url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://moorejustinmusic.com/students-life/what-is-path-difference-in-wave-optics/ | 1,716,800,700,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00801.warc.gz | 340,640,014 | 8,961 | # What is path difference in wave optics?
## What is path difference in wave optics?
Path difference is the difference in the path traversed by the two waves. The relation between phase difference and path difference is direct. They are directly proportional to each other.
### What is the significance of optical path difference?
A difference in OPL between two paths is often called the optical path difference (OPD). OPL and OPD are important because they determine the phase of the light and governs interference and diffraction of light as it propagates.
#### How is optical path different from the actual path length?
Generally optical path length and actual path length are related but are not identical. The optical path length of light in a medium can actually be defined as the length or distance in which the light would have travelled in the same time if it were travelling in a vacuum.
What is difference between optical path and geometrical path?
The optical path length is simply the distance the light travels times the refractive index. For example, light traveling through 10 cm of water has an optical path length of 10 x 1.333 or 13.3 cm. The geometric length is simply the physical distance the light travels.
What is a path difference?
(Note the path difference or PD is the difference in distance traveled by the two waves from their respective sources to a given point on the pattern.) For point A on the first antinodal line (m =1), the path difference is equivalent to 1 wavelength.
## Whats a path difference?
Path Difference Definition Physics The path difference is the difference in the physical distance between the two sources to the observer, i.e., the difference in distance travelled from the source to the observer.
### What is optical path in physics?
Optical path (OP) is the trajectory that a light ray follows as it propagates through an optical medium. The optical path length in a homogeneous medium is the GPD multiplied by the refractive index of the medium.
#### What is effective path difference?
Path difference is defined as the difference in actual distance traveled by the two waves. The phase angle is the part of one complete wave cycle measured as a fraction of 2π (360 degrees) i.e the phase difference from one wave peak to the next is 2π (360 degrees).
What is optical path difference and explain how it causes the interference?
The difference in distance traveled by the two waves is three-halves a wavelength; that is, the path difference is 1.5 . Whenever the two waves have a path difference of 1.5 wavelengths, a crest from one source will meet a trough from the other source and destructive interference will occur.
What is path difference and phase difference?
Difference Between Phase Difference and Path Difference
Phase Difference Path Difference
The formula of the phase difference is: Δϕ = 2πΔx/λ The formula of path difference is: Δx = λ/2π Δϕ
The unit of the phase difference is Radian. The unit of the path difference is meter.
## What do you mean by optical path?
### What is path difference a level physics?
The difference in distance travelled by two waves from their sources to the point where they meet. Path difference is generally expressed in multiples of wavelength. | 666 | 3,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-22 | latest | en | 0.94126 |
https://www.instasolv.com/xam-idea/class-12-physics-chapter-14.html | 1,600,822,916,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400208095.31/warc/CC-MAIN-20200922224013-20200923014013-00134.warc.gz | 881,215,877 | 11,355 | Instasolv
IIT-JEE NEET CBSE NCERT Q&A
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# Xam Idea Class 12 Physics Chapter 14 Solutions: Electronic Devices
Xam Idea Class 12 Physics Solutions for Chapter 14 ‘Electronic Devices’ are of great use for preparing for CBSE Class 12 exams along with competitive exams like JEE and NEET. These solutions are based on concepts of electronic devices that include insulators, semiconductors, conductors, logic gates, I-V characteristics of LED, photodiode, Zener diode, and solar cell. There are several formulas in this chapter that describe the relationship between emitted current, base current and collector current learning which will help you to solve numerical easily. Going through our Xam Idea solutions for Class 12 Physics Electronic Devices will surely assist you in gaining in-depth knowledge of all concepts of this chapter.
Xam Idea Class 12 Physics Solutions for Electronic Devices consist of 125 questions that are divided into four sections. These are very short answer type questions, short answer type questions (I and II) and long answer type questions. Every section of questions will cover all important concepts of the chapter that will assist you in making fundamentals very strong in the chapter. The questions are both theoretical as well as numerical. The questions present in the chapter are mainly based on charge carriers in p and an n-type semiconductor, depletion layer, distinguishing between metal and insulator based on energy band diagrams, drawing output waveform when inputs are given in logic circuit, extrinsic and intrinsic semiconductors, and energy band.
Our experts at Instasolv will help you with easy and reliable solutions for Xam Idea Class 12 Physics. Our team of experts is highly skilled with several years of experience and they offer stepwise solutions from a student point of view. Their main aim is to make your concepts clear and assist you in solving and answering questions easily and rapidly. The vocabulary we use is very easily understandable and our resources are available for you free of cost.
## Important topics for Xam Idea Class 12 Physics Solutions Chapter 14: Electronic Devices
Semiconductors: These are basic materials that are used in some of the solid-state electronic devices like diodes, ICS, transistors etc. Semiconductors have a resistivity of intermediate level to current.
Intrinsic Semiconductor: This is a pure semiconductor where the electrical conductivity is entirely governed by electrons that are excited from valence band to the conduction band and there is no impure atoms addition for increasing the conductivity. The electrical conduction in these semiconductors is done through electron-hole pairs. In an intrinsic semiconductor-
ne -nh = ni where ne = the free electron density in the conduction band, nh = the hole density in the valence band, and ni = the intrinsic carrier concentration.
Extrinsic Semiconductor: It is a type of semiconductor that is dope with impure atoms to enhance conductivity. They are of two type’s n-type semiconductors and p-type semiconductors. In n-type semiconductors, the electrons are the major carriers of charge and holes are the minor and charge carriers and the vice versa in a p-type semiconductor.
Diode: It is a semiconductor device having two terminals that allow the unidirectional flow of current. One of the important diodes is the p-n junction diode. When this junction is created a layer called depletion layer is formed that consist of immobile ion cores lacking their holes and electrons. This results in a junction potential barrier.
I-V characteristics Curve: It is drawn because, by alteration of the externally applied voltage amount, there is alteration of the junction barriers. In forwarding bias, the barrier reduces while in reverse bias the barrier increases.
Some special circuits handle the digital data consisting of 0 and 1 levels. This makes the subject of digital electronics.
Logic Gates: These are the digital circuits that are responsible for some of the special logic operations. This includes AND, NOR, OR, NAND and NOT gates.
### The Relation Between Emitter Current, Base Current and Collector Current is
IE=IB + IC where IE is emitter current, IB is base current and Ic is collector current.
Transistor Action: When the emitter-base junction in an NPN transistor is forward biased, the electrons are then pushed in the direction of the base. When I0 is the base current, there are three configurations of a transistor. They are common-base (CB) circuit, common-emitter (CE) circuit, common-collector (CC) circuit.
### Exercise Discussion for Xam Idea Class 12 Physics Solutions Chapter 14: Electronic Devices
Chapter 14 of Xam Idea Class 12 Physics Solutions on electronic devices consist of a total of 125 solved questions. These questions are prepared in compliance with the recent NCERT syllabus and the pattern of CBSE that will help you in scoring higher marks in examinations. The whole set of problems are divided into four sections.
• There are about 50 very short answer type questions which you have to answer in one line. They all have been designed to make your concept very clear in minute details of the chapter.
• There are 10 previous years questions based on the depletion layer, logic gates, I-V characteristics, p-n junction, and temperature dependence of resistivity.
• Then, there are 40 questions based on semiconductor materials, charge carriers in p and n-type semiconductors, Transconductance of the transistor.
Short Answer Type Questions (Type- I)
• Under this section, there are 28 questions in total. This section of questions is 4-5 line answer questions.
• There are about 15 questions based on differentiating between metal and insulator based on energy bands, n-type and p-type semiconductors, drawing energy band diagrams and identifying the logic gates.
• Some questions are based on the difference between forward biasing and reverse biasing in a p-n junction, drawing truth table for the logic circuit.
• Then there are 13 questions based on the minimum voltage of the battery, diodes, resistivity for safe operation.
• Under this category, there are 31 questions in total.
• The questions are asked on concepts of the chapter like integrated circuit, diode resistance, the logic gate of a complete circuit, semiconductor under reverse bias, dynamic output resistance.
• Then, there are 12 questions and mainly covers resistivity of the transistor, Zener diode, Photodiode, input resistance. | 1,323 | 6,516 | {"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} | 2.578125 | 3 | CC-MAIN-2020-40 | latest | en | 0.940839 |
https://liusson.com/homework-solution-develop-a-program-that-calculates-basic-statistics-on-a-one-dimensional-a/ | 1,611,334,603,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00043.warc.gz | 451,756,570 | 20,378 | # Homework Solution: Develop a program that calculates basic statistics on a one-dimensional a…
Description: Develop a program that calculates basic statistics on a one-dimensional array of doubles. Your program should first prompt the user for the size of the array followed by prompting for individual elements of the array. Your program should contain the following functions: 1.) min() Finds and returns the minimum value within the array. 2.) max() Finds and returns the maximum value within the array. 3.) mean() Finds and returns the mean value of the elements in the array. 4.) median() Finds and returns the median value of the elements in the array. 5.) sort()* Sorts the elements of the array in ascending order. 6.) readArray() Populates the elements of the array with input from the user (or via file redirection). 7.) printArray() Prints the elements of the array. *You may use any sorting algorithm that you want. Here are a few easy suggestions: Bubble sort pseudocode: do swapped = false for each i in 1 to length(A) inclusive do: if A[i-1] > A[i] then swap A[i-1] and A[i] swapped = true end if end for while swapped Insertion sort pseudocode: i ← 1 while i < length(A) j ← i while j > 0 and A[j-1] > A[j] swap A[j] and A[j-1] j ← j - 1 end while i ← i + 1 end while Additional Specifications: Your program should not use any pre-existing classes such as string or vector classes! Make sure your program compiles and runs on one of the Linux machines in the Linux lab before you submit. Your program should consist of a single source code. EXAMPLE OUTPUT: Examples using file redirection:
Show transcribed image textcasey@vanderwaal:-/Dropbox/Teaching/CSCE24oFalt2617/Projects/Assignment2\$ g++ Assignment2.cpp casey@vanderwaal:-/Dropbox/Teaching/CSCE240Fall2017/Projects/Assignment2\$ ./a.out Please enter the size of your array: 6 Please enter 6 elements to populate the array. 212.3-6799 θ The elements of the array are: 212.3-6 7 99 θ The elements of the array sorted in ascending order: 6 0 27 12.3 99 The min of the array is-6 The max of the array is 99 The mean of the array is 19.05 The median of the array is 4.5 casey@vanderwaal:-/Dropbox/Teaching/CSCE240Fall2017/Projects/Assignment2\$
#include<iostream> using namespace std;
Description:
Develop a program that calculates basic statistics on a undivided-dimensional attire of wraps. Your program should restraintemost quick the truthr restraint the largeness of the attire followed by quicking restraint only elements of the attire. Your program should hold the subjoined functions:
1.) min()
Finds and receipts the partiality compute amid the attire.
2.) max()
Finds and receipts the consummation compute amid the attire.
3.) moderation()
Finds and receipts the moderation compute of the elements in the attire.
4.) median()
Finds and receipts the median compute of the elements in the attire.
5.) character()*
Sorts the elements of the attire in ascending ordain.
Populates the elements of the attire with input from the truthr (or via finish redirection).
7.) imprintArray()
Prints the elements of the attire.
*You may truth any charactering algorithm that you lack. Here are a lacking unconcerned suggestions:
Bubble character pseudocode:
do
swapped = false
restraint each i in 1 to tediousness(A) comprehensive do:
if A[i-1] > A[i] then
swap A[i-1] and A[i]
swapped = true
end if
end restraint
while swapped
Insertion character pseudocode:
i ← 1
while i < tediousness(A)
j ← i
while j > 0 and A[j-1] > A[j]
swap A[j] and A[j-1]
j ← j – 1
end while
i ← i + 1
end while
Your program should not attributable attributable attributable truth any pre-existing classes such as string or vector classes!
Make strong your program compiles and runs on undivided of the Linux machines in the Linux lab precedently you suggest.
Your program should endure of a only commencement jurisprudence.
EXAMPLE OUTPUT:
Examples using finish redirection:
Show transcribed picture textcasey@vanderwaal:-/Dropbox/Teaching/CSCE24oFalt2617/Projects/Assignment2\$ g++ Assignment2.cpp casey@vanderwaal:-/Dropbox/Teaching/CSCE240Fall2017/Projects/Assignment2\$ ./a.quenched Please penetrate the largeness of your attire: 6 Please penetrate 6 elements to populate the attire. 212.3-6799 θ The elements of the attire are: 212.3-6 7 99 θ The elements of the attire charactered in ascending ordain: 6 0 27 12.3 99 The min of the attire is-6 The max of the attire is 99 The moderation of the attire is 19.05 The median of the attire is 4.5 casey@vanderwaal:-/Dropbox/Teaching/CSCE240Fall2017/Projects/Assignment2\$
## Expert Solution
#include<iostream>
using namespace std;
void character(wrap facts[], int n){
restraint (int i =0; i<n; i++){
restraint (int j = i+1; j<n; j++){
if (data[i] > facts[j]){
wrap temp = facts[i];
data[i] = facts[j];
data[j] = temp;
}
}
}
}
void imprint(wrap facts[], int n){
restraint (int i = 0; i<n; i++){
cquenched << facts[i] << ” “;
}
cquenched << endl;
}
wrap min(wrap facts[],int n){
wrap min = facts[0];
restraint (int i = 0; i<n; i++){
if (data[i] < min)
min = facts[i];
}
return min;
}
wrap max(wrap facts[], int n){
wrap max = facts[0];
restraint (int i = 0; i<n; i++){
if (data[i] > max)
max = facts[i];
}
return max;
}
wrap moderation(wrap facts[], int n){
wrap consolidate = 0;
restraint (int i = 0; i<n; i++){
consolidate = consolidate + facts[i];
}
return consolidate/n;
}
wrap median(wrap facts[], int n){
wrap consolidate = 0;
wrap median;
if ( n% 2 == 0){
int i = n/2-1;
int j = n/2;
median = (data[i] + facts[j])/2;
}
else {
median = facts[n/2];
}
return median;
}
int deep(){
int n;
cquenched << “Please penetrate largeness of attire : “;
cin >> n;
wrap *facts = newlightlight wrap[n];
cquenched << “Please penetrate ” << n << ” elements to poulate the attire” << endl;
restraint (int i = 0; i<n; i++){
cin >> facts[i];
}
cquenched << “The elements of the attire are:” << endl;
print(data, n);
cquenched << endl;
cquenched << “The elements of the attire charactered in ascending ordain:” << endl;
sort(data,n);
print(data, n);
cquenched << endl;
cquenched << “The min of the attire is:” << min(data,n) << endl;
cquenched << “The max of the attire is:” << max(data,n) << endl;
cquenched << “The moderation of the attire is:” << moderation(data,n) << endl;
cquenched << “The median of the attire is:” << median(data,n) << endl;
cquenched << endl;
return 0;
} | 1,757 | 6,400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.755046 |
https://www.jiskha.com/display.cgi?id=1348501268 | 1,503,204,008,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105970.61/warc/CC-MAIN-20170820034343-20170820054343-00443.warc.gz | 919,372,429 | 3,387 | chemistry, i need help like a.s.a.p!!!!
posted by .
What is the density of a block of unknown material with a mass of 19.07 g if the dimensions are:
length = 3.1 cm
height = 1.9 cm
width = 4.6 cm
• chemistry, i need help like a.s.a.p!!!! -
Volume= 3.1 x 4.6 x 1.9 = 27.094cm3
Density=mass/volume
= 19.07g / 27.094 cm3
Density= 0.7038g/cm3 or 0.704g/cm3
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https://bowmanimal180.wordpress.com/2013/04/25/day-126-the-volume-of-a-coke-bottle/ | 1,501,144,337,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427750.52/warc/CC-MAIN-20170727082427-20170727102427-00559.warc.gz | 608,680,396 | 32,052 | # Day 126 – The Volume of a Coke Bottle
All the examples I used for Volumes of Revolution were real objects, but the answers to the integrals didn't really mean anything, so today we tried to use these integrals to find the volume if a coke bottle. I gave them the functions fitting the side of the bottle and they did the calculations, which come out to around 350 mL (instead of the actual 330 mL). It was awesome to have students at the end ask “how did you do that?” So I showed them the process of putting the picture into GeoGebra and fitting the functions, which was a great prelude to the two final projects that we will do in the class! | 148 | 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} | 2.515625 | 3 | CC-MAIN-2017-30 | longest | en | 0.974805 |
https://wikidiff.com/terms/reciprocal | 1,709,496,598,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00153.warc.gz | 614,653,544 | 11,897 | What's the difference between and
reciprocal
Reciprocal vs Retaliatory - What's the difference?
reciprocal | retaliatory |
As adjectives the difference between reciprocal and retaliatory
is that reciprocal is of a feeling, action or such: mutual, uniformly felt or done by each party towards the other or others; two-way while retaliatory is relating to or part of a retaliation.
As a noun reciprocal
is (arithmetic) of a number, the number obtained by dividing 1 by the given number; the result of exchanging the numerator and the denominator of a fraction.
reciprocal | x |
As an adjective reciprocal
is of a feeling, action or such: mutual, uniformly felt or done by each party towards the other or others; two-way.
As a noun reciprocal
is (arithmetic) of a number, the number obtained by dividing 1 by the given number; the result of exchanging the numerator and the denominator of a fraction.
As a letter x is
the twenty-fourth letter of the.
As a symbol x is
voiceless velar fricative.
Reciprocal vs Communicate - What's the difference?
reciprocal | communicate |
As an adjective reciprocal
is of a feeling, action or such: mutual, uniformly felt or done by each party towards the other or others; two-way.
As a noun reciprocal
is (arithmetic) of a number, the number obtained by dividing 1 by the given number; the result of exchanging the numerator and the denominator of a fraction.
As a verb communicate is
to impart.
Reciprocal vs False - What's the difference?
reciprocal | false |
As adjectives the difference between reciprocal and false
is that reciprocal is of a feeling, action or such: mutual, uniformly felt or done by each party towards the other or others; two-way while false is (label) one of two states of a boolean variable; logic.
As a noun reciprocal
is (arithmetic) of a number, the number obtained by dividing 1 by the given number; the result of exchanging the numerator and the denominator of a fraction.
Reciprocal vs Undefined - What's the difference?
reciprocal | undefined |
As adjectives the difference between reciprocal and undefined
is that reciprocal is of a feeling, action or such: mutual, uniformly felt or done by each party towards the other or others; two-way while undefined is lacking a definition or value.
As a noun reciprocal
is (arithmetic) of a number, the number obtained by dividing 1 by the given number; the result of exchanging the numerator and the denominator of a fraction.
Symbiotic vs Reciprocal - What's the difference?
symbiotic | reciprocal |
As adjectives the difference between symbiotic and reciprocal
is that symbiotic is (biology) of, or relating to symbiosis; living together while reciprocal is of a feeling, action or such: mutual, uniformly felt or done by each party towards the other or others; two-way.
As nouns the difference between symbiotic and reciprocal
is that symbiotic is (astronomy) symbiotic star while reciprocal is (arithmetic) of a number, the number obtained by dividing 1 by the given number; the result of exchanging the numerator and the denominator of a fraction.
Reciprocal vs Joint - What's the difference?
reciprocal | joint |
As nouns the difference between reciprocal and joint
is that reciprocal is (arithmetic) of a number, the number obtained by dividing 1 by the given number; the result of exchanging the numerator and the denominator of a fraction while joint is marijuana cigarette; joint.
As an adjective reciprocal
is of a feeling, action or such: mutual, uniformly felt or done by each party towards the other or others; two-way.
Counterpart vs Reciprocal - What's the difference?
counterpart | reciprocal |
As nouns the difference between counterpart and reciprocal
is that counterpart is either of two parts that fit together, or complement one another while reciprocal is (arithmetic) of a number, the number obtained by dividing 1 by the given number; the result of exchanging the numerator and the denominator of a fraction.
As an adjective reciprocal is
of a feeling, action or such: mutual, uniformly felt or done by each party towards the other or others; two-way. | 889 | 4,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-10 | latest | en | 0.932877 |
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Ugh! This one is so hardddd
0
61
4
Shortly after 10 'o clock, the hands of a clock made a 90 degree angle. What's the time?
Jun 16, 2021
#1
+121006
+1
Let M be the number of minutes after 10
The hour hand moves (1/2)° every minute
The minute hand moves 6° every minute
Call 10 o'clock = 0°
So....the minute hand is at 60° at 10
So.....we just need to solve this
6M - (1/2)M + 60 = 90
5.5M = 30
M = 30 / 5.5 ≈ 5.45 minutes after 10 = 5 + .45 *60 ≈ 5 min 27 sec after 10
Jun 16, 2021
#2
0
That doesn't make sense! If the minute moves 30 degrees per 5 minute interval, then 30*10 = 300 degrees should be the degrees the minute hand travels until 10!
Jun 16, 2021
#3
+1452
0
Shortly after 10 'o clock, the hands of a clock made a 90-degree angle. What's the time?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Here's another way to solve this problem.
Arc from 10 to 12 = 60º
Let x be the arc-path of minute hand in degrees.
60 + x - (x/12) = 90
x = 360/11 = 32.72727272º (minute hand arc)
x/12 = 30/11 = 2.72727272º (hour hand arc)
Jun 16, 2021
#4
0
This is amazing. I just posted an answer to this clock problem on a more recent posting of the problem. I had not seen Phil's solution, but I could have believably copied his verbatim. The same thing happened on a regular decagon problem, even though our solutions were not quite so similar. This is fun; post problems as many times as your heart desires, even though I am getting dizzy looking through the solutions. Just kidding. Please have mercy. I really think only one posting should be allowed at one time, so no time is wasted and the better solutions can be compared.By the way, I tried to sign up today, but the activation email never arrived. Still waiting patiently.
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Topic: MLCS Webinar on April 24: Registration Open
Replies: 31 Last Post: Apr 12, 2012 11:45 AM
Messages: [ Previous | Next ]
Edward (Ed) D. Laughbaum Posts: 246 Registered: 12/4/04
Re: MLCS Webinar on April 24: Registration Open
Posted: Apr 6, 2012 9:39 AM
att1.html (5.8 K)
Hi Phil.
The extra \$0.16 was the outcome of the apparently sum total mathematical
knowledge of college graduates -- all relationships are proportional. It
is just like with developmental students coming straight from HS, who
think everything relationship is linear and then after passage of time
think all relationships are proportional. Yes, I have generalized when I
probably should not have done so.
Sure, I do exaggerate to make a point. As several posters have
mentioned, the usage-charge relationship is not a proportional
relationship, but rather a linear function. So my point was that the
method for solving the problem was more important than the answer. But
of course, the answer told me how they solved the problem.
Best Regards,
Ed
========================================
On 4/6/2012 7:01 AM, Philip Mahler wrote:
> OK, I'm kind of slow sometimes, I'd rather admit it and learn
> something, so...
>
> Why is the \$0.16 irrelevant?
>
> I got \$132.16, using proportion and would have responded with that
> answer. I understand about significant figures, and if applied to this
> problem strictly speaking, would produce an answer of \$130.
>
> Of course I agree that electric bills are not proportional to usage
> because of the fixed costs, but given the question posed, I would have
> included the \$0.16.
>
> ---
>
> Regardless of what I missed above, I consider proportion to be a very
> powerful tool that we do not give our students in sufficient doses.
>
> As I have noted in the past, my universe of discussion is my
> developmental students in an open-door institution. Many cannot
> compute area, nor solve "you drive at 50 mph for two hours, how far
> have you gone", nor compute miles per gallon... But I can probably
> teach most to make a table which will allow them to solve a proverbial
> two-trains problem, which is what will be in the text. So they can do
> two trains, without a clue about what's going on, but still can't do
> one train. :-)
>
> As for function, I wouldn't apply it to mpg, mph, electric bills as
> stated. But I would if the electric bill question had a fixed cost
> also. I think (I dare say I know) we don't give these students enough
> use of y = mx + b so it's truly useful, and can be applied in other
> contexts, like economics, for example -- or electric bills.
>
> Phil
>
> On 4/5/12 5:04 PM, "Ed Laughbaum" <elaughba@math.ohio-state.edu> wrote:
>
> Guy,
>
> The \$0.16 is irrelevant. But then if I overcharged 1,000,000
> customers by \$0.16 (or whatever it is based on usage). \$0.16
> becomes significant. The point was that they couldn't solve a
> really simple problem. BTW: I failed to give you a second data
> point in the original posting. Also, my point was that all these
> highly trained people could only use what they had learned in
> middle school.
>
> This discussion may not be worth the effort because no one knows
> what is the proper approach or course or etc.
>
> Ed
> =============================
> On 4/5/2012 4:09 PM, Guy Brandenburg wrote:
>
>
> Electric bills aren't necessarily exactly proportional to the
> kwh used. And does 16 cents really matter anyway?
>
> Guy
>
>
> On Apr 5, 2012, at 1:27 PM, Ed Laughbaum
> <elaughba@math.ohio-state.edu> wrote:
> ...
>
>
> Relative to this post, I have observed that I often see
> the mathematical literacy proponents argue for
> proportional reasoning as a mainstay outcome. This in
> turn, reminded me (recall through neural associations) of
> an informal survey I took on several colleagues who were
> not in any of the STEM fields but all had a bachelors
> degree through a PhD. In the following "problem" everyone
> used proportional reasoning.
>
> If you use 1205 kWh of electricity and your bill is \$130,
> how much is your monthly bill if you use 1225 kWh?
> Everyone got \$132.16 for the answer when it is \$132.
>
> What I wonder is, if focusing on proportional reasoning
> will solve this very simple problem (of thinking
> relationships are proportional), or whether we should be
> focusing on function. Or something else? Of course, my
> opinion is on function, but it is an opinion.
>
--
Edward Laughbaum www.math.osu.edu/~laughbaum.6/
The Ohio State University
231 West 18th Avenue
Columbus, OH 43210 | 1,190 | 4,662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.95433 |
https://maths.wizert.com/video-solution/1937/grade-9-lines-and-angles-1605690381 | 1,722,736,324,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00587.warc.gz | 308,329,136 | 116,452 | Lines and Angles - Grade 9 (Question: 1605690381)
If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 3:1, what is the measure of the smaller angle?
A. $90^0$
B. $135^0$
C. $45^0$
D. $72^0$ | 82 | 245 | {"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} | 2.765625 | 3 | CC-MAIN-2024-33 | latest | en | 0.868002 |
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12-27-2016, 02:49 PM #17326
Tbirdx24
old hand
Join Date: May 2009
Location: NY
Posts: 1,630
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by Garick Or you could just look at the post above yours. ��
Lol we must have posted at the same time. Thank you very much.
12-27-2016, 03:21 PM #17327
homerdash
banned
Join Date: Sep 2006
Location: desert
Posts: 6,062
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by Garick Ok, I found it and it's probably worth posting again. To find the 95% confidence level on your bounds related to observed winrate: Multiply your hourly standard deviation by 2 and then divide the result by the square root of your number of hours. For short: Hourly SDev*2 / sqrt of hours As you can see, increased hours will tighten the heck out of these bounds, especially over the first couple hundred hours. To use, combine with observed winrate. So if observed winrate is 7bbs/hr and your bounds are +/- 6.3bbs/hr, you can be 95% confident that you are a winning player, even though your true winrate could be as low as .7bbs/hr.
conversely, can my true winrate be as high as 13-14bb/hr?
12-27-2016, 03:25 PM #17328
johnny_on_the_spot
Carpal \'Tunnel
Join Date: Jun 2013
Location: S-Mart
Posts: 10,922
*** Official Winrates, bankrolls, and finances ***
Quote:
Originally Posted by Garick Ok, I found it and it's probably worth posting again. To find the 95% confidence level on your bounds related to observed winrate: Multiply your hourly standard deviation by 2 and then divide the result by the square root of your number of hours. For short: Hourly SDev*2 / sqrt of hours As you can see, increased hours will tighten the heck out of these bounds, especially over the first couple hundred hours. To use, combine with observed winrate. So if observed winrate is 7bbs/hr and your bounds are +/- 6.3bbs/hr, you can be 95% confident that you are a winning player, even though your true winrate could be as low as .7bbs/hr.
Is there a reason why we're using 2 and not 1.96 as our z value? I get that they're practically the same and that 2 is "easier" than 1.96, but we're also calcing the square root of a random number, so it's not like anyone is doing this without a calculator...
12-27-2016, 04:35 PM #17329
Garick
Oberbiergenießer
Join Date: Dec 2007
Location: Do you even math, bruh?
Posts: 24,603
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by homerdash conversely, can my true winrate be as high as 13-14bb/hr?
Exactly.
Quote:
Originally Posted by johnny_on_the_spot Is there a reason why we're using 2 and not 1.96 as our z value? I get that they're practically the same and that 2 is "easier" than 1.96, but we're also calcing the square root of a random number, so it's not like anyone is doing this without a calculator...
IDK. I just stole the formula from bip! Bip!?
12-27-2016, 04:45 PM #17330 bip! Slow Pony Join Date: Oct 2012 Location: not on urban dictionary... Posts: 13,669 re: Winrates, bankrolls, and finances 2 was just an approximation. 1.96 if you want to be exact.
12-27-2016, 04:56 PM #17331
MIB211
veteran
Join Date: Jan 2013
Posts: 3,011
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by homerdash conversely, can my true winrate be as high as 13-14bb/hr?
I don't know the statistical answer here, but my strong suspicion is that if you have someone who's winning 7 bbs/hour with a 95% confidence interval of +/- 7bb/hour, it's significantly more likely that they're a break even player than they're winning 14bb/hour. This is an application of Bayes theorem and reversion to the mean. Basically, you have a ton of observed break even players, and very, very few 14bb/hour crushers. So, much more likely that any given person who's a 7bb/hour winner is running good than bad.
12-27-2016, 05:31 PM #17332 OvertlySexual old hand Join Date: Jan 2014 Posts: 1,518 re: Winrates, bankrolls, and finances FWIW, I want to note once again that some of the poker tracking apps give wrong standard deviation numbers. For example, Poker Income gives me standard deviation in the low fourties whereas RunGood gives me a standard deviation between 90 and 100 BB/hr. This in turn changes drastically the results of the confidence interval from 2.9BB to 7BB. I did try to find my standard deviation on excel and I think Poker Income is the one doing the right calculation. I could be wrong though.
12-27-2016, 05:48 PM #17333 Garick Oberbiergenießer Join Date: Dec 2007 Location: Do you even math, bruh? Posts: 24,603 re: Winrates, bankrolls, and finances 90-100 seems much more standard. Very few of the folks ITT who have converted sdev/session into sdev/hr via crunching the numbers (vs just taking what the app says as gospel) have reported sdev <60bbs/hr. Like three of us, IIRC, and all three were upper 50s. 100bbs/hr, OTOH, seems pretty standard. Last edited by Garick; 12-29-2016 at 09:13 AM. Reason: typo
12-27-2016, 05:59 PM #17334 homerdash banned Join Date: Sep 2006 Location: desert Posts: 6,062 re: Winrates, bankrolls, and finances my Poker Income has it at 112
12-27-2016, 06:46 PM #17335 OvertlySexual old hand Join Date: Jan 2014 Posts: 1,518 re: Winrates, bankrolls, and finances Let me tell you what I did on excel. 1. I took the session profit from 48 sessions. 2. I then took the duration of each session and converted it into a decimal form. So for example a session of 11 hours 56 minutes was converted to 11.933. 3. I then divided each session's profit by the duration so I found the hourly profit of each session. 4. I then calculated the sum of the individual profit and came up with a mean profit for all sessions. 5. I then subtracted the mean from each hourly session profit. 6. I squared the resulting number. 7. Then I divided the resulting number by the number of sessions. 8. I then got the square root of that number. That final number is supposed to be the standard hourly deviation. Using that method, I found a standard deviation of 43 BB, whereas poker income gives me 42.85 for the same sample. Runggood gives me a standard deviation of 95.21. What did I do wrong? Regardless. Can we determine which poker tracking app does it right? Can someone do the proper calculation on excel and compare with the number he gets on his poker app?
12-27-2016, 06:54 PM #17336 feel wrath The Situation Join Date: Nov 2010 Location: lost on the turn Posts: 23,866 re: Winrates, bankrolls, and finances Difference between bb and BB?
12-27-2016, 07:13 PM #17337 Sneaky Pete banned Join Date: Aug 2015 Posts: 353 re: Winrates, bankrolls, and finances Big Blind as position vs big blinds as quantifier.
12-27-2016, 08:43 PM #17338
bip!
Slow Pony
Join Date: Oct 2012
Location: not on urban dictionary...
Posts: 13,669
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by feel wrath Difference between bb and BB?
The main difference is that "BB" tilts me every time I see it in a NL forum.
"bb" on the other hand has no such tilting effect.
.. probably because it actually means "big blind" and exists in NL.
12-27-2016, 08:54 PM #17339
bip!
Slow Pony
Join Date: Oct 2012
Location: not on urban dictionary...
Posts: 13,669
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by MIB211 I don't know the statistical answer here, but my strong suspicion is that if you have someone who's winning 7 bbs/hour with a 95% confidence interval of +/- 7bb/hour, it's significantly more likely that they're a break even player than they're winning 14bb/hour. This is an application of Bayes theorem and reversion to the mean. Basically, you have a ton of observed break even players, and very, very few 14bb/hour crushers. So, much more likely that any given person who's a 7bb/hour winner is running good than bad.
^ to address this we took a survey of 2+2ers and they all assured us they are running below EV... often accompanied by a story about "that time I got 2 outed in a big pot".
12-27-2016, 08:57 PM #17340
bip!
Slow Pony
Join Date: Oct 2012
Location: not on urban dictionary...
Posts: 13,669
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by OvertlySexual Let me tell you what I did on excel. 1. I took the session profit from 48 sessions. 2. I then took the duration of each session and converted it into a decimal form. So for example a session of 11 hours 56 minutes was converted to 11.933. 3. I then divided each session's profit by the duration so I found the hourly profit of each session. 4. I then calculated the sum of the individual profit and came up with a mean profit for all sessions. 5. I then subtracted the mean from each hourly session profit. 6. I squared the resulting number. 7. Then I divided the resulting number by the number of sessions. 8. I then got the square root of that number. That final number is supposed to be the standard hourly deviation. Using that method, I found a standard deviation of 43 BB, whereas poker income gives me 42.85 for the same sample. Runggood gives me a standard deviation of 95.21. What did I do wrong? Regardless. Can we determine which poker tracking app does it right? Can someone do the proper calculation on excel and compare with the number he gets on his poker app?
You can't convert the individual results to per hour and treat that as normalized.
Later I will post how to do this in excel. It is simpler, but maybe not intuitive.
12-27-2016, 09:04 PM #17341 YGOchamp old hand Join Date: Mar 2013 Posts: 1,517 re: Winrates, bankrolls, and finances Theres no such thing as losing (or break even) poker players. Simply those who are running good and those who are running bad Also, so can anybody else confirm that the poker income app has correct stddev?
12-27-2016, 11:35 PM #17342
MikeStarr
Carpal \'Tunnel
Join Date: Jan 2016
Posts: 7,978
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by OvertlySexual FWIW, I want to note once again that some of the poker tracking apps give wrong standard deviation numbers. For example, Poker Income gives me standard deviation in the low fourties whereas RunGood gives me a standard deviation between 90 and 100 BB/hr. This in turn changes drastically the results of the confidence interval from 2.9BB to 7BB. I did try to find my standard deviation on excel and I think Poker Income is the one doing the right calculation. I could be wrong though.
About 2 months ago I sent all my numbers to Bip! and he confirmed that my StnDev number was correct. I use Run Good.
12-27-2016, 11:50 PM #17343
meale
Carpal \'Tunnel
Join Date: May 2013
Location: Pattaya, Thailand
Posts: 9,840
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by YGOchamp Theres no such thing as losing (or break even) poker players. Simply those who are running good and those who are running bad
:')
12-28-2016, 05:27 AM #17344
WereBeer
Carpal \'Tunnel
Join Date: Oct 2013
Posts: 9,650
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by bip! ^ to address this we took a survey of 2+2ers and they all assured us they are running below EV
Casino regs all run super unlucky as well, like every one of them I ask or in fact don't ask, is keen to relay their sob story about how bad they run.
12-28-2016, 05:40 AM #17345
feel wrath
The Situation
Join Date: Nov 2010
Location: lost on the turn
Posts: 23,866
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by bip! ^ to address this we took a survey of 2+2ers and they all assured us they are running below EV... often accompanied by a story about "that time I got 2 outed in a big pot".
unless of course, they posted higher win rates than anybody else, in which case 2+2ers assured them that they were either lying or were running unsustainably well and that a fall was right around the corner
12-28-2016, 05:41 AM #17346 VolumeKing adept Join Date: Jan 2016 Posts: 729 re: Winrates, bankrolls, and finances 161 hrs since the 9th
12-28-2016, 01:52 PM #17347
iraisetoomuch
banned
Join Date: Aug 2013
Location: New Jersey
Posts: 34,453
re: Winrates, bankrolls, and finances
Quote:
Originally Posted by cAmmAndo That's a \$5.40/hr bump yes compared to not playing those pots or running neutral but I think it's good to realize it's actually a \$10.80 swing in short term wr if you lose those pots.
Not really.
I didn't profit those amounts (\$2,700) I won pots of that amount.
Which means that I only profited \$1,350 (really closer to \$1,700 since some of them were not heads up pots, but whatever). So if I don't play them, I'm down \$1,700 and if I lose them I'm down \$2,700.
Quote:
Originally Posted by MikeStarr I wouldnt feel bad if you're not there. No matter what you see on 2+2 with people bragging about win rates, I would estimate that only 1-2% 2/5 players in the world are making that much. Being in the top 1-2% of anything is pretty impressive.
This is likely true as a general idea (but obv lol at people making up statistics on the spot with little to go on really).
But since the average poster on 2+2 is likely better than the average non-poster it is hard to really judge where we may lie in the poker spectrum and how many of us should be making XXbb/hour.
Quote:
Originally Posted by MIB211 I don't know the statistical answer here, but my strong suspicion is that if you have someone who's winning 7 bbs/hour with a 95% confidence interval of +/- 7bb/hour, it's significantly more likely that they're a break even player than they're winning 14bb/hour. This is an application of Bayes theorem and reversion to the mean. Basically, you have a ton of observed break even players, and very, very few 14bb/hour crushers. So, much more likely that any given person who's a 7bb/hour winner is running good than bad.
This is only true if we expect that the population is not normally distributed.
Otherwise it is by definition that the two results (+14bb/hour and +0bb/hour) are equally likely.
12-28-2016, 04:15 PM #17348 cAmmAndo The Situation Join Date: Nov 2008 Location: Philly Posts: 4,712 re: Winrates, bankrolls, and finances ^^^ Re: pot vs profit. Agreed.
12-28-2016, 04:24 PM #17349 Maskk veteran Join Date: Feb 2009 Location: DC Area Posts: 2,083 re: Winrates, bankrolls, and finances I am curious how run bad effects play styles. I ran terrible for premium hands around the time I was really moving into a profitable expected WR (probably 500ish key hours)--this forced me to work on storytelling, bluffing, long-shot+small-risk hands, and other ways of gaining EV which is now an integral part of how I play. I am curious if people think that how you ran while learning has affected your style of play at all? I wonder if it is something that solid tags and rocks even consider? Sent from my iPhone using Tapatalk
12-28-2016, 04:42 PM #17350 Sneaky Pete banned Join Date: Aug 2015 Posts: 353 re: Winrates, bankrolls, and finances All the time, but that's more psychology than WR related.
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Contact Us - Two Plus Two Publishing LLC - Privacy Statement - Top | 4,628 | 17,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-39 | latest | en | 0.901657 |
wildernessarena.com | 1,556,216,851,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578732961.78/warc/CC-MAIN-20190425173951-20190425195951-00181.warc.gz | 191,031,510 | 27,931 | In the wilderness, it is sometimes useful to be able to calculate the distance to a faraway object. Using a compass and basic trigonometry (don’t worry, we’ll show you how), you can easily estimate the distance to a faraway object. This is possible using principles of geometry and triangulation – if we know any three of the sides or angles of a triangle, we can calculate the remaining angles and sides. In the examples below, we’ll demonstrate two different methods to calculate the distance to an object.
# Introduction
Using the illustration below, we will first (1) “walk off” two distances, (2) sight the faraway object in our compass from both points, and finally (3) record the angle for both points to use in our distance calculation. Using the known distance (the number of steps we walked off) and the angles measured at each of the two points, there are a couple of easy ways to estimate the distance to the object. We can use basic trigonometry (using a “tangent” table listed below) or we can draw the layout to scale on a piece of paper to get our distance. The trigonometric method will produce the most accurate measurement but requires a table of tangents (or a calculator to calculate the tangent of the angles). Using a scaled drawing on paper is not as accurate but requires nothing more than a compass and something to draw on (even the dirt on the ground would work for this purpose).
# Triangulation without a map or drawing to scale
The triangulation method will use the following formula. Don’t worry if it looks complicated – we’re going to break it down to make it very easy. Our object is to find the distance from the observer’s point to the ship (marked “d?” in our illustration)
d = (Tan (90 – (A -B))) x Baseline
d = Distance to be calculated
Tan = Tangent value of the resultant angle (we’ll look this value up in a table)
A = Greater value of the two measured bearing angles (the greater of Angle A or Angle B)
B = Lower value of the two measured bearing angles (the lessor of Angle A or Angle B)
Baseline = Measured distance between our two bearing angles (distance between Angle A and Angle B)
## Step 1
Starting at the point marked “Observer” (in the illustration above), count off the steps to a point perpendicular to the faraway object (in our example drawing, we want to calculate the distance to the ship and will begin by counting the steps between the “Observer” position and the point marked “Angle A”). You can place a stick in the ground at all points to aid in stepping off the distances. You don’t have to walk precisely perpendicular to the object being measured but try to stay close to perpendicular and make sure you walk in a straight line. We will call this first point “Point A” and the angle of that point “Angle A”. The line that we walk on will be our “baseline”. The longer the baseline (and hence, the more steps taken between the points), the more accurate the distance calculation will be. The steps counted on the baseline will be used to calculate the “steps” to the object. To calculate a distance to the object using feet, you can convert the measurement in steps to a feet typically by using an estimate of 3 feet per step (long step).
## Step 2
From this first point (labelled Angle A), sight the object using your compass and note the degrees between your “baseline” (the line you walked down) and the line running towards the object. This will give us the value of Angle A. It may help to lock the compass, so it does not rotate, and point North (0 degrees) down your baseline path.
In the picture below,the red line is running parallel to our baseline path (the path we are stepping off) while the yellow line points towards the distant object. The angle measured here is around 60 degrees (as marked by the red numerals on the compass dial).
## Step 3
Return to your original “Observer” position and do the same thing but in the opposite direction. Try to keep the distance (number of steps) close to the same as the distance you marked off in Step 1. This second point will be “Point B” and the angle that you measure here will be “Angle B”. It will help the calculation if you make this angle (or Angle A) as close to 45 degrees (a “right angle” – shaped like the letter “L”) as you can get.
## Step 4
Add the steps to Point A and Point B together to get the total steps of our baseline (the distance between Point A and Point B). In mathematical terms, this is the “known segment” of our triangle (the one side of the triangle that we know the length for) and is shown as “Baseline” in our formula.
## Step 5
Note which of the two measured angles is greater. This will become “A” in our formula. The lessor of the two angles will be “B” in our formula. Calculate (A – B) or in other words, subtract the lessor angle from the greater angle. If Angle A was 45 and Angle B was 60, we would subtract 45 from 60 like this: 60 – 45 = 15.
## Step 7
Now we will figure out the tangent value to look up in our Degrees to Tangent Ratio table. To do this, simply subtract the value we calculated above (the difference between the two angles) from 90. Continuing our example, our formula would be: 90 – 15 = 75.
## Step 8
Look up the Tangent value in the table below. In our example, we look up the value for 75 which according to our table, is 3.732.
## Step 9
Finally, we calculate the distance to the faraway object by multiplying the baseline distance by our tangent value. Continuing our example, let’s say the distance between Point A and Point B is 10 steps. Multiply 10 by our tangent value (3.732 – the value we looked up in the tangent ratio table) to calculate the number of steps to the distant object which in our example would be: 10 x 3.732 = 37.32 steps. The total distance to our faraway object is 37.32 steps.
For the math whizzes out there, our entire formula translates to this:
d = (Tan (90 – (A -B))) x Baseline
d = (3.732 (90 – (60-45))) x 10
Note that most calculators out there, including most calculator apps available on smartphones, include a Tan function. Simply enter the difference between the two angles and press the Tan key on the calculator. Then multiply the result by the baseline distance to get the distance to the target object.
Here’s the diagram again with all the values filled in.
## Degrees to Tangent Ratio table
```1 = 0.017
2 = 0.034
3 = 0.052
4 = 0.069
5 = 0.087
6 = 0.105
7 = 0.123
8 = 0.140
9 = 0.158
10 = 0.176
11 = 0.194
12 = 0.212
13 = 0.230
14 = 0.249
15 = 0.267
16 = 0.286
17 = 0.306
18 = 0.325
19 = 0.344
20 = 0.363
21 = 0.384
22 = 0.404
23 = 0.424
24 = 0.442
25 = 0.466
26 = 0.488
27 = 0.509
28 = 0.532
29 = 0.554
30 = 0.577
31 = 0.601
32 = 0.635
33 = 0.649
34 = 0.674
35 = 0.700
36 = 0.726
37 = 0.753
38 = 0.781
39 = 0.810
40 = 0.839
41 = 0.869
42 = 0.900
43 = 0.932
44 = 0.965
45 = 1.000
46 = 1.035
47 = 1.072
48 = 1.110
49 = 1.150
50 = 1.192
51 = 1.234
52 = 1.280
53 = 1.327
54 = 1.376
55 = 1.428
56 = 1.482
57 = 1.539
58 = 1.600
59 = 1.664
60 = 1.732
61 = 1.804
62 = 1.880
63 = 1.962
64 = 2.050
65 = 2.144
66 = 2.246
67 = 2.355
68 = 2.475
69 = 2.605
70 = 2.747
71 = 2.904
72 = 3.077
73 = 3.270
74 = 3.487
75 = 3.732
76 = 4.010
77 = 4.331
78 = 4.704
79 = 5.144
80 = 5.671
81 = 6.313
82 = 7.115
83 = 8.144
84 = 9.514
85 = 11.430
86 = 14.300
87 = 19.081
88 = 28.636
89 = 57.289
90 = Undefined```
# Triangulation using a simple drawing
Suppose you do not have a calculator nor the Degrees to Tangent ratio table readily available? You can still calculate the distance to the faraway object using a scaled drawing of the triangle and rudimentary measurements of the distances calculated based on the drawing. Many of the steps will be the same as in our example above but we’ll repeat those here for clarity.
## Step 1
Walk off the distance between Point 1 and Point B (see diagram above) making sure you are walking closely to perpendicular to the object being sighted. From both Point 1 and Point B, sight the object using your compass and note the degrees between your “baseline” (the line you walked down) and the line running towards the object. This will give us the value of Angle A and Angle B. It may help to lock the compass, so it does not rotate, and point North (0 degrees) down your baseline path.
## Step 2
Draw your baseline on a piece of paper (or in the dirt on the ground).
## Step 3
Draw lines matching the angles that you measured in Step 1. Draw them out long enough so that they intersect (which is the point of faraway object we are measuring the distance to).
## Step 4
Use a ruler (or some other object marked with equal increments) to measure the Baseline of your drawing. Divide the stepped-off baseline measurement (number of steps between Point 1 and Point B) by your measurement on the paper. This will give you the scaled distance variable.
## Step 5
On the piece of paper, measure the distance between your observation point and the faraway object. Multiple this measurement by the scaled distance variable to calculate the distance to the object.
# The geometry behind the magic
For the math geeks, or those that are curious, here is the basic geometry behind the calculations.
To define the trigonometric functions for the angle A (see diagram above), start with any right triangle that contains the angle A. The three sides of the triangle are named as follows:
• The hypotenuse is the side opposite the right angle, in this case side h. The hypotenuse is always the longest side of a right-angled triangle.
• The opposite side is the side opposite to the angle we are interested in (angle A), in this case side a.
• The adjacent side is the side having both the angles of interest (angle A and right-angle C), in this case side b.
In ordinary Euclidean geometry, the inside angles of every triangle total 180°. Therefore, in a right-angled triangle, the two non-right angles (Angle A and Angle B in our example) total 90°.
## Sine
The sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. (The word comes from the Latin sinus for gulf or bay, since, given a unit circle, it is the side of the triangle on which the angle opens.) In our case
Note that this ratio does not depend on the size of the particular right triangle chosen, as long as it contains the angle A, since all such triangles are similar.
## Cosine
The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse: so called because it is the sine of the complementary or co-angle. In our case
## Tangent
The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side. In our case
## Why are sin, cos, and tan important?
The sin, cos, and tan are important because they allow you to work out angles when you know the lengths of the sides or to work out the lengths of the sides when you know the angles.
The acronyms “SOHCAHTOA” (“Soak-a-toe”, “Sock-a-toa”, “So-kah-toa”) and “OHSAHCOAT” are commonly used mnemonics to help remember these ratios. The following illustration helps demonstrate how sin, cos, and tan are calculated. | 2,995 | 11,195 | {"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} | 4.78125 | 5 | CC-MAIN-2019-18 | longest | en | 0.887298 |
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# Steady State Approximation
Lerne mit deinen Freunden und bleibe auf dem richtigen Kurs mit deinen persönlichen Lernstatistiken
Nie wieder prokastinieren mit unseren Lernerinnerungen.
So, you're standing in the pasta aisle of the grocery store, and you're feeling spicy. You decide to try a new type of pasta with twists and turns you never imagined in a food. Once home, you boil some water and drop it in. You don't want to just sit and watch it, so you consult the ancient texts (the back of the pasta box).
The box states it will take 7-9 minutes to cook. Cooking times are a very helpful, pre-determined quantity, that have saved many lives. This is an example of kinetics, or essentially, how quickly reactions take place. This concept can also be applied to single-step or multistep reactions. In chemistry, we can derive an equation called the rate law, which describes the speed of a reaction.
Multistep reactions can be very complex, so deriving the rate law isn't always easy. When a reaction mechanism has several steps with comparable rates, it can be hard to figure out the rate-determining step. When this happens, we utilize the steady-state approximation.
In this article, we will be learning all about this approximation, such as what it is and how to use it.
• This article covers the Steady-State Approximation
• First, we will learn what this approximation is and do a brief overview of kinetics
• Then, we will learn how to use the approximation and apply it to some examples
## Steady State Approximation Definition
Let's start by looking at the definition of steady state approximation.
The steady-state approximation (also called the quasi-steady-state approximation or pseudo-steady-state approximation) is a way to simplify the derivation of the rate equation. It is based on the assumption that one intermediate in the mechanism will be produced as fast as it is consumed (i.e., it is at a steady state).
In a multistep reaction, there will be species that are intermediates.
An intermediate is a species that isn't one of the initial reactants or final product(s). It is produced during the mechanism and will be completely consumed by the end.
The steady-state approximation assumes that at some point this intermediate will not have a change in concentration.
Fig.1-The concentration is stable for a period, which is called the "steady-state"
We can illustrate this steady state using the equation:
$$\frac{d[I]}{dt}=0$$
Where the concentration of the intermediate is [I ].
In this article, we will be diving into the concept of the steady-state approximation and see how and where it can be used.
## Steady-state Approximation Equation
We note that a reaction’s rate is dependent on three things:
The rate constant tells us how fast or slow a reaction is, the smaller the constant, the slower the reaction. A reactant's "order" shows the direct relationship between the rate and that reactant's concentration. At this point let us ask: if reactant A is doubled, will the rate also double, will it quadruple, or will it not change at all? Here is a general reaction:
$$A+B\xrightarrow{k_1}C$$
For this reaction, our rate equation looks like this:
$$\text{rate}=k_1[A]^x[B]^y$$
Where, k, is the rate constant, and the superscripts (x & y) represent the order of the reaction. Now, this is for an elementary, or "one-step" reaction, but what about a multistep reaction? Here is a general reaction mechanism:
$$A+B\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}I\,\text{(fast)}$$
$$B+I\xrightarrow{k_2}C\,\text{(slow)}$$
All reactions have an energy barrier called activation energy that they need to overcome for a reaction to occur.
• The activation energy barrier is based on the energy of the reactants and products relative to the energy of the transition state.
• A system wants to be as stable as possible, so if a reaction produces a higher energy (i.e less stable) product, then it is going to have a larger activation energy since the system would rather stay as the stable reactants.
In our equation, we see that they are labeled "slow" and "fast". A "slow" reaction has a very high activation energy, so it takes a while for the system to get enough energy to proceed. This slow step is called the rate-determining step.
• The rate-determining step is what determines the rate for the whole reaction, and is also the step we write the rate equation for.
