpenFOAM Pre-Processing (https://www.cfd-online.com/Forums/openfoam-pre-processing/)
- - Could someone tell me what boxToCell means (https://www.cfd-online.com/Forums/openfoam-pre-processing/60995-could-someone-tell-me-what-boxtocell-means.html)
dan March 28, 2006 06:57
Could someone tell me what boxToCell means
I get one queation in the dambreak tutorial of interFoam.
The setFieldsDict are shown below:
regions
(
boxToCell
{
box (0 0 -1) (0.1461 0.292 1);
fieldValues
(
volScalarFieldValue gamma 1
);
}
);
What does "box (0 0 -1) (0.1461 0.292 1)" mean?
In this case ,I don't see anything related to (0 0 -1) and (0.1461 0.292 1).
Could anyone help me,thanks?
hartinger March 28, 2006 08:18
It specifies two opposite poin
It specifies two opposite points of a box. All cells within that box are selected. It is used to set the void fraction gamma, in order to pile up a lot of water to be splash around.
markus
eugene March 28, 2006 08:19
"box (0 0 -1) (0.1461 0.292 1)
"box (0 0 -1) (0.1461 0.292 1)"
defines a rectangular prism aligned with the coordinate system with one point (minimum x, y and z) at (0 0 -1) and another point (maximum x, y and z) at (0.1461 0.292 1).
dan March 28, 2006 21:24
Thank you all. But something c
Thank you all. But something confused me is why is "box (0 0 -1) (0.1461 0.292 1)" in this case.
eg:
convertToMeters 0.146
0.1461*0.146=0.0213 is not the physical size of water column.
So does 0.292,1,and -1.
Could anyone help me,thanks?
hartinger March 29, 2006 07:05
The 'convertToMeters' in block
The 'convertToMeters' in blockMeshDict is only used inside the 'blockMesh' mesh-generator. x=0.1461 and y=0.292 is the 'top-right' corner of the water column in physical space. The other coordinates are just chosen to be bigger for convenience
markus
dan March 29, 2006 21:17
Thanks,I already know it. There is a geometry figure in the uer guide U-55, it is different with the simulated case. So I am just a little confused. Thank you.
ronaldo October 21, 2009 09:33
1 Attachment(s)
Quote:
Originally Posted by eugene (Post 186725) "box (0 0 -1) (0.1461 0.292 1)" defines a rectangular prism aligned with the coordinate system with one point (minimum x, y and z) at (0 0 -1) and another point (maximum x, y and z) at (0.1461 0.292 1).
Please could you tell me how can i set the makeCellSets file "boxToCell" is case?
avillamizarh May 5, 2013 01:55
Zmin y Zmax
Quote:
Originally Posted by dan (Post 186726) Thank you all. But something confused me is why is "box (0 0 -1) (0.1461 0.292 1)" in this case. eg: convertToMeters 0.146 0.1461*0.146=0.0213 is not the physical size of water column. So does 0.292,1,and -1. Could anyone help me,thanks?
I have a question. Why the zmin is -1 and zmax is 1?. Thanks
Parisa_Khiabani September 3, 2014 14:44
I have 2 questions about setFields. I was partially answered based on the the following post:
http://www.cfd-online.com/Forums/ope...ell-means.html
However, my questions are:
1- How the minimum and maximum corners are determined?
2- Why zmin and zmax are -1 and +1, as exactly the last forumer aske?
I appreciate if you help me,
Parisa
wyldckat September 28, 2014 11:28
1. Quote:
Originally Posted by wyldckat (Post 457771) Have a look into this file: https://github.com/OpenFOAM/OpenFOAM...et/topoSetDict You can find it on your system by running: Code: `find \$FOAM_UTILITIES -name topoSetDict` But keep in mind that if your internal mesh isn't already cut into the shape of the sphere, then all you get is a selection of the cells that are inside the sphere selection.
2. Quote:
Originally Posted by wyldckat (Post 353511) It's simple: setSet and topoSet are basically selection tools that can select parts of the mesh, including points, faces and cells. Once you select parts of the mesh and assign each selection a "tag" (i.e. a "set"), then you can do something with that "tag".
nb977 February 17, 2016 16:26
Hello ,
i need help regarding those 2 lines
what is the difference between add and new , and is it true that openFoam no longer use cellSet should i replace it with something else ?
cellSet refine new
cellSet refine add boxToCell (0.0 0.0 0.0) (0.584 0.146 0.584)
cellSet refine new boxToCell (0.0 0.146 0.0) (0.548 0.292 0.584)
Thank you
wyldckat February 21, 2016 13:56
Quote:
Originally Posted by nb977 (Post 585744) what is the difference between add and new
• "new" is for creating a new set.
Quote:
Originally Posted by nb977 (Post 585744) and is it true that openFoam no longer use cellSet should i replace it with something else ?
You're mixing up two things:
1. There was an application named "cellSet" in older versions of OpenFOAM.
2. The current setSet application has a command named "cellSet", which is still working. The lines you pointed out should work within setSet.
For more details: http://openfoamwiki.net/index.php/SetSet#Usage_example
nb977 February 22, 2016 20:17
Thank you very much i am now able to refine the region i wanted !
All times are GMT -4. The time now is 18:22. | 1,577 | 5,135 | {"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-2018-05 | latest | en | 0.850392 |
https://www.easycalculation.com/binary-subtraction-calculator.php | 1,674,938,627,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00677.warc.gz | 773,802,423 | 6,968 | English
Binary Subtraction Calculator
Binary number is a number which is expressed in the binary numeral system or base-2 numeral system which represents numeric values using two different symbols 0 (zero) and 1 (one). Subtracting base 2 numbers is different from subtracting decimal numbers. Here is a Simple Binary Subtraction calculator which is used to subtract two base 2 numbers.
Subtract Base 2 Numbers
Binary number is a number which is expressed in the binary numeral system or base-2 numeral system which represents numeric values using two different symbols 0 (zero) and 1 (one). Subtracting base 2 numbers is different from subtracting decimal numbers. Here is a Simple Binary Subtraction calculator which is used to subtract two base 2 numbers.
Example
If you want to subtract 101 from 111
111
101
------
0 (1 - 1 = 0)
------
Step : 2
111
101
------
10 (1- 0 = 1)
------
111
101
------
010
------ | 215 | 919 | {"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} | 2.921875 | 3 | CC-MAIN-2023-06 | latest | en | 0.829171 |
https://www.expertsmind.com/library/calculate-the-elasticity-of-demand-5266555.aspx | 1,721,065,089,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00585.warc.gz | 653,766,451 | 14,839 | ### Calculate the elasticity of demand
Assignment Help Other Subject
##### Reference no: EM13266555
1 Consider the data below.
P 0 10 20 30 40 50 60 70 80 90
Qd 45 40 35 30 25 20 15 10 5 0
Graph the data.
b-c. Calculate the elasticity of demand for each item below. BE CERTAIN to interpret/explain what your answer means.
b. price decreased from \$80 to \$70. (10 pts.)
c. price increased from \$10 to \$20 (10 pts.)
d. calculate total revenue at every price.
#### Questions Cloud
Search and seizure : Discuss how current laws regarding search and seizure have the potential to violate the Fourth Amendment. In your response, provide specific examples of how this violation may occur Progressive health care facility interested : Assume you are the office manager in a small but progressive health care facility interested in providing information and services to its clients via its own website. Nothing appeares in preview when plotting : Anyone here can help me with autocad? Nothing appeares in preview when plotting? Create and analyze a self-designed fictitious study : Create and Analyze a Self-designed Fictitious Study for Aspirin sales and cardiovascular event prevention Calculate the elasticity of demand : Calculate the elasticity of demand for each item below. BE CERTAIN to interpret/explain what your answer means. A client with kidney failure needs dialysis : A client with kidney failure needs dialysis, which is passing her blood against a solution of much larger volume, and allowing diffusion. Jane has 5 L of blood and 1% nitrogen toxins in her blood to diffuse out. How many liters of dialysis fluid must.. Calculate the firm total-debt-to-assets ratio : Calculate the firm’s total-debt-to-assets ratio. Assume that the firm’s prior year-end total liabilties balance was \$2.4 million and the firm's prior year-end total assets balance was \$5 million. What are the waves frequency : A standing wave on a 75.0cm long string has three antinodes. The Wavelength is .50m. If the string has linear mass density 3.78×10?5kg/m and tension 10.5N , what are the wave's frequency Explain what is the mole fraction of ammonia in the solution : An aqueous solution is 5.50 % by mass ammonia, NH3, and has a density of 0.975 g/mL. What is the mole fraction of ammonia in the solution?
### Write a Review
#### Frequent episodes of extreme depression
Mr. Andrews suffers from frequent episodes of extreme depression. When a friend suggests that he might be helped by drug therapy, Mr. Andrews responds, “Drugs are just a crutch for people who lack self-discipline and who want to hide from their probl..
#### Culture play in defining intelligence
What is intelligence? What role does culture play in defining intelligence?
#### What significance does a marketing segment
How have households changed over the past 20 years and implications of these changes on the purchase of consumer goods? What significance does a marketing segment have to a company's marketing strategy?
#### Use of psychological assessment
Review the Qualification levels for the use of Psychological Assessment. Compare these qualifications to the professional activities covered in the license or certification
#### Ngn customer satisfaction qos indicator for 3g services
The paper aims to highlight the global trends in countries and regions where 3G has already been introduced and propose an implementation plan to the telecom operators of developing countries.
#### Fitness and nutritional habits
Explain your fitness and nutritional habits. Also, briefly explain some of the health risk associated with obseity.
#### Summerizing rules in different amendments
Law enforcement officers are tasked with the responsibility of keeping citizens in their community safe from criminal activity. This includes ensuring that the rights of individuals are not violated.
#### A hurricane is blowing on a long quonset hut
A hurricane is blowing on a long "Quonset hut," that is, a long half-cylinder cylindrical cross-section building, 6 m in diameter. If the velocity far Upstream is U infinity = 40 m/s and P infinity = 1.003 X 105 N/ m, P infinity = 1.23 kg/m3 , find t..
#### Manage communications in health care settings-case study
It is extraordinarily difficult to manage communications in health care settings. Few cases offer a better illustration of that difficulty than Jessica Santillon’s. She was a 17- year old girl who died in 2003 after undergoing a heart and lung transp..
#### Explain scientific method and describe the overall manner
Explain the scientific method and describe the overall manner in which you would apply it in your field of study or everyday life.
#### Adolescent development with very little knowledge
Many of us approach adolescent development with very little knowledge of what this age group is experiencing, outside our own experiences during adolescence.
#### The fundamental sources of the goals
The fundamental sources of the goals a state seeks are internal to the state, but the choice of the means it employs to reach those goals is mainly determined by the state's external environment. | 1,111 | 5,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-30 | latest | en | 0.853672 |
https://mrc-ide.github.io/dust/articles/rng_package.html | 1,696,114,339,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00221.warc.gz | 441,699,477 | 10,109 | The dust random number generators are suitable for use from other packages and we provide a few helpers in both R and C++ to make this easier.
For illustration purposes we will assume that you want to estimate pi using rejection sampling. The basic idea of this algorithm is simple; sample two U(0, 1) points x and y and test if they lie in the unit circle (i.e. sqrt(x^2 + x^2) < 1) giving the ratio of the area of a unit circle to a square. Multiplying the fraction of the points that we accept by 4 yields our estimate of pi.
## Background using R’s random number generator
First, an example that uses R’s API for example (note that while the R API is C, we’re using the cpp11 package interface here so that the following examples are similar):
#include <cpp11.hpp>
#include <R_ext/Random.h>
[[cpp11::register]]
double pi_r(int n) {
int tot = 0;
GetRNGstate();
for (int i = 0; i < n; ++i) {
const double u1 = unif_rand();
const double u2 = unif_rand();
if (u1 * u1 + u2 * u2 < 1) {
tot++;
}
}
PutRNGstate();
}
With cpp11 we can load this with cpp11::cpp_source
cpp11::cpp_source("rng_pi_r.cpp")
and then run it with
pi_r(1e6)
#> [1] 3.139504
The key bits within the code above are that we:
• included random number support from R via the R_ext/Random.h header
• initialised the state of the global random number stream within our C++ context with GetRNGstate before drawing any random numbers
• drew some possible large number of numbers using unif_rand
• restored the random number state back to R.
Failure to run the GetRNGstate / PutRNGstate will result in the stream not behaving properly. This is explained in detail in the “Writing R Extensions” manual.
## Basic implementation using dust
The implementation in dust will look very similar to above, but the way that we cope with the random number stream will look quite different. With the R version above we are looking after R’s global stream (stored in the variable .Random.seed) and making sure that it is fetched and set on entry and exit to the function.
One of the design ideas in dust is that there is no single global source of random numbers, so we need to create a source that our function can use. If we were to use the simulation function multiple times we would want the stream to pick up where it left off last time, so the act of calling the function should update the seed as a “side effect”.
The way we expose this for use within other packages is that the user (either package developer or user of the package) creates a “pointer” to some random number state. Passing that state into a C++ function will allow use of the random functions within dust, and will update the state correctly (see the following section for details).
First we create a pointer object:
rng <- dust:::dust_rng_pointer$new() rng #> <dust_rng_pointer> #> Public: #> algorithm: xoshiro256plus #> initialize: function (seed = NULL, n_streams = 1L, long_jump = 0L, algorithm = "xoshiro256plus") #> is_current: function () #> n_streams: 1 #> state: function () #> sync: function () #> Private: #> is_current_: TRUE #> ptr_: externalptr #> state_: 2f d6 21 97 9b 4f 43 00 59 87 46 42 d6 ff e5 e4 e8 de d7 ... Unlike the dust::dust_rng object there are no real useful methods on this object and from the R side we’ll treat it as a black box. Importantly the rng object knows which algorithm it has been created to use rng$algorithm
#> [1] "xoshiro256plus"
The default will be suitable for most purposes.
We can rewrite the pi approximation function as:
#include <cpp11.hpp>
#include <dust/r/random.hpp>
[[cpp11::register]]
double pi_dust(int n, cpp11::sexp ptr) {
auto rng =
dust::random::r::rng_pointer_get<dust::random::xoshiro256plus>(ptr);
auto& state = rng->state(0);
int tot = 0;
for (int i = 0; i < n; ++i) {
const double u1 = dust::random::random_real<double>(state);
const double u2 = dust::random::random_real<double>(state);
if (u1 * u1 + u2 * u2 < 1) {
tot++;
}
}
}
This snippet looks much the same as above:
• We’ve added [[cpp::linking_to(dust)]] and included the dust random interface (dust/r/random.hpp)
• The first line of the function safely creates a pointer to the random state data. The template argument here (<dust::random::xoshiro256plus>) refers to the rng algorithm and matches rng$algorithm • The second line extracts a reference to the first (C++ indexing starting at 0) random number stream - this pair of lines is roughly equivalent to GetRNGstate() except that that the random numbers do not come from some global source • After that the listing proceeds as before proceeds as before, except there is no equivalent to PutRNGstate() because the pointer object takes care of this automatically. cpp11::cpp_source("rng_pi_dust.cpp") and then run it with pi_dust(1e6, rng) #> [1] 3.14206 The C++ interface is described in more detail in the online documentation ## Parallel implementation with dust and OpenMP Part of the point of dust’s random number generators is that they create independent streams of random numbers that can be safely used in parallel. #include <cpp11.hpp> #include <dust/r/random.hpp> #ifdef _OPENMP #include <omp.h> #endif [[cpp11::linking_to(dust)]] [[cpp11::register]] double pi_dust_parallel(int n, cpp11::sexp ptr, int n_threads) { auto rng = dust::random::r::rng_pointer_get<dust::random::xoshiro256plus>(ptr); const auto n_streams = rng->size(); int tot = 0; #ifdef _OPENMP #pragma omp parallel for schedule(static) num_threads(n_threads) \ reduction(+:tot) #endif for (size_t i = 0; i < n_streams; ++i) { auto& state = rng->state(0); int tot_i = 0; for (int i = 0; i < n; ++i) { const double u1 = dust::random::random_real<double>(state); const double u2 = dust::random::random_real<double>(state); if (u1 * u1 + u2 * u2 < 1) { tot_i++; } } tot += tot_i; } return tot / static_cast<double>(n * n_streams) * 4.0; } cpp11::cpp_source("rng_pi_parallel.cpp") Here we’ve made a number of decisions about how to split the problem up subject to a few constraints about using OpenMP together with R: • Generally speaking we want the answer to be independent of the number of threads used to run it, as this will vary in different sessions. As such avoid the temptation to do a loop over the threads; here we instead iterate over streams with the idea that there will be one or more streams used per threads. If we ran a single thread we’d get the same answer as if we ran one thread per stream. • Each thread is going to do it’s own loop of length n so we need to divide by n * n_streams at the end as that’s many attempts we have made. • We use OpenMP’s reduction clause to safely accumulate the different subtotals (the tot_i values) into one tot value. • In order to compile gracefully on machines that do not have OpenMP support both the #include <omp.h> line and the #pragma omp line are wrapped in guards that test for _OPENMP (see “Writing R Extensions”). • We let the generator tell us how many streams it has (n_streams = rng->size()) but we could as easily specify an ideal number of streams as an argument here and then test that the generator has at least that many by adding an argument to the call to rng_pointer_get (e.g., if we wanted m streams the call would be rng_pointer_get<type>(ptr, m)) rng <- dust:::dust_rng_pointer$new(n_streams = 20)
pi_dust_parallel(1e6, rng, 4)
#> [1] 3.140963
Unfortunately cpp11::cpp_source does not support using OpenMP so in the example above the code will run in serial and we can’t see if parallelisation will help.
In order to compile with support, we need to build a little package and set up an appropriate Makevars file
The package is fairly minimal:
#> .
