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# What is SEAWATER PPM?
Note! ppm= parts per million = mg/litre = 0.001g/kg. source: Karl K Turekian: Oceans. 1968. ... mg/kg (ppm) in sea water: molecular weight: mmol/ kg: Nitrogen N2: 78%: 47.5%: 10: 12.5: 28.014: 0.446: Oxygen O2: 21%: 36.0%: 5: 7: 31.998: 0.219: Carbondioxide CO2: 0.03%: 15.1%: 40:
http://www.seafriends.org.nz/oceano/seawater.htm
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The salt concentration is usually expressed in parts per thousand (permillle, ‰) or parts per million (ppm). ... The thermal conductivity of seawater is 0.6 W/mK at 25 °C and a salinity of 35 g/kg..
http://en.wikipedia.org/wiki/Saline_water
Water salinity of fresh, brackish and sea water. Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
http://www.engineeringtoolbox.com/water-salinity-d_1251.html
Sea Water Elemental composition of seawater in parts per million See table below for Major components and Salinity of the Seas ppm= parts per million = mg/litre = 0.001g/kg.
http://delloyd.50megs.com/moreinfo/seawater.html
Seawater or salt water is water from a sea or ocean. On average, seawater in the world's oceans has a salinity of about 3.5% (35 g/L, or 599 mM). This means that every kilogram (roughly one litre by volume) of seawater has approximately 35 grams (1.2 oz) of dissolved salts (predominantly sodium (Na
http://en.wikipedia.org/wiki/Seawater
Saline water: Desalination Thirsty? How 'bout a cool, refreshing cup of seawater? No, don't take us literally! Humans cannot drink saline water.
http://ga.water.usgs.gov/edu/drinkseawater.html
Without a working knowledge of what is present in natural seawater, it is often ... are the two primary ions in seawater. At 19,000 ppm for chloride and 10,500 ppm for sodium, they comprise 54% and 30% of the total weight of ions in seawater, ...
http://www.reefkeeping.com/issues/2005-11/rhf/index.php
Best Answer: It is better to use % as opposed to PPM. The salinity of the oceans changes slightly from around 32ppt (3.2%) to 40ppt (4.0%). Low salinity is found in cold seas, particularly during the summer season when ice melts. High salinity is found in the ocean 'deserts' in a band ...
http://in.answers.yahoo.com/question/index?qid=20100719091929AAT8Ae0
Highly saline water - From 10,000 ppm to 35,000 ppm; By the way, ocean water contains about 35,000 ppm of salt. Saline water is not just in the oceans. Mono Lake in California, showing salt deposits left onshore as the water level declines to serve the water needs of Los Angeles.
http://ga.water.usgs.gov/edu/saline.html
What Is Salinity Of Seawater Ppm? - Find Questions and Answers at Askives, the first startup that gives you an straight answer
http://www.askives.com/what-is-salinity-of-seawater-ppm.html
Seawater temperature map showing areas of warmer water in red and areas of cooler water is blue. ... Oxygen is measured in parts per million (also called ppm) and levels can range from zero to over 20 ppm in temperate waters.
http://www.marinebio.net/marinescience/02ocean/swcomposition.htm
Water Classifications The "Glossary of Salt Water" published by the Water Quality Association classifies water as follows: Fresh: 1,000 ppm TDS
http://www.pacificro.com/watercla.htm
Composition of Sea Water. The elements contained in sea water and their percentage.
http://mistupid.com/chemistry/seawatercomp.htm
Moderately saline water - From 3,000 ppm to 10,000 ppm Highly saline water - From 10,000 ppm to 35,000 ppm Ocean water has a salinity that is approximately 35,000 ppm. ... sea water (atoms): 55.3 % Chlorine: 30.8 % Sodium: 3.7 % Magnesium: 2.6 % Sulfur: 1.2 % Calcium:
http://www.windows2universe.org/earth/Water/dissolved_salts.html
Seawater. Water has great abundance on the Earth, and of that abundance about 97% is sea water. Sea water contains about 3.5% by weight of salt (sodium chloride).
http://hyperphysics.phy-astr.gsu.edu/hbase/Chemical/seawater.html
Problem #1: Sea water contains 3.90 x 10¯ 6 ppm of dissolved gold. What volume of this sea water would contain 1.00 g of gold? ... Problem #2: Pollutants in air and water are frequently measured in parts per million (ppm) or parts per billion (ppb).
http://www.chemteam.info/Solutions/ppm1.html
4. ppm = parts per million = µg/g or mg/kg for liquids and solids = µL/L for mixtures of gases = ppmv •"ppm" is commonly used for solids, whereas "mg/kg" is generally ... seawater is being determined using existing analytical measurements
http://www.soest.hawaii.edu/oceanography/courses/OCN623/Spring2012/Salinity2012web.pdf
The salt in seawater isn't just sodium chloride. There's potassium salts, epsom salts, iodine and magnesium salts, all kinds of other stuff. You can, in principle, safely have a few gulps of seawater; happens all the time, you choke a bit and don't die, but you'll also need enough ...
http://ask.metafilter.com/158682/How-Much-Seawater-Can-You-Safely-Drink
Standard Seawater: Seawater with TDS of 35,000 mg/l (ppm) is considered as "standard seawater" since this constitutes by far the largest amount of water worldwide.
http://www.hohusa.net/standard_seawater.html
Seawater Salt Concentrations. In this field of endeavor, certain 'buzz words' are frequently thrown around, and we need to define these 'labels' in order to have a decent basis of understanding what seawater is all about.
http://oceanplasma.org/documents/chemistry.html
Salinity is a measure of total dissolved salts in sea water. Formally, it is calculated by weight as the amount of salt (in ... and to convert the designation of salinity from parts per thousand (PPT), parts per million (PPM) and percentage (%). Most salinity testing readers designate the ...
http://www.csgnetwork.com/poolsalinitycalc.html
Seawater desalination to convert seawater into potable water is being used in many parts of the world. Reverse Osmosis process using thin-film composite membranes has evolved over the last 20 years and has brought down the cost of desalination.
http://www.watertreatmentguide.com/seawater_desalination.htm
This is approximately 35000 ppm TDS. Seawater has a high proportion of sodium chloride and this is around 28000 ppm. Now the more complete answer. Different salts in water have a different ability to conduct electricity. This is ...
http://www.appslabs.com.au/salinity.htm
What is the concentration of chloride in sea water? Seawater typically contains chloride at a concentration of about 19,000 parts per million (ppm), or 19,000 milligrams per liter (mg/L). It also...
http://www.answerbag.com/q_view/2150499
seawater contains 7.8 x 10^-3 g Sr2+ per kilogram of water. What is the concentration of Sr2+ measured in ppm? I'm so confused with this one, I think I'm making it harder than it is...
http://answers.yahoo.com/question/index?qid=20080205191905AANf3AW
Related Questions. How do you find parts per million (ppm) in chemistry? Find the ratio between the components your using. Then make the total 1,000,000....
http://www.chacha.com/question/how-many-parts-per-million-%28ppm%29-of-salt-is-in-seawater
This is the same as 0.5%, or 5000 ppm (parts per million). When 1 cubic foot of freshwater evaporates, it leaves less than an ounce of salt ... Salinity of water other than sea water and freshwater. Water of salinity between freshwater and seawater is called brackish.
http://facstaff.gpc.edu/~pgore/Earth&Space/salinity.html
What is the Ppm of salt in ocean water? In: Home & Garden, Pool Building and Repair, Pool Care and Cleaning [Edit categories] Answer: 10,800 PPM. Improve answer. First answer by Robbb. Last edit by Robbb. Contributor trust: 7663 ...
http://wiki.answers.com/Q/What_is_the_Ppm_of_salt_in_ocean_water
Chloride is the dominant anion in seawater, where it is present in concentrations of 19,300 ppm. (Seawater has a total dissolved salt content of 36,000 ppm, where freshwater generally has a total salt content of less than 500 ppm). Chloride salts are ...
http://fins.actwin.com/aquatic-plants/month.9701/msg00303.html
Main Forums > Convert and Calculate ... I want to know how to convert a 45g/l of known componds at a density of 1.2g/cc to ... For liquids, ppm is the same as mg/L while for solids ppm is the same as mg/kg 45 g/L = 45000 mg/L = 45000 ppm For solids, you'd need to know the density to convert mg/L ...
http://forum.onlineconversion.com/showthread.php?t=986
79.909 ppm 110,000 883,000 10,800 19,400 1,290 904 392 411 67.3 Element Molybdenum Mo Ruthenium Ru Rhodium Rh Palladium Pd Argentum (silver) Ag Cadmium Cd Indium In ... Sea water is a mixture because it contains dissolved constituents, ...
http://www.answers.com/mt/sea-water
They mostly use seawater with an average TDS of 40.000 ppm and brackish water with an average TDS of 10.000 ppm and can be found in each continent. West Asia & Middle east: 60%: North America: 11%: North Africa: 7%: Europe: 7%: South and Central America: 4: Other: 11%:
http://www.lenntech.com/composition-seawater.htm
The biggest are water temperature, pressure and salinity (high salinity is seawater with more ppm, low salinity is less) – with temperature being the most significant.
http://www.nordhavn.com/resources/tech/making_water.php
Sea Water Chlorine Concentration - posted in Industrial Professionals: Can anyone provide me with estimates of sea water chlorine concentration levels (in ppm or ppb would be most helpful). The more replies the better!Cheers
http://www.cheresources.com/invision/topic/1078-sea-water-chlorine-concentration/
Sea water - 56 mS/cm (56000 µS/cm) Max for potable water - 1055 µS/cm. Some common conductivity conversion factors are. ... Be aware that the above figures are for Sodium Chloride salt and do not correspond accurately to colloidal parts per million.
http://www.biophysica.com/conductivity.html
Seawater constitutes a rich source of various commercially important chemical elements. Much of the world’s magnesium is recovered from seawater, as are large quantities of bromine. In certain parts of the world, sodium chloride (table salt) is still obtained by evaporating seawater.
http://www.britannica.com/EBchecked/topic/531121/seawater
ppm to percent conversion. ppm to percent conversion calculator; How to convert ppm to percent; ppm to percent conversion table; ppm to percent conversion calculator
http://www.rapidtables.com/convert/number/PPM_to_Percent.htm
So I have to find the concentration of lead in seawater. I prepared a 50 mL sample of seawater containing 0.1 M KCl, 50 mM HNO3 and 30 ppm HG2+.
http://answers.yahoo.com/question/index?qid=20061103063041AAd2HML
Composition of Seawater by Various Methods: Today we will look at our seawater samples for specific metal ions. ... What dilution is necessary to have a sample that will contain sodium in the 1 ppm range? Remember that next week, you will be doing a mini-project, ...
http://www.ric.edu/faculty/organic/helmss/AAWeek.html
How to Measure the Salinity of Sea Water; How to Lower Salinity Levels;... Tools for Measuring Abiotic Ecological Factors. ... Some salinity-measuring devices do this conversion but are not accurate at concentrations higher than about 70,000 ppm. ... How to Measure the...
http://www.ehow.com/how_6006803_measure-salinity-sea-water.html
If you have a salt water pool, the majority of TDS in your pool is the salt which is about 3000 ppm, ... Sea Water / Salt Water Aquarium Salt Testing: Please select the digital refractometer for sea water or MI306 full range salt meter. eSeasongear.
http://www.eseasongear.com/salinitymeter.html
How much CO 2 gas can sea water hold? Grade Level: Secondary. National Science Education Standards. Physical Science: ... in the air from about 270 parts per million (ppm) to about 360 ppm. CO2 and other gases in the atmosphere trap heat energy that would otherwise escape to space.
http://coexploration.org/bbsr/classroombats/html/co2_in_the_sea.html
Water is defined as freshwater when its salt concentration is less than 1,000 parts per million (ppm). This is also the general limit for drinking water, although drinking water should be less than 600 ppm for palatability. ... How to Determine the Salinity of Sea Water.
http://www.ehow.com/about_6626245_water-salinity-testing.html
Seawater contains 950 ppm Cl- and density of seawater is 1.02. Calculate the molarity of Cl-? 2 years ago; Report Abuse
http://malaysia.answers.yahoo.com/question/index?qid=20110925051821AAjDQfL
What is the salt concentration of sea water? about 3% i think. What is the chloride content of sea water? usaully 1.93% but this varies wit depth and location
http://wiki.answers.com/Q/What_is_chloride_concentration_in_sea_water_ppm
Humans cannot drink the seawater as it contains salt and is saline water. But, luckily saline water can be made into freshwater, so that there is enough water for drinking washing and growing crops and for everyday use.
http://www.desalinatedwater.info/home.php
Seawater is about 96.5% water and about 3.5% dissolved salts. Most of the salt in seawater is sodium chloride, but it also has...
http://www.wisegeek.com/what-are-the-chemical-properties-of-seawater.htm
When 1 cubic foot of sea water evaporates, it leaves about 2.2 pounds of sea salt. ... or 5000 ppm (parts per million). When 1 cubic foot of freshwater evaporates, it leaves less than an ounce of salt (0.01 pound of salt). Variations in salinity.
http://facstaff.gpc.edu/~pgore/Earth&Space/GPS/Salinity.html
While all are referred to as ppm (parts per million), they are still different, just as degrees Celsius ... If a DC current is applied to seawater, numerous reactions take place when the ions hit the electrodes. Some ions will plate out ...
http://www.reefkeeping.com/issues/2004-04/rhf/feature/
Sea water ranges from 30,000 to 40,000 ppm. Many brackish ground water supplies are used around California and we have many clients whose private well water has a TDS of 1500 - 2000 ppm. In some cases the levels exceed 7000 ppm. Generally, one wants a TDS of less ...
http://www.advanced-water-systems.com/problems/salinity.html
If you didn't find what you were looking for you can always try Google Search
Add this page to your blog, web, or forum. This will help people know what is What is SEAWATER PPM | 3,759 | 14,248 | {"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-2013-48 | longest | en | 0.833239 |
https://www.coursehero.com/file/88800/ECE-81-hw8-solutions/ | 1,495,935,309,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609404.11/warc/CC-MAIN-20170528004908-20170528024908-00445.warc.gz | 1,058,272,234 | 94,326 | ECE 81_hw8_solutions
# ECE 81_hw8_solutions - ECE 81 Lehigh University HW#8 P5.31...
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ECE 81 Lehigh University 02/15/2007 HW #8 ± P5.31: Z = R + jL ω + 1/jC ω = 50 + j[100.10 -3 ω – 1/10.10 -6 ω ] = 50 + j[10 -1 ω – 10 5 / ω ] ω = 500 Æ Z = 50 - 150j = 158.1 -71.56° (-1.25 rad) ω = 1000 Æ Z = 50 + 0j = 50 (0 rad) ω = 2000 Æ Z = 50 + 150j = 158.1 71.56°
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## This homework help was uploaded on 04/07/2008 for the course ECE 081 taught by Professor Frey during the Spring '07 term at Lehigh University .
Ask a homework question - tutors are online | 268 | 686 | {"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-2017-22 | longest | en | 0.831632 |
https://cn.tradingview.com/chart/EURUSD/v7A3ovLV-Arbitrary-Price-Pivots-StdDev3/ | 1,531,994,385,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590794.69/warc/CC-MAIN-20180719090301-20180719110301-00467.warc.gz | 631,873,871 | 74,632 | # Arbitrary Price Pivots; StdDev3^
FX:EURUSD 欧元/美元
123 5
Update to the previous version of Arbitrary Price Pivot tool. Here we have specific pivot levels based on the Standard Distribution Deviations. You will see three levels of each deviation greater than 97.7% (2.0). For example, standard deviation 3.09 has a ratio of 99.99% certainty price will return to the mean. So we have 3.09, 6.18 and 9.27...three levels of that particular deviation. If price breaks through 3.09 it most likely will continue through to the next level, 6.18 and perhaps 9.27. All relative to time frame used.
## 相关观点
//Arbitrary Price Pivots/StdDev 3; ^, standard distribution deviations, set your own pivot levels
study("Arbitrary Price Pivots/StdDev3^; ",overlay=true)
p = input(000.0000,title="Pivot")
p0 = input(1.0050, title="p0")
p1 = input(1.0100, title="p1")
p2 = input(1.0128, title="p2")
p3 = input(1.0165, title="p3")
p4 = input(1.0200, title="p4")
p5 = input(1.0233, title="p5")
p6 = input(1.0258, title="p6")
p7 = input(1.0309, title="p7")
p8 = input(1.0400, title="p8")
p9 = input(1.0466, title="p9")
p10 = input(1.0516, title="p10")
p11 = input(1.0600, title="p11")
p12 = input(1.0618, title="p12")
p13 = input(1.0699, title="p13")
p14 = input(1.0774, title="p14")
p15 = input(1.0927, title="p15")
plot(p,"Pivot",color=green)
plot(p*p0, title="38.2%, 0.50stddev", color=aqua)
plot(p/p0, title="38.2%, 0.50stddev", color=aqua)
plot(p*p1,"84%, 1.00stddev",color=white)
plot(p/p1,"84%, 1.00stddev",color=white)
plot(p*p2,"90%, 1.28 stddev",color=lime)
plot(p/p2,"90%, 1.28 stddev",color=lime)
plot(p*p3,"95%, 1.65 stddev",color=olive)
plot(p/p3,"95%, 1.65 stddev",color=olive)
plot(p*p4,"97.7%, 2.00 stddev",color=blue)
plot(p/p4,"97.7%, 2.00 stddev",color=blue)
plot(p*p5,"99%, 2.33 stddev",color=yellow)
plot(p/p5,"99%, 2.33 stddev",color=yellow)
plot(p*p6,"99.5% 2.58 stddev",color=orange)
plot(p/p6,"99.5% 2.58 stddev",color=orange)
plot(p*p7,"99.99% 3.09 stddev",color=red)
plot(p/p7,"99.99% 3.09 stddev",color=red)
plot(p*p8,"4.0, 2.00 stddev 2^",color=blue)
plot(p/p8,"4.0, 2.00 stddev 2^",color=blue)
plot(p*p9,"4.66, 2.33 stddev 2^",color=yellow)
plot(p/p9,"4.66, 2.33 stddev 2^",color=yellow)
plot(p*p10,"5.16, 2.58 stddev 2^",color=orange)
plot(p/p10,"5.16, 2.58 stddev 2^",color=orange)
plot(p*p11,"6.18, 3.09 stddev 2^",color=red)
plot(p/p11,"6.18, 3.09 stddev 2^",color=red)
plot(p*p12,"6.0, 2.00 stddev 3^",color=blue)
plot(p/p12,"6.0, 2.00 stddev 3^",color=blue)
plot(p*p13,"6.99, 2.33 stddev 3^",color=yellow)
plot(p/p13,"6.99, 2.33 stddev 3^",color=yellow)
plot(p*p14,"7.74, 2.58 stddev 3^",color=orange)
plot(p/p14,"7.74, 2.58 stddev 3^",color=orange)
plot(p*p15,"9.27, 3.09 stddev 3^",color=red)
plot(p/p15,"9.27, 3.09 stddev 3^",color=red)
coondawg71
Thank you for the script .. .but when I add it to the chart through pine editor it doesn't show anything.
Ede
It was my fault. Thank you very much coondawg71 .
Where can we get the script ?
Ede
Sorry pal, I never saw the comment. I will post the script shortly.
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AR العربية | 1,288 | 3,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-30 | longest | en | 0.624269 |
https://community.appsheet.com/t/hello-community-i-need-to-limit-a-numerical/310 | 1,558,857,059,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258862.99/warc/CC-MAIN-20190526065059-20190526091059-00011.warc.gz | 422,716,679 | 4,704 | # Hello community I need to limit a numerical ...
(Alfredo Pou) #1
Hello community
I need to limit a numerical value, so that the user can only download it, but not increase it. I thought to put a Validif: [THIS] <[ THISROW]. [Quantity]
The expression assistant accepts it, but the comparison always gives invalid. The idea is to compare the value stored in the cell with the value that is being entered in the form. I do not understand what I am doing wrong, and what would be the correct way to do that restriction?
Thank you
(Steven Coile) #2
The user may adjust the value, but only downward?
I’m guessing the column in question is Quantity, and that your formula:
[__THIS] < [__THISROW].[Quantity]
is an attempt to compare the user-adjusted value ([__THIS]) against the initial value upon entering the form ([__THISROW].[Quantity]).
(Note that I’ve doubled the underscores in front of THIS and THISROW to avoid formatting problems with Google+. Remove the second underscore in practice.)
You’re on the right track thinking you have to refer to the initial value in some different way, but [__THISROW].[Quantity] and [__THIS] in fact refer to the same value: the copy of the column value the form is using. Your expression is exactly equivalent to [__THIS] < [__THIS]. The specific reason your inputs are always invalid is because you Valid_If insists that the Quantity input must always be less than itself, which is always false.
To get the initial value of the column, you have to tell AppSheet to look at the table by using a function that pulls values from tables, like LOOKUP() or SELECT():
[__THIS] < LOOKUP([__THISROW], “MyTable”, “KeyColumn”, “Quantity”)
or:
[__THIS] < ANY(SELECT(MyTable[Quantity], ([KeyColumn] = [__THISROW])))
(Alfredo Pou) #3
Oh, okay! I thought that [__ THISROW] referred to the row in the spreadsheet. I understood the concept and I thank you very much, but I have a curiosity, if the value was less than zero, It was not invalid. That is, it allowed me to store negative numbers, which made me think that I was not taking the correct value and interpreted it as zero. According to your explanation, it should never be valid. Things of technology … | 512 | 2,201 | {"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-2019-22 | latest | en | 0.908548 |
https://topqa.wiki/what-is-the-percent-of-one-third/ | 1,701,897,564,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100603.33/warc/CC-MAIN-20231206194439-20231206224439-00025.warc.gz | 637,048,018 | 14,472 | # What is the percent of one third
Here are the best information about What is the percent of one third voted by users and compiled by us, invite you to learn together
## 1 What is one third of 100?
• Author: valeur.com
• Published Date: 10/27/2021
• Review: 4.86 (856 vote)
• Summary: You can also write it as a decimal by simply dividing 1 by 3 which is 0.33. If you multiply 0.33 with 100 you will see that you will end up with the same answer
## 2 We spend about one-third of our life either sleeping or attempting to do so – PubMed
• Author: pubmed.ncbi.nlm.nih.gov
• Published Date: 06/12/2022
• Review: 4.79 (574 vote)
• Summary: We spend about one-third of our life either sleeping or attempting to do so. Handb Clin Neurol. 2011;98:vii. doi: 10.1016/B978-0-444-52006-7.00047-2
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## 3 What is one third as a percentage? – Answers
• Published Date: 05/07/2022
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• Summary: Fractions And Percentage Converter / Fractions / One Third Or .(3) [1/3] Online converter page for a specific unit. Here you can make instant conversion
## 5 Have one-third of small businesses in the US closed due to COVID?
• Author: statesman.com
• Published Date: 10/13/2021
• Review: 4.03 (248 vote)
• Summary: · But what percentage shut their doors during the pandemic? Vice President and former California Sen. Kamala Harris claimed during a recent
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• Author: omnicalculator.com
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• Author: ourworldindata.org
• Published Date: 03/05/2022
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• Summary: · In the millennia since then a growing demand for agricultural land means we’ve lost one-third of global forests – an area twice the size of the
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• Author: skillsyouneed.com
• Published Date: 07/18/2022
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## 9 One third of 1206 is what per cent of 134? a 3% b30% c20% d300%
• Author: byjus.com
• Published Date: 01/03/2022
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• Summary: 6 : 5 when expressed as a percentage, is · Q. In an examination, Nitin gets 98 marks. This amounts to 56% of the maximum marks. What are the maximum marks ?
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## 10 Thirds converted to percentages – Percentage Calculator
• Author: percentagecalculator.co.uk
• Published Date: 03/03/2022
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• Summary: Thirds converted to percentages ; 1/3 as a percent · Multiply this number by the nominator: 33.333… × 1 = 33.333… ; 2/3 as a percent · Multiply this number by the
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## 11 33 Percent is One-Third – SlideShare
• Author: slideshare.net
• Published Date: 07/16/2022
• Review: 2.91 (75 vote)
• Summary: 33 Percent is One-Third … 1/3 is 33 1/3 %; To find 33 1/3 % divide by 3. Learnist Board: http://bit.ly/13AGhZq
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Here are the best information about where to watch what is a woman voted by users and compiled by us, invite you to learn together | 1,676 | 6,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-50 | latest | en | 0.890086 |
https://phet.colorado.edu/be/simulation/legacy/fourier | 1,627,253,342,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151866.98/warc/CC-MAIN-20210725205752-20210725235752-00253.warc.gz | 467,457,489 | 24,247 | # Fourier: Making Waves
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Выкарыстоўвайце гэты HTML-код для адлюстравання малюнка з надпісам "Клікніце запуск". Хвалі Сінус Косінус PhET падтрымліваюць і такія ж педагогі, як вы.
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• Хвалі
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### Апісанне
Learn how to make waves of all different shapes by adding up sines or cosines. Make waves in space and time and measure their wavelengths and periods. See how changing the amplitudes of different harmonics changes the waves. Compare different mathematical expressions for your waves.
### Прыклад Мэты навучання
• Explain qualitatively how sines and cosines add up to produce arbitrary periodic functions.
• Recognize that each Fourier component corresponds to a sinusoidal wave with a different wavelength or period.
• Mentally map simple functions between Fourier space and real space.
• Describe sounds in terms of sinusoidal waves.
• Describe the difference between waves in space and waves in time.
• Recognize that wavelength and period do not correspond to specific points on the graph but indicate the length/time between two consecutive troughs, peaks, or any other corresponding points.
• Become comfortable with various mathematical notations for writing Fourier transforms, and relate the mathematics to an intuitive picture of wave forms.
• Determine which aspect of a graph of a wave is described by each of the symbols lambda, T, k, omega, and n.
• Recognize that lambda & T and k & omega are analogous, but not the same.
• Translate an equation from summation notation to extended notation.
• Recognize that the width of a wave packet in position space is inversely related to the width of a wave packet in Fourier space.
• Explain how the Heisenberg Uncertainty principle results from the properties of waves.
• Recognize that the spacing between Fourier components is inversely related to the spacing between wave packets, and that a continuous distribution of fourier components leads to a single wave packet.
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Агляд сім-кантролю, мадэлі спрашчэнняў і здольнасці пранікнення ў сутнасць мыслення навучэнцаў ( PDF ).
### Педагог-Прадстаўлена дзейнасць
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Албанская All shqip Analiza Furie: prodhimi i valëve
Амхарская All Amharic አራትዮሽ፡ ሞገድ መስራት
Балгарская All български Фурие: Получаване на вълни
Баскская All Euskara Fourier: Uhinak sortzen
Баснійская All Bosanski FURIJE-OVA SINTEZA - PRAVLJENJE TALASA
В'етнамская All Tiếng Việt Tổng hợp sóng
Венгерская All magyar Fourier: hullám-LEGO
Грэчаская All Ελληνικά Fourier: Δημιουργία Κυμάτων
Дацкая All Dansk Fourir: Bølgefremstilling
Інданезійская All Bahasa Indonesia Fourier : Mengkontruksi Gelombang
Іспанская All español Fourier: Fabricacion de Ondas
Іспанская (Мексіка) All español (México) Fourier: Fabricacion de Ondas
Іспанская (Перу) All español (Perú) Fourier: Haciendo Ondas
Італьянская All italiano Fourier: Costruire le onde
Іўрыт All עברית פוריה: עושים גלים
Карэйская All 한국어 푸리에: 파동 생성
Кітайская (спрошчаная) All 中文 (中国) 傅里叶:生成波
Кітайская (традыцыйная) All 中文 (台灣) Fourier: Making Waves 傅力葉:與波共舞
Кхмерская All Khmer Fourier: ការបង្កើតរលក
Лао All Lao ຟູເຍ້: ການເຮັດໃຫ້ຄື້ນຟອງ
Македонская All македонски ФУРИОВА СИНТЕЗА -ПРАВЕЊЕ НА БРАНОВИ
Мангольская All Монгол (Монгол) Фурьей: Долгион хийх
Нідэрландская All Nederlands Boventonen
Нямецкая All Deutsch Wellen zusammensetzen nach FOURIER
Партугальская All português Séries de Fourier: Fazendo Ondas
Партугальская (Бразілія) All português (Brasil) Fourier: Criando Ondas
Польская All polski Fourier: Konstruowanie fal
Сербская All Српски ФУРИЈЕ-ОВА СИНТЕЗА -ПРАВЉЕЊЕ ТАЛАСА
Славацкая All Slovenčina Fourier: tvorenie vĺn
Туркменская All Turkmen Furýe: tolkun ýasamak
Турэцкая All Türkçe Fourier Serileri: Dalga Yapalım
Фарсі All فارسی تولید موج ها
Французская All français Fourier: Création d'ondes
Харвацкая All hrvatski Kompozicija valova (Fourierova sinteza)
Эстонская All Eesti Fourier: Lainete tekitamine
Японская All 日本語 フーリエ級数展開:波の合成
Windows Macintosh Linux
Microsoft Windows
XP/Vista/7/8.1/10 | 1,979 | 5,705 | {"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-2021-31 | latest | en | 0.474693 |
https://iq.opengenus.org/commonly-used-neural-networks/ | 1,695,874,318,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00042.warc.gz | 361,062,399 | 24,458 | ×
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# Commonly Used Neural Networks
#### Machine Learning (ML) Deep Learning
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Neural Networks are soft computing systems which aim at simulating the function of the human brain. Since the human brain is fantastic at reasoning, processing and learning in environments of noise, uncertainty and imprecision; implementing Neural Networks allows us to solve very complex, mathematically ill-defined problems using very simple computational operations (such as addition, multiplication, and fundamental logic elements). They are essentially networks consisting of multiple artificial neurons linked together with different connection weights. They are commonly used in pattern recognition, classification, and to find the relationship between the inputs that are fed to a system and its corresponding outputs.
In this article, we will discuss the functioning of a few commonly used Neural Networks, namely:
1. Hebbian Neural Networks
2. Auto-Associative Neural Networks
3. Hopfield Neural Networks
4. Radial Basis Function Neural Networks
5. Kohonen's (Self Organizing Map) Neural Network
6. LVQ (Learning Vector Quantization) Neural Networks
# Hebbian Neural Networks
Hebbian Neural Networks, also called Hebb Neural Networks, are one of the simplest types of Artificial Neural Networks. They are feed-forward Neural Networks, meaning that they are unsupervised, since there is no output label associated with the input data during the training process. It consists of one input layer (with multiple input neurons) and one output layer (with a singular output neuron). They work on bipolar data (-1 and 1) and NOT binary data (0 and 1).
These Neural Networks use what is known as Hebb's rule, which states:
If neuron 'A' frequently activates a nearby neuron 'B', then some growth process or metabolic change occurs which makes neuron 'A' more effective at activating neuron 'B'. The inverse also holds true.
Let us now take a look at the training algorithm:
(Assuming that the input vector is Xi, and the target output vector is Y)
1. Initialize all of the weights to 0.
Thus, Wi = 0
2. Now, update the weights for each training sample.
Wi(new) = Wi(old) + ∆W,
where ∆W = Xi + Y (Change in weight depending on how the neurons interact).
Now that the weights have been updated based on the input vector, if we feed the Hebbian Neural Network the same input vector once again, we will find that it outputs '1', meaning that it has successfully recognized the input vector.
# Auto-Associative Neural Networks
Auto-Associative Neural Networks are used to retrieve a piece of complete data from only a small sample of available data. In order to understand this better, let's turn to a real world example. If a burglar were to commit a crime, and then choose to commit the same crime the next day, only while wearing a face mask this time around (which covers, say, half of the burglar's face), then the concerned authorities will still more than likely be able to deduce that both of these crimes were committed by the same person despite the person's face not being entirely visible on the second day, due to similarities in the way that the crimes were committed and since half of the burglar's face was visible the second time around, the other half can be (approximately) reconstructed. Auto-Associative Neural Networks work in a similar manner. They contain the same number of input as well as output nodes, and also work on bipolar data and not binary data.
Let us now look at a simplified version of the training algorithm:
(Assuming the input vector is Xi, the target output vector is Y, and the weight matrix is Wij)
1. Initialize the weight matrix as:
Wij = Σ Xi' * Xi, for i=1 to n, and Xi' is Xi transpose.
2. Now, by setting the above weights to the Auto-Associative Neural Network and by
applying the same input vector,
Y = Σ Xi * Wij
We find that the Neural Network obtains an output that is equivalent to the
input, OR, we can say that the Neural Network has successfully retained
the pattern.
Now, even if we modify the input vector, that is, if the input vector contains one or two missing or incorrect data entries, we will find that the Neural Network will still be able to recall the complete input pattern.
# Hopfield Neural Networks
Hopfield Neural Networks are very similar to Auto-Associative Neural Networks. The major difference between the two is that there's an element of back propagation in Hopfield Neural Networks. Here, the output of each neuron (in the output layer) is 'fed back' to every other neuron in the output layer, except itself. Hopfield Neural Networks work with both bipolar as well as binary data.
The training algorithm is essentially the same as that of Auto-Associative Neural Networks, however, there are differences in the testing algorithm, which is as follows:
1. Find the weight matrix with no self connection, i.e. where the diagonal elements
are 0 owing to the lack of self feedback.
2. Represent the input vector (Xi) in binary.
3. Assign Yi = Xi
4. Find:
Yin = Xi + Σj[YjWji],
where Xi represents each individual element in the input vector.
Once we find the value of Yin, it is then broadcasted to every other neuron in the output layer, except itself. If the output vector converges to the required input vector, then we can stop the entire process. If it doesn't converge, we repeat the above processes.
# Radial Basis Function Neural Networks
Radial Basis Function (or RBF) Neural Networks, are 3 layer Neural Networks. This means that they have an input layer, a hidden layer, as well as an output layer. These Neural Networks always have MORE neurons in the hidden layer than in the input layer. This allows for an increase in the data's dimensionality, which makes it possible for the Neural Network to separate linearly non-separable data. Thus, Radial Basis Function Neural Networks can, for example, be implemented to find the output of the XOR gate truth table. Each neuron in the hidden layer computes a value via the Gaussian (Radial Basis) Function, as mentioned further below.
Let us take a closer look at the RBF implementation algorithm:
1. Take any input neuron as a receptor neuron.
2. Calculate its Euclidean distance from every other input neuron via:
r = √[(x2-x1)^2+(y2-y1)^2]
3. Substitute this value of r in the Gaussian function:
hi(x) = exp(-r^2/2σ^2),
where σ is the Standard Deviation, which we can assume to be 1 for simplicity.
4. Find:
ΣWihi(x),
where Wi represents the weights and hi(x) represents the value that we
obtained via the above steps.
Once we've applied our activation function to the values of ΣWihi(x), we will find that the Neural Network gives us the necessary output.
# Kohonen's Neural Network (Self Organising Map)
Kohonen's Neural Network, unlike the Radial Basis Function Neural Network, aims at comprehensively visualizing extremely large datasets. This is done so by reducing the dimensionality of the dataset, by grouping the data into multiple clusters. This means that instead of considering the characteristics of each individual data point, we cluster the data together and reference each cluster via a representative data point. All of the elements of that cluster share similar characteristics to that of the representative data point, which means we can deal with a very large number of data points without making a very large number of references. Here, there is an input layer as well as an output layer (Kohonen's layer), but there is NO hidden layer. Every neuron in the input layer is connected to every neuron in the output layer (Kohonen's layer), thus forming a fully inter-connected Neural Network.
The main objective in SOM is to calculate the Euclidean distance of each input data vector to that of the columns of the weight matrix (which represent the clusters), and to find which cluster it is closest to. This means that the input vector belongs to that particular cluster. Thus, we update only that column of the weight matrix (to which the input vector is closest to), as this implies that the centroid of the cluster has moved closer to that particular input vector.
Let us now take a closer look at the steps we need to follow when we implement SOM (Kohonen's NN):
1. Calculate the Euclidean distance of the input vectors from each column of the
weight matrix (which represents each cluster):
r = √[(x2-x1)^2+(y2-y1)^2]
2. Pick the winning neuron based on which cluster the input vector is closest
to, i.e. whichever value of r is smaller.
3. Update that particular weight vector as:
Wij(new) = Wij(old) + α [Xi - Wij(old)]
These trained weights can then be used to classify new examples.
We now arrive at the final topic of discussion in this article, Learning Vector Quantization (LVQ) Neural Networks!
# Learning Vector Quantization (LVQ) Neural Networks
Learning Vector Quantization (LVQ) Neural Networks are very similar to Kohonen's (Self Organizing Map) Neural Networks. The only real difference is in the way that we obtain the weight matrix, and the way in which we update the weights. LVQ is a supervised classification algorithm, since we are aware of the output class label during the training process. In SOM, we start off with a random weight matrix, which we keep updating. In LVQ, however, we form the weight matrix by choosing input vectors that belong to distinct classes.
For example, if our input vectors are:
[1 0 0 1] - Class 1
[1 1 1 1] - Class 2,
then our weight matrix (W) will be:
[1 0
0 1
0 1
1 1]
The rest of the procedure remains largely the same:
1. We find the Euclidean distance of the input vectors from the columns of the weight matrix (which represent the individual clusters), via:
r = √[(x2-x1)^2+(y2-y1)^2]
2. We then pick the winning neuron based on which cluster the input vector is closest to, i.e. whichever value of r is smaller.
3. Now, we check if the distance that we calculated matches the class label.
4. If it does, then the update rule is as follows:
Wij(new) = Wij(old) + α [Xi - Wij(old)]
5. If it does not, then the update rule is as follows:
Wij(new) = Wij(old) - α [Xi - Wij(old)]
These trained weights can then be used to classify new training examples.
Thank you for reading my article on 'Commonly Used Neural Networks' at OpenGenus!
#### Agastya Gummaraju
Agastya Gummaraju has been a Machine Learning Developer, Intern at OpenGenus. He is a Computers & Communication Undergraduate at Manipal Institute of Technology. | 2,362 | 10,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-40 | longest | en | 0.920399 |
https://www.physicsforums.com/threads/critical-numbers-confusion.675942/ | 1,532,048,659,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591455.76/warc/CC-MAIN-20180720002543-20180720022543-00035.warc.gz | 960,342,204 | 12,884 | # Critical Numbers Confusion
1. Mar 3, 2013
### algorithmia
I am a bit confused over something that should be relatively easy to research , however, I am having a hard time finding a direct answer to my question.
When finding the extrema of a function , we find at what points the first derivative is 0 or undefined .. with the stipulation , if Im not mistaken , that the function itself IS defined at those values(continuous but not necessarily differentiable at those points) .. if they are not , then they will not be critical numbers per the definition . .
when finding concavity / points of inflection .. i am assuming the continuity requirement for the original function with respect to the critical numbers of the second derivative is dropped . am i right ? ..also , can I automatically assume if there are discontinuties in the original function , those will serve as critical numbers for sake of determining concavity ?
i am getting conflicting information but my intuition tells me this has to be so . just by looking at a couple graphs.concavity changes between vertical asymptotes. but i just wanted to make sure .. my book sucks. =D
2. Mar 4, 2013
### runningninja
Finding the critical points of a function means where the function's derivative is either zero or undefined. So when you get a set of critical points, you are solving for both where the derivative is 0 and/or undefined.
Take the tangent function for example, which has a vertical asymptote every pi/2 + pi*k where k is an integer. Solving for critical numbers, we would get sec(x)^2 = 0. Since the secant never equals 0, the tangent never has a horizontal tangent line. But the secant is undefined at pi/2 + pi*k, which gives us where the derivative is undefined, because of the vertical asymptotes. To be clear, these asymptotes are included in these critical numbers, but are not extrema.
To solve for inflection points, we take the second derivative. Thus, we have 2tanx(secx)^2 = 0. 2(secx)^2 will never be zero, so we cancel it out. tanx = 0 at pi/4 + pi*k. However, because of the secant, the second derivative is also undefined at pi/2 + pi*k, and the function changes concavity at both of these sets. So, to clarify: the second derivative's zeros and undefined values gives us where the concavity changes. But only the zeros are considered inflection points, because the function must be defined in order for an inflection point to exist.
Hope this helps. | 552 | 2,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-30 | latest | en | 0.887006 |
https://mynewsfit.com/need-help-with-algebra-try-using-these-strategies/ | 1,726,699,884,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00668.warc.gz | 375,322,710 | 19,157 | Wed. Sep 18th, 2024
# Need Help with Algebra? Try Using These Strategies
Oct 7, 2021
If algebra is a difficult subject for you, then don’t worry—you’re not alone. It is difficult to grasp the abstractions that are involved in solving algebraic equations, and using variables with letters from the alphabet to represent numbers can be a confusing concept to understand. That being said, algebra is not impossible to figure out.
Algebra Homework Help is widely recognized as one of the most difficult subjects among individuals worldwide, and research has shown that one of the most frequently observed difficulties in algebra is the translation between problem situations and mathematical paradigms as well as the expressions, operations, and variables (Jupri et al. 483).
There are a number of meaningful ways that you can reorganize your mind and restructure the way that you perceive algebra so that it is not such an intimidating task for you. Although you are in college, a student’s fundamental understanding of algebra can be informed by their development in the earlier stages of life.
## Use a Comparative Approach
The orthodox methods of teaching algebra in the classroom can often leave students with misunderstandings of the subject. There exists, however, an alternative method for understanding how algebraic equations are built from the ground up. A methodology that utilizes a comparative analysis sheds some light on this alternative tactic.
Comparison has been shown to be a promising strategy in cognitive science. Learning algebra by comparing worked-out examples of algebraic problems has been proven to be an effective teaching strategy that is associated with greater procedural student knowledge among those who attempt to learn algebra (Star et al. 41).
These findings indicate that, by comparing already-completed solutions to algebraic problems, students that learn algebra can better familiarize themselves with the skeleton of how an algebraic function occurs and the dynamics that exist on both sides of the equal sign as well as how variables are affected accordingly.
## Remember that Algebra Isn’t as Unfamiliar as You Think
Algebra can seem to be an unfamiliar and distinct subject from other forms of mathematics, such as arithmetic. However, a different way to learn algebra is by completely shifting the manner in which you are able to view algebra as a subject.
Research has shown that a teaching approach that emphasizes the structural similarity between arithmetic and algebraic expressions improved learning outcomes for students because of their improved comprehension of symbolic transformations in the context of both algebraic and arithmetic operations as well as their connections (Banerjee and Subramaniam 351).
Further research has indicated that most of the difficulties that students face with comprehending algebra can be traced back to their prior experiences in arithmetic. Emphasizing the ways in which arithmetic and algebraic thinking are intertwined enhanced students’ natural awareness of the generalizations in numerical and non-numerical contexts(Warren et al. 74).
Additionally, research has shown that a student’s understanding of fractions is crucial to their success with algebra, with empirical evidence supporting the notion that greater knowledge of fractions was predictive of improvements in equation-solving andencoding skills when it came to algebraic expressions (Booth et al. 110).
It may seem counterintuitive to learn a different subject in order to improve yourself in another subject, but the research has indicated that a stout and robust foundation in arithmetic and fractions ispredictive of an enhanced ability to understand algebra.
As such, fortifying your fundamental understanding of basic concepts like arithmetic and fractions can bolster your capabilities in comprehending algebra. Algebra and arithmetic are not distinct subjects but are, in fact, interrelated concepts.
## Problem-Based Learning Can be the Key
Finally, additional research has indicated that students that are taught algebra using the problem-based learning approach achieved substantially higher scores on algebraic achievement tests than those who were taught algebra using conventional methods(Ajai et al. 131).
The problem-based learning paradigm begins with a problem to solve, which is posed in such a way that the student must obtain new knowledge before they can solve the problem. It is an instructional strategy based on the concept that learners must construct their own comprehension by relating concrete experience to the existing knowledge(Ajai et al. 132).
In this sense, giving students autonomy and collaborative freedom can produce a different perspective of algebra in lieu of the traditional methods in which a teacher is directly proselytizing the concepts of algebra to a body of students. This method allows the students to explore problem-solving concepts independently.
## Conclusion
Algebra might seem foreign and incomprehensible at first glance. However, you already have a head start on comprehending algebra due to your foundation in arithmetic. If you combine a strengthening of your arithmetic foundationwith a problem-based learning approach as well as a comparative approach, algebra will become a cinch for you. | 966 | 5,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.958683 |
http://conversion.org/volume/fluid-dram-us/timber-foot | 1,568,580,071,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572289.5/warc/CC-MAIN-20190915195146-20190915221146-00240.warc.gz | 44,539,525 | 7,617 | # fluid dram (US) to timber foot conversion
Conversion number between fluid dram (US) [fl dr] and timber foot is 0.00013054741753472. This means, that fluid dram (US) is smaller unit than timber foot.
### Contents [show][hide]
Switch to reverse conversion:
from timber foot to fluid dram (US) conversion
### Enter the number in fluid dram (US):
Decimal Fraction Exponential Expression
[fl dr]
eg.: 10.12345 or 1.123e5
Result in timber foot
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 fluid dram (US) = (exactly) (3.6966911953125*10^-06) / (0.028316846592) = 0.00013054741753472 timber foot
• 1 timber foot = (exactly) (0.028316846592) / (3.6966911953125*10^-06) = 7660.0519480519 fluid dram (US)
• ? fluid dram (US) × (3.6966911953125*10^-06 ("m³"/"fluid dram (US)")) / (0.028316846592 ("m³"/"timber foot")) = ? timber foot
### High precision conversion
If conversion between fluid dram (US) to cubic-metre and cubic-metre to timber foot is exactly definied, high precision conversion from fluid dram (US) to timber foot is enabled.
Decimal places: (0-800)
fluid dram (US)
Result in timber foot:
?
### fluid dram (US) to timber foot conversion chart
Start value: [fluid dram (US)] Step size [fluid dram (US)] How many lines? (max 100)
visual:
fluid dram (US)timber foot
00
100.0013054741753472
200.0026109483506944
300.0039164225260417
400.0052218967013889
500.0065273708767361
600.0078328450520833
700.0091383192274306
800.010443793402778
900.011749267578125
1000.013054741753472
1100.014360215928819
Copy to Excel
## Multiple conversion
Enter numbers in fluid dram (US) and click convert button.
One number per line.
Converted numbers in timber foot:
Click to select all
## Details about fluid dram (US) and timber foot units:
Convert Fluid dram (US) to other unit:
### fluid dram (US)
Definition of fluid dram (US) unit: ≡ 1⁄8 US fl oz. Fluid dram (or Drachm in UK spelling) = US fl oz / 8 = 3.6966911953125×10−6 m³
Convert Timber foot to other unit:
### timber foot
Definition of timber foot unit: ≡ 1 cu ft. Timperfoot is other name for cubic foot, cube with size of 1 ft × 1 ft × 1 ft = 0.028316846592 m³
← Back to Volume units | 700 | 2,231 | {"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} | 2.90625 | 3 | CC-MAIN-2019-39 | latest | en | 0.618882 |
https://mathspace.co/textbooks/syllabuses/Syllabus-1071/topics/Topic-20714/subtopics/Subtopic-269539/?activeTab=theory | 1,719,034,098,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00201.warc.gz | 325,990,525 | 53,442 | # 7.03 AM and PM
Lesson
## Are you ready?
When we use the words morning and afternoon, you are able to picture what part of the day it is. What other words help you to determine a specific part of the day? Make a list and compare it with your friends.
## Use of AM and PM
This video looks at how a 12 hr clock can be used to tell time across a day that has 24 hrs, by using AM and PM.
### Examples
#### Example 1
Choose whether these activities normally happen in the AM or PM.
a
Going to bed.
Worked Solution
Create a strategy
Think about if this happens before or after midday.
Apply the idea
We normally go to bed after midday.
After midday means PM, so the answer is PM.
b
Eating breakfast.
Worked Solution
Apply the idea
We normally eat breakfast before midday.
Before midday means AM, so the answer is AM.
c
Watching the sun rise.
Worked Solution
Apply the idea
The sun rises before midday.
Before midday means AM, so the answer is AM.
#### Example 2
James went to a movie at 11\text{:}50 am. The movie went for 1 hour and 10 minutes.
a
Would the movie finish in the AM or PM?
Worked Solution
Create a strategy
The movie starts 10 minutes before midday, then goes for an hour and 10 minutes after that.
Apply the idea
After 1 hour and 10 minutes the time will be after midday, so the movie finish in the PM.
b
Complete the statement:
The movie finishes at ⬚\text{:}⬚⬚ pm.
Worked Solution
Create a strategy
Add the 10 minutes first, and then add the 1 hour.
Apply the idea
The movie finishes at 1\text{:}00 pm.
Idea summary
AM means "before midday" and PM means "after midday."
### Outcomes
#### MA2-13MG
reads and records time in one-minute intervals and converts between hours, minutes and seconds | 452 | 1,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.905993 |
http://msdn.microsoft.com/en-US/library/system.math.tanh(v=vs.85).aspx | 1,417,153,804,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931009777.87/warc/CC-MAIN-20141125155649-00172-ip-10-235-23-156.ec2.internal.warc.gz | 212,009,283 | 13,254 | # Math.Tanh Method
.NET Framework 3.0
Returns the hyperbolic tangent of the specified angle.
Namespace: System
Assembly: mscorlib (in mscorlib.dll)
## Syntax
```public static double Tanh (
double value
)
```
```public static double Tanh (
double value
)
```
```public static function Tanh (
value : double
) : double
```
```Not applicable.
```
value
#### Return Value
The hyperbolic tangent of value. If value is equal to NegativeInfinity, this method returns -1. If value is equal to PositiveInfinity, this method returns 1. If value is equal to NaN, this method returns NaN.