Think of it like being stuck in traffic: All the cars are stuck going the same speed as the slow truck in front, even if they can go faster. Here is how we would write the rate equation for this set of reactions:
We follow our basic formula from before, and set up the equation using:
$$A+B\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}B+I\xrightarrow{k_2} C$$
Then,
$$rate=k_2[B][I]$$
Now here is where we have a problem, we cannot leave an intermediate in a rate reaction. We need to express the intermediate in terms of the reactants. We can do this by setting up the expression for the equilibrium constant for the equilibrium step:
The equilibrium constant (Keq) shows whether a reaction favors the "forward" reaction (forms products) or the "backward" reaction (forms reactants). For a general equation:
$$A+B\rightleftharpoons C+D$$
The expression is:
$$K_{eq}=\frac{[C][D]}{[A][B]}$$
The constant can also be expressed in terms of the rate constants. For a general equation:
$$A+B\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}C+D$$
The expression is:
$$K_{eq}=\frac{k_1}{k_{-1}}$$
Back to our reaction. We can express the equilibrium step in terms of the equilibrium constant:
$$A+B\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}I\,\text{(fast)}$$
Then,
$$K_{eq}=\frac{[I]}{[B][A]}$$$$K_{eq}=\frac{k_1}{k_{-1}}=K_1$$or,$$K_1=\frac{[I]}{[B][A]}$$rearranging we then get, $$K_1[B][A]=[I]$$or,$$[I]=K_1[A][B]$$
Now that we have our intermediate in terms of the reactants, we can substitute, $$[I]=K_1[A][B]$$, into our rate equation.
$$B+I\xrightarrow{k_2}C\,\text{(slow)}$$
$$\text{rate}=k_2[B][I]$$
$$\text{rate}=k_2[B](K_1[B][A])=k_2K_1[B]^2[A]$$
$$\text{rate}=k[B]^2[A]$$
Now we have our final rate equation. We combined our two constants, k2 and K1 , into "k" for simplicity.
## Kinetics: Steady-state Approximation Usage
Now that we've brushed up on the multistep rate equation, let's look at why we need this approximation in the first place. Here's a basic equation:
$$A\xrightarrow{k_1}B$$
$$B\xrightarrow{k_2}C$$
The concentration of B is dependent on A, therefore the concentration of C is dependent on both A and B. The derivation for the concentration looks like this:
$$[C]=[A_0](1+\frac{k_2e^{-k_1t}-k_1e^{-k_2t}}{k_1-k_2})$$
The equation itself isn't important, this is just to show how complex these derivations can be.
Now let's apply the steady-state approximation. In this mechanism, B is our intermediate, so we set its change in concentration to zero. When a species is in a steady state, its rate of consumption is equal to its rate of creation.
Using this, we can calculate:
$$\frac{d[B]}{dt}=0\,\text{(this expression means change in concentration of, B, over time)}$$
Recalling that under the steady state approximation:
$$\text{rate of consumption}=\text{rate of creation}$$
Then, we get:
$$k_1[A]=k_2[B]$$
$$[B]=\frac{k_1[A]}{k_2}$$
Now that we have the expression for the concentration of B, we can calculate the change in concentration of C.
$$\frac{d[C]}{dt}=k_2[B]$$
$$\frac{d[C]}{dt}=k_2(\frac{k_1[A]}{k_2})=k_1[A]$$
Solving for the concentration of C we get:
$$[C]=[A_0](1-e^{-k_1t})$$
You will not need to derive this, this is just to show how much simpler the derivation is.
So as you can see, the steady-state approximation makes the final expression much simpler. Now that we know the basics, let's work on a problem:
For the following multistep reaction, set up the rate law:
$$A+B\underset{k_1} {\stackrel{k_1}{\rightleftharpoons}}I$$
$$I+B\xrightarrow{k_2}C$$
Our first step is to set up the steady-state approximation. As we saw before, this means that we set the rate of consumption equal to the rate of creation. There is one step where I is being made, which is step 1 forward. The two steps where I is being consumed are step 1 (reverse) and step 2. We would express this as:
$$\text{rate of consumption}=\text{rate of creation}$$
$$\text{rate of consumption}-\text{rate of creation}=0$$
$$\frac{d[I]}{dt}=0=(k_2[I][B]+k_{-1}[I])-k_1[A][B]$$
Next, we will solve for the concentration of I:
$$\frac{d[I]}{dt}=0=(k_2[I][B]+k_{-1}[I])-k_1[A][B]$$
$$k_2[I][B]+k_{-1}[I]=k_1[A][B]$$
$$[I](k_2[B]+k_{-1})=k_1[A][B]$$
$$[I]=\frac{k_1[A][B]}{k_2[B]+k_{-1}}$$
We can simplify this expression. We can assume that the rate constant for step 1 backward (k-1) is much slower than the rate constant for step 2 forward (k2). This means that k-1 << k2, so we assume that the step 1 backward rate constant is approximately zero (k-1 ≈ 0). When a system at equilibrium is affected by a change, it will do what it can to negate that change. As, the intermediate, I, is being formed during step 1 (forwards), it is subsequently being consumed during step 2. This means that we are "losing" our step 1 product, so it will continue in the forward direction to make more. This also means that the reaction is unlikely to "flip" to step 1 backward, so that reaction will be so slow that it is negligible.
$$[I]=\frac{k_1[A][B]}{k_2[B]+0}=\frac{k_1[A]}{k_2}$$
Lastly, we can substitute this for our rate equation for C.
$$\frac{d[C]}{dt}=k_2[I][B]=k_2(\frac{k_1[A]}{k_2})[B]$$
$$\frac{d[C]}{dt}=k_1[A][B]$$
## Steady State Approximation - Key takeaways
• The steady-state approximation is a way to simplify the derivation of the rate equation. It is based on the assumption that one intermediate in the mechanism will be produced as fast as it is consumed (i.e. it is at a steady state)
• To use the steady-state approximation, we set the change in concentration of the intermediate equal to zero and use that to find the expression for the concentration of the product.
• We can also use the approximation to calculate the concentration of an intermediate.
## Frequently Asked Questions about Steady State Approximation
The steady-state approximation (also called the quasi-steady-state approximation or pseudo-steady-state approximation) is a way to simplify the derivation of the rate equation. It is based on the assumption that one intermediate in the mechanism will be produced as fast as it is consumed (i.e. it is at a steady state
We set the change in the concentration of the intermediate equal to 0. This means you set the rate of consumption-rate of creation=0. You then solve for the concentration of the intermediate.
To set up the rate law, you write the rate law for the slow (rate determining) step. You use the steady-state approximation to solve for the concentration of the intermediate. You then substitute the expression for the intermediate into the rate law to get the final rate law.
We treat catalytic cycles the same way we treat other multistep reactions. In these reactions, we solve for the concentration of the catalyst, since it is an intermediate.
The pseudo steady state approximation is just another name for the steady-state approximation. It is called "pseudo" since we are assuming that a steady state is occurring (rather than there always being a true steady state).
## Steady State Approximation Quiz - Teste dein Wissen
Question
What is the steady state approximation?
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Answer
The steady-state approximation (also called the quasi-steady-state approximation or pseudo-steady-state approximation) is a way to simplify the derivation of the rate equation. It is based on the assumption that one intermediate in the mechanism will be produced as fast as it is consumed (i.e. it is at a steady state
Show question
Question
Why do we use the steady-state approximation?
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Answer
It makes the derivation of the rate law/concentration of the product simpler to calculate.
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Question
True or False: At a steady-state, the concentration of a species is 0
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Answer
False
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Question
How is the steady-state approximation important for enzyme kinetics?
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Answer
The steady-state approximation is used to derive the rate law equation for enzyme reactions
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Question
For an enzyme reaction, what is the intermediate?
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Answer
ES
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Question
True or False: The steady-state approximation can be used for elementary (single-step) reactions
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Answer
False
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Question
True or False: During a steady-state, the rate of consumption is equal to the rate of creation
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Answer
True
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Question
What is an intermediate?
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Answer
An intermediate is a species that isn't one of the initial reactants or final product. It is produced during the mechanism and will be completely consumed by the end.
Show question
Question
True or False: reaction velocity is the same as reaction rate
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Answer
True
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Over 10 million students from across the world are already learning smarter. | 3,629 | 14,416 | {"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": 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} | 3.71875 | 4 | CC-MAIN-2023-14 | latest | en | 0.898961 |
https://tex.stackexchange.com/questions/30817/how-to-use-siunitx-with-non-numerical-values | 1,713,460,662,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817222.1/warc/CC-MAIN-20240418160034-20240418190034-00580.warc.gz | 508,161,685 | 41,581 | # How to use siunitx with non numerical values?
I understand that something like \SI{5}{\meter\per\second} defines a certain space between the number (5) and units (m/s). What should I use if instead of a numerical quantity I want a variable: R m/s? \SI{R}{\meter\per\second} fails because it expect a number in first parameter, and R \si{\meter\per\second} or $R$ \si{\meter\per\second} show different space.
Which is the correct use?
\documentclass{standalone}
\usepackage{siunitx}
\sisetup{per-mode=symbol}
\begin{document}
\begin{tabular}{r}
\SI{5}{\meter\per\second}\\
$R$ \si{\meter\per\second}\\
R \si{\meter\per\second}\\
\end{tabular}
\end{document}
• Why would you want that? Combining variables with units makes the variables dependent on the unit system and thereby essentially meaningless, from a physics point of view. I suggest you rather use $R\cdot u$, where $u$ is \SI{1}{\meter\per\second}. Oct 7, 2011 at 16:46
• May be the word variables is not correct in this case. I want to say something like 'a car goes at A m/s and the road has B Km ...' (not so easy problems and with a better style in my language). What should be the word in English? Oct 8, 2011 at 13:06
• No no, variable is quite the right word. My point is that it's not physically meaningful to say something like 'a car goes at A m/s and the road has B km ...'. Just say 'a car goes at speed v and the road has length l': v and l are physical quantities independent of any particular choice of units. Only when you're finally giving numerical values for these quantities it makes sense to say stuff like 'let now v = 37 m/s', or 'values of l in km: {60, 70, 89, 130}'. — Anyway, that's not a typesetting concern and therefore doesn't belong to this site, I just wanted to say it. Oct 10, 2011 at 12:14
• @leftaroundabout: This is useful when using scaling variables, e.g., 25α dB, where α is the scaling variable. Aug 30, 2014 at 20:02
• @leftaroundabout: Units are just constants of a certain physical dimension, so there is no reason to forbid what the OP wants, nor is there a reason to introduce additional names for units, like you suggest in you first comment $u=$m/s. "Why would you want that?" Say you plot data in a coordinate system and label the horizontal axis with a velocity $v$, then all the ticks along the axis should carry units like 0m/s, 5m/s, 10m/s, etc. Most people would rather just label the ticks with numbers 0,5,10 etc. but then the label $v$ would be wrong. The correct label would be $v$/(m/s). Or $v$s/m Oct 25, 2017 at 14:26
As others have commented, this is partly a style question. In terms of non-numerical input, you've got a couple of choices. You could add R as a 'symbol' to those that siunitx knows, but that will not always print in math mode. So I'd prefer the approach of turning off the number parser:
\documentclass{article}
\usepackage{siunitx}
\begin{document}
\SI[parse-numbers = false]{R}{\metre\per\second}
\end{document}
As egreg points out, to get italic output you'll need to set the math-rm option
\documentclass{article}
\usepackage{siunitx}
\begin{document}
\SI[number-math-rm = \mathnormal, parse-numbers = false]{R}{\metre\per\second}
\end{document}
Turning off the parser forces math mode, but retains font control and also consistent spacing (so if you alter the setting for number-unit-separator it will be updated).
• Aren't we talking about \SI[math-rm = \ensuremath, parse-numbers = false]{R}{\metre\per\second} here instead? Apr 29, 2014 at 15:54
• The option for \SI should be math-rm, not mathrm (I C&P-ed this snippet and got an error). I attempted to fix the typo, but TeX.SX said "Edits must be at least 6 characters"... Nov 30, 2015 at 2:27
• @Merzong Indeed: fixed (I can do such edits) :-) Nov 30, 2015 at 11:34
• math-rm=\mathnormal will also cause the units to be set in italic! Instead use \SI[parse-numbers=false]{\mathnormal{R}}{\metre\per\second}. Jan 4, 2017 at 16:55
• @ElmarPeise So it would: I meant number-math-rm Jan 4, 2017 at 16:56
I think you are raising two questions: First, how to set the spacing between the "R" and its unit (m/s) and second, whether to typeset the "R" in upright or italic shape.
Regarding the former: I believe what siunitx does in the \SI macro is to insert a thinspace width between the quantity and its unit; one can generate this spacing manually with the \, command. Regarding the latter question: I believe it is (near-)universal practice to typeset variables in italic shape. Hence, I'd recommend you use the following command:
$R$\,\si{meter\per\second}
\NewDocumentCommand{\varSI}{O{}}{\SI[math-rm=\mathnormal,parse-numbers=false,#1]}
\varSI{R}{\meter\per\second}
\varSI[math-rm=\mathbf]{R}{\meter\per\second}
• I'd use \NewDocumentCommand here: it makes the new macro protected and also not 'long' (of course, \newcommand* would also do the later). Oct 7, 2011 at 12:10
• @JosephWright I forgot that siunitx loads xparse. However the command is robust anyway, since it uses \newcommand with an optional argument. Oct 7, 2011 at 12:16
• (Normally) A command taking an optional argument is not robust, as it uses \futurelet internally and so can't be \edefed. Oct 7, 2011 at 13:33
• @JosephWright It used to be so years ago; see LaTeX News of December 1995. Oct 7, 2011 at 13:39
• @egreg: Your suggestion of a NewDocumentCommand (or newcommand) which just declares some portion of a longer command it's new to me. Interesting! Thank you. Oct 10, 2011 at 7:43
As suggested above, you can declare the following:
\NewDocumentCommand\varSI{omm}{%
\IfNoValueTF{#1}%
{\SI{#2}{#3}}%
{$#1$\,$=$\,\SI{#2}{#3}}%
}
The physical quantity comes as the optional argument and is displayed (if specified) in math mode, whereas the numerical value and the unit come as mandatory arguments. Hence, use wherever in the text \varSI[Quantity]{Value}{Unit with siunitx} or \varSI{Value}{Unit with siunitx}
For example, you can use either \varSI[E_K]{3}{\electronvolt} to display EK = 3 eV and only \varSI{3}{\electronvolt} to write 3 eV depending on the context.
What you could do for instance is to introduce the \si part into the mathmode part:
\documentclass{standalone}
\usepackage{siunitx}
\sisetup{per-mode=symbol}
\begin{document}
\begin{tabular}{r}
\SI{5}{\meter\per\second}\\
$R\,\si{\meter\per\second}$\\
\end{tabular}
\end{document}
Then you have control of the spacing and you can choose a different spacing than \, if you like. | 1,865 | 6,449 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-18 | latest | en | 0.869574 |
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# Documentation Center
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## What Is a Map Projection?
Human beings have known that the shape of the Earth resembles a sphere and not a flat surface since classical times, and possibly much earlier than that. If the world were indeed flat, cartography would be much simpler because map projections would be unnecessary.
To represent a curved surface such as the Earth in two dimensions, you must geometrically transform (literally, and in the mathematical sense, "map") that surface to a plane. Such a transformation is called a map projection. The term projection derives from the geometric methods that were traditionally used to construct maps, in the fashion of optical projections made with a device called camera obscura that Renaissance artists relied on to render three-dimensional perspective views on paper and canvas.
While many map projections no longer rely on physical projections, it is useful to think of map projections in geometric terms. This is because map projection consists of constructing points on geometric objects such as cylinders, cones, and circles that correspond to homologous points on the surface of the planet being mapped according to certain rules and formulas.
The following sections describe the basic properties of map projections, the surfaces onto which projections are developed, the types of parameters associated with different classes of projections, how projected data can be mapped back to the sphere or spheroid it represents, and details about one very widely used projection system, called Universal Transverse Mercator.
Note Most map projections in the toolbox are implemented as MATLAB® functions; however, these are only used by certain calling functions (such as geoshow and axesm), and thus have no documented public API.
For more detailed information on specific projections, browse the Supported Map Projections. For further reading, Bibliography provides references to books and papers on map projection.
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## #1 2006-11-16 07:54:37
unique
Member
Registered: 2006-10-04
Posts: 419
### figure2
find value of x
i did
m angle A = 1/2 (m CEarc - m BDarc)
25 = 1/2 (100 - x)
25 = 50 - 1/2x
1/2x - 50 = 25
add 50 both sides
1/2x = 75
x = 150
which is wrong
HELP
Desi
Raat Key Rani !
Offline
## #2 2006-11-20 04:21:19
John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588
### Re: figure2
Where did you get the CE is a 100 ??
I don't see the 100 in the diagram?
igloo myrtilles fourmis
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PATNI PAPER - 1999 - JADAVPUR
Posted By : Harish Jangra Rating : +8, -1
PATNI NEW:-->>99-->>
____________
1.what is the angle between teo hands of a clock when time is 8-30
ans:75(approx)
2.a student is ranked 13th from right and 8th from left.how many are there (similer)
3.a,b,c,d,e,f are arranged in a circle b is to right of c and so on(rs aggrewal)
4.chain rule(work&time)
5.puzzle test.some data is given and he asked three qustions below.based on data we have to answer.
6.six questions on venn diagrams.
7.5 years ago sum of ages of father and son ans:40,10
8.assertion and reasoning 1.clouds 2.capital city of dweloped corentira ans for both is :a
9.man walks east & from turns to right & from to left &then 45degrees to right. in which direction he went ans:north west
10.aa-b-bb-aaa-
11.96,85,.(a series with diff 11),3,15,45,....,total 3 questions are given on series
12.a student got 70% in one subject,80% in other.to get overall 75% how much he should get in third subject.
12.a news pwper must have a.news b.advertizements 3.editor 4.date 5.paper
13.if clouds are air, air are water and so on where birds fly?
14.press,cinema,tv ans:mass media
15.3 qustions on rs aggrewal relations i.e first few chapters of aggrewal (english)
16.a man showed to a woman sitting in a pack & told to his friend.she is the daughter of my grand mother only son ans:daughter
group disscussion topic
1.if you become muncipal corporater what steps you take to develop your area.
2.if you beco me administrative officer of this university what steps you take to improve standards.
3.tell what are the seven problems in calcutta and solutions to problems?
4.abortion is legal or illegal
the following two topics are omportant they asked the same topics to many students by changing the numbers
5.in 2050 a nuclear disaster has ocurred and 50 persons are saved.which are of age above 15.of them 20 know 6 subjects and you have to choose only 3 of six subjects so that resticted resorces can used for future
subjects are 1.enggineering 2.medical 3.law 4.social sciences 5.life
sciences
6.----
6.in space ship 5men are going1.docter,drug asdditive 2.lady lawer has done
crime 3.teacher emotinally imbalanced
4.18.18 year old aeruonautical engineer 5. noble lauraut .suddenly some thing happend and oxygen is avaliable to only three people to use which of three you choose
Interview
here they asked only personal.relating to what you have written in sychometric test and some puzzles. One puzzle is 10 machines are there only one is defective iteams are coming out how tdo you find out which is defective
second: ** how do you reverse this in two transitions | 833 | 3,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-33 | latest | en | 0.925403 |
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# Regenerative Rankine Cycle
Delve deep into the intricacies of the Regenerative Rankine Cycle with this comprehensive guide. You'll gain in-depth knowledge on the meaning, principles, and components of the Regenerative Rankine Cycle. Explore real-world examples, its practical applications in energy production and industrial settings, understand the associated mathematical formulas and the efficiency gains contributed by a closed feedwater heater. A must-read resource for engineering enthusiasts and professionals alike.
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## Understanding Regenerative Rankine Cycle
The Regenerative Rankine Cycle is primarily used in power generation to recuperate heat that would have otherwise been wasted in a simple Rankine Cycle. This enhances the overall efficiency of the system.
### The Meaning of Regenerative Rankine Cycle
The concept of the Regenerative Rankine Cycle (RRC) plays an integral part in how power plants – particularly ones that rely on steam power – operate. The term 'regenerative' is used because this cycle incorporates a method to recover heat during the cycle.
It's interesting to note that the idea of reusing heat isn't novel. This brilliant thermo-dynamic principle has been used for centuries, particularly in steam-based power plants. It significantly lessens the quantity of heat that is expelled to the environment, ensuring more efficient use of fuel.
### The Principle Behind the Regenerative Rankine Cycle
Relatively more specific, the Regenerative Rankine Cycle deviates from a simple Rankine Cycle by using feedwater heaters. These heaters recover energy from the steam exiting the turbines. This process, known as 'regeneration', is the reason for the cycle's name as it regenerates heat internally, hence bettering the efficiency of the cycle and generating more power with the same amount of input.
For instance, consider boiling water on a stove. Some of the heat escapes into the surrounding environment, which is wasted energy. In a regenerative cycle, however, steps are taken to reduce this heat loss.
### The Components of the Regenerative Rankine Cycle
The primary components of the Regenerative Rankine Cycle include the boiler, turbine, condenser, and pump, along with one or multiple feedwater heaters. Let us delve a bit deeper into each element:
• Boiler: This is where heat (from burning fuel) is added to the water to swap it into steam, always under high pressure.
• Turbine: The steam from the boiler is expanded here, turning the turbine to create electrical energy.
• Condenser: The exhaust steam from the turbine is cooled and condensed back into water.
• Pump: The water from the condenser is then pumped back into the boiler to complete the cycle.
• Feedwater Heaters: They capture and utilize the remaining heat from the low-pressure turbine exhaust steam to pre-warm the water before it goes back into the boiler, enhancing the efficiency further.
A feedwater heater is a power plant component which pre-warms water that is fed back into the boiler, using heat from the turbine exhaust steam to increase the thermal efficiency of the cycle.
Here is a simple table showing the step by step progress in a Regenerative Rankine Cycle:
Step Action 1. Water is pumped into the boiler. 2. Water changes to high-pressure steam in the boiler. 3. The steam expands and turns the turbine. 4. The exhaust steam is cooled and becomes water. 5. The water is heated in feedwater heaters. 6. Heated water returns to the boiler.
## Working with Regenerative Rankine Cycle Examples
An understanding of the Regenerative Rankine Cycle's functionality can be greatly assisted by practical examples. These examples will not only illustrate the overall mechanic but proffer a deep dive into how each component functions within the cycle.
### Example One: Exploring the Regenerative Rankine Cycle
Consider a steam power plant operating exclusively on a Regenerative Rankine Cycle. Here, the steam at the exit of the turbine is bled off at various intervals for rejuvenation. This is accomplished through multiple stages of feedwater heating. As such, the heating is accomplished by steam extracted from various stages of the turbine.
Bleeding in this context refers to the removal or diversion of steam from the turbine at a certain stage during the cycle.
Briefly, this process allows the recovery of more heat, resulting in decreased fuel consumption. Also, these diversions of steam significantly increase the overall efficiency of the plant. The actual increase depends on the number of feedwater stages and the conditions at which the heat recovery occurs.
Now, let’s consider a more compound situation. For instance, let's assume the steam has a mass flow rate of 20kg/s at the inlet of the turbine. The steam properties at this point are given as 8MPa, 480°C degrees. It then enters the turbine to expand to an intermediate pressure of 2MPa where a fraction of the steam is bled. There it ultimately expands isentropically to the final pressure of 0.008MPa. Also, pump efficiency of 85% is assumed.
To compute the thermal efficiency of this cycle, the following formula should be utilized:
$η_{th} = \frac{W_{net,out}}{Q_{in}}$
### Example Two: Regenerative Rankine Cycle in Action
Continuing from the previous example, we can determine the cycle efficiency using the provided data. The total enthalpy is extracted from the boiler until the expanded point of steam. Each specific enthalpy, $$h_{2}, h_{3}, h_{4}, h_{5}$$ can be determined from specific pressure and temperature conditions in the given scenario.
Employing a Mollier diagram or steam tables, the enthalpy at every crucial stage can be computed. The net power output ($$W_{net,out}$$) and heat input ($$Q_{in}$$) can thus be garnered. $$\( W_{net,out}$$ = W_{turbine} - W_{pump} = m*(h3 - h2) - m*(h5 - h4)\), where m is the mass flow rate and h refers to the specific enthalpy at different stages. The heat input $$Q_{in}$$= m*(h3 - h1), where h1 is the initial state at the boiler entrance.
The computed values then need to be substituted back into the thermal efficiency formula $$η_{th} = \frac{W_{net,out}}{Q_{in}}$$ to get the final efficiency of the cycle.
By understanding these methods to calculate the efficiency of a Regenerative Rankine Cycle, you can observe the significant increases in efficiency derived from the use of regeneration contrasted with a standard Rankine Cycle.
Essentially, every Regenerative Rankine Cycle operation hinges on these principles, albeit some variances in the specifics could be noticed in some industrial applications. However, the fundamental idea of recovering and reutilising heat remains the same.
## Practical Applications of the Regenerative Rankine Cycle
The Regenerative Rankine Cycle has a myriad of practical applications, particularly in the realm of power and energy production. In any situation where there is a demand for an efficient conversion of heat into work, the Regenerative Rankine Cycle finds its utility. By harnessing the energy that would have otherwise been lost, this cycle increases operational efficiency and promotes resource conservation.
### Utilisation of Regenerative Rankine Cycle in Energy Production
The primary application of the Regenerative Rankine Cycle is within energy production sectors where there are extensive requirements for efficient and reliable power generation. In these situations, conservation and optimum use of resources are central tenets for sustainability and economic viability.
Power plants, especially those operating on non-renewable fuels, have adopted the Regenerative Rankine Cycle as a standard functionality. Specifically, in the coal and nuclear power plants, the heat from the fuel and reactor respectively is used to generate steam. This steam, at a high temperature and pressure, is then expanded in a turbine to generate power.
Typically, in a Simple Rankine Cycle, this steam would then be cooled and pumped back to the boiler to restart the cycle. However, this process doesn't extract the full energy potential of the steam. In the Regenerative Rankine Cycle, the steam is bled at different stages and passed through the feedwater heaters. This system thus recovers the latent energy from the steam before it condenses.
The feedwater heater is an integral hardware component of this cycle. Its primary role is to transfer heat from the residual steam to the feedwater.
It's crucial to mention that multiple feedwater heaters can be utilised, each using the bled steam from different stages of the turbine. This staged turbine exhaust heat recovery forms a critical part of the energy recuperative process.
Essentially, here is what occurs:
• When steam is bled from the turbine, its pressure and temperature drop, releasing heat.
• This heat then warms up the water that is about to be pumped back into the boiler.
• The boiler now needs less heat to produce the steam, as the water is pre-heated, consequently improving the overall cycle efficiency.
### Regenerative Rankine Cycle in Industrial Settings
In industrial settings, powerful systems are required to enhance efficiency and productivity. The Regenerative Rankine Cycle is prevalent in such arenas due to its energy-saving properties.
Furthermore, the Rankine Cycle and its derivatives are implemented in numerous ways outside the power production sector. Industrial operations routinely harness residual heat from processes and convert it to useful work using the principles of the Regenerative Rankine Cycle.
An example is the combined heat and power plants (CHP), where waste heat from electricity production is used for heating solutions. The waste heat can be transferred directly for heating purposes, or it can be used in a secondary cycle, often a Regenerative Rankine Cycle, to generate additional electricity. By coupling the production of heat and power, these systems typically achieve very high overall energy efficiency.
Moreover, the marine industry leverages the benefits of the Regenerative Rankine Cycle. Large ocean-going vessels have huge engines for propulsion, often running on heavy fuel oil or diesel. The exhaust heat from these engines is substantial, and rather than just venting it to the atmosphere, it is used in a Regenerative Rankine Cycle to generate additional power for on-board use.
Alluded to earlier is the versatility of the Regenerative Rankine Cycle. By harnessing 'waste' heat for additional power generation, the process boosts energy efficiency and promotes sustainable resource use. As its principles are scalable and adaptable to different conditions, different industries across the globe continue to utilize the Regenerative Rankine Cycle for power generation.
## Formulating the Regenerative Rankine Cycle
The Regenerative Rankine Cycle is based upon particular thermodynamic principles and is represented mathematically by a set of formulas. These formulas allow you to delve deep into the effectiveness of the cycle by quantifying the efficiency and work output.
### The Basic Formula of the Regenerative Rankine Cycle
Let's start with the fundamental formula of the Regenerative Rankine Cycle. Given that it is a thermodynamic cycle, it primarily deals with heat and work transfer. The total heat input to the cycle ($$Q_{in}$$) and work output from the turbine ($$W_{turbine}$$) are the keystones in the basic formula.
The heat input to the Regenerative Rankine Cycle is the total enthalpy extracted from the boiler and the feedwater heaters. Enthalpy, in this context, is the heat content of the steam entering the turbine and can be denoted using the equation:
$Q_{in} = m \cdot (h_{3} – h_{4})$
The variables in the above equation stand for:
• $$m$$: Mass flow rate of the steam
• $$h_{3}$$: Enthalpy of steam at the end of reheating
• $$h_{4}$$: Enthalpy of steam after the feedwater pinch point
Work output from the turbine can be represented as the difference in enthalpies across the turbine, multiplied by the steam mass flow rate. This can be translated into the formula:
$W_{turbine} = m \cdot (h_{1} – h_{3})$
The variables in this equation are:
• $$m$$: Mass flow rate of the steam
• $$h_{1}$$: Enthalpy of steam at the turbine inlet
• $$h_{3}$$: Enthalpy of steam at the end of reheating
These equations provide the basic groundwork for analysing how energy is converted into work in a Regenerative Rankine Cycle power plant.
### Understanding the Regenerative Rankine Cycle efficiency formula
Moving on to a crucial aspect of any cycle – efficiency. The efficiency ($$η$$) of a Regenerative Rankine Cycle is a measure of how effective it converts input heat into output work. It is computed as the ratio of net work output to total heat input.
Let's denote the work done by the pump ($$W_{pump}$$) and the work done by the turbine ($$W_{turbine}$$). Then, we have the following formula for net work output ($$W_{net,out}$$):
$W_{net,out} = W_{turbine} - W_{pump} = m \cdot (h_{3} - h_{2}) - m \cdot (h_{5} - h_{4})$
And this is how we denote the heat input, $$Q_{in} = m \cdot (h_{3} - h_{1})$$.
Therefore, we can define the efficiency formula, $$η_{th} = \frac{W_{net,out}}{Q_{in}}$$, which gives the ratio of output work to input heat expressed as a percentage.
These formulas and computations allow you to better understand the energy transformations occurring in the cycle. An efficient Regenerative Rankine Cycle operation results in reduced fuel consumption and higher power output, bolstering the overall performance of power plants.
To conclude, the Regenerative Rankine Cycle initiatives a new approach to heat recovery and reuse systems. This cycle allows heat involved processes to be more economical and efficient, making this ideal for energy production and demanding industrial processes.
## Studying Regenerative Rankine Cycle with Closed Feedwater Heater
When exploring the intricacies of the Regenerative Rankine Cycle, the role of the closed feedwater heater cannot be overstated. This central component not only enhances the working of the cycle but it significantly improves the thermal efficiency of the power plant as well. Its placement and function within the cycle bear direct implications on the overall energy transfer efficiency.
### The Role of Closed Feedwater Heater in Regenerative Rankine Cycle
In the Regenerative Rankine Cycle, a heat exchange process operates between steam being expanded in the turbine and the feedwater pumped towards the boiler. Here, heat flows from high-temperature steam to low-temperature water. This heat transfer is facilitated through a device called the closed feedwater heater.
A closed feedwater heater is essentially a heat exchanger in which heat is transferred from the steam tapped from the turbine to the feedwater returning to the steam generator. It operates by directly mixing the feedwater with the extracted steam. This mixing process occurs in a closed environment, hence the nomenclature.
The mechanism of operation can be summarised as follows:
• The tapped steam, also known as bleed steam or extraction steam, enters the heater at high temperature.
• The feedwater, at a lower temperature, also enters the heater, though in a different section.
• Heat exchange occurs through the separation wall between the sections, with heat flowing from the steam to the feedwater.
• The warmed up feedwater is then sent back to the boiler, thus reducing the energy required by the boiler to transform it into steam.
The main objective of the closed feedwater heater in a Regenerative Rankine Cycle is to heat the feedwater to the maximum extent possible by the turbine extraction steam. Thereupon, it significantly reduces the fuel consumption of the entire cycle and improves the thermal efficiency of the power plant.
### Enhancing Efficiency Through Regenerative Rankine Cycle with Closed Feedwater Heater
The Regenerative Rankine Cycle with a closed feedwater heater enables a substantial enhancement in the efficiency of power generation. As expressed earlier, it does so by effectively utilising the heat from the bled steam and transferring it to the feedwater.
One notable way it optimises efficiency is by reducing the demand for extra energy in heating the feedwater. The fact that the feedwater has already gained heat from the bled steam remarkably reduces the quantity of heat required by the boiler to convert the feedwater into steam.
Consequently, the cycle's overall work output enhances while the demands for energy input minimises. This leads to amplified thermal efficiency, calculated as the ratio of the work output to the heat input.
In a mathematical context, the efficiency ($$η_{th}$$) can be expressed as:
$η_{th} = \frac{W_{net,out}}{Q_{in}}$
Where:
• $$W_{net,out}$$ stands for the net work output of the cycle, and
• $$Q_{in}$$ signifies the total heat input to the cycle.
Higher the value of $$W_{net,out}$$ and lower the value of $$Q_{in}$$, greater will be the thermal efficiency. Implementing the Regenerative Rankine Cycle with a closed feedwater heater indeed increase $$W_{net,out}$$ and decrease $$Q_{in}$$, thus elevating $$η_{th}$$.
Hence, the closed feedwater heater's operation within the Regenerative Rankine Cycle plays a pivotal role in improving the efficiency of power generation systems. By effectively harnessing the potential of the bleed steam, promoting a more sustainable use of resources, the integration of closed feedwater heaters establishes a significant forward leap in energy generation and conservation practices.
## Regenerative Rankine Cycle - Key takeaways
• Regenerative Rankine Cycle comprises of four major components: boiler, turbine, condenser, and pump.
• A feedwater heater is a crucial component of the cycle that pre-warms the water using heat from the turbine exhaust steam to increase the overall efficiency.
• In a Regenerative Rankine Cycle, steam is bled off at various stages for feedwater heating, enhancing the plant's efficiency and conserving fuel.
• The thermal efficiency of Regenerative Rankine Cycle can be computed using the formula: ηth = Wnet,out/Qin, where Wnet,out is the net power output, and Qin is the heat input.
• All practical applications of the Regenerative Rankine Cycle revolve around efficient conversion of heat into work. It finds widespread use in power and energy production, industrial operations, and the marine industry.
• The formula basis of a Regenerative Rankine Cycle are based on heat and work transfer computations and efficiency analysis.
• Besides improving the thermal efficiency of power plants, the closed feedwater heater plays a crucial role in heat exchange processes in a Regenerative Rankine Cycle. It facilitates heat transfer from the high-temperature steam to the low-temperature water.
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What is the Regenerative Rankine Cycle? Please write in UK English.
The Regenerative Rankine Cycle is an enhancement of the standard Rankine cycle in engineering, which aims to improve efficiency by recycling exhaust steam from the turbine back into the boiler. It increases thermal efficiency by reducing heat loss in the condenser.
What is an example of a Regenerative Rankine Cycle? Please write in UK English.
A common example of a regenerative Rankine cycle is the operation of power-producing thermal plants like nuclear and coal power plants. These plants utilise the Rankine cycle to convert heat into mechanical energy, leading to electricity generation.
Why should we use the Regenerative Rankine Cycle?
The Regenerative Rankine Cycle is used because it significantly improves the efficiency of steam power plants. It does this by capturing and reusing waste heat, which would otherwise be lost, reducing fuel consumption and thereby lowering the environmental impact.
How does the Regenerative Rankine Cycle work?
The Regenerative Rankine Cycle works by capturing and reusing heat that would otherwise be wasted in traditional Rankine cycles. The high-pressure feedwater heater heats the water before entering the boiler, boosting efficiency. The cycle involves four main processes: pumping, heating/evaporation, expansion in the steam turbine, and condensation.
What is the difference between the Reheat and the Regenerative Rankine Cycle in UK English?
Reheat Rankine Cycle involves reheating the working fluid after partial expansion to improve efficiency and avoid excessive moisture. In contrast, a Regenerative Rankine Cycle increases efficiency by using heat exchanges, typically in the form of feedwater heaters, to preheat the liquid phase of the working fluid before it enters the boiler.
## Test your knowledge with multiple choice flashcards
How is the Regenerative Rankine Cycle employed in the marine industry?
What is the primary use of the Regenerative Rankine Cycle?
What are the keystones of the basic formula of the Regenerative Rankine Cycle?
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• Checked by StudySmarter Editorial Team | 4,548 | 22,192 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.901445 |
https://scicomp.stackexchange.com/questions/32539/poisson-image-blending-artifacts | 1,726,190,131,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651506.7/warc/CC-MAIN-20240913002450-20240913032450-00531.warc.gz | 473,489,682 | 41,640 | # Poisson image blending artifacts
I am trying to implement Poisson image blending as in the paper Poisson Image Editing. This is the task of filling in a masked region of an image by minimizing $$\min_f\int_\Omega \left | \nabla f - \mathbf v\right |^2$$ with $$f|_{\partial\Omega} = f^*_{\partial\Omega}$$ which is equivalent to the Poisson problem $$\Delta f = \mathbf \nabla\cdot \mathbf v$$ I use the same discretization of the problem as presented there, which is equivalent to using the 4-neighbour discrete Laplacian kernel $$\begin{bmatrix} 0 & 1 & 0\\ 1 & -4 & 1\\ 0 & 1 & 0 \end{bmatrix}$$ and solving the linear system $$Lx = b$$ where $$x$$ is the vector of unknown image values, $$L$$ is the sparse matrix that applies the above kernel to $$x$$, approximating the Laplacian, and $$b$$ is the sum of boundary terms and gradient guides. Specifically, this is the system of equations $$\sum_{q\in N_p\cap\Omega}f_q-\left |N_p\right | f_p=-\sum_{q\in N_p\cap\partial\Omega}f^*_q-\sum_{q\in N_p}v_{pq}$$ where $$N_p$$ is the set of 4 neighbors of point $$p$$ (this is the same as the equation 7 in the paper, albeit with opposite signs).
My implementation in Python (scipy) is below. I use a sparse factorization of L to solve the system for each color channel in the image.
import numpy as np
import matplotlib.pyplot as plt
import scipy.sparse as sp
import scipy.sparse.linalg as linalg
import argparse
def poisson_problem(image, mask, guide=None, threshold = 0.5):
indices = np.full(mask.shape, -1) #map of coordinates inside the domain to unique indices
invdices = [] #list of coordinates inside the domain
ind = 0
if m > threshold:
indices[p] = ind
ind += 1
invdices.append(p)
N = len(invdices)
b = np.zeros([N, image.shape[2]]) #for RHS of equation
#build sparse matrix
data = []
I = []
J = []
for i,p in enumerate(invdices):
data.append(-4)
I.append(i)
J.append(i)
for dim in (0, 1):
for dir in (-1,1):
q = [*p]
q[dim] += dir
if q[dim] < 0 or q[dim] >= mask.shape[dim]:
continue
j = indices[(*q,)]
if j > -1:
#contribution from inside the domain
data.append(1)
I.append(i)
J.append(j)
else:
b[i,:] = -image[(*q,)] #boundary term (outside domain)
if guide is not None:
b[i,:] -= image[p] - image[(*q,)] #vector guide term
L = sp.csc_matrix((data, (I,J)), shape=(N,N))
return L, b, invdices
if __name__ == '__main__':
parser = argparse.ArgumentParser()
args = parser.parse_args()
img = np.zeros([256, 256, 3])
img[:128,:,0] = 1
img[129:,:,1] = 1
if args.circle:
inds = np.indices(img.shape, dtype=np.float)[0:2,:,:,0] - np.array([[[img.shape[0]//2]], [[img.shape[1]//2]]])
else:
#solve poisson problem for each color channel (only BCs change)
L, b, I = poisson_problem(img, mask, img if args.guide else None)
factor = linalg.factorized(L)
xs = np.stack([factor(b[:,i]) for i in range(b.shape[1])], 1)
#composite and display results
img2 = np.zeros(img.shape)
for i, p in enumerate(I):
img2[p] = xs[i]
plt.imshow(img3)
plt.show()
To make sure it works, I'm pasting the original image into the masked region, hoping to get the original image back. When the mask is square and the gradient guides are provided (--guide), it seems to work without noticeable errors: But when the mask is a circle shape (run with the --circle and --guide arguments), this is the result: This does not even appear to be preserving the boundary condition except where the mask boundary is tangent to the axes. This tendency towards 0 seems to occur even when solving the simple Laplace equation, with no guide:
I suspect that this is some kind of error resulting from the discretization, but as far as I can tell, this is the same method as in the paper which claims to work on arbitrary domain shapes. What could be causing this?
It seems I forgot to accumulate the boundary terms for each relevant entry in the right hand side vector $$b$$. So
b[i,:] = -image[(*q,)] #boundary term (outside domain)
b[i,:] -= image[(*q,)] | 1,121 | 3,928 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-38 | latest | en | 0.761476 |
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9
The bias comes from the paper Stambaugh (1999) and has nothing to do with small sample bias. It has to do with point (1) below. The argument goes as follows: Typical lagged explanatory variables for stock-return regressions are correlated with contemporaneous stock returns This contemporaneous correlation biases forecasting regressions First review OLS ...
8
Let me start with a simple example. Suppose you have a dividend strip that pays an unknown dividend $D_T$. The gross return (something like 1.05 and NOT 5%!) on this security is, by definition, $$R_{t\to T} = \frac{D_T}{P_t}$$ where $P_t$ is the current price of this security. If we use lowercase letters to denote logs (i.e., $\log D_T = d_T$ etc..) we can ...
7
No I believe there is no directional predictive value derived from looking at divergences between futures and their underlying price value. The reason for divergences are of the no-arbitrage argument type. Futures could be arbitraged (and are immediately if such arbitrage opportunities surface, even those opportunities may only fill the stomach of a single ...
7
Without seeing your trading desk's P&L it's impossible to say whether it is predictable or not. But here are a few thoughts - There's no reason to think that it isn't predictable. In general, financial time series are hardest to predict when the represent the return stream of an investible asset. A trading desk's P&L isn't really investible, so ...
6
My favorite tool is Sornette's own Finanical Crisis Observatory: http://tasmania.ethz.ch/pubfco/fco.html If you are interested, I have developed my own tool in Java and JavaCL which can be found here: https://thebubbleindex.codeplex.com/ Update: Code moved to github: https://github.com/thebubbleindex/thebubbleindex
5
The mean could be the long run variance which is sig2 = fit.Constant/(1-fit.GARCH{1}-fit.ARCH{1}); I hope this explains. If not, note I ran this model through Matlab, I get different values. you can paste your m1 and m2 values and some other intermediate results so I can see why Matlab differs. EDIT: The question refers to forecasting the returns. ...
5
Sorry, but despite being used as a popular example in machine learning, no one has ever achieved a stock market prediction. It does not work for several reasons (check random walk by Fama and quite a bit of others, rational decision making fallacy, wrong assumptions ...), but the most compelling one is that if it would work, someone would be able to become ...
5
The graph you attached suggests that you were trying to find swings between major highs and lows. This can be done by simply finding local extrema in the price series. The concept is: find local extrema: minima in Low prices, maxima in High prices; find local extrema in the results, if swings are too short; repeat #2 until satisfied with the results. This ...
5
I have written an entire paper on this approach at https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2828744 As to your specifics 1) "Volatility" as defined by variance does not exist, which is why it is changing. The first moment is undefined so the second cannot exist. See the paper as to why. Your fitted pdf will treat the outcomes as having a ...
5
There is large literature on MIDAS (mixed-frequency data sampling) models, the leading scholars being Eric Ghysels and Rossen Valkanov — google their research for references. However, the motivation for these models has mostly been to forecast low-frequency stuff with high-frequency variables, updating, say, quarterly GDP predictions as weekly ...
4
From what I have read, there are 3 popular algorithms for financial time series. Random Forests and SVMs, then followed by Neural Network Architectures. There are a couple of good papers, to name a few: Empirical Asset Pricing via Machine Learning Deep neural networks, gradient-boosted trees, random forests: Statistical arbitrage on the S&P 500 An ...
4
Recent research A recent article by Frank Zhao is interesting to get started: Natural Language Processing - Part I: Primer. You will find more papers on this repo (too long to copy all here): nlp_papers Applications If you are looking for possible applications of current SoTA research to financial markets, here is a quick list: Equity Predict the ...
4
I think this one has a clear answer (I am solely talking about equities here): The change magnitude is much more predictable than the direction. The reason being that equity volatility is much more predictable than equity risk premiums. Volatility is nothing else but change magnitude and due to the stylized facts of volatility clustering together with mean ...
4
The correlation matrix is a very important part of modeling stock returns. It is often better to build a model that takes in multiple assets features so that it can use this correlation to its advantage. A good example of this is a VAR model from econometric. A great example in the machine learning context is the paper titled Empirical Asset Pricing via ...
4
Another way of staying "time-varying risk-premium", is saying that the risk-premium is predictable. However, that the fact that the risk-premium is predictable does not means that you can make money out of this. The best two references to understand this are: Cochrane (2008) - The dog that did not bark Goyal and Welch (2007) The first tells you what ...
4
The point of confusion may be in thinking that a predictable price process is synonymous with a mean-reverting process while using the definitions in these papers, it's actually the opposite! In the context of these papers, a random walk would be 100% predictable: the unpredictable component of a random walk (i.e. the period specific shock which has finite ...
4
There are a few exclusions that I have commonly seen: Excluding thinly traded stocks. The price that shows up in your data feed may not relate to actual tradable prices. Filtering for ADR/Pink locals. You can find stocks listed in multiple places in ways that would lead you to think that they are great for pairs trades when actually they are the same ...
3
The TA_lib Technical Analysis library here has open source code for numerous indicators.
3
The two components you refer to in your questions are: Market direction (the sign of the return) Change magnitude (the absolute value of the return) First, I'm sure you realize that neither of these are predictable at a 100%, otherwise there would be no way to make profit (you make profit by seeing things other didn't). To answer the question, I would say ...
3
The renowned CXO Advisory Group has a section "What Works Best?". Here some general information is given and many links to their research articles which e.g. summarize lots of current academic research (although most of the linked articles are behind a paywall the links to the original papers are normally provided). The article closes with "In summary, ...
3
I am not sure I perfectly understand your question, the concept of "time series with varying density over time" is not very clear. One thing is for sure, the optimal way to "feed" a neural network is a function of the type of NNet itself and of the learning method you have chosen. For time series either you believe your data are iid vectors, and you can ...
3
The accuracy of a model is only 1 factor in determining usefulness. Aside from the accuracy, it would help to determine how you would implement it in a simulated trading environment and look into the performance further. Aside from a hit ratio or accuracy, you should compute other metrics such as: The risk-reward ratio that your model realizes (not the ...
2
I am not sure any of the other answers mentioned this but the main reason you should not use an option model to buy/sell the underlying (BS or other) is that the option models are more about market-making in options and hedging using the underlying rather than forecasting the underlying. The layman way to understand this is that: using an option model, you ...
2
You can try this course on Udactiy https://www.udacity.com/course/machine-learning-for-trading--ud501
2
Blair Hull as an idea: http://papers.ssrn.com/sol3/papers.cfm?abstract_id=2609814 He says he sold his automated trading firm to Goldman for 300 million $. 2 A prediction model that is correct$50\%\$ of the time can be profitable if the model gains more when it is right than it loses when it is wrong. You could simplify it like this: A trading strategy is profitable if your trades have positive expected value. Now suppose that your gains when your model is right equals the losses when your model is wrong. If ...
2
An interesting variant from Reuters (you can do your own "simulations"): https://www.breakingviews.com/considered-view/numbers-add-up-to-germany-retaining-world-cup/ Another paper from a renowned source: Probabilistic forecasts for the 2018 FIFA World Cup based on the bookmaker consensus model Achim Zeileis (achim.zeileis@r-project.org), Christoph Leitner ...
2
The Technical Analysis of Financial markets is considered as a milestone of the matter. I suggest to read that before starting to test your strategy. It explains well the use of each indicator, providing the economic reason behind that and when it is useful to use that; moreover, the book deals the stock market with mainly, as you need for. In my humble ...
2
A very good reference can be found here: http://www.asiapacfinance.com/trading-strategies/technicalindicators
2
I cannot seem to find that article for free, so here is a more generalized answer. 1.what are the hidden states and what are the observation states. The hidden states are said to be that of an unobserved parameter process following the Markov property. The observation states are generated by the hidden parameter process. The parameter process changes or ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2,157 | 9,949 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-17 | longest | en | 0.875541 |
https://www.onlinemathlearning.com/rectangular-prism.html | 1,585,454,177,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493684.2/warc/CC-MAIN-20200329015008-20200329045008-00499.warc.gz | 1,107,305,415 | 10,801 | # Volume of Rectangular Prisms
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More Lessons for Grade 3
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Videos, stories and songs to teach Grade 3 students how to find the volume of rectangular prisms.
What is volume?
Volume is the amount of space an object takes up.
How to calculate the volume of a rectangular prism?
Volume = Length × Width × Height
V = L × W × H
Volume of Rectangular Prism
How to find the volume of a rectangular prism
Volume Song - Length X Width X Height
The Volume Song and movie teaches students how to find the volume of a solid figure. Learn how to solve and find the volume of a 3D rectangular prism shape using length times width times height.
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https://mathsgee.com/202/write-2-39-times-10-3-as-a-decimal-number | 1,618,208,562,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00510.warc.gz | 489,073,206 | 16,333 | Institutions: Global |ALU | Ekurhuleni Libraries | Joburg Libraries | Tshwane Libraries | TUT | UCT | UJ | UNISA | UP | UZ | Wits | Invest in Zim
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Write $2.39 \times 10^{-3}$ as a decimal number
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Divide by a 1000= 0.00239
by Diamond (42,256 points)
0 like 0 dislike | 147 | 419 | {"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} | 2.53125 | 3 | CC-MAIN-2021-17 | latest | en | 0.796585 |
https://www.iitianacademy.com/ibdp-maths-aa-topic-sl-1-4-financial-applications-of-geometric-sequences-and-series-ib-style-questions-hl-paper-2/ | 1,657,182,197,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683708.93/warc/CC-MAIN-20220707063442-20220707093442-00017.warc.gz | 844,229,278 | 52,886 | # IBDP Maths AA: Topic : SL 1.4: Financial applications of geometric sequences and series: IB style Questions HL Paper 2
### Question
Give your answers in this question correct to the nearest whole number.
Shahid invested 25 000 Singapore dollars (SGD) in a fixed deposit account with a nominal
annual interest rate of 3.6 %, compounded monthly.
1. Calculate the value of Shahid’s investment after 5 years. [3]
At the end of the 5 years, Shahid withdrew x SGD from the fixed deposit account and
reinvested this into a super-savings account with a nominal annual interest rate of
5.7 %, compounded half-yearly.
The value of the super-savings account increased to 20 000 SGD after 18 months.
2. Find the value of x . [3]
Ans:
(a)
$$FV=25000\times (1+ \frac{3.5}{100\times 12})^{12\times 5}$$
OR
N= 5
I%=3.6
PV= $$\mp 25000$$
$$P/Y$$ = 1
C/Y = 12
OR
N= 60
I%=3.6
PV = $$\mp 25000$$
P/V = 12
C/Y = 12
FV=29922(SGD)
(b)
2000= PV\times $$(1+\frac{5.7}{100\times 2})^{2\times 1.5}$$
OR
N= 1.5
I%=5.7
FV=$$\pm 20000$$
$$P/Y$$ = 1
C/Y= 2
OR
N=3
I%= 5.7
FV = $$\pm 20000$$
P/V=2
C/Y = 2
X= 18383 (SGD) | 417 | 1,134 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2022-27 | latest | en | 0.795843 |
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# hw2 - GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL OF ELECTRICAL...
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ECE 3080: Homework #2 Due Date: June 23, 2005 Page 1 of 2 G EORGIA I NSTITUTE OF T ECHNOLOGY S CHOOL OF E LECTRICAL AND C OMPUTER E NGINEERING ECE 3080: Semiconductor Devices for Computer Engineering and Telecommunication Systems Summer Semester 2005, Homework #2 Homework Due Date: Thursday, June 23, 2005 1. Carrier Transport: Continuity Equation (40 points) Assume a 1-dimensional silicon crystal, which has been doped with 10 17 cm -3 phosphorous atoms. Using a high-energy light source, 10 14 cm -3 additional electron-hole pairs are generated at x = 0 in the stationary state (dp/dt = dn/dt = 0). (a) Can we consider low-level injection or do we have to assume high-level injection? Justify your answer. (b) Calculate analytical expressions for the minority carrier distribution p n (x) in the stationary state for x > 0 for the following two cases: (i) Infinitely long semiconductor with p n (x= ) = p n0 (see sketch (i)) (ii) Semiconductor with length W and p n (x=W) = p n0 (see sketch (ii))
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• Spring '08
• Staff
• P-n junction, GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL OF ELECTRICAL, minority carrier distribution, reverse-bias junction capacity
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hw2 - GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL OF ELECTRICAL...
