url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://www.epa.gov/riskassessment/expobox/routes/inh-cal.htm | 1,438,222,425,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042987034.19/warc/CC-MAIN-20150728002307-00308-ip-10-236-191-2.ec2.internal.warc.gz | 434,790,236 | 6,047 | EPA-Expo-Box (A Toolbox for Exposure Assessors)
# Inhalation
#### Calculations
Using EPA’s current methodology, it is unnecessary to calculate an inhaled dose when using dose-response factors from IRIS in a risk assessment. However, inhalation risk assessments may require that an adjusted air concentration be used to represent continuous exposure. The adjusted air concentration (Cair-adj) may be estimated as shown below. As described under the Methods tab, for noncarcinogens, the air concentration is adjusted based on the time over which exposure occurs (i.e., the exposure duration). For carcinogens, the concentration is averaged over the lifetime of the exposed individual (often assumed to be 70 years).
Cair-adj=Cair x ET x 1day/24 hours x EF x ED/AT
Where:
Cair = Concentration of contaminant in air (mg/m3)
ET = Exposure time (hours/day)
EF = Exposure frequency (days/year)
ED = Exposure duration (years)
AT = Averaging time (days)
• The concentration in air (Cair) is either a measured or modeled value. Air concentrations may be measured in the breathing zone of individuals using personal monitoring equipment or in indoor or outdoor air using stationary or portable monitoring devices. Air measurements may represent gas phase or particulate-phase contaminants, or both.
• Temporal parameters in the equation include the following:
• Exposure time (ET) and exposure frequency (EF) refer to the frequency with which the exposure occurs and might be provided in hours per day and days per year, respectively.
• Exposure duration (ED) is the amount of time that an individual or population is exposed to the contaminant being evaluated and is typically given in years.
• Averaging time (AT) is the amount of time over which exposure is averaged and is equal to ED for assessing non-cancer risks. For chronic assessments (e.g., cancer), potential lifetime average daily dose (LADD) is calculated in which lifetime (LT, in days) is substituted for AT.
In some cases, it may be necessary to calculate an inhalation dose using the equation below. This algorithm can be used to calculate the average daily potential dose from inhalation of a contaminant in air. The potential dose of a contaminant is the product of the contaminant concentration, inhalation rate, exposure time, exposure frequency, and exposure duration divided by the product of averaging time and body weight. The equation parameters below must be defined for each inhalation exposure scenario, and all parameters must be expressed in consistent units; in some cases, unit conversion factors may be necessary. Average Daily Dose (ADD) is generally expressed as mass of contaminant per unit body weight over time (e.g., mg/kg-day).
ADD = Cair x InhR x ET x EF x ED/BW x AT
Where:
ADD = Average daily dose (mg/kg-day)
Cair = Concentration of contaminant in air (mg/m3)
InhR = Inhalation rate (m3/hour)
ET = Exposure time (hours/day)
EF = Exposure frequency (days/year)
ED = Exposure duration (years)
BW = Body weight (kg)
AT = Averaging time (days)
See above for descriptions for many of the relevant equation parameters. The additional parameters are described below.
• Inhalation rate (InhR) represents the volume of air inhaled over a specified timeframe. Long-term inhalation rates are typically expressed in units of m3/day. Short-term inhalation rates are typically indexed to activity levels and are expressed in units of m3/hour or m3/minute. Assessors should choose inhalation rate data that best represent the population for which exposures are being assessed. For example, some assessments might focus on certain subsets of the general population (e.g., older adults) whose inhalation rates might vary from those of the general population. Chapter 6 of the Exposure Factors Handbook: 2011 Edition (U.S. EPA, 2011) provides inhalation rate data for various age groups (see the Factors tab in this module).
• Body weight (BW) of an individual, typically expressed in kilograms (kg), is also included so that the dose is normalized to that value. Sometimes the inhalation rate is already normalized to body weight (e.g., in units of m3/kg-day). In this case, a separate term for body weight would not be necessary.
Additional information on exposure scenarios involving the inhalation route can be found in the Indirect Estimation Module in the Approaches Tool Set of EPA-Expo-Box. The Exposure Calculation Spreadsheet (XLSX) (360 K) estimates the inhalation exposure concentration or inhalation dose when user-defined values are entered for the various exposure parameters that are indicated in BOLD.
Top of Page | 985 | 4,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | latest | en | 0.913117 |
https://kmmiles.com/355-1-km-in-miles | 1,719,068,865,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00355.warc.gz | 301,654,505 | 5,941 | kmmiles.com
Search
# 355.1 km in miles
## Result
355.1 km equals 220.5171 miles
You can also convert 355.1 km to mph.
## Conversion formula
Multiply the amount of km by the conversion factor to get the result in miles:
355.1 km × 0.621 = 220.5171 mi
## How to convert 355.1 km to miles?
The conversion factor from km to miles is 0.621, which means that 1 km is equal to 0.621 miles:
1 km = 0.621 mi
To convert 355.1 km into miles we have to multiply 355.1 by the conversion factor in order to get the amount from km to miles. We can also form a proportion to calculate the result:
1 km → 0.621 mi
355.1 km → L(mi)
Solve the above proportion to obtain the length L in miles:
L(mi) = 355.1 km × 0.621 mi
L(mi) = 220.5171 mi
The final result is:
355.1 km → 220.5171 mi
We conclude that 355.1 km is equivalent to 220.5171 miles:
355.1 km = 220.5171 miles
## Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case three hundred fifty-five point one km is approximately two hundred twenty point five one seven miles:
355.1 km ≅ 220.517 miles
## Conversion table
For quick reference purposes, below is the kilometers to miles conversion table:
kilometers (km) miles (mi)
356.1 km 221.1381 miles
357.1 km 221.7591 miles
358.1 km 222.3801 miles
359.1 km 223.0011 miles
360.1 km 223.6221 miles
361.1 km 224.2431 miles
362.1 km 224.8641 miles
363.1 km 225.4851 miles
364.1 km 226.1061 miles
365.1 km 226.7271 miles
## Units definitions
The units involved in this conversion are kilometers and miles. This is how they are defined:
### Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.
### Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada. | 647 | 2,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-26 | latest | en | 0.817916 |
https://web2.0calc.com/questions/domain_120 | 1,642,798,928,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303709.2/warc/CC-MAIN-20220121192415-20220121222415-00317.warc.gz | 682,564,305 | 5,295 | +0
# Domain
0
94
1
What is the sum of all values of y for which the expression (y + 6)/(y^2 - 5) is undefined?
Aug 9, 2021
#1
0
y^2 - 5 = 0
y^2 = 5
y = +- √5
√5 + -(√5) = 0.
Aug 9, 2021 | 96 | 195 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-05 | latest | en | 0.837317 |
https://byjus.com/question-answer/the-velocity-associated-with-a-proton-moving-in-a-potential-difference-of-1000-v-is-3/ | 1,643,433,327,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00208.warc.gz | 202,742,722 | 22,546 | Question
# The velocity associated with a proton moving in a potential difference of 1000 V is $$\displaystyle 4.37\times { 10 }^{ 5 }\ { ms }^{ -1 }$$. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Solution
## The associated wavelength is obtained from de-Broglie equation.$$\displaystyle \lambda = \frac {h}{mv} = \frac {6.626 \times 10^{-34} }{0.1 \times 4.37 \times 10^{5} } = 1.516 \times 10^{-28} m$$ChemistryNCERTStandard XI
Suggest Corrections
0
Similar questions
View More
Same exercise questions
View More
People also searched for
View More | 184 | 632 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-05 | latest | en | 0.770094 |
https://www.vrcbuzz.com/allcalculators/statistics-calculators/page/2/ | 1,695,620,328,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506686.80/warc/CC-MAIN-20230925051501-20230925081501-00163.warc.gz | 1,153,857,932 | 23,351 | ## Moment coefficient of skewness Calculator for ungrouped data
Skewness Skewness is a measure of symmetry, or more precisely, the lack of symmetry. A data set is symmetric if it looks the same to …
## Mean absolute deviation calculator with examples
Mean absolute deviation Calculator Use this calculator to find the Mean Absolute Deviation using frequency distribution (uniform or discrete),frequencies and grouped data Mean absolute deviation …
## Five (5) number summary calculator with examples
Five number summary for ungrouped data A five number summary is a quick and easy way to determine the the center, the spread and outliers …
## Deciles Calculator for Grouped Data with Examples
Deciles for grouped data Deciles are the values which divide whole distribution into ten equal parts. They are 9 in numbers namely \$D_1, D_2, \cdots, …
## Plus Four Confidence Interval for Proportion Calculator
Plus Four Confidence Interval for Proportion Calculator Use this calculator to compute the plus four confidence interval for population proportion. Plus Four Confidence Interval Calculator …
## Deciles Calculator for Ungrouped Data with Examples
Deciles for ungrouped data Deciles are the values of arranged data which divide whole data into ten equal parts. They are 9 in numbers namely …
## Confidence Interval For Population Variance Calculator
Confidence Interval For Population Variance Calculator Use below Confidence interval for population variance calculator to calculate degees of freedom,chi-square critical values, confidence limits. Calculator Confidence …
## Bowley’s Coefficient of Skewness Calculator for grouped data
Bowley’s Coefficient of Skewness for grouped data Bowley’s coefficient of skewness is based on quartiles of the data. It is based on the middle 50 …
## One Mean Z test Calculator with Examples
One mean z test Calculator with examples In this tutorial we will discuss z test calculator for testing population mean with step by step numerical …
## Paasche Price Index Number Calculator
Paasche Price Index Number Calculator Use below calculator to calculate Paasche Price Index Number based on input value of Based Year Prices,Base Year Quantities,Current year … | 434 | 2,218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.686222 |
https://socratic.org/questions/how-do-you-find-the-x-and-y-intercepts-of-6y-3x-6 | 1,638,757,733,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363229.84/warc/CC-MAIN-20211206012231-20211206042231-00495.warc.gz | 588,173,782 | 5,877 | # How do you find the x and y intercepts of 6y=3x+6?
Mar 6, 2018
$x$-intercept is $- 2$ and $y$-intercept is $1$
#### Explanation:
To find $x$-intercept of any line (actually, I should say any curve) just put $y = 0$
and to $x$-intercept of any line (again any curve) just put $x = 0$.
Hence $x$-intercept of $6 y = 3 x + 6$ is given by $6 \cdot 0 = 3 x + 6$ or $3 x = - 6$ i.e. $x = - 2$
and $y$-intercept of $6 y = 3 x + 6$ is given by $6 y = 3 \cdot 0 + 6$ or $6 y = 6$ i.e. $y = 1$.
graph{6y=3x+6 [-2.958, 2.042, -1, 1.5]} | 245 | 534 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-49 | latest | en | 0.853656 |
https://news.ycombinator.com/item?id=21182982 | 1,576,213,757,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540548544.83/warc/CC-MAIN-20191213043650-20191213071650-00239.warc.gz | 472,110,654 | 13,232 | Elliptic Curve Cryptography Explained 471 points by fangpenlin 66 days ago | hide | past | web | favorite | 44 comments
So, to summarise:* Choose an Elliptic Curve (EC) and a large prime;* The rational points on the EC modulo the prime give you a group;* Your private key is a large integer N;* Perform Diffie-Hellman-Merkle-Williamson (DHMW) in the group;* You now have a shared point in the EC;* Use a symmetric cipher using the shared point as the key.Performing DHMW goes as follows:* A and B agree on a point P in the EC;* A and B each choose a large number as their secret key;* For simplicity, or confusion, we'll say that X's secret key is X;* They compute AP and BP as their public keys and exchange them;* When A receives BP they compute A(BP), B does likewise;* Now they share a point: (AB)P.If you have the language of groups, and understand DHMW in that context, ECDHMW is trivial, as is (this version of) EC PK-Cryptography.TL;DR : ECC is simply Symmetric Cryptography using a key that's been agreed by using DHMW in an Elliptic Curve group.
For those like me, who are confused by the acronym DHMW, it's another name for 'Diffie-Helman'.Diffie and Helman felt that the entire scheme was based on work that Ralph Merkle did (same guy as Merkle Trees), and so should be included in the scheme name, and I think that Williamson refers to Malcolm J. Williamson, a GHCQ employee, who had secretly developed the same system 7 years before Diffie and Helman. What I don't know is why Williamson and not Cocks and Ellis are part of the acronym.
Cocks and Ellis developed what we know as RSA, Malcolm Williamson developed what most people call Diffie-Hellman.I did give the full version before the acronym on the fifth line of my comment, so if people are confused it might be why I'm referring to DHMW and not just DH. Reading up about it all it's pretty clear that Merkle and Williamson should get more credit than they do.
Nobody should get credit for inventing something in secret. You make your decision — invent for corporations or the military industrial complex and get the big bucks or the jingoistic frisson, or for academia and get the accolades. No way a person should double dip.
so you would revoke credit from Turing?
Credit for what? The theoretical work for which Turing is most known - the concept of Turing machines and undecidability - was published publicly and before the war. He did not contribute to the design of a physical computer until after the war. Colossus, the code-breaking "computer", was not of his design (and the man who did design it is not at all famous for it).
Did anyone else discover what he did before his work became public?
yep, eniac and univac were the the first publicly known computers. it became known later that turing’s work predated eniac.
It's not a question of credit, it's a question of attribution and historical accuracy.
Tooting my own horn just a little bit, I spend the first 3 chapters of my book on exactly this topic: https://www.amazon.com/Programming-Bitcoin-Learn-Program-Scr...There are exercises that can help you learn it from scratch.
I've led a number of people through these exercises at this point. They're good for making the math concrete.
I can recommend https://www.amazon.com/Elliptic-Tales-Curves-Counting-Number.... It goes in-depth into the motivating use cases for elliptic curves (as well as their group law) and explains some very elegant theorems (Bézout's theorem) as well as open problems in Maths (the Birch-Swinnerton-Dyer conjecture).
> even elliptic curves with no rational points on them still have points in finite fieldsThat’s interesting. How is that possible?
Take the elliptic curve y^2 = x^3 - 5 that has no rational solutions [1]. But it's easy to see (from brute force) that, for example, 3^2 - 5 = 22 = 27^2 (mod 101).
This [0] was also a very good introduction to elliptic curve cryptography written by a friend of mine in 2015. Highly recommended.
Yes absolutely, I have Andrea's whole series on the topic bookmarked, it is very well done.Another, slightly more topical / simple explanation [0] was posted by Cloudflare almost exactly 6 years ago, I still point people to it as well.
That's a better introduction, imo
I don't have much experience in cryptography, so this may be a stupid question, but I've always wondered about whether Elliptic Curve cryptography opens up some possibilities for partially decentralizing encryption standards.My understanding is that end-users of ECC have to decide which curves to use, and constructing a curve de-novo isn't a choice laymen or crypto end-users should ever make, so a set of standardized curves are issued by standards bodies, such as NIST.Cryptographers endorse the math of ECC as not known to be decipherable, provided that the chosen curve (defined eg by 4 points) isn't specifically constructed to be easily compromised; and the problem becomes whether the curve-issuing standards bodies, such as NIST, are acting in the interests of state security agencies (for argument's sake, lets say NSA) who have vested interests in crypto users adopting curves that NSA can break.Could an international, decentralized curve be constructed by standards bodies from several geopolitical adversaries such as US, China, Russia and Turkey all simultaneously issuing 1 point each to create a combined 4-point curve, so that no single standards body has opportunity to purposely make the result insecure?
> Could an international, decentralized curve be constructed by standards bodies from several geopolitical adversaries such as US, China, Russia and Turkey all simultaneously issuing 1 point each to create a combined 4-point curve, so that no single standards body has opportunity to purposely make the result insecure?The process would be a little more complicated than simply choosing 4 points, but yes you could do something close enough to this in theory.In actual practice there's really only two sets of curves most people [0] implement, and just about everyone agrees to use:- The NIST p curves- Curve25519/Curve448 by Bernstein &c.Which you use tends to fall on what side of the crypto divide you fall on:- NIST p curves if you care about governmental compliance such as FIPS- Bernstein &c's curves if you care about security and distrust NIST created curves.I personally fall on the latter side but spend most of my time doing software for the former. :) A promised later revision to FIPS will standardize Bernstein &c's curves.Almost nobody implements negotiating arbitrary curves. That's really unsafe.So, as 'tptacek would say... just use Curve25519.[0]: I'm talking about major software such as TLS, VPNs, SSH, Kerberos... etc.
I don’t follow. How can the curve itself be insecure? Isn’t the security generated by the random points on the curve?
There are all sorts of pitfalls and potential attacks even if the curves are truly random[1]. Some curves also require you to verify whether public keys are valid points on the curve (and their security breaks if you don't do so). So they're harder to implement safely. Others are hard to implement in a way that avoids timing attacks.This is one of the reasons more paranoid people have generally preferred Curve25519 over the NIST curves -- the NIST curves have very arbitrary base point values which (in theory) could have been backdoored. NIST later published a proof that if you hashed some other arbitrary values, you get the base points -- but then the follow up question is where did the other arbitrary values come from.
It's misleading to claim that 25519 pubkeys don't require validation: In some applications validation isn't required, in other applications validation is simpler but still required.Marketing bullet points aren't a replacement for careful cryptographic review.
https://en.wikipedia.org/wiki/Dual_EC_DRBGThat's the case I know of.Basically you mathematically design a curve that makes it look like it's random numbers but the numbers aren't actually random.
While I don't know enough about this stuff to answer your main question, I will point out that there are some popular curves which aren't tied to any one nation or agency.Ed25519 specifically has major contributions from 5 different people from multiple nationalities.It's not quite the decentralized ideal you talk about, but it's somewhat close!
ed25519 is the signature system based on curve25519
There have been attempts in this direction, it's absolutely doable to use a shared seed where many people contribute a random value.However ultimately a different approach was chosen that solves the same problem: You don't choose an arbitrary curve, instead you define a set of properties that you want your curve to have, based on security, speed and ease of implementation. Then you end up picking the very first curve that fulfils that property.That's how Curve25519 was created. There's very little wiggle room in there.Also it should be said that the hypothesies of choosing a "bad" curve that noone can spot are very hypothetical. We know these NIST curves have an unexplained random seed, but noone has an idea how this could've been used for a backdoor.
This is really helpful!I find RSA really simple to understand, but ECC has always seemed a lot harded to grok - I think I finally get it!
This is the best explanation I have seen so far. Some aspects are still left in the shadow, but most points are now connected.
second that! great article.
It's great that people enjoy this-- but I've never seen explanations that centered on the group law as actually being especially useful for building insight producing understandings complex cryptosystems.I think it's a lot more useful to start from a generic group: e.g. you have some group G generated by g, and a group operator ... plus a way to (de)serialize members, a method to identify the additive identity, maybe a hash function and not much else.From that point you can build a pretty solid understanding of real cryptosystems.Knowledge of how to implement a particular group law is necessary to build one from scratch-- which people should almost never be doing, esp as it requires a serious amount of number theory to do well-- but it doesn't give any real insight into the cryptography, not even any intuition as to why the DL problem in some groups is believed to be hard.I think that understanding ECC starting at group law is kind of like understanding quick sort by studying CMOS logic: you can't implement quicksort in practice without (someone) building some digital logic, but if you're down in the trees it's not particularly easy to see the forest.When I've trained people on the subject I've enjoyed using a sequence that goes like:1. Generic group, how we can construct efficient 'multiplication' from only addition (I prefer to call the group operator addition, but I talk about both notations and how both are equally valid). General intuition about how algebra essentially JustWorks(tm) in finite fields.2. CDH and Diffe-hellman, the fact that the hardness of the DL problem (in particular groups) is a strong assumption.3. Constructing a pedersen commitment using the additive homomorphism 'in the exponent', establishing the security assumption in pedersen commitments that the two generators have an unknown DL between them.4. Constructing a chameleon hash function-- a trap-door hash function where someone knowing a secret can freely construct collisions at will-- by violating the pedersen security assumption.5. Constructing a schnorr digital signature by feeding the output of a chameleon hash back into itself, creating an impossible causal loop that can only be resolved using the trapdoor.6. Extending the causal loop: creating an OR proof (A signature that shows I know at least one of two secrets).[FWIW, we cover 4,5,6 in this paper https://pdfs.semanticscholar.org/4160/470c7f6cf05ffc81a98e8f... for the purpose of describing a more efficient than typical OR proof.]7. From there extending into other more elaborate protocols like range proofs (built from OR proofs of related secrets), private set-intersection (built from polynomial evaluation in-the-exponent), etc.8. And critically, interesting attacks on these systems, with a particular focus on how fragile they can be:8a. Linearly related nonce attacks against signatures. (unfortunately, these attacks are easier taught if the signature algorithm is understood as a hidden linear system, then the attack is just a 'as many knows as unknowns attack', but I strongly prefer the causal loop understanding of signatures for the intuition it creates for more complex protocols).8b. Random-modular-subset sum attacks on naive blind signatures and naive multi-signature.8c. At least one attack that break the generic group abstraction. I'm fond of invalid key attacks on DH with twist-insecure curves. (This is probably the only example that I think is important to teach that requires some group law understanding).If I do get into group law, I think it's interesting to talk about optimizations. Particularly, projective coordinates, precomputation, addition/subtraction ladders and signed digit representations, doubling sharing for multi-exp.Bos-coster for sum-of-many products is an especially fun and easy algorithm to implement (similar kind of fun as implementing the peeling algorithm for decoding sparse linear codes in F(2)).I'd like to have a 9 on that list that goes into the building blocks of proof techniques for these systems e.g. programmable oracles/forking lemma... but that is enough out of my area of expertise that I don't know how to teach it effectively.
This is a fine comment but is a little like saying "the best way to understand elliptic curves is not to study elliptic curves", which, sure, a first course in abstract algebra will be more profitable for designing entirely new constructions (or evaluating them from first principles) --- which most practitioners don't actually do.Some people are just interested in curves themselves. That's the point of an article like this. If you want to learn how to build a signature scheme, there are good sources for that as well.
I think it would be more precise to say "The best way to understand elliptic curve cryptography is to study discrete log cryptography, not to study elliptic curves".There is a tremendous amount of ECC tutorials that walk people mechanically through the group law, I think I've seen a dozen new ones posted on HN in the past year. I don't think I've ever see an article on actual elliptic curve cryptography (beyond briefly explaining DH, as this one does).Like-- say the article were instead explaining elliptic curves as part of factoring, how would it differ? (This one would omit the DH part, but many don't even have that.)Personally, I think they're not that useful from the perspective of cryptography. This level of article doesn't bring you to number theory insight or anything like that. They are useful for slavishly coding your own ECC thing without any understanding, but the results of doing that are inevitably totally insecure toys. Hopefully that isn't something "most practitioners" are doing-- hopefully they're linking libsodium or something. :)If you like it, great! I think I provided a pretty useful curriculum that someone could self-study if they wanted to actually dive deep on the cryptography.
They're not meant to be useful from the perspective of cryptography, in the same way as learning textbook RSA is not useful in that sense. They're useful in the sense that people want to have some faint clue of what's happening.
Maybe it's just a difference in how I look at it, but knowing what group law is doing is not a useful clue to me.From being able to apply group law you couldn't predict the properties or (most of) the vulnerabilities. (or even performance, since these things don't generally cover projective coordinates).At least the way I see it is if you asked about how a spell checker worked, and then I set out and explain how a digital adder circuit works from the gate level. You can't build a spell checker without adders yet you wouldn't be informed. :)This sort of thing also strikes me as the sort of thing that more "feeling of understanding" without creating much understanding. I seldom think it's good to do that, though it isn't always harmful.But learning styles is certainly something that differs a lot between people, so I certainly don't think my experience is universal by any means.I can say that personally learning to ignore the machinery made my understanding grow 100x faster and directly resulted in constructing more than a few publishable (and widely deployed) designs.
UWO?
Yup! | 3,571 | 16,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.951402 |
http://mathematica.stackexchange.com/questions/4030/are-there-advantages-to-using-generalized-part-extraction-instead-of-specialized | 1,469,528,366,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824757.8/warc/CC-MAIN-20160723071024-00164-ip-10-185-27-174.ec2.internal.warc.gz | 154,167,225 | 20,113 | # Are there advantages to using generalized Part extraction instead of specialized functions like First, Last?
There are many ways to extract different parts of lists in Mathematica. For example, the first part of a list v can be accessed either as v[[1]] or First[v] or Take[v,1]. Likewise the last element can be accessed as v[[-1]], Last[v] or Take[v,-1]. Similarly v[[2;;]] is equivalent to Rest[v] and to Drop[v,1], and v[[;;-2]] is equivalent to Most[v] and Drop[v,-1].
The more specialized functions First, Last, Rest and Most are marginally more efficient. On the other hand, it makes sense to use the more general Part ([[ ]]) or Take/Drop if it is part of a calculation that also requires parts to be accessed that do not have a specialist function.
Beyond these considerations, are there any reasons to prefer Part over Take/Drop or the more specialized functions, other than personal preferences over coding style?
-
You cannot make assignments to First, Last, Rest, or Most the way you can with Part. Therefore, there is greater consistency in using Part for all operations. See this answer for an example of and argument for this consistency.
Also, you must change functions if you need to update your code to index a different element or change an element to a Span. By using Part from the beginning you have the greatest flexibility without changing functions. This can make comparing different code revisions easier. It can also make a user defined function more flexible, e.g.: f[x_, part__ : 1] := foo @ x[[part]]
Less importantly x[[1]] is more terse than First@x, especially in the FrontEnd using 〚 〛, and I am a fan of terse coding.
-
+1, for discussing the maintenance cost. – rcollyer Apr 9 '12 at 0:29
In a sea of brackets, x[[1]] is less comprehensible than First@x and as a result probably less maintainable, albeit one character longer :) But whilst developing x[[1]] is easier to modify. Conclusion, develop with Part and deploy with First. – image_doctor Apr 9 '12 at 8:07
Am the only one with Second, Third, Fourth, Fifth and Sixth defined too? :) – image_doctor Apr 9 '12 at 8:10
@image_doctor I find that using \[LeftDoubleBracket] and \[RightDoubleBracket] greatly helps readability compared to [[ and ]] in the frontend – acl Apr 9 '12 at 9:25
First/Last are so elitist. Why not FirstAmongEquals and LastButNotLeast? – Jens Apr 9 '12 at 15:20
Putting aside any minor performance differences, the specialized functions look cleaner and clear in intent in cases where you know for sure that you will not be needing any arbitrary part. Some examples that come to mind are
1: expression//Timing//First (* Get only the timing *)
2: (... Sow[x]...)//Reap//Last (* Get only sowed values *)
3: Most@#/Last@#&@CoefficientList[poly] (* normalize by coefficient of highest degree term *)
4: Rest@#/First@#&@Eigenvalues[matrix] (* normalize by largest eigenvalue *)
5: Most@#/Rest@#&@list (* divide each element by the next *)
I'm not very into golfing, but I do like minimizing my keystrokes that are spent doing nothing (and avoid the mouse). If I'm at the end of a long expression and I need to take the first element, I'd simply use Part to extract it rather than move the cursor all the way back to insert a First@. In other words:
something + (insert very long expression here)[[1]]
something + First@(insert very long expression here)
If I change my mind and decide I need a different element, I have the flexibility to do so and also include all elements with [[;;]]. In such cases, you can't simply stick in a //First at the end, because of the very low precedence of Postfix — the expression returned might not be what you intended to get.
In the case of something + (insert very long expression here) (already bracketed as shown) you can avoid the mouse or excessive left-cursor-key by pressing: Ctrl + . + . and then left-cursor a single time. Four keystrokes total. – Mr.Wizard Apr 9 '12 at 14:11
@Mr.Wizard Not a very big fan of Ctrl+. — never been able to wrap my head around what it's going to select next. Of course, that might be just due to unconventional learning and a reluctance to use it, but I might get around to taming it like I did for Infix :) – R. M. Apr 9 '12 at 14:53 | 1,037 | 4,275 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-30 | latest | en | 0.885475 |
https://mail-block.com/need-979 | 1,674,900,867,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499541.63/warc/CC-MAIN-20230128090359-20230128120359-00265.warc.gz | 399,600,301 | 4,562 | Here, we will show you how to work with Hard math questions and answers. Our website can solve math word problems.
The Best Hard math questions and answers
Hard math questions and answers can be found online or in math books. What is log x? In mathematics, log (also written logarithm) is a way of expressing the natural (base 10) logarithm of a number. It is used to show how much one number is raised to another. The logarithm of a number with base 10 is equal to the power to which that number must be raised to equal its logarithm with base e (the natural logarithm). For example: The base 10 logarithm of 12 is 2, whereas the base e logarithm of 12 is 2. This means that 12 must be multiplied by 2^e to equal its base 10 logarithm, or 2. Similarly, the base 10 logarithm of 100 is 3 and its base e logarithm of 100 is 3. This means that 100 must be multiplied by e^3 to equal its base 10 logarithm, or 6.
Solving exponential equations can be a bit tricky. Most of the time you will need to use an inverse function to get from one number to the other. However, it is possible to solve some equations without using such techniques. Here are some examples: One way to solve an exponential equation is to use a logarithm table. For example, if you have an equation of the form y = 4x^2 + 32, then you would use the logarithm table found here. Then, you would find that log(y) = -log(4) = -2 and log(32) = 2. These values would be used in the original equation to obtain the solution: 4*y = -2*4 + 32 = -16 + 32 = 16. This value is the desired answer for y in this problem. Another way to solve an exponential equation is by using a combination of substitution and elimination. You can start by putting x into both sides of the equation and simplifying: ax + b c where a c if and only if b c/a . Then, once this is done, you can eliminate b from each side (using square roots or taking logs if necessary) to obtain a single solution that does not involve x . c if and only if , then you can substitute for y in both sides, thus eliminating x
There are many online resources that can help you solve algebra equations. For example, you can use an online calculator to solve your math equations. You can also use an online equation solver to get answers for some of the most common algebra equations. Another option is to use an online calculator app on your smartphone or tablet. These apps allow you to do basic calculations and solve simple algebra equations. They also give you step-by-step instructions on how to solve complex equations. However, keep in mind that these tools are not perfect and may not always give accurate results. It’s also important to be careful when using them because they could lead to incorrect answers if you don’t follow the instructions carefully.
When you are working with a y-axis, it is important to know the purpose of your data. What do you need the y-axis to tell you? Are you trying to show the relationship between two variables? Are you trying to show the relationship between one variable and another? Is it simply a visual representation of your data? Once you know what your y-axis is for, then you can start working on how to display it. For example, if you are showing the relationship between two variables and want to use a line graph, then you can use a line graph. If you are showing the relationship between one variable and another and want to use a scatter plot, then you can use a scatter plot. And if it’s a visual representation of your data, then you can display the data in any way that works best for your situation. With some experience, you will be able to figure out which type of visual is needed for your specific situation.
The best math experts are those who can teach math with clarity and ease. They have a good grasp of the subject matter and can explain it in an organized and insightful way. As well, they have the patience to work with students who might not be up to snuff when it comes to understanding certain concepts. Finally, they have a passion for the subject that shows in their teaching. Clearly, no two people are alike, so finding someone that you click with is key if you want your lessons to be memorable and effective. However, there are some qualities that all great math teachers should share. First and foremost is enthusiasm—it’s one of the most important factors in any lesson. You need to be engaged in order to make sure your students are too; otherwise, you’re just wasting your time. Next is patience—you need to be able to sit back, listen to what your students have to say, and take their feedback seriously. Lastly, you need to know the subject matter inside and out since this is one area where mistakes will happen more often than not.
It’s really helpful. I never do my homework cut I either forget or I run out of time. I can do it at school though. Again, I rarely do. This app allows me to get the answers and add the work later. It’s amazing because it helped me bring my C of 71% in Algebra 1 to an A of 106%. Thanks whoever made this app. You have helped me and my future.
Quina Long
This app is so helpful and amazing. I just think there should be a geometry future to find the area and perimeter of a shape and maybe some more features. but otherwise, if you are reading this to find out if you should get this you really should and it not only solves the problem but explains how you can do it and it shows many different solutions to the problem for whatever the question is asking for you can always find the answer you are looking for.
Kamila James | 1,240 | 5,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-06 | latest | en | 0.934463 |
https://www.ehow.com/how_8077169_calculate-much-lumber-need.html | 1,545,211,192,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831933.96/warc/CC-MAIN-20181219090209-20181219112209-00158.warc.gz | 854,204,850 | 15,547 | # How to Calculate How Much Lumber I Need
Lumber is usually sold either in bulk by the board foot or in 16 foot long boards. To bid on some jobs you may need to estimate board feed. The formula to calculate board feet is length in inches multiplied by width in inches multiplied by height in inches divided by 144. Math phobic carpenters and contractors might make an estimate of the board feet they will need then add 20 percent to that. A more detailed approach, considering exactly what you are going to make and how you are going to make it, is more accurate.
### Things You'll Need
• Pencil
• Paper
• Calculator (optional)
## Video of the Day
Draw your project with every board in that project on a piece of paper. Identify the width, thickness and length of every board. Make separate lists for every plank of different widths and every stud, strip or timber you will need. Think of planks of different widths and thicknesses, studs, strips and timbers as separate categories.
Add up all the lengths in all those categories. Divide the total length in feet in each category by 16 and round up. You might need, for example, three 16 foot lengths of 1 x 8 plank, two of 1 x 6 and one length of 4 x 4 timber.
Calculate the cuts you will need to make in all those 16 foot lengths of wood.
Add the end cuts for wood that has been slightly damaged in a lumber yard and allow for the kerfs, or the width of the cuts, you will make. For example, a 16 foot long plank will never yield two, true eight foot planks because some of the wood, the kerf, will be destroyed in cutting.
Draw representations of all the 16 foot lengths of each plank, stud, strip or timber with representations of the cuts you will make. Add the number of boards at your lumber yard that you expect to be cupped (warped horizontally) or bowed (warped vertically) from poor storage or handling. | 425 | 1,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.933637 |
https://www.jiskha.com/search?query=what+is+the+biggest+planet+mercury+venus+earth+jupiter | 1,603,252,414,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107875980.5/warc/CC-MAIN-20201021035155-20201021065155-00154.warc.gz | 797,452,872 | 14,360 | # what is the biggest planet mercury venus earth jupiter
10,040 results
1. ## Science
Can someone please check my answers? I would greatly appreciate it. Thanks! 1. Which of the following inner planets has virtually no atmosphere? a. Mercury*** b. Venus c. Mars d. Earth 2. Which of the following statements about the inner planets is true?
2. ## Science
A student uses a diagram to show the scale sizes of the planets. Which planet's diameter would be approximately the distance of the West Coast to the East Coast of the United States? Uranus Jupiter**** Mercury Neptune Check Please
3. ## Physics
A bag of sugar weighs 4.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixth that on Earth? 1 N Repeat for Jupiter, where g is 2.64 times that on Earth. 2 N Find the mass of the bag of sugar in kilograms
4. ## earth
The weight (V) of an object on Venus varies directly as its weight (E) on Earth. A person weighing 120 lb on Earth would weigh 106 lb on Venus. How much would a person weighing 180 lb on Earth weigh on Venus?
5. ## science
Test name: Moon and Planets Quick Check How does the mass of the Earth compare to the mass of the moon? (1 point) A. The mass of the Earth is about 40 times the mass of the moon. B. The mass of the Earth is about 50 times the mass of the moon. C. The mass
6. ## Math
One of Kepler's three laws of planetary motion states that the square of the period, P, of a body orbiting the sun is proportional to the cube of its average distance, d, from the sun. The Earth has a period of 365 days and its distance from the sun is
7. ## Science
A model shows that the moon has grown to twice its size, yet it has remained in the same place. In one to two sentences, explain how this would impact the gravity between Earth and the moon.(2 points)
8. ## science/astronomy
4. Which of the following is a common feature among the four inner planets? (1 point) atmosphere composition distance from the sun rocky surfaces**** temperature range 5. Which of the following is the smallest terrestrial planet? (1 point) Mars Mercury**
9. ## Science
Scientists have discovered a new planet. The planet is in a solar system whose star is similar in size to the sun and is about as far from the star as Venus is from the sun. The planet is similar in size to Earth, and its atmosphere is similar in thickness
10. ## english
:Select the correct excerpt. Which excerpt from Robert Stawell Ball's Great Astronomers is best supported by this diagram? The diagram show the earth in the middle the sun circling it then the moon circling earth but farther away with mars circling the
11. ## english
Which excerpt from Robert Stawell Ball's Great Astronomers is best supported by this diagram? Excerpt 1 These movements were wholly incompatible with the supposition that the journeys of Venus were described by a single motion of the kind regarded as
12. ## algebra
please review the questions and table and check my answers. tank you tutors. table planets' orbital velocity planet orbital velocity (mi/s) mercury 29.74 venus 21.76 earth 18.5 mars 14.99 jupiter 8.12 saturn 6.02 uranus 4.23 solve. show your work. express
13. ## Physics
the radius of the planet venus is nearly the same to that of the earth, but its mass is only 80% that of the earth. if an objects weighs we on the earth, what does it weigh on venus? calculate the value of g on venus.
14. ## Physics
The radius of the planet venus is nearly the same of that of the eart but it's mass is only 80% that of the earth. If an object weighs We on the earth (a) what does it weigh on venus (b) calculate the value of g in venus
15. ## Science help ms sue.
Multiple Choice 1. Which of the following terms refers to an object that orbits the sun and has enough gravity to be spherical but has not cleared the area of its orbit? (1 point) comet asteroid dwarf planet planet 2. Which of the following statements
16. ## science
The gravity of the sun has helped create and maintain the solar system we know today. Which planet does the sun exert the greatest gravitational pull upon?(1 point) Neptune Earth Mercury Jupiter plz help
17. ## Science
In a drawing that represents a scale model of the solar system, the sun is placed at the center and the planets are shown orbiting in circles around it. While in reality Mars is 1.5 AU away from the sun, in the model Mars appears at 30 cm from it. This
18. ## physics
The radius of the planet Venus is nearly the same as that of the earth, but its mass is only 80% that of the earth. If an object weighs W on the earth, what does it weigh on Venus? Calculate the value of 'g' on Venus
19. ## PHYSICS
the radius of the planet Venus is nearly the same as that of the earth.if an object weighs We on the earth,what is valu
20. ## physics
Calculate the mass of Jupiter knowing that a person who weighs 490 N on the earth's surface would weight 1293 N on the surface of that planet. Jupiter has a radius of 7.15 × 10^7 m.
21. ## Science
Multiple Choice 1. Which of the following terms refers to an object that orbits the sun and has enough gravity to be spherical but has not cleared the area of its orbit? (1 point) comet asteroid dwarf planet planet 2. Which of the following statements
22. ## PHYSICS
THE MASS OF VENUS IS 81.5% OF THE EARTH AND ITS RADIUS IS 94.9% THAT OF THE EARTH. A)COMPUTE THE ACCELERATION DUE TO GRAVITY ON THE SURFACE OF VENUS FROM THESE DATA.B)IF A ROCK WEIGHS 75.0N ON EARTH,WHAT WOULD IT WEIGH ON THE SURFACE OF VENUS?
23. ## science
Which planet by itself contains the majority of mass of all the planets? saturn jupiter uranus venus earth
24. ## physics
mercury rotates about its own axis once every 58.6 Earth days. a satellite orbits mercury once every murcurian day. how high is this satellite above mercury's surface? mass of mercury = 3.2x10^23 kg. the radius of mercury is 2400km.
25. ## astronomy 101
Given that the angular size of Venus is 53ŒŒ'' when the planet is 47000000kmkm from Earth, calculate Venus's diameter (in kilometers).
26. ## Science
Which of these statements suggests that the orbital period of Venus is shorter than that of Saturn? A) Saturn is much larger in size than Venus. B) Saturn is a gaseous planet; Venus is not. C) Venus is closer to the Sun than Saturn is. D) Saturn takes less
27. ## MATH
Earth, Jupiter, and Uranus all revolve around the sun. Earth takes 1 yr, Jupiter 12 yr, Saturn 30 yr, and Uranus 84 yr to make a complete revolution. One night, you look at those three distant planets and wonder how many years it will take before they have
28. ## physics
A bag of sugar weighs 3.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixth that on Earth? N Repeat for Jupiter, where g is 2.64 times that on Earth. N Find the mass of the bag of sugar in kilograms at
29. ## math
The volume of Jupiter is about 1.43*10^15 cubic kilometers. The volume of Earth is about 1.09*10^12 cubic kilometers. The number of Earths that can fit inside Jupiter can be found by dividing Jupiter's volume by Earth's volume. Find this quotient and
30. ## Physics
I have a hard time solving this problem. The weight of an object at the surface of a planet is proportional to the planet's mass and inversely proportional to the square of the radius of the planet. Jupiter's radius is 11 times Earth's and its mass is 320
31. ## physics help 3!! **
Based on the following data about planet X (which orbits around the Sun): Planet X's distance from Sun = 3.6*1012 m Planet X's radius = 2*106 m Planet X's mass = 8.2*1022 kg a.) Find gx, the size of the acceleration due to gravity on the surface of Planet
32. ## Physics
The mass of Venus is 0.8140 times that of earth, and its radius is 0.950 times that of earth. Calculate the acceleration of gravity at the surface of Venus. Express your answer in g's (i.e. as a ratio to our acceleration of gravity). 9.01×10-1 is the
33. ## physics
A bag of sugar weighs 3.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixth that on Earth? N Repeat for Jupiter, where g is 2.64 times that on Earth. N Find the mass of the bag of sugar in kilograms at
35. ## AP Physics
At their closest approach, Venus and Earth are 4.2x10^10m apart. The mass of Venus is 4.87x10^24kg, the mass of Earth is 5.97x10^24kg, and G=6.67x10^-11Nm2/kg2. What is the force exerted by Venus on Earth at that point?
36. ## physics
with what orbital speed will a satellite circle Jupiter if placed at a height of 7.80 * 10 ^ 6 m above the surface of the planet? The mass of Jupiter is 1.90 * 10 ^ 27 kg and the radius of Jupiter is 7.14 * 10 ^ 7 * m .
37. ## college algebra
Kepler’s third law. According to Kepler’s third law of planetary motion, the ratio t^2/r^3 has the same value for every planet in our solar system. R is the average radius of the orbit of the planet measured in astronomical units (AU), and T is the
38. ## Astronomy
Why is the greenhouse effect on Earth not as drastic as on Venus? -There is no carbon dioxide in Earth’s atmosphere. -The oceans dissolved much of the carbon dioxide. -Earth is smaller than Venus so it can’t hold the gases in the atmosphere. -There is
39. ## Physics
The radius of the planet Venus is nearly thesame as that of the earth but it's mass is only 80% that of the earth. If an object weighs 0.8 on the earth what does it weigh on Venus? Calculate the value of g on venus
40. ## Math
an object’s weight on a planet is directly proportional to the mass of that planet and inversely proportional to the square of the radius of the planet. jupiter is 318 times as massive as Earth and has a radius 11 times as large as that of Earth. if
41. ## jupiter and earth
The planet of Jupiter is more than 300 times more massive than earth.however,an object scarcely weighs 3 times as much on the surface of Jupiter as it does on the surface of the earth. How can that be? Determine the radius of Jupiter in terms of earth
43. ## physics
How many minutes would it take a radio wave to travel from the planet Venus to earth?
44. ## Physics for Life science
The planet Jupiter is more than 300 times more massive than Earth. However, an object scarcely weighs 3 times as much on the surface of Jupiter as it does on the surface of the Earth. How can that be? determine the radius of Jupiter in terms of Earth
45. ## Astronomy
Assume that right now the Moon is full, Mercury is at superior conjunction, Venus is at greatest western elongation, and Mars is at opposition. Draw a simple sketch of the solar system, showing the relative positions of the Sun, Earth, the Moon, Mercury,
46. ## earth
how long does it take for light from Jupiter to come to earth. for example tonight the planet Jupiter can be seen by the moon, 2/25/12 and we see the light from the planet and how old is that light. thanks. jerry in colorado USA
47. ## science help pls pls
The table below shows the average distances of Jupiter and Earth from the sun. Name of Planet Average distance from the sun (in AU) Jupiter 5.2 Earth 1.0 What is the difference between the orbital periods of Jupiter and Earth? 4.20 years 6.20 years 8.86
48. ## Math :(
Jupiter is 9.28 x 108 km from Earth when they are farthest apart. Mars is 410,000,000 km from Earth when they are farthest apart. a) How much farther away from Earth is Jupiter? b) When Jupiter is closest to Earth, it is only 629 million km away? What is
49. ## math
1. Number Sense: write three numbers that are greater than 1,543,000 and less than 1,544,000 2. Put the planets in order from the one closest to the sun to the one farthest from the sun. Earth 93,000,000 Jupiter 483,000,000 Mars 142,000,000 Mercury
50. ## physics
A laboratory centrifuge on earth makes n rpm (rev/min) and produces an acceleration of 3.30 g at its outer end. This centrifuge is now used in a space capsule on the planet Mercury, where g mercury is 0.378 what it is on earth. How many rpm (in terms of n)
51. ## Physics
Use the following numbers for this question. MEarth 5.98e+24 kg MJupiter 1.9e+27 kg MSun 2e+30 kg Earth-Sun distance 150000000000 m Jupiter-Sun distance 778000000000 m rE 6380000 m rJ 69000000 m G = 6.67e-11 Nm2kg-2 the acceleration due to gravity at the
52. ## geololgy
Earth and Venus are so similar in size and overall composition that they are almost “twins.” Why did these two planets evolve so differently? Why is Earth's atmosphere rich in oxygen and poor in carbon dioxide, whereas the reverse is true on Venus?
53. ## college
A spaceship lands on the planet whose radius is 5 times that of planet earth and whose mass is 25 times that of planet earth. Find the apparent weight of the astronaut on the second planet if the astronaut weighs 150lbs on planet earth
54. ## Astronomy
Why is the greenhouse effect on Earth not as drastic as on Venus? There is no carbon dioxide in Earth’s atmosphere. The oceans dissolved much of the carbon dioxide. Earth is smaller than Venus so it can’t hold the gases in the atmosphere. There is more
55. ## Third Side, Reply to Jen
Hi Jen, I've seen the drawing. My site is down at the moment. Draw a line from Venus (either of the two positions) to the line Earth-Sun at right angles. Let's call the distance Earth Venus d. Then you have a right angle triangle with a hypotenuse d, the
56. ## science
what is the biggest planet mercury venus earth jupiter
57. ## science
1. Scientists hypothesize that the sun formed ____. (1 point) less than a billion years ago from a cloud of gas and dust less than a million years ago from a chunk of Jupiter 2. Two of the inner planets are ____. (1 point) Venus and Saturn Mars and Venus
58. ## social
if the planet was to collapse where would everybody go. a.mars b.mercury c.venus d.jupiter
59. ## Math
There are 61 known moons. Mercury and Venus do not have moons, but each of the 7 other planets has at least 1. How do I find the % of total moons for each planet? Earth - 1 Mars - 2 Jupiter- 16 Saturn - 18 Uranus - 15 Neptune - 8 Pluto - 1 Thanks for your
60. ## Science
Please check my answers? My textbook won't load, so i've had to use google and guess. 1. Scientists hypothesize that the sun formed ____. (1 point) less than a billion years ago **from a cloud of gas and dust less than a million years ago from a chunk of
61. ## Science
Which of the following would be an appropriate use for a probe? (Choose two of the following) A. a study of the planet Saturn, it's rings, on to Jupiter and some of its moons B. military surveillance of enemy troops C. telecommunications D. a mission to
62. ## Physics
The moon Europa, of the planet Jupiter, has an orbital period of 3.55 days and an average distance from the center of the planet equal to 671,000 km. If the magnitude of the gravitational acceleration at the surface of Jupiter is 2.36 times greater than
63. ## physical science,
the planet jupiter is more than 300 times more massive than earth. However,an object scarcley weighs 3 times as much on the surface of jupiter as it does on the surface of the earth. how can that be? Determine the radius of jupiter in terms of earth radii
Okay I am really stuck on this question Tip: ^ = exponent Example: 6x10^7 7 is exponent Question: Venus is the second planet from the sun. Earth is the third planet from the sun. The distance from Venus to the sun is 6.72x10^7 and the distance from Earth
65. ## Physics
The radius of the planet Venus is nearly the same as that of the earth, but its mass is only 80% that of the earth, if an object weighs we on the earth, what does it weighs on Venus?
66. ## Social Impact of Technology
Which of these planets is in the so-called “Goldilocks Zone”? A. Earth B. Mars C. Jupiter D. Venus Ans: Earth
67. ## physics
An astronaut in full gear has a weight of 2900 N on Earth. Jupiter; weight on planet relative to earth is 2.53 and weight of a 150-lb person is 380. How much will the astronaut weigh on the Jupiter? W_{Jupiter} =________________ N Neptune; weight on planet
68. ## physics
The planet Jupiter is more than 300 times as massive as the Earth But it so happens that a body would scarcely weigh three times as much on the surface of Jupiter as it would on the surface of Earth. Can you think of an explanation for why this is so?
69. ## physics
The planet Jupiter is more than 300 times as massive as the Earth But it so happens that a body would scarcely weigh three times as much on the surface of Jupiter as it would on the surface of Earth. Can you think of an explanation for why this is so?
70. ## science with planets
What was the planet last to be discovered out of these? 1.Mars 2.Mercury 3.Neptune 4.Pluto 5.Saturn 6.Uranus 7.Venus
71. ## physics
Calculate the mass of Jupiter knowing that a person who weighs 490 N on the earth's surface would weight 1293 N on the surface of that planet. Jupiter has a radius of 7.15 × 107 m.
72. ## Physics
Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The only known force a planet exerts on objects on Earth is gravitational. (a) Calculate the gravitational force exerted on a 4.20
73. ## science
how are mercury and venus similar to earth?
74. ## help help help
The asteroid belt between Mars and Jupiter consists of many fragments (which some space scientists think came from a planet that once orbited the Sun but was destroyed). (a) If the center of mass of the asteroid belt is about 2 times farther from the Sun
75. ## science
how do jupiter's moons stay in orbit around jupiter They rotate around jupiter. If you rotate around an object you are accelerating toward it, even though the distance can stay the same. This so-called centripetal acceleration multiplied by the mass must
76. ## Science question
for my science project I have to chose a planet that I would like to visit. I choose "Venus." I know it's not possible to travel to Venus, but I don't believe man has visited in other planet in our solar system, besides the moon. and the moon isn't really
77. ## algebra
I need help badly with this assignment. If Venus is 33.3 million miles farther from the sun than Mercury, then Earth is 25.6 million miles farther from the sum than venus. When the total of the ditances for these three planets from the sum is 191.2 million
78. ## science
Multiple Choice 1. Which of the following terms refers to an object that orbits the sun and has enough gravity to be spherical but has not cleared the area of its orbit? (1 point) comet asteroid dwarf planet planet 2. Which of the following statements
79. ## ms. sue
Multiple Choice 1. Which of the following terms refers to an object that orbits the sun and has enough gravity to be spherical but has not cleared the area of its orbit? (1 point) comet asteroid dwarf planet planet 2. Which of the following statements
80. ## Science
Compare the nature greenhouse effect process with the human induced. 2. What are the greenhouse gases? Which one is the biggest concern? 3. What are scientists doing to collect data? 4. What role does the carbon cycle play in this problem? 5. What are the
81. ## Space craft
How will you overcome these challenges with the design of your spacecraft? (I decided to choose the inner planet mercury) I put: Mercury is very different than Earth. It’s far rougher than earth; the main problem is the temperature. Humans can’t just
82. ## Science
Which terrestrial planet has an atmosphere similar to Earth's? Mars* Venus None of the above
83. ## Astronomy
Jupiter is farther from the sun than is the Earth. In the course of several earth years what phases would you expect jupiter to exhibit to an observer on earth. explain
84. ## English
I have to list the verb or verbs. Could someone please check. 1. The plant Venus seems more like Earth in appearance than any other planet does. seems like
85. ## Physical Science
the planet Jupiter is more than 300 as massive as earth. But a body would scarecly weighs 3 times as much on the surface of Jupiteras it would on the surface of the earth. Can you think of an explanation for why this is so?
86. ## Math
Sunitha is interested to find the Relation between the diameters of Jupiter and the earth.From the given data,it is known that the volume of Jupiter is 729 times the volume of earth.she concluded that the diameter of Jupiter is 9 times the diameter of the
87. ## Science
1> What 2 types of electromagnetic energy have wavelengths of 0.76 cm? 2> What is the pressure at a depth of 3000 km inside the Earth? 3> How many inches of mercury are equivalent to 974.5 millibars of air pressure? 4> Jupiter's period of revolution is how
88. ## Math
The weight of an object on the planet Mercury is about 2/5 the weight of that object on the Earth. An object that weighs 30 pounds on Earth would weigh how many pounds on Mercury? Need solution because I think it is 7.5 but my parents think the answer is
89. ## 3rd grade - Solor System
Two Questions: If it takes Jupiter 9.9 hours to make one rotation around the earth. This means that one day on Jupiter is actually how long? This is as many as how many Earth days?
90. ## space
gowing 10,00 miles per hour how long will it take to get to jupiter from earth with earth the sun and jupiter alined like a straight line or just an estimation
91. ## Science
the force of gravity on jupiter is much stronger than the force of gravity on earth. which of the following explains why this is true. A. Jupiter's orbit is farther away from the Sun and Earths orbit B. Jupiter has more mass than Earth C. Jupiter's orbit
92. ## Science
How are earth and Venus similar? How is Venus different from Earth? Provide two similarities and two differences.
93. ## Science HELP PLS
how are earth and venus similar? how is venus different from earth? provide two similarities and two differences
94. ## Science
How are earth and Venus similar? How is Venus different from earth? (Provide two similarities and two differences)
95. ## Rabboni christian school(natural science)
writing a letter asking financial assistance on products to sell at jupiter planet asking to a manager from as you are in jupiter
96. ## Astronomy
A simple temperature model for the solar nebula disk is Tdisk(a) = 20 � SQRT 100 AU/ a K, (1) where a should be expressed in AU in this equation. (a) Find the temperature in the solar nebula disk at the orbit of Mer- cury, Earth and Jupiter. (b)
97. ## physics / astronomy
A simple temperature model for the solar nebula disk is Tdisk(a) = 20 × SQRT 100 AU/ a K, (1) where a should be expressed in AU in this equation. (a) Find the temperature in the solar nebula disk at the orbit of Mer- cury, Earth and Jupiter. (b)
98. ## Planets
Why is Venus hotter than Mercury even though Mercury is closest to the Sun?
99. ## physics
A bag of sugar weighs 3.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixth that on Earth? N Repeat for Jupiter, where g is 2.64 times that on Earth. N Find the mass of the bag of sugar in kilograms at
100. ## physics
The planet Jupiter has a satellite, Io, which travels in an orbit of radius 4.220×108 m with a period of 1.77 days. Calculate the mass of Jupiter from this information. | 5,806 | 23,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-45 | latest | en | 0.929464 |
https://oeis.org/A323424 | 1,642,854,202,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303845.33/warc/CC-MAIN-20220122103819-20220122133819-00090.warc.gz | 476,371,735 | 4,368 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A323424 Number of cycles (mod n) under Collatz map. 1
1, 1, 2, 1, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 3, 2, 2, 2, 3, 2, 3, 3, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 3, 2, 3, 2, 2, 2, 4, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 3, 2, 2, 2, 3, 3, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS This sequence is likely to be unbounded. LINKS Rémy Sigrist, Illustration for n = 13 FORMULA a(n) >= 2 for any n > 4 (as we have at least the cycles (0) and (1, 4, 2)). EXAMPLE The initial terms, alongside the corresponding cycles, are: n a(n) cycles -- ---- -------------------- 1 1 (0) 2 1 (0) 3 2 (0), (1) 4 1 (0) 5 2 (0), (1, 4, 2) 6 2 (0), (1, 4, 2) 7 3 (0), (1, 4, 2), (3) 8 2 (0), (1, 4, 2) 9 2 (0), (1, 4, 2) 10 2 (0), (1, 4, 2) 11 3 (0), (1, 4, 2), (5) 12 2 (0), (1, 4, 2) 13 3 (0), (1, 4, 2), (3, 10, 5) 14 2 (0), (1, 4, 2) 15 3 (0), (1, 4, 2), (7) 16 2 (0), (1, 4, 2) 17 2 (0), (1, 4, 2) 18 2 (0), (1, 4, 2) 19 3 (0), (1, 4, 2), (9) 20 2 (0), (1, 4, 2) PROG (PARI) a(n, f = k -> if (k%2, 3*k+1, k/2)) = { my (c=0, s=0); for (k=0, n-1, if (!bittest(s, k), my (v=0, i=k); while (1, v += 2^i; i = f(i) % n; if (bittest(s, i), break, bittest(v, i), c++; break)); s += v)); return (c) } CROSSREFS See A000374, A023135, A023153, A233521 for similar sequences. Cf. A006370. Sequence in context: A304486 A188550 A064122 * A334098 A263922 A057526 Adjacent sequences: A323421 A323422 A323423 * A323425 A323426 A323427 KEYWORD nonn AUTHOR Rémy Sigrist, Jan 14 2019 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified January 22 05:19 EST 2022. Contains 350481 sequences. (Running on oeis4.) | 1,108 | 2,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-05 | latest | en | 0.316447 |
https://secondsminutes.com/32577-seconds-in-minutes | 1,708,918,523,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474650.85/warc/CC-MAIN-20240226030734-20240226060734-00424.warc.gz | 513,414,343 | 6,736 | # 32577 seconds in minutes
## Result
32577 seconds equals 542.95 minutes
You can also convert 32577 seconds to minutes and seconds or to hours and minutes
## Conversion formula
Multiply the amount of seconds by the conversion factor to get the result in minutes:
32577 s × 0.0166667 = 542.95 min
## How to convert 32577 seconds to minutes?
The conversion factor from seconds to minutes is 0.0166667, which means that 1 seconds is equal to 0.0166667 minutes:
1 s = 0.0166667 min
To convert 32577 seconds into minutes we have to multiply 32577 by the conversion factor in order to get the amount from seconds to minutes. We can also form a proportion to calculate the result:
1 s → 0.0166667 min
32577 s → T(min)
Solve the above proportion to obtain the time T in minutes:
T(min) = 32577 s × 0.0166667 min
T(min) = 542.95 min
The final result is:
32577 s → 542.95 min
We conclude that 32577 seconds is equivalent to 542.95 minutes:
32577 seconds = 542.95 minutes
## Result approximation:
For practical purposes we can round our final result to an approximate numerical value. In this case thirty-two thousand five hundred seventy-seven seconds is approximately five hundred forty-two point nine five minutes:
32577 seconds ≅ 542.95 minutes
## Conversion table
For quick reference purposes, below is the seconds to minutes conversion table:
seconds (s) minutes (min)
32578 seconds 542.967753 minutes
32579 seconds 542.984419 minutes
32580 seconds 543.001086 minutes
32581 seconds 543.017753 minutes
32582 seconds 543.034419 minutes
32583 seconds 543.051086 minutes
32584 seconds 543.067753 minutes
32585 seconds 543.08442 minutes
32586 seconds 543.101086 minutes
32587 seconds 543.117753 minutes
## Units definitions
The units involved in this conversion are seconds and minutes. This is how they are defined:
### Seconds
The second (symbol: s) (abbreviated s or sec) is the base unit of time in the International System of Units (SI). It is qualitatively defined as the second division of the hour by sixty, the first division by sixty being the minute. The SI definition of second is "the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom". Seconds may be measured using a mechanical, electrical or an atomic clock. SI prefixes are combined with the word second to denote subdivisions of the second, e.g., the millisecond (one thousandth of a second), the microsecond (one millionth of a second), and the nanosecond (one billionth of a second). Though SI prefixes may also be used to form multiples of the second such as kilosecond (one thousand seconds), such units are rarely used in practice. The more common larger non-SI units of time are not formed by powers of ten; instead, the second is multiplied by 60 to form a minute, which is multiplied by 60 to form an hour, which is multiplied by 24 to form a day. The second is also the base unit of time in other systems of measurement: the centimetre–gram–second, metre–kilogram–second, metre–tonne–second, and foot–pound–second systems of units.
### Minutes
The minute is a unit of time or of angle. As a unit of time, the minute (symbol: min) is equal to 1⁄60 (the first sexagesimal fraction) of an hour, or 60 seconds. In the UTC time standard, a minute on rare occasions has 61 seconds, a consequence of leap seconds (there is a provision to insert a negative leap second, which would result in a 59-second minute, but this has never happened in more than 40 years under this system). As a unit of angle, the minute of arc is equal to 1⁄60 of a degree, or 60 seconds (of arc). Although not an SI unit for either time or angle, the minute is accepted for use with SI units for both. The SI symbols for minute or minutes are min for time measurement, and the prime symbol after a number, e.g. 5′, for angle measurement. The prime is also sometimes used informally to denote minutes of time. In contrast to the hour, the minute (and the second) does not have a clear historical background. What is traceable only is that it started being recorded in the Middle Ages due to the ability of construction of "precision" timepieces (mechanical and water clocks). However, no consistent records of the origin for the division as 1⁄60 part of the hour (and the second 1⁄60 of the minute) have ever been found, despite many speculations. | 1,082 | 4,423 | {"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} | 4.15625 | 4 | CC-MAIN-2024-10 | latest | en | 0.764509 |
https://sciencedocbox.com/Physics/112961909-Storage-tank-or-the-hull-of-a-ship-at-rest-is-subjected-to-fluid-pressure-distributed-over-its-surface.html | 1,632,718,678,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00023.warc.gz | 531,980,761 | 29,274 | # storage tank, or the hull of a ship at rest, is subjected to fluid pressure distributed over its surface.
Size: px
Start display at page:
Download "storage tank, or the hull of a ship at rest, is subjected to fluid pressure distributed over its surface."
Transcription
1 Hydrostatic Forces on Submerged Plane Surfaces Hydrostatic forces mean forces exerted by fluid at rest. - A plate exposed to a liquid, such as a gate valve in a dam, the wall of a liquid storage tank, or the hull of a ship at rest, is subjected to fluid pressure distributed over its surface. - Hydrostatic forces form a system of parallel forces, - We need to determine the magnitude of the force; - and its point of application, Centre of Pressure Note: When analyzing hydrostatic forces on submerged surfaces; the atmospheric pressure can be subtracted for simplicity when it acts on both sides of the structure. Consider the plane surface shown in the figure below. The total area is made up of many elemental areas. The force on each elemental area is always normal to the surface but, in general, each force is of different magnitude as the pressure usually varies.
2 Forces on Horizontal Submerged surface: For a horizontal plane submerged in a liquid (or a plane experiencing uniform pressure over its surface), the pressure, p, will be equal at all points of the surface. Thus the resultant force will be given by P = γh : Uniform on the entire plane. Resultant force, F R = P. A = γ. A. h (A: the bottom area of container) Forces on Inclined Submerged surface: This plane surface is totally submerged in a liquid of density ρ and inclined at an angle of θ to the horizontal. Taking pressure as zero at the surface and measuring down from the surface, the pressure on an element γa, submerged distance z, is given by P = ρgz And therefore, the force on the element is F = γzδa
3 The resultant force can be found by summing all of these forces: F R = γ zδa Where, zδa known as 1 st Moment of Area which is equal to AZ. zδa = AZ = 1 st Moment of Area about the line of free surface. Where A is the area of the plane and Z is the depth (distance from the free surface) to the centroid, G. Steps to solve the exerted bodies problems: 1- First, we have the main equation to find the resultant forces: - Location of the force: F R = γ. A. h c 3- h c, for the vertical surface is the distance from the surface to the centre, for example if the depth.5, then the hc = h c, for inclined surface, will be h c = distance from the surface of the fluid to the up edge of the submerged surface + length of the surface/ sinθ 5- The Area term in the resultant force equation can be calculated directly, depend on the shape of submerged surface, for example, if the gate is square, the area will be (Width * Length) as so on. 6- Now we find the distance where the force F acts from the hinge: h c y c =, for inclined surface. sinθ y c = h c, for vertical surface. 7- I x,c, can be calculated based on the shape of the submerged body wich can be shown for common bodies:
4 Shape Dimensions Area I x,c, nd moment of area Rectangle bd bd 3 1 Triangular bh bh 3 36 Circle πr πr 4 4 Semi-Circle πr 0.11R 4 Example 1: A swimming pool is 18m long and 7m wide. Determine the magnitude and the location of the resultant force of the water on the vertical end of the pool where the depth is.5m.
5 The force is The location of force: F R = γ. A. h c F R = = 14.6 kn y c = h c, for vertical surface y c =.5 = 1.5 m The second moment of area can be calculated from the table for the rectangular: I x,c = bd3 1 I x,c = 7.53 = m y R = = 1.66 m Example : A square 3m*3m gate is located in 45 sloping side of a dam. Some measurements indicate that the resultant force of the water on the gate is 500 kn. 1- Determine the pressure at the bottom of the gate. - Show where this force acts. 1- The force is F R = γ. A. h c = h c h c = 5.66 m P = γ (h c + 3 sin45) P = 65.9 kn - The location of force:
6 y c = h c sinθ y c = 5.66 sin45 = 8 m The second moment of area can be calculated from the table for the rectangular: I x,c = bd3 1 I x,c = 3 33 = 6.75 m4 1 y R = = m Example 3: the vertical cross section of a 7m long closed storage tank is shown below. The tank contains ethyl alcohol and air pressure is 40 kpa. Determine the magnitude of the resultant fluid force acting on one end of the tank. Break the figure to three parts as shown below: 1- For part 1: - For part : F 1 = γ. A. h c F R = P A F R = 40 = 160 kn
7 F R = P air A + γ ethyl h c A F = ( 4 ) 4 = 444 kn 3- For part 3: F R = P air A + γ ethyl h c A Area of part 3: F 3 = ( 3 4) 1 4 = 43 kn A = 1 bottom height, h c = 3 height Now, the resultant force can be evaluated by: F R = F 1 + F + F 3 = = 847 kn Example 4: the rectangular gate CD shown in the below figure is 1.8m wide and m long. Assuming the material of the gate to be homogeneous and neglecting friction at the hinge C, determine the resultant and the location of this force of the gate necessary to keep it shut until the water level rises to m above the hinge. To start solving this, we need to find the angle first: θ = tan 1 ( 4 3 ) = F R = γ. A. h c F R = ( + sin (53.13) = kn Now, the location can be found by: Where,
8 y c = h c sin53.13 = + sin53.13 = 3.5m sin53.13 And from table the second moment of area can be calculated by: I x,c = bd3 1 y R = = = 1. m = 3.595m Example 5: A 60-cm square gate (a = 0.6) has its top edge 1 m below the water surface. It is on a 45 o angle and its bottom edge is hinged as shown in figure below. What force P is needed to just open the gate? the force can be calculated by: F R = γ. A. h c F R = ( Now we find the distance d where the force F acts from the hinge: Where, y c = h c sin45 = = 17.7m sin45 sin45 And from table the second moment of area can be calculated by: y R = The force P can be calculated: I x,c = bd3 1 = = m = 17.7m d = 0.3 (y R y c ) = 0.3 ( ) = 0.98m sin (45) = kn
9 P = d F R a = 1.04kN Example 6: An inclined rectangular gate (1.5m wide) contains water on one side. Determine the total resultant force acting on the gate and the location. the force can be calculated by: F R = γ. A. h c F R = ( Now we find the distance d where the force F acts from the hinge: Where, y c = h c sin30 = sin30 sin30 = 5.4m And from table the second moment of area can be calculated by: I x,c = bd3 1 y R = = = 0.16 m = 5.4m sin (30) = kn Example 7: An inclined circular with water on one side is shown in the fig. Determine the total resultant force acting on the gate and the location.
10 Area of the circular gate, A = π 4 d A = m the force can be calculated by: F R = γ. A. h c F R = ( sin(60)) = 17. kn Now we find the distance where the force F acts: Where, y c = h c sinθ = sin60 sin60 =.578m And from table the second moment of area can be calculated by: I x,c = πr4 4 y R = = π = 0.049m =.6m Example 8: Gate AB in the figure below is 1m long and 0.7m wide. Calculate force F on the gate and position X of c.p. (specific gravity of oil is 0.81)
11 Area of the rectangular gate, A = length width A = 0.7 m The density of oil can be calculated by: And h c can be evaluated as below: the force can be calculated by: ρ oil = = 810 kg m 3 h c = sin sin50 = 4.15 m F R = γ. A. h c F R = = kn Now we find the distance where the force F acts: Where, y c = h c sinθ = 4.15 sin50 = 5.4m And from table the second moment of area can be calculated by: I x,c = bd3 1 = = m 4 y R = = 5.435m Example 9: The gate in figure below is 5 m wide, is hinged at point B, and rests against a smooth wall at point A. Compute: a) The force on the gate due to the water pressure.
12 b) The distance between the centre of gate and location of force act. c) The reactions at hinge B. To start solving this, we need to find the angle first: Now, the location can be found by: Where, θ = tan 1 ( 6 ) = h c = 15 3 = 1m F R = γ. A. h c F R = = 5886 kn y c = h c sin37 = 1 sin37 = 19.94m And from table the second moment of area can be calculated by: I x,c = bd3 1 = = m 4 1 y R = = 0.4m Note: the distance between the centre of the gate and centre of pressure can be calculated by: d = y R y c = = 0.46m Summing moments counter clockwise about B gives: P Lsinθ = F(5 d)
13 P 10sin37 = 5886(5 0.46) P = kn With F and P known, the reactions Bx and Bz are found by summing forces on the gate. F x = 0 P = Bx + Fsin37 Bx = 898kN Example 10: A tank holding water has a triangular gate, hinged at the top, in one wall. Find the moment at the hinge required to keep this triangular gate closed. The force can be calculated by: h c = = 1.7m A = bh = 1.5 = 1.5 m F R = γ. A. h c F R = = 5 kn Now we find the distance where the force F acts: Where, y c = h c = 1.7m And from table the second moment of area can be calculated by: I x,c = bh3 36 = 1.53 = m 4 36
14 y R = = m The moment on the hinge from the water is: Moment = 5 ( ) = 14.35kN Submerged Curved surface: The easiest way to determine the resultant hydrostatic force FR acting on a twodimensional curved surface is to determine the horizontal and vertical components FH and FV separately. Horizontal force component on curved surface: F H = F x Horizontal force component on curved surface: F V = F y + W where the summation F y + W is a vector addition (i.e., add magnitudes if both act in the same direction and subtract if they act in opposite directions). Thus, we conclude that: 1- The horizontal component of the hydrostatic force acting on a curved surface is equal (in both magnitude and the line of action) to the hydrostatic force acting on the vertical projection of the curved surface. - The vertical component of the hydrostatic force acting on a curved surface is equal to the hydrostatic force acting on the horizontal projection of the curved surface, plus (minus, if acting in the opposite direction) the weight of the fluid block. The magnitude of the resultant hydrostatic force acting on the curved surface is: F R = F H + F V And the tangent of the angle it makes with the horizontal is α = F V F H
15 Example 1: The water is on the right side of the curved surface AB, which is one quarter of a circle of radius 1.3m. The tank s length is.1m. Find the horizontal and vertical component of the hydrostatic acting on the curved surface. Horizontal force, F H = γ. A. h c F H = ( ) = 84.4 kn Vertical force, F V = γ. volume F V = ( π ) = 94.97kN Example : The gate AB shown is hinged at A and is in the form of quarter-circle wall of radius 1m. If the width of the gate is 30m, calculate the horizontal and vertical force. Horizontal force, F H = γ. A. h c F H = ( 1 ) 1 30 = kn Vertical force, F V = γ. volume F V = ( π ) = kN
16 Example 3: Three gates of negligible weight are used to hold back water in the channel of width b as shown in the figure below. The force of the gate against the block for gate (b) is R. determine in term of R, the force against the blocks for the other two gates. For gate (b) F R = γ water h c A F R = 9810 h h b F R = 4905h b Where, y c = h c y c = h And from table the second moment of area can be calculated by: Thus: y R = I x,c = bd3 1 = bh3 1 bh h h b h = h 6 + h = 4h 6 = h 3 M = 0 R = y R F R R = h h b R = 370 h b eq. 1 For case (a): The weight can be calculated by: W = 9810 volume W = 9810 ( π 4 (h ) b) W = 196. h b
17 This lead to: M H = 0 W ( h 4h 6π ) + F R ( h 3 ) = F B h F B = 385.9h b With equation 1: F B = 1.17R For case (c): The force FB on the curve section passes through the hinge and therefore does not contribute to the moment around H. on the bottom part of the gate: F R = γ water h c A F R = h 4 h b F R = h b Where, y c = h c y c = 3h 4 And from table the second moment of area can be calculated by: And with equation 1: y R = I x,c = bd3 1 = bh3 1 bh 3 1 3h + 3h 4 h b 4 = 0.777h M H = 0 F R 0.777h = F B h F B = 0.875R
18 Example 3: A 4m long curved gate is located in the side of a reservoir containing water as shown in figure below. Determine the magnitude of the horizontal and vertical components of the force of the water on the gate. For equilibrium: Based on the figure: Similarly: Now: F x = 0 F H = F F H = γ water h c A = 9810 (6 + 3 ) 3 4 = 88.9 kn F y = 0 F V = F 1 + W F 1 = (3 4) = kn W = γ water Volume W = 9810 ( π 4 3 4) = 77.4 kn F V = F 1 + W = = kn Buoyancy and Stability It is a common experience that an object feels lighter and weighs less in a liquid than it does in air. This can be demonstrated easily by weighing a heavy object in water by a waterproof spring scale. Also, objects made of wood or other light materials float on water. These and other observations suggest that a fluid exerts an upward force on a
19 body immersed in it. This force that tends to lift the body is called the buoyant force and is denoted by FB. The buoyant force is caused by the increase of pressure in a fluid with depth. Consider, for example, a flat plate of thickness h submerged in a liquid of density ρ f parallel to the free surface, as shown in figure below. The area of the top (and also bottom) surface of the plate is A, and its distance to the free surface is s. The pressures at the top and bottom surfaces of the plate are ρ f. g. s and ρ f. g. (s + h), respectively. Then the hydrostatic force F top = ρ f. g. s. A acts downward on the top surface, and the larger force F top = ρ f. g. (s + h). A acts upward on the bottom surface of the plate. The difference between these two forces is a net upward force, which is the buoyant force. F buoyant = F bottom F top = F top = ρ f. g. (s + h). A ρ f. g. s. A F buoyant = ρ f. g. h. A = ρ f. g. V, where, V = A. h, volume of the plate we conclude that the buoyant force acting on the plate is equal to the weight of the liquid displaced by the plate. Archimedes principle: The buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume. For floating bodies, the weight of the entire body must be equal to the buoyant force, which is the weight of the fluid whose volume is equal to the volume of the submerged portion of the floating body. That is, F B = w ρ f. g. V submerged = ρ average,body. g. V total V sub V total = ρ average,body ρ f Notes:
20 1- The submerged volume fraction of a floating body is equal to the ratio of the average density of the body to the density of the fluid. Note that when the density ratio is equal to or greater than one, the floating body becomes completely submerged. - A body immersed in a fluid (1) remains at rest at any point in the fluid when its density is equal to the density of the fluid, () sinks to the bottom when its density is greater than the density of the fluid, and (3) rises to the surface of the fluid and floats when the density of the body is less than the density of the fluid. 3- The buoyant force is proportional to the density of the fluid, and thus we might think that the buoyant force exerted by gases such as air is negligible. This is certainly the case in general, but there are significant exceptions. For example, the volume of a person is about 0.1 m 3, and taking the density of air to be 1. kg/m 3, the buoyant force exerted by air on the person is: F buoyant = ρ f. g. h. A = N The weight of an 80kg person is 80 * 9.81 = 788 N. Therefore, ignoring the buoyancy in this case results in an error in weight of just 0.15 percent, which is negligible. Example 1: A crane is used to lower weights into the sea (density =105 kg/m 3 ) for an underwater construction project as shown below. Determine the tension in the rope of the crane due to a rectangular 0.4m * 0.4m * 3m concrete block
21 (density = 300 kg/m 3 ) when it is (a) suspended in the air and (b) completely immersed in water. The volume of the rectangle, V = = 0.48 m 3 F T,air = ρ concret. g. V F T,air = = 10.8 kn And the force when the body immersed in water is: Now the force is: F B = ρ water. g. V F B = = 4.8 kn F water = w F B = = 6 kn Note: that the weight of the concrete block, and thus the tension of the rope, decreases by ( )/10.8 = 55% percent in water. Example : A spherical buoy has a diameter of a 1.5m, weight 8.5KN and is anchored to the seafloor with a cable as shown in figure below. Although, the buoy normally floats on the surface, at certain times the water depth increases so that the buoy is
22 completely immersed as illustrated. For this conditions what is the tension of the cable? The volume of the spherical shape, V = π 6 D3 = π 6 (1.5)3 = 1.77m 3 γ seawater = 10.1 kn m 3, given F B = γ seawater V F B = = kn The tension in the cable can be calculated by: T = F B w = = kn
### Static Forces on Surfaces-Buoyancy. Fluid Mechanics. There are two cases: Case I: if the fluid is above the curved surface:
Force on a Curved Surface due to Hydrostatic Pressure If the surface is curved, the forces on each element of the surface will not be parallel (normal to the surface at each point) and must be combined
### CHARACTERISTIC OF FLUIDS. A fluid is defined as a substance that deforms continuously when acted on by a shearing stress at any magnitude.