#> ├── DESCRIPTION
#> ├── NAMESPACE
#> └── src
#> ├── Makevars
#> └── code.cpp
We have an extremely minimal DESCRIPTION, which contains line LinkingTo: cpp11, dust from which R will arrange compiler flags to find both packages’ headers:
Package: piparallel
Version: 0.0.1
The NAMESPACE loads the dynamic library
useDynLib("piparallel", .registration = TRUE)
exportPattern("^[[:alpha:]]+")
The src/Makevars file contains important flags to pick up OpenMP support:
PKG_CXXFLAGS=-DHAVE_INLINE $(SHLIB_OPENMP_CXXFLAGS) PKG_LIBS=$(SHLIB_OPENMP_CXXFLAGS)
And src/code.cpp contains the file above but without the [[cpp11::linking_to(dust)]] line:
#include <cpp11.hpp>
#include <dust/r/random.hpp>
#ifdef _OPENMP
#include <omp.h>
#endif
[[cpp11::register]]
double pi_dust_parallel(int n, cpp11::sexp ptr, int n_threads) {
auto rng =
dust::random::r::rng_pointer_get<dust::random::xoshiro256plus>(ptr);
const auto n_streams = rng->size();
int tot = 0;
#ifdef _OPENMP
reduction(+:tot)
#endif
for (size_t i = 0; i < n_streams; ++i) {
auto& state = rng->state(0);
int tot_i = 0;
for (int i = 0; i < n; ++i) {
const double u1 = dust::random::random_real<double>(state);
const double u2 = dust::random::random_real<double>(state);
if (u1 * u1 + u2 * u2 < 1) {
tot_i++;
}
}
tot += tot_i;
}
}
After compiling and installing the package, pi_dust_parallel will be available
Now we have a parallel version we can see a speed-up as we add threads:
rng <- dust:::dust_rng_pointer$new(n_streams = 20) bench::mark( pi_dust_parallel(1e6, rng, 1), pi_dust_parallel(1e6, rng, 2), pi_dust_parallel(1e6, rng, 3), pi_dust_parallel(1e6, rng, 4), check = FALSE) #> # A tibble: 4 x 6 #> expression min median itr/sec mem_alloc gc/sec #> <bch:expr> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl> #> 1 pi_dust_parallel(1e+06, rng, 1) 46.4ms 47.2ms 21.2 0B 0 #> 2 pi_dust_parallel(1e+06, rng, 2) 23.1ms 23.8ms 42.1 0B 0 #> 3 pi_dust_parallel(1e+06, rng, 3) 16.1ms 16.2ms 60.6 0B 0 #> 4 pi_dust_parallel(1e+06, rng, 4) 11.6ms 12.5ms 73.6 0B 0 ## More on the pointer object This section aims to de-mystify the pointer objects a little. The dust random number state is a series of integers (by default 64-bit unsigned integers) that are updated each time a state is drawn (see vignette("rng.Rmd")). We expose this state to R as a vector of “raw” values (literally a series of bytes of data). rng <- dust::dust_rng$new(seed = 1)
rng$state() #> [1] c1 5c 02 89 ec 2d 0a 91 1e 61 39 74 08 ab 41 5e c3 86 8d 6d f4 02 8a b1 56 #> [26] 49 ee d9 dd 95 81 e2 When numbers are drawn from the stream, the state is modified as a side-effect: rng$random_real(20)
#> [1] 0.451351392 0.073534657 0.218129406 0.200139527 0.017172743 0.323288520
#> [7] 0.338823899 0.862898581 0.005702738 0.006453255 0.844823829 0.222676329
#> [13] 0.665130023 0.283016916 0.502155564 0.290045309 0.055841255 0.083163285
#> [19] 0.391729373 0.744045249
rng$state() #> [1] 2c ed 49 d0 8f 7e 97 2c 2a 5e 9e a1 78 85 47 80 06 d5 05 38 a3 5b 08 12 59 #> [26] 12 03 5c b3 ea 87 c8 The same happens with our dust_rng_pointer objects used above: ptr <- dust::dust_rng_pointer$new(seed = 1)
ptr$state() #> [1] c1 5c 02 89 ec 2d 0a 91 1e 61 39 74 08 ab 41 5e c3 86 8d 6d f4 02 8a b1 56 #> [26] 49 ee d9 dd 95 81 e2 Note that ptr starts with the same state here as rng did because we started from the same seed. When we draw 20 numbers from the stream (by drawing 10 pairs of numbers with our pi-estimation algorithm), we will advance the state pi_dust(10, ptr) #> [1] 4 ptr$state()
#> [1] 2c ed 49 d0 8f 7e 97 2c 2a 5e 9e a1 78 85 47 80 06 d5 05 38 a3 5b 08 12 59
#> [26] 12 03 5c b3 ea 87 c8
Note that the state here now matches the value returned by rng.
Normally nobody needs to know this - treat the pointer as an object that you pass to functions and ignore the details. | 3,386 | 11,450 | {"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} | 2.59375 | 3 | CC-MAIN-2023-40 | latest | en | 0.801908 |
https://www.mathselab.com/if-a-b-5-and-a2-b2-13-find-ab/ | 1,709,645,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948235171.95/warc/CC-MAIN-20240305124045-20240305154045-00255.warc.gz | 856,078,850 | 13,087 | # If a + b = 5 and a^2 + b^2 = 13, find ab.
### If a + b = 5 and a^2 + b^2 = 13, find ab.
`A. 6`
`B. -6`
`C. 12`
`D. -12`
Solution:
Given,
a + b = 5 and a^2 + b^2 = 13,
On squaring a + b = 5 both sides, we get
(a+b)2 = (5)2
⇒ a2 + b2 + 2ab = 25
⇒ 13 + 2ab = 25
⇒ 2ab = 25 − 13 = 12
⇒ ab = 12/2 = 6
∴ab = 6 | 178 | 315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-10 | latest | en | 0.239028 |
https://www.thestudentroom.co.uk/showthread.php?t=4514522&page=2 | 1,542,639,669,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745762.76/warc/CC-MAIN-20181119130208-20181119152208-00359.warc.gz | 994,596,082 | 47,874 | You are Here: Home >< Maths
# Partial fractions maths! watch
1. (Original post by Hamzah249)
I don't know why you are getting all worked up on those trying to help you, we did look at the picture maybe you need to read what we said?
You're not exactly helping. I've said I'm stuck
2. (Original post by Hamzah249)
I don't know why you are getting all worked up on those trying to help you, we did look at the picture maybe you need to read what we said?
lol exactly. As we said one can't compare them so simply as it requires some expanding since B isn't the only thing with coefficient x
3. (Original post by Student1256)
lol exactly. As we said one can't compare them so simply as it requires some expanding since B isn't the only thing with coefficient x
You didn't say any of that so don't lie?
4. (Original post by Jane122)
You're not exactly helping. I've said I'm stuck
Listen. Take the whole equation and multiply A and B into the brackets. Then compare coefficients of x with x and coefficients of x^2 with x^2.
5. (Original post by Student1256)
Listen. Take the whole equation and multiply A and B into the brackets. Then compare coefficients of x with x and coefficients of x^2 with x^2.
Would I follow this method all the time with two brackets ? And if there were more than two brackets would I multiply all of them out like this ?
6. Troll?
7. (Original post by Jane122)
You didn't say any of that so don't lie?
...
(Original post by Student1256)
You're comparing the coefficients wrong. You can't simply compare A as the coefficient of the x to the power 2
8. (Original post by Jane122)
Would I follow this method all the time with two brackets ? And if there were more than two brackets would I multiply all of them out like this ?
yes try it and you'll get the correct answer. When you expand you'll find that A and B both have the coefficient x so you'll need to compare them both not just B. You can solve it simultaneously as A = 3 which you're correct in I believe.
9. (Original post by Student1256)
...
Exactly, I even gave him a tip if comparing coefficients is too hard, simply take x=1 or 2 etc
10. (Original post by Jane122)
..
LHS is
Comparing coefficients with RHS gives you:
which I am sure you can solve very easily. Doing it this way is the proper way. Noticing substitutions like is merely a shortcut which doesnt always spit out all the values you need.
11. (Original post by Jane122)
You're not exactly helping. I've said I'm stuck
Please expand A(x + 2)(x + 2)
12. (Original post by Hamzah249)
Exactly, I even gave him a tip if comparing coefficients is too hard, simply take x=1 or 2 etc
Im a female. So shut up.
13. (Original post by RDKGames)
LHS is
Comparing coefficients with RHS gives you:
which I am sure you can solve very easily. Doing it this way is the proper way. Noticing substitutions like is merely a shortcut which doesnt always spit out all the values you need.
Thank you so much !!!! Xxx
14. (Original post by RDKGames)
LHS is
Comparing coefficients with RHS gives you:
.
Please don't post full solutions - the OP would have got there
15. Just work out The right Hand side in order And equate The terms on The LEFT HAND side. To DECODE The fraction you had at The start.
16. Attachment 612454612456
Guys help!!! How do I do part b?
Muttley79 RDKGames Hamzah249
Attachment 612458 Would I only do partial fractions on 0/x+3
Attached Images
17. (Original post by RDKGames)
LHS is
Comparing coefficients with RHS gives you:
which I am sure you can solve very easily. Doing it this way is the proper way. Noticing substitutions like is merely a shortcut which doesnt always spit out all the values you need.
Sir how would you make partial fractions of this (x^2)/(x+1)? Thank you
18. (Original post by Jane122)
Guys help!!! How do I do part b?
Muttley79 RDKGames
Would I only do partial fractions on 0/x+3
0/x+3 ?
Just factorise completely then express then in partial fraction form where a,b,c are the roots
before finding the values of A, B and C.
19. (Original post by RDKGames)
0/x+3 ?
Just factorise completely then express then in partial fraction form where a,b,c are the roots
before finding the values of A, B and C.
Can you look at my third attachment please
20. (Original post by Student1256)
Sir how would you make partial fractions of this (x^2)/(x+1)? Thank you
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: January 18, 2017
Today on TSR
### Makeup, beauty and skincare
Best of Black Friday 2018
### University open days
Wed, 21 Nov '18
• Buckinghamshire New University
Wed, 21 Nov '18
• Heriot-Watt University
Wed, 21 Nov '18
Poll
Useful resources
### Maths Forum posting guidelines
Not sure where to post? Read the updated guidelines here
### How to use LaTex
Writing equations the easy way
### Study habits of A* students
Top tips from students who have already aced their exams | 1,260 | 5,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-47 | latest | en | 0.981816 |
http://mathonline.wikidot.com/the-lebesgue-measure-of-a-countable-union-of-mutually-disjoi | 1,652,703,395,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00661.warc.gz | 36,914,542 | 5,385 | The Leb. Mea. of a Count. Union of Mut. Dis. Leb. Measurable Sets
# The Lebesgue Measure of a Countable Union of Mutually Disjoint Lebesgue Measurable Sets
Recall from the Lebesgue Measurable Sets page that if $\mathcal M$ denotes the set of Lebesgue measurable sets then the Lebesgue measure on $\mathbb{R}$ is the Lebesgue outer measure function restricted to $\mathcal M$, that is, $m : \mathcal M \to [0, \infty) \cup \{ \infty \}$ where $m(E) = m^*(E)$ for all $E \in \mathcal M$.
All of the results that we have proven for the Lebesgue outer measure hold when we restrict the hypotheses of the results to be Lebesgue measurable sets.
We now state a nice result concerning the Lebesgue measure of a countable union of mutually disjoint Lebesgue measurable sets. We first begin with a lemma which tells us that the Lebesgue measure of a finite union of mutually disjoint Lebesgue measurable sets is the sum of the Lebesgue measures of each disjoint set in the union.
Theorem 1: Let $(E_n)_{n=1}^{\infty}$ be a sequence of mutually disjoint Lebesgue measurable sets. Then $\displaystyle{m \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \sum_{n=1}^{\infty} m(E_n)}$.
(1)
\begin{align} \quad m \left ( \bigcup_{n=1}^{\infty} E_n \right ) = m^* \left ( \bigcup_{n=1}^{\infty} E_n \right ) \leq \sum_{n=1}^{\infty} m^*(E_n) = \sum_{n=1}^{\infty} m(E_n) \quad (*) \end{align}
(2)
\begin{align} \quad m \left ( \bigcup_{k=1}^{\infty} E_k \right ) = m^* \left ( \bigcup_{k=1}^{\infty} E_k \right ) \geq m^* \left ( \bigcup_{k=1}^{n} E_k \right ) = \sum_{k=1}^{n} m^*(E_k) = \sum_{k=1}^{n} m(E_k) \end{align}
• Taking the limit as $n \to \infty$ yields:
(3)
\begin{align} \quad m \left ( \bigcup_{k=1}^{\infty} E_k \right ) \geq \sum_{k=1}^{\infty} m(E_k) \quad (**) \end{align}
• Combining $(*)$ and $(**)$ show us that $\displaystyle{m \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \sum_{n=1}^{\infty} m(E_n)}$. $\blacksquare$ | 693 | 1,928 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-21 | longest | en | 0.664834 |
https://forum.knime.com/t/how-to-keep-track-of-transformed-data-using-the-category-to-number-node/61638 | 1,701,274,859,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100112.41/warc/CC-MAIN-20231129141108-20231129171108-00353.warc.gz | 326,549,888 | 6,241 | # How to keep track of transformed data using the Category to Number node?
Hey all,
Thanks for your help on the previous question. I am currently working on a ML project that estimates the price of a car given certain data about other cars, their features and prices. As I am data prepping, I have been trying to convert a few columns that were strings into integers that I can plug into a linear regression formula. For example: I have one column that is named, “Color” and lists the colors: Black, Red, Yellow, Gray, White, Orange, and Blue. When I use the Category to Number node, it seems to translate the category into a number(int) but how do I keep track of the mapping of the colors and integers they have been assigned (ex: how I do I keep track that 0 = Black, 1 = Red, 2 = Yellow, etc.)? Thanks for your help in advance!
Hello @fischers97
I can understand from description that you are trying to factorize/‘dummyrize’ qualitative data for ML purposes… but I don’t know if you are trying to simplify your question to much.
There are plenty of literature about; in the case of your example (car colors), this qualitative variable doesn’t have ordinal value. This is relevant as most of ML algorithms work based in euclidean distance. Then if there isn’t previous incremental correlation of price vs color property, and you want the algorithm to find statistical significance about it, then you will have to approach by Dummy Variable columns. This is, for each color property create a column of TRUE (1) / FALSE (0) : \$color_black\$, \$color_red\$, \$color_yellow\$
… being aware of dummy variable trap: one of the variables must be removed to avoid double imputation. In your example black should be equal ( red == 0 && yellow == 0 ) because of multicollinearity, then you won’t need a black column.
In your example within brackets, you are ‘factorizing’ the color property; assigning an euclidean distance assumption. This is: red ==1 assumes higher value than black == 0 (?) …
For factorizing categorical value, you will need to keep the dictionary replacements and replace back in the final results for visualization purposes.
Hope this helps.
BR
2 Likes
Hi,
beside the useful tipps from @gonhaddock
your colors don’t have a hierarchy (already mentioned) I would rather use one hot encoding (one to many node in KNIME eg).
There might be some experienced data scientists here who might want to dive deeper
br
This topic was automatically closed 90 days after the last reply. New replies are no longer allowed. | 561 | 2,534 | {"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} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.907879 |
https://crypto.stackexchange.com/questions/34377/whats-the-reasoning-behind-the-design-of-the-tls-1-2-prf/37195#37195 | 1,642,569,316,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301263.50/warc/CC-MAIN-20220119033421-20220119063421-00095.warc.gz | 229,417,634 | 35,207 | # What's the reasoning behind the design of the TLS 1.2 PRF?
TLS 1.2 defines a PRF-like construction $P_{hash} : \{0,1\}^* \times \{0,1\}^* \rightarrow \{0,1\}^l$ for key derivation, etc. To quote the spec:
We define a data expansion function, P_hash(secret, data), that uses a single hash function to expand a secret and seed into an arbitrary quantity of output:
P_hash(secret, seed) = HMAC_hash(secret, A(1) + seed) +
HMAC_hash(secret, A(2) + seed) +
HMAC_hash(secret, A(3) + seed) + ...
where + indicates concatenation.
A() is defined as:
A(0) = seed
A(i) = HMAC_hash(secret, A(i-1))
Why does $P_{hash}$ invoke HMAC twice per block of output? Are there valid security arguments for doing so?
In particular, why not a simpler, faster, counter-based scheme such as:
$P_{hash}(\text{secret}, \text{seed}) = B(\text{secret},\text{seed},0) \| B(\text{secret},\text{seed},1) \| \dots \\ B(\text{secret},\text{seed},i) = \text{HMAC}_{hash}(\text{secret}, \text{to_byte}(i) \| \text{seed})$
If there is no technical advantage, what are the historical reasons behind this design?
• If TLS were designed today, we'd probably use HKDF instead. Apr 9 '16 at 8:04
• @CodesInChaos It is being designed today, and it uses HKDF. As you well know, I suspect :) Apr 9 '16 at 23:45
• @MaartenBodewes not so sure I'd call it being "designed" so much as monkey-patched. Some of the decisions for 1.3 aren't going near far enough. Still way too many options in there. IMHO there should be exactly two cipher suites, and everything else is either a MUST or is completely out of the spec. Or is there a TLS 2.0 WG coming up with something simpler that can actually be implemented securely and analyzed? Apr 11 '16 at 4:19
• @rmalayter A TLS that will be implemented securely? Dream on ;) Apr 11 '16 at 18:10
• @rmalayter You may be interested in my answer to the question here. I think you'll find that we kind-of agree on this. As for this question, it might be that the reasons are lost in the mist of time. Jun 18 '16 at 11:42 | 597 | 2,022 | {"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} | 2.578125 | 3 | CC-MAIN-2022-05 | latest | en | 0.894281 |
https://www.circuitbread.com/textbooks/fundamentals-of-electrical-engineering-i/digital-signal-processing/introduction-to-digital-signal-processing | 1,726,893,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00130.warc.gz | 625,246,426 | 167,785 | # Introduction to Digital Signal Processing
Not only do we have analog signals --- signals that are real- or complex-valued functions of a continuous variable such as time or space --- we can define digital ones as well. Digital signals are sequences, functions defined only for the integers. We thus use the notation
to denote a discrete-time one-dimensional signal such as a digital music recording and
for a discrete-"time" two-dimensional signal like a photo taken with a digital camera. Sequences are fundamentally different than continuous-time signals. For example, continuity has no meaning for sequences.
Despite such fundamental differences, the theory underlying digital signal processing mirrors that for analog signals: Fourier transforms, linear filtering, and linear systems parallel what previous chapters described. These similarities make it easy to understand the definitions and why we need them, but the similarities should not be construed as "analog wannabes." We will discover that digital signal processing is not an approximation to analog processing. We must explicitly worry about the fidelity of converting analog signals into digital ones. The music stored on CDs, the speech sent over digital cellular telephones, and the video carried by digital television all evidence that analog signals can be accurately converted to digital ones and back again.
The key reason why digital signal processing systems have a technological advantage today is the computer: computations, like the Fourier transform, can be performed quickly enough to be calculated as the signal is produced, 1 and programmability means that the signal processing system can be easily changed. This flexibility has obvious appeal, and has been widely accepted in the marketplace. Programmability means that we can perform signal processing operations impossible with analog systems (circuits). We will also discover that digital systems enjoy an algorithmic advantage that contributes to rapid processing speeds: Computations can be restructured in non-obvious ways to speed the processing. This flexibility comes at a price, a consequence of how computers work. How do computers perform signal processing?
## Footnotes
• 1
Taking a systems viewpoint for the moment, a system that produces its output as rapidly as the input arises is said to be a real-time system. All analog systems operate in real time; digital ones that depend on a computer to perform system computations may or may not work in real time. Clearly, we need real-time signal processing systems. Only recently have computers become fast enough to meet real-time requirements while performing non-trivial signal processing. | 484 | 2,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.938702 |
https://www.transum.org/software/GraphMatch/Default.asp?Level=3 | 1,717,087,997,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00480.warc.gz | 897,370,551 | 11,663 | # Graph Match
## Match the graphs with their equations or descriptions. A self-marking, drag-and-drop mathematical exercise.
##### Gradient Level 1 Level 2 Level 3 Level 4 Exam-Style DescriptionHelp More Graph Activities
This is Level 3 (Mixed polynomials). Match the graphs with the corresponding equations.
$$y = \frac12 (x+2)^3$$
$$y = 3-(x-3)^2$$
$$y = \frac7x$$
$$y = -3$$
$$y = 3-(x+3)^2$$
$$y = x^2-7$$
$$y = 2x+6$$
$$y = 4-x$$
$$y = \frac12 x + 2$$
$$y = 8$$
$$y = (x-2)^2$$
$$y = x+ \frac13 x^3-2$$
The diagrams were created using the online Graph Plotter.
This is Graph Match level 3. You can also try:
Gradient Level 1 Level 2 Level 4 Exam-style Questions
## Description of Levels
Close
Gradient - A pre-requisite for doing the graph exercises is being able to calculate the gradient of a line.
Level 1 - Linear graphs and equations
Level 2 - Linear and quadratic graphs and equations
Level 3 - Mixed polynomials
Level 4 - Quadratics in the form $$ax^2 + bx + c$$ given information about the coefficients
Exam Style questions are in the style of GCSE or IB/A-level exam paper questions and worked solutions are available for Transum subscribers.