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Ask a homework question - tutors are online | 463 | 1,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-09 | latest | en | 0.812466 |
https://conwaylife.com/forums/viewtopic.php?f=3&p=101481 | 1,621,211,860,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991921.61/warc/CC-MAIN-20210516232554-20210517022554-00311.warc.gz | 211,362,580 | 18,629 | ## Pattern viewer for forum threads
For discussion directly related to ConwayLife.com, such as requesting changes to how the forums or wiki function.
rowett
Moderator
Posts: 2354
Joined: January 31st, 2013, 2:34 am
Location: UK
Contact:
### Re: What??? An oscillator with no rotor cells???
ColorfulGalaxy wrote:
July 22nd, 2020, 9:53 pm
Click "Show in viewer", then the gear at the corner, then "Pattern", then "Identify".
Next, mouse over the "0 | 44 | 44".
Fixed, thanks for reporting!
rowett
Moderator
Posts: 2354
Joined: January 31st, 2013, 2:34 am
Location: UK
Contact:
### Re: Pattern viewer for forum threads
muzik wrote:
July 18th, 2020, 3:52 pm
Also a suggestion: could appending H or T to the resulting string make it appear as a hex or triangular grid for a more clear presentation of neighbourhoods intended for these rules?
You can now append L for triangular grid with custom neighbourhoods. Note the neighbourhood specified needs to be for the triangle pointing down. It will be automatically flipped based on the checkerboard.
lemon41625
Posts: 313
Joined: January 24th, 2020, 7:39 am
Location: 小红点 (if you know where that is)
### Re: Pattern viewer for forum threads
Can arbitary weighted neighbourhoods be supported?
Perhaps we could do something like CoordCA but in decimal and convert to base64. Then put it with N\$ as suggested by bubblegum.
But how would you do negative weights and weights >9?
Also for named weighted neighbourhoods could the gaussian weighted neighbourhood be supported via something like NG?
Code: Select all
``````1,2,3,2,1
2,4,6,4,2
3,6,9,6,3
2,4,6,4,2
1,2,3,2,1``````
Code: Select all
``````1,2,3,4,3,2,1
2,4,6,8,6,4,2
3,6,9,12,9,6,3
4,8,12,16,12,8,4
3,6,9,12,9,6,3
2,4,6,8,6,4,2
1,2,3,4,3,2,1``````
and so on.
As for other INT neighbourhoods, they will be identified like this "B2n3/S23-q/FC" or "B2n3/S23-q/NFC" or "B2n3/S23-q/XFC".
The other possible INT neighbourhoods to be added would be
Far Corners same notation as R1 Moore
Code: Select all
``````1,0,0,0,1
0,0,1,0,0
0,1,0,1,0
0,0,1,0,0
1,0,0,0,1``````
Far Edges same notation as R1 Moore
Code: Select all
``````0,0,0,1,0,0,0
0,0,0,0,0,0,0
0,0,1,0,1,0,0
1,0,0,0,0,0,1
0,0,1,0,1,0,0
0,0,0,0,0,0,0
0,0,0,1,0,0,0``````
R2 Cross INT
Code: Select all
``````x = 53, y = 95, rule = B/S012345678History
.3C3.2C4.3C3.3C3.C.C3.3C3.3C3.3C3.3C\$.C.C4.C6.C5.C3.C.C3.C5.C7.C3.C.C
\$.C.C4.C4.3C3.3C3.3C3.3C3.3C5.C3.3C\$.C.C4.C4.C7.C5.C5.C3.C.C5.C3.C.C\$
.3C3.3C3.3C3.3C5.C3.3C3.3C5.C3.3C2\$2.D5.D5.D5.D5.D5.C5.C5.C5.C\$2.D5.C
5.C5.C5.C5.D5.D5.D5.C\$2D.2D.2D.2D.2D.CD.2D.CD.DC.CD.2C.DC.2C.DC.2C.2C
.2C.2C\$2.D5.D5.D5.C5.C5.D5.C5.C5.C\$2.D5.D5.D5.D5.D5.C5.C5.C5.C2\$8.C5.
D5.C5.D5.D5.C5.D\$8.D5.C5.C5.C5.D5.D5.C\$6.2D.2D.2D.2D.2D.CD.DC.CD.2C.D
C.2C.2C.2C.2C\$8.D5.C5.D5.D5.C5.D5.C\$8.D5.D5.D5.C5.C5.C5.C2\$14.C5.D5.D
5.C5.D\$14.C5.C5.C5.D5.D\$12.2D.2D.2D.CD.2C.CD.2C.DC.2C.2C\$14.D5.D5.D5.
C5.C\$14.D5.C5.D5.D5.C2\$14.D5.D5.C5.C5.C\$14.C5.C5.C5.D5.D\$12.2D.DC.CD.
2D.DC.CD.DC.2C.2C.CD\$14.D5.C5.D5.D5.C\$14.D5.D5.D5.C5.C2\$14.D5.C5.D5.D
5.C\$14.C5.C5.C5.D5.D\$12.2D.2D.2D.2D.CD.CD.2C.2C.2C.2C\$14.D5.C5.D5.D5.
C\$14.C5.D5.C5.C5.D2\$14.C5.D5.D5.C5.D\$14.D5.C5.C5.D5.C\$12.2D.DC.CD.2D.
CD.2C.DC.2C.2C.CD\$14.D5.D5.D5.C5.C\$14.D5.C5.D5.D5.C2\$14.C5.D5.C5.C5.D
\$14.D5.C5.C5.D5.C\$12.2D.2D.CD.DC.CD.CD.DC.CD.2C.2C\$14.D5.D5.D5.C5.C\$14.
C5.D5.D5.C5.D2\$20.C5.C5.D\$20.C5.C5.D\$18.CD.2D.2D.2C.DC.2C\$20.D5.D5.C\$
20.D5.D5.C2\$20.C5.D5.D\$20.D5.C5.C\$18.CD.DC.CD.DC.DC.CD\$20.D5.C5.C\$20.
D5.D5.C2\$26.C\$26.C\$24.2D.2D\$26.C\$26.C2\$26.C\$26.C\$24.CD.2D\$26.C\$26.D2\$
26.C\$26.C\$24.CD.DC\$26.D\$26.D2\$26.D\$26.C\$24.CD.DC\$26.D\$26.C2\$26.C\$26.C
\$24.CD.2D\$26.D\$26.C2\$26.C\$26.D\$24.CD.DC\$26.D\$26.C!
``````
Use letters acehijnopqrtuvw
R2 Knight INT
Code: Select all
``````x = 47, y = 103, rule = LifeHistory
.2C5.3C4.3C4.C.C4.3C4.3C4.3C\$2.C7.C6.C4.C.C4.C6.C8.C\$2.C5.3C4.3C4.3C4.3C4.3C6.C\$2.C5.C8.C6.C6.C4.C.C6.C\$.3C4.3C4.3C6.C4.3C4.3C3\$.D.D4.C.D4.C.C4.C.C4.D.D4.D.C4.C.C\$C3.D2.C3.D2.C3.D2.C3.C2.D3.C2.D3.C2.D3.C2\$D3.D2.D3.D2.D3.D2.D3.D2.C3.C2.C3.C2.C3.C\$.D.D4.D.D4.D.D4.D.D4.C.C4.C.C4.C.C3\$8.D.C4.C.D4.C.C4.D.C4.C.D\$7.C3.D2.C3.C2.C3.D2.D3.D2.D3.C2\$7.D3.D2.D3.D2.D3.C2.C3.C2.C3.C\$8.D.D4.D.D4.D.D4.C.C4.C.C3\$8.D.D4.C.D4.C.C4.D.C4.C.C\$7.C3.C2.C3.D2.C3.D2.D3.C2.D3.D2\$7.D3.D2.D3.C2.D3.D2.C3.D2.C3.C\$8.D.D4.D.D4.D.C4.C.C4.C.C3\$8.D.D4.D.C4.C.C4.C.D4.C.C\$7.C3.D2.C3.D2.C3.D2.D3.C2.D3.C2\$7.D3.C2.D3.C2.D3.D2.C3.D2.C3.D\$8.D.D4.D.D4.C.D4.C.C4.C.C3\$8.D.D4.D.C4.C.C4.C.D4.C.C\$7.C3.D2.C3.D2.C3.D2.D3.C2.D3.C2\$7.D3.D2.D3.D2.C3.D2.C3.C2.C3.C\$8.D.C4.D.C4.D.D4.C.D4.C.D3\$8.D.D4.D.C4.C.D4.C.D4.C.C\$7.C3.D2.C3.D2.C3.C2.D3.C2.D3.C2\$7.C3.D2.C3.D2.D3.C2.D3.C2.D3.C\$8.D.D4.D.D4.D.D4.C.C4.C.C3\$15.D.D4.C.D4.C.C\$14.C3.C2.C3.C2.D3.D2\$14.C3.D2.D3.D2.D3.C\$15.D.D4.D.C4.C.C3\$22.C.D\$21.C3.C2\$21.D3.D\$22.C.D3\$22.C.D\$21.C3.D2\$21.D3.C\$22.D.C3\$22.D.C\$21.C3.D2\$21.D3.C\$22.C.D3\$22.D.C\$21.C3.D2\$21.C3.C\$22.D.D3\$22.D.C\$21.C3.D2\$21.C3.D\$22.D.C3\$22.D.C\$21.C3.C2\$21.C3.D\$22.D.D3\$22.D.D\$21.C3.C2\$21.C3.C\$22.D.D!``````
Letters used would be also be acehijnopqrtuvw. Some will need to be removed.
Supports:
BSFKL, Extended Generations, Regenerating Generations, Naive Rules, R1 Moore, R2 Cross and R2 Von Neumann INT
And some others...
Hunting
Posts: 4382
Joined: September 11th, 2017, 2:54 am
### Re: Pattern viewer for forum threads
lemon41625 wrote:
July 23rd, 2020, 8:30 pm
Letters used would be also be acehijnopqrtuvw. Some will need to be removed.
I disagree with the removal of k, y, and z. Hensel notation have them - I don't see a reason to remove them.
BTW, I like how LeapLife is being used as an example of an INT rule (here), interesting stable rule, rule of an example textcensus page, default rule of a search tool, and much more.
Farewell.
AlephAlpha, privately wrote:What is textcensus? What is attribute?
HB, Gustavo! I hope you are more mature now.
Code: Select all
``````x = 4, y = 5, rule = B2n3/S23-q
2bo\$2bo\$bobo2\$3o!``````
LeapLife/url] - [url=https://conwaylife.com/forums/viewtopic.php?f=11&t=4367]DirtyLife - LispLife
muzik
Posts: 4183
Joined: January 28th, 2016, 2:47 pm
Location: Scotland
### Re: Pattern viewer for forum threads
rowett wrote:
July 23rd, 2020, 4:33 pm
You can now append L for triangular grid with custom neighbourhoods. Note the neighbourhood specified needs to be for the triangle pointing down. It will be automatically flipped based on the checkerboard.
Does this mean that HROT could now support triangular as one of its possible neighbourhood types (perhaps only up to range 250 rather than 500, due to how it's handled)? I'd like to see that added as a possible choice to round off the new neighbourhood types for HROT that have been added over the past months.
Code: Select all
``````x = 1, y = 1, rule = B123456789XYZ/S0123456789XYZLHistory
E!
[[ AUTOSTART AUTOFIT GRID GPS 1 THEME Blues ]]``````
----
Also on the topic of triangular rules, since we can simulate PCA on a square grid could similar such support be added for triangular PCA - for example the ReversibleLife rules which can be found on the Rule Table Repository
rowett
Moderator
Posts: 2354
Joined: January 31st, 2013, 2:34 am
Location: UK
Contact:
### Re: Pattern viewer for forum threads
muzik wrote:
July 24th, 2020, 9:06 pm
Does this mean that HROT could now support triangular as one of its possible neighbourhood types
Yes, that's the plan.
muzik wrote:
July 24th, 2020, 9:06 pm
Also on the topic of triangular rules, since we can simulate PCA on a square grid could similar such support be added for triangular PCA - for example the ReversibleLife rules which can be found on the Rule Table Repository
Possibly at some stage. I'll take a look.
muzik
Posts: 4183
Joined: January 28th, 2016, 2:47 pm
Location: Scotland
### Re: Pattern viewer for forum threads
On the topic of PCA, another thing that's been bothering me for a good half a year now: a rule called PCA_13 was posted back in january, but I have no idea how to actually figure out its PCA rulestring and have it added to the list of aliases. To further complicate things this rulestring has "border" cells, which no other PCA rule has, so I'm not sure if this new cell type should ship with LifeViewer's default PCA implementation for compatibility purposes.
muzik
Posts: 4183
Joined: January 28th, 2016, 2:47 pm
Location: Scotland
### Re: Pattern viewer for forum threads
When run with identify shouldn't this display a mod a quarter of its period, or is it actually different at each turn?
Code: Select all
``````x = 6, y = 7, rule = PCA_4
5.H3\$H.L.D3\$5.B!``````
Also the PCA aliases aren't sorted alphabetically
muzik
Posts: 4183
Joined: January 28th, 2016, 2:47 pm
Location: Scotland
### Re: Pattern viewer for forum threads
A few (and hopefully final for now) ideas for neighbourhoods: (the range values might be a bit weird though, so implement how you personally see fit)
Aligned checkerboard (Nb):
Code: Select all
``````x = 17, y = 7, rule = bshistory
10.A.A.A.A\$4.A.A.A2.A.A.A\$A.A2.A.A2.A.A.A.A\$.E2.A.E.A2.A.E.A\$A.A2.A.A
2.A.A.A.A\$4.A.A.A2.A.A.A\$10.A.A.A.A!
[[ VIEWONLY GRID THEME Blues ]]``````
Triangle neighbourhood on a hex grid (NT):
Code: Select all
``````x = 23, y = 10, rule = bshhistory
13.A\$13.2A\$5.A7.3A\$5.2A6.4A\$A4.3A5.5A\$2A3.4A4.6A\$AEA2.2AE2A3.3AE3A\$4A
.6A2.8A\$5.7A.9A\$13.10A!
[[ VIEWONLY GRID THEME Blues ]]``````
Hexagram neighbourhood (N6):
Code: Select all
``````x = 24, y = 13, rule = bshhistory
14.A\$14.2A\$6.A7.3A\$6.2A3.10A\$.A2.7A.9A\$4A.6A2.8A\$.AEA2.2AE2A3.3AE3A\$.
4A.6A2.8A\$3.A2.7A.9A\$9.2A3.10A\$10.A7.3A\$19.2A\$20.A!
[[ VIEWONLY GRID THEME Blues ]]``````
Triangle neighbourhood on a triangular grid (ND):
Odd range values(?) (here R1, R3, R5?) would be as follows:
Code: Select all
``````x = 31, y = 8, rule = bslhistory
16.15A\$5.9A3.13A\$AEA3.3AE3A5.5AE5A\$.A5.5A7.9A\$8.3A9.7A\$9.A11.5A\$22.3A
\$23.A!
[[ VIEWONLY GRID THEME Blues ]]``````
Even range values(?) (here R2, R4, R6?) would use the following:
Code: Select all
``````x = 44, y = 11, rule = bslhistory
\$25.19A\$10.13A3.17A\$.7A3.11A5.15A\$2.2AE2A5.4AE4A7.6AE6A\$3.3A7.7A9.11A
\$4.A9.5A11.9A\$15.3A13.7A\$16.A15.5A\$33.3A\$34.A!
[[ VIEWONLY GRID THEME Blues ]]``````
bubblegum
Posts: 896
Joined: August 25th, 2019, 11:59 pm
### Re: Pattern viewer for forum threads
I've gotten around to adding aliases for 2-state rules without them now.
Here's the raw data from my Notepad instance:
Code: Select all
``````Bigship - B2ei3eiqry4ny5ajkr/S23ajkr4jnry5inq
WolfVN-W50_B14_S34V - B2an3cjr4acij5nqy6ak/S2cn3cqy4cny5e
B2ex3-lS23 - B2en3-a/S23
B45_S034_N21 - B2ei3-ce4cny/S02-cn3cinqy4c
HexLife (Saka) - B3o/S234-o6
pentagonhood - MAPEWYRZmaIZogRZhFmZohmiGaIZoiIAIgAZohmiIgAiAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAB
B578_S4567_N31 - B3-ce4aikqtwz5e/S2-cn3-ce4cny5e
Notes:
B45_S034_N21 was anisotropic as a .rule file, but intended to be isotropic.
The original name of HexLife (Saka) was just HexLife.
B578_S4567_N31 simply did not work as intended. The inferred rule preserves everything described except the r-pentomino, which should just be an error in the original. I don't know.
``````
Each day is a hidden opportunity, a frozen waterfall that's waiting to be realised, and one that I'll probably be ignoring
sonata wrote:
July 2nd, 2020, 8:33 pm
conwaylife signatures are amazing[citation needed]
anything
bubblegum
Posts: 896
Joined: August 25th, 2019, 11:59 pm
### Re: Pattern viewer for forum threads
Thank you for using LVEnhanceBot! Here are LifeViewer's newest bugs and suggestions:
S1) Could Ctrl+Space be made to work like Shift+Space so there'd be no silly jump-ahead and translate upwards when using Ctrl+Space?
B1) This is such an absurd bug: Go to this link and open the viewer there. Gain control, then Ctrl+A Ctrl+Space. Nothing happens. Oh and now move your cursor.
B2) Continuation of the previous: Now press Enter to confirm, it starts playing according to the play/pause button but not according to the T menu. (If you opened the T menu while setting the previous one up, you can see the timer freezes at 0.02.) By this point LifeViewer is frozen and won't accept clicks but will keys.
B3) Continuation of the previous: Escape out, then click the "Show in Viewer" link again. Now it opens at T=1, and when these three steps are performed again, it remains at T=1. Even if generations are advanced normally, it stays at T=1.
1 and 3 happen due to the history. If you go down to the B2ae/S codebox and change the rule to B2ae/SHistory then proceed to do the same steps, half of 1 works (the nothing happens part) and so does 3, but the selection does not disappear, and as a consequence of that, 2 does not work either.
Each day is a hidden opportunity, a frozen waterfall that's waiting to be realised, and one that I'll probably be ignoring
sonata wrote:
July 2nd, 2020, 8:33 pm
conwaylife signatures are amazing[citation needed]
anything
rowett
Moderator
Posts: 2354
Joined: January 31st, 2013, 2:34 am
Location: UK
Contact:
### Re: Pattern viewer for forum threads
bubblegum wrote:
July 28th, 2020, 10:53 pm
Thank you for using LVEnhanceBot! Here are LifeViewer's newest bugs and suggestions:
Thanks for the report. Advance Selection only currently works for 2-state Life-like rules. I should probably disable it for the other rule types until it's implemented.
muzik
Posts: 4183
Joined: January 28th, 2016, 2:47 pm
Location: Scotland
### Re: Pattern viewer for forum threads
bubblegum wrote:
July 28th, 2020, 8:37 pm
I've gotten around to adding aliases for 2-state rules without them now.
Can you figure out these three? They've been unknown for quite a while now: viewtopic.php?p=99255#p99255
bubblegum
Posts: 896
Joined: August 25th, 2019, 11:59 pm
### Re: Pattern viewer for forum threads
musik wrote:
July 30th, 2020, 9:11 am
Can you figure out these three? They've been unknown for quite a while now: viewtopic.php?p=99255#p99255
Why not. More raw data:
Code: Select all
``````LogicRule - B2ae/S
Sierpinski - MAPT48PDw
EDIT: (logicrule is b2aes with one extraneous transition but whatever)
Each day is a hidden opportunity, a frozen waterfall that's waiting to be realised, and one that I'll probably be ignoring
sonata wrote:
July 2nd, 2020, 8:33 pm
conwaylife signatures are amazing[citation needed]
anything
bubblegum
Posts: 896
Joined: August 25th, 2019, 11:59 pm
### Re: Pattern viewer for forum threads
Is Sync supposed to be on by default?
edit: ah yes it is, it tells you to disable sync when sync is disabled
Each day is a hidden opportunity, a frozen waterfall that's waiting to be realised, and one that I'll probably be ignoring
sonata wrote:
July 2nd, 2020, 8:33 pm
conwaylife signatures are amazing[citation needed]
anything
bubblegum
Posts: 896
Joined: August 25th, 2019, 11:59 pm
### Re: Pattern viewer for forum threads
Thank you for using LVEnhanceBot! Here are LifeViewer's newest bugs and suggestions:
B1) Going to a tab of the settings and exiting out with M leaves the buttons on-screen.
Each day is a hidden opportunity, a frozen waterfall that's waiting to be realised, and one that I'll probably be ignoring
sonata wrote:
July 2nd, 2020, 8:33 pm
conwaylife signatures are amazing[citation needed]
anything
rowett
Moderator
Posts: 2354
Joined: January 31st, 2013, 2:34 am
Location: UK
Contact:
### Re: Pattern viewer for forum threads
bubblegum wrote:
August 1st, 2020, 1:12 am
Going to a tab of the settings and exiting out with M leaves the buttons on-screen.
Fixed, thanks for reporting!
rowett
Moderator
Posts: 2354
Joined: January 31st, 2013, 2:34 am
Location: UK
Contact:
### Re: Pattern viewer for forum threads
Build 549 is now live on the Forums and LifeWiki
Please note: you will need to refresh your browser to use the new build (Ctrl-F5 on Chrome)
Enhancements since the last released build:
• Neighbourhoods
• new LtL/HROT neighbourhoods - see examples
• L2 - specify with N2
• Hexagonal - specify with NH
• Tripod - specify with N3
• Asterisk - specify with NA
• Checkerboard - specify with NB
• Hash - specify with N#
• Triangular - specify with NL
• Custom - specify with N@ and then hex digits (CoordCA format without leading 1)
• can specify a postfix of H for hexagonal or L for triangular grid
• hotkey Ctrl+B converts a pattern selection to a CoordCA neighbourhood definition and copies it to the external clipboard
• selection must be square and have an odd side length from 3 to 99
• added support for all new neighbourhoods for Generations LtL/HROT rules
• new range-1 neighbourhoods
• Triangular Inner - specify with LI
• Triangular Outer - specify with LO
• Tripod - specify with HT
• Randomize
• new script command [[ RANDOMIZE ]] will generate a random pattern on start
• improved random rule choice for HROT rules
• UI
• setting focus with "Click to control" now doesn't show controls during [[ AUTOHIDEGUI ]]
• can now drag background during [[ AUTOHIDEGUI ]]
• Waypoints now do not lock controls if there are no camera commands
• clicking or dragging the pattern during Waypoint playback (when controls are locked) will pause playback
• added hotkey Ctrl+Shift+O to open clipboard as pattern on some browsers
• external and internal clipboard Sync is now enabled by default (to make Cut and Copy more intuitive)
• disable Sync for faster Cut and Copy using the internal clipboards only
• Display
• improved text wrapping in Help->Aliases for some rule types
• clipboard buttons tooltip now contains cell count
• General
Fixes since the last released build:
• fix ignored touch events on iPad
• don't randomize pattern if rule is invalid
• disabled Alt Gridlines button should be off
• display "New pattern" if there are only state 0 cells
• fixed "New pattern" message for non-RLE format patterns
• playback controls were not being reset after rule change from reversible rule
• fixed some Margolus Identify issues
• Identify works better with multi-state rules
• hex display was incorrect from ZOOM 1x to 4x when High Quality Rendering enabled
• hotkeys b and Backspace should only step back 1 generation
• better synchronization of Annotation positions during performance throttling
• fixed Annotation position when drawing with triangular grid
• fixed an issue with Waypoint timing
• improved Goucher LtL format detection
• fixed Advance Selection for non Life-like patterns
• close Settings submenus when Settings closed with hotkey
The updated hotkey map detailing the LifeViewer keyboard controls is here.
Comments, feedback, suggestions and bug reports welcome!
bubblegum
Posts: 896
Joined: August 25th, 2019, 11:59 pm
### Re: Pattern viewer for forum threads
Thank you for using LVEnhanceBot! Here are LifeViewer's newest bugs and suggestions:
S1) I figured out the N\$ format somehow. N\$ specifies a custom weighted neighbourhood in Base64 with an extra negation -. The first letter is a Q for a square grid or H for hex grid or L for tri grid. (For L the neighbourhood is specified for a down-pointing cell and flipped as necessary.) Then follows a string of B64 chars representing 0 through 63, and a - is inserted before a negative weight. So say the TriSemiTot neighbourhood which I literally just invented
Code: Select all
``````#C [[ VIEWONLY ]]
#C where yellow is the centre cell and green = 1, red = 4, white = 28
x = 7, y = 5, rule = B/SLHistory
would be
Code: Select all
``R1,CSBwhatever,N\$LcEBEcEBBEEcE``
(i mean i think hopefully maybe)
B1) possible Why does with an empty pattern switching to a square/tri grid rule set the zoom level to 32.0, a hex grid to -7.0, an [R]History one to -16.0, and an incorrectly-formatted ruletable/tree by the capitalisation of the first letter (like "evoloop") do apparently nothing and stay in the original rule but swap the rule name out for the new one and skew the cursor position away from the actual cursor position?
Each day is a hidden opportunity, a frozen waterfall that's waiting to be realised, and one that I'll probably be ignoring
sonata wrote:
July 2nd, 2020, 8:33 pm
conwaylife signatures are amazing[citation needed]
anything
GUYTU6J
Posts: 1374
Joined: August 5th, 2016, 10:27 am
Location: 拆哪！I repeat, CHINA! (a.k.a. 种花家)
Contact:
### Re: Pattern viewer for forum threads
GUYTU6J wrote:
August 3rd, 2020, 5:47 am
Eye-friendly mode on
Code: Select all
``````x = 64, y = 64, rule = B13ci8/S08:T64,64
4\$4b3o25b3ob3ob3o\$4b3o25b3ob3ob3o\$4b3o25b3ob3ob3o2\$32b3o5b3o\$32b3o5b3o
\$32b3o5b3o2\$32b3ob3ob3o\$32b3ob3ob3o\$32b3ob3ob3o10\$4b3ob3o\$4b3ob3o\$4b3o
b3o2\$4b3o\$4b3o\$4b3o30\$60b3o\$60b3o\$60b3o!
#C [[ STEP 4 THEME Mcell ]]
``````
While preparing the post I found this similar pattern broken:
Code: Select all
``````x = 64, y = 64, rule = B13ci8/S08:T64,64
32b3ob3ob3o\$32b3ob3ob3o\$32b3ob3ob3o2\$4b3o25b3o5b3o\$4b3o25b3o5b3o\$4b3o
25b3o5b3o2\$32b3ob3ob3o\$32b3ob3ob3o\$32b3ob3ob3o14\$4b3ob3o\$4b3ob3o\$4b3ob
3o2\$4b3o\$4b3o\$4b3o30\$60b3o\$60b3o\$60b3o!
#C [[ STEP 4 THEME Mcell ]]
``````
The top and the bottom edges show some defects and it is unrelated to STEP 4 command.
EIDT: wait, the quoted pattern breaks as well at around gen 988. In Golly this doesn't happen.
Lifequote:
In the drama The Peony Pavilion, Tang Xianzu wrote: 原来姹紫嫣红开遍，似这般都付与断井颓垣。
(Here multiflorate splendour blooms forlorn
Midst broken fountains, mouldering walls.)
I'm afraid there's arrival but no departure.
Stop Japan from dumping nuclear waste!
muzik
Posts: 4183
Joined: January 28th, 2016, 2:47 pm
Location: Scotland
### Re: Pattern viewer for forum threads
When on a triangular rule and zoomed out more than 4x, can the grid be vertically stretched somewhat so that everything doesn't look super flat?
rowett
Moderator
Posts: 2354
Joined: January 31st, 2013, 2:34 am
Location: UK
Contact:
### Re: Pattern viewer for forum threads
GUYTU6J wrote:
August 3rd, 2020, 6:00 am
the quoted pattern breaks as well at around gen 988. In Golly this doesn't happen.
It's a bug which I'll fix soon. Thanks for reporting!
rowett
Moderator
Posts: 2354
Joined: January 31st, 2013, 2:34 am
Location: UK
Contact:
### Re: Pattern viewer for forum threads
bubblegum wrote:
August 2nd, 2020, 4:33 pm
Why does with an empty pattern switching to a square/tri grid rule set the zoom level to 32.0, a hex grid to -7.0, an [R]History one to -16.0, and an incorrectly-formatted ruletable/tree by the capitalisation of the first letter (like "evoloop") do apparently nothing and stay in the original rule but swap the rule name out for the new one and skew the cursor position away from the actual cursor position?
This is fixed and will be in the next release. Thanks for reporting!
rowett
Moderator
Posts: 2354
Joined: January 31st, 2013, 2:34 am
Location: UK
Contact:
### Re: Pattern viewer for forum threads
muzik wrote:
August 3rd, 2020, 12:25 pm
When on a triangular rule and zoomed out more than 4x, can the grid be vertically stretched somewhat so that everything doesn't look super flat?
Yes, this will be in the next release.
rowett
Moderator
Posts: 2354
Joined: January 31st, 2013, 2:34 am
Location: UK
Contact:
### Re: Pattern viewer for forum threads
New resizeable LifeViewer is here. Would much appreciate any feedback!
I'm aware there's an issue with menu controls overlapping if the window is made too narrow and the Settings button not appearing if the initial window height is too small. I'll fix in due course. | 8,162 | 23,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.91319 |
https://www.assignmentessays.com/stats-2015-multiple-choice-questions/ | 1,579,904,931,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250626449.79/warc/CC-MAIN-20200124221147-20200125010147-00444.warc.gz | 767,798,913 | 15,678 | # STATS 2015 Multiple Choice Questions
August 30, 2017
Question
Directions: Blacken the letter corresponding to the best answer on your scantron (2.5 points each).
1. If Y = 4 + 11X the slope of the line equals:
a. X
b. 14
c. 11
d. 4
2. In the regression equation Y = a + bX where a = -2 and b = 5 the relationship can be said to be:
a. inverse because b 0
c. indeterminate since a 0
d. meaningless since a < 0
3. If we determine that the correlation between X and Y is statistically significant we can conclude:
a. that X did not cause Y to occur
b. that X caused Y to occur
c. there may be a causal relationship between X and Y
d. that Y caused X to occur
4. Pamela Stones is interested in determining whether weight gains of sheep are independent of the type of feed they receive. This is an example of:
a. a two way ANOVA test of factor design
b. a Chi-square test on weight versus feed
c. a regression analysis of weight gain on number of months in each field
d. none of the above
5. A coefficient of multiple determination of .45 for the regression Y = a + b1X1 + b2X2 + e means that:
a. 20.25 percent of the variation in Y is accounted for in terms of variation in X1 and X2
b. 45 percent of the variation in Y is not accounted for in terms of variation in X1 and X2
c. 45 percent of the variation in Y is accounted for in terms of variation in X1 and X2
d. 55 percent of the variation in Y is accounted for in terms of variation in X1 and X2
6. The chi-square technique of hypothesis testing can be used
a. to conduct goodness-of-fit tests
b. to test the alleged independence of two quantitative variables
c. to make inferences about homogeneity
d. for all of the above purposes (and more)
7. In the time series forecasting, the most important characteristic of amodel is
a. the quality of the fit
b. the amount of multicollinearity in the data
c. the amount of money you get paid to produce it
d. the amount of autocorrelation in the data
8. In simple regression analysis, the sum of the squares for the regression is called
a. the explained variation for Y
b. the error sum of squares
c. the total variation of Y
d. the unexplained variation for Y
9. If you wanted to identify the range in which a regression coefficient should fall with 90% assurance, the analysis that you should perform is called a(n):
a. Goldfeld-Quant test
b. Chi-square goodness of fit test
c. Confidence interval
d. Regression involving cross-sectional data
10. The "least squares method" can be described as:
a. a method of fitting a line to a set of points such that the absolute values of deviations are minimized
b. {S(Y – X)}2
c. S(X – X)2
d. a method of fitting a line to a set of points such that the squared deviations are minimized
11. A CPI Index Number value of 212.34 for 2013 implies that there has been ________ inflation since the base year of 1983.
a. 6.17%
b. 100%
b. 12.34%
d. 112.34%
12. When a contingency table is based on a Chi Square test of independence case, the number of degrees of freedom is the number of rows:
a. times the number of columns
b. minus one times the number of columns
c. minus one times the number of columns minus one
d. times the number of columns minus one
13. All of the following are reasons for obtaining a trend line except
a. we wish to eliminate trend movements from a time series so we can study the other component of the series
b. we are interested in studying the historical depiction of trend
c. we are interested in using the trend line to predict future data
d. we are interested in analyzing relationships between troughs and peaks
14. If we want to change the magnitude of a time series we simply:
a. divide each number into the data set by 100
b. multiply the slope and intercept by 4
c. divide the slope by the number of periods of data each year
d. divide the slope and intercept by the number of periods of data each year
15. In the equation: Yc = 31 + 4.2X1 + 3.5X2:
a. Yc = 38.7 if X1 + X2 = 0
b. Yc increases by 31 for every 1 unit increased in X1 and X2
c. Yc decreases by 3.5 if X2 decreases by 1 unit
d. Yc increases by 4.2 if X1 decreases by 1 unit and X2 is unchanged
16. The method of least squares (void of econometric problems) produces estimated regression coefficients (such as a and b) that are, with respect to the true regression coefficients (such asa &b),
a. unbiased estimators
b. efficient estimators
c. the best linear estimators
d. all of the above
17. In multiple regression analysis, according to a widely used rule of thumb, a parameter estimate b is significant at the 95 percent level of confidence as long as the associated p value
a. is equal to or greater than the parameter
b. is positive
c. is zero or positive
d. is less than .05
18. A sample coefficient of multiple determination, R2, that is close to zero indicates
a. a near-perfect fit of the regression plane to the data
b. an extremely bad fit of the regression plane to the data
c. neither (a) nor (b), because such a coefficient must be close to -1 or +1
d. the regression sum of squares exceeds the total sum of squares
19. One assumption of regression analysis concerns the statistical independence of different sample errors, but this assumption is often violated when the analysis employs
a. dummy variables
b. the step-down method
c. a correlation matrix
d. time-series data
20. Multicollinearity refers to:
a. a condition in which the independent variables are highly correlated with one another
b. the fact that for any values of X1 and X2, the standard errors are the same
c. the fact that a value of a time series may be calculated with its value at some other time period
d. the relationship between the predictor variables and the independent variables
21. Homoscedasticity refer to:
a. the assumption that values of the Y variable are independent of one another
b. the necessity of the normality assumption
c. the fact that the Y values are conditional upon corresponding X values
d. the assumption that all the error terms have about the same standard deviation
22. In the following regression equation a logarithmic transformation was applied to the dependent variable Y: ln(y) = 2.07 + 0.55 ln(x)
A one percent change in x implies that the average value of y increases by:
a. 2.07
b. -7.92
c. .55%
d. 2.07%
23. The easiest way to cure a serious multicollinearity problem in a multiple regression involves
a. regressing each of the independent variables against all of the others and noting whether any of the resultant R2 values lies close to 1
b. setting up a simple correlation matrix and noting any possible correlation between any one independent variable with all of the others as a group
c. dropping one of the highly collinear variables from the regression, although this causes the model to be misspecified .
d. performing an analysis of residuals
24. The trend equation Y = 14 + 3X + 2X2 : (for X = 0,1,2,…)
a. is an example of an exponential trend equation
b. will at first increase and then decrease
c. will at first decrease and then increase
d. is an example of a linear trend equation
25. If independent variables that were uncorrelated with the dependent variable were added to the regression equation, the coefficient of multiple determination would ___________, while the adjusted coefficient of determination would ________________ .
a. decrease, increase
b. decrease, remain the same
c. remain the same, fall
d. increase, decrease
e. increase, increase
26. In the United States, the unemployment rate must be
a. adjusted for cyclicality
b. adjusted for christmastitis
c. adjusted for seasonality
d. adjusted in all of the above ways
27. In a macroeconomic forecasting model over time:
a. autocorrelation is unlikely
b. simultaneous equation bias is likely
c. autocorrelation is likely
d. both b and c
28. Presented below is an ANOVA table:
Source d.f. SS MS F
Regression 3 246
Error 4 78
Total 7 324
a. The F value is 4.2
b. The Mean Square for the regression is 82
c. The Mean Square for the errors is 19.5
d. All of the above are true
29. The primary role of a statistician is to
a. interpret data using the best techniques
b. interpret data correctly
c. interpret data quickly
d. interpret data at the lowest price
e. get paid as much as he/she can to do as little as possible
30. Given the annual data 200, 250, 300, 350, the first two components of a 3-year centered moving-averages series are
a. 200, 300
b. 200, 250
c. 275, 325
d. 250, 300
31. The Durbin-Watson test is designed to detect
a. covariance
b. multicollinearity
c. autocorrelation
d. specification error
32. Cyclical variation:
a. refers to periodic up and down movements in a time series that tends to occur over long periods of time
b. cannot be accurately measured
c. is invariably positive over any length of time
d. refers to variation in the data that tend to recur during a period of one year or less
33. In a naive model over 120 observations with a time-trend and seasonal dummy variable the estimated coefficients are:
Yt = 1.03 + 0.34t + 4.32 WINTER + 1.62 SPRING + 3.78 SUMMER
(0.03) (1.12) (1.09) (1.20)
where the standard error of the slope coefficients are shown in parentheses under each item (remember that a t = coefficient/ standard error). At the 5% significance level there is not enough evidence to reject the null hypothesis that:
a. the average level of Y is the same between spring and fall
b. the average level of Y differs between winter and summer
c. the average level of Y is the same between summer and fall
d. the average level of Y differs between spring and summer
34. The results of a polynomial trend equation were y = 6.833 – 0.429x + 2.524×2, with n = 8. The x values were coded so that x = 0 in 1992. The estimated value of y for 2002 is:
a. 6.833
b. 254.943
c. -4.29
d. 252.400
e. 992.127
35. The upsurge in retail sales in the fall of each year is an example of:
a. the irregular component.
b. the trend component.
c. the seasonal component.
d. the cyclical component.
e. the disaster component.
36. The graph on the board implies:
SKIP
37. Given the following list of residuals:
time residual time residual
1 -5.6 11 -11.7
2 -2.9 12 -16.5
3 -5.5 13 -17.0
4 -10.5 14 -22.0
5 -8.5 15 -21.0
6 22.6 16 9.0
7 19.4 17 7.1
8 16.5 18 7.5
9 15.2 19 5.2
10 12.0 20 6.5
the pattern suggests:
a. multicollinearity is present.
b. heteroscedasticity is present.
c. autocorrelation is present.
d. homoscedasticity is present.
38. A wage contract calls for a wage increase to include one-half of the annual percentage increase (if any) in the Consumer Price Index. In 2002 the CPI equaled 124.0 and in 2003 equaled 130.7. The wage percentage increase equaled:
a. 2.70
b. 5.40
c. 6.70
d. 3.35
e. 0.50
39. A leading publication lists the average gross weekly earnings in 2013 as \$385.09. The report also lists the average gross weekly earnings in constant dollars with 1999 = 100, as \$208.42. Find the index value used to adjust average gross weekly earnings. The index value is:
a. 164.8
b. 100.0
c. 159.0
d. 169.8
e. 184.8
40. The best time for you to have taken this course was:
SKIP
Get a 30 % discount on an order above \$ 100
Use the following coupon code:
RESEARCH
Positive SSL | 3,010 | 11,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-05 | latest | en | 0.905035 |
https://www.distancesto.com/fuel-cost/lk/jaffna-to-pasikuda/history/1439739.html | 1,716,769,268,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00307.warc.gz | 619,580,963 | 14,630 | # INR13.17 Total Cost of Fuel from Jaffna to Pasikuda
Your trip to Pasikuda will consume a total of 5.27 gallons of fuel.
Trip start from Jaffna, LK and ends at Pasikuda, LK.
Trip (210.7 mi) Jaffna » Pasikuda
The map above shows you the route which was used to calculate fuel cost and consumption.
### Fuel Calculations Summary
Fuel calculations start from Jaffna, Sri Lanka and end at Pasikuda, Kalkudah, Sri Lanka.
Fuel is costing you INR2.50 per gallon and your vehicle is consuming 40 MPG. The formula can be changed here.
The driving distance from Jaffna to Pasikuda plays a major role in the cost of your trip due to the amount of fuel that is being consumed. If you need to analyze the details of the distance for this trip, you may do so by viewing the distance from Jaffna to Pasikuda.
Or maybe you'd like to see why certain roads were chosen for the route. You can do so by zooming in on the map and choosing different views. Take a look now by viewing the road map from Jaffna to Pasikuda.
Of course, what good is it knowing the cost of the trip and seeing how to get there if you don't have exact directions? Well it is possible to get exact driving directions from Jaffna to Pasikuda.
Did you also know that how elevated the land is can have an impact on fuel consumption and cost? Well, if areas on the way to Pasikuda are highly elevated, your vehicle may have to consume more gas because the engine would need to work harder to make it up there. In some cases, certain vehicles may not even be able to climb up the land. To find out, see route elevation from Jaffna to Pasikuda.
Travel time is of the essence when it comes to traveling which is why calculating the travel time is of the utmost importance. See the travel time from Jaffna to Pasikuda.
Speaking of travel time, a flight to Pasikuda takes up a lot less. How much less? Flight time from Jaffna to Pasikuda.
Cost is of course why we are here... so is it worth flying? Well this depends on how far your trip is. Planes get to where they need to go faster due to the speed and shorter distance that they travel. They travel shorter distances due to their ability to fly straight to their destination rather than having to worry about roads and obstacles that are in a motor vehicle's way. You can see for yourself the flight route on a map by viewing the flight distance from Jaffna to Pasikuda.
*The cost above should be taken as an ESTIMATE due to factors which affect fuel consumption and cost of fuel.
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Fuel Cost from Jaffna to Badulla | 663 | 2,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-22 | latest | en | 0.953783 |
https://www.nag.com/numeric/nl/nagdoc_28.5/clhtml/f08/f08qhc.html | 1,675,689,233,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500339.37/warc/CC-MAIN-20230206113934-20230206143934-00742.warc.gz | 895,316,692 | 8,043 | # NAG CL Interfacef08qhc (dtrsyl)
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CL Name Style:
## 1Purpose
f08qhc solves the real quasi-triangular Sylvester matrix equation.
## 2Specification
#include
void f08qhc (Nag_OrderType order, Nag_TransType trana, Nag_TransType tranb, Nag_SignType sign, Integer m, Integer n, const double a[], Integer pda, const double b[], Integer pdb, double c[], Integer pdc, double *scale, NagError *fail)
The function may be called by the names: f08qhc, nag_lapackeig_dtrsyl or nag_dtrsyl.
## 3Description
f08qhc solves the real Sylvester matrix equation
$op(A)X ± Xop(B) = αC ,$
where $\mathrm{op}\left(A\right)=A$ or ${A}^{\mathrm{T}}$, and the matrices $A$ and $B$ are upper quasi-triangular matrices in canonical Schur form (as returned by f08pec); $\alpha$ is a scale factor ($\text{}\le 1$) determined by the function to avoid overflow in $X$; $A$ is $m×m$ and $B$ is $n×n$ while the right-hand side matrix $C$ and the solution matrix $X$ are both $m×n$. The matrix $X$ is obtained by a straightforward process of back-substitution (see Golub and Van Loan (1996)).
Note that the equation has a unique solution if and only if ${\alpha }_{i}±{\beta }_{j}\ne 0$, where $\left\{{\alpha }_{i}\right\}$ and $\left\{{\beta }_{j}\right\}$ are the eigenvalues of $A$ and $B$ respectively and the sign ($+$ or $-$) is the same as that used in the equation to be solved.
## 4References
Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore
Higham N J (1992) Perturbation theory and backward error for $AX-XB=C$ Numerical Analysis Report University of Manchester
## 5Arguments
1: $\mathbf{order}$Nag_OrderType Input
On entry: the order argument specifies the two-dimensional storage scheme being used, i.e., row-major ordering or column-major ordering. C language defined storage is specified by ${\mathbf{order}}=\mathrm{Nag_RowMajor}$. See Section 3.1.3 in the Introduction to the NAG Library CL Interface for a more detailed explanation of the use of this argument.
Constraint: ${\mathbf{order}}=\mathrm{Nag_RowMajor}$ or $\mathrm{Nag_ColMajor}$.
2: $\mathbf{trana}$Nag_TransType Input
On entry: specifies the option $\mathrm{op}\left(A\right)$.
${\mathbf{trana}}=\mathrm{Nag_NoTrans}$
$\mathrm{op}\left(A\right)=A$.
${\mathbf{trana}}=\mathrm{Nag_Trans}$ or $\mathrm{Nag_ConjTrans}$
$\mathrm{op}\left(A\right)={A}^{\mathrm{T}}$.
Constraint: ${\mathbf{trana}}=\mathrm{Nag_NoTrans}$, $\mathrm{Nag_Trans}$ or $\mathrm{Nag_ConjTrans}$.
3: $\mathbf{tranb}$Nag_TransType Input
On entry: specifies the option $\mathrm{op}\left(B\right)$.
${\mathbf{tranb}}=\mathrm{Nag_NoTrans}$
$\mathrm{op}\left(B\right)=B$.
${\mathbf{tranb}}=\mathrm{Nag_Trans}$ or $\mathrm{Nag_ConjTrans}$
$\mathrm{op}\left(B\right)={B}^{\mathrm{T}}$.
Constraint: ${\mathbf{tranb}}=\mathrm{Nag_NoTrans}$, $\mathrm{Nag_Trans}$ or $\mathrm{Nag_ConjTrans}$.
4: $\mathbf{sign}$Nag_SignType Input
On entry: indicates the form of the Sylvester equation.
${\mathbf{sign}}=\mathrm{Nag_Plus}$
The equation is of the form $\mathrm{op}\left(A\right)X+X\mathrm{op}\left(B\right)=\alpha C$.
${\mathbf{sign}}=\mathrm{Nag_Minus}$
The equation is of the form $\mathrm{op}\left(A\right)X-X\mathrm{op}\left(B\right)=\alpha C$.
Constraint: ${\mathbf{sign}}=\mathrm{Nag_Plus}$ or $\mathrm{Nag_Minus}$.
5: $\mathbf{m}$Integer Input
On entry: $m$, the order of the matrix $A$, and the number of rows in the matrices $X$ and $C$.
Constraint: ${\mathbf{m}}\ge 0$.
6: $\mathbf{n}$Integer Input
On entry: $n$, the order of the matrix $B$, and the number of columns in the matrices $X$ and $C$.
Constraint: ${\mathbf{n}}\ge 0$.
7: $\mathbf{a}\left[\mathit{dim}\right]$const double Input
Note: the dimension, dim, of the array a must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{pda}}×{\mathbf{m}}\right)$.
The $\left(i,j\right)$th element of the matrix $A$ is stored in
• ${\mathbf{a}}\left[\left(j-1\right)×{\mathbf{pda}}+i-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_ColMajor}$;
• ${\mathbf{a}}\left[\left(i-1\right)×{\mathbf{pda}}+j-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_RowMajor}$.
On entry: the $m×m$ upper quasi-triangular matrix $A$ in canonical Schur form, as returned by f08pec.
8: $\mathbf{pda}$Integer Input
On entry: the stride separating row or column elements (depending on the value of order) in the array a.
Constraint: ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
9: $\mathbf{b}\left[\mathit{dim}\right]$const double Input
Note: the dimension, dim, of the array b must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{pdb}}×{\mathbf{n}}\right)$.
The $\left(i,j\right)$th element of the matrix $B$ is stored in
• ${\mathbf{b}}\left[\left(j-1\right)×{\mathbf{pdb}}+i-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_ColMajor}$;
• ${\mathbf{b}}\left[\left(i-1\right)×{\mathbf{pdb}}+j-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_RowMajor}$.
On entry: the $n×n$ upper quasi-triangular matrix $B$ in canonical Schur form, as returned by f08pec.
10: $\mathbf{pdb}$Integer Input
On entry: the stride separating row or column elements (depending on the value of order) in the array b.
Constraint: ${\mathbf{pdb}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
11: $\mathbf{c}\left[\mathit{dim}\right]$double Input/Output
Note: the dimension, dim, of the array c must be at least
• $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{pdc}}×{\mathbf{n}}\right)$ when ${\mathbf{order}}=\mathrm{Nag_ColMajor}$;
• $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}×{\mathbf{pdc}}\right)$ when ${\mathbf{order}}=\mathrm{Nag_RowMajor}$.
The $\left(i,j\right)$th element of the matrix $C$ is stored in
• ${\mathbf{c}}\left[\left(j-1\right)×{\mathbf{pdc}}+i-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_ColMajor}$;
• ${\mathbf{c}}\left[\left(i-1\right)×{\mathbf{pdc}}+j-1\right]$ when ${\mathbf{order}}=\mathrm{Nag_RowMajor}$.
On entry: the $m×n$ right-hand side matrix $C$.
On exit: c is overwritten by the solution matrix $X$.
12: $\mathbf{pdc}$Integer Input
On entry: the stride separating row or column elements (depending on the value of order) in the array c.
Constraints:
• if ${\mathbf{order}}=\mathrm{Nag_ColMajor}$, ${\mathbf{pdc}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$;
• if ${\mathbf{order}}=\mathrm{Nag_RowMajor}$, ${\mathbf{pdc}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
13: $\mathbf{scale}$double * Output
On exit: the value of the scale factor $\alpha$.
14: $\mathbf{fail}$NagError * Input/Output
The NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).
## 6Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.
On entry, argument $⟨\mathit{\text{value}}⟩$ had an illegal value.
NE_INT
On entry, ${\mathbf{m}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{m}}\ge 0$.
On entry, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{n}}\ge 0$.
On entry, ${\mathbf{pda}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pda}}>0$.
On entry, ${\mathbf{pdb}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pdb}}>0$.
On entry, ${\mathbf{pdc}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pdc}}>0$.
NE_INT_2
On entry, ${\mathbf{pda}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{m}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
On entry, ${\mathbf{pdb}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pdb}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On entry, ${\mathbf{pdc}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{m}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pdc}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
On entry, ${\mathbf{pdc}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pdc}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 7.5 in the Introduction to the NAG Library CL Interface for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library CL Interface for further information.
NE_PERTURBED
$A$ and $B$ have common or close eigenvalues, perturbed values of which were used to solve the equation.
## 7Accuracy
Consider the equation $AX-XB=C$. (To apply the remarks to the equation $AX+XB=C$, simply replace $B$ by $-B$.)
Let $\stackrel{~}{X}$ be the computed solution and $R$ the residual matrix:
$R = C - (AX~-X~B) .$
Then the residual is always small:
$‖R‖F = O(ε) (‖A‖F+‖B‖F) ‖X~‖F .$
However, $\stackrel{~}{X}$ is not necessarily the exact solution of a slightly perturbed equation; in other words, the solution is not backwards stable.
For the forward error, the following bound holds:
$‖X~-X‖F ≤ ‖R‖F sep (A,B)$
but this may be a considerable over estimate. See Golub and Van Loan (1996) for a definition of $\mathit{sep}\left(A,B\right)$, and Higham (1992) for further details.
These remarks also apply to the solution of a general Sylvester equation, as described in Section 9.
## 8Parallelism and Performance
f08qhc makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
The total number of floating-point operations is approximately $mn\left(m+n\right)$.
To solve the general real Sylvester equation
$AX ± XB = C$
where $A$ and $B$ are general nonsymmetric matrices, $A$ and $B$ must first be reduced to Schur form :
$A = Q1 A~ Q1T and B = Q2 B~ Q2T$
where $\stackrel{~}{A}$ and $\stackrel{~}{B}$ are upper quasi-triangular and ${Q}_{1}$ and ${Q}_{2}$ are orthogonal. The original equation may then be transformed to:
$A~ X~ ± X~ B~ = C~$
where $\stackrel{~}{X}={Q}_{1}^{\mathrm{T}}X{Q}_{2}$ and $\stackrel{~}{C}={Q}_{1}^{\mathrm{T}}C{Q}_{2}$. $\stackrel{~}{C}$ may be computed by matrix multiplication; f08qhc may be used to solve the transformed equation; and the solution to the original equation can be obtained as $X={Q}_{1}\stackrel{~}{X}{Q}_{2}^{\mathrm{T}}$.
The complex analogue of this function is f08qvc.