CHARACTERISTIC OF FLUIDS A fluid is defined as a substance that deforms continuously when acted on by a shearing stress at any magnitude. In a fluid at rest, normal stress is called pressure. 1 Dimensions,
### hapter 13 Archimedes Up-thrust
hapter 13 Archimedes Up-thrust In science, buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed object. The buoyant force is also called Archimedes Up-thrust force. Proof
### Upthrust and Archimedes Principle
1 Upthrust and Archimedes Principle Objects immersed in fluids, experience a force which tends to push them towards the surface of the liquid. This force is called upthrust and it depends on the density
### Physics 107 HOMEWORK ASSIGNMENT #9
Physics 07 HOMEORK ASSIGNMENT #9 Cutnell & Johnson, 7 th edition Chapter : Problems 6, 8, 33, 40, 44 *6 A 58-kg skier is going down a slope oriented 35 above the horizontal. The area of each ski in contact
### Chapter 3 Fluid Statics
Chapter 3 Fluid Statics 3.1 Pressure Pressure : The ratio of normal force to area at a point. Pressure often varies from point to point. Pressure is a scalar quantity; it has magnitude only It produces
### Engineering Mechanics. Equivalent force systems: problems
Engineering Mechanics Equivalent force systems: problems A 36-N force is applied to a wrench to tighten a showerhead. Knowing that the centerline of the wrench is parallel to the x axis. Determine the
### MULTIPLE-CHOICE PROBLEMS:(Two marks per answer) (Circle the Letter Beside the Most Correct Answer in the Questions Below.)
MULTIPLE-CHOICE PROLEMS:(Two marks per answer) (Circle the Letter eside the Most Correct Answer in the Questions elow.) 1. The absolute viscosity µ of a fluid is primarily a function of: a. Density. b.
### P = ρ{ g a } + µ 2 V II. FLUID STATICS
II. FLUID STATICS From a force analysis on a triangular fluid element at rest, the following three concepts are easily developed: For a continuous, hydrostatic, shear free fluid: 1. Pressure is constant
### Chapter 1 INTRODUCTION
Chapter 1 INTRODUCTION 1-1 The Fluid. 1-2 Dimensions. 1-3 Units. 1-4 Fluid Properties. 1 1-1 The Fluid: It is the substance that deforms continuously when subjected to a shear stress. Matter Solid Fluid
### Fluid Statics. Pressure. Pressure
Pressure Fluid Statics Variation of Pressure with Position in a Fluid Measurement of Pressure Hydrostatic Thrusts on Submerged Surfaces Plane Surfaces Curved Surfaces ddendum First and Second Moment of
### Fluid Mechanics. Forces on Fluid Elements. Fluid Elements - Definition:
Fluid Mechanics Chapter 2: Fluid Statics Lecture 3 Forces on Fluid Elements Fluid Elements - Definition: Fluid element can be defined as an infinitesimal region of the fluid continuum in isolation from
### Eric G. Paterson. Spring 2005
Eric G. Paterson Department of Mechanical and Nuclear Engineering Pennsylvania State University Spring 2005 Reading and Homework Read Chapter 3. Homework Set #2 has been posted. Due date: Friday 21 January.
### Chapter 15. m. The symbolic equation for mass density is: ρ= m V. Table of Densities
Chapter 15 Density Often you will hear that fiberglass is used for racecars because it is lighter than steel. This is only true if we build two identical bodies, one made with steel and one with fiberglass.
### PHYS 185 Practice Final Exam Fall You may answer the questions in the space provided here, or if you prefer, on your own notebook paper.
PHYS 185 Practice Final Exam Fall 2013 Name: You may answer the questions in the space provided here, or if you prefer, on your own notebook paper. Short answers 1. 2 points When an object is immersed
### CHAPTER 2 Fluid Statics
Chapter / Fluid Statics CHPTER Fluid Statics FE-type Eam Review Problems: Problems - to -9. (C). (D). (C).4 ().5 () The pressure can be calculated using: p = γ h were h is the height of mercury. p= γ h=
### Hydrostatics. ENGR 5961 Fluid Mechanics I: Dr. Y.S. Muzychka
1 Hydrostatics 2 Introduction In Fluid Mechanics hydrostatics considers fluids at rest: typically fluid pressure on stationary bodies and surfaces, pressure measurements, buoyancy and flotation, and fluid
### MULTIPLE-CHOICE PROBLEMS :(Two marks per answer) (Circle the Letter Beside the Most Correct Answer in the Questions Below.)
Test Midterm 1 F2013 MULTIPLE-CHOICE PROBLEMS :(Two marks per answer) (Circle the Letter Beside the Most Correct nswer in the Questions Below.) 1. The absolute viscosity µ of a fluid is primarily a function
### ME 262 BASIC FLUID MECHANICS Assistant Professor Neslihan Semerci Lecture 4. (Buoyancy and Viscosity of water)
ME 262 BASIC FLUID MECHANICS Assistant Professor Neslihan Semerci Lecture 4 (Buoyancy and Viscosity of water) 16. BUOYANCY Whenever an object is floating in a fluid or when it is completely submerged in
### 1 Fluid Statics. 1.1 Fluid Properties. Fluid
1 Fluid Statics 1.1 Fluid Properties Fluid A fluid is a substance, which deforms when subjected to a force. A fluid can offer no permanent resistance to any force causing change of shape. Fluid flow under
### Chapter 2 Hydrostatics Buoyancy, Floatation and Stability
Chapter 2 Hydrostatics uoyancy, Floatation and Stability Zerihun Alemayehu Rm. E119 AAiT Force of buoyancy an upward force exerted by a fluid pressure on fully or partially floating body Gravity Archimedes
### Chapter 15: Fluid Mechanics Dynamics Using Pascal s Law = F 1 = F 2 2 = F 2 A 2
Lecture 24: Archimedes Principle and Bernoulli s Law 1 Chapter 15: Fluid Mechanics Dynamics Using Pascal s Law Example 15.1 The hydraulic lift A hydraulic lift consists of a small diameter piston of radius
### Fluid Mechanics-61341
An-Najah National University College of Engineering Fluid Mechanics-61341 Chapter [2] Fluid Statics 1 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics Fluid Statics Problems Fluid statics refers to
### 2.6 Force reacts with planar object in fluid
2.6 Force reacts with planar object in fluid Fluid surface Specific weight (γ) => Object sinks in fluid => C is center of gravity or Centroid => P is center of pressure (always under C) => x axis is cross
### ME 262 BASIC FLUID MECHANICS Assistant Professor Neslihan Semerci Lecture 3. (Hydrostatic forces on flat and curved surfaces)
ME 262 BSIC FLUID MECHNICS ssistant Professor Neslihan Semerci Lecture 3 (Hydrostatic forces on flat and curved surfaces) 15. HYDROSTTIC PRESSURE 15.1 Hydrostatic pressure force on flat surfaces When a
### ME-B41 Lab 1: Hydrostatics. Experimental Procedures
ME-B41 Lab 1: Hydrostatics In this lab you will do four brief experiments related to the following topics: manometry, buoyancy, forces on submerged planes, and hydraulics (a hydraulic jack). Each experiment
### CIVE HYDRAULIC ENGINEERING PART I Pierre Julien Colorado State University
CIVE 401 - HYDRAULIC ENGINEERING PART I Pierre Julien Colorado State University Problems with and are considered moderate and those with are the longest and most difficult. In 2018 solve the problems with
### Pressure in stationary and moving fluid Lab- Lab On- On Chip: Lecture 2
Pressure in stationary and moving fluid Lab-On-Chip: Lecture Lecture plan what is pressure e and how it s distributed in static fluid water pressure in engineering problems buoyancy y and archimedes law;
### Fluid Mechanics Discussion. Prepared By: Dr.Khalil M. Al-Astal Eng.Ahmed S. Al-Agha Eng.Ruba M. Awad
Discussion Prepared By: Dr.Khalil M. Al-Astal Eng.Ahmed S. Al-Agha Eng.Ruba M. Awad 2014-2015 Chapter (1) Fluids and their Properties Fluids and their Properties Fluids (Liquids or gases) which a substance
### 11.1 Mass Density. Fluids are materials that can flow, and they include both gases and liquids. The mass density of a liquid or gas is an
Chapter 11 Fluids 11.1 Mass Density Fluids are materials that can flow, and they include both gases and liquids. The mass density of a liquid or gas is an important factor that determines its behavior
### Ch. 5: Distributed Forces
234 5.0 Outline 234 Introduction 235 Center of Mass 236 Centroids (Line, Area, Volume) 241 Composite Bodies and Figures 257 Theorem of Pappus 265 Fluid Statics 272 5.0 Outline 5.0 Introduction Ch. 5: Distributed
### Introduction to Marine Hydrodynamics
1896 1920 1987 2006 Introduction to Marine Hydrodynamics (NA235) Department of Naval Architecture and Ocean Engineering School of Naval Architecture, Ocean & Civil Engineering Shanghai Jiao Tong University
### Pressure in stationary and moving fluid. Lab-On-Chip: Lecture 2
Pressure in stationary and moving fluid Lab-On-Chip: Lecture Fluid Statics No shearing stress.no relative movement between adjacent fluid particles, i.e. static or moving as a single block Pressure at
### Nicholas J. Giordano. Chapter 10 Fluids
Nicholas J. Giordano www.cengage.com/physics/giordano Chapter 10 Fluids Fluids A fluid may be either a liquid or a gas Some characteristics of a fluid Flows from one place to another Shape varies according
### Chapter 9: Solids and Fluids
Chapter 9: Solids and Fluids State of matters: Solid, Liquid, Gas and Plasma. Solids Has definite volume and shape Can be crystalline or amorphous Molecules are held in specific locations by electrical
### Engineering Mechanics: Statics in SI Units, 12e
Engineering Mechanics: Statics in SI Units, 12e 5 Equilibrium of a Rigid Body Chapter Objectives Develop the equations of equilibrium for a rigid body Concept of the free-body diagram for a rigid body
### BTECH MECHANICAL PRINCIPLES AND APPLICATIONS. Level 3 Unit 5
BTECH MECHANICAL PRINCIPLES AND APPLICATIONS Level 3 Unit 5 FORCES AS VECTORS Vectors have a magnitude (amount) and a direction. Forces are vectors FORCES AS VECTORS (2 FORCES) Forces F1 and F2 are in
### Forces on a curved surface
Civil Engineering Hydraulics Forces on Curved Surfaces There is a theory which states that if ever anybody discovers exactly what the Universe is for and why it is here, it will instantly disappear and
### The general rules of statics (as applied in solid mechanics) apply to fluids at rest. From earlier we know that:
ELEMENTARY HYDRAULICS National Certificate in Technology (Civil Engineering) Chapter 2 Pressure This section will study the forces acting on or generated by fluids at rest. Objectives Introduce the concept
### CHAPTER 13. Liquids FLUIDS FLUIDS. Gases. Density! Bulk modulus! Compressibility. To begin with... some important definitions...
CHAPTER 13 FLUIDS Density! Bulk modulus! Compressibility Pressure in a fluid! Hydraulic lift! Hydrostatic paradox Measurement of pressure! Manometers and barometers Buoyancy and Archimedes Principle! Upthrust!
### JNTU World. Subject Code: R13110/R13
Set No - 1 I B. Tech I Semester Regular Examinations Feb./Mar. - 2014 ENGINEERING MECHANICS (Common to CE, ME, CSE, PCE, IT, Chem E, Aero E, AME, Min E, PE, Metal E) Time: 3 hours Max. Marks: 70 Question
### 1. A sphere with a radius of 1.7 cm has a volume of: A) m 3 B) m 3 C) m 3 D) 0.11 m 3 E) 21 m 3
1. A sphere with a radius of 1.7 cm has a volume of: A) 2.1 10 5 m 3 B) 9.1 10 4 m 3 C) 3.6 10 3 m 3 D) 0.11 m 3 E) 21 m 3 2. A 25-N crate slides down a frictionless incline that is 25 above the horizontal.
### CEE 342 Aut Homework #2 Solutions
CEE 4 Aut 005. Homework # Solutions.9. A free-body diagram of the lower hemisphere is shown below. This set of boundaries for the free body is chosen because it isolates the force on the bolts, which is
### Linear Transport Relations (LTR)
Linear Transport Relations (LTR) Much of Transport Phenomena deals with the exchange of momentum, mass, or heat between two (or many) objects. Often, the most mathematically simple way to consider how
### Homework Assignment on Fluid Statics
AMEE 0 Introduction to Fluid Mecanics Instructor: Marios M. Fyrillas Email: m.fyrillas@fit.ac.cy Homework Assignment on Fluid Statics --------------------------------------------------------------------------------------------------------------
### TALLER DE HIDROSTÁTICA
TALLER DE HIDROSTÁTICA 1) Substance A has a density of 3 g/cm 3 and substance B has a density of 4 g/cm 3. In order to obtain equal masses of these two substances, what must be the ratio of the volume
### Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur. Lecture - 9 Fluid Statics Part VI
Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 9 Fluid Statics Part VI Good morning, I welcome you all to this session of Fluid
### Announcements. Equilibrium of a Rigid Body
Announcements Equilibrium of a Rigid Body Today s Objectives Identify support reactions Draw a free body diagram Class Activities Applications Support reactions Free body diagrams Examples Engr221 Chapter
### Hydrostatic. Pressure distribution in a static fluid and its effects on solid surfaces and on floating and submerged bodies.
Hydrostatic Pressure distribution in a static fluid and its effects on solid surfaces and on floating and submerged bodies. M. Bahrami ENSC 283 Spring 2009 1 Fluid at rest hydrostatic condition: when a
### SOLUTION According to Equation 11.3, pressure is defined as P= F/ A; therefore, the magnitude of the force on the lid due to the air pressure is
PHYS 3 Fall 07 Week Recitation: Chapter :, 7, 40, 44, 64, 69.. ssm An airtight box has a remoable lid of area.3 0 m and negligible weight. The box is taken up a mountain where the air pressure outside
### TUTORIAL SHEET 1. magnitude of P and the values of ø and θ. Ans: ø =74 0 and θ= 53 0
TUTORIAL SHEET 1 1. The rectangular platform is hinged at A and B and supported by a cable which passes over a frictionless hook at E. Knowing that the tension in the cable is 1349N, determine the moment
### MTE 119 STATICS FINAL HELP SESSION REVIEW PROBLEMS PAGE 1 9 NAME & ID DATE. Example Problem P.1
MTE STATICS Example Problem P. Beer & Johnston, 004 by Mc Graw-Hill Companies, Inc. The structure shown consists of a beam of rectangular cross section (4in width, 8in height. (a Draw the shear and bending
### (You may need to make a sin / cos-type trigonometric substitution.) Solution.
MTHE 7 Problem Set Solutions. As a reminder, a torus with radii a and b is the surface of revolution of the circle (x b) + z = a in the xz-plane about the z-axis (a and b are positive real numbers, with
### Mechanics Answers to Examples B (Momentum) - 1 David Apsley
TOPIC B: MOMENTUM ANSWERS SPRING 2019 (Full worked answers follow on later pages) Q1. (a) 2.26 m s 2 (b) 5.89 m s 2 Q2. 8.41 m s 2 and 4.20 m s 2 ; 841 N Q3. (a) 1.70 m s 1 (b) 1.86 s Q4. (a) 1 s (b) 1.5
### Chapter 6 Some Applications of the Integral
Chapter 6 Some Applications of the Integral Section 6.1 More on Area a. Representative Rectangle b. Vertical Separation c. Example d. Integration with Respect to y e. Example Section 6.2 Volume by Parallel
### Liquids CHAPTER 13 FLUIDS FLUIDS. Gases. Density! Bulk modulus! Compressibility. To begin with... some important definitions...
CHAPTER 13 FLUIDS FLUIDS Liquids Gases Density! Bulk modulus! Compressibility Pressure in a fluid! Hydraulic lift! Hydrostatic paradox Measurement of pressure! Manometers and barometers Buoyancy and Archimedes
### TEST REPORT. Question file: P Copyright:
Date: February-12-16 Time: 2:00:28 PM TEST REPORT Question file: P12-2006 Copyright: Test Date: 21/10/2010 Test Name: EquilibriumPractice Test Form: 0 Test Version: 0 Test Points: 138.00 Test File: EquilibriumPractice
### ENGI Multiple Integration Page 8-01
ENGI 345 8. Multiple Integration Page 8-01 8. Multiple Integration This chapter provides only a very brief introduction to the major topic of multiple integration. Uses of multiple integration include
### Figure 3: Problem 7. (a) 0.9 m (b) 1.8 m (c) 2.7 m (d) 3.6 m
1. For the manometer shown in figure 1, if the absolute pressure at point A is 1.013 10 5 Pa, the absolute pressure at point B is (ρ water =10 3 kg/m 3, ρ Hg =13.56 10 3 kg/m 3, ρ oil = 800kg/m 3 ): (a)
### Q1 Give answers to all of the following questions (5 marks each):
FLUID MECHANICS First Year Exam Solutions 03 Q Give answers to all of the following questions (5 marks each): (a) A cylinder of m in diameter is made with material of relative density 0.5. It is moored
### EQUILIBRIUM OBJECTIVES PRE-LECTURE
27 FE3 EQUILIBRIUM Aims OBJECTIVES In this chapter you will learn the concepts and principles needed to understand mechanical equilibrium. You should be able to demonstrate your understanding by analysing
### 1. Please complete the following short problems.
Name 1. Please complete the following short problems. For parts 1A and 1B, we will consider three M88 recovery vehicles pulling an M1 tank back onto the road as shown below. F2 F1 50 M88 #1 50 M88 #2 y
### All questions are of equal value. No marks are subtracted for wrong answers.
(1:30 PM 4:30 PM) Page 1 of 6 All questions are of equal value. No marks are subtracted for wrong answers. Record all answers on the computer score sheet provided. USE PENCIL ONLY! Black pen will look
### Figure 1 Answer: = m
Q1. Figure 1 shows a solid cylindrical steel rod of length =.0 m and diameter D =.0 cm. What will be increase in its length when m = 80 kg block is attached to its bottom end? (Young's modulus of steel
### DIMENSIONS AND UNITS
DIMENSIONS AND UNITS A dimension is the measure by which a physical variable is expressed quantitatively. A unit is a particular way of attaching a number to the quantitative dimension. Primary Dimension
### TOPIC B: MOMENTUM EXAMPLES SPRING 2019
TOPIC B: MOMENTUM EXAMPLES SPRING 2019 (Take g = 9.81 m s 2 ). Force-Momentum Q1. (Meriam and Kraige) Calculate the vertical acceleration of the 50 cylinder for each of the two cases illustrated. Neglect
Answers: Review questions 25 ANSERS O REVIE QUESIONS IMPORAN NOE: READ HIS FIRS. here are three different kinds of answer here. he usual form for quantitative questions is a short entry giving the final
### Apex Grammar School O & A Level Evening Classes. Physics EVALUATION TEST PAPER. REAL EXAMINATION QUESTIONS for Secondary 4
Apex Grammar School O & A Level Evening Classes O Level Power Revision Series EVALUATION TEST PAPER REAL EXAMINATION QUESTIONS for Secondary 4 Name: Time Start: Date: Time End: Total Marks : / 40 40 questions
### FE Fluids Review March 23, 2012 Steve Burian (Civil & Environmental Engineering)
Topic: Fluid Properties 1. If 6 m 3 of oil weighs 47 kn, calculate its specific weight, density, and specific gravity. 2. 10.0 L of an incompressible liquid exert a force of 20 N at the earth s surface.
### TOPICS. Density. Pressure. Variation of Pressure with Depth. Pressure Measurements. Buoyant Forces-Archimedes Principle
Lecture 6 Fluids TOPICS Density Pressure Variation of Pressure with Depth Pressure Measurements Buoyant Forces-Archimedes Principle Surface Tension ( External source ) Viscosity ( External source ) Equation
### TOPIC D: ROTATION EXAMPLES SPRING 2018
TOPIC D: ROTATION EXAMPLES SPRING 018 Q1. A car accelerates uniformly from rest to 80 km hr 1 in 6 s. The wheels have a radius of 30 cm. What is the angular acceleration of the wheels? Q. The University
### b) (5) Find the tension T B in the cord connected to the wall.
General Physics I Quiz 6 - Ch. 9 - Static Equilibrium July 15, 2009 Name: Make your work clear to the grader. Show formulas used. Give correct units and significant figures. Partial credit is available
### 2016 ENGINEERING MECHANICS
Set No 1 I B. Tech I Semester Regular Examinations, Dec 2016 ENGINEERING MECHANICS (Com. to AE, AME, BOT, CHEM, CE, EEE, ME, MTE, MM, PCE, PE) Time: 3 hours Max. Marks: 70 Question Paper Consists of Part-A
### Chapter 10. Solids & Liquids
Chapter 10 Solids & Liquids Next 6 chapters use all the concepts developed in the first 9 chapters, recasting them into a form ready to apply to specific physical systems. 10.1 Phases of Matter, Mass Density
### Formulae that you may or may not find useful. E v = V. dy dx = v u. y cp y = I xc/a y. Volume of an entire sphere = 4πr3 = πd3
CE30 Test 1 Solution Key Date: 26 Sept. 2017 COVER PAGE Write your name on each sheet of paper that you hand in. Read all questions very carefully. If the problem statement is not clear, you should ask
### Created by T. Madas WORK & ENERGY. Created by T. Madas
WORK & ENERGY Question (**) A B 0m 30 The figure above shows a particle sliding down a rough plane inclined at an angle of 30 to the horizontal. The box is released from rest at the point A and passes
### Physics 202 Homework 2
Physics 202 Homework 2 Apr 10, 2013 1. An airplane wing is designed so that the speed of the air across the top of the 192 kn wing is 251 m/s when the speed of the air below the wing is 225 m/s. The density
### 2.8 Hydrostatic Force on a Plane Surface
CEE 3310 Hydrostatics, Sept. 7, 2018 35 2.7 Review Pressure is a scalar - independent of direction. Hydrostatics: P = γ k. pressure, vertical motion: z P, z P. Horizontal motion in the same fluid - no
### Chapter 7 Applications of Integration
Chapter 7 Applications of Integration 7.1 Area of a Region Between Two Curves 7.2 Volume: The Disk Method 7.3 Volume: The Shell Method 7.4 Arc Length and Surfaces of Revolution 7.5 Work 7.6 Moments, Centers
### PHYS 101 Previous Exam Problems. Kinetic Energy and
PHYS 101 Previous Exam Problems CHAPTER 7 Kinetic Energy and Work Kinetic energy Work Work-energy theorem Gravitational work Work of spring forces Power 1. A single force acts on a 5.0-kg object in such
### The online of midterm-tests of Fluid Mechanics 1
The online of midterm-tests of Fluid Mechanics 1 1) The information on a can of pop indicates that the can contains 460 ml. The mass of a full can of pop is 3.75 lbm while an empty can weights 80.5 lbf.
### PHYS 101 Previous Exam Problems. Force & Motion I
PHYS 101 Previous Exam Problems CHAPTER 5 Force & Motion I Newton s Laws Vertical motion Horizontal motion Mixed forces Contact forces Inclines General problems 1. A 5.0-kg block is lowered with a downward
### TOPIC E: OSCILLATIONS EXAMPLES SPRING Q1. Find general solutions for the following differential equations:
TOPIC E: OSCILLATIONS EXAMPLES SPRING 2019 Mathematics of Oscillating Systems Q1. Find general solutions for the following differential equations: Undamped Free Vibration Q2. A 4 g mass is suspended by
### Types of Forces. Pressure Buoyant Force Friction Normal Force
Types of Forces Pressure Buoyant Force Friction Normal Force Pressure Ratio of Force Per Unit Area p = F A P = N/m 2 = 1 pascal (very small) P= lbs/in 2 = psi = pounds per square inch Example: Snow Shoes
### 24/06/13 Forces ( F.Robilliard) 1
R Fr F W 24/06/13 Forces ( F.Robilliard) 1 Mass: So far, in our studies of mechanics, we have considered the motion of idealised particles moving geometrically through space. Why a particular particle
### Practice Exam 1 Solutions
Practice Exam 1 Solutions 1a. Let S be the region bounded by y = x 3, y = 1, and x. Find the area of S. What is the volume of the solid obtained by rotating S about the line y = 1? Area A = Volume 1 1
### ME3560 Tentative Schedule Spring 2019
ME3560 Tentative Schedule Spring 2019 Week Number Date Lecture Topics Covered Prior to Lecture Read Section Assignment Prep Problems for Prep Probs. Must be Solved by 1 Monday 1/7/2019 1 Introduction to
### (a) On the dots below that represent the students, draw and label free-body diagrams showing the forces on Student A and on Student B.
2003 B1. (15 points) A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown above. One end of the rope is held by Student A of mass 70 kg, who is at rest on
### R09. d water surface. Prove that the depth of pressure is equal to p +.
Code No:A109210105 R09 SET-1 B.Tech II Year - I Semester Examinations, December 2011 FLUID MECHANICS (CIVIL ENGINEERING) Time: 3 hours Max. Marks: 75 Answer any five questions All questions carry equal
### Definition. is a measure of how much a force acting on an object causes that object to rotate, symbol is, (Greek letter tau)
Torque Definition is a measure of how much a force acting on an object causes that object to rotate, symbol is, (Greek letter tau) = r F = rfsin, r = distance from pivot to force, F is the applied force
### Fluid Mechanics. du dy
FLUID MECHANICS Technical English - I 1 th week Fluid Mechanics FLUID STATICS FLUID DYNAMICS Fluid Statics or Hydrostatics is the study of fluids at rest. The main equation required for this is Newton's
### Force 10/01/2010. (Weight) MIDTERM on 10/06/10 7:15 to 9:15 pm Bentley 236. (Tension)
Force 10/01/2010 = = Friction Force (Weight) (Tension), coefficient of static and kinetic friction MIDTERM on 10/06/10 7:15 to 9:15 pm Bentley 236 2008 midterm posted for practice. Help sessions Mo, Tu
### b) (6) What is the volume of the iron cube, in m 3?
General Physics I Exam 4 - Chs. 10,11,12 - Fluids, Waves, Sound Nov. 14, 2012 Name Rec. Instr. Rec. Time For full credit, make your work clear to the grader. Show formulas used, essential steps, and results
### Wiley Plus. Final Assignment (5) Is Due Today: Before 11 pm!
Wiley Plus Final Assignment (5) Is Due Today: Before 11 pm! Final Exam Review December 9, 009 3 What about vector subtraction? Suppose you are given the vector relation A B C RULE: The resultant vector
### Approximate physical properties of selected fluids All properties are given at pressure kn/m 2 and temperature 15 C.
Appendix FLUID MECHANICS Approximate physical properties of selected fluids All properties are given at pressure 101. kn/m and temperature 15 C. Liquids Density (kg/m ) Dynamic viscosity (N s/m ) Surface
### Fluid Mechanics. If deformation is small, the stress in a body is proportional to the corresponding
Fluid Mechanics HOOKE'S LAW If deformation is small, the stress in a body is proportional to the corresponding strain. In the elasticity limit stress and strain Stress/strain = Const. = Modulus of elasticity.
### Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur. Lecture - 8 Fluid Statics Part V
Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 8 Fluid Statics Part V Good morning, I welcome you all to the session of fluid mechanics.
### UNIVERSITY OF MANITOBA
PAGE NO.: 1 of 6 + Formula Sheet Equal marks for all questions. No marks are subtracted for wrong answers. Record all answers on the computer score sheet provided. USE PENCIL ONLY! Black pen will look | 11,730 | 43,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-39 | latest | en | 0.9352 |
http://www.convertit.com/Go/Maps/Measurement/Converter.ASP?From=mile&To=width | 1,610,787,657,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703505861.1/warc/CC-MAIN-20210116074510-20210116104510-00283.warc.gz | 132,763,211 | 5,790 | Search Maps.com
Channel Features
Travel Deals Travel Alert Bulletin Travel Tools Business Traveler Guide Student Traveler Guide Map & Travel Store
Get travel news, alerts, tips, deals, and trivia delivered free to your email in-box. Email Address: tell me more
Site Tools
Site Map About Maps.com Contact Maps.com Advertise with Maps.com Affiliate Program Order Tracking View Cart Check Out Help
Site Map | Help | Home
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```mile = 1609.344 length (length) ``` Related Measurements: Try converting from "mile" to agate (typography agate), angstrom, archin (Russian archin), arpentlin, bolt (of cloth), cable length, chain (surveyors chain), earth to moon (mean distance earth to moon), fathom, line, link (surveyors link), mil, nail (cloth nail), naval shot, rod (surveyors rod), shaku (Japanese shaku), skein, soccer field, vara (Mexican vara), yard, or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: mile = 44 bolt (of cloth), 2,534,400 bottom measure, 7.33 cable length, 17.6 city block (informal), 86,794.52 digitus (Roman digitus), 1.61E+18 fermi, 8 furlong (surveyors furlong), 2,172.84 gradus (Roman gradus), 20,864.98 Greek palm, 2,906.42 Israeli cubit, .86897624 nautical mile, 58.67 naval shot, 2,112 pace, 5.22E-14 parsec, 3,620.57 Roman cubit, 17.6 soccer field, 7,040 span (cloth span), 5,279.99 survey foot, 1,920.58 vara (Mexican vara), 1,760 yard.
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 564 | 2,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-04 | longest | en | 0.676949 |
https://www.mathworks.com/matlabcentral/profile/authors/1320466 | 1,695,452,053,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.35/warc/CC-MAIN-20230923062631-20230923092631-00699.warc.gz | 980,056,294 | 21,573 | Community Profile
# Colin Warwick
### Agilent Technologies
Active since 2003
Professional Interests: signal integrity, electronic design automation, computational electromagnetism
All
#### Content Feed
View by
Solved
Find all elements less than 0 or greater than 10 and replace them with NaN
Given an input vector x, find all elements of x less than 0 or greater than 10 and replace them with NaN. Example: Input ...
11 years ago
Solved
Select every other element of a vector
Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...
11 years ago
Solved
Is my wife right?
Regardless of input, output the string 'yes'.
11 years ago
Solved
Given a and b, return the sum a+b in c.
11 years ago
Solved
Find the sum of all the numbers of the input vector
Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
11 years ago
Solved
Make the vector [1 2 3 4 5 6 7 8 9 10]
In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...
11 years ago
Solved
Make a checkerboard matrix
Given an integer n, make an n-by-n matrix made up of alternating ones and zeros as shown below. The a(1,1) should be 1. Example...
11 years ago
Solved
Column Removal
Remove the nth column from input matrix A and return the resulting matrix in output B. So if A = [1 2 3; 4 5 6]; and ...
11 years ago
Solved
Determine if input is odd
Given the input n, return true if n is odd or false if n is even.
11 years ago
Solved
Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
11 years ago
Submitted
Analog Circuit Effects in Signal Processing and Communication System Models
Demos used in webinar "Analog Circuit Effects in Signal Processing and Communication System Models"
Submitted
erlangb
These are M-file examples related to the Erlang B formula for blocking probability.
Submitted
Frequency Domain to Baseband-complex time-domain Modeling in RF Blockset
Illustrates the relationships between time and frequency parameters used in RF Blockset.
Submitted
Random source using the Embedded MATLAB Function Block
This is a demonstration of the Embedded MATLAB Function Block, using a portable random number genera | 589 | 2,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.752074 |
https://ch.mathworks.com/help/stats/work-with-multinomial-probability-distribution-objects.html | 1,679,459,035,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943749.68/warc/CC-MAIN-20230322020215-20230322050215-00572.warc.gz | 203,646,429 | 18,698 | Main Content
# Multinomial Probability Distribution Objects
This example shows how to generate random numbers, compute and plot the pdf, and compute descriptive statistics of a multinomial distribution using probability distribution objects.
### Step 1. Define the distribution parameters.
Create a vector `p` containing the probability of each outcome. Outcome 1 has a probability of 1/2, outcome 2 has a probability of 1/3, and outcome 3 has a probability of 1/6. The number of trials `n` in each experiment is 5, and the number of repetitions `reps` of the experiment is 8.
```p = [1/2 1/3 1/6]; n = 5; reps = 8;```
### Step 2. Create a multinomial probability distribution object.
Create a multinomial probability distribution object using the specified value `p` for the `Probabilities` parameter.
`pd = makedist('Multinomial','Probabilities',p)`
```pd = MultinomialDistribution Probabilities: 0.5000 0.3333 0.1667 ```
### Step 3. Generate one random number.
Generate one random number from the multinomial distribution, which is the outcome of a single trial.
```rng('default') % For reproducibility r = random(pd)```
```r = 2 ```
This trial resulted in outcome 2.
### Step 4. Generate a matrix of random numbers.
You can also generate a matrix of random numbers from the multinomial distribution, which reports the results of multiple experiments that each contain multiple trials. Generate a matrix that contains the outcomes of an experiment with `n = 5` trials and `reps = 8` repetitions.
`r = random(pd,reps,n)`
```r = 8×5 3 3 3 2 1 1 1 2 2 1 3 3 3 1 2 2 3 2 2 2 1 1 1 1 1 1 2 3 2 3 2 1 3 1 1 3 1 2 1 1 ```
Each element in the resulting matrix is the outcome of one trial. The columns correspond to the five trials in each experiment, and the rows correspond to the eight experiments. For example, in the first experiment (corresponding to the first row), one of the five trials resulted in outcome 1, one of the five trials resulted in outcome 2, and three of the five trials resulted in outcome 3.
### Step 5. Compute and plot the pdf.
Compute the pdf of the distribution.
```x = 1:3; y = pdf(pd,x); bar(x,y) xlabel('Outcome') ylabel('Probability Mass') title('Trinomial Distribution')```
The plot shows the probability mass for each $k$ possible outcome. For this distribution, the pdf value for any `x` other than 1, 2, or 3 is 0.
### Step 6. Compute descriptive statistics.
Compute the mean, median, and standard deviation of the distribution.
`m = mean(pd)`
```m = 1.6667 ```
`med = median(pd)`
```med = 1 ```
`s = std(pd)`
```s = 0.7454 ``` | 705 | 2,581 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-14 | longest | en | 0.82145 |
http://www.nag.com/numeric/MB/manual64_24_1/html/F07/f07asf.html | 1,386,863,374,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164643547/warc/CC-MAIN-20131204134403-00004-ip-10-33-133-15.ec2.internal.warc.gz | 450,440,679 | 5,848 | Integer type: int32 int64 nag_int show int32 show int32 show int64 show int64 show nag_int show nag_int
Chapter Contents
Chapter Introduction
NAG Toolbox
# NAG Toolbox: nag_lapack_zgetrs (f07as)
## Purpose
nag_lapack_zgetrs (f07as) solves a complex system of linear equations with multiple right-hand sides,
AX = B , ATX = B or AHX = B , $AX=B , ATX=B or AHX=B ,$
where A$A$ has been factorized by nag_lapack_zgetrf (f07ar).
## Syntax
[b, info] = f07as(trans, a, ipiv, b, 'n', n, 'nrhs_p', nrhs_p)
[b, info] = nag_lapack_zgetrs(trans, a, ipiv, b, 'n', n, 'nrhs_p', nrhs_p)
## Description
nag_lapack_zgetrs (f07as) is used to solve a complex system of linear equations AX = B$AX=B$, ATX = B${A}^{\mathrm{T}}X=B$ or AHX = B${A}^{\mathrm{H}}X=B$, the function must be preceded by a call to nag_lapack_zgetrf (f07ar) which computes the LU$LU$ factorization of A$A$ as A = PLU$A=PLU$. The solution is computed by forward and backward substitution.
If trans = 'N'${\mathbf{trans}}=\text{'N'}$, the solution is computed by solving PLY = B$PLY=B$ and then UX = Y$UX=Y$.
If trans = 'T'${\mathbf{trans}}=\text{'T'}$, the solution is computed by solving UTY = B${U}^{\mathrm{T}}Y=B$ and then LTPTX = Y${L}^{\mathrm{T}}{P}^{\mathrm{T}}X=Y$.
If trans = 'C'${\mathbf{trans}}=\text{'C'}$, the solution is computed by solving UHY = B${U}^{\mathrm{H}}Y=B$ and then LHPTX = Y${L}^{\mathrm{H}}{P}^{\mathrm{T}}X=Y$.
## References
Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore
## Parameters
### Compulsory Input Parameters
1: trans – string (length ≥ 1)
Indicates the form of the equations.
trans = 'N'${\mathbf{trans}}=\text{'N'}$
AX = B$AX=B$ is solved for X$X$.
trans = 'T'${\mathbf{trans}}=\text{'T'}$
ATX = B${A}^{\mathrm{T}}X=B$ is solved for X$X$.
trans = 'C'${\mathbf{trans}}=\text{'C'}$
AHX = B${A}^{\mathrm{H}}X=B$ is solved for X$X$.
Constraint: trans = 'N'${\mathbf{trans}}=\text{'N'}$, 'T'$\text{'T'}$ or 'C'$\text{'C'}$.
2: a(lda, : $:$) – complex array
The first dimension of the array a must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The LU$LU$ factorization of A$A$, as returned by nag_lapack_zgetrf (f07ar).
3: ipiv( : $:$) – int64int32nag_int array
Note: the dimension of the array ipiv must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
The pivot indices, as returned by nag_lapack_zgetrf (f07ar).
4: b(ldb, : $:$) – complex array
The first dimension of the array b must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array must be at least max (1,nrhs)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{nrhs}}\right)$
The n$n$ by r$r$ right-hand side matrix B$B$.
### Optional Input Parameters
1: n – int64int32nag_int scalar
Default: The first dimension of the arrays a, b The second dimension of the arrays a, ipiv.
n$n$, the order of the matrix A$A$.
Constraint: n0${\mathbf{n}}\ge 0$.
2: nrhs_p – int64int32nag_int scalar
Default: The second dimension of the array b.
r$r$, the number of right-hand sides.
Constraint: nrhs0${\mathbf{nrhs}}\ge 0$.
lda ldb
### Output Parameters
1: b(ldb, : $:$) – complex array
The first dimension of the array b will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array will be max (1,nrhs)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{nrhs}}\right)$
ldbmax (1,n)$\mathit{ldb}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
The n$n$ by r$r$ solution matrix X$X$.
2: info – int64int32nag_int scalar
info = 0${\mathbf{info}}=0$ unless the function detects an error (see Section [Error Indicators and Warnings]).
## Error Indicators and Warnings
info = i${\mathbf{info}}=-i$
If info = i${\mathbf{info}}=-i$, parameter i$i$ had an illegal value on entry. The parameters are numbered as follows:
1: trans, 2: n, 3: nrhs_p, 4: a, 5: lda, 6: ipiv, 7: b, 8: ldb, 9: info.
It is possible that info refers to a parameter that is omitted from the MATLAB interface. This usually indicates that an error in one of the other input parameters has caused an incorrect value to be inferred.
## Accuracy
For each right-hand side vector b$b$, the computed solution x$x$ is the exact solution of a perturbed system of equations (A + E)x = b$\left(A+E\right)x=b$, where
|E| ≤ c(n)εP|L||U| , $|E|≤c(n)εP|L||U| ,$
c(n)$c\left(n\right)$ is a modest linear function of n$n$, and ε$\epsilon$ is the machine precision.
If $\stackrel{^}{x}$ is the true solution, then the computed solution x$x$ satisfies a forward error bound of the form
( ‖x − x̂‖∞ )/( ‖x‖∞ ) ≤ c(n)cond(A,x)ε $‖x-x^‖∞ ‖x‖∞ ≤c(n)cond(A,x)ε$
where cond(A,x) = |A1||A||x| / x cond(A) = |A1||A| κ (A) $\mathrm{cond}\left(A,x\right)={‖|{A}^{-1}||A||x|‖}_{\infty }/{‖x‖}_{\infty }\le \mathrm{cond}\left(A\right)={‖|{A}^{-1}||A|‖}_{\infty }\le {\kappa }_{\infty }\left(A\right)$.
Note that cond(A,x)$\mathrm{cond}\left(A,x\right)$ can be much smaller than cond(A)$\mathrm{cond}\left(A\right)$, and cond(AH)$\mathrm{cond}\left({A}^{\mathrm{H}}\right)$ (which is the same as cond(AT)$\mathrm{cond}\left({A}^{\mathrm{T}}\right)$) can be much larger (or smaller) than cond(A)$\mathrm{cond}\left(A\right)$.
Forward and backward error bounds can be computed by calling nag_lapack_zgerfs (f07av), and an estimate for κ(A)${\kappa }_{\infty }\left(A\right)$ can be obtained by calling nag_lapack_zgecon (f07au) with norm = 'I'${\mathbf{norm}}=\text{'I'}$.
The total number of real floating point operations is approximately 8n2r$8{n}^{2}r$.
This function may be followed by a call to nag_lapack_zgerfs (f07av) to refine the solution and return an error estimate.
The real analogue of this function is nag_lapack_dgetrs (f07ae).
## Example
```function nag_lapack_zgetrs_example
trans = 'N';
a = [ -1.34 + 2.55i, 0.28 + 3.17i, -6.39 - 2.2i, 0.72 - 0.92i;
-0.17 - 1.41i, 3.31 - 0.15i, -0.15 + 1.34i, 1.29 + 1.38i;
-3.29 - 2.39i, -1.91 + 4.42i, -0.14 - 1.35i, 1.72 + 1.35i;
2.41 + 0.39i, -0.56 + 1.47i, -0.83 - 0.69i, -1.96 + 0.67i];
b = [ 26.26 + 51.78i, 31.32 - 6.7i;
6.43 - 8.68i, 15.86 - 1.42i;
-5.75 + 25.31i, -2.15 + 30.19i;
1.16 + 2.57i, -2.56 + 7.55i];
% Factorize a
[a, ipiv, info] = nag_lapack_zgetrf(a);
% Compute solution
[bOut, info] = nag_lapack_zgetrs(trans, a, ipiv, b)
```
```
bOut =
1.0000 + 1.0000i -1.0000 - 2.0000i
2.0000 - 3.0000i 5.0000 + 1.0000i
-4.0000 - 5.0000i -3.0000 + 4.0000i
0.0000 + 6.0000i 2.0000 - 3.0000i
info =
0
```
```function f07as_example
trans = 'N';
a = [ -1.34 + 2.55i, 0.28 + 3.17i, -6.39 - 2.2i, 0.72 - 0.92i;
-0.17 - 1.41i, 3.31 - 0.15i, -0.15 + 1.34i, 1.29 + 1.38i;
-3.29 - 2.39i, -1.91 + 4.42i, -0.14 - 1.35i, 1.72 + 1.35i;
2.41 + 0.39i, -0.56 + 1.47i, -0.83 - 0.69i, -1.96 + 0.67i];
b = [ 26.26 + 51.78i, 31.32 - 6.7i;
6.43 - 8.68i, 15.86 - 1.42i;
-5.75 + 25.31i, -2.15 + 30.19i;
1.16 + 2.57i, -2.56 + 7.55i];
% Factorize a
[a, ipiv, info] = f07ar(a);
% Compute solution
[bOut, info] = f07as(trans, a, ipiv, b)
```
```
bOut =
1.0000 + 1.0000i -1.0000 - 2.0000i
2.0000 - 3.0000i 5.0000 + 1.0000i
-4.0000 - 5.0000i -3.0000 + 4.0000i
0.0000 + 6.0000i 2.0000 - 3.0000i
info =
0
``` | 3,116 | 7,491 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 77, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2013-48 | longest | en | 0.538789 |
https://pdfhall.com/numerical-optimization-of-a-hysteresis-model_5b73e067097c47aa6f8b457b.html | 1,656,544,515,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00434.warc.gz | 500,057,727 | 13,674 | ## Numerical optimization of a hysteresis model
Elektra program. References. [1] A. Bergqvist, G. Engdahl, A thermodynamic representation of pseudoparticles with hysteresis, IEEE Trans. Magn. 31. (6) (1995) ...
ARTICLE IN PRESS
Physica B 343 (2004) 35–38
Numerical optimization of a hysteresis model J.H. Kraha,*, A.J. Bergqvistb a
Royal Institute of Technology, Electrical Engineering, Teknikringen 22, Stockholm S-100 44, Sweden b ( S-721 78, Sweden ABB Corporate Research, Vaster as .
Abstract An equation system for the anhysteretic curve and the distribution function required by a hysteresis model is suggested. A least-square fitting confirms that the solution is unique and gives optimal agreement with the measurements. Moreover, the used fitting function with three parameters replaces the measured vector, which increases simulation performance. r 2003 Elsevier B.V. All rights reserved. PACS: 75.60.Ej; 02.60.Pn Keywords: Anhysteretic curve; Pinning strength; Distribution function; Approximation
1. Introduction Since 1996, a new hysteresis model based on thermodynamic considerations has been developed [1,2]. For a given material, the model requires only the measurement of the major loop and the virgin curve. The unknown functions are the anhysteretic curve and the distribution function of the involved volume fractions lumped together with respect to magnetic pinning strength, so called pseudo particles. So far the anhysteretic curve was obtained by trial and error, because the middle curve between upper and lower branches of the major loop [3] did not work well. This paper suggests an equation system with the two unknown functions and the two measured curves. A numerical way is to perform least-square fitting, which at the same time reduces the required *Corresponding author. Fax: +46-8-205268. E-mail address: [email protected] (J.H. Krah).
measured curves for a given material to a simple function with a few parameters. This technique reduces the amount of data in a material database and a function is easier and faster to handle during simulations, which is needed, for instance, for lumped element modeling of transformer cores.
2. The model The hysteresis model calculates the magnetization M½H of the magnetic material B ¼ m0 ðH þ M½HÞ; M½H ¼ c Man ðHÞ þ
ð1Þ m X
Man ðPli k ½HÞBðli Þ;
ð2Þ
i¼1
where Pli k ; i ¼ 1; 2; y; m are the play operators of the involved pseudo particles li with pinning strengths li k as backlash values and volume fractions zðli Þ: k is the mean pinning strength
ARTICLE IN PRESS J.H. Krah, A.J. Bergqvist / Physica B 343 (2004) 35–38
and Man the anhysteretic curve. The parameter c takes into account some magnetic reversibility occurring in real magnetic materials. The virgin curve and the major loop can be reproduced uniquely by measurements and they are the easiest cases of magnetizing process. The backlashes of the pseudo particles Pli k ½H are activated one by one according to the distribution function zðli Þ: For H > 0 Eq. (2) can then be rewritten as Z H=k Man ðH lkÞBðlÞ dl ¼ V ðHÞ; cMan ðHÞ þ 0
ð3Þ where V ðHÞ is the measured virgin curve. Similarly, the lower half of the major loop Ml ðHÞ gives Eq. (4), where the pseudo particles are initially saturated to MS and the backlashes start with negative pinning forces li k instead of zero. Z ðHHmin Þ=2k Man ðH 2lkÞBðlÞ dl 0 Z N BðlÞ dl ¼ Ml ðHÞ: ð4Þ MS ðHHmin Þ=2k
Eqs. (3) and (4) contain two unknown and two measured functions. After elimination of zðli Þ the anhysteretic curve can be determined. One numerical way is to insert an approximation function with a few parameters for the anhysteretic curve and then to perform a least-square fitting. Linearizing Man for small H; Eq. (3) gives Z H=k 0 0 Man ð0ÞðH lkÞBð0Þ dl: MEcMan ð0ÞH þ 0
ð5Þ 0 Replacing the slope Man (0) of the anhysteretic curve for small H by the anhysteretic susceptibility w; Eq. (5) reads
MEcwH þ
1 Bð0ÞwH 2 : 2k
1
0
-1
-2 -5000
-2500
0
2500
5000
H [A/m]
Fig. 1. Measurement of major loop and virgin curve.
where w0 þ m0 and k are the material-specific socalled Rayleigh parameters.
3. The measurement In order to capture a smooth virgin curve without edges, it is important to demagnetize with sufficiently many loops. Another problem was a considerable DC offset of the magnetic field through the sample sheet due to small remanence in the used yoke system [4] and a small DC offset of the electric amplifier generating the magnetomotive force. This required a program that first applies an alternating H field with slowly decreasing amplitude and then compares the obtained final B-field value with the mean value of the extreme B-field values, which correspond to positive and negative saturation. If the difference is larger than a specified tolerance, a PI controller calculates a new compensation offset for the drive current generating the H-field. Fig. 1 shows a successful measurement of the virgin curve.
ð6Þ
Eq. (6) confirms Rayleigh’s [3] observation in 1887 that the virgin curve for small magnetization levels has the approximate form BðHÞ ¼ ðw0 þ m0 Þ H þ kH 2 ;
2
B [T]
36
ð7Þ
4. The virgin curve The beginning of the virgin curve for the above measurement was fit by a parabola, see Fig. 2.
ARTICLE IN PRESS J.H. Krah, A.J. Bergqvist / Physica B 343 (2004) 35–38 1.5
37
B -7 2
-4
y = 6.4·10 ·x + 1.4·10 ·x
B [T]
Hc
H
Fig. 3. One way to determine the middle curve of the major loop.
0 -500
H H-Hc H+Hc
0
500
1000
1500
H [A/m]
2
Fig. 2. Fitting of begin of virgin curve by a parabola.
1
Initially, the anhysteretic curve was assumed as a middle curve between upper and lower halves of the major loop [3]. One way using the coercivity HC is illustrated in Fig. 3. The resulting vector for the expected anhysteretic curve for the measured material was fit by a parameterized Sigmoid function. The Sigmoid function given by Eq. (5) has the asymptotic values 0 in the left-half plane and 1 in the righthalf plane. Around the point of origin, it increases smoothly: 1 : 1 þ ex
0
-1
-2 -5000
-2500
0
2500
5000
H [A/m]
Fig. 4. Middle curve of the measured major loop (dotted) for M700 and fitting by a Sigmoid function with a ¼ 1:81; b ¼ 395 and c ¼ 38:2: 2
ð5Þ
Shifting fs ðxÞ down by 0.5 and introducing parameters as amplitude a; a scale factor b for the argument and finally superposing a line with slope c gives Eq. (6), which was fit to the expected anhysteretic curve, see Fig. 4. 2 Ban ðHÞ ¼ a 1 þ cm0 H: ð6Þ 1 þ eH=b Then, a demagnetization example was simulated with 25 pseudo particles distributed according to a Gauss curve, see Fig. 5. The maximal slope of the calculated major loop is lower than that of the measured one, because the backlashes of the pseudo particles involve delays. This means that for small B-field levels the slope of the anhysteretic curve has to be higher than the slope of the middle curve of the major loop.
1
B [T]
fs ðxÞ ¼
B [T]
5. The anhysteretic curve
0
-1
-2 -5000
-2500
0
2500
5000
H [A/m] Fig. 5. Measured virgin curve and major loop and a demagnetization simulation with the original model implementation with 25 pseudo particles.
Furthermore, the used Gauss distribution of the pseudo particles involves negative pinning strengths. A more physical approach is to relate
ARTICLE IN PRESS J.H. Krah, A.J. Bergqvist / Physica B 343 (2004) 35–38
38
curve. The parameters a and c changed slightly to 1.85 and 34.5, respectively. The updated simulated demagnetization curve in Fig. 7 shows that the new calculated major loop fits better than the original in Fig. 5. However, the low number of the 12 used pseudo particles in Fig. 6 obviously involves discretization errors.
0.3
Volume fraction [%]
0.25 0.2 0.15 0.1 0.05 0
6. Conclusions 0
1000
2000
3000
4000
H [A/m]
Fig. 6. Optimal distribution of hysteretic pseudo particles for M700 (‘ ’) and a rough estimation related to measurement (‘+’).
2
B [T]
1
0
-1
-2 -5000
The hysteresis model works accurately with about 30 pseudo particles and an approximation of the anhysteretic curve with a maximum slope higher than that of the major loop. The anhysteretic curve could be fit well by a Sigmoid function. This reduced the required memory from a measured vector to three parameters and increased the simulation speed by a factor of 8.5. The solution of the equation system for the anhysteretic curve and the distribution function is unique for a measured material.
Acknowledgements
-2500
0
2500
5000
This work was financed by ABB within the Elektra program.
H [A/m] Fig. 7. Measured virgin curve and major loop and updated simulation with 12 pseudo particles.
References
the difference between upper and lower halves of the measured major loop to the total remaining backlash. The normalized derivative with respect to H in Fig. 6 can be used as a qualitative starting point for an approximation of the distribution function for the studied material. The optimum was obtained by least-square fitting of the calculated to the measured major loop. The Sigmoid parameter b decreased to around 200 involving a clearly higher initial slope of the anhysteretic
[1] A. Bergqvist, G. Engdahl, A thermodynamic representation of pseudoparticles with hysteresis, IEEE Trans. Magn. 31 (6) (1995) 3539. [2] A. Bergqvist, Magnetic vector hysteresis model with dry friction-like pinning, Phys. B 233 (1997) 342. [3] R.M. Bozorth, Ferromagnetism, Bell Telephone Laboratires, D. Van Nostrand Company, Princeton, NJ, 1951, p. 476. [4] J.H. Krah, G. Engdahl, Experimental verification of alternative audio frequency 2D magnetization set-up for soft magnetic sheets and foils, Proc. Seventh International Workshop on 1&2-Dimensional Magnetic Measurement and Testing, Germany, 2002, Technische Bundesanstalt, Braunschweig, 2003, p. 103–108. | 2,779 | 9,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-27 | latest | en | 0.856862 |
http://www.jiskha.com/display.cgi?id=1294104420 | 1,496,010,322,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463611569.86/warc/CC-MAIN-20170528220125-20170529000125-00401.warc.gz | 681,642,488 | 3,571 | # math
posted by on .
write the steps you would follow to find 8 x 5 x 9 using the multiplication properties and multiples of 10
• math - ,
8*5*9 = 4*2*5*9 = 4*10*9 = 36*10 = 360 | 67 | 182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-22 | latest | en | 0.835742 |
https://www.jiskha.com/questions/1440915/A-true-false-test-consists-of-15-items-b-What-is-the-probability-that-Chris-gets | 1,542,478,778,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743717.31/warc/CC-MAIN-20181117164722-20181117190722-00067.warc.gz | 928,889,026 | 5,749 | # math
A true-false test consists of 15 items.
(b) What is the probability that Chris gets 80% or more for the test?
(c) If it is a 25 item true-false test, would you think it is easier or more difficulty for Chris to get 80% or more? (Hint: please calculate the corresponding probability that Chris gets 80% or more for the test, and compare your results with the one obtained in (b) to answer this question.)
b) he must get 13 right or 14 right or 15 right
= C(15,13)(1/2)^13 (1/2)^2 + C(15,14)(1/2)^14(1/2) + C(15,15) (1/2)^15
= 0.01922%
c) he must get 23 right or 24 right or 25 right
= C(25,23)(1/2)^23 (1/2)^2 + C(25,24)(1/2)^24(1/2) + C(25,25) (1/2)^25
= 5.36441803e-0.5
1. 0
2. 0
3. 6
1. The assumption would be that he would be randomly guessing at the answer.
Since 12/15 = .8, why did you not include the probability of getting 12 correct out of the 15 ?
Try adding that to your other cases.
For b) wouldn't getting 20 out of 25 correct yield 80% ?
Same for 21 and 22
So you have a few more cases to include
1. 0
2. 0
posted by Reiny
2. maybe speak to your tutor if you're having issues. very easy to get done for plagiarism for this when it shows up as the first result on google. QUT is coming for you
1. 0
2. 0
posted by lol
## Similar Questions
1. ### math
A true-false test consists of 15 items. (a) If Chris does not study at all and guesses each and every item in the test, describe the probability model for the number of correct guesses. (b) What is the probability that Chris gets
asked by susan on August 24, 2016
2. ### Math
A true-false test consists of 10 items. What is the probability that Chris gets 80% or more for the test?
asked by Denzil on April 8, 2016
3. ### math
A true-false test consists of 15 items. (a) If Chris does not study at all and guesses each and every item in the test, describe the probability model for the number of correct guesses. each individual question has a probability
asked by susan on August 24, 2016
4. ### MATH Prob.
Jack is taking a four-item true–false test. He has no knowledge about the subject of the test and decides to flip a coin to answer the items. What is the probability that he receives a perfect score? What is the probability on a
asked by Twg on August 6, 2009
5. ### math
the question is: If you know that 4 of the 10 items on a true-false test are false, how many different ways could you select 4 items to mark as being false? I think the answer is 10P4=10*9*8*7/4*3*2*1=5040/24=210 ways
asked by ginger on February 4, 2011
6. ### math
Could someone please tell me if I am on track with my answer? Jack is taking a four-item true-false test. He has no knowledge about the subject of the test and decides to flip a coin to answer the items. What is the probability
asked by Punkie on November 19, 2009
7. ### statistics
a test consists of true/false questions .to pass the test a student must answer at least 6questions correctly .if a student guesses on each question,what is the probability that the student will pass the test
asked by AILY on December 6, 2009
8. ### Math
Jack is taking a four-item true-false test. He has no knowledge about the subject of the test and decided to flip a coin to answer the items. What is the probability that he receives a perfect score? What is the probability on a
asked by B.B. on August 4, 2009
9. ### MATH 157
Can anyone please help me understand these math problems. #4. How many three-symbol codes (letter-number-number)can be made from the letters S, P, Y, and two-digits from the set of (0, 1, 2,..., 9) without repetition? #16. A pair
asked by Rayna on August 4, 2010
10. ### Statistics
A research article reports the results of a drug test. This medicine used to reduce vision loss in people suffering from macular degeneration. The article state p = 0,04 section findings. The following issues are of three
asked by Andrew on July 19, 2013
More Similar Questions | 1,138 | 3,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | longest | en | 0.937971 |
https://www.coursemerit.com/solution-details/35241/MAT-543-Week-6-Homework--Chapter-8Chapter-9-Exercises-9-1-and-9-2-Exercises-8-1-through-8-5- | 1,558,842,266,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00027.warc.gz | 765,473,176 | 10,910 | Question DetailsNormal
\$ 30.00
MAT 543 Week 6 Homework Chapter 8:Chapter 9: Exercises 9-1 and 9-2 Exercises 8-1 through 8-5,
Question posted by
MAT 543 Week 6 Homework
Chapter 8: Exercises 8-1 through 8-5 (page 155 of the text) Chapter 9: Exercises 9-1 and 9-2 (page 174 of the text)
8-1 Two types of visits are provided by the Durham Health Clinic, first-time visits and return visits. Table 8-5 provides the processing time for each work station and the available staff hours per week. Determine the production frontiers for this clinic and indicate which station should be expanded to increase the overall capacity of the clinic. Which service station could be reduced?
8-2 Durham Health Clinic has a contribution margin of \$35 per visit. Calculate the break-even point in visits with fixed costs at \$4000, \$6500, and \$8500 per week. Given this analysis, as a manager, what would you recommend and why?
8-3 Durham Health Clinic is considering signing a contract to perform 50 pre-employment physicals per week for a specific corporation. In terms of staff time, a pre-employment physical requires 0.20 hours in Reception/Discharge, 0.45 hours in Nursing and Testing, and 0.20 hours in Medical Examination. By workstation, determine how many work hours per week will be needed to perform these physicals.
8-4 Currently the clinic does 250 visits per week, with 50% of all visits as return visits. Each employee (physician, nurse, and receptionist) is scheduled to work 35 hours per week.
a. How many employees by type does the clinic currently need?
b. How many employees by type will the clinic need if it signs the contract for pre-employment physicals?
c. If return visits shift to 10% of all regular visits, how many employees by type will the clinic need with and without the contract for pre-employment physicals?
d. How will the answers to “b” and “c” change if the number of physicals is modified to 35 pre employment physicals per week?
Throughout these analyses, specify all assumptions, including assumptions concerning worker productivity.
8-5 How would your answers change for problem 8-1 if nursing and testing time was increased to 0.50 hours for both first and repeat visits, and medical exam and treatment time was reduced to 0.30 hours for a first visit and 0.20 hours for a return visit?
9-1 Alpha Walk-in Clinic operates as a single channel single server system. On Tuesdays, its average arrival rate (λ) per hour is 7.0. Analysis indicates that its service rate (μ) is 8.5 patients per hour. Using queuing theory, describe this service system. What is:
a. The probability that the clinic is idle—no patients waiting or being served?
b. The average number of patients in the system?
c. The average time (hours) a patient spends in the system (waiting + service time)?
d. The average number of patients in the queue waiting for service?
e. The average time (hours) a patient spends in the queue waiting?
f. The probability that the patient, upon arrival, must wait?
9-2 The following data have been collected from a hospital pharmacy. This service system operates as a single server, single channel system.
7–3 pm 3–11 pm 11–7 am
Service rate per hour 200 100 50
Arrival rate per hour 60 50 40
The service rate can be increased or decreased in increments of 50 prescriptions per hour. The expense associated with each 50-prescription increment is \$100. In other words, to be able to process 50 additional prescriptions will cost an additional \$100 per hour. If the current rate of processing or service is lowered by 50 prescriptions per hour, the savings are \$100 per hour. Using queuing theory, describe this service system. What is:
a. The probability that the clinic is idle—no patients waiting or being served?
b. The average number of patients in the system?
c. The average time (hours) a patient spends in the system (waiting + service time)?
d. The average number of patients in the queue waiting for service?
e. The average time (hours) a patient spends in the queue waiting?
f. The probability that a patient, upon arrival, must wait?
Given the associated costs, should the service rate be changed? What are the financial implications associated with your recommendations?
Available Solution
\$ 30.00
MAT 543 Week 6 Homework Chapter 8:Chapter 9: Exercises 9-1 and 9-2 Exercises 8-1 through 8-5,
• This solution has not purchased yet.
• Submitted On 15 Nov, 2017 11:31:56
Solution posted by
MAT 543 Week 6 Homework Chapter 8: Exercises 8-1...
Buy now to view full solution.
Attachment
\$ 629.35 | 1,103 | 4,679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.949125 |
https://www.thestudentroom.co.uk/showthread.php?t=5237078 | 1,531,753,045,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589350.19/warc/CC-MAIN-20180716135037-20180716155037-00242.warc.gz | 1,016,210,773 | 42,114 | Turn on thread page Beta
You are Here: Home >< Maths
proportion watch
1. I don't know how to work part b out
Posted on the TSR App. Download from Apple or Google Play
2. (Original post by Wildnatxox)
I don't know how to work part b out
Basically asks you for the amount of cookies that use at most 1kg of the flour.
3. (Original post by RDKGames)
Basically asks you for the amount of cookies that use at most 1kg of the flour.
how are you meant to know
Posted on the TSR App. Download from Apple or Google Play
4. (Original post by Wildnatxox)
how are you meant to know
30 biscuits = 230g
60 biscuits = 460g
90 biscuits = 690g
...
What's the highest can you go on the biscuits so that the grams on the RHS stay below 1000g?
5. work out how much flour is needed for 1 bikkie. then see how many of that amount will fit into 1000g.
6. (Original post by the bear)
work out how much flour is needed for 1 bikkie. then see how many of that amount will fit into 1000g.
how do you find out how much there is needed for one biscuit
Posted on the TSR App. Download from Apple or Google Play
7. (Original post by Wildnatxox)
how do you find out how much there is needed for one biscuit
so you are told that 230g will make 30 biscuits. from there you can find how much for 1 biscuit.
8. (Original post by the bear)
so you are told that 230g will make 30 biscuits. from there you can find how much for 1 biscuit.
230 divided by 30?
Posted on the TSR App. Download from Apple or Google Play
9. (Original post by Wildnatxox)
230 divided by 30?
yh
10. (Original post by the bear)
work out how much flour is needed for 1 bikkie. then see how many of that amount will fit into 1000g.
But then how can you have 1/15th of an egg for a single cookie? Seems impractical to me!
11. (Original post by RDKGames)
But then how can you have 1/15th of an egg for a single cookie? Seems impractical to me!
that takes the ermm biscuit
Reply
Submit reply
Turn on thread page Beta
Related university courses
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: March 7, 2018
Today on TSR
Top unis in Clearing
Tons of places at all these high-ranking unis
Poll
Useful resources
Make your revision easier
Maths Forum posting guidelines
Not sure where to post? Read the updated guidelines here
How to use LaTex
Writing equations the easy way
Study habits of A* students
Top tips from students who have already aced their exams
Create your own Study Planner
Never miss a deadline again
Thinking about a maths degree?
Chat with other maths applicants
Can you help? Study help unanswered threads
Groups associated with this forum:
View associated groups
The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.
Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE
Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice. | 821 | 3,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-30 | latest | en | 0.937651 |
https://www.physicsforums.com/threads/electricity-questions.292408/ | 1,532,307,830,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594675.66/warc/CC-MAIN-20180722233159-20180723013159-00241.warc.gz | 976,215,424 | 14,042 | # Homework Help: Electricity questions
1. Feb 14, 2009
### Jnblose
1. The problem statement, all variables and given/known data
f a voltage of 40.7 V is applied between points a and b, find the current in each resistor.
The drawing looks like this (idk how to add a picture here) A-[ ]-/\/-[ ]-B In between brackets (space at top and bottom are resistors)... top of the first (A) is 12, bottom 6, middle there is one (designated by /\/) at 5. On top and bottom of B are 4 and 8 respectively.
2. Relevant equations
3. The attempt at a solution First question is Find the equivalent resistance between point a and b. So I tried to break them down and do Rab (hypothetically) as 1/r1 + 1/r2 so 1/12 + 1/6 = .25 then I made a Rcd (hypothetically) and did 1/r4 + 1/r5 which is 1/4 + 1/8 = .375 then I added R3 to get the total R which would be +5 so answer would be 5.625 This was marked incorrect. Then it says if a voltage 40.7 is applied b/t a and B find current in each resistor. There I am completely lost but I attempted something random and got of course a wrong answer.
2. Feb 14, 2009
### Delphi51
I can't follow your description of the diagram. If you can scan one in, you can attach it to your message. Or post it to a site like photobucket.com and give us a link. Some are putting (IMG) and (/IMG), with square brackets instead of round, around the link so it shows up as an image in the post.
Perhaps you could just describe it better. The current goes through the 12 ohm resister . . .
3. Feb 14, 2009
### Jnblose
Last edited by a moderator: Apr 24, 2017
4. Feb 14, 2009
### Jnblose
Ehhh fail wrong one sorry... 1 second.
5. Feb 14, 2009
### Jnblose
6. Feb 14, 2009
### kron
Hi,
you have to calculate for the 12 and 6 Ohm the following:
1/12 + 1/6 = 1/r_1 so r_1 = 6*12/(6 + 12) = 4 = 1/0.25.
So 0.25 is wrong. Do the same for 1/4 + 1/8 = 1/r_3 -> r_3.
r_2 = 5 Ohm.
So R = r_1 + r_2 + r_3.
Now you have I = U/R (between a and b) with U = 40.7 V.
Try to figure out the other currents by yourself. | 649 | 2,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.93395 |
https://www.12manage.com/description_black_scholes_model.html | 1,624,383,279,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488519183.85/warc/CC-MAIN-20210622155328-20210622185328-00606.warc.gz | 538,102,625 | 7,623 | # Black-Scholes Model
Knowledge Center
## Summary, forum, best practices, expert tips and information sources.
12 items • 12.449 visits
Summary
### What is the Black-Scholes Model? Meaning.
Black-Scholes Model is a pricing model of financial instruments, and in particular stocks and options, derived by Fischer Black and Myron Scholes in 1973. It is based on arbitrage arguments that uses the stock price, the exercise price, the risk-free interest rate, the time to expiration, and the standard deviation (volatility) of the stock return.
The key assumptions of the model are:
• The price of the underlying instrument is a geometric Brownian motion, in particular with constant drift and volatility.
• It is possible to short sell the underlying stock.
• There are no arbitrage opportunities.
• Trading in the stock is continuous.
• There are no transaction costs or taxes.
• All securities are perfectly divisible (e.g. it is possible to buy 1/100th of a share).
• The risk-free interest rate is constant, and the same for all maturity dates.
Special Interest Group Join
Black-Scholes Model Special Interest Group. Special Interest Group (43 members)
Forum New Topic
Forum about the Black-Scholes Model.
Comparing 2 Mutual Funds using Standard Deviation How can standard deviation be used to compare the rate of return on two mutual funds? (...) 5 1 comments
Explanation of Black-Scholes Option Valuation Equation The main equation for the Black-Scholes Option Valuation is as follows: C = S.N(d1. - X.e-RT. N(d2. (12) where (...) 3
🔥 NEW Problems with Using Standard Deviation Standard deviation is the most widely used measure of investment risk. It assumes all investors agree on the degree of r (...) 2
Start a new forum topic about the Black-Scholes ModelExchanging your ideas stimulates your personal and professional development. And you can help other people! More info.
Best Practices Sign up
The top-rated topics about the Black-Scholes Model. Here you will find the most valuable ideas and practical suggestions.
Expert Tips Sign up
Advanced insights about the Black-Scholes Model. Here you will find professional advices by experts.
Information Sources Sign up
Various sources of information regarding the Black-Scholes Model. Here you will find powerpoints, videos, news, etc. to use in your own lectures and workshops.
Black-Scholes Option ValuationBlack-Scholes Model, Put Valuation, Call Valuation, Option ValuationThis presentation aims to briefly and clearly explain the Black-Scholes model and the formula’s that are used within thi (...)
Research Links Sign up
Automatically jump to further useful sources regarding the Black-Scholes Model.
News Videos Presentations Books More
Compare with: CAPM | APT | European-Style Option
Special Interest Group
Are you interested in the Black-Scholes Model? Sign up for free
Notify your students
Copy this into your study materials: Knowledge Center about Black-Scholes Model (12manage) and add a hyperlink to: https://www.12manage.com/description_black_scholes_model.html
Link to this knowledge center
Copy this HTML code to your web site: Knowledge Center about Black-Scholes Model (12manage)
Return to Management Hub: Decision-making & Valuation | Finance & Investing More on Management | Return to Management Dictionary
This ends our Black-Scholes Model summary and forum.
About 12manage | Advertising | Link to us / Cite us | Privacy | Suggestions | Terms of Service
© 2021 12manage - The Executive Fast Track. V15.8 - Last updated: 22-6-2021. All names ™ of their owners. | 796 | 3,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-25 | longest | en | 0.859128 |
https://lists.nongnu.org/archive/html/axiom-developer/2005-07/msg00203.html | 1,571,007,528,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986648343.8/warc/CC-MAIN-20191013221144-20191014004144-00493.warc.gz | 690,179,862 | 3,247 | axiom-developer
[Top][All Lists]
## [Axiom-developer] [#193 Symbolic values (without variables) are not orde
From: wyscc Subject: [Axiom-developer] [#193 Symbolic values (without variables) are not ordered properly in 'EXPR INT'] Two different orderings mixed up Date: Wed, 20 Jul 2005 17:48:32 -0500
Changes
http://page.axiom-developer.org/zope/mathaction/193SymbolicValuesWithoutVariablesAreNotOrderedProperlyInEXPRINT/diff
--
Hi Martin:
is clear that the meaning of the operator '<' in 'EXPR INT' (or any domain that
includes symbolic expressions like 'POLY INT') is not meant to be an ordering
of the ground domain (here 'INT'). In polynomial rings, '<' is a term-ordering
to facilitate polynomial arithmetic and variable elimination methods. If I read
'expr.spad' correctly, when 'R' is an integral domain, 'EXPR R' has 'FRAC
SMP(R, Kernel EXPR R)' as its representation ('Rep'). You can explain better
what that means than I (I would much appreciate your writing a page explaining
kernels and operators in Axiom). In any case, the ordering is inherited from
the 'Rep' and so I believe '%e' or 'exp(1)' is treated as a symbol. Indeed that
may be the reason we have:
\begin{axiom}
exp(1)^2 - exp(2)
(exp(1)^2 - exp(2) = 0)::Boolean
\end{axiom}
The sign of a polynomial expression is the sign of the coefficient of the
highest monomial term.