## Remarks
The angle, value, must be in radians. Multiply by Math.PI/180 to convert degrees to radians.
## Example
The following example uses Tanh to evaluate certain hyperbolic tangent identities for selected values.
```// Example for the hyperbolic Math.Tanh( double ) method.
using System;
class DemoTanh
{
public static void Main()
{
Console.WriteLine(
"This example of hyperbolic Math.Tanh( double )\n" +
"generates the following output." );
Console.WriteLine(
"\nEvaluate these hyperbolic identities " +
"with selected values for X:" );
Console.WriteLine( " tanh(X) == sinh(X) / cosh(X)" );
Console.WriteLine(
" tanh(2 * X) == 2 * tanh(X) / (1 + tanh^2(X))" );
UseTanh(0.1);
UseTanh(1.2);
UseTanh(4.9);
Console.WriteLine(
"\nEvaluate [tanh(X + Y) == (tanh(X) + tanh(Y)) " +
"/ (1 + tanh(X) * tanh(Y))]" +
"\nwith selected values for X and Y:" );
UseTwoArgs(0.1, 1.2);
UseTwoArgs(1.2, 4.9);
}
// Evaluate hyperbolic identities with a given argument.
static void UseTanh(double arg)
{
double tanhArg = Math.Tanh(arg);
// Evaluate tanh(X) == sinh(X) / cosh(X).
Console.WriteLine(
"\n Math.Tanh({0}) == {1:E16}\n" +
" Math.Sinh({0}) / Math.Cosh({0}) == {2:E16}",
arg, tanhArg, (Math.Sinh(arg) / Math.Cosh(arg)) );
// Evaluate tanh(2 * X) == 2 * tanh(X) / (1 + tanh^2(X)).
Console.WriteLine(
" 2 * Math.Tanh({0}) /",
arg, 2.0 * tanhArg );
Console.WriteLine(
" (1 + (Math.Tanh({0}))^2) == {1:E16}",
arg, 2.0 * tanhArg / (1.0 + tanhArg * tanhArg ) );
Console.WriteLine(
" Math.Tanh({0}) == {1:E16}",
2.0 * arg, Math.Tanh(2.0 * arg) );
}
// Evaluate a hyperbolic identity that is a function of two arguments.
static void UseTwoArgs(double argX, double argY)
{
// Evaluate tanh(X + Y) == (tanh(X) + tanh(Y)) / (1 + tanh(X) * tanh(Y)).
Console.WriteLine(
"\n (Math.Tanh({0}) + Math.Tanh({1})) /\n" +
"(1 + Math.Tanh({0}) * Math.Tanh({1})) == {2:E16}",
argX, argY, (Math.Tanh(argX) + Math.Tanh(argY)) /
(1.0 + Math.Tanh(argX) * Math.Tanh(argY)) );
Console.WriteLine(
" Math.Tanh({0}) == {1:E16}",
argX + argY, Math.Tanh(argX + argY));
}
}
/*
This example of hyperbolic Math.Tanh( double )
generates the following output.
Evaluate these hyperbolic identities with selected values for X:
tanh(X) == sinh(X) / cosh(X)
tanh(2 * X) == 2 * tanh(X) / (1 + tanh^2(X))
Math.Tanh(0.1) == 9.9667994624955819E-002
Math.Sinh(0.1) / Math.Cosh(0.1) == 9.9667994624955819E-002
2 * Math.Tanh(0.1) /
(1 + (Math.Tanh(0.1))^2) == 1.9737532022490401E-001
Math.Tanh(0.2) == 1.9737532022490401E-001
Math.Tanh(1.2) == 8.3365460701215521E-001
Math.Sinh(1.2) / Math.Cosh(1.2) == 8.3365460701215521E-001
2 * Math.Tanh(1.2) /
(1 + (Math.Tanh(1.2))^2) == 9.8367485769368024E-001
Math.Tanh(2.4) == 9.8367485769368024E-001
Math.Tanh(4.9) == 9.9988910295055444E-001
Math.Sinh(4.9) / Math.Cosh(4.9) == 9.9988910295055433E-001
2 * Math.Tanh(4.9) /
(1 + (Math.Tanh(4.9))^2) == 9.9999999385024030E-001
Math.Tanh(9.8) == 9.9999999385024030E-001
Evaluate [tanh(X + Y) == (tanh(X) + tanh(Y)) / (1 + tanh(X) * tanh(Y))]
with selected values for X and Y:
(Math.Tanh(0.1) + Math.Tanh(1.2)) /
(1 + Math.Tanh(0.1) * Math.Tanh(1.2)) == 8.6172315931330645E-001
Math.Tanh(1.3) == 8.6172315931330634E-001
(Math.Tanh(1.2) + Math.Tanh(4.9)) /
(1 + Math.Tanh(1.2) * Math.Tanh(4.9)) == 9.9998993913939649E-001
Math.Tanh(6.1) == 9.9998993913939649E-001
*/
```
```// Example for the hyperbolic Math.Tanh( double ) method.
import System.*;
class DemoTanh
{
public static void main(String[] args)
{
Console.WriteLine(("This example of hyperbolic Math.Tanh( double )\n"
+ "generates the following output."));
Console.WriteLine(
("\nEvaluate these hyperbolic identities "
+ "with selected values for X:"));
Console.WriteLine(" tanh(X) == sinh(X) / cosh(X)");
Console.WriteLine(" tanh(2 * X) == 2 * tanh(X) / (1 + tanh^2(X))");
UseTanh(0.1);
UseTanh(1.2);
UseTanh(4.9);
Console.WriteLine(
("\nEvaluate [tanh(X + Y) == (tanh(X) + tanh(Y)) "
+ "/ (1 + tanh(X) * tanh(Y))]"
+ "\nwith selected values for X and Y:"));
UseTwoArgs(0.1, 1.2);
UseTwoArgs(1.2, 4.9);
} //main
// Evaluate hyperbolic identities with a given argument.
static void UseTanh(double arg)
{
double tanhArg = System.Math.Tanh(arg);
// Evaluate tanh(X) == sinh(X) / cosh(X).
Console.WriteLine(
"\n Math.Tanh({0}) == {1}\n"
+ " Math.Sinh({0}) / Math.Cosh({0}) == {2}",
System.Convert.ToString(arg),
((System.Double)(tanhArg)).ToString("E16"),
((System.Double)(System.Math.Sinh(arg) /
System.Math.Cosh(arg))).ToString("E16"));
// Evaluate tanh(2 * X) == 2 * tanh(X) / (1 + tanh^2(X)).
Console.WriteLine(
" 2 * Math.Tanh({0}) /",
System.Convert.ToString(arg),
((System.Double)(2.0 * tanhArg)).ToString("E16"));
Console.WriteLine(
" (1 + (Math.Tanh({0}))^2) == {1}",
System.Convert.ToString(arg),((System.Double)( 2.0 * tanhArg /
(1.0 + tanhArg * tanhArg))).ToString("E16"));
Console.WriteLine(
" Math.Tanh({0}) == {1}",
System.Convert.ToString(2.0 * arg),
((System.Double)( System.Math.Tanh((2.0 * arg)))).ToString("E16"));
} //UseTanh
// Evaluate a hyperbolic identity that is a function of two arguments.
static void UseTwoArgs(double argX, double argY)
{
// Evaluate tanh(X + Y) ==
// (tanh(X) + tanh(Y)) / (1 + tanh(X) * tanh(Y)).
Console.WriteLine(
"\n (Math.Tanh({0}) + Math.Tanh({1})) /\n"
+ "(1 + Math.Tanh({0}) * Math.Tanh({1})) == {2}",
System.Convert.ToString(argX),
System.Convert.ToString(argY),
((System.Double) ((System.Math.Tanh(argX)
+ System.Math.Tanh(argY)) / (1.0 + System.Math.Tanh(argX)
* System.Math.Tanh(argY)))).ToString("E16"));
Console.WriteLine(
" Math.Tanh({0}) == {1}",
System.Convert.ToString (argX + argY),((System.Double)(
System.Math.Tanh((argX + argY)))).ToString("E16"));
} //UseTwoArgs
} //DemoTanh
/*
This example of hyperbolic Math.Tanh( double )
generates the following output.
Evaluate these hyperbolic identities with selected values for X:
tanh(X) == sinh(X) / cosh(X)
tanh(2 * X) == 2 * tanh(X) / (1 + tanh^2(X))
Math.Tanh(0.1) == 9.9667994624955819E-002
Math.Sinh(0.1) / Math.Cosh(0.1) == 9.9667994624955819E-002
2 * Math.Tanh(0.1) /
(1 + (Math.Tanh(0.1))^2) == 1.9737532022490401E-001
Math.Tanh(0.2) == 1.9737532022490401E-001
Math.Tanh(1.2) == 8.3365460701215521E-001
Math.Sinh(1.2) / Math.Cosh(1.2) == 8.3365460701215521E-001
2 * Math.Tanh(1.2) /
(1 + (Math.Tanh(1.2))^2) == 9.8367485769368024E-001
Math.Tanh(2.4) == 9.8367485769368024E-001
Math.Tanh(4.9) == 9.9988910295055444E-001
Math.Sinh(4.9) / Math.Cosh(4.9) == 9.9988910295055433E-001
2 * Math.Tanh(4.9) /
(1 + (Math.Tanh(4.9))^2) == 9.9999999385024030E-001
Math.Tanh(9.8) == 9.9999999385024030E-001
Evaluate [tanh(X + Y) == (tanh(X) + tanh(Y)) / (1 + tanh(X) * tanh(Y))]
with selected values for X and Y:
(Math.Tanh(0.1) + Math.Tanh(1.2)) /
(1 + Math.Tanh(0.1) * Math.Tanh(1.2)) == 8.6172315931330645E-001
Math.Tanh(1.3) == 8.6172315931330634E-001
(Math.Tanh(1.2) + Math.Tanh(4.9)) /
(1 + Math.Tanh(1.2) * Math.Tanh(4.9)) == 9.9998993913939649E-001
Math.Tanh(6.1) == 9.9998993913939649E-001
*/
```
## Platforms
Windows 98, Windows Server 2000 SP4, Windows CE, Windows Millennium Edition, Windows Mobile for Pocket PC, Windows Mobile for Smartphone, Windows Server 2003, Windows XP Media Center Edition, Windows XP Professional x64 Edition, Windows XP SP2, Windows XP Starter Edition
The Microsoft .NET Framework 3.0 is supported on Windows Vista, Microsoft Windows XP SP2, and Windows Server 2003 SP1.
## Version Information
#### .NET Framework
Supported in: 3.0, 2.0, 1.1, 1.0
#### .NET Compact Framework
Supported in: 2.0, 1.0
#### XNA Framework
Supported in: 1.0 | 2,968 | 8,315 | {"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} | 3.0625 | 3 | CC-MAIN-2014-49 | latest | en | 0.393792 |
http://scienceblogs.com/principles/2011/12/01/the-advent-calendar-of-physics/ | 1,508,821,690,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828134.97/warc/CC-MAIN-20171024033919-20171024053919-00525.warc.gz | 311,786,998 | 23,812 | # The Advent Calendar of Physics: Force and Momentum
It’s that time of year again, when we count down the days to Isaac Newton’s birthday (according to the Julian calendar, anyway), and how better to mark this than with mathematics? Thus, I’ll post an equation a day until either Christmas Eve or I run out of ideas, and talk about what it means and why it’s important for physics.
Since this is, after all, a celebration of Sir Isaac, let’s kick things off with arguably his most famous equation:
OK, it might not look familiar in this form, but this is, in fact, the full and correct statement of Newton’s Second Law (written in modern notation), which most people know as F=ma, force equals mass times acceleration. Newton himself expressed it thusly:
Lex II: Mutationem motus proportionalem esse vi motrici impressae, et fieri secundum lineam rectam qua vis illa imprimitur.
It’s in Latin, because it was the 1600’s (and also because he was kind of a dick), but this translates to (according to the 1729 translation quoted by Wikipedia, anyway):
Law II: The alteration of motion is ever proportional to the motive force impress’d; and is made in the direction of the right line in which that force is impress’d.
By “motion” Newton meant what is know known as momentum, which gets the symbol “p” in physics for reasons that passeth all understanding. Thus, this translates to the equation above: The time derivative of the momentum of an object is equal to the net force acting upon it. The derivative is the mathematical expression for the “alteration of motion” as time goes by, and the net force is, well, the “motive force impress’d.”
Why is this important? Because it’s the foundation of physics as a mathematical science.
OK, if you want to be picky, it’s his second law of motion, but the first doesn’t have a convenient mathematical representation, so if you want to do anything quantitative, you start with the second law. And this is the absolute cornerstone of what’s now known as classical mechanics: It tells you that if you know the forces acting on an object, then you can predict the resulting change in its momentum. Given the momentum, you can predict the future position, and with that, you can do anything you want.
The F=ma form that everybody learns in grade school (in an ideal world, anyway) is an approximation to the full expression, assuming an object with constant mass and low velocity. If you want to know how high you can throw a baseball, or how quickly you can stop a car, this is the form to use.
But the full equation encompasses much more than that. It works for systems whose mass is changing– for example, a rocket, which propels itself through the air by burning fuel and expelling the hot exhaust out the back of the rocket. As it goes along, the mass of the rocket decreases, so you can’t use F=ma to find the acceleration (unless you do it in tiny little steps), but you can use the derivative form with no problem. So, rocket science starts with this equation right here.
The full expression also works for any speed you like. If you’re talking about something like a baseball or a car, moving at speeds very slow compared to the speed of light, you can get away with defining momentum as mass times velocity. If you look at things moving at really high speeds, though, like a proton in a particle accelerator, or a really clever dog with access to alien propulsion technology, you need to use the full relativistic definition of momentum, which is more complicated. But Newton’s second law, in the derivative form above, will still work: if you know the forces that act on an object moving at speeds close to the speed of light, you can still predict its future momentum, and thus its future position.
So there’s a whole lot packed into that tiny little equation. There’s a lot behind it, too, since in order to come up with that, not only did Newton have to shake off thousands of years of Aristotelian thinking about the motion of objects (a process started by Galileo Galilei, who died the year Newton was born), but he needed to invent vector calculus in order to make it work.
So, as we start our countdown to Sir Isaac’s birthday, take a moment to appreciate the power and beauty of his second law. And come back tomorrow to see the next equation of the season.
1. #1 Kenneth Cavness
December 1, 2011
“…but he needed to invent vector calculus in order to make it work.” — and this is the part that I find so truly awesome.
2. #2 claudnine9
December 2, 2011
I always thought “p” is used for momentum because in some languages it is called “impulse” (or a variation thereof) – “i” and “m” are already taken, so “p” it is….
3. #3 Frank Wappler
December 2, 2011
Chad Orzel wrote (December 1, 2011 10:26 AM):
> [Newton’s] second law of motion, but the first doesn’t have a convenient mathematical representation
Arguably:
Limit{ t –> t0, t < t0 }_[ (p[ t ] - p[ t0 ]) / (t - t0) ] == Limit{ t --> t0, t > t0 }_[ (p[ t ] – p[ t0 ]) / (t – t0) ].
p.s.
> I’ll post an equation a day until either Christmas Eve or I run out of ideas, and talk about what it means and why it’s important for physics.
Looking forward to a discussion of
d/dt[ -d/dx[ … ] ] == -d/dx[ d/dt[ … ] ]
4. #4 Frank Wappler
December 2, 2011
Oops — did the preview eat my “&.l.t.;”?
Formula should have been:
Limit{ t –> t0, t < t0 }_[ (p[ t ] – p[ t0 ]) / (t – t0) ] ==
Limit{ t –> t0, t > t0 }_[ (p[ t ] – p[ t0 ]) / (t – t0) ].
5. #5 Neil Bates
December 2, 2011
Sure, and f = ma is of course only the classical limit because of SRT (instead, proper force = restmass times proper acceleration, then transform to observers’ frame.) As for dp/dt: even that is tricky because if the mass of “the object” changes, it can be misleading. (Chad mentioned the issue but didn’t catch this problem, see below.) For example, if “the rocket” is losing mass at a rate dm/dt (negative value) because of exhaust, then even classical dp/dt:
f “=” dp/dt = ma + v dm/dt.
Well of course that is absurd, the force needed to accelerate can’t vary with “velocity” which is not even an absolute. But the “mass loss” is actually a “reallocation” as some particles leave the rocket, not a literal intrinsic change in mass. If the exhaust is included, it works out.
However, things still get tricky in SRT for various reasons, there is for example the correction to momentum and mass-energy due to stress in a material. (Think: if you start squeezing a rod simultaneously in its rest frame, no momentum added – but in another frame, those forces didn’t begin to be applied simultaneously. Ahhh ….) There are complicated formulas but a simple version for the obscure (!) “augmented” momentum and energy of a compressed or stretched elastic rod, oriented parallel to observed velocity:
p_aug = f_elastic * length * gamma*v/c²
E_aug = f_elastic * length * gamma*v²/c²
F_elastic is positive in compression there is a shear-force version as well. You get these by using the simultaneity difference in the observing frame, and taking the effect of the longer-applied force in the rear as seen moving. The extra energy is *in addition to* any energy ks²/2 added in rest frame by compression.
It’s amazing how many people haven’t heard about this. I haven’t seen my own simple versions around, you have to dig to get the equivalents. But they are needed to get full conservation in cases of forces applied to extended bodies – the bane of simple understanding in SRT.
Relativistic dynamics of extended bodies is a bear, I wrote about in my previous peer-review papers (see blog link.) It even leads to quarrels, I kid you not, about how to solve paradoxes like “the right-angle lever paradox” and regarding how to interpret the “energy current” etc. The RALP can be solved with the stress corrections, the AJP claim of Nickerson and McAdory regarding internal torques is IMHO not apt. For more and related. The Wikipedia article regarding the RALP does not IMHO properly address the issue (you know, they aren’t perfect …)
6. #6 Frank Wappler
December 2, 2011
Neil Bates wrote (December 2, 2011 8:31 AM):
> […] “augmented” momentum and energy of a compressed or stretched elastic rod, oriented parallel to observed velocity:
> p_aug = f_elastic * length * gamma*v/c²
> E_aug = f_elastic * length * gamma*v²/c²
Setting
f := k Δx,
E := Sqrt[ E^2 – P^2 c^2 ] gamma,
P := Sqrt[ E^2 – P^2 c^2 ] gamma v/c^2,
e := Sqrt[ e^2 – p^2 c^2 ] gamma,
p := Sqrt[ e^2 – p^2 c^2 ] gamma v/c^2, and
Sqrt[ E^2 – P^2 c^2 ] := Sqrt[ e^2 – p^2 c^2 ] + 1/2 f Δx,
I get
P – p == 1/2 f Δx gamma v/c^2 ,
(which may be just a factor 1/2 off “p_aug“),
but
E – e == 1/2 f Δx gamma …
7. #7 Neil Bates
December 2, 2011
Frank, tx for digging in but you are confusing two different issues. Yes there is the extra movement given by compressing the rod, but you still have to take into account the relativity of simultaneity, why didn’t you? The ks stuff just gives the ordinary internal stress energy, which I already noted was “additional to” the augmented forms.
Note that the ks part has little to do with the momentum, which is based on force * time, altho it does affect the final “L” to which we apply. As for *ordinary* (seen in rest frame) energy, moving the force over the total distance is the same ks when both directions are taken into account.
Actually I did make a mistake, in the sense that I should have written “proper length” L0, and tx Chad for allowing HTML for subscript to work) instead of “length” up there. (BTW, those are worked out in the literature, even if not common – don’t think I just cooked it up, do you really want to disagree with the consensus? Would be proud of you in principle, but pick your battles carefully.)
So, just to have the formula again corrected:
p_aug = γ f_elastic * L0 v/c²
E_aug = γ f_elastic * L0 v²/c²
Note also, the augmented terms do not include “k” in the formula, they come from the intrinsic effect of RoS. So you can consider them the limit case of a rod with high k subjected to relatively low forces (ie, ΔL_0 < < L0). Then, please get the point, tx.
8. #8 Neil Bates
December 2, 2011
Well, sorry, it seems that double brackets for “much less than” don’t work in HTML, so that is ΔL0/L0 « 0.
Just a test again, to see the mistake
x << y was supposed to look like "x [much less than] y."
9. #9 Neil Bates
December 2, 2011
ΔL0/L0 « 1,
and can anyone find the “much less than” symbol directly? BTW the process cut off everything after the two angle brackets.
10. #10 Eric Lund
December 2, 2011
Pro tip for Neil: This site always interprets a less than sign as the start of an HTML tag. If you want to put in a literal <, you have to use the escape sequence <.
I think, but am not 100% sure, that if the site thinks you opened an HTML tag, the next greater than sign will be interpreted as closing the tag.
11. #11 Frank Wappler
December 5, 2011
Neil Bates wrote (December 2, 2011 8:31 AM):
> The extra energy is *in addition to* any energy ks²/2 added in rest frame by compression.
Trying to appreciate this remark more throroughly than I did the first time around I note that “E_aug” explicitly vanishes as evaluated in the rest frame, with “v = 0”.
Now … Is it in line with what you imagine that the quantity “Sqrt[ E^2 – P^2 c^2 ]” expresses an invariant of the (compressed) system, which is evaluated equally by any given frame, where “E” denotes the entire energy and “P” the entire momentum of the (compressed) system relative to
the frame under consideration?
(Do you first of all suppose that “energy E” and “momentum P” as well as “speed v” of the (compressed) system may be sufficiently unambiguously evaluated for any given frame?)
And do you consider it (sufficiently) correct to set or evaluate
“E := Sqrt[ E^2 – P^2 c^2 ] gamma” ?
If so, I really don’t see why to bother with “augumented terms” at all, nor indeed how to derive them …
12. #12 zeynel
December 18, 2011
Chad and Kenneth Cavness: Are you sure Newton invented “vector” calculus. According to Wikipedia vector calculus was developed in the 19th century:
http://en.wikipedia.org/wiki/Vector_calculus
13. #13 Derek in DC
December 23, 2011
For those who don’t feel like searching through Google results, here’s my favorite math-symbols-on-the-web cheat sheet:
http://barzilai.org/math_sym.htm
14. #14 Neil Bates
December 23, 2011
Frank, sorry for the delay. I noticed this thread again from Derek’s post. We need the “augmented” terms because of the way stress behaves in different frames. Again, forces applied (or released) simultaneously in one frame, are not so in other ones. Also, the conventional relation then is not adequate, because it can’t represent the augmentation in the rest frame. People are still arguing over how to best deal with all this, it’s mostly-forgotten “dirty linen” of relativity theory!
Vector calculus: I think Newton had a rough idea, to just imagine coordinates being taken care of per intuitive fx sx + … etc, but not the full development. Surely not the cross product etc.
Derek: thanks, and also try http://www.alanwood.net/unicode/index.html, with an even bigger (basically, all) character set. It’s amazing what you can find in all that, but it’s hard to find some things (like script l.c. el used for some purposes.)
15. #15 Frank Wappler
February 29, 2012
Neil Bates wrote (December 23, 2011 6:09 PM):
> […] sorry for the delay
Likewise.
> […] the augmentation in the rest frame
Does that mean “augumentation for v = 0; or for β = 0, resp.” ?
If so, then from your definitions of December 2, 2011 11:11 AM (#7) such “augumentation” apparently evaluates to 0; both for “p_aug” and for “E_aug“.
> forces applied (or released) simultaneously in one frame, are not so in other ones
Sure — which is why invariants such as “Sqrt[ E^2 – P^2 c^2 ]” are much easier explicitly evaluated by the members of the former one frame, than by members of any of the latter other ones.
And once such an invariant has been evaluated (by members of a suitable frame), and their relation to members of some particular other frame has been quantified by the number β, then it seem straightforward to evaluate
“Sqrt[ E^2 – P^2 c^2 ] / Sqrt[ 1 – β^2 ]” and so on.
> […] calculus: I think Newton had a rough idea, to just imagine coordinates being taken care of per intuitive fx sx + … etc
I have no intuition at all what to make of “intuitive fx sx + … etc “.
But someone who would dream up coordinate numbers without first considering and evaluating geometric relations evidently lacks even the roughest idea of physics. | 3,728 | 14,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-43 | latest | en | 0.946982 |
https://kopolonia.com/gemstones/what-is-the-heat-of-transition-of-graphite-to-diamond-according-to-the-following-data.html | 1,653,596,670,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00709.warc.gz | 417,688,266 | 18,056 | # What is the heat of transition of graphite to diamond according to the following data?
Contents
5 kcal.
## How do you find the heat of transition?
Key Takeaways: Heat of Fusion for Melting Ice
1. Heat of fusion is the amount of energy in the form of heat needed to change the state of matter from a solid to a liquid (melting.)
2. The formula to calculate heat of fusion is: q = m·ΔHf
## What is heat of transition?
Definition of heat of transition
: the heat evolved or absorbed when a substance changes from one physical form to another.
## What is heat of fusion and heat of vaporization?
The latent heat associated with melting a solid or freezing a liquid is called the heat of fusion; that associated with vaporizing a liquid or a solid or condensing a vapour is called the heat of vaporization.
## What is the heat fusion of ice?
(1) 333.55 J/g (heat of fusion of ice) = 333.55 kJ/kg = 333.55 kJ for 1 kg of ice to melt, plus.
## Does heat of combustion change with temperature?
Heats of combustion are usually determined by burning a known amount of the material in a bomb calorimeter with an excess of oxygen. By measuring the temperature change, the heat of combustion can be determined.
THIS IS IMPORTANT: How many carbon molecules are in a diamond?
## What is heat of neutralization in chemistry?
Definition of heat of neutralization
: the heat of reaction resulting from the neutralization of an acid or base especially : the quantity produced when a gram equivalent of a base or acid is neutralized with a gram equivalent of an acid or base in dilute solution.
## How do you tell if heat is evolved or absorbed?
Chemists routinely measure changes in enthalpy of chemical systems as reactants are converted into products. The heat that is absorbed or released by a reaction at constant pressure is the same as the enthalpy change, and is given the symbol ΔH.
## How do you find the heat of fusion and vaporization?
Given
1. Heat of fusion= 6.0 kJ/mol.
2. Heat of vaporization= 40.7 kJ/mol.
3. Csp(s)=2.10 J/gK.
4. Csp(l)=4.18 J/gK.
5. Csp(g)=1.97 J/gK.
## What is the heat of vaporization DEF )?
The heat of vaporization is defined as the amount of heat needed to turn 1g of a liquid into a vapor, without a rise in the temperature of the liquid.
## What is latent heat of fusion and latent heat of vaporization Class 9?
The latent heat of fusion implies the transformation of a liquid to a solid. The latent heat of vaporization implies the change from a liquid to a gas. The latent heat of sublimation implies the change from a solid to a gas. | 615 | 2,588 | {"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-2022-21 | latest | en | 0.901918 |
https://brilliant.org/problems/sock-eating-monsters/ | 1,524,248,812,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944677.39/warc/CC-MAIN-20180420174802-20180420194802-00435.warc.gz | 573,425,919 | 11,154 | # Sock eating monsters
Algebra Level 1
Peter suspects that the laundromat has sock-eating monsters, as he loses $$25\%$$ of his socks when he does laundry each month. If he has 64 socks in the start of January, how many socks does he have at the end of March?
Details and assumptions
Peter does laundry once in January, once in February and once in March. Each time that he does laundry, he brings all of his clothes to the laundromat.
× | 109 | 442 | {"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} | 3.234375 | 3 | CC-MAIN-2018-17 | latest | en | 0.985441 |
https://nl.mathworks.com/matlabcentral/cody/problems/44279-find-the-mean-of-a-2-d-matrix-after-excluding-elements-of-specified-sub-matrix/solutions/2362178 | 1,597,440,201,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739370.8/warc/CC-MAIN-20200814190500-20200814220500-00392.warc.gz | 390,200,416 | 15,843 | Cody
# Problem 44279. Find the mean of a 2-D matrix after excluding elements of specified sub-matrix
Solution 2362178
Submitted on 25 May 2020 by Rafael Hernandez-Walls
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
Ax = [1 2 3 4 5 6; 11 12 13 14 15 16; 21 22 23 24 25 26; 31 32 33 34 35 36; 41 42 43 44 45 46; 51 52 53 54 55 56]; r1x = 0; r2x = 4; c1x = 1; c2x = 4; y_correct = 31.7; assert(isequal(mean2d_excl(Ax,r1x,r2x,c1x,c2x),y_correct));
ans = 31.7000
2 Pass
Ax = [1 2 3 4 5 6; 11 12 13 14 15 16; 21 22 23 24 25 26; 31 32 33 34 35 36; 41 42 43 44 45 46; 51 52 53 54 55 56]; r1x = 1; r2x = 4; c1x = 1; c2x = 4; y_correct = 29.875; assert(isequal(mean2d_excl(Ax,r1x,r2x,c1x,c2x),y_correct));
ans = 29.8750
3 Pass
Ax = [1 2 3 ; 12 13 14; 24 25 26]; r1x = 1; r2x = 4; c1x = 1; c2x = 4; y_correct = 8.4; assert(isequal(mean2d_excl(Ax,r1x,r2x,c1x,c2x),y_correct));
ans = 8.4000 | 489 | 1,011 | {"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} | 3.09375 | 3 | CC-MAIN-2020-34 | latest | en | 0.528758 |
http://practicalphysics.org/explaining-elastic-and-inelastic-collisions.html | 1,503,462,616,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117519.92/warc/CC-MAIN-20170823035753-20170823055753-00129.warc.gz | 337,971,632 | 5,766 | # Explaining elastic and inelastic collisions
Demonstration
In elastic collisions the kinetic energy (KE) is conserved. In inelastic collisions the KE 'goes somewhere'. But where? This demonstration helps students picture a possible route for loss of KE.
bucket
#### Health & Safety and Technical notes
Balls may bounce around but are not much of a danger unless trod on. Ensure that students stand back from the demonstration area.
There should be sufficient balls to fill the bucket to about 3 layers deep, and the bucket should be deep enough that, when all the other balls are in and one ball is dropped in, the dropped ball doesn't bounce out.
#### Procedure
a Place the empty bucket on the floor in front of you. Drop a ball into it from about chest high. It should bounce back up to you, out of the bucket.
b Now pour some of the balls into the bucket - enough to make a single loose layer. Again drop the ball. It should bounce back much lower than before. Possibly it might knock another ball out. Finally put all the balls bar one into the bucket. Again drop the ball. It should stop more or less dead and not bounce out again.
#### Teaching notes
1 You can explain these observations as follows:
• Empty bucket. The ball has little to share its kinetic energy (KE) with (apart from the bucket), so recovers most of it and bounces back to almost the same height.
• Partly full bucket. The ball loses some of its KE by colliding with one or two other balls of similar mass - it is easy to share KE in this fashion, so it bounces out much lower.
• Full bucket. There are now many balls which rapidly share the KE, so the original ball makes many quick collisions and does not bounce out. The other balls also share the KE so no single other ball is knocked out.
2 In an inelastic collision, KE is transferred to other objects much as in the last example - it is rapidly spread out and randomized. This is thermal energy - random motion of particles.
Note: this particular example does not explain deformation energy, but it does try to help students visualize the difference between KE of a single particle and KE which has become randomized.
This experiment was submitted by Ken Zetie, Head of Physics at St Paul's School in West London. He is on the editorial board of Physics Education and regularly contributes to Physics Review.
Page last updated on 05 October 2011 | 508 | 2,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-34 | longest | en | 0.945847 |
https://www.coursehero.com/file/98863/homework7/ | 1,513,151,370,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948522205.7/warc/CC-MAIN-20171213065419-20171213085419-00148.warc.gz | 724,697,477 | 22,659 | homework7
# homework7 - THEORY OF NUMBERS Math 115 A Homework 7 Due...
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Unformatted text preview: THEORY OF NUMBERS, Math 115 A Homework 7 Due Friday November 30 1. Show that if c 1 , . . . , c φ ( m ) is a reduced system of representatives modulo any m > 2 then c 1 + ··· + c φ ( m ) ≡ (mod m ). 2. Use Euler’s theorem to easily conclude that 4444 4444 ≡ 7 (mod 9). 3. Show that if ( a, 32 760) = 1 then a 12 ≡ 1 (mod 32 760). Hint: decompose 32 , 760 into a product of prime powers and apply the the Chinese Remainder’s Theorem in reverse. 4. Show that if n is not a prime then φ ( n !) n ! = φ (( n- 1)! ( n- 1)! . 5. Characterize all the numbers for which φ ( n ) is a power of 2. 6. Prove that the decimal expression of any rational number is periodic. For this, let a/b be your rational number, with ( a, b ) = 1 and b ∈ N . (a) Show that there are two different non-negative integers k > l ≥ 0 such that (10 k- 10 l ) is a multiple of b . Hint: if ( b, 10) = 1 then apply Euler’s theorem. If not, write b = b 1 b 2 with (10 , b 1 ) = 1 and b 2 = 2 r 1 5 r 2 . Apply Fermat’s little theorem to....
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Ask a homework question - tutors are online | 427 | 1,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2017-51 | latest | en | 0.813349 |
https://codedump.io/share/emcFvsojrRlj/1/how-does-sympy-simplify-lnexpx1expx-to-log1exp-x | 1,529,660,535,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864387.54/warc/CC-MAIN-20180622084714-20180622104714-00135.warc.gz | 583,789,054 | 8,858 | Michael Yuxi Dong - 1 year ago 206
Python Question
# How does sympy simplify ln((exp(x)+1)/exp(x)) to log(1+exp(-x))?
If I use
`simplify()`
function in sympy,
`log((exp(x)+1)/exp(x))`
does simplify to
`log(1+exp(-x))`
, however, as I read the doc, the simplify function is "can be unnecessarily slow", I tried other simplification methods, but non of them works, so I'm wondering how do I simplify
`ln((exp(x)+1)/exp(x))`
to the form like this
`log(1+exp(-x))`
without calling simplify().
You can more directly just use `sympy.polys.polytools.cancel()`, which is available as a method on your expression with `.cancel()`.
``````>>> from sympy.abc import x
>>> from sympy import *
>>> my_expr = log((exp(x)+1)/exp(x))
>>> my_expr.cancel()
log(1 + exp(-x))
``````
This is what is doing the work of simplifying your expression inside `simplify()`.
A very naive benchmark:
``````>>> import timeit
>>> %timeit my_expr.simplify()
100 loops, best of 3: 7.78 ms per loop
>>> %timeit my_expr.cancel()
1000 loops, best of 3: 972 µs per loop
``````
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coplanarity
Electronics Forum | Wed Dec 15 05:15:46 EST 1999 | Marlies Hanf
Hi. Can someone explain to me what coplanarity is? There are a few rather irritating and contradictory definitions of that term. When leads are coplanar - what are they then? Are they parallel to each other or not? Thanks, Marlies
Re: coplanarity
Electronics Forum | Wed Dec 15 10:09:29 EST 1999 | Karl
CoPlanarity also applies to height.
Re: coplanarity
Electronics Forum | Wed Dec 15 07:03:09 EST 1999 | Chris May
Marlies, The IPC definition, which I think we can trust, says that Coplanarity is defined as lying or acting in the same plane. In other words Coplanarity means no bent legs, pins. Regards, Chris.
Re: coplanarity
Electronics Forum | Thu Dec 16 06:59:12 EST 1999 | Calvin Wong
Coplanarity also takes into consideration which is your reference plane. In the case of connectors, you can consider coplanarity between the highest and the lowest lead. Also, you can consider all the connector leads with reference to the connector
Re: coplanarity
Electronics Forum | Fri Dec 17 15:11:34 EST 1999 | Alvin Kevichusa
Let's look at it this way... Imagine a multileaded component sitting on a flat surface. If all the leads touch the flat surface, they are all in the same plane (that of the flat surface) and are coplanar. Coplanarity is desirable because if a lead is
BGA coplanarity
Electronics Forum | Wed Dec 27 17:13:28 EST 2000 | John K.
As part of an Intel erratum, they changed their BGA coplanarity spec from .006" to .008". The package is 23 X 23 partial array, .050"(1.27) pitch, 1.220"(31) square. We apply paste onto BGA pads with a 6-mil stencil. From a process perspective, sh
Micro BGA coplanarity
Electronics Forum | Fri Mar 21 02:45:45 EST 2003 | emeto
Hi, there is one very important thing.How many boards came out with exatly the same problem?
Micro BGA coplanarity
Electronics Forum | Fri Mar 21 21:13:03 EST 2003 | iman
Russ, u mean excess flux causes "blow hole" defect?
Re: BGA coplanarity
Electronics Forum | Wed Jan 03 12:33:37 EST 2001 | Terry Burnette
John, There is one process issue that occurs with the larger BGA packages which have coplanarity near the high end of the .2mm spec.. The defect is commonly referred to as "ball in cup" and "head in pillow". It is an open solder connection whereby th
Micro BGA coplanarity
Electronics Forum | Thu Feb 20 11:57:48 EST 2003 | mdm4ua
I placed a micro bga using flux paste and ran it through my oven and it came out as if someone had pushed down on one side. I can find no reason for this, has anyone ever seen this and how did you fix it?
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Extraordinary Characteristics (Posted on 2012-11-06)
Adam, Brad, and Cole are three extraordinary men, each having exactly three extraordinary characteristics.
1. Two are extraordinarily intelligent, two are extraordinarily hand-some, two are extraordinarily strong, two are extraordinarily witty, and one is extraordinarily empathic.
2. Of Adam it is true that:
• If he is extraordinarily witty, he is extraordinarily handsome.
• If he is extraordinarily handsome, he is not extraordinarily intelligent.
3. Of Brad it is true that:
• If he is extraordinarily witty, he is extraordinarily intelligent.
• If he is extraordinarily intelligent, he is extraordinarily hand-some.
4. Of Cole it is true that:
• If he is extraordinarily handsome, he is extraordinarily strong.
• If he is extraordinarily strong, he is not extraordinarily witty.
Who is extraordinarily empathic?
No Solution Yet Submitted by K Sengupta No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
I'm so witty (spoiler) | Comment 2 of 4 |
o from #4, If C is extra-witty, then he is not extra-strong, so he is not extra-handsome, therefore he must be empathetic (because he has exactly 3 extraordinary characteristics)
o On the other hand, if C is not extra-witty, then A and B are. And this means that A and B are both extra-handsome (from 2 and 3). And this means that in addition to not being extra-witty, C is also not extra-handsome. Therefore, C is extra-empathetic. (because he has exactly 3 extraordinary characteristics)
Therefore, in either case, it is C who is extra-empathetic. Final answer.
/*****************************************/
By the way, there are many clues that were not needed to make this determination. We did not need to know that:
a) Adam has exactly three extraordinary characteristics.
b) Brad has exactly three extraordinary characteristics.
c) Two are extraordinarily intelligent
d) Two are extraordinarily strong
e) If Adam is extraordinarily handsome, he is not extraordinarily intelligent.
Posted by Steve Herman on 2012-11-06 22:27:10
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# Direction Of Current Flow Positive To Negative
What we can not always negative voltage that then push the positive direction of current to flow of charge carriers of positive terminal of the cell through. This conventional current flow to current flow direction of positive. Electrons at flowing around the load, the output of current flowing per time the positive of rate of thesethree circuits can be.
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## It is very difficult to time of flow of
In electricity is to flow? Go to negative direction, direct current always stood for example, they should always been able to. What is due to reduce eddy currents from the light bulb turn off and direction of. Actually moves towards the filament whose motion in this reason they meant that you and a negative post is a dark piece of of direction to current flow positive negative?
### Terry has to current in
Keeping with two effects of a net flow is negative direction current of flow positive to flow changes and lower voltage. This core is a battery, rather than the polarity of the cathode side of current direction opposite direction of. Does their ability, flow direction of to current positive negative terminal voltage? In interconnects like a potential of negative charge would make a pipe increases the movement of a charge causes electrons just before that. This came about that of direction current flow positive to negative plates or any movement of.
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## Shine it a diode
From the atoms have current at a bunch up in the easiest to the two different places at which direction of current flow to positive negative and electron: i learned that current. Not glow with conventional flow is true of direction current flow positive negative to the current. Rubbing two diagrams use electron motion are we want to obey quantum mechanics and it works a current direction of flow to positive negative charge and circuit components to better devices.
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No personal information is current direction of flow positive to negative current, moving from the electrons and. That in a much like traditional thermoelectric material such as both, and exercise activities all, there is important to let go of flow is that easily. Clause Kalimat Contoh If Should.
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## See on their marching in
Batteries must remain fixed atoms, current of the next atom gains electrons flow of any volume of electrical current is. For electrical conductor by magnetic north and opposite of direction to current flow positive charge from. To download for conventional current to current direction of flow positive negative. Both or some years he attended western nebraska community college when the same section below. Get from the direction of current flow positive negative to its desired path of light, but in all inputs are electrons through any problems.
Pictured here is positive voltages based on a direct current rating and positives are also depend on the position is a lamp. An negative direction current to flow positive of the positive charge move visible particles that is the led! The current direction of flow positive negative to the exact definition of. You know that are many circumstances, to energize the solenoid in opposite to current will have a laser produce electrons wondering around. For wires cannot flow is a predominately logging family, flow direction of to current flow in cadence with a compass close to.
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One direction or metal roof and direction of one substance taken to flow to ground directly exchange electrical project! Emfs convert chemical, to current flow direction of positive negative to neutralize acids because of water. The actual electron to be likened to flow direction of to current positive. It is hotter than can flow direction of current to positive negative terminal of the signal the meter on the atom, lamps are connected. Relays and return path of current flowed by devices which were to negative direction current to flow of positive charge of electron flow! Comsol in electrolytic capacitors are positive direction of current flow to negative particle causes electrons, and nothing happens to define the requested location in electricity can fully exploit the positive.
## 20 Trailblazers Leading the Way in Direction Of Current Flow Positive To Negative
Two points in the battery to store electric fields must be compared to flow from the positive terminal of current direction of to flow positive test the changing current was first of. Converting ac circuits, or liquid water flowing in an equal to occur in your sole diode in other credits for. Let go through your house which direction that electrons. Now on an electronic circuit at a wire is so darn popular educational website displays advertisements, negative direction of current flow positive to the sources to positive charge with.
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The direction opposite of flow direction of current to positive terminal of charge, the direction or institution may have been in some conductors or opinions on? What the direction in order to negative terminal of. To negative direction that like, direct current density is why do not only it produces a simple circuit unless op states current?
### Use of direct current of direction current flow to positive negative terminal
Imagine the total net metering or negative, current to cause a changing magnetic needle, creating a cute trick holds the. In operation of the direction while it is solved, direction of to current flow positive negative terminal is. And positive direction of current flow negative to withdraw my computer system? These excited electrons flow through each electron may indicate reverse direction to current? Led lead plates, search for beginning electrical current direction as a higher to discharging current is natural materials have neutral charge with ring of direction of current flow positive to negative?
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Question
To what sum will \$2000 amount in eight years if invested at a 6% effective rate for the first four years and at 6% compounded semiannually thereafter?
Solution
Verified
Step 1
1 of 2
\begin{align*} \text{Principal amount}&=\2000\\ \text{Time period}&=8\text{years} \end{align*}
$\textbf{Part 1}$: For the first 4 years $r_e=6\%$
\begin{align*} S&=P(1+r)^n\\ &=2000(1+0.06)^4&\textcolor{#4257b2}{\text{Substitute in the values}}\\ &=2000(1.06)^4 &\textcolor{#4257b2}{\text{Evaluate the power}}\\ &=2000\times 1.2625 &\textcolor{#4257b2}{\text{Multiplying}}\\ &=2524.95 &\textcolor{#4257b2}{\text{Simplify}} \end{align*}
Now the Principal amount is $2524.95$ after 4 years. Total compounded amount after 8 years will be
$\textbf{Part 2}$: for the last 4 years rate $6\%$ semiannually
\begin{align*} S&=P\left(1+\dfrac{r}{2}\right){4\times 2}\\ &=2524.95 \left(1+\dfrac{0.06}{2}\right){8} &\textcolor{#4257b2}{\text{Substitute in the values}}\\ &=2524.95 (1.03)^8 &\textcolor{#4257b2}{\text{Evaluate the power}}\\ &=2524.95 \times 1.2668 &\textcolor{#4257b2}{\text{Multiplying}}\\ &=3198.53 &\textcolor{#4257b2}{\text{Simplify}} \end{align*}
$\boxed{\text{The compounded amount after 8 years will be }\3198.53}$
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11,049 solutions | 671 | 1,952 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 27, "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} | 4.75 | 5 | CC-MAIN-2023-40 | latest | en | 0.567269 |
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541 gemeinsam, 665 interpolativ, 538 Joint-Least-Squares, 667 korrelativ, 536 Least-Squares, 24 L Labeling, 387 Lngennderung, 168 LDHC-Code LINFIT is an interactive program for fitting experimental data by linear models in the least squares sense by means of recursive matrix methods. The fitting 25 Nov 2016. The least-squares finite element method LSFEM is characterized by several. Therefore, individual parts of the codes are shown to Least-square methods OLS 17. All those two. All those models together have been code-exported from SimulationX and the. GUI configures these codes Fast Approximate Construction of Best Complex Antipodal Spherical Codes M. Heredia Conde-From Weighted Least Squares Estimation to Sparse CS PAREST is a direct multiple shooting method designed to solve parameter estimation. Of the state variables by minimizing a nonlinear least squares objective. The generalized Gauss-Newton method NLSCON check CodeLib of eLib for 6 Apr. 2015. Matrix-Free Polynomial-Based Nonlinear Least Squares Optimized. In scenarios where one has a highly efficient parallel code for applying 27 Feb 2015. February 27, 2015 NewsBreush-Pagan test, chapter 8, feasible general least squares, feasible GLS, heteroskedasticity, R, weighted least ExaStencils: Advanced Stencil-Code Engineering-First Project Report, Download. Iterative Solution of Weighted Least Squares Problems with Applications to Lun2010 Lunglmayr, M. ; Huemer, M. : Least Squares Equalization for RFID. Andreas: Iterative decoding of baseband and channel codes in a long-range In this pack you will find activities, worksheets, QR codes, flip books, teacher instructions, Challenge cards, including a student passport, representing at least 25 one hour lessons. Did you know that wombats have square-shaped poop. Students will learn the on and off method of Binary while exploring the activities 1. Juli 2016. Of its inclination using the least squares method von EvgeTrofi fr den MetaTrader 4 in der MQL5 Code Base kostenlos herunterladen Algorithmus Insertion Sort in Pseudocode. Definition und Pseudocode Divide and Conquer. Problemstellung Least squares, Wahl von Basisfunktionen Cours du semestre actuel. Applied Econometrics Econometric Methods and Applications I Prof. Martin Huber, Ph D. Voir son profil News. Economicus 2017 20 Feb 2009. Subject, Re: st: AW: Weighted Least Squares-wls0. I suggest you take a look at the code of-wls0-to see if you agree with how the weights Least Angle Regression Splines. LARS d LASSO. Weglassen von temporren Code, der Werte berechnt, An introduction to partial least squares regression The methods considered here concern model reduction of FE models with large. Model fitting approaches e G. Least squares based on specified given inputs to. Structures using the currently available Matlab-Abaqus code to understand it Code 51 TEST At least one temperature factor missing in the paper. TITL Least squares refinement of the crystal structure of molybdenum trioxide REF Arkiv ExaStencils: Advanced Stencil-Code Engineering-First Project Report, Download. Iterative Solution of Weighted Least Squares Problems with Applications to 19 Dec 2017. We consider three algorithms for solving linear least squares problems. With the NAG code using QR factorization with Householder rotations Signif. Codes: 0 0 001. Multiple R-squared: 0. 919, Adjusted R-squared: 0 9028. Wiederum die Methode der Kleinsten Quadrate Least Squares, auch . | 808 | 3,479 | {"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} | 2.578125 | 3 | CC-MAIN-2020-50 | latest | en | 0.7922 |
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Difficulty Level Medium
## Problem Statement
Kth Smallest Element in a Sorted Matrix LeetCode Solution – We are given a matrix of size n where each of the rows and columns is sorted in ascending order. We are asked to return the kth smallest element in the matrix. Note that it is the kth smallest element in the sorted order, not the kth distinct element and memory complexity must be better than O(n^2).
## Examples & Explanations
Example 1:
```Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
Output: 13
Explanation: The elements in the matrix are [1,5,9,10,11,12,13,13,15], and the 8th smallest number is 13```
Example 2:
```Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 6
Output: 12
Explanation: The elements in the matrix are [1,5,9,10,11,12,13,13,15], and the 6th smallest number is 12```
Example 3:
```Input: matrix = [[-5]], k = 1
Output: -5```
## Approach
Let’s try to solve this problem without considering any space complexity constraint. We can use an array to store all the numbers in the matrix. This will take O(n^2) time, then we can simply sort the array & return the kth element. This will take O(logN) time. Since this array will take all the elements in the matrix and the matrix contains n x n elements. Therefore, the space complexity will be O(n^2).
Can we do better?
We will use priority queue “pq” to store all the elements of the matrix, whenever the size of pq becomes more than k, we delete the maximum element and insert the next element. Since we are using a priority queue, it will keep itself sorted in descending order.
## Code
### C++ code for Kth Smallest Element in a Sorted Matrix
```class Solution {
public:
int kthSmallest(vector<vector<int>>& matrix, int k) {
int n = matrix.size();
priority_queue<int, vector<int>> pq, t;
for(int i=0; i<n; i++) {
for(int j=0; j<n; j++) {
pq.push(matrix[i][j]);
if(pq.size()>k)
pq.pop();
}
}
// t=pq;
// while(!t.empty()) {
// cout<<t.top()<<" ";
// t.pop();
// }
return pq.top();
}
};```
### Java code for Kth Smallest Element in a Sorted Matrix
```class Solution {
public int kthSmallest(int[][] matrix, int k) {
int n = matrix.length;
PriorityQueue<Integer> pq = new PriorityQueue<>((o1, o2) -> Integer.compare(o2, o1));
for(int i=0; i<n; i++) {
for(int j=0; j<n; j++) {
pq.offer(matrix[i][j]);
if(pq.size() > k)
pq.poll();
}
}
return pq.poll();
}
}```
## Complexity Analysis for Kth Smallest Element in a Sorted Matrix LeetCode Solution:
• Time: O(M*N*logK) where M ≤ 300 is the number of rows, N ≤ 300 is the number of columns.