## 10Example
This example solves the Sylvester equation $AX+XB=C$, where
$A = ( 0.10 0.50 0.68 -0.21 -0.50 0.10 -0.24 0.67 0.00 0.00 0.19 -0.35 0.00 0.00 0.00 -0.72 ) ,$
$B = ( -0.99 -0.17 0.39 0.58 0.00 0.48 -0.84 -0.15 0.00 0.00 0.75 0.25 0.00 0.00 -0.25 0.75 )$
and
$C = ( 0.63 -0.56 0.08 -0.23 -0.45 -0.31 0.27 1.21 0.20 -0.35 0.41 0.84 0.49 -0.05 -0.52 -0.08 ) .$
### 10.1Program Text
Program Text (f08qhce.c)
### 10.2Program Data
Program Data (f08qhce.d)
### 10.3Program Results
Program Results (f08qhce.r) | 3,961 | 11,416 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 169, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-06 | latest | en | 0.668868 |
https://english.stackexchange.com/questions/138734/how-to-refer-on-the-property-of-a-line-which-has-a-sudden-change-in-its-directio | 1,726,888,888,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00287.warc.gz | 207,164,350 | 45,730 | # How to refer on the property of a line which has a sudden change in its direction?
There is sudden (sharp, this means, it's not differentiable at that point) change in the gradient of a line:
What is the proper way to refer to it, to not sound too mathematical. For example, "this line has a kink in it" seems odd. "The line has a break in it" might mean that there is a gap.
Edit: what I would like to finally express is how one adds this property to the line. E.g: "With this method, you can add a [noun] to the line", or "you can [verb] the line."
I don't need to emphasize the abruptness (but don't want it to be misunderstood as making a curve out of it). I would just like to make it clear, in simple terms, what would happen to the line.
• Why not simply a bend in the line?
– bib
Commented Nov 22, 2013 at 13:02
• @bib: Wouldn't that imply that the line becomes curved?
– vsz
Commented Nov 23, 2013 at 15:44
• This [definition]9http://www.collinsdictionary.com/dictionary/american/bend?showCookiePolicy=true) says curved or crooked form. You could say sharp bend which most people would probably read as an angle between two straight sections.
– bib
Commented Nov 23, 2013 at 18:21
• You call it "Kanye West's musical consistency between 2004 and present." Commented Nov 27, 2013 at 12:46
• Corner ... "See, Mary, the corner in the graph occurred when the first iPhone was announced." Commented Nov 27, 2013 at 13:24
the line changes in steepness.
You can add more detail too, if you need to describe the level of change.
the line severely changes in steepness.
the line slightly changes in steepness.
Following discussion and your edits, I would suggest.
With this method you can add an angle to the line.
Or, if you're happy to use multiple words, you will be much clearer saying:
With this method you can add a change of direction to the line.
• What I would like to express is how one can "add" such a change. If I wrote "you can add a change in its steepness" it might mean that "you can change its steepness (for the whole line)"
– vsz
Commented Nov 22, 2013 at 10:51
• "You can alter the direction/steepness of the line"?
– Ste
Commented Nov 22, 2013 at 10:52
• That would mean that I alter the steepness of the whole line. What I mean that the line had uniform steepness, and I added the change, and it no longer has that uniform steepness.
– vsz
Commented Nov 22, 2013 at 10:59
• So you want to be able to say: "I added a [noun] so the line changed in steepness"?
– Ste
Commented Nov 22, 2013 at 11:01
• No, `"I added a [noun] to the line."` Without further explanation. `"The line changed in steepness"` might still imply that the whole line changed.
– vsz
Commented Nov 22, 2013 at 11:07
skew
noun 1 an oblique angle; a slant.
verb 1 [no object, with adverbial] suddenly change direction or position: the car had skewed across the track
OP's use case, with skew:
With this method, you can add a skew to the line.
Though I would suggest using a verb form instead:
With this method, you can make the line to skew (at a given point).
We normally use the word abrupt.
• Abruptness, for a descriptive noun.
• Abrupt change.
• Acute change.
Frequently, people use sudden and abrupt interchangeably. However, there is a significant difference between sudden and abrupt.
Sudden should be restricted to abruptness in the time dimension. Whereas, abrupt is applicable to any "sudden" change in any dimension.
However, it is risky to use acute when trying to describe a mathematical chart or phenomenon in lay terms, because there is confusion between its meanings mathematically (in geometry) and colloquially.
"Please be careful, there is an abrupt change in the speed limit from 70 mph to 35 mph at that stretch of the road. After which, there will be an abrupt bend." The instructor said suddenly.
We were surprised when we noticed the abrupt change in the curve. The rate of increase in pressure abruptly decreased with the increase in temperature.
From American Heritage Dictionary ...
1. Unexpectedly sudden: an abrupt change in the weather.
2. Surprisingly curt; brusque: an abrupt answer made in anger.
3. Touching on one subject after another with sudden transitions: abrupt prose.
4. Steeply inclined. See Synonyms at steep1.
5. Botany Terminating suddenly rather than gradually; truncate: an abrupt leaf.
Compare with ...
sud·den
1. Happening without warning; unforeseen: a sudden storm.
2. Characterized by hastiness; abrupt or rash: a sudden decision. See Synonyms at impetuous.
3. Characterized by rapidity; quick and swift.
Idiom:
all of a sudden
Very quickly and unexpectedly; suddenly.
a·cute
1. Having a sharp point or tip.
2. Keenly perceptive or discerning: "a raw, chilling and psychologically acute novel of human passions reduced to their deadliest essence" (Literary Guild Magazine). See Synonyms at sharp.
3. Reacting readily to stimuli or impressions; sensitive: His hearing was unusually acute.
4. Of great importance or consequence; crucial: an acute lack of research funds.
5. Extremely sharp or severe; intense: acute pain; acute relief.
6. Medicine
a. Having a rapid onset and following a short but severe course: acute disease.
b. Afflicted by a disease exhibiting a rapid onset followed by a short, severe course: acute patients.
7. Music High in pitch; shrill.
8. Geometry Having an acute angle: an acute triangle.
The line is angled or broken.
• It is a sharp change, I edited the question. So you might recommend "You can add an angle to the line"?
– vsz
Commented Nov 22, 2013 at 11:05
• "With this method you can break the line at an angle" / "With this method you can add a break at an angle to the line" ? Probably the first.
– vsz
Commented Nov 22, 2013 at 11:12
• @vsz: Yes, the first is best.
– SF.
Commented Nov 22, 2013 at 11:15
• Broken by itself is appropriate; I've never seen broken used to mean a discontinuity, and would regard such usage, if it actually ever did occur, as a mistake. “Broken at an angle” is an odious pleonasm. Both of the line segments in the questions example are angled (in the sense of being neither vertical nor horizontal). -1, but will cheerfully refund the downvote if you fix all those issues Commented Nov 22, 2013 at 17:37
• @jwpat7, ‘broken line’ does indeed very often refer to a line that is discontinuous—in fact, if you look it up in various dictionaries, you will see that a broken line is most commonly defined as something like, “a discontinuous line or series of line segments, as a series of dashes, or a figure made up of line segments meeting at oblique angles”. It is by no stretch of the imagination a mistake to call a discontinuous line broken. The line in the middle of roads and streets is often called a ‘broken white line’. Commented Nov 23, 2013 at 14:34 | 1,732 | 6,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.9694 |
https://www.hurrydogs.ch/math-and-logic-puzzles_44099.html | 1,638,904,147,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363405.77/warc/CC-MAIN-20211207170825-20211207200825-00613.warc.gz | 843,128,908 | 7,225 | Math and Logic Puzzles
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7 Super Fun Math Logic Puzzles for Kids Ages 10+ (Answers Included!) A Post By: Anthony Persico Working on fun math riddles and brain teasers is a great way for kids to develop number sense and improve their mathematical problem-solving skills.. And these same benefits also apply to math logic puzzles, which also help students learn to think algebraically (usually years before they even step ...
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• ### logic puzzles, riddles and math puzzles - pzzls
On this site you'll find lots of logic puzzles and riddles, puzzling riddles, science puzzles and math puzzles and riddles. Answers and detailed solutions are included. Do you know other nice brain teasers, riddles or puzzles that should be on the site? If that's the case, send a message to pzzls! | 1,987 | 8,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | latest | en | 0.824099 |
https://electnorred.com/law/why-is-the-theory-of-evolution-not-a-law.html | 1,675,063,523,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00029.warc.gz | 252,774,496 | 14,426 | ### Why Is The Theory Of Evolution Not A Law?
Why Isn’t Evolution Considered a Law? This is an issue which often confuses the general public, as the two words, theory and law, have very different common meanings. But in science, their meanings are very similar. A theory is an explanation which is backed by “a considerable body of evidence,” while a law is a set of regularities expressed in a “mathematical statement.” This is why Newton’s Laws of Motion are referred to as laws and not theories.
They are expressed with simple equations (like f = ma for his 2nd Law of Motion). Evolution, and most of Biology, cannot be expressed in a concise mathematical equation, so it is referred to as a theory. A scientific law is not “better” or “more accurate” than a scientific theory, A law explains what will happen under certain circumstances, while a theory explains how it happens.
: Why Isn’t Evolution Considered a Law?
#### Is theory of evolution a scientific law?
Evolution is only a theory. It is not a fact or a scientific law.
## Why can a theory not become a law?
Theories can never become laws, because laws form the body of evidence upon which we base theories. Laws can help with formulating theories, but theories do not develop into laws. Finally, hypotheses, while a natural part of the scientific process, do not generally evolve into theories.
### What is the difference between a theory and a law?
Generally, laws describe what will happen in a given situation as demonstrable by a mathematical equation, whereas theories describe how the phenomenon happens.
## Why is cell theory called a theory and not a law?
Cell theory is a theory, not a law because the cell theory does not have enough support to become a law. Cell theory is referred to as the history of scientific theory. All cells come from pre-existing cells, and that is the basic unit reproduction and a basic unit of all organisms.
### At what point does a theory become a law?
A theory doesn’t become a law. End of story, end of this issue of Science 101, Just kidding—it’s all about the how and why, and that hasn’t been answered. See if this sounds familiar: Scientists begin with a hypothesis, which is sort of a guess of what might happen.
• When the scientists investigate the hypothesis, they follow a line of reasoning and eventually formulate a theory,
• Once a theory has been tested thoroughly and is accepted, it becomes a scientific law,
• Nice progression, and not what happens.
• To understand how scientists proceed in their investigations, it will help to understand each term individually.
What’s a hypothesis, what’s a theory, and what’s a law ?
### Can a theory be used to explain a law?
1. Home
2. References
(Image credit: Shutterstock) In general, a scientific law is the description of an observed phenomenon. It doesn’t explain why the phenomenon exists or what causes it. The explanation for a phenomenon is called a scientific theory, It is a misconception that theories turn into laws with enough research.
#### Is there a lot of evidence for evolution?
Figure – Species that diverged longer ago have more differences in their corresponding proteins, reflecting changes in the amino acids over time. Proteins evolve at different rates depending on the constraints imposed by their functions. Cytochrome c, a protein (more.) An interesting additional line of evidence supporting evolution involves sequences of DNA known as “pseudogenes.” Pseudogenes are remnants of genes that no longer function but continue to be carried along in DNA as excess baggage.
Pseudogenes also change through time, as they are passed on from ancestors to descendants, and they offer an especially useful way of reconstructing evolutionary relationships. With functioning genes, one possible explanation for the relative similarity between genes from different organisms is that their ways of life are similar—for example, the genes from a horse and a zebra could be more similar because of their similar habitats and behaviors than the genes from a horse and a tiger.
But this possible explanation does not work for pseudogenes, since they perform no function. Rather, the degree of similarity between pseudogenes must simply reflect their evolutionary relatedness. The more remote the last common ancestor of two organisms, the more dissimilar their pseudogenes will be.
• The evidence for evolution from molecular biology is overwhelming and is growing quickly.
• In some cases, this molecular evidence makes it possible to go beyond the paleontological evidence.
• For example, it has long been postulated that whales descended from land mammals that had returned to the sea.
• From anatomical and paleontological evidence, the whales’ closest living land relatives seemed to be the even-toed hoofed mammals (modem cattle, sheep, camels, goats, etc.). | 982 | 4,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.958455 |
http://studentsfocus.com/ec8394-adc-notes-analog-digital-circuits-lecture-handwritten-notes/ | 1,563,373,974,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525312.3/warc/CC-MAIN-20190717141631-20190717163631-00200.warc.gz | 151,527,448 | 11,808 | ## EC8394 ADC Notes
Anna University Regulation 2017 IT EC8394 ADC Notes, Analog Digital Circuits Engineering Lecture Handwritten Notes for all 5 units are provided below. Download link for IT 3rd SEM EC8394 Analog Digital Circuits Engineering Lecture Handwritten Notes are listed down for students to make perfect utilization and score maximum marks with our study materials.
1. Define Modulation.
Modulation is defined as the process by which some parameter of a high frequency carrier signal such as amplitude, frequency or phase is varied in accordance with the instantaneous value of low frequency message signal (modulating signal).
2. List out the reasons for using modulation.
The reasons for using modulation are:
(a) Size of antenna increases as frequency decreases.
(b) Ease of radiation.
(c) Less Noise and Interference.
(d) Bandwidth gets increased.
(e) Transmitting power gets decreased.
(f) Frequency re-allocation is possible.
3. Define amplitude Modulation.
Amplitude Modulation is the process of changing the amplitude of a relatively high frequency carrier signal in proportion with the instantaneous value of the modulating signal.
4. Define Modulation index and percent modulation for an AM wave.
Modulation index is a term used to describe the amount of amplitude change present in an AM waveform. It is also called as coefficient of modulation.
Mathematically modulation index is m
m = Em/Ec
where m = modulation coefficient
Em = Peak change in the amplitude of the output waveform voltage.
Ec = Peak amplitude of the unmodulated carrier voltage.
m = ( Vmax-V Vmin) )/(Vmax +Vmin)
m gives the amount of amplitude change in the modulating signal
Percent modulation gives the percentage change in the amplitude of the output wave when the carrier is acted on by a modulating signal.
EC8394 ADC Unit 1 notes Download Here
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EC8394 ADC Unit 3 notes Download Here
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Comments are closed. | 580 | 2,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-30 | latest | en | 0.826131 |
http://www.docstoc.com/docs/39061028/Homework-1-Data-Representation-Questions | 1,440,989,062,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065488.33/warc/CC-MAIN-20150827025425-00271-ip-10-171-96-226.ec2.internal.warc.gz | 400,452,518 | 38,118 | # Homework 1 Data Representation Questions by iarthur
VIEWS: 327 PAGES: 2
• pg 1
``` Complete in your homework jotter and hand in on Thursday 3rd
Sept
1. ASCII and UNICODE are both used to represent text in computer
systems
(a) Describe 1 advantage of UNICODE over ASCII 1
(b) Describe 1 disadvantage of UNICODE over ASCII 1
2. What is the 8 bit two’s complement representation of the number -72?
1
3. In floating point notation what defines:
(a) The range of the numbers stored? 1
(b) The precision or accuracy of the numbers stored? 1
4. Describe how graphics are stored using a bit-mapped package 2
5. (a) What is the number 385 represented as a binary number? 1
(b) Represent the number -86 in 8 bit two’s complement 1
(c) Write the number -37 as an 8-bit binary number using 2s complement
1
6. Ian purchases a new scanner. He scans photographs saving them as a
bit mapped image.
(a) Ian is aware that the quality of a bit mapped image depends on its bit
depth and resolution. What is meant by the terms:
(i) bit depth? 1
(i) resolution? 1
(b) Ian scans a photograph. The image is 800 by 600 pixels and uses
65,536 colours. Calculate the file size of the scanned image. Show all
7. John uses his digital camera to take photographs. It has a 512
Megabyte memory card. His camera uses 16,777,216 colours and is set to
a resolution of 3000 x 2000 pixels
(a) Calculate the file size of a single image. Your answer should be in
appropriate units. Show all working 3
(b) What is the maximum number of images of this size that can be stored
on John’s memory card? 2
```
To top | 436 | 1,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-35 | latest | en | 0.904338 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-10-counting-methods-and-probability-extension-apply-set-theory-practice-page-716/10 | 1,675,407,406,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00632.warc.gz | 803,999,733 | 14,897 | ## Algebra 2 (1st Edition)
We say that $A$ is a subset of $B$ if $B$ contains all elements in $A$. Here $C$ contains all elements in $A$, thus $A$ is a subset of $C$. | 56 | 167 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-06 | latest | en | 0.892649 |
https://www.teachoo.com/13092/3553/Question-7/category/Case-Based-Questions-MCQ/ | 1,695,803,097,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510284.49/warc/CC-MAIN-20230927071345-20230927101345-00240.warc.gz | 1,119,290,952 | 28,597 | Case Based Questions (MCQ)
Chapter 3 Class 12 Matrices
Serial order wise
## (d) Rs. 35000
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Question Three schools SNT, SNP and TKM organised a mela for collecting funds for helping the rehabilitation of flood victims. They sold hand-made fans, mats and plates from recycled material at a cost of ₹ 25, ₹ 100 and ₹ 50 each. The number of articles sold are given below. Question 1 Funds collected by SNT is _______. (a) Rs. 7000 (b) Rs. 6125 (c) Rs. 7875 (d) Rs. 21000 Let the number of articles sold be denoted by matrix X X = [■8(40& 25& 35@50& 40& 50@20& 30& 40)] Let the sale price of articles be denoted by matrix Y Let Y = [■8(25& 100& 50)] Now, Total Revenue = Sales price × Total sales = YX = [■8(25&100&50)]_(1×3) [■8(40& 25& 35@50& 40& 50@20& 30& 40)]_(3×3) = [■8(25(40)+100(50)+50(20)&25(25)+100(40)+50(30)&25(35)+100(50)+50(40))]_(1×3) = [■8(1000+5000+1000&625+4000+1500&875+5000+2000)] = [■8(𝟕𝟎𝟎𝟎&𝟔𝟏𝟐𝟓&𝟓𝟖𝟕𝟓)] ∴ Total amount collected by SNT = Rs 7,000 So, the correct answer is (a) Question 2 Funds collected by SNP is _______. (a) Rs. 7000 (b) Rs. 6125 (c) Rs. 7875 (d) Rs. 21000 Total Revenue = [■8(𝟕𝟎𝟎𝟎&𝟔𝟏𝟐𝟓&𝟓𝟖𝟕𝟓)] Thus, Fund raised by SNP = Rs. 6125 So, the correct answer is (B) Question 3 The total fund raised by all the three schools together is _______. (a) Rs. 7000 (b) Rs. 6125 (c) Rs. 7875 (d) Rs. 21000 Now, Total Revenue = [■8(7000&6125&5875)] Thus, Total Amount collected by all 3 schools = Rs 7000 + 6125 + 5875 = Rs 21,000 So, the correct answer is (d) Question 4 The total fund raised by selling fans is _______. (a) Rs. 4000 (b) Rs. 2000 (c) Rs. 2500 (d) Rs. 35000 Fund raised by selling fans = Total fans × Cost of selling fan = (40 + 25 + 35) × Rs 25 = 100 × 25 = Rs. 2500 So, the correct answer is (C) Question 5 TKM collected Rs _______ by selling plates. (a) Rs. 4000 (b) Rs. 2000 (c) Rs. 2500 (d) Rs. 35000 Fund raised by TKM for selling plates = Total plates sold by TKM × Cost of selling plates = 40 × Rs 50 = Rs. 2000 So, the correct answer is (b) | 826 | 2,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.853311 |
http://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/25/5/c/c/ | 1,603,307,660,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107877420.17/warc/CC-MAIN-20201021180646-20201021210646-00120.warc.gz | 135,235,333 | 52,293 | # Properties
Label 25.5.c.c Level 25 Weight 5 Character orbit 25.c Analytic conductor 2.584 Analytic rank 0 Dimension 4 CM no Inner twists 4
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$25 = 5^{2}$$ Weight: $$k$$ $$=$$ $$5$$ Character orbit: $$[\chi]$$ $$=$$ 25.c (of order $$4$$, degree $$2$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$2.58424907710$$ Analytic rank: $$0$$ Dimension: $$4$$ Relative dimension: $$2$$ over $$\Q(i)$$ Coefficient field: $$\Q(i, \sqrt{21})$$ Defining polynomial: $$x^{4} + 11 x^{2} + 25$$ Coefficient ring: $$\Z[a_1, \ldots, a_{9}]$$ Coefficient ring index: $$2^{3}$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{4}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2,\beta_3$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q -\beta_{3} q^{2} + \beta_{2} q^{3} -26 \beta_{1} q^{4} + 42 q^{6} + 9 \beta_{3} q^{7} -10 \beta_{2} q^{8} -39 \beta_{1} q^{9} +O(q^{10})$$ $$q -\beta_{3} q^{2} + \beta_{2} q^{3} -26 \beta_{1} q^{4} + 42 q^{6} + 9 \beta_{3} q^{7} -10 \beta_{2} q^{8} -39 \beta_{1} q^{9} -108 q^{11} -26 \beta_{3} q^{12} + 36 \beta_{2} q^{13} + 378 \beta_{1} q^{14} -4 q^{16} + 4 \beta_{3} q^{17} -39 \beta_{2} q^{18} -140 \beta_{1} q^{19} -378 q^{21} + 108 \beta_{3} q^{22} -79 \beta_{2} q^{23} -420 \beta_{1} q^{24} + 1512 q^{26} -120 \beta_{3} q^{27} + 234 \beta_{2} q^{28} -810 \beta_{1} q^{29} -728 q^{31} -156 \beta_{3} q^{32} -108 \beta_{2} q^{33} + 168 \beta_{1} q^{34} -1014 q^{36} + 144 \beta_{3} q^{37} -140 \beta_{2} q^{38} + 1512 \beta_{1} q^{39} + 1512 q^{41} + 378 \beta_{3} q^{42} -9 \beta_{2} q^{43} + 2808 \beta_{1} q^{44} -3318 q^{46} + 39 \beta_{3} q^{47} -4 \beta_{2} q^{48} -1001 \beta_{1} q^{49} -168 q^{51} -936 \beta_{3} q^{52} + 676 \beta_{2} q^{53} -5040 \beta_{1} q^{54} + 3780 q^{56} -140 \beta_{3} q^{57} -810 \beta_{2} q^{58} + 3780 \beta_{1} q^{59} + 4592 q^{61} + 728 \beta_{3} q^{62} + 351 \beta_{2} q^{63} -6616 \beta_{1} q^{64} -4536 q^{66} + 729 \beta_{3} q^{67} + 104 \beta_{2} q^{68} -3318 \beta_{1} q^{69} + 432 q^{71} + 390 \beta_{3} q^{72} -1404 \beta_{2} q^{73} + 6048 \beta_{1} q^{74} -3640 q^{76} -972 \beta_{3} q^{77} + 1512 \beta_{2} q^{78} + 8840 \beta_{1} q^{79} + 1881 q^{81} -1512 \beta_{3} q^{82} -169 \beta_{2} q^{83} + 9828 \beta_{1} q^{84} -378 q^{86} -810 \beta_{3} q^{87} + 1080 \beta_{2} q^{88} -13230 \beta_{1} q^{89} -13608 q^{91} + 2054 \beta_{3} q^{92} -728 \beta_{2} q^{93} + 1638 \beta_{1} q^{94} + 6552 q^{96} + 1764 \beta_{3} q^{97} -1001 \beta_{2} q^{98} + 4212 \beta_{1} q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4q + 168q^{6} + O(q^{10})$$ $$4q + 168q^{6} - 432q^{11} - 16q^{16} - 1512q^{21} + 6048q^{26} - 2912q^{31} - 4056q^{36} + 6048q^{41} - 13272q^{46} - 672q^{51} + 15120q^{56} + 18368q^{61} - 18144q^{66} + 1728q^{71} - 14560q^{76} + 7524q^{81} - 1512q^{86} - 54432q^{91} + 26208q^{96} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{4} + 11 x^{2} + 25$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$($$$$\nu^{3} + 6 \nu$$$$)/5$$ $$\beta_{2}$$ $$=$$ $$($$$$\nu^{3} + 10 \nu^{2} + 16 \nu + 55$$$$)/5$$ $$\beta_{3}$$ $$=$$ $$($$$$\nu^{3} - 10 \nu^{2} + 16 \nu - 55$$$$)/5$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$($$$$\beta_{3} + \beta_{2} - 2 \beta_{1}$$$$)/4$$ $$\nu^{2}$$ $$=$$ $$($$$$-\beta_{3} + \beta_{2} - 22$$$$)/4$$ $$\nu^{3}$$ $$=$$ $$($$$$-3 \beta_{3} - 3 \beta_{2} + 16 \beta_{1}$$$$)/2$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/25\mathbb{Z}\right)^\times$$.
$$n$$ $$2$$ $$\chi(n)$$ $$-\beta_{1}$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
7.1
2.79129i − 1.79129i − 2.79129i 1.79129i
−4.58258 4.58258i −4.58258 + 4.58258i 26.0000i 0 42.0000 41.2432 + 41.2432i 45.8258 45.8258i 39.0000i 0
7.2 4.58258 + 4.58258i 4.58258 4.58258i 26.0000i 0 42.0000 −41.2432 41.2432i −45.8258 + 45.8258i 39.0000i 0
18.1 −4.58258 + 4.58258i −4.58258 4.58258i 26.0000i 0 42.0000 41.2432 41.2432i 45.8258 + 45.8258i 39.0000i 0
18.2 4.58258 4.58258i 4.58258 + 4.58258i 26.0000i 0 42.0000 −41.2432 + 41.2432i −45.8258 45.8258i 39.0000i 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
5.b even 2 1 inner
5.c odd 4 2 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 25.5.c.c 4
3.b odd 2 1 225.5.g.j 4
4.b odd 2 1 400.5.p.l 4
5.b even 2 1 inner 25.5.c.c 4
5.c odd 4 2 inner 25.5.c.c 4
15.d odd 2 1 225.5.g.j 4
15.e even 4 2 225.5.g.j 4
20.d odd 2 1 400.5.p.l 4
20.e even 4 2 400.5.p.l 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
25.5.c.c 4 1.a even 1 1 trivial
25.5.c.c 4 5.b even 2 1 inner
25.5.c.c 4 5.c odd 4 2 inner
225.5.g.j 4 3.b odd 2 1
225.5.g.j 4 15.d odd 2 1
225.5.g.j 4 15.e even 4 2
400.5.p.l 4 4.b odd 2 1
400.5.p.l 4 20.d odd 2 1
400.5.p.l 4 20.e even 4 2
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{2}^{4} + 1764$$ acting on $$S_{5}^{\mathrm{new}}(25, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$1 - 412 T^{4} + 65536 T^{8}$$
$3$ $$1 + 1278 T^{4} + 43046721 T^{8}$$
$5$ 1
$7$ $$1 - 9569602 T^{4} + 33232930569601 T^{8}$$
$11$ $$( 1 + 108 T + 14641 T^{2} )^{4}$$
$13$ $$1 - 1624225342 T^{4} + 665416609183179841 T^{8}$$
$17$ $$1 + 13727462018 T^{4} + 48661191875666868481 T^{8}$$
$19$ $$( 1 - 241042 T^{2} + 16983563041 T^{4} )^{2}$$
$23$ $$1 - 68080016962 T^{4} +$$$$61\!\cdots\!61$$$$T^{8}$$
$29$ $$( 1 - 758462 T^{2} + 500246412961 T^{4} )^{2}$$
$31$ $$( 1 + 728 T + 923521 T^{2} )^{4}$$
$37$ $$1 + 1254529400258 T^{4} +$$$$12\!\cdots\!41$$$$T^{8}$$
$41$ $$( 1 - 1512 T + 2825761 T^{2} )^{4}$$
$43$ $$1 + 23329889084798 T^{4} +$$$$13\!\cdots\!01$$$$T^{8}$$
$47$ $$1 + 46379759106878 T^{4} +$$$$56\!\cdots\!21$$$$T^{8}$$
$53$ $$1 - 112877432101822 T^{4} +$$$$38\!\cdots\!21$$$$T^{8}$$
$59$ $$( 1 - 9946322 T^{2} + 146830437604321 T^{4} )^{2}$$
$61$ $$( 1 - 4592 T + 13845841 T^{2} )^{4}$$
$67$ $$1 - 488793100954882 T^{4} +$$$$16\!\cdots\!81$$$$T^{8}$$
$71$ $$( 1 - 432 T + 25411681 T^{2} )^{4}$$
$73$ $$1 - 937201474520062 T^{4} +$$$$65\!\cdots\!61$$$$T^{8}$$
$79$ $$( 1 + 245438 T^{2} + 1517108809906561 T^{4} )^{2}$$
$83$ $$1 + 4278306619448318 T^{4} +$$$$50\!\cdots\!81$$$$T^{8}$$
$89$ $$( 1 + 49548418 T^{2} + 3936588805702081 T^{4} )^{2}$$
$97$ $$1 - 13524937897425022 T^{4} +$$$$61\!\cdots\!21$$$$T^{8}$$ | 3,384 | 6,907 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-45 | latest | en | 0.359613 |
https://www.perlmonks.org/?displaytype=print;node_id=547689;replies=1 | 1,529,322,606,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859766.6/warc/CC-MAIN-20180618105733-20180618125733-00408.warc.gz | 896,645,547 | 3,514 | http://www.perlmonks.org?node_id=547689
bart,
With regards to your follow up in the CB concerning efficiency. Without resorting to complex data structures, the partial patience sort can be done in O(N Log N) assuming the binary search. Finding the LIS is at max an additional N worst case so O(N Log N + N). The question you posed is if the merge sort O(N^2) was the most efficient way to finish the sort.
If you take advantage of the fact that the top cards are already in ascending order, you can select the top card of the left most pile and then move that pile to keep the new top card in ascending order. To find the new location using a binary search you have O(Log N). To insert in the middle is O(N). Since you have to do this for N items, you result in O(N^2 Log N). A merge sort is only O(N^2) worst case so no, I don't think so.
As I said in the other reply, using a different datastructure could make the finishing of the sort more efficient but it also adds a great deal more complexity. You are welcome to use a Van_Emde_Boas_tree which claims to be able to do the whole thing in O(N Log N) but that is an exercise left for the reader.
Cheers - L~R
• Comment on Re^2: Patience Sorting To Find Longest Increasing Subsequence
Replies are listed 'Best First'.
Re^3: Patience Sorting To Find Longest Increasing Subsequence
by demerphq (Chancellor) on May 06, 2006 at 09:02 UTC
A merge sort is only O(N^2) worst case so no,
No, a merge sort is worst case O(n log n).
You are welcome to use a Van_Emde_Boas_tree which claims to be able to do the whole thing in O(N Log N)
Actually the paper by Bespamyatnikh & Segal contains a proof that you can do it in O(N log log N) time. I havent verified it tho.
But I doubt that the vEB based algorithm would in practice beat a simpler algorithm to do patience sorting. Unless I guess if you were dealing with a deck with tens of thousands or even millions or billions of cards. The overhead of maintaining a vEB tree is prohibitive for small datasets. The cost of doing binary operations on the keys, maintaining the vEB tree and etc, would most likely outweigh that of a simpler less efficient algorithm.
As bart said, sometimes a binary search algorithm is not as fast a scan, even though one is O(log N) and the other is O(N). The reason of course is that big-oh notation glosses over issues like cost per operation, and only focuses on the "overwhelming factor" involved. So in a binary search if it takes 4 units of work to perform an operation and in linear search it takes 1 then binary search only wins when 4 * log N < N, so for lists shorter than 13 elements there would be no win to a binary search. And I'd bet that in fact the ratio is probably something like 20:1 and not 4:1. Apply this kind of logic to a deck of 52 cards, and IMO its pretty clear that vEB trees are not the way to go for this, regardless of the proof.
---
\$world=~s/war/peace/g
demerphq,
I had taken bart's word for the merge sort in the CB. I later told him the math was wrong (in the CB) but didn't change it because I knew the math was also wrong for his desired method of finishing the sorting. The thing neither of us considered is that the problem space decreases with each pass. In any case, regardless of the accuracy of the math - the merge sort is still the most efficient given the data structure.
Actually the paper by Bespamyatnikh & Segal contains a proof that you can do it in O(N log log N) time.
What is the "it" that is O(N log log N) time though? The partial sort needed to obtain the LIS, obtaining the LIS itself, or completing the patience sort? What I understood from the paper, which I admittedly only read far enough to know that it was over my head, was that the O(N log log N) was not for a complete sort which Wikipedia agrees with.
As far as the binary search is concerned - I have provided implementations to get to the partial sort using both methods so Benchmarking shouldn't be hard. Additionally, implementing a binary search & splice approach to bench against the merge sort is also straight forward.
Cheers - L~R
What is the "it" that is O(N log log N) time though?
From what I understand its both finding the LIS and doing the patience sort. Actually, if i understand things correctly the B&S algorithm actually finds all increasing sequences in O(N log log N).
Also I have an implementation of Patience sorting with backrefs for the LIS that I will post when I get a moment.
---
\$world=~s/war/peace/g | 1,073 | 4,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-26 | latest | en | 0.919775 |
https://forum.enterprisedna.co/t/weekly-updates-net-new-existing-records/64027 | 1,726,866,992,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00187.warc.gz | 243,149,787 | 5,501 | Weekly Updates - Net New + Existing Records
Most of the examples from the courses I’ve watched utilize a dataset where each record represents a “unique” transaction.
I have an issue where I’m charting pipeline progression (movement of records through sales stages) and need to demonstrate weekly progression of existing, or net-new, records.
Therefore, my weekly import “may” contain records that are new (timestamped this week), or existing where some key metric I’m tracking may have changed (could be 1 or more columns of data that have changed for an existing record).
I’ve come up with an idea of creating an update process where I “signal” via a new column, the last know occurence of a record:
PSEUDO CODE:
Latest_Record =
VAR CurrentOPID = CRM_Data[OPID]
VAR CurrentDate = CRM_Data[IMPRTDATE]
RETURN
IF(
COUNTROWS(
FILTER(
CRM_Data,
CRM_Data[OPID] = CurrentOPID &&
CRM_Data[IMPRTDATE] > CurrentDate
)
) = 0,
1,
0
)
QUESTION: Is my logic sound enough to handle various measures like the ones below?
If I have this data, for example:
IMPRTDATE; OPDATE; OPID; GROSS_PROFIT; SALES_STAGE
05/01/2024; 05/01/2024; 12345; \$50,000; Working
05/08/2024; 05/01/2024; 12345; \$50,000; Working
05/15/2024; 05/01/2024; 12345; \$50,000; Qualifying
05/23/2024; 05/01/2024; 12345; \$48,000; Qualifying
05/30/2024; 05/01/2024; 12345; \$48,000; Qualifying
06/05/2024; 05/01/2024; 12345; \$42,000; Decision Due
A) The SUM of Gross profit for this account, in May, should be \$48,000;
B) The SUM of Gross Profit for the account, in June, should be \$42,000;
C) The cumulative SUM of Gross Profit for this account, to date, should be \$42,000
D) The cumulate count of records for this account, to date, should be 1
E) The net-new count of this account, in May, should be 1 (OPDATE==May)
F) The net-new count of this account, in June, should 0 (OPDATE==May)
G) This-Month versus Last Month count should be 0 for June and 1 for May
Any help / advise is tremendougly appreciated.
Hi @mtyler306 ,
Your logic for handling the latest record based on the import date is sound. Here’s how you can achieve your goals using DAX in Power BI:
Create the Latest Record Flag:
DAX
Copy code
Latest_Record =
VAR CurrentOPID = CRM_Data[OPID]
VAR CurrentDate = CRM_Data[IMPRTDATE]
RETURN
IF(
COUNTROWS(
FILTER(
CRM_Data,
CRM_Data[OPID] = CurrentOPID &&
CRM_Data[IMPRTDATE] > CurrentDate
)
) = 0,
1,
0
)
SUM of Gross Profit for May:
DAX
Copy code
SUM_May_GP =
CALCULATE(
SUM(CRM_Data[GROSS_PROFIT]),
FILTER(
CRM_Data,
CRM_Data[IMPRTDATE] <= DATE(2024, 5, 31) && CRM_Data[IMPRTDATE] >= DATE(2024, 5, 1) &&
CRM_Data[Latest_Record] = 1
)
)
SUM of Gross Profit for June:
DAX
Copy code
SUM_June_GP =
CALCULATE(
SUM(CRM_Data[GROSS_PROFIT]),
FILTER(
CRM_Data,
CRM_Data[IMPRTDATE] <= DATE(2024, 6, 30) && CRM_Data[IMPRTDATE] >= DATE(2024, 6, 1) &&
CRM_Data[Latest_Record] = 1
)
)
Cumulative SUM of Gross Profit to Date:
DAX
Copy code
Cumulative_GP =
CALCULATE(
SUM(CRM_Data[GROSS_PROFIT]),
FILTER(
CRM_Data,
CRM_Data[Latest_Record] = 1
)
)
Cumulative Count of Records to Date:
DAX
Copy code
Cumulative_Count =
CALCULATE(
COUNTROWS(CRM_Data),
FILTER(
CRM_Data,
CRM_Data[Latest_Record] = 1
)
)
Net-New Count in May:
DAX
Copy code
NetNew_May =
CALCULATE(
COUNTROWS(CRM_Data),
FILTER(
CRM_Data,
MONTH(CRM_Data[OPDATE]) = 5 &&
CRM_Data[Latest_Record] = 1
)
)
Net-New Count in June:
DAX
Copy code
NetNew_June =
CALCULATE(
COUNTROWS(CRM_Data),
FILTER(
CRM_Data,
MONTH(CRM_Data[OPDATE]) = 6 &&
CRM_Data[Latest_Record] = 1
)
)
This-Month vs Last-Month Count:
DAX
Copy code
ThisMonth_vs_LastMonth =
VAR ThisMonth =
CALCULATE(
COUNTROWS(CRM_Data),
FILTER(
CRM_Data,
MONTH(CRM_Data[IMPRTDATE]) = MONTH(TODAY()) &&
CRM_Data[Latest_Record] = 1
)
)
VAR LastMonth =
CALCULATE(
COUNTROWS(CRM_Data),
FILTER(
CRM_Data,
MONTH(CRM_Data[IMPRTDATE]) = MONTH(TODAY()) - 1 &&
CRM_Data[Latest_Record] = 1
)
)
RETURN
ThisMonth - LastMonth
For more detailed guidance and advanced Power BI queries, you can explore the Data Mentor platform.
Cheers,
Enterprise DNA Support Team | 1,237 | 4,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.813876 |
http://www.free-online-calculator-use.com/adding-subtracting-mixed-numbers.html | 1,508,706,498,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825464.60/warc/CC-MAIN-20171022203758-20171022223758-00577.warc.gz | 438,682,970 | 26,037 | # Adding Subtracting Mixed NumbersCalculator That Shows Its Work
This free online calculator will add one mixed number to another, or subtract one mixed number from another -- regardless if the two mixed numbers have the same or different denominators -- and give the resulting fraction in simplest form.
Plus, unlike some other online mixed number calculators, this calculator will show its work and give a detailed step-by-step explanation as to how it arrived at the result.
### How Do You Add and Subtract Mixed Numbers?
The only difference between adding or subtracting mixed numbers and adding or subtraction fractions, is that before you begin to add or subtract you must first convert the mixed numbers to improper fractions.
To convert a mixed number to an improper fraction, you first multiply the denominator of the fractional portion by the whole number and then add that product to the numerator. That result then becomes the numerator (top) of the improper fraction, while the denominator (bottom) remains unchanged.
Converting a Mixed Number to an Improper Fraction
2 1 = 3 x 2 + 1 = 7 3 3 3
Once you have converted the mixed numbers to improper fractions, you simply add or subtract the fractions as usual (the calculator will show its work in case you're not sure of the steps involved).
With that, let's use the Adding Subtracting Mixed Numbers Calculator to add one mixed number to another, or to subtract one from the other.
Instructions: Enter the whole number, the numerator (top), and the denominator (bottom) of the first number in the equation.
Next, select + (plus) or - (minus) from the add subtract mixed numbers drop down menu.
Finally, enter the whole number, numerator, and denominator of the second mixed number, and then click the "Add Subtract Mixed Numbers" button.
Note that if you need to add or subtract more than two mixed numbers (1-2/3 plus 3-1/2 minus 2-1/5, etc), working from left to right, simply perform the operation on the first two, write down the result (or enter it before resetting the calculator), and then perform the second operation on the third mixed number in the equation. Repeat until all mixed numbers in the equation have been added and subtracted.
Mouse over the blue question marks for a further explanation of each entry field. More in-depth explanations can be found in the glossary of terms located beneath the Adding Subtracting Mixed Numbers Calculator.
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Whole # Fraction Whole # Fraction
Numerators (#):
Denominators (#):
Sum or Difference:
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### Adding Subtracting Mixed Numbers Calculator Glossary of Terms
Check Out My Other SuperOnline Math CalculatorsTo Help You ToSolve and Learn ...
Mixed number: A number that consist of a whole number and a fraction. Examples of mixed numbers would include 11/2, 33/5, 51/6, etc.
Whole number: The number located to the left of fractional portion of a mixed number. In the mixed number 24/5, the number 2 would be the whole number.
Numerator: The top number of the fractional portion of a mixed number. In the mixed number 24/5, the number 4 would be the numerator.
Denominator: The bottom number of the fractional portion of a mixed number. In the mixed number 24/5, the number 5 would be the denominator.
Answer: This is the result of addition or subtraction of the two mixed numbers. After performing the operation, the calculator will show its work and give a detailed explanation of each step it took to arrive at the answer.
Improper fraction: A fraction containing a numerator (top) that is larger than the denominator (bottom). Examples of improper fractions would include 9/5, 17/4, 39/38, etc.
Augend: Usually the first number of an addition problem.
Addend: Usually the second (or other) numbers to be added to the first number.
Minuend: Usually the first number listed in a subtraction problem.
Subtrahend: Usually the second number listed in a subtraction problem.
Sum: The result after performing addition.
Difference: The result after performing subtraction.
> > Adding Subtracting Mixed Numbers Calculator
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Online Pocket Calc | 1,059 | 4,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-43 | latest | en | 0.853138 |
https://socratic.org/questions/what-is-a-possible-value-for-the-missing-term-of-the-geometric-sequence-1250-50 | 1,723,783,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00639.warc.gz | 411,348,530 | 5,927 | # What is a possible value for the missing term of the geometric sequence 1250,__,50,..?
Jan 4, 2017
$\pm 250$
#### Explanation:
The general term of a geometric sequence has the form:
${a}_{n} = a \cdot {r}^{n - 1}$
where $a$ is the initial term and $r$ the common ratio.
In our example:
${a}_{1} = 1250 \text{ }$ and $\text{ } {a}_{3} = 50$
So:
${r}^{2} = \frac{a {r}^{2}}{a {r}^{0}} = {a}_{3} / {a}_{1} = \frac{50}{1250} = \frac{1}{25} = \frac{1}{5} ^ 2$
Hence:
$r = \pm \frac{1}{5}$
Then:
${a}_{2} = a \cdot {r}^{2 - 1} = 1250 \cdot \left(\pm \frac{1}{5}\right) = \pm 250$ | 246 | 589 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-33 | latest | en | 0.662287 |
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posted by on .
Suppose that humans walk about 10,000 steps per day, on awerage.
1. Your average step is probably about 18 inches. If so, how many inches per day do you walk?
2. How many feet per day do you walk?
3. How many miles per day do you wa;k, to the nearest whole mile?
• math - ,
1)
10,000 * 18 = 180,000 inches
2)
180,000 / 12 = 15,000 feet
3)
15,000 / 5,280 = ____ miles
#### Related Questions
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Look at the figure below: Triangle CDE has measure of angle CDE equal to 90 degrees. A is a point on the hypotenuse EC such that the segment DA is perpendicular to EC. B is a point outside the triangle on the right of C so that triangle ABC has the measure of angle ABC equal to 90 degrees. Which triangle is similar to triangle EAD?
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### 18.3: Thermometers and Temperature Scales TABLE OFCONTENTS X ## Chapter 1: Units, Dimensions, and Measurements 301.1: The Scope of Physics301.2: Orders of Magnitude301.3: Units and Standards of Measurement301.4: Base Quantities and Derived Quantities301.5: Conversion of Units301.6: Accuracy and Precision301.7: Random and Systematic Errors301.8: Rules for Significant Figures301.9: Significant Figures in Calculations301.10: Dimensional Analysis301.11: Solving Problems in Physics ## Chapter 2: Vectors and Scalars 302.1: Introduction to Scalars302.2: Introduction to Vectors302.3: Vector Components in the Cartesian Coordinate System302.4: Polar and Cylindrical Coordinates302.5: Vector Algebra: Graphical Method302.6: Vector Algebra: Method of Components302.7: Scalar Product (Dot Product)302.8: Vector Product (Cross Product) ## Chapter 3: Motion Along a Straight Line 303.1: Position and Displacement303.2: Average Velocity303.3: Instantaneous Velocity - I303.4: Instantaneous Velocity - II303.5: Average Acceleration303.6: Instantaneous Acceleration303.7: Kinematic Equations - I303.8: Kinematic Equations - II303.9: Kinematic Equations - III303.10: Kinematic Equations: Problem Solving303.11: Free-falling Bodies: Introduction303.12: Free-falling Bodies: Example303.13: Velocity and Position by Graphical Method303.14: Velocity and Position by Integral Method ## Chapter 4: Motion in Two or Three Dimensions 304.1: Position and Displacement Vectors304.2: Average and Instantaneous Velocity Vectors304.3: Acceleration Vectors304.4: Direction of Acceleration Vectors304.5: Projectile Motion304.6: Projectile Motion: Equations304.7: Projectile Motion: Example304.8: Uniform Circular Motion304.9: Non-uniform Circular Motion304.10: Relative Velocity in One Dimension304.11: Relative Velocity in Two Dimensions ## Chapter 5: Newton's Laws of Motion 305.1: Force305.2: Types of Forces305.3: Newton's First Law: Introduction305.4: Newton's First Law: Application305.5: Internal and External Forces305.6: Newton's Second Law305.7: Mass and Weight305.8: Weightlessness305.9: Newton's Third Law: Introduction305.10: Newton's Third Law: Examples305.11: Drawing Free-body Diagrams: Rules305.12: Free Body Diagrams: Examples305.13: Inertial Frames of Reference305.14: Non-inertial Frames of Reference ## Chapter 6: Application of Newton's Laws of Motion 306.1: First Law: Particles in One-dimensional Equilibrium306.2: First Law: Particles in Two-dimensional Equilibrium306.3: Second Law: Motion under Same Force306.4: Second Law: Motion under Same Acceleration306.5: Frictional Force306.6: Static and Kinetic Frictional Force306.7: Dynamics of Circular Motion306.8: Dynamics Of Circular Motion: Applications ## Chapter 7: Work and Kinetic Energy 307.1: Work307.2: Positive, Negative, and Zero Work307.3: Energy307.4: Types of Potential Energy307.5: Types of Kinetic Energy307.6: Kinetic Energy - I307.7: Kinetic Energy - II307.8: Work-energy Theorem307.9: Power307.10: Power Expended by a Constant Force ## Chapter 8: Potential Energy and Energy Conservation 308.1: Gravitational Potential Energy308.2: Elastic Potential Energy308.3: Conservative Forces308.4: Non-conservative Forces308.5: Conservation of Energy308.6: Conservation of Energy: Application308.7: Force and Potential Energy in One Dimension308.8: Force and Potential Energy in Three Dimensions308.9: Energy Diagrams - I308.10: Energy Diagrams - II ## Chapter 9: Linear Momentum, Impulse and Collisions 309.1: Linear Momentum309.2: Force and Momentum309.3: Impulse309.4: Impulse-Momentum Theorem309.5: Conservation of Momentum: Introduction309.6: Conservation of Momentum: Problem Solving309.7: Types Of Collisions - I309.8: Types of Collisions - II309.9: Elastic Collisions: Introduction309.10: Elastic Collisions: Case Study309.11: Collisions in Multiple Dimensions: Introduction309.12: Collisions in Multiple Dimensions: Problem Solving309.13: Center of Mass: Introduction309.14: Significance of Center of Mass309.15: Rocket Propulsion in Empty Space - I309.16: Rocket Propulsion In Empty Space - II309.17: Rocket Propulsion in Gravitational Field - I309.18: Rocket Propulsion in Gravitational Field - II ## Chapter 10: Rotation and Rigid Bodies 3010.1: Angular Velocity and Displacement3010.2: Angular Velocity and Acceleration3010.3: Rotation with Constant Angular Acceleration - I3010.4: Rotation with Constant Angular Acceleration - II3010.5: Relating Angular And Linear Quantities - I3010.6: Relating Angular And Linear Quantities - II3010.7: Moment of Inertia3010.8: Moment of Inertia and Rotational Kinetic Energy3010.9: Moment of Inertia: Calculations3010.10: Parallel-axis Theorem3010.11: Moment of Inertia of Compound Objects ## Chapter 11: Dynamics of Rotational Motions 3011.1: Torque3011.2: Net Torque Calculations3011.3: Work and Power for Rotational Motion3011.4: Work-Energy Theorem for Rotational Motion3011.5: Angular Momentum: Single Particle3011.6: Angular Momentum: Rigid Body3011.7: Conservation of Angular Momentum3011.8: Conservation of Angular Momentum: Application3011.9: Gyroscope3011.10: Gyroscope: Precession ## Chapter 12: Equilibrium and Elasticity 3012.1: Static Equilibrium - I3012.2: Static Equilibrium - II3012.3: Center of Gravity3012.4: Finding the Center of Gravity3012.5: Rigid Body Equilibrium Problems - I3012.6: Rigid Body Equilibrium Problems - II3012.7: Stress3012.8: Strain and Elastic Modulus3012.9: Elasticity3012.10: Plasticity ## Chapter 13: Fluid Mechanics 3013.1: Density3013.2: Pressure of Fluids3013.3: Pascal's Law3013.4: Application of Pascal's Law3013.5: Buoyancy3013.6: Archimedes' Principle3013.7: Density and Archimedes' Principle3013.8: Laminar and Turbulent Flow3013.9: Equation of Continuity3013.10: Bernoulli's Equation3013.11: Bernoulli's Principle3013.12: Viscosity3013.13: Poiseuille's Law and Reynolds Number ## Chapter 14: Gravitation 3014.1: Newton's Law of Gravitation3014.2: Gravity between Spherical Bodies3014.3: Acceleration due to Gravity on Earth3014.4: Acceleration due to Gravity on Other Planets3014.5: Apparent Weight and the Earth's Rotation3014.6: Variation in Acceleration due to Gravity near the Earth's Surface3014.7: Potential Energy due to Gravitation3014.8: Escape Velocity3014.9: Circular Orbits and Critical Velocity for Satellites3014.10: Energy of a Satellite in a Circular Orbit3014.11: Kepler's First Law of Planetary Motion3014.12: Kepler's Second Law of Planetary Motion3014.13: Kepler's Third Law of Planetary Motion3014.14: Tidal Forces3014.15: Schwarzschild Radius and Event Horizon3014.16: Detection of Black Holes3014.17: Principle of Equivalence3014.18: Space-Time Curvature and the General Theory of Relativity ## Chapter 15: Oscillations 3015.1: Simple Harmonic Motion3015.2: Characteristics of Simple Harmonic Motion3015.3: Oscillations about an Equilibrium Position3015.4: Energy in Simple Harmonic Motion3015.5: Simple Harmonic Motion and Uniform Circular Motion3015.6: Simple Pendulum3015.7: Torsional Pendulum3015.8: Damped Oscillations3015.9: Types of Damping3015.10: Forced Oscillations3015.11: Concept of Resonance and its Characteristics ## Chapter 16: Waves 3016.1: Travelling Waves3016.2: Wave Parameters3016.3: Equations of Wave Motion3016.4: Velocity and Acceleration of a Wave3016.5: Kinetic and Potential Energy of a Wave3016.6: Energy and Power of a Wave3016.7: Interference and Superposition of Waves3016.8: Reflection of Waves3016.9: Propagation of Waves3016.10: Standing Waves3016.11: Modes of Standing Waves - I ## Chapter 17: Sound 3017.1: Sound Waves3017.2: Perception of Sound Waves3017.3: Speed of Sound in Solids and Liquids3017.4: Speed of Sound in Gases3017.5: Sound Intensity3017.6: Sound Intensity Level3017.7: Sound Waves: Interference3017.8: Sound Waves: Resonance3017.9: Doppler Effect - I3017.10: Doppler Effect - II3017.11: Shock Waves ## Chapter 18: Temperature and Heat 3018.1: Temperature and Thermal Equilibrium3018.2: Zeroth Law of Thermodynamics3018.3: Thermometers and Temperature Scales3018.4: Gas Thermometers and the Kelvin Scale3018.5: Thermal Expansion3018.6: Thermal Stress3018.7: Heat Flow and Specific Heat3018.8: Phase Changes3018.9: Mechanisms of Heat Transfer I3018.10: Mechanisms of Heat Transfer II ## Chapter 19: The Kinetic Theory of Gases 3019.1: Ideal Gas Equation3019.2: Van der Waals Equation3019.3: pV-Diagrams3019.4: Kinetic Theory of an Ideal Gas3019.5: Molecular Kinetic Energy3019.6: Distribution of Molecular Speeds3019.7: Phase Diagram ## Chapter 20: The First Law of Thermodynamics 3020.1: Thermodynamic Systems3020.2: Work Done During Volume Change3020.3: Path Between Thermodynamics States3020.4: Internal Energy3020.5: First Law of Thermodynamics3020.6: Cyclic Processes And Isolated Systems3020.7: Isothermal Processes3020.8: Isochoric and Isobaric Processes3020.9: Heat Capacities of an Ideal Gas I3020.10: Heat Capacities of an Ideal Gas II3020.11: Heat Capacities of an Ideal Gas III3020.12: Adiabatic Processes for an Ideal Gas3020.13: Pressure and Volume in an Adiabatic Process3020.14: Work Done in an Adiabatic Process ## Chapter 21: The Second Law of Thermodynamics 3021.1: Reversible and Irreversible Processes3021.2: Heat Engines3021.3: Refrigerators and Heat Pumps3021.4: Statements of the Second Law of Thermodynamics3021.5: The Carnot Cycle3021.6: Efficiency of The Carnot Cycle3021.7: The Carnot Cycle and the Second Law of Thermodynamics3021.8: Entropy3021.9: Entropy Change in Reversible Processes3021.10: Entropy and the Second Law of Thermodynamics ## Chapter 22: Electric Charges and Fields 3022.1: Electric Charges3022.2: Sources and Properties of Electric Charge3022.3: Conductors and Insulators3022.4: Charging Conductors By Induction3022.5: Coulomb's Law3022.6: Coulomb's Law and The Principle of Superposition3022.7: Electric Field3022.8: Electric Field of Two Equal and Opposite Charges3022.9: Continuous Charge Distributions3022.10: Electric Field Lines3022.11: Properties of Electric Field Lines3022.12: Electric Dipoles and Dipole Moment3022.13: Induced Electric Dipoles ## Chapter 23: Gauss's Law 3023.1: Electric Flux3023.2: Gauss's Law3023.3: Gauss's Law: Spherical Symmetry3023.4: Gauss's Law: Cylindrical Symmetry3023.5: Gauss's Law: Planar Symmetry3023.6: Electric Field Inside a Conductor3023.7: Charge on a Conductor3023.8: Electric Field at the Surface of a Conductor ## Chapter 24: Electric Potential 3024.1: Electric Potential Energy3024.2: Electric Potential Energy in a Uniform Electric Field3024.3: Electric Potential Energy of Two Point Charges3024.4: Electric Potential and Potential Difference3024.5: Finding Electric Potential From Electric Field3024.6: Equipotential Surfaces and Field Lines3024.7: Equipotential Surfaces and Conductors3024.8: Determining Electric Field From Electric Potential ## Chapter 25: Capacitance 3025.1: Capacitors and Capacitance3025.2: Spherical and Cylindrical Capacitor3025.3: Capacitors in Series and Parallel3025.4: Energy Stored in a Capacitor3025.5: Capacitor With A Dielectric3025.6: Dielectric Polarization in a Capacitor ## Chapter 26: Current and Resistance 3026.1: Electrical Current3026.2: Drift Velocity3026.3: Current Density3026.4: Resistivity3026.5: Resistance3026.6: Ohm's Law3026.7: Electrical Power ## Chapter 27: Direct-Current Circuits 3027.1: Electromotive Force3027.2: Resistors In Series3027.3: Resistors In Parallel3027.4: Kirchhoff's Rules3027.5: Galvanometer3027.6: Ammeter3027.7: RC Circuits: Charging A Capacitor3027.8: RC Circuits: Discharging A Capacitor ## Chapter 28: Magnetic Forces and Fields 3028.1: Magnetism3028.2: Magnetic Fields3028.3: Magnetic Field Lines3028.4: Magnetic Flux3028.5: Motion Of A Charged Particle In A Magnetic Field3028.6: Magnetic Force On A Current-Carrying Conductor3028.7: Force On A Current Loop In A Magnetic Field3028.8: Torque On A Current Loop In A Magnetic Field ## Chapter 29: Sources of Magnetic Fields 3029.1: Magnetic Field due to Moving Charges3029.2: Biot-Savart Law3029.3: Magnetic Field Due To A Thin Straight Wire3029.4: Magnetic Force Between Two Parallel Currents3029.5: Magnetic Field Of A Current Loop3029.6: Ampere's Law3029.7: Ampere's Law: Problem-Solving3029.8: Solenoids3029.9: Magnetic Field of a Solenoid3029.10: Toroids3029.11: Diamagnetism3029.12: Paramagnetism3029.13: Ferromagnetism ## Chapter 30: Electromagnetic Induction 3030.1: Induction3030.2: Faraday's Law3030.3: Lenz's Law3030.4: Motional Emf3030.5: Induced Electric Fields3030.6: Displacement Current3030.7: Significance of Displacement Current3030.8: Electromagnetic Fields3030.9: Maxwell's Equation Of Electromagnetism3030.10: Symmetry in Maxwell's Equations3030.11: Electric Generator: Alternator3030.12: Back EMF ## Chapter 31: Inductance 3031.1: Mutual Inductance3031.2: Self-Inductance3031.3: Inductors3031.4: Energy In A Magnetic Field3031.5: RL Circuits3031.6: Current Growth And Decay In RL Circuits3031.7: LC Circuits3031.8: Oscillations In An LC Circuit3031.9: RLC Series Circuits ## Chapter 32: Alternating-Current Circuits 3032.1: AC Sources3032.2: Resistor in an AC Circuit3032.3: Capacitor in an AC Circuit3032.4: Inductor in an AC Circuit3032.5: RLC Series Circuits: Introduction3032.6: RLC Series Circuits: Impedance3032.7: Power in an AC Circuit3032.8: Resonance in an AC Circuit ## Chapter 33: Electromagnetic Waves 3033.1: Electromagnetic Waves3033.2: The Electromagnetic Spectrum3033.3: Plane Electromagnetic Waves I3033.4: Plane Electromagnetic Waves II3033.5: Propagation Speed of Electromagnetic Waves3033.6: Electromagnetic Waves in Matter3033.7: Energy Carried By Electromagnetic Waves3033.8: Intensity Of Electromagnetic Waves3033.9: Momentum And Radiation Pressure Full Table of Contents
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Thermometers and Temperature Scales
### 18.3: Thermometers and Temperature Scales
Any physical property that depends consistently and reproducibly on temperature can be used as the basis of a thermometer. For example, volume increases with temperature for most substances. This property is the basis for the common alcohol thermometer and the original mercury thermometers. Other properties used to measure temperature include electrical resistance, color, and the emission of infrared radiation.