The way that '4-%e < 0' can be evaluated they way you intend is to substitute a
numerical value for '%e'. So users should be aware that in Axiom, symbolic
entries are not the same as their numerical counterpart. What they should do is:
\begin{axiom}
4.0 - %e < 0
\end{axiom}
or
\begin{axiom}
(4-%e)::(EXPR Float) < 0
\end{axiom}
(This is reminescent of FORTRAN: type promotion from INTEGER to REAL). In other
words, if one wants to test numerical inequalities involving symbolic
constants, one should work in 'EXPR Float', not 'EXPR Integer' since most
symbolic constants are not integers. To provide an automatic type promotion
from 'EXPR INT' to 'EXPR Float' would interfere with the ordering in polynomial
expressions, unless a symbolic constant can be distinguished from a polynomial
variable. We cannot overload the operator '<' for two distinct orderings in the
same domain. I think the suggestion I posted in #47 of a 'SymbolicFloat' domain
may be worth exploring. In 'SymbolicFloat', we can retain symbolic manipulation
of expresions with symbolic constants by using the term-ordering of the
underllying representation, and still use the numerical ordering of the domain
'SymbolicFloat'.
William
-- | 667 | 2,564 | {"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} | 2.71875 | 3 | CC-MAIN-2019-43 | latest | en | 0.842174 |
https://www.coursehero.com/tutors-problems/Math/8216134-001-part-1-of-2-100-points-Patty-is-only-7-years-old-but-she-insi/ | 1,550,294,286,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479885.8/warc/CC-MAIN-20190216045013-20190216071013-00244.warc.gz | 789,570,206 | 21,176 | View the step-by-step solution to:
# 001 (part 1 of 2) 10.0 points Patty is only 7 years old, but she insists she can play basketball with a regulation-height hoop.
001 (part 1 of 2) 10.0 points
Patty is only 7 years old, but she insists she
can play basketball with a regulation-height
hoop. She can throw the ball with an initial
velocity of 10 feet per second.
The equation for the height of the ball is
y = −16t^2+ vt + s
where v is the initial velocity, t is in seconds,
and s is the initial height.
How high can she throw a ball?
### Why Join Course Hero?
Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.
### -
Educational Resources
• ### -
Study Documents
Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.
Browse Documents | 227 | 919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-09 | latest | en | 0.927337 |
http://www.chegg.com/homework-help/statistical-reasoning-for-everyday-life-3rd-edition-chapter-1.2-solutions-9780321286727 | 1,475,028,976,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661289.57/warc/CC-MAIN-20160924173741-00107-ip-10-143-35-109.ec2.internal.warc.gz | 406,885,932 | 16,810 | # Statistical Reasoning for Everyday Life (3rd Edition) View more editions Solutions for Chapter 1.2
• 1445 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
Census and Sample. What is a census, what is a sample, and what is the difference between them?
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 1
Census is the process of collecting data from each individual of the population. Sample is part of population and its characteristics are same as that of population. The difference between census and sample is that census involves data collection from whole population whereas sample involves data collection from part of population.
Corresponding Textbook
Statistical Reasoning for Everyday Life | 3rd Edition
9780321286727ISBN-13: 0321286723ISBN: Authors:
Alternate ISBN: 9780205646425, 9780321287069, 9780321287113, 9780321505736, 9780321505743, 9780321656117, 9780321656612, 9780321831057 | 269 | 1,058 | {"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} | 2.640625 | 3 | CC-MAIN-2016-40 | latest | en | 0.864707 |
https://oeis.org/A088197 | 1,627,649,567,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00135.warc.gz | 445,344,395 | 3,657 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A088197 First differences of A088196. 7
1, 3, 4, 1, 3, 4, 4, 5, 3, 5, 3, 4, 4, 5, 7, 1, 7, 4, -2, 10, 4, 4, 6, 7, 3, 4, 1, 3, 16, 4, 4, 4, 9, 3, 5, 7, 4, 5, 7, 1, 11, -2, 7, 3, 12, 12, 4, 1, 3, 8, -4, 16, 4, 8, 5, 3, 5, 3, 4, 9, 15, 4, -2, 7, 15, 2, 14, 1, 3, 8, 8, 5, 7, 4, 5, 8, 3, 4, 16, 1, 11, -2, 10, 4, 4, 6, 7, 3, 4, 12, 8, 4, 8, 4, 5, 11, 4, 17 (list; graph; refs; listen; history; text; internal format)
OFFSET 2,2 LINKS PROG (PARI) qnrp_d(n)= { /* The difference sequence of the sequence with the largest QnR modulo the primes */ local(k=1, m, p, fl, jj, j, v=[]); for(i=3, n, m=0; p=prime(i); jj=0; fl=2^p-1; j=2; while((j<=(p-1)/2), jj=(j^2)%p; fl-=2^jj; j++); j=p-1; while(m==0, if(bitand(2^j, fl), m=j); j--); v=concat(v, m-k); k=m); print(v)} CROSSREFS Cf. A088191, A088196, A088198, A088199, A088200, A088201. Sequence in context: A101667 A117378 A278518 * A087517 A128529 A205547 Adjacent sequences: A088194 A088195 A088196 * A088198 A088199 A088200 KEYWORD easy,sign AUTHOR Ferenc Adorjan (fadorjan(AT)freemail.hu), Sep 23 2003 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified July 30 08:52 EDT 2021. Contains 346355 sequences. (Running on oeis4.) | 710 | 1,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-31 | latest | en | 0.5608 |
http://encyclopedia2.thefreedictionary.com/Computer+algorithm | 1,521,865,150,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649683.30/warc/CC-MAIN-20180324034649-20180324054649-00053.warc.gz | 93,487,927 | 23,039 | # algorithm
(redirected from Computer algorithm)
Also found in: Dictionary, Thesaurus, Medical, Financial.
Related to Computer algorithm: Pseudocode, Quantum algorithm
## algorithm
(ăl`gərĭth'əm) or
## algorism
(–rĭz'əm) [for Al-KhowarizmiAl-Khowarizmi
, fl. 820, Arab mathematician of the court of Mamun in Baghdad. His treatises on Hindu arithmetic and on algebra made him famous. He is said to have given algebra its name, and the word algorithm is said to have been derived from his name.
], a clearly defined procedure for obtaining the solution to a general type of problem, often numerical. Much of ordinary arithmetic as traditionally taught consists of algorithms involving the fundamental operations of addition, subtraction, multiplication, and division. An example of an algorithm is the common procedure for division, e.g., the division of 1,347 by 8, in which the remainders of partial divisions are carried to the next digit or digits; in this case the remainder of 5 in the division of 13 by 8 is placed in front of the 4, and 8 is then divided into 54. The software that instructs modern computers embodies algorithms, often of great sophistication.
## algorithm
any method, procedure, or set of instructions for carrying out a task by means of a precisely specified series of steps or sequence of actions, e.g. as in long division, the hierarchical sequence of steps in a typical computer program, or the steps in a manufacturing process.
## Algorithm
one of the basic concepts (categories) of mathematics devoid of a formal definition in terms of simpler concepts and abstracted directly from experience. Examples of algorithms are the familiar rules of addition, subtraction, multiplication, and long division taught in elementary school. In general, the term “algorithm” denotes any precise procedure specifying a calculational process (called algorithmic in this case) that begins with an arbitrary initial datum (drawn from a certain set of initial data possible for the given algorithm) and that is directed toward a result fully determined by the initial datum. For example, in the case of the algorithms of arithmetical operations mentioned above, the possible results may be natural numbers expressed in the decimal system, while the possible initial data may consist of ordered pairs of natural numbers. Thus, besides directions for carrying out the algorithmic process, the procedure must include (1) an indication of the set of possible initial data and (2) a rule by which the process is recognized as completed, when the desired result is attained. It is not assumed that the result must be achieved: the process of applying the algorithm to a specific possible initial datum (that is, an algorithmic process that develops from this datum onward) may also be terminated without a result or not terminated at all. If the process terminates (or does not terminate) by achieving a result, the algorithm is said to be applicable (or inapplicable) to the possible initial datum under consideration. It is possible to construct an algorithm II for which there exists no algorithm that discerns, on the basis of the possible initial datum for II, whether II is applicable to it or not; for instance, such an algorithm II can be constructed so that the set of positive integers serves as the set of its possible initial data.
The algorithm concept occupies a central position in modern mathematics, especially in computational mathematics. Thus, the problem of a numerical solution to equations of a given type reduces to finding an algorithm that will convert any pair, consisting of an arbitrary equation of the given type and an arbitrary rational number €, into a number (or an n-tuple of numbers) which is less than € and different from the root or roots of the equation. Improvements in computers offer the possibility of realizing increasingly complex algorithms with their use. However, the term “computational process” used in defining the algorithm concept must not be understood in the restricted meaning of digital calculations. Thus, even in school algebra courses one speaks of literal calculations, to say nothing of such nondigital symbols as brackets, equals signs, and the signs of arithmetical operations used in arithmetical calculations. It is possible to go further and consider calculations involving arbitrary symbols and their combinations; such precisely is the broad approach used to describe the algorithm concept. In this sense, one may speak of an algorithm for translation from one language to another, of an algorithm for train dispatching (which transforms information on train movements into orders), and of other such examples involving algorithmic descriptions of control processes. It is precisely for this reason that the algorithm is one of the central concepts of cybernetics. In general, the most varied constructive entities can serve as the initial data and results of algorithms. To take one example, the results of so-called recognition algorithms, are the words “yes” and “no.”
Example of an algorithm. Let the possible initial datum and the possible results consist of all possible finite sequences (including the empty sequence) of the letters a and b—“words in the alphabet {a, b}.” We shall agree to call the transition from word X to word Y “permissible” in the following two cases (P will denote an arbitrary word):(l) X has the form aP and Y has the form Pb and (2) X has the form baP and Y has the form Paba. An instruction is formulated: “starting with an arbitrary word, make permissible transitions until a word of the form aaP is obtained, then stop; the word P is the result.” This instruction forms an algorithm which we denote as I. We take the word babaa as the initial datum and obtain after one transition baaaba, after two, aabaaba. By virtue of the instruction we must stop, since the result is baaba. We take the word baaba as the initial datum and obtain, successively, abaaba, baabab, abababa, bababab, babababa, .... It can be demonstrated that this process will never end—that is, there will never appear a word beginning with aa, and for every word obtained it will always be possible to perform a permissible transition. Let us now take the word abaab as the initial datum. We obtain baabb, abbaba, bbabab. At this point, however, no further permissible transition is possible, yet there is no signal to stop. This is what is called the resultless stop. Thus, I is applicable to the word babaa and inapplicable to the words baaba and abaab.
Significance of algorithms. Algorithms abound in science; the ability to solve a problem “in the general form” always means, essentially, a knowledge of some algorithm. In speaking, for example, of a person’s ability to add numbers, one has in mind not the fact that he can sooner or later find the sum of any two numbers, but rather the fact that he possesses a unified method of addition applicable to any two specific notations of numbers—in other words, an addition algorithm (the familiar rule for addition of numbers by column is such an algorithm). The notion of a problem “in the general form” is explicated with the help of the concept of a mass problem. A mass problem is specified by a series of separate, individual problems and consists in the requirement to find a general method—that is, an algorithm—for their solution. Thus, the problem of the numerical solution of equations of a given type and the problem of automatic translation are mass problems: the individual problems constituting them are, in the first case, problems of the numerical solutions of individual equations of a given type and, in the second, problems of the translation of individual phrases. The role of mass problems determines both the significance and the sphere of application of the algorithm concept. Mass problems are extremely characteristic of and important in mathematics: for example, in algebra mass problems arise in the verification of algebraic equations of various types; in mathematical logic, we find mass problems for recognizing the derivability of propositions from given axioms; and so on. In the case of mathematical logic, the concept of the algorithm is all the more so essential because it is the basis of calculus—the central concept of mathematical logic. This concept serves as a generalization and explication of the intuitive concepts of “derivation” and “proof.” Establishing the unsolvability of some mass problem (say the problem of recognizing the truth or demonstrability of some logicomathematical language)—that is, the absence of a unified algorithm that permits finding solutions to all individual problems of a given set—is an important cognitive act which shows that in the solution of concrete individual problems it is fundamentally necessary to have specific methods for each such problem. The existence of unsolvable mass problems is thus a sign of the inexhaustibility of the cognitive process.
Substantive phenomena which underlay the formation of the algorithm concept long occupied an important position in science. From very ancient times, many problems of mathematics consisted in the search for constructive methods of one kind or another. This search, especially intensified with the advent of convenient symbolism and with the realization that certain sought-for methods cannot, in principle, be found (the problem of squaring the circle and the like), was a powerful factor in the development of scientific knowledge. Realization of the impossibility of solving problems by direct calculation led to the creation, in the 19th century, of the set-theoretic concept. Only after a period of turbulent development of this concept, during which the question of constructive methods in the modern sense of the term did not arise at all, did it become possible, in the middle of the 20th century, to turn once again to questions of constructivity, this time at a new level, enriched by the emergent concept of the algorithm. It was this concept which formed the basis of a special constructive trend in mathematics.
The very word algorithm derives from algorithmic a Latin transliteration of the Arabic name of al-Khwarizmi, a ninth-century mathematician from the district of Khorezm. In medieval Europe, the term algorism (algorithm) was used for the decimal positional number system and the art of calculating in it, since it was through a 12th-century Latin translation of al-Khwarizmi’s treatise that Europeans first became acquainted with the positional system of notation.
Structure of the algorithmic process. The algorithmic process is one of sequential transformation of constructive entities; it proceeds in discrete steps, each one of which consists in the replacement of a given constructive entity with another. Thus, in applying the algorithm I to the word baaba, we get a succession of words baaba, abaaba, baabab, and so forth. If, say, we apply the algorithm of subtraction by column to the pair <307, 49>, the following succession of constructive entities appears:
In this series of sequential constructive objects, each succeeding constructive object is fully determined, within the limits of the given algorithm, by its immediate predecessor. In a stricter approach, it is also assumed that the transition from every constructive object to the one immediately following it is sufficiently “elementary” in the sense that the transformation, in one step, of the preceding constructive object into the following one is of local character. The transformation does not embrace the whole constructive object, but only a portion of it delineated beforehand for the given algorithm, and the transformation itself is determined not by the whole preceding constructive object, but only by this limited portion.
Thus, along with sets of possible initial data and possible results, there exists for any algorithm a set of intermediate results which make up the working medium in which the algorithmic process develops. For I, all three sets coincide, but not for the subtraction-by-column algorithm: the possible initial data are pairs of numbers, the possible results are numbers (all in the decimal system), while intermediate results are complex fractions of the type
where q is the notation of the number in the decimal system, r is a similar notation or empty word, and ρ is the notation of a number in the decimal system with an allowance for dots over certain digits.
The functioning of the algorithm begins with a preparatory step in which the possible initial datum is transformed into the initial member of the sequence of intermediate results; this transformation takes place on the basis of a special “rule of beginning” which forms part of the algorithm under consideration. This rule, for I, consists in the application of an identity transformation and, for the subtraction algorithm, in the replacement of the pair <a, b> with the expression
Then the “rule of direct processing” is applied, which effects the successive transformation of each arising intermediate result into its successor. These transformations continue until a certain test, to which each intermediate result is subjected as it appears, indicates that a given intermediate result is conclusive; this test is applied on the basis of a special “rule of completion.” For example, for I, the rule of completion consists in verifying whether the intermediate result begins with aa. If the rule of completion does not produce the stop signal for any intermediate result, then the rule of direct processing is either applicable to every arising intermediate result and the algorithmic process continues indefinitely or it is inapplicable to a certain intermediate result and the process is terminated without result. Finally, the final result is extracted from the conclusive intermediate result also on the basis of a special rule; for Entity, this extraction consists in discarding the first two a’s and, for the subtraction algorithm, in discarding everything except the bottom line of digits. In many important cases, the rule of beginning and the rule of extraction of result both assign identical transformations and therefore are not formulated separately. Thus, for every algorithm it is possible to isolate seven (not independent!) parameters that characterize it: (1) the set of possible initial data, (2) the set of possible results, (3) the set of intermediate results, (4) the rule of beginning, (5) the rule of direct processing, (6) the rule of completion, and (7) the rule of extraction of result.
“Refinement” of the concept of the algorithm. Further “refinements” of the concept of the algorithm are possible, and these, strictly speaking, lead to a certain narrowing of the concept. Every such refinement consists in a precise description of a certain class for each of the seven parameters mentioned above—a class within which the given parameter can change. The selection of these classes is what distinguishes one refinement from another. In many refinements, all classes except two—the class of sets of intermediate results and the class of rules of direct processing—are chosen individually; that is, all parameters, except the two exceptions mentioned, are rigidly fixed. Since the seven parameters determine a certain algorithm unambiguously, the choice of the seven classes of variation of these parameters determines a certain class of algorithms. However, such a choice is properly referred to as a “refinement” only if we are convinced that for an arbitrary algorithm having permissible (by the given choice) sets of possible initial data and possible results it is possible to designate an equivalent algorithm taken from the class of algorithms defined by the given choice. This conviction is formulated for each refinement as a basic hypothesis, which, at the present level of our ideas on the matter, cannot be the subject of mathematical proof.
The first refinements of the type described were proposed in 1936 by the American mathematician E. L. Post and the English mathematician A. M. Turing. Also well known are the refinements formulated by the Soviet mathematicians A. A. Markov and A. N. Kolmogorov. The latter proposed treating constructive entities as topological complexes of a specific type; this offered the possibility of explicating the property of “localness” of a transformation For each of the proposed refinements, the corresponding main hypothesis is in good agreement with practice. Favoring this hypothesis is the fact that, as can be demonstrated, all proposed refinements are in a certain natural sense equivalent to one another.
As an example (in modernized form), we may take the refinement proposed by Turing. In order to specify a Turing algorithm, we must indicate (1) pairwise nonintersecting alphabets B, D, C, with letter ɑ isolated in D and letters and ω in C and (2) a set of pairs of the form <, ƞT̲q> where p, qC, ξ, η∊∪D, and T is one of the signs −, 0, +, assuming that in this set (called a program) there are no two pairs with identical first members. The parameters of the algorithm are assigned as follows: possible initial data and possible results are words in B; possible intermediate results are words in BDC containing not more than one letter from C. The rule of beginning: the initial word P is translated into the word λαPλ. The rule of completion: the final result is the intermediate result containing ω. The rule of extraction of result: the result is decreed to be the sequence of all those letters of the conclusive intermediate result which follows ω and precedes the first letter not contained in B. The rule of direct processing, which translates A into A’ consists in the following; we adjoin the letter λ to A on the right and on the left; then in the word thus formed, we replace the portion of form ∊ρξ, where ρ∊C, with the word Q by the following rule: in the program, we seek the pair having the first member ρξ; let the second member of this pair be η Tq; if T is −, then Q = q∊η; if T is 0. then Q-∊qy; if T is+, then Q=€ηg. The word appearing after this replacement is A’.
V. A. USPENSKII
## algorithm
[′al·gə‚rith·əm]
(mathematics)
A set of well-defined rules for the solution of a problem in a finite number of steps.
## Algorithm
A well-defined procedure to solve a problem. The study of algorithms is a fundamental area of computer science. In writing a computer program to solve a problem, a programmer expresses in a computer language an algorithm that solves the problem, thereby turning the algorithm into a computer program. See Computer programming
#### Operation
An algorithm generally takes some input, carries out a number of effective steps in a finite amount of time, and produces some output. An effective step is an operation so basic that it is possible, at least in principle, to carry it out using pen and paper. In computer science theory, a step is considered effective if it is feasible on a Turing machine or any of its equivalents. A Turing machine is a mathematical model of a computer used in an area of study known as computability, which deals with such questions as what tasks can be algorithmically carried out and what cannot. See Automata theory
Many computer programs deal with a substantial amount of data. In such applications, it is important to organize data in appropriate structures to make it easier or faster to process the data. In computer programming, the development of an algorithm and the choice of appropriate data structures are closely intertwined, and a decision regarding one often depends on knowledge of the other. Thus, the study of data structures in computer science usually goes hand in hand with the study of related algorithms. Commonly used elementary data structures include records, arrays, linked lists, stacks, queues, trees, and graphs.
#### Applications
Many algorithms are useful in a broad spectrum of computer applications. These elementary algorithms are widely studied and considered an essential component of computer science. They include algorithms for sorting, searching, text processing, solving graph problems, solving basic geometric problems, displaying graphics, and performing common mathematical calculations.
Sorting arranges data objects in a specific order, for example, in numerically ascending or descending orders. Internal sorting arranges data stored internally in the memory of a computer. Simple algorithms for sorting by selection, by exchange, or by insertion are easy to understand and straightforward to code. However, when the number of objects to be sorted is large, the simple algorithms are usually too slow, and a more sophisticated algorithm, such as heap sort or quick sort, can be used to attain acceptable performance. External sorting arranges stored data records.
Searching looks for a desired data object in a collection of data objects. Elementary searching algorithms include linear search and binary search. Linear search examines a sequence of data objects one by one. Binary search adopts a more sophisticated strategy and is faster than linear search when searching a large array. A collection of data objects that are to be frequently searched can also be stored as a tree. If such a tree is appropriately structured, searching the tree will be quite efficient.
A text string is a sequence of characters. Efficient algorithms for manipulating text strings, such as algorithms to organize text data into lines and paragraphs and to search for occurrences of a given pattern in a document, are essential in a word processing system. A source program in a high-level programming language is a text string, and text processing is a necessary task of a compiler. A compiler needs to use efficient algorithms for lexical analysis (grouping individual characters into meaningful words or symbols) and parsing (recognizing the syntactical structure of a source program). See Software engineering
A graph is useful for modeling a group of interconnected objects, such as a set of locations connected by routes for transportation. Graph algorithms are useful for solving those problems that deal with objects and their connections—for example, determining whether all of the locations are connected, visiting all of the locations that can be reached from a given location, or finding the shortest path from one location to another.
Mathematical algorithms are of wide application in science and engineering. Basic algorithms for mathematical computation include those for generating random numbers, performing operations on matrices, solving simultaneous equations, and numerical integration. Modern programming languages usually provide predefined functions for many common computations, such as random number generation, logarithm, exponentiation, and trigonometric functions.
In many applications, a computer program needs to adapt to changes in its environment and continue to perform well. An approach to make a computer program adaptive is to use a self-organizing data structure, such as one that is reorganized regularly so that those components most likely to be accessed are placed where they can be most efficiently accessed. A self-modifying algorithm that adapts itself is also conceivable. For developing adaptive computer programs, biological evolution has been a source of ideas and has inspired evolutionary computation methods such as genetic algorithms. See Genetic algorithms
Certain applications require a tremendous amount of computation to be performed in a timely fashion. An approach to save time is to develop a parallel algorithm that solves a given problem by using a number of processors simultaneously. The basic idea is to divide the given problem into subproblems and use each processor to solve a subproblem. The processors usually need to communicate among themselves so that they may cooperate. The processors may share memory, through which they can communicate, or they may be connected by communication links into some type of network such as a hypercube. See Concurrent processing, Multiprocessing, Supercomputer
## algorithm
1. a logical arithmetical or computational procedure that if correctly applied ensures the solution of a problem
2. Logic Maths a recursive procedure whereby an infinite sequence of terms can be generated
## algorithm
(algorithm, programming)
A detailed sequence of actions to perform to accomplish some task. Named after the Iranian mathematician, Mohammed Al-Khawarizmi.
Technically, an algorithm must reach a result after a finite number of steps, thus ruling out brute force search methods for certain problems, though some might claim that brute force search was also a valid (generic) algorithm. The term is also used loosely for any sequence of actions (which may or may not terminate).
Paul E. Black's Dictionary of Algorithms, Data Structures, and Problems.
## algorithm
A set of ordered steps for solving a problem, such as a mathematical formula or the instructions in a program. The terms algorithm and "program logic" are synonymous as both refer to a sequence of steps to solve a problem. However, an algorithm often implies a more complex problem rather than the input-process-output logic of typical business software. See encryption algorithm.
References in periodicals archive ?
For both hospitals combined, the number of hospital-acquired, primary, CVC-associated bloodstream infections varied by method, investigator review (n = 48), infection control professional review (n = 56), positive culture plus manual CVC determination (n = 86), computer algorithm (n = 64), and computer algorithm plus manual CVC determination (n = 48).
5 million grant is received by Computer scientists at Brown University from the National Science Foundation and National Institutes of Health to develop new computer algorithms and statistical methods to analyze large, complex datasets.
At EPFL, a team from the Audiovisual Communications Laboratory (LCAV), under the direction of Professor Martin Vetterli developed the computer algorithm.
And several top-ranked players outdid state-of-the-art computer algorithms that tackle the same tasks.
When Kranskopfasked his question, she suddenly realized that the computer algorithm could be interpreted as crochet instructions.
In the experiment, the researchers used approximately1, 000 paintings of 34 well-known artists, and let the computer algorithm analyze the similarity between them based solely on the visual content of the paintings, and without any human guidance.
Using a computer algorithm, mathematician and computer scientist Benjamin N.
The team used a computer algorithm to create a set of 20 pairs of faces at opposing ends of the trustworthiness scale.
Fifth-place winner Ryan Marques Harrison, 17, of Baltimore Polytechnic Institute developed a computer algorithm to predict how proteins interact at various acidities.
According to an analysis by Kanjoya, a San Francisco company that uses a computer algorithm to detect emotion in social media posts, differing opinions dominated Twitter.
Washington, Oct 14 (ANI): A group of Israeli researchers has built a computer algorithm that analyses biblical text to decipher its different authors.
But now, using advances in neuroimaging techniques, researchers including one of Indian-origin from the Stanford University School of Medicine trained a computer algorithm to interpret magnetic resonance imaging (MRI) data of the brain and determine whether someone is in pain.
Site: Follow: Share:
Open / Close | 5,412 | 27,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | longest | en | 0.9318 |
https://www.mathworks.com/matlabcentral/cody/problems/52-what-is-the-next-step-in-conway-s-life/solutions/475692 | 1,607,220,243,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141753148.92/warc/CC-MAIN-20201206002041-20201206032041-00339.warc.gz | 763,653,248 | 18,812 | Cody
# Problem 52. What is the next step in Conway's Life?
Solution 475692
Submitted on 20 Jul 2014 by Alvery
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
A = [ ... 1 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0]; B = [ ... 1 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0]; assert(isequal(life(A),B))
2 Pass
%% A = [ ... 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0]; B = [ ... 1 0 1 1 1 0 0 0 0 0 1 1 0 1 1 0]; assert(isequal(life(A),B))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 270 | 651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.723272 |
https://www.convert-measurement-units.com/convert+Square+foot+US+to+Square+mile+US.php | 1,675,756,096,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00569.warc.gz | 711,876,469 | 13,290 | Convert sqft to sqmi (Square foot (US) to Square mile (US))
## Square foot (US) into Square mile (US)
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Square+foot+US+to+Square+mile+US.php
## How many Square mile (US) make 1 Square foot (US)?
1 Square foot (US) [sqft] = 0.000 000 035 870 064 283 58 Square mile (US) [sqmi] - Measurement calculator that can be used to convert Square foot (US) to Square mile (US), among others.
# Convert Square foot (US) to Square mile (US) (sqft to sqmi):
1. Choose the right category from the selection list, in this case 'Area'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Square foot (US) [sqft]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Square mile (US) [sqmi]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '833 Square foot (US)'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Square foot (US)' or 'sqft'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Area'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '86 sqft to sqmi' or '57 sqft into sqmi' or '43 Square foot (US) -> Square mile (US)' or '12 sqft = sqmi' or '29 Square foot (US) to sqmi' or '13 sqft to Square mile (US)' or '65 Square foot (US) into Square mile (US)'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(88 * 91) sqft'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '833 Square foot (US) + 2499 Square mile (US)' or '67mm x 99cm x 97dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.092 954 076 473 9×1032. For this form of presentation, the number will be segmented into an exponent, here 32, and the actual number, here 1.092 954 076 473 9. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.092 954 076 473 9E+32. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 109 295 407 647 390 000 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 1,017 | 4,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.854213 |
https://socratic.org/questions/a-charge-of-2-c-is-at-2-1-and-a-charge-of-1-c-is-at-5-3-if-both-coordinates-are- | 1,575,953,379,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525821.56/warc/CC-MAIN-20191210041836-20191210065836-00484.warc.gz | 543,531,143 | 6,449 | # A charge of -2 C is at (-2 , 1 ) and a charge of -1 C is at (5, 3) . If both coordinates are in meters, what is the force between the charges?
Jan 27, 2017
According to Coulumbs Law: $m a t h b f F = \frac{k {Q}_{1} {Q}_{2}}{{\left\mid \setminus m a t h b f {r}_{1} - \setminus m a t h b f {r}_{2} \right\mid}^{2}} m a t h b f {e}_{{r}_{1} - {r}_{2}}$
Coulumb's Constant is k approx 8.99×10^9 Nm^2/${C}^{2}$
The magnitude of the force evaluates as:
| mathbf F |= ( 8.99×10^9 * (-2) * (-1))/(abs(((5),(3)) - ((3),(1)))^2) = 2.2475 * 10^9 N
....with ${\left\mid \setminus m a t h b f {r}_{1} - \setminus m a t h b f {r}_{2} \right\mid}^{2} = 8$.
[The answer may look big but note that a charge of "only" 1C is a lot of charge.] | 303 | 734 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-51 | latest | en | 0.734229 |
http://fins.actwin.com/aquatic-plants/month.200204/msg00871.html | 1,511,561,797,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808972.93/warc/CC-MAIN-20171124214510-20171124234510-00090.warc.gz | 117,258,219 | 2,276 | # spray bar
```> Hi list members, I don't post often.I read the list every day and have
> learned a lot from this list. Now I would like to add a spray bar to my
> planted tank. I have a125 gal tank with a 20 gal sump powered by a Mag
> driven 250 gal per hour pump. The piping is 11/4 in. I think that I want
> the bar to be 5 ft. it will fit between the return and the overflow at the
> other end of the tank. The return is about 5 in below the water line. What I
> need to know is should I make the holes in the spray bar different sizes or
> all the size? I will have to decide what size holes that I will need. Are
> there any formulas for this? Thanks for any help.
> Blendia
You can make them all the same size. Hydraulic pressure will force more or
less the same volume through all the holes. Same way a garden sprinkler hose
works. With air it's a different matter, since air is more compressible, but
water behaves hydraulically.
I don't know of any hole size formulas, but basically you'll want to make
the total hole size of your spray bar (area of all the holes added together)
at least the size of your return pipe's opening area. Assuming a return pipe
diameter of 11/4ths inches:
1.25/2 = 0.625 (radius of pipe)
0.625 squared = 0.390625
0.390625 x pi (3.1415) = 1.227148437 square inches
If you drill one hole every 6 inches (10 holes total) you need holes with an
area of at least 0.1227148437 square inches each. Reversing the circle area
formula we get a hole radius of 0.1976423537, or a diameter of about 0.4 of
an inch. Use a 7/16 inch drill bit.
If you drill one hole every 3 inches (4 per foot, 20 total) you need holes
with a diameter of about 0.14 inch each, or about a 3/16 inch drill bit.
Etc, etc.
Just always err on the high side (round up) and you won't restrict your
pump.
Dan Dixon
PS. Someone be sure to check my math!!! :)
``` | 526 | 1,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.916072 |
https://electronics.stackexchange.com/questions/369972/diode-only-speed-control-of-dc-motor | 1,702,187,403,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101195.85/warc/CC-MAIN-20231210025335-20231210055335-00855.warc.gz | 184,899,334 | 44,218 | # Diode only speed control of DC motor
I was salvaging a motor / gearbox from a breast pump and noticed something I didn't understand at all. The control panel consists of a 6 position switch (5 speeds and off) and then a series of diodes. No resistors, no capacitors, and no 555 timer. It doesn't seem to be a PWM situation, so what am I not understanding about diodes that let them act as a speed controller? I thought their only function was to limit power flow in a single direction.
I also got a couple 1/2 horsepower DC permanent magnet motors for a go kart project and wondering if this information could be useful at all on a larger motor.
• Possible that the switch adjusts how many diodes are in series. Each diode would drop the voltage by around 0.5-1V, so it could be used for a rather odd and crude speed controller. Can you post a pic of the back of the board and tell us what the input voltage is from the DC adapter? Apr 22, 2018 at 0:59
• Thanks Phil, I didn't realize that you would get a voltage drop through a diode, that may be just what is happening. Would that create the same inefficiency that would exist with resistors in generating heat? I could just use a PWM, but more than anything I am trying to understand what I was looking at. The supply is a 12V 2A source.
– Alan
Apr 22, 2018 at 1:19
• draw the schematic ... it may help you understand the circuit Apr 22, 2018 at 3:14
• @Alan The most common application of standard diodes is to control the direction of current flow, but that's not all they're good for and there's no such thing as a component without some level of voltage loss and heat dissipation. Check out this article if you want to get a better introduction to how diodes work. Apr 22, 2018 at 4:43
It's a very cheap way to create a speed controller that has a predictable voltage drop without the same level of current limiting created by a resistor based voltage divider.
The switch puts a variable number of diodes in series to create a combined voltage drop.
The larger diodes (D1-D5) seem to read "1N593" from what I can make out. If that's correct, they would be zener diodes in the forward biased mode. That family of zeners have a 15-39V zener voltage and a 1.5V forward voltage. They're using the 1.5V forward voltage to limit the motor input.
The smaller diode (D6) is probably a standard rectifying diode. Hopefully one with a relatively high reverse voltage, because they don't seem to have anything else to deal with back EMF when that motor stops. This would add another 0.6V voltage drop.
If that's all correct, you should expect to see about 10V on high and about 4.5V on low with 1.5V increments between. Regardless of exact numbers, that's the basic theory of operation.
It's not very efficient and is only useful for small increments, so I would not be trying to adapt this for larger motors.
Here's a simulated version of the circuit (based on my assumptions for exact parts used and with some equivalent components for the rotary switch and motor)
• So are they actually 2 zeners in series inside then? Apr 22, 2018 at 3:43
• @HenryCrun I'm sorry, I don't understand the question. This board has as many as 5 or 6 zeners in series (assuming I'm right about it having any). If you're asking about the zeners themselves, none of the datasheet I've seen have shown an equivalent circuit diagram showing 2 diodes in series to explain the voltage drops. It's possible they are made of 2 (or more) PN junctions in series, but I'd think it far more likely that it is a single PN junction with a specific doping method to create the characteristics. I'm not going to claim to be a zener diode physics expert though. Apr 22, 2018 at 4:02
• Can't find much information, only a few seem to give Vf, BZX79C 0.9V@10mA (cf. 1n4148=0.7), 1N593X 1.2V@200mA (cf 1n4001=0.8V). Nothing about it varying with Vz. Guess its just an effect of the doping levels. Apr 22, 2018 at 10:17
A series of diodes will create a series of stepped voltage drops. The voltage drops will not change proportionally with current as the voltage drops across resistors would. That means that there will be less speed variation with changing load. That is an advantage over series resistors. Using diodes in this way still causes heat and inefficiency similarly to series resistors. If the motor requires a maximum of 2 amps and the voltage drop is 1 volt per diode, six diodes would dissipate 12 watts and the motor would dissipate another two watts or so in winding resistance, friction etc. leaving about 10 watts converted to mechanical energy with an efficiency of about 42%. That is a pretty low efficiency, but only 14 watts dissipated as heat.
The same technique could be used with a larger motor, but the energy lost as heat would increase proportionally. With a 1/2 Hp motor, a similar setup would use 888 watts and dissipate more than 500 watts of power as heat to provide 373 watts of mechanical power.
• Thanks for your excellent response Charles, I marked Phil as correct as the diagram and explanation helped with the issue, though yours gave great insight on what would happen as it scaled up.
– Alan
Apr 23, 2018 at 20:43 | 1,272 | 5,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | longest | en | 0.962123 |
https://www.ademcetinkaya.com/2022/09/should-you-buy-now-or-wait-ip-stock.html | 1,685,306,654,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644506.21/warc/CC-MAIN-20230528182446-20230528212446-00259.warc.gz | 688,962,935 | 60,154 | Stock market also called as equity market is the aggregation of the sellers and buyers. It is concerned with the domain where the shares of various public listed companies are traded. For predicting the growth of economy, stock market acts as an index. Due to the nonlinear nature, the prediction of the stock market becomes a difficult task. But the application of various machine learning techniques has been becoming a powerful source for the prediction. We evaluate International Paper prediction models with Transfer Learning (ML) and Factor1,2,3,4 and conclude that the IP stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy IP stock.
Keywords: IP, International Paper, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.
## Key Points
1. Stock Rating
2. How do you know when a stock will go up or down?
3. Game Theory
## IP Target Price Prediction Modeling Methodology
This study presents financial network indicators that can be applied to global stock market investment strategies. We propose to design both undirected and directed volatility networks of global stock market based on simple pair-wise correlation and system-wide connectedness of stock date using a vector auto-regressive model. We consider International Paper Stock Decision Process with Factor where A is the set of discrete actions of IP stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Factor)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Transfer Learning (ML)) X S(n):→ (n+3 month) $\begin{array}{l}\int {r}^{s}\mathrm{rs}\end{array}$
n:Time series to forecast
p:Price signals of IP stock
j:Nash equilibria
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## IP Stock Forecast (Buy or Sell) for (n+3 month)
Sample Set: Neural Network
Stock/Index: IP International Paper
Time series to forecast n: 11 Sep 2022 for (n+3 month)
According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy IP stock.
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Yellow to Green): *Technical Analysis%
## Conclusions
International Paper assigned short-term Ba3 & long-term Ba3 forecasted stock rating. We evaluate the prediction models Transfer Learning (ML) with Factor1,2,3,4 and conclude that the IP stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy IP stock.
### Financial State Forecast for IP Stock Options & Futures
Rating Short-Term Long-Term Senior
Outlook*Ba3Ba3
Operational Risk 8881
Market Risk4667
Technical Analysis7186
Fundamental Analysis6535
Risk Unsystematic5336
### Prediction Confidence Score
Trust metric by Neural Network: 93 out of 100 with 473 signals.
## References
1. Bottou L. 2012. Stochastic gradient descent tricks. In Neural Networks: Tricks of the Trade, ed. G Montavon, G Orr, K-R Müller, pp. 421–36. Berlin: Springer
2. Bessler, D. A. S. W. Fuller (1993), "Cointegration between U.S. wheat markets," Journal of Regional Science, 33, 481–501.
3. Arora S, Li Y, Liang Y, Ma T. 2016. RAND-WALK: a latent variable model approach to word embeddings. Trans. Assoc. Comput. Linguist. 4:385–99
4. Abadie A, Diamond A, Hainmueller J. 2015. Comparative politics and the synthetic control method. Am. J. Political Sci. 59:495–510
5. D. Bertsekas. Dynamic programming and optimal control. Athena Scientific, 1995.
6. Tibshirani R. 1996. Regression shrinkage and selection via the lasso. J. R. Stat. Soc. B 58:267–88
7. S. Bhatnagar, R. Sutton, M. Ghavamzadeh, and M. Lee. Natural actor-critic algorithms. Automatica, 45(11): 2471–2482, 2009
Frequently Asked QuestionsQ: What is the prediction methodology for IP stock?
A: IP stock prediction methodology: We evaluate the prediction models Transfer Learning (ML) and Factor
Q: Is IP stock a buy or sell?
A: The dominant strategy among neural network is to Buy IP Stock.
Q: Is International Paper stock a good investment?
A: The consensus rating for International Paper is Buy and assigned short-term Ba3 & long-term Ba3 forecasted stock rating.
Q: What is the consensus rating of IP stock?
A: The consensus rating for IP is Buy.
Q: What is the prediction period for IP stock?
A: The prediction period for IP is (n+3 month) | 1,305 | 5,092 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-23 | latest | en | 0.897717 |
https://education.ti.com/en/activity/detail?id=89C197C4541C475A967A94A00E1080A3 | 1,726,323,496,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.38/warc/CC-MAIN-20240914125424-20240914155424-00104.warc.gz | 202,327,754 | 11,289 | Education Technology
# Walking the Line
Math: Algebra I: Linear Functions
9-12
60 Minutes
TI-30X IIS
This is Activity 4 from the EXPLORATIONS Book:
Math Investigations with the TI-30X IIS: Activities for Secondary Mathematics
The following material is required for this activity:
• Coordinate graph paper
Lessons
# Walking the Line
Activity Overview
Students use linear functions to model and solve problems in situations with slope and a constant rate of change. They learn to represent situations with variables in expressions, equations, and inequalities and use tables and graphs as tools to interpret them.
Before the Activity
• See the attached PDF file for detailed instructions for this activity
• Print pages 38 - 42 from the attached PDF file for your class
• During the Activity
Distribute the pages to the class.
• Given the coordinates of two points, use the slope formula to find the slope of the line passing through the two points
• Change the decimal result to a fraction
• For each given ordered pair of numbers, plot the points, draw a line through the points, and find the slope of the line
• Solve a real world problem and understand the meaning of the slope in the given situation
• Make a plot of the data pairs from the given problem, draw a line through the pairs, and use the linear model to predict results
• Write an equation to model given data, use the equation to generate several data pairs, plot the points, and find answers
• After the Activity
Students complete the Student Activity pages.
• Review student results
• As a class, discuss questions that appeared to be more challenging
• Re-teach concepts as necessary | 350 | 1,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-38 | latest | en | 0.885395 |
https://math.stackexchange.com/questions/3001116/showing-the-power-series-sum-n-0-infty-frac13-cosx-has-a-rad | 1,563,814,935,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528141.87/warc/CC-MAIN-20190722154408-20190722180408-00120.warc.gz | 478,111,959 | 37,651 | # Showing the power series $\sum_{n = 0}^{\infty} \frac{1}{3 - \cos(x)}$ has a radius of convergence $R > 1$
I want to show that the power series
$$\sum_{n = 0}^{\infty} a_{n}x^{n} = \frac{1}{3 - \cos(x)}$$
has a radius $$R > 1$$? I don't think the ratio test will work here. From Googling, I found the Cauchy-Hadamard Theorem, and I was wondering if I could somehow apply it to this. Can someone please help me? I'm not so familiar with $$\limsup$$.
This can be seen as a heuristic way which may not be necessarily correct.
There is a slightly easier way using the expansion for $$\dfrac{1}{1-x}$$.
$$\dfrac{1}{3 - \cos(x)}=\dfrac{1}{3}\left(\dfrac{1}{1-\dfrac{\cos(x)}{3}}\right)=\dfrac{1}{3}\displaystyle\sum_{n=0}^\infty\left(\dfrac{\cos(x)}{3}\right)^n$$ exists if $$\left|\dfrac{\cos(x)}{3}\right|<1 \implies |\cos(x)|<3$$ which is true for all $$x\in\mathbb{R}$$.
$$\rule{17cm}{1pt}$$
To find the power series we calculate using Taylor series
$$f(0)=\dfrac{1}{2}, \: f'(x)=\dfrac{-\sin x}{(3-\cos x)^2}\implies f'(0)=0, \: \\f''(x)=\dfrac{(3-\cos x)(-\cos x)+2\sin^2 x(3-\cos x)}{(3-\cos x)^4}\implies f''(0)=\dfrac{1}{4}$$.
Continuing this way we get $$\quad$$ $$\dfrac{1}{3 - \cos(x)}=\dfrac{1}{2}-\dfrac{x^2}{8}+\dfrac{x^4}{24}+\cdots$$
$$\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}=3\implies e^{2iz}-6e^{iz}+1=0 \implies e^{iz}=\dfrac{6\pm\sqrt{36-4}}{2}=3\pm2\sqrt{2}$$ $$\implies z=2\pi n -i\ln(3\pm2\sqrt{2}),\: n\in\mathbb{Z}$$.
Now since radius of curvature is defined such that the function is analytic in $$\{z: |z - z_0| < r\}$$, so the radius of convergence is at least $$r$$. We infer that if $$f$$ has a pole at $$c$$ then it is analytic for $$|z-z_o|, so radius of curvature is exactly the distance to the closest pole.
Thus $$|z|=|\ln(3+2\sqrt{2})|$$. Thus we see that $$R=|\ln(3+2\sqrt{2})|>1$$.
• that is very clever. – joseph Nov 16 '18 at 13:21
• I don't think this works. The power series of $1/(3- \cos x)$ is not $\frac{1}{3} \sum_n (\frac{\cos x}{3})^n$, because it is not a power series. If you expand it, it does not converge uniformly on the whole real line. Actually the power series has a finite radius of convergence. That's because otherwise it would converge uniformly on the whole $\Bbb C$: but there is the problem that on complex numbers there is a solution for $\cos z=3$. – Crostul Nov 16 '18 at 13:26
• @Crostul: Isnt this true for $x\in\mathbb{R}$ ? How does the power series look like in the case $x\in\mathbb{C}$? – Yadati Kiran Nov 16 '18 at 13:31
• The radius of convergence of that power series is bounded by $\ln (3+2\sqrt 2) = 1.76...$. I think that is exactly the radius of convergence, though I'm not sure. You can see it graphically at wolframalpha.com/input/?i=power+series+of+(3-cos+z)%5E(-1) . – Crostul Nov 16 '18 at 13:33
• @Crostul: I shall check in the case for $x\in\mathbb{C}$. But what about when $x\in\mathbb{R}$ – Yadati Kiran Nov 16 '18 at 13:38
The poles of the function occur at
$$\cos z=3$$ or $$z=\pm i\text{ arcosh}(3)+2k\pi=\pm i\log(3+\sqrt8)+2k\pi.$$
As the radius of convergence is the distance to the closest pole,
$$R=\log(3+\sqrt8)\approx1.76274717.$$
You may also invoke Cauchy's integral theorem. $$f(z)=\frac{1}{3-\cos(z)}$$ is holomorphic over $$|z|\leq\frac{3}{2}$$, since
$$\left|3-\cos(z)\right|\geq 2-\sum_{n\geq 1}\frac{|z|^{2n}}{(2n)!}=3-\cosh(|z|)$$ and $$K=3-\cosh\left(\tfrac{3}{2}\right)>0$$. In particular
$$|a_n|=\left|\frac{1}{2\pi i}\oint_{|z|=\frac{3}{2}}\frac{dz}{z^{n+1}(3-\cos z)}\right|\leq \frac{1}{K\left(\frac{3}{2}\right)^n}$$ by the triangle inequality, and the radius of convergence of $$\sum a_n z^n$$ is at least $$\frac{3}{2}$$.
The same argument works also by replacing $$\frac{3}{2}$$ with $$\frac{7}{4}$$, since $$\cosh\frac{7}{4}$$ still is less than three. | 1,439 | 3,786 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 33, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-30 | latest | en | 0.763115 |
http://www.evi.com/q/convert_85_mm_to_inches | 1,394,483,044,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011005264/warc/CC-MAIN-20140305091645-00013-ip-10-183-142-35.ec2.internal.warc.gz | 316,090,634 | 14,269 | # convert 85 mm to inches
• 3.34645669291339 inches
the height 3.34645669291339 inches
• tk10npubl tk10ncanl
## TrueKnowledge.com is now Evi.com
Evi, is our best selling mobile app that turns your phone into a mobile assistant. Over the next few months we will be adding all of Evi's power including local information on shopping, restaurants and more... to this site.
Until then, to experience all of the power of Evi now, download the Evi app for iOS or Android here.
## Top ways people ask this question:
• convert 85 mm to inches (56%)
• 85mm in inches (11%)
• 85mm converted to inches (9%)
• convert 85mm to inches (5%)
• 85mm converted into inches (3%)
• convert 85 mm into inches (2%)
• 85 mm is how many inches (1%)
• 85mm equals how many inches (1%)
• 85 mm to inches (1%)
• 85 mm in inches (1%)
## Other ways this question is asked:
• 85 mms in inches
• 85 millimeters to inches
• 85 millimeter equals how many inches
• 85mm conversion to inches
• how many inchs is 85mm
• convert 85mm into inches
• 085 mm to inches
• 85 mm converted to inches
• 85 milimeters equals how many inches
• 85 mm to inch
• what is 85 millimeters in inches
• convert 85 mm into inces
• 85mm is equal to how many inches?
• what is the equivalent of 85mm to inches
• 85mm = how many inch
• how much is 85mm in inches
• convert 085mm to inches
• 85milimeters equals how many inches
• how many inches is 85.0mm
• whats 85 mm in inches
• 85.00mm is how many inches
• 85mm in ''
• 85mm in "
• what's 85mm in inches
• 85mm converts to how many inchs
• convert "85mm" to inches | 473 | 1,567 | {"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} | 2.53125 | 3 | CC-MAIN-2014-10 | longest | en | 0.872548 |
http://rgtusyllabus.blogspot.com/2011/12/ce503-fluid-mech-ii-ce-5th-sem-civil.html | 1,485,159,877,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282202.61/warc/CC-MAIN-20170116095122-00131-ip-10-171-10-70.ec2.internal.warc.gz | 243,482,648 | 18,629 | ### CE503 Fluid Mech. – II CE 5th sem CIVIL Engineering(CE) 5th Semester RGTU/RGPVSyllabus
CE503 Fluid Mech. – II Syllabus
RGTU/RGPV Fluid Mech. – II
CIVIL Engineering(CE) 5th - Fifth Semester Syllabus
CE503 Fluid Mech. – II Course Contents:
Unit-I
Turbulent flow : Laminar and turbulent boundary layers and laminar sublayer, hydrodynamically smooth andrough boundaries, velocity distribution in turbulent flow, resistance of smooth and artificially roughened pipes,commercial pipes, aging of pipes.
Pipe flow problems : Losses due to sudden expansion and contraction, losses in pipe fittings and valves,concepts of equivalent length, hydraulic and energy gradient lines, siphon, pipes in series, pipes in parallel,branching of pipes.
Pipe Network : *Water Hammer (only quick closure case). transmission of power. *Hardy Cross Method
Unit-II
Uniform flow in open channels : Channel geometry and elements of channel section, velocity distribution,energy in open channel flow, specific energy, types of flow, critical flow and its computations, uniform flow and its computations, Chezy’s and Manning’s formulae, determination of normal depth and velocity, Normal and critical slopes, Economical sections, Saint Vegnet equation.
Unit-III
Non uniform flow in open channels : Basic assumptions and dynamic equations of gradually varied flow,characteristics analysis and computations of flow profiles, rapidly varied flow hydraulic jump in rectangular channels and its basic characteristics, surges in open channels & channel flow routing, venturi flume.
Unit-IV
Forces on immersed bodies: Types of drag, drag on a sphere, a flat plate, a cylinder and an aerofoil development of lift, lifting vanes, magnus effect.
Unit-V
Fluid Machines:
Turbines : Classifications, definitions, similarity laws, specific speed and unit quantities, Pelton turbine-their construction and settings, speed regulation, dimensions of various elements, Action of jet, torque, power and efficiency for ideal case, characteristic curves.
Reaction turbines: construction & settings, draft tube theory, runaway speed, simple theory of design and characteristic curves, cavitation.
Pumps:
Centrifugal pumps : Various types and their important components, manometric head, total head, net positive suction head, specific speed, shut off head, energy losses, cavitation, principle of working and characteristic curves.
Reciprocating pumps: Principle of working, Coefficient of discharge, slip, single acting and double acting pump, Manometric head, Acceleration head.
List of Experiment
1. Study the performances characteristics of Pelton Wheel
2. Study the performances characteristics of Francis Turbine
3. Study the performances charactristics of Kaplan Turbine
4. Caliration of multistage (Two) Pump & Study of characteristic of variable speed pump
5. To study the performance & details of operation of Hyd. Ram
6. Determination of coefficient of discharge for a broad crested weir & to plot water surface profile over weir 7. Study of the characteristic of the Reciprocating pump
Suggested Books & Study Material:
1. Fluid Mechanics - Modi & Seth - Standard Book house, Delhi
2. Open Channel Flow by Rangaraju - Tata Mc Graw - Hill Publishing Comp. Ltd., New Delhi
3. Fluid Mechanics - A.K. Jain - Khanna Publishers, Delhi
4. Fluid Mechanics, Hydraulics & Hydraulic Machanics - K.R. Arora - Standard Publishers Distributors 1705- B, Nai Sarak, Delhi-6
5. Hyd. of open channels By Bakhmetiff B.A. (McGraw Hill, New York)
6. Open Channel Hyd. By Chow V.T. (McGraw Hill, New York)
7. Engineering Hydraulics By H. Rouse
8. Centrifugal & Axial Flow Pump By Stempanoff A.J. New York
9. Relevant IS codes.
CIVIL Engineering(CE) 5th - Fifth Semester other Syllabus
CE501 Transportation Engineering – II Syllabus | 871 | 3,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.823757 |
https://www.math.uni-hamburg.de/home/khomskii/forcing2021/note.html | 1,620,547,185,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988961.17/warc/CC-MAIN-20210509062621-20210509092621-00593.warc.gz | 907,315,946 | 1,354 | A note on Kunen
Both of Kunen's textbooks use nonstandard notation. The most important deviations are:
• R(α) denotes the levels of the cumulative hierarchy of sets, normally denoted by Vα
• ON denotes the class of Ordinals
• HF denotes the set of "hereditarily finite sets", usually denoted by Vω
• ZC denotes ZFC − Replacement. Z denotes ZF − Replacement. ZFC is ZFC − Foundation and ZF is ZF − Foundation.
Kunen sometimes puts emphasis on proving results in ZFC without the Axiom of Foundation first (later showing that the cumulative hierarchy of sets V, which doesn't require Foundation to be defined, satisfies ZFC with Foundation). We will not be concerned with this so you can always assume that Foundation is a ZFC axiom and skip any discussion of Kunen that relates to Foundation. Whenever Kunen assumes ZFC or ZF, you can exchange it with ZFC or ZF, respectively. | 207 | 875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-21 | latest | en | 0.898507 |
https://es.mathworks.com/help/risk/bias-mitigation-for-credit-scoring-model-by-reweighting.html | 1,725,934,491,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00660.warc.gz | 212,531,048 | 23,153 | # Bias Mitigation in Credit Scoring by Reweighting
Bias mitigation is the process of removing bias from a data set or a model in order to make it fair. Bias mitigation usually follows a bias detection step, where a series of metrics are computed based on a data set or model predictions. Bias mitigation has three stages: pre-processing, in-processing, and post-processing. This example demonstates a pre-processing method to mitigate bias in a credit scoring workflow. The example uses bias detection and bias mitigation functionality from the Statistics and Machine Learning Toolbox™. For a detailed example on bias detection, see the following example: Explore Fairness Metrics for Credit Scoring Model.
The bias mitigation method in this example is Reweighting which essentially reweights observations within a data set to guarantee fairness between different subgroups within a sensitive attribute. As a result of reweighting, the Statistical Parity Difference (SPD) of all subgroups goes to 0 and the Disparate Impact metric becomes 1. This example demonstrates how reweighting works in a credit scoring workflow.
Load the CreditCardData data set and discretize the 'CustAge' predictor.
AgeGroup = discretize(data.CustAge,[min(data.CustAge) 30 45 60 max(data.CustAge)], ...
'categorical',{'Age < 30','30 <= Age < 45','45 <= Age < 60','Age >= 60'});
CustID CustAge AgeGroup TmAtAddress ResStatus EmpStatus CustIncome TmWBank OtherCC AMBalance UtilRate status
______ _______ ______________ ___________ __________ _________ __________ _______ _______ _________ ________ ______
1 53 45 <= Age < 60 62 Tenant Unknown 50000 55 Yes 1055.9 0.22 0
2 61 Age >= 60 22 Home Owner Employed 52000 25 Yes 1161.6 0.24 0
3 47 45 <= Age < 60 30 Tenant Employed 37000 61 No 877.23 0.29 0
4 50 45 <= Age < 60 75 Home Owner Employed 53000 20 Yes 157.37 0.08 0
5 68 Age >= 60 56 Home Owner Employed 53000 14 Yes 561.84 0.11 0
6 65 Age >= 60 13 Home Owner Employed 48000 59 Yes 968.18 0.15 0
7 34 30 <= Age < 45 32 Home Owner Unknown 32000 26 Yes 717.82 0.02 1
8 50 45 <= Age < 60 57 Other Employed 51000 33 No 3041.2 0.13 0
Split the data set into training and testing data. Use the training data to fit the model and the testing data to predict from the model.
rng('default');
c = cvpartition(size(data,1),'HoldOut',0.3);
data_Train = data(c.training(),:);
data_Test = data(c.test(),:);
### Compute Fairness Metrics at Predictor and Model Level
Compute the fairness metrics for the training data by creating a fairnessMetrics object and then generating a metrics report using report. Since you are only working with data and there is no fitted model, only two bias metrics are computed for StatisticalParityDifference and DisparateImpact. The two group metrics computed are GroupCount and GroupSizeRatio. The fairness metrics are computed for two sensitive attributes, Age ('AgeGroup') and Residential Status ('ResStatus').
trainingDataMetrics = fairnessMetrics(data_Train, 'status', 'SensitiveAttributeNames',{'AgeGroup', 'ResStatus'});
tdmReport = report(trainingDataMetrics)
tdmReport=7×4 table
SensitiveAttributeNames Groups StatisticalParityDifference DisparateImpact
_______________________ ______________ ___________________________ _______________
AgeGroup Age < 30 0.039827 1.1357
AgeGroup 30 <= Age < 45 0.096324 1.3282
AgeGroup 45 <= Age < 60 0 1
AgeGroup Age >= 60 -0.19181 0.34648
ResStatus Home Owner 0 1
ResStatus Tenant 0.01689 1.0529
ResStatus Other -0.02108 0.93404
figure
tiledlayout(2,1)
nexttile
plot(trainingDataMetrics,'spd')
nexttile
plot(trainingDataMetrics,'di')
Looking at the DisparateImpact bias metric for both AgeGroup and ResStatus, you can see that there is a much larger variance in the AgeGroup predictor as compared to the ResStatus predictor. This suggests that users are treated more unfairly when it comes to their age as compared to their residential status. This example focuses on the AgeGroup predictor and attempts to reduce bias among its subgroups.
To begin, fit a credit scoring model and compute the model-level bias metrics. This provides a baseline for comparison.
Since CustAge and AgeGroup are essentially the same predictor and this is a sensitive attribute, you can exclude it from the model. Additionally, you can use 'status' as the response variable and 'CustID' as the ID variable.
PredictorVars = setdiff(data_Train.Properties.VariableNames, ...
{'CustAge','AgeGroup','CustID','FairWeights','status'});
sc1 = creditscorecard(data_Train,'IDVar','CustID', ...
'PredictorVars',PredictorVars,'ResponseVar','status');
sc1 = autobinning(sc1);
sc1 = fitmodel(sc1,'VariableSelection','fullmodel');
Generalized linear regression model:
logit(status) ~ 1 + TmAtAddress + ResStatus + EmpStatus + CustIncome + TmWBank + OtherCC + AMBalance + UtilRate
Distribution = Binomial
Estimated Coefficients:
Estimate SE tStat pValue
________ ________ ________ __________
(Intercept) 0.73924 0.077237 9.5711 1.058e-21
ResStatus 1.755 1.295 1.3552 0.17535
EmpStatus 0.88652 0.32232 2.7504 0.0059516
CustIncome 0.95991 0.19645 4.8862 1.0281e-06
TmWBank 1.132 0.3157 3.5856 0.00033637
OtherCC 0.85227 2.1198 0.40204 0.68765
AMBalance 1.0773 0.31969 3.3698 0.00075232
UtilRate -0.19784 0.59565 -0.33214 0.73978
840 observations, 831 error degrees of freedom
Dispersion: 1
Chi^2-statistic vs. constant model: 66.5, p-value = 2.44e-11
pointsinfo1 = displaypoints(sc1)
pointsinfo1=38×3 table
Predictors Bin Points
_______________ _________________ _________
{'ResStatus' } {'Tenant' } -0.017688
{'ResStatus' } {'Home Owner' } 0.11681
{'ResStatus' } {'Other' } 0.29011
{'ResStatus' } {'<missing>' } NaN
{'EmpStatus' } {'Unknown' } -0.097582
{'EmpStatus' } {'Employed' } 0.33162
{'EmpStatus' } {'<missing>' } NaN
{'CustIncome' } {'[-Inf,30000)' } -0.61962
{'CustIncome' } {'[30000,36000)'} -0.10695
{'CustIncome' } {'[36000,40000)'} 0.0010845
{'CustIncome' } {'[40000,42000)'} 0.065532
⋮
pd1 = probdefault(sc1,data_Test);
Set the threshold value that controls the allocation of "goods" and "bads."
threshold = 0.35;
predictions1 = double(pd1>threshold);
Create a fairnessMetrics object to compute fairness metrics at the model level and then generate a metrics report using report.
modelMetrics1 = fairnessMetrics(data_Test, 'status', 'Predictions', predictions1, 'SensitiveAttributeNames','AgeGroup');
mmReport1 = report(modelMetrics1)
mmReport1=4×7 table
ModelNames SensitiveAttributeNames Groups StatisticalParityDifference DisparateImpact EqualOpportunityDifference AverageAbsoluteOddsDifference
__________ _______________________ ______________ ___________________________ _______________ __________________________ _____________________________
Model1 AgeGroup Age < 30 0.54312 2.6945 0.47391 0.5362
Model1 AgeGroup 30 <= Age < 45 0.19922 1.6216 0.35645 0.22138
Model1 AgeGroup 45 <= Age < 60 0 1 0 0
Model1 AgeGroup Age >= 60 -0.15385 0.52 -0.18323 0.16375
Measure accuracy of model using validatemodel.
validatemodel(sc1)
ans=4×2 table
Measure Value
________________________ _______
{'Accuracy Ratio' } 0.33751
{'Area under ROC curve'} 0.66876
{'KS statistic' } 0.26418
{'KS score' } 1.0403
figure
tiledlayout(2,1)
nexttile
plot(modelMetrics1,'spd')
nexttile
plot(modelMetrics1,'di')
### Reweight Data at Predictor and Model Level
Use fairnessWeights to reweight the training data to remove bias for the sensitive attribute 'AgeGroup'.
fairWeights = fairnessWeights(data_Train, 'AgeGroup', 'status');
data_Train.FairWeights = fairWeights;
CustID CustAge AgeGroup TmAtAddress ResStatus EmpStatus CustIncome TmWBank OtherCC AMBalance UtilRate status FairWeights
______ _______ ______________ ___________ __________ _________ __________ _______ _______ _________ ________ ______ ___________
1 53 45 <= Age < 60 62 Tenant Unknown 50000 55 Yes 1055.9 0.22 0 0.95879
2 61 Age >= 60 22 Home Owner Employed 52000 25 Yes 1161.6 0.24 0 0.75407
3 47 45 <= Age < 60 30 Tenant Employed 37000 61 No 877.23 0.29 0 0.95879
4 50 45 <= Age < 60 75 Home Owner Employed 53000 20 Yes 157.37 0.08 0 0.95879
7 34 30 <= Age < 45 32 Home Owner Unknown 32000 26 Yes 717.82 0.02 1 0.82759
8 50 45 <= Age < 60 57 Other Employed 51000 33 No 3041.2 0.13 0 0.95879
9 50 45 <= Age < 60 10 Tenant Unknown 52000 25 Yes 115.56 0.02 1 1.0992
10 49 45 <= Age < 60 30 Home Owner Unknown 53000 23 Yes 718.5 0.17 1 1.0992
Use fairnessMetrics to compute fairness metrics for the training data after reweighting and use report to generate a fairness metrics report..
trainingDataMetrics_AfterReweighting = fairnessMetrics(data_Train, 'status', 'SensitiveAttributeNames','AgeGroup','Weights','FairWeights');
tdmrReport = report(trainingDataMetrics_AfterReweighting)
tdmrReport=4×4 table
SensitiveAttributeNames Groups StatisticalParityDifference DisparateImpact
_______________________ ______________ ___________________________ _______________
AgeGroup Age < 30 -2.9976e-15 1
AgeGroup 30 <= Age < 45 -5.5511e-16 1
AgeGroup 45 <= Age < 60 0 1
AgeGroup Age >= 60 -2.9421e-15 1
By applying the reweighting algorithm to the AgeGroup predictor, you can completely remove the disparate impact for AgeGroup. Then use this debiased data to fit a model to produce predictions with an overall reduced disparate impact at the model level.
Use creditscorecard to fit a new credit scoring model with the new fair weights and compute model-level bias metrics.
sc2 = creditscorecard(data_Train,'IDVar','CustID', ...
'PredictorVars',PredictorVars,'WeightsVar','FairWeights','ResponseVar','status');
sc2 = autobinning(sc2);
sc2 = fitmodel(sc2,'VariableSelection','fullmodel');
Generalized linear regression model:
logit(status) ~ 1 + TmAtAddress + ResStatus + EmpStatus + CustIncome + TmWBank + OtherCC + AMBalance + UtilRate
Distribution = Binomial
Estimated Coefficients:
Estimate SE tStat pValue
________ ________ ________ __________
(Intercept) 0.74055 0.076222 9.7158 2.5817e-22
ResStatus 2.0467 1.7669 1.1584 0.24672
EmpStatus 0.91879 0.32197 2.8536 0.0043222
CustIncome 0.91038 0.33216 2.7407 0.00613
TmWBank 1.1067 0.30826 3.5901 0.0003305
OtherCC 0.42264 3.5078 0.12049 0.9041
AMBalance 1.1347 0.3447 3.2919 0.00099504
UtilRate -0.39861 0.77284 -0.51577 0.60601
840 observations, 831 error degrees of freedom
Dispersion: 1
Chi^2-statistic vs. constant model: 46.6, p-value = 1.85e-07
pointsinfo2 = displaypoints(sc2)
pointsinfo2=34×3 table
Predictors Bin Points
_______________ _________________ ________
{'ResStatus' } {'Tenant' } 0.016048
{'ResStatus' } {'Home Owner' } 0.091092
{'ResStatus' } {'Other' } 0.28326
{'ResStatus' } {'<missing>' } NaN
{'EmpStatus' } {'Unknown' } -0.10352
{'EmpStatus' } {'Employed' } 0.33653
{'EmpStatus' } {'<missing>' } NaN
{'CustIncome' } {'[-Inf,30000)' } -0.37618
{'CustIncome' } {'[30000,40000)'} 0.047483
{'CustIncome' } {'[40000,42000)'} 0.10244
{'CustIncome' } {'[42000,47000)'} 0.14652
{'CustIncome' } {'[47000,Inf]' } 0.40015
⋮
pd2 = probdefault(sc2,data_Test);
predictions2 = double(pd2>threshold);
Use fairnessMetrics to compute fairness metrics at the model level and report to generate a fairness metrics report.
modelMetrics2 = fairnessMetrics(data_Test, 'status', 'Predictions', predictions2, 'SensitiveAttributeNames','AgeGroup');
mmReport2 = report(modelMetrics2)
mmReport2=4×7 table
ModelNames SensitiveAttributeNames Groups StatisticalParityDifference DisparateImpact EqualOpportunityDifference AverageAbsoluteOddsDifference
__________ _______________________ ______________ ___________________________ _______________ __________________________ _____________________________
Model1 AgeGroup Age < 30 0.39394 2.1818 0.37391 0.39377
Model1 AgeGroup 30 <= Age < 45 0.094298 1.2829 0.22947 0.11509
Model1 AgeGroup 45 <= Age < 60 0 1 0 0
Model1 AgeGroup Age >= 60 -0.13333 0.6 -0.18323 0.1511
Measure accuracy of model using validatemodel.
validatemodel(sc2)
ans=4×2 table
Measure Value
________________________ _______
{'Accuracy Ratio' } 0.27735
{'Area under ROC curve'} 0.63868
{'KS statistic' } 0.22702
{'KS score' } 0.90741
figure
tiledlayout(2,1)
nexttile
plot(modelMetrics2,'spd')
nexttile
plot(modelMetrics2,'di')
The process of reweighting removed all the bias from the training data. When you use the new data to fit a model, the overall bias in the model is reduced when compared to a model trained with biased data. As a consequence of this reduction in bias, there is a drop in model accuracy. You can choose to make tradeoff to improve fairness.
### References
[1] Nielsen, Aileen. "Chapter 4. Fairness PreProcessing." Practical Fairness. O'Reilly Media, Inc., Dec. 2020.
[2] Mehrabi, Ninareh, et al. “A Survey on Bias and Fairness in Machine Learning.” ArXiv:1908.09635 [Cs], Sept. 2019. arXiv.org, https://arxiv.org/abs/1908.09635.
[3] Wachter, Sandra, et al. Bias Preservation in Machine Learning: The Legality of Fairness Metrics Under EU Non-Discrimination Law. SSRN Scholarly Paper, ID 3792772, Social Science Research Network, 15 Jan. 2021. papers.ssrn.com, https://papers.ssrn.com/sol3/papers.cfm?abstract_id=3792772. | 4,463 | 16,746 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.785852 |
http://mathhelpforum.com/algebra/97061-easiest-way-solve.html | 1,508,539,861,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824357.3/warc/CC-MAIN-20171020211313-20171020231313-00053.warc.gz | 219,039,663 | 10,977 | # Thread: Easiest way to solve
1. ## Easiest way to solve
What is the simplest way to solve this problem?
$|x+1|^2+3|x+1|-4=0$
2. Hello, Mike9182!
What is the simplest way to solve this equation?
. . $|x+1|^2+3|x+1|-4\:=\:0$
We have a quadratic . . . Solve it accordingly.
Factor: . $\left(|x+1| + 4\right)\left(|x+1| - 1\right) \:=\:0$
Set each factor equal to zero and solve:
. . $|x+1| + 4 \:=\:0 \quad\Rightarrow\quad|x+1| \:=\:-4$ . . . no roots
. . . .
An absolute value is never negative.
. . $|x+1| - 1 \:=\:0 \quad\Rightarrow\quad |x+1| \:=\:1 \quad\Rightarrow\quad x+1 \:=\:\pm1 \quad\Rightarrow\quad x \:=\:0,\:2$
But $x=2$ is extraneous.
. . Therefore, the only solution is: . $x \:=\:0$
3. ## Absolute value
Hello Mike9182
Originally Posted by Mike9182
What is the simplest way to solve this problem?
$|x+1|^2+3|x+1|-4=0$
Let $y = |x+1|$. Then:
$y^2 +3y - 4=0$
$\Rightarrow (y+4)(y-1)=0$
$\Rightarrow y=-4, \,1$
But $y = -4$ is impossible, since $|x+1| \ge 0$
$\Rightarrow y =1= |x+1|$
$\Rightarrow x+1 = \pm 1$
$\Rightarrow x = 0,\, -2$
Grandad
4. Originally Posted by Soroban
$x+1 \:=\:\pm1 \quad\Rightarrow\quad x \:=\:0,\:2$
Am I missing something?
5. Originally Posted by alexmahone
Am I missing something?
probably just a typo ... it happens. | 487 | 1,286 | {"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": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-43 | longest | en | 0.598773 |
https://www.convertunits.com/from/chinese+yard/to/step | 1,695,992,370,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510516.56/warc/CC-MAIN-20230929122500-20230929152500-00825.warc.gz | 783,087,484 | 12,525 | ## Convert chinese yard to step
chinese yard step
How many chinese yard in 1 step? The answer is 0.85470085470085. We assume you are converting between chinese yard and step. You can view more details on each measurement unit: chinese yard or step The SI base unit for length is the metre. 1 metre is equal to 1.1216546649618 chinese yard, or 1.3123359580052 step. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between chinese yards and steps. Type in your own numbers in the form to convert the units!