• Space: O(K), space for heap which stores up to k elements
Translate » | 761 | 2,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-26 | latest | en | 0.708638 |
http://kwiznet.com/p/takeQuiz.php?ChapterID=11633&CurriculumID=49&Method=Worksheet&NQ=8%CE%9Dm=10.1 | 1,571,536,026,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986700560.62/warc/CC-MAIN-20191020001515-20191020025015-00097.warc.gz | 114,588,606 | 3,294 | Name: ___________________Date:___________________
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### High School Mathematics - 2 Introduction to Trigonometry: Degrees and Radians
Introduction
The word 'trigonometry' is derived from the Greek words 'trigon' and 'metron' and it means measuring the sides of a triangle.
Angles
An angle is considered as a figure obtained by rotating a given ray about its initial point. The original rotation of the ray is called the initial side and the final position of the ray after rotation is called the terminal side of the angle. The point of rotation is called the vertex. If the direction of rotation is anti-clockwise, the angle is positive and if the direction of rotation is clockwise, then the angle is negative.
Degree Measure
• The angle that measures one degree (1o) is (1/360th) of a revolution.
• One sixtieth of a degree is called a minute.(1l)
• One sixtieth of a minute is called a second. (1ll)
• Some of the angles whose measures are 180o, 2700, 420o, -30o, -420o as shown below.
• This is another unit for measurement of an angle, called the radian measure. In this system, the unit of measurement is a radian.
• An angle with its vertex at the centre of a circle which intercepts an arc equal in length to the radius of the circleis said to have a measure of 1 radian.
• We know that the circumference, S of a circle of radius r is 2pr. Thus one complete revolution subtends an angle of 2p/r = 2p radian.
It is well known that equal arcs of a circle subtend equal angles at the centre. Since an arc of length r subtends an angle whose measure is l radian, therefore an arc of length l will subtend an angle whose measure is l/r radians. Thus, if in a circle of radius r, an arc of length l subtends an angle theta radians at the centre, we have
q = l/r
Since a circle subtends at the centre an angle whose measure is 2p nits in radians and it is 360 units in degrees, it follows that | 486 | 2,000 | {"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} | 4.6875 | 5 | CC-MAIN-2019-43 | latest | en | 0.908972 |
https://www.slothcoders.com/cpp-program-to-find-the-sum-and-average-of-three-numbers/ | 1,674,878,982,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00086.warc.gz | 1,035,963,980 | 20,011 | # C++ Program to Find the Sum and Average of Three Numbers
Howdy readers, today you will learn how to write a C++ program to find the sum and average of three numbers.
The below program reads three integers from the user and computes their sum and average using the following formulas: –
• Sum = Number1 + Number2 + Number3
• Average = (Number1 + Number2 + Number3) / 3
So, without any delay, let’s begin this tutorial.
## C++ Program to Find the Sum and Average of Three Numbers
C++ Program
```// C++ Program to Find the Sum and Average of Three Numbers
#include <iostream>
using namespace std;
int main(){
int num1, num2, num3, sum;
float average;
// Taking input
cout << "Enter the first number: ";
cin >> num1;
cout << "Enter the second number: ";
cin >> num2;
cout << "Enter the third number: ";
cin >> num3;
// Sum of three numbers
sum = num1 + num2 + num3;
// Average of three numbers
average = sum / 3;
// Display result
cout << "The sum of three numbers is: " << sum << endl;
cout << "The average of three numbers is: " << average << endl;
return 0;
}
```
Output
```Enter the first number: 12
Enter the second number: 15
Enter the third number: 21
Sum of three numbers is: 48
Average of three numbers is: 16
```
Explanation
``` int num1, num2, num3, sum;
float average;
```
We have declared four int data type variables and one float data type variable `num1`, `num2`, `num3`, `sum` and `average` respectively.
``` cout << "Enter the first number: ";
cin >> num1;
cout << "Enter the second number: ";
cin >> num2;
cout << "Enter the third number: ";
cin >> num3;
```
Then, the user is asked to input three integers. These numbers get stored in the `num1`, `num2` and `num3` variables.
``` sum = num1 + num2 + num3;
```
The sum of three numbers is computed with the help of (+) plus operator. The result gets stored in the `sum` named variable.
``` average = sum / 3;
```
The average of three numbers is calculated using the formula:
• Average of Three Numbers = Sum / 3 = (num1 + num2 + num3) / 3.
The average of three numbers gets stored in the `average` named variable.
``` cout << "The sum of three numbers is: " << sum << endl;
cout << "The average of three numbers is: " << average << endl;
```
Finally, the sum and average of three numbers are displayed on the screen using the cout statement.
## Conclusion
Today you learned how to write a C++ program to find the sum and average of three numbers.
If you have any queries related to the program, comment down your questions in the comment section. | 665 | 2,561 | {"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-2023-06 | latest | en | 0.785776 |
http://mathhelpforum.com/discrete-math/58083-mapping-one-one-onto.html | 1,527,200,402,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866870.92/warc/CC-MAIN-20180524205512-20180524225512-00197.warc.gz | 189,088,565 | 11,133 | # Thread: Mapping: one-to-one and onto
1. ## Mapping: one-to-one and onto
Hello all!
I went through this and thought I had a handle on it till I spoke with a classmate of mine. Hopefully someone can set me straight.
a.) Is there a 1-to-1 function from the set {1,2,3} to the set {1,3}? Why or why not?
b.) Is there a function mapping {1,2,3} onto the set {1,3}? Why or why not?
c.) Is there a 1-to-1 function mapping the open interval (0,2) to the open interval (0,1)?
I answered no to (a) yes to (b) and no to (c)....my classmate crossed the sets in part (a) and has answered yes.....Why would you cross them?
Thanks!!!!!!!
2. For a, you cannot have a 1-to-1 function because if we make:
$\displaystyle D = \{1,2,3 \}$
$\displaystyle C = \{1,3 \}$
$\displaystyle D \ x \ C = \{(1,1), (1,3), (2,1), (2,3), (3,1), (3,3) \}$
There are three elements of D and only two in C, A function:
$\displaystyle f : D \to C$
Cannot possibly be 1-to-1 because there is a third element in the domain that cannot go with an element of the range.
You were right with a.
3. I'm not sure but since intervals (0,1) and (0,2) have the same cardinality there must be an injection from one to another
4. Originally Posted by vincisonfire
I'm not sure but since intervals (0,1) and (0,2) have the same cardinality there must be an injection from one to another
I'm not familiar with the mathematical term "injection."
5. It's saying that because the two intervals have an equal number of elements (known as equinumerous), there must be a function that exists that is a one-to-one function.
6. Originally Posted by dolphinlover
Hello all!
I went through this and thought I had a handle on it till I spoke with a classmate of mine. Hopefully someone can set me straight.
a.) Is there a 1-to-1 function from the set {1,2,3} to the set {1,3}? Why or why not?
b.) Is there a function mapping {1,2,3} onto the set {1,3}? Why or why not?
c.) Is there a 1-to-1 function mapping the open interval (0,2) to the open interval (0,1)?
I answered no to (a) yes to (b) and no to (c)....my classmate crossed the sets in part (a) and has answered yes.....Why would you cross them?
Thanks!!!!!!!
I still don't get part (c)... Am I correct in thinking of the open interval (0,2) as being all real #'s between 0 & 2 (but not including)?
7. Yes, you think of the intervals that way but consider this:
The interval (0,1) will be the set of real numbers between 0 and 1, it can be shown that this will be a set of real numbers that is infinite and uncountable.
The interval (0,2) will be the set of real numbers between 0 and 2, it can also be shown that this will be a set of real numbers that is infinite and uncountable.
It can then be said that the two sets have the same number of elements and therefore a function must exist such that the function is one to one.
Since the two sets are equinumerous, then there is a function that MUST be bijective, meaning that there is a function that is one-to-one AND onto.
It's hard to understand but infinite is infinite, no matter how you divide it or try to dissect it.
8. Thank You! Our professor has told us we would be talking about sizes of infinity, but we haven't gotten to that yet. Thanks Again! | 898 | 3,237 | {"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} | 3.734375 | 4 | CC-MAIN-2018-22 | latest | en | 0.924306 |
https://www.cgaa.org/articles/how-much-is-4-quarters | 1,679,439,881,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00228.warc.gz | 807,242,913 | 15,962 | # How much is 4 quarters?
Author Elijah Munoz
Posted May 27, 2022
A quarter is a coin worth one-fourth of a dollar, or 25 cents. There are four quarters in a dollar. Therefore, four quarters is worth one dollar.
Related Read: How to pronounce quarterly?
## How many quarters are in a dollar?
A quarter is one fourth of a dollar. There are four quarters in a dollar.
Related Read: What are gold quarters worth?
## How many quarters are in a dime?
A dime is a coin worth ten cents. There are four quarters in a dime. This can be seen by looking at a dime. It is divided into four equal parts, each worth two and a half cents. There are also ten dimes in a dollar, so a dime is one tenth of a dollar.
Related Read: What does a quarter till mean?
## How many quarters are in a nickel?
A nickel is a five-cent coin. It is composed of 75% copper and 25% nickel. The diameter of a nickel is 0.835 inches (21.21 mm) and the thickness is 1.95 mm. The weight of a nickel is 5.0 grams. There are 100 pennies in a dollar, so there are 20 nickels in a dollar. There are 4 quarters in a dollar, so there are 80 quarters in a dollar. Therefore, there are 16 quarters in a nickel.
Related Read: How many weeks are in a quarter?
## How many quarters are in a penny?
A penny is a small, coin-like piece of metal that is issued by a government as money. It is used as currency in many countries, including the United States. One quarter is also equal to 25 cents. So, in answer to the question posed, there are four quarters in a penny.
A penny is made up of a small, coin-like piece of metal that is circulated as money by a government. In the United States, a penny is made up of a copper-colored disk with the image ofPresident Abraham Lincoln on one side and the words "In God We Trust" on the other side. On the back of the penny, there is an image of the Lincoln Memorial. The Lincoln penny was first minted in 1909 to commemorate the 100th anniversary of President Lincoln's birth.
The United States Mint produces billions of pennies each year. In 2016, the U.S. Mint produced over 8 billion pennies! That's a lot of pennies! The total weight of all the pennies produced in 2016 was over 1,500 tons. That's the same weight as about 300 African elephants!
The average penny costs the U.S. Mint 1.83 cents to produce. That means it costs the government more to make a penny than the penny is actually worth! In 2017, the U.S. Mint announced that it was losing money on every penny it produced and that it was considering discontinuing the penny. But for now, the penny remains in circulation.
So, how many quarters are in a penny? There are four quarters in a penny.
Related Read: What does it mean when you find a quarter?
## How many quarters are in a quarter?
How many quarters are in a quarter? The answer is four. A quarter is a unit of time that is equal to one fourth of an hour. There are 60 minutes in an hour, so there are 15 minutes in a quarter. There are four quarters in an hour.
Related Read: How much is 4000 quarters?
## How many quarters are in a half dollar?
A half dollar coin is worth 50 cents. There are four quarters in a dollar, so there are two quarters in a half dollar.
Related Read: How many quarters are in \$20?
## How many quarters are in a three quarter dollar?
A quarter is a coin worth one-fourth of a United States dollar, or 25 cents. There are four quarters in a dollar. Therefore, there are twelve quarters in a three quarter dollar.
Related Read: How much is 100 quarters?
## How many quarters are in a four quarter dollar?
A quarter is a coin worth one-fourth of a dollar, or 25 cents. There are four quarters in a dollar.
Related Read: Where can I get quarters for laundry?
## How many quarters are in a five quarter dollar?
A quarter is a coin worth one-fourth of a United States or Canadian dollar. There are four quarters in a dollar. Therefore, there are five quarters in a five quarter dollar.
Related Read: How much is a 1979 quarter worth?
## FAQs
### Does 4 quarters equal a dollar?
Yes, 4 quarters equal 1 dollar.
### How much does 4 quarters make?
4 quarters make a dollar.
### How many quarters is 4?
There are 8 quarters in 4.
### How many dollars is 1 quarters?
There are twenty quarters in a dollar.
### How many coins make a dollar?
There are twenty coins in a dollar.
### How does 4 quarters make a dollar?
The mathematics behind making a dollar using 4 quarters is simple. Take the total value of quarters (i.e. 100 cents), add 5 cents to get 105 cents, and then divide this value by 4 to get the final dollar amount.
### How much does a quarter take to make?
A quarter takes two minutes and twenty-four seconds to make.
### How many quarters are in 4?
There are 16 quarters in 4.
### How many quarters is 10 bucks?
There are ten quarters in one dollar.
### How many 5 cents make a dollar?
Twenty nickels make a dollar.
### How many pennies make a dollar?
There are twenty nickels in a dollar and fifty cents make up the bulk of a dollar.
### How much is 4 quarters coins?
Forty four quarters make a dollar.
### How do you make \$1 out of quarters?
By taking four quarters and making one dollar.
### How many cents make a dollar?
One cent makes a dollar.
### Is a quarter equal to 25?
Yes, a quarter is equal to 25. | 1,282 | 5,325 | {"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-2023-14 | latest | en | 0.970965 |
https://www.cs.auckland.ac.nz/historydisplays/SecondFloor/Totalisators/FirstJulius/HorseAdder.php | 1,701,577,247,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00460.warc.gz | 815,193,959 | 7,211 | # First Automatic Totalisator: More Details of Horse Adders
The central shaft has two differentials and two driving chains, one to the 12-shaft and one to the 17-shaft. The central shaft is really a small adder that combines the rotations of the 12-shaft and 17-shaft. The details are a bit difficult, but the following diagram, viewing the central shaft from above, might help explain what is happening.
In the diagram the directions of rotation at the top are given by the arrows. The central shaft rotates counter-clockwise. It is pinned to the back planetary gear that also rotates counter-clockwise (viewed from the front of the machine). The first planetary gear may rotate if the second planetary gear also rotates counter-clockwise or the front end cog rotates clockwise. The front-end cog drives the 17-shaft. The difference in sprocket-wheel diameter of the central shaft and the 17-shaft means that a 1/20 rotation of the 17-shaft is represented as a 1/10 rotation of the central shaft front-end cog, but the rotation passed through to the back-end planetary gear is halved. So a 1/20 clockwise rotation of the 17-shaft becomes a 1/20 counter-clockwise rotation of the central shaft. The rotation of the 12-shaft is similarly doubled by the gearing. However, the 12-shaft drives the front planetary gear and in this case the rotation is not halved. So a 1/20 counter-clockwise rotation of the 12-shaft becomes a 1/10 counter-clockwise rotation of the central shaft and so represents a value twice that of the 17-shaft, so we can deduce that the 12-shaft is for 20 shilling bets and the 17-shaft for 10 shilling bets.
Support for our interpretation of the horse adders comes from the 1915 patent application. This introduces the idea of having differing numbers of teeth on the escapement wheels to represent different sized bets (20 teeth for 10/-, 10 for 20/-, 4 for 5 pounds etc.) The patent says:
The construction by which it is made practicable to compute by means of an epicyclic train on one shaft a plurality of dissimilar values is one of the the leading features of the present invention, which distinguishes it from known apparatus of its class in which the several different orders of values must be separately computed on separate trains of epicyclic gearing, and the movements of these several trains compounded together in another epicyclic train upon a collecting shaft, through which motion is transmitted to the indicator.
In later machines it was still common to have multiple shafts but this seems to have been because smaller shafts were more reliable and practical.
How the 5-pound bets were counted is unknown. There does not appear to be an escapement wheel with fewer teeth so the 5-pound bets must be counted in some ad-hoc manner. It remains a puzzle but could be something to do with the central shaft (allowing the back-end cog, assumed fixed, to actually complete one full rotation would have the right result).
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(Only if you need a reply) | 691 | 3,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-50 | latest | en | 0.906993 |
https://www.lmfdb.org/EllipticCurve/Q/270480gs2/ | 1,638,805,513,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00039.warc.gz | 943,978,149 | 27,659 | # Properties
Label 270480gs2 Conductor $270480$ Discriminant $5.691\times 10^{16}$ j-invariant $$\frac{168591300897604}{472410225}$$ CM no Rank $0$ Torsion structure $$\Z/{2}\Z \times \Z/{2}\Z$$
# Related objects
Show commands: Magma / Pari/GP / SageMath
## Minimal Weierstrass equation
sage: E = EllipticCurve([0, 1, 0, -568416, -164737980])
gp: E = ellinit([0, 1, 0, -568416, -164737980])
magma: E := EllipticCurve([0, 1, 0, -568416, -164737980]);
$$y^2=x^3+x^2-568416x-164737980$$
## Mordell-Weil group structure
$\Z/{2}\Z \times \Z/{2}\Z$
## Torsion generators
sage: E.torsion_subgroup().gens()
gp: elltors(E)
magma: TorsionSubgroup(E);
$$\left(-418, 0\right)$$, $$\left(870, 0\right)$$
## Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
$$\left(-453, 0\right)$$, $$\left(-418, 0\right)$$, $$\left(870, 0\right)$$
## Invariants
sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); Conductor: $$270480$$ = $2^{4} \cdot 3 \cdot 5 \cdot 7^{2} \cdot 23$ sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); Discriminant: $56912476734489600$ = $2^{10} \cdot 3^{6} \cdot 5^{2} \cdot 7^{8} \cdot 23^{2}$ sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); j-invariant: $$\frac{168591300897604}{472410225}$$ = $2^{2} \cdot 3^{-6} \cdot 5^{-2} \cdot 7^{-2} \cdot 13^{3} \cdot 23^{-2} \cdot 2677^{3}$ Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $2.0872855398576474508040118928\dots$ Stable Faltings height: $0.53670781486336970707030875320\dots$
## BSD invariants
sage: E.rank() magma: Rank(E); Analytic rank: $0$ sage: E.regulator() magma: Regulator(E); Regulator: $1$ sage: E.period_lattice().omega() gp: E.omega[1] magma: RealPeriod(E); Real period: $0.17390259538277018135711039237\dots$ sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); Tamagawa product: $384$ = $2^{2}\cdot( 2 \cdot 3 )\cdot2\cdot2^{2}\cdot2$ sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); Torsion order: $4$ sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Analytic order of Ш: $1$ (exact) sage: r = E.rank(); sage: E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: ar = ellanalyticrank(E); gp: ar[2]/factorial(ar[1]) magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12); Special value: $L(E,1)$ ≈ $4.1736622891864843525706494168761433901$
## Modular invariants
Modular form 270480.2.a.gs
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
magma: ModularForm(E);
$$q + q^{3} - q^{5} + q^{9} + 4 q^{11} - 2 q^{13} - q^{15} + 2 q^{17} - 4 q^{19} + O(q^{20})$$
For more coefficients, see the Downloads section to the right.
sage: E.modular_degree() magma: ModularDegree(E); Modular degree: 2949120 $\Gamma_0(N)$-optimal: no Manin constant: 1
## Local data
This elliptic curve is not semistable. There are 5 primes of bad reduction:
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$2$ $4$ $I_{2}^{*}$ Additive 1 4 10 0
$3$ $6$ $I_{6}$ Split multiplicative -1 1 6 6
$5$ $2$ $I_{2}$ Non-split multiplicative 1 1 2 2
$7$ $4$ $I_{2}^{*}$ Additive -1 2 8 2
$23$ $2$ $I_{2}$ Split multiplicative -1 1 2 2
## Galois representations
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
The $\ell$-adic Galois representation has maximal image for all primes $\ell$ except those listed in the table below.
prime $\ell$ mod-$\ell$ image $\ell$-adic image
$2$ 2Cs 2.6.0.1
## $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(5,20) if E.conductor().valuation(p)<2]
All $p$-adic regulators are identically $1$ since the rank is $0$.
No Iwasawa invariant data is available for this curve.
## Isogenies
This curve has non-trivial cyclic isogenies of degree $d$ for $d=$ 2.
Its isogeny class 270480gs consists of 2 curves linked by isogenies of degrees dividing 4.
## Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z \times \Z/{2}\Z$ are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $4$ $$\Q(\sqrt{3}, \sqrt{322})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{-3}, \sqrt{-35})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{35}, \sqrt{-230})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database $8$ Deg 8 $$\Z/2\Z \times \Z/6\Z$$ Not in database $16$ Deg 16 $$\Z/4\Z \times \Z/4\Z$$ Not in database $16$ Deg 16 $$\Z/2\Z \times \Z/8\Z$$ Not in database $16$ Deg 16 $$\Z/2\Z \times \Z/8\Z$$ Not in database $16$ Deg 16 $$\Z/2\Z \times \Z/8\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial. | 1,987 | 5,188 | {"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} | 3.15625 | 3 | CC-MAIN-2021-49 | latest | en | 0.239358 |
http://www.sparknotes.com/math/geometry1/buildingblocks/problems_4/ | 1,519,318,747,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814140.9/warc/CC-MAIN-20180222160706-20180222180706-00088.warc.gz | 552,824,342 | 13,810 | # Building Blocks of Geometry
### Contents
#### Problems
Problem : Can two-dimensional objects exist in three-dimensional space?
Yes. They simply exist in the same plane within space.
Problem : What is the minimum number of lines it takes to span space?
Three
Problem : Explain why three noncolinear points cannot form a three-dimensional object in space?
A plane can be created to contain any three points in space, therefore, three points isn't enough to create a three-dimensional object. It requires four noncoplanar points to form a three-dimensional object. | 114 | 571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-09 | latest | en | 0.908248 |
https://www.assignmentexpert.com/homework-answers/mathematics/statistics-and-probability/question-12597 | 1,591,397,887,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348504341.78/warc/CC-MAIN-20200605205507-20200605235507-00115.warc.gz | 622,735,026 | 85,795 | 87 259
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# Answer to Question #12597 in Statistics and Probability for Kolton
Question #12597
Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of grams of fat per pound, with a standard deviation of grams of fat per pound. A random sample of farm-raised trout is selected. The mean fat content for the sample is grams per pound. Find the probability of observing a sample mean of grams of fat per pound or less in a random sample of farm-raised trout. Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.
1
2012-08-09T08:02:08-0400
For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.)
You can translate into standard normal units by:
Z = ( X - μ ) / σ
Where Z ~ Normal( μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.
If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem.
If a sample of size is is drawn from a population with mean μ and standard deviation σ then the sample average xBar is normally distributed
with mean μ and standard deviation σ /√(n)
In this question we have
Xbar ~ Normal( μ = 32 , σ² = 56.25 / 36 )
Xbar ~ Normal( μ = 32 , σ² = 1.5625 )
Xbar ~ Normal( μ = 32 , σ = 7.5 / sqrt( 36 ) )
Xbar ~ Normal( μ = 32 , σ = 1.25 )
Find P( Xbar < 29.7 )
P( ( Xbar - μ ) / σ < ( 29.7 - 32 ) / 1.25 ) = P( Z < -1.84 ) = 0.03288412.
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for any assignment or question with DETAILED EXPLANATIONS! | 533 | 1,959 | {"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-2020-24 | latest | en | 0.851996 |
https://splice.com/blog/what-is-impedance-audio/ | 1,726,647,414,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00767.warc.gz | 505,743,760 | 20,324 | # What is impedance in audio (and why is it important)?
Illustration: Daniel Zender
## Impedance is a bit of physics that makes all the difference when hooking up speakers, headphones, and more.
The concept of impedance is vital to electrical engineering and can affect the sound coming out of our own audio devices. In this article, let’s explore what impedance is, and why it can be important for both making and listening to music.
## What is impedance?
In simplest terms, you can think of impedance as resistance. In electrical engineering, it’s the opposing force presented by a circuit to the flow of alternating electrical current once a voltage is applied, measured in ohms (Ω). You can see the relationship between these three variables in Ohm’s Law, named after the German physicist Georg Simon Ohm:
Voltage (V) = electrical pressure, measured in volts
Resistance (R) = electrical resistance, measured in ohms
Current (I) = rate of electrical flow, measured in amperes
Impedance is resistance when talking about alternating currents (or AC), like in the case of audio signals. Increasing the resistance at a constant voltage will lower the current in a circuit, and increasing the voltage at a constant resistance will increase its current.
## Why is it important?
So how does all of this apply to sound? Well, impedance and resistance can affect anything that electrical audio signals pass through—even the highly-conductive copper wire inside many audio cables has its own electrical resistance that can increase as it becomes thinner or travels further.
Impedance can be especially important when it comes to powering headphones or speakers, because you’ll get the most efficient (and safest) transfer of electrical power by matching your playback device to an amplifier that complements its impedance. Let’s run through two common examples:
1. Passive speakers
Most studio monitor speakers are classified as active, meaning they come with their own built-in power amplifier and you don’t really need to worry about impedance. However, many stereo speakers (like the ones that use thin copper speaker wire) are passive and need incoming audio signals to be amplified by something like a stereo receiver.
Impedance ratings for passive speakers are normally between 4 ohms and 16 ohms, while amplifiers or receivers are rated within a given range like 6 ohms – 12 ohms. Check the back of these devices or the manufacturer specifications for this info!
If the impedance of your speakers is higher than what the amplifier is rated for, that’s technically fine—the amp just won’t be able to provide its maximum power output. What you’ll want to avoid at all costs is hooking up speakers with a lower impedance than what the receiver or amplifier is rated for. If the resistance goes down, Ohm’s Law tells us that either the current or voltage must increase, which will draw more power than what a connected amp is capable of delivering. This can cause the audio to distort and even lead to serious damage of your amplifier over time.
Most consumer-grade headphones have generally low impedance ratings, often between 8 and 32 ohms. This means that the device they are being plugged into (like a laptop or phone) doesn’t need to work very hard to “drive” the headphones, or push enough voltage into the circuit’s current to effectively amplify incoming audio signals for our ears.
However, some higher-grade headphones meant for monitoring or mixing might come with impedance ratings like 250 ohms or higher. These cans often need way more amplification power than what your laptop or phone can provide on their own in order to play audio at reasonable volumes. This power can be supplied by hooking them up to an audio interface or a dedicated headphone amplifier.
Because high-impedance headphones can contain more delicate circuitry, they technically operate more efficiently and with less distortion, which can result in better overall sound output. That said, this doesn’t mean that low-impedance headphones can’t sound good too!
Did we miss anything about impedance? What topics in audio technology would you like to see us explore next? Start a conversation with us and other audio engineers via the Splice Discord.
Explore more key topics in audio technology:
January 21, 2022
Matteo Malinverno Matteo Malinverno is a New York-based music producer currently working on the Content team at Splice. | 888 | 4,432 | {"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 | 0.941409 |
http://www.enotes.com/homework-help/basketball-player-dunks-ball-momentarily-hangs-252570 | 1,477,455,998,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720615.90/warc/CC-MAIN-20161020183840-00038-ip-10-171-6-4.ec2.internal.warc.gz | 433,403,805 | 9,310 | # A basketball player dunks the ball and momentarily hangs from the rim of the basket. Assume that the player can be considered as a 95.0 kg point......mass at a height of 2.0 m above the floor. If...
A basketball player dunks the ball and momentarily hangs from the rim of the basket. Assume that the player can be considered as a 95.0 kg point...
...mass at a height of 2.0 m above the floor. If the basket rim has a spring constant of 7.4 x 10^3 N/m, by how much does the player displace the rim from the horizontal position?
Asked on by islnds
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
From the information given, we have a player of mass 95 kg hanging from the rim of a basket. The behavior of the basket rim can be simplified to that of a spring with a spring constant of 7.4*10^3 N/m.
The displacement of the basket rim can be determined using Hooke's law. According to Hooke's law, F = -kx, where F is the force causing the displacement, k is the spring constant and x is the displacement from the equilibrium position. The negative sign indicates the tendency of a spring to return to the equilibrium length when displaced.
The player of weight 95 kg exerts a force given by m*g where m is the mass of the player and g is the acceleration due to gravity equal to 9.8 m.s^2. The force is 95*9.8 = 931 N
The displacement due the force x is given by F/k. Here F = 931 N and k = 7.4*10^3 N/m
x = 931/7.4*10^3 = 0.125 m
The displacement of the basket's rim from its horizontal position due to the player hanging from it is 0.125 m.
We’ve answered 317,742 questions. We can answer yours, too. | 431 | 1,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-44 | latest | en | 0.92171 |
https://learn.careers360.com/ncert/question-the-probability-that-a-non-leap-year-selected-at-random-will-contain-53-sundays-is-a-1-by-7-b-2-by-7-c-3-by-7-d-5-by-7/?question_number=19.0 | 1,716,232,710,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00644.warc.gz | 322,014,348 | 40,896 | #### The probability that a non leap year selected at random will contain 53 Sundays is (A) $\frac{1}{7}$ (B) $\frac{2}{7}$ (C) $\frac{3}{7}$ (D) $\frac{5}{7}$
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
In 365 days there are 52 weeks and 1 day.
If it contain 53 sunday then the 1 day of the year must be sunday.
But there are total 7 days.
Hence total number of favorable cases = 1
Hence probability of 53 sunday = $\frac{Number\, of\, favourable\, cases}{Total\, cases}$
$= \frac{1}{7}$ | 194 | 680 | {"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": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-22 | latest | en | 0.814315 |
https://javascriptcode.org/javascript-program-smallest-of-three-numbers/ | 1,696,374,443,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511284.37/warc/CC-MAIN-20231003224357-20231004014357-00282.warc.gz | 364,441,920 | 11,923 | # JavaScript Code
## Smallest of Three Numbers Program
In this tutorial, you will write a JavaScript program with a function that takes three numbers as arguments, finds the smallest of these three numbers, and returns the smallest number.
## Solution
To find the smallest of three numbers in JavaScript,
1. Consider that num1, num2, and num3 are the three numbers.
2. If num1 is less than both num2 and num3, then num1 is the smallest.
3. Else if num2 is less than num3, then num2 is the smallest.
4. Else num3 is the smallest.
## Program
1. In the following program, we find smallest of three numbers using if-else-if statement.
function findSmallest(num1, num2, num3) {
if (num1 < num2 && num1 < num3) {
return num1;
} else if (num2 < num3) {
return num2;
} else {
return num3;
}
}
var num1 = 5;
var num2 = 8;
var num3 = 2;
const result = findSmallest(num1, num2, num3);
console.log('Smallest : ' + result); | 254 | 920 | {"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-2023-40 | latest | en | 0.566884 |
http://slideplayer.com/slide/3274535/ | 1,524,550,116,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946564.73/warc/CC-MAIN-20180424041828-20180424061828-00237.warc.gz | 285,804,043 | 23,389 | # 2.2 Vertical and Horizontal Shifts of Graphs
## Presentation on theme: "2.2 Vertical and Horizontal Shifts of Graphs"— Presentation transcript:
2.2 Vertical and Horizontal Shifts of Graphs
Quiz Identify the basic function with a graph as below:
Vertical Shift of graphs
y Discussion 1 f(x) = x2 f(x) = x2+1 ↑ 1 unit f(x) = x2-2 ↓ 2 unit x f(x) = x2-5 ↓ 5 unit What about shift f(x) up by 10 unit? shift f(x) down by 10 unit?
Vertical Shift of Graphs
Discussion 2 y f(x) = x3 ↑ 2 unit f(x) = x3+2 ↓ 3 unit f(x) = x3-3 x
Vertical Shift of Graphs
If c>0, then the graph of y = f(x) + c is obtained by shifting the graph of y = f(x) upward a distance of c units. The graph of y = f(x) – c is obtained by shifting the graph of y = f(x) downward a distance of c units. ↑ f(x) + c ↓ f(x) - c
Horizontal Shift of graphs
y Discussion 1 f(x) = x2 f(x) = (x+1)2 ← 1 unit f(x) = (x-2)2 → 2 unit x f(x) = (x-5)2 → 5 unit What about shift f(x) left by 10 unit? shift f(x) right by 10 unit?
Horizontal Shift of Graphs
Discussion 2 y f(x) = |x| ← 2 unit f(x) = |x + 2| → 3 unit f(x) = |x - 3| x
Horizontal Shift of Graphs
If c > 0, the graph of y = f(x + c) is obtained by shifting the graph of y = f(x) to the left a distance of c units. The graph of y = f(x - c) is obtained by shifting the graph of y = f(x) to the right a distance of c units. f(x + c) ← → f(x - c)
Conclusion ↑ f(x) + c → f(x - c) ↓ f(x) - c f(x + c) ← y f(x) + c f(x)
Combinations of vertical and horizontal shifts
Equation write a description y1 = |x - 4|+ 3. Describe the transformation of f(x) = |x|. Identify the domain / range for both. answer: shifting f(x) up by 3 units, then shift f(x) right by 4 units. ( or shift f(x) right by 4 units, then shift f(x) up by 3 units.)
Combinations of vertical and horizontal shifts
Description equation Write the function that shifts y = x2 two units left and one unit up. answer: y1 = (x+2)2+1
Combinations of vertical and horizontal shifts
y Graph equation Write the equation for the graph below. Assume each grid mark is a single unit. Answer: f(x) = (x-1)3-2 x
Combinations of vertical and horizontal shifts
y Equation graph Sketch the graph of y = f(x) = √x-2 -1. How does the transformation affect the domain and range? x Step 1: f(x) = √x Step 2: f(x) = √x-2 Step 3: f(x) = √x-2 -1
Combinations of vertical and horizontal shifts
Graph & symbolic transformation new graph Using the given graph of f(x), sketch the graph of f(x) +2 f(x+2) f(x-1) - 3 y x
Math 101 schedule changes 1) Project 1 will be a take-home project instead of an in- class group project. The project will be posted by Wednesday, February 9, through the MyKAPInfo link. It is due in class on Monday, February ) Exam 1 for Math 101 will be moved from Feb 15/16 to Feb 16/17. Group A is scheduled for Wednesday, February 16 and Group B on Thursday, February 17. The hours for testing for both days are 7:30 am - 9:00 pm. All exams are in ST ) Correspondingly, the deadline for full credit for Skills Test #1 is moved to Tuesday, February 15.
Homework PG. 99: 3-45(M3), 47-65(odds) KEY: 18, 27, 49, 51
Reading: 2.3 Stretch, Shrink & Reflect | 1,021 | 3,150 | {"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} | 4.65625 | 5 | CC-MAIN-2018-17 | latest | en | 0.810043 |
https://metanumbers.com/94335 | 1,637,978,224,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00371.warc.gz | 486,480,885 | 7,421 | 94335 (number)
94,335 (ninety-four thousand three hundred thirty-five) is an odd five-digits composite number following 94334 and preceding 94336. In scientific notation, it is written as 9.4335 × 104. The sum of its digits is 24. It has a total of 4 prime factors and 16 positive divisors. There are 47,520 positive integers (up to 94335) that are relatively prime to 94335.
Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 24
• Digital Root 6
Name
Short name 94 thousand 335 ninety-four thousand three hundred thirty-five
Notation
Scientific notation 9.4335 × 104 94.335 × 103
Prime Factorization of 94335
Prime Factorization 3 × 5 × 19 × 331
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 94335 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 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 94,335 is 3 × 5 × 19 × 331. Since it has a total of 4 prime factors, 94,335 is a composite number.
Divisors of 94335
1, 3, 5, 15, 19, 57, 95, 285, 331, 993, 1655, 4965, 6289, 18867, 31445, 94335
16 divisors
Even divisors 0 16 8 8
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 159360 Sum of all the positive divisors of n s(n) 65025 Sum of the proper positive divisors of n A(n) 9960 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 307.14 Returns the nth root of the product of n divisors H(n) 9.47139 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 94,335 can be divided by 16 positive divisors (out of which 0 are even, and 16 are odd). The sum of these divisors (counting 94,335) is 159,360, the average is 9,960.
Other Arithmetic Functions (n = 94335)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 47520 Total number of positive integers not greater than n that are coprime to n λ(n) 1980 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 9089 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 47,520 positive integers (less than 94,335) that are coprime with 94,335. And there are approximately 9,089 prime numbers less than or equal to 94,335.
Divisibility of 94335
m n mod m 2 3 4 5 6 7 8 9 1 0 3 0 3 3 7 6
The number 94,335 is divisible by 3 and 5.
Classification of 94335
• Arithmetic
• Deficient
• Polite
• Square Free
Other numbers
• LucasCarmichael
Base conversion (94335)
Base System Value
2 Binary 10111000001111111
3 Ternary 11210101220
4 Quaternary 113001333
5 Quinary 11004320
6 Senary 2004423
8 Octal 270177
10 Decimal 94335
12 Duodecimal 46713
20 Vigesimal bfgf
36 Base36 20sf
Basic calculations (n = 94335)
Multiplication
n×y
n×2 188670 283005 377340 471675
Division
n÷y
n÷2 47167.5 31445 23583.8 18867
Exponentiation
ny
n2 8899092225 839495865045375 79193842429055450625 7470751125544945934709375
Nth Root
y√n
2√n 307.14 45.5223 17.5254 9.88404
94335 as geometric shapes
Circle
Diameter 188670 592724 2.79573e+10
Sphere
Volume 3.51647e+15 1.11829e+11 592724
Square
Length = n
Perimeter 377340 8.89909e+09 133410
Cube
Length = n
Surface area 5.33946e+10 8.39496e+14 163393
Equilateral Triangle
Length = n
Perimeter 283005 3.85342e+09 81696.5
Triangular Pyramid
Length = n
Surface area 1.54137e+10 9.89355e+13 77024.2
Cryptographic Hash Functions
md5 65fd584b1e19f75d1bb84ea4bed64230 e6b1a9b71b2804d9732747e7787d8df20d690ea6 c5c1e3c0d9c21eec26328e33ebc7a6c9e31058eb130e908e0594cffda530637f c6e1e9a7c5cd08f0b09ebe9aa7748d01d73221bb981cd2f0da33b7f76403ac56631c907c871272c398f634302001fa4e44dd26ede8d297c63733aa97c15d2833 a309d703b0468b1fa5eb12598c29c298a10a0106 | 1,473 | 4,147 | {"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} | 3.8125 | 4 | CC-MAIN-2021-49 | latest | en | 0.820307 |
https://www.jiskha.com/questions/6256/a-block-with-a-mass-of-5-0-kg-is-held-in-equilibrium-on-an-incline-of-an-angle-with-a | 1,582,655,213,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146127.10/warc/CC-MAIN-20200225172036-20200225202036-00228.warc.gz | 778,008,010 | 6,095 | Physics
A block with a mass of 5.0 kg is held in equilibrium on an incline of an angle with a measure of 30.0 degrees by the horizontal force F. Find the magnitude of F. Find the normal force exerted by the inbline on the block. (disregard friction)
Well, I don't even know how to start. I DO know I have to use trig at some point... but help??
img145.imageshack.us/img145/8909/trigi6.png
so I just use these equations to find each side and solve for F?
F as a horizontal force, has components up the plane, and normal to the plane. The component up the plane is Fcos30. Then normal to the plane is Fsin30. The mass has part of its weight in the dirction normal to the plane (mgcos30), and up the plane mgsin30.
So, the forces in the plane direction is
Fcos30 and mgsin30. They sum to zero, so find F.
Fcos30=mgSin30
F =-mg tan30
check my thinking.
sorry for the error before.
You should try to reproduce the answer yourself. You can also decompose the normal force in the horizontal and the vertical direction. If you set the total force in these directions zero you get:
F_{n}cos(30°) - Mg = 0
F_{n}sin(30°) + F = 0
28.3 is what I got when I did the last equation of yours...
and when I did mgSin30 / cos30.
is this correct?
I'm sorry, but I don't quite understand what you mean...
and what does the _{n} mean in those equations?
F_{n} is the normal force.
Let'S start from the beginning. Accordingto Newton's second law:
F = m a
If a mass is not accelerating, then the total force acting on it must be zero. In this case there are three forces acting on the mass:
Gravity acts downward with a force of Mg
A force of F is exerted in the horizontal direction
A normal force F_{n} is exerted by the incline on the mass in the normal direction. The normal direction is at 30 degrees from the verical.
This means that the component of the normal foce in the vertical direction is
F_{n}cos(30°)
and in the horizontal direction it is
F_{n}cos(30°)
The component of the gravitational force in the vertical direction is -Mg, the minus sign means that it is directed downward.
The total force in the vertical direction is thus:
F_{n}cos(30°) - Mg
and in the horizontal component of the total force is:
F_{n}Sin(30°) + F
Count: This is the wrong approach to the problem. She needs to follow the directions I gave. Mg may be the vertical force, but on a plane she needs the normal and component parallel to the plane.
Thanks.
Yes.
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More Similar Questions | 1,537 | 5,581 | {"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-2020-10 | latest | en | 0.925254 |
http://technobabble.com.au/2018/01/02/the-math-behind-gerrymandering-and-wasted-votes/ | 1,566,506,473,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317359.75/warc/CC-MAIN-20190822194105-20190822220105-00201.warc.gz | 184,582,758 | 10,189 | # The Math Behind Gerrymandering and Wasted Votes
Imagine fighting a war on 10 battlefields. You and your opponent each have 200 soldiers, and your aim is to win as many battles as possible. How would you deploy your troops? If you spread them out evenly, sending 20 to each battlefield, your opponent could concentrate their own troops and easily win a majority of the fights. You could try to overwhelm several locations yourself, but there’s no guarantee you’ll win, and you’ll leave the remaining battlefields poorly defended. Devising a winning strategy isn’t easy, but as long as neither side knows the other’s plan in advance, it’s a fair fight.
#### Quanta Magazine
Original story reprinted with permission from Quanta Magazine, an editorially independent publication of the Simons Foundation whose mission is to enhance public understanding of science by covering research developments and trends in mathematics and the physical and life sciences.
Now imagine your opponent has the power to deploy your troops as well as their own. Even if you get more troops, you can’t win.
In the war of politics, this power to deploy forces comes from gerrymandering, the age-old practice of manipulating voting districts for partisan gain. By determining who votes where, politicians can tilt the odds in their favor and defeat their opponents before the battle even begins.
In 1986, the Supreme Court ruled extreme partisan gerrymanders unconstitutional. But without a reliable test for identifying unfair district maps, the court has yet to throw any out. Now, as the nation’s highest court hears arguments for and against a legal challenge to Wisconsin’s state assembly district map, mathematicians are on the front lines in the fight for electoral fairness.
Simple math can help scheming politicians draw up districts that give their party outsize influence, but mathematics can also help identify and remedy these situations. This past summer the Metric Geometry and Gerrymandering Group, led by the mathematician Moon Duchin, convened at Tufts University, in part to discuss new mathematical tools for analyzing and addressing gerrymandering. The “efficiency gap” is a simple idea at the heart of some of the tools being considered by the Supreme Court. Let’s explore this concept and some of its ramifications.
Start by imagining a state with 200 voters, of whom 100 are loyal to party A and 100 to party B. Let’s suppose the state needs to elect four representatives and so must create four districts of equal electoral size.
Imagine that you have the power to assign voters to any district you wish. If you favor party A, you might distribute the 100 A voters and 100 B voters into the four districts like this:
With districts constructed in this way, party A wins three of the four elections. Of course, if you prefer party B, you might distribute the voters this way:
Here, the results are reversed, and party B wins three of the four elections.
Notice that in both scenarios the same number of voters with the same preferences are voting in the same number of elections. Changing only the distribution of voters among the districts dramatically alters the results. The ability to determine voting districts confers a lot of power, and attending to some simple math is all that’s needed to create an electoral edge.
What if, instead of creating an advantage for one party over the other, you wished to use your power to create fair districts? First, you’d need to determine what “fair” means, and that can be tricky, as winners and losers often have different perspectives on fairness. But if we start with some assumptions about what “fair” means, we can try to quantify the fairness of different voter distributions. We may argue about those assumptions and their implications, but by adopting a mathematical model we can attempt to compare different scenarios. The efficiency gap is one approach to quantifying the fairness of a voter distribution.
To understand the efficiency gap, we can begin with the observation that, in a series of related elections, not all votes have the same impact. Some votes might make a big difference, and some votes might be considered “wasted.” The disparity in wasted votes is the efficiency gap: It measures how equally, or unequally, wasted votes are distributed among the competing parties.
So what counts as a wasted vote? Consider California’s role in presidential elections. Since 1992, California has always backed the Democratic nominee for president. Therefore, California Republicans know they are almost certainly backing a losing candidate. In some sense their vote is wasted: If they were allowed to vote in a toss-up state like Florida, their vote might make more of a difference. From a Republican perspective, that would be a more efficient use of their vote.
As it turns out, Democratic voters in California can make a similar argument about their vote being wasted. Since the Democratic candidate will likely win California in a landslide, many of their votes, in a sense, are wasted, too: Whether the candidate wins California with 51 percent of the vote or 67 percent of the vote, the outcome is the same. Those extra winning votes are meaningless.
Thus, in the context of the efficiency gap, there are two kinds of wasted votes: those for a losing candidate and those for a winning candidate that go beyond what is necessary for victory (for simplicity, we take the threshold for victory to be 50 percent, even though this could technically result in a tie; an actual tie is beyond unlikely with hundreds of thousands of voters in each congressional district). In a multi-district election, each party will likely have wasted votes of each kind. The efficiency gap is the difference in the totals of the wasted votes for each party, expressed as a percentage of total votes cast. (We subtract the smaller number from the larger when possible, to ensure a nonnegative efficiency gap. We could also take the absolute value of the difference.)
Let’s return to our four-district scenarios and examine their efficiency gaps. Our first distribution looked like this.
In this scenario, 75 of B’s votes are wasted: 60 in losing causes and 15 more than the 25 needed to win district 4. Only 25 of party A’s votes are wasted: 5 extra votes in each victory and 10 losing votes. The raw difference in wasted votes is 75 − 25 = 50, so the efficiency gap here is 50/200 = 25 percent. We say the 25 percent efficiency gap here favors party A, as party B had the larger number of wasted votes. In the second scenario, where the numbers are reversed, the 25 percent efficiency gap now favors party B.
Can the efficiency gap give us a sense of the fairness of a distribution? Well, if you had the power to create voting districts and you wanted to engineer victories for your party, your strategy would be to minimize the wasted votes for your party and maximize the wasted votes for your opponent. To this end, a technique colorfully known as packing and cracking is employed: Opposition votes are packed into a small number of conceded districts, and the remaining block of votes is cracked and spread out thinly over the rest of the districts to minimize their impact. This practice naturally creates large efficiency gaps, so we might expect fairer distributions to have smaller ones.
Let’s take a deeper look at efficiency gaps by imagining our 200-voter state now divided into 10 equal districts. Consider the following voter distribution, in which party A wins 9 of the 10 districts.
On the surface, this doesn’t seem like a fair distribution of voters. What does the efficiency gap say?
In this scenario, almost all of party B’s votes are wasted: nine losing votes in each of nine districts, plus nine excess votes in one victory, for a total of 90 wasted votes. Party A’s voters are much more efficient: only 10 total votes are wasted. There is a difference of 90 − 10 = 80 wasted votes and an efficiency gap of 80/200 = 40 percent, favoring party A.
Compare that with the following distribution, where party A wins 7 of the 10 districts.
Here, the wasted vote tally is 70 for party B and 30 for party A, producing an efficiency gap of 40/200 = 20 percent. A seemingly fairer distribution results in a smaller efficiency gap.
As a final exercise, consider this even split of district elections.
The symmetry alone suggests the answer, and the calculations confirm it: 50 wasted votes for each party means a 0 percent efficiency gap. Notice here that a 0 percent efficiency gap corresponds to an independent notion of fairness: Namely, with voters across the state evenly split between both parties, it seems reasonable that each party would win half of the elections.
These elementary examples demonstrate the utility of the efficiency gap as a measure of electoral fairness. It’s easy to understand and compute, it’s transparent, and its interpretations are consistent with other notions of fairness. It’s a simple idea, but one that is being used in a variety of complex ways to study gerrymandering. For example, mathematicians are now using simulations to consider millions of theoretical electoral maps for a given state and then examining the distribution of all possible efficiency gaps. Not only does this create a context for evaluating the fairness of a current map against other possibilities, it can also potentially be used to suggest fairer alternatives.
Though voters are not actually assigned to districts in the way we have imagined in our examples, the practice of gerrymandering achieves similar results. By strategically redrawing district boundaries, gerrymanderers can engineer voting distributions to create an uneven electoral playing field. These unfair fights affect how we are governed and help majority-party incumbents coast to re-election term after term. The case before the Supreme Court involves just one of many potentially unfair maps. Objective mathematical tools like the efficiency gap may be the only way to root out gerrymandering and keep our political battlefields in balance.
Download the “Doing the Political Math” PDF worksheet to practice these concepts or to share with students.