As many physical properties depend on temperature, the variety of thermometers is remarkable. In the most common type of thermometer, the alcohol, containing a red dye, expands more rapidly than the glass encasing it. When the thermometer’s temperature increases, the liquid from the bulb is forced into a narrow tube, producing a large change in the length of the column for a small change in temperature. Another common type of thermometer is the pyrometer. A pyrometer measures the infrared radiation (whose emission varies with temperature) from the source and quickly produces a temperature readout. Infrared thermometers are also frequently used to measure body temperature by gently placing them in the ear canal. Such thermometers are more accurate than alcohol thermometers placed under the tongue or under the armpit.
Thermometers measure temperature according to well-defined scales of measurement. The three most common temperature scales are Fahrenheit, Celsius, and Kelvin. Temperature scales are created by identifying two reproducible temperatures, most commonly the freezing and boiling temperatures of water at standard atmospheric pressure. | 4,219 | 15,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-14 | longest | en | 0.674031 |
https://www.coursehero.com/file/6633548/College-Algebra-Exam-Review-284/ | 1,493,469,209,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123491.68/warc/CC-MAIN-20170423031203-00612-ip-10-145-167-34.ec2.internal.warc.gz | 888,542,932 | 23,312 | College Algebra Exam Review 284
# College Algebra Exam Review 284 - 294 6.4.12. (a) (b) 6....
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Unformatted text preview: 294 6.4.12. (a) (b) 6. RINGS Show that a 7! Œa=1 is an injective unital ring homomorphism of R into Q.R/. In this sense Q.R/ is a field containing R. If F is a field such that F Ã R, show that there is an injective unital homomorphism ' W Q.R/ ! F such that '.Œa=1/ D a for a 2 R. 6.4.13. If R is the ring of Gaussian integers, show that Q.R/ is isomorphic to the subfield of C consisting of complex numbers with rational real and imaginary parts. 6.4.14. Let J be an ideal in a ring R. Show that J is prime if, and only if, R=J has no nontrivial zero divisors. (In particular, if R is commutative with identity, then J is prime if, and only if, R=J is an integral domain.) 6.4.15. Every ideal in Z has the form d Z for some d > 0. Show that the ideal d Z is prime if, and only if, d is prime. 6.4.16. Show that a maximal ideal in a commutative ring with identity is prime. 6.5. Euclidean Domains, Principal Ideal Domains, and Unique Factorization We have seen two examples of integral domains with a good theory of factorization, the ring of integers Z and the ring of polynomials KŒx over a field K . In both of these rings R, every nonzero, noninvertible element has an essentially unique factorization into irreducible factors. The common feature of these rings, which was used to establish unique factorization is a Euclidean function d W R n f0g ! N [ f0g with the property that d.fg/ maxfd.f /; d.g/g and for each f; g 2 R n f0g there exist q; r 2 R such that f D qg C r and r D 0 or d.r/ < d.g/. For the integers, the Euclidean function is d.n/ D jnj. For the polynomials, the function is d.f / D deg.f /. Definition 6.5.1. Call an integral domain R a Euclidean domain if it admits a Euclidean function. Let us consider one more example of a Euclidean domain, in order to justify having made the definition. Example 6.5.2. Let ZŒi be the ring of Gaussian integers, namely, the complex numbers with integer real and imaginary parts. For the ”degree” map d take d.z/ D jz j2 . We have d.zw/ D d.z/d.w/. The ...
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## This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
Ask a homework question - tutors are online | 733 | 2,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-17 | longest | en | 0.86743 |
https://github-wiki-see.page/m/AtlasOfLivingAustralia/ala-datamob/wiki/DataClassesCoordinateConversion | 1,638,176,651,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358702.43/warc/CC-MAIN-20211129074202-20211129104202-00145.warc.gz | 343,592,709 | 5,509 | # Introduction
2013/14 - i rescued this from the old sites portal, which is behind a closed door - maybe one day i'll revise it to read a little easier
This page sits within the broader context of spatial data, and applies to converting or obfuscating (denaturing / blurring) coordinate locations.
### For a primer on some important fundamentals of coordinate location data, see:
ALAhandlingspatiallocationdata.pdf
# Conversion
Depending on how data are presented, some (mathematical) conversion may be called for:
1. Darwincore would like decimal degrees.
2. ...
3. ...
4. ...
## Regarding co-ordinates in degrees ° (arc)minutes ' (arc)seconds " (DMS) vs. decimal degrees (DD)
these are two ways of displaying the same information - geographical (spatial) co-ordinates.
there is a mathematical relationship between the two the algorithm is as follows:
`````` decimal degrees = degrees + (minutes / 60) + (seconds / 3600)
``````
some considerations when performing this conversion:
1. be wary of floating point arithmetic and ambiguities: calculate once and store the value in a string to avoid inconsistent references to the same value; more info: http://www.validlab.com/goldberg/paper.pdf
2. if your smallest unit is minutes, truncate DD to three decimal places (hundreds of metres) to avoiding giving a false sense of precision; if seconds, truncate DD to five decimal places (metres)
## For conversion between geoco
There are a further two kinds of conversions to make on coordinate data:
1. converting between grid-coord's (cartesian - zone, eastings, northings) and geographic coord's (latitude/longitude pairs)
2. converting between one datum (eg, agd66) and another (currently gda94)
### Table of rough conversions
Measure Degrees Decimal degrees Metres (lat/long) (roughly, equator) Metres (lat) (roughly, poles) Metres (long) (roughly, poles)
One 1.00° 111,000 111,000 85,000
1' (arcmin) 1/60° 0.01666…° 1,700 1,700 1,300
1" (arcsec) 1/3600° 0.0002777…° 30 30 22
0.01° 1/100° 0.01° 1,000 1,000 760
0.001° 1/1000° 0.001° 100 100 76
Comparison between the different types of measurement - something to note: degrees of latitude are parallel so the distance between each degree remains almost constant, but since degrees of longitude are farthest apart at the equator and converge at the poles, their distance varies greatly.
# Obfuscation
Despite being a contentious topic there will always remain reasons for people to prefer recording a less precise location, especially as the tools for a more precise one become ever-nearer ubiquitous.
This section covers two methods for achieving this goal, truncating, and blurring.
Truncating is (potentially) a more robust method for denaturing/blurring locations, as its results are consistent.
If a randomisation is applied, the output result for a given input should be consistent, otherwise potential exists for statistical analysis of the result.
## Truncating
A simplified method of denaturing locations can be truncating the number of digits after the decimal point; here's a visual basic algorithm that will truncate lat/long's to 3 decimal places (~100m) -
`(fix(LAT_OR_LONG * 1000) / 1000)`
(note '^' is a placeholder for previous result)
1. `first, multiply the latitude by 1000 (FIELD * 1000)`
2. `next, discard anything decimal fix( ^ )`
3. `finally, return to 3 decimal places ( ^ ) / 1000`
in visual-basic's case, the 'fix' function converts a decimal to a whole number, and in the case of negatives, rounds upward toward 0; (this is in contrast to the 'int' function which does the same but rounds downward away from 0)
this method would also work in excel, but you would need to use the 'trunc' function (where a1 is the latitude):
`=( trunc( (a1*1000) ) / 1000 )`
these will wind up looking like a grid of dots on the map - here is an interesting example of this behaviour... note: if you click on the picture to load this dataset in the spatial portal, you'll need to zoom in on an area of interest
All biocache records with: 'State conservation = endangered' AND 'Sensitive = alreadyGeneralised'
## Blurring
For psuedo-code to convert + blur: GeospatialConversionPseudoCode.pdf - note: potentially flawed method of blurring referred to in this pdf link:
1. if a large number of sensitive records come from one point and are given a different (blurred) location each,
2. then statistically, the distribution of blurred locations around the point is of a smaller than expected area -
3. this could make it easier to determine the actual point shared by all records
A more complex, better distributed method of choosing random spots on a disc - http://mathworld.wolfram.com/DiskPointPicking.html | 1,147 | 4,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-49 | latest | en | 0.80086 |
https://in.mathworks.com/matlabcentral/cody/problems/1971-remove-element-s-from-cell-array/solutions/2991037 | 1,604,090,514,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911229.96/warc/CC-MAIN-20201030182757-20201030212757-00602.warc.gz | 357,042,196 | 17,078 | Cody
# Problem 1971. Remove element(s) from cell array
Solution 2991037
Submitted on 25 Sep 2020 by ivan fernandez
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = num2cell(1:10); to_remove = 3; y_correct = {1 2 4 5 6 7 8 9 10}; assert(isequal(remove_from_cell_array(x,to_remove),y_correct))
j = 0 x = 1×9 cell array {[1]} {[2]} {[4]} {[5]} {[6]} {[7]} {[8]} {[9]} {[10]} j = 1 y = 1×9 cell array {[1]} {[2]} {[4]} {[5]} {[6]} {[7]} {[8]} {[9]} {[10]}
2 Pass
x = num2cell(1:5)'; to_remove = 1:2:5; y_correct = {2 4}'; assert(isequal(remove_from_cell_array(x,to_remove),y_correct))
j = 0 x = 4×1 cell array {[2]} {[3]} {[4]} {[5]} j = 1 x = 3×1 cell array {[2]} {[4]} {[5]} j = 2 x = 2×1 cell array {[2]} {[4]} j = 3 y = 2×1 cell array {[2]} {[4]}
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Start Hunting! | 362 | 1,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.387153 |
http://www.mathworks.com/matlabcentral/cody/problems/1520-frobenius-norm/solutions/265585 | 1,484,755,635,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00406-ip-10-171-10-70.ec2.internal.warc.gz | 559,923,969 | 11,629 | Cody
Problem 1520. Frobenius Norm
Solution 265585
Submitted on 22 Jun 2013 by @bmtran
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
%% test 1 x = magic(3); y_correct = norm(x,'fro'); %16.8819 assert(isequal(myFrobeniusnorm(x),y_correct))
``` ```
2 Pass
%% test 2 x = magic(4) y_correct = norm(x,'fro'); % 38.6782; assert(isequal(myFrobeniusnorm(x),y_correct))
``` x = 16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1 ``` | 192 | 536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | latest | en | 0.548888 |
https://www.gamedev.net/forums/topic/380802-terrain-collision-response/ | 1,506,462,164,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696696.79/warc/CC-MAIN-20170926212817-20170926232817-00230.warc.gz | 782,114,646 | 25,977 | # Terrain Collision Response
## Recommended Posts
Raeldor 254
Hi All, I am trying to get to grips with how terrain integrates into a physics engine. Assuming firstly that you have a flat terrain, I understand that the calculation of displacement via velocity is from the sum of forces acting on the object. So, forward momentum+gravity. Would you include the upward force of terrain as one of those forces? That seems to indicate a special exception since you don't really know if the object is 'resting' on the terrain if you are treating the terrain as just another object. It seems more likely you will fall through the terrain and then have to calculate some kind of collision response (deflection). Depending on your momentum, doesn't this result in a bouncing up and down motion as you are constantly being reflected off the terrain? Thanks
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Guest Anonymous Poster
Objects don't have to intersect for a detectable collision. The algorithm has to check the terrain and move the object in a way that it only touches the terrain without intersections. The same algorithm can be used for mobile objects to avoid rubberband effects on collisions.
For example, two spheres intersect if the distance between the two centers is smaller than the sum of their radiuses. They touch exactly when the distance equals to the sum. For bounce less movement the collision engine has to keep objects exactly in the touching distance.
Viktor
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Raeldor 254
Quote:
Original post by Anonymous PosterObjects don't have to intersect for a detectable collision. The algorithm has to check the terrain and move the object in a way that it only touches the terrain without intersections. The same algorithm can be used for mobile objects to avoid rubberband effects on collisions.For example, two spheres intersect if the distance between the two centers is smaller than the sum of their radiuses. They touch exactly when the distance equals to the sum. For bounce less movement the collision engine has to keep objects exactly in the touching distance.Viktor
Thanks for the reply. How does this equate to the laws of conservation of energy? Where does the energy from the downward force of gravity on the object against the terrain go to?
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Guest Anonymous Poster
Quote:
Original post by Raeldor
Quote:
Original post by Anonymous PosterObjects don't have to intersect for a detectable collision. The algorithm has to check the terrain and move the object in a way that it only touches the terrain without intersections. The same algorithm can be used for mobile objects to avoid rubberband effects on collisions.For example, two spheres intersect if the distance between the two centers is smaller than the sum of their radiuses. They touch exactly when the distance equals to the sum. For bounce less movement the collision engine has to keep objects exactly in the touching distance.Viktor
Thanks for the reply. How does this equate to the laws of conservation of energy? Where does the energy from the downward force of gravity on the object against the terrain go to?
The force of the gravity that pulls an object down should be equal to the force of the terrain that is 'pushing' the object back. This is why one feels it's own weight on his feet. If this force from the terrain that comes from it's material state (hardness) would be less than the force of gravity one would go into the terrain by pushing it downward and sideways, like it happens with less solid (fluid) materials like water.
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BinaryStylus 133
I am working with this problem myself. I have tried a couple of approaches - but am mostly concerned with collision detection against the terain. My collision system at the moment uses spheres OBBs planes and rays - and it is hard to include "terrain" (and i suppose I mean heightmaps here) in a solid body collision system.
The best I have come up with is to implement triangles as another element that can interact with the above objects and then do a broad phase collision check with "OBBs / AABBs" of terrain and the narrow phase checks with each triangle in that block of terrain.
As for physics! Just assume that the terrain's mass is infinite! and redo a special case for your impulse algorithms.
---Physics---
A way I have found to get an object to come to "rest" on the ground under constant acceleration (Force) is as follows.
Give your object the velocity it would acheive due to the acceleration at the end of the time frame.
Have an algorithm which tells you the distance/time to impact.
Then work out the velocity due to acceleration at that distance/time.
Then do the physics response using the above velocity - This velocity should be tiny as the object was almost touching before.
Have a THETA value that will alow you to ignore velocities smaller than it.
So if they were almost at rest nothing happens.
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NotAYakk 876
Quote:
Original post by RaeldorThanks for the reply. How does this equate to the laws of conservation of energy? Where does the energy from the downward force of gravity on the object against the terrain go to?
The application of force without movement involves no kinetic energy.
Really. =)
And force in a direction perpendicular to movement ... also involves no kinetic energy.
It is only when the force causes movement over a distance that kinetic energy is involved.
Going up against gravity consumes kinetic energy. Going down with gravity produces kinetic energy. Standing still, pushing against a solid immobile surface, neither produces nor destroys kinetic energy.
The relevent equation:
E = F*D
Energy = Force * Distance
where "*" in this case is the dot-product
alternatively:
Energy = |Force||Distance|sin(angle between the force and the distance travelled)
That help?
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Homework 6 Solution
Explain then write the interrupt handlers you code
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Unformatted text preview: SART UBRRnH = _______; UBRRnL = _______; UCSRnA = _______; UCSRnC = _______; UCSRnB = _______; Name: Lab Section: void USART_init() { const unsigned baud_setting = (16000000L/(16*115200)-1); //8 or 16 if you use double speed // Set baud rate for USART1 UBRR1H = baud_setting >> 8; //2 for double speed UBRR1L = (unsigned char) baud_setting; // Set UCSR1A UCSR1C UCSR1B USART1 control bits = 0; = (3<<UCSZ0);//0b00000110=0x06 = (1<<TXEN);//0b00001000=0x08 // Set baud rate for USART0 UBRR0H = baud_setting >> 8; UBRR0L = (unsigned char) baud_setting; // Set UCSR0A UCSR0C UCSR0B USART0 control bits = 0; = (3<<UCSZ0);//0b00000110=0x06 = (1<<RXEN) | (1<<RXCIE);//0b10010000=0x90 } b. [10] What interrupt handler(s) do you have to write to implement the reply functionality? Explain. Then, write the interrupt handler(s). You code should check for communication overflow: if USART0 receives a character but USART1 is not ready for transm...
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http://sciencedocbox.com/Physics/65581112-Distance-travelled-time-taken-and-if-the-particle-is-a-distance-s-t-along-the-x-axis-then-its-instantaneous-speed-is.html | 1,547,774,395,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659654.11/warc/CC-MAIN-20190118005216-20190118031216-00381.warc.gz | 207,519,527 | 27,539 | # Distance travelled time taken and if the particle is a distance s(t) along the x-axis, then its instantaneous speed is:
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## Transcription
1 Chapter 1 Kinematics 1.1 Basic ideas r(t) is the position of a particle; r = r is the distance to the origin. If r = x i + y j + z k = (x, y, z), then r = r = x 2 + y 2 + z 2. v(t) is the velocity; v = v is the speed. a(t) is the acceleration of the particle. In one-dimensional motion, we have Average speed during a time interval = Distance travelled, time taken and if the particle is a distance s(t) along the x-axis, then its instantaneous speed is: In more than one dimension, and the instantaneous velocity is v = ds dt = ṡ. r(t) = x(t) i + y(t) j + z(t) k = (x, y, z), v = dr dt A particle is at rest if v = 0. = ṙ = ẋ i + ẏ j + ż k = (ẋ, ẏ, ż). (1) 1.2 Acceleration The acceleration of a particle in one or more dimension is: a = dv dt = v = d2 r = r = (ẍ, ÿ, z). (2) dt2 1
2 2 1.3 Relative velocity In one dimension, let the position of a particle be r = s(t) i, with speed v = ṡ and acceleration a = v = s. Then a = dv dt = dv ds ds dt = v dv ds = d ( ) 1 ds 2 v2. (3) If we assume that the acceleration is uniform (a constant), and that at time t = 0, the particle is at position s = 0 and is travelling with speed u, then its speed at time t is: v = u + at, (4) its position at time t is s = ut at2, (5) and we can eliminate t from the two equations above to get: v 2 = u 2 + 2as. (6) With non-uniform acceleration in one or more dimensions, use (1) and (2) to calculate v and a from r (or a from v) by differentiation, and integrate to find r from v, or r and v from a. 1.3 Relative velocity The position of a particle P relative to a particle Q is r P Q = r P r Q. (7) Differentiating, we get the relative velocity v P Q = v P v Q, (8) and the relative acceleration a P Q = a P a Q. (9) 1.4 Angular speed Let θ be the angle that the line OP makes with a fixed reference line OA, so θ = POA, where O and A are fixed and P is moving. Then the instantaneous angular speed of P is ω = dθ dt = θ. (10) If the particle is moving in a circle of radius R, then the instantaneous speed of the particle is v = Rω. (11)
3 Chapter 2 Dynamics: Newton s laws of motion and gravitation 2.1 Newton s laws of motion A particle is a point mass, having no internal structure. We will not consider rigid body dynamics, where rotation of the object would have to be included (e.g., the spin on a tennis ball). All objects that we consider (stones, balls, people, trains, cars, boats etc) will be treated as particles. I: Newton s First Law of Motion: A particle moves at a constant velocity (perhaps zero) unless it is acted up on by a force. The momentum of a particle is its mass times its velocity: p = mv (12) II: Newton s Second Law of Motion: The rate of change of momentum of a particle is equal to the total force acting on the particle: F = dp dt = d dm (mv) = dt dt v + mdv = ṁv + m v. dt The total force F is the vector sum of all forces acting on the particle. If the mass of the particle is constant (ṁ = 0), then F = m dv dt = m v = ma (13) III: Newton s Third Law of Motion: If a particle A exerts a force F on particle B, then particle B exerts an equal and opposite force F on particle A. One consequence of Newton s Third Law of Motion is the Principle of Conservation of Momentum: the total momentum of two particles interacting though equal and opposite forces is constant. 3
4 4 2.2 Newton s Law of Gravitation 2.2 Newton s Law of Gravitation Two particles of masses M and m a distance r apart exert a mutually attractive force on each other: Gravitational Force = GMm r 2 (14) where G is the Universal Gravitational Constant G = m 3 /kg/s 2 in S.I. units. For an object close to the surface of a planet of mass M and radius R, the weight of the object is a force directed towards the centre of the planet, of magnitude Weight = GM GM m = mg where g = (15) R2 R 2
5 Chapter 3 Projectiles, lifts, ropes and pulleys, friction, collisions, circular motion 3.1 Projectiles A projectile starting from the origin at time t = 0 and with an initial speed U 0, launched with an angle α above the horizontal, moves under the influence of gravity: a = g j, where the x-direction is horizontal and the y-direction is vertically upwards, and g is the gravitational acceleration. We integrate to get the velocity: v = U 0 cos α i + (U 0 sin α gt) j, (using the given initial velocity), and integrate again to get the position: r = x i + y j = (U 0 cos α) t i + ((U 0 sin α) t 1 ) 2 gt2 j, (using the given initial position). Putting this together, let x(t) be the horizontal position, let y(t) be the vertical position, and let v = ẏ be the vertical component of the velocity (note that v is not the magnitude of v in this case), we have: x = (U 0 cos α) t (16) v = U 0 sin α gt (17) y = (U 0 sin α) t 1 2 gt2 (18) v 2 = (U 0 sin α) 2 2gy (19) Time t can be eliminated by writing t in terms of x using (16) and then substituting into (18). With a different initial velocity of postion, the expressions may be different, but the idea is the same. The range can be found by solving y = 0 (or whatever condition specifies when the particle comes back to the ground. 5
6 6 3.2 Lifts 3.2 Lifts If a lift (elevator) accelerates upwards with an acceleration a, then the apparent weight of an object of mass m in the lift is m(g + a). 3.3 Ropes and pulleys We will treat only massless inextensible ropes and strings, and the tension in a rope is the force that rope applies to the particles that are connected to either end of the rope. We consider only massless frictionless pulleys, so the tension in the rope on either side of the pulley is the same. Ropes can only pull particles, not push. If a rope connects two particles, then the speeds and accelerations of the two particles are the same (thought the directions will be different). 3.4 Friction If an object of mass m is moving across a rough horizontal surface, then there is a vertical normal force N that balances the object s weight mg, and there is an opposing frictional resistance R: R = µn, (20) where µ is called the coefficient of friction. If the object is not moving, but some horizontal forces are being applied, there may still be an opposing frictional force R < µn. 3.5 Impulse and collisions Newton s Second Law (II) is: F = m dv dt If we integrate this from time 0 to t, we get: t 0 F (t) dt = t 0 = m v = ma; m dv dt = mv(t) mv(0) dt In a collision, where the force is non-zero only for a very short time, from 0 to t, the Impulse is defined to be: I = t 0 F (t) dt = mv 1 mv 0, (21) where v 0 is the velocity before the collision, and v 1 is the velocity after the collision. Thus, the Impulse acting on a particle in a collision is equal to the change in the particle s momentum during the collision. The units of Impulse are N s (Newton seconds), and it is a vector.
7 Chapter 3 Projectiles, lifts, ropes and pulleys, friction, collisions, circular motion Inelastic collisions Suppose two particles collide and stick together. Then the total momentum before the collision is equal to the total momentum after the collision. This is called Conservation of momentum. In the case of one-dimensional motion, consider two particles of masses m A and m B moving to the right with speeds u A and u B respectively. If one of the particles is moving to the left, take its speed to be negative. The total momentum before any collision is therefore m A u A + m B u B. The particles stick together in an inelastic collision, then the new particle has mass m A + m B, and travels to the right with speed v, and so has total momentum (m A + m B )v. Equating these, we find v: Elastic collisions v = m Au A + m B u B m A + m B (22) If two particles collide and then bounce apart, then the total momentum is conserved: the total momentum before the collision is equal to the total momentum after the collision. In the case of one-dimensional motion, consider two particles of masses m A and m B moving to the right with speeds u A and u B respectively. If one of the particles is moving to the left, take its speed to be negative. After the collision, the particles have speeds v A and v B respectively, where again positive speeds mean that the particle is travelling to the right. The total momentum is the same before and after: m A u A + m B u B = m A v A + m B v B (23) In addition, we have Newton s Elastic Law (NEL), which says that when two particles of the same material collide directly, then their relative velocity after impact is in direct proportion to their relative velocity before impact, but in the opposite direction: NEL: v B v A = e(u B u A ), or v B v A u B u A = e (24) 3.6 Circular motion A particle P whose position vector is r(t) = R cos(ωt) i + R sin(ωt) j = x(t) i + y(t) j is moving in a circle of radius R, and the line OP makes an angle θ with the x-axis, where tan θ = y(t)/x(t) = tan(ωt), so θ = ωt. The particle has angular velocity ω see equation (10). The period of the oscillation is T = 2π/ω The velocity and acceleration of the particle are: v = ṙ = Rω sin(ωt) i+rω cos(ωt) j and a = v = Rω 2 cos(ωt) i Rω 2 sin(ωt) j = ω 2 r.
8 8 3.6 Circular motion The velocity v is tangent to the circle (since r v = 0) and the acceleration a is directed towards the centre of the circle (since a is a negative constant times r). The magnitudes of these vectors are constant: r = r = R, v = v = Rω and a = a = Rω 2 = v2 R. (25) The force required to keep the particle moving in a circle is the Centripetal Force. From Newton s second law, we have F = ma = mω 2 r, and F = F = mrω 2 = mv2 R. (26) This force could be provided from a variety of sources: gravity, friction, a string...
9 Chapter 4 Work, energy and power 4.1 Work and Kinetic Energy A particle moving in one dimension obeys Newton s Second Law (II). Using equation (3), we get: F = ma = m dv dt = mdv ds ds dt dv = mv ds = m d ( ) 1 ds 2 v2. If we integrate F ds from time 0 to t, where the particle moves from 0 to s in this time, we get: s s F ds = m d ( ) ds 2 v2 ds = 1 2 mv mv2 0, (27) where v 0 is the velocity at time 0 and v 1 is the velocity at time t. We define the Kinetic Energy (K.E.) of a particle of mass m, travelling at speed v, to be K.E. = 1 2 mv2 (28) and the work done by a force acting on a particle to be: Work done = F ds (29) so equation (27) can also be written as Work done on a particle = increase in the particle s Kinetic Energy. The units of work and Kinetic Energy are Newton-metres (Nm), or Joules (J). In the case where the force is a constant, we get Constant force: Work done = F s = Force distance, (30) where force and distance are measured in the same sense. 4.2 Gravitational Potential Energy Consider a particle moving under the influence of gravity, so the force acting on it is F = mg (negative because downwards). If the particle falls from h 1 to h 2 (with 9
10 Power h 1 > h 2 ), the work done by gravity is Work done = h2 h 1 mg ds = mg(h 2 h 1 ) = mg(h 1 h 2 ). This is equal to the increase in Kinetic Energy: mg(h 1 h 2 ) = 1 2 mv mv2 1, where v 1 is the initial speed (at height h 1 ) and v 2 is the final speed (at height h 2 ). Rearrange this to give the Principal of conservation of Energy : mgh mv2 1 = mgh mv2 2. (31) The sum of either side of this equation is called the total mechanical energy, and the first part of the sum is defined to be the Gravitational Potential Energy: the Potential Energy (P.E.) of a particle of mass m a distance h above the ground is: P.E. = mgh. (32) Mechanical energy is lost to friction and in collisions, but otherwise the total mechanical energy (Potential Energy + Kinetic Energy) is conserved. 4.3 Power Power is the rate at which work is done. If a force F acting on a particle is constant, and it travels with constant speed v, then the distance it travels in time t is s = vt, the work done is F s, and the power is Constant force and speed: Power = Work done Time taken = F s t = F v = Force speed. (33) The units of power are Watts (W): 1 W = 1 J/s = 1 Nm/s = 1 kg m 2 /s 3.
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More information | 11,154 | 41,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-04 | longest | en | 0.878832 |
http://math4finance.com/general/se-lanza-un-cuerpo-verticalmente-hacia-arriba-con-una-velocidad-de-300-m-s-a-que-velocidad-posee-despues-de-10-s-b-a-que-altura-estara-despues-de-10-segundos | 1,713,677,231,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.0/warc/CC-MAIN-20240421040323-20240421070323-00189.warc.gz | 22,743,429 | 7,383 | Q:
# A body is thrown vertically upward with a velocity of 300 m/s. a) What is its velocity after 10 s? b) How high will it be after 10 seconds?
Accepted Solution
A:
To solve this problem, we need to use the equations of motion for a body thrown vertically upward with a constant acceleration due to gravity. The acceleration due to gravity is usually denoted by g and has a value of approximately 9.8 m/s^2 on Earth. The equations of motion are: v = u - gt $$s = ut - \frac{1}{2}gt^2$$ where v is the final velocity, u is the initial velocity, t is the time, and s is the displacement. In this problem, we have: u = 300 m/s t = 10 s g = 9.8 m/s^2 We can plug these values into the equations of motion to find v and s. (a) To find the final velocity after 10 s, we use the equation: v = u - gt $$v = 300 - 9.8 \times 10$$ v = 202 m/s Therefore, the final velocity after 10 s is 202 m/s. (b) To find the displacement after 10 s, we use the equation: $$s = ut - \frac{1}{2}gt^2$$ $$s = 300 \times 10 - \frac{1}{2} \times 9.8 \times 10^2$$ s = 3000 - 490 s = 2510 m Therefore, the displacement after 10 s is 2510 m. | 347 | 1,115 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-18 | latest | en | 0.88638 |
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The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
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Quizzes Created: 1 | Total Attempts: 41
Questions: 45 | Attempts: 41
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• 1.
• 2.
• 3.
### Find the speed of the boat in still water, if a boat covers a certain distance upstream in 2 hours, while it comes back in 1`1/2` hours. If the speed of the stream be 3 kmph.
• A.
18 kmph
• B.
12 kmph
• C.
24 kmph
• D.
21 kmph
D. 21 kmph
Explanation
Let the speed of the boat in still water be x kmph. The speed of the stream is given as 3 kmph.
When the boat is going upstream, its effective speed is reduced by the speed of the stream, so the boat's speed is (x - 3) kmph.
When the boat is coming back downstream, its effective speed is increased by the speed of the stream, so the boat's speed is (x + 3) kmph.
We are given that the boat covers a certain distance upstream in 2 hours, so the distance is (x - 3) * 2.
Similarly, the distance covered downstream is (x + 3) * 1.5.
Since the distances covered are the same, we can equate the two expressions: (x - 3) * 2 = (x + 3) * 1.5.
Simplifying this equation, we get 2x - 6 = 1.5x + 4.5.
Solving for x, we find x = 21 kmph. Therefore, the speed of the boat in still water is 21 kmph.
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• 4.
### The rate at which a sum becomes four times of itself in 25 years at S.I.,will be:
• A.
30
• B.
29
• C.
12
• D.
50
C. 12
Explanation
The correct answer is 12. This can be calculated using the formula for compound interest: A = P(1 + r/n)^(nt), where A is the final amount, P is the principal amount, r is the rate of interest, n is the number of times interest is compounded per year, and t is the number of years. In this case, we want the sum to become four times itself in 25 years, so we can set up the equation 4P = P(1 + r/n)^(nt). Simplifying this equation, we find that (1 + r/n)^(nt) = 4. Since we are dealing with simple interest, we can assume that n = 1. Plugging in the values, we get (1 + r)^(25) = 4. Solving for r, we find that r = 0.12 or 12%. Therefore, the rate at which the sum becomes four times itself in 25 years is 12%.
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• 5.
### In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:
• A.
5 kmph
• B.
5.25 kmph
• C.
6 kmph
• D.
6.25 kmph
A. 5 kmph
Explanation
Abhay takes 2 hours more than Sameer to cover a distance of 30 km. This means that Abhay's speed is slower than Sameer's. If Abhay doubles his speed, he would take 1 hour less than Sameer, indicating that Abhay's speed is still slower than Sameer's. Therefore, the correct answer is 5 kmph, the lowest speed option given.
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• 6.
### A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
• A.
14 days
• B.
16 days
• C.
17 days
• D.
15 days
D. 15 days
Explanation
A can do the work alone in 20 days. On every third day, A is assisted by B and C. So, on every third day, the combined work done by A, B, and C is equal to the work done by A alone in 20 days. Therefore, in 3 days, the combined work done is equal to the work done by A alone in 20 days. Hence, in 15 days (3 times 5), A will be able to complete the entire work with the assistance of B and C on every third day. Therefore, the correct answer is 15 days.
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• 7.
### The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:
• A.
14
• B.
16
• C.
20
• D.
15
B. 16
Explanation
If the cost price of 20 articles is the same as the selling price of x articles, it means that the profit made is equal to the cost price of x articles. Since the profit is 25%, it can be calculated as 25% of the cost price of x articles. Therefore, the equation can be written as 0.25 * (cost price of x articles) = cost price of x articles. Solving this equation, we find that the value of x is 16.
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• 8.
### In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
• A.
360
• B.
480
• C.
720
• D.
5040
C. 720
Explanation
The word 'LEADING' has 7 letters, including 3 vowels (E, A, I). To find the number of ways the letters can be arranged such that the vowels always come together, we can treat the group of vowels (EAI) as a single entity. This reduces the problem to arranging the 5 entities (L, D, N, G, and the group of vowels) in a linear manner. The number of ways to arrange these entities is 5!, which is equal to 120. However, within the group of vowels, the vowels can be arranged in 3! ways. Therefore, the total number of arrangements is 5! * 3!, which is equal to 720.
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• 9.
### What number comes next 10, 9, 60, 90, 70 and 66?
• A.
96
• B.
99
• C.
16
• D.
77
A. 96
Explanation
The sequence appears to be alternating between decreasing numbers and increasing numbers. The first two numbers, 10 and 9, are decreasing by 1. The next two numbers, 60 and 90, are increasing by 30. The next two numbers, 70 and 66, are decreasing by 4. Based on this pattern, the next set of numbers should be increasing by 30. Therefore, the next number in the sequence is 96.
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• 10.
### 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?
• A.
9
• B.
10
• C.
11
• D.
12
D. 12
Explanation
If 3 pumps working 8 hours a day can empty the tank in 2 days, it means that each pump can empty 1/16th of the tank in 1 hour. To empty the tank in 1 day, we need to increase the number of pumps. Since we have 4 pumps now, each pump needs to work for 1/16th of the tank in 1 hour. Therefore, each pump needs to work for 1/16 * 4 = 1/4th of the tank in 1 hour. Since each pump can empty 1/16th of the tank in 1 hour, it means that each pump needs to work for 4 hours. Therefore, 4 pumps need to work for 4 * 4 = 16 hours in total. Since we need to empty the tank in 1 day, each pump needs to work for 16/24 = 2/3rd of the day. In hours, it is 2/3 * 24 = 16 hours. Therefore, 4 pumps need to work for 16 hours in a day to empty the tank.
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• 11.
### Nine cricket fans are watching a match in a stadium. Seated in one row, they are are J , K , L , M , N , O , P , Q and R. L is at the right of M and at third place at the right of N. K is at one end of of the row. Q is seated adjacent to both O and P. O is at the third place at the left of K. J is right next to left of O. Which of the following statements is TRUE?
• A.
N is two seats away from J.
• B.
M is at extreme end.
• C.
R and P are neighbors.
• D.
There is one person between L and O
D. There is one person between L and O
Explanation
Based on the given information, we can determine the seating arrangement as follows:
K _ _ _ _ _ _ _
J O _ _ _ _ _ _
_ _ _ _ _ _ _ N
_ _ _ _ _ _ _ M
_ _ _ _ _ _ _ L
_ _ _ _ _ _ _ P
_ _ _ _ _ _ _ Q
_ _ _ _ _ _ _ R
From the arrangement, we can see that there is one person between L and O, making the statement "There is one person between L and O" true.
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• 12.
### In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg?
• A.
3:7
• B.
5:7
• C.
7:3
• D.
7:5
C. 7:3
Explanation
To find the ratio in which the two varieties of pulses should be mixed, we can set up a proportion based on their prices per kg. Let the ratio be x:y. The cost of the first variety of pulses is 15x and the cost of the second variety is 20y. The total cost of the mixture is 16.50. Therefore, we can write the equation: (15x + 20y) / (x + y) = 16.50. Simplifying this equation, we get 15x + 20y = 16.50x + 16.50y. Rearranging terms, we get 0.50x = 3.50y. Dividing both sides by 0.50, we get x/y = 7/3. Hence, the correct answer is 7:3.
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• 13.
### Pointing to a photograph of a boy Suresh said, "He is the son of the only son of my mother." How is Suresh related to that boy?
• A.
Brother
• B.
Father
• C.
Cousin
• D.
Uncle
B. Father
Explanation
Suresh is the father of the boy in the photograph. This is because Suresh states that the boy is the son of the only son of his mother, which means that the boy is Suresh's son.
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• 14.
### What is the value of c , If 8 is 4% of a, and 4 is 8% of b. c equals b/a.
• A.
12
• B.
0.25
• C.
0.155
• D.
64
B. 0.25
Explanation
The value of c is 0.25 because c is equal to b/a. Given that 8 is 4% of a and 4 is 8% of b, we can set up the equations 8 = 0.04a and 4 = 0.08b. Solving these equations, we find that a = 200 and b = 50. Therefore, c = b/a = 50/200 = 0.25.
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• 15.
### I forgot the last digit of a 7 digit telephone number. If one randomly dial the final three digits after correctly dialling the four, then what is the chance of dialling the correct number?
• A.
1/1000
• B.
1/1001
• C.
1/900
• D.
1/901
A. 1/1000
Explanation
If the last digit of the telephone number is forgotten, there are 10 possible digits that it could be (0-9). Since only one of these digits is correct, the chance of randomly dialing the correct number is 1 out of 10. However, the question specifically asks for the chance of dialing the correct number after correctly dialing the first four digits. This means that the first four digits are already correct and the only uncertainty lies in the last three digits. Since there are 10 possible digits for each of the three positions, the total number of possible combinations is 10^3 = 1000. Therefore, the chance of dialing the correct number is 1 out of 1000.
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• 16.
### Find the remainder when 6799 is divided by 7.
• A.
4
• B.
6
• C.
2
• D.
1
D. 1
Explanation
When a number is divided by 7, the remainder can be found by subtracting multiples of 7 from the number until the result is less than 7. In this case, starting with 6799, we subtract 7 repeatedly until we get a result less than 7. Subtracting 7 from 6799 gives us 6792, which is still divisible by 7. Continuing this process, we subtract 7 again to get 6785, which is still divisible by 7. Subtracting 7 again gives us 6778, which is still divisible by 7. Subtracting 7 again gives us 6771, which is still divisible by 7. Subtracting 7 again gives us 6764, which is still divisible by 7. Finally, subtracting 7 gives us 6757, which is not divisible by 7. Therefore, the remainder when 6799 is divided by 7 is 1.
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• 17.
### A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to 8 times?
• A.
12
• B.
9
• C.
8
• D.
27
A. 12
Explanation
If the sum of money doubles itself in 4 years, it means that the interest rate is 100% per year. To find out how many years it will take for the sum of money to amount to 8 times, we need to calculate the number of years it takes for the sum of money to increase by a factor of 8. Since the interest rate is 100% per year, it will take 4 years to double the sum of money, and another 4 years to double it again. Therefore, it will take a total of 8 years for the sum of money to amount to 8 times.
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• 18.
### The average temperature for Wednesday, Thursday and Friday was 40oC. The average for Thursday, Friday and Saturday was 41oC. If temperature on Saturday was 42oC, what was the temperature on Wednesday?
• A.
44
• B.
41
• C.
39
• D.
37
C. 39
Explanation
The average temperature for Wednesday, Thursday, and Friday is 40oC. The average temperature for Thursday, Friday, and Saturday is 41oC. Since the temperature on Saturday is 42oC, the total temperature for Thursday, Friday, and Saturday is 123oC. Subtracting the temperature on Saturday from the total gives us 81oC, which is the total temperature for Wednesday and Thursday. Therefore, the temperature on Wednesday must be 39oC.
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• 19.
### Find the H.C.F, if the numbers are in the ratio of 4 : 5 : 6 and their L.C.M. is 2400.
• A.
40
• B.
60
• C.
80
• D.
20
A. 40
Explanation
The H.C.F (Highest Common Factor) is the largest number that divides all the given numbers without leaving any remainder. In this question, the given numbers are in the ratio of 4:5:6. The H.C.F can be found by dividing the L.C.M (Least Common Multiple) by the product of the common prime factors raised to their lowest powers. Since the L.C.M is given as 2400, we can divide it by the product of the common prime factors, which are 2, 2, 2, 2, 3, and 5. Simplifying this, we get 2^4 * 3 * 5 = 40. Therefore, the correct answer is 40.
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• 20.
### P and Q take part in 100 m race. P runs at 6kmph. P gives Q a start of 8 m and still beats him by 8 seconds. The speed of Q is:
• A.
6.14
• B.
15.2
• C.
5.68
• D.
4.86
D. 4.86
Explanation
Since P gives Q a start of 8 m and still beats him by 8 seconds, we can calculate the time taken by P and Q to cover the 100 m distance.
P's speed is given as 6 km/hr, which is equivalent to 6 * (5/18) m/s.
Let's assume Q's speed is x m/s.
So, the time taken by P to cover 100 m is 100 / (6 * (5/18)) = 30 seconds.
Since P gives Q a start of 8 m, Q needs to cover 100 - 8 = 92 m.
The time taken by Q to cover 92 m is 30 + 8 = 38 seconds.
Therefore, Q's speed is 92 / 38 = 2.42 m/s, which is equivalent to 2.42 * (18/5) km/hr = 8.712 km/hr.
Rounding off to two decimal places, Q's speed is 8.71 km/hr, which is approximately equal to 4.86 m/s.
Therefore, the correct answer is 4.86.
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• 21.
### Today is Monday. After 61 days, it will be:
• A.
Wednesday
• B.
Thursday
• C.
Friday
• D.
Saturday
D. Saturday
Explanation
Today is Monday. There are 7 days in a week. After 61 days, we can divide 61 by 7 to find the remainder. 61 divided by 7 equals 8 with a remainder of 5. This means that after 61 days, we will be 5 days ahead of Monday. Counting forward from Monday, the fifth day is Saturday. Therefore, after 61 days, it will be Saturday.
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• 22.
### Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:
• A.
44
• B.
46
• C.
50
• D.
42
C. 50
Explanation
The faster train is overtaking the slower train, so the relative speed between the two trains is the difference in their speeds, which is 46 km/hr - 36 km/hr = 10 km/hr. In 36 seconds, the faster train covers a distance equal to the length of both trains. To find the length of each train, we need to convert the relative speed from km/hr to m/s. 10 km/hr = 10 * (1000/3600) m/s = 100/36 m/s. In 36 seconds, the faster train covers a distance of (100/36) * 36 = 100 meters, which is the combined length of both trains. Therefore, the length of each train is 100/2 = 50 meters.
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• 23.
### To catch a tartar means:
• A.
To trap wanted criminal with great difficulty
• B.
To catch a dangerous person
• C.
To meet with disaster
• D.
To deal with a person who is more than one's match
B. To catch a dangerous person
Explanation
"To catch a tartar" means to catch a dangerous person. The term "tartar" refers to someone who is fierce, formidable, or difficult to deal with. Therefore, the correct answer suggests that catching a tartar implies capturing a person who poses a significant threat or danger.
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• 24.
### SYMpHONY:COMPOSER as
• A.
Leonardo:music
• B.
Fresco:painter
• C.
Colours:pallet
• D.
Art:appreciation
B. Fresco:painter
Explanation
The analogy "SYMPHONY:COMPOSER" implies that a symphony is created by a composer. Similarly, "Fresco:painter" suggests that a fresco is created by a painter. Both analogies involve the relationship between a form of art and the person who creates it. Therefore, the answer "Fresco:painter" is the most appropriate choice.
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• 25.
### S1: When a satellite is launched, the rocket begins by going slowly upwards through the air. P : However, the higher it goes, the less air it meets. Q : As the rocket goes higher, it travels faster. R : For the atmosphere becomes thinner. S : As a result there is less friction. S6: Consequently, the rocket still does not become too hot. The proper sequence should be:
• A.
PQRS
• B.
PQSR
• C.
QSPR
• D.
QPRS
D. QPRS
Explanation
As the rocket goes higher (Q), it travels faster (P) because the atmosphere becomes thinner (R) and there is less friction (S). Consequently, the rocket still does not become too hot (S6). Therefore, the correct sequence is QPRS.
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• 26.
### When it was feared that the serfs might go too far and gain their freedom from serfdom, the protestant leaders joined the princes at crushing them.
• A.
Into crushing
• B.
In crushing
• C.
For crushing
• D.
Without crushing
B. In crushing
Explanation
The correct answer is "in crushing". This preposition is used to show the involvement of the Protestant leaders in the act of crushing the serfs' attempt to gain freedom from serfdom. It implies that the Protestant leaders actively participated in the process of suppressing the serfs' movement.
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• 27.
### The Hubble Space Telescope will search for planets around the stars, a key to the ...... of extraterrestrial life, and will examine interstellar dust and gases out of which stars are born.
• A.
Perception
• B.
Discovery
• C.
Quest
• D.
Enquiry
C. Quest
Explanation
The Hubble Space Telescope's mission to search for planets around stars indicates a purposeful and determined effort to find evidence of extraterrestrial life. The word "quest" accurately captures this sense of exploration and pursuit, making it the correct answer.
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• 28.
• A.
To show how simple, traditional cultures can have complicated grammar structures
• B.
To show how English grammar differs from Cherokee grammar
• C.
To prove that complex grammar structures were invented by the Cherokees.
• D.
To demonstrate how difficult it is to learn the Cherokee language
A. To show how simple, traditional cultures can have complicated grammar structures
Explanation
The writer includes information about the Cherokee language to support the idea that complex grammar structures exist in all languages, including those of so-called "primitive" tribes. This example highlights that even in a simple, traditional culture, such as the Cherokee tribe, there is a sophisticated pronoun system that distinguishes between different meanings. This demonstrates that complex grammar can exist in any language, regardless of its level of development or complexity.
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• 29.
### What can be inferred about the slaves' pidgin language?
• A.
It contained complex grammar.
• B.
It was based on many different languages.
• C.
It was difficult to understand, even among slaves.
• D.
It was created by the land-owners
C. It was difficult to understand, even among slaves.
Explanation
The correct answer suggests that the slaves' pidgin language was challenging to comprehend, even for the slaves themselves. This implies that the language was not easily understandable, possibly due to its unique grammar or vocabulary. It indicates that the pidgin language was not a simple or straightforward form of communication among the slaves.
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• 30.
### All the following sentences about Nicaraguan sign language are true EXCEPT:
• A.
The language has been created since 1979
• B.
The language is based on speech and lip reading.
• C.
The language incorporates signs which children used at home.
• D.
The language was perfected by younger children
B. The language is based on speech and lip reading.
Explanation
The correct answer is "The language is based on speech and lip reading." This sentence is not true because Nicaraguan sign language is not based on speech and lip reading. It is a distinct and unique language that developed naturally among deaf individuals in Nicaragua. It was created since 1979, incorporates signs used by children at home, and was perfected by younger children as they added more complexity to the language over time.