## Quick conversion chart of chinese yard to step
1 chinese yard to step = 1.17 step
5 chinese yard to step = 5.85 step
10 chinese yard to step = 11.7 step
15 chinese yard to step = 17.55 step
20 chinese yard to step = 23.4 step
25 chinese yard to step = 29.25 step
30 chinese yard to step = 35.1 step
40 chinese yard to step = 46.8 step
50 chinese yard to step = 58.5 step
## Want other units?
You can do the reverse unit conversion from step to chinese yard, or enter any two units below:
## Enter two units to convert
From: To:
## 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, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 407 | 1,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-40 | latest | en | 0.887515 |
http://blog.csdn.net/yulinxx/article/details/73522391 | 1,519,248,817,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813803.28/warc/CC-MAIN-20180221202619-20180221222619-00733.warc.gz | 51,880,127 | 15,577 | # 八叉树 C++ 基础 源码
http://blog.csdn.net/pizi0475/article/details/6269060
#include <iostream>
using namespace std;
//定义八叉树节点类
template<class T>
struct OctreeNode
{
T data; //节点数据
T xMin, xMax; //节点坐标，即六面体个顶点的坐标
T yMin, yMax;
T zMin, zMax;
OctreeNode <T> *top_left_front, *top_left_back; //该节点的个子结点
OctreeNode <T> *top_right_front, *top_right_back;
OctreeNode <T> *bottom_left_front, *bottom_left_back;
OctreeNode <T> *bottom_right_front, *bottom_right_back;
OctreeNode //节点类
(T nodeValue = T(),
T xminValue = T(), T xmaxValue = T(),
T yminValue = T(), T ymaxValue = T(),
T zminValue = T(), T zmaxValue = T(),
OctreeNode<T>* top_left_front_Node = NULL,
OctreeNode<T>* top_left_back_Node = NULL,
OctreeNode<T>* top_right_front_Node = NULL,
OctreeNode<T>* top_right_back_Node = NULL,
OctreeNode<T>* bottom_left_front_Node = NULL,
OctreeNode<T>* bottom_left_back_Node = NULL,
OctreeNode<T>* bottom_right_front_Node = NULL,
OctreeNode<T>* bottom_right_back_Node = NULL)
: data(nodeValue),
xMin(xminValue), xMax(xmaxValue),
yMin(yminValue), yMax(ymaxValue),
zMin(zminValue), zMax(zmaxValue),
top_left_front(top_left_front_Node),
top_left_back(top_left_back_Node),
top_right_front(top_right_front_Node),
top_right_back(top_right_back_Node),
bottom_left_front(bottom_left_front_Node),
bottom_left_back(bottom_left_back_Node),
bottom_right_front(bottom_right_front_Node),
bottom_right_back(bottom_right_back_Node) {}
};
//创建八叉树
template <class T>
void createOctree(OctreeNode<T> * &root, int maxdepth, double xMin, double xMax, double yMin, double yMax, double zMin, double zMax)
{
cout << "处理中，请稍候……" << endl;
maxdepth = maxdepth - 1; //每递归一次就将最大递归深度-1
if (maxdepth >= 0)
{
root = new OctreeNode<T>();
root->data = 9; //为节点赋值，可以存储节点信息，如物体可见性。由于是简单实现八叉树功能，简单赋值为。
root->xMin = xMin; //为节点坐标赋值
root->xMax = xMax;
root->yMin = yMin;
root->yMax = yMax;
root->zMin = zMin;
root->zMax = zMax;
double xMind = (xMax - xMin) / 2;//计算节点个维度上的半边长
double yMind = (yMax - yMin) / 2;
double zMind = (zMax - zMin) / 2;
//递归创建子树，根据每一个节点所处（是几号节点）的位置决定其子结点的坐标。
createOctree(root->top_left_front, maxdepth, xMin, xMax - xMind, yMax - yMind, yMax, zMax - zMind, zMax);
createOctree(root->top_left_back, maxdepth, xMin, xMax - xMind, yMin, yMax - yMind, zMax - zMind, zMax);
createOctree(root->top_right_front, maxdepth, xMax - xMind, xMax, yMax - yMind, yMax, zMax - zMind, zMax);
createOctree(root->top_right_back, maxdepth, xMax - xMind, xMax, yMin, yMax - yMind, zMax - zMind, zMax);
createOctree(root->bottom_left_front, maxdepth, xMin, xMax - xMind, yMax - yMind, yMax, zMin, zMax - zMind);
createOctree(root->bottom_left_back, maxdepth, xMin, xMax - xMind, yMin, yMax - yMind, zMin, zMax - zMind);
createOctree(root->bottom_right_front, maxdepth, xMax - xMind, xMax, yMax - yMind, yMax, zMin, zMax - zMind);
createOctree(root->bottom_right_back, maxdepth, xMax - xMind, xMax, yMin, yMax - yMind, zMin, zMax - zMind);
}
}
int i = 1;
//先序遍历八叉树
template <class T>
void preOrder(OctreeNode<T> * & p)
{
if (p)
{
cout << i << ".当前节点的值为：" << p->data << "\n坐标为：";
cout << " xMin: " << p->xMin << " xMax: " << p->xMax;
cout << " yMin: " << p->yMin << " yMax: " << p->yMax;
cout << " zMin: " << p->zMin << " zMax: " << p->zMax;
i += 1;
cout << endl;
preOrder(p->top_left_front);
preOrder(p->top_left_back);
preOrder(p->top_right_front);
preOrder(p->top_right_back);
preOrder(p->bottom_left_front);
preOrder(p->bottom_left_back);
preOrder(p->bottom_right_front);
preOrder(p->bottom_right_back);
cout << endl;
}
}
//求八叉树的深度
template<class T>
int depth(OctreeNode<T> *& p)
{
if (p == NULL)
return -1;
int h = depth(p->top_left_front);
return h + 1;
}
//计算单位长度，为查找点做准备
int cal(int num)
{
int result = 1;
if (1 == num)
result = 1;
else
{
for (int i = 1; i < num; i++)
result = 2 * result;
}
return result;
}
//查找点
int maxdepth = 0;
int times = 0;
static double xMin = 0, xMax = 0, yMin = 0, yMax = 0, zMin = 0, zMax = 0;
int tmaxdepth = 0;
double txm = 1, tym = 1, tzm = 1;
template<class T>
void find(OctreeNode<T> *& p, double x, double y, double z)
{
double xMind = (p->xMax - p->xMin) / 2;
double yMind = (p->yMax - p->yMin) / 2;
double zMind = (p->yMax - p->yMin) / 2;
times++;
if (x > xMax || x<xMin || y>yMax || y<yMin || z>zMax || z < zMin)
{
cout << "该点不在场景中！" << endl;
return;
}
if (x <= p->xMin + txm && x >= p->xMax - txm && y <= p->yMin + tym &&
y >= p->yMax - tym && z <= p->zMin + tzm && z >= p->zMax - tzm)
{
cout << endl << "找到该点！" << "该点位于" << endl;
cout << " xMin: " << p->xMin << " xMax: " << p->xMax;
cout << " yMin: " << p->yMin << " yMax: " << p->yMax;
cout << " zMin: " << p->zMin << " zMax: " << p->zMax;
cout << "节点内！" << endl;
cout << "共经过" << times << "次递归！" << endl;
}
else if (x < (p->xMax - xMind) && y < (p->yMax - yMind) && z < (p->zMax - zMind))
{
cout << "当前经过节点坐标：" << endl;
cout << " xMin: " << p->xMin << " xMax: " << p->xMax;
cout << " yMin: " << p->yMin << " yMax: " << p->yMax;
cout << " zMin: " << p->zMin << " zMax: " << p->zMax;
cout << endl;
find(p->bottom_left_back, x, y, z);
}
else if (x < (p->xMax - xMind) && y<(p->yMax - yMind) && z>(p->zMax - zMind))
{
cout << "当前经过节点坐标：" << endl;
cout << " xMin: " << p->xMin << " xMax: " << p->xMax;
cout << " yMin: " << p->yMin << " yMax: " << p->yMax;
cout << " zMin: " << p->zMin << " zMax: " << p->zMax;
cout << endl;
find(p->top_left_back, x, y, z);
}
else if (x > (p->xMax - xMind) && y < (p->yMax - yMind) && z < (p->zMax - zMind))
{
cout << "当前经过节点坐标：" << endl;
cout << " xMin: " << p->xMin << " xMax: " << p->xMax;
cout << " yMin: " << p->yMin << " yMax: " << p->yMax;
cout << " zMin: " << p->zMin << " zMax: " << p->zMax;
cout << endl;
find(p->bottom_right_back, x, y, z);
}
else if (x > (p->xMax - xMind) && y<(p->yMax - yMind) && z>(p->zMax - zMind))
{
cout << "当前经过节点坐标：" << endl;
cout << " xMin: " << p->xMin << " xMax: " << p->xMax;
cout << " yMin: " << p->yMin << " yMax: " << p->yMax;
cout << " zMin: " << p->zMin << " zMax: " << p->zMax;
cout << endl;
find(p->top_right_back, x, y, z);
}
else if (x<(p->xMax - xMind) && y>(p->yMax - yMind) && z < (p->zMax - zMind))
{
cout << "当前经过节点坐标：" << endl;
cout << " xMin: " << p->xMin << " xMax: " << p->xMax;
cout << " yMin: " << p->yMin << " yMax: " << p->yMax;
cout << " zMin: " << p->zMin << " zMax: " << p->zMax;
cout << endl;
find(p->bottom_left_front, x, y, z);
}
else if (x<(p->xMax - xMind) && y>(p->yMax - yMind) && z > (p->zMax - zMind))
{
cout << "当前经过节点坐标：" << endl;
cout << " xMin: " << p->xMin << " xMax: " << p->xMax;
cout << " yMin: " << p->yMin << " yMax: " << p->yMax;
cout << " zMin: " << p->zMin << " zMax: " << p->zMax;
cout << endl;
find(p->top_left_front, x, y, z);
}
else if (x > (p->xMax - xMind) && y > (p->yMax - yMind) && z < (p->zMax - zMind))
{
cout << "当前经过节点坐标：" << endl;
cout << " xMin: " << p->xMin << " xMax: " << p->xMax;
cout << " yMin: " << p->yMin << " yMax: " << p->yMax;
cout << " zMin: " << p->zMin << " zMax: " << p->zMax;
cout << endl;
find(p->bottom_right_front, x, y, z);
}
else if (x > (p->xMax - xMind) && y > (p->yMax - yMind) && z > (p->zMax - zMind))
{
cout << "当前经过节点坐标：" << endl;
cout << " xMin: " << p->xMin << " xMax: " << p->xMax;
cout << " yMin: " << p->yMin << " yMax: " << p->yMax;
cout << " zMin: " << p->zMin << " zMax: " << p->zMax;
cout << endl;
find(p->top_right_front, x, y, z);
}
}
//main函数
int main()
{
OctreeNode<double> * rootNode = NULL;
int choiced = 0;
while (true)
{
system("cls");
cout << "请选择操作：\n";
cout << "1.创建八叉树 2.先序遍历八叉树\n";
cout << "3.查看树深度 4.查找节点 \n";
cout << "0.退出\n\n";
cin >> choiced;
if (choiced == 0)
{
return 0;
}
else if (choiced == 1)
{
system("cls");
cout << "请输入最大递归深度：" << endl;
cin >> maxdepth;
cout << "请输入外包盒坐标，顺序如下：xMin,xMax,yMin,yMax,zMin,zMax" << endl;
cin >> xMin >> xMax >> yMin >> yMax >> zMin >> zMax;
if (maxdepth >= 0 || xMax > xMin || yMax > yMin || zMax > zMin || xMin > 0 || yMin > 0 || zMin > 0)
{
tmaxdepth = cal(maxdepth);
txm = (xMax - xMin) / tmaxdepth;
tym = (yMax - yMin) / tmaxdepth;
tzm = (zMax - zMin) / tmaxdepth;
createOctree(rootNode, maxdepth, xMin, xMax, yMin, yMax, zMin, zMax);
}
else
{
cout << "输入错误！";
return 0;
}
}
else if (choiced == 2)
{
system("cls");
cout << "先序遍历八叉树结果：\n";
i = 1;
preOrder(rootNode);
cout << endl;
system("pause");
}
else if (choiced == 3)
{
system("cls");
int dep = depth(rootNode);
cout << "此八叉树的深度为" << dep + 1 << endl;
system("pause");
}
else if (choiced == 4)
{
system("cls");
cout << "请输入您希望查找的点的坐标，顺序如下：x,y,z\n";
double x, y, z;
cin >> x >> y >> z;
times = 0;
cout << endl << "开始搜寻该点……" << endl;
find(rootNode, x, y, z);
system("pause");
}
else
{
system("cls");
cout << "\n\n错误选择！\n";
system("pause");
}
}
}
• 本文已收录于以下专栏：
## 八叉树Octree
• huapenguag
• 2016年03月09日 19:28
• 1432
## 八叉树Octree
• Augusdi
• 2014年06月30日 16:41
• 12780
## 场景管理：八叉树算法C++实现
• u012234115
• 2015年07月30日 21:06
• 2945
## 八叉树三维数据结构
（一）基本原理 用八叉树来表示三维形体，并研究在这种表示下的各种操作及应用是在进入80年代后才比较全面地开展起来的。这种方法，既可以看成是四叉树方法在三维空间的推广，也可以认为是用三维体素阵列...
• Chinamming
• 2013年11月24日 13:15
• 3309
## 空间八叉树剖分
• Augusdi
• 2014年06月30日 16:49
• 4652
## 四叉树与八叉树
• hjwang1
• 2016年09月03日 01:36
• 1196
## 八叉树及K-D树的应用和实现
1. 八叉树、k-d树的原理2. 八叉树、k-d树的应用、优缺点3. 八叉树、k-d树的实现八叉树和k-d树都经常用来处理三维空间数据，k-d树的使用范围更宽泛些，适用于k维空间的数据，在Sift算法...
• Augusdi
• 2014年06月30日 20:46
• 2264
## 基于八叉树的网格生成算法剖析
• Kaitiren
• 2016年03月11日 11:40
• 2392
## 四叉树与八叉树
• zhanxinhang
• 2011年08月21日 15:34
• 38888
## 叉树Octree原理及简单实现（C++版）
1、对Octree的描述 Octree的定义是：若不为空树的话，树中任一节点的子节点恰好只会有八个，或零个，也就是子节点不会有0与8以外的数目。那么，这要用来做什么？想象一个立方体，我们最少可以切成...
• Chinamming
• 2013年11月24日 13:22
• 2094
举报原因： 您举报文章：八叉树 C++ 基础 源码 色情 政治 抄袭 广告 招聘 骂人 其他 (最多只允许输入30个字) | 4,060 | 9,530 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-09 | latest | en | 0.250537 |
https://www.expertsmind.com/library/what-is-annual-equivalent-worth-of-this-marketing-campaign-51853619.aspx | 1,709,264,005,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00077.warc.gz | 772,621,417 | 12,276 | ### What is annual equivalent worth of this marketing campaign
Assignment Help Financial Management
##### Reference no: EM131853619
1. A company is considering entering into a new marketing campaign. If it engages in this marketing campaign, it must pay \$9,000 immediately and \$7,000 each at the end of year 1 and year 2. The company believes its annual revenues due to the marketing campaign will be \$11,000 at the end of year 1, \$9,000 at the end of year 2, and \$6,000 at the end of year 3. What is the annual equivalent worth of this marketing campaign over the next three years? The interest rate is 6.8% compounded annually."
2. "Motors R Us requires a specific painting process for one its automobiles that will be manufactured for exactly eight years. Three options are available. Neither option 2 nor option 3 can be repeated after its process life. However option 1 will always be available at the same cost during the period of eight years. Here are the options: -Option 1 is to subcontract out the process at a cost of \$116,000 per year. -Option 2 costs \$151,000 immediately and has annual operating and labor costs of \$85,000 and a useful service life of eight years with a salvage value of \$41,000. -Option 3 costs \$118,000 immediately and has annual operating and labor costs of \$75,000 and a useful service life of five years with a salvage value of \$23,000. For years six through eight, the firm will follow option 1. Enter the ""net present cost"" for the option that you would recommend if i = 14.8%. Enter your answer as positive number."
### Write a Review
#### What was the dollar-based return on the security
Over the holding period the euro moves from \$1.1/euro to \$1.15/euro. What was the U.S. dollar-based return on the security?
#### Suppose that the installation of low-loss thermal windows
Suppose that the installation of low-loss thermal windows is expected to save 450 per year on bills. If you live in your home for 40 years and could earn 6% per year on other investments, how much could you afford to pay now to have the windows insta..
#### How much money per year will the bank provide
Your rich uncle has promised to invest \$500,000 in your startup 3 years from today. How much money per year will the bank provide?
#### Determining profit or loss from an investment
Determining Profit or Loss from an Investment. Three years ago, you purchased 150 shares of IBM stock for \$88 a share. Today, you sold your IBM stock for \$103 a share. For this problem, ignore commissions that would be charged to buy and sell your IB..
#### Firm cost of capital is affected by firm capital structure
A firm’s cost of capital is affected by the firm’s capital structure. A bond that sells at a price higher than its par value is called a face value bond.
#### European call option contracts
You take a short position in 100 European call option contracts, with strike price \$50 and maturity three months, on a stock that is trading at \$52. The annual volatility of the stock is constant and equal to 22%. The annual, continuous risk-free int..
#### Discount rate-calculate the present value of annuity
You will receive \$5,000 per year, every year for the next five (5) years, beginning at the end of this year. If you use 6% as your discount rate, calculate the present value of this annuity.
#### No-arbitrage band for the stock-futures price relationship
The one-year futures price on a particular stock-index portfolio is 406, the stock index currently is 400, the one-year risk-free interest rate is 3%, and the year-end dividend that will be paid on a \$400 investment in the index portfolio is \$5. Give..
#### Share in commissions for each transaction
You paid 50 cents per share in commissions for each transaction. What is the balance of your account? Annual interest rate is 10%.
#### In a period of heightened price volatility-what is a buyback
In a period of heightened price volatility, the beta of the market will: What is a buyback? What is a bank?
#### Establish delta-neutral position
how many shares of stock must you buy to establish a delta-neutral position?
#### The analysis of the chosen type of trace evidence
Describe materials that can be transferred from a person to another person or to an inanimate object during a crime. | 968 | 4,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.949354 |
http://www.slideserve.com/sylvester-potts/evaluating-hypothesis | 1,506,415,569,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695113.88/warc/CC-MAIN-20170926070351-20170926090351-00232.warc.gz | 560,118,824 | 16,229 | 1 / 25
# Evaluating Hypothesis - PowerPoint PPT Presentation
Evaluating Hypothesis. Introduction Estimating Accuracy Sampling Theory and Confidence Intervals Differences in Error Comparing Learning Algorithms. Introduction. Given some training data a learning algorithm produces a hypothesis.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about ' Evaluating Hypothesis' - sylvester-potts
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
• Introduction
• Estimating Accuracy
• Sampling Theory and Confidence Intervals
• Differences in Error
• Comparing Learning Algorithms
Given some training data a learning algorithm produces a hypothesis.
The next step is to estimate the accuracy of the hypothesis on
testing data:
Data Learning Hypothesis
Algorithm
Performance Assessment
• How do we know how precise is our estimation?
• There are two difficulties:
• Bias in the estimate
• Variance in the estimate
• Bias in the estimate. Normally overoptimisitc.
• To avoid it we use a separate set of data.
• Variance in the estimate. The estimation varies from
• sample to sample. The smaller the sample the larger the variance.
Estimated Accuracy
Variance
accuracy
True accuracy
Bias
sample size
• Introduction
• Estimating Accuracy
• Sampling Theory and Confidence Intervals
• Differences in Error
• Comparing Learning Algorithms
Examples in the input space are randomly distributed according
to some probability distribution D.
p(X)
Input Space X
• Questions:
• Given a hypothesis h and a dataset with n examples randomly
• obtained based on D, what is the accuracy of h in future examples?
• 2. What is the error in this (accuracy) estimate?
Classification of mushrooms.
Some mushrooms are more likely to show up than others.
Example:
More likely to appear
frequency
size
Sample error:
errorS (h) = 1/n Σ δ (f(X),h(X))
Where f is the true target function, h is the hypothesis, and
δ(a,b) = 1 if a = b, 0 otherwise.
True Error:
errorD (h) = P[ f(X) = h(x)]D
How good is errorS (h) in estimating errorD (h) ?
There are 4 mushrooms in our dataset: {X1, X2, X3, X4}
out of a space of 6 possible mushrooms.
The probability distribution is such that
P(X1) = 0.2 P(X4) = 0.1
P(X2) = 0.1 P(X5) = 0.2
P(X3) = 0.3 P(X6) = 0.1
Our hypothesis classifies correctly X1, X2, and X3 but not X4.
The sample error is ¼ (0 + 0 + 0 + 1) = ¼ = 0.25
Our hypothesis also classifies correctly X6 but not X5.
The true error is
0.2(0) + 0.1(0) + 0.3(0) + 0.1(1) + 0.2(1) + 0.1(1) = 0.3
• Introduction
• Estimating Accuracy
• Sampling Theory and Confidence Intervals
• Differences in Error
• Comparing Learning Algorithms
• Assume the following conditions are present:
• The sample has n examples drawn according to probability D.
• n > 30
• Hypothesis h has made r errors in the n examples.
• Then with probability of 95%, the true error lies in the interval:
• errorS(h) +- 1.96 errorS(h) (1 - errorS(h)) / n
For example if n = 40 and r =12 then with 95% confidence the
interval lies in 0.30 +- 0.14
How much does the size of the dataset affect the difference between
the sample error and the true error?
We have a sample of size n obtained according to distribution D.
Instances are drawn independently from each other.
This probability can be modeled through the binomial distribution
• Combinations:
• Assume we wish to select r objects from n objects.
• In this case we do not care about the order in which
we select the r objects.
• The number of possible combinations of r objects
from n objects is
n ( n-1) (n-2) … (n –r +1) / r! = n! / (n-r)! r!
• We denote this number as C(n,r)
• Let X be a discrete random variable that takes the
following values:
x1, x2, x3, …, xn. Let P(x1), P(x2), P(x3),…,P(xn) be
their respective probabilities. Then the expected value of X,
E(X), is defined as
E(X) = x1P(x1) + x2P(x2) + x3P(x3) + … + xnP(xn)
E(X) = Σi xi P(xi)
• Let X be a discrete random variable that takes the
following values:
x1, x2, x3, …, xn. Let u be the mean. Then the variance is:
variance(X) = E[ X – u] 2
• What is the probability of getting x successes in n trials?
• Assumption: all trials are independent and the probability of
success remains the same.
Let p be the probability of success and let q = 1-p
then the binomial distribution is defined as
P(x) = nCx p x q n-x for x = 0,1,2,…,n
The mean equals n p
The variance equals n p q
The sampling error can be modeled using a binomial distribution.
For example. Let’s suppose we have a dataset of size n = 40 and
that the true probability of error is 0.3.
Then the expected number of errors is np = 40(0.3) = 12
Plotting the binomial distribution:
0 10 20 30 40
If r is the number of errors in our dataset of size n, then
our estimation of the sample error is r/n.
The true error is p.
The bias of an estimator Y is E[Y] – p.
The sample error is an unbiased estimator of the true error
because the expected value of r/n is p.
The standard deviation is approx.
errorS(h) (1 - errorS(h)) / n
• Introduction
• Estimating Accuracy
• Sampling Theory and Confidence Intervals
• Differences in Error
• Comparing Learning Algorithms
Suppose we are comparing two hypothesis from two
algorithms, say a decision tree and a neural network:
h1
h2
Type B
Type A
We would like to compute the difference in error
d = error(h1) – error(h2).
Variable d can be considered approx. normal.
(because the difference of two normal distributions is also normal).
The variance is the sum of the variances of both errors.
So the interval confidence is defined as:
d +- Zn error(h1)(1 – error(h1)) / n1 + error(h2)(1 – error(h2)) / n2
• Introduction
• Estimating Accuracy
• Sampling Theory and Confidence Intervals
• Differences in Error
• Comparing Learning Algorithms
We have two algorithms LA and LB.
How do we know which is one is better on average for
learning some target function?
We want to compute the expected value of their different
performance according to distribution D:
E[ error(LA) – error(LB) ] D
With a limited sample what we can do is the following:
For i = 1 to k
Separate the data into a training set and a testing set
Train LA and LB on the testing set
Compute the difference in error:
di = error(LA) – error(LB)
End
Return dmean = 1/k Σ di
The standard deviation for this statistic is:
standard dev = [1 / k(k-1)] Σ (di – dmean)2
So the confidence intervals are:
dmean +- t(n,k-1) (standard dev)
• Estimating the accuracy of a hypothesis may have some error.
• Errors in estimation are the bias and variance factors.
• One can compute confidence intervals using statistical theory.
• The sampling error can be modeled using a Binomial distribution.
• Differences in error can be computed using multiple subsampling. | 1,926 | 7,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-39 | latest | en | 0.799184 |
https://www.scribd.com/document/351357473/Indiabix-With-Explanations | 1,560,883,915,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998813.71/warc/CC-MAIN-20190618183446-20190618205446-00011.warc.gz | 914,220,044 | 85,006 | You are on page 1of 46
# MICROWAVE COMMS Therefore, some electrons are accelerated and some
## INDIABIX are retarded. This process is called velocity
SECTION 1 modulation.
The velocity modulation causes bunching of electrons.
1. The velocity factor of a transmission line depends on This bunching effect converts velocity modulation into
A. temperature density modulation of beam.
B. skin effect The input is fed at buncher cavity and output is taken
C. relative permittivity of dielectric at catcher cavity.
D.none of the above In a two cavity klystron only buncher and catcher
C cavity are used. In multi cavity klystron one or more
intermediate cavities are also used.
Answer: Option C The features of a multicavity klystron are :
Explanation: 1. Frequency range - 0.25 GHz to 100 GHz
r of a dielectric changes with humidity. 2. Power output - 10 kW to several hundred kW
3. Power gain - 60 dB (nominal value)
2. A loss less line of characteristic impedance Z0 is 4. Efficiency - about 40%.
terminated in pure reactance of -jZ0 value. VSWR is A multicavity klystron is used in UHF TV transmitters,
A. 10 Radar transmitter and satellite communication.
B. 2
C. 1 4. In a circular waveguide the dominant mode is
D.infinity A. TE01
D B. TE11
Explanation: D.TE21
A pure reactance does not absorb any power. B
Therefore VSWR is infinite because |rv| = 1. Answer: Option B
Explanation:
3. In a klystron amplifier the input cavity is called Some applications require dual polarization capability.
A. buncher Circular waveguide has this capability.
B. catcher These analysis uses cylindrical coordinates.
C. Pierce gun In circular waveguide TE11 mode has the lowest cut off
D.collector frequency and is the dominant mode.
A If D is diameter of waveguide
Answer: Option A c = 1.706 D for TE11 mode
Explanation: c = 1.029 D for TE21 mode
A Klystron is a vacuum tube used for c = 0.82 D for TE01 mode
generation/amplification of microwaves. c = 1.306 D for TM01 mode.
An electron beam is produced by oxide coated
indirectly heated cathode and is focussed and 5. The reflection coefficient on a line is 0.2 45. The
accelerated by focussing electrode. SWR is
This beam is transmitted through a glass tube. The A. 0.8
input cavity where the beam enters the glass tube is B. 1.1
called buncher. C. 1.2
As electrons move ahead they see an accelerating field D.1.5
for half cycle and retarding field for the other half D
Explanation:
4. Noise - about 5 dB for low power TWT 25 dB for
high power TWT
TWT is used as RF amplifier in broadband microwave
6. Microwave resonators are used in receivers, repeater amplifier in broad band
A. microwave oscillators communication systems, communication satellites etc.
B. microwave narrow band amplifier
C. microwave frequency metres 8. In a TWT the amplitude of resultant wave travelling
D.all of the above down the helix
A. increases exponentially
D
B. increases linearly
Answer: Option D C. decreases exponentially
Explanation: D.is almost constant
They are used in all the devices. A
## 7. Assertion (A): TWT uses a focussing mechanism to Answer: Option A
prevent the electron beam from spreading. Explanation:
Reason (R): In a TWT the electron beam has to travel a In a klystron the resonant structure limits the
much longer distance than in klystron. bandwidth.
Both A and R are correct and R is correct explanation A TWT is a broadband device. Its main components are
A. electron gun (to produce the electron beam) and a
of A
Both A and R are correct but R is not correct structure supporting the slow electromagnetic wave.
B. The velocity of wave propagation along the helix
explanation of A
C. A is correct but R is wrong structure is less than velocity of light.
D.A is wrong but R is correct The beam and wave travel along the structure at the
same speed.
A
Thus interaction occurs between beam and wave and
Answer: Option A the beam delivers energy to the RF wave.
Explanation: Therefore the signal gets strengthened and amplified
In a klystron the resonant structure limits the output is delivered at the other end of tube.
bandwidth. The main features of TWT are :
A TWT is a broadband device. Its main components are 1. Frequency range - 0.5 GHz to 90 GHz
electron gun (to produce the electron beam) and a 2. Power output - 5 mW at low frequencies(less than
structure supporting the slow electromagnetic wave. 20 GHz) 250 kW (continuous wave) at 3 GHz 10 MW
The velocity of wave propagation along the helix (pulsed) at 3 GHz
structure is less than velocity of light. 3. Efficiency - about 5 to 20%
The beam and wave travel along the structure at the 4. Noise - about 5 dB for low power TWT 25 dB for
same speed. high power TWT
Thus interaction occurs between beam and wave and TWT is used as RF amplifier in broadband microwave
the beam delivers energy to the RF wave. receivers, repeater amplifier in broad band
Therefore the signal gets strengthened and amplified communication systems, communication satellites etc.
output is delivered at the other end of tube.
The main features of TWT are : 9. Which of the following is not a travelling wave?
1. Frequency range - 0.5 GHz to 90 GHz A. e = Em sin (x - t)
2. Power output - 5 mW at low frequencies(less than B. e = Em cos (x - t)
20 GHz) 250 kW (continuous wave) at 3 GHz 10 MW C. e = Em sin (t - x)
(pulsed) at 3 GHz D.e = Em sin (x)
3. Efficiency - about 5 to 20% D
Explanation: C. directivity (dB) equals isolation plus coupling
In a travelling wave both x and t increase D.isolation (dB) equals (coupling) (directivity)
simultaneously so that a constant phase point moves A
in the direction of positive (or negative) x.
Explanation:
Isolation equals coupling plus directivity.
10. Both Impatt and Trapatt devices use avalanche effect
A.True 12. As the frequency is increased, the charging MVAR in
B. False a cable
A. decreases
A B. increases
Answer: Option A C. remain the same
Explanation: D.decreases or remains the scheme
An Impatt diode has n+ - p - i - p + structure and is B
used with reverse bias.
It exhibits negative resistance and operates on the Answer: Option B
principle of avalanche breakdown. Explanation:
Impatt diode circuits are classified as broadly tunable Charging current and MVAR are proportional to
circuit, low Q circuit and high Q circuit. frequency.
The impedance of Impatt diode is a few ohms. The
word Impatt stands for Impact Avalanche Transit 13. Loss angle of a good quality cable is about
Time diode. A. 1
The features of Impatt diode oscillator are : B. 30
frequency 1 to 300 GHz, Power output (0.5 W to 5 W C. 70
for single diode circuit and upto 40 W for D.90
combination of several diodes), efficiency about 20%. A
power microwave transmitter etc. Explanation:
Avalanche diode can also be operated in large signal For ideal dielectric loss angle is zero. For good quality
high efficiency mode called Trapped Avalanche cables, loss angle is very small, about 1 only.
Transit Time mode.
The Trapatt oscillations depend on the delay in the 14. In a three cavity klystron amplifier, the oscillations
current caused by avalanche process. are excited in
The avalanche delay makes it possible to increase the A. input cavity
diode voltage well above the breakdown voltage. B. output cavity
Therefore a very rapid multiplication of charge C. intermediate cavity
carriers occurs. A Trapatt diode is also a negative D.both (a) and (b)
resistance device.
C
The features of Trapatt diode oscillator are :
Frequency 3 to 50 GHz, Power output 1-3 W, Answer: Option C
Its applications are low power doppler radar, A Klystron is a vacuum tube used for
microwave beacon landing system etc. generation/amplification of microwaves.
An electron beam is produced by oxide coated
11. In a directional coupler indirectly heated cathode and is focussed and
A. isolation (dB) equals coupling plus directivity accelerated by focussing electrode.
B. coupling (dB) equals isolation plus directivity
This beam is transmitted through a glass tube. The B. Coaxial
input cavity where the beam enters the glass tube is C. Both
called buncher. D.None of the above
As electrons move ahead they see an accelerating B
field for half cycle and retarding field for the other
Therefore, some electrons are accelerated and some Explanation:
are retarded. This process is called velocity Since fields are confirmed within a coaxial cable it is
The velocity modulation causes bunching of
electrons. This bunching effect converts velocity 17. Assertion (A): The impedance of a matched load is
modulation into density modulation of beam. equal to characteristic impedance of line.
The input is fed at buncher cavity and output is taken Reason (R): A matched termination absorbs all the
at catcher cavity. power incident on it.
In a two cavity klystron only buncher and catcher Both A and R are correct and R is correct
A.
cavity are used. In multi cavity klystron one or more explanation of A
intermediate cavities are also used. Both A and R are correct but R is not correct
B.
The features of a multicavity klystron are : explanation of A
1. Frequency range - 0.25 GHz to 100 GHz C. A is correct but R is wrong
2. Power output - 10 kW to several hundred kW D.A is wrong but R is correct
3. Power gain - 60 dB (nominal value) A
A multicavity klystron is used in UHF TV transmitters, Explanation:
15. In the given figure reflection coefficient at load is 18. Skin effect is more pronounced at high frequencies.
A.True
B. False
A
Explanation:
Skin effect increases as frequency increases.
## 19. Assertion (A): The phenomenon of differential
A. 0.6 mobility is called transferred electron effect.
B. - 0.6 Reason (R): GaAs exhibits transferred electron effect.
C. 0.4 Both A and R are correct and R is correct
A.
D.- 0.4 explanation of A
Both A and R are correct but R is not correct
B B.
explanation of A
Answer: Option B C. A is correct but R is wrong
Explanation: D.A is wrong but R is correct
B
Explanation:
16. Which of the following lines is non-radiating?
A. Open two wire
A Gunn diode uses GaAs which has a negative
differential mobility, i.e., a decrease in carrier 21. Which of the following parameters is negligible in
velocity with increase in electric field. transmission lines?
This effects is called transferred electron effect. The A. R
impedance of a Gunn diode is tens of ohms. B. L
A Gunn diode oscillator has a resonant cavity, an C. C
arrangement to couple Gunn diode to cavity, biasing D.G
arrangement for Gunn diode and arrangement to D
Applications of Gunn diode oscillator include Answer: Option D
receivers. Shunt conductance can be neglected in most of
calculations.
20. Which of the following devices uses a helix?
A. Klystron amplifier 22. The width of a radio beam from a 1 m diameter
B. Klystron oscillator parabolic antenna at 10 GHz is about
C. TWT A. 100
D.Both (a) and (b) B. 50
C. 5
C D.1
Explanation:
In a klystron the resonant structure limits the Answer: Option C
bandwidth. Explanation:
A TWT is a broadband device. Its main components
are electron gun (to produce the electron beam) and .
a structure supporting the slow electromagnetic
wave. 23. The diagram to show distance time history of
The velocity of wave propagation along the helix electrons in klystron amplifier is called
structure is less than velocity of light. A. apple gate diagram
The beam and wave travel along the structure at the B. asynchronous diagram
same speed. C. bunching diagram
Thus interaction occurs between beam and wave and D.velocity modulation diagram
the beam delivers energy to the RF wave.
A
Therefore the signal gets strengthened and amplified
output is delivered at the other end of tube. Answer: Option A
The main features of TWT are : Explanation:
1. Frequency range - 0.5 GHz to 90 GHz Applegate diagram is distance time plot.
2. Power output - 5 mW at low frequencies(less than
20 GHz) 250 kW (continuous wave) at 3 GHz 10 MW 24. Impedance level of Impatt diodes is generally lower
(pulsed) at 3 GHz than that of Gunn diodes
3. Efficiency - about 5 to 20% A.True
4. Noise - about 5 dB for low power TWT 25 dB for B. False
high power TWT A
TWT is used as RF amplifier in broadband microwave
Explanation:
communication systems, communication satellites
An Impatt diode has n+ - p - i - p + structure and is
etc.
used with reverse bias.
It exhibits negative resistance and operates on the Both A and R are correct but R is not correct
B.
principle of avalanche breakdown. explanation of A
Impatt diode circuits are classified as broadly tunable C. A is correct but R is wrong
circuit, low Q circuit and high Q circuit. D.A is wrong but R is correct
The impedance of Impatt diode is a few ohms. The B
word Impatt stands for Impact Avalanche Transit
The features of Impatt diode oscillator are : Explanation:
frequency 1 to 300 GHz, Power output (0.5 W to 5 W It is somewhat similar to TWT and can deliver
for single diode circuit and upto 40 W for microwave power over a wide frequency band.
combination of several diodes), efficiency about 20%. It has an electron gun and a helix structure. However
Its applications include police radar systems, low the interaction between electron beam and RF wave
power microwave transmitter etc. is different than in TWT.
The growing RF wave travels in opposite direction to
25. A quarter wave line open circuited at far end behaves the electron beam.
as The frequency of wave can be changed by changing
A. inductance the voltage which controls the beam velocity.
B. L and C in parallel Moreover the amplitude of oscillations can be
C. capacitance decreased continuously to zero by changing the
D.L and C in series beam current.
It features are:
D
1. Frequency range - 1 GHz to 1000 GHz.
Answer: Option D 2. Power output - 10 mV to 150 mW (continuous
Explanation: wave) 250kW (pulsed).
A quarter wave line o.c. at far end behaves as a series It is used as signal source in transmitters and
tuned circuit. instruments.
26. A line has Z0 = 300 0 . If ZL = 150 0 , 28. In a microstrip transmission the signal can be easily
reflection coefficient is redirected by changing the centre strip.
A. 0.5 A.True
B. 0.3333 B. False
C. -0.3333 A
D.-0.5
C
Explanation:
Answer: Option C A Microstrip line has a single dielectric substratc with
Explanation: ground plane on one side and a strip on the other
Reflection coefficient face.
fabrication.
The high dielectric constant of the substrate reduces
27. Assertion (A): A backward wave oscillator has an guide wavelength and circuit dimensions.
internal positive feedback. A microstrip line is the most commomly used
Reason (R): A positive feedback is necessary for transmission structure for microwave integrated
sustained oscillations. circuits.
Both A and R are correct and R is correct
A.
explanation of A 29. The frequency of oscillation in a backward wave
oscillator can be changed by
A. varying the voltage which controls beam velocity 31. In the given figure the time taken by the wave to make one ro
B. varying the beam current
both by varying the beam current and by light
C.
varying the voltage which controls beam velocity
D.changing the rate of thermionic emission
A
Explanation:
It is somewhat similar to TWT and can deliver
microwave power over a wide frequency band.
A. 10 ns
It has an electron gun and a helix structure. However
B. 20 ns
the interaction between electron beam and RF wave
C. 30 ns
is different than in TWT.
D.40 ns
The growing RF wave travels in opposite direction to
the electron beam. D
The frequency of wave can be changed by changing Answer: Option D
the voltage which controls the beam velocity. Explanation:
Moreover the amplitude of oscillations can be
decreased continuously to zero by changing the
beam current.
It features are: 32. For a 50 resistor for 3 GHz application, the stray
1. Frequency range - 1 GHz to 1000 GHz. capacitance should be less than
2. Power output - 10 mV to 150 mW (continuous A. 1 F
wave) 250kW (pulsed). B. 1 nF
It is used as signal source in transmitters and C. 1 pF
instruments. D.0.1 pF
D
30. A line has a phase constant of 29.8 rad/m. At 1000 MHz the wavelength is
A. 29.8 m Answer: Option D
B. 2.98 m Explanation:
C. 2.1 m XC should be at least ten times greater than R.
D.0.21 m Therefore for 50 resistor at 3 GHz, C should be less
D than 0.1 pF.
## Answer: Option D 33. In a travelling wave tube distributed interaction
Explanation: between an electron beam and a travelling wave
takes place
. A.True
B. False
A
Explanation:
In a klystron the resonant structure limits the
bandwidth.
A TWT is a broadband device. Its main components
are electron gun (to produce the electron beam) and
a structure supporting the slow electromagnetic A. RL = 1
wave. B. R = 0
The velocity of wave propagation along the helix
C.
structure is less than velocity of light.
The beam and wave travel along the structure at the D.
same speed.
Thus interaction occurs between beam and wave and D
the beam delivers energy to the RF wave.
Therefore the signal gets strengthened and amplified
Explanation:
output is delivered at the other end of tube.
The condition for minimum distortion is found by
The main features of TWT are :
1. Frequency range - 0.5 GHz to 90 GHz
2. Power output - 5 mW at low frequencies(less than equating The result is .
20 GHz) 250 kW (continuous wave) at 3 GHz 10 MW 36. The action of backward wave oscillator is similar to
(pulsed) at 3 GHz that of
3. Efficiency - about 5 to 20% A. klystron amplifier
4. Noise - about 5 dB for low power TWT 25 dB for B. reflex klystron oscillator
high power TWT C. TWT
TWT is used as RF amplifier in broadband microwave D.magnetron
communication systems, communication satellites C
Explanation:
34. The fabrication of microstrip line is done by It is somewhat similar to TWT and can deliver
A. photo etching microwave power over a wide frequency band.
B. printed circuit technique It has an electron gun and a helix structure. However
C. oxidation the interaction between electron beam and RF wave is
B The growing RF wave travels in opposite direction to
the electron beam.
Answer: Option B The frequency of wave can be changed by changing
Explanation: the voltage which controls the beam velocity.
Printed circuit techniques is used for micro strip line. Moreover the amplitude of oscillations can be
decreased continuously to zero by changing the beam
35. The condition of minimum distortion in a transmission current.
line is It features are:
1. Frequency range - 1 GHz to 1000 GHz.
2. Power output - 10 mV to 150 mW (continuous wave)
250kW (pulsed).
It is used as signal source in transmitters and
instruments.
37. In the given figure the reflection coefficient at source Explanation:
end is It is somewhat similar to TWT and can deliver
microwave power over a wide frequency band.
It has an electron gun and a helix structure. However
the interaction between electron beam and RF wave
is different than in TWT.
The growing RF wave travels in opposite direction to
the electron beam.
The frequency of wave can be changed by changing
the voltage which controls the beam velocity.
Moreover the amplitude of oscillations can be
A. 1 decreased continuously to zero by changing the
B. -1 beam current.
C. 1/3 It features are:
D.-1/3 1. Frequency range - 1 GHz to 1000 GHz.
C
2. Power output - 10 mV to 150 mW (continuous
wave) 250kW (pulsed).
Answer: Option C It is used as signal source in transmitters and
Explanation: instruments.
## . 40. A line is excited by a 100 V dc source. If reflection
coefficients at both ends are 1 each then
38. Assertion (A): Artificial transmission lines are A. there will be no oscillations on line
frequently used in laboratories. B. there will be only 1 or 2 oscillations on line
Reason (R): An artificial transmission line can be used C. there will be a finite number of oscillations on line
to represent an actual line and can also be used as a D.the oscillations will continue indefinitely
delay circuit, as attenuator, as filter network etc. D
Both A and R are correct and R is correct Answer: Option D
A.
explanation of A Explanation:
Both A and R are correct but R is not correct Because rv = 1 at each end, the line voltage will not
B.
explanation of A reach a steady value and oscillations will continue
C. A is correct but R is wrong indefinitely.
D.A is wrong but R is correct
A 41. Assertion (A): A coaxial line is a non-radiating line.
Answer: Option A Reason (R): In a coaxial line the electric and magnetic
Explanation: fields are confined to the region between the
Artificial transmission lines have many applications. concentric conductors.
One of the applications is to simulate an actual line in Both A and R are correct and R is correct
A.
the laboratory. explanation of A
Both A and R are correct but R is not correct
B.
39. In a backward wave oscillator the wave travelling explanation of A
along the line winds itself back and forth C. A is correct but R is wrong
A.True D.A is wrong but R is correct
B. False A
Since the fields are confined, there is no radiation.
42. Reflex klystron oscillator is essentially a low power Explanation:
device Since all ports matched, input SWR = 1. No power is
A.True reflected and Pin = P4 = 1 W.
B. False This power splits between ports 1 and 2. Therefore P1
A = P2 = 0.5 , P3 = 0.
Answer: Option A 45. Which TM mode in rectangular waveguide has lowest
Explanation: cutoff frequency?
It uses a single cavity resonator for generating A. TM11
microwave oscillations. B. TM01
Its parts are electron gun, resonator, repeller and C. TM10
output coupling. D.TM21
It operates on the principle of positive feed back.
A
The repeller electrode is at negative potential and
sends the partially bunched electron beam back to Answer: Option A
resonator cavity. Explanation:
This positive feedback supports oscillations. Its Hz = 0
feature are:
1. Frequency range - 2 to 100 GHz
2. Power output - 10 MW to about 2 W
3. Efficiency - 10 - 20 %
oscillator in microwave devices, oscillator for
microwave measurements in laboratories etc.
Ex = ZTM Hy
43. A 10 km long line has a characteristic impedance of Ey = ZTM Hx
400 ohms. If line length is 100 km, the characteristic where E0 is the amplitude of the wave.
impedance is The expressions for , c, fc, g, a are the same as for
A. 4000 TE waves.
B. 400 For TM wave the lowest cut off frequency in
C. 40 rectangular wave guide is for TM11 mode.
D.4
B If , the cutoff frequency for TM11 mode is
about 12% more than that for TE20 mode.
Explanation:
46. A resistive microwave load with ZL = 150 is
connected to 50 coaxial line. SWR is
and is independent of line length. A. more than 3
B. less than 3
44. A matched generator (Zg = Z0) with 1 available. If all C. equal to 3
ports are matched, the power delivered at ports 1, 2, D.either (a) or (c)
3 respectively are
C
A. 0.5 W, 0.5 W, 0 W
B. 0 W, 0 W, 1 W Answer: Option C
C. 0.33 W each Explanation:
D.0 W, 0.25 W, 0.25 W
A
Since the two wire radiators are at 90 to each other,
47. Which of the following devices uses a slow wave they have to be excited 90 out of phase with each
structure? other.
A. Klystron two cavity amplifier
B. Klystron multicavity amplifier 49. A waveguide section in a microwave circuit acts as
C. Reflex klystron oscillator A. LP filter
D.TWT B. Bandpass filter
D C. HP filter
D.Band stop filter
Explanation: C
In a klystron the resonant structure limits the Answer: Option C
bandwidth. Explanation:
A TWT is a broadband device. Its main components It allows high frequencies.
are electron gun (to produce the electron beam) and
a structure supporting the slow electromagnetic 50. In a klystron amplifier dc electron velocity is v0 and dc
wave. electron charge density is r0 the dc beam current is
The velocity of wave propagation along the helix A. r0v0
structure is less than velocity of light. B. r0/v0
The beam and wave travel along the structure at the C. v0/r0
same speed. D.(v0/r0)2
Thus interaction occurs between beam and wave and A
the beam delivers energy to the RF wave.
Therefore the signal gets strengthened and amplified Answer: Option A
output is delivered at the other end of tube. Explanation:
The main features of TWT are :
1. Frequency range - 0.5 GHz to 90 GHz = coulomb |sec|m2 = current
2. Power output - 5 mW at low frequencies(less than density.
20 GHz) 250 kW (continuous wave) at 3 GHz 10 MW
(pulsed) at 3 GHz
3. Efficiency - about 5 to 20%
4. Noise - about 5 dB for low power TWT 25 dB for
high power TWT
TWT is used as RF amplifier in broadband microwave
communication systems, communication satellites
etc.
## 48. In a turnstile antenna two-half wave resonant wire
radiators are placed at 90 to each other in the same
plane and are excited
A. in phase
B. in phase opposition
C. 90 out of phase with each other
D.45 out of phase with each other
C
Explanation:
As electrons move ahead they see an accelerating field
for half cycle and retarding field for the other half
cycle.
Therefore, some electrons are accelerated and some
are retarded. This process is called velocity
SECTION 2 modulation.
The velocity modulation causes bunching of electrons.
1. The directivity in a receiving antenna This bunching effect converts velocity modulation into
increases the intercept area in forward direction density modulation of beam.
reduces the noise picked up from other sources The input is fed at buncher cavity and output is taken
provides a means of discriminating against undesired at catcher cavity.
signals originating in directions other than in which the In a two cavity klystron only buncher and catcher
desired transmitter lies. cavity are used. In multi cavity klystron one or more
Which of the above statements are correct? intermediate cavities are also used.
A. 1 only The features of a multicavity klystron are :
B. 1 and 2 only 1. Frequency range - 0.25 GHz to 100 GHz
C. 1, 2 and 3 2. Power output - 10 kW to several hundred kW
D.2 and 3 only 3. Power gain - 60 dB (nominal value)
C 4. Efficiency - about 40%.
A multicavity klystron is used in UHF TV transmitters,
2. Assertion (A): PIN diode can be used as attenuator and Radar transmitter and satellite communication.
limiter.
Reason (R): PIN diode has a thin intrinsic layer. 4. The cut off wavelength in circular waveguide
Both A and R are correct and R is correct explanation A. guide diameter
A.
of A B. square of guide diameter
Both A and R are correct but R is not correct C. cube of guide diameter
B.
explanation of A D.square root of guide diameter
C. A is correct but R is wrong
A
D.A is wrong but R is correct
Explanation:
3. The bunching action which occurs in multicavity Some applications require dual polarization capability.
klystron amplifier can be represented by Applegate Circular waveguide has this capability.
diagram These analysis uses cylindrical coordinates.
A.True In circular waveguide TE11 mode has the lowest cut off
B. False frequency and is the dominant mode.
A If D is diameter of waveguide
Answer: Option A c = 1.706 D for TE11 mode
Explanation: c = 1.029 D for TE21 mode
A Klystron is a vacuum tube used for c = 0.82 D for TE01 mode
generation/amplification of microwaves. c = 1.306 D for TM01 mode.
An electron beam is produced by oxide coated
indirectly heated cathode and is focussed and 5. In Reflex Klystron oscillator the focussing electrode is
accelerated by focussing electrode. at a high potential
This beam is transmitted through a glass tube. The A. True
input cavity where the beam enters the glass tube is B. False
called buncher. B
Explanation: It exhibits negative resistance and operates on the
It uses a single cavity resonator for generating principle of avalanche breakdown. Impatt diode
microwave oscillations. circuits are classified as broadly tunable circuit, low
Its parts are electron gun, resonator, repeller and Q circuit and high Q circuit.
output coupling. The impedance of Impatt diode is a few ohms. The
It operates on the principle of positive feed back. word Impatt stands for Impact Avalanche Transit
The repeller electrode is at negative potential and Time diode.
sends the partially bunched electron beam back to The features of Impatt diode oscillator are :
resonator cavity. Frequency 1 to 300 GHz, Power output (0.5 W to 5
This positive feedback supports oscillations. Its feature W for single diode circuit and upto 40 W for
are: combination of several diodes), efficiency about
1. Frequency range - 2 to 100 GHz 20%.
2. Power output - 10 MW to about 2 W Its applications include police radar systems, low
3. Efficiency - 10 - 20 % power microwave transmitter etc.
in microwave devices, oscillator for microwave 7. In high frequency circuits impedance matching can
measurements in laboratories etc. be done by
open circuited stubs
6. Consider the following statements short circuited stubs
Impedance of Gunn diode is about tens of ohms.
Impedance of Impatt diode is a few ohms. transformer
Impedance of Impatt diode are of the same order. Which of the above correct?
Impedance of Impatt diode is more than that of A. 1, 2 and 3
Gunn diode. B. 1 and 2 only
Which of the above statement are correct? C. 1 and 3 only
A. all D. 2 and 3 only
B. 1 and 2 only
A
C. 1, 2 and 3
D. 1, 2 and 4 Answer: Option A
B Explanation:
All the three can be used for impedance matching in
Answer: Option B high frequency circuits.
Explanation:
A Gunn diode uses GaAs which has a negative 8. The directive gain of a transmitting antenna is
differential mobility, i.e., a decrease in carrier proportional to
velocity with increase in electric field. A. its cross-sectional area
This effects is called transferred electron effect. The B. square of cross-sectional area
impedance of a Gunn diode is tens of ohms. C. square root of cross-sectional area
A Gunn diode oscillator has a resonant cavity, an D. cube root of cross-sectional area
arrangement to couple Gunn diode to cavity,
A
biasing arrangement for Gunn diode and
Applications of Gunn diode oscillator include Explanation:
An Impatt diode has n+ - p - i - p + structure and is and is thus proportional
used with reverse bias. to cross sectional area.
9. In a loss line RL < Z0, then 11. If a line having Z0 = 300 0 W is open circuited at far
A. Reflection coefficient is zero end, VSWR is
B. Reflection coefficient is A. 0
C. Reflection coefficient is negative B. 1
D. Reflection coefficient is positive C.
C D. 2
Explanation:
## Since RL < Z0, rv is negative.
If line is o.c. |rv| = 1 and VSWR = .
10. Assertion (A): A backward wave oscillator can be
used as a sweep generator. 12. If f is the frequency of electromagnetic wave, fc is
Reason (R): The frequency of oscillation of a cutoff frequency, then in a rectangular waveguide
backward wave oscillator can be changed by A. attenuation is low when f > fc
varying the voltage which controls the beam B. attenuation is low when f < fc
velocity. C. attenuation is high when f < fc
Both A and R are correct and R is correct D.either (b) or (c)
A.
explanation of A A
Both A and R are correct but R is not correct
explanation of A
C. A is correct but R is wrong Explanation:
D. A is wrong but R is correct Wavelength should be less than cutoff wavelengths.
Therefore frequency should be higher than cutoff
A
frequency for minimum attenuation.
Explanation: 13. Assertion (A): The condition of minimum distortion in
It is somewhat similar to TWT and can deliver a transmission line is L = RC/G.
It has an electron gun and a helix structure. inductance to satisfy the condition L = RC/G.
However the interaction between electron beam Both A and R are correct and R is correct
A.
and RF wave is different than in TWT. explanation of A
The growing RF wave travels in opposite direction Both A and R are correct but R is not correct
B.
to the electron beam. explanation of A
The frequency of wave can be changed by C. A is correct but R is wrong
changing the voltage which controls the beam D.A is wrong but R is correct
velocity. B
Moreover the amplitude of oscillations can be
decreased continuously to zero by changing the Answer: Option B
beam current. Explanation:
It features are: Oscillator circuits using vacuum tubes (triodes,
1. Frequency range - 1 GHz to 1000 GHz. pentodes) have the following limitations at very high
2. Power output - 10 mV to 150 mW (continuous frequencies (microwave region).
wave) 250kW (pulsed). The stray capacitances and inductances become
It is used as signal source in transmitters and important and affect the operation of the circuit.
instruments. At low frequencies the transit time between cathode
and anode is a small fraction of period of oscillation.
However, at microwave frequencies this transit time C
becomes comparable to time period of oscillations.
Answ er: Option C
14. A magnetron has a cylindrical cathode surrounded by Explanation:
an anode structure having cavities opening into
interaction space by means of slots
If line is o.c. |rv| = 1 and VSWR = .
A.True
B. False 12. If f is the frequency of electromagnetic wave, fc is
A cutoff frequency, then in a rectangular waveguide
Answer: Option A A. attenuation is low when f > fc
Explanation: B. attenuation is low when f < fc
It is somewhat similar to TWT and can deliver C. attenuation is high when f < fc
microwave power over a wide frequency band. D.either (b) or (c)
It has an electron gun and a helix structure. However A
the interaction between electron beam and RF wave Answer: Option A
is different than in TWT. Explanation:
The growing RF wave travels in opposite direction to Wavelength should be less than cutoff wavelengths.
the electron beam. Therefore frequency should be higher than cutoff
The frequency of wave can be changed by changing frequency for minimum attenuation.
the voltage which controls the beam velocity.
Moreover the amplitude of oscillations can be 13. Assertion (A): The condition of minimum distortion in
decreased continuously to zero by changing the a transmission line is L = RC/G.
It features are: inductance to satisfy the condition L = RC/G.
1. Frequency range - 1 GHz to 1000 GHz. Both A and R are correct and R is correct
2. Power output - 10 mV to 150 mW (continuous A.
explanation of A
wave) 250kW (pulsed). Both A and R are correct but R is not correct
It is used as signal source in transmitters and B.
explanation of A
instruments. C. A is correct but R is wrong
D.A is wrong but R is correct
15. The correct sequence of parts in klystron amplifier
are B
A. anode, catcher cavity, cathode, buncher cavity Answer: Option B
B. cathode, buncher cavity, catcher cavity, cavity Explanation:
C. anode, buncher cavity, catcher cavity, cathode Oscillator circuits using vacuum tubes (triodes,
D. cathode, catcher cavity, anode, buncher cavity pentodes) have the following limitations at very high
B frequencies (microwave region).
The stray capacitances and inductances become
Answer: Option B important and affect the operation of the circuit.
Explanation: At low frequencies the transit time between cathode
Cathode is the first part and anode is the last. and anode is a small fraction of period of oscillation.
However, at microwave frequencies this transit time
11. If a line having Z0 = 300 0 W is open circuited at far becomes comparable to time period of oscillations.
end, VSWR is
A. 0 14. A magnetron has a cylindrical cathode surrounded by
B. 1 an anode structure having cavities opening into
C. interaction space by means of slots
D. 2
A.True A. 1.265 - 18.43
B. False B. 1.01 - 10
A C. 1.14 66.68
D. 1.09 66.68
Explanation: A
It is somewhat similar to TWT and can deliver Answer: Option A
microwave power over a wide frequency band. Explanation:
It has an electron gun and a helix structure. However
the interaction between electron beam and RF wave
is different than in TWT. Transmission coefficient
The growing RF wave travels in opposite direction to = 1.265 - 18.43.
the electron beam.
The frequency of wave can be changed by changing 17. In a reflex klystron oscillator, repeller electrode is at
the voltage which controls the beam velocity. A
Moreover the amplitude of oscillations can be low positive potential
.
decreased continuously to zero by changing the B
beam current. high positive potential
.
It features are: C
1. Frequency range - 1 GHz to 1000 GHz. negative potential
.
2. Power output - 10 mV to 150 mW (continuous D
wave) 250kW (pulsed). zero potential
.
It is used as signal source in transmitters and C
instruments.
15. The correct sequence of parts in klystron amplifier Explanation:
are It uses a single cavity resonator for generating
A. anode, catcher cavity, cathode, buncher cavity microwave oscillations.
B. cathode, buncher cavity, catcher cavity, cavity Its parts are electron gun, resonator, repeller and
C. anode, buncher cavity, catcher cavity, cathode output coupling.
D. cathode, catcher cavity, anode, buncher cavity It operates on the principle of positive feed back.
The repeller electrode is at negative potential and
B
sends the partially bunched electron beam back to
Explanation: This positive feedback supports oscillations. Its
Cathode is the first part and anode is the last. feature are:
1. Frequency range - 2 to 100 GHz
16. A transmission line has Z0 = 300 and ZL = (300 - 2. Power output - 10 MW to about 2 W
j300) ohm. The transmission coefficient is 3. Efficiency - 10 - 20 %
oscillator in microwave devices, oscillator for
microwave measurements in laboratories etc.
## 18. A quarter wave line short circuited at load end
behaves as
A. an inductance
B. an inductance and capacitance in series
C. a capacitance C
D.an inductance and capacitance in parallel
D Explanation:
Answer: Option D A Gunn diode uses GaAs which has a negative
Explanation: differential mobility, i.e., a decrease in carrier
A quarter wave line short-circuited at far end velocity with increase in electric field.
behaves as a parallel tuned circuit. This effects is called transferred electron effect. The
impedance of a Gunn diode is tens of ohms.
19. A (75 - J50) ohm load is connected to a coaxial of Z0 = A Gunn diode oscillator has a resonant cavity, an
75 ohm at 10 GHz. The best method of matching is to arrangement to couple Gunn diode to cavity, biasing
connect arrangement for Gunn diode and arrangement to
B. a short circuited stub at load Applications of Gunn diode oscillator include
C. a short circuited stub at specific distance from load continuous wave radar, pulsed radar and microwave
C
22. A magnetron requires an external magnetic field with
Answer: Option C flux lines parallel to axis of cathode
Explanation: A.True
A short-circuited stub is a suitable method of B. False
matching.
A
20. The width of a radio beam from a 1 m diameter Answer: Option A
parabolic antenna at 1 GHz is about Explanation:
A It is somewhat similar to TWT and can deliver
100
. microwave power over a wide frequency band.
B It has an electron gun and a helix structure. However
50
. the interaction between electron beam and RF wave
C is different than in TWT.
5
. The growing RF wave travels in opposite direction to
D the electron beam.
2
. The frequency of wave can be changed by changing
B the voltage which controls the beam velocity.
Moreover the amplitude of oscillations can be
Answer: Option B decreased continuously to zero by changing the
Explanation: beam current.
Beam width between nulls of a parabolic antenna It features are:
1. Frequency range - 1 GHz to 1000 GHz.
. 2. Power output - 10 mV to 150 mW (continuous
Therefore 50 is the nearest value. wave) 250kW (pulsed).
It is used as signal source in transmitters and
21. Which one of the following is transferred locking instruments.
A. BARITT diode
B. IMPATT diode 23. When a line short circuited at far end, the minimum
C. Gunn diode voltage occurs at
D. Step recovery diode A. far end
B. source end
C. midway between source and far end A. 20 dB more
D.none of the above B. 20 dB less
A C. 700 dB more
D.100 dB less
Explanation: A
When a line is short-circuited at far end, the voltage Answer: Option A
at far end is zero. Explanation:
As frequency increases, transmission loss increases
24. In a TWT the axial component of electric field slightly.
A. at a velocity that is almost equal to speed of light
B. at a velocity that is a small fraction of speed of light 26. For a circular wave guide
C. at a velocity that is about 50% of speed of light A. cutoff frequency for TE10 and TE01 modes are same
at a velocity that may be even more than speed of cutoff frequency for TE10 and TE01 modes are
D. B.
light different
B cutoff frequency for TE10 mode = twice the cutoff
C.
frequency for TE20 mode
Answer: Option B cutoff frequency for TE10 mode = haf the cutoff
Explanation: D.
frequency for TE20 mode
In a klystron the resonant structure limits the
bandwidth. A
A TWT is a broadband device. Its main components Answer: Option A
are electron gun (to produce the electron beam) and Explanation:
a structure supporting the slow electromagnetic When a circular waveguide is rotated by 90, the
wave. configuration remain the same.
The velocity of wave propagation along the helix
structure is less than velocity of light. 27. In the given figure the reflected voltage wave after
The beam and wave travel along the structure at the first reflection is
same speed.
Thus interaction occurs between beam and wave and
the beam delivers energy to the RF wave.
Therefore the signal gets strengthened and amplified
output is delivered at the other end of tube.
The main features of TWT are :
1. Frequency range - 0.5 GHz to 90 GHz
2. Power output - 5 mW at low frequencies(less than
20 GHz) 250 kW (continuous wave) at 3 GHz 10 MW
(pulsed) at 3 GHz A. 18 V
3. Efficiency - about 5 to 20% B. -18 V
4. Noise - about 5 dB for low power TWT 25 dB for C. 30 V
high power TWT D.-30 V
TWT is used as RF amplifier in broadband microwave B
communication systems, communication satellites Answer: Option B
etc. Explanation:
## 25. If other parameter are constant, transmission loss at 5
GHz as compared to that at 0.5 GHz is .
31. Which of the following terminations makes the
28. Assertion (A): A half wavelength line can be used as a input impedance of a line equal to characteristic
1 : 1 transformer. impedance Z0?
Reason (R): The input impedance of a half A. line open circuited at far end
wavelength line is equal to load impedance. B. line terminated in Z0
Both A and R are correct and R is correct C. line short circuited at far end
A.
explanation of A D. line terminated in inductance
Both A and R are correct but R is not correct B
B.
explanation of A
C. A is correct but R is wrong Answer: Option B
D.A is wrong but R is correct Explanation:
If ZL is load impedance and Z0 is characteristic
A impedance.
Explanation:
Since a half wavelength line has an input impedance
Input impedance =
equal to load impedance, the impedance
. If ZL = Z0 the input impedance = Z0.
transformation ratio is 1:1.
32. In microwave system the function of mode filter is
29. The magnitudes of OC and SC input impedances of a
A. to suppress modes with lower cut off frequencies
transmission line an 100 and 25 . The
B. to suppress modes with higher cut off frequencies
characteristic impedance is
C. to change mode of wave transmission
A. 25
D.both (b) and (c)
B. 50
C. 75 D
B Explanation:
Mode filter suppress modes with lower cutoff
frequencies.
Explanation:
33. For the strip line in the given figure the capacitance
per unit length =
30. A loss less line is terminated in a circular lines are E
lines
A. SWR = 0
B. SWE = 0
C. SWR is finite
D. SWR =
C
Explanation:
A.
= finite number and VSWR
B.
= finite number. C.
Reason (R): A quarter wave transformer is a
D. transmission line of quarter wave length.
Both A and R are correct and R is correct
A A.
explanation of A
Answer: Option A Both A and R are correct but R is not correct
B.
Explanation: explanation of A
C. A is correct but R is wrong
D.A is wrong but R is correct
B
34. The total field developed by an antenna array at a Answer: Option B
distant point is Explanation:
phasor sum of fields produced by individual
A.
antennas of the array A line of length is called quarter wavelength line.
algebraic sum of fields produced by individual Such a line is used for impedance matching.
B.
antennas of the array
C. either (a) or (b) depending on type of array If
D.neither (a) nor (b)
A
Answer: Option A A quarter wave line can match a source impedance
Explanation: Zin with load impedance ZL by selecting a proper value
Since field is a phasor quantity we have to take of Z0 so as to satisfy equation.
phasor sum.
Such a line is also called transformer.
35. High speed logic circuits use
A. pulses with very small width 38. Assertion (A): Magnetron is generally used in n mode.
B. pulses with very large width Reason (R): Frequency for p mode can be easily
C. pulses whose width is neither small nor large separated from adjacent modes.
D.either (b) or (c) Both A and R are correct and R is correct
A.
A explanation of A
Both A and R are correct but R is not correct
explanation of A
Explanation:
C. A is correct but R is wrong
High speed means time period is small. Therefore pulses should also have small width.
D.A is wrong but R is correct
36. To couple a coaxial line to a parallel wire line it is best to use A
A. slotted line Answer: Option A
B. balun Explanation:
C. directional coupler It is somewhat similar to TWT and can deliver
D./4 transformer microwave power over a wide frequency band.
B It has an electron gun and a helix structure. However
the interaction between electron beam and RF wave
is different than in TWT.
Explanation:
The growing RF wave travels in opposite direction to
A balun gives 4 : 1 impedance transformation.
the electron beam.
The frequency of wave can be changed by changing
37. Assertion (A): A quarter wave transformer is used to
the voltage which controls the beam velocity.
match a resistive load to a transmission line.
Moreover the amplitude of oscillations can be Explanation:
decreased continuously to zero by changing the TE01 mode is unaffected because its horizontally oriented elect
beam current. to the conducting strips.
It features are:
1. Frequency range - 1 GHz to 1000 GHz. 42. The directive gain of a transmitting antenna is
2. Power output - 10 mV to 150 mW (continuous A. proportional to wavelength
wave) 250kW (pulsed). B. inversely proportional to wavelength
It is used as signal source in transmitters and C. proportional to square of wavelength
instruments. D. inversely proportional to square of wavelength
D
39. A balun should have
A. low SWR at both ports Answer: Option D
B. high SWR at both ports Explanation:
C. high SWR at input port and low SWR at output port
D.low SWR at input port and high SWR at output
A
Answer: Option A 43. A duplexer is used to
Explanation: A. couple two antennas to a transmitter
Turn ratio is so selected as to give low SWR at both B. isolate the antenna from the local oscillator
ports. prevent interference between two antennas
C.
use an antenna for reception or transmission
D.
40. Assertion (A): The velocity factor of a line is the ratio of wave velocity
withoutoninterference
the line to speed of light.
Reason (R): If the conductors of a line are immersed in a non magnetic
D insulating liquid, the wave
velocity increases.
A. Both A and R are correct and R is correct explanation of A Answer: Option D
B. Both A and R are correct but R is not correct explanation of Explanation:
A
C. A is correct but R is wrong Same antenna is used for transmission and reception.
D.A is wrong but R is correct
44. A branched duplexer requires
C
A. TR tubes
Answer: Option C B. ATR tube
Explanation: C. both TR and ATR tube
R is wrong because wave velocity decreases. D.none of the above
C
41. In mode filter of the given figure which of the following modes is unaffected
Explanation:
A branched duplexer funds use in narrow bandwidth
applications.
TR tube stands for transmit-receive tube and ATR
A. TE01 tube stands for antitransmit receive tube.
B. TE10 Both these tubes are used in branched duplexer.
C. TE11
D.all TM 45. In a backward wave oscillator the wave
A. travelling along the line winds itself back and forth
A
B. progresses only in forward direction
Answer: Option A C. progresses only in backward direction
D.either (a) or (c) Both A and R are correct but R is not correct
B.
A explanation of A
C. A is correct but R is wrong
Answer: Option A D.A is wrong but R is correct
Explanation:
It is somewhat similar to TWT and can deliver microwave powerBover a wide frequency band.
It has an electron gun and a helix structure. However the interaction
Option Belectron beam and RF
wave is different than in TWT. Explanation:
The growing RF wave travels in opposite direction to the electron
A PINbeam.
diode has an intrinsic (i) layer between p and n
The frequency of wave can be changed by changing the voltage layers.
which
When
controls
reverse
the beam
bias isvelocity.
applied depletion layers
Moreover the amplitude of oscillations can be decreased continuously
are formed to at
zero
p-i by
andchanging
i-n junctions.
the beam
current. The effective/width of depletion layer increases by
It features are: the width of i layer. It can be used as a voltage
1. Frequency range - 1 GHz to 1000 GHz. controlled attenuator.
2. Power output - 10 mV to 150 mW (continuous wave) 250kW At(pulsed).
high frequencies the rectification effect ceases
It is used as signal source in transmitters and instruments. and impedance of diode is effectively that of i layer.
This impedance varies with the applied bias. It is used
46. A coaxial line has L = 500 nH/m and C = 50 pF/m. The characteristic
in highimpedance
frequency isswitching circuits, limiters,
A. 500 modulators etc.
B. 250
C. 100 49. In a vacuum tube, the transit time of electron
D.50 between cathode and anode is important at
C A. low frequencies
B. high frequencies
Answer: Option C C. both (a) and (b)
Explanation: frequencies which are neither very low nor very
D.
high
B
.
47. The main feature of a parametric amplifier is Explanation:
A. low noise At high frequencies transit time is large as compared
B. very high gain to the period of microwave signal.
D.both (b) and (c) 50.
C Assertion (A): A line of length and short circuited
Answer: Option C at far end has an input impedance of infinity.
Explanation:
Since reactance does not contribute thermal noise to Reason (R): A line of length and short circuited at
the circuit, it is a low noise device. far end behaves as a parallel resonant circuit.
Both A and R are correct and R is correct
A.
48. Assertion (A): PIN diode is commonly used for explanation of A
microwave control. Both A and R are correct but R is not correct
B.
Reason (R): A PIN diode uses heavily doped p and n explanation of A
materials. C. A is correct but R is wrong
Both A and R are correct and R is correct D.A is wrong but R is correct
A.
explanation of A A
Answer: Option A 1. A cavity resonator is
Explanation: A. a hollow metallic enclosure
A parallel tuned circuit has an infinite impedance if R a hollow enclosure having magnetic material as its
B.
is zero. walls
a hollow enclosure having dielectric material as its
C.
walls
D.either (b) or (c)
A
Explanation:
## Hollow metallic enclosures exhibit resonance
behaviour when excited by electromagnetic field.
## 2. If antenna diameter is increased four times, the
maximum range is increased by a factor of
A. 2
B. 2
C. 4
D.0.2
C
Explanation:
device.
## Reason (R): A Gunn oscillator uses the phenomenon
of transferred electron effect.
Both A and R are correct and R is correct
A.
explanation of A
Both A and R are correct but R is not correct
B.
explanation of A
C. A is correct but R is wrong
D.A is wrong but R is correct
MICROWAVE COMMS
INDIABIX B
SECTION 3
5. The component in the given figure is
Explanation:
## A Gunn diode uses GaAs which has a negative
differential mobility, i.e., a decrease in carrier velocity
with increase in electric field.
## This effects is called transferred electron effect. The
impedance of a Gunn diode is tens of ohms.
A. attenuator
B. T type low pass filter
A Gunn diode oscillator has a resonant cavity, an
C. T type high pass filter
arrangement to couple Gunn diode to cavity, biasing
D. phase shifter
arrangement for Gunn diode and arrangement to
couple RF power to load. B
## Applications of Gunn diode oscillator include Answer: Option B
4. Atomic and molecular resonance is observed in many Z01 provides series inductances and teflon filled low impedance
substances line Z02 provides shunt capacitance.
A. at microwave frequencies
B. at low frequencies
C. at frequencies used in AM broadcast 6. The semiconductor diode which can be used in switching circui
D.at both (b) and (c) microwave range is
A A. PIN diode
B. Varactor diode
Answer: Option A C. Tunnel diode
D. Gunn diode
Explanation: A
## Atomic and molecular distances are very small. Answer: Option A
Therefore resonance can occur only at microwave
frequencies. Explanation:
## A PIN diode has an intrinsic (i) layer between p and n layers.
When reverse bias is applied depletion layers are formed at p-i
and i-n junctions.
## The effective/width of depletion layer increases by the width o
layer. It can be used as a voltage controlled attenuator.
At high frequencies the rectification effect ceases and impedance
B. False
of diode is effectively that of i layer. B
## This impedance varies with the applied bias. It is used in high
frequency switching circuits, limiters, modulators etc. Answer: Option B
## 7. In the given figure the reflected current wave after Explanation:
first reflection is
The quantities required to be measured in microwave
circuits are frequency, power and impedance.
## Microwave frequency measurement is done by slotted
line, resonant cavities and transfer oscillator.
## Microwave power measurement uses bolometers and
micromave power meters.
## A. 0.18 A A self balancing bridge working on the principle of
B. -0.18 A power substitution is commomly used.
C. 0.3 A
D.-0.3 A Microwave impedance measurement is done by
measurement of reflection coefficient and VSWR.
B
## A slotted line and probe is a basic tool for these
Answer: Option B measurements.' Network analyser enables rapid
impedance measurement over a broad frequency
Explanation: range.
## 10. A reflex klystron oscillator uses
A. one cavity resonator
B. two cavity resonators
8. A line has an attenuation of 0.054 Np/m. The C. three cavity resonators
attenuation in decibels is D.none of the above
A. 4.7 dB/m A
B. 0.47 dB/m
C. 0.54 dB/m
B
Explanation:
Answer: Option B It uses a single cavity resonator for generating microwave osci
Explanation: Its parts are electron gun, resonator, repeller and output coup
0.054 x 8.68 = 0.47 dB/m. It operates on the principle of positive feed back.
9. If a line is open circuited Zin = Z0 tanh (gl). The repeller electrode is at negative potential and sends the p
A.True
This positive feedback supports oscillations. Its feature are: Explanation:
1. Frequency range - 2 to 100 GHz Power handling capacity (voltage)2 and is inversely
proportional to f2 max.