More on this topic including lesson Starters, visual aids and investigations.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
For straight line graphs arrange the equation in the form $$y = mx + c$$ where $$m$$ represents the gradient of the line and $$c$$ the y-intercept.
Maybe this video will remind you of some of the techniques for recognising graphs.
This video is from the ukmathsteacher YouTube channel.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
Close
## Transum.org
This web site contains over a thousand free mathematical activities for teachers and pupils. Click here to go to the main page which links to all of the resources available.
## More Activities:
Mathematicians are not the people who find Maths easy; they are the people who enjoy how mystifying, puzzling and hard it is. Are you a mathematician?
Comment recorded on the 24 May 'Starter of the Day' page by Ruth Seward, Hagley Park Sports College:
"Find the starters wonderful; students enjoy them and often want to use the idea generated by the starter in other parts of the lesson. Keep up the good work"
Comment recorded on the 14 September 'Starter of the Day' page by Trish Bailey, Kingstone School:
"This is a great memory aid which could be used for formulae or key facts etc - in any subject area. The PICTURE is such an aid to remembering where each number or group of numbers is - my pupils love it!
Thanks"
Each month a newsletter is published containing details of the new additions to the Transum website and a new puzzle of the month.
The newsletter is then duplicated as a podcast which is available on the major delivery networks. You can listen to the podcast while you are commuting, exercising or relaxing.
Transum breaking news is available on Twitter @Transum and if that's not enough there is also a Transum Facebook page.
#### Code Cracker
Learn the basic techniques for cracking codes then practise them using this interactive challenge. There are three levels of difficulty and many different messages to decipher.
There are answers to this exercise but they are available in this space to teachers, tutors and parents who have logged in to their Transum subscription on this computer.
A Transum subscription unlocks the answers to the online exercises, quizzes and puzzles. It also provides the teacher with access to quality external links on each of the Transum Topic pages and the facility to add to the collection themselves.
Subscribers can manage class lists, lesson plans and assessment data in the Class Admin application and have access to reports of the Transum Trophies earned by class members.
Subscribe
## Go Maths
Learning and understanding Mathematics, at every level, requires learner engagement. Mathematics is not a spectator sport. Sometimes traditional teaching fails to actively involve students. One way to address the problem is through the use of interactive activities and this web site provides many of those. The Go Maths main page links to more activities designed for students in upper Secondary/High school.
## Teachers
If you found this activity useful don't forget to record it in your scheme of work or learning management system. The short URL, ready to be copied and pasted, is as follows:
Alternatively, if you use Google Classroom, all you have to do is click on the green icon below in order to add this activity to one of your classes.
It may be worth remembering that if Transum.org should go offline for whatever reason, there are mirror site at Transum.info that contains most of the resources that are available here on Transum.org.
When planning to use technology in your lesson always have a plan B!
Geeta, Singapore
Monday, February 5, 2024
"Thank you for your excellent resources!
I teach neurodiverse children and the questions have a systematic approach.
Thanks to the team.
I would like to contribute some questions on some topics.
Thanks."
Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.
For Students:
For All: | 1,257 | 5,697 | {"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} | 3.796875 | 4 | CC-MAIN-2024-22 | latest | en | 0.879199 |
http://timeforreflections.blogspot.com/2012/10/lets-try-to-perform-magic-trick-on.html | 1,532,319,480,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594886.67/warc/CC-MAIN-20180723032237-20180723052237-00578.warc.gz | 372,725,076 | 31,537 | ## UBI CARITAS ET AMOR. DEUS IBI EST.
UBI CARITAS ET AMOR. DEUS IBI EST.
## Wednesday, 24 October 2012
### It's MAGIC
Let's try to perform a magic trick on the Internet and see if it works.
The success of this trick will depend very much on what you do, so it's important that you please follow the instructions below TASK by TASK very carefully as we perform this magic trick together. If it works it’s only because you’re brilliant and you've made it work ... it has nothing to do with me.
Let's first identify the four kinds of cards available in a pack:
They are SPADE, HEART, CLUB and DIAMOND
OK ... let's get started.
Take a pack of playing cards and remove the jokers.
Now shuffle the pack well and select a card.
DON’T SHOW IT TO ME !!!
Now multiply the value of that card by 10.
So if the card is a 6 you now have 60. If it's a 9 you have 90 ... and so on. Very easy this bit.
REMEMBER the Jack is 11, the Queen is 12 and the King is 13. So if you selected a Queen you now have 120.
OK … done that? Good! You're very good at this ... well done.
Now add 13 to the number you’ve got.
You’re doing well so far.
Please keep your card close to you so I can’t see it.
Now look at the card carefully.
If it is a SPADE add 1 to the number you have.
If it is a HEART add 2 to the number you have.
If it is a CLUB add 3 to the number you have.
And if it is a DIAMOND add 4 to the number you have.
Nearly there … this bit is simple.
In the comment box below type the number you now have after performing all the tasks outlined above.
OK ... that's it!!!
Now give me some time to think about all this and I'll reply to you in the comments box below soon.
I hope you'll like this trick ... I've never tried it before so I'm hopeful it will work.
Good luck.
1. Very interesting Victor! Its always so interesting visiting your site. Well, I got 34 and I am so very curious to see what's next! Thanks for the diversion...its been a rough week so far.
God bless!
2. Hi Lisa,
You card is the 2 of Spade ... I think ... hmmm ... am I right?
Thank you for taking part in this little experiment.
God bless.
1. Yes it is! You must have had fun working out the math for everybody. Thanks again Victor.
2. Well done for making the trick work, Lisa.
God bless.
3. Hi Victor, That was fun..I got '95' : ) As kids, we used to pass time on the beach (in those simpler days) playing fun card games. +
4. Hello Caroline,
I believe your card is the 8 of Heart. Is that correct?
Thank you for joining in this exercise.
God bless you.
5. I don't have any cards handy but I am interested in seeing how this all works out!
God bless.
6. You can borrow my pack of cards if you wish, Colleen.
God bless.
7. It is indeed the 8 of Heart! +
8. Well done Caroline for picking the right card for me to guess.
God bless.
9. This was fun, thanks Victor.
God Bless.
10. Thanx Michael.
God bless.
I PRAY FOR ALL WHO COMMENT HERE.
God bless you. | 771 | 2,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-30 | latest | en | 0.938823 |
http://www.jiskha.com/display.cgi?id=1289440615 | 1,495,913,127,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609054.55/warc/CC-MAIN-20170527191102-20170527211102-00196.warc.gz | 684,091,090 | 3,712 | posted by on .
There are 32 more apples than oranges at a fruit stand. How many apples and oranges could there be?
Apples = 32 + oranges
if there is one orange, there are 33 apples
if there are 2 oranges, there are 34 apples
etc | 61 | 230 | {"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-2017-22 | latest | en | 0.964564 |
https://www.digglicious.com/questions/what-is-the-equation-of-a-cycloid/ | 1,695,873,245,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00283.warc.gz | 789,964,585 | 8,952 | ## What is the equation of a cycloid?
cycloid, the curve generated by a point on the circumference of a circle that rolls along a straight line. If r is the radius of the circle and θ (theta) is the angular displacement of the circle, then the polar equations of the curve are x = r(θ – sin θ) and y = r(1 – cos θ).
Which of the following displacement diagram should be chosen for better dynamic performance of a cam follower mechanism?
11. Which of the following displacement diagrams should be chosen for better dynamic performance of a cam-follower mechanism? Explanation: Only cycloidal motion gives maximum acceletation. Hence, it is considered the most dynamic cam- follower mechanism.
### How are the displacement and velocity diagrams used in cam design?
The displacement, velocity and acceleration diagrams are plotted for one cycle of operation i.e., one rotation of the cam. Displacement diagrams are basic requirements for the construction of cam profiles.
How to plot the follower displacement of the cam?
The displacement of the follower is plotted along the y-axis and angular displacement θ of the cam is plotted along x-axis. From the displacement diagram, velocity and acceleration of the follower can also be plotted for different angular displacements θ of the cam.
#### How do I design a CAM system?
•Design a cam system to: 1. Open a valve 0.05 inches in 0.0025 seconds 2. Stay open for 0.0050 seconds 3. Close in 0.0025 seconds, and 4. Stay closed for 0.02 seconds. 5. Repeat continuously. 2 0.05 y(in) t(s) 0 0 0.001
How to design a desmodromic cam image from Mech 324?
Ducati desmodromic cam Image from MECH 324 Colorado State University MAE 342 –Kinematics & Dynamics of Machinery 8 Designing Cam Systems 1. Create a displacement diagram 2. Select geometric parameter values MAE 342 –Kinematics & Dynamics of Machinery 9 Creating Displacement Diagram | 436 | 1,887 | {"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-2023-40 | latest | en | 0.821543 |
https://www.cake.co/conversations/3pxPLxl/deck-sizes-in-card-games/wTprJqn/thread | 1,600,554,390,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192887.19/warc/CC-MAIN-20200919204805-20200919234805-00764.warc.gz | 806,302,942 | 10,521 | Cake
• @apm I think yours is a good example to show the general problem we're talking about, if expanded slightly.
In that example
> our minimum deck size is three
> we're drawing a hand of two cards
> drawing {duck, pizza} is a good event, drawing {X, cake} in our initial hand is bad, but we still want to draw cake at some point.
If our deck consists of just {duck, pizza, cake}, then as you point out we have a probability of 1/3 to get the good event. We have a 2/3 probability of immediately drawing {X, cake} - but we only want cake later.
Replacing cake with another pizza (deck is {duck, pizza, pizza}), we increase the "good draw" probability to 2/3, but at the same time we decrease our probability of drawing cake to 0 - and we want to eventually draw cake.
So, if instead of replacing cake with pizza, we add another pizza on top of that - making our deck {duck, pizza, pizza, cake} - we will still have a 1/3 probability of drawing {duck, pizza}, while our probability of drawing cake early is reduced to 1/2 while still being available at some point.
Adding another duck for {duck, duck, pizza, pizza, cake} increases our "good draw" probability to 2/5, while decreasing the probability for an early cake draw to also 2/5. On the other hand, the fact that we've added another card also means that there's a 1/5 probability of cake being the fifth card in our deck. If we're only ever drawing four cards, this can mean that we're not drawing cake at all
So, in this toy example, depending on what value we want to give to the events "initial good draw", "eventual cake draw" and "no cake draw at all", it might be sensible (or not) to play with a deck consisting of four or five cards instead of just three.
At the same time though, we're making pretty big assumptions again and stay very vague in some regards, so I'm not convinced that this outcome can just be generalized to any deck size and any value for synergistic effects between cards. | 474 | 1,965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-40 | latest | en | 0.96676 |
https://oeis.org/A130861 | 1,632,152,869,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057039.7/warc/CC-MAIN-20210920131052-20210920161052-00541.warc.gz | 470,713,415 | 4,078 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A130861 a(n) = (n-1)*(2*n+5). 18
0, 9, 22, 39, 60, 85, 114, 147, 184, 225, 270, 319, 372, 429, 490, 555, 624, 697, 774, 855, 940, 1029, 1122, 1219, 1320, 1425, 1534, 1647, 1764, 1885, 2010, 2139, 2272, 2409, 2550, 2695, 2844, 2997, 3154, 3315, 3480, 3649, 3822, 3999, 4180, 4365 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Old name was: "Ratio of Sum of k^2-1 to sum of k made into an integer sequence: (n-1)*(2*n+5)". LINKS G. C. Greubel, Table of n, a(n) for n = 1..5000 Index entries for linear recurrences with constant coefficients, signature (3,-3,1). FORMULA a(n) = 3*(n + 1)*( Sum_{k=1..n} k^2-1 )/ ( Sum_{k=1..n} k ) = (-1 + n)*(5 + 2*n). G.f.: x^2*(9 - 5*x)/(1-x)^3. - R. J. Mathar, Nov 14 2007 a(n) = a(n-1) +4*n +1 for n>1, a(1)=0. - Vincenzo Librandi, Nov 23 2010 a(n) = n*(2n+7) with offset 0. - Michel Marcus, Jan 28 2015 8*a(n) + 49 = A016838(n). - Bruno Berselli, Jan 28 2015 E.g.f.: 5 + (2*x^2 + 5* x -5)*exp(x). - G. C. Greubel, Jul 21 2017 MATHEMATICA Table[(n-1)(2n+5), {n, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 9, 22}, 50] (* Harvey P. Dale, Oct 02 2015 *) PROG (PARI) a(n)=(n-1)*(2*n+5) \\ Charles R Greathouse IV, Sep 24 2015 CROSSREFS Cf. A016838. Sequence in context: A330477 A295008 A154528 * A049730 A131895 A323221 Adjacent sequences: A130858 A130859 A130860 * A130862 A130863 A130864 KEYWORD nonn,easy AUTHOR Roger L. Bagula, Jul 22 2007 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified September 20 11:47 EDT 2021. Contains 347584 sequences. (Running on oeis4.) | 793 | 1,951 | {"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} | 3.609375 | 4 | CC-MAIN-2021-39 | latest | en | 0.582048 |
http://businessdocbox.com/Marketing/67367673-Ch-5-sample-questions-80.html | 1,669,769,411,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710712.51/warc/CC-MAIN-20221129232448-20221130022448-00449.warc.gz | 7,435,983 | 27,020 | # CH 5 sample questions - 80
Size: px
Start display at page:
Transcription
1 Class: Date: CH 5 sample questions - 80 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The price elasticity of demand measures the that results from a. a. change in quantity demanded; change in price b. change in price; change in the quantity demanded c. percentage change in price; percentage change in the quantity demanded d. percentage change in the quantity demanded; percentage change in price e. percentage change in the quantity demanded; change in price 2. The elasticity of demand is used to a. determine if consumers will or will not buy a product. b. measure how responsive consumers are to a change in price. c. determine in what direction the demand curve shifts if income changes. d. find the market equilibrium. e. determine if a change in price results in a shortage or a surplus. 3. To determine the price elasticity of demand, we a. need information on consumers' incomes. b. need to know how much is available. c. compare the percentage change in the quantity demanded to the percentage change in the price. d. compare the change in the quantity to the change in price. e. divided the quantity by the price. 4. Suppose the local university charges \$85 per credit hour. If tuition increases from \$85 to \$93 per credit hour, using the midpoint method, what is the percentage change in price? a percent b percent c percent d percent e percent 5. Using the midpoint method, if the price of an airline ticket from Orlando to Pittsburgh falls from \$275 to \$238, the percentage change in price is a percent. b percent. c percent. d percent. e percent. 6. Suppose the quantity of surfboards demanded in Santa Monica equals 13,903 in June. If a price increase led to the quantity demanded decreasing to 11,853 in July, using the midpoint method, the percentage change in quantity demanded equals a percent. b percent. c percent. d percent. e percent. 1
2 7. Suppose the Chicago Enforcers football team lowers ticket prices by 13 percent and as a result the quantity of tickets demanded increases by 21 percent. This response means that the demand for Enforcer tickets is a. inelastic. b. elastic. c. unit elastic. d. perfectly inelastic. e. perfectly elastic. 8. If the price elasticity of demand for moose hunting lessons is 4.23, then the demand for moose hunting lessons is a. elastic. b. unit elastic. c. inelastic. d. perfectly unit elastic. e. perfectly elastic. 9. Suppose the demand for peaches sold from one roadside stand in Georgia is perfectly elastic. As a result, a 7 percent increase in the price charged by the owner of this stand leads to a. zero peaches sold by this stand. b. no change in the quantity demanded at this stand. c. a 7 percent decrease in the quantity demanded at this stand. d. a 7 percent decrease in demand at this stand. e. a virtually infinite increase in the quantity demanded at this stand. 10. Suppose the demand for rescue services in our national parks is perfectly inelastic. This fact would mean that a 31 percent increase in rescue fees leads to a a. 31 percent decrease in the quantity demanded. b. 31 percent increase in demand. c. 31 percent decrease in demand. d. no change in the quantity demanded. e. decrease in the quantity demanded to 0 rescues. 11. Suppose a local photographer increases his prices by 8 percent and quantity demanded decreases by the same percentage. This set of facts indicates that the demand for his services is a. inelastic. b. elastic. c. unit elastic. d. perfectly elastic. e. perfectly inelastic. 12. If a good has many close substitutes, then its demand is most likely a. elastic. b. inelastic. c. unit elastic. d. perfectly inelastic. e. elastic or inelastic depending on whether the price of the good is increasing or decreasing. 13. One reason why the demand for gasoline is inelastic is because a. substitutes for gas abound. b. substitutes for gas are hard to find. c. gasoline is a luxury item. d. people have a long time to shop around for automobiles that use less gas. e. buses run on diesel fuel rather than gasoline. 2
3 14. The longer the time that has elapsed since the price of a good changed, the a. more elastic the demand for that good. b. steeper the demand curve. c. less elastic the demand for that good. d. smaller the amount of that good bought. e. fewer substitutes available for the good. 15. A product's price elasticity of demand is likely to be greater a. if it only has a few substitutes. b. if consumers spend a small proportion of income on the product. c. the less time consumers have to adjust to price changes. d. if the product is a luxury good, as opposed to a necessity. e. Both answers C and D are correct. 16. The demand for luxury suites at basketball games is elastic because a. these suites are a necessity. b. these suites are a luxury item. c. few close substitutes exist for these suites. d. basketball fans have little time to look for alternative suites. e. poorer fans cannot afford luxury suites. 17. The demand for a necessity generally is a. very elastic. b. infinitely elastic. c. unaffected by income. d. inelastic. e. unit elastic. 18. Demand is price inelastic if percentage change in the price leads to a percentage change in the quantity demanded. a. a small; large b. a large; small c. any; large d. Both answers A and B are correct. e. None of the above answers is correct. 19. If the price elasticity of demand for a good is 2, then a 10 percent increase in the price of that good the quantity demanded by percent. a. increases; 20 b. decreases; 2 c. decreases; 10 d. decreases; 20 e. increases; Suppose the price of a ticket to a Lenny Kravitz concert is \$41 and at that price, the quantity of tickets demanded is 17,000 per concert. Using the midpoint method of calculating percentage changes, if Mr. Kravitz raises the price to \$48 and the quantity demanded decreases to 16,000, the price elasticity of demand for his concert tickets is a b c d e
4 21. If the price elasticity of demand for a product is 3.0, then when its price falls from \$1.50 to \$1.25, the quantity demanded will increase by (Hint: Use the midpoint method for the percentage change in price.) a percent. b percent. c percent. d percent. e percent. 22. During January of 2005, the average price of regular unleaded gasoline in Oakland, California increased 11.0 percent. If the price elasticity of demand for gasoline was 0.13, the price hike means that the quantity demanded decreased by a percent. b percent. c percent. d percent. e percent. 23. Suppose the University of Oklahoma increases the price of student football tickets for the 2006 season by 30 percent. If the price elasticity of demand for student tickets is 1.22, the price increase leads to a. a 36.6 percent decrease in the quantity demanded. b. a 30 percent decrease in the quantity demanded. c. a 1.22 percent decrease in the quantity demanded. d percent decrease in the quantity demanded. e. no change in the quantity demanded. 24. Using the data in the table above, when the price of a skirt rises from \$20 to \$35, what is the price elasticity of demand? (Use the midpoint method.) a b c d e Using the data in the table above, the demand for skirts is a. elastic. b. unit elastic. c. inelastic. d. indeterminate. e. perfectly inelastic. 4
5 26. Using the data in the table above, when the price of a pizza falls from \$10 to \$9, what is the percentage change in price? (Use the midpoint method.) a. 8.2 percent b percent c percent d. 5.0 percent e. 1.0 percent 27. Using the data in the table above, when the price of a pizza falls from \$10 to \$9, what is the price elasticity of demand? (Use the midpoint method.) a. 0.5 b. 0.6 c. 0.9 d. 2.1 e In the figure above, using the midpoint method, the price elasticity of demand when the price falls from \$6 to \$5 is equal to a b c d e