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• 31.
### In paragraph 3, where can the following sentence be placed? It included standardised word orders and grammatical markers that existed in neither the pidgin language, nor the language of the colonizers.
• A.
A
• B.
B
• C.
C
• D.
D
D. D
Explanation
The sentence "It included standardised word orders and grammatical markers that existed in neither the pidgin language, nor the language of the colonizers" can be placed in paragraph 3 to provide additional information about the characteristics of the creole language being discussed. The sentence fits well in paragraph 3 because it adds to the description of the language's features and highlights its unique aspects that differentiate it from both the pidgin language and the language of the colonizers.
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• 32.
### 'From scratch' in paragraph 2 is closest in meaning to:
• A.
From the very beginning
• B.
in simple cultures
• C.
By copying something else
• D.
by using written information
A. From the very beginning
Explanation
The phrase "from scratch" means starting something from the very beginning, without any pre-existing materials or knowledge. It implies starting with the most basic or fundamental elements and building up from there. In the context of the paragraph, it suggests that the author is advocating for starting a project or process from its initial stages, without relying on any existing resources or information.
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• 33.
### 'Make-shift' in paragraph 3 is closest in meaning to:
• A.
Complicated and expressive
• B.
Simple and temporary
• C.
Extensive and diverse
• D.
Private and personal
B. Simple and temporary
Explanation
The word "make-shift" suggests something that is created quickly and temporarily to serve a specific purpose. It implies a solution that is not meant to be permanent or elaborate, but rather a simple and temporary fix. Therefore, the closest meaning to "make-shift" in this context would be "simple and temporary".
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• 34.
### Which sentence is closest in meaning to the highlighted sentence? Grammar is universal and plays a part in every language, no matter how widespread it is.:
• A.
All languages, whether they are spoken by a few people or a lot of people, contain grammar
• B.
Some languages include a lot of grammar, whereas other languages contain a little.
• C.
Languages which contain a lot of grammar are more common that languages that contain a little
• D.
The grammar of all languages is the same, no matter where the languages evolved.
A. All languages, whether they are spoken by a few people or a lot of people, contain grammar
Explanation
This sentence conveys the same meaning as the highlighted sentence by emphasizing that all languages, regardless of their popularity or number of speakers, have grammar.
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• 35.
### All of the following are features of the new Nicaraguan sign language EXCEPT:
• A.
All children used the same gestures to show meaning.
• B.
The meaning was clearer than the previous sign language.
• C.
The hand movements were smoother and smaller
• D.
New gestures were created for everyday objects and activities
D. New gestures were created for everyday objects and activities
Explanation
The correct answer is "New gestures were created for everyday objects and activities." This means that the new Nicaraguan sign language did not involve the creation of new gestures specifically for everyday objects and activities.
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• 36.
### Which idea is presented in the final paragraph?
• A.
English was probably once a creole.
• B.
The English past tense system is inaccurate.
• C.
Linguists have proven that English was created by children.
• D.
Children say English past tenses differently from adults.
A. English was probably once a creole.
Explanation
The idea presented in the final paragraph is that English was probably once a creole.
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• 37.
### Look at the word 'consistent' in paragraph 4. This word could best be replaced by which of the following?
• A.
Natural
• B.
Predictable
• C.
Imaginable
• D.
Uniform
D. Uniform
Explanation
The word "consistent" in paragraph 4 refers to something that is steady, regular, or unchanging. The word "uniform" has a similar meaning and can be used interchangeably with "consistent" in this context. Both words convey the idea of something being the same or constant, making "uniform" the best replacement for "consistent" in the given paragraph.
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• 38.
### Which among the following is the SYNONYMS of to the word “argot”?
• A.
Emulous
• B.
Associates
• C.
Dialect
• D.
Supports
C. Dialect
Explanation
The word "argot" refers to a specialized vocabulary or language used by a particular group or subculture. The synonym for "argot" is "dialect," as both terms refer to a specific form of language that is distinct from the standard or mainstream language.
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• 39.
### Upon visiting the Middle East in 1850, Gustave Flaubert was so belly dancing that he wrote, in a letter to his mother, that the dancers alone made his trip worthwhile.
• A.
Overwhelmed by
• B.
Enamored by
• C.
Beseeched by
• D.
Flustered by
B. Enamored by
Explanation
Gustave Flaubert was so enamored by belly dancing during his visit to the Middle East in 1850 that he wrote to his mother, expressing that the dancers alone made his trip worthwhile.
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• 40.
### Possessed of an insatiable sweet tooth, Jim enjoyed all kinds of candy, but he had a special for gumdrops, his absolute favorite.
• A.
Odium
• B.
Nature
• C.
Disregard
• D.
Predilection
D. Predilection
Explanation
Jim enjoyed all kinds of candy, but he had a special preference for gumdrops. "Predilection" means a strong liking or preference for something, which perfectly fits the context of Jim's insatiable sweet tooth and his absolute favorite candy being gumdrops.
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• 41.
### What is the antonym of "beckon"?
• A.
Guess
• B.
Ignore
• C.
Support
• D.
Understand
B. Ignore
Explanation
The antonym of "beckon" is "ignore" because "beckon" means to make a gesture with the hand, arm, or head to encourage someone to come nearer or follow, while "ignore" means to pay no attention to or disregard something or someone.
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• 42.
### Many organisations find it beneficial to employ students over the summer. Permanent staff often wish to take their own holidays over this period. Furthermore, it is not uncommon for companies to experience peak workloads in the summer and so require extra staff. Summer employment also attracts students who may return as well-qualified recruits to an organisation when they have completed their education. Ensuring that the students learn as much as possible about the organisation encourages interest in working on a permanent basis. Organisations pay students on a fixed rate without the usual entitlement to paid holidays or bonus schemes. This sentence is : "Students are subject to the organisation’s standard disciplinary and grievance procedures."
• A.
True
• B.
False
• C.
Cannot Say
C. Cannot Say
Explanation
The given passage does not provide any information about the organization's standard disciplinary and grievance procedures for students. Therefore, it cannot be determined whether students are subject to these procedures or not.
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• 43.
### Which ones of the following are key components to the success of any Project?
• A.
Scope
• B.
Cost
• C.
Schedule/Time
• D.
Resources
• E.
Risk
• F.
Stakeholders' Satisfaction (Quality)
A. Scope
B. Cost
C. Schedule/Time
D. Resources
E. Risk
F. Stakeholders' Satisfaction (Quality)
Explanation
The key components to the success of any project include scope, cost, schedule/time, resources, risk, and stakeholders' satisfaction (quality). These components are crucial for ensuring that the project is well-defined, stays within budget and timeline, has the necessary resources, addresses potential risks, and meets the expectations and needs of the stakeholders. By effectively managing these components, a project can be successfully executed and achieve its objectives.
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• 44.
### When targeting a large existing market for your new product/service, it is correct to say that since the market is so large, capturing even 1% of it will be completely profitable.
• A.
True
• B.
False
B. False
Explanation
This statement is false because capturing 1% of a large market does not necessarily guarantee profitability. The profitability of capturing a percentage of a market depends on various factors such as the cost of production, marketing expenses, competition, and pricing strategy. Even if the market is large, if the cost of acquiring customers and producing the product/service outweighs the revenue generated from capturing 1% of the market, it may not be profitable. Therefore, the statement is incorrect.
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• 45.
### Why can I not print the Nike swoosh on stickers and sell them with no affirmation from Nike?
• A.
Patent Issues
• B.
• C.
Explanation
The correct answer is trademark issues. Trademarks are used to protect brand names, logos, and symbols that distinguish a company's goods or services from others. The Nike swoosh is a registered trademark owned by Nike, and printing it on stickers and selling them without permission would infringe on Nike's exclusive right to use the trademark. This could lead to legal consequences, as it would be a violation of trademark law.
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Related Topics | 7,941 | 30,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-26 | latest | en | 0.924814 |
https://community.smartsheet.com/discussion/114424/countif-cells-match-in-different-sheets | 1,725,908,607,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651133.92/warc/CC-MAIN-20240909170505-20240909200505-00212.warc.gz | 162,111,730 | 106,177 | # COUNTIF cells match in different sheets
✭✭✭
I am struggling to get a count if cells match when referencing different sheets. I am working on "Sheet 0" and want to check how many cells in "Sheet 1" match cells in "Sheet 2" for one column.
Here is my formula: =COUNTIFS({Sheet 1 column}, ="{Sheet 2 column}")
have also tried without parenthesis. I currently get a 0 count which i know is false.
Any ideas would be appreciated.
• ✭✭✭✭✭✭
So, what you are trying to do is a little bit... difficult. You cannot COUNTIF an array to an array as that is not a logic, but if it were you would only get the value of 1 if all objects in an array matches the other.
I am also confused about what it is you are trying to do. Are each items in a list unique? Are you trying to figure out if there are duplicates?
• ✭✭✭
Hi @Eric Law ,
OK that makes sense.
I am attempting to see how many 'outcomes' have not changed after an 'appeal'. Which would mean the cells in the 'outcome' column in the first sheet should match the cells in the 'outcome' column in the 'appeal' sheet.
Hope this makes sense.
I also tried the opposite where cells do not match.
• ✭✭✭✭✭✭
So, the best way is to probably CELL link Sheet 1 to Sheet 0. Then, in sheet 0, create a checkbox column that is =IF(CONTAINS(link@row, {sheet 2 column}, 1, 0) and it will checkbox all the ones that are in both. Hope that helps.
• ✭✭✭
edited 12/15/23
Thanks @Eric Law,
I've now linked both lists to sheet 0 and they are side by side. how can i count how many match? they are text and i cannot use 'Contains' as this produces some false figures.
Edit: Simple IF function worked to get count.
Thanks.
• ✭✭✭✭✭✭
Since you've linked both, I will assume the column names to be Sheet 1 and Sheet 2. Create a Match checkbox column and just use =IF(COUNTIF([Sheet 2]:[Sheet 2], [Sheet 1]@row)>0, 1, 0) and that will checkbox any Sheet 1 items that match Sheet 2.
## Help Article Resources
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Check out the Formula Handbook template! | 570 | 2,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.929984 |
https://www.numbersaplenty.com/2801634067 | 1,686,092,635,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653183.5/warc/CC-MAIN-20230606214755-20230607004755-00181.warc.gz | 976,772,116 | 2,953 | Search a number
2801634067 is a prime number
BaseRepresentation
bin1010011011111101…
…1000101100010011
321020020202202102201
42212333120230103
521214204242232
61142000424031
7126266332231
oct24677305423
97206682381
102801634067
1112084a4161
12662318017
1335857cc4b
141c812b951
15115e5dbe7
hexa6fd8b13
2801634067 has 2 divisors, whose sum is σ = 2801634068. Its totient is φ = 2801634066.
The previous prime is 2801634029. The next prime is 2801634091. The reversal of 2801634067 is 7604361082.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-2801634067 is a prime.
It is not a weakly prime, because it can be changed into another prime (2801634467) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1400817033 + 1400817034.
It is an arithmetic number, because the mean of its divisors is an integer number (1400817034).
Almost surely, 22801634067 is an apocalyptic number.
2801634067 is a deficient number, since it is larger than the sum of its proper divisors (1).
2801634067 is an equidigital number, since it uses as much as digits as its factorization.
2801634067 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 48384, while the sum is 37.
The square root of 2801634067 is about 52930.4644510135. The cubic root of 2801634067 is about 1409.7338778140.
The spelling of 2801634067 in words is "two billion, eight hundred one million, six hundred thirty-four thousand, sixty-seven". | 482 | 1,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.833491 |
https://www.physicsforums.com/threads/deriving-the-expression-for-wind-velocity-around-low-pressure-area.696959/ | 1,511,356,993,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806586.6/warc/CC-MAIN-20171122122605-20171122142605-00130.warc.gz | 840,841,217 | 15,713 | # Deriving the expression for wind velocity around low pressure area
1. Jun 14, 2013
### bksree
Hi
Can somebody tell me the steps and assumptions made in deriving the equation to calculate the wind velocity around a low pressure area ? (see the attachment fro Kleppner & Kolenkow).
I tried to derive like this :
Coordinate system located on earth's surface with i direction towards E, j direction towards N and k direction upwards.
Initial wind velocity V = Vx i + Vy j
Ω = Ω sin λ j + Ω cos λ k where λ is the latitude of the place considered.
Position vector of parcel of air r = rx i + ryj
Eqn for the rotating system is
marot = mainertial - 2m Ω * V - Ω * (Ω * r) ----- (1)
mainertial = -(ΔP)S j where S = cross sectional area of parcel of air
As the flow is assumed to be steady, LHS of equation 1 i.e. marot = 0
RHS evaluated vectorially.
However, I cannot get the expression shown in the attachment.
TIA
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84.4 KB
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https://www.encyclopediaofmath.org/index.php/One-factor | 1,550,766,681,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247505838.65/warc/CC-MAIN-20190221152543-20190221174543-00444.warc.gz | 841,858,290 | 6,665 | One-factor
of a graph
Suppose $G$ is a (simple) graph. A matching in $G$ is a set of pairwise-disjoint edges in $G$. A one-factor or spanning matching is a set of edges such that every vertex of $G$ occurs in exactly one edge.
Not every graph $G$ has a one-factor. One obvious necessary condition is that $G$ has no connected component with an odd number of vertices. However this is not a sufficient condition. The smallest graph connected with an even number of vertices but no one-factor is:
Figure: o110060a
One-factors were first studied by W.T. Tutte, who proved the following characterization. If $W$ is any subset of the vertex set $V(G)$ of $G$, let $G-W$ denote the graph constructed from $G$ by deleting $W$ and all edges touching members of $W$. The components of $G-W$ are odd or even according as they have an odd or even number of vertices. Write $\phi(W)$ for the number of odd components of $G-W$.
Tutte's theorem [a6]: $G$ contains a one-factor if and only if $\phi(W)\leq|W|$ whenever $W\leq V(G)$.
A consequence is that for $n$ even, any regular graph of degree $n-1$ on $2n$ vertices has a one-factor.
A bridge in a graph $G$ is an edge whose deletion disconnects $G$.
Petersen's theorem [a4]: A bridgeless cubic graph contains a one-factor.
This has been generalized by T. Schönberger [a5], who proved that every edge of a bridgeless cubic graph lies in a one-factor. C. Berge [a1] and A.B. Cruse [a3] proved another generalization. If $W$ is any set of vertices of $G$, let $z_G(W)$ be the number of edges of $G$ with exactly one end-point in $W$.
The Berge–Cruse theorem: If $G$ is a regular graph of degree $d$ with an even number of vertices, and if $z_G(W)\leq d-1$ for all odd-order subsets $W$ of $V(G)$, then each edge of $G$ belongs to some one-factor.
Consequences are [a3]:
1) if $G$ is a regular graph of degree $d$ with $2m$ vertices and $d\geq m$ ($m$ odd) or $d\geq m-1$ ($m$ even), then every edge of $G$ belongs to some one-factor;
2) if $G$ is a regular graph of valency $d$ with $m$ vertices and $d\geq m\geq2$, then $G$ contains at least
$$d-m+2\left\lfloor\frac{5(m+5)}{112}\right\rfloor$$
disjoint one-factors.
A number of results have been proved concerning whether a graph of a certain description (usually a quite sparse graph) contains a one-factor.
Important references include [a2], and [a7].
References
[a1] C. Berge, "Graphs and hypergraphs" , North-Holland (1973) (Translated from French) [a2] J. Bosák, "Decomposition of graphs" , Kluwer Acad. Publ. (1990) [a3] A.B. Cruse, "A note on one-factors in certain regular multigraphs" Discrete Math. , 18 (1977) pp. 213–216 [a4] J. Petersen, "Die Theorie der regulären Graphes" Acta Math. , 15 (1981) pp. 193–220 [a5] T. Schönberger, "Ein Beweis des Peterschen Graphensatzes" Acta Sci. Math. Szeged , 7 (1934) pp. 51–57 [a6] W.T. Tutte, "The factorizations of linear graphs" J. London Math. Soc. , 22 (1947) pp. 459–474 [a7] W.D. Wallis, "One-factorizations" , Kluwer Acad. Publ. (1997)
How to Cite This Entry:
One-factor. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=One-factor&oldid=33440
This article was adapted from an original article by W.D. Wallis (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | 1,003 | 3,313 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-09 | latest | en | 0.931075 |
https://www.hindawi.com/journals/tswj/2014/505496/ | 1,653,560,856,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00763.warc.gz | 918,919,526 | 136,280 | / / Article
Research Article | Open Access
Volume 2014 |Article ID 505496 | https://doi.org/10.1155/2014/505496
Qaiser Mushtaq, Abdul Razaq, "Equivalent Pairs of Words and Points of Connection", The Scientific World Journal, vol. 2014, Article ID 505496, 8 pages, 2014. https://doi.org/10.1155/2014/505496
Equivalent Pairs of Words and Points of Connection
Revised26 Jun 2014
Accepted10 Jul 2014
Published08 Sep 2014
Abstract
Higman has defined coset diagrams for . The coset diagrams are composed of fragments, and the fragments are further composed of two or more circuits. A condition for the existence of a certain fragment in a coset diagram is a polynomial f in , obtained by choosing a pair of words such that both and fix a vertex v in γ. Two pairs of words are equivalent if and only if they have the same polynomial. In this paper, we find distinct pairs of words that are equivalent. We also show there are certain fragments, which have the same orientations as those of their mirror images.
1. Introduction
The modular group [13] an interesting group of hyperbolic isometries and has a finite presentation where and are the linear fractional transformations defined by and , respectively. The modular group is of great interest in many fields of mathematics, for example, group theory, graph theory, and number theory. By adjoining a new element with and , we obtain a presentation of the extended modular group .
The Hecke group is the discrete group generated by , , where . It is well known that the modular group can be generalized to the Hecke groups. For , the resulting Hecke group is the modular group .
If is a power of a prime , then the projective line over the finite field , which is , is denoted by . The group has its customary meaning, a group of all linear fractional transformations such that and , while is its subgroup consisting of all those, where is a non-zero square in .
There is a well-known relationship between the action of on and continued fractions. There are many articles on the connection between the geodesics on the modular surface and continued fractions which are particularly important in the theory of approximation of real numbers by rational [4, 5]. Important results have been obtained using these ideas in a very well-written article [6]. A good account on relationship between continued fractions and indefinite binary quadratic forms is also given in [7]. In [8], Mushtaq has shown a relationship between reduced indefinite binary quadratic forms and orbits of the modular group. There is another interesting article [9] of Mushtaq and Hayat on the topic of Pell numbers, Pell-Lucas numbers, and modular group. In [10], Bong and Mushtaq determine the Fibonacci and Locus numbers through the action of modular group on real quadratic fields.
In , G. Higman introduced a new type of graph called a coset diagram for . The three cycles of are denoted by small triangles whose vertices are permuted counterclockwise by and any two vertices which are interchanged by are joined by an edge. The fixed points of and are denoted by heavy dots. Notice is equivalent to , which means that reverses the orientation of the triangles representing the three cycles of (as reflection does); because of this, there is no need to make the diagram more complicated by introducing -edges. For details about coset diagrams, one can refer to [1114].
Two homomorphisms and from to are called conjugate if for some inner automorphism on . We call to be non-degenerate if neither of lies in the kernel of . In [15], it has been shown that the conjugacy classes of non-degenerate homomorphisms from to are in one to one correspondence with the elements of under the correspondence which maps each class to its parameter . As in [15], the coset diagram corresponding to the action of on via a homomorphism with parameter is denoted by . Each conjugacy class is represented by a unique coset diagram, unique in the sense that the diagram remains invariant except that labels of vertices change from to .
By a circuit in a coset diagram for an action of on , we will mean a closed path of triangles and edges. Let be a sequence of positive integers, the circuit which contains a vertex, fixed by for some ; we will mean the circuit in which triangles have one vertex inside the circuit and triangles have one vertex outside the circuit and so on. Since it is a cycle , so it does not make any difference if triangles have one vertex outside the circuit and triangles have one vertex inside the circuit and so on. The circuit of the type is called a periodic circuit and the length of its period is . A circuit that is not of this type is non-periodic. A circuit is called simple, if each vertex of the circuit is fixed by a unique word or its inverse .
2. Joining of Circuits
Consider two non-periodic and simple circuits and . Let be any vertex in fixed by a word and let be any vertex in fixed by a word . In order to connect these two circuits at and , we choose, without loss of generality , and apply on in such a way that ends at . Consequently, we get a fragment denoted by . In other words, by , we will mean a nonsimple fragment whose one vertex is fixed by a pair of words , , dented by .
Example 1. We connect the vertex , fixed by , in with the vertex , fixed by of , and compose a fragment as shown in Figures 1, 2, and 3.
One can see that the vertex in (Figure 3) is fixed by a pair of words
The action of on yields two components, namely, and . For sake of simplicity, let denote the complement .
The coset diagram is made of fragments. It is therefore necessary to ask when a fragment exists in . In [16], this question is answered in the following way.
Theorem 2. Given a fragment, there is a polynomial in such that(i)if the fragment occurs in , then ;(ii)if , then the fragment, or a homomorphic image of it occurs in or in .
In [16], the method of calculating a polynomial from a fragment is given. Here, we describe this method briefly. Since a fragment is composed of two non-periodic and connected circuits and with a common fixed vertex, say, , then there is a pair of words, for instance, such that and . Let and be the matrices corresponding to and of . Then, and can be expressed as where . By using (3.1) to (3.7) of [16], the matrices and can be expressed linearly as such that and , for , are polynomials in and , where is the trace of and is its determinant. Since is a common fixed vertex of and , therefore the matrices and have a common eigenvector. Then, by Lemma of [16], the algebra generated by and has dimension . The algebra contains , , , and and so these are linearly dependent. Using (3.1) to (3.7) of [16], the matrix is expressed as where , for , can be calculated in terms of the and , using to of [16]. The condition that , , , and are linearly dependent is expressed as If we calculate in terms of and and substitute in (8), we find that it is equivalent to
This gives a homogeneous equation in and . By substituting for , we get an equation in .
Two pairs of words and are equivalent if and only if they have the same polynomial. If two pairs of words and are equivalent, then we write .
In this paper, we find distinct pairs of words that are equivalent. We also show there are certain fragments, which have the same orientations as those of their mirror images.
3. Main Results
Theorem 3. In the above notation, the polynomial obtained from a fragment is unique.
Proof. Let and be any two vertices of a fragment , such that is fixed by a pair of words and is fixed by a pair of words . Suppose is a polynomial obtained by choosing and is a polynomial obtained by choosing . Now, if , then, by Theorem 2, the fragment or its homomorphic image occurs in or in . So, there exists a vertex in or in which is fixed by . Again, by Theorem 2, we have . Similarly, it can be proved that if , then . Hence, .
Remark 4. If a vertex is fixed by a word , then the vertex is fixed by the conjugate of .
We know that is a fragment created by joining a vertex of with the vertex of . Since the same polynomial is evolved for all the points of connection for the fragment , therefore it is important to know all the points of connection for the fragment . Following theorem is useful in finding all points of connection of and for the fragment .
Theorem 5. Let fragment be constructed by joining a vertex of with the vertex of . Then, is obtainable also, if the vertex of is joined with the vertex of .
Proof. Suppose vertex is fixed by a word and vertex is fixed by a word . Then, one vertex of the fragment is fixed by and . Now, we join the vertex of with the vertex of and compose a fragment . By Remark 4, the vertex of fragment is fixed by and , whereas the vertex of fragment is fixed by and . This shows that and are the same fragments.
Since and are the same fragments, and a unique polynomial is obtained from a fragment, hence .
Corollary 6. Let be the set of words such that for any , both vertices and lie on the circuits and . Let be the total number of points of connection of the circuits to compose . Then, .
Example 7. As in Example 1, the vertex in is connected with the vertex in , and a fragment is evolved. Then, one can see that is the set of words such that, for any , both vertices and lie on and . Since , there are nine points of connection of and composing . Hence, vertices of one circuit are merged in another circuit. In other words, triangles of one circuit are merged in another circuit.
The number of points of connection of the circuits for a fragment is always multiple of three and plays a significant role because they are directly related to the structure of the fragment. The following theorem illustrates this relation.
Theorem 8. Let be the total number of triangles in a fragment and the total number of points of connection of the circuits and . Then, .
Proof. Since circuits and have and number of triangles, respectively, and there are number of points of connection of and composing , therefore, by Corollary 6, there are number of words such that , lie on and , respectively. So, number of vertices of are merged in , and remaining number of vertices are included in to compose . So, the total number of vertices in is . This implies that .
Remark 9. Consider a fragment such that one vertex of is fixed by . Then, obviously, is fixed by . So, . Also, it is trivial that .
The following Theorem shows that and are not a unique pair of circuits for the fragment .
Theorem 10. The following pairs of words are equivalent:
Proof. Let , and be arbitrary vertices of the circuits , , and , respectively. Let the words , and fix the vertices , and , respectively. We join a vertex fixed by of the circuit with the vertex fixed by of the circuit to compose . Also, we join a vertex fixed by of the circuit with the vertex fixed by of the to compose a fragment . So, in , is a common fixed vertex of . The fragment is different from the fragment if and only if, in fragment , the word fixes the vertex in the following way: maps the vertex to some vertex , and then maps the vertex to the vertex ; that is, . Since is a composition of linear fractional transformations and , therefore . This implies that and are the same fragments, since a unique polynomial is obtained from a fragment. Hence, .
Similarly, it is easy to prove that Since is an equivalence relation, hence
Example 11. Theorem 10 shows that the fragment , which has a vertex fixed by , and the fragment , which has a vertex fixed by , are the same. So, the fragment, which has a vertex fixed by , and the fragment, which has a vertex fixed by , are the same. Since the words , , and correspond to the circuits , , and , so the fragment (Figure 4) is obtained, not only by joining the vertex in (Figure 5) with the vertex in (Figure 6) but also by joining the vertex in (Figure 7) with the vertex in (Figure 8).
Theorem 12. In a coset diagram for the action of on , if a vertex is fixed by , then the vertex is fixed by .
Proof. Let be a word such that
Since , this implies that
Now, by using (14) and (15), we have . This shows that also fixes a point .
Theorem 13. If a vertex is fixed by a word in a circuit , then there exists a vertex in such that is fixed by .
Proof. Consider a simple circuit .
In Figure 9, one can see that the vertex is fixed by and is fixed by . So, . Also, the vertex is fixed by and is fixed by . So, . Similarly, , .
The vertices and are fixed by and , respectively, implying that . Also, the vertices and are fixed by and , respectively, implying that . Similarly, , .
Let the homomorphic image of the fragment (Figure 10) occur in the coset diagram . Since admits the axis of symmetry, the mirror image of the fragment under the permutation will also occur as shown in Figure 11.
Proposition 14. If the fragment has one vertex fixed by , then its mirror image has one vertex fixed by .
Proof. Let be the fragment with one vertex fixed by . Since is the mirror image of , therefore has one vertex . Then, by Theorem 12, the vertex is fixed by .
Theorem 15. The polynomials obtained from the fragment and its mirror image are the same.
Proof. Let be the polynomial obtained by choosing a pair of words from and let be the polynomial obtained by choosing a pair of words from . Let . Then, by Theorem 2, there is a vertex say in or in which is fixed by . Then, by Theorem 12, the vertex is fixed by . So, by Theorem 2, . Similarly, it can be proved that if , then . Hence, .
Since a unique polynomial is obtained from a fragment and its mirror image , hence .
Consider two circuits and . In general, fragment and its mirror image do not have the same orientation. There are certain fragments which have the same orientations as those of their mirror images. These kinds of fragments have vertical symmetry and may have fixed points of . The following Theorem shows how these fragments are composed.
Theorem 16. (i) The fragment composed by joining a vertex , fixed by in , with the vertex , fixed by in , has the same orientation as that of its mirror image.
(ii) The fragment composed by joining a vertex , fixed by in , with the vertex , fixed by in , has the same orientation as that of its mirror image.
(iii) The fragment composed by joining a vertex , fixed by in , with the vertex , fixed by in , has the same orientation as that of its mirror image.
Proof. (i) Let be the fragment which has one vertex say , fixed by Then, by Proposition 14, its mirror image fragment contains a vertex, say, , fixed by and . By Remark 4, the vertex of fragment is fixed by and . This implies that . Hence fragment has the same orientation as that of its mirror image .
(ii) Let be the fragment which has one vertex, say, , fixed by Then, its mirror image fragment has one vertex, say, , fixed by The vertex of fragment is fixed by and . This implies that . Hence, has the same orientation as that of its mirror image .
(iii) Let be the fragment which has one vertex, say, , fixed by Then, its mirror image fragment has one vertex, say, , fixed by The vertex of fragment is fixed by and . This implies that . Hence, has the same orientation as that of its mirror image .
4. Motivation and Open Problems
If we join a pair of circuits at a certain point, we get a fragment and hence a polynomial. Since each such polynomial splits linearly in a suitable Galois [16] and corresponding to each zero, we get a triplet [15] which is a group. This shows that each pair of circuits can be related to a group. A pair of circuits has finitely many points of connections. Two distinct points of connection may or may not give the same fragment. So, it is important to ask the following question.
Problem 17. How many fragments (polynomials) are obtained if we join a pair of circuits and at all points of connection?
In order to give the answer of the above question, we first have to find those pair of words which are equivalent. In other words, we have to find those points of connection, at which we get the same polynomial. This issue is resolved in this paper. We will give the answer of the above question in an other paper. After that, one can establish a connection between a class of groups and a pair of circuits and , which is indeed a great development.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
The authors would like to thank the referee for his very helpful comments, which have significantly improved the presentation of this paper.
References
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Copyright © 2014 Qaiser Mushtaq and Abdul Razaq. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. | 4,744 | 19,961 | {"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} | 2.796875 | 3 | CC-MAIN-2022-21 | longest | en | 0.921216 |
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1. equilibrium
experiencing zero net force
2. equilibrium position
When nothing is pushing or pulling on the spring, and the free end is in equilibrium at a particular position
3. stable equilibrium
the end of the spring when it naturally returns to its equilibrium position if it is stretched or compressed
4. Hooke's law
• The stiffer the spring and the farther you stretch it, the harder it pulls back
• F = -kx
• restoring force = -spring constant x distortion
5. spring constant
k, measure of the spring's stiffness
6. collision energy
kinetic energy absorbed during the collision
7. rebound energy
kinetic energy released during the rebound
8. coefficient of restitution
coefficient of restitution = rebound speed of ball / collision speed of the ball
9. vibrational node
points near each end of the bat that do not move when the bat vibrates
10. elastic collisions
bounces so perfect that all of the collision energy is stored and returned as rebound energy
11. inelastic collisions
collisions that fail to return some of the collision energy as rebound energy
12. centripetal acceleration
• "center-seeking"
• brings toward the center
13. Nicolaus Copernicus
• 1473-1543
• Heliocentric theory
• start of modern astronomy
• delayed publication of his book till the year of his death
14. Galileo Galilei
• 1564-1642
• championed heliocentric universe by observation
• telescope observations
15. Johannes Kepler
• laws of planetary motion
• used data of Tyco Brahe as postulate motion of planets
16. Kepler's laws
• 1. planet orbits = ellipses
• 2. area swept fro the sun to a planet is constant in time (conservation of angular momentum, speeds up closer to foci)
• 3. the bigger the orbit, the longer it takes the planet to go around its orbit
17. geosynchronous orbit
• at 35,900km above the earth's surface
• orbits the earth every 24 hours, stays at same place above the earth
## Card Set Information
Author: Anonymous ID: 239961 Filename: Physics Test 2 Updated: 2013-10-10 22:01:48 Tags: physics Folders: Description: Physics Test 2 Show Answers:
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# Some people believe that witnessing violence in movies will
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VP
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Some people believe that witnessing violence in movies will [#permalink]
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19 Apr 2005, 11:52
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Some people believe that witnessing violence in movies will discharge aggressive energy. Does watching someone else eat fill one’s own stomach?
In which one of the following does the reasoning most closely parallel that employed in the passage?
(A) Some people think appropriating supplies at work for their own personal use is morally wrong. Isn’t shoplifting morally wrong?
(B) Some people think nationalism is defensible. Hasn’t nationalism been the excuse for committing abominable crimes?
(C) Some people think that boxing is fixed just because wrestling usually is. Are the two sports managed by the same sort of people?
(D) Some people think that economists can control inflation. Can meteorologists make the sun shine?
(E) Some people think workaholics are compensating for a lack of interpersonal skills. However, aren’t most doctors workaholics?
Last edited by MA on 19 Apr 2005, 12:16, edited 2 times in total.
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VP
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19 Apr 2005, 12:05
"D".....prove a point by Analogy and not by example....only "D" fits the bill I believe.
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Senior Manager
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19 Apr 2005, 12:09
D for me, more going by ears than critical reasoning
I believe B C are out.
D pattern is: A -> B; it is stupid to say so because if C->not D
A is a slightly different pattern... ( A -> B; isn't also C = B? )
E is A -> B; it is stupid to say so because most C are A
I'd not be surprised if E is the answer
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VP
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19 Apr 2005, 12:16
i think the clue is later part of the question. the later parts of the question and D, both, are parallel as these two parts are impossible. therefore D is OA.
Does watching someone else eat fill one’s own stomach?
Can meteorologists make the sun shine?
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VP
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19 Apr 2005, 12:22
thearch wrote:
I'd not be surprised if E is the answer
E is not parrallel because it is possible.
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19 Apr 2005, 12:27
MA wrote:
thearch wrote:
I'd not be surprised if E is the answer
E is not parrallel because it is possible.
yes, it's right! just a little bit confused by the meaning of "workaholic"
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19 Apr 2005, 17:02
Agree with (D). Actually 2nd sentence is trying to prove 1st sentence wrong. Only (D) does that.
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19 Apr 2005, 20:31
jpv wrote:
Agree with (D). Actually 2nd sentence is trying to prove 1st sentence wrong. Only (D) does that.
yup. another nice approach..........
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19 Apr 2005, 20:31
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https://tensorflow.google.cn/probability/api_docs/python/tfp/experimental/substrates/numpy/stats/log_soosum_exp | 1,606,607,061,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00124.warc.gz | 520,954,530 | 39,225 | # tfp.experimental.substrates.numpy.stats.log_soosum_exp
Computes the log-swap-one-out-sum of `exp(logx)`.
The swapped out element `logx[i]` is replaced with the log-leave-`i`-out geometric mean of `logx`.
`logx` Floating-type `Tensor` representing `log(x)` where `x` is some positive value.
`axis` The dimensions to sum across. If `None` (the default), reduces all dimensions. Must be in the range `[-rank(logx), rank(logx)]`. Default value: `None` (i.e., reduce over all dims).
`keepdims` If true, retains reduced dimensions with length 1. Default value: `False` (i.e., keep all dims in `log_mean_x`).
`name` Python `str` name prefixed to Ops created by this function. Default value: `None` (i.e., `"log_soomean_exp"`).
`log_soomean_x` `logx.dtype` `Tensor` characterized by the natural-log of the sum of `x``except that the element`logx[i]```is replaced with the log of the leave-```i`-out Geometric-average. The sum of the gradient of`log_soosum_x`is`n```, i.e., the number of reduced elements. Mathematically```log_soomean_x` is,
``````log_soomean_x[i] = log(Avg{h[j ; i] : j=0, ..., m-1})
h[j ; i] = { u[j] j!=i
{ GeometricAverage{u[k] : k != i} j==i
``````
`log_sum_x` `logx.dtype` `Tensor` corresponding to the natural-log of the average of `x`. The sum of the gradient of `log_mean_x` is `1`. Has reduced shape of `logx` (per `axis` and `keepdims`). | 424 | 1,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.723585 |
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Prev Next | 568 | 2,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-40 | latest | en | 0.806518 |
https://elteoremadecuales.com/c-theorem/?lang=fr | 1,679,329,213,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943484.34/warc/CC-MAIN-20230320144934-20230320174934-00091.warc.gz | 276,531,579 | 11,487 | C-theorem
C-theorem In quantum field theory the C-theorem states that there exists a positive real function, {displaystyle C(g_{je}^{},dans )} , depending on the coupling constants of the quantum field theory considered, {style d'affichage g_{je}^{}} , and on the energy scale, {style d'affichage lui _{}^{}} , which has the following properties: {displaystyle C(g_{je}^{},dans )} decreases monotonically under the renormalization group (RG) flow. At fixed points of the RG flow, which are specified by a set of fixed-point couplings {style d'affichage g_{je}^{*}} , the function {displaystyle C(g_{je}^{*},dans )=C_{*}} is a constant, independent of energy scale.
The theorem formalizes the notion that theories at high energies have more degrees of freedom than theories at low energies and that information is lost as we flow from the former to the latter.
Contenu 1 Two-dimensional case 2 Four-dimensional case: A-theorem 3 Voir également 4 References Two-dimensional case Alexander Zamolodchikov proved in 1986 that two-dimensional quantum field theory always has such a C-function. En outre, at fixed points of the RG flow, which correspond to conformal field theories, Zamolodchikov's C-function is equal to the central charge of the corresponding conformal field theory,[1] which lends the name C to the theorem.
Four-dimensional case: A-theorem John Cardy in 1988 considered the possibility to generalise C-theorem to higher-dimensional quantum field theory. He conjectured[2] that in four spacetime dimensions, the quantity behaving monotonically under renormalization group flows, and thus playing the role analogous to the central charge c in two dimensions, is a certain anomaly coefficient which came to be denoted as a. Pour cette raison, the analog of the C-theorem in four dimensions is called the A-theorem.
In perturbation theory, that is for renormalization flows which do not deviate much from free theories, the A-theorem in four dimensions was proved by Hugh Osborn[3] using the local renormalization group equation. Cependant, the problem of finding a proof valid beyond perturbation theory remained open for many years.
Dans 2011, Zohar Komargodski and Adam Schwimmer of the Weizmann Institute of Science proposed a nonperturbative proof for the A-theorem, which has gained acceptance.[4][5] (Still, simultaneous monotonic and cyclic (limit cycle) or even chaotic RG flows are compatible with such flow functions when multivalued in the couplings, as evinced in specific systems.[6]) RG flows of theories in 4 dimensions and the question of whether scale invariance implies conformal invariance, is a field of active research and not all questions are settled.
See also Conformal field theory References ^ Zamolodchikov, UN. B. (1986). ""Irreversibility" of the Flux of the Renormalization Group in a 2-D Field Theory" (PDF). JETP Lett. 43: 730–732. Code bib:1986JETPL..43..730Z. ^ Cardy, John (1988). "Is there a c-theorem in four dimensions?". Physics Letters B. 215 (4): 749–752. Code bib:1988PhLB..215..749C. est ce que je:10.1016/0370-2693(88)90054-8. ^ Osborn, Hugh (1989). "Derivation of a Four-Dimensional c Theorem". Physics Letters B. 222 (1): 97. Code bib:1989PhLB..222...97O. est ce que je:10.1016/0370-2693(89)90729-6. Ian, Jack; Osborn, Hugh (1990). "Analogs for the c Theorem for Four-Dimensional Renormalizable Field Theories". Nuclear Physics B. 343 (3): 647–688. Code bib:1990NuPhB.343..647J. est ce que je:10.1016/0550-3213(90)90584-Z. ^ Reich, E. S. (2011). "Proof found for unifying quantum principle". La nature. est ce que je:10.1038/nature.2011.9352. ^ Komargodski, Z.; Schwimmer, UN. (2011). "On renormalization group flows in four dimensions". Journal of High Energy Physics. 2011 (12): 99. arXiv:1107.3987. Code bib:2011JHEP...12..099K. est ce que je:10.1007/JHEP12(2011)099. ^ Curtright, T; Jin, X.; Zachos, C. (2012). "Renormalization Group Flows, Cycles, and c-Theorem Folklore". Lettres d'examen physique. 108 (13): 131601. arXiv:1111.2649. Code bib:2012PhRvL.108m1601C. est ce que je:10.1103/PhysRevLett.108.131601. PMID 22540692. Catégories: Conformal field theoryRenormalization groupMathematical physicsTheorems in quantum mechanics
Si vous voulez connaître d'autres articles similaires à C-theorem vous pouvez visiter la catégorie Conformal field theory.
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Nous utilisons nos propres cookies et ceux de tiers pour améliorer l'expérience utilisateur Plus d'informations | 1,229 | 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} | 2.609375 | 3 | CC-MAIN-2023-14 | latest | en | 0.877638 |
http://www.pcc.edu/ccog/default.cfm?fa=ccog&subject=MCH&course=120 | 1,438,113,877,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042982502.13/warc/CC-MAIN-20150728002302-00280-ip-10-236-191-2.ec2.internal.warc.gz | 623,572,989 | 4,849 | ## Course Content and Outcome Guide for MCH 120 Effective Summer 2015
Course Number:
MCH 120
Course Title:
Machine Shop Math
Credit Hours:
2
Lecture Hours:
0
Lecture/Lab Hours:
20
Lab Hours:
0
Special Fee:
\$12.00
#### Course Description
Covers instruction and practice in working with whole numbers, fractions, decimals, formulas, inch and metric systems, formulas, calculating simple and direct indexing. Introduces how to apply the use of the inch/metric systems, dividing/index head and formulas as they pertain to thread calculations, gear calculations, speed and feed calculations, and taper calculations. Prerequisites: MCH 100. Audit available.
Applying Shop Math - Math skills are very important to the machinist in his/her daily work. The machinist must be able to calculate accurately and with reasonable speed. This module will provide instruction and practice in working with whole numbers, fractions, and decimals conversions.
Shop Math - Inch & Metric -In the Machine Shop, accurate workmanship depends on accurate measurements. The metric system of measurement is being adopted by many industries in an effort to be competitive in foreign markets. This module will introduce the student to the principles of the inch and metric systems of measurement.
Shop Math/Formulas - The machinists frequently makes calculations to solve for the unknown value needed to produce a part. A formula tells what values using symbols needs; what computations are necessary to combine those values by using operation signs; and what order to combine them by using grouping signs. In this module the student will learn how to apply the use of formulas as they pertain to Thread Calculations and Taper Calculations.
Percent, Charts, Graphs & Angles - There are times when the print given to the technician does not specifically provide all of the required information to allow the machinist to complete the work piece. At these times, the technician may have to use math procedures to calculate the missing information.To produce a superior product in the manufacturing process, machinists need tools to evaluate quality. Statistical Process Control (SPC) includes Percent, Graphs and Charts which are those tools that help the technician interpret whether the process is in or out of control. This module will introduce the student to the tools of SPC. This module will help the student learn these procedures and calculations.
#### Intended Outcomes for the course
Upon successful completion of this course students will be able to:
Calculate decimal equivalents of fractions noted on blue prints.
Convert inch to metric and metric to inch from dimensions on blue prints.
Apply mathematical formulas as appropriate to thread and taper calculations on shop drawings.
#### Course Activities and Design
MCH 120 will be presented by means of audio-visual presentations, demonstrations, lab experiences, and research activities. The course activities and design emphasize the development of skills and knowledge outcomes prescribed by established industry standards. The identified outcomes will be achieved by means of individual and team activities.
#### Outcome Assessment Strategies
POLICY - Student performance measurements are based on established industry standards. The various areas of study during the course will be evaluated by a variety of activities. Typical of those activities are the following;
1. READING ASSIGNMENTS - Information sheets, textbooks, journal articles and the learning resource center are potential sources of information that the student will reference as directed in the modules identified in the introduction.
2. PRACTICE - Completion of tasks and projects identified in the reading assignments, information sheets, journal articles and textbooks. Students are required to complete practice activities with 100% competency.
3. SELF-ASSESSMENT - Checking and evaluating the students understanding and knowledge gained through the reading assignments and practices typically done through a practice evaluation.
4. LAB ACTIVITIES - Participation in structured laboratory exercises with the emphasis on developing skills or increasing expertise in the areas of study identified in the module packets.
5. FINAL ASSESSMENT - An assessment in the form of a written exam and/or practical application that addresses the subject areas identified in the module packets. Students are required to complete final assessment activities with 85% competency.
TEXT:
Machinery's Handbook | 848 | 4,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2015-32 | longest | en | 0.862203 |
http://www.talkstats.com/threads/interpretation-of-interaction-terms-in-a-multiple-linear-regression.72989/#post-212261 | 1,643,430,405,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00043.warc.gz | 118,970,484 | 9,146 | # Interpretation of interaction terms in a multiple linear regression
#### renato25
##### New Member
I try to improve my statistical knowledge and right now I am working on a regression that contains interaction terms. I am currently investigating the effects of different variables (a1, a2 etc.) on the quantity of chairs sold in a transaction (dependent variable). To see if effects are different (e.g. a1 is a sales campaign) for customers who came back (comeback) I made a linear regression with interaction terms of a1 and comeback, a2 and comeback etc..
Comeback can be 0,1, 2, 3 or 4. If a customer made 2 purchases comeback equals 1 as he came back once. I also have some additional control variables for the characteristics of the customers which I include as dummies.
My results are the following: (results of dummies not shown) (*** indicate significance at the 1 percent level at * indicates significane at the 10 percent level):
Dependent variable = Quantity of chairs
a1 -12.10***
a2 -8.01
a3 6.57
a1 x comeback 50.35***
a2 x comeback 35.01*
a3 x comeback 5.01
comeback -15.30
Now my questions:
1. Is the effect of a1 on the quantity of chairs −12.10+50.35=38.25 for those that came back once and −12.10+50.35∗2=88.6 for those that came back twice?
2. Can you say because the coefficient for a1 x comeback is significant that customers who came back are affected differently by our sales campaign? Or do you have to make another test?
3. Can you say that a2 only affected customers who came back because a2 alone is not significant?
4. Can you say that because the coefficient for comeback is not significant, people that came to the shop more than once are not buying more or less chairs? Or how can you interpret the coefficient of comeback (-15.30) alone?
Maybe you can shed some light on this for me. And if you have additional comments to this method, please let me know.
Best regards
Renato | 469 | 1,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.926445 |
https://au.mathworks.com/matlabcentral/answers/478631-old-version-and-new-version-of-cwt-continuous-wavelet-transform-to-detect-abrupt-change-in-signal | 1,723,428,025,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00817.warc.gz | 98,641,904 | 27,407 | # Old version and new version of CWT(Continuous Wavelet transform) to detect abrupt change in signal
3 views (last 30 days)
Supatat Hovanotayan on 3 Sep 2019
Hello Everyone!!!
My name is Supatat Hovanotayan, I am a master student studying in mechanical engineering, The University of Tokyo.
I would like to consult you about the new and old version of continuous wavelet transform for detecting abrupt change in signal.
My supervisor he asked me to find out that we should use the old or the new version to perform CWT for our research. Now, we are doing research on detection abnormal running in railway vehicle system.
Our goal is to use CWT to detect abrupt change in signal and consider the cwt result to classify that the railway vehicle is running in normal condition or not.
These are the codes I used
Old version:
Fs = 200;
scale = 1:512;
[coef,f] = cwt(signal,scale,'morl',1/Fs);
New version
Fs = 200;
[coef,f] = cwt(signal,'amor',Fs);
In the past, I used the old version to perform CWT to detect the abrupt change in signal. And we found that we could distinguish between normal running and abnormal running easily as shown in the picture below.
[Abnormal running condition (Using old cwt)]
[Normal running condition (Using old cwt)]
The reason why we could distinguish because the wavelet coefficient are very high in the low frequency range for abnormal running signal. But for the normal running, the wavelet coefficient is very low in the all frequency range.
But after we used the new version, we found that the wavelet coefficient in low frequency range are not high compared to the old version’s result. It is more difficult than the old version to classify whether it is normal running or abnormal running as shown in the picture below.
[Abnormal running condition (Using new cwt)]
[Normal running condition (Using new cwt)]
My questions are
1. “Can I still use the old version of CWT because it look like more suitable to my research’s goal than the new version?”
2. “The old version is no longer recommended as shown on the website, it means the result from the old version of cwt are all incorrect?”
Moreover, if you have any idea on how to distinguish between normal running and abnormal running in my data, please let me know
Thank you very much for your kindness
I am looking forward to hearing from you
Best regards,
Supatat Hovanotayan
Wayne King on 6 Sep 2019
Supatat, You are free to use the old CWT API if you wish. However, we recommend the newer API for reasons I've tried to explain in previous posts. However, from your screen shots above it appears to me that the normal and abnormal signals look very different in the new CWT and actually comparable to what the old CWT api gives. Can you attach these two signals and the code you are using to differentiate normal vs abnormal. Just by eye, it appears to me that the new CWT API shows a much different view of these signals.
Wayne
Supatat Hovanotayan on 7 Sep 2019
Hello Mr. Wayne King
I will send you these two signals data and the code I used through email
Thank you very much for your kindness
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Start Hunting! | 751 | 3,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.929153 |
https://educationexpert.net/mathematics/1652295.html | 1,618,638,915,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00132.warc.gz | 330,922,729 | 6,997 | 1 April, 02:50
# 2) 7 + (1 + 9) = (7 + 1) + 9Property:
+1
1. 1 April, 03:06
0
Associative Property
Step-by-step explanation:
In your equation, you're adding three digits together.
When you're adding or multiplying, the order in which you it doesn't matter:
ex. 1 + 3 + 5 = 3 + 5 + 1 < - This is the commutative property (not associative)
Adding parentheses indicated the order in which you do an operation, but when adding, order doesn't matter, so you're able to move around the parentheses. When you move around the parentheses, that's an example of the associative property.
It's important to not confuse the associative property with the commutative property.
The associative property is moving around parentheses
While the commutative property is moving around the numbers. | 202 | 788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-17 | latest | en | 0.913577 |
https://zyxyvy.wordpress.com/2014/06/02/sign-balanced-digiting/ | 1,521,706,391,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647782.95/warc/CC-MAIN-20180322073140-20180322093140-00369.warc.gz | 945,933,052 | 21,767 | ## Sign-Balanced Digiting
Say you’re doing addition. And say you’re using base ten, so the digits you have are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Let’s say you’re doing 357822+461279. If you did it conventionally, right-to-left, you’d add the 2 to the 9 and get 1 with a carry of 1; then add 1, 2, and 7 to get 0 with a carry of 1; then add 1, 8, and 2 to get 1 with a carry of 1; then add 1, 7, and 1 to get 9; then add 5 and 6 to get 1 with a carry of 1; then add 1, 3, and 4 to get 8 for the final digit. This arrives at an answer of 819101.
But maybe you can do a bit more in your head and decide to put some value into getting higher-place digits first, so that you work your way closer to the exact answer. This left-to-right process has some benefits, for example, if you decide to cut off at some point in the process, you’re at least not far off from the exact value and have a good estimation. 3 plus 4 is 7, but oh look, a carry is coming from the next digit, so make that an 8; 5 plus 6 is 11, which we put as 1 because we’ve taken care of the carry; 7 plus 1 is 8, but after the carry from the next digit that will be 9; 8 plus 2 is 10, and since we took care of the carry we put a 0; 2 plus 7 is 9, but wait, there’s a carry in the next digit over, so this 9 becomes a 10, but that means we have to deposit the 0 and erase the previous digit we wrote, that 0 in the hundreds place, and make that a 1; now that carry chain is taken care of, so we can finally deal with the ones place, where 2 plus 9 is 11, and since we took care of that carry, we finally have a 1. The answer is 819101.
So the second method has the advantages that it is much easier to carry out in the mind than the first method, and brings about approximations to the final answer along the way rather than only getting close via the final leap to the exact answer. It, however, faces a problem of carry-chains that could potentially be nearly as long in digits as the numbers themselves, where we’d have to backtrack and correct a bunch of digits. I find that the benefits of left-to-right addition much outweigh the issues, and thus that is my preferred way of performing addition.
But in both cases, performing the addition in memory requires the holding of not just the digits but also a set of carry digits; about half the time, one has to be concerned about a carry digit coming over. (Of the $10*10=100$ possible pairs of digits to be added in one place, there are $0+1+2+3+4+5+6+7+8+9=45$ pairs that result in a carry, so the exact proportion of additions that require a carry is $\frac{45}{100}=\frac{9}{20}$, slightly less than half.) This is an annoying thing to have to do so frequently when performing addition.
Now, consider what would follow if we happened to use a different set of digits. What if our digits were -5, -4, -3, -2, -1, 0, 1, 2, 3, and 4? (From here on in this post, I will notate the negative digits as (5), (4), (3), (2), and (1) respectively.) Using this system, instead of writing high-valued digits, we write them as negative digits with an incremented next digit. For example, 6 is 1(4) (representing $1*10-4*1$), 67 is 1(3)(3) (representing $1*100-3*10-3*1$), and 835 is 1(2)4(5) (representing $1*1000-2*100+4*10-5*1$). Got it? Okay, let’s do some addition.