2. Power output - 10 MW to about 2 W
13. To couple a coaxial line to a parallel wire line it is best
3. Efficiency - 10 - 20 % to use
A. slotted line
B. balundevices, oscillator for microwave measurements in
laboratories etc. C. directional coupler
D./4 transformer
B
11. A coaxial RF cable has a characteristic impedance of 50 and C equal to 40 pF/m. The inductance is
A. 1 H/m
B. 10 H/m
C. 0.1 H/m
Explanation:
D.0.01 H/m
C A balun gives 4 : 1 impedance transformation.
## Answer: Option C 14. In a multicavity klystron amplifier the signal to be
amplified develops an ac voltage of signal frequency
Explanation: across the gap in buncher cavity
A.True
B. False
A
.
## 12. In a microwave coaxial line, the maximum operating Answer: Option A
frequency is f max and breakdown strength of
dielectric is Ed' Then maximum power handling Explanation:
capacity is proportional to
A Klystron is a vacuum tube used for
A. generation/amplification of microwaves.
## B. An electron beam is produced by oxide coated
indirectly heated cathode and is focussed and
accelerated by focussing electrode.
C.
This beam is transmitted through a glass tube. The
D. input cavity where the beam enters the glass tube is
called buncher.
B
As electrons move ahead they see an accelerating
field for half cycle and retarding field for the other
Therefore, some electrons are accelerated and some 16. In a circular waveguide TE21 mode has lowest cutoff
are retarded. This process is called velocity frequency
modulation. A.True
B. False
The velocity modulation causes bunching of B
electrons. This bunching effect converts velocity
modulation into density modulation of beam.
The input is fed at buncher cavity and output is taken
at catcher cavity. Explanation:
In a two cavity klystron only buncher and catcher Some applications require dual polarization
cavity are used. In multi cavity klystron one or more capability. Circular waveguide has this capability.
intermediate cavities are also used.
These analysis uses cylindrical coordinates.
The features of a multicavity klystron are :
In circular waveguide TE11 mode has the lowest cut
1. Frequency range - 0.25 GHz to 100 GHz off frequency and is the dominant mode.
## A multicavity klystron is used in UHF TV transmitters, c = 0.82 D for TE01 mode
c = 1.306 D for TM01 mode.
15. If VSWR is infinite, the transmission line is terminated
in 17. The number of TV channels which can be
A. short circuit accommodated in a spectrum of 300 MHz is about
B. complex impedance A. 5
C. open circuit B. 10
D.either (a) or (c) C. 50
D.200
D
C
Explanation:
Explanation:
## A TV channel has a bandwidth about 5 MHz. Hence
and . Since VSWR = 300 MHz spectrum can accommodate about 50
, |rv| = 1. This can happen if ZL = 0 or ZL = , i.e., channels.
line is o.c or s.c.
18. The noise figure of multicavity klystron amplifier is
very low
A.True 4. Efficiency - about 40%.
B. False
B A multicavity klystron is used in UHF TV transmitters,
Answer: Option B 19. Assertion (A): The velocity of electromagnetic waves
on overhead lines and coaxial cables is the same.
Explanation:
Reason (R): Free space has an intrinsic impedance of
A Klystron is a vacuum tube used for 377 ohms.
generation/amplification of microwaves. Both A and R are correct and R is correct
A.
explanation of A
An electron beam is produced by oxide coated Both A and R are correct but R is not correct
indirectly heated cathode and is focussed and B.
explanation of A
accelerated by focussing electrode. C. A is correct but R is wrong
D.A is wrong but R is correct
This beam is transmitted through a glass tube. The
D
input cavity where the beam enters the glass tube is
called buncher.
As electrons move ahead they see an accelerating
field for half cycle and retarding field for the other Explanation:
half cycle.
Velocity of em waves on caoxial cables is less than
Therefore, some electrons are accelerated and some that on overhead lines.
are retarded. This process is called velocity
modulation. 20. In the given figure the E and H lines in a coaxial cable
## The velocity modulation causes bunching of
electrons. This bunching effect converts velocity
modulation into density modulation of beam.
## The input is fed at buncher cavity and output is taken
at catcher cavity.
In a two cavity klystron only buncher and catcher A. The radial lines are E lines and circular lines are H lines
cavity are used. In multi cavity klystron one or more B. Radial lines are H lines and circular lines are E lines
intermediate cavities are also used. C. The directions of E and H lines are wrong
D.Radial lines may be E and H lines depending an direction of c
The features of a multicavity klystron are : A
## 1. Frequency range - 0.25 GHz to 100 GHz
2. Power output - 10 kW to several hundred kW
Explanation:
3. Power gain - 60 dB (nominal value)
H lines are concentric circles around conductor and E lines are
In circular waveguide TE11 mode has the lowest cut
21. Roughly the time required for microwave cooking as compared
offtofrequency
conventional
and cooking
is the dominant
is mode.
B. about half If D is diameter of waveguide
D.about one-tenth c = 1.706 D for TE11 mode
D
c = 1.029 D for TE21 mode
c = 0.82 D for TE01 mode
Explanation:
c = 1.306 D for TM01 mode.
Time required for microwave cooking is much less than the time required for conventional cooking.
24. Assertion (A): Klystron amplifiers use one or more
intermediate cavities in addition to buncher and
22. If the minimum range of a radar is to be doubled, the
catcher cavity.
peak power has to be increased by a factor of
A. 2
Reason (R): When one or more intermediate cavities
B. 4
are used the bandwidth can be increased.
C. 8
Both A and R are correct and R is correct
D.16 A.
explanation of A
D Both A and R are correct but R is not correct
B.
explanation of A
Answer: Option D C. A is correct but R is wrong
D.A is wrong but R is correct
Explanation: A
Range = (power)025.
23. Which mode has the lowest cut off frequency in
Explanation:
circular wave guides?
A. TE01
A Klystron is a vacuum tube used for
B. TE11
generation/amplification of microwaves.
C. TE20
D.TE21
An electron beam is produced by oxide coated
B indirectly heated cathode and is focussed and
accelerated by focussing electrode.
This beam is transmitted through a glass tube. The
Explanation: input cavity where the beam enters the glass tube is
called buncher.
Some applications require dual polarization
capability. Circular waveguide has this capability. As electrons move ahead they see an accelerating
field for half cycle and retarding field for the other
These analysis uses cylindrical coordinates. half cycle.
Therefore, some electrons are accelerated and some
are retarded. This process is called velocity Answer: Option A
modulation.
Explanation:
The velocity modulation causes bunching of
electrons. This bunching effect converts velocity It uses a single cavity resonator for generating microwave osci
modulation into density modulation of beam.
Its parts are electron gun, resonator, repeller and output coup
The input is fed at buncher cavity and output is taken
at catcher cavity. It operates on the principle of positive feed back.
In a two cavity klystron only buncher and catcher The repeller electrode is at negative potential and sends the p
cavity are used. In multi cavity klystron one or more
intermediate cavities are also used. This positive feedback supports oscillations. Its feature are:
## 2. Power output - 10 kW to several hundred kW 3. Efficiency - 10 - 20 %
3. Power gain - 60 dB (nominal value) Its applications include radar receivers, local oscillator in micro
laboratories etc.
27. Consider the following applications
A multicavity klystron is used in UHF TV transmitters,
Radar transmitter and satellite communication. 1. TV tuning
2. Active filter
25. A wave Em cos (bx - t) is a backward wave. 3. Microwave frequency multiplication
A.True
B. False In which of above can a varactor diode be used?
B A. 1 2 and 3
B. l and 2 only
C. 1 and 3 only
D.2 and 3 only
Explanation: A
## It is a forward wave because as t increases, x increases. Answer: Option A
Explanation:
26. A reflex klystron oscillator is a
A. low power device Varactor diode is used in all the three applications.
B. high power device
C. high efficiency device 28. Consider the following statements
D.both (a) and (b)
A
1. Dissipative attenuator has a fixed value of
2. Reflective attenuator has a fixed value of
attenuation. Explanation:
3. Both dissipative and reflective attenuators
are available only with fixed attenuation. A Klystron is a vacuum tube used for generation/amplification
4. Both dissipative and reflective attenuators
are available with either fixed or variable An electron beam is produced by oxide coated indirectly heate
attenuation. electrode.
Which of the above are correct? This beam is transmitted through a glass tube. The input cavity
A. 1 only
B. 1 and 2 As electrons move ahead they see an accelerating field for hal
C. 1, 2 and 3
D.4 only Therefore, some electrons are accelerated and some are retar
D
The velocity modulation causes bunching of electrons. This bu
Answer: Option D modulation of beam.
Explanation: The input is fed at buncher cavity and output is taken at catche
Both dissipative and reflective attenuators are In a two cavity klystron only buncher and catcher cavity are us
available with either fixed or variable attenuation. cavities are also used.
29. Impedance inversion may be obtained with The features of a multicavity klystron are :
A. a short circuited stub
B. a quarter wave line 1. Frequency range - 0.25 GHz to 100 GHz
C. an open circuited stub
D.a half wave line 2. Power output - 10 kW to several hundred kW
B
3. Power gain - 60 dB (nominal value)
## Impedance inversion occurs when load impedance is
not matched with characteristic impedance of line. 31. If a short line is terminated in characteristic impedance it beha
A.True
30. Power gain of a multicavity klystron amplifier is B. False
A. more than 30 dB
A
B. causes retardation of all electrons
C. causes acceleration of some electrons and retardation of others
D.none of the above Answer: Option A
A
Explanation:
In a line terminated by Z0, there are no reflections. Explanation:
32. Ga As exhibits negative differential mobility A Klystron is a vacuum tube used for
A.True generation/amplification of microwaves.
B. False
A An electron beam is produced by oxide coated
indirectly heated cathode and is focussed and
accelerated by focussing electrode.
This beam is transmitted through a glass tube. The
Explanation:
input cavity where the beam enters the glass tube is
called buncher.
A Gunn diode uses GaAs which has a negative
differential mobility, i.e., a decrease in carrier
As electrons move ahead they see an accelerating
velocity with increase in electric field.
field for half cycle and retarding field for the other
half cycle.
This effects is called transferred electron effect. The
impedance of a Gunn diode is tens of ohms.
Therefore, some electrons are accelerated and some
are retarded. This process is called velocity
A Gunn diode oscillator has a resonant cavity, an modulation.
arrangement to couple Gunn diode to cavity, biasing
arrangement for Gunn diode and arrangement to
The velocity modulation causes bunching of
electrons. This bunching effect converts velocity
modulation into density modulation of beam.
Applications of Gunn diode oscillator include
The input is fed at buncher cavity and output is taken
at catcher cavity.
33. Consider the following statements
In a two cavity klystron only buncher and catcher
cavity are used. In multi cavity klystron one or more
1. Bunching of electrons occurs in two cavity
intermediate cavities are also used.
klystron amplifier.
2. Bunching of electrons occurs in multi cavity
The features of a multicavity klystron are :
klystron amplifier.
3. Bunching of electrons occurs in reflex cavity
1. Frequency range - 0.25 GHz to 100 GHz
klystron amplifier.
2. Power output - 10 kW to several hundred kW
Which of the above statements are correct?
A. 1, 2, and only
3. Power gain - 60 dB (nominal value)
B. 1, 2 and 3 only
C. 1 and 3 only
D.2 and 3 only 4. Efficiency - about 40%.
B
A multicavity klystron is used in UHF TV transmitters,
34. Assertion (A): For high frequency lines inductive 36. Which one of the following is not a negative resistance device
reactance is very high as compared to ac resistance. A. Gunn diode
B. Tunnel diode
Reason (R): Due to skin effect ac resistance of line is C. Impatt diode
higher than dc resistance. D. Varactor diode
Both A and R are correct and R is correct D
A.
explanation of A
Both A and R are correct but R is not correct
explanation of A
C. A is correct but R is wrong
D.A is wrong but R is correct Explanation:
B
Gunn diode, Tunnel diode and Impatt diodes are negative
resistance devices.
The limitation of solid state devices at high frequencies includ
Explanation: those associated with transit time and junction capacitances.
XL = L. As frequency increases line resistance The devices used are : Transferred electron oscillators (Gunn
increases slightly but XL increases directly as per diode), Avalanche diode oscillators (Impatt diode, Trapatt dio
frequency. Masters, Lasers, Tunnel diode, Varactor etc).
35. Consider the following applications 37. Microwave links are typically 50 km apart
A. because of atmospheric attenuation
1. Switch B. because of output tube power limitation
2. Attenuator C. because of earth's curvature
3. Phase shifter D.to ensure that applied dc voltage is not excessive
4. Oscillator C
## In which of the above can a PIN diode be used?
1 2 3 and 4
.
B Explanation:
1 2 and 3 only
.
C Earth's curvature limits the distance between
1 and 2 only microwave links.
.
D
2, 3 and 4 only 38. For matching over a range of frequencies in a
.
transmission line it is best to use
B
A. a balun
B. a broad band directional coupler
Answer: Option B C. double stub
D.a single stub of adjustable position
Explanation: C
## PIN diode is not used in oscillators.
Explanation: 1. Frequency range - 1 GHz to 1000 GHz.
Double stub is useful for matching over a range of 2. Power output - 10 mV to 150 mW (continuous
frequencies. wave) 250kW (pulsed).
39. Assertion (A): In the interaction region of magnetron It is used as signal source in transmitters and
an electron is subjected to three forces: force due to instruments.
electric field, force due to magnetic field and
centrifugal force. 40. Which of the following is wrong for a magic used to tee?
A. E and H arms are decoupled
Reason (R): The mechanism of generation of B. coplanar arms are coupled
microwaves, in a magnetron, involves interaction of C. all ports are perfectly matched
electromagnetic fields with electrons moving in static D.A signal into coplanar arm splits equally between E and H ar
electric and magnetic fields oriented at right angles B
to each other.
Both A and R are correct and R is correct
explanation of A
Both A and R are correct but R is not correct
B. Explanation:
explanation of A
C. A is correct but R is wrong
D.A is wrong but R is correct Coplanar arms are decoupled.
41. The expression HZ = C cos(Bx) cos(Ay)yz is for
A
A. TE wave
B. TM waves
Answer: Option A C. both TE and TM waves
D.some TE and some TM waves
Explanation: A
## It is somewhat similar to TWT and can deliver
microwave power over a wide frequency band. Answer: Option A
## It has an electron gun and a helix structure. However Explanation:
the interaction between electron beam and RF wave
is different than in TWT. In TM mode Hz = 0.
The growing RF wave travels in opposite direction to 42. Assertion (A): Impedance matching can be done by
the electron beam. using stubs.
The frequency of wave can be changed by changing Reason (R): A double stub is used for impedance
the voltage which controls the beam velocity. matching when frequency of signal varies.
Both A and R are correct and R is correct
A.
Moreover the amplitude of oscillations can be explanation of A
decreased continuously to zero by changing the Both A and R are correct but R is not correct
B.
beam current. explanation of A
C. A is correct but R is wrong
D.A is wrong but R is correct
It features are:
A
Therefore the signal gets strengthened and amplified
Answer: Option A output is delivered at the other end of tube.
## Explanation: The main features of TWT are :
Hollow metallic enclosures exhibit resonance 1. Frequency range - 0.5 GHz to 90 GHz
behaviour when excited by electromagnetic field.
2. Power output - 5 mW at low frequencies(less than
These enclosures are called cavity resonators. 20 GHz) 250 kW (continuous wave) at 3 GHz 10 MW
(pulsed) at 3 GHz
3. Efficiency - about 5 to 20%
1. It uses thermionic emission.
2. It uses an attenuater. 4. Noise - about 5 dB for low power TWT 25 dB for
3. It is inherently a resonant device. high power TWT
TWT is used as RF amplifier in broadband microwave
Which of the above are correct? receivers, repeater amplifier in broad band
A. 1, 2, 3 and 4 communication systems, communication satellites
B. 1, 2 and 3 etc.
C. 1, 3 and 4
D.1, 2 and 4 44. If a line having Z0 = 300 0 W is short circuited at far
D end, VSWR is
A. 0
B. 1
D.2
Explanation:
C
## A TWT is a broadband device. Its main components Explanation:
are electron gun (to produce the electron beam) and
a structure supporting the slow electromagnetic If line is s.c |rv| = 1 and VSWR = .
wave.
45. A klystron amplifier generally uses Pierce gun
The velocity of wave propagation along the helix A.True
structure is less than velocity of light. B. False
A
The beam and wave travel along the structure at the
same speed.
Thus interaction occurs between beam and wave and
Explanation:
the beam delivers energy to the RF wave.
It uses a single cavity resonator for generating microwave This impedance varies with the applied bias. It is used in high
oscillations. frequency switching circuits, limiters, modulators etc.
Its parts are electron gun, resonator, repeller and output47. A pulsed radar produces 1 sec pulses at a rate of
coupling. 1000 per second. The duty cycle is
A. 0.001
It operates on the principle of positive feed back. B. 0.01
C. 0.1
The repeller electrode is at negative potential and sends the D.1
partially bunched electron beam back to resonator cavity. A
## This positive feedback supports oscillations. Its feature are:
1. Frequency range - 2 to 100 GHz
Explanation:
2. Power output - 10 MW to about 2 W
1 x 10-6 x 1000 = 0.001.
3. Efficiency - 10 - 20 %
48. Assertion (A): Impedance measurement at
microwave frequencies is done by finding SWR.
microwave devices, oscillator for microwave measurements in
Reason (R): SWR and reflection coefficient depend
laboratories etc.
on the characteristic impedance and load impedance.
46. Assertion (A): PIN diode is used as a fast switch.
Both A and R are correct and R is correct
A.
explanation of A
Reason (R): PIN diode has very high resistance when reverse Both A and R are correct but R is not correct
biased and very low resistance when forward biased. B.
explanation of A
A. Both A and R are correct and R is correct explanation of A C. A is correct but R is wrong
B. Both A and R are correct but R is not correct explanation ofD.AA is wrong but R is correct
C. A is correct but R is wrong
B
D.A is wrong but R is correct
A
Explanation:
Explanation:
## A PIN diode has an intrinsic (i) layer between p and n layers.
When reverse bias is applied depletion layers are formed at p-i
and i-n junctions. 49. When a line is loaded the characteristic impedance is
equal to
The effective/width of depletion layer increases by the widthA. (R + jL)(G + jC)
of i layer. It can be used as a voltage controlled attenuator. B.
## At high frequencies the rectification effect ceases and C.
impedance of diode is effectively that of i layer.
D.LC
B The velocity modulation causes bunching of
electrons. This bunching effect converts velocity
modulation into density modulation of beam.
The input is fed at buncher cavity and output is taken
Explanation: at catcher cavity.
## In a two cavity klystron only buncher and catcher
cavity are used. In multi cavity klystron one or more
intermediate cavities are also used.
or
The features of a multicavity klystron are :
.
1. Frequency range - 0.25 GHz to 100 GHz
## 2. Power output - 10 kW to several hundred kW
50. Which device has internal positive feedback?
A. Two cavity klystron amplifier 3. Power gain - 60 dB (nominal value)
B. Multi-cavity klystron amplifier
C. Reflex klystron amplifier 4. Efficiency - about 40%.
D.All of the above
C A multicavity klystron is used in UHF TV transmitters,
Explanation:
## A Klystron is a vacuum tube used for
generation/amplification of microwaves.
## An electron beam is produced by oxide coated
indirectly heated cathode and is focussed and MICROWAVE COMMS
accelerated by focussing electrode. INDIABIX
SECTION 4
This beam is transmitted through a glass tube. The
input cavity where the beam enters the glass tube is 1. The domain mode in rectangular waveguide is
called buncher. A. TE01
B. TE02
As electrons move ahead they see an accelerating C. TE10
field for half cycle and retarding field for the other D.TM11
half cycle. C
Therefore, some electrons are accelerated and some
are retarded. This process is called velocity
2. The reverse voltage applied to a varactor diode
modulation.
A. is less than avalanche breakdown voltage
B. is more than avalanche breakdown voltage
may be more or less than a avalanche breakdown Answer: Option D
C.
voltage
is very high as compared to avalanche breakdown 6. In a TWT the amplitude of resultant wave travelling
D.
voltage down the helix remains constant
A A.True
B. False
Explanation: B
The action of varactor diode is due to the capacitance Answer: Option B
of depletion layer. The thickness of depletion layer Explanation:
depends on the reverse bias which should not cause The signal gets strengthened.
breakdown.
7. In the given figure forward current wave has a
magnitude of
3. To ensure that only dominant mode TE10 is allowed to
propagate in an air filled rectangular waveguide, the
lower frequency limit and upper frequency limit are
about 25% above fc for TE10 and 5% below fc for TE20
A.
mode respectively
about 50% above fc for TE10 and 25% below fc for
B.
TE20 mode respectively
about 25% above fc for TE10 and 25% below fc for
C.
TE20 mode respectively
A. 0.45 A
D.none of the above
B. 0.3 A
A C. 0.2 A
B
4. The skin depth at 1000 MHz as compared to that at Answer: Option B
500 MHz is
Explanation:
A. 2
B. 12
C. 0.707
D.0.5
8. In an infinite line the input impedance at every point is
C equal to characteristic impedance.
B. False
5. The transit time (in cycles) for electrons in repeller A
space of reflex klystron oscillator for sustaining
oscillations is (n is any integer)
A. 2(n - 1)
9. Consider the following time parameters
B. 2n - 1
Domain growth time constant.
C. Transit time.
Dielectric relaxation time in positive mobility regime.
D. Which of the above are used in connection width
transferred electron device?
D A. 1, 2, and 3
B. 1 and 2 only
C. 1 and 3 only 14. In the given figure shows the equivalent circuit of a
D.2 and 3 only magic tee. If all ports are matched
A
## 10. PIN diode has
A. p+ and n layers separated by i layer
B. p+ and n+ layers separated by i layer
C. p- and n- layers separated by i layer
D.none of the above
B A. nE = nH = 2
B. nE = nH = 0.707
C. nE = 2, nH = 0.707
D.nH = 0.707, nH = 2
11. The two terms used to describe performance of a
directional coupler are A
A. coupling and directivity Answer: Option A
B. gain and coupling
C. gain and directivity
D.gain and isolation 15. Assertion (A): Microstrip is very commonly used in
A microwave integrated circuits.
surface so that active and passive discrete
components can be easily mounted.
12. The number of PM radio channels which can be
Both A and R are correct and R is correct
accommodated in a spectrum of 300 MHz is about A.
explanation of A
A. 10
Both A and R are correct but R is not correct
B. 100 B.
explanation of A
C. 500
C. A is correct but R is wrong
D.1500
D.A is wrong but R is correct
D
A
Explanation:
13. Consider the following statements
A Microstrip line has a single dielectric substratc with
Gunn diode can be used in combinational logic
ground plane on one side and a strip on the other
circuits.
face.
Gunn diode can be used in sequential logic circuits.
Both Gunn diode and GaAs Mesfet can be used in
logic circuits.
fabrication.
Which of the above statements are correct?
The high dielectric constant of the substrate reduces
A. 1, 2, and 3
guide wavelength and circuit dimensions.
B. 1 and 2 only
A microstrip line is the most commomly used
C. 2 and 3 only
transmission structure for microwave integrated
D.1 and 3 only
circuits.
A
16. A 75 ohm line is first short terminated and minima D.(pulse width) + (PRF)
locations are noted. Then the short is replaced by A
resistive load and minima location are again noted. If
minima location are not altered and VSWR is 3, the Answer: Option A
A. 25 20. In a klystron amplifier the bunching effect
B. 50 converts velocity modulation into current
A.
C. 225 modulation of beam
D.250 converts current modulation into velocity
B.
modulation of beam
C C. both (c) and (b)
Answer: Option C D.neither (a) nor (b)
Explanation: A
rv and VSWR . Explanation:
A Klystron is a vacuum tube used for
17. The word 'LORAN' means generation/amplification of microwaves.
A. long range navigation An electron beam is produced by oxide coated
B. long range TV transmission indirectly heated cathode and is focussed and
C. long range cable transmission accelerated by focussing electrode.
D.either (b) or (c) This beam is transmitted through a glass tube. The
A input cavity where the beam enters the glass tube is
called buncher.
field for half cycle and retarding field for the other
18. In the given figure forward voltage wave has a half cycle.
magnitude Therefore, some electrons are accelerated and some
are retarded. This process is called velocity
modulation.
The velocity modulation causes bunching of
electrons. This bunching effect converts velocity
modulation into density modulation of beam.
The input is fed at buncher cavity and output is taken
at catcher cavity.
In a two cavity klystron only buncher and catcher
cavity are used. In multi cavity klystron one or more
A. 90 V intermediate cavities are also used.
B. 60 V The features of a multicavity klystron are :
C. 30 V 1. Frequency range - 0.25 GHz to 100 GHz
D.10 V 2. Power output - 10 kW to several hundred kW
C 3. Power gain - 60 dB (nominal value)
A multicavity klystron is used in UHF TV transmitters,
19. The duty cycle of a radar transmitter is equal to
A. (PRF) (pulse width)
21. Out of modes TE20 and TE30 of propagation of
B. (PRF)/(pulse width)
electromagnetic engery
C. (pulse width)/(PRF)
A. both have the same cutoff frequency
TE20 has lower cut off frequency as compared to The input is fed at buncher cavity and output is taken
B.
TE30 at catcher cavity.
TE30 has lower cutoff frequency as compared to In a two cavity klystron only buncher and catcher
C.
TE20 cavity are used. In multi cavity klystron one or more
D.either (a) or (c) intermediate cavities are also used.
B The features of a multicavity klystron are :
1. Frequency range - 0.25 GHz to 100 GHz
Answer: Option B 2. Power output - 10 kW to several hundred kW
3. Power gain - 60 dB (nominal value)
22. A TE10 rectangular waveguide is to be designed for 4. Efficiency - about 40%.
operation over 25-35 GHz and the band centre is 1.5 A multicavity klystron is used in UHF TV transmitters,
times the cutoff frequency. The dimension of Radar transmitter and satellite communication.
A. 15 mm 24. If load impedance ZL >> Z0, then
B. 10 mm A. Z0 = ZL (VSWR)
C. 9 mm B. ZL = Z0 (VSWR)
D.7.5 mm C. ZL = Z0 (VSWR - 1)
D D.ZL = Z0 VSWR
23. In a klystron amplifier the force on all electrons Explanation:
A. causes acceleration of all electrons
B. causes retardation of all electrons
causes acceleration of some electrons and If ZL >> Z0, |rv| and VSWR .
C.
retardation of others Therefore ZL = Z0(VSWR).
D.none of the above
C 25. The inductance of a twin feeder used to connect
A. 0.2 H/m
Explanation:
B. 0.6 H/m
A Klystron is a vacuum tube used for
C. 1.6 H/m
generation/amplification of microwaves.
D.10 H/m
An electron beam is produced by oxide coated
indirectly heated cathode and is focussed and C
accelerated by focussing electrode. Answer: Option C
This beam is transmitted through a glass tube. The
input cavity where the beam enters the glass tube is 26. When resonant cavities are coupled together the
called buncher. result is
As electrons move ahead they see an accelerating A. one resonant frequency
field for half cycle and retarding field for the other B. n resonant frequencies
half cycle. C. resonant frequencies
Therefore, some electrons are accelerated and some
are retarded. This process is called velocity D.either (b) or (c)
modulation. B
The velocity modulation causes bunching of Answer: Option B
electrons. This bunching effect converts velocity
modulation into density modulation of beam.
27. For a rectangular waveguide having width a and B. 0.01
height b, the cutoff wavelength for TM11 mode is C. 0.05
equal to D.0.25
A. D
B.
30. As wavelength decrease the size of high directivity
C. antenna
A. decreases
D. B. increases
C. is not affected
B D.either (b) or (c)
28. The path length between ports 1 and 2 for the two
waves in a hybrid ring is
31. Which of the following modes of transmission will
A. 1
not be supported by a rectangular waveguide?
B. A. TE15
B. TE12
C. C. TM11
D.TM10
D. D
C
Answer: Option C 32. In a circular waveguide with radius r, the dominant
Explanation: mode is
The quantities required to be measured in A. TM01
microwave circuits are frequency, power and B. TE01
impedance. C. TM11
Microwave frequency measurement is done by D.TE11
slotted line, resonant cavities and transfer oscillator.
B
Microwave power measurement uses bolometers
and microwave power meters. Answer: Option B
A self balancing bridge working on the principle of Explanation:
power substitution is commonly used. Some applications require dual polarization
Microwave impedance measurement is done by capability. Circular waveguide has this capability.
measurement of reflection coefficient and VSWR. These analysis uses cylindrical coordinates.
A slotted line and probe is a basic tool for these In circular waveguide TE11 mode has the lowest cut
measurements.' Network analyser enables rapid off frequency and is the dominant mode.
impedance measurement over a broad frequency If D is diameter of waveguide
range. c = 1.706 D for TE11 mode
c = 1.029 D for TE21 mode
29. The radiation resistance of a circular loop of one turn c = 0.82 D for TE01 mode
is 0.01 ohm. For 5 turn loop the radiation resistance c = 1.306 D for TM01 mode.
is
A. 0.002
33. If 'a' is the width of rectangular wave guide and '' is for single diode circuit and upto 40 W for
the wave length, then combination of several diodes), efficiency about 20%.
A. Its applications include police radar systems, low
power microwave transmitter etc.
B.
C. 36. Which one of the following can be used for
amplification of microwave energy?
D. A. TWT
B
B. Magnetron
C. Reflex klystron
A
34. A parametric amplifier uses
A. non linear resistance Answer: Option A
B. non linear reactance
C. either (a) or (b) 37. If ZTE is wave impedance for TE waves, Ed is maximum
D.neither (a) nor (b) dielectric strength of insulating material, a and b are
B
the width and height of a rectangular wave guide, the
maximum power handling capability Pmax for TE10
Explanation:
In a parametric amplifier the non-linear element is A.
varactor diode or inductor.
B.
35. Assertion (A): Impatt diode can be used in both
amplifiers and oscillators.
C.
Reason (R): Impatt diode has a low resistance.
Both A and R are correct and R is correct
A. D.
explanation of A
Both A and R are correct but R is not correct
B. B
explanation of A
C. A is correct but R is wrong Answer: Option B
D.A is wrong but R is correct
38. At frequencies of 10 MHz and higher the impedance
B
represented by ground is primarity
Explanation: B. inductive
An Impatt diode has n+ - p - i - p + structure and is C. capacitive
used with reverse bias. D.partly resistive and partly capacitive
It exhibits negative resistance and operates on the
C
principle of avalanche breakdown.
circuit, low Q circuit and high Q circuit.
The impedance of Impatt diode is a few ohms. The 39. A disadvantage of microstrip line as the connections
word Impatt stands for Impact Avalanche Transit of compared to strip line is that former
Time diode. A. do not lend themselves to printed circuit
The features of Impatt diode oscillator are : B. are more likely to radiate
frequency 1 to 300 GHz, Power output (0.5 W to 5 W C. are bulky
D.are more expensive and complex
Explanation:
Answer: Option B The operation of magnetron is based on interaction
of electromagnetic fields with electrons moving in
40. As frequency increases, the transmission efficiency static electric and magnetic fields oriented at 90
A. decreases with respect to each other.
B. increases In a magnetron three forces act on electron viz. force
C. is not affected due to electric field (equal to - eE), force due to
D.either (a) or (c) magnetic field [equal to - e (v x B)] and centrifugal
B force (equal to mv2/r).
Answer: Option B The path of the electron can be found by balancing
these three forces. Magnetron can be of three types
41. A circular polarizer converts a linearly polarized wave i.e., negative resistance device, cyclotron frequency
into a circularly polarised wave. device and travelling wave or cavity device.
A.True In negative resistance magnetron use is made of the
B. False negative resistance between two anode segments.
It has low efficiency and is used at frequencies less
A
than 0.5 GHz.
Answer: Option A In cyclotron magnetron a synchronism exists
between ac component of electric field and periodic
42. If D is diameter of circular waveguide the cutoff oscillation of electrons in a direction parallel to the
wave-length for TE11 mode is equal to field.
A. 1.706 D It is used for frequencies higher than 100 MHz.
B. 2.11 D A cavity magnetron has a number of cylindrical
C. 0.82 D cavities in the interaction region.
D.0.41 D The cavity magnetron is the most common type of
A magnetron. Its features are :
1. Frequency range - 500 MHz to 10 GHz
2. Power output - 250 kW (pulses)
Explanation:
Some applications require dual polarization
It is used in radar systems, industrial heating systems
capability. Circular waveguide has this capability.
and microwave ovens.
These analysis uses cylindrical coordinates.
In circular waveguide TE11 mode has the lowest cut
44. Varactor diodes use
off frequency and is the dominant mode.
A. silicon
If D is diameter of waveguide
B. GaAs
c = 1.706 D for TE11 mode
C. either (a) or (b)
c = 1.029 D for TE21 mode
D.neither (a) nor (b)
c = 0.82 D for TE01 mode
c = 1.306 D for TM01 mode. C
43. The external magnetic field in a magnetron is such
that lines are 45. A circular waveguide carries TE11 mode whole radial
A. parallel to the axis of cathode electric field is given by
B. perpendicular to the axis of cathode Er = E0j1(r) sin V/m
C. inclined to the axis of cathode where r is radial distance in cm from the axis
D.either (b) or (c) The cut off wavelength is
A A. 10 cm
B. 3 p cm power transmission is least efficient when there
A.
C. 2 p cm are no standing waves on the line
D.8 cm power transmission is most efficient when there
B.
A are no standing waves on line
C. Load power depends on phase constant
Answer: Option A D.standing waves will always exist on the line
46. Assertion (A): Impatt diode is an avalanche diode. B
Reason (R): Avalanche breakdown phenomenon Answer: Option B
occurs when a p-n junction is reverse biased. Explanation:
Both A and R are correct and R is correct If standing waves are not there, whole of the power
A.
explanation of A is absorbed by load.
Both A and R are correct but R is not correct
B.
explanation of A 49. In the given figure a short circuited transmission line
C. A is correct but R is wrong resonator If n = 1, 2, 3...
D.A is wrong but R is correct
B
Explanation:
An Impatt diode has n+ - p - i - p + structure and is
used with reverse bias.
It exhibits negative resistance and operates on the For series reasonance l = np and for parallel
principle of avalanche breakdown. A.
Impatt diode circuits are classified as broadly tunable resonance
circuit, low Q circuit and high Q circuit.
The impedance of Impatt diode is a few ohms. The
B. For series resonance and for
word Impatt stands for Impact Avalanche Transit
Time diode. parallel resonance l = np
The features of Impatt diode oscillator are : C. l = np for both series and parallel resonance
frequency 1 to 300 GHz, Power output (0.5 W to 5 W
for single diode circuit and upto 40 W for D. for both series and parallel
combination of several diodes), efficiency about 20%. resonance
Its applications include police radar systems, low
A
power microwave transmitter etc.
47. A radar has a maximum range of 120 km. The
maximum allowable pulse repetition frequency for 50. The electric field in a TWT due to applied signal
unambiguous reception is A. is directed along the helix axis
A. 1250 B. is directed radially from helix axis
B. 330 C. is inclined to the helix axis by about 60
C. 2500 D.is inclined to the helix axis by about 45
D.8330 A
In a klystron the resonant structure limits the
bandwidth.
48. In a line with finite attenuation
A TWT is a broadband device. Its main components
are electron gun (to produce the electron beam) and
a structure supporting the slow electromagnetic
wave.
The velocity of wave propagation along the helix
structure is less than velocity of light.
The beam and wave travel along the structure at the
same speed.
Thus interaction occurs between beam and wave and
the beam delivers energy to the RF wave.
Therefore the signal gets strengthened and amplified
output is delivered at the other end of tube.
The main features of TWT are :
1. Frequency range - 0.5 GHz to 90 GHz
2. Power output - 5 mW at low frequencies(less than
20 GHz) 250 kW (continuous wave) at 3 GHz 10 MW
(pulsed) at 3 GHz
3. Efficiency - about 5 to 20%
4. Noise - about 5 dB for low power TWT 25 dB for
high power TWT
TWT is used as RF amplifier in broadband microwave | 24,067 | 93,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.85837 |
https://kat1055.com/algebra-1-staar-test-2022-answer-key/ | 1,695,803,287,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510284.49/warc/CC-MAIN-20230927071345-20230927101345-00475.warc.gz | 357,580,679 | 12,048 | # Algebra 1 Staar Test 2022 Answer Key: Everything You Need To Know
## Introduction
If you are a student who has just taken the Algebra 1 STAAR test in 2022, you must be eagerly waiting for the answer key to be released. This article will provide you with all the information you need to know about the Algebra 1 STAAR test 2022 answer key.
### Why is the Algebra 1 STAAR Test Important?
The Algebra 1 STAAR test is a standardized test taken by students in Texas to assess their knowledge and skills in Algebra 1. This test is an essential part of the Texas Education Assessment program and is used to determine the academic progress of students. The results of the test are also used by schools and districts to evaluate the effectiveness of their teaching methods.
### When Will the Algebra 1 STAAR Test 2022 Answer Key be Released?
The answer key for the Algebra 1 STAAR Test 2022 is expected to be released in the month of June. The exact date of release has not yet been announced by the Texas Education Agency (TEA).
### How to Access the Algebra 1 STAAR Test 2022 Answer Key?
The Algebra 1 STAAR Test 2022 answer key will be available on the TEA website. Students can access the answer key by logging in to their student portal using their unique identification number (UIN) and date of birth.
### How to Interpret the Algebra 1 STAAR Test 2022 Answer Key?
The Algebra 1 STAAR Test 2022 answer key will contain the correct answers to all the questions asked in the test. Students can use the answer key to evaluate their performance and identify areas where they need to improve. The answer key will also provide an idea of how the student has performed in comparison to other students who have taken the test.
### Tips for Using the Algebra 1 STAAR Test 2022 Answer Key
Here are some tips for using the Algebra 1 STAAR Test 2022 answer key: 1. Use the answer key to identify your strengths and weaknesses in Algebra 1. 2. Focus on the questions you got wrong and try to understand why you made the mistake. 3. Use the answer key to practice similar questions and improve your skills. 4. Use the answer key to compare your performance with that of other students who have taken the test.
### Conclusion
The Algebra 1 STAAR Test 2022 answer key is an essential tool for students to evaluate their performance and identify areas where they need to improve. By following the tips mentioned above, students can use the answer key to their advantage and achieve better results in the future. | 551 | 2,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-40 | longest | en | 0.925129 |
https://groups.yahoo.com/neo/groups/aima-talk/conversations/topics/674?o=0&var=1&l=1 | 1,498,605,332,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321938.75/warc/CC-MAIN-20170627221726-20170628001726-00281.warc.gz | 757,404,924 | 18,531 | Sorry, an error occurred while loading the content.
## an error at chapter 15, page 556, equation 15.20
Expand Messages
• �Ut+1 = (I - (Kt+1))(F(�Ut)(F^T) + (�Ux)) == �Ut+1 = (I - (Kt+1)H)(F(�Ut)(F^T) + (�Ux))
Message 1 of 1 , Apr 23, 2006
£Ut+1 = (I - (Kt+1))(F(£Ut)(F^T) + (£Ux))
==>
£Ut+1 = (I - (Kt+1)H)(F(£Ut)(F^T) + (£Ux))
Your message has been successfully submitted and would be delivered to recipients shortly. | 183 | 421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.772042 |
http://www.mywordsolution.com/question/a-random-sample-is-obtained-from-the-normal/916709 | 1,485,219,721,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560283475.86/warc/CC-MAIN-20170116095123-00354-ip-10-171-10-70.ec2.internal.warc.gz | 585,421,425 | 8,169 | +1-415-315-9853
info@mywordsolution.com
## Statistics
A random sample is obtained from the normal population with mean of µ = 80 and standard deviation of σ = 8. Which of following outcomes is more likely? Describe your answer.
A. A sample mean greater than 86 for a sample of n = 4 scores.
B. A sample mean greater than 84 for a sample of n = 16 scores.
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M916709
Have any Question?
## Related Questions in Statistics and Probability
### Spss assignment 02the data set you need to do the
SPSS Assignment 02 The data set you need to do the assignment can be found on Blackboard in the folder "Assignments and due dates" and in the spreadsheet "Data for Assignment 2". This is a group assignment and you can su ...
### Cnsider selecting one card at a time from a 52-card deck
Consider selecting one card at a time from a 52-card deck. What is the probability that the first card is from the suit of hearts and the second card is also from the suit of hearts? (Note there are 13 cards from the sui ...
### Twenty years ago 40 children were born in the same hospital
Twenty years ago 40 children were born in the same hospital in Denmark. Researchers classified the children by whether or not they were breast-fed for at least 3 months. They were interested in whether breast-feeding ten ...
### The annual commissions earned by sales representatives of
The annual commissions earned by sales representatives of Machine Products, Inc. a manufacturer of light machinery, follow the normal probability distribution. The mean yearly amount earned is \$40,000 and the standard d ...
### A caterer is competing for a companys business the caterer
A caterer is competing for a company's business. The caterer randomly selects a past client and submits a tasting menu consisting of all the dishes prepared for the past client. What type of sample is this? A college fac ...
### Question -after collecting the data a forecaster plans on
Question - After collecting the data, a forecaster plans on initially estimating the following model y = β 0 + i=1 ∑ 8 β i X i + β 9 (X 5 *X 7 ) +e, then dropping all of the X variables that are not significant at the 0. ...
### The mean incubation time for a type of fertilized egg kept
The mean incubation time for a type of fertilized egg kept at 100.7° F is 22days. Suppose that the incubation times are approximately normally distributed with a standard deviation of 2 days. (a) The probability that a r ...
### 1 what does the reliability coefficient of a measure
1. What does the reliability coefficient of a measure indicate if it is .60? .OO? 1.00? 2. What does a correlation coefficient tell us? Why are correlation coefficients useful when assessing reliability? 3. What are the ...
### A philosophy professor assigns letter grades on a test
A philosophy professor assigns letter grades on a test according to the following scheme. A: Top 12% of scores B: Scores below the top 12% and above the bottom 63% C: Scores below the top 37% and above the bottom 22% D: ...
### When how and why was israel created discuss the six day
When, how, and why was Israel created? Discuss the Six Day War and the Yom Kipper War from both the Israeli and Arab points of view. Who are the most important leaders and personalities in the Middle East today?How might ...
• 4,153,160 Questions Asked
• 13,132 Experts
• 2,558,936 Questions Answered
## Looking for Assignment Help?
Start excelling in your Courses, Get help with Assignment
Write us your full requirement for evaluation and you will receive response within 20 minutes turnaround time.
Ask Now Help with Problems, Get a Best Answer
### Section onea in an atwood machine suppose two objects of
SECTION ONE (a) In an Atwood Machine, suppose two objects of unequal mass are hung vertically over a frictionless
### Part 1you work in hr for a company that operates a factory
Part 1: You work in HR for a company that operates a factory manufacturing fiberglass. There are several hundred empl
### Details on advanced accounting paperthis paper is intended
DETAILS ON ADVANCED ACCOUNTING PAPER This paper is intended for students to apply the theoretical knowledge around ac
### Create a provider database and related reports and queries
Create a provider database and related reports and queries to capture contact information for potential PC component pro
### Describe what you learned about the impact of economic
Describe what you learned about the impact of economic, social, and demographic trends affecting the US labor environmen | 1,031 | 4,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-04 | longest | en | 0.949079 |
https://aipapa44.com/2020/10/27/casino-angels-bonus-demons/ | 1,606,570,255,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195656.78/warc/CC-MAIN-20201128125557-20201128155557-00171.warc.gz | 181,454,779 | 8,281 | # Casino Angels & Bonus Demons
For example, as a player, can be along with a three and an ace, bringing your cards’ total value to 4, and then you are dealt one more card. To win the game, the total of your cards always be nine in the most. Suppose, you are dealt an ace, a 2 and a three, bringing your card total in order to six whereas the dealership gets an ace using a 4 giving him a card total of five, then the the successful. But, if your total exceeds nine, may well mean damage to you.
Having less decks changes the odds very slightly in favor of household on tie bets. Regarding any 6-deck game the house edge is 1.06% on a banker bet, 1.24% for that player bet, and eighteen.44% on the tie. Whereas in the eight deck version the house edge is 1.06% throughout the banker bet, 1.24% on the player bet, and fourteen inches.36% on the tie.
Here’s a good example of how are you affected if your total surpasses nine. Imagine you have received an ace and a three in they dealt to you, one more card, which might be an eight, thus giving you an absolute of an even dozen. Don’t forget inside of rule is get nearest to being unfaithful. If your total exceeds nine, the first digit is dropped! m88up That is, twelve is treated as only two and in case the dealer gets some thing than two but below nine, he wins the bet.
Rules Of Baccarat To know which value of any hand, we add diet plan the cards in the hand. When the total is a one-digit number, then here is the total value of the part. However, if the total is a two-digit number, the value of the second digit may be the value belonging to the hand. For example, in the event the total from the two-digit number is 18, then the value of the hands is 8, if the total is 17, the associated with the hand is more effective. Please notice that the lowest value a Baccarat hand may have is zero, and the finest is trying to find. Any two-card hand having a associated with 8 or 9 is called natural.
First option to take is regarding acquainted this particular game’s end goal. What is the supposed associated with the credit cards? Are you winning? Ask these questions to yourself before and during play. Here’s the trick: you require a combined value of two cards close to 9 during the banker’s cards. Additionally you need to remember that each card has specific to it corresponding cost. For example, an Ace can be 1 or 11. | 570 | 2,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-50 | latest | en | 0.950153 |
http://www.ck12.org/arithmetic/Decimal-Rounding-on-a-Number-Line/?by=all&difficulty=all | 1,448,529,231,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398446997.59/warc/CC-MAIN-20151124205406-00356-ip-10-71-132-137.ec2.internal.warc.gz | 361,804,958 | 16,695 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Decimal Rounding on a Number Line
## Rounding according to proximity on number line
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Decimal Rounding on a Number Line
Discover how to use a number line to round decimals.
1
## Decimal Rounding on a Number Line
Round decimals using a number line.
0
## Decimal Rounding on a Number Line
Round decimals using a number line.
0
• PLIX
## Decimal Rounding on a Number Line: Rounding a Runner
Decimal Rounding on a Number Line Interactive
0
• Video
## Rounding Decimals
by CK-12 //basic
Learn to round numbers to the nearest decimal value.
0
• Video
## Examples: Rounding Decimals
by CK-12 //basic
This video provides an example of rounding a given decimal to different place values.
0
• Practice
0%
2
## Decimal Rounding on a Number Line Double Entry Diary
Connect your daily life experiences with specific points in a reading using a Double Entry Diary.
0
## Decimal and Whole Number Jeopardy
by Kyle Bardman //basic
1
• Flashcards
## Decimal Rounding on a Number Line
by Tiffany Kwok //basic | 383 | 1,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2015-48 | longest | en | 0.806219 |
https://www.rosettacode.org/wiki/Talk:Special_factorials | 1,653,652,613,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00004.warc.gz | 1,117,735,685 | 6,609 | I'm working on modernizing Rosetta Code's infrastructure. Starting with communications. Please accept this time-limited open invite to RC's Slack.. --Michael Mol (talk) 20:59, 30 May 2020 (UTC)
# Talk:Special factorials
### Reverse factorial algorithm
I took a stab at translating the reverse factorial algorithm used in the Factor entry to Java. It should be almost as efficient as taking the factorial itself.
public static int rf(int n) {
if (n == 1)
return 0; //1 has two answers -- return the lower one
int a = 1;
int b = 1;
while (n > a) {
b++;
a = a * b;
}
if (a == n)
return b;
else return -1; //undefined
}
--Chunes (talk) 17:06, 16 March 2021 (UTC)
Note that the factorial inverse (or reverse factorial) of unity has two possible answers: zero and unity.
It is normal when searching a series (in this case, the series of factorial products) to use the first match found in the series. -- Gerard Schildberger (talk) 17:37, 16 March 2021 (UTC)
Good catch. I revised the algorithm above and will make a note about it in the task description. --Chunes (talk) 17:58, 16 March 2021 (UTC)
### Why is af(0) 0?
Is it that 0 is the additive identity the way that factorial of 0 is 1 is the multiplicative identity? If so why should it be the identity anyway?
I think it's just that the formula literally produces 0 for n = 0.
First run through, i is 1, this produces -1. Second run through, i is 0, this produces 1 and their sum is 0. --Chunes (talk) 17:36, 16 March 2021 (UTC)
So the math summation notation goes backwards automatically when i > n? This may make a good task -- C will do one iteration and stop, etc. --Wherrera (talk) 21:11, 16 March 2021 (UTC) | 481 | 1,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-21 | latest | en | 0.970602 |
http://physicsgoeasy.blogspot.com/2008/04/waves-concept_17.html | 1,490,526,623,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189214.2/warc/CC-MAIN-20170322212949-00148-ip-10-233-31-227.ec2.internal.warc.gz | 282,374,471 | 22,849 | ## Pages
### Waves Concept
PART2
Interference of waves:-
-From principle of superposition we know that overlaping waves algbrically add togather to produce a net wave without altering the way of each other or the individual waves.
-If two sinusoidal waves of the same amplitude and wavelength travell in the same direction they interfere to produce a resultant sinusoidal wave travelling in that direction.
-The resultant wave due to interference of two sinusoidal waves is given by the relation
y′(x,t)=[2Amcos(υ/2)]sin(ωt-kx+υ/2)where υ is the phase difference between two waves.
-If υ=0n then there would be no phase difference between the travelling waves and the interference would be fully constructive.
-If υ=π then waves would be out of phase and there interference would be distructive.
Reflection of waves:-
-When a apulse or travelling wave encounters any boundary it gets reflected.
-If the boundary is not completely rigid then then a part of wave gets reflected and rest of it's part gets transmitted or refracted.
-A travelling wave at a rigid boundary is reflected with a phase reversal but the reflection at open boundary takes place without phase change.
-if an incident wave is represented by
yi(x,t)=A sin(ωt-kx)then reflected wave at rigid boundary is
yr(x,t)=A sin(ωt+kx+π)
=-Asin(ωt+kx)
and for reflections at open boundary reflected wave is given by
yr(x,t)=Asin(ωt+kx)
Standing waves:-
-The interference of two identical waves moving in opposite directions produces standing waves.
-For a string with fixed ends standing wave is given by
y(x,t)=[2Acos(kx)]sin(ωt)above equation does not represent travelling wave since it does not have characterstic form involving (ωt-kx) or (ωt+kx) in the argument of trignometric function.
-In standing waves amplitude of waves is different at different points i.e., at nodes amplitude is zero and at antinodes amplitude is maximumwhich is equal to sum of amplitudes of constituting waves.
-At intermediate points amplitude of wave varies between these two limits of maxima and minima
Normal modes of stretched string:--Frequency of transverse motion of stretched string of length L fixed at both the ends is given by
f=nv/2L
where n=1,2,3,4,.......
-The set of frequencies given by above relation are called normal modes of oscillation of the system.
-The mode with n=1 is called the fundamental mode with frequancy
f1=v/2L-Similarly second harmonic is the oscillation mode with n=2 and so on.
-Thus the string has infinite number of possible frequency of viberation which are harmonics of fundamental frequency f1 such that fn=nf1 | 610 | 2,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-13 | longest | en | 0.890968 |
http://oeis.org/A144226 | 1,550,743,707,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247503844.68/warc/CC-MAIN-20190221091728-20190221113728-00574.warc.gz | 194,659,987 | 4,010 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A144226 Prime numbers containing an equal number of odd and even digits. 6
23, 29, 41, 43, 47, 61, 67, 83, 89, 1009, 1021, 1049, 1061, 1063, 1069, 1087, 1201, 1223, 1229, 1249, 1283, 1289, 1409, 1423, 1427, 1429, 1447, 1481, 1483, 1487, 1489, 1601, 1607, 1609, 1621, 1627, 1663, 1667, 1669, 1801, 1823, 1847, 1861, 1867, 1889 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Can it be proved that this sequence has relative density 0 in the primes? Numbers with equal numbers of even and odd decimal digits have k * n/sqrt(log(n)) members up to n (k varies by upper or lower density). - Charles R Greathouse IV, Nov 12 2010 LINKS T. D. Noe, Table of n, a(n) for n = 1..10000 FORMULA A000040 INTERSECTION A227870. - Jonathan Vos Post, Nov 04 2013 EXAMPLE The prime 1889 contains an equal number of odd and even digits. MATHEMATICA fQ[n_] := Block[{id = IntegerDigits[n]}, Length[Select[id, OddQ]] == Length[Select[id, EvenQ]]]; Select[Prime[Range[300]], fQ] (* Robert G. Wilson v, Sep 24 2008 *) eoQ[n_]:=Module[{idn=IntegerDigits[n]}, Count[idn, _?OddQ]==Count[ idn, _?EvenQ]]; Select[Prime[Range[300]], eoQ] (* Harvey P. Dale, Mar 07 2017 *) CROSSREFS Cf. A000040, A144205, A227870. Sequence in context: A156983 A230456 A174196 * A173709 A225319 A228139 Adjacent sequences: A144223 A144224 A144225 * A144227 A144228 A144229 KEYWORD nonn,base AUTHOR Parthasarathy Nambi, Sep 15 2008 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified February 21 04:18 EST 2019. Contains 320371 sequences. (Running on oeis4.) | 654 | 1,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-09 | latest | en | 0.60669 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/02%3A_Gas_Laws/2.09%3A_Temperature_and_the_Ideal_Gas_Thermometer | 1,643,274,902,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00159.warc.gz | 225,287,981 | 26,107 | # 2.9: Temperature and the Ideal Gas Thermometer
In Section 2.2 we suppose that we have a thermometer that we can use to measure the temperature of a gas. We suppose that this thermometer uses a liquid, and we define an increase in temperature by the increase in the volume of this liquid. Our statement of Charles’ law asserts that the volume of a gas is a linear function of the volume of the liquid in our thermometer, and that the same linear function is observed for any gas. As we note in Section 2.8, there is a problem with this statement. Careful experiments with such thermometers produce results that deviate from Charles’ law. With sufficiently accurate volume measurements, this occurs to some extent for any choice of the liquid in the thermometer. If we make sufficiently accurate measurements, the volume of a gas is not exactly proportional to the volume of any liquid (or solid) that we might choose as the working substance in our thermometer. That is, if we base our temperature scale on a liquid or solid substance, we observe deviations from Charles’ law. There is a further difficulty with using a liquid as the standard fluid on which to base our temperature measurements: temperatures outside the liquid range of the chosen substance have to be measured in some other way.
Evidently, we can choose to use a gas as the working fluid in our thermometer. That is, our gas-volume measuring device is itself a thermometer. This fact proves to be very useful because of a further experimental observation. To a very good approximation, we find: If we keep the pressures in the thermometer and in some other gaseous system constant at low enough values, both gases behave as ideal gases, and we find that the volumes of the two gases are proportional to each other over any range of temperature. Moreover, this proportionality is observed for any choice of either gas. This means that we can define temperature in terms of the expansion of any constant-pressure gas that behaves ideally. In principle, we can measure the same temperature using any gas, so long as the constant operating pressure is low enough. When we do so, our device is called the ideal gas thermometer. In so far as any gas behaves as an ideal gas at a sufficiently low pressure, any real gas can be used in an ideal gas thermometer and to measure any temperature accurately. Of course, practical problems emerge when we attempt to make such measurements at very high and very low temperatures.
The (very nearly) direct proportionality of two low-pressure real gas volumes contrasts with what we observe for liquids and solids. In general, the volume of a given liquid (or solid) substance is not exactly proportional to the volume of a second liquid (or solid) substance over a wide range of temperatures.
In practice, the ideal-gas thermometer is not as convenient to use as other thermometers—like the mercury-in-glass thermometer. However, the ideal-gas thermometer is used to calibrate other thermometers. Of course, we have to calibrate the ideal-gas thermometer itself before we can use it.
We do this by assigning a temperature of 273.16 K to the triple point of water. (It turns out that the melting point of ice isn’t sufficiently reproducible for the most precise work. Recall that the triple point is the temperature and pressure at which all three phases of water are at equilibrium with one another, with no air or other substances present. The triple-point pressure is 611 Pa or $$\mathrm{6.03\times }{\mathrm{10}}^{\mathrm{-3\ }}$$atm. See Section 6.3.) From both theoretical considerations and experimental observations, we are confident that no system can attain a temperature below absolute zero. Thus, the size$${}^{3}$$ of the kelvin (one degree on the Kelvin scale) is fixed by the difference in temperature between a system at the triple point of water and one at absolute zero. If our ideal gas thermometer has volume $$V$$ at thermal equilibrium with some other constant-temperature system, the proportionality of $$V$$ and $$T$$ means that
$\frac{T}{V}=\frac{273.16}{V_{273.16}}$
With the triple point fixed at 273.16 K, experiments find the freezing point of air-saturated water to be 273.15 K when the system pressure is 1 atmosphere. (So the melting point of ice is 273.15 K, and the triple-point is 0.10 C. We will find two reasons for the fact that the melting point is lower than the triple point: In Section 6.3 we find that the melting point of ice decreases as the pressure increases. In Section 16.10 we find that solutes usually decrease the temperature at which the liquid and solid states of a substance are in phase equilibrium.)
If we could use an ideal gas in our ideal-gas thermometer, we could be confident that we had a rigorous operational definition of temperature. However, we note in Section 2.8 that any real gas will exhibit departures from ideal gas behavior if we make sufficiently accurate measurements. For extremely accurate work, we need a way to correct the temperature value that we associate with a given real-gas volume. The issue here is the value of the partial derivative
${\left(\frac{\partial V}{\partial T}\right)}_P$
For one mole of an ideal gas,
${\left(\frac{\partial V}{\partial T}\right)}_P=\frac{R}{P}=\frac{V}{T}$
is a constant. For a real gas, it is a function of temperature. Let us assume that we know this function. Let the molar volume of the real gas at the triple point of water be $$V_{273.16}$$ and its volume at thermal equilibrium with a system whose true temperature is $$V$$ be $$V_T$$. We have
$\int_{273.16}^T \left( \frac{ \partial V}{ \partial T} \right)_P dT = \int_{V_{273.16}}^{V_T} dV = V_T - V_{273.16}$
When we know the integrand on the left as a function of temperature, we can do the integration and find the temperature corresponding to any measured volume, $$V_T$$.
When the working fluid in our thermometer is a real gas we make measurements to find $${\left({\partial V}/{\partial T}\right)}_P$$ as a function of temperature. Here we encounter a circularity: To find $${\left({\partial V}/{\partial T}\right)}_P$$ from pressure-volume-temperature data we must have a way to measure temperature; however, this is the very thing that we are trying to find.
In principle, we can surmount this difficulty by iteratively correcting the temperature that we associate with a given real-gas volume. As a first approximation, we use the temperatures that we measure with an uncorrected real-gas thermometer. These temperatures are a first approximation to the ideal-gas temperature scale. Using this scale, we make non-pressure-volume-temperature measurements that establish $${\left({\partial V}/{\partial T}\right)}_P$$ as a function of temperature for the real gas. [This function is
${\left(\frac{\partial V}{\partial T}\right)}_P=\frac{V+{\mu }_{JT}C_P}{T}$
where $$C_P$$ is the constant-pressure heat capacity and $${\mu }_{JT}$$ is the Joule-Thomson coefficient. Both are functions of temperature. We introduce $$C_P$$ in Section 7.9. We discuss the Joule-Thomson coefficient further in Section 2.10 below, and in detail in Section 10.14. Typically $$V\gg C_P$$, and the value of $${\left({\partial V}/{\partial T}\right)}_P$$ is well approximated by $${V}/{T}={R}/{P}$$. With $${\left({\partial V}/{\partial T}\right)}_P$$ established using this scale, integration yields a second-approximation to the ideal-gas temperatures. We could repeat this process until successive temperature scales converge at the number of significant figures that our experimental accuracy can support.
In practice, there are several kinds of ideal-gas thermometers, and numerous corrections are required for very accurate measurements. There are also numerous other ways to measure temperature, each of which has its own complications. Our development has considered some of the ideas that have given rise to the concept$${}^{4}$$ that temperature is fundamental property of nature that can be measured using a thermodynamic-temperature scale on which values begin at zero and increase to arbitrarily high values. This thermodynamic temperature scale is a creature of theory, whose real-world counterpart would be the scale established by an ideal-gas thermometer whose gas actually obeyed $$PV=nRT$$ at all conditions. We have seen that such an ideal-gas thermometer is itself a creature of theory.
The current real-world standard temperature scale is the International Temperature Scale of 1990 (ITS-90). This defines temperature over a wide range in terms of the pressure-volume relationships of helium isotopes and the triple points of several selected elements. The triple points fix the temperature at each of several conditions up to 1357.77 K (the freezing point of copper). Needless to say, the temperatures assigned at the fixed points are the results of painstaking experiments designed to give the closest possible match to the thermodynamic scale. A variety of measuring devices—thermometers—can be used to interpolate temperature values between different pairs of fixed points. | 2,011 | 9,081 | {"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} | 3.359375 | 3 | CC-MAIN-2022-05 | latest | en | 0.929639 |
https://math.answers.com/algebra/What_is_0_raised_to_the_first_power | 1,670,370,941,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711121.31/warc/CC-MAIN-20221206225143-20221207015143-00350.warc.gz | 417,082,599 | 48,597 | 0
# What is 0 raised to the first power?
Wiki User
2013-02-17 21:27:59
the answer is 0. 0 raised to to any power is 0. When you try to solve these types of problems think of the power the number of times you multiply the base by. So, 0 to the first power is written out as 0 so the answer is 0.
Wiki User
2013-02-17 21:27:59
Study guides
20 cards
➡️
See all cards
3.8
1785 Reviews | 128 | 388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.935535 |
https://www.hometheaterforum.com/community/threads/per-channel-watts-specs-for-home-theater-receivers.365594/ | 1,642,825,181,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00566.warc.gz | 800,198,530 | 26,737 | Forum list
What's new
# Per Channel Watts Specs for Home Theater Receivers (1 Viewer)
#### richatct
##### Auditioning
Here is something I have always wondered about the RMS Watts spec of a home theater receiver.
Any time you view a spec (of any brand) of a multi channel receiver (let's say 7 channels), it is always written in this format:
"140 watts per channel into 8 ohms (20-20,000 Hz) at .08% THD, with 2 channels driven".
And I am like ok.... 90% of my use of the receiver are in watching movies or TV, which will typically be powering all 7 channels (not just two for stereo). But the spec talks about watts per channel with ONLY 2 channels driven.
So, does that mean I can get 140 watts X 7 channels?
Or if that spec is 140 x 2 = 280 watts, then in reality in using 7 channels at the same time I am actually getting: 280 watts / 7 channels = 40 watts for each of the 7 channels?
#### Gary Seven
##### Grand Poo Pah
Senior HTF Member
No, usually. The wpc spec is what it is capable of, not necessarily what it does. It depends on the power supply. Take the power supply total wattage divided by number of channels to get a better idea what it does for those channels.
#### JohnRice
##### Bounded In a Nutshell
Supporter
Senior HTF Member
As Gary alluded to, that specs show the maximum each channel is capable of putting out, but you're completely right that it doesn't represent what can be produced running all channels. With most receivers, the power supply is exhausted driving two channels to their max. Unfortunately, they never give an all channels driven spec, so you can only speculate what it is.
In most cases, your math of 280/7 isn't far off. It probably isn't as low as 40, but it's probably no more than 60.
Keep in mind that surround channels don't require as much power, since they generally don't reproduce low frequencies. The lower the frequency, the more power needed to reproduce it.
Manufacturers have upped the ante on deceptiveness lately with a new rating that's a 1KHz tone at 10% distortion, one channel driven, rather than 20Hz-20KHz at a reasonable distortion level with two channels driven. That spec is completely BS. No connection to reality at all.
This is a major reason I use external amps in my main HT. The amps I have are able to provide full power to all channels, but they weigh a ton. In reality, that's overkill, since surround channels don't need as much power. That's why many amps, including the current versions of the ones I have, no longer are able to run all channels at 100% simultaneously. However, if you're looking at a two channel amp, you want full power from all channels. Same with a three channel amp for the front three channels in a surround system.
Replies
3
Views
674
Replies
22
Views
2K
Replies
0
Views
2K | 678 | 2,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-05 | latest | en | 0.958864 |
https://blog.collegehockeyranked.com/2022/01/17/so-long-and-thanks-for-all-the-fish/ | 1,721,403,600,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00386.warc.gz | 122,622,682 | 9,191 | # So long, and thanks for all the fish!
I’m taking a break (indefinite?) from updating this site. There are a number of reasons now is the right time, but first a look back!
It’s been about 20 years that I’ve been independently calculating PWR, first at SiouxSports.com, then 8 years ago, launching this site. I did it because while there were a few other sources for the PWR table itself, I noticed that when my friends and I were discussing PWR, we were doing a lot of repetitive work that a computer could be doing. I felt that by making that information more accessible, I could add a lot to fans’ understanding of PWR.
What we were all always interested in about PWR was knowing what changes were likely, so my first innovation was the teams’ comparison details, which puts in front of the reader precisely how a teams’ PWR is calculated, making it much easier to guess at likely changes in the future.
About 15 years ago, I took it a step further, realizing that by simulating a weekend’s games, I could forecast likely PWR changes, based on whether a team wins or loses. This truly automated the work we all used to do notepads and the backs of envelopes, providing (somewhat) easy to interpret probability distributions of a teams’ likely PWR based its own performance. This sort of “UND needs to win about half its remaining games to get a bid” prediction was quite revolutionary at the time.
Finally, realizing that the reason we care about forecasting PWR is that we really want to know who’s going to make the tournament at the end of the season, I created the analysis of the wins needed to get a particular PWR rank table, which lets you see at a glance where every team is likely to fall based on its performance in its remaining games. I’m again proud of the simplicity of the presentation which, though it seems obvious in retrospect, was quite groundbreaking in making that sort of information accessible to everyone.
So why stop now? This year really is a perfect storm of conditions for me, as the amount of worked peaked right at a point my interest has fallen. First, because of the pandemic, I just haven’t paid attention to this site or college hockey much in the last two years. Second, over the years other have built on those innovations (imitation being the sincerest form of flattery), so me reviving this work is less necessary for the community. Third, the data source for college hockey scores that I (and the rest of the college hockey community) have been using for 20 years also shut down this year. Finally, the NCAA implemented the first major change to the PWR formula since 2013-14–crediting an OT win as only .55 of a win, thus necessitating a fair amount of work for me to get the rankings accurate again.
So, I’m going to hit the “pause” button on updating this site. Because I don’t want to post inaccurate information, I’ll take down the PWR/RPI tables, but as I mentioned above there are other great sources! I’ll leave the blog here as a historical archive for some time.
So long, and I’m sure I’ll see you around!
## 2 thoughts on “So long, and thanks for all the fish!”
1. Wes in Denver
Thanks for everything you’ve done. I’ve been checking the website every month or so the last two years looking for an update.
I appreciate all the historical data that kept me busy looking and analyzing over the years.
UND ‘07
2. Brandon
I have no idea if you’re still watching these, but like you, I am attempting my own tool for PWR calculations/predictions.
If you see this and could email me in response, I can explain further, but I’m having issues getting OWP/OOWP accurate to get a proper RPI number, my weighted win % are matching all published resources perfectly, so I know my numbers there are good, but the aggregation of OWP/OOWP is where something is going amiss. | 846 | 3,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.98186 |
http://parasys.net/error-propagation/error-propegation.php | 1,521,524,608,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647299.37/warc/CC-MAIN-20180320052712-20180320072712-00105.warc.gz | 215,937,889 | 4,590 | # parasys.net
Home > Error Propagation > Error Propegation
# Error Propegation
## Contents
The results for addition and multiplication are the same as before. The sine of 30° is 0.5; the sine of 30.5° is 0.508; the sine of 29.5° is 0.492. Sprache: Deutsch Herkunft der Inhalte: Deutschland Eingeschränkter Modus: Aus Verlauf Hilfe Wird geladen... Please note that the rule is the same for addition and subtraction of quantities.
Then the error in any result R, calculated by any combination of mathematical operations from data values x, y, z, etc. Let's say we measure the radius of a very small object. The size of the error in trigonometric functions depends not only on the size of the error in the angle, but also on the size of the angle. Suppose n measurements are made of a quantity, Q.
## Error Propagation Calculator
Since both distance and time measurements have uncertainties associated with them, those uncertainties follow the numbers throughout the calculations and eventually affect your final answer for the velocity of that object. The propagation of error formula for $$Y = f(X, Z, \ldots \, )$$ a function of one or more variables with measurements, $$(X, Z, \ldots \, )$$ The measured track length is now 50.0 + 0.5 cm, but time is still 1.32 + 0.06 s as before. JCGM.
Wird geladen... the relative determinate error in the square root of Q is one half the relative determinate error in Q. 3.3 PROPAGATION OF INDETERMINATE ERRORS. It can tell you how good a measuring instrument is needed to achieve a desired accuracy in the results. Error Propagation Chemistry In the first step - squaring - two unique terms appear on the right hand side of the equation: square terms and cross terms.
Why can this happen? Your cache administrator is webmaster. This also holds for negative powers, i.e. Melde dich bei YouTube an, damit dein Feedback gezählt wird.
In both cases, the variance is a simple function of the mean.[9] Therefore, the variance has to be considered in a principal value sense if p − μ {\displaystyle p-\mu } Error Propagation Calculus Example: If an object is realeased from rest and is in free fall, and if you measure the velocity of this object at some point to be v = - 3.8+-0.3 p.37. Since uncertainties are used to indicate ranges in your final answer, when in doubt round up and use only one significant figure.
## Error Propagation Example
It may be defined by the absolute error Δx.
This leads to useful rules for error propagation. Error Propagation Calculator We can also collect and tabulate the results for commonly used elementary functions. Error Propagation Formula General functions And finally, we can express the uncertainty in R for general functions of one or mor eobservables.
Then our data table is: Q ± fQ 1 1 Q ± fQ 2 2 .... GUM, Guide to the Expression of Uncertainty in Measurement EPFL An Introduction to Error Propagation, Derivation, Meaning and Examples of Cy = Fx Cx Fx' uncertainties package, a program/library for transparently The system returned: (22) Invalid argument The remote host or network may be down. For example, the bias on the error calculated for logx increases as x increases, since the expansion to 1+x is a good approximation only when x is small. Error Propagation Physics
Also, if indeterminate errors in different measurements are independent of each other, their signs have a tendency offset each other when the quantities are combined through mathematical operations. It can be written that $$x$$ is a function of these variables: $x=f(a,b,c) \tag{1}$ Because each measurement has an uncertainty about its mean, it can be written that the uncertainty of Since the uncertainty has only one decimal place, then the velocity must now be expressed with one decimal place as well. The derivative, dv/dt = -x/t2.
Simanek. ERROR PROPAGATION RULES FOR ELEMENTARY OPERATIONS AND FUNCTIONS Let R be the result of a calculation, without consideration of errors, and ΔR be the error (uncertainty) in that result. Error Propagation Addition In fact, since uncertainty calculations are based on statistics, there are as many different ways to determine uncertainties as there are statistical methods. But more will be said of this later. 3.7 ERROR PROPAGATION IN OTHER MATHEMATICAL OPERATIONS Rules have been given for addition, subtraction, multiplication, and division. | 995 | 4,361 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | longest | en | 0.818454 |
https://goalofbusines.xyz/what-is-surface-area-formula/ | 1,659,939,996,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570767.11/warc/CC-MAIN-20220808061828-20220808091828-00196.warc.gz | 249,947,530 | 22,355 | Measurements
# What is surface area formula?
Surface area is the sum of the areas of all faces (or surfaces) on a 3D shape. … We can also label the length (l), width (w), and height (h) of the prism and use the formula, SA=2lw+2lh+2hw, to find the surface area.
### Also, What is the area and volume of a cylinder?
A cylinder’s volume is π r² h, and its surface area is 2π r h + 2π r². Learn how to use these formulas to solve an example problem.
### Hereof, Where do we use total surface area?
For a two-dimensional object, that’s also its total surface area. In three dimensions, like a cube, a sphere, or a pyramid, the surfaces can’t all be seen at one time. Total surface area in that case means adding up the areas of all the surfaces. For a cube, that means adding up the surface area of all six sides.
Also to know What is surface area and volume? The surface area of any given object is the area or region occupied by the surface of the object. Whereas volume is the amount of space available in an object. In geometry, there are different shapes and sizes such as sphere, cube, cuboid, cone, cylinder, etc.
### What unit is surface area?
What is surface area? Surface area is the total area of the faces of a three-dimensional shape. Surface area is measured in square units.
## What is the formula of volume?
Whereas the basic formula for the area of a rectangular shape is length × width, the basic formula for volume is length × width × height.
## How do you calculate volume and surface area?
Surface Area Formulas:
1. Volume = (1/3)πr
2
h.
2. Lateral Surface Area = πrs = πr√(r
2
+ h
2
)
3. Base Surface Area = πr
2
4. Total Surface Area. = L + B = πrs + πr
2
= πr(s + r) = πr(r + √(r
2
+ h
2
))
## What’s the difference between total surface area and surface area?
“Area” & “surface area” of a figure mean exactly the same thing. So “total surface area” simply means “total area of the figure, regardless of whether the figure is a 2-D figure (such as a square) or a 3-D figure (such as a cube) or a 4-D figure … at 2018 Jan 16 Tue 1907H.
## What is the difference between surface and total surface area?
The lateral surface of an object is the area of all the faces of the object, excluding the area of its base and top. For a cube, the lateral surface area would be the area of four sides. Total surface area is the area of all the faces including the bases.
## What is difference between curved surface area and total surface area?
Curved Surface Area (CSA) – It includes the area of all the curved surfaces. Lateral Surface Area (LSA) – It includes the area of all the surface excluding the top and bottom areas. Total Surface Area (TSA) – It includes the area of all the surfaces of the object including the bases.
## What is the difference between total surface area and volume?
Surface area is the sum of the areas of all the faces of the solid figure. … Volume is the number of unit cubes that make up a solid figure. Volume is the amount of space inside of the solid figure.
## What is surface area and volume of 3D shapes?
The surface area of a 3D shape is the total area of all its faces. • Volume of a pyramid = 1. 3. × area of base × vertical height.
## What is perimeter formula?
The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides. x is in this case the length of the rectangle while y is the width of the rectangle. The perimeter, P, is: P=x+x+y+y.