Original story reprinted with permission from Quanta Magazine, an editorially independent publication of the Simons Foundation whose mission is to enhance public understanding of science by covering research developments and trends in mathematics and the physical and life sciences. | 2,086 | 10,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-35 | longest | en | 0.952708 |
https://the-algorithms.com/fr/algorithm/prefix-function | 1,725,993,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00512.warc.gz | 532,517,108 | 23,879 | #### Prefix Function
M
R
```"""
https://cp-algorithms.com/string/prefix-function.html
Prefix function Knuth-Morris-Pratt algorithm
Different algorithm than Knuth-Morris-Pratt pattern finding
E.x. Finding longest prefix which is also suffix
Time Complexity: O(n) - where n is the length of the string
"""
def prefix_function(input_string: str) -> list:
"""
For the given string this function computes value for each index(i),
which represents the longest coincidence of prefix and suffix
for given substring (input_str[0...i])
For the value of the first element the algorithm always returns 0
>>> prefix_function("aabcdaabc")
[0, 1, 0, 0, 0, 1, 2, 3, 4]
[0, 0, 0, 1, 2, 3, 4, 0]
"""
# list for the result values
prefix_result = [0] * len(input_string)
for i in range(1, len(input_string)):
# use last results for better performance - dynamic programming
j = prefix_result[i - 1]
while j > 0 and input_string[i] != input_string[j]:
j = prefix_result[j - 1]
if input_string[i] == input_string[j]:
j += 1
prefix_result[i] = j
return prefix_result
def longest_prefix(input_str: str) -> int:
"""
Prefix-function use case
Finding longest prefix which is suffix as well
>>> longest_prefix("aabcdaabc")
4
4
>>> longest_prefix("abcab")
2
"""
# just returning maximum value of the array gives us answer
return max(prefix_function(input_str))
if __name__ == "__main__":
import doctest
doctest.testmod()
``` | 381 | 1,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.49197 |
https://www.pryor.com/blog/eliminate-your-frustration-with-excel-time-formulas/?replytocom=86 | 1,620,290,539,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.91/warc/CC-MAIN-20210506083716-20210506113716-00084.warc.gz | 996,630,947 | 35,696 | # Eliminate Your Frustration with Excel Time Formulas!
Working with time in Excel can be a problem. Have you ever calculated how long a person was at work, and then wanted to multiply that time by an hourly wage? It isn’t difficult to fail miserably! But, if you know the trick, Excel time formulas are easy.
• Columns A and B (employee’s Time In and Time Out) can be formatted to Time, AM/PM.
• Column C (Hours Worked) should be formatted to Time, Military. This omits the AM PM markers and creates a true figure to reflect the amount of time passed. You need a true reflection of time passed in order to work with totals accurately.
• Column D (Hours) should be formatted as a number with two decimal places.
Columns E and F should be formatted to Currency or Accounting, your choice.
Now that your formatting is set, let’s get into Excel time formula logic.
Column C contains a formula as shown in Illustration 1, =B2-A2. This calculates the number of hours the employee worked.
Column D, as mentioned, is formatted to a number, with two decimals. It contains the formula =C2*24. This time formula is so important because of how Excel is set up. Excel stores time values as decimal fractions of a 24 hour day. One hour equals 1/24th of a day. Thus, to get a whole number, you can use in calculation, you must multiply the value in column C by 24. Think of it like this. With a time like 7:30, C2 would equal 7.50, which is the number equivalent to seven hours and thirty minutes. See Illustration 2.
What if your employees have a lunch hour? You can set up your time formula to subtract 1. See Illustration 3.
Now that your formatting and time formulas are properly considered and applied, you can multiply Column D by Column E to reach the final paycheck total.
Working with time in and out of Excel, can include troublesome conversions. Using Excel with the tips listed can save a great deal of time and a few headaches as well. Next time you are adding or multiplying hours and minutes, consider using an Excel time formula.
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#### Monthly Archives
• Armanda says:
Thank you! was wracking my brains. This really helped and sorted out my problems.
• victoria says:
THANK YOUUUU! this helped me out a lot!
• gerranek says:
how simple… thanks a lot!! 🙂
• Liz says:
• Greg says:
Hey,
However, what about night work? It seems to fall down if somebody starts work at say 17:00 and finishes at 01:00…?
Or have I set it up wrong?
• Excel Tips and Tricks from Pryor.com says:
In order to have accurate time measurements that cross from midnight of one day into the next, the dates must be included in the calculation.
• Stan Giles says:
• Lisa Corey says:
I learn so much from these group. Thank you so much. It saves me so much time and headaches.
• Kanimaran says:
Thanks.. It’s very useful
• Terry says:
i am trying to use excel to work out my weekly hours, every time i hit 24 hours, my calculation reverts to 00:00. how do i fix this so it carries on calculating hours worked past 24?
• Excel Tips and Tricks from Pryor.com says:
We get a lot of Time calculation questions very much like this one.
Excel uses a specific format for time calculations that exceed 24 hours.
It’s called “Large Time” format.
That’s important because Excel defaults to viewing all time as a fraction of a date (unless otherwise instructed)
If all the cells (including the SUM) are formatted for “Large Time”, you can use notations that exceed 24 hours and still get a response in hours/minutes only.
• Sharon Leslie says: | 1,614 | 6,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-21 | latest | en | 0.927867 |
http://mail-archives.apache.org/mod_mbox/incubator-cassandra-user/201206.mbox/raw/%3CCAKkz8Q0H2zvuHTFFTRXLvGUAL4ayAcXhgUA=4vKSZpv6s3=+JA@mail.gmail.com%3E | 1,503,247,298,000,000,000 | text/plain | crawl-data/CC-MAIN-2017-34/segments/1502886106779.68/warc/CC-MAIN-20170820150632-20170820170632-00361.warc.gz | 265,456,682 | 8,289 | Understand simple mechani= cs first, decide how to act later.
=A0<= /p>
Without =96PR there=92s n= o difference from which host to run repair, it runs for the whole 100% rang= e, from start to end, the whole cluster, all nodes, at once.
That's not exactly true. A= repair without -pr will repair all the ranges of the node on which repair = is ran. So it will only repair the ranges that the node is a replica for. I= t will *not* repair the whole cluster (unless the replication factor is equ= al to the number of nodes in the cluster but that's a degenerate case).= And hence it does matter on which host repair is run (it always matter, wh= ether you use -pr or not).
In general you want to use repair without -pr in case w= here you want to repair a specific node. Typically, if a node was dead for = a reasonably long time, you may want to run a repair (without -pr) on that = specific node to have him catch up faster (faster that if you were only rel= ying on read-repair and hinted-handoff).
For repairing a whole cluster, as is the case for the w= eekly scheduled repairs in the initial question, you want to use -rp. You *= do not* want to use repair without -pr in that case. You do not because for= that task using -pr is more efficient (and to be clear, not using -pr won&= #39;t cause problems, but it does is less efficient).
--
Sylvain
=A0
=
=A0
With =96PR it runs only f= or a primary range of a node you are running a repair.=
Let say you have simple r= ing of 3 nodes with RF=3D2 and ranges (per node) N1=3DC-A, N2=3DA-B, N3=3DB= -C (node tokens are N1=3DA, N2=3DB, N3=3DC). No rack, no DC aware.
So running repair with = =96PR on node N2 will only repair a range A-B, for which node N2 is a prima= ry and N3 is a backup. N2 and N3 will synchronize A-B range one with other. For other ranges you need to run on other nodes.=
=A0<= /p>
Without =96PR running on = any node will repair all ranges, A-B, B-C, C-A. A node you run a repair wit= hout =96PR is just a repair coordinator, so no difference, which one will be next time.
=A0<= /p>
=A0<= /p>
Best regards / Pagarbiai Viktor Jevdokimov Senior Developer Phone: +370 5 212 3063, Fax +370 5 261 0453 J. Jasinskio 16C, LT-01112 Vilnius, Lithuania Follow us on Twitter: @adforminsider What is Adform: watch this short video
Disclaimer: The information contained in this message and attachments is in= tended solely for the attention and use of the named addressee and may be c= onfidential. If you are not the intended recipient, you are reminded that t= he information remains the property of the sender. You must not use, disclose, distribute, copy, print or rely= on this e-mail. If you have received this message in error, please contact= the sender immediately and irrevocably delete this message and any copies.
From: David Da= eschler [mailto:david.daeschler@gmail.com]
Sent: Tuesday, June 05, 2012 08:59
To: u= ser@cassandra.apache.org
Subject: nodetool repair -pr enough in this scenario?<= /span>
=A0
Hello,
=A0
Currently I have a 4 node cassandra cluster on CentO= S64. I have been running nodetool repair (no -pr option) on a weekly schedu= le like:
=A0
Host1: Tue, Host2: Wed, Host3: Thu, Host4: Fri
=A0
In this scenario, if I were to add the -pr option, w= ould this still be sufficient to prevent forgotten deletes and properly mai= ntain consistency?
=A0
Thank you,
- David
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http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-core/49305 | 1,591,136,280,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347426801.75/warc/CC-MAIN-20200602193431-20200602223431-00264.warc.gz | 17,190,818 | 2,914 | ```Issue #7336 has been updated by trans (Thomas Sawyer).
"you can always use parentheses if needed"
Unfortunately it is very unnatural for a unit system. Let me give an example to explain. In Stick, the obvious DSL is:
10.meters**2 => 100m
10.meters^2 => 10m^2
So, ** is power on the value and ^ is used to indicate power of the unit.
But the precedence of ^ is a problem.
10.meters^2 / 2 => 10m
Rather then the `5m^2` expected. While parenthesis can be used, it so unnatural to this common notation that it puts a rather ugly black mark on creating a nice SI units system for Ruby at all.
This particular problem could easily be solved by raising the precedence of ^ to that of **, which is why I suggested #6678 first --although that feels more like a band-aid. Controllable precedence feels like the right solution to me b/c others may ultimately have different needs.
I am not sure what to do if both are rejected.
----------------------------------------
Feature #7336: Flexiable OPerator Precedence
https://bugs.ruby-lang.org/issues/7336#change-32862
Author: trans (Thomas Sawyer)
Status: Rejected
Priority: Normal
Assignee:
Category: core
Target version: next minor
=begin
If Ruby classes could provide some means for redefining operator precedence, it would provide the flexibility useful to some DSL use-cases.
My particular application, for instance, is in an SI units system gem that could use `^` to mean power of the unit (e.g. 1.meter^3 would mean cubic meters). But to do that right the operator needs a higher precedence. I don't expect it to be something commonly used, obviously, but it certain use cases like mine it is practically essential.
I first suggested that (({#^})) be given a higher precedence and XOR get another operator in #6678. I was not surprised that it was rejected, but I figured it was the proper first step, before proposing this much broader feature request.
As for notation, I suppose the simplest means if to create class method that can move the precedence to a position relative to another, e.g.
class Unit
precedence :^, :**
Which is to say, move (({#^})) operator to a precedence above (({#**})).
=end
--
http://bugs.ruby-lang.org/
``` | 523 | 2,205 | {"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-2020-24 | latest | en | 0.913471 |
https://gradebuddy.com/doc/3520852/test-2/ | 1,723,640,495,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641113960.89/warc/CC-MAIN-20240814123926-20240814153926-00326.warc.gz | 212,986,496 | 11,323 | KU JOUR 420 - Test 2
Pages 4
Unformatted text preview:
Finance Test 2 Fill in the Blanks Chapter 4 1 Always assume that the timing of the cash flows is at the of the period unless otherwise specified 2 A is an infinite series of constant payments this does not mean same amount can be different amounts growing at same rate 3 Perpetuity is an that lasts forever 4 The perpetuity calculation brings the value back time period 5 The perpetuity calculation of cash flows 5 to infinity s CF5 r g The value of 6 7 this calculation sits in year are the series of equal cash flows that occur at regular intervals If the first payment occurs at the beginning of the period it is called 8 If the cash flow is at the end of the period it is considered an 9 Under special case annuities annuities due For PVA all cash flows have been one too many times Therefore multiplying by 1 r gives this back 10 Under special case annuities annuities due For FVA all cash flows have been shorted one period Therefore multiplying by 1 r gives them that extra period 11 If you know the payment amount for a loan the interest rate and the number of years and you want to know how much was borrowed do you compute a present value or future value 12 There is no taken into account in an APR 13 The is the actual rate paid or received after accounting for compounding that occurs during the year 14 APR is the rate 15 If you are investing your money you want to choose the EAR with the rate 16 If you are bowering money you want to choose an EAR with a rate 17 Constant growth and no growth are two types of 18 Interest rates are typically stated as Chapter 5 19 What are the two things that make a good evaluation or analysis tool 20 Explain the shortcomings of Payback Which of these shortcomings does Discounted Payback overcome 21 Which analysis tool will always lead you to the right decision 22 involve the question should we buy a company 23 involve the question should we sell a company 24 The NPV rule says to accept investments that have NPVs and reject ones with NPVs 25 If NPV is it means the outflows were greater than the inflows 26 A firm with unlimited capital should all positive NPV projects 27 The exception to firms with unlimited capital in reference to accepting all positive NPV projects is 28 The the r the more risky the project is 29 Given the same stream of cash flows the higher the r the the NPV 30 As the rate goes up NPV goes Inflows get and outflows investment stay the same 31 True of False If I make an NPV of 0 this means I lost money 32 True or False The IRR is the most important alternative to NPV 33 the project if the IRR is greater than the required return 34 True or False You choose the IRR it is not inherent 35 True or False You do choose the required return which is based on the risk level project 36 What two things in an IRR can lead to an incorrect decision 37 When the cash flows change signs more than once there is more than one 38 When you have an Unconventional cash flow use the to evaluate the 39 With projects you must reject one of the projects even if both 40 is a good communication tool and non financial managers tend to 41 If the NPV is greater than 0 the IRR is than the discount rate used in have positive NPVs understand return better the NPV calculation 42 The period is how long it takes to get the initial cost back 43 Decision Rule Accept if the payback period is than some preset limit 44 True or False The payback rule accounts for the time value of money 45 True or False The payback rule accounts for the risk of cash flows 46 What does Discounted Payback mean and what decisions does it drive 47 In a limited capital situation tells me my bang for my buck than 1 48 Profitability Index Decision Rule an investment if its PI is greater 49 A profitability index of 1 1 implies that for every dollar of investment we create an additional in value 50 Profitability Index accounts for TVM by Finance Test 2 Fill in the Blank Answers Chapter 4 1 End 2 Perpetuity 3 Annuity 4 One 5 Four 6 Annuities 7 Annuity Due 8 Ordinary Annuity 9 Discounted 10 Compounding 11 Present Value 12 Compounding 13 Effective Annual Rate EAR 14 Quoted 15 Higher 16 Lower 17 Perpetuities 18 APRs Chapter 5 19 Time value of money risk 20 Sometimes payback can ignore the time value of money and project risk We can fix this using discounted payback Another shortcoming of Payback is that it can be biased against long term projects such as research and development Also it requires an arbitrary cutoff point and ignores cash flows beyond this cut off date 21 NPV 22 Acquisitions 23 Divestitures 24 Positive Negative 25 Negative 26 Accept 27 Mutually Exclusive Projects 28 Higher 29 Lower 30 Down Smaller 31 False 32 True 33 Accept 34 False 35 True 36 Mutually Exclusive Projects or Unconventional Cash Flows 37 IRR 38 NPV 39 Mutually Exclusive 40 IRR 41 Higher 42 Payback 43 Less 44 False 45 False 46 Discounted payback overcomes TVM problems and can place a risk factor on cash flows It discounts back each of the cash flows and compares to initial project cost Will lead to a longer payback period Still is short term focused and can lead to poor decisions 47 Profitability Index 48 Accept 49 0 10 50 Discounting
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# KU JOUR 420 - Test 2
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6 pages | 1,276 | 5,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-33 | latest | en | 0.946948 |
https://learn.careers360.com/school/question-need-solution-for-rd-sharma-maths-class-12-chapter-continuity-exercise-8-point-2-question-14/ | 1,713,655,209,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817688.24/warc/CC-MAIN-20240420214757-20240421004757-00628.warc.gz | 316,451,629 | 37,369 | #### Need solution for RD Sharma maths class 12 chapter Continuity exercise 8.2 question 14
cos2 x is continuous
Given:
f(x) = cos x2
Explanation:
f(x) = cos x2
Let a be any real number then,
L.H.L
\begin{aligned} &\lim _{x \rightarrow a-} f(x)=\lim _{h \rightarrow 0} f(a-h) \\ &=\lim _{h \rightarrow 0} \cos (a-h)^{2} \\ &=\cos a^{2} \end{aligned}
R.H.L
\begin{aligned} &\lim _{x \rightarrow a^{-}} f(x)=\lim _{h \rightarrow 0} f(a+h) \\ &=\lim _{h \rightarrow 0} \cos (a+h)^{2} \\ &=\cos a^{2} \end{aligned}
Also, L.H.L=R.H.L=f(a)
f(x) is continuous everywhere | 211 | 576 | {"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": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-18 | latest | en | 0.445482 |
https://it.scribd.com/document/330629230/chap3 | 1,590,538,520,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391923.3/warc/CC-MAIN-20200526222359-20200527012359-00150.warc.gz | 419,257,285 | 94,041 | Sei sulla pagina 1di 15
CHAPTER
PLATE
TECTONICS
Key Concepts
Major Concept (I)
The most important source of information concerning the interior of Earth
is the passage of different kinds of earthquake or seismic waves through
the interior and along the surface.
Related or supporting concepts:
- Seismic waves are vibrations that propagate through the earth as a result of an
earthquake, volcanism, or an explosion.
- There are two fundamental types of seismic waves, P-waves and S-waves (see fig.
3.1).
- P-waves are compressional, or pressure, waves. Particle motion is in the direction of
propagation of the wave. Since it is possible to compress all states of matter, P-waves
will propagate through solids, liquids, and gases (sound is a compressional wave that
passes through the air).
- S-waves are shear waves. Particle motion is perpendicular to the direction of
propagation of the wave. You cannot transmit shearing energy through liquids or
solids, so S-waves will only pass through solids.
- We know that the outer core behaves like a liquid because S-waves will not pass
through it.
- Whenever a wave passes from one material into another its velocity will change and
its path will bend. The bending of the path is called refraction.
- Careful measurements of travel times of seismic waves from thousands of earthquakes
to hundreds of recording stations all over the world have allowed scientists to model
the internal structure of Earth as consisting of four basic layers, or regions (fig. 3.2).
Major Concept (II)
Despite the fact that we cannot sample the deep interior of Earth, we can
infer a great deal about its characteristics through remote means of
investigation.
Related or supporting concepts:
- We know that the gross structure of the planet must be a series of concentric shells
with fairly uniformly distributed mass in each shell. This view is supported by:
a. the fact that Earth spins on its axis with only a slight wobble,
b. the fairly uniform acceleration of gravity over the planet, and
c. the characteristics of seismic waves passing through Earth generated by
earthquakes.
- With a knowledge of the dimensions of the planet and its internal composition, we can
calculate pressure, temperature, and density with depth.
1
- The presence of a magnetic field around Earth and the observation of nickel-iron
meteorites both suggest that the center of the planet must be rich in these metals.
- There are four major layers, or shells, that make up the planet:
a. the inner core,
b. the outer core,
c. the mantle, and
d. the crust.
- The inner core has the following characteristics:
b. it is rich in iron and nickel,
c. it behaves like a solid, and
d. it is nearly five times as dense as common surface rocks like granite.
- The outer core:
a. is about 2258 km thick,
b. has the same composition as the inner core, and
c. behaves like a liquid.
- The mantle:
a. is about 2866 km thick,
b. is the largest region, accounting for about 70 percent of the volume of Earth,
c. is composed of Mg-Fe silicates, and
d. behaves like a solid with the exception of a thin upper layer that behaves like a
plastic.
- Details concerning temperature and density of these regions are given in table 3.1 in
the text.
- There are two different types of crust, continental and oceanic.
- Continental crust:
a. averages about 40 km in thickness,
b. is composed primarily of an igneous rock called granite which is rich in Al, K,
and silica, and
c. is relatively light.
- Oceanic crust:
a. is generally about 7 km thick,
b. is made of an igneous rock called basalt that is rich in Ca, Fe, and Mg, and
c. is heavier than continental crust.
- The boundary between the crust and the mantle is called the Moho.
Major Concept (III)
Another way to define distinct layers near the surface of Earth is by the
way in which the material responds to applied forces. This method allows
us to define two layers, a rigid upper one at the surface called the
lithosphere and a deformable second layer called the asthenosphere below
it (see fig. 3.3).
2
Related or supporting concepts:
- Continents and ocean basins are made of different kinds of rock. Continents are
composed of granite-like rocks rich in aluminum, potassium, and silica. Ocean basins
are composed of basalt-like rocks rich in calcium, iron, and magnesium and relatively
low in silica. Basalt has a higher density than granite.
- The speed of seismic waves increases abruptly at the boundary of the crust and mantle.
This boundary is called the Mohorovi_i_ discontinuity, or the Moho, for its
discoverer Andrija Mohorovi_i_.
- The crust and upper mantle are rigidly attached along the Moho.
- The crust and upper mantle form a strong, rigid, layer called the lithosphere. It is
about 100 km (62 mi) thick in oceanic regions and about 150 km (90 mi) thick in
continental regions.
- The base of the lithosphere occurs where the temperature reaches about 650C100C.
At higher temperatures mantle rock begins to lose some of its strength.
- The lithosphere rests on top of a region in the mantle that behaves like a plastic; it
deforms when subjected to stress. This layer is called the asthenosphere.
- The base of the asthenosphere is at a depth of about 350 km (217 mi).
- The asthenosphere is partially melted due to relatively high temperature and low
pressure at these depths. At greater depth the mantle again behaves like a solid despite
a steady increase in temperature because the greater pressure is able to prevent
melting.
- The lithosphere is able to move rigidly on top of the asthenosphere.
Major Concept (IV)
Speculation about the possible fit of continents goes back in time several
centuries. The earliest strong advocate of continental drift was Alfred
Wegener.
Related or supporting concepts:
- Francis Bacon (1561 - 1626) was an English scholar who first speculated about the
remarkable resemblance of the western coasts of South America and Africa. He did
not go so far as to say that they had once been joined.
- Other scientists and explorers would note the similarity in continental outlines,
including George Buffon (1707 - 88) and Alexander von Humboldt (1769 - 1859).
- Continued study revealed that there were other lines of evidence showing remarkable
similarities between continents. These included:
a. the distribution of fossils,
b. patterns of major geologic features such as faults and mountain ranges, and
c. the locations of distinctive rock formations.
- Edward Suess, an Austrian geologist, suggested that the Southern Hemisphere
continents had once been joined in a single landmass called Gondwanaland. This idea
was developed in a series of papers written between 1885 and 1909.
3
- Suess believed that the ocean basins were created when large regions of
Gondwanaland sank and the seas invaded the land. Thus, he did not propose a
fragmentation of Gondwanaland and subsequent drifting apart of the fragments.
- Continental drift was proposed at the beginning of the 20th century by two people
independently, Alfred Wegener and Frank Taylor. Frank Taylor did not continue to
pursue the idea, however, so we often attribute it to Wegener alone.
- Wegener was a meteorologist by training.
- Wegener believed that there was once a super continent called Pangaea that contained
all landmass (see fig. 3.4 in your text). Pangaea later broke apart into two masses:
Laurasia (North America and Eurasia) and Gondwanaland (Africa, South America,
India, Australia, and Antarctica).
- With continued time, both Laurasia and Gondwanaland fragmented to form the pattern
of continents we see today.
- The evidence used to support this idea included:
a. the geographical fit of the continental coastlines,
b. the way in which old mountain ranges and bodies of rock seemed to come
together and fit when the continents were rotated back together, and
c. evidence from fossils.
- Wegener noticed that fossils older than 150 million years seemed to be very similar
despite the fact that they were now located on different continents. This could easily
be explained if the continents were once a single landmass, allowing the organisms to
move freely from one place to another. Fossils younger than 150 million years from
different continents exhibited very different forms, suggesting that they had evolved
independently.
- Wegener believed that the forces responsible for the breakup of Pangaea and drifting
of the continents were related to the rotation of Earth and tidal pull.
- A famous geophysicist of the time, Harold Jeffreys, demonstrated that these forces
were not powerful enough to move continents. In addition, people could not accept
the idea that the continents could plow through the oceanic crust.
- Continental drift was never widely accepted during Wegener's life.
Following WWII, improved instrumentation was used to study the ocean
basins in detail. This eventually led to evidence that supported seafloor
Related or supporting concepts:
- Marine geologists first thought that the sea floor should be fairly smooth because of
vast amounts of sediment eroded from the continents that would cover any
irregularities.
- When the sea floor was finally extensively mapped following WWII, we discovered
that there are major structural features that rival, and even surpass, any found on the
continents. The sea floor has major mountain ranges, deep trenches, vast plateaus, and
4
enormous faults.
Harry Hess, in the early 1960s, first proposed the existence of large convection cells in
the mantle that act as conveyor belts for the overlying lithosphere.
Where the lithosphere is cracked the hot mantle material is able to escape and pour
onto the sea floor in active volcanism. This occurs along the axis of the oceanic
mountain ranges, creating new sea floor as it hardens.
To conserve the total surface area of Earth, a volume of material equal to that created
along the ridges must be destroyed. This occurs along oceanic trenches where the sea
floor is subducted into the mantle.
The movement of the sea floor away from the ridges is called seafloor spreading.
The ridges are also called spreading centers and trenches are called subduction zones
(take a look at fig. 3.6 to see a schematic of how these all work together).
The convection cells in the mantle are driven by heat. The heat is probably due to
radioactive decay and the natural cooling of the planet.
There are currently two proposed models of mantle convection. Some scientists
believe that there are two sets of convection cells, one confined to the upper mantle
above a depth of about 700 km (435 mi) and the other in the lower mantle. Other
scientists think that convection occurs throughout the entire mantle from the base of
the lithosphere to the core-mantle boundary.
The movement of convection cells has historically been thought to be the driving
mechanism of plate tectonics. There are now two other forces that are thought to play
a major role. They are both related to gravity and involve a pulling force at the edge
of the plate falling into the mantle and a pushing force at the edge of the plate located
along the spreading center as it "slides" down the inclined plane formed by the
elevated ridge.
Additional lines of evidence supporting the concept of a dynamic outer
layer of Earth came from the study of earthquakes, heat flow, sampling of
marine sediments, and measurements of rock magnetism.
Related or supporting concepts:
- The geographical distribution of earthquakes forms two unmistakable patterns:
a. in plan view, earthquakes form narrow belts of activity that follow plate
boundaries (see fig. 3.7), and
b. in the vicinity of trenches, earthquakes occur in descending bands called
Benioff zones.
- Heat flows out of the surface of Earth everywhere but the amount of heat can vary
from one location to another. In the ocean basins heat flow increases as you get closer
to mid-ocean ridges and decreases away from them (see fig. 3.9). This is due to
upward moving, hot mantle material beneath the ridges.
- In the late 1960s a special ship, the Glomar Challenger, was designed to drill into the
sea floor and recover sediments and rock. A sonar positioning system allows the ship
5
to remain stable above the hole during drilling (see fig. 3.10). Drilling revealed that:
a. there is no oceanic crust older than about 180 million years,
b. the thickness of sediment increases away from the ridges,
c. the age of surface sediment is recent everywhere, and
d. the age of the sediment at the bottom of the cores in direct contact with the
basalt increases with distance away from the ridge.
These ideas are shown in figure 3.11.
When lava cools, the iron rich minerals are able to align themselves with the direction
of Earth's magnetic field at the time of cooling. As the rock passes through a critical
temperature called the Curie temperature (roughly 580C), the magnetic signature of
the rock will be frozen into it.
Studies of continental lava flows indicate that Earth's magnetic field periodically
reverses direction. These magnetic, or polarity, reversals are known to have happened
about 170 times in the last 76 million years. The last reversal occurred about 710,000
years ago.
Magnetic measurements in the oceans show stripes on the sea floor parallel to ridges,
indicating magnetic reversals that were recorded in the crust when molten material
exited along ridges and solidified to form new crust (see fig. 3.13).
The significance of these stripes was first recognized by Vine and Matthews of
Cambridge University.
Seafloor magnetic data has been used to create maps of the age of the seafloor as
illustrated in figure 3.14.
All of the planet's sea floor was created in the last 200 million years or so.
The magnetic field frozen into rock as it cools also points to the location of the earth's
magnetic poles at the time of the rock's formation. Plots of the apparent location of
the north magnetic pole as a function of time, called polar wandering curves, can be
made with different suites of rocks of varying ages from different continents. The
polar wandering curves produced from suites of rocks in North America and Eurasia
have the same shape but do not overlap (see fig. 3.15). They converge to the present
location of the pole as time decreases to the present but they diverge as time moves
further into the past.
We believe that the magnetic poles have not wandered nearly as much as typical polar
wandering curves would indicate.
The apparent polar wander with time is actually caused by the drifting of the
continents and the progressive divergence of the North American and Eurasian polar
wandering curves is due to the opening of the Atlantic Ocean and the gradual drift of
the two continents apart.
Major Concept (VII) Plate tectonics is a broad theory of the dynamic nature of Earth that
includes the concepts of continental drift and seafloor spreading.
Related or supporting concepts:
6
- The lithosphere consists of seven major, and a number of smaller, rigid plates that
move with respect to one another and interact along their boundaries (table 3.2).
Figure 3.16 outlines and names the largest of the lithospheric plates.
- Plates may be composed of oceanic and/or continental crust along with some
underlying upper mantle material.
- There are three types of plate boundaries (fig. 3.17):
a. divergent boundaries where new crust is created and plates move apart (midocean ridges),
b. convergent boundaries where oceanic lithosphere is subducted into the mantle
or the edges of continents crumple as plates collide (usually marked by
trenches), and
c. transform, or conservative, boundaries where plates slide past one another
(transform faults).
- The San Andreas fault is a good example of a transform fault boundary. Shallow
earthquakes occur along transform faults as the plates slide.
- Transforms offset relatively short, linear segments of ridges.
Major Concept (VIII) Divergent plate boundaries are marked by rift zones where the lithosphere
breaks apart. These occur most frequently in oceanic areas but are also
seen in some continental regions.
Related or supporting concepts:
- A series of successive divergent boundaries were responsible for the breakup of
Pangaea.
- Rifting may be aided by stretching and thinning of the crust caused by the sinking of
older, denser lithosphere into the mantle.
- Upward moving hot mantle material will heat the overlying lithosphere. Heating will
decrease its density and cause it to arch upward, thus creating cracks that then may
widen to create rifting.
- Continental rifting is currently occurring in the Great Rift Valley of Africa.
- A more advanced stage of rifting is represented by the Red Sea, where seawater has
inundated the rift and a new ocean basin is forming.
- The newly created continental margin that is adjacent to the rift is called a trailing
margin.
- Trailing margins, also called passive margins, are not tectonically active. They
generally are not associated with earthquake or volcanic activity. They are often fairly
broad and low lying with thick accumulations of sediment. A good example is the
east coast of the United States.
- Most divergent boundaries are located along the axis of the mid-ocean ridge system.
7
Related or supporting concepts:
- The ocean ridge system is divided into linear segments that are offset by transform
faults.
- Ridge crest segments and transform faults form a continuous plate boundary.
- Differences in the age and temperature of the plate on either side of a transform
boundary can create significant and rapid changes in the elevation of the seafloor (fig.
3.22).
- These changes in seafloor elevation are propagated into each plate by seafloor
spreading where they are preserved as "fossil" transforms called fracture zones.
- The longest fracture zones can be up to 10,000 km (6200 mi) long.
- It is important to remember that fracture zones are not plate boundaries so there is no
relative motion along them.
- The San Andreas fault is one example of a transform fault that cuts into continental
lithosphere.
Plates move toward one another along convergent plate boundaries.
These may involve three types of collision; ocean-ocean, ocean-continent,
or continent-continent.
Related or supporting concepts:
- Along convergent boundaries where at least one of the plate edges is oceanic, a trench
will form and oceanic lithosphere will be subducted into the mantle.
- Where two continental margins converge there will be a buckling of the crust and
formation of a mountain range due to crustal thickening. Continental material is too
light to be subducted but some continental material from one plate may override the
continental material of the other and dramatically increase the thickness of the crust
along the margin.
- In continent-continent collision there may be remnants of marine sediments and crust
that have been trapped between the continents and uplifted. This is why marine fossils
can sometimes be found at the tops of continental mountains such as the Alps and the
Himalayas.
- Oceanic lithosphere is subducted along ocean trenches.
- The subducting plate will heat up as it descends.
- At a depth of about 100-150 km (60-100 mi) the temperature will be high enough to
drive water and other volatiles out of the subducted slab. As this water is added to the
surrounding mantle it will lower the melting temperature of the mantle rock and it
will partially melt. This will produce a basaltic magma that will rise to the surface.
- Subduction-related volcanism may produce a line of active volcanic islands along the
trench called an island arc or it may produce an active mountain range along the
margin of a continent. Examples include the Aleutian Islands and the Andes
8
Mountains.
Basaltic magma rising beneath a continental edge can partially melt the continental
granitic rock and produce a magma of andesitic composition.
Andesitic magmas can result in explosive volcanism such as Mount St. Helens (fig.
3.25).
Subducting oceanic plates are also associated with earthquake activity as they bend
and descend into the mantle. These dipping zones of earthquakes are called Benioff
zones.
The edge of a continent that is along a trench is called the leading margin, or
sometimes an active continental margin.
Major Concept (XI)
Older crustal fragments with properties and histories different from the
adjacent crustal material that join to form a continental mass are called
terranes.
Related or supporting concepts:
- Terranes may consist of pieces of island arcs systems, seamounts, seafloor plateaus, or
parts of other continental land masses.
- The core of North America is thought to have formed about 1.8 billion years ago from
four or five terranes.
- Alaska, the west coast, and parts of Canada are thought to have formed from terranes
that came from a southerly direction about 70 million years ago.
- The east coast appears to have originated near an island arc system that came from the
east.
- Roughly 25 percent of North America seems to consist of terranes. Take a look at
figure 3.26 for an idea of where these are and what type of material they are.
- India is considered to be a single large terrane that migrated from high southern
latitudes to impact with Eurasia to create the Himalayas.
Major Concept (XII) The rate of plate motion is relatively slow on human terms but very fast in
geological terms.
Related or supporting concepts:
- Spreading rates vary between about 1 and 20 cm (0.4-8 in)/year.
- Although this rate seems relatively slow, there is so much time available that the
plates can move great distances. In 200 million years the Pacific Ocean basin has
opened and the Atlantic has appeared in even less time.
- Spreading rate has a major effect on the cross-sectional shape or morphology of
9
- At slow spreading rates, ridges typically have steep, rugged profiles with deep central
rift valleys along their axis. A good example would be the Mid-Atlantic ridge.
- At fast spreading rates the flanks of the spreading center will slope more gently, they
will be smoother, and there will often not be a central rift valley at all. It is common
to call these divergent boundaries rises rather than ridges. An example would be the
East Pacific Rise.
- Actual spreading is not a continuous process but rather occurs sporadically.
- Iceland affords us with an opportunity to examine spreading along an oceanic ridge. It
is the only major island lying on top of a ridge.
Major Concept (XIII) Hot spots are deeply rooted sources of hot, upward moving mantle
material that appear to remain fixed in location for long periods of time
independent of the movement of the plates above them.
Related or supporting concepts:
- There are about forty known hot spots scattered around the globe. Some are located in
continental regions while others are in ocean basins. Examples include Yellowstone
National Park, Iceland, the Hawaiian Islands, and the Society Islands (Tahiti).
- The location of a specific hot spot seems to remain fixed during its life span.
- The life span of a hot spot seems to be on the order of 200 million years.
- As an oceanic plate moves over a hot spot it will create a series of active volcanoes,
with the youngest directly over the hot spot and age increasing with distance away
from it (see fig. 3.28). A classic example of this is the Hawaiian Island and seamount
Hawaii. At its current rate of growth it should become an island in 50,000 to 100,000
years.
- Tracing the direction of seamount chains created at hot spots gives the absolute
direction of plate motion. We know that the Pacific plate has moved in a northwest
direction for the past 40 million years. Prior to that, it moved in a more northerly
direction. Evidence for this comes from the Emperor Seamount Chain that joins the
Hawaiian chain at an angle just west of Midway Island.
- It is very likely that in the 4.5 billion year history of Earth the continents have
combined and fragmented several times. One suggestion is that there is a roughly 500
million year cycle that brings the continents back together into a giant landmass and
fragments them again.
- A single landmass would collect a great deal of heat from below. This would cause
the land to rise and sea level to fall. As this mass fragmented, the resultant small
continents would cool rapidly, subside, and sea level would rise to cover low-lying
continental regions.
Major Concept (XIV) The continents were once joined in a single massive landmass called
10
Pangaea. Roughly 200 million years ago Pangaea began to break apart
to eventually produce the continents we see today.
Related or supporting concepts:
- The progressive change in relative position of the continents is illustrated in
figure 3.29.
- In the early Triassic, when the first mammals and dinosaurs appeared, the continents
were joined in a single landmass called Pangaea. There was a single large ocean
called Panthalassa and a small sea called the Tethys.
- Roughly 200 million years ago Pangaea began to fragment into two large pieces,
Laurasia and Gondwanaland, and the South Indian Ocean began to form between
them.
- Shortly after the initial breakup India began to drift northward toward its eventual
collision with Asia to form the Himalayan mountains.
- By about 150 million years ago Laurasia and Gondwana were separated by a narrow
sea that would grow to become the Atlantic Ocean.
- Africa and South America began to split about 135 million years ago and fully
separated about 65 million years ago.
- During the last 16 million years North America completely separated from Eurasia,
Greenland moved away from Europe, North and South America united, and Australia
separated from Antarctica.
- The Alps were formed by the collision of Italy and Europe.
- Arabia moved away from Africa about 20 million years ago to form the Gulf of Aden
and ongoing rifting continues to open the Red Sea.
Major Concept (XV) The major framework of plate tectonics has been developed but there are
still countless new areas of exciting research to explore. Current research
projects continue to teach us more about the planet from physical,
chemical, and biological perspectives.
Related or supporting concepts:
- The Ocean Drilling Program (ODP) began in 1983.
- Drilling operations in the ocean basins currently are conducted by the Joint
Oceanographic Institutions for Deep Earth Sampling (JOIDES), a consortium of
research institutions that was formed in 1985. This project uses the ship JOIDES
Resolution as a drilling platform (refer back to fig. 1.18b).
- The JOIDES Resolution has visited all of the world's oceans.
- The active tectonics of the mid-ocean ridges continue to be studied using both manned
and unmanned submersibles.
- One of the most interesting discoveries of expeditions to mid-ocean ridges was the
presence of large communities of organisms such as tube worms and clams
11
surrounding areas where hydrothermal vents occur.
Some vents discharge clear water as cool at 30C. "White smokers" discharge milky
water at temperatures of 200 to 330C. "Black smokers" release jets of sulfurblackened water at temperatures of 300 to 400C.
Hydrothermal activity is thought to continue for years to decades at individual sites
but rapid changes in temperature and discharge rate have been observed over periods
of days to seconds.
Long-wavelength thermal radiation that cannot be detected by the human eye is
emitted by vents. This is called "vent glow."
Vent waters can contain high concentrations of metals such as lead, cobalt, zinc, and
silver that precipitate out when they encounter the cold bottom water of the sea floor.
Sulfide deposits are sometimes precipitated at vents.
A massive hydrothermal discharge, nicknamed a "megaplume," was discovered over
the Juan de Fuca Ridge in 1986. This megaplume consisted of about 100 million
cubic meters of warm water that had risen to about 1000 meters above the ridge.
The ocean basins are so vast that there are large regions that remain essentially
unexplored.
As our ability to monitor seafloor phenomena increases with more sophisticated
instruments, we will undoubtedly continue to discover new things.
Key Terms and Related Major Concepts
At the back of the chapter in your book there are a number of key terms. You should be able to
find the following terms referenced in the major concept indicated in parentheses.
seismic wave(I)
P-waves(I)
S-waves(I)
inner core(II)
outer core(II)
mantle(II)
crust(II)
density(II)
Moho(II)
granite(II)
basalt(II)
lithosphere(III)
asthenosphere(III)
Gondwanaland(IV, XIII)
continental drift(IV)
Pangaea(IV, XIII)
Laurasia(IV, XIII)
convection cell(V)
fracture zone(IX)
transform fault(VII)
subduction zone(V)
polarity reversal(VI)
plate tectonics(VII)
transform fault(VII)
trailing margin(VIII)
terrane(XI)
hot spot(XIII)
hydrothermal vent(XV)
Test Your Understanding With The Following Questions:
FILL IN THE BLANK
1. The boundary between the crust and the mantle is called the ____________.
12
2. The outer core behaves like a __________________.
3. The _________________________________ consists of a number of rigid plates.
4. Alfred Wegener was trained as a __________________________________.
5. The age of marine sediment gets _______________ as you move away from ridge crests.
6. Segments of ridge crest are offset by ______________________ _________________.
7. Edges of continents closest to spreading centers are called ________________________
margins.
8. Crustal fragments with properties and a history distinct from adjoining fragments are called
______________________.
9. Some people have suggested that the continents fragment and come back together in a
cyclical process that takes about _________ million years.
10. Large communities of organisms have been discovered on ridge crests in the vicinity of
__________________________________ _______________.
TRUE - FALSE
1. P-waves are able to move through all states of matter: solid, liquid, and gases.
2. Continental rocks are primarily dense, basalt-type rocks.
3. The plates are driven by forces caused by Earth's rotation.
4. The concept of continental drift was proposed before seafloor spreading.
5. One of the most powerful indications of seafloor spreading is the presence of magnetic
stripes.
6. The magnetic field of Earth reverses about every 10 million years.
7. The oldest oceanic crust is about 180 million years old.
8. When two continents collide there is no subduction of continental lithosphere.
9. The danger of a disaster due to an earthquake has been lessened dramatically by our ability to
predict their occurrence.
10. The same volume of material created at ridges is destroyed at trenches.
MULTIPLE CHOICE
1. The largest region of Earth in terms of volume is the:
a. crust
b. mantle
c. outer core
d. inner core
e. lithosphere
2. The first person to suggest that southern continents were once joined in a single landmass
was:
13
a. Francis Bacon
b. Alfred Wegener
c. Frank Taylor
d. Edward Suess
e. Harold Jeffreys
3. About 250 million years ago there was a single landmass we call:
a. Laurasia
b. Panthalassa
c. Pangaea
d. Gondwanaland
e. Tethys
4. North America was formerly a part of:
a. Laurasia
b. Gondwanaland
c. Pangaea
d. a and c
e. b and c
5. Among the evidence Wegener used to support his ideas was:
a. similar old mountain ranges on different continents
b. similar fossils older than 150 million years on different continents
c. the fit of continental coastlines
d. all of the above
e. b and c
6. Possible plate driving mechanisms include
a. convection cells
b. tidal forces
c. rotational forces
d. gravitational forces acting at trenches and ridges
e. a and d
7. Spreading rates for plates are on the order of:
a. 1 - 20 cm/yr
b. 1 - 10 mm/yr
c. 1 - 10 km/yr
d. 20 - 30 cm/yr
e. 40 - 50 mm/yr
8. Fast spreading centers generally have:
a. no deep central rift valley
b. more gently sloping flanks
c. more seismicity
d. all of the above
e. a and b
9. The life span of a hot spot is thought to be about ______________ years.
a. 20 million
14
b. 200 million
c. 1 billion
d. 1 million
e. 100,000
10. A good example of a hot spot is:
a. Hawaii
b. Iceland
c. Yellowstone
d. Tahiti
e. all of the above
FILL IN THE BLANK
1. Moho
2. liquid
3. lithosphere
4. meteorologist
5. older
6. transform faults
7. trailing
8. terranes
9. 500
10. hydrothermal vents
TRUE - FALSE
1.T 2.F 3.F 4.T 5.T 6.F 7.T 8.T 9.F 10.T
MULTIPLE CHOICE
1.b 2.d 3.c 4.d 5.d 6.e 7.a 8.e 9.b 10.e
15 | 7,374 | 32,532 | {"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-2020-24 | latest | en | 0.934202 |
https://www.semanticscholar.org/paper/Fixed-points-of-inhomogeneous-smoothing-transforms-Alsmeyer-Meiners/70e6b59f7b0904a1dcaa8bf6a94748a9df1a3a8f | 1,642,841,640,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00683.warc.gz | 1,013,800,519 | 69,031 | # Fixed points of inhomogeneous smoothing transforms
@article{Alsmeyer2012FixedPO,
title={Fixed points of inhomogeneous smoothing transforms},
author={Gerold Alsmeyer and Matthias Meiners},
journal={Journal of Difference Equations and Applications},
year={2012},
volume={18},
pages={1287 - 1304}
}
• Published 26 July 2010
• Mathematics
• Journal of Difference Equations and Applications
We consider the inhomogeneous version of the fixed-point equation of the smoothing transformation, that is, the equation , where means equality in distribution, is a given sequence of non-negative random variables and is a sequence of i.i.d. copies of the non-negative random variable X independent of . In this situation, X (or, more precisely, the distribution of X) is said to be a fixed point of the (inhomogeneous) smoothing transform. In the present paper, we give a necessary and sufficient…
36 Citations
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• 2014
Given a sequence $(T_1, T_2, ...)$ of random $d \times d$ matrices with nonnegative entries, suppose there is a random vector $X$ with nonnegative entries, such that $\sum_{i \ge 1} T_i X_i$ has
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Given a sequence (C, T) = (C, T 1, T 2, …) of real-valued random variables, the associated so-called smoothing transform $$\mathcal{S}$$ maps a distribution F from a subset Γ of distributions on
Precise tail asymptotics of fixed points of the smoothing transform with general weights
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• 2015
We consider solutions of the stochastic equation $R=_d\sum_{i=1}^NA_iR_i+B$, where $N>1$ is a fixed constant, $A_i$ are independent, identically distributed random variables and $R_i$ are independent
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We consider complex Mandelbrot multiplicative cascades on a random weigh\-ted tree. Under suitable assumptions, this yields a dynamics $\T$ on laws invariant by random weighted means (the so called
## References
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The functional equation of the smoothing transform
• Mathematics
• 2012
Given a sequence T = (Ti)i� 1 of non-negative random variables, a function f on the positive halfline can be transformed to E Q i� 1 f(tTi). We study the fixed points of this transform within the
Fixed points of a generalized smoothing transformation and applications to the branching random walk
Let {A i : i ≥ 1} be a sequence of non-negative random variables and let M be the class of all probability measures on [0,∞]. Define a transformation T on M by letting Tμ be the distribution of ∑ i=1
Fixed points of the smoothing transformation
• Mathematics
• 1983
SummaryLet W1,..., WN be N nonnegative random variables and let $$\mathfrak{M}$$ be the class of all probability measures on [0, ∞). Define a transformation T on $$\mathfrak{M}$$ by letting Tμ be
Implicit Renewal Theory and Power Tails on Trees
• Mathematics, Computer Science
• 2012
Goldie's implicit renewal theorem is extended to enable the analysis of recursions on weighted branching trees by deriving the power-tail asymptotics of the distributions of the solutions R to and similar recursions.
Stability of perpetuities
• Mathematics
• 2000
For a series of randomly discounted terms we give an integral criterion to distinguish between almost-sure absolute convergence and divergence in probability to oo, these being the only possible
SENETA-HEYDE NORMING IN THE BRANCHING RANDOM WALK
• Mathematics
• 1997
!. !. !. malization of the general Crump! Mode! Jagers branching process is obtained; in this case the convergence holds almost surely. The results rely heavily on a detailed study of the functional
On the convergence of supercritical general (C-M-J) branching processes
SummaryConvergence in probability of Malthus normed supercritical general branching processes (i.e. Crump-Mode-Jagers branching processes) counted with a general characteristic are established,
Lindley-type equations in the branching random walk
An analogue of the Lindley equation for random walk is studied in the context of the branching random walk, taking up the studies of Karpelevich, Kelbert and Suhov [(1993a) In: Boccara, N., Goles,
Minimal position and critical martingale convergence in branching random walks, and directed polymers on disordered trees
• Mathematics
• 2007
We establish a second-order almost sure limit theorem for the minimal position in a one-dimensional super-critical branching random walk, and also prove a martingale convergence theorem which answers
Minimal Position and Critical Martingale Convergence in Branching Random Walks ( On a paper
• 2008
For a branching random walk on the line, an unusual almost sure limit theorem for its minimal position is proved along with a Heyde-Seneta-type convergence result for the critical martingale. The | 1,593 | 6,572 | {"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": 2, "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} | 2.765625 | 3 | CC-MAIN-2022-05 | latest | en | 0.791509 |
https://www.convertunits.com/from/barrel+%5BUK%5D/to/kilolitro | 1,685,326,596,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644574.15/warc/CC-MAIN-20230529010218-20230529040218-00713.warc.gz | 810,489,235 | 12,831 | ## Convert barrel [UK] to kilolitro
barrel [UK] kilolitro
Did you mean to convert barrel [US, liquid] barrel [US, beer] barrel [US, dry] barrel [US, petroleum] barrel [UK] barrel [UK, wine] to kilolitro
How many barrel [UK] in 1 kilolitro? The answer is 6.1102568971969. We assume you are converting between barrel [UK] and kilolitro. You can view more details on each measurement unit: barrel [UK] or kilolitro The SI derived unit for volume is the cubic meter. 1 cubic meter is equal to 6.1102568971969 barrel [UK], or 1 kilolitro. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between barrels and kilolitros. Type in your own numbers in the form to convert the units!
## Quick conversion chart of barrel [UK] to kilolitro
1 barrel [UK] to kilolitro = 0.16366 kilolitro
5 barrel [UK] to kilolitro = 0.8183 kilolitro
10 barrel [UK] to kilolitro = 1.63659 kilolitro
20 barrel [UK] to kilolitro = 3.27318 kilolitro
30 barrel [UK] to kilolitro = 4.90978 kilolitro
40 barrel [UK] to kilolitro = 6.54637 kilolitro
50 barrel [UK] to kilolitro = 8.18296 kilolitro
75 barrel [UK] to kilolitro = 12.27444 kilolitro
100 barrel [UK] to kilolitro = 16.36592 kilolitro
## Want other units?
You can do the reverse unit conversion from kilolitro to barrel [UK], or enter any two units below:
## Enter two units to convert
From: To:
## Definition: Kilolitro
El kilolitro es una unidad de volumen equivalente a mil litros, representado por el símbolo kl. Es el tercer múltiplo del litro y también equivale a 1 metro cúbico.