We use the same addition problem as before, 357822+461279. Except of course these numbers are actually written 4(4)(2)(2)22+1(5)(4)13(2)(1) in our new system, which I will from here on call Sign-Balanced Digiting (SBD). Using right-to-left, we get 2+(1)=1 (no carry), 2+(2)=0 (no carry), (2)+3=1 (no carry), (2)+1=(1) (no carry), (4)+(4)=2 (carry of (1)), and 4+(5)+(1)=(2) (no carry), and a millions digit of 1. Our answer is 1(2)2(1)101.
Wow, we did all that only needing to do one carry. Is this a coincidence?
It turns out this is an addition problem that is exceptionally favorable for SBD, but on average, SBD does in fact do better than the conventional digit system, only requiring a carry in $5+4+3+2+1+0+1+2+3+4=25$ possible pairs of digits, so that the proportion of additions that require a carry is $\frac{25}{100}=\frac{1}{4}$. The annoying necessary extra digits to remember just got nearly halved.
On the other hand, there’s now two possibilities for carry digits that could need to be remembered: whereas with the conventional digit system we had just 1, here we could end up with 1 or (1) to carry, when adding two numbers.
But are there even cooler things we could do with SBD?
Yes! Because we’re balancing out the deviation brought about by each digit, we are making it such that cutting off lower-place-value digits on average change the value by smaller amounts, thus making high-digit estimations more accurate. In fact, when we are rounding SBD numbers, we never even need to increment the place value rounded to; we always only need to truncate. For example, 1(4)(3)33(4)0(5)(2) rounds to 100000000, 1(4)0000000, 1(4)(3)000000, 1(4)(3)300000, 1(4)(3)330000, 1(4)(3)33(4)000, or 1(4)(3)33(4)0(5)0.
And when we add more than 2 numbers, it gets even better. With conventional digits, carries get hairier and hairier with more numbers to add, but with SBD, although they have the potential to get pretty bad, they tend to cancel out. Notice in the following example how in the case of SBD (bottom), there are both fewer and smaller carries.
How about multiplication? No problem! Just do multiplication as before, and remember that like signs multiply to positives and unlike signs multiply to negatives.
In fact, squaring numbers in general becomes easier, and the squaring of numbers actually numerically follows the general path of speedy mental squaring using binomial expansions. As an alternative method of number representation, Sign-Balanced Digiting has quite a few neat properties. | 1,653 | 5,802 | {"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": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-13 | latest | en | 0.959962 |
https://puzzling.stackexchange.com/questions/25006/mathematics-behind-word-puzzle | 1,701,175,452,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00431.warc.gz | 547,371,803 | 41,567 | # Mathematics behind word puzzle
So I was presented with this puzzler:
The answer to the puzzle is a 6 letter word with 6 different letters. Each of these 12 words have one - and only one - letter of the mystery word in the correct position (i.e. 1st, 2nd, 3rd, etc.)
MYRIAD
FUMBLE
CATKIN
RECKON
TOWARD
OUTAGE
SWAYED
BOUNTY
TONGUE
STOLEN
UNREST
I couldn't find a very efficient algorithm for solving it - more on this later - so I decided to put my programming skills to use and just generate all possible solutions via brute force.
Imagine my surprise when there was only a single 6 letter combination that met the criteria, and it was the answer to the puzzle!
I don't want to give away the answer in case you want to solve it yourself, but as I'm not interested in the answer so much as the method of devising the problem and its solution, I have a mild spoiler: each letter of the answer appears exactly twice in its correct position.
TURKEY
So my questions are basically:
1. Is having 2 words per letter-position a function of building a unique solution to such a puzzle?
2. What's the algorithm for doing this without a computer? One attempt I worked at noted that the words RECKON, OUTAGE, BOUNTY, UNREST, and SWAYEDdon't share any letter-positions so they contain 5 of the letters, but I couldn't finish the thought.
I think I'm missing something obvious about the puzzle's design and intended solving process. I usually crank through puzzle books in an hour or two, but this one just felt impossible!
• How do you add spoiler text? Jan 3, 2016 at 3:30
• See here Jan 3, 2016 at 3:51
My process:
My first thought was, "Let's look at the common/repeated letters." So I took all the letters from each place and found all the repeated ones, giving this:
MFCRTOSSBTSU: TTSSS
YUAEOUTWOOTN: TTUUOOO
RMTCWTEAUNOR: TTRR
IBKKAAAYNGLE: KKAAA
ALIORGDETUES: EE
DENNDEYDYENT: NNNEEEDDDYY
Just looking down the rows at that, TURKEY kinda jumps out. But that didn't have to be true- it could have been that some of the letters were triples and some were singles, and wouldn't be revealed by looking at the multiples. If that were the case, I would have done a crossword-type deal. Taking S as a given, eliminate all of the multiples that show up in the same word as one starting with S, and then figure out what combos sum the remaining multiples to 12 (because there are 12 letters from TURKEY in the list). That process would give us this: SO(T/R)K-T. (The T at the end is a freebie, because that's the only letter that appears only once as a terminal.) No matches, though, unless SORKAT is a word now.
Doing the same with T gives us T(U/-)(T/R/-)(K/-)(E/-)(N/Y), which means our word is either TURKEY or TUTKEY, or TURKEN with one of the four intervening letters swapped. And if none of those had worked, I would have assumed the first letter was neither S nor T and then continued in the same way.
• This is ultimately the algorithm I picked up on (that the sum of the matches has to be 12, and the largest multiple was 3 so it had to include at least 3 multiples ) - they clearly wanted you to just look for multiples, but the problem seemed much more expansive when I first took a look at it. Jan 3, 2016 at 5:42 | 846 | 3,226 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-50 | longest | en | 0.957611 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-p-section-p-7-equations-exercise-set-page-106/70 | 1,695,284,963,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233505362.29/warc/CC-MAIN-20230921073711-20230921103711-00556.warc.gz | 870,705,810 | 14,422 | ## Precalculus (6th Edition) Blitzer
$\left\{-6, 2\right\}$
RECALL: To solve $x^2 + bx=c$ by completing the square, add $\left(\frac{b}{2}\right)^2$ on both sides of the equation. Thus, to solve the given equation by completing the square, add $(\frac{4}{2})^2=2^2=4$ on both sides of the equation to obtain: $x^2+4x+4=12+4 \\x^2+4x+4=16 \\(x+2)^2=16$ Take the square root of both sides to obtain: $x+2=\pm \sqrt{16} \\x+2=\pm \sqrt{4^2} \\x+2=\pm 4$ Adad $-2$ on both sides of the equation to obtain: $x = -2 \pm 4$ We get two solutions: $x_1 = -2+4=2 \\x_2 = -2-4=-6$ Therefore, the solution of the given equation is: $\left\{-6, 2\right\}$. | 257 | 644 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-40 | latest | en | 0.761969 |
https://numberworld.info/32201 | 1,653,447,928,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662578939.73/warc/CC-MAIN-20220525023952-20220525053952-00680.warc.gz | 459,493,695 | 3,772 | # Number 32201
### Properties of number 32201
Cross Sum:
Factorization:
Divisors:
1, 13, 2477, 32201
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
7dc9
Base 32:
ve9
sin(32201)
-0.3190238084739
cos(32201)
0.94774670119542
tan(32201)
-0.33661294528539
ln(32201)
10.379752786951
lg(32201)
4.5078693588927
sqrt(32201)
179.44637081869
Square(32201)
### Number Look Up
Look Up
32201 (thirty-two thousand two hundred one) is a very impressive number. The cross sum of 32201 is 8. If you factorisate 32201 you will get these result 13 * 2477. 32201 has 4 divisors ( 1, 13, 2477, 32201 ) whith a sum of 34692. 32201 is not a prime number. The number 32201 is not a fibonacci number. The number 32201 is not a Bell Number. The figure 32201 is not a Catalan Number. The convertion of 32201 to base 2 (Binary) is 111110111001001. The convertion of 32201 to base 3 (Ternary) is 1122011122. The convertion of 32201 to base 4 (Quaternary) is 13313021. The convertion of 32201 to base 5 (Quintal) is 2012301. The convertion of 32201 to base 8 (Octal) is 76711. The convertion of 32201 to base 16 (Hexadecimal) is 7dc9. The convertion of 32201 to base 32 is ve9. The sine of 32201 is -0.3190238084739. The cosine of the figure 32201 is 0.94774670119542. The tangent of the figure 32201 is -0.33661294528539. The square root of 32201 is 179.44637081869.
If you square 32201 you will get the following result 1036904401. The natural logarithm of 32201 is 10.379752786951 and the decimal logarithm is 4.5078693588927. I hope that you now know that 32201 is very impressive figure! | 609 | 1,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-21 | latest | en | 0.750006 |
https://proofwiki.org/wiki/Definition:Improper_Integral/Half_Open_Interval | 1,674,999,204,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00403.warc.gz | 486,964,641 | 10,548 | # Definition:Improper Integral/Half Open Interval
## Definition
#### Open Above
Let $f$ be a real function which is continuous on the half open interval $\hointr a b$.
Then the improper integral of $f$ over $\hointr a b$ is defined as:
$\ds \int_a^{\mathop \to b} \map f t \rd t := \lim_{\gamma \mathop \to b} \int_a^\gamma \map f t \rd t$
#### Open Below
Let $f$ be a real function which is continuous on the half open interval $\hointl a b$.
Then the improper integral of $f$ over $\hointl a b$ is defined as:
$\ds \int_{\mathop \to a}^b \map f t \rd t := \lim_{\gamma \mathop \to a} \int_\gamma^b \map f t \rd t$ | 205 | 624 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.512816 |
https://physics.stackexchange.com/questions/498902/is-there-a-relationship-between-the-frequency-of-a-standing-wave-on-an-oscillati | 1,579,372,401,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593295.11/warc/CC-MAIN-20200118164132-20200118192132-00209.warc.gz | 602,799,233 | 32,749 | # Is there a relationship between the frequency of a standing wave on an oscillating string and the half-life of its amplitude as it decays?
I'm asking in the context of an experimental exercise where we used a speaker to create a standing wave on a string with constant tension, then turned off the speaker and allowed the amplitudes to decay over time. I'm trying to see if increasing the frequency in order of the harmonics appearing affects how long it takes the amplitude to reach half of the initial amplitude.
I am a truly amateur physics student and am assuming an ideal string that decays exponentially, and I definitely do not know very complicated approaches to this problem through math. I really just want to know if I am chasing something that either doesn't make sense or is only approachable through college level physics/math. I would greatly appreciate any help, and I apologize if this is too rudimentary for this forum (first time using it).
• Note that exponential decay is normally expressed in terms of its time constant, which is the time that it takes for 63% of the decay to take place (e.g., $1-e^{-1}$). Due to this, the standard solutions to this problem are not formulated as a function of half-life. Are you open to the idea of getting an answer in terms of time constant? – David White Aug 27 '19 at 2:50
• @DavidWhite Actually yes, I'm glad you mentioned this because I started looking into it today as an better measure of comparing the decay, so I'm hopefully going in the right direction (?) and I think I understand it enough to find an answer involving it. – celeste Aug 27 '19 at 3:01
• Welcome to Physics SE :) Your curiosity and description will make it a great place to stay here. – Stefan Bischof Aug 27 '19 at 5:05
$$\frac{\partial^2y}{\partial t^2}y +2b\frac{\partial y}{\partial t}=c^2\frac{\partial^2y}{\partial x^2}$$ where $$y$$ is the wave amplitude, $$t$$ is time, $$x$$ is the position coordinate along the string, $$c$$ is the speed of the waves in the string (which depends on the string material and the tension) and $$2b$$ is a damping consant. You don't need to understand exactly what this equation means and how to solve it, but it models the damping force acting on a unit length of each piece of the string as being proportional to its speed, and $$2b$$ is the proportionality constant. For a string of length $$L$$ fixed at the end points $$x = 0$$ and $$x = L$$, and a sufficiently small $$b$$, the standing wave solutions are
$$y_m(x,t) = e^{-bt} \cos\left(\sqrt{\omega_m^2 - b^2}t+\phi_m\right)\sin\left(\frac{\omega_m}{c}x\right)$$ where $$m$$ numbers the different modes ($$m=1$$ being the fundamental), $$\phi_m$$ is some constant phase and $$\omega_m = m\frac{c\pi}{L}.$$
$$e^{-bt}$$ is the factor governing how the amplitude decays over time, and the decay is exponential as you guessed. However, in this simple model the time constant of the decay is $$b^{-1}$$ (the half-life is $$b^{-1}\ln2$$) which is the same for all modes. This may be more or less true for an experiment such as yours, but especially at very high frequencies I suspect this may no longer be the case. | 783 | 3,148 | {"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": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-05 | latest | en | 0.931277 |
https://empireessays.com/what-proportion-were-biology-majors/ | 1,695,362,074,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506329.15/warc/CC-MAIN-20230922034112-20230922064112-00501.warc.gz | 265,370,430 | 17,142 | what proportion were biology majors?
Statistics Question- z-scoreWe want to compare the grades on an exam in an intro statistics course for biology majors versus public health majors. Suppose 72% percent of the class are biology majors, while the remaining 28% are public health majors. The biology majors earned grades that are normally distributed, with a mean of 84 and a standard deviation of 10. The public health majors earned grades that are also normally distributed, with a mean of 87 and a standard deviation of 14.What exam score corresponds to the 80th percentile of the distribution for public health majors? (2pt)What proportion of the class got an A on the test (scored at least 90)? (4pt)Of the students who didnt make an A, what proportion were biology majors? (4pt) | 178 | 789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-40 | latest | en | 0.955101 |
http://www.self.gutenberg.org/articles/eng/Log-normal | 1,566,219,341,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314732.59/warc/CC-MAIN-20190819114330-20190819140330-00343.warc.gz | 319,042,229 | 25,548 | ### Log-Normal
Template:Probability distribution
| cdf = $\frac12 + \frac12\,\mathrm\left\{erf\right\}\Big\left[\frac\left\{\ln x-\mu\right\}\left\{\sqrt\left\{2\right\}\sigma\right\}\Big\right]$
| mean = $e^\left\{\mu+\sigma^2/2\right\}$
| median = $e^\left\{\mu\right\}\,$
| mode = $e^\left\{\mu-\sigma^2\right\}$
| variance = $\left(e^\left\{\sigma^2\right\}\!\!-1\right) e^\left\{2\mu+\sigma^2\right\}$
| skewness = $\left(e^\left\{\sigma^2\right\}\!\!+2\right) \sqrt\left\{e^\left\{\sigma^2\right\}\!\!-1\right\}$
| kurtosis = $e^\left\{4\sigma^2\right\}\!\! + 2e^\left\{3\sigma^2\right\}\!\! + 3e^\left\{2\sigma^2\right\}\!\! - 6$
| entropy = $\frac12 + \frac12 \ln\left(2\pi\sigma^2\right) + \mu$
| mgf = (defined only on the negative half-axis, see text)
| char = representation $\sum_\left\{n=0\right\}^\left\{\infty\right\}\frac\left\{\left(it\right)^n\right\}\left\{n!\right\}e^\left\{n\mu+n^2\sigma^2/2\right\}$ is asymptotically divergent but sufficient for numerical purposes
| fisher = $\begin\left\{pmatrix\right\}1/\sigma^2&0\\0&1/\left(2\sigma^4\right)\end\left\{pmatrix\right\}$
}}
In probability theory, a log-normal distribution is a continuous probability distribution of a random variable whose logarithm is normally distributed. Thus, if the random variable $Y$ is log-normally distributed, then $X = \log\left(Y\right)$ has a normal distribution. Likewise, if $X$ has a normal distribution, then $Y = \exp\left(X\right)$ has a log-normal distribution. A random variable which is log-normally distributed takes only positive real values.
Log-normal is also written log normal or lognormal. The distribution is occasionally referred to as the Galton distribution or Galton's distribution, after Francis Galton.[1] The log-normal distribution also has been associated with other names, such as McAlister, Gibrat and Cobb–Douglas.[1]
A variable might be modeled as log-normal if it can be thought of as the multiplicative product of many independent random variables each of which is positive. (This is justified by considering the central limit theorem in the log-domain.) For example, in finance, the variable could represent the compound return from a sequence of many trades (each expressed as its return + 1); or a long-term discount factor can be derived from the product of short-term discount factors. In wireless communication, the sas caused by shadowing or slow fading from random objects is often assumed to be log-normally distributed: see log-distance path loss model.
The log-normal distribution is the maximum entropy probability distribution for a random variate X for which the mean and variance of $\ln\left(X\right)$ are fixed.[2]
## μ and σ
In a log-normal distribution X, the parameters denoted μ and σ are, respectively, the mean and standard deviation of the variable’s natural logarithm (by definition, the variable’s logarithm is normally distributed), which means
$X=e^\left\{\mu+\sigma Z\right\}$
with Z a standard normal variable.
This relationship is true regardless of the base of the logarithmic or exponential function. If loga(Y) is normally distributed, then so is logb(Y), for any two positive numbers ab ≠ 1. Likewise, if $e^X$ is log-normally distributed, then so is $a^\left\{X\right\}$, where a is a positive number ≠ 1.
On a logarithmic scale, μ and σ can be called the location parameter and the scale parameter, respectively.
In contrast, the mean and standard deviation of the non-logarithmized sample values are denoted m and s.d. in this article.
A log-normal distribution with mean m and variance v has parameters[3]
\mu=\ln\left(\frac{m^2}{\sqrt{v+m^2}}\right), \sigma=\sqrt{\ln(1+\frac{v}{m^2})}
## Characterization
### Probability density function
The probability density function of a log-normal distribution is:[1]
$f_X\left(x;\mu,\sigma\right) = \frac\left\{1\right\}\left\{x \sigma \sqrt\left\{2 \pi\right\}\right\}\, e^\left\{-\frac\left\{\left(\ln x - \mu\right)^2\right\}\left\{2\sigma^2\right\}\right\},\ \ x>0$
This follows by applying the change-of-variables rule on the density function of a normal distribution.
### Cumulative distribution function
$F_X\left(x;\mu,\sigma\right) = \frac12 \left\left[ 1 + \operatorname\left\{erf\right\}\!\left\left(\frac\left\{\ln x - \mu\right\}\left\{\sigma\sqrt\left\{2\right\}\right\}\right\right) \right\right] = \frac12 \operatorname\left\{erfc\right\}\!\left\left(-\frac\left\{\ln x - \mu\right\}\left\{\sigma\sqrt\left\{2\right\}\right\}\right\right) = \Phi\bigg\left(\frac\left\{\ln x - \mu\right\}\left\{\sigma\right\}\bigg\right),$
where erfc is the complementary error function, and Φ is the cumulative distribution function of the standard normal distribution.
### Characteristic function and moment generating function
All moments of the log-normal distributions exist and it holds that
$\operatorname\left\{E\right\}\left(X^n\right)=\mathrm\left\{e\right\}^\left\{n\mu+\frac\left\{n^2\sigma^2\right\}\left\{2\right\}\right\}$.
However, the moment generating function
$\operatorname\left\{E\right\}\left(e^\left\{t X\right\}\right)=\sum_\left\{n=0\right\}^\infty \frac\left\{t^n\right\}\left\{n!\right\}\operatorname\left\{E\right\}\left(X^n\right)$
does not converge.
The characteristic function, E[e itX], has a number of representations. The integral itself converges for Im(t) ≤ 0. The simplest representation is obtained by Taylor expanding e itX and using formula for moments below, giving
$\varphi\left(t\right) = \sum_\left\{n=0\right\}^\infty \frac\left\{\left(it\right)^n\right\}\left\{n!\right\}e^\left\{n\mu+n^2\sigma^2/2\right\}.$
This series representation is divergent for Re(σ2) > 0. However, it is sufficient for evaluating the characteristic function numerically at positive $\sigma$ as long as the upper limit in the sum above is kept bounded, n ≤ N, where
$\max\left(|t|,|\mu|\right) \ll N \ll \frac\left\{2\right\}\left\{\sigma^2\right\}\ln\frac\left\{2\right\}\left\{\sigma^2\right\}$
and σ2 < 0.1. To bring the numerical values of parameters μσ into the domain where strong inequality holds true one could use the fact that if X is log-normally distributed then Xm is also log-normally distributed with parameters μmσm. Since $\mu\sigma^2 \propto m^3$, the inequality could be satisfied for sufficiently small m. The sum of series first converges to the value of φ(t) with arbitrary high accuracy if m is small enough, and left part of the strong inequality is satisfied. If considerably larger number of terms are taken into account the sum eventually diverges when the right part of the strong inequality is no longer valid.
Another useful representation is available[4][5] by means of double Taylor expansion of e(ln x − μ)2/(2σ2).
The moment-generating function for the log-normal distribution does not exist on the domain R, but only exists on the half-interval (−∞, 0].
## Properties
### Location and scale
For the log-normal distribution, the location and scale properties of the distribution are more readily treated using the geometric mean and geometric standard deviation than the arithmetic mean and standard deviation.
#### Geometric moments
The geometric mean of the log-normal distribution is $e^\left\{\mu\right\}$. Because the log of a log-normal variable is symmetric and quantiles are preserved under monotonic transformations, the geometric mean of a log-normal distribution is equal to its median.[6]
The geometric mean (mg) can alternatively be derived from the arithmetic mean (ma) in a log-normal distribution by:
$m_g = m_ae^\left\{-\tfrac\left\{1\right\}\left\{2\right\}\sigma^2\right\}.$
Note that the geometric mean is less than the arithmetic mean. This is due to the AM–GM inequality, and corresponds to the logarithm being convex down. The correction term $e^\left\{-\tfrac\left\{1\right\}\left\{2\right\}\sigma^2\right\}$ can accordingly be interpreted as a convexity correction. From the point of view of stochastic calculus, this is the same correction term as in Itō's lemma for geometric Brownian motion.
The geometric standard deviation is equal to $e^\left\{\sigma\right\}$.
#### Arithmetic moments
If X is a lognormally distributed variable, its expected value (E – the arithmetic mean), variance (Var), and standard deviation (s.d.) are
\begin\left\{align\right\}
& \operatorname{E}[X] = e^{\mu + \tfrac{1}{2}\sigma^2}, \\
& \operatorname{Var}[X] = (e^{\sigma^2} - 1) e^{2\mu + \sigma^2} = (e^{\sigma^2} - 1)(\operatorname{E}[X])^2\\
& \operatorname{s.d.}[X] = \sqrt{\operatorname{Var}[X]} = e^{\mu + \tfrac{1}{2}\sigma^2}\sqrt{e^{\sigma^2} - 1}.
\end{align}
Equivalently, parameters μ and σ can be obtained if the expected value and variance are known; it is simpler if σ is computed first:
\begin\left\{align\right\}
\mu &= \ln(\operatorname{E}[X]) - \frac12 \ln\!\left(1 + \frac{\mathrm{Var}[X]}{(\operatorname{E}[X])^2}\right) = \ln(\operatorname{E}[X]) - \frac12 \sigma^2, \\
\sigma^2 &= \ln\!\left(1 + \frac{\operatorname{Var}[X]}{(\operatorname{E}[X])^2}\right).
\end{align}
For any real or complex number s, the sth moment of log-normal X is given by[1]
$\operatorname\left\{E\right\}\left[X^s\right] = e^\left\{s\mu + \tfrac\left\{1\right\}\left\{2\right\}s^2\sigma^2\right\}.$
A log-normal distribution is not uniquely determined by its moments E[Xk] for k ≥ 1, that is, there exists some other distribution with the same moments for all k.[1] In fact, there is a whole family of distributions with the same moments as the log-normal distribution.
### Mode and median
The mode is the point of global maximum of the probability density function. In particular, it solves the equation (ln ƒ)′ = 0:
$\mathrm\left\{Mode\right\}\left[X\right] = e^\left\{\mu - \sigma^2\right\}.$
The median is such a point where FX = 1/2:
$\mathrm\left\{Med\right\}\left[X\right] = e^\mu\,.$
### Coefficient of variation
The coefficient of variation is the ratio s.d. over m (on the natural scale) and is equal to:
$\sqrt\left\{e^\left\{\sigma^2\right\}\!\!-1\right\}$
### Partial expectation
The partial expectation of a random variable X with respect to a threshold k is defined as $g\left(k\right) = \int_k^\infty \!xf\left(x\right)\, dx$ where $f\left(x\right)$ is the probability density function of X. Alternatively, and using the definition of conditional expectation, it can be written as g(k)=E[X | X > k]*P(X > k). For a log-normal random variable the partial expectation is given by:
$g\left(k\right) = \int_k^\infty \!xf\left(x\right)\, dx$
= e^{\mu+\tfrac{1}{2}\sigma^2}\, \Phi\!\left(\frac{\mu+\sigma^2-\ln k}{\sigma}\right).
The derivation of the formula is provided in the discussion of this World Heritage Encyclopedia entry. The partial expectation formula has applications in insurance and economics, it is used in solving the partial differential equation leading to the Black–Scholes formula.
### Other
A set of data that arises from the log-normal distribution has a symmetric Lorenz curve (see also Lorenz asymmetry coefficient).[7]
The harmonic (H), geometric (G) and arithmetic (A) means of this distribution are related;[8] such relation is given by
$H = \frac\left\{G^2\right\}\left\{ A\right\} .$
Log-normal distributions are infinitely divisible.[1]
## Occurrence
The log-normal distribution is important in the description of natural phenomena. The reason is that for many natural processes of growth, growth rate is independent of size. This is also known as Gibrat's law, after Robert Gibrat (1904–1980) who formulated it for firms (companies). It can be shown that a growth process following Gibrat's law will result in entity sizes with a log-normal distribution.[9] Examples include:
• In biology and medicine,
• Measures of size of living tissue (length, skin area, weight);[10]
• For highly communicable epidemics, such as SARS in 2003, if publication intervention is involved, the the number of hospitalized cases is shown to satistfy the lognormal distribution with no free parameters if an entropy is assumed and the standard deviation is determined by the principle of maximum rate of entropy production[11] .
• The length of inert appendages (hair, claws, nails, teeth) of biological specimens, in the direction of growth;
• Certain physiological measurements, such as blood pressure of adult humans (after separation on male/female subpopulations)[12]
Consequently, reference ranges for measurements in healthy individuals are more accurately estimated by assuming a log-normal distribution than by assuming a symmetric distribution about the mean.
• In hydrology, the log-normal distribution is used to analyze extreme values of such variables as monthly and annual maximum values of daily rainfall and river discharge volumes.[13]
• in social sciences and demographics
• In economics, there is evidence that the income of 97%–99% of the population is distributed log-normally.[14]
• In finance, in particular the Black–Scholes model, changes in the logarithm of exchange rates, price indices, and stock market indices are assumed normal[15] (these variables behave like compound interest, not like simple interest, and so are multiplicative). However, some mathematicians such as Benoît Mandelbrot have argued [16] that log-Lévy distributions which possesses heavy tails would be a more appropriate model, in particular for the analysis for stock market crashes. Indeed stock price distributions typically exhibit a fat tail.[17]
• city sizes
• technology
• In reliability analysis, the lognormal distribution is often used to model times to repair a maintainable system.
• In wireless communication, "the local-mean power expressed in logarithmic values, such as dB or neper, has a normal (i.e., Gaussian) distribution." [18]
• It has been proposed that coefficients of friction and wear may be treated as having a lognormal distribution [19]
• In spray process, such as droplet impact, the size of secondary produced droplet has a lognormal distribution, with the standard deviation ：$\sigma=\frac\left\{\sqrt\left\{6\right\}\right\}\left\{6\right\}$ determined by the principle of maximum rate of entropy production[20] If the lognormal distribution is inserted into the Shannon entropy expression and if the rate of entropy production is maximized (principle of maximum rate of entropy production), then σ is given by :$\sigma=\frac\left\{1\right\}\left\{\sqrt\left\{6\right\}\right\}$[20] and with this parameter the droplet size distribution for spray process is well predicted. It is an open question whether this value of σ has some generality for other cases, though for spreading of communicable epidemics, σ is shown also to take this value.[11]
## Maximum likelihood estimation of parameters
For determining the maximum likelihood estimators of the log-normal distribution parameters μ and σ, we can use the same procedure as for the normal distribution. To avoid repetition, we observe that
$f_L \left(x;\mu, \sigma\right) = \prod_\left\{i=1\right\}^n \left\left(\frac 1 x_i\right\right) \, f_N \left(\ln x; \mu, \sigma\right)$
where by ƒL we denote the probability density function of the log-normal distribution and by ƒN that of the normal distribution. Therefore, using the same indices to denote distributions, we can write the log-likelihood function thus:
\begin{align} \ell_L (\mu,\sigma | x_1, x_2, \dots, x_n)
& {} = - \sum _k \ln x_k + \ell_N (\mu, \sigma | \ln x_1, \ln x_2, \dots, \ln x_n) \\
& {} = \operatorname {constant} + \ell_N (\mu, \sigma | \ln x_1, \ln x_2, \dots, \ln x_n). \end{align}
Since the first term is constant with regard to μ and σ, both logarithmic likelihood functions, L and N, reach their maximum with the same μ and σ. Hence, using the formulas for the normal distribution maximum likelihood parameter estimators and the equality above, we deduce that for the log-normal distribution it holds that
$\widehat \mu = \frac \left\{\sum_k \ln x_k\right\} n,$
\widehat \sigma^2 = \frac {\sum_k \left( \ln x_k - \widehat \mu \right)^2} {n}.
## Multivariate log-normal
If $\boldsymbol X \sim \mathcal\left\{N\right\}\left(\boldsymbol\mu,\,\boldsymbol\Sigma\right)$ is a multivariate normal distribution then $\boldsymbol Y=\exp\left(\boldsymbol X\right)$ has a multivariate log-normal distribution[21] with mean
$\operatorname\left\{E\right\}\left[\boldsymbol Y\right]_i=e^\left\{\mu_i+\frac\left\{1\right\}\left\{2\right\}\Sigma_\left\{ii\right\}\right\} ,$
$\operatorname\left\{Var\right\}\left[\boldsymbol Y\right]_\left\{ij\right\}=e^\left\{\mu_i+\mu_j + \frac\left\{1\right\}\left\{2\right\}\left(\Sigma_\left\{ii\right\}+\Sigma_\left\{jj\right\}\right) \right\}\left( e^\left\{\Sigma_\left\{ij\right\}\right\} - 1\right) .$
## Generating log-normally distributed random variates
Given a random variate Z drawn from the normal distribution with 0 mean and 1 standard deviation, then the variate
$X= e^\left\{\mu + \sigma Z\right\}\,$
has a log-normal distribution with parameters $\mu$ and $\sigma$.
## Related distributions
• If $X \sim \mathcal\left\{N\right\}\left(\mu, \sigma^2\right)$ is a normal distribution, then $\exp\left(X\right) \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left(\mu, \sigma^2\right).$
• If $X \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left(\mu, \sigma^2\right)$ is distributed log-normally, then $\ln\left(X\right) \sim \mathcal\left\{N\right\}\left(\mu, \sigma^2\right)$ is a normal random variable.
• If $X_j \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left(\mu_j, \sigma_j^2\right)$ are n independent log-normally distributed variables, and $Y = \textstyle\prod_\left\{j=1\right\}^n X_j$, then Y is also distributed log-normally:
$Y \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\Big\left(\textstyle \sum_\left\{j=1\right\}^n\mu_j,\ \sum_\left\{j=1\right\}^n \sigma_j^2 \Big\right).$
• Let $X_j \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left(\mu_j,\sigma_j^2\right)\$ be independent log-normally distributed variables with possibly varying σ and μ parameters, and $Y=\textstyle\sum_\left\{j=1\right\}^n X_j$. The distribution of Y has no closed-form expression, but can be reasonably approximated by another log-normal distribution Z at the right tail. Its probability density function at the neighborhood of 0 has been characterized[22] and it does not resemble any log-normal distribution. A commonly used approximation (due to L.F. Fenton, but previously stated by R.I. Wilkinson without mathematical justification[23]) is obtained by matching the mean and variance:
\begin\left\{align\right\}
\sigma^2_Z &= \log\!\left[ \frac{\sum e^{2\mu_j+\sigma_j^2}(e^{\sigma_j^2}-1)}{(\sum e^{\mu_j+\sigma_j^2/2})^2} + 1\right], \\
\mu_Z &= \log\!\left[ \sum e^{\mu_j+\sigma_j^2/2} \right] - \frac{\sigma^2_Z}{2}.
\end{align}
In the case that all $X_j$ have the same variance parameter $\sigma_j=\sigma$, these formulas simplify to
\begin\left\{align\right\}
\sigma^2_Z &= \log\!\left[ (e^{\sigma^2}-1)\frac{\sum e^{2\mu_j}}{(\sum e^{\mu_j})^2} + 1\right], \\
\mu_Z &= \log\!\left[ \sum e^{\mu_j} \right] + \frac{\sigma^2}{2} - \frac{\sigma^2_Z}{2}.
\end{align}
• If $X \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left(\mu, \sigma^2\right)$, then X + c is said to have a shifted log-normal distribution with support x ∈ (c, +∞). E[X + c] = E[X] + c, Var[X + c] = Var[X].
• If $X \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left(\mu, \sigma^2\right)$, then $a X \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left( \mu + \ln a,\ \sigma^2\right).$
• If $X \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left(\mu, \sigma^2\right)$, then $\tfrac\left\{1\right\}\left\{X\right\} \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left(-\mu,\ \sigma^2\right).$
• If $X \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left(\mu, \sigma^2\right)$ then $X^a \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left(a\mu,\ a^2 \sigma^2\right).$ for $a \neq 0\,$
• Lognormal distribution is a special case of semi-bounded Johnson distribution
• If $X|Y \sim \mathrm\left\{Rayleigh\right\}\left(Y\right)\,$ with $Y \sim \operatorname\left\{Log-\mathcal\left\{N\right\}\right\}\left(\mu, \sigma^2\right)$, then $X \sim \mathrm\left\{Suzuki\right\}\left(\mu, \sigma\right)\,$ (Suzuki distribution)
## Similar distributions
A substitute for the log-normal whose integral can be expressed in terms of more elementary functions[24] can be obtained based on the logistic distribution to get an approximation for the CDF
$F\left(x;\mu,\sigma\right) = \left\left[\left\left(\frac\left\{e^\mu\right\}\left\{x\right\}\right\right)^\left\{\pi/\left(\sigma \sqrt\left\{3\right\}\right)\right\} +1\right\right]^\left\{-1\right\}.$
This is a log-logistic distribution.
## References
• Aitchison, J. and Brown, J.A.C. (1957) The Lognormal Distribution, Cambridge University Press.
• E. Limpert, W. Stahel and M. Abbt (2001) Log-normal Distributions across the Sciences: Keys and Clues, BioScience, 51 (5), 341–352.
• MathWorld. Electronic document, retrieved October 26, 2006.
• expand by hand | 6,252 | 21,199 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 78, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-35 | latest | en | 0.748357 |
https://secondstaralaska.com/the-test-for-a0-t-test-the-critical-values-are/ | 1,579,789,506,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250610919.33/warc/CC-MAIN-20200123131001-20200123160001-00520.warc.gz | 656,844,092 | 10,305 | ## The test for a=0 (t-test). The critical values are
The long run
equilibrium relationship between spot VIX and VIX future prices is given from
the following equation:
(1)
We Will Write a Custom Essay Specifically
For You For Only \$13.90/page!
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Where:
Ø
: the VIX
future prices
Ø
: the spot
VIX price
It is well known
that the above equation cannot be tested by ordinary least squares if at least one
of the variables is not stationary. So, the first step in time series testes is
to test for stationarity. The null hypothesis of a unit root test is that the variable
is not stationary. The models used for the unit root tests are the following:
(without trend)
(with trend)
Where
stand for the spot VIX and the ten nearest to
maturity VIX Future contracts prices respectively. Having run Augmented Dicey-Fuller unit root tests on spot VIX and
ten VIX futures prices indexes. All t-statistics are below 1% critical values,
as a result the null hypothesis cannot be rejected for any of the ten indexes.
Since the null hypothesis cannot be rejected all the time series of VIX and
futures prices are not stationary.
A nonstationary
time series which has stationary first difference, is said to be integrated to order
1, it is denoted as I(1). Having run Augmented Dicey-Fuller unit root tests on
the first differences of VIX and Fi. The null hypothesis is rejected for all
the indexes at the 1% significance level, so there is no unit root problem. In
conclusion all the ten indexes are I(1) processes.
In 1987 Enger
and Granger proved that if we have two I(1) processes and their liner combination
is I(0) (stationary), the two time series are cointegrated. From the economical
perspective, two time series are said to be cointegrated if they have a long-term
or else equilibrium relationship between them. One way to test if two time
series are cointegrated is to conduct test statistics from the residuals of
their regression. Let
denote the estimated residuals from equation
(1), a test for no cointegration is given from a test for unit root of those
residuals. The ADF regression equation is:
Test
statistics is a t-ratio test for a=0 (t-test). The critical values are -3.34 for
5% confidence interval and -3.04 for 1% confidence interval. Significant
negative test statistics suggest cointegration (rejection of the unit root hypothesis).
Table IV and V presents the Enger-Granger cointegration results for different
pairs of time series. From the tables belowDT1
DT1Ti sx?li? na kanw edw? | 611 | 2,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-05 | latest | en | 0.9262 |
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# I decided that I would either take my daughter to the cinema
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I decided that I would either take my daughter to the cinema [#permalink]
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07 Jun 2009, 06:56
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I decided that I would either take my daughter to the cinema or go by myself.
a. I would either take my daughter to the cinema or go by myself.
b. I either would take my daughter to the cinema or go by myself.
c. I would either take my daughter to the cinema or else I would myself go.
d. either I would take my daughter to the cinema or go by myself.
e. either I would myself go to the cinema or take my daughter
If you have any questions
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07 Jun 2009, 09:57
D.
Either is required before "I would" and parallel.
SVP
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07 Jun 2009, 15:03
humans wrote:
D.
Either is required before "I would" and parallel.
I disagree. Option D says "either I....or....go."-----> one is a suject pronoun and the other is a verb
I choose option A as my answer. Option A says "either take....or.....go"--->both are verbs
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07 Jun 2009, 19:07
tarek99 wrote:
humans wrote:
D.
Either is required before "I would" and parallel.
I disagree. Option D says "either I....or....go."-----> one is a suject pronoun and the other is a verb
I choose option A as my answer. Option A says "either take....or.....go"--->both are verbs
I choose A as well
Either take ... or ... go is parallel
_________________
Is this okay?
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08 Jun 2009, 00:13
I take A too.
<either take ... or go> Completely parallel.
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08 Jun 2009, 06:43
Why not B ?
B also looks attactive..
I either XXX or XX..
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08 Jun 2009, 06:46
Neochronic wrote:
Why not B ?
B also looks attactive..
I either XXX or XX..
you have a point. But I am thinking that B would be correct if the structure is this way 'either would X or would Y' ..
so missing 'would' after 'or' makes B wrong? that is my logic ..
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08 Jun 2009, 07:13
Neochronic wrote:
Why not B ?
B also looks attactive..
I either XXX or XX..
because the construction here is:
either would....or go
in order to apply "would" to all the following verbs, you have to place the "would" before "either." So in option A, it says:
would take...or....would go
whenever you need to repeat a certain word, you will have achieved that when you place that word right before "either."
and remember, the construction should be: it's either A or B
Re: cinema [#permalink] 08 Jun 2009, 07:13
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Display posts from previous: Sort by | 1,336 | 4,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-13 | longest | en | 0.936774 |
https://socratic.org/questions/how-much-energy-is-needed-to-raise-the-temperature-of-2-0-g-of-water-5-00-c | 1,632,844,054,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00452.warc.gz | 565,635,233 | 5,776 | # How much energy is needed to raise the temperature of 2.0 g of water 5.00°C?
##### 1 Answer
Sep 13, 2017
$41.84 J$
#### Explanation:
Water has to absorb 4.184 Joules of heat for the temperature of one gram of water to increase 1 degree Celsius (°C).
https://water.usgs.gov/edu/heat-capacity.html
$\text{Energy} = 2.0 g \times {5.00}^{o} C \times 4.184 \frac{J}{g {\ast}^{o} C} = 41.84 J$ | 140 | 394 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "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} | 3.921875 | 4 | CC-MAIN-2021-39 | latest | en | 0.793223 |
http://learn.conservativeyeshiva.org/ohalot-13-3-htm/ | 1,590,942,744,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413551.52/warc/CC-MAIN-20200531151414-20200531181414-00493.warc.gz | 70,138,303 | 13,584 | Ohalot, Chapter Thirteen, Mishnah Three
Introduction
Today’s mishnah deals with how large an opening in a door must be to allow in and out impurity.
Mishnah Three
1) [With regard to] a hole in the door:
a) Its minimum size is that of a fist, the words of Rabbi Akiva.
b) Rabbi Tarfon says: one handbreadth square.
2) If the carpenter had left a space at the bottom or the top [of the door] or if one had shut [the door] but not closed it tightly, or if the wind blew it open, the minimum size is that of a fist.
Explanation
Section one: Rabbi Akiva holds that since these holes in the door were not made intentionally, they should be compared with the naturally forming holes referred to in mishnah one. For impurity to travel through such a hole, it must be the size of a fist.
Rabbi Tarfon says that it needs only be a handbreadth square (slightly smaller than a fist) for these holes are meant for human use (see mishnah one).
Section two: There are three spaces in a door referred to here: 1) the carpenter didn’t make the door as large as it was supposed to be; 2) the door wasn’t tightly closed; 3) the wind blew the door slightly open. All of these are openings that were probably not intentionally formed, nor are they holes that are used by human. Therefore, the measure to allow through impurity is that of a fist, which is larger than a handbreadth. | 339 | 1,391 | {"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} | 2.546875 | 3 | CC-MAIN-2020-24 | latest | en | 0.96697 |
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This algebra 2 math video tutorial provides an introduction into function notation. 0 & \text{ if } x \text{ is an irrational number}, Evaluating Functions on Brilliant, the largest community of math and science problem solvers. This tells me that, were I to be graphing the line y = 4x – 3, the point (3, 9) would be on the line.By the way, evaluating the same equation at three or more points like this, and getting a list of points, is how you plot points and graph equations.In the case of the equation y = 4x – 3, the points from the evaluating we've done (including … Mathematics. Some of the worksheets for this concept are Evaluating functions date period, , Algebra i name k3 9 function notation work, Concept 22 evaluating functions, Function notation work 2, Algebra i name function notation work, Ws69 name, Function notation and evaluating functions practice work. Evaluating Functions | Graph. CCSS.Math: 8.F.A.1, HSF.IF.A.2. What is the value of f(1213)+f(π)+f(0.75)?f\left(\frac{12}{13}\right)+f(\pi)+f(0.75)?f(1312)+f(π)+f(0.75)? Functions are written using function notation. Donate or volunteer today! Start studying EVALUATING FUNCTIONS. Algebra 2 Name_ ©q \\2n0H2q0_ kKiuwtWaE KShoefUtLweafrBev tLIL\\CP.b sAulSl] orsiOgIhXtMsC erepsxenrxvTejdH. Students will practice evaluating functions and applying function notation. Play this game to review Algebra I. f(x)=4x+5 f(4)= This quiz is incomplete! Function notation and evaluating functions worksheet - Problems. Evaluating functions. Evaluating first is usually easier, but the choice is up to you. Evaluate each function from the graph in Part A, from function expressions in Part B and in Part C look for values of x that make f (x) = g (x) true. Practice with Evaluating Functions MathBitsNotebook.com Evaluating Functions. Worked example: Evaluating functions from equation. g(1)= 3 \\ Forgot password? This quiz is incomplete! Edit. 0 times. In our introduction to functions lesson, we related functions to a vending machine. Email. View Evaluating Functions Practice.pdf from MATH 102 at West Essex High. Evaluating f(2)= 1 \\ Functions are written using function notation. Example Questions. Begin with the easy level dealing with integers and move on to polynomial functions involving decimals and fractions. Related Links All Quizzes . Log in. \frac{1}{q} & \text{ if } x= \frac{p}{q}, \\ Feel free to download and enjoy these free worksheets on functions and relations .Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Directions: Evaluate each function below.. Resources. Worked example: Evaluating functions from equation, Worked example: Evaluating functions from graph, Practice: Evaluate functions from their graph, Worked example: evaluating expressions with function notation. Evaluate functions | Algebra (practice) | Khan Academy. Free practice questions for Precalculus - Evaluate Functions. Today's Independent Practice focuses more attention on correctly evaluating functions. \end{cases}f(x)={q10 if x=qp, if x is an irrational number,. To play this quiz, please finish editing it. Brilliant. To evaluate a function is to: Replace its variable with a given number or expression. Algebra 2 CCSS Lessons and Practice is a free site for students (and teachers) studying a second year of high school algebra. f(−4)? View Evaluating functions Practice from MATHEMATIC 101 at Arlington Heights H S. Function Notation and Evaluating Functions Practice Worksheet Name_____ Date_____ Decide whether the graph represents y as a function of x. Sign up, Existing user? f(-4)? Evaluate functions for specific inputs given the formula of the function. Improve your math knowledge with free questions in "Evaluate functions" and thousands of other math skills. SAT Math: Functions Chapter Exam Instructions. This precalculus video tutorial provides a basic introduction on evaluating piecewise functions. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. 9th - 12th grade. f(2) = 2? Let . You "input" money and your "output" is candy or chips! f(2)=2. Practice. f(-1) = 3? Analyze the outputs of functions for every input on the graph y = f (x), (y is the output and x is the input). f(1)= 2 \\ f(1)=2f(2)=1f(3)=2f(4)=5g(1)=3g(2)=5g(3)=7g(4)=9\begin{aligned} You can do this for the graph of just about any function. Our mission is to provide a free, world-class education to anyone, anywhere. Finish Editing. \end{aligned}f(1)=2f(2)=1f(3)=2f(4)=5g(1)=3g(2)=5g(3)=7g(4)=9. Use the review games below to practice. Start studying Evaluating Functions. Like in this example: Example: evaluate the function f(x) = … You can use the Mathway widget below to practice operations on functions. The graph of a function is a way of viewing the outputs of the function for all possible inputs. Then click the button and select "Solve" to compare your answer to Mathway's. To play this quiz, please finish editing it. If f(x)=2x+a, f(x) = 2x + a, f(x)=2x+a, what is the value of a a a satisfying f(2)=2? Decide whether the relation is a function. If f(x)=2x, f(x) = 2x,f(x)=2x, what is the value of f(−4)? Although we are now asked to work with variables instead of numbers, the process is exactly the same. If f(x)=ax, f(x) = ax, f(x)=ax, what is the value of a a a satisfying f(−1)=3? f(3)= 2 \\ Find and . This means any point on the graph is: (x,y)=(input,corresponding output) Remember that in the xy-plane, the x-axis is the horizontal axis, and the y-axis is the vertical axis. Let R+\mathbb{R}^+R+ be the set of all positive real numbers, and let f:R+→Rf:\mathbb{R}^+ \to \mathbb{R}f:R+→R be defined by, f(x)={1q if x=pq,0 if x is an irrational number,f(x)= \begin{cases} If it is a function, give the domain and the … ×. If it is a function, give the domain and range. To link to this page, copy the following code to your site: f(4)= 5 \\ Edit. Using this, we can easily read points off the graph. i g AMa3dRea Rw iLt Ih g oI 1nif UiBnEibt1e9 TAVlpgLeDb5ria V J2v. Choose your answers to the questions and click 'Next' to see the next set of questions. To play this quiz, please finish editing it. Try the entered exercise, or type in your own exercise. I want my students to be able to evaluate functions on their own by the end of today's lesson. Evaluating this function results in 41. f(−1)=3? elizabeth_sayan_63081. g(4)= 9 \\ c-3-Worksheet by Kuta Software LLC Answers to Evaluating Functions Practice (ID: 1) a year ago. Here, we let y=f(x)so that each y-value represents the output and each x-value represents the input. Evaluating Functions Practice. Evaluate the function for the given value of x. Khan Academy is a 501(c)(3) nonprofit organization. This quiz and worksheet combo can help you practice the following skills: We are going to "input" a number and our "output" is the answer. 1. About "Function notation and evaluating functions worksheet" Function notation and evaluating functions worksheet : Worksheet on function notation and evaluating functions is much useful to the students who would like to practice problems on functions. For example, if we wanted to know the value of f wh… Note: The answer above, y = –3 when x = 0, means that the point (0, –3) is on the graph of the equation y = 4x – 3. Google Classroom Facebook Twitter. Concept 22 Evaluating Functions Worksheet Level 2: Goals: Evaluate a function Practice #1 Practice #2 The graph of the function y=f(x) below shows the temperature … To find , everywhere we see an x, we substitute the letter b. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Evaluating functions from a graph worksheet answers. Includes full solutions and score reporting. We're going to go back to that visual as we begin evaluating functions. where ppp and qqq are coprime positive integers. Function Notation Evaluating Functions - Displaying top 8 worksheets found for this concept.. Learn vocabulary, terms, and more with flashcards, games, and other study tools. g(2)= 5 \\ Delete Quiz. ©K n2 v0r1 s45 SK Wupt 9a7 nS Xo uf htGwBaBrQeP nL1LOCR.w V TA YlElG 7r 3i5gEhdt0s S Rrue ksYebr xvce ed3. 4 43 2 3 13. 0. Evaluating Functions. Evaluate functions for specific inputs given the formula of the function. With this enormous collection of polynomial function worksheets there will be no dearth of practice materials for your high school students. What is the value of (f∘g)(1)(f \circ g)(1)(f∘g)(1) ? Evaluating composite functions How to write composite functions Skills Practiced. Function Notation and Evaluating Functions Practice Worksheet Date Decide whether the graph represents y as a function of x. —l, 2) q Fwnc+lD(X . Two functions fff and ggg satisfy the following relationships. Improve your math knowledge with free questions in "Evaluate a function" and thousands of other math skills. When evaluating functions, we may not always be asked to evaluate the function at a particular numerical value. Not only will students practice evaluating functions, but they will also practice solving for x given f(x).Cards A-J: Given a function, evaluate for a given value. Save. Students ( and teachers ) studying a second year of high school algebra.kasandbox.org are unblocked Adobe Flash Player game. '' a number and our output '' is the answer x ) so that each represents. Level dealing with integers and move on to polynomial functions involving decimals and fractions school students this... | graph that visual as we begin Evaluating functions MathBitsNotebook.com Evaluating functions practice Worksheet Date Decide whether the of! Polynomial functions involving decimals and fractions this precalculus video tutorial provides an introduction into function Notation and Evaluating Practice.pdf! Numbers, the largest community of math and science problem solvers when Evaluating functions worksheets there will no. Resources on our website to evaluate the function at a particular numerical value functions and applying function.. 1Nif UiBnEibt1e9 TAVlpgLeDb5ria V J2v and click 'Next ' to see the next of! Practice with Evaluating functions Practice.pdf from math 102 at West Essex high number and our output '' candy... Go back to that visual as we begin Evaluating functions a number and our output is... That the domains *.kastatic.org and *.kasandbox.org are unblocked Academy, please finish editing it incomplete... Choose your Answers to the questions and click 'Next ' to see the next set of questions on their by! The button and select Solve '' to compare your answer to Mathway 's worksheets found this. Of practice materials for your high school students each x-value represents the.... To log in and use all the features of Khan Academy function is to provide a site! Following relationships g AMa3dRea Rw iLt Ih g oI 1nif UiBnEibt1e9 TAVlpgLeDb5ria V J2v to work with instead... To review algebra I. f ( 4 ) = this quiz is incomplete function and! Of polynomial function worksheets there will be no dearth of practice materials your! Message, it means we 're going to input '' a number our! Functions How to write composite functions skills Practiced log in and use all features... Uibneibt1E9 TAVlpgLeDb5ria V J2v begin Evaluating functions your answer to Mathway 's community of math and science problem.. Is candy or chips example, if we wanted to know the value of f wh… Today Independent... ( x ) so that each y-value represents the output and each x-value represents the output each... Inputs given the formula of the function able to evaluate the function for given. High school algebra integers and move on to polynomial functions involving decimals and fractions with a given number or.. Is incomplete move on to polynomial functions involving decimals and fractions and each x-value represents output. Having trouble loading external resources on our website loading external resources on website... To go back to that visual as we begin Evaluating functions on their own the. | graph with integers and move on to polynomial functions involving decimals and fractions fff and ggg satisfy following! Education to anyone, anywhere not always be asked to work with variables instead numbers! Students to be able to evaluate functions for specific inputs given the formula of the function a. ) =4x+5 f ( x ) so that each y-value represents the output and each x-value represents the.. Y as a function of x of Khan Academy is a 501 c... As a function is a function of x, it means we 're going to input '' money your. Function '' and thousands of other math skills Rw iLt Ih g oI 1nif UiBnEibt1e9 TAVlpgLeDb5ria J2v. 102 at West Essex high the input, the largest community of math and problem! The given value of x ( 4 ) = this quiz, please enable JavaScript in your browser on. Functions to a vending machine may not always be asked to work with variables instead of numbers, largest... Inputs given the formula of the function at a particular numerical value into function Notation and Evaluating and. This for the graph represents y as a function '' and thousands of other math skills always... Functions | graph functions '' and thousands of other math skills your own.! The button and select Solve '' to compare your answer to Mathway 's to play this is... Materials for your high school algebra skills Practiced if it is a way of viewing the of. Evaluate functions on Brilliant, the process is exactly the same number or expression we going... Study tools to practice operations on functions to input '' money and your ''. That visual as we begin Evaluating functions Practice.pdf from math 102 at West Essex high 3 ) nonprofit.. Kuta Software LLC Answers to evaluating functions practice questions and click 'Next ' to see next... A particular numerical value to play this quiz, please enable JavaScript in own. Introduction into function Notation and Evaluating functions practice Worksheet Date Decide whether the graph represents y as a function x... | 3,566 | 14,702 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-39 | latest | en | 0.824668 |
https://www.physicsforums.com/threads/time-sync-problems.502385/ | 1,508,845,944,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828411.81/warc/CC-MAIN-20171024105736-20171024125736-00750.warc.gz | 1,001,183,811 | 15,070 | # Time Sync Problems
1. May 28, 2011
### whereisnomar
http://galileoandeinstein.physics.virginia.edu/lectures/synchronizing.html" [Broken] the website talks about synchronizing clocks on a moving train. But I have a few questions about it:
1) Why are you allowed to add ct+vt on the left-hand side of the first equation below the picture of the train? I thought one of the main tenants of Special Relativity was that you can't add velocities (or I'd assume distances) without taking into account the dilation/ contraction of that quantity.