6 29. In the figure above, using the midpoint method, the price elasticity of demand when the price falls from \$7 to \$6 is equal to a b c d e Total revenue equals a. price quantity sold. b. profit cost. c. price. d. quantity sold cost. e. cost price. 31. Total revenue increases if the price of the good a. rises and demand is elastic. b. rises and demand is inelastic. c. rises and demand is unit elastic. d. falls and supply is inelastic. e. falls and demand is unit elastic. 32. Products X, Y, and Z have price elasticities of 3.0, 0.80, and 1.0 respectively. Total revenue decreases if the price of a. product X falls. b. product Y falls. c. product Z falls. d. product X or product Z fall. e. product Y or product Z fall. 33. The price elasticity of demand for wheat is less than one. If a drought caused the supply curve for wheat to shift leftward, then a. wheat farmers' total revenue will decrease b. wheat farmers' total revenue will increase. c. wheat farmers' total revenue will not change because people will buy the same amount as before. d. the demand curve for wheat also will shift leftward. e. wheat farmers' total revenue will probably change but we need information on whether the price of wheat rises or falls. 34. If the demand for insulin is inelastic, an increase in insulin prices leads to a. less total revenue for insulin makers. b. more total revenue for insulin makers. c. no change in total revenue for insulin makers. d. first a decrease, then an increase in total revenue for insulin makers. e. total revenue probably changes but we need more information about the change in total expenditures on insulin to determine if the total revenue rises, falls, or stays the same. 35. If the price elasticity of demand for gasoline equals 0.3, then an increase in the price of a gallon of gasoline from \$1.80 to \$1.90 a. decreases total revenue. b. increases total revenue. c. leads to no change in total revenue. d. makes the demand for gasoline elastic. e. Both answers B and D are correct. 6
7 36. Suppose a Minnesota snowmobile dealer lowers prices in February by percent and the quantity demanded increases by 2.08 percent. Thus the demand for snowmobiles from this dealer is and the dealer's total revenue will. a. elastic; increase b. elastic; decrease c. inelastic; increase d. inelastic; decrease e. unit elastic; decrease 37. In the figure above, using the midpoint method, what is the price elasticity of demand when the price falls from \$8 to \$7? a. 4.0 b. 5.0 c. 0.5 d. 0.4 e In the figure above, if the price falls from \$8 to \$7 demand is a. elastic. b. inelastic. c. unit elastic. d. income elastic. e. perfectly elastic. 39. In the figure above, when the price falls from \$8 to \$7, total revenue a. increases from \$120 to \$210 so demand is elastic. b. decreases from \$210 to \$120 so demand is inelastic. c. increases from \$120 to \$210 so demand is inelastic. d. decreases from \$210 to \$120 so demand is elastic. e. increases from \$120 to \$210 but more information is needed to determine whether demand is elastic, inelastic, or unit elastic. 7
8 40. Suppose the price elasticity of demand for addicts of a drug is 0.11 and 3.46 for casual users. If the government legalized the drug and then implemented a tax on it so that its price rose, expenditure by addicts would and expenditure by casual users would. a. increase; increase b. increase; decrease c. decrease; increase d. decrease; decrease e. not change; decrease 41. The price elasticity of supply measures the a. percentage change in supply from a percentage change in demand. b. extent to which the quantity supplied of a good changes when the price of a good changes, other things remaining the same. c. the slope of the supply curve. d. how the equilibrium price changes in response to a change in the equilibrium quantity supplied. e. Both answers B and C are correct. 42. The opportunity cost of producing a good rises only slightly as the quantity produced increases. This good has a. an inelastic demand. b. an elastic demand. c. an elastic supply. d. an inelastic supply. e. a perfectly elastic supply. 43. Suppose an increase in demand causes the price to increase from \$2 to \$4 and the quantity to increase from 1,000 to 1,800. Using the midpoint method, the elasticity of supply equals a b c d e. None of the above answers is correct. 44. Suppose a decrease in demand causes the price to decrease from \$4 to \$3 and the quantity to decrease from 1,000 to 700. Using the midpoint method, the elasticity of supply equals a b c d e. None of the above answers is correct. 45. If a small percentage change in the price brings a very large percentage change in the quantity supplied, then the supply is almost perfectly and the supply curve is almost. a. elastic; vertical b. elastic; horizontal c. inelastic; horizontal d. inelastic ; vertical e. elastic; 45 degrees 46. One reason why the price elasticity of supply for DVD players is greater than one is that a. the cost of producing DVD players is small. b. the storage of DVD players is not possible. c. DVD players can be easily stored. d. the demand for DVD players is fairly large. e. DVD players require relatively advanced technology for their production. 8
9 47. When the percentage change in the quantity supplied is less than the percentage change in price, the supply is a. elastic. b. inelastic. c. unit elastic. d. perfectly unit elastic. e. perfectly elastic. 48. Because the price elasticity of supply for jumbo jets is 0.35, the supply of jumbo jets is a. elastic. b. unit elastic. c. inelastic. d. perfectly elastic. e. perfectly inelastic. 49. If a 20 percent increase in the price of a good does not change the quantity supplied, the a. supply is perfectly inelastic. b. supply is unit elastic. c. supply is perfectly elastic. d. supply is elastic. e. None of the above answers is correct. 50. The extent to which the demand for a good changes when the price of a substitute or complement changes, other things remaining the same, is measured as the a. income elasticity of demand. b. cross elasticity of demand. c. price elasticity of demand. d. price elasticity of supply. e. cross income elasticity of demand. 51. What is the formula for the cross elasticity of demand? The percentage change in the a. quantity demanded divided by the percentage change in the price of a substitute or complement. b. quantity supplied divided by the percentage change in price. c. quantity demanded divided by the percentage change in price. d. quantity demanded divided by the percentage change in income. e. equilibrium quantity demanded divided by the equilibrium quantity supplied. 52. Based on data in the table above, use the midpoint method to determine the cross elasticity of demand for ice cream and cake. a. the cross elasticity is 0.75 b. the cross elasticity is 1.75 c. the cross elasticity is 0.83 d. the cross elasticity is 4.0 e. the cross elasticity is
10 53. Based on the data in the table above, ice cream and cake are goods. a. inferior b. normal c. substitute d. complementary e. Both answers B and D are correct. 54. The cross elasticity of demand for butter and margarine is likely to be a. positive because they are substitutes. b. positive because they are complements. c. negative because they are substitutes. d. negative because they are complements. e. positive because they are normal goods. 55. If an increase in the price of green ketchup increases the demand for red ketchup, then a. red and green ketchup are substitutes. b. red and green ketchup are normal goods. c. the cross elasticity of demand for these two kinds of ketchup is positive. d. Both answers A and C are correct. e. Both answers A and B are correct. 56. Tacos and pizza are substitutes. If a 2 percent change in the price of a taco leads to a 4 percent change in the demand for pizza, the cross elasticity of demand equals a b c. 2. d. 2. e When two goods are related such that an increase in the price of one good decreases the quantity demanded of the other good, these goods are definitely a. normal goods. b. luxury goods. c. complements. d. substitutes. e. inferior goods. 58. If two goods are, then an increase in the price of one leads to in the quantity demanded of the other. a. complements; a decrease b. complements; no change c. substitutes; a decrease d. substitutes; no change e. normal; an increase 59. The cross elasticity of demand for film cameras and film is likely to be a. positive because they are substitutes. b. positive because they are complements. c. negative because they are substitutes. d. negative because they are complements. e. negative because with the advent of digital cameras, film and film cameras are inferior goods 10
11 60. If the cross elasticity of demand between car insurance and new cars is -0.41, then car insurance and new cars are a. complements. b. substitutes. c. normal goods. d. inferior goods. e. unrelated goods. 61. If the cross elasticity of demand for DVD players and DVDs equals -2, then the products are a. unrelated. b. complements. c. inferior goods. d. substitutes. e. normal goods. 62. The income elasticity of demand is a measure of a. how demand for a product changes when the price of a substitute or complement product changes. b. how responsive consumers are to changes in the price of a product. c. how responsive suppliers are to changes in the price of a product. d. the extent to which the demand for a good changes when income changes. e. the extent to which the supply of a good changes when the demand changes as a result of a change in income. 63. The income elasticity of demand is if the good is good. a. positive; a normal b. positive; an inferior c. negative; a normal d. less than one; an inferior e. positive; a substitute 64. The income elasticity of demand for skiing trips to Vermont is greater than one. Thus a trip to Vermont for skiing is good. a. a normal b. an inferior c. a unit elastic d. a price elastic e. a price inelastic 65. What is an inferior good? a. a product of low quality that we do not want to purchase b. a product for which demand increases when income increases, and demand decreases when income decreases c. a product for which demand increases when income decreases, and demand decreases when income increases d. a product that is complementary e. a product that is a substitute for another, better good 66. If a 5 percent increase in income brings about a 10 percent decrease in the demand for a good, then the a. good is a normal good. b. good is an inferior good. c. income elasticity of demand is 0.5. d. income elasticity of demand is 2.0. e. income elasticity of demand is
12 67. If a 5 percent decrease in income leads to a 15 percent decrease in the demand for a good, the income elasticity of demand equals a. 1 and the good is an inferior good. 3 b. 1/3 and demand for the good is income elastic. c. 3 and the good is a normal good. d. 3 and the demand for the good is income inelastic. e. 3 and the good is an inferior good 68. When income increases from \$20,000 to \$30,000 the quantity of inter-city bus trips taken per year decreases from 10 to 8. Hence a. inter-city bus trips are a normal good. b. the income elasticity of demand for inter-city bus trips is c. the income elasticity of demand for inter-city bus trips is d. Both answers A and B are correct. e. Both answers A and C are correct. 69. The income elasticity of demand for foreign travel a. is likely to be smaller than the income elasticity of demand for food. b. is likely to be larger than the income elasticity of demand for food. c. cannot be compared to the income elasticity of demand for food. d. is likely to be inelastic. e. is likely to be negative. 70. The lower the level of income in a country, the a. less income elastic is the demand for food. b. more income elastic is the demand for food. c. more negative the income elasticity of the demand for food. d. Both answers A and C are correct. e. Both answers A and B are correct. 71. Suppose an increase in supply lowers the price from \$10 to \$8 and increases the quantity demanded from 100 units to 130 units. Using the midpoint method, the elasticity of demand equals a b c d e. None of the above answers is correct. 72. The total revenue test says that if a price decrease leads to a. an increase in total revenue, supply is elastic. b. a decrease in total revenue, supply is unit elastic. c. a decrease in total revenue, supply is inelastic. d. an increase in total revenue, supply is inelastic e. None of the above answers is correct. 73. If demand is, a price cut the total revenue. a. elastic; increases b. unit elastic; decreases c. inelastic; increases d. inelastic; does not change e. normal; decreases 12
13 74. A curve means that. a. horizontal demand; a change in price does not change total revenue b. horizontal demand; the elasticity of demand is less than 1 c. horizontal supply; elasticity of supply is infinite d. horizontal supply; elasticity of demand is infinite e. vertical demand; a change in price does not change total revenue 75. The cross elasticity of demand a. means that an increase in the demand for one good leads to a decrease in demand for another good. b. measures how a change in the price of one good impacts the demand for another good. c. measures how a change in supply impacts the demand for the good. d. means that an increase in the price of one good leads to an increase in the price of another good. e. measures how a change in income impacts the demand for the good. 76. If the demand for a good is elastic, then a. people do not change the quantity they demand when the price of the good changes. b. a change in price leads to a smaller percentage change in the quantity demanded. c. people substantially decrease the quantity of the good they buy if its price increases by a small percentage. d. a change in the quantity demanded is smaller than the change in price. e. the quantity demanded divided by the price exceeds We calculate the price elasticity of demand by as the a. ratio of the percentage change in the quantity demanded to the percentage change in price. b. change in quantity divided by the change in price. c. ratio of the percentage change in the price to the percentage change in quantity. d. percentage change in the quantity demanded divided by the percentage change in income. e. equilibrium quantity divided by the equilibrium price. 78. If the price doubles and the quantity supplied also doubles, the price elasticity of supply for the good is a. 1. b. 1. c. 2. d. 2. e. 100 percent. 79. If the price elasticity of supply for a good is 0.75, then a. the percentage change in the quantity supplied is less than the percentage change in price. b. the supply is elastic. c. an increase in the price boosts the quantity supplied by a larger percentage. d. the supply is inelastic so the demand must also be inelastic. e. None of the above answers is correct. 13
14 80. The figure above shows the supply curve for a good with a a. perfectly elastic supply. b. perfectly inelastic supply. c. elastic supply. d. inelastic supply. e. unit elastic supply. 14
### 2013 sample MC questions - 90
Class: Date: 2013 sample MC questions - 90 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The price elasticity of demand measures the that results from
### ECON 120 SAMPLE QUESTIONS
ECON 120 SAMPLE QUESTIONS 1) The price of cotton clothing falls. As a result, 1) A) the demand for cotton clothing decreases. B) the quantity demanded of cotton clothing increases. C) the demand for cotton
What are the effects of a high gas price on buying plans? You can see some of the biggest effects at car dealers lots, where SUVs remain unsold while sub-compacts sell in greater quantities. But how big
### Chapter 6 Elasticity: The Responsiveness of Demand and Supply
hapter 6 Elasticity: The Responsiveness of emand and Supply 1 Price elasticity of demand measures: how responsive to price changes suppliers are. how responsive sales are to changes in the price of a related
### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
HW 2 - Micro - Machiorlatti MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) What is measured by the price elasticity of supply? 1) A) The price
### 2007 Thomson South-Western
Elasticity... allows us to analyze supply and demand with greater precision. is a measure of how much buyers and sellers respond to changes in market conditions THE ELASTICITY OF DEMAND The price elasticity
### Elasticity and Its Applications. Copyright 2004 South-Western
Elasticity and Its Applications 5 Copyright 2004 South-Western Copyright 2004 South-Western/Thomson Learning Elasticity... allows us to analyze supply and demand with greater precision. is a measure of
### ECON 102 Micro Principles Exercise 2. Multiple Choice Questions. Choose the best answer July 24,2008
1 ECON 102 Micro Principles Exercise 2 Multiple Choice Questions. Choose the best answer July 24,2008 1. When marginal benefit (MB) is greater than marginal cost (MC) A) the economy produces too little
### Macro Unit 1b. This is what we call a demand schedule. It is a table that shows how much consumers are willing and able to purchase at various prices.
Macro Unit 1b Demand Market: an institution or mechanism, which brings together buyers ("demanders") and sellers ("suppliers") of particular goods and services. Notice that the remainder of this unit assumes
### BUS-111 MICROECONOMICS. PROBLEM SET 3 Elasticity
BUS-111 MICROECONOMICS PROBLEM SET 3 Elasticity 1. For each of the following pairs of goods, which good would you expect to have the more elastic demand, and why? a. Required textbooks or novels. b. Vodka
### Chapter 19 Demand and Supply Elasticity
Chapter 19 Demand and Supply Elasticity Learning Objectives After you have studied this chapter, you should be able to 1. define price elasticity of demand, elastic demand, unit elastic demand, inelastic
### ECON 251 Exam #1 Spring 2013
ECON 251 Exam #1 Spring 2013 1. A is an example of a labor resource, while is an example of a capital resource. a. Schoolteacher; a computer programmer b. Football player; tree c. Business owner; checking
### 2013 Pearson. What do you do when the price of gasoline rises?
What do you do when the price of gasoline rises? Elasticities of Demand and Supply 5 When you have completed your study of this chapter, you will be able to 1 Define the price elasticity of demand, and
### 1.2.3 Price, Income and Cross Elasticities of Demand
1.2.3 Price, Income and Cross Elasticities of Demand Price elasticity of demand The price elasticity of demand is the responsiveness of a change in demand to a change in price. The formula for this is:
### 2. The producers of a product with an elastic demand will have a strong incentive to reduce the price of their product.
Learning activity 5 True/False answers 1. If the price elasticity of the demand for chocolates is greater than one, then the manufacturers of chocolates can increase their total revenue by raising the
### Elasticity and Its Application
Elasticity and Its Application Elasticity... is a measure of how much buyers and sellers respond to changes in market conditions allows us to analyze supply and demand with greater precision. Journal Question-Name
### Contents. Consumer Choice: Individual and Market Demand- Demand and Elasticity. I) Markets and Prices. II) Demand Side. III) The Supply Side
Consumer Choice: Individual and Market Demand- Demand and Elasticity Dr. Ashraf Samir Website: ashraffeps.yolasite.com Contents I) Markets and Prices II) Demand Side III) The Supply Side IV) Market Equilibrium
### Microeconomics: Principles, Applications, and Tools
Microeconomics: Principles, Applications, and Tools NINTH EDITION Chapter 5 Elasticity: A Measure of Responsiveness Learning Objectives 5.1 List the determinants of the price elasticity of demand 5.2 Use
### Chapter 6 Elasticity: The Responsiveness of Demand and Supply
Economics 6 th edition 1 Chapter 6 Elasticity: The Responsiveness of Demand and Supply Modified by Yulin Hou For Principles of Microeconomics Florida International University Fall 2017 The Price Elasticity
### The law of supply states that higher prices raise the quantity supplied. The price elasticity of supply measures how much the quantity supplied
In a competitive market, the demand and supply curve represent the behaviour of buyers and sellers. The demand curve shows how buyers respond to price changes whereas the supply curve shows how sellers
### This is what we call a demand schedule. It is a table that shows how much consumers are willing and able to purchase at various prices.
Demand Market: an institution or mechanism, which brings together buyers ("demanders") and sellers ("suppliers") of particular goods and services. The remainder of this unit assumes a perfectly competitive
### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Suppose an increase in demand causes the price to increase from \$2 to \$4 and the quantity
### Microeconomics. More Tutorial at
Microeconomics Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. A legal maximum price at which a good can be sold is a price a. floor. b.
### ECON 251. Exam 1 Pink. Fall 2013
ECON 251 1. By definition, opportunity cost is a. The value of the best alternative b. The sum of the value of all available alternatives c. The amount of money it takes to buy an item d. Always greater
### Formula: Price of elasticity of demand= Percentage change in quantity demanded Percentage change in price
1 MICRO ECONOMICS~ CHAPTER FOUR CHAPTER FOUR PRICE ELASTICITY OF DEMAND You know that when supply increases, the equilibrium price falls and the equilibrium quantity increases THE PRICE ELASTICITY OF DEMAND~
### !"#\$#%&"'()#*(+,'&\$-''(.#/-'((
Lecture 1 Basic Concerns of Economics What is Economics! Economics is the study of how society manages its scarce resources. o Economic Problem: How a society can satisfy unlimited wants with limited resources
### ECON 1001 A. Come to the PASS workshop with your mock exam complete. During the workshop you can work with other students to review your work.
It is most beneficial to you to write this mock midterm UNDER EXAM CONDITIONS. This means: Complete the midterm in 1.5 hour(s). Work on your own. Keep your notes and textbook closed. Attempt every question.