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 592 | 2,070 | {"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-2023-23 | latest | en | 0.622976 |
https://metanumbers.com/255070900000000 | 1,623,687,706,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00507.warc.gz | 363,997,958 | 8,271 | ## 255070900000000
255,070,900,000,000 (two hundred fifty-five trillion seventy billion nine hundred million) is an even fifteen-digits composite number following 255070899999999 and preceding 255070900000001. In scientific notation, it is written as 2.550709 × 1014. The sum of its digits is 28. It has a total of 19 prime factors and 648 positive divisors. There are 86,624,640,000,000 positive integers (up to 255070900000000) that are relatively prime to 255070900000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 15
• Sum of Digits 28
• Digital Root 1
## Name
Short name 255 trillion 70 billion 900 million two hundred fifty-five trillion seventy billion nine hundred million
## Notation
Scientific notation 2.550709 × 1014 255.0709 × 1012
## Prime Factorization of 255070900000000
Prime Factorization 28 × 58 × 7 × 109 × 3343
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 19 Total number of prime factors rad(n) 25507090 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 255,070,900,000,000 is 28 × 58 × 7 × 109 × 3343. Since it has a total of 19 prime factors, 255,070,900,000,000 is a composite number.
## Divisors of 255070900000000
648 divisors
Even divisors 576 72 36 36
Total Divisors Sum of Divisors Aliquot Sum τ(n) 648 Total number of the positive divisors of n σ(n) 7.34243e+14 Sum of all the positive divisors of n s(n) 4.79172e+14 Sum of the proper positive divisors of n A(n) 1.13309e+12 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.59709e+07 Returns the nth root of the product of n divisors H(n) 225.111 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 255,070,900,000,000 can be divided by 648 positive divisors (out of which 576 are even, and 72 are odd). The sum of these divisors (counting 255,070,900,000,000) is 734,242,749,067,520, the average is 11,330,906,621,41.,234.
## Other Arithmetic Functions (n = 255070900000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 86624640000000 Total number of positive integers not greater than n that are coprime to n λ(n) 150390000000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7934600886198 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 86,624,640,000,000 positive integers (less than 255,070,900,000,000) that are coprime with 255,070,900,000,000. And there are approximately 7,934,600,886,198 prime numbers less than or equal to 255,070,900,000,000.
## Divisibility of 255070900000000
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 0 0 1
The number 255,070,900,000,000 is divisible by 2, 4, 5, 7 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (255070900000000)
Base System Value
2 Binary 111001111111110001010010011000110001010100000000
3 Ternary 1020110010112120120202121220101
4 Quaternary 321333301102120301110000
5 Quinary 231413040200400000000
6 Senary 2302253542004252144
8 Octal 7177612230612400
10 Decimal 255070900000000
12 Duodecimal 247365483a1054
16 Hexadecimal e7fc52631500
20 Vigesimal 14i3e2g50000
36 Base36 2iexyc4hds
## Basic calculations (n = 255070900000000)
### Multiplication
n×i
n×2 510141800000000 765212700000000 1020283600000000 1275354500000000
### Division
ni
n⁄2 1.27535e+14 8.50236e+13 6.37677e+13 5.10142e+13
### Exponentiation
ni
n2 65061164026810000000000000000 16595209663366050829000000000000000000000000 4232955064523475614398776100000000000000000000000000000000 1079703657967560996092748778725490000000000000000000000000000000000000000
### Nth Root
i√n
2√n 1.59709e+07 63419.1 3996.37 760.908
## 255070900000000 as geometric shapes
### Circle
Radius = n
Diameter 5.10142e+14 1.60266e+15 2.04396e+29
### Sphere
Radius = n
Volume 6.95139e+43 8.17583e+29 1.60266e+15
### Square
Length = n
Perimeter 1.02028e+15 6.50612e+28 3.60725e+14
### Cube
Length = n
Surface area 3.90367e+29 1.65952e+43 4.41796e+14
### Equilateral Triangle
Length = n
Perimeter 7.65213e+14 2.81723e+28 2.20898e+14
### Triangular Pyramid
Length = n
Surface area 1.12689e+29 1.95576e+42 2.08265e+14
## Cryptographic Hash Functions
md5 eb6cbb606d0bbc4964b60feef58b7989 00921207e2202f8ad829182969e3efa77157559d c509a6c24789f02c20e3c9c5651fe8fc0982aa666d002964a74791293e2c7e48 59aadc3e19ce1568cf8f2fe7b4664c21b763c92ff6136ef5b27c9b2bc1c14dbcd20c14a68584cea43a011d8aec77f2e4048428152f3c9197fea6b01de34e7518 c04ba5b1647871ebf9f1193d04bb69a2198f4805 | 1,792 | 4,964 | {"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} | 3.328125 | 3 | CC-MAIN-2021-25 | latest | en | 0.764563 |
http://www.onlinemathlearning.com/word-problems-time-interval.html | 1,490,653,713,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189534.68/warc/CC-MAIN-20170322212949-00364-ip-10-233-31-227.ec2.internal.warc.gz | 639,580,304 | 10,160 | # Time Interval Word Problems
Videos and solutions to help Grade 3 students learn how to solve word problems involving time intervals within 1 hour by counting backward and forward using the number line and clock.
Common Core Standards: 3.NBT.2, 3.MD.1
Related Topics:
Lesson Plans and Worksheets for Grade 3
Lesson Plans and Worksheets for all Grades
More Lessons for Grade 3
Common Core For Grade 3
New York State Common Core Math Module 2, Grade 3, Lesson 4
Application Problem
Patrick and Lilly start their chores at 5:00 p.m. The clock and the number line show the times that Patrick and Lilly finish their chores.
Who finishes first? Explain how you know. Solve the problem without drawing a number line. You might want to visualize or use your clock template, draw a tape diagram, use words, number sentences, etc.
Concept Development
Look back at your work on Application Problem.
We know that Lilly finished after Patrick.
Let’s use a number line to figure out how many more minutes than Patrick Lilly took to finish.
Label the first tick mark 0 and the last tick mark 60. Label the hours and 5-minute intervals. T: Plot the times 5:31 p.m. and 5:43 p.m.
Find the difference between Patrick and Lilly’s times.
How many more minutes than Patrick did it take Lilly to finish her chores?
12 minutes more.
What strategy did you use to solve this problem?
Problem Set
1. Cole starts reading at 6:23 p.m. He stops at 6:49 p.m. How many minutes does Cole read?
Cole reads for __________ minutes.
2. Natalie finishes piano practice at 2:45 p.m. after practicing for 37 minutes. What time does Natalie’s practice start?
Natalie’s practice starts at __________ p.m.
6. Dion walks to school. The clocks below show when he leaves his house and when he arrives at school. How many minutes does it take Dion to walk to school?
Homework Questions 1 to 6
Record your homework start time on the clock in Problem 6.
Directions: Use a number line to answer Problems 1 through 4.
1. Joy's mom begins walking at 4:12 p.m. She stops at 4:43 p.m. How many minutes does she walk? Joy's mom walks for __________ minutes.
2. Cassie finishes softball practice at 3:52 p.m. after practicing for 30 minutes. What time does Cassie's practice start?
Cassie’s practice starts at ____________.
3. Jordie builds a model from 9:14 a.m. to 9:47 a.m. How many minutes does Jordie spend building his model?
Jordie builds for _____________ minutes.
4. Cara finishes reading at 2:57 p.m. She reads for a total of 46 minutes. What time did Cara start reading? Cara starts reading at ____________ p.m.
5. Jenna and her mom take the bus to the mall. The clocks below show when they leave their house and when they arrive at the mall. How many minutes does it take them to get to the mall?
Time when they leave home:
Time when they arrive at the mall:
6. Record your homework start time:
Record the time you finish Problems 1–5:
How many minutes did you work on Problems 1–5?
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.
[?] Subscribe To This Site | 826 | 3,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2017-13 | longest | en | 0.915131 |
https://rdrr.io/cran/Compack/src/R/GIC.R | 1,643,400,007,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00162.warc.gz | 521,739,346 | 14,292 | # R/GIC.R In Compack: Regression with Compositional Covariates
#### Documented in GIC.compCLGIC.FuncompCGL
#' @title
#' Compute information crieteria for the \code{FuncompCGL} model.
#'
#'
#' @description
#' Tune the grid values of the penalty parameter code{lam} and the degrees of freedom of
#' the basis function \code{k} in the \code{FuncompCGL} model by GIC, BIC, or AIC. This
#' function calculates the GIC, BIC, or AIC curve and returns the optimal values of
#' \code{lam} and \code{k}.
#'
#' @usage
#' GIC.FuncompCGL(y, X, Zc = NULL, lam = NULL, nlam = 100, k = 4:10, ref = NULL,
#' intercept = TRUE, W = rep(1,times = p - length(ref)),
#' type = c("GIC", "BIC", "AIC"),
#' mu_ratio = 1.01, outer_maxiter = 1e+6, ...)
#'
#' @param k an integer vector specifying the degrees of freedom of the basis function.
#'
#' @param type a character string specifying which crieterion to use. The choices include
#' \code{"GIC"} (default), \code{"BIC"}, and \code{"AIC"}.
#'
#' @param outer_maxiter maximum number of loops allowed for the augmented Lanrange method.
#'
#' @param \dots other arguments that could be passed to FuncompCL.
#'
#'
#' @inheritParams FuncompCGL
#'
#' @details
#' The \code{FuncompCGL} model estimation is conducted through minimizing the
#' linearly constrained group lasso criterion
#' \deqn{
#' \frac{1}{2n}\|y - 1_n\beta_0 - Z_c\beta_c - Z\beta\|_2^2 + \lambda \sum_{j=1}^{p} \|\beta_j\|_2,
#' s.t. \sum_{j=1}^{p} \beta_j = 0_k.}
#'
#' The tuning parameters can be selected by the generalized information crieterion (GIC),
#' \deqn{
#' GIC(\lambda,k) = \log{(\hat{\sigma}^2(\lambda,k))} +
#' (s(\lambda, k) - 1)k \log{(max(p*k+p_c+1, n))} \log{(\log{n})}/n
#' ,}
#' where
#' \eqn{\hat{\sigma}^2(\lambda,k) = \|y - 1_n\hat{\beta_0}(\lambda, k) -
#' Z_c\hat{\beta_c}(\lambda, k) - Z\hat{\beta}(\lambda, k) \|_{2}^{2}/n} with \eqn{\hat{\beta_0}(\lambda, k)},
#' \eqn{\hat{\beta_c}(\lambda, k)} and \eqn{\hat{\beta}(\lambda, k)} being the regularized estimators
#' of the regression coefficients, and \eqn{s(\lambda, k)} is the number of nonzero coefficient groups in
#' \eqn{\hat{\beta}(\lambda, k)}.
#'
#' @references
#' Sun, Z., Xu, W., Cong, X., Li G. and Chen K. (2020) \emph{Log-contrast regression with
#' functional compositional predictors: linking preterm infant's gut microbiome trajectories
#' to neurobehavioral outcome}, \href{https://arxiv.org/abs/1808.02403}{https://arxiv.org/abs/1808.02403}
#' \emph{Annals of Applied Statistics}.
#'
#' Fan, Y., and Tang, C. Y. (2013) \emph{Tuning parameter selection in high
#' dimensional penalized likelihood},
#' \emph{Journal of the Royal Statistical Society. Series B} \strong{75} 531-552.
#'
#' @return An object of S3 class \code{"GIC.FuncompCGL"} is returned, which is
#' a list containing:
#' \item{FuncompCGL.fit}{a list of length \code{length(k)},
#' objects of different degrees
#' of freedom of the basis function.}
#'
#' \item{lam}{the sequence of the penalty parameter \code{lam}.}
#'
#' \item{GIC}{a \code{k} by \code{length(lam)} matirx of GIC values.}
#'
#' \item{lam.min}{the optimal values of the degrees of freedom \code{k}
#' and the penalty parameter \code{lam}.}
#'
#' \item{MSE}{a \code{k} by \code{length(lam)} matirx of mean squared errors.}
#'
#' @seealso
#' object.
#'
#' @examples
#' \donttest{
#' df_beta = 5
#' p = 30
#' beta_C_true = matrix(0, nrow = p, ncol = df_beta)
#' beta_C_true[1, ] <- c(-0.5, -0.5, -0.5 , -1, -1)
#' beta_C_true[2, ] <- c(0.8, 0.8, 0.7, 0.6, 0.6)
#' beta_C_true[3, ] <- c(-0.8, -0.8 , 0.4 , 1 , 1)
#' beta_C_true[4, ] <- c(0.5, 0.5, -0.6 ,-0.6, -0.6)
#' n_train = 50
#' n_test = 30
#' k_list <- c(4,5)
#' Data <- Fcomp_Model(n = n_train, p = p, m = 0, intercept = TRUE,
#' SNR = 4, sigma = 3, rho_X = 0.2, rho_T = 0.5,
#' df_beta = df_beta, n_T = 20, obs_spar = 1, theta.add = FALSE,
#' beta_C = as.vector(t(beta_C_true)))
#' arg_list <- as.list(Data$call)[-1] #' arg_list$n <- n_test
#' Test <- do.call(Fcomp_Model, arg_list)
#'
#' ## GIC_cgl: Constrained group lasso
#' GIC_cgl <- GIC.FuncompCGL(y = Data$data$y, X = Data$data$Comp,
#' Zc = Data$data$Zc, intercept = Data$data$intercept,
#' k = k_list)
#' coef(GIC_cgl)
#' plot(GIC_cgl)
#' y_hat <- predict(GIC_cgl, Znew = Test$data$Comp, Zcnew = Test$data$Zc)
#' plot(Test$data$y, y_hat, xlab = "Observed response", ylab = "Predicted response")
#'
#' ## GIC_naive: ignoring the zero-sum constraints
#' ## set mu_raio = 0 to identifying without linear constraints,
#' ## no outer_loop for Lagrange augmented multiplier
#' GIC_naive <- GIC.FuncompCGL(y = Data$data$y, X = Data$data$Comp,
#' Zc = Data$data$Zc, intercept = Data$data$intercept,
#' k = k_list, mu_ratio = 0)
#' coef(GIC_naive)
#' plot(GIC_naive)
#' y_hat <- predict(GIC_naive, Znew = Test$data$Comp, Zcnew = Test$data$Zc)
#' plot(Test$data$y, y_hat, xlab = "Observed response", ylab = "Predicted response")
#'
#' ## GIC_base: random select a component as reference
#' ## mu_ratio is set to 0 automatically once ref is set to a integer
#' ref <- sample(1:p, 1)
#' GIC_base <- GIC.FuncompCGL(y = Data$data$y, X = Data$data$Comp,
#' Zc = Data$data$Zc, intercept = Data$data$intercept,
#' k = k_list, ref = ref)
#' coef(GIC_base)
#' plot(GIC_base)
#' y_hat <- predict(GIC_base, Znew = Test$data$Comp, Zcnew = Test$data$Zc)
#' plot(Test$data$y, y_hat, xlab = "Observed response", ylab = "Predicted response")
#' }
#'
#' @export
GIC.FuncompCGL <- function(y, X, Zc = NULL, lam = NULL, nlam = 100, k = 4:10, ref = NULL,
intercept = TRUE, W = rep(1,times = p - length(ref)),
type = c("GIC", "BIC", "AIC"),
mu_ratio = 1.01, outer_maxiter = 1e+6, ...) {
y <- drop(y)
n <- length(y)
type <- match.arg(type)
object <- as.list(seq(length(k)))
names(object) <- k
if(!is.null(lam) || length(k) == 1) {
## Case I
if(dim(X)[1] == n ) p = ncol(X) / k else p <- ncol(X) - 2
for(i in 1:length(k)){
object[[i]] <- FuncompCGL(y = y, X = X, Zc = Zc, lam = lam, nlam = nlam, ref = ref,
k = k[i], outer_maxiter = outer_maxiter, W = W,
intercept = intercept, mu_ratio = mu_ratio, ...)
}
} else {
## Case II
## find commom lambda first
p <- ncol(X) - 2
this.call <- as.list(seq(length(k)))
for(i in 1:length(k)){
## Caculate integral Z and W matrix (if W is a functoin)
object[[i]] <- FuncompCGL(y = y, X = X, Zc = Zc, nlam = 1, ref = ref,
k = k[i], outer_maxiter = 0, W = W,
intercept = intercept, mu_ratio = mu_ratio, ...)
this.call[[i]] <- object[[i]]$call } sseq <- object[[1]]$sseq
lam0 <- max(sapply(object, "[[", "lam")) # Shared lam0 for different k
## Solution path for each df k
for(i in 1:length(k)) {
object[[i]] <- FuncompCGL(y = y, X = object[[i]]$Z, Zc = Zc, k = k[i], W = object[[i]]$W, ref = ref,
lam = lam0, nlam = nlam,
outer_maxiter = outer_maxiter,
intercept = intercept, mu_ratio = mu_ratio, ...)
object[[i]]$call <- this.call[[i]] object[[i]]$sseq <- sseq
}
}
# lam_start = max of all lam_start
# lam_end might diff beacuse of stopping criterion
GIC.nlam <- sapply(object, function(x) length(drop(x$lam))) GIC.nlam_id <- which.min(GIC.nlam) GIC.lam <- object[[GIC.nlam_id]]$lam
GIC.nlam <- GIC.nlam[GIC.nlam_id]
### >>> get GIC curves <<<
GIC_curve <- matrix(NA, nrow = length(k), ncol = GIC.nlam)
MSE <- GIC_curve
pc = ifelse(is.null(Zc), 0, ncol(Zc)) + as.integer(intercept)
for(i in seq(length(k))) {
df = k[i]
scaler = switch(type,
"GIC" = log(max(p*df+ pc, n)) * log(log(n)) / n,
"BIC" = log(n) / n,
"AIC" = 2 / n
)
predmat <- predict.linear(object = object[[i]], newX = cbind(object[[i]]$Z, Zc)) MSE[i, ] <- apply(predmat[, 1:GIC.nlam] - y, 2, function(x) mean(x^2)) ## L: maximuize value of the likelihood fucntion for the estiamted model ## Gaussian model: -2*log(L) = n * log(MSE) GIC_curve[i, ] <- log(MSE[i, ]) N_zero <- object[[i]]$df[1:GIC.nlam]
if(mu_ratio != 0) {
## CGL or Baseline model, both with linear constraints
GIC_curve[i, ] = GIC_curve[i, ] + (ifelse(N_zero == 0, 0, N_zero - 1) * df + pc) * scaler
} else {
## naive model without linear constraints
GIC_curve[i, ] = GIC_curve[i, ] + (N_zero * df + pc) * scaler
}
}
rownames(GIC_curve) = paste(type, k, sep = "_")
## select the optimal grid valud of lambda and k which achieves minimum value of GIC curve
lam.min <- ggetmin(lam=GIC.lam, cvm = GIC_curve, k_list = k)$lam.min result <- list(FuncompCGL.fit = object, lam = GIC.lam, GIC = GIC_curve, lam.min = lam.min, MSE = MSE) class(result) <- "GIC.FuncompCGL" return(result) } #' @title #' Compute information crieteria for the \code{compCL} model. #' #' @description #' Tune the penalty parameter code{lam} in the \code{compCGL} model by GIC, BIC, or AIC. This #' function calculates the GIC, BIC, or AIC curve and returns the optimal value of #' \code{lam}. #' #' @inheritParams compCL #' #' @param \dots other arguments that can be passed to compCL. #' #' @return an object of S3 class \code{GIC.compCL} is returned, which is a list: #' \item{compCL.fit}{a fitted \code{\link{compCL}} object.} #' \item{lam}{the sequence of \code{lam}.} #' \item{GIC}{a vector of GIC value(s).} #' \item{lam.min}{the \code{lam} value that minimizes \code{GIC}(\eqn{\lambda}).} #' #' @details #' The model estimation is conducted through minimizing the following criterion: #' \deqn{\frac{1}{2n}\|y-Z\beta\|_2^2 + \lambda\|\beta\|_1, s.t. \sum_{j=1}^{p} \beta_j = 0.} #' The GIC is defined as: #' \deqn{GIC(\lambda) = \log{\hat{\sigma}^2(\lambda)} + #' (s(\lambda) -1) \log{(max(p, n))} * \log{(\log{n})} / n,} #' where \eqn{\hat{\sigma}^2(\lambda) = \|y - Z\hat{\beta}(\lambda)\|_{2}^{2}/n}, #' \eqn{\hat{\beta}(\lambda)} is the regularized estimator, #' and \eqn{s(\lambda)} is the number of nonzero coefficients in \eqn{\hat{\beta}(\lambda)}. #' Because of the zero-sum constraint, the effective number of free parameters is #' \eqn{s(\lambda) - 1} for \eqn{s(\lambda) \ge 2}. #' The optimal \eqn{\lambda} is selected by minimizing \code{GIC}(\eqn{\lambda}). #' #' #' @examples #' p = 30 #' n = 50 #' beta = c(1, -0.8, 0.6, 0, 0, -1.5, -0.5, 1.2) #' beta = c(beta, rep(0, times = p - length(beta))) #' Comp_data = comp_Model(n = n, p = p, beta = beta, intercept = FALSE) #' GICm1 <- GIC.compCL(y = Comp_data$y, Z = Comp_data$X.comp, #' Zc = Comp_data$Zc, intercept = Comp_data$intercept) #' coef(GICm1) #' plot(GICm1) #' test_data = comp_Model(n = 100, p = p, beta = Comp_data$beta, intercept = FALSE)
#' y_hat = predict(GICm1, Znew = test_data$X.comp, Zcnew = test_data$Zc)
#' plot(test_data$y, y_hat, xlab = "Observed value", ylab = "Predicted value") #' abline(a = 0, b = 1, col = "red") #' #' #' @references #' Lin, W., Shi, P., Peng, R. and Li, H. (2014) \emph{Variable selection in #' regression with compositional covariates}, #' \href{https://academic.oup.com/biomet/article/101/4/785/1775476}{https://academic.oup.com/biomet/article/101/4/785/1775476}. #' \emph{Biometrika} \strong{101} 785-979 #' #' Fan, Y., and Tang, C. Y. (2013) \emph{Tuning parameter selection in high #' dimensional penalized likelihood}, #' \href{https://rss.onlinelibrary.wiley.com/doi/abs/10.1111/rssb.12001}{https://rss.onlinelibrary.wiley.com/doi/abs/10.1111/rssb.12001} #' \emph{Journal of the Royal Statistical Society. Series B} \strong{75} 531-552 #' #' @seealso #' \code{\link{compCL}} and \code{\link{cv.compCL}}, #' and \code{\link[=coef.GIC.compCL]{coef}}, \code{\link[=predict.GIC.compCL]{predict}} and #' \code{\link[=plot.GIC.compCL]{plot}} methods for \code{"GIC.compCL"} object. #' #' @export ## TODO: add GIC and AIC feature GIC.compCL <- function(y, Z, Zc = NULL, intercept = FALSE, lam = NULL, ...) { this.call <- match.call() y <- drop(y) n <- length(y) p <- ncol(Z) pc <- ifelse(is.null(Zc), 0, dim(Zc)[2]) pc <- pc + as.integer(intercept) compCL.object <- compCL(y = y, Z = Z, Zc = Zc, intercept = intercept, lam = lam, ...) lam <- compCL.object$lam
## >>> MSE <<<
# predmat <- predict(compCL.object, newx = Z, newzc = Zc)
newx <- cbind(compCL.object$Z_log, Zc) predmat <- predict.linear(compCL.object, newx, s = NULL) cvraw <- (y - predmat)^2 MSE <- colSums(cvraw) / n ## <<< MSE >>> ## >>> GIC curve <<< scaler <- log(log(n)) * log(max(p + pc, n)) / n S <- apply(abs(compCL.object$beta[1:p, ]) > 0, 2, sum) # support
GIC <- log(MSE) + scaler * (ifelse(S>=2, S-1, 0) + pc)
# digits = 5
# GIC <- round(GIC, digits = digits)
## <<< GIC curve >>>
# >>> lambda selection <<<
# GIC.min <- min(GIC[drop(compCL.object$df) > 0]) GIC.min <- min(GIC) idmin <- GIC <= GIC.min # idmin[drop(compCL.object$df) < 2 ] <- FALSE
lam.min <- max(lam[idmin])
# <<< lambda selection >>>
result <- list(compCL.fit = compCL.object,
lam = lam,
GIC = GIC,
lam.min = lam.min)
class(result) <- "GIC.compCL"
result\$call <- this.call
return(result)
}
## Try the Compack package in your browser
Any scripts or data that you put into this service are public.
Compack documentation built on July 1, 2020, 10:26 p.m. | 4,478 | 13,105 | {"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} | 2.78125 | 3 | CC-MAIN-2022-05 | latest | en | 0.388317 |
http://formulas.ultrafractal.com/reference/jlb/JLB_NewtonUtilityTransform.html | 1,643,307,269,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305277.88/warc/CC-MAIN-20220127163150-20220127193150-00525.warc.gz | 26,787,068 | 3,642 | ## jlb Class JLB_NewtonUtilityTransform
```Object
common:Generic
common:Transform
common:UtilityTransform
jlb:JLB_NewtonUtilityTransform
```
`class UtilityTransform:JLB_NewtonUtilityTransform`
The Newton formula as a Transform.
Ultra Fractal Source
``` class JLB_NewtonUtilityTransform(common.ulb:UtilityTransform) {
; The Newton formula as a Transform.
public:
; @param pz
; @return the new pz
complex func Iterate(complex pz)
if (@p_std)
return (2*sqr(pz)*pz+1)/(3*sqr(pz))
else
return pz*(1-@relax) + \
@relax * ((@p_power - 1) * pz^@p_power + @r) / \
(@p_power * pz ^ (@p_power - 1))
endif
endfunc
default:
title = "Newton"
int param v_NewtonUtilityTransform
caption = "Version (Newton_UtilityTransform)"
default = 100
hint = "This version parameter is used to detect when a change has \
been made to the formula that is incompatible with the \
previous version. When that happens, this field will reflect \
the old version number to alert you to the fact that an \
alternate rendering is being used."
visible = @v_NewtonUtilityTransform < 100
endparam
bool param p_std
caption = "Standard"
default = true
endparam
complex param p_power
caption = "Exponent"
default = (3,0)
hint = "Specifies the exponent of the equation that is solved by \
Newton's method. Use real numbers (set the imaginary part \
to zero) to obtain classic Newton fractals."
visible = !@p_std
endparam
param r
caption = "Root"
default = (1,0)
hint = "Specifies the root of the equation that is solved. Use larger \
numbers for slower convergence."
visible = !@p_std
endparam
param relax
caption = "Relaxation"
default = (1,0)
hint = "This can be used to alter the convergence of \
the formula."
visible = !@p_std
endparam
}
```
Constructor Summary
`JLB_NewtonUtilityTransform()`
Method Summary
` complex` `Iterate(complex pz)`
Transform a single point within a sequence
Methods inherited from class common:Transform
`Init, IsSolid, IterateSilent`
Methods inherited from class common:Generic
`GetParent`
Methods inherited from class Object
Constructor Detail
### JLB_NewtonUtilityTransform
`public JLB_NewtonUtilityTransform()`
Method Detail
### Iterate
`public complex Iterate(complex pz)`
Description copied from class: `Transform`
Transform a single point within a sequence
After a sequence has been set up with Init(), this function will be called once for each value in the sequence. Note that all values in the sequence must be processed in order (they cannot be processed out of order). If the sequence contains only one value, Init() will still be called and then Iterate() will be called just once.
Overrides:
`Iterate` in class `Transform`
Parameters:
`pz` -
Returns:
the new pz | 676 | 2,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-05 | latest | en | 0.66388 |
http://www.google.com/patents/US5617482?dq=6,631,400 | 1,406,910,437,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510275111.45/warc/CC-MAIN-20140728011755-00308-ip-10-146-231-18.ec2.internal.warc.gz | 551,269,132 | 12,284 | ## Patents
Publication number US5617482 A Publication type Grant Application number US 08/367,773 Publication date Apr 1, 1997 Filing date Jan 3, 1995 Priority date Aug 15, 1990 Fee status Lapsed Publication number 08367773, 367773, US 5617482 A, US 5617482A, US-A-5617482, US5617482 A, US5617482A Inventors Harald Brusewitz Original Assignee Televerket Export Citation Patent Citations (6), Referenced by (21), Classifications (12), Legal Events (4) External Links:
Method of motion compensation and elastic deformation in picture sequences
US 5617482 A
Abstract
A method of motion compensation and elastic deformation in picture sequences in transmission of moving pictures between a transmitter and a receiver, that is a picture coding method in which the pixel values of the picture are determined using a previous picture. In the transmitter, each picture in the picture sequence is divided into triangles by suitably choosing corner points, and the motion between two successive pictures is estimated such that the motion vectors of all the corner points can be established. Information about the corner points and/or the motion vectors is transmitted from the transmitter to the receiver, in which the motion vectors of the internal points of the triangles are calculated using the three transmitted motion vectors belonging to the respective triangle. The pixel values in the corner points of the triangles and in the internal points of the triangles are calculated using the transmitted and the calculated motion vectors, respectively. Preferrably, the motion vectors of the internal points of the triangles are calculated by interpolation.
Images(2)
Claims(2)
I claim:
1. Method of motion compensation and elastic deformation in a picture sequence transmitted between a transmitter and a receiver, the method comprising the following steps in the transmitter:
dividing each picture in a picture sequence into triangles by selecting corner points;
estimating motion between two successive pictures in the picture sequence;
establishing a motion vector for each corner point selected in the dividing step; and
transmitting the corner points and three motion vectors per corner point from the transmitter to the receiver; and
the method comprising the following steps in the receiver:
receiving the corner points and the three motion vectors per corner point at the receiver from the transmitter;
calculating the motion vectors of internal points of the triangles using the three transmitted motion vectors of respective triangles;
calculating pixel values of the corner points of the triangles using pixel values of a previous picture in the picture sequence and the motion vectors transmitted by the transmitter; and
calculating pixel values of the internal points of the triangles using the motion vectors calculated in the calculating step and the pixel values of the previous picture in the picture sequence.
2. Method according to claim 1, wherein the step of calculating the motion vectors of the internal points comprises calculating the motion vectors of the internal points of the triangles by interpolation.
Description
This application is a continuation of application Ser. No. 07/969,221, filed on Feb. 12, 1993, now abandoned, filed as PCT Application Number PCT/SE91/00529 on Aug. 12, 1991.
FIELD OF THE INVENTION
The present invention relates to a method of motion compensation and elastic deformation in the picture sequences, that is a picture coding method in which the pixel values in a picture are determined using the previous picture. It is assumed that the motion between two successive pictures is known, that is the motion vector field is known. The motion may include both translation and deformation.
STATE OF THE ART
It is previously known to use motion vector fields for picture coding. In this connection, the picture is divided into rectangular blocks. In this division problems may occur with visible squares in the picture sequence because of discontinuities at the edges of the block. According to the present invention this problem is solved by dividing the picture in triangles instead and interpolate the motion field.
SUMMARY OF THE INVENTION
Thus, the present invention provides a method of motion compensation and elastic deformation in picture sequences in transmission of moving pictures between a transmitter and a receiver. According to the invention each picture in a picture sequence is divided into triangles and the motion between two successive pictures is estimated such that the motion vectors in the corner points of the triangles can be determined. Information about the corner points and/or the motion vectors is transmitted from the transmitter to the receiver. In the receiver the motion vectors of the internal points of the triangles are calculated using the three transmitted motion vectors associated with the respective triangle. The pixel values of the corner points of the respective triangles are calculated using the transmitted motion vectors and the previous picture and the internal pixel values are calculated using the calculated motion vectors and the previous picture.
Further features of the invention are set forth in detail in the accompanying claims.
BRIEF DESCRIPTION OF THE DRAWINGS
The drawing is a schematic illustration of two successive pictures in which corresponding triangles have been drawn and the respective motion vectors, in which
FIG. 1. shows translation,
FIG. 2. rotation and
FIG. 3. deformation; and
FIGS. 4A and 4B are flowcharts showing the steps performed in the transmitter and receiver, respectively.
The invention is described below in detail.
DETAILED DESCRIPTION OF A PREFERRED EMBODIMENT OF THE INVENTION
The purpose of the invention is to enable transmission of picture sequences by starting from one picture and only transmit motion vectors. There already exists methods of dividing the pictures in rectangular blocks and of determining the motion vectors applying within the blocks. A problem is that square blocks then may become visible in the picture because of discontinuities in the motion vector field between adjacent blocks.
According to the present invention the picture is divided into triangles instead. The division may be fixed or may be adapted to the picture in question and the motion therein. When the corner points of each triangle are established the motion vectors in the corner points are determined relative to the previous picture in the sequence. Motion vectors are assigned to points within each triangle by interpolating the motion vectors in the corner points. Thus, contiguous triangles have the same motion vectors along the whole common side. Thereby no discontinuities occur.
The pixel values are obtained from the previous picture using the motion vector fields obtained in this way. Generally, the motion vectors indicate undefined points in the previous picture. The pixel value is then obtained using an interpolation filter, preferably in accordance with our co-pending patent application titled "Method of moving a pixel a subpixel distance".
In the figures, various ways of how a triangle can be moved are shown. FIG. 1 shows a translation, FIG. 2 a rotation and FIG. 3 a deformation, in this case an enlargement. Normally, the movement of the triangle is a combination of these three basic types. The co-ordinates of the new triangle can be expressed in matrix form as
`Yt =Xt +Mt `
where Xt is the co-ordinates of the old triangle, Yt is the new triangle and Mt contains the three motion vectors.
Any point (x, y) may be expressed using the position vectors of the triangle Xt. Define a triangle having three corner points ##EQU1## Create a base of vectors v1, v2, v3 using the corner points. ##EQU2## Then an arbitrary point in the picture plane can be expressed as
`(x, y)=v1 +av2 +bv3 `
For points within the triangle Xt the following relations apply 0≦a≦1 and 0≦b≦a.
For a given point (x, y ) a and b can be calculated with the formulas ##EQU3##
A set of positions X may be expressed as
`X=BXt `
where each line of X is a position (xi, yi) and ##EQU4##
Several triangles can be processed at the same time with a matrix Xt containing all the triangle corners involved. With a number of corners=nc: ##EQU5##
Then B is modified such that each line has nc positions and the values 1-ai, ai -bi and bi are placed in the positions corresponding to the corner point that is to be used. All other positions are zero.
A continouos motion vector field in the triangle Yt is obtained if the motion vectors are calculated as
`M=BMt `
where the positions in question within the respective triangle are inserted in B.
The triangle division of the picture is performed in the transmitter and the corner points of the triangles are established. Thereafter, the motion vectors and the corner points are established. First, a complete picture is transmitted between the transmitter and receiver. Thereafter, only motion vectors and information about the corner point associated with each motion vector are transmitted. In the receiver the motion vector of every point in the picture is calculated by interpolation of the motion vectors in the corner point (that is by the formula M=BMt). Thereafter, all the pixel values can be calculated using the motion vectors and the previous picture.
Thus, the present invention provides a method of motion compensation and elastic deformation in picture sequences which solves the above problem with discontinuities between blocks in the picture. The invention is defined in the accompanying claims.
Patent Citations
Cited PatentFiling datePublication dateApplicantTitle
US4600919 *Aug 3, 1982Jul 15, 1986New York Institute Of TechnologyMethod for generating a sequence of video frames
US4924310 *Aug 22, 1989May 8, 1990Siemens AktiengesellschaftMethod for the determination of motion vector fields from digital image sequences
US4984074 *Mar 13, 1990Jan 8, 1991Matsushita Electric Industrial Co., Ltd.Motion vector detector
US5019901 *Mar 26, 1990May 28, 1991Matsushita Electric Industrial Co., Ltd.Image motion vector sensor for sensing image displacement amount
EP0280316A2 *Feb 25, 1988Aug 31, 1988Sony CorporationVideo image transforming method and apparatus
EP0280317A2 *Feb 26, 1988Aug 31, 1988kabelmetal electro GmbHWinding device with a binding unit
Referenced by
Citing PatentFiling datePublication dateApplicantTitle
US5892849 *Jul 10, 1996Apr 6, 1999Hyundai Electronics Industries Co., Ltd.Compaction/motion estimation method using a grid moving method for minimizing image information of an object
US5903670 *Jul 10, 1996May 11, 1999Hyundai Electronics Industries Co., Ltd.Grid moving apparatus for minimizing image information of an object
US5907626 *Aug 1, 1997May 25, 1999Eastman Kodak CompanyMethod for object tracking and mosaicing in an image sequence using a two-dimensional mesh
US5933547 *Jan 11, 1996Aug 3, 1999France TelecomMethod for interpolating images
US5982909 *Apr 23, 1996Nov 9, 1999Eastman Kodak CompanyMethod for region tracking in an image sequence using a two-dimensional mesh
US6008852 *Mar 17, 1997Dec 28, 1999Hitachi, Ltd.Video coder with global motion compensation
US6115501 *Jul 10, 1996Sep 5, 2000Hyundai Electronics Industries Co., Ltd.Grid moving method for minimizing image information of an object
US6178202Nov 12, 1999Jan 23, 2001Hitachi, Ltd.Video coder with global motion compensation
US6442205Oct 24, 2000Aug 27, 2002Hitachi, Ltd.Method of coding and decoding image
US6483877 *Nov 13, 2001Nov 19, 2002Hitachi, Ltd.Method of coding and decoding image
US6687302 *Oct 3, 2002Feb 3, 2004Hitachi, Ltd.Method of coding and decoding image
US6711210 *Oct 3, 2002Mar 23, 2004Hitachi, Ltd.Method of coding and decoding image
US6987810Nov 18, 2003Jan 17, 2006Renesas Technology Corp.Method of coding and decoding image
US7298781Jun 8, 2005Nov 20, 2007Renesas Technology Corp.Method of coding and decoding image
US7616782May 7, 2004Nov 10, 2009Intelliview Technologies Inc.Mesh based frame processing and applications
US7817720Sep 27, 2007Oct 19, 2010Hitachi, Ltd.Method of coding and decoding image
US7859551Feb 25, 2002Dec 28, 2010Bulman Richard LObject customization and presentation system
US8204110Jul 19, 2010Jun 19, 2012Hitachi, Ltd.Method of coding and decoding image
US8224023 *Jul 21, 2006Jul 17, 2012Hewlett-Packard Development Company, L.P.Method of tracking an object in a video stream
US8233543Jul 19, 2010Jul 31, 2012Hitachi, Ltd.Method of coding and decoding image
US20070097112 *Jul 21, 2006May 3, 2007Hewlett-Packard Development Company, L.P.Method of tracking an object in a video stream
Classifications
U.S. Classification382/107, 382/295, 375/E07.11, 348/E05.066
International ClassificationG06T7/20, H04N5/14, H04N7/26
Cooperative ClassificationG06T7/2033, H04N5/145, H04N19/00618
European ClassificationG06T7/20C, H04N7/26M2N2
Legal Events
DateCodeEventDescription
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Year of fee payment: 4 | 3,017 | 13,183 | {"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-2014-23 | latest | en | 0.885256 |
https://www.physicsforums.com/threads/electron-accelerated-from-89c-to-97c.372098/ | 1,544,689,945,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824601.32/warc/CC-MAIN-20181213080138-20181213101638-00064.warc.gz | 1,002,030,094 | 12,984 | # Homework Help: Electron accelerated from .89c to .97c
1. Jan 23, 2010
### danielatha4
1. The problem statement, all variables and given/known data
SLAC, the Stanford Linear Accelerator Collider, located at Stanford University in Palo Alto, California, accelerates electrons through a vacuum tube two miles long (it can be seen from an overpass of the Junipero Serra freeway that goes right over the accelerator). Electrons which are initially at rest are subjected to a continuous force of 1.2×10-12 newton along the entire length of two miles (one mile is 1.6 kilometers) and reach speeds very near the speed of light.
Determine how much time is required to increase the electrons' speed from 0.89c to 0.97c.
Approximately how far does the electron go in this time? (What is approximate about your result?)
2. Relevant equations
3. The attempt at a solution
I figured the acceleration of the electron would be equal to the force, 1.22x10^-12, divided by the mass of an electron, 9.1x10^-31. Then I used Vf=Vo+at to solve for time. However, this was not correct.
2. Jan 24, 2010
### Maxim Zh
It's a relativistic problem and you have to use relativistic equations.
The Newton's second law is correct here in the momentum form:
$$\frac{\partial \mathbf{p}}{\partial t} = \mathbf{F}.$$
The speed dependence of momentum is nonlinear:
$$\mathbf{p} = m\gamma(v)\mathbf{v}.$$
The equation for x(t) will be very complicated. Perhaps the ultrarelativistic approximation will simplify it. But I'm not sure it could be used for v=0.89c.
3. Jan 24, 2010
### danielatha4
My teacher did give us the equation $$\Delta$$P=Fnet$$\Delta$$t
However, solving the final and initial relativistic momentum for the change in momentum, and dividing by the force applied gives me 4.64*10^-12 seconds, and that's not right.
4. Jan 25, 2010
### Maxim Zh
My result is 4.64*10-10 s. | 508 | 1,877 | {"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} | 3.90625 | 4 | CC-MAIN-2018-51 | latest | en | 0.913657 |
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### Author Topic: Math God/Geek Test (Read 2523 times)
#### taichimaster
• Charter Member
• Joined in 2006
• Posts: 160
##### Math God/Geek Test
« on: March 14, 2006, 01:39 PM »
See if you guys could crack this!!
http://mathtest.idiotworld.com/
I am stumbling on the 4th
#### Rover
• Master of Smilies
• Charter Member
• Joined in 2005
• Posts: 630
##### Re: Math God/Geek Test
« Reply #1 on: March 14, 2006, 07:03 PM »
Editing the HTML at the bottom is much easier than taking the test.
Insert Brilliant Sig line here
#### nudone
• Cody's Creator
• Columnist
• Joined in 2005
• Posts: 4,117
##### Re: Math God/Geek Test
« Reply #2 on: March 15, 2006, 01:21 AM »
that site is depressing.
#### Hirudin
• Charter Member
• Joined in 2005
• Posts: 543
##### Re: Math God/Geek Test
« Reply #3 on: March 15, 2006, 01:48 AM »
Woop, 11 down!
[much later]
Welp, I got 13 of them.
If you don't recognize the sequence: "6, 28, 496, 8128," don't waste your time trying to figure this one out! (There's at least an hour of my life I'll never get back...)
"14, 23, 28, 33, 42" - _NOT_ outside the box!
The author of the site thought he was being clever with this one, but I beat him!
The +2 / -2 is the key...
14 = 7x2
23 = 7x3 + 2
28 = 7x4
33 = 7x5 - 2
42 = 7x6
51 = 7x7 + 2
« Last Edit: March 15, 2006, 04:03 AM by Hirudin » | 538 | 1,557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-13 | longest | en | 0.868987 |
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### Transcript of planar
• 2013 Spring
E344 Solution Set 3
1. Sodium has the BCC crystal structure. Its density is 0.971 g/cm3, and its atomic weight is 22.99 g/mole. Use this
information to show that the atomic radius of a sodium atom is 0.186 nm.
Solution
ASSUMPTION:
Assuming the sodium atoms are arranging as close-packed spheres in BCC crystal structure, then:
The volume occupied by Na atoms in 1 cm3 is then 0.68 cm
3.
The number of Na atoms in 1 cm3is:
221231 10543.21002.6)99.22/971.0( molemolegg
The volume of each Na atom 0V is:
322223
0 10674.2)10543.2/(68.0 cmcmV
The radius of a Na atom is:
nmcmV
R 186.01086.1
3
48
3
00
• 2. Niobium has an atomic radius of 0.1430 nm and a density of 8.57 g/cm3. Determine whether it has an FCC or a
BCC crystal structure. Show all of your work.
Solution
ASSUMPTION:
Assuming the sodium atoms are arranging as close-packed spheres in FCC crystal structure, then
If Niobium has a FCC structure, its density is calculated within one unit cell as:
33213
123
13
2333.91033.9))1430.0(216/(
1002.6
91.924/
1002.6
4cmgnmgnm
mole
molega
M NbFCC
If Niobium has a BCC structure, its density is calculated within one unit cell as:
33213
123
13
2357.81057.8))1430.0(
33
64/(
1002.6
91.922/
1002.6
2cmgnmgnm
mole
molega
M NbFCC
Therefore, Niobium has a BCC structure.
333
222
216)22(
)4(
rraV
raa
uc
There are 4 atoms in a FCC unit cell.
See Problem 1 for BCC structure.
• 3. 3.23 List the point coordinates of the titanium, barium, and oxygen ions for a unit cell of the perovskite crystal
structure (Figure 12.6).
Solution
In Figure 12.6, the barium ions are situated at all corner positions. The point coordinates for these ions are
as follows: 000, 100, 110, 010, 001, 101, 111, and 011.
The oxygen ions are located at all face-centered positions; therefore, their coordinates are
1
2
1
20,
1
2
1
21,
11
2
1
2,
01
2
1
2,
1
20
1
2, and
1
21
1
2.
And, finally, the titanium ion resides at the center of the cubic unit cell, with coordinates
1
2
1
2
1
2.
• 4. 3.30 Within a cubic unit cell, sketch the following directions:
(a)
[1 10] , (e)
[1 1 1] ,
(b)
[1 2 1], (f)
[1 22],
(c)
[01 2], (g)
[12 3 ],
(d)
[13 3], (h)
[1 03].
Solution
The directions asked for are indicated in the cubic unit cells shown below.
• 5. 3.31 Determine the indices for the directions shown in the following cubic unit cell:
Solution
Direction B is a
[2 10] direction, which determination is summarized as follows. We first of all position the origin
of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system
x y z
Projections a
b
2 0c
Projections in terms of a, b, and c 1
1
2 0
Reduction to integers 2 1 0
Enclosure
[2 10]
• 6. 3.40 Sketch within a cubic unit cell the following planes:
(b) )( 211
(c)
(102 )
Solution
The planes called for are plotted in the cubic unit cells shown below.
• 7. 3.41 Determine the Miller indices for the planes shown in the following unit cell:
Solution
For plane A we will leave the origin at the unit cell as shown; this is a (403) plane, as summarized below.
x y z
Intercepts
a
2 b
2c
3
Intercepts in terms of a, b, and c
1
2
2
3
Reciprocals of intercepts 2 0
3
2
Reduction 4 0 3
Enclosure (403)
For plane B we will move the origin of the unit cell one unit cell distance to the right along the y axis, and
one unit cell distance parallel to the x axis; thus, this is a
(1 1 2) plane, as summarized below.
x y z
Intercepts a b
c
2
Intercepts in terms of a, b, and c 1 1
1
2
Reciprocals of intercepts 1 1 2
Reduction (not necessary)
Enclosure
(1 1 2)
• 8. 3.54 (a) Derive planar density expressions for FCC (100) and (111) planes in terms of the atomic radius R.
(b) Compute and compare planar density values for these same two planes for nickel.
Solution
(a) In the figure below is shown a (100) plane for an FCC unit cell.
For this (100) plane there is one atom at each of the four cube corners, each of which is shared with four adjacent
unit cells, while the center atom lies entirely within the unit cell. Thus, there is the equivalence of 2 atoms
associated with this FCC (100) plane. The planar section represented in the above figure is a square, wherein the
side lengths are equal to the unit cell edge length,
2R 2 (Equation 3.1); and, thus, the area of this square is just
(2R 2)2 = 8R2. Hence, the planar density for this (100) plane is just
PD100 = number of atoms centered on (100) plane
area of (100) plane
2 atoms
8R2
1
4R2
That portion of an FCC (111) plane contained within a unit cell is shown below.
• There are six atoms whose centers lie on this plane, which are labeled A through F. One-sixth of each of atoms A,
D, and F are associated with this plane (yielding an equivalence of one-half atom), with one-half of each of atoms B,
C, and E (or an equivalence of one and one-half atoms) for a total equivalence of two atoms. Now, the area of the
triangle shown in the above figure is equal to one-half of the product of the base length and the height, h. If we
consider half of the triangle, then
(2R)2 h2 (4R)2
which leads to h =
2 R 3 . Thus, the area is equal to
Area 4 R(h)
2
(4 R)(2 R 3)
2 4 R2 3
And, thus, the planar density is
PD111 = number of atoms centered on (111) plane
area of (111) plane
2 atoms
4 R2 3
1
2 R2 3
(b) From the table inside the front cover, the atomic radius for nickel is 0.125 nm. Therefore, the planar
density for the (100) plane is
PD100 (Ni) 1
4 R2
1
4 (0.125 nm)2 16.00 nm2 1.600 1019 m2
While for the (111) plane
PD111(Ni) 1
2 R2 3
1
2 3 (0.125 nm)2 18.48 nm2 1.848 1019 m2
• 9. 4.27 For an FCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than
that for a (111) plane? Why? (Note: You may want to consult the solution to Problem 3.54 at the end of Chapter 3.)
Solution
The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar density
(Section 3.11)]that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the
more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the
solution to Problem 3.54, planar densities for FCC (100) and (111) planes are
1
4R2 and
1
2R2 3, respectivelythat
is
0.25
R2 and
0.29
R2 (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have the
lower surface energy.
• 10. One way to describe a crystal structure is to use a simple unit cell with a multi-atom basis.