## What is cube formula?
So for a cube, the formulas for volume and surface area are V=s3 V = s 3 and S=6s2 S = 6 s 2 .
## Is total surface area and volume the same?
Surface area is a two-dimensional measure, while volume is a three-dimensional measure. Two figures can have the same volume but different surface areas. For example: A rectangular prism with side lengths of 1 cm, 2 cm, and 2 cm has a volume of 4 cu cm and a surface area of 16 sq cm.
## How are surface area and volume related?
The volume is how much space is inside the shape. The surface-area-to-volume ratio tells you how much surface area there is per unit of volume. This ratio can be noted as SA:V. To find this ratio, you divide the formula for surface area by the formula for volume and then you simplify.
## Where is total surface area and curved surface area used?
Surface Area Formula of Cone
If the radius of the base of the cone is “r” and the slant height of the cone is “l”, the surface area of a cone is given as: Total surface area of a cone, T = πr(r + l) The curved surface area of a cone, S = πrl.
## What is the difference between total surface area and curved surface area?
The term surface area is defined as the total area of the surface of the given solid object. … Curved Surface Area (CSA) – It includes the area of all the curved surfaces. Lateral Surface Area (LSA) – It includes the area of all the surface excluding the top and bottom areas.
## What is a volume of 3d shapes?
Volume of a 3-d shape is defined as the total space enclosed/occupied by any 3-dimensional object or solid shape. It also can be defined as the number of unit cubes that can be fit into the shape. The SI unit of volume is cubic meters.
## What are the formulas for 3d shapes?
Formulas for 3D Shapes
Shape Surface Area Terms
Cube 6a
2
a = length of the edge
Rectangular prism 2(wl + hl + hw) l = length w = width h = height
Cylinder 2πr(r + h) r = radius of the circular base h = height of the cylinder
Cone πr(r + l) r = radius of the circular base l = slant height
## How do you find an area of a shape?
The area is calculated by multiplying the length of a shape by its width. In this case, we could work out the area of this rectangle even if it wasn’t on squared paper, just by working out 5cm x 5cm = 25cm² (the shape is not drawn to scale).
## How do you find the area and perimeter?
Divide the perimeter by 4: that gives you the length of one side. Then square that length: that gives you the area. In this example, 14 ÷ 4 = 3.5.
## What is π?
Succinctly, pi—which is written as the Greek letter for p, or π—is the ratio of the circumference of any circle to the diameter of that circle. Regardless of the circle’s size, this ratio will always equal pi. In decimal form, the value of pi is approximately 3.14.
## Are volume and surface area directly proportional?
So for a sphere, the ratio of surface area to volume is given by: S/V = 3/R. … One interesting thing that you should notice is that the surface area to volume ratio is inversely proportional to the size of an object. For example, we found that for a sphere S/V is 3/R.
Measurements
Measurements
Measurements | 1,695 | 6,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | longest | en | 0.897591 |
https://www.beatthegmat.com/mba/2019/04/11/understanding-negation-and-sample-spaces | 1,561,487,405,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999876.81/warc/CC-MAIN-20190625172832-20190625194832-00422.warc.gz | 681,776,638 | 15,304 | # Understanding Negation and Sample spaces:
by on April 11th, 2019
## Introduction
1. Introduction to Negation: In this section, we will briefly talk about what negating a statement means and why is it important.
2. Key skills for negation
3. 4 Questions
If you feel, you lag behind in the planning part of your preparation, we invite you for a free webinar on GMAT Strategy. Learn the best way to prepare for GMAT by taking a free trial of our online GMAT preparation resource. You can write to us at acethegmat@e-gmat.com if you need any GMAT related advice.
# What is negation and negation test?
To understand the above let’s take a simple argument.
Every super-smart GMAT aspirant registers on BeatTheGMAT. Hence, every Harvard admit is a registered student at BeatTheGMAT.
• Assumption: Harvard only admits super-smart GMAT aspirants.
Negation technique can be used to determine if the given answer choice is the true assumption or not. In other words, it a sure shot way to validate the answer (recommended use) or to determine whether an answer choice is an assumption
## Why does negation test work?
The correct answer choice is like a missing premise. Which means that has to be true for the conclusion to hold true. This implies that the negated version of the correct answer choice will shatter, falsify or invalidate the conclusion. Hence, once we have narrowed down our answer choice(s), we can verify it by evaluating whether the negated version really shatters the conclusion or not. If it does, then your selected choice is indeed the conclusion, otherwise not.
In the example above, the negated version of the assumption could be one of the following:
1. Harvard does not only admit super-smart GMAT aspirants.
2. At least one Harvard admit is not super-smart aspirant.
Ok, sounds good. But the next question comes to mind.
## How do we negate?
When you think of negation, it helps to think in terms of sets. Each answer choice is “a part” of a superset. Hence, the negated statement = Super Set – Space occupied by the answer choice.
`Negated Statement = Super set - Space occupied by the answer choice`
Let’s take an example answer choice.
All GMAT test takers who take the exam after 1 month of preparation perform better than they did in their first mock
Q: What is the super set?
• Super set is the set of possibilities for GMAT takers who take the exam after one month of preparation. Let’s assume that there are 100 test takers who take the exam after 1 month of the preparation. Then there are multiple possibilities that exist for this group of test takers
1. All 100 test takers who take the exam after 1 month of preparation perform better than they did on their first mock.
2. Not All test takers who take the exam after 1 month of preparation perform better than they did on their first mock.
3. Some Test Takers who take the exam after 1 month of preparation perform better than they did on their first mock.
4. Most test takers who take the exam after 1 month of preparation perform better than they did on their first mock.
5. Less than 50% of test takers who take the exam after 1 month of preparation perform better than they did on their first mock.
6. No test taker who takes the exam after 1 month of preparation performs better than he/she did on their first mock.
There are a few things to note in the picture above.
1. Understand the Super Set: Note, we are only talking about a segment of the population – the test takers who take the exam 1 month after preparing for their GMAT. This is our Universe or the super set. We are not concerned with test takers who take the exam after 15 days of preparation or 2 months of preparation. Words such as “Who” (user for people), “that” (animals, corporations etc.), “which” and “those” have a segmentation effect. Therefore, any answer choice that deals with choices outside this segment – the segment in which people take the exam after 1 month of preparation – is incorrect.
• Take Away: Words such as Who, that, which, those etc. have a segmentation effect.
2. Understand the Sample space occupied by the choice: Once you have defined the Super set, understand the sample space occupied by the choice that you want to negate. Words such as all, none, some, most, etc. define these sample spacing. It is important to know what these words imply in terms of sample spaces. Taking the “Super Set” in discussion and assuming that it contains 100 test takers, here is what the sample spaces mean:
Super Set = Set of 100 test takers Term Sample Space it occupies All All the 100 Test takers Some 1 to 100 test takers (including both 1 and 100). Note some does not include 0 Not All 0-99 Test Takers (Notice how Not All is different from Some) Most 51-100 test takers None 0 Test takers out of 100
3. Understand what negation means: Negation or logical negation means that you select an answer choice that occupies the sample space that excludes the original choice. If it maps to another term above, then perfect, otherwise you may make the use of word “Not” to find the logical opposite. In the example above, the logical opposite of “All” is “Not All”. Note, that you may be tempted to use “Some” but “some” includes All (Notice, it includes 100) and hence is not the logical opposite. Here is a table of logical opposites.
Term Logical Opposite All Not All Some None (Notice, Some does not include 0) Not All All Most Not more than half (which means half or less, 0-50) None Some Exactly X Not Exactly X (Note that the sample space is both before XX and after X, just not X) Significant Insignificant Never Sometimes Always Not Always Everywhere Not everywhere
If you feel, you lag behind in the planning part of your preparation, we invite you for a free webinar on GMAT Strategy. Learn the best way to prepare for GMAT by taking a free trial of our online GMAT preparation resource. You can write to us at acethegmat@e-gmat.com if you need any GMAT related advice.
Coming back to the example answer choice:
All GMAT test takers who take the exam after 1 month of preparation perform better than they did in their first mock. Let’s now look at the answer choices:
Answer Choice Explanation Some Test Takers who take the exam after 1 month of preparation perform better than they did on their first mock. Incorrect: For the reasons discussed above. Notice “some” = 1-100 whereas we are looking for a sample space of 0-99. Hence, this choice is not correct. Most test takers who take the exam after 1 month of preparation perform better than they did on their first mock Incorrect: Only includes 51 to 100 Not All test takers who take the exam after 1 month of preparation perform better than they did on their first mock Correct: This is the correct choice. Notice, how this occupies the complementary sample space and is the logical opposite. Less than 50% of test takers who take the exam after 1 month of preparation perform better than they did on their first mock. Incorrect: Only includes 0-50
### WHAT IF YOU NEGATE THE VERB?
Let’s bring the Original choice again for clarity purposes.
All GMAT test takers who take the exam after 1 month of preparation perform better than they did in their first mock. Notice, that in all the negated choices above, we did not modify the verb – perform better. Let us see what happens when we negate the verb. Consider the following negated choice
Neg 1: Some GMAT test takers who take the exam after 1 month of preparation do not perform better than they did in their first mock.
What does the above answer choice mean?
The answer choice implies that, out of 100 test takers, between 1 and 100 test takers did not perform better than they did in their first mock. This means that 0 to 99 test takers did perform better on their mock. This is the same is “Not all”. Hence, the above answer choice is Correct. Let’s see how this happened:
Test Taker did not perform better Test taker did perform better 1 test taker did not perform better 99 test takers performed better 100 test takers did not perform better 0 test takers performed better
Take Away: It is sometimes possible to negate by either negating the subject of an answer choice or by negating the Verb.
If you feel, you lag behind in the planning part of your preparation, we invite you for a free webinar on GMAT Strategy. Learn the best way to prepare for GMAT by taking a free trial of our online GMAT preparation resource. You can write to us at acethegmat@e-gmat.com if you need any GMAT related advice.
# Exercise Sentences:
Ex1: Most GMAT test takers who take the exam after 1 month of preparation perform better than they did in their first mock. (Multiple answers may be correct)
1. Most GMAT test takers who take the exam after 1 month of preparation do not perform better than they did in their first mock.
2. All GMAT test takers who take the exam after 1 month of preparation perform better than they did in their first mock.
3. No GMAT test taker who takes the exam after 1 month of preparation performs better than they did in their first mock.
4. At least half of GMAT test takers who take the exam after 1 month of preparation perform better than they did in their first mock.
5. Up to half of GMAT test takers who take the exam after 1 month of preparation perform better than they did in their first mock.
6. Up to half of GMAT test takers who take the exam after 1 month of preparation do not perform better than they did in their first mock.
7. Half or more of GMAT test takers who take the exam after 1 month of preparation do not perform better than they did in their first mock.
Ex2: Every internet dating profile in the world begs for some spontaneity (Multiple answers may be correct)
1. Not every internet dating profile in the world begs for some spontaneity
2. No internet dating profile in the world begs for some spontaneity
3. Most internet dating profiles in the world begs for some spontaneity
4. Some internet dating profiles in the world do not beg for some spontaneity
Ex3: Innovations such as iPhone happen nowhere outside Apple. (Multiple answers may be correct)
1. Innovations such as iPhone do happen outside Apple.
2. Innovations such as iPhone do not happen outside Apple.
3. No other company is as innovative as Apple is.
4. Innovations such as iPhone happen both inside Apple and in other companies.
Ex4: No sane person who has never given a ride to a stranded passenger will give a ride to someone dressed in a hood (Multiple answers may be correct)
1. Some sane people who have never given a ride to a stranded passenger will give a ride to someone dressed in a hood
2. At least one sane person who has never given a ride to a stranded passenger will give a ride to someone dressed in a hood
3. All sane people who have never given a ride to a stranded passenger will give a ride to someone dressed in a hood
4. No sane person who has ever given a ride to a stranded passenger will give a ride to someone dressed in a hood.
5. Some sane people who have given a ride to a stranded passenger will give a ride to someone dressed in a hood
Ex5:Every writer will tell you: first, find a good café. (Multiple answers may be correct)
1. Every writer will not tell you: first, find a good café.
2. Some writers will not tell you: first, find a good café
3. No writer will tell you: first, find a good café.
4. Most writers will not tell you: first, find a good café.
Ex-6: All the people who became sick did not get vaccinated.(Multiple answers may be correct)
1. Not everyone who became sick got vaccinated.
2. Some people who became sick did not get vaccinated
3. At least one person who became sick got vaccinated
4. All people who got vaccinated did not become sick
5. Most people who did not become sick got vaccinated
6. Most people who did not get vaccinated did not become sick.
If you feel, you lag behind in the planning part of your preparation, we invite you for a free webinar on GMAT Strategy. Learn the best way to prepare for GMAT by taking a free trial of our online GMAT preparation resource. You can write to us at acethegmat@e-gmat.com if you need any GMAT related advice. | 2,815 | 12,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-26 | latest | en | 0.881104 |
https://www.mrexcel.com/board/threads/lookup-help.176067/ | 1,695,808,202,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510284.49/warc/CC-MAIN-20230927071345-20230927101345-00144.warc.gz | 1,009,440,455 | 19,246 | # Lookup Help
#### SonicBoomGolf
##### Active Member
I am trying to lookup a value and am having some trouble. Below is a small sample dataset to illustrate my goal.
Column A Column B
Sold To Revenue
1000080 \$-
1000080 \$-
1000080 \$1,000,000.00
1000170 \$145.00
1000196 \$144.00
1004570 \$187.00
1004572 \$198.00
I have a unique list of numbers that have multiple rows of cost data but only one row of revenue data, much like number 1000080. My goal is to find the cell in coulmn B that has a value above 0. A VLOOKUP would work, but it returns only the first match for a given number. Instead, I need the lookup to return the first non-zero match it finds.
So using the model above, I would want \$1,000,000 returned for 1000080. Any ideas?
### Excel Facts
Format cells as currency
Select range and press Ctrl+Shift+4 to format cells as currency. (Shift 4 is the \$ sign).
Book1.xls
ABCD
1Sold toRevenue1000080
21000080\$ -\$ 1,000,000.00
31000080\$ -
41000080\$ 1,000,000.00
51000170\$ 145.00
61000196\$ 144.00
71004570\$ 187.00
81004572\$ 198.00
9
Sheet2
The above is an array formula, which requires entry via Control+Shift+Enter [CSE], not just Enter. Any direct edit of the formula will require re-entry via CSE. You can tell if the formula has been entered correctly, as it will have braces - { } - around it afterward.
Here is the formula I am using (with CRTL+SHIFT+ENTER), but I am getting a zero for every number I try
=MIN(IF((\$A\$1:\$A\$959=B966)*(\$G\$1:\$G\$959>0),\$B\$1:\$B\$959))
Using my previous example, if the first two criteria are met (match for the given Sold-to number and greater than zero in column G) then I want the result from column B brough in. I get zeros back for every number I try. Not sure what is wrong.......
P.S. B966 is the first unique number I am researching (1000080) and column G is the revenue column. Column B is the respective district of the sold-to.
Base on just_jon sample
You could use this formula its a non-array. Put the formula in cell C2 and just hit enter.
=LOOKUP(2,1/((A2:A8=C1)),B2:B8)
vane0326 said:
Base on just_jon sample
You could use this formula its a non-array. Put the formula in cell C2 and just hit enter.
=LOOKUP(2,1/((A2:A8=C1)),B2:B8)
There's a problem with that --
Book1
ABCD
1SoldRevenue1000080
210000800.99
31000080-
410000801,000,000.00
510000800.99
61000196144
71004570187
81004572198
9
Sheet1
Replies
2
Views
2K
Replies
0
Views
1K
Replies
1
Views
576
Replies
1
Views
3K
Replies
3
Views
553
1,203,252
Messages
6,054,385
Members
444,721
Latest member
BAFRA77
### We've detected that you are using an adblocker.
We have a great community of people providing Excel help here, but the hosting costs are enormous. You can help keep this site running by allowing ads on MrExcel.com.
### Which adblocker are you using?
1)Click on the icon in the browser’s toolbar.
2)Click on the icon in the browser’s toolbar.
2)Click on the "Pause on this site" option.
Go back
1)Click on the icon in the browser’s toolbar.
2)Click on the toggle to disable it for "mrexcel.com".
Go back
### Disable uBlock Origin
Follow these easy steps to disable uBlock Origin
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back
### Disable uBlock
Follow these easy steps to disable uBlock
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back | 1,030 | 3,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.84458 |
http://mathhelpforum.com/algebra/72060-simultaneous-equations-print.html | 1,513,008,028,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513611.22/warc/CC-MAIN-20171211144705-20171211164705-00202.warc.gz | 172,969,049 | 2,711 | # simultaneous equations
• Feb 5th 2009, 06:20 PM
crafty
simultaneous equations
A + B = 1 Ax + By = 0 solve for A and B???
• Feb 5th 2009, 08:51 PM
Oiram
The only choice for A and B is for both A and B to equal zero.
• Feb 5th 2009, 10:01 PM
mr fantastic
Quote:
Originally Posted by Oiram
The only choice for A and B is for both A and B to equal zero.
? So A + B = 1 => 0 + 0 = 1 ....??
• Feb 5th 2009, 10:03 PM
mr fantastic
Quote:
Originally Posted by crafty
A + B = 1 Ax + By = 0 solve for A and B???
Substitute A = 1 - B into Ax + By = 0: (1 - B)x + By = 0 => x - Bx + By = 0 => x = B(x + y) => B = x/(x + y).
Now get A.
• Feb 5th 2009, 10:20 PM
Oiram
I'm sorry that misread the statement. I thought the statement was :
A+B = 1Ax + By = 0. I didn't realize that they were two different equations. | 302 | 805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-51 | longest | en | 0.958256 |
https://rightbankwarsaw.com/players/is-ronaldo-both-footed.html | 1,627,768,218,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154126.73/warc/CC-MAIN-20210731203400-20210731233400-00345.warc.gz | 484,306,699 | 19,837 | # Is Ronaldo both footed?
Contents
The Portuguese superstar has incredible technique with both feet and is capable of scoring with virtually any part of his body. Last season, he scored 29 times with his right foot, 16 times with his left and eight times with his head.
## Is Ronaldo right footed?
Cristiano Ronaldo scores 6 of every 10 goals with his right foot, 2 of every 10 goals with his left foot and 2 of every 10 goals with a header.
## Is Messi two footed?
While many top football players can score off their wrong foot, most do not prefer to do so, simply because they are not proficient enough. An example in this regard is Argentine maestro Lionel Messi, who is renowned for his magical left foot but rarely scores with his other foot.
## Is Ronaldo ambidextrous?
The only two players who have a better record in the modern era are Messi and Cristiano Ronaldo – both who are widely considered as the two best athletes to ever play the sport. … He might not be the most ambidextrous athlete out there, but he was simply so good that I couldn’t leave him out.
## What footed is cr7?
– Nacional U15
Full name: Cristiano Ronaldo dos Santos Aveiro
Height: 1,87 m
Citizenship: Portugal
Position: attack – Left Winger
Foot: both
## Who jumps higher Messi or Ronaldo?
Earlier this week, Cristiano Ronaldo reached a height of 2.56 metres to score a header against Sampdoria. … Let’s do some maths: Messi’s leap was estimated to be at 1.04 metres which makes the total height of his jump up to 2.74 metres. We shouldn’t forget that the Argentine is 17 cm shorter than the Portuguese.
## Who has more goals cr7 or Messi?
While Ronaldo currently has more goals overall, Messi has the edge in the scoring department, with a higher season average (40 to 35), having hit a high of 73 goals in 2011-12.
Cristiano Ronaldo vs Lionel Messi: Club goals.
Cristiano Ronaldo Season Lionel Messi
42 2016-17 54
44 2017-18 45
28 2018-19 51
37 2019-20 31
## What do you call someone who can use both feet?
‘Ambidextrous’ is the word. ‘ambi-‘ = both, ‘-dextrous’ = right, so together it is as though both are equally dominant. Usually, your feet (and eyes) follow the same dominance as your hands. … your feet are ambidextrous.
## What is ambidextrous for feet?
The word ambidextrous is familiar to most people: it describes somebody who is be able to use both hands (and sometimes, in the case of soccer, feet) equally well. … This means to be equally clumsy or unskilled with both hands!
## Who is the best left-footed soccer player?
Lionel Messi is considered by many to be the best left-footed soccer player. He plays for FC Barcelona, a Spanish soccer club, and also plays for the Argentina national team. Lionel Messi is considered by many to be the greatest left-footed player of all time (The GOAT).
IT IS INTERESTING: Frequent question: Is football an aerobic exercise?
## Is Messi left handed?
No he isn’t left handed. He uses his right hand to write and do other stuff. But he uses his strong left foot to play football. … Lionel Messi may uses his right hand to write and do other things but It is a known fact that a left footed person is also a left handed…
6′ 2″
## Who is the best player of football right now?
• Thiago Alcântara. Midfield mastermind of Bayern Munich’s treble-winning 2019/20 season (UEFA Champions League, Bundesliga and German Cup).
• Kevin De Bruyne. …
• Kylian Mbappé …
• Neymar. …
• Sergio Ramos. …
• Mohamed Salah. …
• Virgil Van Dijk.
## Why did Ronaldo leave Real?
Ronaldo became the most expensive player above the age of 30 when he moved from Real Madrid to Juventus, while his €100 million transfer fee is also the highest sum paid by an Italian club in history. Upon joining Juventus, Ronaldo revealed that the reason he left Real Madrid was to take on a new challenge.
## What team is Ronaldo in 2020?
Ronaldo set to make first Champions League appearance of 2020-21 season after being named in Juventus squad. Cristiano Ronaldo is set to make his first Champions League appearance of the 2020-21 season after being named in Juventus’ latest European squad.
## Is cr7 retiring?
In a recent interview, he admitted that he has been lucky to enjoy the leadership of the former Manchester United man before claiming that Ronaldo would still be active when he leaves his current position in 2024. … On his retirement he added: “He is 36, we are in 2021 and I will remain here only until 2024.
IT IS INTERESTING: How many national championships does ASU football have? | 1,092 | 4,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.964336 |
http://sudoku.com.au/1V30-9-2017-sudoku.aspx | 1,550,542,561,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489304.34/warc/CC-MAIN-20190219020906-20190219042906-00447.warc.gz | 259,707,028 | 12,448 | Timer 00:00:00 Sudoku Sudoku
## if (sTodaysDate) document.write('Sudoku for ' + sTodaysDate.split('-')[0] + '/' + monthname() + '/' + sTodaysDate.split('-')[2]); else document.write('Enter your own sudoku puzzle.')
Help
Play this pic as a Jigsaw
or Sliding Puzzle
Previous / Next
Choose a number, and place it in the grid above.
1 2 3 4 5 6 7 8 9
This number is a possibility
Automatically remove Possibilities Allow incorrect Moves Clicking the playing grid places the current number Highlight Current Square Grey out Used Numbers Possibilities in Grid Format
Check out the latest post in the Sudoku Forum
BREAKING NEWS
Submitted by: uno hu
Indicate which comments you would like to be able to see GeneralJokesOtherSudoku Technique/QuestionRecipes
Basics to UP = 231. (9=31)h13 - (1)h8 = i8 - i5 = a5 - (1=59)ab4 - (5&9)c46 = HP(95)c31 - (5)i1 = HP(52)ig3 => -9gi2 ; UP = 482. Y wing: (3=89)af9 - (9=3)d8 => -3a8 ; UP = 81 30/Sep/17 1:10 AM | |
Sorry for typo:HP(52)ig3 should read HP(52)ig2 30/Sep/17 1:25 AM | |
Basics to 231. ALS-XY wing (467=5)c189-(5=1369)gi1.h13-(9=4567)c1389 =>-67c6; basics to 482. XY-wing (3=8)a9-(8=9)f9-(9=3)d8 =>-3a8 30/Sep/17 1:48 AM | |
NQ(4567)c1389=9c3-9h3=NQ(1356)h3.ghi1*-5c1=(467)c189 => -{5a1*, (67)b6}; 15 singlesXY-Wing(369)di8.g7-3def7; stte 30/Sep/17 1:57 AM | |
NQ(4567)c1389=9c3-9h3=NQ(1356)h3.ghi1*-5c1=NT(467)c189 => -{5a1*, (67)b6}; 15 singlesXY-Wing(369)di8.g7-3def7; stte 30/Sep/17 1:59 AM | |
Very pretty photo, of a lovely place. 30/Sep/17 3:24 AM | |
Lovely water fall photo! Peaceful environment 30/Sep/17 5:58 AM | |
#24 1?@a5=>b4=9=>c3=9=>13@h13=>g1=6=>i1=5=>a2=5=>a4={}=>a5<>1#49 2@de7=>g7<>2 6?@g7=>i8=9=>d8=3=>a….i7<>3=>g7<>6VHBC to #81V:Only cell left in Vertical column for this candidateH:Only cell More... 30/Sep/17 6:38 AM | |
Waterfall or Spillway? 30/Sep/17 6:43 AM | |
Not a member? Joining is quick and free.
As a member you get heaps of benefits.
You can also try the Chatroom (No one chatting right now - why not start something? )
Check out the Sudoku Blog Subscribe
Easy Medium Hard Tough Or try the Kids Sudokus (4x4 & 6x6) 16x16 or the Parent's Page.
Printer Friendly versions:
Members Get Goodies! Become a member and get heaps of stuff, including: stand-alone sudoku game, online solving tools, save your times, smilies and more!
Welcome our latest MembersJohno from GCRukiya223 from New York cityJohn Maxwell from Eltham
Member's Birthdays TodayNone Today. | 945 | 2,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-09 | latest | en | 0.642048 |
https://wcalculator.com/conversion/time/1-millenium-to-hour | 1,701,228,082,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100047.66/warc/CC-MAIN-20231129010302-20231129040302-00349.warc.gz | 698,791,518 | 5,373 | # How to convert 1 millenium to hour?
1 nameeee = 8,765,812.7777778 nameeee
## How many nameeee in an nameeee?
To convert from nameeee to nameeee, divide by 0.0000001
1 nameeee = 8,765,812.7777778 nameeee,
## How many nameeee in an nameeee?
To convert from nameeee to nameeee, multiply by 0.0000001
1 nameeee = 0.0000001 nameeee,
## Convert 1 nameeee to other units:
nameeee HOUR 1 nameeee = 31,556,926,000,000,000,000 Nameeee 1 nameeee = 31,556,926,000,000,000 Nameeee 1 nameeee = 31,556,926,000,000 Nameeee 1 nameeee = 31,556,926,000 Nameeee 1 nameeee = 525,948,766.6666667 Nameeee 1 nameeee = 8,765,812.7777778 Nameeee 1 nameeee = 365,242.1990741 Nameeee 1 nameeee = 52,177.4570106 Nameeee 1 nameeee = 12,000.0174922 Nameeee 1 nameeee = 1,000.0008239 Nameeee 1 nameeee = 100.0000824 Nameeee 1 nameeee = 10.0000082 Nameeee | 321 | 834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-50 | latest | en | 0.54121 |
https://socratic.org/questions/what-is-40-000-30 | 1,726,081,490,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651400.96/warc/CC-MAIN-20240911183926-20240911213926-00610.warc.gz | 492,842,073 | 5,747 | # What is 40,000 -: 30?
##### 1 Answer
Mar 9, 2018
$1333.333$ or $1333 \frac{1}{3}$
#### Explanation:
$\frac{40000}{30} = \frac{4000}{3}$ This is a simplification to make long division easier (I assume you can't use a calculator... but can use the internet?)
I can't format long division using the functions available but you probably (might?) already know how to do it.
$\frac{4000}{3} = 1333 \frac{1}{3}$ | 131 | 410 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-38 | latest | en | 0.778808 |
https://www.jovoto.com/projects/betterbeta/ideas/13297?page=5&scope2=published&scope=rating | 1,660,105,150,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571147.84/warc/CC-MAIN-20220810040253-20220810070253-00024.warc.gz | 753,386,269 | 59,026 | 1+3+1+3=8
My first Idea, the beta-version so to speak...hrhr
Solution
Idea:
13=B
8=2xB=Better Beta
1+3+1+3=8
rational+abstrakt=Better Beta
(the picture i used is from : http://wvs.topleftpixel.com/archives/photostextures/0604261536.shtml )
Other entries in this project | 96 | 273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-33 | latest | en | 0.636861 |
https://www.wangt.cc/2022/08/leetcode-854-k-similar-strings/ | 1,679,765,691,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945368.6/warc/CC-MAIN-20230325161021-20230325191021-00574.warc.gz | 1,186,902,003 | 7,790 | • 欢迎光临~
# LeetCode 854. K-Similar Strings
Strings `s1` and `s2` are `k`-similar (for some non-negative integer `k`) if we can swap the positions of two letters in `s1` exactly `k` times so that the resulting string equals `s2`.
Given two anagrams `s1` and `s2`, return the smallest `k` for which `s1` and `s2` are `k`-similar.
Example 1:
```Input: s1 = "ab", s2 = "ba"
Output: 1
```
Example 2:
```Input: s1 = "abc", s2 = "bca"
Output: 2```
Constraints:
• `1 <= s1.length <= 20`
• `s2.length == s1.length`
• `s1` and `s2` contain only lowercase letters from the set `{'a', 'b', 'c', 'd', 'e', 'f'}`.
• `s2` is an anagram of `s1`.
The smallest swap, we could use BFS.
For a current string, if cur.equals(s2), return returns current level.
If not, we get all its neighbors.
To get a neighbor, we first find the different char between cur and s2, mark its index as ind.
Then find the next char where cur.charAt(j) == s2.charAt(ind).
Time Complexity: O(k*n^2). n = s1.length(). k = swap count, could be up to n. There could be k level BFS iteration. In each iteration, for each cur, we could find n neighbor, totally there could be n node in the que, thus there are n^2 neighbors.
Space: O(n^2).
AC Java:
``` 1 class Solution {
2 public int kSimilarity(String s1, String s2) {
4 HashSet<String> visited = new HashSet<>();
7 int level = 0;
8
9 while(!que.isEmpty()){
10 int size = que.size();
11 while(size-- > 0){
12 String cur = que.poll();
13 if(cur.equals(s2)){
14 return level;
15 }
16
17 for(String can : getNei(cur, s2)){
18 if(visited.contains(can)){
19 continue;
20 }
21
24 }
25 }
26
27 level++;
28 }
29
30 return -1;
31 }
32
33 private List<String> getNei(String s1, String s2){
34 char [] s1Arr = s1.toCharArray();
35 char [] s2Arr = s2.toCharArray();
36 int n = s1.length();
37 int ind = 0;
38 while(ind < n && s1Arr[ind] == s2Arr[ind]){
39 ind++;
40 }
41
42 List<String> res = new ArrayList<>();
43 for(int j = ind + 1; j < n; j++){
44 if(s1Arr[j] == s2Arr[j] || s1Arr[j] != s2Arr[ind]){
45 continue;
46 }
47
48 swap(s1Arr, ind, j);
50 swap(s1Arr, ind, j);
51 }
52
53 return res;
54 }
55
56 private void swap(char [] arr, int i, int j){
57 char temp = arr[i];
58 arr[i] = arr[j];
59 arr[j] = temp;
60 }
61 }``` | 785 | 2,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-14 | latest | en | 0.545356 |
https://physics.stackexchange.com/questions/600012/meaning-of-the-eigenvalues-and-eigenvectors-of-a-quantum-operator | 1,701,736,434,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00608.warc.gz | 516,200,309 | 40,985 | # Meaning of The Eigenvalues and Eigenvectors of a Quantum Operator
This is more a check to ensure I know the physical meaning of eigenvectors and eigenvalues in quantum mechanics, and to ask the general community if this is wrong:
On some observable, represented by the operator $$\hat{Q}$$ the eigenfunction of this operator is as follows: $$\hat{Q} f = qf$$
Where $$q$$ is the eigenvalue of the operator. In this case, $$q$$ represents values that one could get while attempting to measure $$\hat{Q}$$, and the eigenvector produced by $$q$$ is the state that the system is in at the time of the measurement that produces $$q$$. Is this right?
That's not correct. Eigenvectors aren't "produced" by an eigenvalue: eigenvectors are "produced" by the operator $$\hat Q$$, inasmuch as "produced" is understood to mean "vectors (or functions) on which the operator acts by multiplying the original vectors by numbers". Synonyms for "eigenvectors" are proper vectors (in French vecteurs propres) or sometimes characteristic vectors. Characteristic vectors is good because it implies that the eigenvectors are enough to characterize the operator.
If the system is prepared in an eigenstate or eigenfunction $$f$$ of $$\hat Q$$, then measuring the observable $$Q$$ will always yield the outcome $$q$$, and $$q$$ is the only outcome from a measurement of $$\hat Q$$ with the system prepared in the state $$f$$.
If you make a measurement of $$Q$$ and obtain the result $$q$$, then yes the system will "collapse" to an eigenstate of $$Q$$, and evolve from that collapsed state thereafter. If there is more than one eigenstate with this eigenvalue, things are a little more complicated as you can get linear combinations of those eigenstates with eigenvalue $$q$$. | 415 | 1,758 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-50 | latest | en | 0.916428 |
https://math.answers.com/Q/How_do_you_write_eight_thousandths_in_decimal | 1,719,133,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00687.warc.gz | 325,364,323 | 47,331 | 0
How do you write eight thousandths in decimal?
Updated: 9/17/2023
Wiki User
14y ago
0.008
0.1 = one tenth
0.01 = one hundreth
0.001 = one thousandth
Wiki User
14y ago
Earn +20 pts
Q: How do you write eight thousandths in decimal?
Submit
Still have questions?
Related questions
How do you write eight and five thousandths in a decimal?
Eight and five thousandths = 8.005
How do you write seventy-eight and three thousandths in decimal form?
Seventy-eight and three thousandths in decimal form is 78.003
How would you write eight hundred sixteen thousandths in decimal form?
.816 is eight hundred sixteen thousandths in decimal form .816 is eight hundred sixteen thousandths in decimal form .816 is eight hundred sixteen thousandths in decimal form
How do you write sixty-eight hundred-thousandths in decimal form?
Sixty-eight hundred-thousandths written in decimal form is 0.00068
It is 0.058
.0008
8.002
It is 0.038
It is 8.003
==> 0.008
How do you write nine hundred sixty-eight thousandths in decimal form?
Nine hundred sixty-eight thousandths in decimal form is 0.968
How do you write fifty-one and twenty-eight thousandths in decimal form?
Fifty-one and twenty-eight thousandths in decimal form is 51.028 | 321 | 1,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-26 | latest | en | 0.891343 |
https://en.m.wikiversity.org/wiki/Fundamental_Mathematics/Arithmetic/Arithmetic_Equation/Linear_Equation | 1,638,001,076,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00292.warc.gz | 305,269,616 | 9,169 | # Fundamental Mathematics/Arithmetic/Arithmetic Equation/Linear Equation
## Linear equation
${\displaystyle f(x)=ax+b=0}$
In this equation,
f(x) represents function
x represents the x axis
b represents the y-intercept.
## Finding equation's root
To find value satisfy linear equation above by dividing both sides by a
${\displaystyle x+{\frac {b}{a}}=0}$
${\displaystyle x=-{\frac {b}{a}}}$
Hence, x is the root of equation
## Straight line
${\displaystyle y=ax+b}$
Gradient is worked out by :${\displaystyle rise/run\,}$ , rise being the distance a line travels vertically, and run being the distance it travels horizontally. The y-intercept (or y-int) is the point on a linear graph where a linear equation crosses the y-axis. Similarly, the x-intercept (or x-int) is the point where a linear equation crosses the x-axis.
for x=0
${\displaystyle y=a(0)+b\ =b}$ indicates y-intercept
for y=0
${\displaystyle 0=ax+b}$ indicates x-intercept
${\displaystyle x=-{\frac {b}{a}}}$
For every value of x there is a corresponding value as shown below
x 0 1 2 y b a+b 2a+b | 302 | 1,084 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-49 | latest | en | 0.821614 |
https://www.misostudy.com/question/34th-question-please | 1,721,531,268,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00596.warc.gz | 758,217,599 | 25,156 | Flat 50% off
Ends in
8929 803 804
0
## In the Young's double slit experiment, the intensities at two points $P_{1}$ and $P_{2}$ on the screen are respectively $I_{1}$ and $I_{2}$. If $P_{1}$ is located at the centre of a bright fringe and $P_{2}$ is located at a distance euqal to a quarter of fringe width from $P_{1}$, then $\frac{I_{1}}{I_{2}}$ is(a) 2(b) $\frac{1}{2}$(c) 4(d) 16
### Asked By kona cheran kumar
Updated Mon, 27 May 2019 05:16 pm | 180 | 453 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.683391 |
https://oeis.org/A004745 | 1,653,746,707,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016853.88/warc/CC-MAIN-20220528123744-20220528153744-00732.warc.gz | 483,296,311 | 4,368 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A004745 Numbers whose binary expansion does not contain 001. 8
0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 20, 21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 32, 40, 42, 43, 44, 45, 46, 47, 48, 52, 53, 54, 55, 56, 58, 59, 60, 61, 62, 63, 64, 80, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 104, 106, 107, 108, 109, 110 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008. FORMULA Sum_{n>=2} 1/a(n) = 5.808784664093998434778841785199192904637860758506854276321167162567685504669... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 13 2022 MATHEMATICA Select[Range[0, 110], ! StringContainsQ[IntegerString[#, 2], "001"] &] (* Amiram Eldar, Feb 13 2022 *) PROG (PARI) is(n)=n=binary(n); for(i=4, #n, if(n[i]&&!n[i-1]&&!n[i-2], return(0))); 1 \\ Charles R Greathouse IV, Mar 29 2013 (PARI) is(n)=while(n>8, if(bitand(n, 7)==1, return(0)); n>>=1); 1 \\ Charles R Greathouse IV, Feb 11 2017 (Haskell) a004745 n = a004745_list !! (n-1) a004745_list = filter f [0..] where f x = x < 4 || x `mod` 8 /= 1 && f (x `div` 2) -- Reinhard Zumkeller, Jul 01 2013 CROSSREFS Cf. A007088; A003796 (no 000), A004746 (no 010), A004744 (no 011), A003754 (no 100), A004742 (no 101), A004743 (no 110), A003726 (no 111). Sequence in context: A191845 A274375 A343107 * A158037 A332109 A287519 Adjacent sequences: A004742 A004743 A004744 * A004746 A004747 A004748 KEYWORD nonn,base,easy AUTHOR STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified May 28 09:09 EDT 2022. Contains 354112 sequences. (Running on oeis4.) | 850 | 2,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-21 | latest | en | 0.535216 |
https://www.geeksforgeeks.org/modify-a-binary-array-to-bitwise-and-of-all-elements-as-1/ | 1,568,918,749,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573570.6/warc/CC-MAIN-20190919183843-20190919205843-00237.warc.gz | 882,279,131 | 24,650 | # Modify a binary array to Bitwise AND of all elements as 1
Given an array a[] consisting only 0 and 1. The task is to check if it is possible to transform the array such that the AND value between every pair of indices is 1. The only operation allowed is to:
• Take two indices i and j and replace the a[i] and a[j] with a[i] | a[j] where ‘|’ means bitwise OR operation.
If it is possible then the output is “YES”, otherwise the output is “NO”.
Examples:
```Input: arr[] = {0, 1, 0, 0, 1}
Output: Yes
Choose these pair of indices (0, 1), (1, 2), (3, 4).
Input: arr[] = {0, 0, 0}
Output: No
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach: The main observation is if the array consists at least one 1 then the answer will be YES, otherwise the output will be NO because OR with 1 will give us 1 as the array consists only 0 and 1.
If there is at least one 1 then we will choose all index with 0 value and replace with OR value with the index having 1 and the OR value will be 1 always.
After all operation, the array will consist of only 1 and the AND value between any pair of indices will be 1 as (1 AND 1)=1.
Below is the implementation of above approach:
## C++
`// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` ` `// Function to check if it is possible or not ` `bool` `check(``int` `a[], ``int` `n) ` `{ ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``if` `(a[i]) ` ` ``return` `true``; ` ` ` ` ``return` `false``; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` ` ``int` `a[] = { 0, 1, 0, 1 }; ` ` ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` ` ` ` ``check(a, n) ? cout << ``"YES\n"` ` ``: cout << ``"NO\n"``; ` ` ` ` ``return` `0; ` `} `
## Java
`// Java implementation of the above approach ` `class` `GFG ` `{ ` ` ` ` ``// Function to check if it is possible or not ` ` ``static` `boolean` `check(``int` `a[], ``int` `n) ` ` ``{ ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``if` `(a[i] == ``1``) ` ` ``return` `true``; ` ` ` ` ``return` `false``; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `main (String[] args) ` ` ``{ ` ` ``int` `a[] = { ``0``, ``1``, ``0``, ``1` `}; ` ` ``int` `n = a.length; ` ` ` ` ``if``(check(a, n) == ``true` `) ` ` ``System.out.println(``"YES\n"``) ; ` ` ``else` ` ``System.out.println(``"NO\n"``); ` ` ``} ` `} ` ` ` `// This code is contributed by Ryuga `
## Python3
`# Python 3 implementation of the ` `# above approach ` ` ` `# Function to check if it is ` `# possible or not ` `def` `check(a, n): ` ` ``for` `i ``in` `range``(n): ` ` ``if` `(a[i]): ` ` ``return` `True` ` ` ` ``return` `False` ` ` `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` ` ``a ``=` `[``0``, ``1``, ``0``, ``1``] ` ` ``n ``=` `len``(a) ` ` ` ` ``if``(check(a, n)): ` ` ``print``(``"YES"``) ` ` ``else``: ` ` ``print``(``"NO"``) ` ` ` `# This code is contributed by ` `# Surendra_Gangwar `
## C#
`// C# implementation of the above approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` ` ``// Function to check if it is possible or not ` ` ``static` `bool` `check(``int` `[]a, ``int` `n) ` ` ``{ ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``if` `(a[i] == 1) ` ` ``return` `true``; ` ` ` ` ``return` `false``; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `Main () ` ` ``{ ` ` ``int` `[]a = { 0, 1, 0, 1 }; ` ` ``int` `n = a.Length; ` ` ` ` ``if``(check(a, n) == ``true` `) ` ` ``Console.Write(``"YES\n"``) ; ` ` ``else` ` ``Console.Write(``"NO\n"``); ` ` ``} ` `} ` ` ` `// This code is contributed ` `// by Akanksha Rai `
## PHP
` `
Output:
```YES
```
My Personal Notes arrow_drop_up
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
Article Tags :
Practice Tags :
1
Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 1,637 | 4,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-39 | longest | en | 0.725225 |
http://blog.janaretz.de/2019/11/19/craps-systems-dead-or-alive/ | 1,685,314,798,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00359.warc.gz | 6,969,090 | 6,483 | ## Craps Systems – Dead or Alive?
Craps Systems – Dead or Alive?
Much more Home elevators Craps Systems.Many members use several craps sporting products despite the fact that you should recognize that no craps systems can shift the home edge relating to any specific bet.
Casino craps Solutions: Introduction.Many competitors use some casino craps dissipated devices nevertheless you want to recognise that none of the craps products can go up your place benefit with all bet. Your property benefit is normally resolved through the amount of revenue any gambling house compensates financially people after you win a house game regarding the actual likelihood for ones wager.What some sort of casino craps person anticipates to achieve with the casino craps units is always to a lot more than brief flip scope fluctuations. As an example during snake eyes your rate from a 7 staying thrown is actually five in order to one. Which means that (in the long term in excess of large numbers or maybe even billions of throws) it should average out to your 7 becoming folded 1 time for every 6 rolls. But, this doesn’t actually take place around the short-term run. A player might recede 3 Times 7’s uninterruptedly, or even rotate forty conditions not having rolling whatever 7’s. Available on the market we mean by way of the short term change of craps cube probabilities, and this means that a few casino craps models may also help members revenue in some learning sessions. Your snake eyes systems other I love to make use of is actually that 10/4 Press.
Casino craps Devices: Gaming the actual Exact opposite Point.If I am bets these distribute wire, I enjoy add an area wager for the perpendicular availablility of typically the point. In the event that you appear very closely in certain cube, if you want to your top and bottom results will be telephone number a mixture opposites. The four, like, will be complete opposite of the actual eight. In particular: If your 6 is by your 2 cube by means of 5-1 (six) about 1 section, therefore sleep issues will likely be 6-2 (eight). Your jazz group of five certainly is the the complete opposite of all the 90 years in addition to the mixture of four will be complete opposite of ten.A insert chance around the 4 or 10 will pay 9 money for a 5 cent bet. Generally if the 4 or go with the 10 certainly is the idea, When i make a 5 bill position wager to the variety opposite. I then change my very own spot option up to a “buy” bet.
Snake eyes Methods: Gaming Example.Place some and also 10 meant for 5 dollars. If ever the player with the dice 7s away you are usually out and about 5 dollars. At the initially hit you may be paid out 9 dollars. Show the card dealer to assist you to advertising your wager, along with our velocity guarantee goes to help you 10 money (you can get 4 dollars). In the 2nd hit you could be paid out 18 dollars. Explain the casino dealer for you to effortlessly find Betting house Guides – Chukua Hatua the range, which means the dealership increases ones bet to 25 money and provide you with again 3 dollars.You pay back the latest five percent fee that should be compensated legitimate possibilities on your own guess, when you get yourself a number. When you acquire a 4/10, you will end up spent a couple to 1 as opposed to 9 decades to for a bet. Certain gambling houses will need you to attributes needed compensation for the acquire can guess at first, dissimilar to many people don’t.
Comments are closed. | 765 | 3,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.930625 |
https://www.supportyourtech.com/excel/how-to-do-a-random-sample-in-excel-a-step-by-step-guide/ | 1,723,323,923,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00162.warc.gz | 801,008,532 | 28,407 | # How to Do a Random Sample in Excel: A Step-by-Step Guide
How to Do a Random Sample in Excel
Creating a random sample in Excel is a quick and effective way to analyze a subset of your data without dealing with the entire dataset. To achieve this, you’ll use Excel’s built-in functions to generate random numbers and sort your data accordingly. Here’s a step-by-step guide to help you through the process.
## Step-by-Step Tutorial on How to Do a Random Sample in Excel
This tutorial will walk you through the steps to select a random sample from your data in Excel. By the end, you’ll have a randomized subset of your data ready for analysis.
### Step 1: Open Your Excel File
First, open your Excel file that contains the dataset.
Having your dataset open allows you to work directly with the data you want to analyze.
### Step 2: Insert a New Column
Add a new column next to your data. Label this column "Random" or something similar.
This new column will be used to generate random numbers for each row in your dataset.
### Step 3: Generate Random Numbers
In the first cell of the new column, type `=RAND()`. Press Enter and then copy this formula down through all the rows of your data.
The `=RAND()` function generates a random number between 0 and 1 for each row, creating the basis for your random sample.
### Step 4: Copy and Paste Values
Select the entire column with the random numbers, copy it, and then paste it as values.
Pasting as values ensures that the random numbers don’t change every time Excel recalculates.
### Step 5: Sort by Random Numbers
Highlight your entire dataset, including the new column, and sort it by the "Random" column.
Sorting your data by the random numbers will shuffle your dataset, effectively creating a random order.
### Step 6: Select Your Sample
Choose the top ‘n’ number of rows from your sorted data to form your random sample.
For example, if you need a sample size of 100, select the first 100 rows from the sorted data.
After following these steps, you’ll have a randomized subset of your data.
## Tips for How to Do a Random Sample in Excel
• Backup Your Data: Always make a copy of your original dataset before making changes.
• Consistent Sampling: Use the same random seed if you need to repeat the procedure for consistency.
• Check for Duplicates: Ensure your random sample doesn’t have unwanted duplicates by using the `Remove Duplicates` feature.
• Scalable Method: This method works for datasets of any size, from a few rows to thousands.
• Recalculate Random Numbers: If needed, press `F9` to generate new random numbers without changing your dataset.
### What is the purpose of generating random samples?
Random samples are used to analyze a subset of data, making it easier to detect patterns and make predictions without processing the entire dataset.
### Can I use this method for large datasets?
Yes, Excel handles large datasets efficiently, although the number of rows may be limited by your computer’s memory.
### Does this method work on Excel Online?
Yes, the steps are similar, but the interface might differ slightly. Always ensure your formulas and sorting options are correct.
### Can I automate this process?
Yes, you can use Excel macros to automate the process of generating random samples.
### What if my dataset includes non-numeric data?
No problem. The random number generation and sorting methods work regardless of whether your data is numeric or non-numeric.
## Summary
2. Insert a new column.
3. Generate random numbers.
4. Copy and paste values.
5. Sort by random numbers.
## Conclusion
Creating a random sample in Excel is straightforward and incredibly useful for data analysis. By following these steps, you can easily randomize your dataset and focus on a subset for more detailed analysis. This method not only saves time but also ensures that your sample is truly random, which is crucial for accurate results.
If you’re working with large datasets or need to repeat this process regularly, consider learning about Excel macros to automate the steps. Additionally, always backup your data and double-check for any duplicates to maintain the integrity of your sample.
Whether you’re a student working on a project, a researcher conducting a study, or a business analyst diving into data, mastering the art of random sampling in Excel is a valuable skill. So, dive in and start experimenting with your data today! | 920 | 4,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-33 | latest | en | 0.814296 |
https://stats.stackexchange.com/questions/tagged/linear-dynamical-system | 1,721,936,274,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00829.warc.gz | 454,708,741 | 48,336 | # Questions tagged [linear-dynamical-system]
Dynamic linear models refers to modeling problems where coefficients (as in regression) are allowed to vary with time. This is the so called state-space approach.
36 questions
Filter by
Sorted by
Tagged with
23 views
1 vote
135 views
### Choleskly constraint in mlemodel in statsmodel
I want to constraint the off diagonal terms in the covariance matrix in a dynamic linear model. I tried using Cholesky method but it does not seem to converge. I am trying to fit a multivariate CAPM ...
• 329
1 vote
51 views
63 views
### Help on statistical modeling of pedestrian flow in subways
I'm a New Yorker and take the subways every day. I have a growing interest in understanding the distribution of paths people take on the subways to work every day. I.e. if there are $n$ subway ...
• 151
1 vote
49 views
### Determining state space for a dynamic linear model
Are there any techniques for determining a good state space to use for a dynamic linear model? I'm trying to model ad-clicks with observed values being whether a user clicked on an ad and I'm curious ...
• 517
1 vote
68 views
### What should be the termination criteria for my problem with a closed loop system identification?
I have modelled a dynamic system which needs to be validated against test data. A closed loop system identification process is used for the validation. In this process, the time domain simulation of ...
270 views
### How to add stochastic drift in dynamic linear model?
As I'm not able to comment (yet), my question follows the one raised by @mzuba here I would like to use the DLM R package to model the local linear trend model, which unlike mzuba specified, has a ...
• 207
1 vote
37 views
### Predictions in a control loop like airconditions
I wonder if there are special things to consider with predictions in a control loop, e.g. An airconditioner trys to hold the target temperature at 20 degrees. I want to predict the energy consumption,...
• 1,239
667 views
### How to estimate coefficients of a state space when relevant data is provided?
I have a state space system $\dot{x}$ = $Ax$ + $Bu$ $y$ = $Cx$ I know C matrix exactly. And A matrix looks something like this, and some of the $x_{ij}$ in A are known as well. Same goes with B. \...
• 143
1 vote
317 views
### Help in CRLB for linear model
The model is an FIR (MA) filter $$x(t) = h_1 u(t-1) + h_2 u(t-2) + u(t) \tag{1}$$ $$y(t) = h^T x(t) + v(t) \tag{2}$$ $u(t)$ is a pseudo-random binary signal (PRBS) that excites/ drives the ...
• 787
1 vote
125 views
### Simulating a dynamical system
Basically I need to replicate Hartley's 'A User's Guide to Solving Real Business Cycle Models' . Specifically (to make question relevant to stats.stackexchange), I want to simulate the dynamical ...
• 410
160 views
• 13.9k
316 views
### Learning a mapping from one time series to another with a Kalman Filter
I am interested in finding the relation between two (possibly multi dimensional) time series $x_{1:T}$ and $y_{1:T}$. I wonder how I can do that with a linear dynamical system/Kalman filter. My ...
• 13.9k
1 vote | 789 | 3,131 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-30 | latest | en | 0.909747 |
https://books.google.fi/books?id=7WGBT2K_H6cC&hl=fi&lr= | 1,586,479,566,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371880945.85/warc/CC-MAIN-20200409220932-20200410011432-00244.warc.gz | 364,737,302 | 10,779 | # Mercantile Arith
Richardson, Lord & Holbrook, 1831
### Mitä ihmiset sanovat -Kirjoita arvostelu
Yhtään arvostelua ei löytynyt.
### Sisältö
Numeration 9 Practical Questions in Federal Money 22 freight 23 115 51 Vulgar Fractions 70 Practice 81 Tare and Tret 87 Mode of calculating the interest of notes with endorsements and settlement 96
Permutation 134 Extraction of the Biquadrate Root 143 Tables of Cordage 150 A Table of angles which every point and quarter point of the compass 156 Mensuration 169 Gauging 177 Exchange 184 Account of Sales of United States Eagles 190
### Suositut otteet
Sivu 2 - In conformity to the act of Congress of the United States, entitled, " An act for the encouragement of learning, by securing the copies of maps, charts and books, to the authors and proprietors of such copies, during the times therein mentioned ;
Sivu 170 - District, has deposited in this office the title of a book, the right whereof he claims as proprietor, in the words following, to wit...
Sivu 250 - District Clerk's Office. BE IT REMEMBERED, that on the tenth day of August, AD 1829, in the fifty-fourth year of the Independence of the United States of America, JP Dabney, of the said district, has deposited in this office the title of a book, the right whereof he claims as author, in the words following, to wit...
Sivu 97 - ... interest at that time due : add that interest to the principal, and from the sum subtract the payment made at that time, together with the preceding payments (if any) and the remainder forms a new principal ; on which compute and subtract the interest, as upon the first principal: And proceed in this manner to the time of judgment.
Sivu 322 - States, to me in hand paid, at or before the ensealing and delivery of these presents by of the City of , party of the second part...
Sivu 71 - Multiply the whole numbers by the denominator of the fraction, and to the product add the numerator for a new numerator, and place it over the denominator. NOTE. To express a whole number fraction-wise, set one for a denominator to the given number. EXAMPLES. 1. Reduce 5f to an improper fraction.
Sivu 72 - Multiply all the numerators together for a new numerator, and all the denominators for a new denominator: then reduce the new fraction to its lowest terms.
Sivu 131 - When the number of terms are odd, the middle term multiplred into itself will be equal to the two extremes, or any two means, equally distant from the mean : As 2, 4, 8, 16, 32, where 2X32=4X 16=8X8=64.
Sivu 12 - SIMPLE SUBTRACTION TEACHETH to take a less number from a greater of the same denomination, and thereby show the difference. The greater is called the minuend, and the less the subtrahend. Rule. Place the subtrahend, or less number, under the minuend or greater, and subtract units from units, tens from tens, and so on ; if any figure of the subtrahend be greater than the corresponding one of the minuend, add ten to the upper figure, and.
Sivu 324 - In witness whereof the master or purser of the said ship hath affirmed to three bills of lading... | 782 | 3,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-16 | latest | en | 0.841934 |
https://profound-answers.com/how-do-you-round-to-the-nearest-hundredth/ | 1,680,184,373,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949331.26/warc/CC-MAIN-20230330132508-20230330162508-00423.warc.gz | 535,928,247 | 43,587 | Collection of recommendations and tips
# How do you round to the nearest hundredth?
## How do you round to the nearest hundredth?
To round a number to the nearest hundredth , look at the next place value to the right (the thousandths this time). Same deal: If it’s 4 or less, just remove all the digits to the right. If it’s 5 or greater, add 1 to the digit in the hundredths place, and then remove all the digits to the right.
## What is 0.564 rounded to the nearest hundredth?
Rounding Decimals
A B
Round 0.564 to the nearest hundredth. 0.56
Round 4.084 to the nearest whole. 4
Round 4.81 to the nearest whole. 5
Round 5.961 to the nearest whole. 6
What is the nearest hundredth?
Nearest hundredth is the second digit after the decimal point.
### Where is the hundredth place?
The second digit to the right of the decimal point indicates the number of hundredths.
### What is 2.5 rounded off?
Both 1.5 and 2.5 are rounded to 2 . 3.5 and 4.5 are rounded to 4 .
How do you write 7 100 as a decimal?
7/100 as a decimal is 0.07.
## What place value is hundredth?
The third digit to the left of decimal point is in the hundreds place and so on. The first digit to the right of decimal point is in the tenths place. The second digit to the right of decimal point is in the hundredths place.
## What is hundredth math?
In arithmetic, a hundredth is a single part of something that has been divided equally into a hundred parts. A hundredth is the reciprocal of 100. A hundredth is written as a decimal fraction as 0.01, and as a vulgar fraction as 1/100.
What does 1.5 round to?
2
For example, 1.5 (pronounced as “one point five” or “one and a half”) would be rounded up to 2, and 2.1 would be rounded down to 2.
### What does 4.5 round to?
3.5 and 4.5 are rounded to 4 .
### What is 34 100 as a decimal?
34/100 as a decimal is 0.34.
How much is rounding to the nearest hundred?
Rounding to the nearest hundred is 800. Rounding to the nearest ten is 840. Rounding to the nearest one is 838. Rounding to the nearest tenth is 838.3. Rounding to the nearest hundredth is 838.27.
## What is the result of rounding 13495 to the nearest tenth?
So, the result of rounding the number 13495 to the nearest tenth will be 13495.00. This way, in case you are still a student, you will be showing that you know how to round numbers no matter if you are using a round to the nearest hundredth calculator or doing it manually.
## What happens when you round a number to the nearest integer?
Rounding a number involves replacing the number with an approximation of the number that results in a shorter, simpler, or more explicit representation of said number based on specific rounding definitions. For example, if rounding the number 2.7 to the nearest integer, 2.7 would be rounded to 3.
Is the decimal number 6.1234 rounded to its nearest hundredth?
The decimal number 6.1234 rounded to its nearest hundredth is 6.12, as the digit next to the hundredth digit is less than 5. Round the number 4.1269 to its nearest hundredth. | 800 | 3,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | longest | en | 0.926557 |
https://jueqingsizhe66.github.io/archives/the-second-interpreter-from-one/ | 1,653,704,091,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663011588.83/warc/CC-MAIN-20220528000300-20220528030300-00212.warc.gz | 387,267,038 | 11,841 | # The Second Interpreter from One
1. const-exp
2. diff-exp
3. zero?-exp
4. if-exp
5. var-exp
6. let-exp
## 1. 增加四则运算
### 1.1 lang.scm修改四则运算
;;;new add + * /
(expression ("+" "(" expression "," expression ")") add-exp)
(expression ("*" "(" expression "," expression ")") mult-exp)
(expression ("/" "(" expression "," expression ")") div-exp)
### 1.2 interp.scm修改四则运算
(add-exp (exp1 exp2)
(let ((val1 (value-of exp1 env))
(val2 (value-of exp2 env)))
(let ((num1 (expval->num val1))
(num2 (expval->num val2)))
(num-val
(+ num1 num2)))))
(mult-exp (exp1 exp2)
(let ((val1 (value-of exp1 env))
(val2 (value-of exp2 env)))
(let ((num1 (expval->num val1))
(num2 (expval->num val2)))
(num-val
(* num1 num2)))))
(div-exp (exp1 exp2)
(let ((val1 (value-of exp1 env))
(val2 (value-of exp2 env)))
(let ((num1 (expval->num val1))
(num2 (expval->num val2)))
(num-val
(/ num1 num2)))))
### 1.3 测试四则运算结果
(run "let x = -(4,1) in +(x,1)")
(run "let x = -(4,1) in *(x,2)")
(run "let x = -(4,1) in /(x,2)")
(num-val 4)
(num-val 6)
(num-val 1 1/2)
## 2. 增加逻辑比较
### 2.1 lang.scm修改逻辑比较
;;增加逻辑比较
(expression ("equal?" "(" expression "," expression ")") equal?-exp)
(expression ("less?" "(" expression "," expression ")") less?-exp)
(expression ("greater?" "(" expression "," expression ")") greater?-exp)
(expression ("minus" "(" expression ")") minus-exp)
### 2.2 interp.scm修改逻辑比较
scheme (equal?-exp (exp1 exp2) (let ((val1 (value-of exp1 env)) (val2 (value-of exp2 env))) (let ((num1 (expval->num val1)) (num2 (expval->num val2))) (bool-val (= num1 num2)))))
(less?-exp (exp1 exp2)
(let ((val1 (value-of exp1 env))
(val2 (value-of exp2 env)))
(let ((num1 (expval->num val1))
(num2 (expval->num val2)))
(bool-val
(< num1 num2)))))
(greater?-exp (exp1 exp2)
(let ((val1 (value-of exp1 env))
(val2 (value-of exp2 env)))
(let ((num1 (expval->num val1))
(num2 (expval->num val2)))
(bool-val
(> num1 num2)))))
(minus-exp (body-exp)
(let ((val1 (value-of body-exp env)))
(let ((num (expval->num val1)))
(num-val (- 0 num)))))
<h3 id="2.3"> 2.3 测试增加逻辑比较结果</h3>
scheme
(run "if less?(1, 2) then 1 else 2")
(run "if greater?(2, 1) then minus(1) else minus(2)")
(num-val 1)
(num-val -1)
## 3. 加入列表操作
### 3.1 lang.scm修改列表操作
;;; 增加list比较
;;new stuff
(expression ("cons" "(" expression "," expression ")") cons-exp)
(expression ("car" "(" expression ")") car-exp)
(expression ("cdr" "(" expression ")") cdr-exp)
(expression ("emptylist") emptylist-exp)
(expression ("null?" "(" expression ")") null?-exp)
### 3.2 interp.scm修改列表操作
;;;增加了list操作
(emptylist-exp ()
(emptylist-val))
(cons-exp (exp1 exp2)
(let ((val1 (value-of exp1 env))
(val2 (value-of exp2 env)))
(pair-val val1 val2)))
(car-exp (body)
(expval-car (value-of body env)))
(cdr-exp (body)
(expval-cdr (value-of body env)))
(null?-exp (exp)
(expval-null? (value-of exp env)))
### 3.3 data-structures.scm增加了expval值类型和4个操作
• 修改expval部分
(pair-val
(car expval?)
(cdr expval?))
(emptylist-val)
• 增加了4个expval类型变换操作
;;增加4个expval->操作
(define expval->pair
(lambda (v)
(cases expval v
(pair-val (car cdr)
(cons car cdr))
(else (expval-extractor-error 'pair v)))))
(define expval-car
(lambda (v)
(cases expval v
(pair-val (car cdr) car)
(else (expval-extractor-error 'car v)))))
(define expval-cdr
(lambda (v)
(cases expval v
(pair-val (car cdr) cdr)
(else (expval-extractor-error 'cdr v)))))
(define expval-null?
(lambda (v)
(cases expval v
(emptylist-val () (bool-val #t))
(else (bool-val #f)))))
### 3.4 测试修改列表操作结果
(run "cons(1, 2)")
(run "car (cons (1, 2))")
(run "cdr (cons (1, 2))")
(run "null? (emptylist)")
(run "null? (cons (1, 2))")
(run "let x = 4
in cons(x,
cons(cons(-(x,1),
emptylist),
emptylist))")
(pair-val (num-val 1) (num-val 2))
(num-val 1)
(num-val 2)
(bool-val #t)
(bool-val #f)
(pair-val (num-val 4) (pair-val (pair-val (num-val 3) (emptylist-val)) (emptylist-val)))
> (run "car (3 5 3)")
. . parsing: at line 1: looking for ")", found number 5 in production
((string "car") (string "(") (non-term expression) (string ")") (reduce #<procedure:car-exp>))
> (run "car (3,5)")
. . parsing: at line 1: looking for ")", found literal-string111 "," in production
((string "car") (string "(") (non-term expression) (string ")") (reduce #<procedure:car-exp>))
> (run "null? ()")
. . parsing: at line 1: nonterminal <expression> can't begin with literal-string111 ")"
> (run "car (cons (3,5))")
(num-val 3)
(car-exp (body) (expval-car (value-of body env)))
(pair-val
(car expval?)
(cdr expval?))
(emptylist-val)
(define expval-car
(lambda (v)
(cases expval v
(pair-val (car cdr) car) ;;v在这边是expval类型,更具体来说是pair-val,其他类型没有对应的操作
;;而在当前的情况细 pair-val只能通过cons创建！！
(else (expval-extractor-error 'car v)))))
(define-datatype expval expval?
(num-val
(value number?))
(bool-val
(boolean boolean?))
(pair-val
(car expval?)
(cdr expval?))
(emptylist-val))
(cons-exp (exp1 exp2)
(let ((val1 (value-of exp1 env))
(val2 (value-of exp2 env)))
(pair-val val1 val2)))
### 注意操作符后面不是必须直接跟上括号(空格会被直接忽略) 而是应该保证括号的对称性。
> (run "* (3,4")
. . parsing: at line 1: looking for ")", found end-marker #f in production
((string "*") (string "(") (non-term expression) (string ",") (non-term expression) (string ")") (reduce #<procedure:mult-exp>))
> (run "+(3,4)")
(num-val 7)
> (run "*(3,4)")
(num-val 12)
## 4.list的具体实现
### 4.1 lang.scm修改list具体实现
(expression ("list" "(" (separated-list expression ",") ")" ) list-exp)
### 4.2 interp.scm修改list具体实现
1. 增加一个apply-elm用于list操作。
;; used as map for the list
(define apply-elm
(lambda (env)
(lambda (elem)
(value-of elem env))))
1. 增加了list类型
;;; 增加了list操作
(list-exp (args)
(list-val (map (apply-elm env) args)))
### 4.3 data-structures.scm修改list的具体实现
;;;增加了list的具体实现
(define list-val
(lambda (args)
(if (null? args)
(emptylist-val)
(pair-val (car args)
(list-val (cdr args))))))
### 4.4 测试list具体实现的结果
(run "list(1, 2, 3)")
(run "car(cdr(list(1, 2, 3)))")
(run "let x = 4
in list(x, -(x,1), -(x,3))")
(pair-val (num-val 1) (pair-val (num-val 2) (pair-val (num-val 3) (emptylist-val))))
(num-val 2)
(pair-val (num-val 4) (pair-val (num-val 3) (pair-val (num-val 1) (emptylist-val))))
## 5. cond条件比较
### 5.1 lang.scm修改cond条件比较
;;增加了cond具体语法
;; new stuff
(expression ("cond" (arbno expression "==>" expression) "end") cond-exp)
### 5.2 interp.scm修改cond条件比较
1. 增加一个cond-val，之所以不在类似加入列表实现list具体实现添加val转换，是因为cond-val涉及到value-of 程序，所以需要放在interp.scm中,放在apply-elm之后即可.
;;增加了cond-val的处理
;;new stuff
(define cond-val
(lambda (conds acts env)
(cond ((null? conds)
(error 'cond-val "No conditions got into #t"))
((expval->bool (value-of (car conds) env))
(value-of (car acts) env))
(else
(cond-val (cdr conds) (cdr acts) env)))))
1. 在value-of中增加了具体的实现
;;;增加了cond操作
(cond-exp (conds acts)
(cond-val conds acts env))
### 5.3 测试cond条件比较结果
(run "less?(1, 2)")
(run "cond less?(1, 2) ==> 2 end")
(run "cond less?(2, 1) ==> 1 greater?(2, 2) ==> 2 greater?(3, 2) ==> 3 end")
(bool-val #t)
(num-val 2)
(num-val 3)
## 6. print显示
### 6.1 lang.scm修改print显示
;; new stuff
(expression ("print" "(" expression ")") print-exp)
### 6.2 interp.scm修改print显示
;;;增加了print
(print-exp (arg)
(let ((val (value-of arg env)))
;(print val) ;;编译不通过 改为display
(display val) ;
(num-val 1)))
### 6.3 测试print显示结果
(run "print( less? (1, 2))")
#(struct:bool-val #t)(num-val 1)
## 7. let的改进和let*的加入
### 一个问题
(run "let x = 30
in let x = -(x,1)
y = -(x,2)
in -(x, y)")
. . parsing: at line 3: looking for "in", found identifier y in production
((string "let") (term identifier) (string "=") (non-term expression) (string "in") (non-term expression) (reduce #<procedure:let-exp>))
(let-exp (var exp1 body)
(let ((val1 (value-of exp1 env)))
(value-of body
(extend-env var val1 env))))
### let改进之interp.scm 修改
1. 增加两个let-exp的辅助程序
;; let-exp的嵌套实现
(define value-of-vals
(lambda (vals env)
(if (null? vals)
'()
(cons (value-of (car vals) env)
(value-of-vals (cdr vals) env)))))
(define extend-env-list
(lambda (vars vals env)
(if (null? vars)
env
(let ((var1 (car vars))
(val1 (car vals)))
(extend-env-list (cdr vars) (cdr vals) (extend-env var1 val1 env))))))
1. 改变letexp的具体实现
; (let-exp (var exp1 body)
; (let ((val1 (value-of exp1 env)))
; (value-of body
; (extend-env var val1 env))))
(let-exp (vars vals body)
(let ((_vals (value-of-vals vals env)))
(value-of body (extend-env-list vars _vals env))))
### 7.2 let改进之lang.scm
; (expression
; ("let" identifier "=" expression "in" expression)
; let-exp)
(expression
("let" (arbno identifier "=" expression) "in" expression)
let-exp)
link: bad variable linkage;
reference to a variable that is uninitialized
reference phase level: 0
variable module: "/home/canbetter/let-lang/lang.scm"
variable phase: 0
reference in module: "/home/canbetter/let-lang/interp.scm" in: a-program?
### 7.3 let改进结果
(run "let x = -(4,1) in let x =+(x,1) in -(x,10)")
(num-val -6)
(run "let x = -(4,1) in let x =+(x,1) y=-(x,10) in -(x,y)")
(num-val 11) ;; x=4 y=-(3,10)=-7 所以-(x,y)=11而不是 10
### 7.4 let*的lang.scm具体实现
(expression ("let*" (arbno identifier "=" expression) "in" expression) let*-exp)
### 7.5 let*的interp.scm具体实现
1. 增加了一个extend-env-list-iter操作,由于也是存在value-of所以放入interp.scm中。
(define extend-env-list-iter
(lambda(vars vals env)
(if (null? vars)
env
(let ((var1 (car vars))
(val1 (value-of (car vals) env)))
(extend-env-list-iter (cdr vars) (cdr vals)
(extend-env var1 val1 env))))))
1. let*的类型定义
(let*-exp (vars vals body)
(value-of body (extend-env-list-iter vars vals env)))
### 7.6 let*的测试结果
(run "let x = 30
in let x = -(x,1)
y = -(x,2)
in -(x, y)")
(run "let x = 30
in let* x = -(x,1)
y = -(x,2)
in -(x, y)")
(num-val 1)
(num-val 2)
## 8. unpack列表赋值
### 8.1 unpack的interp.scm修改
1. 增加了extend-env-list-exp的操作，用于针对unpack的特殊的环境拓展,当然可以把它放在data-structures.scm中
;;; 关于unpack的操作
(define extend-env-list-exp
(lambda (vars vals env)
(if (null? vars)
env
(let ((var1 (car vars))
(val1 (expval-car vals)))
(extend-env-list-exp (cdr vars)
(expval-cdr vals)
(extend-env var1 val1 env))))))
1. unpack-exp的具体实现:
(unpack-exp (vars vals body)
(let ((_vals (value-of vals env)))
(value-of body (extend-env-list-exp vars _vals env))))
### 8.2 unpack的lang.scm修改
;;new stuff
(expression ("unpack" (arbno identifier) "=" expression "in" expression) unpack-exp)
### 8.3 测试unpack修改结果
;; new testcase
(run "let u = 7
in unpack x y = cons(u, cons(3,emptylist))
in -(x,y)")
(num-val 4) ;;;(x y)= (7 3) -(7,3)=4
## 结论
1[]:http://jueqingsizhe66.github.io/blog/2016/02/25/first-interpreter-from-eopl/
Related
##### 叶昭良
###### Engineer of offshore wind turbine technique research
My research interests include distributed energy, wind turbine power generation technique , Computational fluid dynamic and programmable matter. | 3,772 | 10,849 | {"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} | 2.875 | 3 | CC-MAIN-2022-21 | longest | en | 0.527728 |
https://www.khipufieldguide.com/catalog/UR1151.html | 1,722,802,475,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00250.warc.gz | 668,087,312 | 6,338 | Fieldmark Fieldmark XRay
UR1151
Colored Pendant Sum
Group Sum Bands
Ascher Decreasing Group
DataFile:
UR1151
Notes:
Ascher Databook Notes:
1. This is one of several khipus acquired by the Museum in 1907 with provenance Pachacamac. For a list of them, see UR1097.
2. By spacing, the khipu is separated into 4 groups of 1, 6, 12, and 4 pendants respectively.
1. The pendant in group 1 and the pendants in group 4 are LB and zero-valued (or blank). The subsidiaries in group 1 are W and non-zero.
2. The 6 pendants of group 2 are one pair of pendants LB, Wand then 4 LB pendants each with a W subsidiary. Each of the values in the first pair is 13; the 4 LB pendant values sum to 13; and 4 W sub_sidiaries sum to 13.
3. The 12 pendants of group 3 are 5 LB pendants with LB subsidiaries, then 5 W pendants with W subsidiaries, and then 1 pair of pendants LB, W each with a subsidiary of the same color. The first 5 LB pendant values sum to 26; then the 5 W pendant values sum to 26; and each of values in the last pair is 26. Excluding subsidiaries suspended from subsidiaries rather than from pendants, the subsidiaries also sum to 26.
4. With the exception of the subsidiaries of subsidiaries of group 3, all the values in groups 2 and 3 are summarized in Table 1.