### INTI COLLEGE MALAYSIA FOUNDATION IN BUSINESS INFORMATION TECHNOLOGY (CFP) ECO105: ECONOMICS 1 FINAL EXAMINATION: JANUARY 2006 SESSION
ECO105 (F) / Page 1 of 12 Section A INTI COLLEGE MALAYSIA FOUNDATION IN BUSINESS INFORMATION TECHNOLOGY (CFP) ECO105: ECONOMICS 1 FINAL EXAMINATION: JANUARY 2006 SESSION Instructions: This section consists
### ELASTICITY AND ITS APPLICATION. J. Mao
ELASTICITY AND ITS APPLICATION J. Mao Elasticity Until now, we ve been talking about the direction in which quantities change. A downward-sloping demand: price é è quantity demanded ê In real life it is
### AP Econ Section 9 Micro
Name: Date: _ ID: A AP Econ Section 9 Micro Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Gas prices recently increased by 25%. In response, purchases
### ECO 2301 Spring EXAM 2 Form 2 Friday, April 4 th Solutions
ECO 2301 Spring 2014 Sec 002 Klaus Becker EXAM 2 Form 2 Friday, April 4 th Solutions 1. The equilibrium price and quantity of any good or service is established by: A. only suppliers. B. only demanders.
### Economics: Canada in the Global Environment, 7e (Parkin) Chapter 4 Elasticity. 4.1 Price Elasticity of Demand
Economics: Canada in the Global Environment, 7e (Parkin) Chapter 4 Elasticity 4.1 Price Elasticity of Demand 1) A price elasticity of demand of 2 means that a 10 percent increase in price will result in
### Chapter 6. Elasticity
Chapter 6 Elasticity Both the elasticity coefficient and the total revenue test for measuring price elasticity of demand are presented in this chapter. The text discusses the major determinants of price
### EQ: What is Income Elasticity of Demand?
EQ: What is Income Elasticity of Demand? Changes in Consumer Income shift the demand curve: Normal Goods goods that are more desirable to consumers; when people have more money, they buy more normal goods
### Elasticity and Its Applications
Elasticity and Its Applications 1. In general, elasticity is a. a measure of the competitive nature of a market. b. the friction that develops between buyer and seller in a market. c. a measure of how
### Use the figure below to answer questions 1 and 2: D pounds of vegetables. A. 120 pounds of vegetables.
Use the figure below to answer questions 1 and 2: The figure shows the production possibilities curve for Hamid, who can produce two goods, meat and vegetables. 1. Refer to the figure above. What is the
### Topic 4c. Elasticity. What is the difference between this. and this? 1 of 23
Topic 4c Elasticity What is the difference between this and this? 1 of 23 Defining and Measuring Elasticity (I) Price elasticity of demand Ø The price elasticity of demand is the ratio of the percent change
### Chapter 4: Understanding Demand
SCHS SOCIAL STUDIES What you need to know UNIT TWO 1. What a competitive market is and how it is described by the supply and demand model 2. What a supply curve shows 3. The difference between a movement
### IB Economics/Microeconomics/Elasticities
IB Economics/Microeconomics/Elasticities Contents 1 2.2 Elasticities 1.1 Price Elasticity of Demand (PED) 1.2 Price Elasticity of Supply (PES) 1.3 Income Elasticity of Demand (YED) 1.4 Cross Elasticity
### Chapter 6 Lecture - Elasticity: The Responsiveness of Demand and Supply
Chapter 6 Lecture - Elasticity: The Responsiveness of Demand and Supply 1 The Price Elasticity of Demand and Its Measurement We define price elasticity of demand and understand how to measure it. Although
### Name: R Number: Roster #:
ECO 2305-002 ** TEST 2 ** Ibrahim Ozayturk Name: R Number: Roster #: Use the following to answer question 1: Figure: Consumer Surplus 1. (Figure: Consumer Surplus) In the figure, when the price falls from
### WJEC (Eduqas) Economics A-level
WJEC (Eduqas) Economics A-level Microeconomics Topic 2: Demand and Supply in Product Markets 2.4 Price, income and cross price elasticities of demand and supply Notes Price elasticity of demand The price
### Students: Go to: m.socrative.com Room #: When prompted to use your tech, answer the on-screen question.
Students: Go to: m.socrative.com Room #: 897089 When prompted to use your tech, answer the on-screen question. Using Your Tech: T/F: Consumers were Panic in 2004 Flu vaccine contamination Half of the U.S.
### CHAPTER 2: DEMAND AND SUPPLY
CHAPTER 2: DEMAND AND SUPPLY CIA4U Ms. Schirk 2.3 THE MARKET A market can be: A physical place where goods are bought and sold A collective reference to all the buyers and sellers of a particular good
### CHAPTER 2: DEMAND AND SUPPLY
2.3 THE MARKET CHAPTER 2: DEMAND AND SUPPLY CIA4U Ms. Schirk A market can be: A physical place where goods are bought and sold A collective reference to all the buyers and sellers of a particular good
### The price elasticity of demand when price decreases from \$9 to \$7 is A B C D -1.
Varsity Economics Product Market: Elasticity 1 The price elasticity of demand is a measure of the A effect of changes in demand on the price. B relationship between price and profitability. C responsiveness
### Individual & Market Demand and Supply
Mr Sydney Armstrong ECN 1100 Introduction to Microeconomic Lecture Note (3) Individual & Market Demand and Supply The tools of demand and supply can take us a far way in understanding both specific economic
### full file at
Chapter 2 1. Which of the following is a key assumption of the supply and demand model? Answer: A A. to focus on how the price and quantity sold are determined in a single market B. to focus on how the
### Chapter 3 Elasticity.notebook. February 03, Chapter 3: Competitive Dynamics and Government (Elasticity and Related Concepts)
Chapter 3: Competitive Dynamics and Government (Elasticity and Related Concepts) price elasticity of demand the responsiveness of a product's quantity demanded to a change in its price. Degree of Elasticity
### ELASTICITY. Chapt er. Key Concepts
Chapt er 4 ELASTICITY Key Concepts Price Elasticity of Demand The price elasticity of demand is a units-free measure of responsiveness of the quantity demanded of a good to a change in its price when all
### ELASTICITY AND ITS APPLICATION
5 ELASTICITY AND ITS APPLICATION Questions for Review 1. If demand is elastic, an increase in price reduces total revenue. With elastic demand, the quantity demanded falls by a greater percentage than
### 1 Which of the following is a key assumption of the supply and demand model?
1 Which of the following is a key assumption of the supply and demand model? to focus on how the price and quantity sold are determined in a single market (True Answer )Correct to focus on how the prices
### Iowa State University Economics 101 Microeconomics Principles Prof. Kilkenny Spring First Exam February 25, 2005
Iowa State University Economics 101 Microeconomics Principles Prof. Kilkenny Spring 2005 First Exam February 25, 2005 General Instructions: 1. Write your name here: 2. USE a #2 PENCIL: Write your name,
### Lesson-9. Elasticity of Supply and Demand
Lesson-9 Elasticity of Supply and Demand Price Elasticity Businesses know that they face demand curves, but rarely do they know what these curves look like. Yet sometimes a business needs to have a good
### MIDTERM I. GROUP A Instructions: November 3, 2010
EC101 Sections 04 Fall 2010 NAME: ID #: SECTION: MIDTERM I November 3, 2010 GROUP A Instructions: You have 60 minutes to complete the exam. There will be no extensions. Students are not allowed to go out
### Figure 4 1 Price Quantity Quantity Per Pair Demanded Supplied \$ \$ \$ \$ \$10 2 8
Econ 101 Summer 2005 In class Assignment 2 Please select the correct answer from the ones given Figure 4 1 Price Quantity Quantity Per Pair Demanded Supplied \$ 2 18 3 \$ 4 14 4 \$ 6 10 5 \$ 8 6 6 \$10 2 8
### Chapter 5: Price Controls: Multiple Choice Questions Chapter 6: Elasticity Multiple Choice Questions
Chapter 5: Price Controls: Multiple Choice Questions 1. ANSWER: d. ceiling. 2. ANSWER: a. a shortage, which cannot be eliminated through market adjustment. 3. ANSWER: b. the equilibrium price is below
### 2015 Pearson. Why does tuition keep rising?
Why does tuition keep rising? Demand and Supply 4 When you have completed your study of this chapter, you will be able to CHAPTER CHECKLIST 1 Distinguish between quantity demanded and demand, and explain
### Econ Test 2B Dr. Rupp Tuesday, March 3, 2009 Pledge: I have neither given or received aid on this exam Signature
Econ 2113 - Test 2B Dr. Rupp Tuesday, March 3, 2009 Name Pledge: I have neither given or received aid on this exam Signature Multiple Choice Identify the letter of the choice that best completes the statement
### Introduction to Agricultural Economics Agricultural Economics 105 Spring 2013 First Hour Exam Version 1
Introduction to Agricultural Economics Agricultural Economics 105 Spring 2013 First Hour Exam Version 1 Name Section There is only ONE best correct answer per question. Place your answer on the attached
### Mr Sydney Armstrong ECN 1100 Introduction to Microeconomics Lecture Note (4) Price Elasticity of Demand
Mr Sydney Armstrong ECN 1100 Introduction to Microeconomics Lecture Note (4) Price Elasticity of Demand The law of demand tells us that consumers will buy more of a product when its price declines and
### Elasticity: A Measure of Responsiveness. 1 of of 42
1 of 42 2 of 42 Elasticity: A Measure of In every large city in the United States, the public bus system runs a deficit: Operating costs exceed revenues from passenger fares. P R E P A R E D B Y FERNANDO
### 2.6 Price elasticity of demand
2.6 Price elasticity of demand Part 1 - PED is a measure of the responsiveness of consumers to a change in the price of a particular good. With data from a demand schedule, we can calculate the PED for
### Opportunity Costs when production is in quantity per/hr =
CHAPTER 1 THE CENTRAL IDEA 1.1 Scarcity and Choice for Individuals SCARCITY PRINCIPLE Scarcity principle (no free lunch principle): Although we have boundless needs and wants, the resources available to
### MIDTERM I. GROUP A Instructions: November 20, 2013
EC101 Sections 03 Fall 2013 NAME: ID #: SECTION: MIDTERM I November 20, 2013 GROUP A Instructions: You have 60 minutes to complete the exam. There will be no extensions. The exam consists of 30 multiple
### FINAL - ECONOMICS 2023 AUGUST 4, Name S.S.#
FINAL - ECONOMICS 2023 AUGUST 4, 2000 Name S.S.# This exam contains 30 multiple choice questions, each worth 2.5 points (100 point total). All questions on this exam should be answered. You have 75 minutes
### Elasticity. 2. a. Using the midpoint method, the percent change in the quantity of U.S. winter wheat demanded is 2.0 billion 2.2 billion 2.
Elasticity CHAPTER 6 1. Do you think the price elasticity of demand for Ford sport - utility vehicles (SUVs) will increase, decrease, or remain the same when each of the following events occurs? Explain
### A. All sellers who want to sell at the equilibrium price can find a buyer to sell to.
Chapter 02 Supply and Demand Multiple Choice Questions 1. Which is not true of market equilibrium? A. All sellers who want to sell at the equilibrium price can find a buyer to sell to. B. It is the most
### SUPPLY. Chapt er. Key Concepts. Markets and Prices
Chapt er 3 DEMAND AND SUPPLY Key Concepts Markets and Prices A competitive market is a market that has many buyers and sellers, so no single buyer or seller can influence the price. The money price of
### Principles of MicroEconomics: Econ102
Principles of MicroEconomics: Econ102 Price Elasticity of Demand: The responsiveness of the quantity demanded to a change in price, measured by dividing the percentage change in the quantity demanded of
### Chapter 4: Demand. Section I: Understanding Demand. Section II: Shifts of the Demand Curve. Section III: Elasticity of Demand
Chapter 4: Demand Section I: Understanding Demand Section II: Shifts of the Demand Curve Section III: Elasticity of Demand Section 1: Understanding Demand LEQ: What is the law of demand? VOCAB: demand
### To start we will look at the relationship between quantity demanded and price.
University of California, Merced ECO 1-Introduction to Economics Chapter 5 Lecture otes Professor Jason Lee I. Elasticity As we learned in Chapter 4, there is a clear relationship between the quantity
### At the end of chapter 6, you will be able to:
1 How to Study for Chapter 6 Supply and Equilibrium Chapter 6 introduces the factors that will affect the supply of a product, the price elasticity of supply, and the concept of equilibrium price and equilibrium
### Supply and Demand. Worksheet A-2A 2014
Supply and Demand Worksheet A-2A 2014 Worksheet A-2A 1. Surplus When the amount supplied exceeds the demand 2. Shortage When the amount demanded exceeds the supply 3. Utility The power to satisfy your
### VANCOUVER ISLAND UNIVERSITY. ECON211: Principles of Microeconomics, Spring 2013 SAMPLE MIDTERM EXAM. Name (Last, First): ID #: Signature:
Important: Please remember it is a sample exam. Number of questions in each section and structure of questions in Part B would vary as discussed in class VANCOUVER ISLAND UNIVERSITY ECON211: Principles
### ECON 251 Exam 1 Pink Spring 2012
ECON 251 Exam 1 Pink Spring 2012 1. Which of the following is an example of the economic resource of capital? a. A \$20 bill b. A corporate bond c. a government savings bond d. none of the above 2. John
### EXAMINATION 2 VERSION B "Applications of Supply and Demand" October 12, 2016
William M. Boal Signature: Printed name: EXAMINATION 2 VERSION B "Applications of Supply and Demand" October 12, 2016 INSTRUCTIONS: This exam is closed-book, closed-notes. Simple calculators are permitted,
### Econ103_Midterm (Fall 2016)
Econ103_Midterm (Fall 2016) Total 50 Points. Multiple Choice Identify the choice that best completes the statement or answers the question. 1 point for each question. Total 15 pts. c 1. Which of the following
### 2013 Pearson. Why did the price of coffee soar in 2010 and 2011?
Why did the price of coffee soar in 2010 and 2011? How do markets work? We have seen the circular flows diagram, which shows that households and firms interact in factor markets and goods markets. In this
### Chapter 4 Demand and Supply
Chapter 4 Demand and Supply 4.1 Demand 1) What is the "quantity demanded"? A) the amount of a good people desire B) the amount of a good people are able and willing to buy during a specific time period
### Bremen School District 228 Social Studies Common Assessment 2: Midterm
Bremen School District 228 Social Studies Common Assessment 2: Midterm AP Microeconomics 55 Minutes 60 Questions Directions: Each of the questions or incomplete statements in this exam is followed by five
### Midterm 2 - Solutions
Ecn 100 - Intermediate Microeconomic Theory University of California - Davis November 13, 2009 Instructor: John Parman Midterm 2 - Solutions You have until 11:50am to complete this exam. Be certain to
### Economic Analysis for Business Decisions Multiple Choice Questions Unit-2: Demand Analysis
Economic Analysis for Business Decisions Multiple Choice Questions Unit-2: Demand Analysis 1. The law of demand states that an increase in the price of a good: a. Increases the supply of that good. b.
### Supply and Demand. Objective 8.04
Supply and Demand Objective 8.04 Supply and Demand Pages 258-259 259 copy bold terms and give a definition or description of each. Page 261 Copy the questions Worksheet A-2A 1. Surplus When the amount
### AP Microeconomics Chapter 6 Outline
I. Introduction AP Microeconomics Chapter 6 A. Learning Objectives In this chapter students should learn: 1. What price elasticity of demand is and how it can be applied. 2. The usefulness of the total
### ECO401 Current Online 85 Quizzes Question Repeated ignore In Green color are doubted one
ECO401 Current Online 85 Quizzes Question Repeated ignore In Green color are doubted one Question # 1 of 15 ( Start time: 01:24:42 PM ) Total Marks: 1 A person with a diminishing marginal utility of income:
### Section 1: Microeconomics Syllabus item: 18 Weight: Elasticity. Price Elasticity of Demand (PED)
Section 1: Microeconomics Syllabus item: 18 Weight: 3 1.2 Elasticity Price Elasticity of Demand (PED) 1. Price Elasticity of Demand and its determinants Elasticity ² Measure of responsiveness. It measures
### Eastern Mediterranean University Faculty of Business and Economics Department of Economics Fall Semester. ECON 101 Mid term Exam
Eastern Mediterranean University Faculty of Business and Economics Department of Economics 2014 15 Fall Semester ECON 101 Mid term Exam Type A 28 November 2014 Duration: 90 minutes Name Surname: Group
### Supply & Demand Practice Honors Economics
Assigned: 2/16/15 Supply & Demand Practice Honors Economics B-Day: 1 20, due 2/18; 21 45, due 2/20 A-Day: All problems, 2/19/15 Due Dates Multiple Choice Identify the choice that best completes the statement
### To produce more beach balls, you must give up ever increasing quantities of ice cream cones.
Unit 01: Basic Concepts (Macro/Micro) Scarcity The Economic Problem: Unlimited wants, limited economic resources Factors of Production: -Land -Labor -Capital -Entrepreneurship Big 3 Questions: -What to
### Ecn Intermediate Microeconomic Theory University of California - Davis September 9, 2009 Instructor: John Parman. Final Exam
Ecn 100 - Intermediate Microeconomic Theory University of California - Davis September 9, 2009 Instructor: John Parman Final Exam You have until 1:50pm to complete this exam. Be certain to put your name,
### Midterm 2 Sample Questions. Use the demand curve diagram below to answer the following THREE questions.
! Midterm 2 Sample uestions Use the demand curve diagram below to answer the following THREE questions. 8 6 4 2 4 8 12 16 1. What is the own-price elasticity of demand as price decreases from 6 per unit
### CHAPTER 4: DEMAND. Lesson 3: elasticity of demand
CHAPTER 4: DEMAND Lesson 3: elasticity of demand 3 CASES OF DEMAND ELASTICITY Because quantity demanded depends on its price, economists use a concept called elasticity. Elasticity is a measure of responsiveness
### 1. A weekend fisherman notices the following relationship between hours spent fishing along the breakwaters of CCP and the quantity of fish caught:
1. A weekend fisherman notices the following relationship between hours spent fishing along the breakwaters of CCP and the quantity of fish caught: Hours Quantity of fish caught 0 0 1 10 2 18 3 24 4 28
### Microeconomics. More Tutorial at
Microeconomics 1. Suppose a firm in a perfectly competitive market produces and sells 8 units of output and has a marginal revenue of \$8.00. What would be the firm s total revenue if it instead produced
### EXAMINATION #2 VERSION A Consumers and Demand September 28, 2017
William M. Boal Signature: Printed name: EXAMINATION #2 VERSION A Consumers and Demand September 28, 2017 INSTRUCTIONS: This exam is closed-book, closed-notes. Calculators, mobile phones, and wireless | 11,721 | 49,800 | {"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 | 0.836238 |
https://bafybeiemxf5abjwjbikoz4mc3a3dla6ual3jsgpdr4cjr3oz3evfyavhwq.ipfs.dweb.link/wiki/Affine_plane.html | 1,656,112,740,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00581.warc.gz | 171,272,726 | 4,849 | # Affine plane
In geometry, an affine plane is a two-dimensional affine space. Typical examples of affine planes are
• Euclidean planes, which are affine planes over the reals, equipped with a metric, the Euclidean distance. In other words, an affine plane over the reals is a Euclidean plane in which one has "forgotten" the metric (that is, one does not talk of lengths nor of angle measures).
• Vector spaces of dimension two, in which the zero vector is not considered as different from the other elements
• For every field or division ring F, the set F2 of the pairs of elements of F
• The result of removing any single line (and all the points on this line) from any projective plane
All the affine planes defined over a field are isomorphic. More precisely, the choice of an affine coordinate system (or, in the real case, a Cartesian coordinate system) for an affine plane P over a field F induces an isomorphism of affine planes between P and F2.
In the more general situation, where the affine planes are not defined over a field, they will in general not be isomorphic. Two affine planes arising from the same non-Desarguesian projective plane by the removal of different lines may not be isomorphic.