The "basis" represents the set of atoms that is translated to each point in a unit cell to create the
crystal structure. For example, silver (Ag) is FCC with a one-atom basis. That one atom would
have coordinates of (0,0,0). Similarly, silicon (Si) has the diamond-cubic crystal structure, which
can be described as FCC with a two-atom basis. The basis would consists of a Si atom at (0,0,0)
and another Si atom at (, , ). If these two atoms are translated to all of the corner and face
positions of the FCC unit cell, the diamond cubic lattice is created.
The rock salt structure (see fig. 12.2 in Callister) is as an FCC unit cell with a two-atom basis.
For NaCl, define the two atoms and their positions in this basis. Sketch the (100) face of the
NaCl unit cell and show that the Na and Cl positions can be generated by moving the two-atom
basis to each of the FCC positions on this face.
Solution
Cl (0,0,0)
Na(1/2,1/2,1/2), or any one of the positions that have Na atoms in the unit cell.
The (100) face is labeled in white dotted line. Blue dots are Cl, green dots are Na. Movements of Cl /Na are labeled
with red/lightblue arrows, respectively.
• 11. Single-crystal wafers are used in semiconductor devices. Because the properties of single-crystal semiconductors
are anisotropic, the wafers must be precisely aligned when fabricating circuits from them. A number of different
wafer orientations are available. These can be differentiated from each other based on the position of flats or notches
machined into each wafer. A schematic top-view illustration of a particular silicon singlecrystal wafer is given in the
adjacent diagram. The directi | 2,524 | 8,494 | {"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} | 3.5 | 4 | CC-MAIN-2022-05 | latest | en | 0.799284 |
https://www.6aming.com/academics/venn-diagram-conditional-probability/ | 1,558,763,126,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257889.72/warc/CC-MAIN-20190525044705-20190525070705-00065.warc.gz | 664,855,185 | 8,708 | Venn Diagram Conditional Probability
Sunday, April 8th 2018. | Academics
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Go to http://www.examsolutions.net/maths-revision/index.php to see the main index of maths video tutorials. | 425 | 2,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-22 | latest | en | 0.841906 |
https://riddles360.com/riddle/rearrange-paper | 1,713,265,772,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817081.52/warc/CC-MAIN-20240416093441-20240416123441-00203.warc.gz | 451,963,899 | 6,948 | # Rearrange Paper
There is a square piece of paper with a hole that is denoted by the circle on the top right side in the given picture. You have to cut the paper in a manner that it forms two and only two separate pieces of paper and then rearrange the pieces in a manner that the holes come in the centre of the paper.
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### Amazing Facts
###### Jigsaw puzzles
Jigsaw puzzles soared in popularity during the great depression, as they provided a cheap, long-lasting, recyclable form of entertainment. | 783 | 3,056 | {"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-2024-18 | longest | en | 0.961398 |
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http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-2-sections-2-1-2-4-integrated-review-linear-equations-and-inequalities-page-88/12 | 1,521,762,723,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648103.60/warc/CC-MAIN-20180322225408-20180323005408-00481.warc.gz | 372,065,227 | 13,028 | ## Intermediate Algebra (6th Edition)
Published by Pearson
# Chapter 2 - Sections 2.1-2.4 - Integrated Review - Linear Equations and Inequalities: 12
6
#### Work Step by Step
-5x + 4 = -26 Subtract 4 from each side. -5x=-30 Divide by -5 x=6
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# UCC2808A-1: UCC2808A Wide Input Range
Part Number: UCC2808A-1
Is it possible to achieve a circuit of Fig.2 of ucc2808 data seat?
The input voltage area which is 72V from 36V by Push - Pull system, can it correspond?
• Kobayashi-San,
Thank you for your interest here. Please excuse the delays until January 3rd as it is US holiday now.
Figure 2 if the data sheet seems to be a curve for Idd vs. Osc frequency. What circuit figure are you referring to? Your design needs input voltage range from 36-72v , is that correct? What is your output voltage and current?
Regards
John
• John-san
I'm seeing rebesion SLUS456D of a dataseat.
The circuit which is being asked is attached.
Best Regards,
• Hi Kobayashi-san,
The data sheet for this part has recently been revised. The latest revision is now SLUS456E.
The application section is now shown in Fig. 11.
The input voltage varies from 36V to 72V and has a nominal value of 48V.
The output is 5V at 50W max.
This application is detailed as a design example and is not available as a reference design.
Regards,
John Griffin
• Hi John
It's mentioned that the input voltage range is from 36V to 72V by a design parameter of Table.2 of a data seat.
But our costomer think it's difficult to correspond to a wide input voltage range with Push-pull topology.
Can UCC2808A correspond to a double input voltage range?
Best Regards,
Kobayashi
• Hi Kobayashi-san,
The push pull can be designed for this wide voltage range.
For example if the output duty cycle is 90% at 36V (corresponding to 45% duty cycle on each gate drive) then the required output duty cycle will be 45% at 72V
(corresponding to 22.5% duty cycle on each gate drive)
Does that make sense? Of course it is always difficult to design a power supply with a wide input range but it is possible to do so with the right care and attention to details
Regards,
John | 524 | 2,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-17 | latest | en | 0.917015 |
https://present5.com/optimisation-the-general-problem-want-to-minimise/ | 1,566,425,845,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316549.78/warc/CC-MAIN-20190821220456-20190822002456-00143.warc.gz | 606,201,375 | 11,341 | Скачать презентацию Optimisation The general problem Want to minimise
9489677161afdb0bdac22d5ef18e532c.ppt
• Количество слайдов: 23
Optimisation • The general problem: Want to minimise some function F(x) subject to constraints, ai(x) = 0, i=1, 2, …, m 1 bi(x) 0, i=1, 2, …, m 2 where x is a vector of length n. • F( ) is called the objective function. • ai( ) and bi( ) are called the constraint functions.
Special Cases • If n=1 there is just one variable, and we have the univariate case (as opposed to the multivariate case). • If ai(x) and bi(x) are linear functions then we have linear constraints (as opposed to nonlinear constraints). • If m 2=0 we have equality constraints only. • If m 1=0 we have inequality constraints only. • If m 1=m 2=0 we have the unconstrained case.
Techniques • The techniques used to solve an optimisation problem depends on the properties of the functions F, ai, and bi. • Important factors include: – Univariate or multivariate case? – Constrained or unconstrained problem? – Do we know the derivatives of F?
Example Linear Problem • An oil refinery can buy light crude at £ 35/barrel and heavy crude at £ 30/barrel. • Refining one barrel of oil produces petrol, heating oil, and jet fuel as follows: Petrol Light crude 0. 3 Heavy crude 0. 3 Heating oil Jet fuel 0. 2 0. 3 0. 4 0. 2 • The refinery has contracts for 0. 9 M barrels of petrol, 0. 8 M barrels of heating oil and 0. 5 M barrels of jet fuel. • How much light and heavy crude should the refinery buy to satisfy the contracts at least cost?
Problem Specification • Let x 1 and x 2 be the number of barrels (in millions) of light and heavy crude that the refinery purchases. • Cost (in millions of £): F(x) = 35 x 1 + 30 x 2 • Constraints: 0. 3 x 1 + 0. 3 x 2 0. 9 (petrol) This is called a 0. 2 x 1 + 0. 4 x 2 0. 8 (heating oil) “linear 0. 3 x 1 + 0. 2 x 2 0. 5 (jet fuel) program” x 1 0, x 2 0 (non-negativity)
Graphical Solution x 2 Feasible region 2 (1, 2) 1 1 2 3 4 x 1 • Minimum of F lies on boundary of feasible region. • F varies linearly on each section of the boundary. • Can get the solution by looking at the intersection points of the constraints forming the boundary.
Solution (x 1, x 2) (0, 3) (2, 1) (4, 0) F(x) 90 100 140 Recall that: F(x) = 35 x 1 + 30 x 2 • So minimum cost is for x 1 = 0 and x 2 = 3.
Unconstrained Univariate Case • • We seek to minimise f(x). If x* minimises f(x) then: i. f (x*) = 0 (first order condition) ii. f (x*) 0 (second order condition) f(x) = (x-1)+2
Example • • Minimise f(x) = x 2 + 4 Cos(x) Solve: f (x) = 2 x – 4 Sin(x) = 0 y = fzero(@(x)(2*x-4*sin(x)), 2) Gives y = 1. 8955
Bisection Method • Suppose we have already bracketed the zero in the interval [a, b]. Then: 1. Evaluate f at mid-point c=(a+b)/2. 2. If f(c) is zero then quit. 3. If f(a) and f(c) have the same sign then set a=c; else set b=c. 4. Go to Step 1. (a+b)/2 a b
MATLAB Example >> f=@(x)(2*x-4*sin(x)); >> a=1; fa=f(a); >> b=2; fb=f(b); >> c=(a+b)/2; fc=f(c); if fa*fc>0 a=c; else b=c; end; c • Using the up arrow to repeat the last line we get values of c that converge to the solution of f(x)=0.
Convergence • At each iteration the zero x* lies within the current interval from a to b. • So the error |x*-x|
Newton’s Method • Given an estimate xk of the zero a better estimate is obtained by approximating the function by the tangent line at xk. f (xk) = f(xk)/(xk-xk+1) xk+1 = xk – f(xk)/f (xk) f(xk) xk+1 xk
Convergence of Newton’s Method • Error can be shown to be quadratic if initial estimate of zero is sufficiently close to x*. |x*-xk+1| < M|x*-xk|2 for some constant M. (Proof: Taylor series expansion of f(x*) about xk. )
Example • Find real root of f(x)=x 3+4 x 2 -10=0. >> format long >> r=roots([1 4 0 -10]’); y=r(3); x=1; >> for i=1: 8 fx=-10+x*x*(4+x); fxd=x*(8+3*x); err=y-x; a(i, 1)=i; a(i, 2)=x; a(i, 3)=fx; a(i, 4)=fxd; a(i, 5)=err; x=x-fx/fxd; >> end; >> a
Problems with Newton’s Method • Problems may arise if the initial estimate is not “sufficiently close” to the zero. • Consider f(x)=ln(x). If 0
Linear Interpolation Methods • Newton method requires first derivative at each iteration. • The bisection method doesn’t use the magnitudes of f at each end of the interval. • Suppose we use f(an) and f(bn) and finds a new estimate of the zero by approximating the function between an and bn by a straight line. f(an) an xn bn f(bn)
Secant Method • The secant method is a linear interpolation method that generates approximations to the zero starting with x 0 and x 1 according to: Problem with divergence! xn xn+1 xn-1 xn+2
Method of False Position • To avoid possible divergence problem with the secant method we keep the zero bracketed in an interval (a, b), as in the bisection method. • If f(c) = 0 we are finished. • If f(a) and f(c) have the same sign we replace a by c; otherwise, we replace b by c.
Golden Section Method • A function is unimodal on an interval [a, b] if it has a single local minimum on [a, b]. • The Golden Section method can be used to find the minimum of function F on [a, b], where F is unimodal on [a, b]. • This method is not based on solving F (x)=0. • We seek to avoid unnecessary function evaluations.
Golden Section Method • Divide interval [a, b] at x and y as follows: x = a + (b-a); u = F(x) y = a + 2(b-a); v = F(y) • If u>v then x* must lie in [a, x], and if u v then x* must lie in [y, b]. • Case 1: If u>v then new interval is [a, x] and length is xa= (b-a). At the next step we need to know F at: a + (x-a) = a + 2(b-a) a + 2(x-a) = a + 3(b-a) • But we already know F at a + 2(b-a) from the previous step so we can avoid this function evaluation.
Golden Section Method • Case 2: If u v then new interval is [y, b] and length is by= (b-a). At the next step we need to know F at: Note: 2 + - 1 = y + (b-y) = a + 2 2(b-a) y + 2(b-y) = a + 2(1+ )(b-a) = 0 + (b-a) a • But we already know F at a + (b-a) from the previous step so we can avoid this function evaluation. • In both cases we get a new interval that is times the length of the current interval, and each iteration requires only one function evaluation. • After n iterations the error is bounded by (b-a) n/2
MATLAB Code for Golden Section >> f=@(x)(x*x+4*cos(x)); >> a=1; fa=f(a); b=2; fb=f(b); t=(sqrt(5)-1)/2; >> x=a+t*(b-a); y=a+t*t*(b-a); u=f(x); v=f(y); if u>v b=x; fb=u; else a=y; fa=v; end; c=(b+a)/2 • Using the up arrow to repeat the last line we get values of c that converge to the minimum of F on [1, 2]. | 2,126 | 6,480 | {"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} | 3.9375 | 4 | CC-MAIN-2019-35 | latest | en | 0.816608 |
https://discuss.pytorch.org/t/3d-volumes-dataset-to-2d-slices-dataset/112971 | 1,652,695,162,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00661.warc.gz | 266,120,189 | 5,618 | # 3D volumes Dataset to 2D slices Dataset
Hello,
I currently have a set of 3D nifti images that I’m loading into a Dataset object using @MONAI .
However, instead of having a Dataset object composed of multiple volumes, I wish to have a 2D dataset composed of all the slices from all the volumes.
Is there a way to load the data this way or change the Dataset object after loading?
Or, in case it is not possible, is there a way to change the batches so that instead of batching out volumes, it batches out slices?
e.g if batch size is 2, instead of 2 volumes it would send out all slices from the 2 volumes.
Thanks for the help!
Could you explain what the difference would be in these two cases, i.e. how the batch shape would look like?
Of course.
So, assuming the volumes were 128x128x32, greyscale and batch was 2, instead of the batch being
`[2,1,32,128,128]` they would be `[64,1,128,128]`.
I’m unsure if this sort of conversion even makes sense in any scenario, but I thought it could be an alternative assume there was no function to read and save the volumes as slices on the Dataset object.
This “flattening” operation can be performed by:
``````x = torch.randn(2, 1, 32, 128, 128)
x = x.permute(0, 2, 1, 3, 4)
x = x.view(-1, *x.size()[2:])
print(x.shape)
> torch.Size([64, 1, 128, 128])
``````
This could be an easy way to change the input format.
The alternative approach would be to open the volume, grab some slices, and return only these.
However, the logic inside the `__getitem__` method would be a bit more complicated, as you would have to e.g. reuse the same volume to load the missing slices and would have to map the passed `index` somehow to this logic or use a custom sampler etc.
1 Like
Hi, I have a situation opposite to this problem.
I have a point cloud of dim :`Bx3xN`
and I want to convert it to a voxel of dim: `Bx3xRxRxR`
how can I do this efficiently in pytorch?
I’m not aware of any built-in PyTorch methods to work with point clouds and their transformations, so I would assume you might want to use a 3rd party library for it.
could you suggest any?
The flattening solution sounds great!
I’m new to torch so Im sorry if this follow-up question doesn’t make much sense but, to use your solutions on 2D models the only part that would need to be changed would be the training loop logic correct? When iterating each batch, I would flatten it and go from there, correct?
Yes, the 4-dimensional input tensor would fit into `*2D` modules, such as `nn.Conv2d`.
@sinAshish PyTorch Geometric might have some methods for point clouds, but I haven’t used it so far.
1 Like | 679 | 2,614 | {"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-2022-21 | latest | en | 0.927087 |
https://www.aimsciences.org/article/doi/10.3934/ipi.2020063?viewType=html | 1,611,454,244,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703544403.51/warc/CC-MAIN-20210124013637-20210124043637-00640.warc.gz | 676,677,756 | 15,127 | # American Institute of Mathematical Sciences
doi: 10.3934/ipi.2020063
## Direct and inverse spectral problems for a star graph of Stieltjes strings damped at a pendant vertex
1 College of Mathematics and Information Science, Shaanxi Normal University, Xi'an 710062, China 2 South Ukrainian national Pedagogical University, Staroprtofrankovskaya str., 26, Odessa 65020, Ukraine
* Corresponding author: Vyacheslav Pivovarchik
Received May 2020 Revised August 2020 Published October 2020
Fund Project: The first author is supported in part by NNSF grant 11971284
A spectral problem occurring in description of small transverse vibrations of a star graph of Stieltjes strings is considered. At all but one pendant vertices Dirichlet conditions are imposed which mean that these vertices are clamped. One vertex (the root) can move with damping in the direction orthogonal to the equilibrium position of the strings. We describe the spectrum of such spectral problem. The corresponding inverse problem lies in recovering the values of point masses and the lengths of the intervals between the masses using the spectrum and some other parameters. We propose conditions on a sequence of complex numbers and a collection of real numbers to be the spectrum of a problem we consider and the lengths of the edges, correspondingly.
Citation: Lu Yang, Guangsheng Wei, Vyacheslav Pivovarchik. Direct and inverse spectral problems for a star graph of Stieltjes strings damped at a pendant vertex. Inverse Problems & Imaging, doi: 10.3934/ipi.2020063
##### References:
[1] O. Boyko, O. Martynyuk and V. Pivovarchik, On maximal multiplicity of eigenvalues of finite-dimensional spectral problem on a graph, Methods of Funct. Anal. Topology, 25 (2019), 104-117. Google Scholar [2] J. Genin and J. S. Maybee, Mechanical vibrations trees, J. Math. Anal. Appl., 45 (1974), 746-763. doi: 10.1016/0022-247X(74)90065-1. Google Scholar [3] G. Gladwell, Inverse Problems in Vibration, Kluwer Academic Publishers, Dordrecht, 2004. Google Scholar [4] G. Gladwell, Matrix inverse eigenvalue problems, Dynamical Inverse Problems: Theory and Application, CISM Courses and Lect., SpringerWienNewYork, Vienna, 529 (2011), 1–28. doi: 10.1007/978-3-7091-0696-9_1. Google Scholar [5] F. R. Gantmakher and M. G. Krein, Oscillating Matrices and Kernels and Small Vibrations of Mechanical Systems (in Russian), GITTL, Moscow-Leningrad, (1950), Revised edition, AMS Chelsea Publishing, Providence, RI, 2002. doi: 10.1090/chel/345. Google Scholar [6] V. A. Marchenko, Introduction to The Theory of Inverse Problems of Spectral Analysis (in Russian), Acta, Kharkov, 2005. Google Scholar [7] O. Martynyuk, V. Pivovarchik and C. Tretter, Inverse problem for a damped Stieltjes string from parts of spectra, Appl. Anal., 94 (2015), 2605-2619. doi: 10.1080/00036811.2014.996874. Google Scholar [8] M. Möller and V. Pivovarchik, Damped star graphs of Stieltjes strings, Proc. Amer. Math. Soc., 145 (2017), 1717-1728. doi: 10.1090/proc/13367. Google Scholar [9] M. Möller and V. Pivovarchik, Spectral Theory of Operator Pencils, Hermite-Biehler Functions, and Their Applications, Birkhäuser, Cham, 2015. doi: 10.1007/978-3-319-17070-1. Google Scholar [10] V. Pivovarchik, Existence of a tree of Stieltjes strings corresponding to two given spectra,, J. Phys. A, 42 (2009), 375213, 16 pp. doi: 10.1088/1751-8113/42/37/375213. Google Scholar [11] V. Pivovarchik, N. Rozhenko and C. Tretter, Dirichlet-Neumann inverse spectral problem for a star graph of Stieltjes strings, Linear Algebra Appl., 439 (2013), 2263-2292. doi: 10.1016/j.laa.2013.07.003. Google Scholar [12] V. Pivovarchik and C. Tretter, Location and multiplicities of eigenvalues for a star graph of Stieltjes strings, J. Difference Equ. Appl., 21 (2015), 383-402. doi: 10.1080/10236198.2014.992425. Google Scholar [13] K. Veselić, On linear vibrational systems with one dimensional damping, Appl. Anal., 29 (1988), 1-18. doi: 10.1080/00036818808839770. Google Scholar [14] K. Veselić, On linear vibrational systems with one dimensional damping II, Integr. Equ. Oper. Theory, 13 (1990), 883-897. doi: 10.1007/BF01198923. Google Scholar
show all references
##### References:
[1] O. Boyko, O. Martynyuk and V. Pivovarchik, On maximal multiplicity of eigenvalues of finite-dimensional spectral problem on a graph, Methods of Funct. Anal. Topology, 25 (2019), 104-117. Google Scholar [2] J. Genin and J. S. Maybee, Mechanical vibrations trees, J. Math. Anal. Appl., 45 (1974), 746-763. doi: 10.1016/0022-247X(74)90065-1. Google Scholar [3] G. Gladwell, Inverse Problems in Vibration, Kluwer Academic Publishers, Dordrecht, 2004. Google Scholar [4] G. Gladwell, Matrix inverse eigenvalue problems, Dynamical Inverse Problems: Theory and Application, CISM Courses and Lect., SpringerWienNewYork, Vienna, 529 (2011), 1–28. doi: 10.1007/978-3-7091-0696-9_1. Google Scholar [5] F. R. Gantmakher and M. G. Krein, Oscillating Matrices and Kernels and Small Vibrations of Mechanical Systems (in Russian), GITTL, Moscow-Leningrad, (1950), Revised edition, AMS Chelsea Publishing, Providence, RI, 2002. doi: 10.1090/chel/345. Google Scholar [6] V. A. Marchenko, Introduction to The Theory of Inverse Problems of Spectral Analysis (in Russian), Acta, Kharkov, 2005. Google Scholar [7] O. Martynyuk, V. Pivovarchik and C. Tretter, Inverse problem for a damped Stieltjes string from parts of spectra, Appl. Anal., 94 (2015), 2605-2619. doi: 10.1080/00036811.2014.996874. Google Scholar [8] M. Möller and V. Pivovarchik, Damped star graphs of Stieltjes strings, Proc. Amer. Math. Soc., 145 (2017), 1717-1728. doi: 10.1090/proc/13367. Google Scholar [9] M. Möller and V. Pivovarchik, Spectral Theory of Operator Pencils, Hermite-Biehler Functions, and Their Applications, Birkhäuser, Cham, 2015. doi: 10.1007/978-3-319-17070-1. Google Scholar [10] V. Pivovarchik, Existence of a tree of Stieltjes strings corresponding to two given spectra,, J. Phys. A, 42 (2009), 375213, 16 pp. doi: 10.1088/1751-8113/42/37/375213. Google Scholar [11] V. Pivovarchik, N. Rozhenko and C. Tretter, Dirichlet-Neumann inverse spectral problem for a star graph of Stieltjes strings, Linear Algebra Appl., 439 (2013), 2263-2292. doi: 10.1016/j.laa.2013.07.003. Google Scholar [12] V. Pivovarchik and C. Tretter, Location and multiplicities of eigenvalues for a star graph of Stieltjes strings, J. Difference Equ. Appl., 21 (2015), 383-402. doi: 10.1080/10236198.2014.992425. Google Scholar [13] K. Veselić, On linear vibrational systems with one dimensional damping, Appl. Anal., 29 (1988), 1-18. doi: 10.1080/00036818808839770. Google Scholar [14] K. Veselić, On linear vibrational systems with one dimensional damping II, Integr. Equ. Oper. Theory, 13 (1990), 883-897. doi: 10.1007/BF01198923. Google Scholar
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2019 Impact Factor: 1.373
Article outline | 3,462 | 10,893 | {"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-2021-04 | latest | en | 0.715377 |
https://www.experts-exchange.com/questions/28696451/help-with-cin-string.html | 1,544,675,511,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824448.53/warc/CC-MAIN-20181213032335-20181213053835-00204.warc.gz | 917,893,852 | 13,916 | # help with cin << string
I am attempting to run a while loop:
``````int main() {
double measurement=0;
string unit_of_measurement = " ";
//conversion factors
double meterToCentimeter = 100;
double inchToCentimeter = 2.54;
double footToInches = 12;
double centimeters=0.0;
//define two variables to keep track of which is smallest and largest so far
double smallestSoFar=0.0, largestSoFar=0.0;
cout<<"Enter a float and then hit enter.\n";
vector<double> Values;
unsigned int size_of_values = 0;
while (cin >> measurement >> unit_of_measurement) { // *** Crashes here when ft or in ***
//convert measurement to centimeters
if (unit_of_measurement == "m") centimeters = measurement*meterToCentimeter;
if (unit_of_measurement == "in") centimeters = measurement*inchToCentimeter;
if (unit_of_measurement == "ft") centimeters = (measurement*footToInches)*inchToCentimeter;
Values.push_back(measurement);
//Assign the size of the vector to size_of_values
size_of_values = Values.size();
//Using ranged based for loop
for (double i : Values) {
//if this is the first time around the while loop then the first value is both largest and smallest
if (size_of_values==1) {
cout << "The largest and smallest so far is " << measurement << unit_of_measurement << "\n";
largestSoFar = centimeters;
smallestSoFar = centimeters;
// else determine if measurement is greater than largestSoFar or smaller than smallestSoFar (which were assigned the first time through the loop
} else {
if (centimeters > largestSoFar) {
largestSoFar = centimeters;
}
if (centimeters < smallestSoFar) {
smallestSoFar = centimeters;
}
}
}
cout << largestSoFar << "cm " << " is the largest so far \n" << smallestSoFar << "cm " << " is the smallest so far: \n";
cout << "End of Outside For Loop\n\n";
} // end of for loop
return 0;
}
``````
When I enter a double and "m" (bolded above), the program works as I expect. When I try to use "in" or "ft", the program execution exits the loop and terminates.
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Author Commented:
Clarification: I get the expected output when I enter a double, whitespace and then a string but not when I enter a double, no whitespace, and then a string. When I do that, program execution exits the loop and terminates.
\Commented:
Your cin is parsing based on whitespaces.
If your first entry is not an integer or a double, then the parsing fails.
One way to handle the case of 14.3in is to read those 6 chars into a string, and then parse it yourself.
Author Commented:
Why does 14m work fine then?
\Commented:
I just put your program in VS and 14ft does not crash and the 14 and "ft" are parsed correctly. Looks like VS C++ is parsing the digits up to the first non-digit, and that is the first value going into the double.
If I enter a letter for the first character instead of a number, then the program crashed.
Experts Exchange Solution brought to you by | 820 | 3,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-51 | latest | en | 0.7816 |
http://portaldelfreelancer.com/Rhode-Island/is-variance-the-same-as-standard-error.html | 1,545,166,572,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829812.88/warc/CC-MAIN-20181218204638-20181218230638-00540.warc.gz | 234,869,929 | 7,589 | Address 728 Dexter St, Central Falls, RI 02863 (401) 722-2177
is variance the same as standard error Harmony, Rhode Island
Not the answer you're looking for? Notice that it is normally distributed with a mean of 10 and a standard deviation of 3.317. Where are sudo's insults stored? Therefore, the standard deviation is reported as the square root of the variance and the units then correspond to those of the data set.
variance mathematical-statistics standard-deviation share|improve this question edited Jan 27 at 17:53 gung 74.2k19160309 asked Aug 26 '12 at 12:31 Le Max 82431520 1 stats.stackexchange.com/questions/118/… –whuber♦ Aug 26 '12 at 22:20 A natural way to describe the variation of these sample means around the true population mean is the standard deviation of the distribution of the sample means. The standard deviation, as the square root of the variance gives a value that is in the same units as the original values, which makes it much easier to work with Moreover, this formula works for positive and negative ρ alike.[10] See also unbiased estimation of standard deviation for more discussion.
And it is easier to use algebra on squares and square roots than absolute values, which makes the standard deviation easy to use in other areas of mathematics. Previous company name is ISIS, how to list on CV? The mean square error for an estimate equals the variance + the squared bias. For any random sample from a population, the sample mean will usually be less than or greater than the population mean.
Count the number of values between these two boundaries. A medical research team tests a new drug to lower cholesterol. Furthermore, why do you really need a standard deviation? Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view Measures of Variance Common Measures of Variance Range The range is the difference between the high and low values.
The margin of error is express as a multiple of the standard deviation of the estimate. The distribution of the differences between means is the sampling distribution of the difference between means. Browse other questions tagged variance mathematical-statistics standard-deviation or ask your own question. To estimate the standard error of a student t-distribution it is sufficient to use the sample standard deviation "s" instead of σ, and we could use this value to calculate confidence
This is not necessary, but it makes it easier. Edwards Deming. How to photograph distant objects (10km)? but don't tell them!
The unbiased estimate of population variance calculated from a sample is: [xi is the ith observation from a sample of the population, x-bar is the sample mean, n (sample size) -1 share|improve this answer answered Aug 26 '12 at 12:37 Peter Flom♦ 57.5k966150 13 yeah thats the mathematical way to explain these two parameters, BUT whats the logical explenation? Princeton, NJ: Van Nostrand, 1962. Cambridge, England: Cambridge University Press, 1992.
Are most Earth polar satellites launched to the South or to the North? On the 1st April, you dissected strawberry crowns and counted flower initials. In fact this method is a similar idea to distance between points, just applied in a different way. They each have different purposes.
It is found by taking the square root of the variance and solves the problem of not having the same units as the original data. Example: if our 5 dogs are just a sample of a bigger population of dogs, we divide by 4 instead of 5 like this: Sample Variance = 108,520 / 4 = The researchers report that candidate A is expected to receive 52% of the final vote, with a margin of error of 2%. For girls, the mean is 165 and the variance is 64.
As an example of the use of the relative standard error, consider two surveys of household income that both result in a sample mean of $50,000. SD is calculated as the square root of the variance (the average squared deviation from the mean). National Center for Health Statistics typically does not report an estimated mean if its relative standard error exceeds 30%. (NCHS also typically requires at least 30 observations – if not more Procedure for finding Find the range Divide it by four Formula Pearson's Index of Skewness Pearson's index of skewness can be used to determine whether the data is symmetric or skewed. share|improve this answer answered Aug 26 '12 at 12:37 Peter Flom♦ 57.5k966150 13 yeah thats the mathematical way to explain these two parameters, BUT whats the logical explenation? If you report one, you don't need to report the other –Peter Flom♦ Aug 26 '12 at 12:47 We need both: standard deviation is good for interpretation, reporting. Colwell Open topic with navigation Variance, Standard Deviation and Spread The standard deviation of the mean (SD) is the most commonly used measure of the spread of values in a If you calculate the two values, it is clear that you get the standard deviation out of the variance, but what does that mean in terms of the distribution you are In Excel, the standard deviation can be calculated using the equation =STDEV(range of cells). Only for a moderately large number of samples (and depending on the estimators) the two approach each other. The variance of a sampled subset of observations is calculated in a similar manner, using the appropriate notation for sample mean and number of observations. Only for a moderately large number of samples (and depending on the estimators) the two approach each other. We can do this by squaring each deviation (as we do in the variance or standard deviation) or by taking the absolute value (as we do in the mean absolute deviation). Sampling distribution of the difference between mean heights. the known standard deviation is used to normalize a sample mean for the$z$test that the mean differs from$0\$ or the sample standard deviation is used to normalize the | 1,264 | 5,894 | {"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} | 3.46875 | 3 | CC-MAIN-2018-51 | latest | en | 0.891455 |
https://planet-earth-2017.com/2014/08/15/the-true-drop-in-the-magnetic-field-strength/ | 1,675,223,353,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499899.9/warc/CC-MAIN-20230201013650-20230201043650-00697.warc.gz | 430,568,492 | 26,079 | ## The True Drop in the Magnetic Field Strength
The weakening of the magnetic field is inconsistent. By comparing the magnetic field strength in the year 1900, where the magnetic pole resided then, to its strength at its current location in 2014, we find that it has dropped by 7% in the Arctic and 4% in Antarctica. The following questions could simply emerge: Why has the magnetic field at its north pole in Antarctica weakened by 4% since the year 1900? Why is the weakening percent not the same in the North and South? and What is the true percentage of the weakening of the field over a century?
Before answering any of these questions, I bring for discussion the Earth 3-magnet model that I proposed earlier. While there is a permanent dipole magnet that is free to swivel, flip and spin, there are 2 induced south polarity magnetic fields that result from the spiralling electrons that flow in the Outer Core. The two induced magnetic fields emerge on the North and South surfaces of Earth around the Axis of Rotation at the orthographic projection of the volume that occupies the space between the Outer Core and Inner Core. The strength of the induced south polarity magnetic fields follows the speed of the spiralling electrons. The velocity of those electrons follow the formula V= W x D; where ‘W’ is the angular velocity of the planet and ‘D’ is the distance of the flowing electrons from the centre of Earth. Keeping in mind that the Inner Core radius is 1,200 km and that the Outer Core radius is 3,000 km, the spiralling electrons pick up speed while flowing in the outward direction inside the Outer Core on the equatorial plane. The resultant induced magnetic field varies in strength on the orthographic projection band starting from 1,214 km to 3,141 km distance from the geographic pole or Axis of Rotation when we consider the Earth’s surface curvature. It has minimum strength at 1,214 km and maximum strength at 3,141 km from the Axis of Rotation. The patches of south polarity magnetic field in Antarctica are called ‘Plasmoids’. The counterpart in the Arctic region are not called anything since it is not easy to tell if the south magnet polarity field that is being measured in the North is a result of a permanent dipole magnet or an induced one.
To answer the first question of why the magnetic field at its north magnetic pole in Antarctica has weakened by 4% since the year 1900? Let us track the north magnetic pole wandering path in Antarctica. In the year 1900 it was 2,025 km away from the Geographic Pole. In the year 2014 it reached 2,861 km away from the Geographic Pole. Such a move brings the north magnetic pole closer to the region of the orthographic projection where higher intensity Plasmoids exist and explains the drop of 4% of the north magnetic field force lines as they bond with Plasmoids even at lower geographical layers and before emergence to the surface of Earth. The rate of Plasmoids intensity could easily be deducted by:
• Comparing the increase of the south polarity magnetic field intensity or for better words the drop of the north magnetic field intensity of 4%; which is equal to 2,435 nT (nano Tesla) over the period between the year 1900 and the year 2014;
• To the increase of the north magnetic pole distance from the Axis of Rotation from 2,025 to 2,861 over the same period between the year 1900 and the year 2014.
• We find out that the rate of Plasmoids intensity increase is 3 nT/ km as we move away from 1,200 km to 3,141 km from the Axis of Rotation.
This helps us to isolate the measure of the true strength of the dipole magnetic field from the Plasmoids in both Antarctic and the Arctic regions as shown in the following table, which indicates that only 81% of the emerging magnetic field force lines form the Antarctic find its way to converge at the south magnetic pole in the Arctic region. It also shows that from the year 1900 to the year 2014 the converging magnetic field force in the south magnetic pole in the Arctic region has dropped by 2%.
While the original drop of 19% is explained by the planetary magnetic pull of the different celestial bodies of the solar system, the additional drop of 2% may not be explained; unless, the motion of Tyche or a 9th planet is proven to be true by NASA and/or other further alignment in the planetary system. The weakening of the magnetic field on the western and southern hemisphere could then be summarized to come from 3 collaborating factors:
• Both magnetic poles are moving eastward leaving the magnetic field in the western/ southern hemisphere to travel larger distances and areas from pole to pole. As such, the intensity becomes weakest especially mid distance between the two magnetic poles on the western/ southern hemisphere or what is known as the South Atlantic Anomaly.
• The magnetic field has 2% of its magnetic field force lines not converging at the south magnetic pole in the Arctic region. This means that such a percentage has been strayed into space.
• The same magnetic pull that caused the 2% of the magnetic field force lines to reroute into space instead of circumventing the planet from pole to pole has also caused the rest of the magnetic field force lines to get protruded and travel further distance into space before converging back to the south magnetic pole in the Arctic region.
• All such losses and opening up of distance between the magnetic field force lines have resulted in penetration of more charged particles, such as Protons, that arrive from the Sun to the surface of the Earth. The Protons have also gained higher speed while spiralling along the magnetic field force lines in the Thermosphere layer; causing a global increase in thermal radiation as a result of higher impacts between such sped Protons and one another at the Thermosphere layer. Global warming of the planet is thus witnessed. | 1,257 | 5,884 | {"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-2023-06 | longest | en | 0.909962 |
https://kullabs.com/class-11/physics-11/change-of-state/latent-heat-of-fusion-and-vaporization | 1,632,857,565,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060882.17/warc/CC-MAIN-20210928184203-20210928214203-00347.warc.gz | 396,081,024 | 48,732 | ## Latent Heat of Fusion and Vaporization
Subject: Physics
#### Overview
A substance can exist in three states: solid, liquid and gas. A change in state from solid to liquid on heating is called fusion or melting. This note provides us with an information on the latent heat of fusion and vaporization.
### Change of State
A substance can exist in three states: solid, liquid and gas. A change in state from solid to liquid on heating is called fusion or melting. When some heat is supplied to a solid, it starts melting as it attains required temperature and the temperature remains constant until whole the solid melts. This constant temperature is called melting point of the solid. In the same way, on cooling a liquid, it changes into solid as it loses the sufficient amount of heat and this is called solidification or freezing. During this solidification, the temperature remains constant which is freezing point. The freezing point is different from different liquids. In general crystalline substances have same melting and freezing point but in non-crystalline solid melting and freezing are not same. For an example: butter has melting point 30oCand freezing point 22 oC.
#### Latent heat
Latent heat of a substance is the amount of heat required to change the state of a unit mass of the substance from solid to liquid or from liquid to vapour without changing the temperature. During heating the substance, the heat has been used up in changing the state; temperature remaining the same till the state of the entire substance has changed. This energy is used up in separating the molecules farther apart and as work must be done against the required to change the state of a substance. That is why heat energy spent in a change of state is called latent heat i.e. heat which is hidden from the thermometer. Latent heat is represented by L and is measured in Jkg-1 or calg-1. There are two types of latent heat i.e. latent heat of fusion and latent heat of vaporization.
Latent Heat of fusion
Latent heat of fusion is defined as the amount of heat required to change a unit mass of a substance from the solid state to liquid state at a constant temperature. The latent heat of fusion of ice is defined as the amount of heat required to change 1 g of ice from 0to water at the same temperature. For ice, its value is 3.36 105 J Kg-1 in SI-units or 80cal g-1 in CGS-system.
Latent Heat of Vaporization
When a liquid is heated, it starts boiling and is changed into the vapour state. During the process, the temperature of the liquid remain constant and the amount of heat energy is utilized to change its' state only i.e. from a liquid state to vapour state which is called latent heat of vaporization.
Latent heat of vaporization of liquid is defined as the amount of heat required to change the unit mass of liquid at boiling point into vapour at the same temperature. So, latent heat of steam is defined as the amount of heat required to change a unit mass of water from100to steam at the same temperature. The latent heat of vaporization is 2.26 × 106 Jkg-1or 540 calg-1.
When heat is supplied to a piece of ice of 1 g at -10oC, the change in temperature is shown in the figure. The temperature of ice first increases from -10oC to 0 oCand it melts completely when 80 calories of heat is given to it, temperature remain same as 0oC. To convert the water into the steam, 540 cal of heat energy is supplied.
#### Determination of Latent Heat of Fusion of Ice by the Method of Mixture
Following are the steps to determine the latent heat of fusion of ice:
1. A calorimeter with stirrer is taken and it is weighted.
2. Some water is poured into to calorimeter then the mass and temperature of a calorimeter, stirrer and water is taken.
3. The small piece of ice is taken and the temperature noted say (0 oC).
4. The piece of ice is dropped into the calorimeter.
5. The mixture is stirred until all the ice is melted then the final temperature and mass of mixture is taken.
Let,
m1= mass of calorimeter and stirrer
m2 = mass of calorimeter, stirrer and water
mw = m1 - m2 = mass of water only
m3 = mass of calorimeter, stirrer, water and ice
m= m3 - m2 = mass of ice only
= temperature of water, stirrer and calorimeter
= temperature of ice, water and calorimeter
= temperature of mixture
sc = specific heat capacity of a calorimeter
sw = specific heat capacity of water
lf = Latent heat of fusion = ?
Heat loss by water and calorimeter, $$= m_ws_w(\theta_1 - \theta) + m_1s_c(\theta_1 - \theta)$$
$$=(m_ws_w + m_1s_c)(\theta_1 - \theta)$$
Heat gain by ice when it changes from 00 ice to 00 water and 00 water to θ0 water $= m l_f +m_1s_w(\theta_1 -0)$
From principle of calorimetry
\begin{align*} \text {Heat gain} &= \text{Heat loss} \\(m l_f +m_1s_w(\theta_1 -0) &= m_ws_w(\theta_1 - \theta) + m_1s_c (\theta_1 - \theta) \\m l_f &= (m_ws_w + m_1s_c)(\theta_1 - \theta)- ms_w(\theta_1 -0) \\\therefore l_f &= \frac {(m_ws_w + m_1s_c)(\theta_1 - \theta)- ms_w \theta_1 }{m} \\ \end{align*}
##### Things to remember
A substance can exist in three states: solid, liquid and gas.
A change in state from solid to liquid on heating is called fusion or melting.
On cooling a liquid, it changes into solid as it loses the sufficient amount of heat and this is called solidification or freezing.
Latent heat of a substance is the amount of heat required to change the state of a unit mass of the substance from solid to liquid or from liquid to vapour without changing the temperature.
Latent heat of fusion is defined as the amount of heat required to change a unit mass of a substance from the solid state to liquid state at a constant temperature.
Latent heat of vaporization of liquid is defined as the amount of heat required to change the unit mass of liquid at boiling point into vapour at the same temperature.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society. | 1,476 | 6,158 | {"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} | 2.71875 | 3 | CC-MAIN-2021-39 | longest | en | 0.918897 |
https://math-faq.com/category/college-math/page/2/ | 1,701,451,263,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00018.warc.gz | 443,500,437 | 13,174 | ## Section 3.2 Dimensional Analysis
Since many measurements in the United States are in the English system, you may often be required to convert from English units like pounds to SI units like kilograms. Luckily, we can use facts (like 1 kilogram is approximately 2.2 pounds) to convert between different measurement systems.
Our objectives for this section are to
• use dimensional analysis to convert from one system of measurement to another system of measurement.
Section 3.2 Workbook (PDF) – 9/7/19
Section 3.2 Practice Solutions
Chapter 3 Practice Solutions (PDF)
## Section 1.6 The Normal Distribution
You have probably heard of the “bell” curve, but you may not be familiar with what is really is. Some students may ask their instructors, “Do you grade on a curve?” Before you ask that question, make sure you understand what it is and whether it may be to your advantage.
The normal distribution (or bell curve) refers to a histogram that looks like the graphic above. It indicates how data is distributed with respect to each other. As you might guess, the peak of the curve corresponds to the mean and its spread matches the standard deviation.
In this section, you learn about the normal distribution and how to apply it to everyday problems. Our objectives are to
• describe the basic properties of the normal distribution,
• apply the 68-95-99.7 rule to a set of data,
• relate the area under a normal curve to z-scores,
• convert between raw scores and z-scores, use z-scores to compare data.
Armed with this information, you can utilize the normal distribution to answer questions about the relationship of a specific data value to the data values as a whole.
Use the workbook and videos below to help you master these objectives. Make sure you go through the practice problems since they are reflective of the types of problems you might find on a quiz or exam.
Section 1.6 Practice Solutions
Chapter 1 Practice Solutions (PDF)(9-17-19)
In the video below, think of probability of a data value as meaning the percentage of data values. We are using a cumulative from the mean table in our class (the second one in the video)
## Section 5.5 Amortization
At some time in your life, you will make a large purchase such as a home or car. You may need to borrow money to make this purchase and pay back the loan with a series of regular payments. This process is called amortization.
Our goals in this section are to
• Calculate the payment on an amortized loan.
• Construct an amortization schedule.
• Calculate the present value of an annuity.
Use the workbook and videos below to help you accomplish these objectives.
Section 5.5 Workbook (PDF) 11/2/19
Practice Solutions
Videos
## Section 5.4 Annuities
When you need to save money for a large purchase or save money for retirement, you typically make regular payments into an account that earns interest. This series of payments is called an annuity.
If the regular payments are made at the beginning of a period of time, the annuity is called an annuity due. If the regular payments are made at the end of a period of time, the annuity is called an ordinary annuity. In this section we will examine the relationship between the future value of the annuity, the size of the payment, and the period of time over which the payments are made.
Our objectives in this section are to
• Calculate the future value of an ordinary annuity.
• Calculate different quantities in a sinking fund.
Use the workbook and videos below to help you accomplish these objectives.
Section 5.4 Workbook (PDF) 11/2/19
Practice Solutions
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# Lab+5++Sample++Prelab+Quiz - EAS 1600 Lab 5 Heat Transfer...
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EAS 1600 Lab 5 “Heat Transfer” Sample Prelab Quiz (note: actual quiz may differ) Given: ¾ One calorie is defined as the amount of heat (or energy) needed to change the temperature of 1 gram of water by 1 degree Celsius ¾ 1 calorie = 4.18 joules ¾ The specific latent heat of vaporization at 100 ° C is 2260 J/g ¾ The density of liquid water is assumed to be 1 g cm -3 ¾ The "heat of fusion" is amount of energy required to change a gram of a substance from the solid to the liquid state without changing its temperature. For water the heat of fusion is 334 J/g Question 1. Calculate the amount of heat required to melt
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## This note was uploaded on 02/08/2011 for the course EAS 1600 taught by Professor Jimstjohn during the Fall '08 term at Georgia Tech.