Table 1
Group Color LB Color W
Group 2
P21 = 13
$\sum\limits_{i=3}^{6} P_{2i}\;=\;13$
P22 = 13
$\sum\limits_{i=3}^{6} P_{2i}\;subsidiaries\;=\;13$
Group 3
P3,11 = 26
$\sum\limits_{i=1}^{5} P_{3i}\; = \;26$
$\sum\limits_{i=1}^{5} P_{3i}\;\;subsidiaries\; = \;P_{3,11}\;subsidiary$
P3,12 = 26
$\sum\limits_{i=6}^{10} P_{3i}\; = \;26$
$\sum\limits_{i=1}^{5} P_{3i}\;\;subsidiaries\; = \;P_{3,12}\;subsidiary$
Group 3
$\sum\limits_{i=1}^{10} P_{3i}\;\;subsidiaries\; = \;P_{3,11}\;subsidiary + P_{3,12}\;subsidiary = 26$ | 630 | 1,785 | {"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} | 3.421875 | 3 | CC-MAIN-2024-33 | latest | en | 0.725651 |
https://www.studystack.com/flashcard-854659 | 1,513,066,662,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948515311.25/warc/CC-MAIN-20171212075935-20171212095935-00198.warc.gz | 823,569,747 | 18,159 | or
or
taken
why
Make sure to remember your password. If you forget it there is no way for StudyStack to send you a reset link. You would need to create a new account.
Don't know
Know
remaining cards
Save
0:01
Flashcards Matching Hangman Crossword Type In Quiz Test StudyStack Study Table Bug Match Hungry Bug Unscramble Chopped Targets
Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.
Normal Size Small Size show me how
# 1 step equation esms
### One Step Equations
x + 1 = 10 Minus 1
x + 5 = 25 Minus 5
6 + x = 24 Minus 6
7 + x = 100 Minus 7
x + 2 = 1000 Minus 2
x - 3 = 535 Plus 3
-4 + x = 210 Plus 4
-8 + x = 91 Plus 8
x - 9 = 100 Plus 9
x - 10 = 35 Plus 10
y / 1 = 23 Multiply 1
y / 2 = 56 Multiply 2
y / 3 = 75 Multiply 3
y / 4 = 11 Multiply 4
y / 5 = 13 Multiply 5
6y = 90 Divide by 6
7y = 105 Divide by 7
8y = 120 Divide by 8
9y = 135 Divide by 9
10y = 2000 Divide by 10
Created by: malos | 360 | 980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.756022 |
https://www.convertunits.com/from/hectare+meter/hour/to/acre+foot/second | 1,660,075,174,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571086.77/warc/CC-MAIN-20220809185452-20220809215452-00159.warc.gz | 637,878,927 | 16,962 | ››Convert hectare metre/hour to acre foot/second
hectare meter/hour acre foot/second
Did you mean to convert hectare meter/hour to acre foot/second acre foot/second [survey]
How many hectare meter/hour in 1 acre foot/second? The answer is 444.0534624.
We assume you are converting between hectare metre/hour and acre foot/second.
You can view more details on each measurement unit:
hectare meter/hour or acre foot/second
The SI derived unit for volume flow rate is the cubic meter/second.
1 cubic meter/second is equal to 0.36 hectare meter/hour, or 0.00081071319217801 acre foot/second.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between hectare meters/hour and acre feet/second.
Type in your own numbers in the form to convert the units!
››Quick conversion chart of hectare meter/hour to acre foot/second
1 hectare meter/hour to acre foot/second = 0.00225 acre foot/second
10 hectare meter/hour to acre foot/second = 0.02252 acre foot/second
50 hectare meter/hour to acre foot/second = 0.1126 acre foot/second
100 hectare meter/hour to acre foot/second = 0.2252 acre foot/second
200 hectare meter/hour to acre foot/second = 0.4504 acre foot/second
500 hectare meter/hour to acre foot/second = 1.12599 acre foot/second
1000 hectare meter/hour to acre foot/second = 2.25198 acre foot/second
››Want other units?
You can do the reverse unit conversion from acre foot/second to hectare meter/hour, or enter any two units below:
Enter two units to convert
From: To:
››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! | 533 | 2,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-33 | latest | en | 0.855937 |
http://www.cse.chalmers.se/~nad/listings/up-to/Equivalence.html | 1,552,982,171,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201922.85/warc/CC-MAIN-20190319073140-20190319095140-00172.warc.gz | 255,425,843 | 80,886 | ```------------------------------------------------------------------------
-- Equivalences
------------------------------------------------------------------------
{-# OPTIONS --without-K --safe #-}
-- Partly based on Voevodsky's work on so-called univalent
-- foundations.
open import Equality
module Equivalence
{reflexive} (eq : ∀ {a p} → Equality-with-J a p reflexive) where
open import Bijection eq as Bijection hiding (id; _∘_; inverse)
open Derived-definitions-and-properties eq
open import Equality.Decidable-UIP eq using (propositional-identity⇒set)
open import Equality.Decision-procedures eq
open import Groupoid eq
open import H-level eq as H-level
open import H-level.Closure eq
open import Injection eq using (_↣_; module _↣_; Injective)
open import Logical-equivalence as L-eq hiding (id; _∘_; inverse)
open import Nat eq
open import Preimage eq as Preimage using (_⁻¹_)
open import Prelude as P hiding (id) renaming (_∘_ to _⊚_)
open import Surjection eq as Surjection using (_↠_; module _↠_)
------------------------------------------------------------------------
-- Is-equivalence
-- A function f is an equivalence if all preimages under f are
-- contractible.
Is-equivalence : ∀ {a b} {A : Set a} {B : Set b} →
(A → B) → Set (a ⊔ b)
Is-equivalence f = ∀ y → Contractible (f ⁻¹ y)
abstract
-- Is-equivalence f is a proposition, assuming extensional equality.
propositional :
∀ {a b} → Extensionality (a ⊔ b) (a ⊔ b) →
{A : Set a} {B : Set b} (f : A → B) →
Is-proposition (Is-equivalence f)
propositional {a} ext f =
Π-closure (lower-extensionality a lzero ext) 1 λ _ →
Contractible-propositional ext
-- If the domain is contractible and the codomain is propositional,
-- then Is-equivalence f is contractible.
sometimes-contractible :
∀ {a b} → Extensionality (a ⊔ b) (a ⊔ b) →
{A : Set a} {B : Set b} {f : A → B} →
Contractible A → Is-proposition B →
Contractible (Is-equivalence f)
sometimes-contractible {a} ext A-contr B-prop =
Π-closure (lower-extensionality a lzero ext) 0 λ _ →
cojoin ext (Σ-closure 0 A-contr (λ _ → B-prop _ _))
-- Is-equivalence f is not always contractible.
not-always-contractible₁ :
∀ {a b} →
∃ λ (A : Set a) → ∃ λ (B : Set b) → ∃ λ (f : A → B) →
Is-proposition A × Contractible B ×
¬ Contractible (Is-equivalence f)
not-always-contractible₁ =
⊥ ,
↑ _ ⊤ ,
const (lift tt) ,
⊥-propositional ,
↑-closure 0 ⊤-contractible ,
λ c → ⊥-elim (proj₁ (proj₁ (proj₁ c (lift tt))))
not-always-contractible₂ :
∀ {a b} →
∃ λ (A : Set a) → ∃ λ (B : Set b) → ∃ λ (f : A → B) →
Contractible A × Is-set B ×
¬ Contractible (Is-equivalence f)
not-always-contractible₂ =
↑ _ ⊤ ,
↑ _ Bool ,
const (lift true) ,
↑-closure 0 ⊤-contractible ,
↑-closure 2 Bool-set ,
λ c → Bool.true≢false (cong lower
(proj₂ (proj₁ (proj₁ c (lift false)))))
-- Is-equivalence respects extensional equality.
respects-extensional-equality :
∀ {a b} {A : Set a} {B : Set b} {f g : A → B} →
(∀ x → f x ≡ g x) →
Is-equivalence f → Is-equivalence g
respects-extensional-equality f≡g f-eq = λ b →
H-level.respects-surjection
(_↔_.surjection (Preimage.respects-extensional-equality f≡g))
0
(f-eq b)
abstract
-- If Σ-map id f is an equivalence, then f is also an equivalence.
drop-Σ-map-id :
∀ {a b} {A : Set a} {B C : A → Set b} (f : ∀ {x} → B x → C x) →
Is-equivalence {A = Σ A B} {B = Σ A C} (Σ-map P.id f) →
∀ x → Is-equivalence (f {x = x})
drop-Σ-map-id {b = b} {A} {B} {C} f eq x z =
H-level.respects-surjection surj 0 (eq (x , z))
where
map-f : Σ A B → Σ A C
map-f = Σ-map P.id f
to-P : ∀ {x y} {p : ∃ C} → (x , f y) ≡ p → Set b
to-P {y = y} {p} _ = ∃ λ y′ → f y′ ≡ proj₂ p
to : map-f ⁻¹ (x , z) → f ⁻¹ z
to ((x′ , y) , eq) = elim¹ to-P (y , refl (f y)) eq
from : f ⁻¹ z → map-f ⁻¹ (x , z)
from (y , eq) = (x , y) , cong (_,_ x) eq
to∘from : ∀ p → to (from p) ≡ p
to∘from (y , eq) = elim¹
(λ {z′} (eq : f y ≡ z′) →
_≡_ {A = ∃ λ (y : B x) → f y ≡ z′}
(elim¹ to-P (y , refl (f y)) (cong (_,_ x) eq))
(y , eq))
(elim¹ to-P (y , refl (f y)) (cong (_,_ x) (refl (f y))) ≡⟨ cong (elim¹ to-P (y , refl (f y))) \$
cong-refl (_,_ x) {x = f y} ⟩
elim¹ to-P (y , refl (f y)) (refl (x , f y)) ≡⟨ elim¹-refl to-P _ ⟩∎
(y , refl (f y)) ∎)
eq
surj : map-f ⁻¹ (x , z) ↠ f ⁻¹ z
surj = record
{ logical-equivalence = record { to = to; from = from }
; right-inverse-of = to∘from
}
------------------------------------------------------------------------
-- _≃_
-- Equivalences.
infix 4 _≃_
record _≃_ {a b} (A : Set a) (B : Set b) : Set (a ⊔ b) where
constructor ⟨_,_⟩
field
to : A → B
is-equivalence : Is-equivalence to
-- Equivalent sets are isomorphic.
from : B → A
from y = proj₁ (proj₁ (is-equivalence y))
right-inverse-of : ∀ x → to (from x) ≡ x
right-inverse-of x = proj₂ (proj₁ (is-equivalence x))
abstract
left-inverse-of : ∀ x → from (to x) ≡ x
left-inverse-of x =
cong (proj₁ {B = λ x′ → to x′ ≡ to x}) (
proj₁ (is-equivalence (to x)) ≡⟨ proj₂ (is-equivalence (to x)) (x , refl (to x)) ⟩∎
(x , refl (to x)) ∎)
bijection : A ↔ B
bijection = record
{ surjection = record
{ logical-equivalence = record
{ to = to
; from = from
}
; right-inverse-of = right-inverse-of
}
; left-inverse-of = left-inverse-of
}
open _↔_ bijection public
hiding (from; to; right-inverse-of; left-inverse-of)
abstract
-- All preimages of an element under the equivalence are equal.
irrelevance : ∀ y (p : to ⁻¹ y) → (from y , right-inverse-of y) ≡ p
irrelevance y = proj₂ (is-equivalence y)
-- The two proofs left-inverse-of and right-inverse-of are
-- related.
left-right-lemma :
∀ x → cong to (left-inverse-of x) ≡ right-inverse-of (to x)
left-right-lemma x =
lemma₁ to _ _ (lemma₂ (irrelevance (to x) (x , refl (to x))))
where
lemma₁ : {x y : A} (f : A → B) (p : x ≡ y) (q : f x ≡ f y) →
refl (f y) ≡ trans (cong f (sym p)) q →
cong f p ≡ q
lemma₁ f = elim
(λ {x y} p → ∀ q → refl (f y) ≡ trans (cong f (sym p)) q →
cong f p ≡ q)
(λ x q hyp →
cong f (refl x) ≡⟨ cong-refl f ⟩
refl (f x) ≡⟨ hyp ⟩
trans (cong f (sym (refl x))) q ≡⟨ cong (λ p → trans (cong f p) q) sym-refl ⟩
trans (cong f (refl x)) q ≡⟨ cong (λ p → trans p q) (cong-refl f) ⟩
trans (refl (f x)) q ≡⟨ trans-reflˡ _ ⟩∎
q ∎)
lemma₂ : ∀ {f : A → B} {y} {f⁻¹y₁ f⁻¹y₂ : f ⁻¹ y}
(p : f⁻¹y₁ ≡ f⁻¹y₂) →
proj₂ f⁻¹y₂ ≡
trans (cong f (sym (cong (proj₁ {B = λ x → f x ≡ y}) p)))
(proj₂ f⁻¹y₁)
lemma₂ {f} {y} =
let pr = proj₁ {B = λ x → f x ≡ y} in
elim {A = f ⁻¹ y}
(λ {f⁻¹y₁ f⁻¹y₂} p →
proj₂ f⁻¹y₂ ≡
trans (cong f (sym (cong pr p))) (proj₂ f⁻¹y₁))
(λ f⁻¹y →
proj₂ f⁻¹y ≡⟨ sym \$ trans-reflˡ _ ⟩
trans (refl (f (proj₁ f⁻¹y))) (proj₂ f⁻¹y) ≡⟨ cong (λ p → trans p (proj₂ f⁻¹y)) (sym (cong-refl f)) ⟩
trans (cong f (refl (proj₁ f⁻¹y))) (proj₂ f⁻¹y) ≡⟨ cong (λ p → trans (cong f p) (proj₂ f⁻¹y)) (sym sym-refl) ⟩
trans (cong f (sym (refl (proj₁ f⁻¹y)))) (proj₂ f⁻¹y) ≡⟨ cong (λ p → trans (cong f (sym p)) (proj₂ f⁻¹y))
(sym (cong-refl pr)) ⟩∎
trans (cong f (sym (cong pr (refl f⁻¹y)))) (proj₂ f⁻¹y) ∎)
right-left-lemma :
∀ x → cong from (right-inverse-of x) ≡ left-inverse-of (from x)
right-left-lemma x = subst
(λ x → cong from (right-inverse-of x) ≡ left-inverse-of (from x))
(right-inverse-of x)
(let y = from x in
cong from (right-inverse-of (to y)) ≡⟨ cong (cong from) \$ sym \$ left-right-lemma y ⟩
cong from (cong to (left-inverse-of y)) ≡⟨ cong-∘ from to _ ⟩
cong (from ⊚ to) (left-inverse-of y) ≡⟨ cong-roughly-id (from ⊚ to) (λ _ → true) (left-inverse-of y)
_ _ (λ z _ → left-inverse-of z) ⟩
trans (left-inverse-of (from (to y)))
(trans (left-inverse-of y) (sym (left-inverse-of y))) ≡⟨ cong (trans _) \$ trans-symʳ _ ⟩
trans (left-inverse-of (from (to y))) (refl _) ≡⟨ trans-reflʳ _ ⟩∎
left-inverse-of (from (to y)) ∎)
-- Equivalences are isomorphic to pairs.
≃-as-Σ : ∀ {a b} {A : Set a} {B : Set b} →
A ≃ B ↔ ∃ λ (f : A → B) → Is-equivalence f
≃-as-Σ = record
{ surjection = record
{ logical-equivalence = record
{ to = λ { ⟨ f , is ⟩ → f , is }
; from = uncurry ⟨_,_⟩
}
; right-inverse-of = refl
}
; left-inverse-of = refl
}
-- Bijections are equivalences.
↔⇒≃ : ∀ {a b} {A : Set a} {B : Set b} → A ↔ B → A ≃ B
↔⇒≃ A↔B = record
{ to = to
; is-equivalence = λ y →
(from y , right-inverse-of y) , irrelevance y
}
where
open _↔_ A↔B using (to; from)
is-equivalence : Is-equivalence to
is-equivalence = Preimage.bijection⁻¹-contractible A↔B
right-inverse-of : ∀ x → to (from x) ≡ x
right-inverse-of = proj₂ ⊚ proj₁ ⊚ is-equivalence
irrelevance : ∀ y (p : to ⁻¹ y) → (from y , right-inverse-of y) ≡ p
irrelevance = proj₂ ⊚ is-equivalence
-- There is a logical equivalence between A ↔ B and A ≃ B.
↔⇔≃ :
∀ {a b} {A : Set a} {B : Set b} →
(A ↔ B) ⇔ (A ≃ B)
↔⇔≃ = record
{ to = ↔⇒≃
; from = _≃_.bijection
}
-- The function subst is an equivalence family.
subst-as-equivalence :
∀ {a p} {A : Set a} (P : A → Set p) {x y : A} (x≡y : x ≡ y) →
P x ≃ P y
subst-as-equivalence P {y = y} x≡y = ↔⇒≃ (record
{ surjection = record
{ logical-equivalence = record
{ to = subst P x≡y
; from = subst P (sym x≡y)
}
; right-inverse-of = subst-subst-sym P x≡y
}
; left-inverse-of = λ p →
subst P (sym x≡y) (subst P x≡y p) ≡⟨ cong (λ eq → subst P (sym x≡y) (subst P eq _)) \$ sym \$ sym-sym _ ⟩
subst P (sym x≡y) (subst P (sym (sym x≡y)) p) ≡⟨ subst-subst-sym P _ _ ⟩∎
p ∎
})
abstract
subst-is-equivalence :
∀ {a p} {A : Set a} (P : A → Set p) {x y : A} (x≡y : x ≡ y) →
Is-equivalence (subst P x≡y)
subst-is-equivalence P x≡y =
_≃_.is-equivalence (subst-as-equivalence P x≡y)
------------------------------------------------------------------------
-- Equivalence
-- Equivalences are equivalence relations.
id : ∀ {a} {A : Set a} → A ≃ A
id = ⟨ P.id , singleton-contractible ⟩
inverse : ∀ {a b} {A : Set a} {B : Set b} → A ≃ B → B ≃ A
inverse A≃B = ⟨ from , (λ y → (to y , left-inverse-of y) , irr y) ⟩
where
open _≃_ A≃B
abstract
irr : ∀ y (p : from ⁻¹ y) → (to y , left-inverse-of y) ≡ p
irr y (x , from-x≡y) =
Σ-≡,≡→≡ (from-to from-x≡y) (elim¹
(λ {y} ≡y → subst (λ z → from z ≡ y)
(trans (cong to (sym ≡y))
(right-inverse-of x))
(left-inverse-of y) ≡ ≡y)
(let lemma =
trans (cong to (sym (refl (from x))))
(right-inverse-of x) ≡⟨ cong (λ eq → trans (cong to eq) (right-inverse-of x)) sym-refl ⟩
trans (cong to (refl (from x)))
(right-inverse-of x) ≡⟨ cong (λ eq → trans eq (right-inverse-of x)) \$ cong-refl to ⟩
trans (refl (to (from x)))
(right-inverse-of x) ≡⟨ trans-reflˡ (right-inverse-of x) ⟩∎
right-inverse-of x ∎
in
subst (λ z → from z ≡ from x)
(trans (cong to (sym (refl (from x))))
(right-inverse-of x))
(left-inverse-of (from x)) ≡⟨ cong₂ (subst (λ z → from z ≡ from x))
lemma (sym \$ right-left-lemma x) ⟩
subst (λ z → from z ≡ from x)
(right-inverse-of x)
(cong from \$ right-inverse-of x) ≡⟨ subst-∘ (λ z → z ≡ from x) from _ ⟩
subst (λ z → z ≡ from x)
(cong from \$ right-inverse-of x)
(cong from \$ right-inverse-of x) ≡⟨ cong (λ eq → subst (λ z → z ≡ from x) eq
(cong from \$ right-inverse-of x)) \$
sym \$ sym-sym _ ⟩
subst (λ z → z ≡ from x)
(sym \$ sym \$ cong from \$ right-inverse-of x)
(cong from \$ right-inverse-of x) ≡⟨ subst-trans _ ⟩
trans (sym \$ cong from \$ right-inverse-of x)
(cong from \$ right-inverse-of x) ≡⟨ trans-symˡ _ ⟩∎
refl (from x) ∎)
from-x≡y)
infixr 9 _∘_
_∘_ : ∀ {a b c} {A : Set a} {B : Set b} {C : Set c} →
B ≃ C → A ≃ B → A ≃ C
f ∘ g = record
{ to = to
; is-equivalence = λ y →
(from y , right-inverse-of y) , irrelevance y
}
where
f∘g = ↔⇒≃ \$ Bijection._∘_ (_≃_.bijection f) (_≃_.bijection g)
to = _≃_.to f∘g
from = _≃_.from f∘g
abstract
right-inverse-of : ∀ x → to (from x) ≡ x
right-inverse-of = _≃_.right-inverse-of f∘g
irrelevance : ∀ y (p : to ⁻¹ y) → (from y , right-inverse-of y) ≡ p
irrelevance = _≃_.irrelevance f∘g
-- Equational reasoning combinators.
infix -1 finally-≃
infixr -2 step-≃
-- For an explanation of why step-≃ is defined in this way, see
-- Equality.step-≡.
step-≃ : ∀ {a b c} (A : Set a) {B : Set b} {C : Set c} →
B ≃ C → A ≃ B → A ≃ C
step-≃ _ = _∘_
syntax step-≃ A B≃C A≃B = A ≃⟨ A≃B ⟩ B≃C
finally-≃ : ∀ {a b} (A : Set a) (B : Set b) → A ≃ B → A ≃ B
finally-≃ _ _ A≃B = A≃B
syntax finally-≃ A B A≃B = A ≃⟨ A≃B ⟩□ B □
abstract
-- Some simplification lemmas.
right-inverse-of-id :
∀ {a} {A : Set a} {x : A} →
_≃_.right-inverse-of id x ≡ refl x
right-inverse-of-id {x = x} = refl (refl x)
left-inverse-of-id :
∀ {a} {A : Set a} {x : A} →
_≃_.left-inverse-of id x ≡ refl x
left-inverse-of-id {x = x} =
left-inverse-of x ≡⟨⟩
left-inverse-of (P.id x) ≡⟨ sym \$ right-left-lemma x ⟩
cong P.id (right-inverse-of x) ≡⟨ sym \$ cong-id _ ⟩
right-inverse-of x ≡⟨ right-inverse-of-id ⟩∎
refl x ∎
where open _≃_ id
right-inverse-of∘inverse :
∀ {a b} {A : Set a} {B : Set b} →
∀ (A≃B : A ≃ B) {x} →
_≃_.right-inverse-of (inverse A≃B) x ≡
_≃_.left-inverse-of A≃B x
right-inverse-of∘inverse A≃B = refl _
left-inverse-of∘inverse :
∀ {a b} {A : Set a} {B : Set b} →
∀ (A≃B : A ≃ B) {x} →
_≃_.left-inverse-of (inverse A≃B) x ≡
_≃_.right-inverse-of A≃B x
left-inverse-of∘inverse {A = A} {B} A≃B {x} =
subst (λ x → _≃_.left-inverse-of (inverse A≃B) x ≡
right-inverse-of x)
(right-inverse-of x)
(_≃_.left-inverse-of (inverse A≃B) (to (from x)) ≡⟨ sym \$ _≃_.right-left-lemma (inverse A≃B) (from x) ⟩
cong to (_≃_.right-inverse-of (inverse A≃B) (from x)) ≡⟨ cong (cong to) \$ right-inverse-of∘inverse A≃B ⟩
cong to (left-inverse-of (from x)) ≡⟨ left-right-lemma (from x) ⟩∎
right-inverse-of (to (from x)) ∎)
where open _≃_ A≃B
------------------------------------------------------------------------
-- One can replace either of the functions with an extensionally equal
-- function
with-other-function :
∀ {a b} {A : Set a} {B : Set b}
(A≃B : A ≃ B) (f : A → B) →
(∀ x → _≃_.to A≃B x ≡ f x) →
A ≃ B
with-other-function ⟨ g , is-equivalence ⟩ f g≡f =
⟨ f , respects-extensional-equality g≡f is-equivalence ⟩
with-other-inverse :
∀ {a b} {A : Set a} {B : Set b}
(A≃B : A ≃ B) (f : B → A) →
(∀ x → _≃_.from A≃B x ≡ f x) →
A ≃ B
with-other-inverse A≃B f from≡f =
inverse \$ with-other-function (inverse A≃B) f from≡f
private
-- The two functions above compute in the right way.
to∘with-other-function :
∀ {a b} {A : Set a} {B : Set b}
(A≃B : A ≃ B) (f : A → B)
(to≡f : ∀ x → _≃_.to A≃B x ≡ f x) →
_≃_.to (with-other-function A≃B f to≡f) ≡ f
to∘with-other-function _ _ _ = refl _
from∘with-other-function :
∀ {a b} {A : Set a} {B : Set b}
(A≃B : A ≃ B) (f : A → B)
(to≡f : ∀ x → _≃_.to A≃B x ≡ f x) →
_≃_.from (with-other-function A≃B f to≡f) ≡ _≃_.from A≃B
from∘with-other-function _ _ _ = refl _
to∘with-other-inverse :
∀ {a b} {A : Set a} {B : Set b}
(A≃B : A ≃ B) (g : B → A)
(from≡g : ∀ x → _≃_.from A≃B x ≡ g x) →
_≃_.to (with-other-inverse A≃B g from≡g) ≡ _≃_.to A≃B
to∘with-other-inverse _ _ _ = refl _
from∘with-other-inverse :
∀ {a b} {A : Set a} {B : Set b}
(A≃B : A ≃ B) (g : B → A)
(from≡g : ∀ x → _≃_.from A≃B x ≡ g x) →
_≃_.from (with-other-inverse A≃B g from≡g) ≡ g
from∘with-other-inverse _ _ _ = refl _
------------------------------------------------------------------------
-- The two-out-of-three property
-- If two out of three of f, g and g ∘ f are equivalences, then the
-- third one is also an equivalence.
record Two-out-of-three
{a b c} {A : Set a} {B : Set b} {C : Set c}
(f : A → B) (g : B → C) : Set (a ⊔ b ⊔ c) where
field
f-g : Is-equivalence f → Is-equivalence g → Is-equivalence (g ⊚ f)
g-g∘f : Is-equivalence g → Is-equivalence (g ⊚ f) → Is-equivalence f
g∘f-f : Is-equivalence (g ⊚ f) → Is-equivalence f → Is-equivalence g
two-out-of-three :
∀ {a b c} {A : Set a} {B : Set b} {C : Set c}
(f : A → B) (g : B → C) → Two-out-of-three f g
two-out-of-three f g = record
{ f-g = λ f-eq g-eq →
_≃_.is-equivalence (⟨ g , g-eq ⟩ ∘ ⟨ f , f-eq ⟩)
; g-g∘f = λ g-eq g∘f-eq →
respects-extensional-equality
(λ x → let g⁻¹ = _≃_.from ⟨ g , g-eq ⟩ in
g⁻¹ (g (f x)) ≡⟨ _≃_.left-inverse-of ⟨ g , g-eq ⟩ (f x) ⟩∎
f x ∎)
(_≃_.is-equivalence
(inverse ⟨ g , g-eq ⟩ ∘ ⟨ _ , g∘f-eq ⟩))
; g∘f-f = λ g∘f-eq f-eq →
respects-extensional-equality
(λ x → let f⁻¹ = _≃_.from ⟨ f , f-eq ⟩ in
g (f (f⁻¹ x)) ≡⟨ cong g (_≃_.right-inverse-of ⟨ f , f-eq ⟩ x) ⟩∎
g x ∎)
(_≃_.is-equivalence
(⟨ _ , g∘f-eq ⟩ ∘ inverse ⟨ f , f-eq ⟩))
}
------------------------------------------------------------------------
-- f ≡ g and ∀ x → f x ≡ g x are isomorphic (assuming extensionality)
private
module Separate-abstract-block where
abstract
-- Functions between contractible types are equivalences.
function-between-contractible-types-is-equivalence :
∀ {a b} {A : Set a} {B : Set b} (f : A → B) →
Contractible A → Contractible B → Is-equivalence f
function-between-contractible-types-is-equivalence f cA cB =
Two-out-of-three.g-g∘f
(two-out-of-three f (const tt))
(lemma cB)
(lemma cA)
where
-- Functions from a contractible type to the unit type are
-- contractible.
lemma : ∀ {b} {C : Set b} → Contractible C →
Is-equivalence (λ (_ : C) → tt)
lemma (x , irr) _ = (x , refl tt) , λ p →
(x , refl tt) ≡⟨ Σ-≡,≡→≡
(irr (proj₁ p))
(subst (λ _ → tt ≡ tt)
(irr (proj₁ p)) (refl tt) ≡⟨ elim (λ eq → subst (λ _ → tt ≡ tt) eq (refl tt) ≡ refl tt)
(λ _ → subst-refl (λ _ → tt ≡ tt) (refl tt))
(irr (proj₁ p)) ⟩
refl tt ≡⟨ elim (λ eq → refl tt ≡ eq) (refl ⊚ refl) (proj₂ p) ⟩∎
proj₂ p ∎) ⟩∎
p ∎
-- ext⁻¹ is an equivalence (assuming extensionality).
ext⁻¹-is-equivalence :
∀ {a b} {A : Set a} →
({B : A → Set b} → Extensionality′ A B) →
{B : A → Set b} {f g : (x : A) → B x} →
Is-equivalence (ext⁻¹ {f = f} {g = g})
ext⁻¹-is-equivalence ext {f = f} {g} =
let surj : (∀ x → Singleton (g x)) ↠ (∃ λ f → ∀ x → f x ≡ g x)
surj = record
{ logical-equivalence = record
{ to = λ f → proj₁ ⊚ f , proj₂ ⊚ f
; from = λ p x → proj₁ p x , proj₂ p x
}
; right-inverse-of = refl
}
lemma₁ : Contractible (∃ λ f → ∀ x → f x ≡ g x)
lemma₁ =
H-level.respects-surjection surj 0 \$
_⇔_.from Π-closure-contractible⇔extensionality
ext (singleton-contractible ⊚ g)
lemma₂ : Is-equivalence (Σ-map P.id ext⁻¹)
lemma₂ = function-between-contractible-types-is-equivalence
_ (singleton-contractible g) lemma₁
in drop-Σ-map-id ext⁻¹ lemma₂ f
open Separate-abstract-block public
-- f ≡ g and ∀ x → f x ≡ g x are isomorphic (assuming extensionality).
extensionality-isomorphism :
∀ {a b} →
Extensionality a b →
{A : Set a} {B : A → Set b} {f g : (x : A) → B x} →
(∀ x → f x ≡ g x) ≃ (f ≡ g)
extensionality-isomorphism ext =
inverse ⟨ _ , ext⁻¹-is-equivalence (apply-ext ext) ⟩
-- Note that the isomorphism gives us a really well-behaved notion of
-- extensionality.
good-ext : ∀ {a b} → Extensionality a b → Extensionality a b
apply-ext (good-ext ext) = _≃_.to (extensionality-isomorphism ext)
abstract
good-ext-is-equivalence :
∀ {a b} (ext : Extensionality a b) →
{A : Set a} {B : A → Set b} {f g : (x : A) → B x} →
Is-equivalence {A = ∀ x → f x ≡ g x} (apply-ext (good-ext ext))
good-ext-is-equivalence ext =
_≃_.is-equivalence (extensionality-isomorphism ext)
good-ext-refl :
∀ {a b} (ext : Extensionality a b)
{A : Set a} {B : A → Set b} (f : (x : A) → B x) →
apply-ext (good-ext ext) (λ x → refl (f x)) ≡ refl f
good-ext-refl ext f =
_≃_.to (extensionality-isomorphism ext) (λ x → refl (f x)) ≡⟨ cong (_≃_.to (extensionality-isomorphism ext)) \$ sym \$
apply-ext ext (λ _ → ext⁻¹-refl f) ⟩
_≃_.to (extensionality-isomorphism ext) (ext⁻¹ (refl f)) ≡⟨ _≃_.right-inverse-of (extensionality-isomorphism ext) _ ⟩∎
refl f ∎
cong-good-ext :
∀ {a b} (ext : Extensionality a b)
{A : Set a} {B : A → Set b} {f g : (x : A) → B x}
(f≡g : ∀ x → f x ≡ g x) {x} →
cong (_\$ x) (apply-ext (good-ext ext) f≡g) ≡ f≡g x
cong-good-ext ext f≡g {x} =
cong (_\$ x) (apply-ext (good-ext ext) f≡g) ≡⟨⟩
ext⁻¹ (apply-ext (good-ext ext) f≡g) x ≡⟨ cong (_\$ x) \$ _≃_.left-inverse-of (extensionality-isomorphism ext) f≡g ⟩∎
f≡g x ∎
subst-good-ext :
∀ {a b p} (ext : Extensionality a b)
{A : Set a} {B : A → Set b} {f g : (x : A) → B x} {x}
(P : B x → Set p) {p}
(f≡g : ∀ x → f x ≡ g x) →
subst (λ f → P (f x)) (apply-ext (good-ext ext) f≡g) p ≡
subst P (f≡g x) p
subst-good-ext ext {f = f} {g} {x} P {p} f≡g =
subst (λ f → P (f x)) (apply-ext (good-ext ext) f≡g) p ≡⟨ subst-∘ P (_\$ x) _ ⟩
subst P (cong (_\$ x) (apply-ext (good-ext ext) f≡g)) p ≡⟨ cong (λ eq → subst P eq p) (cong-good-ext ext f≡g) ⟩∎
subst P (f≡g x) p ∎
elim-good-ext :
∀ {a b p}
(ext : Extensionality a b)
{A : Set a} {B : A → Set b} {x : A}
(P : B x → B x → Set p)
(p : (y : B x) → P y y)
{f g : (x : A) → B x}
(f≡g : ∀ x → f x ≡ g x) →
elim (λ {f g} _ → P (f x) (g x)) (p ⊚ (_\$ x))
(apply-ext (good-ext ext) f≡g) ≡
elim (λ {x y} _ → P x y) p (f≡g x)
elim-good-ext ext {x = x} P p f≡g =
elim (λ {f g} _ → P (f x) (g x)) (p ⊚ (_\$ x)) (apply-ext (good-ext ext) f≡g) ≡⟨ sym \$ elim-cong _ _ _ ⟩
elim (λ {x y} _ → P x y) p (cong (_\$ x) (apply-ext (good-ext ext) f≡g)) ≡⟨ cong (elim (λ {x y} _ → P x y) p) (cong-good-ext ext f≡g) ⟩
elim (λ {x y} _ → P x y) p (f≡g x) ∎
-- I based the statements of the following three lemmas on code in
-- the Lean Homotopy Type Theory Library with Jakob von Raumer and
-- Floris van Doorn listed as authors. The file was claimed to have
-- been ported from the Coq HoTT library. (The third lemma has later
-- been generalised.)
good-ext-sym :
∀ {a b} (ext : Extensionality a b)
{A : Set a} {B : A → Set b} {f g : (x : A) → B x}
(f≡g : ∀ x → f x ≡ g x) →
apply-ext (good-ext ext) (sym ⊚ f≡g) ≡
sym (apply-ext (good-ext ext) f≡g)
good-ext-sym ext f≡g =
apply-ext (good-ext ext) (sym ⊚ f≡g) ≡⟨ cong (apply-ext (good-ext ext) ⊚ (sym ⊚_)) \$ sym \$
_≃_.left-inverse-of (extensionality-isomorphism ext) _ ⟩
apply-ext (good-ext ext)
(sym ⊚ ext⁻¹ (apply-ext (good-ext ext) f≡g)) ≡⟨⟩
apply-ext (good-ext ext) (λ x →
sym \$ cong (_\$ x) (apply-ext (good-ext ext) f≡g)) ≡⟨ cong (apply-ext (good-ext ext)) \$
apply-ext ext (λ _ → sym \$ cong-sym _ _) ⟩
apply-ext (good-ext ext) (λ x →
cong (_\$ x) (sym \$ apply-ext (good-ext ext) f≡g)) ≡⟨⟩
apply-ext (good-ext ext)
(ext⁻¹ (sym \$ apply-ext (good-ext ext) f≡g)) ≡⟨ _≃_.right-inverse-of (extensionality-isomorphism ext) _ ⟩∎
sym (apply-ext (good-ext ext) f≡g) ∎
good-ext-trans :
∀ {a b} (ext : Extensionality a b)
{A : Set a} {B : A → Set b} {f g h : (x : A) → B x}
(f≡g : ∀ x → f x ≡ g x) (g≡h : ∀ x → g x ≡ h x) →
apply-ext (good-ext ext) (λ x → trans (f≡g x) (g≡h x)) ≡
trans (apply-ext (good-ext ext) f≡g) (apply-ext (good-ext ext) g≡h)
good-ext-trans ext f≡g g≡h =
apply-ext (good-ext ext) (λ x → trans (f≡g x) (g≡h x)) ≡⟨ sym \$ cong₂ (λ f g → apply-ext (good-ext ext) (λ x → trans (f x) (g x)))
(_≃_.left-inverse-of (extensionality-isomorphism ext) _)
(_≃_.left-inverse-of (extensionality-isomorphism ext) _) ⟩
apply-ext (good-ext ext) (λ x →
trans (ext⁻¹ (apply-ext (good-ext ext) f≡g) x)
(ext⁻¹ (apply-ext (good-ext ext) g≡h) x)) ≡⟨⟩
apply-ext (good-ext ext) (λ x →
trans (cong (_\$ x) (apply-ext (good-ext ext) f≡g))
(cong (_\$ x) (apply-ext (good-ext ext) g≡h))) ≡⟨ cong (apply-ext (good-ext ext)) \$ apply-ext ext (λ _ → sym \$
cong-trans _ _ _) ⟩
apply-ext (good-ext ext) (λ x →
cong (_\$ x) (trans (apply-ext (good-ext ext) f≡g)
(apply-ext (good-ext ext) g≡h))) ≡⟨⟩
apply-ext (good-ext ext)
(ext⁻¹ (trans (apply-ext (good-ext ext) f≡g)
(apply-ext (good-ext ext) g≡h))) ≡⟨ _≃_.right-inverse-of (extensionality-isomorphism ext) _ ⟩∎
trans (apply-ext (good-ext ext) f≡g)
(apply-ext (good-ext ext) g≡h) ∎
cong-post-∘-good-ext :
∀ {a b c} {A : Set a} {B : A → Set b} {C : A → Set c}
{f g : (x : A) → B x} {h : ∀ {x} → B x → C x}
(ext₁ : Extensionality a b)
(ext₂ : Extensionality a c)
(f≡g : ∀ x → f x ≡ g x) →
cong (h ⊚_) (apply-ext (good-ext ext₁) f≡g) ≡
apply-ext (good-ext ext₂) (cong h ⊚ f≡g)
cong-post-∘-good-ext {f = f} {g} {h} ext₁ ext₂ f≡g =
cong (h ⊚_) (apply-ext (good-ext ext₁) f≡g) ≡⟨ sym \$ _≃_.right-inverse-of (extensionality-isomorphism ext₂) _ ⟩
apply-ext (good-ext ext₂)
(ext⁻¹ (cong (h ⊚_) (apply-ext (good-ext ext₁) f≡g))) ≡⟨⟩
apply-ext (good-ext ext₂) (λ x →
cong (_\$ x) (cong (h ⊚_) (apply-ext (good-ext ext₁) f≡g))) ≡⟨ cong (apply-ext (good-ext ext₂)) \$ apply-ext ext₂ (λ _ →
cong-∘ _ _ _) ⟩
apply-ext (good-ext ext₂) (λ x →
cong (λ f → h (f x)) (apply-ext (good-ext ext₁) f≡g)) ≡⟨ cong (apply-ext (good-ext ext₂)) \$ apply-ext ext₂ (λ _ → sym \$
cong-∘ _ _ _) ⟩
apply-ext (good-ext ext₂) (λ x →
cong h (cong (_\$ x) (apply-ext (good-ext ext₁) f≡g))) ≡⟨⟩
apply-ext (good-ext ext₂)
(cong h ⊚ ext⁻¹ (apply-ext (good-ext ext₁) f≡g)) ≡⟨ cong (apply-ext (good-ext ext₂) ⊚ (cong h ⊚_)) \$
_≃_.left-inverse-of (extensionality-isomorphism ext₁) _ ⟩∎
apply-ext (good-ext ext₂) (cong h ⊚ f≡g) ∎
cong-pre-∘-good-ext :
∀ {a b c} {A : Set a} {B : Set b} {C : B → Set c}
{f g : (x : B) → C x} {h : A → B}
(ext₁ : Extensionality a c)
(ext₂ : Extensionality b c)
(f≡g : ∀ x → f x ≡ g x) →
cong (_⊚ h) (apply-ext (good-ext ext₂) f≡g) ≡ apply-ext (good-ext ext₁) (f≡g ⊚ h)
cong-pre-∘-good-ext {f = f} {g} {h} ext₁ ext₂ f≡g =
cong (_⊚ h) (apply-ext (good-ext ext₂) f≡g) ≡⟨ sym \$ _≃_.right-inverse-of (extensionality-isomorphism ext₁) _ ⟩
apply-ext (good-ext ext₁)
(ext⁻¹ (cong (_⊚ h) (apply-ext (good-ext ext₂) f≡g))) ≡⟨⟩
apply-ext (good-ext ext₁) (λ x →
cong (_\$ x) (cong (_⊚ h) (apply-ext (good-ext ext₂) f≡g))) ≡⟨ cong (apply-ext (good-ext ext₁)) \$ apply-ext ext₁ (λ _ →
cong-∘ _ _ _) ⟩
apply-ext (good-ext ext₁) (λ x →
cong (_\$ h x) (apply-ext (good-ext ext₂) f≡g)) ≡⟨ cong (apply-ext (good-ext ext₁)) \$ apply-ext ext₁ (λ _ →
cong-good-ext ext₂ _) ⟩
apply-ext (good-ext ext₁) (λ x → f≡g (h x)) ≡⟨⟩
apply-ext (good-ext ext₁) (f≡g ⊚ h) ∎
------------------------------------------------------------------------
-- Groupoid
abstract
-- Two proofs of equivalence are equal if the function components
-- are equal (assuming extensionality).
--
lift-equality :
∀ {a b} → Extensionality (a ⊔ b) (a ⊔ b) →
{A : Set a} {B : Set b} {p q : A ≃ B} →
_≃_.to p ≡ _≃_.to q → p ≡ q
lift-equality {a} {b} ext {p = ⟨ f , f-eq ⟩} {q = ⟨ g , g-eq ⟩} f≡g =
elim (λ {f g} f≡g → ∀ f-eq g-eq → ⟨ f , f-eq ⟩ ≡ ⟨ g , g-eq ⟩)
(λ f f-eq g-eq →
cong (⟨_,_⟩ f)
(_⇔_.to {To = Proof-irrelevant _}
propositional⇔irrelevant
(propositional ext f) f-eq g-eq))
f≡g f-eq g-eq
-- Two proofs of equivalence are equal if the /inverses/ of the
-- function components are equal (assuming extensionality).
--
lift-equality-inverse :
∀ {a b} → Extensionality (a ⊔ b) (a ⊔ b) →
{A : Set a} {B : Set b} {p q : A ≃ B} →
_≃_.from p ≡ _≃_.from q → p ≡ q
lift-equality-inverse ext {p = p} {q = q} f≡g =
p ≡⟨ lift-equality ext (refl _) ⟩
inverse (inverse p) ≡⟨ cong inverse \$ lift-equality ext f≡g ⟩
inverse (inverse q) ≡⟨ lift-equality ext (refl _) ⟩∎
q ∎
-- _≃_ comes with a groupoid structure (assuming extensionality).
groupoid : ∀ {ℓ} → Extensionality ℓ ℓ → Groupoid (lsuc ℓ) ℓ
groupoid {ℓ} ext = record
{ Object = Set ℓ
; _∼_ = _≃_
; id = id
; _∘_ = _∘_
; _⁻¹ = inverse
; left-identity = left-identity
; right-identity = right-identity
; assoc = assoc
; left-inverse = left-inverse
; right-inverse = right-inverse
}
where
abstract
left-identity : {X Y : Set ℓ} (p : X ≃ Y) → id ∘ p ≡ p
left-identity _ = lift-equality ext (refl _)
right-identity : {X Y : Set ℓ} (p : X ≃ Y) → p ∘ id ≡ p
right-identity _ = lift-equality ext (refl _)
assoc : {W X Y Z : Set ℓ} (p : Y ≃ Z) (q : X ≃ Y) (r : W ≃ X) →
p ∘ (q ∘ r) ≡ (p ∘ q) ∘ r
assoc _ _ _ = lift-equality ext (refl _)
left-inverse : {X Y : Set ℓ} (p : X ≃ Y) → inverse p ∘ p ≡ id
left-inverse p =
lift-equality ext (apply-ext ext \$ _≃_.left-inverse-of p)
right-inverse : {X Y : Set ℓ} (p : X ≃ Y) → p ∘ inverse p ≡ id
right-inverse p =
lift-equality ext (apply-ext ext \$ _≃_.right-inverse-of p)
-- Inverse is involutive (assuming extensionality).
--
-- This property is more general than
-- Groupoid.involutive (groupoid …), because A and B do not have to
-- have the same size.
inverse-involutive :
∀ {a b} {A : Set a} {B : Set b} →
Extensionality (a ⊔ b) (a ⊔ b) →
(p : A ≃ B) →
inverse (inverse p) ≡ p
inverse-involutive ext p = lift-equality ext (refl _)
-- Inverse is an isomorphism (assuming extensionality).
--
-- This property is more general than
-- Groupoid.⁻¹-bijection (groupoid …), because A and B do not have to
-- have the same size.
inverse-isomorphism :
∀ {a b} {A : Set a} {B : Set b} →
Extensionality (a ⊔ b) (a ⊔ b) →
A ≃ B ↔ B ≃ A
inverse-isomorphism ext = record
{ surjection = record
{ logical-equivalence = record
{ to = inverse
; from = inverse
}
; right-inverse-of = inverse-involutive ext
}
; left-inverse-of = inverse-involutive ext
}
------------------------------------------------------------------------
-- A surjection from A ↔ B to A ≃ B, and related results
private
abstract
-- ↔⇒≃ is a left inverse of _≃_.bijection (assuming extensionality).
↔⇒≃-left-inverse :
∀ {a b} {A : Set a} {B : Set b} →
Extensionality (a ⊔ b) (a ⊔ b) →
(A≃B : A ≃ B) →
↔⇒≃ (_≃_.bijection A≃B) ≡ A≃B
↔⇒≃-left-inverse ext _ = lift-equality ext (refl _)
-- When sets are used ↔⇒≃ is a right inverse of _≃_.bijection
-- (assuming extensionality).
↔⇒≃-right-inverse :
∀ {a b} {A : Set a} {B : Set b} →
Extensionality (a ⊔ b) (a ⊔ b) →
Is-set A → (A↔B : A ↔ B) →
_≃_.bijection (↔⇒≃ A↔B) ≡ A↔B
↔⇒≃-right-inverse {a} {b} {B = B} ext A-set A↔B =
cong₂ (λ l r → record
{ surjection = record
{ logical-equivalence = _↔_.logical-equivalence A↔B
; right-inverse-of = r
}
; left-inverse-of = l
})
(apply-ext (lower-extensionality b b ext) λ _ →
_⇔_.to set⇔UIP A-set _ _)
(apply-ext (lower-extensionality a a ext) λ _ →
_⇔_.to set⇔UIP B-set _ _)
where
B-set : Is-set B
B-set = respects-surjection (_↔_.surjection A↔B) 2 A-set
-- There is a surjection from A ↔ B to A ≃ B (assuming
-- extensionality).
↔↠≃ :
∀ {a b} {A : Set a} {B : Set b} →
Extensionality (a ⊔ b) (a ⊔ b) →
(A ↔ B) ↠ (A ≃ B)
↔↠≃ ext = record
{ logical-equivalence = ↔⇔≃
; right-inverse-of = ↔⇒≃-left-inverse ext
}
-- When A is a set A ↔ B and A ≃ B are isomorphic (assuming
-- extensionality).
↔↔≃ :
∀ {a b} {A : Set a} {B : Set b} →
Extensionality (a ⊔ b) (a ⊔ b) →
Is-set A → (A ↔ B) ↔ (A ≃ B)
↔↔≃ ext A-set = record
{ surjection = ↔↠≃ ext
; left-inverse-of = ↔⇒≃-right-inverse ext A-set
}
-- When B is a set A ↔ B and A ≃ B are isomorphic (assuming
-- extensionality).
↔↔≃′ :
∀ {a b} {A : Set a} {B : Set b} →
Extensionality (a ⊔ b) (a ⊔ b) →
Is-set B → (A ↔ B) ↔ (A ≃ B)
↔↔≃′ ext B-set = record
{ surjection = ↔↠≃ ext
; left-inverse-of = λ A↔B →
↔⇒≃-right-inverse
ext
(H-level.respects-surjection
(_↔_.surjection \$ Bijection.inverse A↔B) 2 B-set)
A↔B
}
-- For propositional types there is a split surjection from
-- equivalence to logical equivalence.
≃↠⇔ : ∀ {a b} {A : Set a} {B : Set b} →
Is-proposition A → Is-proposition B → (A ≃ B) ↠ (A ⇔ B)
≃↠⇔ {A = A} {B} A-prop B-prop = record
{ logical-equivalence = record
{ to = _≃_.logical-equivalence
; from = ⇔→≃
}
; right-inverse-of = refl
}
where
⇔→≃ : A ⇔ B → A ≃ B
⇔→≃ A⇔B = ↔⇒≃ record
{ surjection = record
{ logical-equivalence = A⇔B
; right-inverse-of = to∘from
}
; left-inverse-of = from∘to
}
where
open _⇔_ A⇔B
abstract
to∘from : ∀ x → to (from x) ≡ x
to∘from _ = _⇔_.to propositional⇔irrelevant B-prop _ _
from∘to : ∀ x → from (to x) ≡ x
from∘to _ = _⇔_.to propositional⇔irrelevant A-prop _ _
-- For propositional types logical equivalence is isomorphic to
-- equivalence (assuming extensionality).
⇔↔≃ : ∀ {a b} → Extensionality (a ⊔ b) (a ⊔ b) →
{A : Set a} {B : Set b} →
Is-proposition A → Is-proposition B → (A ⇔ B) ↔ (A ≃ B)
⇔↔≃ ext {A} {B} A-prop B-prop = record
{ surjection = record
{ logical-equivalence = L-eq.inverse \$ _↠_.logical-equivalence \$
≃↠⇔ A-prop B-prop
; right-inverse-of = λ _ → lift-equality ext (refl _)
}
; left-inverse-of = refl
}
-- If there is a propositional, reflexive relation on A, and related
-- elements are equal, then A is a set, and (assuming extensionality)
-- the relation is equivalent to equality.
--
-- (This is more or less Theorem 7.2.2 from "Homotopy Type Theory:
-- Univalent Foundations of Mathematics" (first edition).)
propositional-identity≃≡ :
∀ {a b} {A : Set a}
(B : A → A → Set b) →
(∀ x y → Is-proposition (B x y)) →
(∀ x → B x x) →
(∀ x y → B x y → x ≡ y) →
Is-set A ×
(Extensionality (a ⊔ b) (a ⊔ b) → ∀ {x y} → B x y ≃ (x ≡ y))
propositional-identity≃≡ B B-prop B-refl f =
A-set
, λ ext →
_↔_.to (⇔↔≃ ext (B-prop _ _) (A-set _ _))
(record
{ to = f _ _
; from = λ x≡y → subst (B _) x≡y (B-refl _)
})
where
A-set = propositional-identity⇒set B B-prop B-refl f
------------------------------------------------------------------------
-- Closure, preservation
abstract
-- All h-levels are closed under the equivalence operator (assuming
-- extensionality).
h-level-closure :
∀ {a b} →
Extensionality (a ⊔ b) (a ⊔ b) →
∀ {A : Set a} {B : Set b} n →
H-level n A → H-level n B → H-level n (A ≃ B)
h-level-closure {a} {b} ext {A = A} {B} n hA hB =
H-level.respects-surjection
(_↔_.surjection \$ Bijection.inverse ≃-as-Σ) n lemma₂
where
lemma₁ : ∀ n {to : A → B} →
H-level n A → H-level n B →
H-level n (Is-equivalence to)
lemma₁ zero cA cB = sometimes-contractible ext cA (mono₁ 0 cB)
lemma₁ (suc n) _ _ = mono (m≤m+n 1 n) (propositional ext _)
lemma₂ : H-level n (∃ λ (to : A → B) → Is-equivalence to)
lemma₂ = Σ-closure n
(Π-closure (lower-extensionality b a ext) n (λ _ → hB))
(λ _ → lemma₁ n hA hB)
-- For positive h-levels it is enough if one of the sides has the
-- given h-level.
left-closure :
∀ {a b} →
Extensionality (a ⊔ b) (a ⊔ b) →
∀ {A : Set a} {B : Set b} n →
H-level (1 + n) A → H-level (1 + n) (A ≃ B)
left-closure ext {A = A} {B} n hA =
H-level.[inhabited⇒+]⇒+ n λ (A≃B : A ≃ B) →
h-level-closure ext (1 + n) hA \$
H-level.respects-surjection (_≃_.surjection A≃B) (1 + n) hA
right-closure :
∀ {a b} →
Extensionality (a ⊔ b) (a ⊔ b) →
∀ {A : Set a} {B : Set b} n →
H-level (1 + n) B → H-level (1 + n) (A ≃ B)
right-closure ext {A = A} {B} n hB =
H-level.[inhabited⇒+]⇒+ n λ (A≃B : A ≃ B) →
left-closure ext n \$
H-level.respects-surjection
(_≃_.surjection (inverse A≃B)) (1 + n) hB
-- This is not enough for level 0.
¬-left-closure :
∀ {a b} →
Extensionality (a ⊔ b) (a ⊔ b) →
∃ λ (A : Set a) → ∃ λ (B : Set b) →
Contractible A × Is-proposition B × ¬ Contractible (A ≃ B)
¬-left-closure ext =
↑ _ ⊤ ,
⊥ ,
↑-closure 0 ⊤-contractible ,
⊥-propositional ,
λ c → ⊥-elim (_≃_.to (proj₁ c) _)
¬-right-closure :
∀ {a b} →
Extensionality (a ⊔ b) (a ⊔ b) →
∃ λ (A : Set a) → ∃ λ (B : Set b) →
Is-proposition A × Contractible B × ¬ Contractible (A ≃ B)
¬-right-closure ext =
⊥ ,
↑ _ ⊤ ,
⊥-propositional ,
↑-closure 0 ⊤-contractible ,
λ c → ⊥-elim (_≃_.from (proj₁ c) _)
-- ⊥ ≃ ⊥ is contractible (assuming extensionality).
⊥≃⊥-contractible :
∀ {ℓ₁ ℓ₂} →
Extensionality (ℓ₁ ⊔ ℓ₂) (ℓ₁ ⊔ ℓ₂) →
Contractible (⊥ {ℓ = ℓ₁} ≃ ⊥ {ℓ = ℓ₂})
⊥≃⊥-contractible {ℓ₁} {ℓ₂} ext =
↔⇒≃ ⊥↔⊥ , λ ⊥↔⊥′ →
lift-equality ext \$
apply-ext (lower-extensionality ℓ₂ ℓ₁ ext) λ x → ⊥-elim x
where
⊥↔⊥ : ⊥ {ℓ = ℓ₁} ↔ ⊥ {ℓ = ℓ₂}
⊥↔⊥ = record
{ surjection = record
{ logical-equivalence = record
{ to = ⊥-elim
; from = ⊥-elim
}
; right-inverse-of = λ x → ⊥-elim x
}
; left-inverse-of = λ x → ⊥-elim x
}
-- Equalities are closed, in a strong sense, under applications of
-- equivalences.
≃-≡ : ∀ {a b} {A : Set a} {B : Set b} (A≃B : A ≃ B) {x y : A} →
let open _≃_ A≃B in
(to x ≡ to y) ≃ (x ≡ y)
≃-≡ A≃B {x} {y} = ↔⇒≃ record
{ surjection = surjection′
; left-inverse-of = left-inverse-of′
}
where
open _≃_ A≃B
surjection′ : (to x ≡ to y) ↠ (x ≡ y)
surjection′ =
Surjection.↠-≡ \$
_↔_.surjection \$
Bijection.inverse \$
_≃_.bijection A≃B
abstract
left-inverse-of′ :
∀ p → _↠_.from surjection′ (_↠_.to surjection′ p) ≡ p
left-inverse-of′ = λ to-x≡to-y →
cong to (
trans (sym (left-inverse-of x)) \$
trans (cong from to-x≡to-y) \$
left-inverse-of y) ≡⟨ cong-trans to _ _ ⟩
trans (cong to (sym (left-inverse-of x))) (
cong to (trans (cong from to-x≡to-y) (
left-inverse-of y))) ≡⟨ cong₂ trans (cong-sym to _) (cong-trans to _ _) ⟩
trans (sym (cong to (left-inverse-of x))) (
trans (cong to (cong from to-x≡to-y)) (
cong to (left-inverse-of y))) ≡⟨ cong₂ (λ eq₁ eq₂ → trans (sym eq₁) \$
trans (cong to (cong from to-x≡to-y)) \$
eq₂)
(left-right-lemma x)
(left-right-lemma y) ⟩
trans (sym (right-inverse-of (to x))) (
trans (cong to (cong from to-x≡to-y)) (
right-inverse-of (to y))) ≡⟨ _↠_.right-inverse-of (Surjection.↠-≡ \$ _≃_.surjection A≃B) to-x≡to-y ⟩∎
to-x≡to-y ∎
abstract
private
-- We can push subst through certain function applications.
push-subst :
∀ {a₁ a₂ b₁ b₂} {A₁ : Set a₁} {A₂ : Set a₂}
(B₁ : A₁ → Set b₁) {B₂ : A₂ → Set b₂}
{f : A₂ → A₁} {x₁ x₂ : A₂} {y : B₁ (f x₁)}
(g : ∀ x → B₁ (f x) → B₂ x) (eq : x₁ ≡ x₂) →
subst B₂ eq (g x₁ y) ≡ g x₂ (subst B₁ (cong f eq) y)
push-subst B₁ {B₂} {f} g eq = elim
(λ {x₁ x₂} eq → ∀ y → subst B₂ eq (g x₁ y) ≡
g x₂ (subst B₁ (cong f eq) y))
(λ x y → subst B₂ (refl x) (g x y) ≡⟨ subst-refl B₂ _ ⟩
g x y ≡⟨ sym \$ cong (g x) \$ subst-refl B₁ _ ⟩
g x (subst B₁ (refl (f x)) y) ≡⟨ cong (λ eq → g x (subst B₁ eq y)) (sym \$ cong-refl f) ⟩∎
g x (subst B₁ (cong f (refl x)) y) ∎)
eq _
push-subst′ :
∀ {a₁ a₂ b₁ b₂} {A₁ : Set a₁} {A₂ : Set a₂}
(A₁≃A₂ : A₁ ≃ A₂) (B₁ : A₁ → Set b₁) (B₂ : A₂ → Set b₂) →
let open _≃_ A₁≃A₂ in {x₁ x₂ : A₁} {y : B₁ (from (to x₁))}
(g : ∀ x → B₁ (from (to x)) → B₂ (to x)) (eq : to x₁ ≡ to x₂) →
subst B₂ eq (g x₁ y) ≡ g x₂ (subst B₁ (cong from eq) y)
push-subst′ A₁≃A₂ B₁ B₂ {x₁} {x₂} {y} g eq =
subst B₂ eq (g x₁ y) ≡⟨ cong (subst B₂ eq) \$ sym \$ g′-lemma _ _ ⟩
subst B₂ eq (g′ (to x₁) y) ≡⟨ push-subst B₁ g′ eq ⟩
g′ (to x₂) (subst B₁ (cong from eq) y) ≡⟨ g′-lemma _ _ ⟩∎
g x₂ (subst B₁ (cong from eq) y) ∎
where
open _≃_ A₁≃A₂
g′ : ∀ x′ → B₁ (from x′) → B₂ x′
g′ x′ y = subst B₂ (right-inverse-of x′) \$
g (from x′) \$
subst B₁ (sym \$ cong from \$ right-inverse-of x′) y
g′-lemma : ∀ x y → g′ (to x) y ≡ g x y
g′-lemma x y =
let lemma = λ y →
let gy = g (from (to x)) \$
subst B₁
(sym \$ cong from \$ cong to (refl _)) y in
subst B₂ (cong to (refl _)) gy ≡⟨ cong (λ p → subst B₂ p gy) \$ cong-refl to ⟩
subst B₂ (refl _) gy ≡⟨ subst-refl B₂ gy ⟩
gy ≡⟨ cong (λ p → g (from (to x)) \$ subst B₁ (sym \$ cong from p) y) \$ cong-refl to ⟩
g (from (to x))
(subst B₁ (sym \$ cong from (refl _)) y) ≡⟨ cong (λ p → g (from (to x)) \$ subst B₁ (sym p) y) \$ cong-refl from ⟩
g (from (to x))
(subst B₁ (sym (refl _)) y) ≡⟨ cong (λ p → g (from (to x)) \$ subst B₁ p y) sym-refl ⟩
g (from (to x)) (subst B₁ (refl _) y) ≡⟨ cong (g (from (to x))) \$ subst-refl B₁ y ⟩∎
g (from (to x)) y ∎
in
subst B₂ (right-inverse-of (to x))
(g (from (to x)) \$
subst B₁ (sym \$ cong from \$
right-inverse-of (to x)) y) ≡⟨ cong (λ p → subst B₂ p (g (from (to x)) \$ subst B₁ (sym \$ cong from p) y)) \$
sym \$ left-right-lemma x ⟩
subst B₂ (cong to \$ left-inverse-of x)
(g (from (to x)) \$
subst B₁ (sym \$ cong from \$ cong to \$
left-inverse-of x) y) ≡⟨ elim¹
(λ {x′} eq →
(y : B₁ (from (to x′))) →
subst B₂ (cong to eq)
(g (from (to x)) \$ subst B₁ (sym \$ cong from \$ cong to eq) y) ≡
g x′ y)
lemma
(left-inverse-of x) y ⟩∎
g x y ∎
-- If the first component is instantiated to id, then the following
-- lemmas state that ∃ preserves functions, logical equivalences,
-- injections, surjections and bijections.
∃-preserves-functions :
∀ {a₁ a₂ b₁ b₂} {A₁ : Set a₁} {A₂ : Set a₂}
{B₁ : A₁ → Set b₁} {B₂ : A₂ → Set b₂}
(A₁≃A₂ : A₁ ≃ A₂) → (∀ x → B₁ x → B₂ (_≃_.to A₁≃A₂ x)) →
Σ A₁ B₁ → Σ A₂ B₂
∃-preserves-functions A₁≃A₂ B₁→B₂ = Σ-map (_≃_.to A₁≃A₂) (B₁→B₂ _)
∃-preserves-logical-equivalences :
∀ {a₁ a₂ b₁ b₂} {A₁ : Set a₁} {A₂ : Set a₂}
{B₁ : A₁ → Set b₁} {B₂ : A₂ → Set b₂}
(A₁≃A₂ : A₁ ≃ A₂) → (∀ x → B₁ x ⇔ B₂ (_≃_.to A₁≃A₂ x)) →
Σ A₁ B₁ ⇔ Σ A₂ B₂
∃-preserves-logical-equivalences {B₂ = B₂} A₁≃A₂ B₁⇔B₂ = record
{ to = ∃-preserves-functions A₁≃A₂ (_⇔_.to ⊚ B₁⇔B₂)
; from =
∃-preserves-functions
(inverse A₁≃A₂)
(λ x y → _⇔_.from
(B₁⇔B₂ (_≃_.from A₁≃A₂ x))
(subst B₂ (sym (_≃_.right-inverse-of A₁≃A₂ x)) y))
}
∃-preserves-injections :
∀ {a₁ a₂ b₁ b₂} {A₁ : Set a₁} {A₂ : Set a₂}
{B₁ : A₁ → Set b₁} {B₂ : A₂ → Set b₂}
(A₁≃A₂ : A₁ ≃ A₂) → (∀ x → B₁ x ↣ B₂ (_≃_.to A₁≃A₂ x)) →
Σ A₁ B₁ ↣ Σ A₂ B₂
∃-preserves-injections {A₁ = A₁} {A₂} {B₁} {B₂} A₁≃A₂ B₁↣B₂ = record
{ to = to′
; injective = injective′
}
where
open _↣_
to′ : Σ A₁ B₁ → Σ A₂ B₂
to′ = ∃-preserves-functions A₁≃A₂ (_↣_.to ⊚ B₁↣B₂)
abstract
injective′ : Injective to′
injective′ {x = (x₁ , x₂)} {y = (y₁ , y₂)} =
_↔_.to Σ-≡,≡↔≡ ⊚
Σ-map (_≃_.injective A₁≃A₂) (λ {eq₁} eq₂ →
let lemma =
to (B₁↣B₂ y₁) (subst B₁ (_≃_.injective A₁≃A₂ eq₁) x₂) ≡⟨ refl _ ⟩
to (B₁↣B₂ y₁)
(subst B₁ (trans (sym (_≃_.left-inverse-of A₁≃A₂ x₁)) \$
trans (cong (_≃_.from A₁≃A₂) eq₁)
(_≃_.left-inverse-of A₁≃A₂ y₁))
x₂) ≡⟨ cong (to (B₁↣B₂ y₁)) \$ sym \$ subst-subst B₁ _ _ _ ⟩
to (B₁↣B₂ y₁)
(subst B₁ (trans (cong (_≃_.from A₁≃A₂) eq₁)
(_≃_.left-inverse-of A₁≃A₂ y₁)) \$
subst B₁ (sym (_≃_.left-inverse-of A₁≃A₂ x₁))
x₂) ≡⟨ cong (to (B₁↣B₂ y₁)) \$ sym \$ subst-subst B₁ _ _ _ ⟩
to (B₁↣B₂ y₁)
(subst B₁ (_≃_.left-inverse-of A₁≃A₂ y₁) \$
subst B₁ (cong (_≃_.from A₁≃A₂) eq₁) \$
subst B₁ (sym (_≃_.left-inverse-of A₁≃A₂ x₁)) x₂) ≡⟨ sym \$ push-subst′ A₁≃A₂ B₁ B₂
(λ x y → to (B₁↣B₂ x)
(subst B₁ (_≃_.left-inverse-of A₁≃A₂ x) y))
eq₁ ⟩
subst B₂ eq₁
(to (B₁↣B₂ x₁) \$
subst B₁ (_≃_.left-inverse-of A₁≃A₂ x₁) \$
subst B₁ (sym (_≃_.left-inverse-of A₁≃A₂ x₁)) x₂) ≡⟨ cong (subst B₂ eq₁ ⊚ to (B₁↣B₂ x₁)) \$ subst-subst B₁ _ _ _ ⟩
subst B₂ eq₁
(to (B₁↣B₂ x₁) \$
subst B₁ (trans (sym (_≃_.left-inverse-of A₁≃A₂ x₁))
(_≃_.left-inverse-of A₁≃A₂ x₁))
x₂) ≡⟨ cong (λ p → subst B₂ eq₁ (to (B₁↣B₂ x₁) (subst B₁ p x₂))) \$
trans-symˡ _ ⟩
subst B₂ eq₁ (to (B₁↣B₂ x₁) \$ subst B₁ (refl _) x₂) ≡⟨ cong (subst B₂ eq₁ ⊚ to (B₁↣B₂ x₁)) \$
subst-refl B₁ x₂ ⟩
subst B₂ eq₁ (to (B₁↣B₂ x₁) x₂) ≡⟨ eq₂ ⟩∎
to (B₁↣B₂ y₁) y₂ ∎
in
subst B₁ (_≃_.injective A₁≃A₂ eq₁) x₂ ≡⟨ _↣_.injective (B₁↣B₂ y₁) lemma ⟩∎
y₂ ∎) ⊚
Σ-≡,≡←≡
∃-preserves-surjections :
∀ {a₁ a₂ b₁ b₂} {A₁ : Set a₁} {A₂ : Set a₂}
{B₁ : A₁ → Set b₁} {B₂ : A₂ → Set b₂}
(A₁≃A₂ : A₁ ≃ A₂) → (∀ x → B₁ x ↠ B₂ (_≃_.to A₁≃A₂ x)) →
Σ A₁ B₁ ↠ Σ A₂ B₂
∃-preserves-surjections {A₁ = A₁} {A₂} {B₁} {B₂} A₁≃A₂ B₁↠B₂ = record
{ logical-equivalence = logical-equivalence′
; right-inverse-of = right-inverse-of′
}
where
open _↠_
logical-equivalence′ : Σ A₁ B₁ ⇔ Σ A₂ B₂
logical-equivalence′ =
∃-preserves-logical-equivalences A₁≃A₂ (logical-equivalence ⊚ B₁↠B₂)
abstract
right-inverse-of′ :
∀ p →
_⇔_.to logical-equivalence′ (_⇔_.from logical-equivalence′ p) ≡ p
right-inverse-of′ = λ p → Σ-≡,≡→≡
(_≃_.right-inverse-of A₁≃A₂ (proj₁ p))
(subst B₂ (_≃_.right-inverse-of A₁≃A₂ (proj₁ p))
(to (B₁↠B₂ _) (from (B₁↠B₂ _)
(subst B₂ (sym (_≃_.right-inverse-of A₁≃A₂ (proj₁ p)))
(proj₂ p)))) ≡⟨ cong (subst B₂ _) \$ right-inverse-of (B₁↠B₂ _) _ ⟩
subst B₂ (_≃_.right-inverse-of A₁≃A₂ (proj₁ p))
(subst B₂ (sym (_≃_.right-inverse-of A₁≃A₂ (proj₁ p)))
(proj₂ p)) ≡⟨ subst-subst-sym B₂ _ _ ⟩∎
proj₂ p ∎)
∃-preserves-bijections :
∀ {a₁ a₂ b₁ b₂} {A₁ : Set a₁} {A₂ : Set a₂}
{B₁ : A₁ → Set b₁} {B₂ : A₂ → Set b₂}
(A₁≃A₂ : A₁ ≃ A₂) → (∀ x → B₁ x ↔ B₂ (_≃_.to A₁≃A₂ x)) →
Σ A₁ B₁ ↔ Σ A₂ B₂
∃-preserves-bijections {A₁ = A₁} {A₂} {B₁} {B₂} A₁≃A₂ B₁↔B₂ = record
{ surjection = surjection′
; left-inverse-of = left-inverse-of′
}
where
open _↔_
surjection′ : Σ A₁ B₁ ↠ Σ A₂ B₂
surjection′ = ∃-preserves-surjections A₁≃A₂ (surjection ⊚ B₁↔B₂)
abstract
left-inverse-of′ :
∀ p → _↠_.from surjection′ (_↠_.to surjection′ p) ≡ p
left-inverse-of′ = λ p → Σ-≡,≡→≡
(_≃_.left-inverse-of A₁≃A₂ (proj₁ p))
(subst B₁ (_≃_.left-inverse-of A₁≃A₂ (proj₁ p))
(from (B₁↔B₂ _)
(subst B₂ (sym (_≃_.right-inverse-of A₁≃A₂
(_≃_.to A₁≃A₂ (proj₁ p))))
(to (B₁↔B₂ _) (proj₂ p)))) ≡⟨ push-subst B₂ (λ x → from (B₁↔B₂ x))
(_≃_.left-inverse-of A₁≃A₂ (proj₁ p)) ⟩
from (B₁↔B₂ _)
(subst B₂ (cong (_≃_.to A₁≃A₂)
(_≃_.left-inverse-of A₁≃A₂ (proj₁ p)))
(subst B₂ (sym (_≃_.right-inverse-of A₁≃A₂
(_≃_.to A₁≃A₂ (proj₁ p))))
(to (B₁↔B₂ _) (proj₂ p)))) ≡⟨ cong (λ eq → from (B₁↔B₂ _)
(subst B₂ eq
(subst B₂ (sym (_≃_.right-inverse-of A₁≃A₂ _))
(to (B₁↔B₂ _) (proj₂ p))))) \$
_≃_.left-right-lemma A₁≃A₂ _ ⟩
from (B₁↔B₂ _)
(subst B₂ (_≃_.right-inverse-of A₁≃A₂
(_≃_.to A₁≃A₂ (proj₁ p)))
(subst B₂ (sym (_≃_.right-inverse-of A₁≃A₂
(_≃_.to A₁≃A₂ (proj₁ p))))
(to (B₁↔B₂ _) (proj₂ p)))) ≡⟨ cong (from (B₁↔B₂ _)) \$ subst-subst-sym B₂ _ _ ⟩
from (B₁↔B₂ _) (to (B₁↔B₂ _) (proj₂ p)) ≡⟨ left-inverse-of (B₁↔B₂ _) _ ⟩∎
proj₂ p ∎)
-- Σ preserves equivalence.
Σ-preserves : ∀ {a₁ a₂ b₁ b₂} {A₁ : Set a₁} {A₂ : Set a₂}
{B₁ : A₁ → Set b₁} {B₂ : A₂ → Set b₂}
(A₁≃A₂ : A₁ ≃ A₂) → (∀ x → B₁ x ≃ B₂ (_≃_.to A₁≃A₂ x)) →
Σ A₁ B₁ ≃ Σ A₂ B₂
Σ-preserves A₁≃A₂ B₁≃B₂ = ↔⇒≃ \$
∃-preserves-bijections A₁≃A₂ (_≃_.bijection ⊚ B₁≃B₂)
-- A part of the implementation of ≃-preserves (see below) that does
-- not depend on extensionality.
≃-preserves-⇔ :
∀ {a₁ a₂ b₁ b₂}
{A₁ : Set a₁} {A₂ : Set a₂} {B₁ : Set b₁} {B₂ : Set b₂} →
A₁ ≃ A₂ → B₁ ≃ B₂ → (A₁ ≃ B₁) ⇔ (A₂ ≃ B₂)
≃-preserves-⇔ A₁≃A₂ B₁≃B₂ = record
{ to = λ A₁≃B₁ → B₁≃B₂ ∘ A₁≃B₁ ∘ inverse A₁≃A₂
; from = λ A₂≃B₂ → inverse B₁≃B₂ ∘ A₂≃B₂ ∘ A₁≃A₂
}
-- Equivalence preserves equivalences (assuming extensionality).