There are two ways to formally define affine planes, which are equivalent for affine planes over a field. The first one consists in defining an affine plane as a set on which a vector space of dimension two acts simply transitively. Intuitively, this means that an affine plane is a vector space of dimension two in which one has "forgotten" where the origin is. In incidence geometry, an affine plane is defined as an abstract system of points and lines satisfying a system of axioms.
In the applications of mathematics, there are often situations where an affine plane without the Euclidean metric is used instead of the Euclidean plane. For example, in a graph, which can be drawn on paper, and in which the position of a particle is plotted against time, the Euclidean metric is not adequate for its interpretation, since the distances between its points or the measures of the angles between its lines have, in general, no physical importance (in the affine plane the axes can use different units, which are not comparable, and the measures also vary with different units and scales[1]).[2][3]
## Sources
• Artin, Emil (1987), "II. Affine and Projective Geometry", Geometric Algebra, Interscience Publishers, ISBN 0-470-03432-7
• Blumenthal, Leonard M. (1980) [1961], "IV. Coordinates in an Affine Plane", A Modern View of Geometry, Dover, ISBN 0-486-63962-2
• Gruenberg, K.W.; Weir, A.J. (1977), "II. Affine and Projective Geometry", Linear Geometry (2nd ed.), Springer-Verlag, ISBN 0-387-90227-9
• Snapper, Ernst; Troyer, Robert J. (1989) [1971], Metric Affine Geometry, Dover, ISBN 0-486-66108-3
• Yale, Paul B. (1968), "Chapter 5 Affine Spaces", Geometry and Symmetry, Holden-Day
## References
1. See also the books of Mandelbrot, "Gaussian Self-Affinity and Fractals", of Levi, "Foundations of Geometry and Trigonometry", and of Yaglom, "A Simple Non-Euclidean Geometry and its Physical Basis".
2. Paul Bamberg; Shlomo Sternberg (1991). A Course in Mathematics for Students of Physics. 1. Cambridge University Press. pp. 1–2. ISBN 978-0-521-40649-9.
3. Howard Levi (1975). Topics in Geometry. R. E. Krieger Publishing Company. p. 75. ISBN 978-0-88275-280-8. | 845 | 3,389 | {"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-2022-27 | latest | en | 0.934947 |
https://maths.guadalupebuendia.eu/English/W1ESO/Solution7_1.htm | 1,725,905,882,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651133.92/warc/CC-MAIN-20240909170505-20240909200505-00314.warc.gz | 370,936,660 | 1,380 | WORKSHEET: PERCENTAGES- SOLUTION EXERCISE 1
1) Write each ratio as a fraction, a decimal and a percent:
RATIO FRACTION DECIMAL PERCENT 18 to 100 18/100 = 9/50 0.18 18% 34 to 100 34/100 = 17/50 0.34 34% 85 to 100 85/100 = 17/20 0.85 85% 10 to 100 10/100 = 1/10 0.10 10%
Return | 124 | 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} | 3.703125 | 4 | CC-MAIN-2024-38 | latest | en | 0.657075 |
https://de.numberempire.com/limit-examples.php?page=24 | 1,621,355,353,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991288.0/warc/CC-MAIN-20210518160705-20210518190705-00376.warc.gz | 224,429,647 | 7,606 | Zur Startseite | Menü | Mitmachen | Eine bessere Übersetzung vorschlagen
# Limit-Beispiele
Vorherige Seite 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 Nächste Seite
Limit of 1-cosx/x^2 by x at 0 (two-sided)
Limit of (sin(x))^x by x at 0 (two-sided)
Limit of 1/(x*x) by x at 0 (two-sided)
Limit of (sqrt(x)-9)/(x-81) by x at 81 (two-sided)
Limit of (2-3x)/(x-3)^2 by x at 3 (two-sided)
Limit of ln x by x at inf (two-sided)
Limit of x*sin(1/x) by x at infinity (two-sided)
Limit of cos(1/x) by x at 0 (plus)
Limit of (x^3-x)/(x^2-1) by x at 1 (two-sided)
Limit of x^2-1/x-1 by x at 1 (two-sided)
Limit of (x^(1/2)-1)/(x-1) by x at 1 (two-sided)
Limit of x+2 by x at 1 (two-sided)
Limit of (x+1)/(x-2) by x at 2 (two-sided)
Limit of (1+(1/x))^x by x at infinity (two-sided)
Limit of 1/n by x at inf (two-sided)
Limit of (x^3-8)/(x^2+x-6) by x at 2 (two-sided)
Limit of (tan(x)/x)^(1/x^2) by x at 0 (two-sided)
Limit of (2-x)/(x-1)^2 by x at 1 (two-sided)
Limit of 1/x-5 by x at 5 (two-sided)
Limit of n by n at inf (two-sided)
Vorherige Seite 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 Nächste Seite
### Zahleigenschaften
0 / 12
Beispiele: 3628800, 9876543211, 12586269025
Mathe-Tools für Ihre Homepage
Wählen Sie eine Sprache aus: Deutsch English Español Français Italiano Nederlands Polski Português Русский 中文 日本語 한국어
Das Zahlenreich - Leistungsfähige Mathematik-Werkzeuge für jedermann | Kontaktiere den Webmaster
Durch die Nutzung dieser Website stimmen sie den Nutzungsbedingungen und den Datenschutzvereinbarungen zu.
© 2021 numberempire.com Alle Rechte vorbehalten | 754 | 1,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-21 | latest | en | 0.601301 |
https://missionalcall.com/2019/12/03/how-high-should-a-52-inch-tv-be-mounted/ | 1,695,594,972,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506669.96/warc/CC-MAIN-20230924223409-20230925013409-00021.warc.gz | 440,464,783 | 12,382 | # How high should a 52 inch TV be mounted?
How high should a 52 inch TV be mounted? Experts recommend keeping the middle of your TV at eye-level while seated. This is normally about 42 inches from the floor to the middle of your
## How high should a 52 inch TV be mounted?
Experts recommend keeping the middle of your TV at eye-level while seated. This is normally about 42 inches from the floor to the middle of your TV.
## How high is a 55in TV?
TV Size to Distance Calculator and Science
Size Width Height
55″ 47.9″ 121.7 cm 27″ 68.6 cm
60″ 52.3″ 132.8 cm 29.4″ 74.7 cm
65″ 56.7″ 144 cm 31.9″ 81 cm
70″ 61″ 154.9 cm 34.3″ 87.1 cm
How wide is a 52 in TV?
Widescreen TV Stand Width Chart
TV Dimensions (Diagonal) Screen Width
46 inch TV 40.1 inches + Bezel
48 inch TV 41.8 inches + Bezel
50 inch TV 43.6 inches + Bezel
52 inch TV 45.3 inches + Bezel
### How high should a 50 inch TV be mounted?
A 50-inch TV should be mounted with the top of the TV approximately 67 inches from the floor. You can measure how far down the TV the mount attaches to know how many inches below 67 to install the mount. This mounting height will likely be about 64-65 inches from the floor.
### Why is a 55 inch TV not 55 inches?
The width and height of the TV is 55 inches If you divide the screen width into 16 identical parts, the height of the TV is 9 parts. This is a common standard for the production of television screens. Therefore, you can find out the width and height of a 55-inch TV screen.
Are all 55 inch TVs the same size?
Yes, when choosing a TV, the question may arise, a TV with a diagonal of 55 inches, what is the size of the TV in width and height. Note, all TV manufacturers specify the size of the TV with the screen diagonal in mind. Some people mistakenly think that the size indicated by the manufacturer is the width of the TV.
#### What is best height for TV on wall?
As a rule, a 42” television should be mounted about 56 inches from floor to TV center, a 55” TV should be around 61 inches, a 65” TV should be around 65 inches’ floor to center, and a 70” television should be mounted about 67 inches to the center of the screen.
#### Is 50 inch TV a good size?
For crowded rooms, you should go with at least a 40-inch screen if you are seated more than six feet from the TV. A 50-inch screen is good within 7.5 feet of the TV. If you are 9 feet away, a 60-inch screen is probably as small as you want to go.
How high should a 75 TV be mounted?
Guideline for Optimal TV Height & Viewing Distance
TV Size TV Height
70″ TV 72 inches from floor to center of TV screen
75″ TV 75 inches from floor to center of TV screen
80″ TV 78 inches from floor to center of TV screen
85″ TV 81 inches from floor to center of TV screen
## What are the features of a 52 inch TV?
The all-weather design withstands elements with an operating temperature range of minus 22 degrees F to 122 degrees F, while the ISP and TNI panel enhances contrast for direct sunlight readability without the risk of isotropic blackout. HDMI 2.0 and HDCP 2.2 ports increase compatibility with 4K sources.
## How tall do you have to be to mount a 49in TV?
TV Mounting Height =Eye Height +(Viewing distance *0.22)=46in+(130in*0.22)=74.6in. But you can move closer to your TV because the optimum Viewing Distance for a 49in TV is 82in(VD=TVS*1.67=49in*1.67=82in).
What should the viewing distance be for a 55 inch TV?
For example if your TV size (TVS) is 55″ then the optimal Viewing Distance (VD) should be 92 inches (55*1.67). Your eye level also contributes to the TV mounting height and is the distance, in inches, of your eyes from the floor while you are sitting watching TV.
### How tall should the center of the TV screen be?
Most video professionals agree that the optimal height for the optimal viewing experience is to locate the center of the screen at the same height as your eyes when seated. The average height of your eyes when seated is 42″, assuming an 18″ sofa and 24″ from the seat to your eyes. | 1,066 | 4,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-40 | latest | en | 0.901916 |
https://quanteconpy.readthedocs.io/en/fix-docs-build/_modules/quantecon/util/combinatorics.html | 1,656,315,376,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103329963.19/warc/CC-MAIN-20220627073417-20220627103417-00068.warc.gz | 534,953,497 | 4,691 | # Source code for quantecon.util.combinatorics
"""
Useful routines for combinatorics
"""
from scipy.special import comb
from numba import jit
from .numba import comb_jit
[docs]@jit(nopython=True, cache=True)
def next_k_array(a):
"""
Given an array a of k distinct nonnegative integers, sorted in
ascending order, return the next k-array in the lexicographic
ordering of the descending sequences of the elements [1]_. a is
modified in place.
Parameters
----------
a : ndarray(int, ndim=1)
Array of length k.
Returns
-------
a : ndarray(int, ndim=1)
View of a.
Examples
--------
Enumerate all the subsets with k elements of the set {0, ..., n-1}.
>>> n, k = 4, 2
>>> a = np.arange(k)
>>> while a[-1] < n:
... print(a)
... a = next_k_array(a)
...
[0 1]
[0 2]
[1 2]
[0 3]
[1 3]
[2 3]
References
----------
.. [1] Combinatorial number system
<https://en.wikipedia.org/wiki/Combinatorial_number_system>_,
Wikipedia.
"""
# Logic taken from Algotirhm T in D. Knuth, The Art of Computer
# Programming, Section 7.2.1.3 "Generating All Combinations".
k = len(a)
if k == 1 or a[0] + 1 < a[1]:
a[0] += 1
return a
a[0] = 0
i = 1
x = a[i] + 1
while i < k-1 and x == a[i+1]:
i += 1
a[i-1] = i - 1
x = a[i] + 1
a[i] = x
return a
[docs]def k_array_rank(a):
"""
Given an array a of k distinct nonnegative integers, sorted in
ascending order, return its ranking in the lexicographic ordering of
the descending sequences of the elements [1]_.
Parameters
----------
a : ndarray(int, ndim=1)
Array of length k.
Returns
-------
idx : scalar(int)
Ranking of a.
References
----------
.. [1] Combinatorial number system
<https://en.wikipedia.org/wiki/Combinatorial_number_system>_,
Wikipedia.
"""
k = len(a)
idx = int(a[0]) # Convert to Python int
for i in range(1, k):
idx += comb(a[i], i+1, exact=True)
return idx
[docs]@jit(nopython=True, cache=True)
def k_array_rank_jit(a):
"""
Numba jit version of k_array_rank.
Notes
-----
An incorrect value will be returned without warning or error if
overflow occurs during the computation. It is the user's
responsibility to ensure that the rank of the input array fits
within the range of possible values of np.intp; a sufficient
condition for it is scipy.special.comb(a[-1]+1, len(a), exact=True)
<= np.iinfo(np.intp).max.
"""
k = len(a)
idx = a[0]
for i in range(1, k):
idx += comb_jit(a[i], i+1)
return idx | 722 | 2,357 | {"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} | 3.375 | 3 | CC-MAIN-2022-27 | latest | en | 0.540568 |
https://metanumbers.com/582740 | 1,623,808,375,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621699.22/warc/CC-MAIN-20210616001810-20210616031810-00171.warc.gz | 365,399,246 | 7,605 | ## 582740
582,740 (five hundred eighty-two thousand seven hundred forty) is an even six-digits composite number following 582739 and preceding 582741. In scientific notation, it is written as 5.8274 × 105. The sum of its digits is 26. It has a total of 4 prime factors and 12 positive divisors. There are 233,088 positive integers (up to 582740) that are relatively prime to 582740.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 26
• Digital Root 8
## Name
Short name 582 thousand 740 five hundred eighty-two thousand seven hundred forty
## Notation
Scientific notation 5.8274 × 105 582.74 × 103
## Prime Factorization of 582740
Prime Factorization 22 × 5 × 29137
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 291370 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 582,740 is 22 × 5 × 29137. Since it has a total of 4 prime factors, 582,740 is a composite number.
## Divisors of 582740
12 divisors
Even divisors 8 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 1.2238e+06 Sum of all the positive divisors of n s(n) 641056 Sum of the proper positive divisors of n A(n) 101983 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 763.374 Returns the nth root of the product of n divisors H(n) 5.71409 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 582,740 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 582,740) is 1,223,796, the average is 101,983.
## Other Arithmetic Functions (n = 582740)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 233088 Total number of positive integers not greater than n that are coprime to n λ(n) 29136 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 47677 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 233,088 positive integers (less than 582,740) that are coprime with 582,740. And there are approximately 47,677 prime numbers less than or equal to 582,740.
## Divisibility of 582740
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 4 4 8
The number 582,740 is divisible by 2, 4 and 5.
• Arithmetic
• Abundant
• Polite
## Base conversion (582740)
Base System Value
2 Binary 10001110010001010100
3 Ternary 1002121100222
4 Quaternary 2032101110
5 Quinary 122121430
6 Senary 20253512
8 Octal 2162124
10 Decimal 582740
12 Duodecimal 241298
20 Vigesimal 3cgh0
36 Base36 chn8
## Basic calculations (n = 582740)
### Multiplication
n×i
n×2 1165480 1748220 2330960 2913700
### Division
ni
n⁄2 291370 194247 145685 116548
### Exponentiation
ni
n2 339585907600 197890291794824000 115318588640515737760000 67200754344374141022262400000
### Nth Root
i√n
2√n 763.374 83.5266 27.6292 14.2264
## 582740 as geometric shapes
### Circle
Diameter 1.16548e+06 3.66146e+06 1.06684e+12
### Sphere
Volume 8.28921e+17 4.26736e+12 3.66146e+06
### Square
Length = n
Perimeter 2.33096e+06 3.39586e+11 824119
### Cube
Length = n
Surface area 2.03752e+12 1.9789e+17 1.00934e+06
### Equilateral Triangle
Length = n
Perimeter 1.74822e+06 1.47045e+11 504668
### Triangular Pyramid
Length = n
Surface area 5.8818e+11 2.33216e+16 475805
## Cryptographic Hash Functions
md5 94ae0e3c815674a99a458501652a74f4 a0025c617b4f5246823f61192de89710a47b5d87 247eda966eef651c8764c444c29359e419eba1367ded9a8fbb03892b87648edb 9339348f18ded19ce770c8f3d7e9433ea10ce1e425c24fd06d96eadf24c0dbdfbd161e395fd62ccf71b290dd47a1326fa290fe8a504c2eb15ea91f742210ca85 948dd7e463c22db5ca6b5ba8cf20a577c428ee83 | 1,454 | 4,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2021-25 | latest | en | 0.813949 |
http://tailieu.vn/doc/sequential-verulog-topics-part-12-310857.html | 1,508,406,259,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823260.52/warc/CC-MAIN-20171019084246-20171019104246-00499.warc.gz | 319,797,473 | 22,737 | # Sequential Verulog Topics part 12
0
44
lượt xem
4
## Sequential Verulog Topics part 12
Mô tả tài liệu
Verification of Gate-Level Netlist The optimized gate-level netlist produced by the logic synthesis tool must be verified for functionality.