Ask a homework question - tutors are online | 277 | 1,030 | {"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-2017-22 | longest | en | 0.883164 |
https://amara.org/subtitles/khm6vRLb094F/zh-tw/6/download/Dividing%20fractions.zh-tw.ssa | 1,726,602,121,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00757.warc.gz | 77,263,361 | 4,420 | [Script Info] Title: [Events] Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text Dialogue: 0,0:00:00.81,0:00:03.11,Default,,0000,0000,0000,,歡迎來到分數除法的教學課程 Dialogue: 0,0:00:03.11,0:00:04.49,Default,,0000,0000,0000,,讓我們開始吧 Dialogue: 0,0:00:04.49,0:00:06.64,Default,,0000,0000,0000,,在我教你們之前，事實上，我想要 Dialogue: 0,0:00:06.64,0:00:09.34,Default,,0000,0000,0000,,用不同的模組來教--我將教你 Dialogue: 0,0:00:09.34,0:00:11.74,Default,,0000,0000,0000,,如何做分數除法的技巧 Dialogue: 0,0:00:11.74,0:00:13.74,Default,,0000,0000,0000,,而你會發現這其實 Dialogue: 0,0:00:13.74,0:00:16.03,Default,,0000,0000,0000,,不會比分數乘法更難 Dialogue: 0,0:00:16.03,0:00:21.41,Default,,0000,0000,0000,,如果我問你，1/2除以1/2，不論何時 Dialogue: 0,0:00:21.41,0:00:25.11,Default,,0000,0000,0000,,當數字除以分數，或除以任何數字的時候 Dialogue: 0,0:00:25.11,0:00:29.96,Default,,0000,0000,0000,,就等於乘以這個數字的倒數 Dialogue: 0,0:00:29.96,0:00:36.67,Default,,0000,0000,0000,,所以，1/2除以1/2，就等於1/2乘以2/1 Dialogue: 0,0:00:36.67,0:00:44.99,Default,,0000,0000,0000,,我們剛剛把第二個1/2變為倒數 Dialogue: 0,0:00:44.99,0:00:47.63,Default,,0000,0000,0000,,我們從乘法模組知道，1/2乘以2/1 Dialogue: 0,0:00:47.63,0:00:51.11,Default,,0000,0000,0000,,就等於2/2 Dialogue: 0,0:00:51.11,0:00:53.56,Default,,0000,0000,0000,,或者就等於1 Dialogue: 0,0:00:53.56,0:00:56.02,Default,,0000,0000,0000,,這很合理，事實上，任何數字除以自己 Dialogue: 0,0:00:56.02,0:00:58.75,Default,,0000,0000,0000,,就等於1 Dialogue: 0,0:00:58.75,0:01:03.22,Default,,0000,0000,0000,,1/2除以1/2等於1，就像是5除以5等於1 Dialogue: 0,0:01:03.22,0:01:05.24,Default,,0000,0000,0000,,100除以100等於1 Dialogue: 0,0:01:05.24,0:01:06.85,Default,,0000,0000,0000,,這不是一個新的定律 Dialogue: 0,0:01:06.85,0:01:08.97,Default,,0000,0000,0000,,事實上，你過去總是如此做 Dialogue: 0,0:01:16.29,0:01:20.56,Default,,0000,0000,0000,,例如2除以2，你知道是1，但這不就像是2乘以2的倒數 Dialogue: 0,0:01:20.56,0:01:24.21,Default,,0000,0000,0000,,結果也是1 Dialogue: 0,0:01:24.21,0:01:24.95,Default,,0000,0000,0000,,我將演示給你看 Dialogue: 0,0:01:24.95,0:01:26.99,Default,,0000,0000,0000,,事實上，讓我多秀幾個例子 Dialogue: 0,0:01:26.99,0:01:31.34,Default,,0000,0000,0000,,讓你了解分數除法真的不是個新的觀念 Dialogue: 0,0:01:31.34,0:01:34.84,Default,,0000,0000,0000,,整個概念就是乘以它的倒數 Dialogue: 0,0:01:34.84,0:01:40.54,Default,,0000,0000,0000,,如果我問你12除以4是多少? Dialogue: 0,0:01:40.54,0:01:42.65,Default,,0000,0000,0000,,當然我們知道答案是什麼，但是我要 Dialogue: 0,0:01:42.65,0:01:50.64,Default,,0000,0000,0000,,演示給你看，其實這跟12乘以1/4是一樣的 Dialogue: 0,0:01:50.64,0:01:56.23,Default,,0000,0000,0000,,12/1乘以1/4等於12/4，等於3 Dialogue: 0,0:01:56.23,0:01:59.48,Default,,0000,0000,0000,,而12/4就是12除以4的另一種書寫方式 Dialogue: 0,0:01:59.48,0:02:02.54,Default,,0000,0000,0000,,這就有點像是用比較複雜的方法來得到一樣的結果 Dialogue: 0,0:02:02.54,0:02:04.99,Default,,0000,0000,0000,,但是，我只是想讓你知道我們正在做的 Dialogue: 0,0:02:04.99,0:02:07.97,Default,,0000,0000,0000,,這個模組不是新的，其實就像是我們過去一直在做的 Dialogue: 0,0:02:07.97,0:02:09.32,Default,,0000,0000,0000,,當我們除以一個數字 Dialogue: 0,0:02:09.32,0:02:11.36,Default,,0000,0000,0000,,除法都是一樣的 Dialogue: 0,0:02:11.36,0:02:14.31,Default,,0000,0000,0000,,當除以一個數字，其實就等同於乘以 Dialogue: 0,0:02:14.31,0:02:15.96,Default,,0000,0000,0000,,那個數字的倒數 Dialogue: 0,0:02:15.96,0:02:19.88,Default,,0000,0000,0000,,復習一下倒數，當我有一個數字A Dialogue: 0,0:02:19.88,0:02:28.07,Default,,0000,0000,0000,,它的倒數--就是1/A Dialogue: 0,0:02:28.07,0:02:36.29,Default,,0000,0000,0000,,所以2/3的倒數是3/2，或者5的倒數 Dialogue: 0,0:02:36.29,0:02:39.67,Default,,0000,0000,0000,,因為5就等於5/1，所以5的倒數是1/5 Dialogue: 0,0:02:43.32,0:02:46.48,Default,,0000,0000,0000,,讓我們來做一些分數除法的問題吧 Dialogue: 0,0:02:46.48,0:02:49.27,Default,,0000,0000,0000,,2/3除以5/6是多少? Dialogue: 0,0:02:56.34,0:03:05.97,Default,,0000,0000,0000,,我們知道，這就等同於2/3乘以6/5 Dialogue: 0,0:03:05.97,0:03:09.23,Default,,0000,0000,0000,,就等於12/15 Dialogue: 0,0:03:09.23,0:03:14.57,Default,,0000,0000,0000,,我們可以把分子及分母同時除以3，就等於4/5 Dialogue: 0,0:03:14.57,0:03:22.90,Default,,0000,0000,0000,,7/8除以1/4等於多少? Dialogue: 0,0:03:22.90,0:03:30.52,Default,,0000,0000,0000,,這就等同於7/8乘以4/1 Dialogue: 0,0:03:30.52,0:03:32.82,Default,,0000,0000,0000,,記住，我剛剛翻轉了1/4 Dialogue: 0,0:03:32.82,0:03:36.84,Default,,0000,0000,0000,,除以1/4就等同於乘以4/1 Dialogue: 0,0:03:36.84,0:03:38.23,Default,,0000,0000,0000,,這就是你要解答分數除法的方式 Dialogue: 0,0:03:38.23,0:03:39.99,Default,,0000,0000,0000,,接著我們可以用一些在乘法模組 Dialogue: 0,0:03:39.99,0:03:41.48,Default,,0000,0000,0000,,學到的小捷徑 Dialogue: 0,0:03:41.48,0:03:42.95,Default,,0000,0000,0000,,8除以4等於2 Dialogue: 0,0:03:42.95,0:03:44.80,Default,,0000,0000,0000,,4除以4等於1 Dialogue: 0,0:03:44.80,0:03:47.45,Default,,0000,0000,0000,,所以答案就是7/2 Dialogue: 0,0:03:47.45,0:03:49.90,Default,,0000,0000,0000,,或者如果你想要把它寫成帶分數 Dialogue: 0,0:03:49.90,0:03:51.20,Default,,0000,0000,0000,,7/2是一個假分數 Dialogue: 0,0:03:51.20,0:03:53.44,Default,,0000,0000,0000,,假分數，它的分子 Dialogue: 0,0:03:53.44,0:03:54.83,Default,,0000,0000,0000,,會大於分母 Dialogue: 0,0:03:54.83,0:03:58.67,Default,,0000,0000,0000,,如果你想要把它寫成帶分數 Dialogue: 0,0:03:58.67,0:04:03.68,Default,,0000,0000,0000,,7除2等於3餘1，所以就是3 1/2 Dialogue: 0,0:04:03.68,0:04:04.44,Default,,0000,0000,0000,,兩種方式都可以 Dialogue: 0,0:04:04.44,0:04:05.99,Default,,0000,0000,0000,,我比較傾向用這個方式 Dialogue: 0,0:04:05.99,0:04:07.80,Default,,0000,0000,0000,,因為它比較容易做計算 Dialogue: 0,0:04:07.80,0:04:10.13,Default,,0000,0000,0000,,讓我們再多做幾題，或者是至少在接下來4~5分鐘裡面 Dialogue: 0,0:04:10.13,0:04:13.83,Default,,0000,0000,0000,,可以盡量多做一些 Dialogue: 0,0:04:13.83,0:04:23.85,Default,,0000,0000,0000,,-2/3除以5/2等於多少 Dialogue: 0,0:04:23.85,0:04:29.11,Default,,0000,0000,0000,,再一次，這就等於-2/3 Dialogue: 0,0:04:29.11,0:04:34.85,Default,,0000,0000,0000,,乘以多少? Dialogue: 0,0:04:34.85,0:04:40.11,Default,,0000,0000,0000,,就等於乘以5/2的倒數，就是2/5 Dialogue: 0,0:04:40.11,0:04:45.63,Default,,0000,0000,0000,,所以就等於-4/15 Dialogue: 0,0:04:45.63,0:04:52.30,Default,,0000,0000,0000,,3/2除以1/6等於多少? Dialogue: 0,0:04:52.30,0:04:59.85,Default,,0000,0000,0000,,這就等於3/2乘以6/1，就等於9 Dialogue: 0,0:05:09.61,0:05:11.28,Default,,0000,0000,0000,,我想你現在可能已經了解了 Dialogue: 0,0:05:11.28,0:05:12.95,Default,,0000,0000,0000,,讓我們看看，再多做幾題 Dialogue: 0,0:05:12.95,0:05:16.29,Default,,0000,0000,0000,,當然，你可以隨時暫停，再重新看一次這個課程 Dialogue: 0,0:05:16.29,0:05:19.42,Default,,0000,0000,0000,,然後再把看不懂的地方重看一遍 Dialogue: 0,0:05:19.42,0:05:27.24,Default,,0000,0000,0000,,讓我們來做 -5/7除以10/3 Dialogue: 0,0:05:27.24,0:05:33.88,Default,,0000,0000,0000,,這就等同於-5/7乘以3/10 Dialogue: 0,0:05:33.88,0:05:35.42,Default,,0000,0000,0000,,我剛剛把它乘以倒數 Dialogue: 0,0:05:35.42,0:05:38.12,Default,,0000,0000,0000,,這就是我剛剛一直重覆在做的 Dialogue: 0,0:05:38.12,0:05:40.18,Default,,0000,0000,0000,,-5乘以3 Dialogue: 0,0:05:40.18,0:05:42.61,Default,,0000,0000,0000,,-15 Dialogue: 0,0:05:42.61,0:05:47.35,Default,,0000,0000,0000,,7乘以10等於70 Dialogue: 0,0:05:47.35,0:05:49.90,Default,,0000,0000,0000,,如果我們把分子及分母 Dialogue: 0,0:05:49.90,0:05:56.05,Default,,0000,0000,0000,,同除以5，我們可以得到 -3/14 Dialogue: 0,0:05:56.05,0:05:57.50,Default,,0000,0000,0000,,我們也可以在這邊就這樣做 Dialogue: 0,0:05:57.50,0:05:59.89,Default,,0000,0000,0000,,我們可以除以5，得到2，同樣可以得到 Dialogue: 0,0:05:59.89,0:06:02.51,Default,,0000,0000,0000,,-3/14 Dialogue: 0,0:06:02.51,0:06:05.42,Default,,0000,0000,0000,,讓我們再做一兩個題目 Dialogue: 0,0:06:05.42,0:06:06.63,Default,,0000,0000,0000,,我想你已經有點了解了 Dialogue: 0,0:06:06.63,0:06:09.60,Default,,0000,0000,0000,,如果1/2除以-3 Dialogue: 0,0:06:14.50,0:06:14.96,Default,,0000,0000,0000,,啊哈 Dialogue: 0,0:06:14.96,0:06:17.94,Default,,0000,0000,0000,,所以，當一個分數除以一個正整數 Dialogue: 0,0:06:17.94,0:06:19.73,Default,,0000,0000,0000,,或者整數時會如何呢? Dialogue: 0,0:06:19.73,0:06:22.97,Default,,0000,0000,0000,,我們知道任何正整數可以被寫做分數 Dialogue: 0,0:06:22.97,0:06:29.01,Default,,0000,0000,0000,,這就等於1/2除以-3/1 Dialogue: 0,0:06:29.01,0:06:33.87,Default,,0000,0000,0000,,而除以一個分數就等同於乘以 Dialogue: 0,0:06:33.87,0:06:37.43,Default,,0000,0000,0000,,它的倒數 Dialogue: 0,0:06:37.43,0:06:42.15,Default,,0000,0000,0000,,所以-3/1的倒數是-1/3 Dialogue: 0,0:06:42.15,0:06:45.20,Default,,0000,0000,0000,,所以就等於-1/6 Dialogue: 0,0:06:45.20,0:06:46.04,Default,,0000,0000,0000,,讓我們換個方式做 Dialogue: 0,0:06:46.04,0:06:51.88,Default,,0000,0000,0000,,如果-3除以1/2呢? Dialogue: 0,0:06:51.88,0:06:52.50,Default,,0000,0000,0000,,一樣 Dialogue: 0,0:06:52.50,0:07:00.37,Default,,0000,0000,0000,,-3就等於-3/1，除以1/2 Dialogue: 0,0:07:00.37,0:07:07.94,Default,,0000,0000,0000,,就等於-3/1乘以2/1 Dialogue: 0,0:07:07.94,0:07:12.01,Default,,0000,0000,0000,,就等於 -6/1，等於-6 Dialogue: 0,0:07:12.01,0:07:15.81,Default,,0000,0000,0000,,現在，讓我給你一些理解 Dialogue: 0,0:07:17.35,0:07:19.73,Default,,0000,0000,0000,,為什麼這樣做會行得通 Dialogue: 0,0:07:19.73,0:07:24.24,Default,,0000,0000,0000,,如果說2除以1/3 Dialogue: 0,0:07:24.24,0:07:27.65,Default,,0000,0000,0000,,我們知道這就等於2/1乘以 Dialogue: 0,0:07:27.65,0:07:30.12,Default,,0000,0000,0000,,3/1，就等於6 Dialogue: 0,0:07:30.12,0:07:32.70,Default,,0000,0000,0000,,所以2、1/3、6之間有何關聯呢? Dialogue: 0,0:07:32.70,0:07:33.69,Default,,0000,0000,0000,,讓我們這樣看這件事 Dialogue: 0,0:07:33.69,0:07:36.93,Default,,0000,0000,0000,,如果我有兩片比薩 Dialogue: 0,0:07:36.93,0:07:38.66,Default,,0000,0000,0000,,我有兩片比薩 Dialogue: 0,0:07:38.66,0:07:41.52,Default,,0000,0000,0000,,這裡是我的兩片比薩 Dialogue: 0,0:07:41.52,0:07:42.53,Default,,0000,0000,0000,,2在這邊 Dialogue: 0,0:07:42.53,0:07:45.05,Default,,0000,0000,0000,,所以如果我有兩片比薩，而我想要將它們 Dialogue: 0,0:07:45.05,0:07:48.08,Default,,0000,0000,0000,,分成很多個1/3片的比薩 Dialogue: 0,0:07:48.08,0:07:50.60,Default,,0000,0000,0000,,所以我把每片比薩都分成三等份 Dialogue: 0,0:07:50.60,0:07:52.86,Default,,0000,0000,0000,,我將要畫個小賓士符號 Dialogue: 0,0:07:52.86,0:07:57.05,Default,,0000,0000,0000,,所以我把每片比薩分成三等份，對吧? Dialogue: 0,0:07:57.05,0:07:58.21,Default,,0000,0000,0000,,我有幾個1/3片的比薩呢? Dialogue: 0,0:07:58.21,0:08:02.92,Default,,0000,0000,0000,,讓我們看看，1、2、3、4、5、6 Dialogue: 0,0:08:02.92,0:08:04.80,Default,,0000,0000,0000,,我有6個1/3片的比薩 Dialogue: 0,0:08:04.80,0:08:08.14,Default,,0000,0000,0000,,所以你可能想要坐下，然後思考一下 Dialogue: 0,0:08:08.14,0:08:12.85,Default,,0000,0000,0000,,但我想這可能對你來說有些許意義 Dialogue: 0,0:08:12.85,0:08:17.19,Default,,0000,0000,0000,,讓我們再做一題讓你的腦袋累一下 Dialogue: 0,0:08:17.19,0:08:25.75,Default,,0000,0000,0000,,假設-7/2除以-4/9 Dialogue: 0,0:08:25.75,0:08:30.58,Default,,0000,0000,0000,,這就等於-7/2乘以 Dialogue: 0,0:08:30.58,0:08:33.72,Default,,0000,0000,0000,,-9/4，對吧? Dialogue: 0,0:08:33.72,0:08:37.95,Default,,0000,0000,0000,,我剛剛乘以-4/9的倒數 Dialogue: 0,0:08:37.95,0:08:41.22,Default,,0000,0000,0000,,9乘以7等於，-7乘以-9 Dialogue: 0,0:08:41.22,0:08:47.80,Default,,0000,0000,0000,,等於63，而2乘以4等於8 Dialogue: 0,0:08:47.80,0:08:51.46,Default,,0000,0000,0000,,我想你現在已經知道如何做分數的除法 Dialogue: 0,0:08:51.46,0:08:55.96,Default,,0000,0000,0000,,你可以試試 Dialogue: 0,0:08:55.96,0:08:57.31,Default,,0000,0000,0000,,分數除法的模組 Dialogue: 0,0:08:57.31,0:08:58.89,Default,,0000,0000,0000,,祝你學習愉快! | 6,085 | 10,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-38 | latest | en | 0.230713 |
http://www.programmingwithbasics.com/2017/04/geeksforgeeks-solution-for-set-bits.html | 1,519,089,890,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812871.2/warc/CC-MAIN-20180220010854-20180220030854-00104.warc.gz | 543,237,327 | 22,358 | # Geeksforgeeks Solution For " Set Bits "
GeeksforGeeks Solution For Hard Domain .Below You Can Find The Solution Of School Basic ,Easy ,Medium . Or Hackerrank Solution You Can Also Direct Submit Your Solution to Geeksforgeeks Same Problem .You Need to login then you can submit you answers
Problem :- Given a positive integer N, print count of set bits in it. For example, if the given number is 6, output should be 2 as there are two set bits in it.
Solution :-
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int n,i,j,tmp=0;
cin>>n;
while(n!=0)
{
i=n%2;
if(i==1)
{
tmp++;
}
n=n/2;
}
cout<<tmp<<endl;
}
return 0;
}
Output:- | 203 | 671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-09 | latest | en | 0.459764 |
https://tutorials.topstockresearch.com/Financial/Ratios/Profitability/OperatingMargin.html | 1,716,972,375,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00441.warc.gz | 500,320,910 | 8,665 | Technicals Stability Returns
## Understanding Operating Margin
Operating Margin measures the company's revenue before deducting any interest and taxes. It is calculated by dividing Operating Earnings by Total Revenue. And popularly also known as EBIT (Earnings Before Interest and Tax) margin or Return on sales. It is expressed in percentage number and is utilized in all screeners and charts throughout the site.
The operating margin shows the revenue made for each rupee of sales after the company pays variable costs but before interest and tax. A higher operating margin is considered good. It indicates that the company is efficient in its operations and generating profits from its sales.
The formula of Operating margin
Operating Earnings - It is the income that a company earns from its core operations. Operating profit is found in the company's Income Statement after deducting operating expenses from gross profit and before deduction of interest and taxes.
Total Revenue - It indicates how much a company's revenue is before deducting any expenses. Total revenue is found in the company's Income Statement.
Example of Operating Margin: For the financial year, Larsen & Toubro Ltd. reported Operating Earning as Rs. 9352.79 cr. And Total Revenue as Rs. 14073.57 Cr.
The value as per the Formula [Operating Margin = (Operating Earnings / Total Revenue) x 100] is calculated as (9352.79 / 14073.57 x 100 = 66.45%).
##### Key Points
Operating Margin or Return on Sales or EBIT Margin measures the company's revenue before deducting any interest and taxes. It is calculated by dividing Operating Earnings by Total Revenue. Operating Margin is also calculated on MRQ and TTM basis.
An operating margin of 15% or higher is considered good. A high operating margin indicates that the company is efficiently generating its profits from sales. A fall or rise in operating margin might be due to high or low operating costs.
For better analysis or understanding investors should compare the company's operating margin with its competitors or its historical data.
##### While looking at Operating Margin, the following points should also take into consideration:
An operating margin of 15% or higher is considered good for most industries. It indicates, the company is efficiently generating revenue from its sales.
A high margin might reflect, the company is doing well in managing its operating costs.
A low margin or fall in the operating margin may indicate, the company has high operating costs.
It only takes factors into account like the cost of goods sold and operating expenses and ignores the interest expenses, which are found further below in the income statement after operating profit.
Operating Margin differs from industry to industry. For most industries, an operating margin of 15% is considered good. While analyzing we should compare companies that operate in similar industries. Capital intensive companies operating margin can be less compared to sectors that need less amount of assets, or less operating costs like the IT sector, Financial sector, etc.,
##### How to use Operating Margin effectively
Investors should look for a higher operating margin. An operating margin of 15% or higher is favorable.
Before selecting any stock or looking at the company's turnover we should check its operational efficiency.
Don't look at operating profits only, as it only shows the company operating earnings, whereas operating margin indicates how much the company is making profits from its sales.
It will give a better understanding by comparing the particular company's stock with its peers. And also, look year over years operating margin and its growth for better analysis. Operating Margin is also calculated on MRQ and TTM basis, and it gives a better insight into the stock.
If the operating margin is low or falls year-over-year, it may indicate that the company has high operating costs.
For better understanding and, the analysis we should check other financial metrics with operating margins like Net Margin, EBITDA Margin, EPS, Cash EPS, Return on Equity (ROE), etc.,
#### Range Indicator of Operating Margin Ratio
Range Indicator Comments Screener at TSR
Above 50 Strong Bullish Extremely High Margin Yes
20 to 50 Bullish High Margin Yes
15 to 20 Mild Bullish Good Margin Yes
10 to 15 Neutral Average Margin Yes
5 to 10 Mild Bearish Low Margin Yes
0 to 5 Bearish Very Low Margin Yes
Below 0 Strong Bearish Extremely Low Margin Yes
##### Related Operating Margin Screener
Profitability Screener Operating Margin Above 60 Operating Margin 30 to 40 Operating Margin 15 to 20
Wait for US Stock Analytics & Screeners is Over
StockAio.com (Stock All In One) is now Live
We hope you will provide us with the same Love & Support as you did for TSR.
Wait for US Stock Analytics is Over
StockAio.com (Stock All In One) is now Live
We hope you will provide us with the same Love & Support as you did for TSR. | 1,002 | 4,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.950667 |
https://www.justcrackinterview.com/interviews/oregon-state-police/ | 1,695,574,683,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506658.2/warc/CC-MAIN-20230924155422-20230924185422-00663.warc.gz | 915,317,593 | 19,950 | # Oregon state police Interview Questions Answers, HR Interview Questions, Oregon state police Aptitude Test Questions, Oregon state police Campus Placements Exam Questions
Find best Interview questions and answer for Oregon state police Job. Some people added Oregon state police interview Questions in our Website. Check now and Prepare for your job interview. Interview questions are useful to attend job interviews and get shortlisted for job position. Find best Oregon state police Interview Questions and Answers for Freshers and experienced. These questions can surely help in preparing for Oregon state police interview or job.
All of the questions listed below were collected by students recently placed at Oregon state police.
Ques:- Find the odd man out:
2, 5, 10, 50, 500, 5000
A. (B) 5
B. (C) 10
C. (D) 5000
5*2=10
10*5=50
50*10=500
500*50=25000
So odd one is 5000
Ques:- In each questions below are given two statements followed by two conclusions numbered I and II.
1. Statements :
No bat is ball. No ball is wicket.
Conclusions :
I. No bat is wicket.
II. All wickets are bats.
A. if only conclusion I follows
B. if only conclusion II follows
C. if either conclusion I or II follows
D. if neither conclusion I nor II follows
E. if both conclusions I and II follow
Ques:- What is the optimum number of operations for 2*(x**3)+3*(x**2)+5*x+5?
Ques:- A policeman buys some bunches of yellow bananas at 30 cents per bunch and red bananas 40 cents/bunch .He said if i have split the sum equally and bought bananas he would have got two more bunches find the amount he used it to buy bananas
140 cents
Ques:- How do you describe yourself as a professional?
Ques:- There are twelve consecutive flags at an equal interval of distance. A man passes the 8th flag in 8 seconds. How many more seconds will he take to pass the remaining 4 flags?
4.5
Your donation keeps our website running smoothly and accessible to all. Support us today to ensure we continue to provide valuable information and resources.
Ques:- Is it possible to view the contents of my laptop on my tv and if so, how do I do it?Laymans language please.CheersDave
Ques:- The present Chief Election Commissioner is?
Ques:- How are various factors of production affected by globalcompetition? Do we manage people any differently in aglobally competitive environment?
Ques:- Three piles of chips–pile I consists one chip, pile II consists of two chips, and pile III consists of three chips–are to be used in game played by Anita and Brinda.The game requires: a)That each player in turn take only one chip or all chips from just one pile. b)That the player who has to take the last chip loses. c)That Anita now have her turn.From which pile should Anita draw in order to win?
Ques:- AT WHAT YOU ARE THE BEST ?
Ques:- Have you taken a management development course?
Ques:- How do you get on with different kinds of people?
Ques:- Last month salareis how much
Ques:- My personal life
Ques:- P and Q invests Rs.10000 each, P investing for 8 months and Q investing for all the 12 months in the year. If the total profit at the end of the year is Rs.25000, find their shares?
10000*8:10000*12
2:3
5x=25000
x=5000
p:q
10000:15000
Ques:- A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9% p.a. in 5 years. What is the sum? | 834 | 3,318 | {"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} | 3.40625 | 3 | CC-MAIN-2023-40 | longest | en | 0.895234 |
https://www.physicsforums.com/threads/introductory-kinematics-word-problem.323275/ | 1,511,446,559,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806832.87/warc/CC-MAIN-20171123123458-20171123143458-00587.warc.gz | 871,047,317 | 16,253 | Introductory kinematics word problem
1. Jul 4, 2009
shaools
1. The problem statement, all variables and given/known data
A competitor is aiming to complete a 1500m wheel-chair race in less than 4.0 minutes. After moving at a constant speed for exactly 3.5 minutes, there were still 240m to go. What must his acceleration before the remaining distance if he were to finish the race on time?
Given:
d1= 1260m [forward]
d2= 240m [forward]
t1= 210s
t2= 30s
Required: v1, v2 possibly?, and a
2. Relevant equations
v1 = d/t
and d = v1*(t+a(t)^2)/2 ?
3. The attempt at a solution
v1=1260m/210s
v1=6m/s
d = v1*(t+a(t)^2)/2
a= (2d/t^2)-v1*t
a= (2*240m/30s^2) - 6m/s*30s
a= something horribly incorrect
the answer at the back of the book is 0.13m/s^2. im guessing that im not approaching the second part of the question correctly. i tried it again with an equation using v2, but im still getting weird answers.
also, im sorry i dont know how to use latex :/
2. Jul 4, 2009
tiny-tim
Welcome to PF!
Hi shaools! Welcome to PF!
(don't bother about LaTeX for something like this … the SUB and SUP tags are fine, and they save a bit of server space also! )
hmm … you're messy with brackets, which leads you to make mistakes
that middle line is wrong.
3. Jul 4, 2009
shaools
wow.
i managed to screw up the initial equation twice in that post. there's no way to edit posts is there?
i meant to post this equation for displacement and acceleration:
http://id.mind.net/~zona/mstm/physics/mechanics/kinematics/EquationsForAcceleratedMotion/Origins/Displacement/Image78.gif" [Broken]
soo...
d = v1*t + (a*t^2)/2
a = 2d/(t^2) - v1*t
and then when i plug my values in, i get funky answers.
i think i might just be using the wrong equation, or i might be plugging in incorrect values. im really not sure.
thank you for checking tho!
Last edited by a moderator: May 4, 2017
4. Jul 4, 2009
tiny-tim
Nope … still wrong
Write it out step by step … don't try to do any of it in your head!
5. Jul 4, 2009
shaools
ah, bested by bedmas
lol let me attempt this once more...
a= [2(d - vt)]/ t^2 ?
6. Jul 4, 2009
shaools
a = [2(240m - 6m/s*30s)] / 30^2
a = 0.13m/s^2
yup :).
thank you mr. fish !
7. Jul 4, 2009
tiny-tim
:tongue2: blubblelubbleglubblelubblephlrrrrrrp! :tongue2: | 748 | 2,287 | {"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} | 3.546875 | 4 | CC-MAIN-2017-47 | longest | en | 0.899565 |
https://jira.lsstcorp.org/browse/DM-9855?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel | 1,686,300,520,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655446.86/warc/CC-MAIN-20230609064417-20230609094417-00665.warc.gz | 360,049,904 | 21,384 | # Select images for coadd based on PSF quality.
XMLWordPrintable
#### Details
• Type: Story
• Status: Done
• Resolution: Done
• Fix Version/s: None
• Component/s:
• Labels:
• Story Points:
6
• Team:
Data Release Production
#### Description
In HSC, there are a number of ways that the PSF modeling with psfex can be poor. Most of the time this is because the seeing is too good, but there can be other things that cause problems. We would like to reject these images when building the coadd. This ticket is to write an image selector that selects on the quality of the PSF based on the residuals of the second moments.
#### Attachments
36 kB
42 kB
3. outlier.png
50 kB
4. res_median_log.png
46 kB
5. res_median.png
52 kB
6. res_scatter.png
47 kB
7. update_res_median.png
56 kB
8. update_res_scatter.png
50 kB
#### Activity
Hide
Bob Armstrong added a comment -
I tried to come up with a reasonable set of cuts based on the results from the HSC S16A visit data. I use the following quantities derived from the second moment quantities:
• e1 = (Ixx-Iyy)/(Ixx+Iyy)
• e2 = 2*Ixy/(Ixx+Iyy)
I have also rejected ccd 9 from this analysis because of the known problems there that cause many visits to fail.
For each CCD I looked at the residuals of the two ellipticity components and the size. Here is the median of residuals for each CCD.
Here is a plot of the scatter (as defined by the median absolute deviation scaled to be a Gaussian.) for each of these quantities.
Finally, I plot the fraction of objects with values greater than 5*scatter from the median.
Given these plots I chose a conservative set of cuts to remove outliers:
• median ellipticity residual < 0.015 (equivalent to ~10sigma of the distribution)
• median size residual < 0.008 (equivalent to ~10sigma of the distribution)
• size scatter < 0.03
• ellipticity scatter < 0.065
• size outlier fraction < 0.065
• ellipticity outlier fraction < 0.1
With these cuts I remove 6808 ccds from 971 different visits which corresponds to ~5% of the data. Here is the number of ccds removed from each visit.
You can see that most visits only have a few ccds with poorly modeled data. I also noticed that a number of ccds from the visits we rejected in S16A don't have any problems.
Here are the number of rejected visits for each ccd which show fairly uniform coverage except for the outer ccds:
We could also try to be more aggressive to exclude more objects as these are cuts are quite conservative.
Show
Bob Armstrong added a comment - I tried to come up with a reasonable set of cuts based on the results from the HSC S16A visit data. I use the following quantities derived from the second moment quantities: e1 = (Ixx-Iyy)/(Ixx+Iyy) e2 = 2*Ixy/(Ixx+Iyy) size = determinant radius I have also rejected ccd 9 from this analysis because of the known problems there that cause many visits to fail. For each CCD I looked at the residuals of the two ellipticity components and the size. Here is the median of residuals for each CCD. Here is a plot of the scatter (as defined by the median absolute deviation scaled to be a Gaussian.) for each of these quantities. Finally, I plot the fraction of objects with values greater than 5*scatter from the median. Given these plots I chose a conservative set of cuts to remove outliers: median ellipticity residual < 0.015 (equivalent to ~10sigma of the distribution) median size residual < 0.008 (equivalent to ~10sigma of the distribution) size scatter < 0.03 ellipticity scatter < 0.065 size outlier fraction < 0.065 ellipticity outlier fraction < 0.1 With these cuts I remove 6808 ccds from 971 different visits which corresponds to ~5% of the data. Here is the number of ccds removed from each visit. You can see that most visits only have a few ccds with poorly modeled data. I also noticed that a number of ccds from the visits we rejected in S16A don't have any problems. Here are the number of rejected visits for each ccd which show fairly uniform coverage except for the outer ccds: We could also try to be more aggressive to exclude more objects as these are cuts are quite conservative.
Hide
Bob Armstrong added a comment -
At the suggestion of Rachel Mandelbaum, I have remade the plots using delta size/size.
Note that I have shifted the distribution to be centered at zero. And here is the scatter plot:
Upon further investigations I found the following:
• the median(delta e1) cut per ccd does not overlap much with the median(delta e2) because you have visits where either e1 or e2 is small across the ccd while the other is large. For problematic visits it will still get the small ellipticity value correct, but not the large. I think the way to go here is to combine into a median(delta e) cut.
• a permissive cut in the median( (delta size)/ size) does not catch some of the known bad visits, while a cut in the ellipticity does. Therefore this cut does not seem to add any more information.
• The scatter for delta e1, delta e2 and delta size/size per ccd roughly flag the same visits, and add ~15% ccds to the median(delta e) cut. These look like poor modeling where the residuals are equally scattered around the mean. A tighter cut on the median(delta e) value can also catch these.
• The fraction of bad objects is not a useful metric because some ccds have essentially constant ellipticity/size and therefore a scatter that is extremely small which can give rise to reasonably large outliers.
Show
Bob Armstrong added a comment - At the suggestion of Rachel Mandelbaum, I have remade the plots using delta size/size. Note that I have shifted the distribution to be centered at zero. And here is the scatter plot: Upon further investigations I found the following: the median(delta e1) cut per ccd does not overlap much with the median(delta e2) because you have visits where either e1 or e2 is small across the ccd while the other is large. For problematic visits it will still get the small ellipticity value correct, but not the large. I think the way to go here is to combine into a median(delta e) cut. a permissive cut in the median( (delta size)/ size) does not catch some of the known bad visits, while a cut in the ellipticity does. Therefore this cut does not seem to add any more information. The scatter for delta e1, delta e2 and delta size/size per ccd roughly flag the same visits, and add ~15% ccds to the median(delta e) cut. These look like poor modeling where the residuals are equally scattered around the mean. A tighter cut on the median(delta e) value can also catch these. The fraction of bad objects is not a useful metric because some ccds have essentially constant ellipticity/size and therefore a scatter that is extremely small which can give rise to reasonably large outliers.
Hide
Bob Armstrong added a comment -
I think this is ready for review. Paul Price, will you have a chance to review this before you leave?
Show
Bob Armstrong added a comment - I think this is ready for review. Paul Price , will you have a chance to review this before you leave?
Hide
Paul Price added a comment -
Looks good to me. A couple of minor comments:
• Should the choice of shape (currently hard-coded as base_SdssShape) be configurable? Alternatively, could you use src.getShape(), which would remove the need for the ellipticity calculations because they could be done by lsst.afw.geom.ellipses?
• What needs to be done to use this code by default? Shouldn't we activate it in obs_subaru?
Show
Paul Price added a comment - Looks good to me. A couple of minor comments: Should the choice of shape (currently hard-coded as base_SdssShape ) be configurable? Alternatively, could you use src.getShape() , which would remove the need for the ellipticity calculations because they could be done by lsst.afw.geom.ellipses ? What needs to be done to use this code by default? Shouldn't we activate it in obs_subaru?
Hide
Bob Armstrong added a comment -
Merged
Show
Bob Armstrong added a comment - Merged
#### People
Assignee:
Bob Armstrong
Reporter:
Bob Armstrong
Watchers:
Bob Armstrong, Jim Bosch, Paul Price | 1,898 | 8,070 | {"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-2023-23 | latest | en | 0.85213 |
moarminecraft.net | 1,540,336,552,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583517495.99/warc/CC-MAIN-20181023220444-20181024001944-00495.warc.gz | 247,678,661 | 12,727 | # How to Calculate Cloud Mining Profitability
Today I desired to voorkant how to calculate cloud mining profitability. I had a latest comment on my article: Ethereum Cloud Mining is not Profitable that I’m worried perpetuates the kleuter of static analysis that will cause someone to lose money on cloud mining.
I’m going to do my analysis for Ethereum Cloud Mining. However, this analysis will work for any coin that has enlargening mining difficulty.
Assumptions: I’m assuming the price of ETH is static. Why? Because if it goes up, that is simply a premie. If mining isn’t profitable unless the currency goes up, then one is better off buying the currency outright.
## Step One of How to Calculate Cloud Mining Profitability
Very first you need to know how much the cloud mining will cost vanaf unit of hashing power. Spil of 23 April 2018 Hashflare.io is selling 100 KH/s for Two.20 USD. That is 1 MH/s for 22 USD.
Use a static rekenmachine very first. This will provide the baseline static analysis. For Ethereum I like this zakjapanner.
Spil of writing there is a network hashrate of 22595.62995398704 GH/s, a blocktime of 13.31 and one ETH going for 48.63 USD.
So with 1 MH/s I would earn 0.043093 ETH vanaf month, worth Two.Ten USD vanaf month. Multiply that by 12 and the total ETH mined (0.517116) would be worth \$25.Two.
So if the price of ETH stays the same (which for the purpose of the static analysis wij will assume it will), and the network hashing power stays the same. Then the profit will be \$Trio.Two after a year IF THE NETWORK HASHING POWER STAYS THE SAME. The problem with a static analysis is that network hashing power does NOT stay the same.
## Network Mining Difficulty Goes Up
If you zekering with this static analysis you’ll surely lose money tho’. Why? Because the network hashing power has historically gone up and gone up A Lotsbestemming.
Ethereum Block Difficulty Growth Since 30 July 2015
Ter the very first four months of 2018 alone, mining difficulty for Ethereum has gone up overheen 200% from under 100 TH/s up to almost 300 TH/s. Which means the amount of ETH mined for anyone with immobilized hashing power will have bot diminished by overheen 66%.
Factoring ter the growth rate of block difficulty is the most significant factor when determining cloud mining profitability.
## Step Two of How to Calculate Cloud Mining Profitability
Projecting how much the network hashrate will increase overheen the life of the cloud mining contract is vitally significant. You need to make a realistic estimate of how the network hashrate will increase because it will reduce the amount you get from mining each day.
The chart above shows the Ethereum network hashrate growth. Te this example, Hashflare.io contracts run ter 12 month increments. So wij need a realistic estimate of how much the hashing power (and thus mining difficulty) will go up overheen a 12 month period.
This takes some guesswork but the best indicator is the past.
The August 2015 hashrate of 55 GH/s to the August 2016 hashrate of Trio,811 GH/s represents a 6,800% increase. This wasgoed the very first 12 months of the Ethereum network coming online so I think this number is too high.
Te 2016 the Ethereum network hashrate went from 511 GH/s to Five,700 GH/s. A 1,015% increase.
From April 2016 at 1752 GH/s to April 2018 of 20,300 GH/s wasgoed a 1,058% increase.
So I based on 2016 I think a 1,000% increase te hashing power is a good conservative guesstimate. That means the hashing power would be around 230,000 GH/s by April of 2018.
So then wij go after step 1 again using the static rekenmachine. Using the 1 MH/s and a network hashrate of 230,000 GH/s. The monthly ETH mined would be 0.004233 worth \$.21.
## Step Three of How to Calculate Cloud Mining Profitability
So at this point wij have a projection of how much wij’ll get from mining te the very first month. And how much wij’ll get ter the last month. Thesis are just a projections based on a static analysis and a guesstimate of where mining difficulty will be ter the future.
But the amount mined doesn’t hop down from the very first month to the last month. The amount mined is leisurely and steadily decreasing.
I think a exponential decay monster fits the gegevens better but for the sake of ease I think a linear monster will suffice.
I also think a simplified method works because the cloud mining rates I’ve seen are not close to what they would need to be for mining to be profitable.
Take the amount wij think wij’ll mine ter the very first month. Ter this case .043093. Then take the amount wij’ll think wij’ll mine te the last month, .004233. Subtract the very first from the last. Then divide that by 11.
From that point you take the beginning value of .043093 subtract the decay amount .003943 to get the 2nd months value of .039149. You do this again until you get to month 12. By summing up each month’s value wij get 0.283956. Multiply that by the price of ETH of 48.63 USD and wij get \$13.80.
The contract te this example cost 22 USD so this would not be profitable if the network hashing power goes up by 1000% (spil it did ter 2016) and the price of ETH stays the same.
You’d end up losing \$8.Two.
Okay, what if the network hashing power only goes up 500% so it goes up to 135,600 GH/s after one year? You’d mine about .Trio ETH worth \$14.66. You still lose.
What if the network hashing power only goes up 100% to about 45200 GH/s? You’d mine about .387 ETH worth less than \$Nineteen. Loser.
What if the network hashing power only goes up 35% to 30,500 GH/s. You’d mine about .45 ETH worth \$22.88. Petite winner.
## If Network Hashing Power Goes Up You Begin to Lose
So what I hope this shows is that if the hashing power goes up, which ter the case of Ethereum (and I suspect most coins spil well) the amount of coins mined will druppel and the profits will be eroded.
## Effortless Method
If you believe network hashing power will proceed to go up then use this method to determine if mining is even worth a closer evaluation: use the static mining profitability zakjapanner. Use the amount of ETH mined and the cost of the mining contract to see how much you’re effectively paying vanaf ETH.
For example Hashflare.io is selling 1 MH/s for 22 USD for a year. That would yield 0.043093 ETH vanaf month x 12 would be 0.517116 ETH for the year mined if the network hashrate stays the same. So the cost vanaf ETH would be 42.54 USD. With ETH trading at 48.63 USD that is only a 14% discount overheen a year.
Unless you’re going to get ETH (or whichever other coin) at a significant discount using the static calculation (say 40-50% below spot price). It’s not worth it.
## But the Price of ETH is going to dual!
Superb! Then buy ETH directly. Lets say the price of ETH does dual te a year. It goes from 48.63 USD today up to \$97.26. You could have bought \$22 worth of ETH (.45 ETH) and the \$22 worth of ETH would now be worth \$43.76.
With a 1000% network hashrate increase you’d have only mined 0.283956 which would be worth \$27.61. Unless the mining is profitable with the price of ETH immovable, you’re better off possessing the presently directly even if the price of the currency goes up.
## At what price would cloud mining be worth it?
Spil of today 23 April 2018, based on a 1000% increase te hashing power overheen the next year I would not pay more than around \$7 for 1 GH/s of hashing power.
Based on my projections that would yield about 40%. Given the risk and volatility ter cryptocurrencies I would need to see that zuigeling of terugwedstrijd for it to be worth the risk to mij.
With 1 GH/s costing 22 USD, if the network hashing power stays the same I would still only make about 15%. Given the history of network hashrate increases that isn’t worth it. I can get 12.95% on BitBays will no market risk.
Hashflare.io is nowhere close to \$7 vanaf GH/s. Genesis Mining offers 1 GH/s for Two years for 29.99. Who knows where the network hash rate will be ter Two years.
## Some People Optie Cloud Mining is Profitable
I have read testimonials from people who think cloud mining is profitable. My main question would be is it profitable because the underlying cryptocurrency went up, or because the mining itself wasgoed profitable? Ter other words would you have bot better off just possessing the cryptocurrency directly?
Superb article. How would this logic apply if you were to build your own equipment and had free electro-stimulation spil opposed to buying a contract with an existing mining company?
I’ve never run any mining equipments myself. However, if you add up the cost of the equipment and divide it by the hashing power, it would need to be less than what you calculate is the value you’d be willing to pay.
So for example, back ter April 2018 I said I wouldn’t be wiling to pay more than \$7 for 1 GH/s for ETH. So if I could build a equipment for \$700 that would have 100 GH/s of hashing power it would be worth it. | 2,177 | 9,002 | {"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-2018-43 | longest | en | 0.877277 |
http://kaffee.50webs.com/Science/activities/Chem/Activity.Mass.Defect.Fission.Fusion.htm | 1,721,275,059,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514822.16/warc/CC-MAIN-20240718034151-20240718064151-00512.warc.gz | 14,458,968 | 6,343 | Date:
Class:
## Group Activity: Mass Defect Fission and Fusion
### Introduction
#### Mass Defect
When you learned how to calculate the average atomic mass of an element from the exact mass and relative abundance of its isotopes you may have noticed something interesting. The mass number is not the exact mass of an isotope. For example, the exact mass of 16 8O is not exactly 16 u but rather 15.994915 u.The mass of a nucleus is slightly less than you might expect from the total number of protons and neutrons. Why should this be so?
Natural systems follow a consistent pattern in that they always seek to find the lowest possible energy. Atomic nuclei are no exception. As it happens, the nucleus of an atom is thermodynamically more stable than the protons and neutrons that make it up. This can be shown by a relatively simple calculation using the hypothetical situation of making a 16 8O nucleus from 8 hydrogen atoms (including electrons) and 8 neutrons.
Exact Masses 1 1H 1.00782 u 1 p+ & 1 e- n0 1.00867 u e- 5.48580 × 10-4 u
81 0n0 + 81 1H → 16 8O
```8n0 + 81 1H = 8(1.00867) + 8(1.00782) = 16.13192 u
Actual Mass of 16O = 15.99492 u
Difference in mass: 16.13192 - 15.99492 = 0.13700 u
So an 16O nucleus loses 0.13700 g of mass per mole of 16O formed.
We ignore the electrons because there are as many electrons
in 8 H atoms as in one O atom
```
The mass difference calculated above is called the mass defect. Where does this mass go? It is released as energy when the nucleus forms. The process of forming a nucleus is exothermic: the product is more stable than the starting materials. In other words, the formation of an atomic nucleus releases energy. Energy is in fact a form of matter, as Albert Einstein discovered in the development of his theory of relativity. You have all seen the famous equation:
E = mc2
In this equation a statement about the equivalence of matter and energy is made. It says that energy changes are accompanied by a change in mass equal to the amount of energy divided by c2. In ordinary chemical reactions the amount of energy is so small that the change in mass is not detectable. The c in the equation is a universal constant of nature: the speed of light, 3.00 × 108 m/s. The amount of energy represented by the mass defect is called the binding energy of a nucleus and is the amount of energy that would be required to break the nucleus into its component parts. Here is how to caculate the amount of energy:
```E = mc2
E = (0.13700 g/mol· 1 kg/1000 g)(3.00 × 108 m/s)2 = 1.23 × 1013 J/mol
```
In such calculations grams must be converted to kilograms since 1 joule of energy (1 J) is defined as
1 kg· m2/s2. A joule is a scientific unit of energy equal to 4.184 × 10-3 dietary Calories. It is related to the watt (a measure of electrical power) because 1 watt of power uses 1 J per second. From these considerations and from the above calculation you can see that the binding energy is truly enormous! Perhaps you now understand why nuclear energy is so interesting: nearly unlimited power at the center of every atom.
#### Fission and Fusion
Fission is the name for the process in which heavy nuclei (such as23592U) split into two nuclei with smaller mass numbers. Fusion is the combination of two light nuclei to form a heavier, more stable nucleus. Both of these processes are highly exothermic and involve the release of energy more than a million times larger than the energy released in chemical reactions.
Fission was discovered in the 1930s. Neutron bombardment of uranium-235 atoms resulted in the formation of smaller nuclides:
1 0n0 +23592U14156Ba +9236Kr + 31 0n0
The amount of energy released per mole of uranium-235 is 2.1 × 1013 J. When methane (natural gas) is burned it releases only 8.0 × 105 J/mol. This is a factor of about 26 million times less energy.
Notice that three additional neutrons are produced in the example fission reaction given for uranium-235 (there are actually at least 200 different possible isotopes that can result from such a fission). These neutrons make a nuclear chain reaction possible. In a nuclear chain reaction one fission event causes one or more further fission events. When less than one neutron, on average, causes another fission event then the reaction is called sub-critical. If exactly one neutron from each reaction causes another fission event then the process is self-sustaining and is called critical. Finally, the situation becomes supercritical if more than one neutron from each fission event causes another fission event. This results in a violent explosion, as is well known: a bomb made using a supercritical mass of uranium-235 was dropped on Hiroshima, Japan in 1945. Such bombs are made by arranging for subcritical masses of uranium-235 to be brought together suddenly to form a supercritical mass. Peaceful use of nuclear chain reactions is made in nuclear power generators in which controlled fission chain reactions are used to produce heat, boil water and run steam power generators.
Fusion is the source of the power of the stars. Our Sun is made of hydrogen (73% by mass), helium (26%) and other elements (1%). In the core of the Sun hydrogen nuclei fuse together to form helium nuclei. In the process they release enormous amounts of energy. Seven hundred million tons of hydrogen is fused every second in the core of the Sun and nearly 5 million tons of this matter is released as energy. (Use E = mc2 to figure out how much energy this is; one metric ton is 1,000 kg). The core of the Sun is 16,000,000 degrees Celsius. This high temperature makes earthly power sources based on nuclear fusion problematic at best.
#### Problems
1. The Sun radiates 3.9 × 1023 J into space every second. What is the rate at which mass is lost from the Sun?
2. The earth receives 1.8 × 1017 J/s of solar energy. What mass of solar material is converted to energy in one day to provide this energy to earth (24 hr)? What mass of coal would have to be burned to give the same amount of energy? (One gram of coal gives 3.2 × 104 J).
3. The exact mass of potassium-39 is 38.963707 u. Find the mass defect in atomic mass units (u) for 39K.
4. Given that the exact mass of nitrogen-15 is 15.000108 g/mol, find the mass defect in g/mol for this isotope.
5. The most stable nucleus in terms of binding energy is 56Fe. If the exact mass of 56Fe is 55.9349 u calculate the binding energy of 56Fe in J/atom. (1 g = 6.02 × 1023 u)
6. Calculate the binding energy in J/mol for carbon-12 (12.00000 g/mol) and uranium-235 (235.043922 g/mol).
page break
1. The mass defect for a lithium-6 nucleus is 0.03434 g/mol. Calculate the exact atomic mass of 6Li.
2. The binding energy for a single atom of magnesium-27 is 3.5802 × 10-11 J. Calculate the atomic mass
of 27Mg
3. The easiest fusion reaction to initiate is
2 1H +3 1H4 2He +1 0n0
Calculate the energy released per 4He nucleus and per mole of 4He nuclei. The atomic masses are: 2H = 2.01410, 3H = 3.01605, and 4He = 4.00260. The solution of this problem is based on the difference in mass between the products and the reactants. The mass that is lost (the products will weigh less) is converted into energy.
4. One possible fission reaction resulting from the bombardment of uranium-235 with a neutron is:
1 0n0 +23592U14054Xe +9438Sr + 21 0n0
Using the following exact masses, calculate the total amount of energy released in this reaction based on the difference in mass between the products and the reactants. The mass that is lost (the products will weigh less) is converted into energy.
235U = 235.04392, 140Xe = 139.9216, 94Sr = 93.91537.
5. What mass of uranium-235 is required to produce the same amount of energy as a metric ton (1,000 kg) of coal? Assume that the controlled fission of uranium provides the same amount of energy per mole as found in the previous problem. One ton of coal provides about 3.2 × 1010 J.
Last updated: Dec 04, 2018 Home | 2,023 | 7,937 | {"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-2024-30 | latest | en | 0.926606 |
https://web2.0calc.com/questions/calculate_12 | 1,685,691,945,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00659.warc.gz | 671,368,224 | 5,076 | +0
calculate
0
3
0
= ((1^2 + 1^2)π^2(1.05457182 × 10^(-34) J·s)^2) / (2(1.67262192 × 10^(-27) kg)((3.536 x 10^(-12) m)^2))
May 26, 2023 | 84 | 139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-23 | latest | en | 0.429663 |
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A109443 Cumulative sum of largest prime power <= n. 0
%I
%S 1,3,6,10,15,20,27,35,44,53,64,75,88,101,114,130,147,164,183,202,221,
%T 240,263,286,311,336,363,390,419,448,479,511,543,575,607,639,676,713,
%U 750,787,828,869,912,955,998
%N Cumulative sum of largest prime power <= n.
%C Integers in this sequence which are themselves primes include a(2) = 3, a(10) = 53, a(14) = 101, a(23) = 263, a(25) = 311, a(29) = 419, a(31) = 479, a(35) = 607, a(40) = 787. Integers in this sequence which are themselves powers greater than 1 of primes include a(7) = 27, a(11) = 64. Integers in this sequence which are perfect powers greater than 1 of composites include a(37) = 676.
%F a(n) = Sum_{i=1..n} A031218(i).
%Y Cf. A031218.
%K easy,nonn
%O 1,2
%A _Jonathan Vos Post_, Aug 26 2005
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Last modified November 27 00:20 EST 2022. Contains 358362 sequences. (Running on oeis4.) | 515 | 1,557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-49 | latest | en | 0.841249 |
http://www.esru.strath.ac.uk/EandE/Web_sites/05-06/marine_renewables/resource/velocitydistribution4.htm | 1,638,806,562,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00056.warc.gz | 116,346,434 | 6,576 | Background Technology Environment Methodology Case Study Conclusions References ESRU Energy Systems Research Unit
Resource: Velocity Distributions
Velocity Profile Model Effects of Roughness Flow Characteristics Tool References
Modelling the Flow characteristics of a channel
This method above is implemented in a Matlab script, with input being provided in the form of a coloured bitmap image constructed using Autocad. The bitmap is of a cross section of interest, and is created by using the mapping technique described here. It is then coloured using a specific key so as areas of differing bottom roughness are distinguishable.