≃-preserves :
∀ {a₁ a₂ b₁ b₂} →
Extensionality (a₁ ⊔ a₂ ⊔ b₁ ⊔ b₂) (a₁ ⊔ a₂ ⊔ b₁ ⊔ b₂) →
{A₁ : Set a₁} {A₂ : Set a₂} {B₁ : Set b₁} {B₂ : Set b₂} →
A₁ ≃ A₂ → B₁ ≃ B₂ → (A₁ ≃ B₁) ≃ (A₂ ≃ B₂)
≃-preserves {a₁} {a₂} {b₁} {b₂} ext {A₁} {A₂} {B₁} {B₂} A₁≃A₂ B₁≃B₂ =
↔⇒≃ (record
{ surjection = record
{ logical-equivalence = ≃-preserves-⇔ A₁≃A₂ B₁≃B₂
; right-inverse-of = to∘from
}
; left-inverse-of = from∘to
})
where
open _≃_
ext₁ = lower-extensionality (a₁ ⊔ b₁) (a₁ ⊔ b₁) ext
ext₂ = lower-extensionality (a₂ ⊔ b₂) (a₂ ⊔ b₂) ext
abstract
to∘from :
(A₂≃B₂ : A₂ ≃ B₂) →
B₁≃B₂ ∘ (inverse B₁≃B₂ ∘ A₂≃B₂ ∘ A₁≃A₂) ∘ inverse A₁≃A₂ ≡ A₂≃B₂
to∘from A₂≃B₂ =
lift-equality ext₁ \$
apply-ext (lower-extensionality b₂ a₂ ext₁) λ x →
to B₁≃B₂ (from B₁≃B₂ (to A₂≃B₂ (to A₁≃A₂ (from A₁≃A₂ x)))) ≡⟨ right-inverse-of B₁≃B₂ _ ⟩
to A₂≃B₂ (to A₁≃A₂ (from A₁≃A₂ x)) ≡⟨ cong (to A₂≃B₂) \$ right-inverse-of A₁≃A₂ _ ⟩∎
to A₂≃B₂ x ∎
from∘to :
(A₁≃B₁ : A₁ ≃ B₁) →
inverse B₁≃B₂ ∘ (B₁≃B₂ ∘ A₁≃B₁ ∘ inverse A₁≃A₂) ∘ A₁≃A₂ ≡ A₁≃B₁
from∘to A₁≃B₁ =
lift-equality ext₂ \$
apply-ext (lower-extensionality b₁ a₁ ext₂) λ x →
from B₁≃B₂ (to B₁≃B₂ (to A₁≃B₁ (from A₁≃A₂ (to A₁≃A₂ x)))) ≡⟨ left-inverse-of B₁≃B₂ _ ⟩
to A₁≃B₁ (from A₁≃A₂ (to A₁≃A₂ x)) ≡⟨ cong (to A₁≃B₁) \$ left-inverse-of A₁≃A₂ _ ⟩∎
to A₁≃B₁ x ∎
-- Equivalence preserves bijections (assuming extensionality).
≃-preserves-bijections :
∀ {a₁ a₂ b₁ b₂} →
Extensionality (a₁ ⊔ a₂ ⊔ b₁ ⊔ b₂) (a₁ ⊔ a₂ ⊔ b₁ ⊔ b₂) →
{A₁ : Set a₁} {A₂ : Set a₂} {B₁ : Set b₁} {B₂ : Set b₂} →
A₁ ↔ A₂ → B₁ ↔ B₂ → (A₁ ≃ B₁) ↔ (A₂ ≃ B₂)
≃-preserves-bijections ext A₁↔A₂ B₁↔B₂ =
_≃_.bijection \$ ≃-preserves ext (↔⇒≃ A₁↔A₂) (↔⇒≃ B₁↔B₂)
-- Is-equivalence preserves equality, if we see _≃_ as a form of
-- equality (assuming extensionality).
Is-equivalence-preserves :
∀ {a b} {A : Set a} {B : Set b} {f g : A → B} →
Extensionality (a ⊔ b) (a ⊔ b) →
(∀ x → f x ≡ g x) →
Is-equivalence f ≃ Is-equivalence g
Is-equivalence-preserves ext f≡g =
_↠_.from (≃↠⇔ (propositional ext _) (propositional ext _)) (record
{ to = respects-extensional-equality f≡g
; from = respects-extensional-equality (sym ⊚ f≡g)
})
------------------------------------------------------------------------
-- Another property
abstract
-- As a consequence of extensionality-isomorphism and ≃-≡ we get a
-- strengthening of W-≡,≡↠≡.
W-≡,≡≃≡ : ∀ {a b} {A : Set a} {B : A → Set b} →
Extensionality b (a ⊔ b) →
∀ {x y} {f : B x → W A B} {g : B y → W A B} →
(∃ λ (p : x ≡ y) → ∀ i → f i ≡ g (subst B p i)) ≃
(sup x f ≡ sup y g)
W-≡,≡≃≡ {a} {A = A} {B} ext {x} {y} {f} {g} =
(∃ λ p → ∀ i → f i ≡ g (subst B p i)) ≃⟨ Σ-preserves id lemma ⟩
(∃ λ p → subst (λ x → B x → W A B) p f ≡ g) ≃⟨ ↔⇒≃ Σ-≡,≡↔≡ ⟩
((x , f) ≡ (y , g)) ≃⟨ ≃-≡ (↔⇒≃ W-unfolding) ⟩□
(sup x f ≡ sup y g) □
where
lemma : (p : x ≡ y) →
(∀ i → f i ≡ g (subst B p i)) ≃
(subst (λ x → B x → W A B) p f ≡ g)
lemma p = elim
(λ {x y} p → (f : B x → W A B) (g : B y → W A B) →
(∀ i → f i ≡ g (subst B p i)) ≃
(subst (λ x → B x → W A B) p f ≡ g))
(λ x f g →
(∀ i → f i ≡ g (subst B (refl x) i)) ≃⟨ subst (λ h → (∀ i → f i ≡ g (h i)) ≃ (∀ i → f i ≡ g i))
(sym (apply-ext (lower-extensionality lzero a ext) (subst-refl B)))
id ⟩
(∀ i → f i ≡ g i) ≃⟨ extensionality-isomorphism ext ⟩
(f ≡ g) ≃⟨ subst (λ h → (f ≡ g) ≃ (h ≡ g))
(sym \$ subst-refl (λ x' → B x' → W A B) f)
id ⟩□
(subst (λ x → B x → W A B) (refl x) f ≡ g) □)
p f g
``` | 20,946 | 50,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | longest | en | 0.437068 |
https://hg.ucc.asn.au/dropbear/file/DROPBEAR_2020.80/libtommath/bn_s_mp_sqr_fast.c | 1,642,954,323,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304287.0/warc/CC-MAIN-20220123141754-20220123171754-00010.warc.gz | 363,440,160 | 4,408 | ### view libtommath/bn_s_mp_sqr_fast.c @ 1734:73646de50f13DROPBEAR_2020.80
version 2020.80
author Matt Johnston Fri, 26 Jun 2020 21:45:59 +0800 1051e4eea25a
line wrap: on
line source
```#include "tommath_private.h"
#ifdef BN_S_MP_SQR_FAST_C
/* LibTomMath, multiple-precision integer library -- Tom St Denis */
/* the jist of squaring...
* you do like mult except the offset of the tmpx [one that
* starts closer to zero] can't equal the offset of tmpy.
* So basically you set up iy like before then you min it with
* (ty-tx) so that it never happens. You double all those
* you add in the inner loop
After that loop you do the squares and add them in.
*/
mp_err s_mp_sqr_fast(const mp_int *a, mp_int *b)
{
int olduse, pa, ix, iz;
mp_digit W[MP_WARRAY], *tmpx;
mp_word W1;
mp_err err;
/* grow the destination as required */
pa = a->used + a->used;
if (b->alloc < pa) {
if ((err = mp_grow(b, pa)) != MP_OKAY) {
return err;
}
}
/* number of output digits to produce */
W1 = 0;
for (ix = 0; ix < pa; ix++) {
int tx, ty, iy;
mp_word _W;
mp_digit *tmpy;
/* clear counter */
_W = 0;
/* get offsets into the two bignums */
ty = MP_MIN(a->used-1, ix);
tx = ix - ty;
/* setup temp aliases */
tmpx = a->dp + tx;
tmpy = a->dp + ty;
/* this is the number of times the loop will iterrate, essentially
while (tx++ < a->used && ty-- >= 0) { ... }
*/
iy = MP_MIN(a->used-tx, ty+1);
/* now for squaring tx can never equal ty
* we halve the distance since they approach at a rate of 2x
* and we have to round because odd cases need to be executed
*/
iy = MP_MIN(iy, ((ty-tx)+1)>>1);
/* execute loop */
for (iz = 0; iz < iy; iz++) {
_W += (mp_word)*tmpx++ * (mp_word)*tmpy--;
}
/* double the inner product and add carry */
_W = _W + _W + W1;
/* even columns have the square term in them */
if (((unsigned)ix & 1u) == 0u) {
_W += (mp_word)a->dp[ix>>1] * (mp_word)a->dp[ix>>1];
}
/* store it */
/* make next carry */
W1 = _W >> (mp_word)MP_DIGIT_BIT;
}
/* setup dest */
olduse = b->used;
b->used = a->used+a->used;
{
mp_digit *tmpb;
tmpb = b->dp;
for (ix = 0; ix < pa; ix++) { | 703 | 2,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-05 | longest | en | 0.59203 |
https://proelectricacademy.com/hydro-power-plant/transmission-lines-parameters-types-and-theory/ | 1,618,496,044,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038085599.55/warc/CC-MAIN-20210415125840-20210415155840-00385.warc.gz | 487,789,619 | 10,764 | # Transmission Lines
### What is Transmission Lines?
Transmission lines is the long conductor with special design to carry bulk amount of generated power at very high voltage from one station to another as per variation of the voltage level.
### Types of Transmission Line
In transmission line determination of voltage drop, transmission efficiency, line loss etc. are important things to design. These values are affect by line parameter R, L and C of the transmission line. Length wise transmission lines are three types.
### Short Transmission Line
• A length less than 80km (50 miles)
• Voltage level less than 69 kV
• Capacitance effect is negligible
• Only resistance and inductance are taken in calculation capacitance is neglected.
### Medium Transmission Line
A medium transmission line is classified as a transmission line with:
• A length more than 80 km (50 miles) but less than 250 km (150 miles)
• Operational voltage level is from 69 kV to approx 133 kV
• Capacitance effect is present
• Distribute capacitance form is use for calculation purpose.
### Long Transmission Line
A long transmission line is classified as a transmission line with:
• A length more than 250 km (150 miles)
• Voltage level is above 133 kV
• Line constants are consider as distribute over the length of the line
### Efficiency of Transmission Line
Transmission efficiency is define as the ration of receiving end power PR to the sending end power PS and it is express in percentage value.
cosθs is the sending end power factor.
cosθR is the receiving end power factor.
Vs is the sending end voltage per phase.
VR is the receiving end voltage per phase.
### Voltage Regulation of Transmission Line
Voltage regulation of transmission line is define as the ratio of difference between sending and receiving end voltage to receiving end voltage of a transmission line between conditions of no load and full load. It is also express in percentage.
Where, Vs is the sending end voltage per phase and VR is the receiving end voltage per phase.
XL is the reactance per phase.
R is the resistance per phase.
cosθR is the receiving end power factor.
Effect of load power factor on regulation of transmission line:
Now
• Power factor is lagging or unity, and then VR is increased and goes to be positive.
• Power factor is leading, and then VR is decreased and goes to be negative.
### Load Power Factor on Efficiency of Transmission Line
We know efficiency of transmission line is
Now, for short transmission line, IR = IS = I
So, considering three phase short transmission line,
So,
Now it is clear that to transmit given amount of power, the load current is inversely proportional to receiving end power factor.
Again in case of medium and long transmission line,
Here it is clear that transmission efficiency depends on the receiving end power factor.
### Nominal T Method in Medium Transmission Line
In the nominal T method the capacitance of the line is assumed to be concentrated at the middle point of the line, and at both side half of line resistance and inductance is lumped.
Here,
IR is the receiving end load current per phase,
R is the resistance per phase,
XL is the inductive reactance per phase,
C is the capacitance per phase,
cosΦR is the receiving end lagging power factor,
VS is the sending end voltage.
V1 is the voltage across the capacitor.
Voltage across Capacitor C,
Capacitive current
Sending end current
Sending end voltage
### Nominal π Method in Medium Transmission Line
In the nominal pi method total line capacitance is assumed to be lumped and divided into two halves to be connected across sending end and receiving end respectively. Total line resistance and inductance are assumed to be present in middle of the line.
Here IR is the receiving end load current per phase,
R is the resistance per phase,
XL is the inductive reactance per phase,
C is the capacitance per phase,
cosΦR is the receiving end lagging power factor,
VS is the sending end voltage.
Let us assume,
as the reference phasor,
The capacitive current at load end
Line current
Sending end voltage,
Charging current at the sending end is
Sending End current is
### What is Nominal T Method in Medium Transmission Line?
In this method the whole line capacitance is assumed to be concentrated at the middle point of the line and half the line resistance and reactance are lumped on its either side. Therefore in this arrangement full charging current flows over half the line. | 941 | 4,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-17 | latest | en | 0.900786 |
https://www.physicsforums.com/threads/slow-us-down.187783/ | 1,477,186,697,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719136.58/warc/CC-MAIN-20161020183839-00064-ip-10-171-6-4.ec2.internal.warc.gz | 988,319,307 | 20,470 | # Slow us down
1. Sep 29, 2007
### Mackay1011
Hi, is it possible in (theory) to somehow combine lots of atoms, protons ect and make a shield that pushes gravity outwards in every direction so the aircraft or whatever was the weightless? if that were possible, would we be able to travel near the speed of light or even 1/4 the speed of light? please anser in detail, thanks
2. Sep 29, 2007
### malawi_glenn
no you cant sheild gravity, gravity is attractive to all particles, there exist no "negative gravity" as you have in electric charge etc. The only way to not "feel" gravity is to accelerate in opposite direction with equal magnitude (Principle of equivalence).
In order to travel fast, we need much energy per mass of space craft, so there should exist some limt on travel speed.
3. Sep 30, 2007
### Mackay1011
I dont really understand much about science in general, but what is light made of? and why cant we use the atoms or whatever lights made of and use that to help us travel fast? if you understand what I mean please explain on my level lol. no complicated words :P
4. Sep 30, 2007
### malawi_glenn
Light has wave and particle properties, but what light IS.. is perhaps noone who can answer.
But your question "why cant we use the atoms or whatever lights made of and use that to help us travel fast" is really funny and very strange from a scientists point of view.
We need fuel to travel, and the more power per mass of space craft we can travel fast. According to newtons law: Force = mass*acceleration; the more force per mass -> greater Acceleration and hence greater velocity (now Iam very simplified so that you can understand..) And the most effective power source we know of is fusion (the thing that the Sun gets its energy from). And there is also restrictions on how small the space crafts could be. Due to size of humans, and the container to have the fuel in, and magnets etc that is needed to perform fusion. Fusion on earth has not been made yet, but maybe in 30-50 years, the first fusion reactor is build.
now back to your question; "cant we use the atoms or whatever lights made of and use that to help us travel fast?" Specify it first, tell me what kind of device you had in mind.
Well, we can use light as power to travel, if we had a huge sail that uses the momenta of photons to push us forward (not solar cells). But that has been made.. and when we are a certain distance from the sun, the light intesity gets too low to use this effect effective. So best we can use is fusion. As trivia, the space crafts that have been traveling (without humans) to neptune and further, have used nuclear decays for traveling power ;)
5. Sep 30, 2007
### Mackay1011
ok I understand (sort of) I have a perfect example, a ship carrys fuel, the ship has to stop every few weeks or so to re-fuel, but a submarine has a nuclear thingy so the sub dont have to stop for 20 years or so? I think scientists need to stop thing of fuel that runs out quickly, if you think about when you turn on a light what propels the light forward? speed your money inventing NEW forms or propulsion.
Example: I was thinking of something that is either already in space and isnt going to run out, or an engine of some kind that uses an extreme about of thrust and when it shoots out the back it dosent go into the air but can be used again and again? This might seem like a stupid question but to me it seems a possibility.
6. Sep 30, 2007
### malawi_glenn
but we are looking for new energy sources, the best we know of is fusion, and we are trying to control it. Fission has certain troubles, that it quite dangerous, when we use Uran as fission fuel, but there is reaserch in fission too, how to fission other nucleis than uran more effetive. And you can't just invent ANYTHING, there are constraints that nature gives us.
What do you mean by "if you think about when you turn on a light what propels the light forward?" ??????????
A thing that aleady is in space that already exists but does not run out? hehe now you are just guessing, you mean that we can use energy from stars etc? In order to get there, we must first spend approx 1 000 000 years in space travelling there, if you dont mean the sun. And how would you controll that energy from black holes etc? You can't, its just science fiction.
And an enginge that reuses its thrust over and over again sounds like an "eternal machine" and there is no such, was proved in 19th century by laws of thermodynamics and entropy.. there is always losses. And if you gather all the thrust, what would make the space craft move forward?..
You have to motivate WHY you think something is possible, othervise it is not worth anything. And you "complained" on scienteist
"speed your money inventing NEW forms or propulsion" (you must have made a typo); you demanding things that are in contradiction with nature, if you dont know so much about science, why making all these ideas? And complaining? You dont even know what it is you are complaining about.
7. Sep 30, 2007
### Mackay1011
lmao, I have an idea and I have just been searching on the net and I didnt belive what I was seeing. this is the sort of idea I had. http://video.google.co.uk/videoplay...=18&start=0&num=10&so=0&type=search&plindex=1 in this video the black thing is floating, cant you make an entire aircraft like this? and if that was possible then the fuel we use would be really good because in the video he says it is extremely easy to accelerate :D
And when I said what propels light forward, I ment like... when we speek we force the air out, what forces light to move?
Last edited: Sep 30, 2007
8. Sep 30, 2007
### malawi_glenn
That is called Meissner effect, and we can use it to move trains without friction. But we need to cool the material to very low temperatures. Or invent superconductors in room temperature... But you cant accelerate things with it. And yeah, the "black thing" is an ordinary ferromagnet.
In space, you have no friction, so the Meissner effect is just good here on earth ;)
and i dont know what is meant by "force the air out". Light is light, energy conservation from atomic transitions etc creates light.. light is among the most strange things in nature...
9. Sep 30, 2007
### Mackay1011
ok, sorry to keep throwing things at you but im extremely intrested, I found another thing on the net about anti-gravity ect http://www.ufoevidence.org/documents/doc1064.htm theres a few lines in it that says..... "examine possible uses for such a technology". Applications, the company says, could include space launch systems, artificial gravity on spacecraft, aircraft propulsion and ‘fuelless’ electricity generation — so-called ‘free energy’.
10. Sep 30, 2007
### malawi_glenn
ufoevidence.org... not a source I would trust;) I mean, does this page or the guy the refer to present their methods etc? No.. because it is secret, very secret.. hehe.. just to fool peaplo that is not into physics, so that they will not trust real phycisists nomore. Because real physics is relly complicated formulas etc, but in this pseduoscience, you can just write some nice words so that people can understand and be admired.. I say crap :)
11. Sep 30, 2007
### Mackay1011
just one more question, is there "anything" anywhere in the universe that we know about, that would repel gravity... ie inject these "things" into a golf ball for example and the throw the golf ball up into the air and it would just keep going and keep gaining speed and never stop or loose speed? if the answer is yes, I have faith in travel to diffrent solor systems, if the answer is no, we are never going to get diffrent solor systems in a decent amount of time, the one thing that is stoping us is speed, lol i no you new that :P any ideas
12. Sep 30, 2007
### malawi_glenn
look at my first post regarding "negative"
Gravity is the conection between space and time, as inteprented by Einsteins General Relativity, behaves very strange around and inside black holes.
13. Sep 30, 2007
### Mackay1011
ok, malawi glenn id like to thank you for answering all my stupid questions, but i understand a little, just one more question, were does anti-matter and or dark matter exsist? what is it, thanks for you time, ;)
14. Sep 30, 2007
### malawi_glenn
What Dark matter is, no one knows yet. It is matter that dont interact with light in ordinary manner. It is proven that dark matter is sourronding galaxies and hopes of galaxies.
Anti-matter exists only a fraction of a second, in laboratories and in processes such as decays of nuclei and particles, but are almost directly beeing annihilated (meet corresponding particle) and disapear and leaving photons left.
15. Sep 30, 2007
### Mackay1011
no, I dont really understand when I read it on the net anyway, well im going to stop thinking about this sort of stuff now, and hopefully in the next 50 years we will see how much progress we have made with speed and distance travel.
Cya malawi glenn and thanks for answer all my q's
16. Sep 30, 2007
### malawi_glenn
okay, is this way better to let a guy who meets non-physicts maybe once a month to explain to a guy who himself says he dont understand much about science? :P
17. Sep 30, 2007
### Mackay1011
oh, i just remembered my mate told me that light has been slowed down to just 60mph before???? true? and once it reached 60 mph when it was finished being slowed down would it resume its normal speed 186,000 miles per second lol, still cant get my head round that, or would it stay at 60 mph?
18. Sep 30, 2007
### Mackay1011
oh, i just remembered my mate told me that light has been slowed down to just 60mph before???? true? and once it reached 60 mph when it was finished being slowed down would it resume its normal speed 186,000 miles per second lol, still cant get my head round that, or would it stay at 60 mph?
19. Sep 30, 2007
### malawi_glenn
20. Oct 1, 2007
### JeffKoch
Didn't you already have one thread locked due to a link to a crackpot UFO web site on "anti-gravity" and a complete lack of actual physics? | 2,466 | 10,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | longest | en | 0.957026 |
https://nonsmooth.gricad-pages.univ-grenoble-alpes.fr/siconos/reference/python/siconos_kernel/LagrangianR_pyclass.html | 1,545,009,754,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828018.77/warc/CC-MAIN-20181216234902-20181217020902-00284.warc.gz | 687,665,344 | 10,683 | # siconos.kernel.LagrangianR (Python class)¶
class siconos.kernel.LagrangianR(lagType: RELATION::SUBTYPES)[source]
Bases: siconos.kernel.Relation
Lagrangian (Non Linear) Relation (generic interface)
Relations for Lagrangian Dynamical Systems. This class is only an interface for specific (Linear, scleronomic, rheomomic …) Lagrangian Relations (see derived classes).
Class name = type+subType.
If $$y = h(t,q,\dot q,\ldots)$$ describes the constraint (the relation) , all the gradients of h are handled by the following SimpleMatrix and SiconosVector objects.
• The Jacobian of the constraints with respect to the coodinates $$q$$
i.e. $$\nabla^T_q h(t,q,\dot q,\ldots)$$ is stored in SP::SimpleMatrix _jachq .
This Jacobian is mainly used for Newton linearization and to compute the time- derivative of the constraint $$y = h(q,\ldots)$$ that is $$\dot y (t) = \nabla^T_q h(t,q,\dot q,\ldots) (q) \dot q +\ldots$$ This object can also store more general linearized part of the gap function. If $$y=h(q)$$ models a gap function, then the time–derivative can be generically written as $$\dot y (t) = H(q,\ldots) \dot q +\ldots.$$ The matrix $$H(q,\ldots)$$ is also stored in SP::SimpleMatrix _jachq
• The Jacobian of the constraints with respect to the generalized velocities $$\dot q$$ i.e. $$\nabla^\top_{\dot q} h(t,q,\dot q,\ldots)$$ is stored in SP::SimpleMatrix _jachqDot
• The time-derivative of Jacobian of the constraints with respect to the generalized coordinates $$q$$ i.e. $$\frac{d}{dt} \nabla^\top_{q} h(t,q,\dot q,\ldots).$$. This value is useful to compute the second-order time–derivative of the constraints with respect to time.
In corresponding derived classes, h and Jacobians are connected to plug-in functions (user-defined).
Generated class (swig), based on C++ header Program listing for file kernel/src/modelingTools/LagrangianR.hpp.
C() -> array_like (np.float64, 2D)[source]
checkSize(inter: Interaction) → void[source]
checkSize(Interaction inter)=0 -> None
check sizes of the relation specific operators.
Parameters: inter – an Interaction using this relation
computeJacg(time: double, inter: Interaction) → void[source]
computeJacg(double time, Interaction inter)=0 -> None
computeJach(time: double, inter: Interaction) → void[source]
computeJach(double time, Interaction inter)=0 -> None
compute all the H Jacobian
Parameters: time – the current time inter – the interaction using this relation
display() → None[source]
main relation members display
dotJachq() -> array_like (np.float64, 2D)[source]
initialize(Interaction inter) → None[source]
initialize components specific to derived classes.
Parameters: inter – the interaction using this relation
jachlambda() -> array_like (np.float64, 2D)[source]
jachq() -> array_like (np.float64, 2D)[source]
get a pointer on matrix Jach[index]
Returns: a pointer on a SimpleMatrix
jachqDot() -> array_like (np.float64, 2D)[source]
setJachqPtr(array_like (np.float64, 2D) newPtr) → None[source]
set Jach[index] to pointer newPtr (pointer link)
Parameters: newPtr – the new matrix | 837 | 3,081 | {"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": 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} | 2.796875 | 3 | CC-MAIN-2018-51 | latest | en | 0.755543 |
https://lmlib.ch/2.0/html/_gallery_examples/12-filtering/example-ex122.0-polynomial-filters.html | 1,670,240,830,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711016.32/warc/CC-MAIN-20221205100449-20221205130449-00201.warc.gz | 408,433,885 | 5,298 | # Symmetric and Non-Symmetric Polynomial Filters with ALSSMs [ex122.0]#
Applies Composite Costs of polynomials of degrees N=0..6.
Out:
```Badly Conditioned Steady State Matrix W: Use larger boundaries or lower g.
Badly Conditioned Steady State Matrix W: Use larger boundaries or lower g.
Badly Conditioned Steady State Matrix W: Use larger boundaries or lower g.
Badly Conditioned Steady State Matrix W: Use larger boundaries or lower g.
Badly Conditioned Steady State Matrix W: Use larger boundaries or lower g.
Badly Conditioned Steady State Matrix W: Use larger boundaries or lower g.
Badly Conditioned Steady State Matrix W: Use larger boundaries or lower g.
Badly Conditioned Steady State Matrix W: Use larger boundaries or lower g.
```
```import matplotlib.pyplot as plt
import numpy as np
import lmlib as lm
from lmlib.utils.generator import gen_rect
import time
# --- Generating test signal ---
K = 2000
k = np.arange(K)
y = gen_rect(K, 500,250)
# --- ALSSM Filtering ---
y_hats_sym = []
y_hats_left = []
for i in range(0,6):
# Polynomial ALSSM
alssm_poly = lm.AlssmPoly(poly_degree=i)
# Segments
segment_left = lm.Segment(a=-np.inf, b=-1, direction=lm.FORWARD, g=25)
segment_right = lm.Segment(a=0, b=np.inf, direction=lm.BACKWARD, g=25)
# -- Symmetric Filter --
# CompsiteCost
costs = lm.CompositeCost((alssm_poly,), (segment_left, segment_right), F=[[1, 1]])
# filter signal and take the approximation
xs = rls.filter_minimize_x(y)
# extracts filtered signal
y_hats_sym.append(costs.eval_alssm_output(xs, alssm_weights=[1]))
# -- Left-Sided Filter --
# CompsiteCost
costs = lm.CompositeCost((alssm_poly,), (segment_left, segment_right), F=[[1, 0]])
# filter signal and take the approximation
xs = rls.filter_minimize_x(y)
# extracts filtered signal
y_hats_left.append(costs.eval_alssm_output(xs, alssm_weights=[1]))
# --- Plotting ----
STYLES = ['tab:blue','tab:orange','tab:green','tab:red','tab:purple','tab:brown']
fig, ax = plt.subplots(2, sharex='all', figsize=(10,6))
ax[0].plot(k, y, lw=0.6, c='gray', label=rf'\$y\$')
for (i, y_hat) in enumerate(y_hats_sym):
ax[0].plot(k, y_hat, STYLES[i], lw=1, label=r'\$N='+str(i)+'\$')
ax[0].legend(loc='upper right')
ax[0].set_title('Left- and Right-Sided CostSegment (Symmetric)')
ax[1].plot(k, y, lw=0.6, c='gray', label=rf'\$y\$')
for (i, y_hat) in enumerate(y_hats_left):
ax[1].plot(k, y_hat, STYLES[i], lw=1, label=r'\$N='+str(i)+'\$')
ax[1].legend(loc='upper right')
ax[1].set_title('Left-Sided CostSegment only (non-symmetric)')
ax[1].set_xlabel('k')
plt.show()
```
Total running time of the script: ( 0 minutes 0.691 seconds)
Gallery generated by Sphinx-Gallery | 794 | 2,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | latest | en | 0.600554 |
http://www.brainbashers.com/showanswer.asp?ref=ZQPD | 1,481,120,712,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542213.61/warc/CC-MAIN-20161202170902-00465-ip-10-31-129-80.ec2.internal.warc.gz | 367,335,870 | 3,250 | [OK] We use cookies to personalise content and adverts. We also share information about your use of our site with our advertising partners who may combine it with other information you’ve provided to them or they’ve collected from your use of their services. You also agree to our terms and conditions (T&C).
Peter throws two darts at a dartboard, aiming for the centre. The second dart lands farther from the centre than the first. If Peter now throws another dart at the board, aiming for the centre, what is the probability that this third throw is also worse (i.e., farther from the centre) than his first? Assume Peter's skillfulness is constant.
[Ref: ZQPD]
Answer: Since the three darts are thrown independently, they each have a 1/3 chance of being the best throw. As long as the third dart is not the best throw, it will be worse than the first dart. Therefore the answer is 2/3.
Back to the puzzles...
Our Favourite Illusions
Shadow Illusion What Am I? Hidden Faces Impossible Waterfall? Same Eyes? Impossible Prongs?
Duck Or Rabbit? Spinning Dancer Who Turned To? Blind Spot The Butterfly Parallel Cafe Wall Lines? | 253 | 1,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-50 | latest | en | 0.936454 |
https://number.academy/124488 | 1,660,611,469,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00281.warc.gz | 407,707,037 | 13,186 | # Number 124488
Number 124,488 spell 🔊, write in words: one hundred and twenty-four thousand, four hundred and eighty-eight . Ordinal number 124488th is said 🔊 and write: one hundred and twenty-four thousand, four hundred and eighty-eighth. Color #124488. The meaning of number 124488 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 124488. What is 124488 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 124488.
## What is 124,488 in other units
The decimal (Arabic) number 124488 converted to a Roman number is (C)(X)(X)(IV)CDLXXXVIII. Roman and decimal number conversions.
#### Weight conversion
124488 kilograms (kg) = 274446.2 pounds (lbs)
124488 pounds (lbs) = 56467.4 kilograms (kg)
#### Length conversion
124488 kilometers (km) equals to 77354 miles (mi).
124488 miles (mi) equals to 200345 kilometers (km).
124488 meters (m) equals to 408421 feet (ft).
124488 feet (ft) equals 37945 meters (m).
124488 centimeters (cm) equals to 49011.0 inches (in).
124488 inches (in) equals to 316199.5 centimeters (cm).
#### Temperature conversion
124488° Fahrenheit (°F) equals to 69142.2° Celsius (°C)
124488° Celsius (°C) equals to 224110.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
124488 seconds equals to 1 day, 10 hours, 34 minutes, 48 seconds
124488 minutes equals to 3 months, 2 days, 10 hours, 48 minutes
### Codes and images of the number 124488
Number 124488 morse code: .---- ..--- ....- ....- ---.. ---..
Sign language for number 124488:
Number 124488 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 124488
### Multiplications
#### Multiplication table of 124488
124488 multiplied by two equals 248976 (124488 x 2 = 248976).
124488 multiplied by three equals 373464 (124488 x 3 = 373464).
124488 multiplied by four equals 497952 (124488 x 4 = 497952).
124488 multiplied by five equals 622440 (124488 x 5 = 622440).
124488 multiplied by six equals 746928 (124488 x 6 = 746928).
124488 multiplied by seven equals 871416 (124488 x 7 = 871416).
124488 multiplied by eight equals 995904 (124488 x 8 = 995904).
124488 multiplied by nine equals 1120392 (124488 x 9 = 1120392).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 124488
Half of 124488 is 62244 (124488 / 2 = 62244).
One third of 124488 is 41496 (124488 / 3 = 41496).
One quarter of 124488 is 31122 (124488 / 4 = 31122).
One fifth of 124488 is 24897,6 (124488 / 5 = 24897,6 = 24897 3/5).
One sixth of 124488 is 20748 (124488 / 6 = 20748).
One seventh of 124488 is 17784 (124488 / 7 = 17784).
One eighth of 124488 is 15561 (124488 / 8 = 15561).
One ninth of 124488 is 13832 (124488 / 9 = 13832).
show fractions by 6, 7, 8, 9 ...
### Calculator
124488
#### Is Prime?
The number 124488 is not a prime number. The closest prime numbers are 124477, 124489.
#### Factorization and factors (dividers)
The prime factors of 124488 are 2 * 2 * 2 * 3 * 3 * 7 * 13 * 19
The factors of 124488 are
1 , 2 , 3 , 4 , 6 , 7 , 8 , 9 , 12 , 13 , 14 , 18 , 19 , 21 , 24 , 26 , 28 , 36 , 38 , 39 , 42 , 52 , 56 , 57 , 63 , 72 , 76 , 78 , 84 , 91 , 104 , 114 , 117 , 126 , 133 , 152 , 156 , 168 , 171 , 182 , 228 , 234 , 247 , 252 , 266 , 273 , 312 , 342 , 364 , 399 , 456 , 468 , 494 , 504 , 532 , 546 , 684 , 728 , 741 , 798 , 819 , 936 , 988 , 1064 , 1092 , 1197 , 1368 , 1482 , 1596 , 1638 , 1729 , 1976 , 2184 , 2223 , 2394 , 2964 , 3192 , 3276 , 3458 , 4446 , 4788 , 5187 , 5928 , 6552 , 6916 , 8892 , 9576 , 124488 show more factors ...
Total factors 96.
Sum of factors 436800 (312312).
#### Powers
The second power of 1244882 is 15.497.262.144.
The third power of 1244883 is 1.929.223.169.782.272.
#### Roots
The square root √124488 is 352,82857.
The cube root of 3124488 is 49,93164.
#### Logarithms
The natural logarithm of No. ln 124488 = loge 124488 = 11,731965.
The logarithm to base 10 of No. log10 124488 = 5,095127.
The Napierian logarithm of No. log1/e 124488 = -11,731965.
### Trigonometric functions
The cosine of 124488 is 0,731354.
The sine of 124488 is -0,681998.
The tangent of 124488 is -0,932514.
### Properties of the number 124488
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 124488 in Computer Science
Code typeCode value
PIN 124488 It's recommendable to use 124488 as a password or PIN.
124488 Number of bytes121.6KB
CSS Color
#124488 hexadecimal to red, green and blue (RGB) (18, 68, 136)
Unix timeUnix time 124488 is equal to Friday Jan. 2, 1970, 10:34:48 a.m. GMT
IPv4, IPv6Number 124488 internet address in dotted format v4 0.1.230.72, v6 ::1:e648
124488 Decimal = 11110011001001000 Binary
124488 Decimal = 20022202200 Ternary
124488 Decimal = 363110 Octal
124488 Decimal = 1E648 Hexadecimal (0x1e648 hex)
124488 BASE64MTI0NDg4
124488 MD5ca8edc2492854e01c1e1a17df3e329f0
124488 SHA153c72680049f0e913d34c9bb1f01ee7067533e7a
124488 SHA256df00376f07262747289bf5bb15b693c8fdca1a80a90e2c1d176321f62802a781
124488 SHA3848ae9a1957f3223f20b999dd7b56d37be83116b7c5c59e9dbcb0a486b0b6a376359c1e5fb14fd10f9a2d95ff7b19a50b7
More SHA codes related to the number 124488 ...
If you know something interesting about the 124488 number that you did not find on this page, do not hesitate to write us here.
## Numerology 124488
### Character frequency in number 124488
Character (importance) frequency for numerology.
Character: Frequency: 1 1 2 1 4 2 8 2
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 124488, the numbers 1+2+4+4+8+8 = 2+7 = 9 are added and the meaning of the number 9 is sought.
## Interesting facts about the number 124488
### Asteroids
• (124488) 2001 RU35 is asteroid number 124488. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 9/8/2001.
## № 124,488 in other languages
How to say or write the number one hundred and twenty-four thousand, four hundred and eighty-eight in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 124.488) ciento veinticuatro mil cuatrocientos ochenta y ocho German: 🔊 (Anzahl 124.488) einhundertvierundzwanzigtausendvierhundertachtundachtzig French: 🔊 (nombre 124 488) cent vingt-quatre mille quatre cent quatre-vingt-huit Portuguese: 🔊 (número 124 488) cento e vinte e quatro mil, quatrocentos e oitenta e oito Chinese: 🔊 (数 124 488) 十二万四千四百八十八 Arabian: 🔊 (عدد 124,488) مائة و أربعة و عشرون ألفاً و أربعمائةثمانية و ثمانون Czech: 🔊 (číslo 124 488) sto dvacet čtyři tisíce čtyřista osmdesát osm Korean: 🔊 (번호 124,488) 십이만 사천사백팔십팔 Danish: 🔊 (nummer 124 488) ethundrede og fireogtyvetusindfirehundrede og otteogfirs Dutch: 🔊 (nummer 124 488) honderdvierentwintigduizendvierhonderdachtentachtig Japanese: 🔊 (数 124,488) 十二万四千四百八十八 Indonesian: 🔊 (jumlah 124.488) seratus dua puluh empat ribu empat ratus delapan puluh delapan Italian: 🔊 (numero 124 488) centoventiquattromilaquattrocentottantotto Norwegian: 🔊 (nummer 124 488) en hundre og tjue-fire tusen, fire hundre og åtti-åtte Polish: 🔊 (liczba 124 488) sto dwadzieścia cztery tysiące czterysta osiemdziesiąt osiem Russian: 🔊 (номер 124 488) сто двадцать четыре тысячи четыреста восемьдесят восемь Turkish: 🔊 (numara 124,488) yüzyirmidörtbindörtyüzseksensekiz Thai: 🔊 (จำนวน 124 488) หนึ่งแสนสองหมื่นสี่พันสี่ร้อยแปดสิบแปด Ukrainian: 🔊 (номер 124 488) сто двадцять чотири тисячi чотириста вiсiмдесят вiсiм Vietnamese: 🔊 (con số 124.488) một trăm hai mươi bốn nghìn bốn trăm tám mươi tám Other languages ...
## News to email
Privacy Policy.
## Comment
If you know something interesting about the number 124488 or any natural number (positive integer) please write us here or on facebook.
#### Comment (Maximum 2000 characters) *
The content of the comments is the opinion of the users not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illicit or harmful to third parties. Number.academy reserves the right to remove or not make public any inappropriate comment. It also reserves the right to post a comment on another topic. Privacy Policy.
For this topic there are no comments. | 2,909 | 8,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.66716 |
http://conceptmap.cfapps.io/wikipage?lang=en&name=GHz | 1,553,035,554,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202161.73/warc/CC-MAIN-20190319224005-20190320010005-00044.warc.gz | 49,090,621 | 15,849 | # Hertz
(Redirected from GHz)
The hertz (symbol: Hz) is the derived unit of frequency in the International System of Units (SI) and is defined as one cycle per second.[1] It is named for Heinrich Rudolf Hertz, the first person to provide conclusive proof of the existence of electromagnetic waves. Hertz are commonly expressed in multiples: kilohertz (103 Hz, kHz), megahertz (106 Hz, MHz), gigahertz (109 Hz, GHz), terahertz (1012 Hz, THz), petahertz (1015 Hz, PHz), and exahertz (1018 Hz, EHz).
Hertz
Unit systemSI derived unit
Unit ofFrequency
SymbolHz
Named afterHeinrich Hertz
In SI base unitss−1
Top to bottom: Lights flashing at frequencies f = 0.5 Hz, 1.0 Hz and 2.0 Hz, i.e. at 0.5, 1.0 and 2.0 flashes per second, respectively. The time between each flash – the period T – is given by 1f (the reciprocal of f ), i.e. 2, 1 and 0.5 seconds, respectively.
Some of the unit's most common uses are in the description of sine waves and musical tones, particularly those used in radio- and audio-related applications. It is also used to describe the speeds at which computers and other electronics are driven.
## Definition
The hertz is defined as one cycle per second. The International Committee for Weights and Measures defined the second as "the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom"[2][3] and then adds: "It follows that the hyperfine splitting in the ground state of the caesium 133 atom is exactly 9 192 631 770 hertz, ν(hfs Cs) = 9 192 631 770 Hz." The dimension of the unit hertz is 1/time (1/T). Expressed in base SI units it is 1/second (1/s).
In English, "hertz" is also used as the plural form.[4] As an SI unit, Hz can be prefixed; commonly used multiples are kHz (kilohertz, 103 Hz), MHz (megahertz, 106 Hz), GHz (gigahertz, 109 Hz) and THz (terahertz, 1012 Hz). One hertz simply means "one cycle per second" (typically that which is being counted is a complete cycle); 100 Hz means "one hundred cycles per second", and so on. The unit may be applied to any periodic event—for example, a clock might be said to tick at 1 Hz, or a human heart might be said to beat at 1.2 Hz. The occurrence rate of aperiodic or stochastic events is expressed in reciprocal second or inverse second (1/s or s−1) in general or, in the specific case of radioactive decay, in becquerels.[5] Whereas 1 Hz is 1 cycle per second, 1 Bq is 1 aperiodic radionuclide event per second.
Even though angular velocity, angular frequency and the unit hertz all have the dimension 1/s, angular velocity and angular frequency are not expressed in hertz,[6] but rather in an appropriate angular unit such as radians per second. Thus a disc rotating at 60 revolutions per minute (rpm) is said to be rotating at either 2π rad/s or 1 Hz, where the former measures the angular velocity and the latter reflects the number of complete revolutions per second. The conversion between a frequency f measured in hertz and an angular velocity ω measured in radians per second is
${\displaystyle \omega =2\pi f\,}$ and ${\displaystyle f={\frac {\omega }{2\pi }}\,}$ .
This SI unit is named after Heinrich Hertz. As with every International System of Units (SI) unit named for a person, the first letter of its symbol is upper case (Hz). However, when an SI unit is spelled out in English, it is treated as a common noun and should always begin with a lower case letter (hertz)—except in a situation where any word in that position would be capitalized, such as at the beginning of a sentence or in material using title case.
## History
The hertz is named after the German physicist Heinrich Hertz (1857–1894), who made important scientific contributions to the study of electromagnetism. The name was established by the International Electrotechnical Commission (IEC) in 1930.[7] It was adopted by the General Conference on Weights and Measures (CGPM) (Conférence générale des poids et mesures) in 1960, replacing the previous name for the unit, cycles per second (cps), along with its related multiples, primarily kilocycles per second (kc/s) and megacycles per second (Mc/s), and occasionally kilomegacycles per second (kMc/s). The term cycles per second was largely replaced by hertz by the 1970s. One hobby magazine, Electronics Illustrated, declared their intention to stick with the traditional kc., Mc., etc. units.[8]
## Applications
A sine wave with varying frequency
A heartbeat is an example of a non-sinusoidal periodic phenomenon that may be analyzed in terms of frequency. Two cycles are illustrated.
### Vibration
Sound is a traveling longitudinal wave which is an oscillation of pressure. Humans perceive frequency of sound waves as pitch. Each musical note corresponds to a particular frequency which can be measured in hertz. An infant's ear is able to perceive frequencies ranging from 20 Hz to 20,000 Hz; the average adult human can hear sounds between 20 Hz and 16,000 Hz.[9] The range of ultrasound, infrasound and other physical vibrations such as molecular and atomic vibrations extends from a few femtohertz[10] into the terahertz range[11] and beyond.
Electromagnetic radiation is often described by its frequency—the number of oscillations of the perpendicular electric and magnetic fields per second—expressed in hertz.
Radio frequency radiation is usually measured in kilohertz (kHz), megahertz (MHz), or gigahertz (GHz). Light is electromagnetic radiation that is even higher in frequency, and has frequencies in the range of tens (infrared) to thousands (ultraviolet) of terahertz. Electromagnetic radiation with frequencies in the low terahertz range (intermediate between those of the highest normally usable radio frequencies and long-wave infrared light) is often called terahertz radiation. Even higher frequencies exist, such as that of gamma rays, which can be measured in exahertz (EHz). (For historical reasons, the frequencies of light and higher frequency electromagnetic radiation are more commonly specified in terms of their wavelengths or photon energies: for a more detailed treatment of this and the above frequency ranges, see electromagnetic spectrum.)
### Computers
In computers, most central processing units (CPU) are labeled in terms of their clock rate expressed in megahertz (106 Hz) or gigahertz (109 Hz). This specification refers to the frequency of the CPU's master clock signal. This signal is a square wave, which is an electrical voltage that switches between low and high logic values at regular intervals. As the hertz has become the primary unit of measurement accepted by the general populace to determine the performance of a CPU, many experts have criticized this approach, which they claim is an easily manipulable benchmark. Some processors use multiple clock periods to perform a single operation, while others can perform multiple operations in a single cycle.[12] For personal computers, CPU clock speeds have ranged from approximately 1 MHz in the late 1970s (Atari, Commodore, Apple computers) to up to 6 GHz in IBM POWER microprocessors.
Various computer buses, such as the front-side bus connecting the CPU and northbridge, also operate at various frequencies in the megahertz range.
## SI multiples
Submultiples Multiples Value SI symbol Name Value 10−1 Hz dHz decihertz 101 Hz daHz decahertz 10−2 Hz cHz centihertz 102 Hz hHz hectohertz 10−3 Hz mHz millihertz 103 Hz kHz kilohertz 10−6 Hz µHz microhertz 106 Hz MHz megahertz 10−9 Hz nHz nanohertz 109 Hz GHz gigahertz 10−12 Hz pHz picohertz 1012 Hz THz terahertz 10−15 Hz fHz femtohertz 1015 Hz PHz petahertz 10−18 Hz aHz attohertz 1018 Hz EHz exahertz 10−21 Hz zHz zeptohertz 1021 Hz ZHz zettahertz 10−24 Hz yHz yoctohertz 1024 Hz YHz yottahertz Common prefixed units are in bold face.
Higher frequencies than the International System of Units provides prefixes for are believed to occur naturally in the frequencies of the quantum-mechanical vibrations of high-energy, or, equivalently, massive particles, although these are not directly observable and must be inferred from their interactions with other phenomena. By convention, these are typically not expressed in hertz, but in terms of the equivalent quantum energy, which is proportional to the frequency by the factor of Planck's constant.
## Notes and references
1. ^ "hertz". (1992). American Heritage Dictionary of the English Language (3rd ed.), Boston: Houghton Mifflin.
2. ^ "SI brochure: Table 3. Coherent derived units in the SI with special names and symbols".
3. ^ "[Resolutions of the] CIPM, 1964 – Atomic and molecular frequency standards" (PDF). SI brochure, Appendix 1.
4. ^ NIST Guide to SI Units – 9 Rules and Style Conventions for Spelling Unit Names, National Institute of Standards and Technology
5. ^ "(d) The hertz is used only for periodic phenomena, and the becquerel (Bq) is used only for stochastic processes in activity referred to a radionuclide." "BIPM – Table 3". BIPM. Retrieved 2012-10-24.
6. ^ "SI brochure, Section 2.2.2, paragraph 6". Archived from the original on 1 October 2009.
7. ^ "IEC History". Iec.ch. 1904-09-15. Retrieved 2012-04-28.
8. ^ Cartwright, Rufus (March 1967). Beason, Robert G., ed. "Will Success Spoil Heinrich Hertz?" (PDF). Electronics Illustrated. Fawcett Publications, Inc. pp. 98–99. Retrieved 2016-03-29.
9. ^ Ernst Terhardt (20 February 2000). "Dominant spectral region". Mmk.e-technik.tu-muenchen.de. Archived from the original on 26 April 2012. Retrieved 28 April 2012.
10. ^ "Black Hole Sound Waves - Science Mission Directorate". science.nasa.go.
11. ^ Atomic vibrations are typically on the order of tens of terahertz
12. ^ Asaravala, Amit (2004-03-30). "Good Riddance, Gigahertz". Wired.com. Retrieved 2012-04-28. | 2,467 | 9,792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2019-13 | latest | en | 0.860047 |
https://www.ukessays.com/essays/computer-science/web-personalization-using-feedforward-2063.php | 1,606,723,225,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141211510.56/warc/CC-MAIN-20201130065516-20201130095516-00595.warc.gz | 884,522,379 | 19,199 | Web Personalization Using Feedforward Backpropagation
4033 words (16 pages) Essay
26th Mar 2018 Computer Science Reference this
Disclaimer: This work has been submitted by a university student. This is not an example of the work produced by our Essay Writing Service. You can view samples of our professional work here.
Any opinions, findings, conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of UKEssays.com.
WEB PERSONALIZATION USING FEEDFORWARD BACKPROPAGATION NEURAL NETWORK
Chapter 4: Methodology
Chapter 4 shows the methodology of the present work. Section 4.1 present the methodology, section 4.2 includes flow chart of the present work. Section 4.3 present the proposed algorithm.
1.1 METHODOLOGY
• Start
• Configure search engine
• Training on data according to user’s context.
• Testing on data
• ANN optimization for search optimization.
• Stop
1.1
Training: – Data is trained using Feedforward Backpropagation Neural network. Before testing and searching the data is trained. Data training is required for optimal results.
Testing: – Testing of data is performed using Feedforward Backpropagation neural network and Using SVM (Support Vector Machine).
User Query: – User can enter the Query for find the information. Some Web sites name are suggested to the users according the query of user.
1.2 FLOWCHART
The simple flowchart of the designed algorithm is depicted in the figure 4.2
Figure 4.2: flowchart of the designed algorithm
1.3 ALGORITHM DESIGN
The Algorithm for the present work is discussed in this section. The various steps used in algorithm are explained. The pseudo Code for the algorithm is also discussed in this section.
Table4.1 Proposed Algorithm
Proposed Algorithm STEP 1: (GENERATION OF WT(TRAINING AND TESTING) Begin Initialize CAPS chars(‘A’,’Z’) Initialize small chars(‘a’ to ‘z’) For each char in training/testing file Find(pos~char) T(pos.char) STEP 2: (SUPPORT VECTOR MACHINE) Begin test_set=gettest data; For each category in cat Test_Set=generator(Test_Set) Training _set(category, :)=wt(category, :); Test_Set=unique(Test_Set); For j=1:wt.count Kernal.function=linear; Group(j)=j; result=MultiSVM(Ts,Gruop,Test_Set) End Ts=find(unique(training.set)); STEP 3: Begin Initialize Training Set As SVM Group=GroupSVM; Neural.Initialization=newt(TrainSet,Group); Train.hiddenNeurons=10; (hidden neurons may vary according to size) Train.epocs=50; Result_train=train(net,trainset,group); Result_class=simulate(result_train,test_data); STEP 4: (ERROR CALCULATION) Error=sqrt((sum(result_class)-sum(original_class))^2/length(result)
1. Pseudo code of proposed algorithm:-The Pseudo code for proposed algorithm is shown below.
Table 4.2 proposed algorithm in pseudo code
Algorithm_Design Start globaltesting_datauser_found extracting the words from the paragraph result=generator(current_word) load architecture Select A Training File To Upload collecting data from the excel sheet for setting up the target; fori=1:rows for k=1:cols Target(j)=p; j=j+1; end p=p+1; end initiating the neural network loadsvm_group loadtraining_data; fortempind=1:itrind tst=test(tempind,:); C=Cb; T=Tb; u=unique(C); N=length(u); c4=[]; c3=[]; j=1; k=1; if(N>2) itr=1; classes=0; cond=max(C)-min(C); while((classes~=1)&&(itr<=length(u))&& size(C,2)>1 &&cond>0) if you increase the data you will have to adjust the groups also Database Updated Draw Plots End
Chapter-5
RESULT AND PERFORMANCE ANALYSIS
In this chapter results of the present is explained. The figures of result, comparison, comparison tables and graphs of the present work are shown in this chapter.
1.1 TOOLS USED
To implement my work I used Matlab. Matlab Stands for MATrix LABoratory. MATLAB has a modern programming language environment: it has refined data structures, contains built-in editing and debugging tools, and supports object-oriented programming.
Table 5.1: Tools Used
Computer Core i 3 or higher RAM 32 MB Platform Windows xp/ 7/8 Other hardware Keyboard, mouse Software Matlab 2010a
MATLAB
The name MATLAB stands for MATrix LABoratory. MATLAB was written originally to provide easy access to matrix software developed by the LINPACK (linear system package) and EISPACK (Eigen system package) projects MATLAB is a high-performance language for technical computing. It integrates computation, visualization, and programming environment. Furthermore, MATLAB is a modern programming language environment: it has refined data structures, contains built-in editing and debugging tools, and supports object-oriented programming. These factors make MATLAB an outstanding tool for education and research. MATLAB has many advantages compared to conventional computer languages (e.g., C, FORTRAN) for solving technical problems. MATLAB is an interactive system whose basic data element is an array that does not require dimensioning. The software package has been commercially available since 1984 and is now considered as a standard tool at most universities and industries worldwide. It has powerful built-in routines that enable a very wide variety of computations. It also has easy to use graphics commands that make the visualization of results immediately available. Specification applications are collected in packages referred to as toolbox. There are toolboxes for signal processing, symbolic computation, control theory, simulation, and optimization.
If you need assistance with writing your essay, our professional essay writing service is here to help!
After logging into your account, you can enter MATLAB by double-clicking on the MATLAB shortcut icon (MATLAB 7.0.4) on your Windows desktop. When you start MATLAB, a special window called the MATLAB desktop appears. The desktop is a window that contains other windows. The major tools within or accessible from the desktop are:
• The Command Window
• The Command History
• Workspace
• The Current directory
• Help browser
• Start button
5.1.1 MATLAB CHARACTERISTICS
• Developed first and foremost by Cleve Molar in the 1970’s
• Derived from FORTRAN subroutines LINPACK and EISPACK, linear and Eigen value systems.
• Developed principally as an interactive system to access LINPACK and EISPACK.
• Gained its esteem through word of mouth, because it was not authoritatively dispersed.
• Rewritten in C in the 1980’s with more functionality, which include plotting routines.
• The Math Works Inc. was produced (1984) to marketplace and go on with expansion Of MATLAB.
• MATLAB may behave as a calculator or as a programming language
• MATLAB combine adequately calculation and graphic plotting.
• MATLAB is moderately easy to learn
• MATLAB is interpreted (not compiled), errors are easy to fix.
• MATLAB is optimized to be relatively fast when performing matrix operations
• MATLAB does have some object-oriented elements
5.1.3 RESULTS
In this section Screen Shots of the present work are shown. Firstly, Data Set is uploaded after that Neural Network and SVM are used for training and testing of the data. User can create their account and if user has already account then he can sign in for the Personalization. Three parameters are taken for the comparison between the SVM (support vector machine) and Neural Network. Accuracy, Precision and Recall are the three parameters used for the comparison. Neural Network gives the best results.
Figure 5.1: Proposed Flowchart
Fig. 5.1 shows the main working window of the personalization. The above figure has all the training and testing window components in w
+hich the personalized data can be trained through the Neural Network and Support Vector Machine.
Training Model for SVM as well as Neural Network.
Inputs: examples, a set of examples, each with input x = x1; x2; : : : ; xn and output y
Inputs: network, a perceptron with weights Wj ; j = 0; : : : ; n and activation function g
Repeat for each e in examples do
inPnj = 0Wj xj [e] Err y[e] – g(in)
WjWj + _ _ Err _ g0(in) _ xj [e]
End
Until all examples correctly predicted or stopping criterion is reached Return network
Figure 5.2: represents the architecture of the Neural Network
Neural network contains of input and hidden layers. Each and every layer has weight and bandwidth of the data. Hidden Layer contains epochs that means iteration. The maximum iteration provided over here is 50 but it is not necessary that the neural will run till 50. It would cross check the validations and would provide the results required. The results can also be checked by the following graphs.
Figure 5.3: Representing detailed neural architecture
The above figure represents the architecture over which the neural has been tested and trained. There is one validation denoted by the pink line and has been achieved on the 4th Iteration.
Figure 5.4: Personalizing Option
The above figure provides the option to personalize the system according to the choice of the user. Here the user can banned those website link which he or she does not want to see in the future.
Figure 5.6: Results after testing data
The above figure represents results after testing the data. User can test data after fill the data in the box.
Figure 5.7. Different parameters
The above figure shows the different parameter after click on result neural button. Accuracy, Precision, and recall parameters can be calculated. The same parameter can be calculated by SVM also.
5.2Comparison Tables and Graphical Representation
The experiment was conducted for computing Accuracy, Precision and Recall. The experiment has been performed to compare the performance of both Neural Network and SVM (Support Vector Machine). The Accuracy, Precision and Recall for both approaches was different. Given tables and graphs proves the performance of the algorithms.
Table 5.2: Accuracy Comparison
ACCURACY COMPARISON CATEGORY ACCURACY NEURAL ACCURACY SVM POLITICS 97.9348 90.3484 ENTERTAINMENT 99.1023 96.4148 SPORTS 97.9348 89.3067 HOUSE 99.5281 96.76 EDUCATION 95.696 91.742 GOOD 93.7454 98.558 BEAUTIFUL 96.641 94.1688 HAPPY 90.7599 99.0201 FOOD 96.7325 88.1569 FRUITS 90.3681 95.5261 VEGETABLES 85.4341 92.3119 COLORS 95.8125 89.3325 CLOTHES 97.1374 99.1023 TOOLS 96.4148 94.4251 SHAPES 99.5209 89.3036
Figure 5.8: Graph of accuracy comparison
Table 5.3: Precision Comparison
PRECISION COMPARISON CATEGORY PRECISION NEURAL PRECISION SVM POLITICS 1.2828 1.3548 ENTERTAINMENT 1.1053 1.124 SPORTS 1.2828 1.2674 HOUSE 1.891 1.376 EDUCATION 1.0934 1.1997 GOOD 1.164 1.1472 BEAUTIFUL 1.588 1.3946 HAPPY 1.0874 1.2957 FOOD 1.1206 1.0738 FRUITS 1.1985 1.2635 VEGETABLES 1.1878 1.2608 COLORS 1.0489 1.0409 CLOTHES 1.4406 1.1053 TOOLS 1.124 1.1562 SHAPES 1.4564 1.2803
Figure 5.9: graph of Precision Comparison
Table5.4: Recall Comparison
CATEGORY RECALL NEURAL RECALL SVM POLITICS 0.792 0.67773 ENTERTAINMENT 0.89813 0.84409 SPORTS 0.792 0.68945 HOUSE 0.67537 0.89335 EDUCATION 0.85077 0.74268 GOOD 0.95213 0.86386 BEAUTIFUL 0.81413 0.84414 HAPPY 0.77884 0.80129 FOOD 0.85091 0.74772 FRUITS 0.72527 0.9208 VEGETABLES 0.66707 0.87509 COLORS 0.88184 0.78242 CLOTHES 0.73272 0.89813 TOOLS 0.84409 0.96881 SHAPES 0.68564 0.89048
Figure 5.10: Graph of Recall Comparison
Chapter 6: CONCLUSION AND FUTURE SCOPE
Chapter 6 includes conclusion and future scope of the present work. Future scope means that what enhancement can be done in the future. Section 6.1 covers the Conclusion and Section 6.2 covers the Future scope.
6.1 CONCLUSION
Web personalization is an answer for data over-burden issue on World Wide Web .The web personalization assemble the accuracy of web hunt apparatus, streamlines the looking process and reduce the time customer needs to spend for looking for. Today for both Web-based affiliations and for the end customers the web personalization has transformed into a key gadget.
Our academic experts are ready and waiting to assist with any writing project you may have. From simple essay plans, through to full dissertations, you can guarantee we have a service perfectly matched to your needs.
Web utilization mining is the methodology of recognizing delegate patterns and scanning examples depicting the movement in the site, by investigating the clients’ conduct. Site directors can then use this information to redesign or change the site according to the side interests and behaviour of its visitors, or upgrade the execution of their systems. Also, the supervisors of e-trade destinations can procure profitable business brainpower, making buyer profiles and accomplishing business sector division. There exists number of techniques yet none has been accomplished great amount.
This postulation introduced a methodology taking into account neural system for web personalization of web substance. Firstly, in the pre-processing stage the information must be gathered from the better places it is put away (customer side, server side, and intermediary servers). In the wake of recognizing the customers, the snap surges of each customer must be part into sessions. The last venture of the entire web utilization mining methodology is to dissect the examples found amid the example disclosure step. Web Usage Mining attempt to comprehend the examples identified in before step. The most well-known systems is information visualization applying channels High dimensional information stream contains a huge colossal measure of information. Such huge sum information contains a vast information with high measurements with information many-sided quality. A valid example remote sensor framework data, web logs, Google look for, et cetera. Standard strategies are not suitable over high dimensional data as they obliged high figuring expense for taking care of data that is the reason this technique has been realized with some change highlights.
6.2 FUTURE SCOPE
Future misleads examine the half breed utilization structure positioning that can be connected to a bound together web/navigational diagram which extends out of the breaking points of a solitary site. Such approach would empower a “worldwide” significance positioning over the web, improving both web query items and the suggestion process.
Now, if the user wants to revisit URL P3, she would not be able to do that using just the BackButton navigation Stack. If she resorts to the history list to get some help, she will be disappointed to see that its list based textual representation gives no idea about the structure of the navigation pattern. Moreover, even for a modestly sized navigation session, the history list gets cluttered to an extent so that renders it ineffective in searching for a specific page. The bookmark facility is of little help in this case, as the user cannot bookmark each and every page due to overhead associated with the very process of bookmarking. Moreover, even selected bookmarking is of no help as, in most cases, the user does not know at the time of visiting a web page whether it is important enough to be bookmarked.
One thing that has long been acknowledged by the research community is the use of graphical overview diagrams in assisting user navigation through complex information spaces.
The visualization scheme employed should be efficient enough to give a graphical representation of user session history in real time. Computationally and graphically intensive application may cause undue delays in the visualization generation process, especially when the session history grows large. Most of the past work done for WWW subspace visualization is plagued by these delays therefore is inefficient for the ordinary use.
The solution must be designed keeping in mind that it has to replace WWW browser stack based navigation structure and its history list. Therefore it must provide all those facilities that were provided by these browser components. Users who are familiar with the facilities provided by the browser may find it very difficult to adjust to a new scheme that does not provide these facilities.
The visualization scheme should be designed more on an aesthetic rather than a scientific basis. Humans tend to get confused when presented with a large amount of data jumbled up in front of them. It is, therefore, highly recommended that session history data be divided into small and easily manageable groups, neatly knitted together through an elegant link structure.
1
Related Services
View all
DMCA / Removal Request
If you are the original writer of this essay and no longer wish to have your work published on the UKDiss.com website then please:
Related Services | 3,762 | 16,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.863001 |
https://www.physicsforums.com/threads/solving-kinematic-equations-acceleration-from-100m-in-9-77s.470669/ | 1,708,540,797,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00516.warc.gz | 970,645,409 | 16,040 | # Solving Kinematic Equations: Acceleration from 100m in 9.77s
• 7yler
In summary, the person ran 100 m in 9.77 seconds with a maximum speed of 10.23 m/s reached in 3 seconds. The acceleration during the first 3 seconds can be found by using the equation (final velocity - initial velocity)/time, resulting in an acceleration of 3.41 m/s^2. To find the displacement and velocity at 3 seconds and 9.77 seconds, equations containing the acceleration variable can be used. However, it is not possible to solve the equations with two missing variables without additional information.
#### 7yler
A person runs 100 m in 9.77 s. Maximum speed was reached in 3 s, then maintained for the rest of the time. What was the acceleration during the first 3 seconds?
I know that I can find the average velocity, which is 100 m/9.77 s, and i know that initial velocity is 0 m/s. I don't know how to find the acceleration because I don't know what the velocity is. Could anyone guide me through the methodology of this problem? I don't want the numerical answer.
it can be like this.
initial velocity=0
velocity at the end of 3 sec is100/9.77m/s i.e. 10.23m/s
now acc. = (final vel. - initial vel.)/time=(10.23-0)/3=3.41m/sec.sq
so,acc. after 3 sec is 3.41m/sec.sq
The system is telling me that that is incorrect.
Find the displacement and velocity at 3 seconds. Find the displacement and velocity at 9.77 seconds.
How do I find displacement and velocity if I don't know acceleration?
I meant, find equations for them (the equations will contain "a").
I don't understand how I can solve the equations with two missing variables.
## 1. How do you calculate acceleration from a given distance and time?
Acceleration can be calculated using the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. In this case, the given distance of 100m can be used as the displacement (vf - vi) in the equation.
## 2. Why is acceleration important in kinematics?
Acceleration is important in kinematics because it represents the rate of change of velocity. It provides information about how quickly an object is speeding up or slowing down, and is necessary for accurately describing an object's motion.
## 3. What is the difference between acceleration and velocity?
Velocity is a vector quantity that describes the speed and direction of an object's motion, while acceleration is a vector quantity that describes the rate of change of an object's velocity. In other words, velocity tells us how fast and in which direction an object is moving, while acceleration tells us how much an object's velocity is changing over time.
## 4. Can acceleration be negative?
Yes, acceleration can be negative. A negative acceleration, also known as deceleration, means that an object is slowing down. This can happen when the initial velocity is greater than the final velocity, or when the direction of the acceleration is opposite to the direction of the velocity.
## 5. How can kinematic equations be applied in real-world situations?
Kinematic equations can be applied in various real-world situations, such as calculating the acceleration of a roller coaster, determining the speed of a car in a car chase, or predicting the trajectory of a basketball shot. These equations allow us to analyze and understand the motion of objects in our everyday lives. | 777 | 3,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.936144 |
https://lists.umanitoba.ca/pipermail/epidata-list/2005q4/000322.html | 1,624,392,598,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488519735.70/warc/CC-MAIN-20210622190124-20210622220124-00205.warc.gz | 313,766,549 | 2,027 | # [EpiData-list] Howto: Group a date variable by calendar year
epidata-list at lists.umanitoba.ca epidata-list at lists.umanitoba.ca
Wed Dec 14 14:44:06 CST 2005
```* Example pgm file to group a date variable V1
* created by: Jens Lauritsen, EpiData Association
set show info=off
generate 50 // generate 50 empty observations for testing
define v1 <dd/mm/yyyy>
define y ####
y = 1900 + recnumber // could also be 1900 + _n
v1 = dmy(1,7,y)
* so now we have a date variable with dates from 1900 to 1950
* three different strategies to group.
* Strategy 1: extract year from the date and recode to a string:
define y1 ####
define ys ________________________
y1 = year(v1)
recode y1 to ys by 10
tab ys
* Strategy 2: round the number into 10 year groups
define y2 ####
y2 = year(v1)
y2 = trunc(y2/10)*10 // creates groups by 10 years
tab y2
* If you use 5 instead of 10 you get 5 year groups
* y2 = trunc(y2/5)*5 // creates groups by 5 years
* strategy 3: split on specific dates:
define y3 #
let y3 = 0
if (v1 < dmy(1,10,1930) ) then y3 = 1
if (v1 >= dmy(1,10,1930)) and (v1 < dmy(1,10,1939) ) then y3 = 2
if (v1 >= dmy(1,10,1939)) and (v1 < dmy(1,10,1960) ) then y3 = 3
tab y3
* run the above to get the tabulations.
----------------------------------------------------------------------------------
Output:
E.g. for strategy 1:
No. % Cum %
1900 - 1909 9 18.00 18.00
1910 - 1919 10 20.00 38.00
1920 - 1929 10 20.00 58.00
1930 - 1939 10 20.00 78.00
1940 - 1949 10 20.00 98.00
1950 - 1959 1 2.00 100.00
Total 50 100%
* strategy 2:
No. % Cum %
1900 9 18.00 18.00
1910 10 20.00 38.00
1920 10 20.00 58.00
1930 10 20.00 78.00
1940 10 20.00 98.00
1950 1 2.00 100.00
Total 50 100%
* strategy 3
No. % Cum %
1 30 60.00 60.00
2 9 18.00 78.00
3 11 22.00 100.00
Total 50 100%
----------------------------------------------------------------------------------
Regards Jens Lauritsen
EpiData Association
``` | 783 | 2,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-25 | latest | en | 0.488991 |
http://oeis.org/A218532 | 1,591,434,402,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348511950.89/warc/CC-MAIN-20200606062649-20200606092649-00316.warc.gz | 88,528,153 | 3,685 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A218532 Number of times n consecutively appears in A218461. 0
2, 3, 5, 10, 21, 44, 92, 216, 446, 1018, 2356, 5408, 12559, 29709, 70446, 169008, 407539, 988057, 2415086, 5927972 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(n) is roughly e^n*(e-1)/n. - Charles R Greathouse IV, Aug 20 2013 LINKS PROG (PARI) a(nn) = my(lpn=1, lfn=1); for (i=2, nn, p = prime(i); fn = prime(p)\p; if (fn != lfn, print1(i - lpn, ", "); lfn = fn; lpn = i)) \\ Michel Marcus, Aug 16 2013 (PARI) a(n)=my(p=2, t=p, s, quot); forprime(q=2, , if(t--, next); quot=q\p; if(quot>=n, if(quot>n, return(s), s++)); t=nextprime(p+1)-p; p+=t) \\ Charles R Greathouse IV, Aug 20 2013 CROSSREFS Cf. A218461 (floor(prime(prime(n))/prime(n))). Sequence in context: A024494 A131708 A002991 * A022861 A001646 A103595 Adjacent sequences: A218529 A218530 A218531 * A218533 A218534 A218535 KEYWORD nonn AUTHOR Tyler Carrico, Oct 31 2012 EXTENSIONS a(13) from Michel Marcus, Aug 16 2013 a(14)-a(20) from Charles R Greathouse IV, Aug 20 2013 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified June 6 04:15 EDT 2020. Contains 334859 sequences. (Running on oeis4.) | 551 | 1,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.613041 |
https://www.physicsforums.com/threads/fluid-mechanics-problem-oil-pressure-calculations-in-pipe-flow.974022/#post-6199234 | 1,685,277,653,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643784.62/warc/CC-MAIN-20230528114832-20230528144832-00557.warc.gz | 1,060,143,241 | 18,276 | # Fluid Mechanics problem: Oil pressure calculations in pipe flow
• Motorbiker
#### Motorbiker
Homework Statement
An oil with density 900 kg/m3 and viscosity 0.18 Ns/m2 flows through a circular pipe which inclines upwards
at 40° to the horizontal. The length of the pipe is 10 m and the diameter is 6 cm. The fluid pressure at the
lower end of the pipe is 350 kPa and the pressure at the upper end is 250 kPa. (i) Confirm that flow is upward
through the pipe
Relevant Equations
I think Steady flow equation
Problem Statement: An oil with density 900 kg/m3 and viscosity 0.18 Ns/m2 flows through a circular pipe which inclines upwards
at 40° to the horizontal. The length of the pipe is 10 m and the diameter is 6 cm. The fluid pressure at the
lower end of the pipe is 350 kPa and the pressure at the upper end is 250 kPa. (i) Confirm that flow is upward
through the pipe
Relevant Equations: I think Steady flow equation
I have been trying really hard to start this question, but I don't know what equations are relevant or how to start. I would be very grateful if someone could kindly explain the problem to me in basic terms. This way I will hopefully be able to attempt the question and post my attempt here if I need further help.
Thank you.
## Answers and Replies
If the fluid were not flowing, what would the pressure at the top end be if the pressure at the lower end were 350 kPa?
If the fluid were not flowing, what would the pressure at the top end be if the pressure at the lower end were 350 kPa?
The pressure would still be 250kPa at the top end because pressure decreases with altitude.
Oh yeah? OK, let's see your calculation to prove it.
Oh yeah? OK, let's see your calculation to prove it.
Am I correct in my understanding?
For the calculation, I will need to use the bernoulli's equation, is that right?
Am I correct in my understanding?
For the calculation, I will need to use the bernoulli's equation, is that right?
Your understanding was incorrect (quantitatively).
For the calculation without fluid flowing, the Bernoulli equation can be used.
Your understanding was incorrect (quantitatively).
For the calculation without fluid flowing, the Bernoulli equation can be used.
Thank you, can the sfee can be used instead?
Thank you, can the sfee can be used instead?
What is sfee?
What is sfee?
Ah sorry, I meant steady flow energy equation.
Ah sorry, I meant steady flow energy equation.
What does the steady flow energy equation reduce to if the flow is zero?
What does the steady flow energy equation reduce to if the flow is zero?
It reduces to zero?
It reduces to zero?
Let's stick with the Bernoulli equation (at least for now). So let's see your Bernoulli equation calculation.
Let's stick with the Bernoulli equation (at least for now). So let's see your Bernoulli equation calculation.
Sorry, I think misunderstood you. The steady flow energy equation should reduce to:
p1/pg + z1=p2/pg +z2+hf
I think this is correct for cases where the flow is zero.
Sorry, I think misunderstood you. The steady flow energy equation should reduce to:
p1/pg + z1=p2/pg +z2+hf
I think this is correct for cases where the flow is zero.
Well, hf is zero if there is no flow. Otherwise, OK.
Well, hf is zero if there is no flow. Otherwise, OK.
Okay great, I note that we have been given the viscosity in this problem, are we supposed to use it in this section?
Am I correct in thinking that the incline at 40 degrees is somehow related to z1 and z2?
Last edited:
Okay great, I note that we have been given the viscosity in this problem, are we supposed to use it in this section?9
if we are first doing the calculation for a case with no flow, why would we need the viscosity?
Am I correct in thinking that the incline at 40 degrees is somehow related to z1 and z2?
Correct. | 918 | 3,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-23 | latest | en | 0.956722 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.