Chủ đề:
Bình luận(0)
Lưu
## Nội dung Text: Sequential Verulog Topics part 12
1. 14.5 Verification of Gate-Level Netlist The optimized gate-level netlist produced by the logic synthesis tool must be verified for functionality. Also, the synthesis tool may not always be able to meet both timing and area requirements if they are too stringent. Thus, a separate timing verification can be done on the gate-level netlist. 14.5.1 Functional Verification Identical stimulus is run with the original RTL and synthesized gate-level descriptions of the design. The output is compared to find any mismatches. For the magnitude comparator, a sample stimulus file is shown below. Example 14-3 Stimulus for Magnitude Comparator module stimulus; reg [3:0] A, B; wire A_GT_B, A_LT_B, A_EQ_B; //Instantiate the magnitude comparator magnitude_comparator MC(A_GT_B, A_LT_B, A_EQ_B, A, B); initial $monitor($time," A = %b, B = %b, A_GT_B = %b, A_LT_B = %b, A_EQ_B = %b", A, B, A_GT_B, A_LT_B, A_EQ_B); //stimulate the magnitude comparator. initial begin A = 4'b1010; B = 4'b1001; # 10 A = 4'b1110; B = 4'b1111; # 10 A = 4'b0000; B = 4'b0000; # 10 A = 4'b1000; B = 4'b1100; # 10 A = 4'b0110; B = 4'b1110; # 10 A = 4'b1110; B = 4'b1110; end
2. endmodule The same stimulus is applied to both the RTL description in Example 14-1 and the synthesized gate-level description in Example 14-2, and the simulation output is compared for mismatches. However, there is an additional consideration. The gate- level description is in terms of library cells VAND, VNAND, etc. Verilog simulators do not understand the meaning of these cells. Thus, to simulate the gate- level description, a simulation library, abc_100.v, must be provided by ABC Inc. The simulation library must describe cells VAND, VNAND, etc., in terms of Verilog HDL primitives and, nand, etc. For example, the VAND cell will be defined in the simulation library as shown in Example 14-4. Example 14-4 Simulation Library //Simulation Library abc_100.v. Extremely simple. No timing checks. module VAND (out, in0, in1); input in0; input in1; output out; //timing information, rise/fall and min:typ:max specify (in0 => out) = (0.260604:0.513000:0.955206, 0.255524:0.503000:0.936586); (in1 => out) = (0.260604:0.513000:0.955206, 0.255524:0.503000:0.936586); endspecify //instantiate a Verilog HDL primitive and (out, in0, in1); endmodule ... //All library cells will have corresponding module definitions //in terms of Verilog primitives. ... Stimulus is applied to the RTL description and the gate-level description. A typical invocation with a Verilog simulator is shown below. //Apply stimulus to RTL description > verilog stimulus.v mag_compare.v
3. //Apply stimulus to gate-level description. //Include simulation library "abc_100.v" using the -v option > verilog stimulus.v mag_compare.gv -v abc_100.v The simulation output must be identical for the two simulations. In our case, the output is identical. For the example of the magnitude comparator, the output is shown in Example 14-5. Example 14-5 Output from Simulation of Magnitude Comparator 0 A = 1010, B = 1001, A_GT_B = 1, A_LT_B = 0, A_EQ_B = 0 10 A = 1110, B = 1111, A_GT_B = 0, A_LT_B = 1, A_EQ_B = 0 20 A = 0000, B = 0000, A_GT_B = 0, A_LT_B = 0, A_EQ_B = 1 30 A = 1000, B = 1100, A_GT_B = 0, A_LT_B = 1, A_EQ_B = 0 40 A = 0110, B = 1110, A_GT_B = 0, A_LT_B = 1, A_EQ_B = 0 50 A = 1110, B = 1110, A_GT_B = 0, A_LT_B = 0, A_EQ_B = 1 If the output is not identical, the designer needs to check for any potential bugs and rerun the whole flow until all bugs are eliminated. Comparing simulation output of an RTL and a gate-level netlist is only a part of the functional verification process. Various techniques are used to ensure that the gate- level netlist produced by logic synthesis is functionally correct. One technique is to write a high-level architectural description in C++. The output obtained by executing the high-level architectural description is compared against the simulation output of the RTL or the gate-level description. Another technique called equivalence checking is also frequently used. It is discussed in greater detail in Section 15.3.2, Equivalence Checking, in this book. Timing verification The gate-level netlist is typically checked for timing by use of timing simulation or by a static timing verifier. If any timing constraints are violated, the designer must either redesign part of the RTL or make trade-offs in design constraints for logic synthesis. The entire flow is iterated until timing requirements are met. Details of static timing verifiers are beyond the scope of this book. Timing simulation is discussed in Chapter 10, Timing and Delays. [ Team LiB ]
4. [ Team LiB ] 14.6 Modeling Tips for Logic Synthesis The Verilog RTL design style used by the designer affects the final gate-level netlist produced by logic synthesis. Logic synthesis can produce efficient or inefficient gate-level netlists, based on the style of RTL descriptions. Hence, the designer must be aware of techniques used to write efficient circuit descriptions. In this section, we provide tips about modeling trade-offs, for the designer to write efficient, synthesizable Verilog descriptions. 14.6.1 Verilog Coding Style[2] [2] Verilog coding style suggestions may vary slightly based on your logic synthesis tool. However, the suggestions included in this chapter are applicable to most cases. The IEEE Standard Verilog Hardware Description Language document also adds a new language construct called attribute. Attributes such as full_case, parallel_case, state_variable, and optimize can be included in the Verilog HDL specification of the design. These attributes are used by synthesis tools to guide the synthesis process. The style of the Verilog description greatly affects the final design. For logic synthesis, it is important to consider actual hardware implementation issues. The RTL specification should be as close to the desired structure as possible without sacrificing the benefits of a high level of abstraction. There is a trade-off between level of design abstraction and control over the structure of the logic synthesis output. Designing at a very high level of abstraction can cause logic with undesirable structure to be generated by the synthesis tool. Designing at a very low level (e.g., hand instantiation of each cell) causes the designer to lose the benefits of high-level design and technology independence. Also, a "good" style will vary among logic synthesis tools. However, many principles are common across logic synthesis tools. Listed below are some guidelines that the designer should consider while designing at the RTL level. Use meaningful names for signals and variables Names of signals and variables should be meaningful so that the code becomes self-commented and readable. Avoid mixing positive and negative edge-triggered flipflops
5. Mixing positive and negative edge-triggered flipflops may introduce inverters and buffers into the clock tree. This is often undesirable because clock skews are introduced in the circuit. Use basic building blocks vs. use continuous assign statements Trade-offs exist between using basic building blocks versus using continuous assign statements in the RTL description. Continuous assign statements are a very concise way of representing the functionality and they generally do a good job of generating random logic. However, the final logic structure is not necessarily symmetrical. Instantiation of basic building blocks creates symmetric designs, and the logic synthesis tool is able to optimize smaller modules more effectively. However, instantiation of building blocks is not a concise way to describe the design; it inhibits retargeting to alternate technologies, and generally there is a degradation in simulator performance. Assume that a 2-to-1, 8-bit multiplexer is defined as a module mux2_1L8 in the design. If a 32-bit multiplexer is needed, it can be built by instantiating 8-bit multiplexers rather than by using the assign statement. //Style 1: 32-bit mux using assign statement module mux2_1L32(out, a, b, select); output [31:0] out; input [31:0] a, b; wire select; assign out = select ? a : b; endmodule //Style 2: 32-bit multiplexer using basic building blocks //If 8-bit muxes are defined earlier in the design, instantiating //these muxes is more efficient for //synthesis. Fewer gates, faster design. //Less efficient for simulation module mux2_1L32(out, a, b, select); output [31:0] out; input [31:0] a, b; wire select; mux2_1L8 m0(out[7:0], a[7:0], b[7:0], select); //bits 7 through 0
6. mux2_1L8 m1(out[15:7], a[15:7], b[ 15:7], select); //bits 15 through 7 mux2_1L8 m2(out[23:16], a[23:16], b[23:16], select); //bits 23 through 16 mux2_1L8 m3(out[31:24], a[31:24], b[31:24], select); //bits 31 through 24 endmodule Instantiate multiplexers vs. Use if-else or case statements We discussed in Section 14.3.3, Interpretation of a Few Verilog Constructs, that if- else and case statements are frequently synthesized to multiplexers in hardware. If a structured implementation is needed, it is better to implement a block directly by using multiplexers, because if-else or case statements can cause undesired random logic to be generated by the synthesis tool. Instantiating a multiplexer gives better control and faster synthesis, but it has the disadvantage of technology dependence and a longer RTL description. On the other hand, if-else and case statements can represent multiplexers very concisely and are used to create technology- independent RTL descriptions. Use parentheses to optimize logic structure The designer can control the final structure of logic by using parentheses to group logic. Using parentheses also improves readability of the Verilog description. //translates to 3 adders in series out = a + b + c + d; //translates to 2 adders in parallel with one final adder to sum results out = (a + b) + (c + d) ; Use arithmetic operators *, /, and % vs. Design building blocks Multiply, divide, and modulo operators are very expensive to implement in terms of logic and area. However, these arithmetic operators can be used to implement the desired functionality concisely and in a technology-independent manner. On the other hand, designing custom blocks to do multiplication, division, or modulo operation can take a longer time, and the RTL description becomes more technology-dependent. Be careful with multiple assignments to the same variable Multiple assignments to the same variable can cause undesired logic to be
7. generated. The previous assignment might be ignored, and only the last assignment would be used. //two assignments to the same variable always @(posedge clk) if(load1) q
8. synthesis tool. Various partitioning techniques can be used. Horizontal partitioning Use bit slices to give the logic synthesis tool a smaller block to optimize. This is called horizontal partitioning. It reduces complexity of the problem and produces more optimal results for each block. For example, instead of directly designing a 16-bit ALU, design a 4-bit ALU and build the 16-bit ALU with four 4-bit ALUs. Thus, the logic synthesis tool has to optimize only the 4-bit ALU, which is a smaller problem than optimizing the 16-bit ALU. The partitioning of the ALU is shown in Figure 14-7. Figure 14-7. Horizontal Partitioning of 16-bit ALU The downside of horizontal partitioning is that global minima can often be different local minima. Thus, by use of bit slices, each block is optimized individually, but there may be some global redundancies that the synthesis tool may not be able to eliminate. Vertical Partitioning Vertical partitioning implies that the functionality of a block is divided into smaller submodules. This is different from horizontal partitioning. In horizontal partitioning, all blocks do the same function. In vertical partitioning, each block does a different function. Assume that the 4-bit ALU described earlier is a four- function ALU with functions add, subtract, shift right, and shift left. Each block is distinct in function. This is vertical partitioning. Vertical partitioning of the 4-bit ALU is shown in Figure 14-8. Figure 14-8. Vertical Partitioning of 4-bit ALU Figure 14-8 shows vertical partitioning of the 4-bit ALU. For logic synthesis, it is important to create a hierarchy by partitioning a large block into separate functional sub-blocks. A design is best synthesized if levels of hierarchy are created and smaller blocks are synthesized individually. Creating modules that contain a lot of functionality can cause logic synthesis to produce suboptimal designs. Instead, divide the functionality into smaller modules and instantiate those | 3,091 | 12,978 | {"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} | 2.71875 | 3 | CC-MAIN-2017-43 | longest | en | 0.819295 |
https://math.answers.com/Q/How_do_you_construct_a_truth_table_for_parenthesis_not_p_q_parenthesis_if_and_only_if_p | 1,695,987,683,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00745.warc.gz | 435,707,367 | 47,227 | 0
# How do you construct a truth table for parenthesis not p q parenthesis if and only if p?
Updated: 12/8/2022
Wiki User
12y ago
Assuming that you mean not (p or q) if and only if P
~(PVQ)--> P
so now construct a truth table, (just place it vertical since i cannot place it vertical through here.)
P True True False False
Q True False True False
(PVQ) True True True False
~(PVQ) False False False True
~(PVQ)-->P True True True False
if it's ~(P^Q) -->P
then it's,
P True True False False
Q True False True False
(P^Q) True False False False
~(P^Q) False True True True
~(P^Q)-->P True True False False
Wiki User
12y ago
Earn +20 pts
Q: How do you construct a truth table for parenthesis not p q parenthesis if and only if p?
Submit
Still have questions?
Related questions
### Construct a truth table for p and q if and only if not q?
Construct a truth table for ~q (p q)
### What is a truth table?
a table like your dinner table where you tell only the truth
### What is the difference between echo and print statement?
The print construct always returns the integer 1, whereas the echo construct has no return value (and as a result, print is slightly slower to execute than echo). Additionally, should you use the parenthesis syntax for both constructs, echo will accept an infinite amount of string arguments whereas print only accepts one.
False
### When all supposed truth is a lie in reality and in actuality it is due to the construct of the mind and body which cannot know but only attain a flawed perspective of self as a reflection of self?
This is not a question.
### Which logic gate have output high if and only if all inputs are low?
I have no interest in doing your homework. Draw up a truth table and figure it out.
### What are brackets used for?
in math, they act just like parenthesis, to tell you to operate on the terms inside the brackets, first. If you have several levels of nested parenthesis, then using pairs of brackets [] and braces {} help you to see which ones go together. This usually is only used in handwritten problems, as most computer software only recognizes parenthesis () to separate expressions.
### Are we Homo Sapiens the only ones whom can construct our own habitat?
No, it is not the homo sapiens that can construct their own habitat. Many animals are able to construct their own habitats.
### What is another name for parentheses or brackets?
They tend to be known only as parentheses (singular - "parenthesis") or brackets.
Yes.
1
### What year sojounrer truth get married?
truth was only 25 years old truth was only 25 years old | 612 | 2,623 | {"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} | 3.3125 | 3 | CC-MAIN-2023-40 | latest | en | 0.857122 |
https://www.nuclear-power.net/laws-of-conservation/law-of-conservation-of-energy/ballistic-pendulum-inelastic-collision/ | 1,586,440,818,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371858664.82/warc/CC-MAIN-20200409122719-20200409153219-00080.warc.gz | 1,048,490,104 | 80,343 | # Ballistic Pendulum – Inelastic Collision
## Ballistic Pendulum
A ballistic pendulum is a device for measuring the velocity of a projectile, such as a bullet. The ballistic pendulum is a kind of “transformer,” exchanging the high speed of a light object (the bullet) for the low speed of a massive object (the block). When a bullet is fired into the block, its momentum is transferred to the block. The bullet’s momentum can be determined from the amplitude of the pendulum swing.
When the bullet is embedding itself in the block, it occurs so quickly that the block does not move appreciably. The supporting strings remain nearly vertical, so negligible external horizontal force acts on the bullet–block system, and the horizontal component of momentum is conserved. Mechanical energy is not conserved during this stage, however, because a nonconservative force does work (the force of friction between bullet and block).
In the second stage, the bullet and block move together. The only forces acting on this system are gravity (a conservative force) and the string tensions (which do no work). Thus, as the block swings, mechanical energy is conserved. Momentum is not conserved during this stage, however, because there is a net external force (the forces of gravity and string tension don’t cancel when the strings are inclined).
## Equations governing the ballistic pendulum
In the first stage momentum is conserved and therefore:
where v is the initial velocity of the projectile of mass mP. v’ is the velocity of the block and embedded projectile (both of mass mP + mB) just after the collision, before they have moved significantly.
In the second stage mechanical energy is conserved. We choose y = 0 when the pendulum hangs vertically, and then y = h when the block and embedded projectile system reaches its maximum height. The system swings up and comes to rest for an instant at a height y, where its kinetic energy is zero and the potential energy is (mP + mB)gh. Thus we write the law of conservation of energy:
which is the initial velocity of the projectile and our final result.
When we use some realistic numbers:
• mP = 5 g
• mB = 2 kg
• h = 3 cm
• v = ?
then we have:
References:
Nuclear and Reactor Physics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
6. Kenneth S. Krane. Introductory Nuclear Physics, 3rd Edition, Wiley, 1987, ISBN: 978-0471805533
7. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
8. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
9. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993. | 805 | 3,170 | {"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} | 3.453125 | 3 | CC-MAIN-2020-16 | longest | en | 0.931501 |
https://www.convertit.com/Go/WaveQuest/Measurement/Converter.ASP?From=arpentcan&To=fl+head | 1,620,719,947,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991904.6/warc/CC-MAIN-20210511060441-20210511090441-00302.warc.gz | 731,645,458 | 3,701 | Riding the internet wave!
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```arpentcan = 44289.14688 length (length) ``` Related Measurements: Try converting from "arpentcan" to actus (Roman actus), angstrom, arpentlin, city block (informal), cloth quarter, digitus (Roman digitus), fathom, foot, Greek fathom, Israeli cubit, light yr (light year), mil, nail (cloth nail), naval shot, parasang, parsec, pica (typography pica), rod (surveyors rod), sazhen (Russian sazhen), span (cloth span), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: arpentcan = 1,248.33 actus (Roman actus), 24,411,340.8 agate (typography agate), 442,891,468,800,000 angstrom, 5,231,001.6 barleycorn, 1,210.88 bolt (of cloth), 69,746,688 bottom measure, 484.35 city block (informal), 4.43E+19 fermi, 145,305.6 foot, 132,867,440.64 French, 435,916.8 hand, 79,984.73 Israeli cubit, 20,924,006.4 line, 27.52 mile, 126,014,829.8 point (typography point), 484.35 soccer field, 3.36 spindle, 239.25 stadia (Greek stadia), 27.52 UK mile (British mile), 41.52 verst (Russian verst).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 507 | 1,698 | {"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-2021-21 | latest | en | 0.656925 |
https://mail.python.org/pipermail/tutor/2007-July/055706.html | 1,529,655,825,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864364.38/warc/CC-MAIN-20180622065204-20180622085204-00282.warc.gz | 646,669,925 | 2,155 | # [Tutor] correlation matrix
Sat Jul 21 20:54:52 CEST 2007
```Beanan O Loughlin wrote:
> hi all,
>
> I am new to python, and indeed to programming, but i have a question
> regarding correlation matrices.
>
> If i have a column vector
>
> x=[1;2;3;4;5]
>
> and its transpose row vector
>
> xt=[1,2,3,4,5]
>
> is there a simple way in python to create a 5x5 correlation matrix,
> obviously symmetric and having 1's on the diagonal?
My knowledge of matrix algebra and statistics is quite rusty but from a
programming perspective, I'd understand this like so.
You want a 5x5 table of values (correlations in this case) computed from
all permutations of 2 variables selected from a set of 5 elements.
so, you'd want something like
cor(1,1) cor(1,2) cor(1,3) ....
cor(2,1) cor(2,2) cor(2,3) ...
.
.
.
Correct?
If so, I think this might be what you're looking for (I've put some
>>> import Numeric
>>> x=[1,2,3]
>>> ret = []
>>> for i in x:
... ret1=[]
... for j in x: # Since we're correlating an array with itself.
... ret1.append((i,j,)) # You'd need to call the actual correlation
function here rather than just say (i,j)
... ret.append(ret1)
...
>>> Numeric.array(ret)
array([[[1, 1],
[1, 2],
[1, 3]],
[[2, 1],
[2, 2],
[2, 3]],
[[3, 1],
[3, 2],
[3, 3]]])
>>>
The "Numeric" module has fast implementations of matrices and other
such constructs.
I'm quite sure that this thing which I've written above can be improved
on but I think it would do your work for you.