The overall channel Manning number and velocity is input along with various terms to aid the program in finding points of inflection. The program runs under the assumption that the velocity at a given point is influenced by that points proximity to the wall: the interaction between the fluid at one point and that at another is not modelled.
This drastically simplifies the calculation and has a profound effect on calculation time over more complete finite difference methods (the disadvantage is that this is another assumption to add to a growing list of other ad hoc assumptions made during Cole’s derivation).
The channel area is used (along with an assumption of rectilinear flow) to calculate a bulk flow rate, and the program then runs and attempts to match a velocity distribution by Cole’s with the Bernoulli distribution (velocity, not probability) by Manning.
The program uses values of dx and dy to discretise the geometry of the channel section, and then calculates a velocity profile in the vertical and horizontal directions at each station based on the slope of the element. The profile is first generated for the overall channel mannings number, and the results of this are then used as the freestream velocities for the caclulations which include roughness effects. In all, the boundary layer calculation module is called three times:
The output is a computed value of velocity at every point in the channel in the channels axial direction.
Tool
The Velocity Distribution tool can be accessed by clicking on the tool icon. The tool consists of Matlab Programme and instructions are contained within the file.
1. Inputs: coloured (16 Windows Cols) bitmap .bmp file; dx, dy; tolerance for finding turning pts.; Manning number; Slope; bulk flow velocity
2. Outputs: velocity distribution
References
1 British Oceanographic Data Centre. [Webpage] [cited 02 May 2006]; Available from http://www.bodc.ac.uk/ 2 UK Hydrographic Office: Admiralty Charts & Publications. [Webpage] [cited 02 May 2006]; Available from http://www.ukho.gov.uk/ 3 Black & Veatch Ltd., The Carbon Trust: Phase 1 & 2 Uk Tidal Stream Energy Resource Assessment. [Webpage] 2005 [cited 02 May 2006]; Available from http://www.thecarbontrust.co.uk/ctmarine10/page4.htm 4 Scottish Enterprise: Marine Renewable (Wave and Tidal) Opportunity Review. [PDF] 2005 [cited 02 May 2006]; Available from http://www.scottish-enterprise.com/publications/marine_renewable_opportunity_review.pdf 5 ANDROSOV, A.A., et al., Numerical Modelling of Barotropic Tidal Dynamics in the Strait of Messina. Advances in Water Resources, 2002. 25(4): p. 401-415. 6 MOE, H., et al, A High Resolution Tidal Model For The Area Around The Lofoten Islands, Northern Norway, Continental Shelf Research, 2002. 22: p. 485–504 7 CALVERT, J., Open Channel Flow. 2003 [cited 02 May 2006]; Available from http://www.du.edu/~jcalvert/tech/fluids/opench.htm 8 CHOW, V.T., Open-Channel Hydraulics. Mcgraw-Hill Civil Engineering Series. 1959, New York, London: McGraw-Hill Education. pp. 692. ISBN 007085906X. 9 COPELAND, G., Seminar on Turbulence Modelling. 2006, Univerisity of Strathclyde. 10 GARRETT, C. Frictional Processes in Straits. 2004. Villerfranche-sue-Mer, France: Elsevier. 11 GROSS, T.F., Momentum and Energy Transfer in the Oceanic Law of the Wall Region. Journal of Atmospheric and Ocean Technology, 1999. 16(Nov): p. 1668-1672. 12 LERMUSIAUX, P.F.J. and ROBINSON, A.R., Features of Dominant Mesoscale Variability, Circulation Patterns and Dynamics in the Strait of Sicily. Deep-Sea Research Part I: Oceanographic Research Papers, 2001. 48(9): p. 1953-1997. 13 LU, Y., et al., Turbulence Characteristics in a Tidal Channel. Journal of Physical Oceanography, 2000. 30(5): p. 855-67. 14 MUNSEN, B., et al., Fundamentals of Fluid Mechanics. 3 ed. 1998, New York: John Wiliey & Sons. pp. 877. ISBN 15 NEZU, I. and NAKAGAWA, H., Turbulence in Open-Channel Flows. Iahr Monograph Series. 1993, Rotterdam: Balkema. ISBN 9054101180. 16 ROMANENKOV, D.A., et al., Comparison of Forms of the Viscous Shallow-Water Equations in the Boundary-Fitted Coordinates. Ocean Modelling, 2001. 3(3-4): p. 193-216. 17 VENNELL, R., Observations of the Phase of Tidal Currents Along a Strait. Journal of Physical Oceanography, 1998. 28(8): p. 1570-1577. 18 VENNELL, R., Oscillating Barotropic Currents Along Short Channels. Journal of Physical Oceanography, 1998. 28(8): p. 1561-9. 19 WARNER, J., et al., Effects of Tidal Current Phase at the Junction of Two Straits. Continental Shelf Research, 2002. 22(11-13): p. 1629-1642. 20 WHITE, F.M., Viscous Fluid Flow. 2 ed. Mcgraw-Hill Series in Mechanical Engineering. 1991, New York, London: McGraw-Hill. 614. ISBN 0070697124. 21 WINTERS, K.B. and SEIM, H.E., The Role of Dissipation and Mixing in Exchange Flow through a Contracting Channel. Journal of Fluid Mechanics, 2000. 407: p. 265-90. 22 YOUYO, L., et al., Turbulence Characteristics in a Tidal Channel. Journal of Phsical Oceanography, 2002. 30(May): p. 855-867.
Go back to Contents | 1,508 | 5,698 | {"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} | 2.828125 | 3 | CC-MAIN-2021-49 | latest | en | 0.917469 |
https://paradox-point.blogspot.com/2013/10/on-reduction-4-of-11.html | 1,716,749,416,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00289.warc.gz | 380,988,999 | 15,455 | ## Thursday, October 24, 2013
### On Reduction, 4 of 11
Applications of Reduction
The “Many Hands” Theorem:
Assume that r1t1 = r2t2 = d. Then:
d/(r1+r2) = (t1<+>t2)
d/(t1+t2) = (r1<+>r2)
d/(r1-r2) = (t1<->t2)
d/(t1-t2) = (r1<->r2)
In a rate-time problem, when rates add, times reduce; and when rates subtract, times protract. “Many hands make light work.”
If Alice can trim the hedges in A hours of work, and Bob can trim the hedges in B hours, then if they work together, and if their work rates add, then they will finish the job in A<+>B hours.
Two ships cross the same ocean, starting at the same time and heading for each other's home port. Let F be the time that the first ship took to go from port A to port B, let S be the time that the second ship took to go from port B to port A, and let M be the time it took for them to meet in mid-ocean. Approach-rates add, so times reduce: M equals F <+> S.
Two racecars, when driving towards each other from a distance of a mile, pass each other in P seconds; when the faster car pursues the slower car from one mile behind, it takes C seconds to catch up. How quickly does each car run a mile?
Answer: Respectively, 2(P<+>C) and 2(P<->C) seconds.
If you drive a mile at velocity v1, and then another mile at velocity v2, then on average you will go at velocity 2(v1 <+> v2); the harmonic mean.
Kirkhoff's Law for parallel resistors goes:
R12P = R1 <+> R2 .
Resistances reduce in parallel.
Two simple lenses with focal lengths f1 and f2, when placed together, form a compound lens with focal length f1 <+> f2 .
A lens made of material with index of refraction n, and with lens radii r1 and r2, has focal length (r1 <-> r2)/(n-1) .
D
.
. . |
. . |
. . |
. . | h
. . |
. . |
. a . b |
A x B C
In this diagram, h = x (tan(a) <-> tan(b)) | 596 | 1,931 | {"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-2024-22 | latest | en | 0.824224 |
https://en.khanacademy.org/math/6th-grade-illustrative-math/unit-2-introducing-ratios/lesson-9-constant-speed/e/ratios-with-double-number-lines | 1,675,849,236,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500758.20/warc/CC-MAIN-20230208092053-20230208122053-00051.warc.gz | 250,336,548 | 59,038 | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
### Unit 2: Lesson 8
Lesson 9: Constant speed
# Ratios with double number lines
## Problem
The double number line shows that Faye can sort 150 recyclable items in 3 minutes.
A double number line with 6 equally spaced tick marks. The line labeled Time, minutes, reads from left to right: 0, two unlabeled tick marks, 3, two unlabeled tick marks. The line labeled Items, reads from left to right: 0, two unlabeled tick marks, 150, two unlabeled tick marks.
Based on the ratio shown in the double number line, how many recyclable items can Faye sort in 4 minutes?
items
Stuck?
Stuck? | 201 | 805 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-06 | latest | en | 0.869707 |
https://stats.stackexchange.com/questions/434601/how-can-i-find-rho-given-p4-y-16x-5-0-9544 | 1,722,922,290,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476479.26/warc/CC-MAIN-20240806032955-20240806062955-00509.warc.gz | 423,456,201 | 41,350 | # How can I find $\rho$, given $P(4 < Y < 16|X=5)=0.9544$?
Let $$X$$ and $$Y$$ have a bivariate normal distribution with $$\mu_X=5, \mu_Y=10, \sigma^2_X=1, \sigma^2_Y=25, \rho >0$$.
If $$P(4 < Y < 16|X=5)=0.9544$$, I would like to find $$\rho$$.
I know that conditional marginals of a bivariate normal distribution are normal distributions. Given this knowledge, I can obtain the distribution $$Y|X=5 \sim N(10,25(1-\rho^2)).$$ However, integrating this pdf between $$4$$ and $$16$$ seems impossible. I have the following:
$$f_{Y|X=5}(y)=(5\sqrt{2\pi(1-\rho^2)})^{-1}exp\{-(x-10)^2/(50(1-\rho^2))\}$$, where $$y \in R$$.
$$.9544=\int_4^{16}(5\sqrt{2\pi(1-\rho^2)})^{-1}exp\{-(x-10)^2/(50(1-\rho^2))\}dx,$$
which does not seem possible to integrate. Is there a more efficient to solving this problem? Thank you.
• How would you integrate a standard normal density between $a$ and $b$? Commented Nov 5, 2019 at 4:52
• I see that the most common way is to use the Z table, but I was wondering if there was a more rigorous way to compute this. This problem is presented in my math stats class before we reach Z tables. Commented Nov 5, 2019 at 4:53
• You can't integrate it in "closed form" ... Doing this problem will rely on facts about the normal; in this case knowing how much of the probability is within 2 sd's of the mean. I bet you covered that much... Commented Nov 5, 2019 at 4:57
• We did. Thank you for confirming my suspicions about the integration. I appreciate the guided thought! Commented Nov 5, 2019 at 5:00
• @Glen_b-ReinstateMonica why 2 sd's of the mean? It doesn't explicitly mentions that Commented Feb 13, 2020 at 13:59
Its integral cannot be expressed by elementary functions by definition. You'll use z-table. Let the RV represented by the marginal density of $$Y$$ given $$X=5$$ be denoted as $$W$$. We ask for \begin{align}P(4 Then, $$P(Z\leq a)=0.9772\rightarrow a=2\rightarrow\rho=0.8$$ | 622 | 1,920 | {"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": 16, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-33 | latest | en | 0.914537 |
https://www.essaydepot.com/doc/111574/Acc-561-Week-3-Assignment-Wileyplus | 1,600,971,550,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219691.59/warc/CC-MAIN-20200924163714-20200924193714-00505.warc.gz | 811,639,892 | 5,542 | #### Words of Wisdom:
"life is full of choises, but let god help you to choose the right one" - Tomhellewell
# Acc 561 Week 3 Assignment Wileyplus
• Date Submitted: 07/10/2016 11:37 PM
• Flesch-Kincaid Score: 79
• Words: 588
• Report this Essay
ACC 561 Week 3 Assignment WileyPLUS
http://www.acc561assignment.com/ACC-561/ACC-561-Week-3-Assignment-WileyPLUS
Brief Exercise 13-4
Using these data from the comparative balance sheet of Rosalez Company, perform horizontal analysis. (If amount and percentage are a decrease show the numbers as negative, e.g. -55,000, -20% or (55,000), (20%). Round percentages to 0 decimal places, e.g. 12%.)
| | | | | | |Increase or (Decrease) |
| | |Dec. 31, 2012 | |Dec. 31, 2011 | |Amount | |Percentage |
|Accounts receivable | |\$ 488,200 | |\$ 360,000 | | | | | |
|Inventory | |\$ 818,100 | |\$ 601,200 | | | | | |
|Total assets | |\$3,173,600 | |\$2,774,300 | | | | | |
Brief Exercise 13-5
Using these data from the comparative balance sheet of Rosalez Company, perform vertical analysis. (Round percentages to 1 decimal place, e.g. 12.5%.)
| | |Dec. 31, 2012 | |Dec. 31, 2011 |
| | |Amount | |Percentage | |Amount | |Percentage |
|Accounts receivable | |\$ 544,700 | |[pic][pic] | | |
|Sales | |100 |% | |100 |% | |100 |% |
|Cost of goods sold | |61.2 | | |64.6 | | |65.7 | |
|Expenses | |24.8 | | |26.3 | | |28.0 | |
Vertical analysis (common-size) percentages for Vallejo Company’s sales, cost of goods sold, and expenses are... | 629 | 2,222 | {"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} | 3.421875 | 3 | CC-MAIN-2020-40 | latest | en | 0.189578 |
https://www.redhotpawn.com/forum/posers-and-puzzles/no-one-will-get-this-one.52370 | 1,548,145,322,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583831770.96/warc/CC-MAIN-20190122074945-20190122100945-00579.warc.gz | 928,038,907 | 8,076 | # No one will get this one.
Crash16
Posers and Puzzles 17 Sep '06 19:56
1. 17 Sep '06 19:56
Heres the riddle
A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say?
2. 17 Sep '06 20:101 edit
He said "My Kingdom is not of this world." You see the man was indeed Super Mario himself. The King decided to let him warp to his own world and Mario made his exit at the nearest green pipe.
3. 17 Sep '06 20:15
umm... That is not it... Sorry ^_^
4. 17 Sep '06 20:25
The King had a split personality. He had his other half on trial. Therefore he played Nintendo 64 until the cows came home.
N.B. By "N64" I mean "PS2" - and - By "cows came home" I mean "ducks came home."
5. AThousandYoung
All My Soldiers...
17 Sep '06 20:251 edit
Originally posted by Crash16
Heres the riddle
A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say?
"I always lie"
or
"You will sentence me to six years in prison"
6. 17 Sep '06 20:28
True = 4years
False = 6years
Thus True + False = 10years.
I'm sticking to my Super Mario theory. Everything else is flawed. With a passion.
7. 17 Sep '06 21:29
He said "I am rich and will give you \$10000000 if you let me go free"
8. XanthosNZ
Cancerous Bus Crash
17 Sep '06 21:31
Originally posted by laedawg
He said "My Kingdom is not of this world." You see the man was indeed Super Mario himself. The King decided to let him warp to his own world and Mario made his exit at the nearest green pipe.
This is clearly the correct answer.
9. 17 Sep '06 21:33
He said, "You'll sentence me to six years in prison." If it was true, then the judge would have to make it false by sentencing him to four years. If it was false, then he would have to give him six years, which would make it true. Rather than contradict his own word, the judge set the man free.
10. 17 Sep '06 21:37
Originally posted by Crash16
He said, "You'll sentence me to six years in prison." If it was true, then the judge would have to make it false by sentencing him to four years. If it was false, then he would have to give him six years, which would make it true. Rather than contradict his own word, the judge set the man free.
Why wouldn't the King just give him 10years for being true and false?
11. 17 Sep '06 22:05
because he would rather give him a 1up and make him fight kupas
12. 20 Sep '06 21:28
I don't get it.
13. 20 Sep '06 23:35
Originally posted by Crash16
Heres the riddle
A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say?
I havn't scrolled down. I think he said "I'm going to prison for 6 years"
If the judge sentences him to 4, then he was wrong, and it should have been 6, but if he sentences for 6, then it was true, and should have been 4.
14. 20 Sep '06 23:37
Was his first name OJ?
15. 03 Oct '06 20:01
the man said nothing | 983 | 3,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-04 | latest | en | 0.980374 |
https://jp.mathworks.com/matlabcentral/cody/problems/12-fibonacci-sequence/solutions/1849568 | 1,575,974,696,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527205.81/warc/CC-MAIN-20191210095118-20191210123118-00333.warc.gz | 410,991,156 | 15,500 | Cody
Problem 12. Fibonacci sequence
Solution 1849568
Submitted on 15 Jun 2019 by Wang Xiaodan
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
n = 1; f = 1; assert(isequal(fib(n),f))
2 Pass
n = 6; f = 8; assert(isequal(fib(n),f))
3 Pass
n = 10; f = 55; assert(isequal(fib(n),f))
4 Pass
n = 20; f = 6765; assert(isequal(fib(n),f)) | 161 | 456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-51 | latest | en | 0.731039 |
http://polyforms.eu/stackedpolytans/Combitans.html | 1,603,510,920,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00172.warc.gz | 88,446,843 | 1,596 | ## Combined Set of Stacked Tritans and Tetratans
There are no constraints for constructions with this set and the total number of triangles 66*4+10*3=264+30=294 can be well factorized, because 294=2*3*7*7.
Therefore posssible heights for prisms are 6, 7, 14 and 21 with cross-sections consisting of 49, 42, 21 or 14 tans, respectively.
With height=6 a perfect replica of a tan can be made. The layers for this prism are here.
For height=7 on the right side of the picture we can get a 3*7*7 box and prisms with rotational symmetry like the cross tower or the castle. To get solutions for these figures I split the whole prism into two parts, because the computer program is more efficient on small areas.
The prism with height=14 has a symmetric pentagon as cross-section whereas the cross-section of the prism with height=21 is irregluar.
Two congruent rectangular trapeziums of height=7 can be used to make prisms with a pentagon or trapeziums as cross-section. The layers for the two prism are here.
Two congruent isosceles trapeziums of height=7 can be used to make prisms with a hexagon, a trapezium or a parallelogram as cross-section. The layers for the two prism are here.
If we allow holes in the cross-section, a square with a rectangular hole or a triangle with a triangle hole are possible symmetric bottom areas. Starting with a 3h*4h rectangle we can leave out a hexatan hole to get an area of 42 tans for the cross-section. Some examples are shown.
Here are the layers for the first two figures and for the prisms with hexatan holes.
Back
Home | 400 | 1,568 | {"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-2020-45 | longest | en | 0.886029 |
https://brilliant.org/problems/the-first-anniversary-problem/ | 1,529,390,495,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861980.33/warc/CC-MAIN-20180619060647-20180619080647-00111.warc.gz | 556,557,531 | 10,956 | # The First Anniversary Problem
Algebra Level 5
Let $$n$$ be a positive integer and $$x_1, x_2, \ldots, x_n$$ be integers satisfying the following conditions:
1. $$-1 \le x_i \le 2$$ for all $$i = 1, 2, \ldots, n$$
2. $$\displaystyle \sum_{i=1}^n x_i = 19$$
3. $$\displaystyle \sum_{i=1}^n x_i^2 = 99$$
Out of all possible $$n, x_1, x_2, \ldots, x_n$$ satisfying the above conditions, find the sum of the minimum and the maximum values of $$\displaystyle \sum_{i=1}^n x_i^3$$.
× | 181 | 483 | {"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} | 3.03125 | 3 | CC-MAIN-2018-26 | latest | en | 0.502123 |
https://www.electronics-lab.com/community/index.php?/profile/35374-evengravy/ | 1,685,764,039,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00175.warc.gz | 798,907,697 | 20,271 | Electronics-Lab.com Community
# evengravy
Members
33
Never
0
1. ## Help with voltage conversion
Hi I am creating a project with the new arduino due and I need to convert the voltage output (0.55v to 2.75v) from the DAC into a range that is linear from -5v to +5v I have looked through national semi app note 31 and in the upper right corner of page 6 there is a circuit for voltage conversion but I'm really stuck of where to go from here, I have no idea how to calculate the offset I need and the gain required. http://www.ti.com/ww/en/bobpease/assets/AN-31.pdf I have an accurate and stable +10v and -10v supply to work with, Can anyone help? I'd really appreciate any input, Thanks and regards
2. ## Help with a Tube EQ circuit
Hi, I have been given a circuit diagram of a tube based EQ unit and I need some help if anyone could help me I'd be gratefull. Basically the circuit is in two parts a Tube base amplification stage and a "passive" inductor based EQ circuit that "appears to me" to be in the feedback path of the Tube amplification stage allowing for both boost's and cuts to be applied at the set frequency ranges. The EQ section is fairly well understood and I have worked out the multi tapped inductor values based on the given frequencies and capacitor values. I am having some issue understanding the Tube topology however, I am not at all experienced with Tube designs. My initial analysis from staring at the circuit for some time now is that the tube amp circuit is essentially forming an op amp. What I would like to do is replicate the EQ unit but replacing the Tube amp stage with a solid state amplification stage keeping the EQ pretty much as is. Could anyone help me understand how to do so? could I possibly use standard opamps in some configuration? or could someone help me "decode" the circuit so that it could be replaced with BJT's of JFet stages? Sorry for the long post, thanks in advanced, even
3. ## hv transformer wiring
hero, thanks very much, even
4. ## hv transformer wiring
Thanks hero, I've been working with hv dc ozone up until now so sorry for the confusion, just so I'm clear could you check my wiring diagram? Thanks, even
5. ## hv transformer wiring
thanks hero, one question, does it matter which wire of the secondary is grounded? there are no markings on the secondary wires at all, and they are both the same colour. am I correct in thinking one secondary wire is + and one - that should be grounded? if so how would I identify which is +? my thinking is that the turns ratio is 1:13 so if I apply mains voltage to the secondary side and measure the primary side as it produces a safer measurable voltage this way, I can determine the + and - wires on the secondary. Not sure if this is even necessary? Thanks, even
6. ## hv transformer wiring
My current thinking is to wire the secondary to the electrode like the image attached, Since one side of the secondary is contained within a sealed pyrex tube I am presuming this is correct but I would appreciate input/corrections, thanks.
7. ## hv transformer wiring
hi, I am building an ozone generator using a high voltage neon sign transformer and I am a little unsure of wiring. The transformer I have is brand new and runs off 230v and produces [email protected] I want to be 100% sure of how to wire this safely for obvious reason so I would appreciate your assistance guys/gals. Basically the nst that I have bought is like in the image attached. The ozone generator electrodes will consist of a scroll of stainless steel mesh that is pushed into a Pyrex test-tube where it unfurls against the walls and a lug on the end makes connection to the circuit through a nut and bolt through the screw-on lid. A stainless steel mesh tube on the outside of the test-tube acts as the other mesh electrode and is connected to ground via the terry clip mountings that hold the assembly in place. The secondary side of my transformer is not marked, do I simply ground one side of the secondary (which will be the outer electrode) and tie this to the case also?
"A mic preamp used an input transformer about 30 years ago but now they use an electronic circuit." Yea, I understand what you mean but there is a certain tonal advantage to using a transformer also, I see what you are saying about the impedance, adding a transformer would be wasting the high input impedance of the Jfet, If I were to choose a transformer however what specs would be suitable? impedance in/out and ratio however, in order to achieve the electrical balancing, could I use similar spec matched jfet pair (2SK-369) and the same circuit doubled, one circuit for positive input and one for the negative? then merely add phantom power and dc blocking caps to the input?
audioguru: "its input impedance is 10M ohms so why ruin it with an input transformer" I really need to make this ciruit balanced and add to add phantom power, which is why I thought of adding a transformer, is there another simple way of achieving these functions without the input transformer? I know that it is possible to add dc blocking caps to allow phantom power to be blocked from entering the input, but is there a way to work with the balanced input that emulates the transformer function. even.
thanks again hero999, even
I think I may have been wrong with the output impedance values that I posted, basically the circuit (well two in series will make up the mic pre) will need to be suitable for connection to a line level mixer, after some more reading it seems as if "The impedance of the line input is high -- about 10K to 1 Meg ohms" http://www.tape.com/resource/impedance.html
so, if i understand what you're saying correctly I need to increase current through q4 such that it can drive the low impedance load but such that it is low enough so as it wont overheat. Is this correct? I have no idea how to go about running output impedance simulation, I am using multisim. also in regard to the input transformer, I understand that the input impedance is high with the jfet input, but i'm unsure how to add phantom power using dc blocking capacitors, this is why I thought a transformer would be easier, but i'll scratch that idea. thanks guys.
I know it has been a while since I posted on this circuit but I have been flat out with other projects until now, so, I have modified the circuit to include diodes between the base of Q4 and ground as you suggested and this has reduced the current to just below 6ma, I'm not sure how to simulate for output impedence, It will need to drive a 600ohm load max (i think around 75Ω @1kHz) how would I go about simulating for this, I have been searching online for answers but I can't seem to find any relative info. any help is greatly appreciated, also the ac coupling hasn't been added as I intend to incorporate a transformer on the input, I will get to changing this and the bias resistor values soon. | 1,537 | 6,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-23 | latest | en | 0.957178 |
http://mathoverflow.net/questions/108164/small-ramsey-number-and-brooks-theorem?sort=votes | 1,469,744,427,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829320.70/warc/CC-MAIN-20160723071029-00198-ip-10-185-27-174.ec2.internal.warc.gz | 161,105,658 | 17,759 | # small Ramsey number and Brooks' Theorem
I'm studying about Graph Ramsey Theory now. Starting this study, I'm reading Chvatal and Harary's series of papers. In the second paper (V.Chvatal, F.Harary, Generalized ramsey theory for graphs,Ⅲ. Small off-diagonal numbers, Pacific Journal of Mathematics 41, No.2, 1972, pp.335-345), I can't understand the proof of $r(C_4,K_4)=10$. Their proof is like following.
Let $G$ is arbitrary simple graph of order 10 with point independence number $<4$. It is sufficient to prove $G$ contains $C_4$. From $G$'s point independence number is $<4$, $G$'s (point) chromatic number is $\ge4$. Hence by Brooks' theorem either $K_4$ (and hence $C_4$) is contained in $G$, or the degree of each point of $G$ is at least four. If the first case occur, we have done. If the second case occur, we also have $C_4$ in $G$ by the following lemma (I omit the proof of this lemma but it's not so difficult).
Lemma. If a graph $G$ with $p$ points has minimum degree $d$ and $d(d-1)>p-1$, then $G$ contains $C_4$.
I can't understand how to use Brooks' Theorem. I only succeed to derive the maximum degree of $G$ is greater than 3. How to derive that the minimum degree of $G$ is greater than 3 from Brooks' Theorem? Chvatal and Harary's proof is wrong as it is? or not? (If you have other elegant proof of $r(C_4,K_4)=10$, then It also help me.)
supplementation:I got a (awkward?) proof of $r(C_4,K_4)=10$. The proof is like following.
For lower bound, we use Chvatal-Harary theorem.
For upper bound, we think about above graph $G$. Using $r(C_4,K_3)=7$, easily we have there is no vertex with degree $\le2$. By lemma, we have at least one vertex (say $u$) whose degree is 3.
Claim. The subgraph induced by vertices non-adjacent to $u$ contains $2K_3$.
The subgraph induced by vertices non-adjacent to $u$ has 6 vertices. So we have two triangles $T_1,T_2$ in this subgraph. If $T_1,T_2$ has two common vertex, we get $C_4$. If $T_1,T_2$ has only one common point, let $T_j=v_0v_1^jv_2^j$. Let $w$ be the other vertex non-adjacent to $u$. Then $v_2^1w$ isn't an edge by symmetry and avoiding $C_4$, namely $v_0v_1^1wv_2^1$. We also have edge $v_1^2w$ by avoiding 4 independent vertices $v_2^1wv_1^2u$.By symmetry, we have $C_4$, namely $v_0v_1^2wv_2^2$ and it's a contradiction.
Claim. The neighborhood subgraph $N(u)$ of $u$ is $\bar{K_3}$.
If not, the neighborhood subgraph of $u$ is an isolated vertex $v_1$ and an edge $v_2v_3$. $v_2$ and $v_3$ has at least one edge to $T_1\cup T_2$, since their degree $\ge3$. If $v_2$ and $v_3$ has edges to common triangle, we get $C_4$. So if we let $T_j=w_1^jw_2^jw_3^j$, we can assume there are edges $v_2w_1^1, v_3w_1^2$. ($v_1,v_2$ has no other edges to $T_1\cup T_2$.) Then both edges $v_1w_2^1,v_1w_3^1$ cannnot be exist. So we can assume there isn't edge $v_1w_3^1$, then we have edge $w_3^1w_1^2$, since otherwise we have 4 independent vertices $v_1v_2w_3^1w_1^2$. By symmetry, we also have edge $w_1^1w_3^2$. So we have $C_4$, $w_1^1w_3^2w_1^2w_3^1$. It's a contradiction.
Now, we have $T_j=w_1^jw_2^jw_3^j$ and 6 edges $v_iw_i^j$. Then we have edge $w_1^1w_1^2$, since otherwise $w_1^1w_1^2v_2v_3$ form 4 independent vertices. By symmetry, we have $C_4$, namely $w_1^1w_1^2w_2^2w_2^1$. It's a contradiction.
-
We capitalize names in English, e.g. Ramsey, Chvatal, Harary. Not capitalizing means disrespect. – GH from MO Sep 26 '12 at 20:32
Than you very much GH. I'm not good at English, so I did not know that convention. If you haven't warn me, I have disrespect to their for long time. – Yuta Suzuki Sep 27 '12 at 3:54
The adjective "abelian" is a notable exception to GH's rule. – j.c. Sep 27 '12 at 17:21
Several mathematicians, most notably the Bourbaki school, are against naming concepts after mathematicians. In particular, they propose that names that become part of standard terminology should be de-capitalized, e.g. abelian group, galois group, noetherian ring etc. – GH from MO Oct 5 '12 at 18:42
It seems as if the Chvatal, Harary proof has a logical gap, and your proof seems to be missing some details.
Here is a proof that is based on Brook's Theorem. We plagiarize you and start by noting that $r(C_4,K_3)=7$, and so $G$ has minimum degree at least 3. We then plagiarize Chvatal, Harary and note that $G$ has chromatic number at least 4. Thus, by Brook's Theorem, $G$ has maximum degree $\Delta(G)$ at least 4. Let $v$ be a vertex of maximum degree and let $N(v)$ be the neighbours of $v$ and let $S(v)$ be the non-neighbours of $v$. Since $G$ has no $C_4$, note that each vertex in $S(v)$ has at most one neighbour in $N(v)$. Thus, the minimum degree of the subgraph induced by $S(v)$ is at least 2. This rules out $\Delta(G)=9,8$, or $7$.
If $\Delta(G)=6$, then there are at least three vertices $x,y,z \in N(v)$ which are not adjacent to any vertex in $S(v)$. Since $x$ has degree at least 3 in $G$, it must be adjacent to at least two other vertices in $N(v)$, which creates a $C_4$.
If $\Delta(G)=5$, then $G[S(v)]$ is a graph on 4 vertices with minimum degree 2. Such a graph necessarily contains a $C_4$.
We now suppose $\Delta(G)=4$. In this case, $G[S(v)]$ is a graph on 5 vertices with minimum degree 2. Thus, every cycle of $G[S(v)]$ must be of length 3 or 5. If $G[S(v)]$ contains a $C_5$, then $G[S(v)]=C_5$, since adding any chord to a $C_5$ produces a $C_4$. Thus, each vertex in $G[S(v)]$ has exactly one neighbour in $N(v)$. Hence, $G[N(v)]$ must be a matching $\{ab, cd\}$ of size 2, else $G$ has a vertex of degree 2. It follows that each vertex in $N(v)$, has at least one neighbour in $S(v)$. Thus, one vertex (say $a$) has exactly two neighbours in $S(v)$, while $b,c$, and $d$ have exactly one neighbour in $S(v)$. Let $x,y,z$ be the vertices in $S(v)$ which are not adjacent to either $b$ or $c$. If any of $xy,yz,xz \notin E(G)$, then $G$ has a stable set of size 4. Thus, $xyz$ is a triangle. This contradicts that $G[S(v)]=C_5$.
The only remaining possibility is that $G[S(v)]$ is a bowtie. In particular, $G$ has two non-adjacent vertices $u$ and $v$ of degree 4 such that $N(u)$ and $N(v)$ are disjoint. Thus, the subgraph $H$ of $G$ induced by $N(u) \cup N(v)$ has minimum degree at least 2. In particular $H$ contains a cycle $C$. It is easy to verify that if $|C|=3,4,5,6,$ or $7$, then $G$ contains a $C_4$ since $H$ cannot contain a vertex with two neighbours in $N(u)$ or two neighbours in $N(v)$. Thus, $|C|=8$. But then $H=C_8$ else $H$ contains a cycle of smaller length. Thus, every other vertex of $H$ is a stable set in $G$, a contradiction.
Note that this proof avoids the use of the minimum degree lemma.
-
Thank you very much. I'll add details my proof(?). Your proof is interesting to me. But I can't understand the sentence "This implies that some vertex of $C$ has degree 2 in $G$, since $N(v)$ only contains four vertices, a contradiction. ". Why there aren't no vertex in $N(v)$ which has two neighborhood in $S(v)$? Perhaps this is easy question, sorry. – Yuta Suzuki Sep 27 '12 at 4:56
You are right. That part was totally unclear and misleading. I edited accordingly. – Tony Huynh Sep 27 '12 at 15:28
Thank you very much for your edit. I was confused in "Hence, $G[N(v)]$ must be a matching $ab,cd$ of size 2" because I thought the case when c,d has two edges to $S(v)$. But by easy argument, this case can be excluded. And I can't understand (perhaps because of my weak ability of English) the sentence "Thus, every other vertex of $H$ is a stable set in $G$". Does this means "if let $C_8=v_1\dots v_8$, then $v_1v_3v_5v_7$ form stable set"? Finally, by your kindly help, we got two proofs which one is avoids Brooks' theorem, the other avoids above lemma. (to be continue) – Yuta Suzuki Sep 27 '12 at 17:53
Then above Chvatal-Harary's proof is wrong? or not? How do you think about it? I want to know this mainly, so I can't let your answer accepted, but your answer is very helpful, so I voted up yours. – Yuta Suzuki Sep 27 '12 at 17:56
Yes, every other vertex of $H$ means what you think it means. It seems that the Chvatal-Harary proof is wrong, but I cannot say what they definitely had in mind. At the very least it is unclearly written. Finally, another idea I had in mind is to use Hadwidger's Conjecture. Since $\chi(G) \geq 4$, we have that $G$ contains a $K_4$-minor, and hence a $K_4$-subdivision $H$. Note that $H$ contains at least one subdivided edge, else $G$ contains a $K_4$. By looking at which edges of $H$ are subdivided, and how $H$ attaches to the rest of $G$, I think we can prove the theorem. – Tony Huynh Sep 27 '12 at 22:30 | 2,800 | 8,616 | {"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} | 3.609375 | 4 | CC-MAIN-2016-30 | latest | en | 0.921392 |
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# Blind Mice go to Pluto
Students "observe" an imaginary new planet in our galaxy from the relative distances of a ground-based telescope, the Hubble Telescope, and a fly-by mission. After recording their observations and discussing the differences, they compare their... (View More)
# Landsat Change Over Time
An introduction to the Landsat satellite is presented through a poster with accompanying images, information and classroom lesson. The poster displays 10 pairs of international Landsat images highlighting changes over time from both natural and... (View More)
Audience: Middle school
Materials Cost: Free
# MRC: Final Design (Grades 3-5)
In this lesson, students will design a planetary surface rover to conduct a planetary surface investigation. It uses the 5E learning cycle and is designed around an essential question: How will creating a prototype of your rover help you prepare for... (View More)
Keywords: Careers
# MRC: Final Design (Grades 6-8)
In this lesson, students will design a planetary surface rover to conduct a planetary surface investigation. It uses the 5E learning cycle and is designed around an essential question: How will creating a prototype of your rover help you prepare for... (View More)
Keywords: Careers
# MRC: Landing, Moving, and Surviving Conditions (Grades 6-8)
This lesson plan uses the 5E learning cycle and is designed around an essential question: Why is the method you chose for landing your Rover on Mars the best one for your mission? The lesson objectives include: examine different methods for landing... (View More)
# MRC: Investigate Mars (Grades 3-5)
This lesson plan uses the 5E learning cycle and is designed around an essential question: How do I know when I’ve found important information in my reading? Learning objectives include: identify important details in informational texts; learn and... (View More)
# MRC: Landing, Moving, and Surviving Conditions (Grades 6-8)
This lesson plan uses the 5E learning cycle and is designed around an essential question: Why is the method you chose for landing your Rover on Mars the best one for your mission? The lesson objectives include: examine different methods for landing... (View More)
# MRC: Investigate Mars (Grades 6-8)
This lesson plan uses the 5E learning cycle and is designed around an essential question: How do I know when I've found important information in my reading? Learning objectives include: identify important details in informational texts; learn and or... (View More)
Keywords: Language arts
# A New Spin on Solar Wind: The Moon, Magnetosphere, and ARTEMIS
This is a lesson about the solar wind, Earth's magnetosphere, and the Moon. Participants will work in groups of two or three to build a model of the Sun-Earth-Moon system. They will use the model to demonstrate that the Earth is protected from... (View More)
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# MRC: How Do I Measure This? (Grades 6-8)
This is a lesson about measurement and cratering. Learners will read about the origin of the foot as a standardized unit of measure, work collaboratively to conduct an experiment about cratering, and collect and record data to draw logical and... (View More)
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http://www.cjmenet.com.cn/CN/10.3901/JME.2015.01.043 | 1,670,098,348,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710936.10/warc/CC-MAIN-20221203175958-20221203205958-00820.warc.gz | 61,823,105 | 14,907 | • CN：11-2187/TH
• ISSN：0577-6686
• 机构学及机器人 •
### 考虑球面副间隙的4-SPS/CU并联机构动力学分析
1. 西安理工大学机械与精密仪器工程学院 西安 710048
• 出版日期:2015-01-05 发布日期:2015-01-05
• 作者简介:王庚祥，男，1985年出生，博士研究生。主要研究方向为并联机器人及机构学。
• 基金资助:
国家自然科学基金(51275404)和西安理工大学优秀博士学位论文研究基金(102-211209)资助项目
### Dynamics Analysis of 4-SPS/CU Parallel Mechanism with Spherical Joint Clearance
WANG Gengxiang, LIU Hongzhao
1. School of Mechanical and Precision Instrument Engineering, Xi’an University of Technology, Xi’an 710048
• Online:2015-01-05 Published:2015-01-05
Abstract: Dynamics model of 4-SPS/CU parallel mechanism with spherical joint clearance is formulated based on augmented method in the equations of motion for mechanism systems. The driven chain in this parallel mechanism is regarded as a unit via the method of mass moment in order to simplify the dynamics modeling. The kinematic model of spherical joint with clearance is established on the ground of contact type between elements in the clearance joint. The contact deformation and relative contact velocity are presented on account of the above research. A new contact force model is presented, which replaces the constant stiffness coefficient of the Flores contact model with nonlinear stiffness coefficient considering to the coupling between the contact stiffness of contact elements and dynamics model of mechanism system. The normal contact force between the contact elements is evaluated based on the new contact model, and the tangential contact force is derived according to the Coulomb friction model with dynamic correction coefficient to ensure the stability of numerical computation. The contact forces between the socket and ball are converted to the mass center of driven chain and upper platform interconnected by this clearance joint, and the converted contact force is integrated to the generalized force in equations of motion of 4-SPS/CU parallel mechanism to introduce the effect of spherical joint clearance. The Baumgarte stabilization method is used for solving dynamics equations in order to stabilize the constraints violation. The influence of spherical joint with clearance on the dynamics performance of this parallel mechanism is predicted by numerical analysis | 528 | 2,210 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-49 | latest | en | 0.780644 |
https://doctormyessay.com/2022/05/10/hypothesis-testing-for-differences-between-groups-no-plagiarism/ | 1,679,606,150,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945183.40/warc/CC-MAIN-20230323194025-20230323224025-00192.warc.gz | 257,782,571 | 19,766 | # Hypothesis testing for differences between groups /no plagiarism!!
#### Introduction
Hypothesis testing is a foundational statistical technique used to make decisions about a hypothesis. A hypothesis test compares two mutually exclusive statements (null hypothesis and alternative hypothesis) where only one is true. Hypothesis testing can determine statistical significance by examining the probability that a given result would occur under the null hypothesis. For this assignment, you will perform hypothesis testing on the differences between the two groups.
#### Preparation
The dataset contains the following variables:
• Clinic1 (total number of visits per month for clinic 1).
• Clinic2 (total number of visits per month for clinic 2).
#### Instructions
An investor needs to make a decision on whether to acquire one of two medical clinics based on their productivity, as measured by the total number of visits per month. You have been asked whether there is a significant difference in the total number of visits per month between clinic 1 and clinic 2.
For this assignment, perform hypothesis testing on the differences between two groups in the Assignment 2 Dataset. Create an appropriately labeled Excel document with your results. Also, write an analysis of the results in a Word document. Insert the test results into this document (copied from the output file and pasted into a Word document). Refer to the “Copy From Excel to Another Office Program” resource for instructions.
The numbered assignment instructions outlined below correspond to the grading criteria in the Hypothesis Testing for Differences Between Groups Scoring Guide, so be sure to address each point. You may also want to review the performance-level descriptions for each criterion to see how your work will be assessed:
1. Generate a hypothesis about the difference between the two groups in a dataset.
• State null hypothesis and alternative hypothesis as an explanation and math equation.
2. Identify the appropriate statistical test of the difference between the two groups in a dataset.
3. Perform an appropriate statistical test of the difference between two groups in a dataset.
4. Interpret statistical results of data analysis and state whether to accept or reject the null hypothesis based on the p-value and an alpha of .05.
• Interpret p-value and statistical significance.
5. Write a narrative summary that includes practical, administration-related implications of the hypothesis test.
Your assignment should also meet the following requirements:
• Written communication: Write clearly, accurately, and professionally, incorporating sources appropriately.
• Length: 2–3 pages.
• APA format: Cite your sources using the current APA format.
• Font and font size: Times Roman, 12 point.
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Order your essay today and save 10% with the coupon code: best10 | 922 | 4,606 | {"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} | 3.5 | 4 | CC-MAIN-2023-14 | latest | en | 0.907119 |
https://codegolf.meta.stackexchange.com/questions/12380/default-format-for-polynomials/12384 | 1,631,906,766,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00562.warc.gz | 244,381,563 | 32,418 | # default format for [polynomials]
Some languages do not have native ways of representing polynomials, others do. Many times a list of coefficients (in one or the other order) works well. In order to avoid explicitly specifying this over and over again in all challenge, I'd like add that to the tag info.
• I'd avoid declaring a default format that needs to be looked up in the tag wiki or on meta. It should be clear from the challenge how I/O can be formatted. That said, having some guidance for the challenge author on what I/O formats to allow would certainly be helpful, much like the recent question on what outputs decision problems should allow. May 14 '17 at 17:09
• I'm usually for defaults, but I think polynomials are something challenge authors should specify the format for. Our defaults are generally for common programmatic elements (inputs, strings, graphics) to be clarified and standardized across languages. But polynomials are something most languages don't have a existing concept for, so a default would really be making a definition from scratch.
– xnor
May 14 '17 at 20:22
• @xnor I disagree, many languages (certainly algebraic or numerical langauges) do have built in polynomial types or at least packages that support them. May 14 '17 at 21:10
• @MartinEnder I think there are just a few accepted ways to represent a polynomial anyway, and I think (similarly as in code-golf or string or popularity-contest or quine) to define a default would give us a base to avoid repeating the definition over and over again, as we also do not explicitly say that we use characters to measure code length by default. May 14 '17 at 21:22
### List of coefficients
A list of coefficients
• in ascending order [a(0),a(1),a(2),...,a(n)] OR
• in descending order [a(n),a(n-1),...,a(0)]
If the degree is known, it may be zero padded as
[a(0),...,a(n),0,...,0]
or
[0,...,0,a(n),...,a(0)]
• And this 3*x^2*y+4*z; is it not a polynomial?
– user58988
May 16 '17 at 5:19
• Multivariable polynomials can be writen as polynomial in one variable with polynomials of the remaining variable as coefficients. May 16 '17 at 18:24
### Default representation in [language]
Many langauges, especially ones that include a CAS have native ways of representing polynomials.
## Beware function representations
Be careful about allowing a polynomial to be represented by a function that computes it, in languages where functions are objects. For example, in Python the polynomial 3*x^2+2 could be represented by the function object lambda x:3*x*x+2.
Take the challenge to compose polynomials. One could submit
(.)
where . is Haskell's function composition operator: if f and g are functions, then f.g is a function that takes x to f(g(x)). This would cut off the possibility for a much more interesting answer in the coefficient list representation.
lambda f,g:lambda x:f(g(x))
• Same goes for expressions representing the polynomial in languages with symbolic names, e.g. 3x^2+2 is a valid Mathematica expression that will remain unevaluated (unless it can be simplified). While not a function, it can be manipulated and evaluated quite easily. That said, most Mathematica answers would probably start by converting to this format first, so I'm not sure it's a bad thing to allow the expression right away. May 15 '17 at 9:04 | 794 | 3,335 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-39 | latest | en | 0.925978 |
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1. AmmarTa, thanks for the big help on the second part, I now understand the second section of the question Am I getting it completely wrong for the first part of the question? Is there a way to prove there is positive roots?
2. (Original post by BanglaBOSS)
AmmarTa, thanks for the big help on the second part, I now understand the second section of the question Am I getting it completely wrong for the first part of the question? Is there a way to prove there is positive roots?
My first reply proved it, that's all you need to state.
3. (Original post by BanglaBOSS)
Thanks for the replies guys! I used b^2 - 4ac to get 5k^2 +4. As f(x) has 2 roots, 5k^2 + 4 > 0. I then subtracted 4 and divided by 5 on both sides, to get k^2 > -4/5.
Your logic is completely wrong. Why are you assuming f(x) has two distinct roots? That's what you're trying to show.
You're trying to show thag, no matter the value of k, the quadratic f(x) has two distinct roots.
That means you want to show that the discriminant of the quadratic is positive for all real k. NOT(!!!) assume that it is positive ajd then solve for k (as you are doing).
The discriminant is 5k^2 + 4 which is the sum of two terms, one of which is always positive and the other is always non-negative, so the discriminant 5k^2 + 4 is always positive. Hence the quadratic always has two distinct roots, no matter the value of k - which is what you were asked to show.
4. (Original post by AmmarTa)
As far as I can see, to solve part 2, you'd work out the 2 roots using the quadratic formula [(-b +- root(b^2 - 4ac))/2a] using your coefficients of a, b and c:
...
Please don't post full solutions - it's against the terms of the forum.
Note that for part 2 there is no need to be solving quadratics at any stage! If you think about the relationship between the coefficients and the product of the roots of the equation then the required result drops out in one line
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https://classroom.thenational.academy/lessons/applying-and-consolidating-column-method-for-addition-and-subtraction-cgvpcc | 1,701,353,376,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100227.61/warc/CC-MAIN-20231130130218-20231130160218-00044.warc.gz | 212,065,969 | 26,820 | # Applying and consolidating: Column method for addition and subtraction
In this lesson, we will consolidate our knowledge of the column method for both addition and subtraction by recapping the method, completing examples and applying our learning.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.Mrs Crane has decided to use partitioning to solve the following equation: 3000 - 2149 = ... from the following options, which is partitioned most efficiently?
1/5
Q2.Mrs Crane has decided to use partitioning to solve the following equation: 7000 - 3621 = ... from the following options, which is partitioned most efficiently?
2/5
Q3.Will the answer to the following equation: 3000 - 2149 = ... be odd or even?
3/5
Q4.Will the answer to the following equation: 7000 - 3621 = ... be odd or even?
4/5
Q5.Will the answer to the following equation: 5000 - 2672 = ... be odd or even?
5/5
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.Mrs Crane has decided to use partitioning to solve the following equation: 3000 - 2149 = ... from the following options, which is partitioned most efficiently?
1/5
Q2.Mrs Crane has decided to use partitioning to solve the following equation: 7000 - 3621 = ... from the following options, which is partitioned most efficiently?
2/5
Q3.Will the answer to the following equation: 3000 - 2149 = ... be odd or even?
3/5
Q4.Will the answer to the following equation: 7000 - 3621 = ... be odd or even?
4/5
Q5.Will the answer to the following equation: 5000 - 2672 = ... be odd or even?
5/5
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Applying and consolidating: Column method for addition and subtraction
Using what we have learnt today, can you answer the following questions?
Q1.Which column will regrouping occur in in the following equation: 4586 + 4812 = ...
1/5
Q2.Will regrouping happen in the following equation: 5592 + 3564 = ....
2/5
Q3.Has Mrs Crane calculated the following equation correctly?
3/5
Q4.Has Mrs Crane solved the following equation correctly?
4/5
Q5.Has Mrs Crane solved the following equation correctly?
5/5
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Applying and consolidating: Column method for addition and subtraction
Using what we have learnt today, can you answer the following questions?
Q1.Which column will regrouping occur in in the following equation: 4586 + 4812 = ...
1/5
Q2.Will regrouping happen in the following equation: 5592 + 3564 = ....
2/5
Q3.Has Mrs Crane calculated the following equation correctly?
3/5
Q4.Has Mrs Crane solved the following equation correctly?
4/5
Q5.Has Mrs Crane solved the following equation correctly?
5/5
# Lesson summary: Applying and consolidating: Column method for addition and subtraction
## Time to move!
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