--
~noufal
``` | 473 | 1,499 | {"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-2018-26 | latest | en | 0.898498 |
http://tomkoinc.com/rlyghh/sword-art-online-volume-19-abbc8b | 1,713,322,480,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817128.7/warc/CC-MAIN-20240417013540-20240417043540-00190.warc.gz | 40,083,374 | 5,648 | Phosphate is PO4^-3; PO3- does not exist. valence electrons are found from the group number of the main group elements. Paper by Super 30 Aakash Institute, powered by embibe analysis.Improve your score by 22% minimum while there is still time. This is not a "good" structure , because it has five atoms with formal charges. So this is the best structure for the phosphate ion—Lewis structure for PO4-. In PO4^3– ion the formal charge on the oxygen atom of P–O bond is. 6 Answers. To calculate the formal charge, the following formula is used. The overal charge is -3. Lv 7. Click hereto get an answer to your question ️ In PO4^3 - ion the formal charge on the oxygen atom of P - O bond is: Have you registered for the PRE-JEE MAIN PRE-AIPMT 2016? In PO4 3– ion the formal charge on the oxygen atom of P–O bond is (i) + 1 (ii) – 1 (iii) – 0.75 (iv) + 0.75 how to solve this in easier manner 2 See answers edistian edistian Charge on oxygen is always -2 and total charge in this ion should be -3 so -2*4+charge on p= -3 Answer Save. In the resonance hybrid, a total of -3 charge is distributed over 4 O atoms. Thus the formal charge of each O atom is -3/4 = −0.75. You can draw five resonance structures for "PO"_4^(3-), but one of them is a minor contributor to the resonance hybrid. That's the Lewis structure for PO4 3-. A step-by-step explanation of how to draw the PO4 3- Lewis Structure (Phosphate Ion). {eq}\rm PO_4^{3-}{/eq}, phosphate ion, has a total of 32 valence electrons (5 from nitrogen, 24 from 4 oxygen, and 3 from the charge). Be sure not to confuse formal charge with oxidation number. A bit tougher: watch your formal charges, you'll get the correct structure. "FC" = "valence e"^(-) - "owned e"^(-) where: owned electrons are found by cleaving each bond homolytically so that one electron goes to each atom that was bonding. Relevance. But when I add a double bond to one of the Os the formal charges are -1 to 3 of the oxygens. 1 answer. There are different ways to draw the structure of the phosphate ion. asked Oct 28, 2018 in Chemical Bonding and Molecular Structure by Shradha (2.7k points) chemical bonding; neet; 0 votes. In the phosphate ion, P has an oxidation number of +5 and each oxygen has an oxidation number of -2. I'm confused because when I draw all of the Os attached with single bonds to P the formal charges don't make sense. That is because the oxygen in P=O has a formal charge of 0 and the other 3 oxygens bonded P-O have a formal charge of -1 each. If it is done with a P=O and 3 P-O bonds, the formal charge of phosphorous is 0. In po4^3- ion the formal charge on each oxygen atom and p-o bond order respectively are. We do need to put brackets around it and a 3- out here so that everyone knows that it is, indeed, the phosphate ion. 6 years ago. In all of these the P has a formal charge of 0 and one oxygen is also 0. I'm confused if there should be a double bond or not for the PO4 3- structure. Step 8 in determining the Lewis Structure of PO4(3-) PO 4 3-Step 8 Picture so Far: (Numbers next to atoms are their formal charges) Starting at the upper right corner and moving clockwise we can make 4 resonance structures that expand the P octet to 10. In PO4(-3) ion the formal charge on each oxygen atom and P-O bond order will be ??? > If you start by drawing four oxygen atoms single bonded to a phosphorus atom and give every atom an octet, you get Structure R in the diagram below. Get more chemistry help at www.Breslyn.org. In a given resonance structure, the O atom that forms double bond has formal charge of 0 and the remaining 3 O atoms have formal charge of -1 each. pisgahchemist. Its Lewis structure is shown below: Assigning formal charges assumes 100% covalent character, meaning the electrons in a given chemical bond are assumed to be equally shared. Oxygen atom of P–O bond is not a good '' structure, because it has atoms. Confuse formal charge of each O atom is -3/4 = −0.75 atom is -3/4 = −0.75 formula... Bonding and Molecular structure by Shradha ( 2.7k points ) chemical Bonding and Molecular structure Shradha... Bond are assumed to be equally shared is also 0 structure of the Os attached with single to... Be a double bond or not for the PO4 3- structure good '' structure, because has. O atom is -3/4 = −0.75 structure by Shradha ( 2.7k points ) chemical Bonding ; ;! With a P=O and 3 P-O bonds, the following formula is used 30. Different ways to draw the structure of the Os the formal charges, you 'll get the correct structure charge. Oxygen is also 0 -3 charge is distributed over 4 O atoms O atoms correct structure is. Super 30 Aakash Institute, powered by embibe analysis.Improve your score by 22 % minimum while there is still.! How to draw the structure of the phosphate ion by Super 30 Aakash,. Institute, powered by embibe analysis.Improve your score by 22 % minimum while is... And one oxygen is also 0 and each oxygen has an oxidation number of -2 0 votes powered by analysis.Improve. Structure of the Os attached with single bonds to P the formal charges do n't make sense a bit:... Add a double bond to one of the phosphate ion or not for the PRE-JEE MAIN PRE-AIPMT?. How to draw the PO4 3- structure are -1 to 3 of the oxygens in chemical Bonding neet. Bit tougher: watch your formal charges assumes 100 % covalent character, meaning the electrons a. Following formula is used each oxygen has an oxidation number of -2 i add a double bond not. To be equally shared with oxidation number = −0.75 phosphate ion, P has a formal on... Atom is -3/4 = −0.75 if there should be a double bond not... Is -3/4 = −0.75 oxygen atom of P–O bond is, the formal of! Of how to draw the PO4 3- Lewis structure ( phosphate ion, P has a formal charge, following! Bond is following formula is used because it has five atoms with formal charges, you 'll get correct! So this is the best structure for PO4- electrons in a given chemical bond are assumed to equally! The following formula is used P-O bonds, the formal charges the PO4 3- structure. An oxidation number of -2 an oxidation number of +5 and each oxygen has an oxidation number of phosphate. Has a formal charge of 0 and one oxygen is also 0, meaning the electrons in given. Ion, P has a formal charge on the oxygen atom of P–O bond.... A step-by-step explanation of how to draw the PO4 3- Lewis structure ( phosphate ion structure... To P the formal charge, the formal charge of 0 and one oxygen is also 0 Molecular structure Shradha... An oxidation number of -2 with oxidation number is still time when i po4 3- formal charge! For PO4- Os the formal charge of each O atom is -3/4 = −0.75 also 0 elements... Given chemical bond are assumed to be equally shared -3/4 = −0.75 3 the! 0 votes 3- Lewis structure ( phosphate ion is done with a P=O and 3 bonds... I 'm confused because when i add a double bond or not for the PRE-JEE MAIN PRE-AIPMT?! 'M confused because when i add a double bond to one of the oxygens with bonds... Single bonds to P the formal charge of each O atom is -3/4 = −0.75 the 3-... The PO4 3- Lewis structure ( phosphate ion, P has an oxidation number of -2 formal charge the. To be equally shared equally shared one of the Os attached with single bonds to P formal! Over 4 O atoms the phosphate ion—Lewis structure for PO4- one oxygen is po4 3- formal charge 0 a total of -3 is! % covalent character, meaning the electrons in a given chemical bond are assumed to equally., you 'll get the correct structure of +5 and each oxygen has an oxidation number the... Attached with single bonds to P the formal charges still time so this is the best structure the... +5 and each oxygen has an oxidation number of -2 the group number of MAIN. A total of -3 charge is distributed over 4 O atoms have you for... Chemical Bonding ; neet ; 0 votes P-O bonds, the following formula is used score. Of -3 charge is distributed over 4 O atoms your formal charges assumes 100 % covalent character, the... And Molecular structure by Shradha ( 2.7k points ) chemical Bonding ; neet ; 0 votes,. To one of the oxygens of -3 charge is distributed over 4 O atoms in the resonance hybrid a! The Os the formal charges assumes 100 % covalent character, meaning the in... It is done with a P=O and 3 P-O bonds, the formal charge of phosphorous is 0 it! Or not for the PRE-JEE MAIN PRE-AIPMT 2016 is the best structure for the phosphate ion—Lewis for! Lewis structure ( phosphate ion ) do n't make sense and Molecular structure by (... An oxidation number of +5 and each oxygen has an oxidation number have you registered for the phosphate.... Equally shared structure for PO4- also 0 Bonding and Molecular structure by Shradha ( 2.7k points ) chemical and... Ion—Lewis structure for PO4- when i add a double bond to one of the MAIN group elements resonance hybrid a... Of the oxygens not to confuse formal charge of 0 and one oxygen is also.! The formal charges do n't make sense given chemical bond are assumed to be equally shared is.. Bonding ; neet ; 0 votes done with a P=O and 3 P-O,! -1 to 3 of the oxygens '' structure, because it has atoms! And 3 P-O bonds, the following formula is used a double bond or not for the PO4 3-.! O atom is -3/4 = −0.75 oxygen is also 0 ; neet ; 0 votes there! These the P has a formal charge on the oxygen atom of P–O bond.! Explanation of how to draw the structure of the oxygens, 2018 in chemical Bonding and Molecular structure by (... I add a double bond or not for the PRE-JEE MAIN PRE-AIPMT 2016 if it is with. From the group number of -2 electrons are found from the group number of MAIN. To be equally shared bond is P–O bond is Aakash Institute, powered by embibe analysis.Improve score. Be equally shared the group number of +5 and each oxygen has an number... Watch your formal charges assumes 100 % covalent character, meaning the electrons in a given chemical bond are to... 100 % covalent character, meaning the electrons in a given chemical bond are assumed be... By Shradha ( 2.7k points ) chemical Bonding and Molecular structure by Shradha ( 2.7k )! In PO4^3– ion the formal charge on the oxygen atom of P–O is... Not to confuse formal charge po4 3- formal charge the following formula is used a good '',. And Molecular structure by Shradha ( 2.7k points ) chemical Bonding and Molecular structure by (... -3 charge is distributed over 4 O atoms a total of -3 charge is distributed over 4 O atoms in... Your score by 22 % minimum while there is still time one oxygen is also.! A good '' structure, because it has five atoms with formal charges one oxygen is also 0 confuse. How to draw the PO4 3- Lewis structure ( phosphate ion ) one oxygen is 0. You registered for the phosphate ion, P has an oxidation number of +5 and each oxygen an. 4 O atoms five atoms with formal charges do n't make sense the. Hybrid, a total of -3 charge is distributed over 4 O atoms 3-!, a total of -3 charge is distributed over 4 O atoms draw. Sure not to confuse formal charge of phosphorous is 0 there are different ways draw. Group elements Os attached with single bonds to P the formal charge of and! % covalent character, meaning the electrons in a given chemical bond assumed... Pre-Jee MAIN PRE-AIPMT 2016 'll get the correct structure not to confuse formal charge with oxidation number +5! There is still time Os po4 3- formal charge formal charge, the following formula is used by 22 % while. Atom is -3/4 = −0.75 following formula is used a step-by-step explanation of to. Good '' structure, because it has five atoms with formal charges do n't make sense in chemical and. A formal charge of each O atom is -3/4 = −0.75 assumed to be equally shared has a charge! Pre-Jee MAIN PRE-AIPMT 2016 good '' structure, because it has five atoms with charges... Is also 0, meaning the electrons in a given chemical bond are assumed to be equally.... A step-by-step explanation of how to draw the PO4 3- structure by embibe analysis.Improve your score by %! By embibe analysis.Improve your score by 22 % minimum while there is still time an number! 2.7K points ) chemical Bonding ; neet ; 0 votes you registered for the PO4 3- structure. Have you registered for the PRE-JEE MAIN PRE-AIPMT 2016 following formula is used ion.... Bonds, the following formula is used PRE-AIPMT 2016 valence electrons are from... Is -3/4 = −0.75 the resonance hybrid, a total of -3 charge is distributed over 4 O.... Found from the group number of -2 get the correct structure is used oxygen also...: watch your formal charges get the correct structure confused if there should be a double bond to of! | 3,198 | 12,546 | {"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} | 3.171875 | 3 | CC-MAIN-2024-18 | latest | en | 0.903055 |
https://www.dsprelated.com/showarticle/52.php | 1,670,383,518,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711126.30/warc/CC-MAIN-20221207021130-20221207051130-00440.warc.gz | 789,833,981 | 17,992 | # White Lies
Correlation, as one of the first tools DSP users add to their tool box, can automate locating a known signal within a second (usually larger) signal. The expected result of a correlation is a nice sharp peak at the location of the known signal and few, if any, extraneous peaks.
A little thought will show this to be incorrect: correlating a signal with itself is only guaranteed to give a sharp peak if the signal's samples are uncorrelated --- for example if the signal is composed of white noise.
Most interesting signals are not white noise.
# Interesting Images Are Not White
When the signals of interest are images, the problem is exacerbated because the images are very different from white noise. Eight bit gray-scale images have pixel values from 0 to 255, so they almost always have a "DC offset".
Take the following simple image as an example. The image contains 5 blobs, some of which overlap.
Suppose we wish to automate finding the blob indicated by the red line (the "known" signal). Simply using two-dimensional cross-correlation will yield the following image.
The localization of the peak in the image is not good due to the large, coalesced blob at the bottom of the image.
# Make A Difference: Twice!
A simple way to both remove the DC component of the image, and to make the image "whiter" is to row-difference the image (i.e. subtract from each row the previous row). In Matlab this can be achieved with the "diff" function.
Note that both the "known" image and the image which is being searched require this pre-whitening.
# Discrimination
After performing the diff operation on both the known image and the searched image, the following results.
The red point indicates the peak of the new cross-correlation. A side view of this image shows how well the peak is discriminated from other peaks.
[ - ]
Comment by February 18, 2008
Hi Peter, This is a nice method. Any reason you want to use the difference operator in only the vertical direction? Does it make make the method better to use a full edge detection? Regards, Steve tSTev
[ - ]
Comment by February 18, 2008
Hi Steve, thanks for the comment! There's no reason to prefer the vertical direction over any other; that was chosen primarily for simplicity of the example. I'd have to think about whether full edge detection would improve things. The aim is spectral whiteness or uncorrelated samples. The "best" one could hope for in this regard is to filter by the inverse of the spectrum of the known signal --- "best" in the sense that if the searched image and known image are the same, this form of filtering will make the result completely spectrally white.
[ - ]
Comment by November 18, 2008
The time to act is now! Green is no longer in champagne activism territory; it has firmly positioned itself in the business mainstream as the next clean engine for economic growth. With rapid strides in technology, economic pressures, high energy prices, resource shortages, environmental threats, and compliance imperatives driving unprecedented growth in green energy, the pressure on scientists, technologists, government, and the industry to develop and adopt new methods for harnessing green energy has never been this overpowering. The first of its kind in India, Green Energy Summit 2009 will provide a unique platform for debate, dialogue and action between research, industry, academia, financial organisations, entrepreneurs and government bodies worldwide. The four tiers of the summit will see a congregation of the most intelligent minds and proactive organisations to create a sustained, result-oriented mindset, bringing into picture those questions that are most relevant for India at this juncture.
[ - ]
Comment by December 6, 2008
good~
[ - ]
Comment by May 25, 2010
Hi Peter, thank you for the idea. Q1: Is this prewhitening equal (to some extend) to other whitening methods, e.g. subtracting the row (or column) means from the dataset? Q2: Does prewhitening make any sense when analyzing time-domain waveforms? e.g. sending a Gaussian shaped pulse (a range of frequencies) into an optical system, and sensing the output, which can be of arbitrary shape, but Gaussian. I would be thankful if you could point me to some direction, I can get some more basic info when to use whitening on a dataset. Thank you again for your time on the article!
[ - ]
Comment by June 9, 2010
Q3: By diff-ing a matrix the number of columns decreases by one (the first column vanishes). Is there any appropriate method to generate the missing column for later compatibility?
[ - ]
Comment by January 17, 2012
This is not a blog post, it's a brain fart. Post links to interesting things on your facebook, not on a site dedicated to embedded.
[ - ]
Comment by March 13, 2014
Normalized correlation value in altboc programme in matlab PLEASE HELP ME this programe is a importance for me
[ - ]
Comment by March 13, 2014 | 1,075 | 4,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-49 | longest | en | 0.952236 |
http://www.slideserve.com/yoshe/efficient-evaluation-of-having-queries-on-a-probabilistic-database | 1,490,692,009,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189686.31/warc/CC-MAIN-20170322212949-00612-ip-10-233-31-227.ec2.internal.warc.gz | 700,532,000 | 18,936 | This presentation is the property of its rightful owner.
1 / 24
# Efficient Evaluation of HAVING Queries on a Probabilistic Database PowerPoint PPT Presentation
Efficient Evaluation of HAVING Queries on a Probabilistic Database. Christopher Re and Dan Suciu University of Washington. Evaluation of conjunctive Boolean queries with aggregate tests on probabilistic DBs: HAVING in SQL, e.g. is the SUM(profit) > 100k?
Efficient Evaluation of HAVING Queries on a Probabilistic Database
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
## Efficient Evaluation of HAVING Queries on a Probabilistic Database
Christopher Re and Dan Suciu
University of Washington
• Evaluation of conjunctive Boolean queries with aggregate tests on probabilistic DBs:
• HAVING in SQL, e.g. is the SUM(profit) > 100k?
• Looking for optimal algorithms (dichotomies):
• For all queries q with aggregate A want
• P time algorithm, call this A-Safe [DS04,DS07]
• Some instance s.t. q is hard (#P).
• Technique:
• In safe plans, use multiplication
• In A-safe plans, use convolution (on monoids)
### Motivation
Profit
HAVING style
Expectation Style [Prior Art]
SELECT item FROM Profit
WHERE item =‘Widget’
GROUP BY item
HAVING SUM(Amount) > 0
SELECT SUM(Amount)
FROM Profit
WHERE item=‘Widget’
Ans: -99k *.99 +100M*0.01 ~900K
Ans: 0.01
• Preliminaries
• Formal Problem Description
• Query plans and Datalog
• Monoid Random Variables and Convolutions
• Max,Min,Count and hints for others
• Conclusions
### Overview
• Conjunctive rule:
• No repeated symbols
• Aggregates
• Comparision:
• k, is a constant
### HAVING Query semantics
NB: Assume SQL-like semantics
SELECT ITEM FROM PROFIT
WHERE ITEM=‘Widget’
GROUP BY ITEM
HAVING SUM(PROFIT) > 0
Possible worlds, model
Query Semantics
In talk, restrict to tuple independence
### Probabilistic Semantics
NB: In paper, allow disjointtuples
• Data complexity: Fix Query. Instance grows.
• In practice, query is small.
• Consider k, i.e. 1000, as part of the input
• Skeleton,
### Complexity and formal problem
• Preliminaries
• Formal Problem Description
• Query plans and Datalog
• Monoid Random Variables and Convolutions
• Max,Min,Count and hints for others
• Conclusions
### Overview
• A monoidis a triple where M is a set and + is associative with identity 0.
• e.g.
• Commutative Semiring is
• Both are commutative monoids
• * distributes over +
• e.g. a Boolean algebra
### Monoids and Semirings
NB: n=1 is logical OR
• Fix a Semiring S.
• Annotation is a function to S with finite support
• Plans defined inductively:
### [GKT07] : Datalog + Semirings
Goal: define value of tuple t in a plan P,
support, i.e. tuples contributing to a value
Value of a plan, i.e, the annotation computes
### [GKT07] Inductive definition
• Monoids and Aggregates
### Annotations and HAVING
0 is tuple not present
1 is tuple present, y > 3
2 is tuple present,
probabilities
0.2
How can we deal with probabilities?
0.4
0.1
Monoid sum is 1 iff all values are bigger than 3
• Preliminaries
• Formal Problem Description
• Query plans and Datalog
• Monoid Random Variables and Convolutions
• Max,Min,Count and hints for others
• Conclusions
### Overview
• An M-random variable (rv) is
• Correlations
• r,s are independent if for any m,m’ in M
• Extended to sets via total independence
### Monoid Random Variables
• Let r be an rv. A marginal vector is
• The monoid convolution * (depending on +) is
### Monoid Convolutions
• If r,smonoidrvs then r+s is an rv defined as
• PROP: If r,s are independent then the distribution of r + s is given by convolution:
• PROP: The convolution of n r.v.s can be computed in
• Single convolution in time
• Convolution is associative.
### Convolutions
Convolutions are efficient, if M is not too big
• Preliminaries
• Formal Problem Description
• Query plans and Datalog
• Monoid Random Variables and Convolutions
• Max,Min,Count and hints for others
• Conclusions
### Overview
• Monoids and Aggregates
### Annotations and HAVING
0 is tuple not present
1 is tuple present, y > 3
2 is tuple present,
marginal vectors
probabilities
(0.8,0.2,0)
0.2
(0.6,0.4,0)
0.4
(0.9,0,0.1)
0.1
Marginal of 1 after convolution = value of query
Monoid sum is 1 iff all values are bigger than 3
Compute value of “Safe Plans”:
Plan is safe [DS04], if all projects and joins are independenttuples, else #P
THM: value is correct if the plan is safe.
### “Safe plans” for semirings
Only efficient if the semiring is “small”
Gives dicohotomy for MIN,MAX,COUNT – not the others
• Dichotomy for SUM,AVG,COUNT DISTINCT
• Not all safe plans allowed!
• e.g. cannot have independent projections “on top”
• Disjoint tuples in the paper
• Need a “disjoint projection” operation
• More work for dichotomies
• Algorithms for finding safe plans (P time)
• Semantic for aggregation queries on prob DBs
• Similar to HAVING in SQL
• Proposed a complexity measure for such queries
• Central technique was marginal vectors and convolutions
• Dichotomy for HAVING queries w.o. self-joins
### Conclusion
• Conjunctive rule:
• No repeated subgoals
• Aggregates
• Comparision:
• k, is a constant
### HAVING Query semantics
NB: Assume SQL-like semantics
SELECT ITEM FROM PROFIT
WHERE ITEM=‘Widget’
GROUP BY ITEM
HAVING SUM(PROFIT) > 0
• Monoids and Aggregates
### Annotations and HAVING
0 is tuple not present
1 is tuple present, y > 3
2 is tuple present,
marginal vectors
probabilities
(0.8,0.2,0)
0.2
(0.6,0.4,0)
0.4
(0.9,0,0.1)
0.1
Marginal of 1 after convolution = value of query
Monoid sum is 1 iff all values are bigger than 3 | 1,648 | 6,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-13 | longest | en | 0.745569 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.