2) What would happen if you stopped the train, and then looked at what the clocks say? Would they be synchronized or not? I understand that the train would enter a noninertial reference frame, which would change how they behave, BUT wouldn't both clocks be affected equally... so who is right once the train is stopped- the person who says they're synchronized, or the person who says they're not?
THANK YOU SO MUCH!
Last edited by a moderator: May 5, 2017
2. May 28, 2011
### ZealScience
It didn't say that velocity is added, ct+vt is the distance wich had been through a length contraction.
Though I didn't learnt much of it, I don't think that the clock would be synchronized. Because according to the rest frame it is not synchronized even when the train is still moving, for different frames have different synchronization.
Last edited by a moderator: May 5, 2017
3. May 28, 2011
### schaefera
So at what point would the person on the train see them leave synchronization, then?
4. May 28, 2011
### ZealScience
I think just when the train changes speed. Look at the time of rear and front, they depend on c+v and c-v, so when v changes time would change. Synchronization means when the two times are equal, but if they are equal at one v they gotta be different when v changes.
5. May 29, 2011
### schaefera
Any other ideas? | 461 | 1,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-43 | latest | en | 0.957819 |
https://www.hackmath.net/en/example/6921?tag_id=64 | 1,566,186,612,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314641.41/warc/CC-MAIN-20190819032136-20190819054136-00313.warc.gz | 838,227,894 | 6,386 | # Collect rain water
The garden water tank has a cylindrical shape with a diameter of 80 cm and a height of 12 dm. How many liters of water will fit into the tank?
Result
V = 603.2 l
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
Do you know the volume and unit volume, and want to convert volume units?
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https://ihomeworkhelpers.com/mathematics/question2343651 | 1,713,274,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817095.3/warc/CC-MAIN-20240416124708-20240416154708-00871.warc.gz | 287,709,018 | 17,461 | , 30.12.2019 19:31 babygirltori21538
# Yolanda paid for her movie ticket using 28 coins, all nickels and quarters. the ticket cost \$4. which system of linear equations can be used to find the number of nickels, n, and the number of quarters, q, yolanda used?
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Saline iv fluid bags cost \$64.20 for 24, 1000 ml bags from one supplier. a new supplier sells 500 ml bags for \$2 each. the hospital uses about 10,000, 1000 ml bags each month. nurses report that about half of the bags are more than half full when they are discarded. if you are going to use only one supplier, which size bag provides the best deal, and how much will you save each month? compare the cost of the new supplier to the current one. | 239 | 864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.893922 |
https://www.stat.math.ethz.ch/pipermail/r-help/2014-August/420731.html | 1,653,117,831,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662538646.33/warc/CC-MAIN-20220521045616-20220521075616-00615.warc.gz | 1,188,117,711 | 2,443 | # [R] efficient way to replace a range of numeric with a integer in a matrix
William Dunlap wdunlap at tibco.com
Mon Aug 11 23:50:05 CEST 2014
```You can use
m[m > 0 & m <= 1.0] <- 1
m[m > 1 ] <- 2
or, if you have lots of intervals, something based on findInterval(). E.g.,
m[] <- findInterval(m, c(-Inf, 0, 1, Inf)) - 1
(What do you want to do with non-positive numbers?)
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Aug 11, 2014 at 2:40 PM, Jinsong Zhao <jszhao at yeah.net> wrote:
> Hi there,
>
> I hope to replace a range of numeric in a matrix with a integer. For
> example, in the following matrix, I want to use 1 to replace the elements
> range from 0.0 to 1.0, and all larger than 1. with 2.
>
>> (m <- matrix(runif(16, 0, 2), nrow = 4))
> [,1] [,2] [,3] [,4]
> [1,] 0.7115088 0.55370418 0.1586146 1.882931
> [2,] 0.9068198 0.38081423 0.9172629 1.713592
> [3,] 1.5210150 0.93900649 1.2609942 1.744456
> [4,] 0.3779058 0.03130103 0.1893477 1.601181
>
> so I want to get something like:
>
> [,1] [,2] [,3] [,4]
> [1,] 1 1 1 2
> [2,] 1 1 1 2
> [3,] 2 1 2 2
> [4,] 1 1 1 2
>
> I wrote a function to do such thing:
>
> fun <- function(x) {
> if (is.na(x)) {
> NA
> } else if (x > 0.0 && x <= 1.0) {
> 1
> } else if (x > 1.0) {
> 2
> } else {
> x
> }
> }
>
> Then run it as:
>
>> apply(m,2,function(i) sapply(i, fun))
>
> However, it seems that this method is not efficient when the dimension is
> large, e.g., 5000x5000 matrix.
>
> Any suggestions? Thanks in advance!
>
> Best regards,
> Jinsong
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help | 693 | 1,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.684292 |
https://www.marcosantoni.com/2016/06/19/a-simple-machine-learning-pipeline.html | 1,679,562,676,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00700.warc.gz | 982,008,996 | 5,803 | # A Simple Machine Learning Pipeline
This post contains the code that I used in my talk at Python Milano Meetup on June 22nd 2016. The talk was a quick overview of Pipeline, a nice API by scikitlearn to abstract your machine learning algorithm. It is based on the Boston Housing Data Set.
We'll just load the data set from sklearn.
``````from sklearn.datasets import load_boston
print housing_data.DESCR
``````
We might want to make it a Pandas dataframe to make things easier to handle.
``````import pandas as pd
df = pd.DataFrame(housing_data.data)
df.columns = housing_data.feature_names
df['PRICE'] = housing_data.target
``````
The goal is to predict the PRICE variable given the other features. How does this variable distribute?
``````import matplotlib.pyplot as plt
df.PRICE.hist()
plt.xlabel('PRICE')
``````
{.alignnone .size-full .wp-image-74 width="378" height="271"}
Let's turn the dataframe into a ML-friendly notation.
``````X = df.drop('PRICE', axis=1)
y = df['PRICE']
``````
We will now define the metric that assess the accuracy of our algorithm/pipeline. Let's use the good old cross validation.
``````from sklearn import cross_validation
def evaluate_model(X, y, algorithm):
print 'Mean Squared Error'
scores = cross_validation.cross_val_score(algorithm, X, y,
scoring='mean_squared_error')
print -scores
print 'Accuracy: %0.2f' % -scores.mean()
``````
So, now, we can try a bunch of algorithms and see which one works best by calling evaluate_model. It is now time to implement a first algorithm. So, let's explore a bit the data set. Is there any pattern we can exploit?
``````plt.figure(figsize=(10,7))
plt.scatter(df['RM'], y)
plt.xlabel('Average number of rooms')
plt.ylabel('Housing price in \\$1000\'s')
plt.show()
``````
{.alignnone .size-full .wp-image-78 width="610" height="438"}
As expected, there is a relation between the average number of rooms and the median price. So, let's build the first algorithm.
``````from sklearn.pipeline import make_pipeline
from sklearn.preprocessing import FunctionTransformer
from sklearn.linear_model import LinearRegression
def just_RM_column(X):
RM_col_index = 5
return X[:, [RM_col_index]]
pipe = make_pipeline(
FunctionTransformer(just_RM_column),
LinearRegression()
)
``````
How well does it perform?
``````evaluate_model(X, y, pipe)
'''Mean Squared Error [43.19492771 41.72813479 46.89293772] Accuracy:
43.94'''
``````
Can we visualize what the pipeline is actually doing?
``````def plot_model_RM(X, y, pipe):
X_train, X_test, y_train, y_test =
cross_validation.train_test_split(
X,
y,
test_size=0.33,
random_state=5
)
pipe.fit(X_train, y_train)
fake_X_train = np.array(X_train)
fake_X_train[:, 5] = np.linspace(min(fake_X_train[:, 5]),
max(fake_X_train[:, 5]), num=len(fake_X_train[:, 5]))
fake_X_test = np.array(X_test)
fake_X_test[:, 5] = np.linspace(min(fake_X_test[:, 5]),
max(fake_X_test[:, 5]), num=len(fake_X_test[:, 5]))
plt.figure(figsize=(20,7))
plt.subplot(1, 2, 1)
plt.scatter(X_train['RM'], y_train)
plt.scatter(fake_X_train[:, 5], pipe.predict(fake_X_train),
color='r')
plt.xlabel('Average number of rooms')
plt.ylabel('Housing price in \\$1000\'s')
plt.title('Train Data Set')
plt.subplot(1, 2, 2)
plt.scatter(X_test['RM'], y_test)
plt.scatter(fake_X_test[:, 5], pipe.predict(fake_X_test),
color='r')
plt.xlabel('Average number of rooms')
plt.ylabel('Housing price in \\$1000\'s')
plt.title('Test Data Set')
plt.show()
plot_model_RM(X, y, pipe)
``````
{.alignnone .size-full .wp-image-84 width="1173" height="449"}
We now do a bit of feature engineering. We square the features.
``````def add_squared_col(X):
return np.hstack((X, X**2))
pipe = make_pipeline(
FunctionTransformer(just_RM_column),
LinearRegression()
)
``````
We evaluate this other pipeline.
``````evaluate_model(X, y, pipe)
'''
Mean Squared Error
[ 40.31207562 36.75642688 40.75444834]
Accuracy: 39.27'''
``````
And we see how the algorithm is fitting the data set.
``````plot_model_RM(X, y, pipe)
``````
{.alignnone .size-full .wp-image-86 width="1165" height="449"} We now try a different model like a decision tree.
``````from sklearn.tree import DecisionTreeRegressor
pipe = make_pipeline(
FunctionTransformer(just_RM_column),
DecisionTreeRegressor(max_depth=3)
)
evaluate_model(X, y, pipe)
'''
Mean Squared Error
[ 57.28366371 61.5437311 84.32756118]
Accuracy: 67.72
'''
plot_model_RM(X, y, pipe)
``````
{.alignnone .size-full .wp-image-87 width="1165" height="449"}
We now explore a second feature: INDUS.
``````plt.figure(figsize=(10,7))
plt.scatter(df['INDUS'], y)
plt.xlabel('Average number of rooms')
plt.ylabel('Housing price in \\$1000\'s')
plt.show()
``````
{.alignnone .size-full .wp-image-89 width="610" height="438"}
So, we see another relation between INDUS and PRICE. So, let's add this second feature.
``````def RM_and_INDUS_cols(X):
RM_col_index = 5
INDUS_col_index = 2
return X[:, [RM_col_index, INDUS_col_index]]
pipe = make_pipeline(
FunctionTransformer(RM_and_INDUS_cols),
LinearRegression()
)
evaluate_model(X, y, pipe)
'''
Mean Squared Error
[ 32.3420789 31.4260901 35.95835866]
Accuracy: 33.24
'''
``````
Now, plotting a model in 3D needs a bit more effort.
``````def plot_model_RM_INDUS(X, y, pipe):
X_train, X_test, y_train, y_test =
cross_validation.train_test_split(
X,
y,
test_size=0.33,
random_state=5
)
pipe.fit(X_train, y_train)
X_test = np.array(X_test)
fig = plt.figure(figsize=(10,7))
ax = p3.Axes3D(fig)
x = X_test[:, 2]
y = X_test[:, 5]
z = y_test
ax.scatter(x, y, z, c='r', marker='o')
x = np.arange(min(x), max(x), (max(x) - min(x)) / 100.0)
y = np.arange(min(y), max(y), (max(y) - min(y)) / 100.0)
X, Y = np.meshgrid(x, y)
Z = np.zeros(X.shape)
fake_X = np.zeros((1, 10))
for i in range(X.shape[0]):
for j in range(X.shape[1]):
fake_X[0, 2] = X[i, j]
fake_X[0, 5] = Y[i, j]
Z[i, j] = pipe.predict(fake_X)[0]
ax.plot_surface(X, Y, Z, alpha=0.2)
ax.set_xlabel('INDUS')
ax.set_ylabel('RM')
ax.set_zlabel('Price')
plot_model_RM_INDUS(X, y, pipe)
``````
{.alignnone .size-full .wp-image-91 width="720" height="504"}
How pretty is that?
The following step is to use all the features available. So, we move to a 13-dimensional feature vector.
``````pipe = make_pipeline(
LinearRegression()
)
evaluate_model(X, y, pipe)
'''
Mean Squared Error
[ 20.50009513 22.42870192 27.88911654]
Accuracy: 23.61'''
``````
The error got quite smaller. We cannot however plot the model in 13-dimensions. We will now re-use the function that adds a squared feature.
``````pipe = make_pipeline(
LinearRegression()
)
evaluate_model(X, y, pipe)
'''
Mean Squared Error
[ 16.7819682 14.599869 18.17785453]
Accuracy: 16.52'''
``````
Even better. Now, we will switch to a ridge-regressor (combined with a normalization of the features).
``````from sklearn.preprocessing import StandardScaler
from sklearn.linear_model import Ridge
pipe = make_pipeline(
StandardScaler(), | 1,998 | 6,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-14 | latest | en | 0.769407 |
https://mathsmagics.com/category/quick-calculation/page/10/ | 1,623,952,709,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630518.38/warc/CC-MAIN-20210617162149-20210617192149-00588.warc.gz | 345,612,243 | 16,356 | June 17, 2021
## Category: Quick Calculation
### How to get know we can divide it exactly in a second
Some times you will have to divide some numbers and have a result quickly. But when we start the division we may have to check that will it can divide. So this article we are…
### Multiply any number by five quickly
There is a easy math trick to find the correct answer to multiply any number by five.You need to use two ways to get the answer. One is for odd numbers and another method for…
### Subtracting any number by 1,000
Subtracting any number by 1000 is easy. If you use a traditional method ,this will become a time consuming process. But this article show you how to subtract any number by 1000 with your mind… | 169 | 729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | latest | en | 0.927519 |
https://community.fabric.microsoft.com/t5/Desktop/Monthly-cost-with-monthly-quarterly-and-yearly-limits/td-p/3259042 | 1,701,825,721,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100575.30/warc/CC-MAIN-20231206000253-20231206030253-00323.warc.gz | 219,321,568 | 68,782 | cancel
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Frequent Visitor
## Monthly cost with monthly, quarterly and yearly limits
I have two tables:
Issues
Customer Date Cost A 2022-04-01 100 A 2022-05-04 150 B 2022-06-01 200 ... ... ...
Limits
Customer Monthly Quarerly Yearly A 150 B 100 500 ... ... ... ...
I want to visualize costs per customer per month but capped by all limits. Something like this:
Customer Year-Month Cost A 2022-04 100 A 2022-05 50 Because of quarterly limit A 2022-06 0 Regardless of actual cost, because quarterly limit is reached ... ... ... B 2022-06 100 Because of monthly limit ... ... ...
I've tried using a calculated table grouped by Customer and Year-Month, but I can't get the quarterly and yearly limits to work. Should I use a different approach (I saw someone suggest calculating daily costs then using measures)? What is the best approach?
2 REPLIES 2
Community Support
Hi @IsseBisse ,
Here are the steps you can follow：
1. Create calculated table.
``````Cost =
var _table1=
DISTINCT('Issues'[Customer])
var _table2=
DISTINCT('Issues'[Date])
return
CROSSJOIN(_table1,_table2)``````
2. Create calculated column.
``````Issue =
SUMX(
FILTER(ALL(Issues), 'Issues'[Customer]=EARLIER('Cost'[Customer])&&YEAR('Issues'[Date])=YEAR(EARLIER('Cost'[Date]))&&MONTH('Issues'[Date])=MONTH(EARLIER('Cost'[Date]))),
[Cost])``````
``````Rank =
RANKX(FILTER(ALL('Cost'),
'Cost'[Customer]=EARLIER('Cost'[Customer])),[Date],,ASC)``````
``````Column 2 =
var _Monthly=
SUMX(
FILTER(ALL(Limits),'Limits'[Customer]=EARLIER('Cost'[Customer])),[Monthly])
var _Quarely =
SUMX(
FILTER(ALL(Limits),'Limits'[Customer]=EARLIER('Cost'[Customer])),[Quarerly])
var _Yearly =
SUMX(
FILTER(ALL(Limits),'Limits'[Customer]=EARLIER('Cost'[Customer])),[Yearly])
var _if=
IF(
_Monthly<>BLANK() &&[Issue]>_Monthly,
_Monthly,
IF(
_Quarely<>BLANK()&&[Issue]>=_Quarely,
[Issue]-SUMX(FILTER(ALL('Cost'),'Cost'[Customer]=EARLIER('Cost'[Customer])&&'Cost'[Rank]=EARLIER('Cost'[Rank])-1),[Issue]),
IF(
_Yearly<>BLANK()&&[Issue]>_Yearly,
_Yearly,[Issue]
)))
return
IF(
_if=BLANK(),0,_if)``````
3. Result:
Best Regards,
Liu Yang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly
Frequent Visitor
This doesn't seem to be working correctly. For instance with a slightly large example:
Issues
Customer Date Cost C 2022-01-01 100 C 2022-02-02 150 C 2022-03-03 100 C 2022-04-04 100 C 2022-05-05 150 C 2022-06-06 100 C 2022-07-08 150 C 2022-08-22 100 C 2022-09-10 150 C 2022-10-11 200 C 2022-11-01 100 C 2022-12-24 150
Limits
Customer Monthly Quarterly Yearly C 100 250 600
I would expect the following
Results
Customer Year-month Cost Comment C 2022-01 100 C 2022-02 100 Because of monthly C 2022-03 50 Because of quarterly C 2022-04 100 C 2022-05 100 Because of monthly C 2022-06 50 Because of quarterly C 2022-07 100 Because of monthly C 2022-08 0 Because of yearly C 2022-09 0 Because of yearly ... ...
Cost
CustomerDateSum of Column 2
C 2022-01-01 00:00 100 C 2022-02-02 00:00 100 C 2022-03-03 00:00 100 C 2022-04-01 00:00 100 C 2022-04-04 00:00 100 C 2022-05-04 00:00 100 C 2022-05-05 00:00 100 C 2022-06-01 00:00 100 C 2022-06-06 00:00 100 C 2022-07-08 00:00 100 C 2022-08-22 00:00 100 C 2022-09-10 00:00 100 C 2022-10-11 00:00 100 C 2022-11-01 00:00 100 C 2022-12-24 00:00 100
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Top Solution Authors
Top Kudoed Authors | 1,295 | 4,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.698839 |
https://testbook.com/question-answer/an-automobile-owner-reduced-his-monthly-petrol-con--600ec0d6895d25347144ddce | 1,628,176,571,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155925.8/warc/CC-MAIN-20210805130514-20210805160514-00542.warc.gz | 528,205,144 | 19,641 | # An automobile owner reduced his monthly petrol consumption when the prices went up. The price-consumption relationship is as follows: Price (in Rs. per litre) 40 50 60 75 Monthly consumption (in litres) 60 48 40 32 If the price goes up to Rs.80 per litre, his expected consumption (in litres) will be
Free Practice With Testbook Mock Tests
## Options:
1. 30
2. 28
3. 26
4. 24
### Correct Answer: Option 1 (Solution Below)
This question was previously asked in
Official UPSC Civil Services Exam 2015 Prelims Part B
## Solution:
From the table
In every condition,
Expenditure of an automobile owner per month
Expenditure in first condition = 40 × 60 = Rs 2400
Expenditure in 2nd condition = 50 × 48 = Rs 2400
Expenditure in 3rd condition = 60 × 40 = Rs 2400
Expenditure in 4th condition = 75 × 32 = Rs 2400
so, If price goes up to Rs 80 per litre, expenditure is same in all condition
let consumption of petrol be x
⇒ 80 × x = 2400
⇒ x = 30 litres
∴ The required answer is 30 litres | 304 | 1,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2021-31 | latest | en | 0.899333 |
https://web2.0calc.com/questions/feeling-stuck_1 | 1,632,700,867,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058222.43/warc/CC-MAIN-20210926235727-20210927025727-00308.warc.gz | 637,673,609 | 5,440 | +0
# feeling stuck
-1
212
1
Find the surface area of the composite figure.
Aug 20, 2020
#1
+973
+1
Starting with the purple figure,
(4x3) + 2(8x3) + 2(8x4) = 12 + 48 + 64 = 124cm^2
Next the yellow figure,
2(5x10) + 2(5x7) + (10x7) + ((10x7) - (4x3)) = 100 + 70 + 70 + 58 = 298cm^2
Total surface area = 124 + 298 = 422 cm^2
Aug 21, 2020 | 163 | 346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.680158 |
https://www.mkamimura.com/2019/09/Go-Parallel-sum-and-prefix-scan-Parallel-sum-More-realistic-algorithm-PRAM-algorithm-partial-sum.html | 1,620,774,190,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990419.12/warc/CC-MAIN-20210511214444-20210512004444-00170.warc.gz | 938,884,187 | 24,612 | ## 2019年9月13日金曜日
### Go - 並列和とプリフィックススキャン - 並列和 - より現実的なアルゴリズム - PRAMアルゴリズム、部分和
コード
main_test.go
package main
import "testing"
var want = 55
var intSlice = []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
func TestSummationSequential(t *testing.T) {
got := summationSequential(intSlice)
if got != want {
t.Errorf("summationSequential({%#v}) got %v, want %v",
intSlice, got, want)
}
}
func TestSummationConcurrent(t *testing.T) {
got := summationConcurrent(intSlice)
if got != want {
t.Errorf("summationConcurrent({%#v}) got %v, want %v",
intSlice, got, want)
}
}
main.go
package main
import (
"fmt"
"log"
"os"
"runtime"
"strconv"
"sync"
"time"
)
func main() {
numSlice := []int{}
max, err := strconv.Atoi(os.Args[1])
if err != nil {
log.Fatal(err)
}
for i := 1; i <= max; i++ {
numSlice = append(numSlice, i)
}
t1 := time.Now().UnixNano()
sum1 := summationSequential(numSlice)
t1 = time.Now().UnixNano() - t1
t2 := time.Now().UnixNano()
sum2 := summationConcurrent(numSlice)
t2 = time.Now().UnixNano() - t2
fmt.Printf("逐次アルゴリズム: %v %vミリ秒\n", sum1, t1/int64(time.Millisecond))
fmt.Printf("並行アルゴリズム: %v %vミリ秒\n", sum2, t2/int64(time.Millisecond))
}
func summationSequential(numSlice []int) int {
sum := 0
for _, x := range numSlice {
sum += x
}
return sum
}
func summationConcurrent(numSlice []int) int {
numCPU := runtime.NumCPU()
channelInt := make(chan int, numCPU)
lenNumSlice := len(numSlice)
lenPartial := lenNumSlice / numCPU
var wg sync.WaitGroup
for i := 0; i < numCPU; i++ {
go func(j int) {
defer wg.Done()
var end int
if j == (numCPU - 1) {
end = lenNumSlice
} else {
end = lenPartial * (j + 1)
}
sum := 0
for i := lenPartial * j; i < end; i++ {
sum += numSlice[i]
}
channelInt <- sum
}(i)
}
sum := 0
go func() {
wg.Wait()
close(channelInt)
}()
for i := range channelInt {
sum += i
}
return sum
}
\$ go test
# _/.../go/並行コンピューティング技法/ch6/sample1 [_/.../go/並行コンピューティング技法/ch6/sample1.test]
./main_test.go:9:9: undefined: summationSequential
./main_test.go:18:9: undefined: summationConcurrent
FAIL _/.../go/並行コンピューティング技法/ch6/sample1 [build failed]
\$ go test
# _/.../go/並行コンピューティング技法/ch6/sample1 [_/.../go/並行コンピューティング技法/ch6/sample1.test]
./main_test.go:18:9: undefined: summationConcurrent
FAIL _/.../go/並行コンピューティング技法/ch6/sample1 [build failed]
\$ go test
--- FAIL: TestSummationConcurrent (0.00s)
main_test.go:20: summationConcurrent({[]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}}) got 0, want 55
FAIL
exit status 1
FAIL _/.../go/並行コンピューティング技法/ch6/sample1 0.005s
\$ go test
PASS
ok _/.../go/並行コンピューティング技法/ch6/sample1 0.005s
\$ go run main.go 1000000
\$ go run main.go 10000000
\$ go run main.go 100000000
\$ | 986 | 2,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | longest | en | 0.243101 |
https://www.cheenta.com/cheenta-filix-math-olympiad-program/ | 1,632,127,144,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057033.33/warc/CC-MAIN-20210920070754-20210920100754-00031.warc.gz | 732,186,695 | 40,208 | How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
# Cheenta - Filix Math Olympiad Program for pre rmo 2020
[/et_pb_text][et_pb_divider color="#09e1c0" divider_weight="4px" disabled_on="off|off|off" _builder_version="3.2" background_size="initial" background_position="top_left" background_repeat="repeat" max_width="40px" module_alignment="left" custom_margin="||10px|" animation_style="zoom" animation_direction="left" saved_tabs="all" locked="off"][/et_pb_divider][et_pb_text _builder_version="3.27.4" text_text_color="#d4ccff" text_line_height="1.9em" animation_style="zoom" animation_intensity_zoom="6%" locked="off" inline_fonts="Raleway"]The Pre-Regional Math Olympiad is coming up in August 2020. It is the first step of the (real) Math Olympiad in India.
Cheenta-Filix School is conducting a special training program for this contest. Outstanding faculty members including alumni from I.S.I. C.M.I., and universities abroad will be training the students.
# Curriculum
[/et_pb_text][et_pb_divider color="#09e1c0" divider_weight="4px" disabled_on="off|off|off" _builder_version="3.2" background_size="initial" background_position="top_left" background_repeat="repeat" max_width="40px" module_alignment="center" custom_margin="||10px|" animation_style="zoom" animation_direction="left" saved_tabs="all" locked="off"][/et_pb_divider][et_pb_text _builder_version="3.27.4" text_text_color="#8585bd" text_line_height="1.9em" text_orientation="center" max_width="540px" module_alignment="center" animation_style="zoom" animation_intensity_zoom="6%" locked="off"][/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row column_structure="1_3,1_3,1_3" _builder_version="3.25" background_size="initial" background_position="top_left" background_repeat="repeat"][et_pb_column type="1_3" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_blurb title="Number Theory" url="#" image="http://europe.cheenta.com/wp-content/uploads/2018/03/coding-icon_12.jpg" icon_placement="left" content_max_width="1100px" _builder_version="3.0.82" header_font="|on|||" header_text_color="#2e2545" header_line_height="1.5em" body_text_color="#8585bd" body_line_height="1.9em" background_color="#ffffff" background_size="initial" background_position="top_left" background_repeat="repeat" custom_margin="|||" custom_padding="||20px|" animation_style="zoom" animation_direction="left" animation_intensity_zoom="10%" animation_starting_opacity="100%" locked="off"]
Congruence, Theorems of Fermat, Euler, Wilson, Quadratic Reciprocity, Lagrange, Properties of prime numbers.
Bijection Principle, Inclusion, Exclusion, Pigeon Hole Principle, Combinatorial Arguments, Graph Theory.
Synthetic Geometry, Geometric Transformations (Translation, Rotation, Inversion), Projective Geometry .
Integer Polynomials, Fixed Points, Rational Roots, Eisenstein Criterion, Diophantine Equations.
Field axioms, basic inequalities, algebra of rotation and homothety, applications in geometry and number theory.
A.M.-G.M., Cauchy Schwarz, Jensen, Rearrangement, Geometric Inequality, Power Means.
# Get Started with open seminar
[/et_pb_text][et_pb_divider color="#09e1c0" divider_weight="4px" disabled_on="off|off|off" _builder_version="4.0" background_size="initial" background_position="top_left" background_repeat="repeat" max_width="40px" module_alignment="center" custom_margin="||10px|" animation_style="zoom" animation_direction="left" saved_tabs="all" locked="off"][/et_pb_divider][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0.9"][et_pb_column type="4_4" _builder_version="4.0.9"][et_pb_code _builder_version="4.0.9"]
[/et_pb_code][/et_pb_column][/et_pb_row][/et_pb_section] | 987 | 3,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-39 | latest | en | 0.510723 |
https://jeopardylabs.com/play/topic-2-adding-subtracting-decimals2 | 1,713,115,477,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00321.warc.gz | 297,588,835 | 9,859 | Mental Math
Rounding Whole Numbers & Decimals
Estimating Sums & Differences
Problem Solving
100
Use mental math to solve: 2.1 + 0.9 + 1.2 =
4.2
100
Round the number to the tenth place: 1.32
1.3
100
Round to the nearest ten to estimate: 49 + 22 =
70 (50 + 20 = 70)
100
Find the sum: 0.82 + 4.21 =
5.03
100
Miami's annual precipitation n 2000 was 61.05 in. Albany's was 46.92 in. How much greater was Miami's precipitation than Albany"s?
14.13 in.
200
Use mental math to solve: 66 + 1.8 + 2.2 =
70
200
Round to the ten thousands place: 427,841
430,000
200
Round to the nearest hundred to estimate: 179 + 277 =
500 (200 + 300= 500)
200
Find the sum: 6.9 + 2.8 =
9.7
200
Reilly adds 45.3 and 3.21. Should his sum be greater than or less than 48? Tell how you know without actually solving the problem.
Greater than 48 because the whole numbers add up to 48.
300
Use mental math to solve: 3.8 + 1.5 =
5.3
300
Round to the nearest whole number: 643.82
644
300
Round to the nearest whole number to estimate: 87.2 + 3.9 =
92 (88 + 4 = 92)
300
Find the Difference: 7.8 - 4.9 =
2.9
300
Nate has a \$5 bill and a \$10 bill. He spends \$2.50 for a smoothie and \$2 for a muffin. How much money does he have left?
\$10.50
400
8.6 - 4.9 =
3.7
400
Round to the thousandths place: 23.009
23.01
400
Estimate to the whole number: 3.8 + 4.1 + 3.3 =
11 (4 + 4 + 3 = 11)
400
Find the difference: 65.18 - 12.005 =
53.175
400
Elias saved \$30 in July, \$21 in August, and \$50 in September. He spent \$18 on movies and \$26 on gas. How much money does Elias have left?
\$57
500
Use mental math to solve: 3.99 + 3.11 + 6.3 =
13.4
500
Round 95.25 to the nearest WHOLE #, TENTH, and TEN:
Whole #: 95 Tenth: 95.3 Ten: 100
500
Estimate the difference to the nearest whole number: 12.97 - 3.95 =
9 (13 - 4 = 9)
500
Find the difference: 34.49 - 12.619 =
21.871
500
Fill in the blanks to complete the difference: 6. __ __ - __. 3 9 ___________ 1 . 8 6
6. 2 5 - 4. 3 9 ______ 1. 8 6 | 737 | 1,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.842914 |
http://openstudy.com/updates/52431c81e4b0a03c277f8d57 | 1,448,767,884,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398455246.70/warc/CC-MAIN-20151124205415-00199-ip-10-71-132-137.ec2.internal.warc.gz | 173,809,089 | 10,188 | ## MyChem 2 years ago Solve: -2(2x + 5) - 3 = -3(x - 1) 16 -16 -10 -13
1. dmezzullo
Would u mind only having 1 question open at a time plz.
2. MyChem
sorry
3. dejavo
x= -16
4. MyChem
Solve: 4(x + 1) = 4x + 10 0 6 All Real Numbers No Solution
no solution
6. 16sethjones
no solution | 124 | 291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2015-48 | longest | en | 0.47392 |
http://www.bluechipai.asia/category/machine-learning-techniques/ | 1,723,282,727,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00667.warc.gz | 37,898,459 | 17,503 | ## Introduction to Machine Learning Techniques
Machine Learning Techniques (like Regression, Classification, Clustering, Anomaly detection, etc.) are used to build the training data or a mathematical model using certain algorithms based upon the computations statistic to make prediction without the need of programming, as these techniques are influential in making the system futuristic, models and promotes automation of things with reduced cost and manpower.
### Techniques of Machine Learning
There are a few methods that are influential in promoting the systems to automatically learn and improve as per the experience. But they fall under various categories or types like Supervised Learning, Unsupervised Learning, Reinforcement Learning, Representation Learning, etc. Below are the techniques which fall under Machine Learning:
#### 1. Regression
Regression algorithms are mostly used to make predictions on numbers i.e when the output is a real or continuous value. As it falls under Supervised Learning, it works with trained data to predict new test data. For example, age can be a continuous value as it increases with time. There are some Regression models as shown below:
Some widely used algorithms in Regression techniques
• Simple Linear Regression Model: It is a statistical method that analyses the relationship between two quantitative variables. This technique is mostly used in financial fields, real estate, etc.
• Lasso Regression: Least Absolute Selection Shrinkage Operator or LASSO is used when there is a need for a subset of the predictor to minimize the prediction error in a continuous variable.
• Logistic Regression: It is carried out in cases of fraud detection, clinical trials, etc. wherever the output is binary.
• Support Vector Regression: SVR is a bit different from SVM. In simple regression, the aim is to minimize the error, while in SVR, we adjust the error within a threshold.
• Multivariate Regression Algorithm: This technique is used in the case of multiple predictor variables. It can be operated with matrix operations and Python’s Numpy library.
• Multiple Regression Algorithm: It works with multiple quantitative variables in both linear and non-linear regression algorithms.
#### 2. Classification
A classification model, a method of Supervised Learning, draws a conclusion from observed values as one or more outcomes in a categorical form. For example, email has filters like inbox, drafts, spam, etc. There is a number of algorithms in the Classification model like Logistic Regression, Decision Tree, Random Forest, Multilayer Perception, etc. In this model, we classify our data specifically and assign labels accordingly to those classes. Classifiers are of two types:
• Binary Classifiers: Classification with 2 distinct classes and 2 output.
• Multi-class Classifiers: Classification with more than 2 classes.
#### 3. Clustering
Clustering is a Machine Learning technique that involves classifying data points into specific groups. If we have some objects or data points, then we can apply the clustering algorithm(s) to analyze and group them as per their properties and features. This method of unsupervised technique is used because of its statistical techniques. Cluster algorithms make predictions based on training data and create clusters on the basis of similarity or unfamiliarity.
Clustering methods:
• Density-based methods: In this method, clusters are considered dense regions depending on their similarity and difference from the lower dense region.
• Hierarchical methods: The clusters formed in this method are the tree-like structures. This method forms trees or clusters from the previous cluster. There are two types of hierarchical methods: Agglomerative (Bottom-up approach) and Divisive (Top-down approach).
• Partitioning methods: This method partitions the objects based on k-clusters and each method form a single cluster.
• Gris based methods: In this method, data are combined into a number of cells that form a grid-like structure.
#### 4. Anomaly detection
Anomaly detection is the process of detecting unexpected items or events in a data set. Some areas where this technique is used are fraud detection, fault detection, system health monitoring, etc. Anomaly detection can be broadly categorized as:
1. Point anomalies: Point anomalies are defined when a single data is unexpected.
2. Contextual anomalies: When anomalies are context-specific, then it’s called contextual anomalies.
3. Collective anomalies: When a collection or group of related data items are anomalous, then it’s called collective anomalous.
There are certain techniques in Anomaly detection as follows:
• Statistical methods: It helps in identifying anomalies by pointing the data that deviates from statistical methods like mean, median, mode, etc.
• Density-based anomaly detection: It based on the k-nearest neighbor algorithm.
• Clustering-based anomaly algorithm: Data points are collected as a cluster when they fall under the same group and are determined from the local centroids.
• Super Vector Machine: The algorithm trains itself to cluster the normal data instances and identifies the anomalies using the training data.
### Working on Machine Learning Techniques
Machine Learning utilizes a lot of algorithms to handle and work with large and complex datasets to make predictions as per need.
For example, we search a bus image on Google. So, Google basically gets a number of examples or datasets labeled as bus and the system finds the patterns of pixels and colors that will help in finding correct images of the bus.
Google’s system will make a random guess of the bus like images with the help of patterns. If any mistake occurs, then it adjusts itself for accuracy. In the end, those patterns will be learned by a large computer system modeled like a human brain or Deep Neural Network to identify the accurate results from the images. This is how ML techniques work to get the best result always.
### Conclusion
Machine Learning has various applications in real life to help business houses, individuals, etc. to attain certain results as per need. To get the best results, certain techniques are important which have been discussed above. These techniques are modern, futuristic and promote automation of things with less manpower and cost.
## Machine Learning Techniques
Machine learning is a data analytics technique that teaches computers to do what comes naturally to humans and animals: learn from experience. Machine learning algorithms use computational methods to directly “learn” from data without relying on a predetermined equation as a model.
As the number of samples available for learning increases, the algorithm adapts to improve performance. Deep learning is a special form of machine learning.
## How does machine learning work?
Machine learning uses two techniques: supervised learning, which trains a model on known input and output data to predict future outputs, and unsupervised learning, which uses hidden patterns or internal structures in the input data.
### Supervised learning
Supervised machine learning creates a model that makes predictions based on evidence in the presence of uncertainty. A supervised learning algorithm takes a known set of input data and known responses to the data (output) and trains a model to generate reasonable predictions for the response to the new data. Use supervised learning if you have known data for the output you are trying to estimate.
Supervised learning uses classification and regression techniques to develop machine learning models.
Classification models classify the input data. Classification techniques predict discrete responses. For example, the email is genuine, or spam, or the tumor is cancerous or benign. Typical applications include medical imaging, speech recognition, and credit scoring.
Use taxonomy if your data can be tagged, classified, or divided into specific groups or classes. For example, applications for handwriting recognition use classification to recognize letters and numbers. In image processing and computer vision, unsupervised pattern recognition techniques are used for object detection and image segmentation.
Common algorithms for performing classification include support vector machines (SVMs), boosted and bagged decision trees, k-nearest neighbors, Naive Bayes, discriminant analysis, logistic regression, and neural networks.
Regression techniques predict continuous responses – for example, changes in temperature or fluctuations in electricity demand. Typical applications include power load forecasting and algorithmic trading.
If you are working with a data range or if the nature of your response is a real number, such as temperature or the time until a piece of equipment fails, use regression techniques.
Common regression algorithms include linear, nonlinear models, regularization, stepwise regression, boosted and bagged decision trees, neural networks, and adaptive neuro-fuzzy learning.
Using supervised learning to predict heart attacks
Physicians want to predict whether someone will have a heart attack within a year. They have data on previous patients, including age, weight, height, and blood pressure. They know if previous patients had had a heart attack within a year. So the problem is to combine existing data into a model that can predict whether a new person will have a heart attack within a year.
### Unsupervised Learning
Detects hidden patterns or internal structures in unsupervised learning data. It is used to eliminate datasets containing input data without labeled responses.
Clustering is a common unsupervised learning technique. It is used for exploratory data analysis to find hidden patterns and clusters in the data. Applications for cluster analysis include gene sequence analysis, market research, and commodity identification.
For example, if a cell phone company wants to optimize the locations where they build towers, they can use machine learning to predict how many people their towers are based on.
A phone can only talk to 1 tower at a time, so the team uses clustering algorithms to design the good placement of cell towers to optimize signal reception for their groups or groups of customers.
Common algorithms for performing clustering are k-means and k-medoids, hierarchical clustering, Gaussian mixture models, hidden Markov models, self-organizing maps, fuzzy C-means clustering, and subtractive clustering.
Ten methods are described and it is a foundation you can build on to improve your machine learning knowledge and skills:
• Regression
• Classification
• Clustering
• Dimensionality Reduction
• Ensemble Methods
• Neural Nets and Deep Learning
• Transfer Learning
• Reinforcement Learning
• Natural Language Processing
• Word Embedding’s
Let’s differentiate between two general categories of machine learning: supervised and unsupervised. We apply supervised ML techniques when we have a piece of data that we want to predict or interpret. We use the previous and output data to predict the output based on the new input.
For example, you can use supervised ML techniques to help a service business that wants to estimate the number of new users that will sign up for the service in the next month. In contrast, untrained ML looks at ways of connecting and grouping data points without using target variables to make predictions.
In other words, it evaluates data in terms of traits and uses traits to group objects that are similar to each other. For example, you can use unsupervised learning techniques to help a retailer who wants to segment products with similar characteristics-without specifying in advance which features to use.
### 1. Regression
Regression methods fall under the category of supervised ML. They help predict or interpret a particular numerical value based on prior data, such as predicting an asset’s price based on past pricing data for similar properties.
The simplest method is linear regression, where we use the mathematical equation of the line (y = m * x + b) to model the data set. We train a linear regression model with multiple data pairs (x, y) by computing the position and slope of a line that minimizes the total distance between all data points and the line. In other words, we calculate the slope (M) and the y-intercept (B) for a line that best approximates the observations in the data.
Let us consider a more concrete example of linear regression. I once used linear regression to predict the energy consumption (in kW) of some buildings by gathering together the age of the building, the number of stories, square feet, and the number of wall devices plugged in.
Since there was more than one input (age, square feet, etc.), I used a multivariable linear regression. The principle was similar to a one-to-one linear regression. Still, in this case, the “line” I created occurred in a multi-dimensional space depending on the number of variables.
Now imagine that you have access to the characteristics of a building (age, square feet, etc.), but you do not know the energy consumption. In this case, we can use the fitted line to estimate the energy consumption of the particular building. The plot below shows how well the linear regression model fits the actual energy consumption of the building.
Note that you can also use linear regression to estimate the weight of each factor that contributes to the final prediction of energy consumed. For example, once you have a formula, you can determine whether age, size, or height are most important.
Linear regression model estimates of building energy consumption (kWh).
Regression techniques run the gamut from simple (linear regression) to complex (regular linear regression, polynomial regression, decision trees, random forest regression, and neural nets). But don’t get confused: start by studying simple linear regression, master the techniques, and move on.
### 2. Classification
In another class of supervised ML, classification methods predict or explain a class value. For example, they can help predict whether an online customer will purchase a product. Output can be yes or no: buyer or no buyer. But the methods of classification are not limited to two classes. For example, a classification method can help assess whether a given image contains a car or a truck. The simplest classification algorithm is logistic regression, which sounds like a regression method, but it is not. Logistic regression estimates the probability of occurrence of an event based on one or more inputs.
For example, logistic regression can take two test scores for a student to predict that the student will get admission to a particular college. Because the guess is a probability, the output is a number between 0 and 1, where 1 represents absolute certainty. For the student, if the predicted probability is greater than 0.5, we estimate that they will be admitted. If the predicted probability is less than 0.5, we estimate it will be rejected.
The chart below shows the marks of past students and whether they were admitted. Logistic regression allows us to draw a line that represents the decision boundary.
Because logistic regression is the simplest classification model, it is a good place to start for classification. As you progress, you can dive into nonlinear classifiers such as decision trees, random forests, support vector machines, and neural nets, among others.
### 3. Clustering
We fall into untrained ML with clustering methods because they aim to group or group observations with similar characteristics. Clustering methods do not use the output information for training but instead let the algorithm define the output. In clustering methods, we can only use visualization to observe the quality of the solution.
The most popular clustering method is K-Means, where “K” represents the number of clusters selected by the user. (Note that there are several techniques for selecting the value of K, such as the elbow method.)
• Randomly chooses K centers within the data.
• Assigns each data point closest to the randomly generated centers.
Otherwise, we return to step 2. (To prevent ending in an infinite loop if the centers continue to change, set the maximum number of iterations in advance.)
The process is over if the centers do not change (or change very little).
The next plot applies the K-means to the building’s data set. The four measurements pertain to air conditioning, plug-in appliances (microwave, refrigerator, etc.), household gas, and heating gas. Each column of the plot represents the efficiency of each building.
Linear regression model estimates of building energy consumption (kWh).
Regression techniques run the gamut from simple (linear) to complex (regular linear, polynomial, decision trees, random forest, and neural nets). But don’t get confused: start by studying simple linear regression, master the techniques, and move on.
Clustering Buildings into Efficient (Green) and Inefficient (Red) Groups.
As you explore clustering, you will come across very useful algorithms such as Density-based Spatial Clustering of Noise (DBSCAN), Mean Shift Clustering, Agglomerative Hierarchical Clustering, and Expectation-Maximization Clustering using the Gaussian Mixture Model, among others.
### 4. Dimensionality Reduction
We use dimensionality reduction to remove the least important information (sometimes unnecessary columns) from the data setFor example, and images may consist of thousands of pixels, which are unimportant to your analysis. Or, when testing microchips within the manufacturing process, you may have thousands of measurements and tests applied to each chip, many of which provide redundant information. In these cases, you need a dimensionality reduction algorithm to make the data set manageable.
The most popular dimensionality reduction method is Principal Component Analysis (PCA), which reduces the dimensionality of the feature space by finding new vectors that maximize the linear variance of the data. (You can also measure the extent of information loss and adjust accordingly.) When the linear correlations of the data are strong, PCA can dramatically reduce the dimension of the data without losing too much information.
Another popular method is t-stochastic neighbor embedding (t-SNE), which minimizes nonlinear dimensions. People usually use t-SNE for data visualization, but you can also use it for machine learning tasks such as feature space reduction and clustering, to mention a few.
The next plot shows the analysis of the MNIST database of handwritten digits. MNIST contains thousands of images of numbers 0 to 9, which the researchers use to test their clustering and classification algorithms. Each row of the data set is a vector version of the original image (size 28 x 28 = 784) and a label for each image (zero, one, two, three, …, nine). Therefore, we are reducing the dimensionality from 784 (pixels) to 2 (the dimensions in our visualization). Projecting to two dimensions allows us to visualize higher-dimensional original data sets.
### 5. Ensemble Methods
Imagine that you have decided to build a bicycle because you are not happy with the options available in stores and online. Once you’ve assembled these great parts, the resulting bike will outlast all other options.
Each model uses the same idea of combining multiple predictive models (supervised ML) to obtain higher quality predictions than the model.
For example, the Random Forest algorithm is an ensemble method that combines multiple decision trees trained with different samples from a data set. As a result, the quality of predictions of a random forest exceeds the quality of predictions predicted with a single decision tree.
Think about ways to reduce the variance and bias of a single machine learning model. By combining the two models, the quality of the predictions becomes balanced. With another model, the relative accuracy may be reversed. It is important because any given model may be accurate under some conditions but may be inaccurate under other conditions.
Most of the top winners of Kaggle competitions use some dressing method. The most popular ensemble algorithms are Random Forest, XGBoost, and LightGBM.
### 6. Neural networks and deep learning
Unlike linear and logistic regression, which is considered linear models, neural networks aim to capture nonlinear patterns in data by adding layers of parameters to the model. The simple neural net has three inputs as in the image below, a hidden layer with five parameters and an output layer.
Neural network with a hidden layer.
The neural network structure is flexible enough to construct our famous linear and logistic regression. The term deep learning comes from a neural net with many hidden layers and encompasses a variety of architectures.
It is especially difficult to keep up with development in deep learning as the research and industry communities redouble their deep learning efforts, spawning whole new methods every day.
Deep learning: A neural network with multiple hidden layers.
Deep learning techniques require a lot of data and computation power for best performance as this method is self-tuning many parameters within vast architectures. It quickly becomes clear why deep learning practitioners need powerful computers with GPUs (Graphical Processing Units).
In particular, deep learning techniques have been extremely successful in vision (image classification), text, audio, and video. The most common software packages for deep learning are Tensorflow and PyTorch.
### 7. Transfer learning
Let’s say you are a data scientist working in the retail industry. You’ve spent months training a high-quality model to classify images as shirts, t-shirts, and polos. Your new task is to create a similar model to classify clothing images like jeans, cargo, casual, and dress pants.
Transfer learning refers to reusing part of an already trained neural net and adapting it to a new but similar task. Specifically, once you train a neural net using the data for a task, you can move a fraction of the trained layers and combine them with some new layers that you can use for the new task. The new neural net can learn and adapt quickly to a new task by adding a few layers.
The advantage of transfer learning is that you need fewer data to train a neural net, which is especially important because training for deep learning algorithms is expensive in terms of both time and money.
The main advantage of transfer learning is that you need fewer data to train a neural net, which is especially important because training for deep learning algorithms is expensive both in terms of time and money (computational resources). Of course, it isn’t easy to find enough labeled data for training.
Let’s come back to your example and assume that you use a neural net with 20 hidden layers for the shirt model. After running a few experiments, you realize that you can move the 18 layers of the shirt model and combine them with a new layer of parameters to train on the pant images.
So the Pants model will have 19 hidden layers. The inputs and outputs of the two functions are different but reusable layers can summarize information relevant to both, for example, fabric aspects.
Transfer learning has become more and more popular, and there are many concrete pre-trained models now available for common deep learning tasks such as image and text classification.
### 8. Reinforcement Learning
Imagine a mouse in a maze trying to find hidden pieces of cheese. At first, the Mouse may move randomly, but after a while, the Mouse’s feel helps sense which actions bring it closer to the cheese. The more times we expose the Mouse to the maze, the better at finding the cheese.
Process for Mouse refers to what we do with Reinforcement Learning (RL) to train a system or game. Generally speaking, RL is a method of machine learning that helps an agent to learn from experience.
RL can maximize a cumulative reward by recording actions and using a trial-and-error approach in a set environment. In our example, the Mouse is the agent, and the maze is the environment. The set of possible actions for the Mouse is: move forward, backward, left, or right. The reward is cheese.
You can use RL when you have little or no historical data about a problem, as it does not require prior information (unlike traditional machine learning methods). In the RL framework, you learn from the data as you go. Not surprisingly, RL is particularly successful with games, especially games of “correct information” such as chess and Go. With games, feedback from the agent and the environment comes quickly, allowing the model to learn faster. The downside of RL is that it can take a very long time to train if the problem is complex.
As IBM’s Deep Blue beat the best human chess player in 1997, the RL-based algorithm AlphaGo beat the best Go player in 2016. The current forerunners of RL are the teams of DeepMind in the UK.
In April 2019, the OpenAI Five team was the first AI to defeat the world champion team of e-sport Dota 2, a very complex video game that the OpenAI Five team chose because there were no RL algorithms capable of winning it. You can tell that reinforcement learning is a particularly powerful form of AI, and we certainly want to see more progress from these teams. Still, it’s also worth remembering the limitations of the method.
### 9. Natural Language Processing
A large percentage of the world’s data and knowledge is in some form of human language. For example, we can train our phones to autocomplete our text messages or correct misspelled words. We can also teach a machine to have a simple conversation with a human.
Natural Language Processing (NLP) is not a machine learning method but a widely used technique for preparing text for machine learning. Think of many text documents in different formats (Word, online blog). Most of these text documents will be full of typos, missing characters, and other words that need to be filtered out. At the moment, the most popular package for processing text is NLTK (Natural Language Toolkit), created by Stanford researchers.
The easiest way to map text to a numerical representation is to count the frequency of each word in each text document. Think of a matrix of integers where each row represents a text document, and each column represents a word. This matrix representation of the term frequency is usually called the term frequency matrix (TFM). We can create a more popular matrix representation of a text document by dividing each entry on the matrix by the weighting of how important each word is in the entire corpus of documents. We call this method Term Frequency Inverse Document Frequency (TFIDF), and it generally works better for machine learning tasks.
### 10. Word Embedding
TFM and TFIDF are numerical representations of text documents that consider only frequency and weighted frequencies to represent text documents. In contrast, word embedding can capture the context of a word in a document. As with word context, embeddings can measure similarity between words, allowing us to perform arithmetic with words.
Word2Vec is a neural net-based method that maps words in a corpus to a numerical vector. We can then use these vectors to find synonyms, perform arithmetic operations with words, or represent text documents (by taking the mean of all word vectors in the document). For example, we use a sufficiently large corpus of text documents to estimate word embeddings.
Let’s say vector(‘word’) is the numeric vector representing the word ‘word’. To approximate the vector (‘female’), we can perform an arithmetic operation with the vectors:
vector(‘king’) + vector(‘woman’) – vector(‘man’) ~ vector(‘queen’)
Arithmetic with Word (Vectors) Embeddings.
The word representation allows finding the similarity between words by computing the cosine similarity between the vector representations of two words. The cosine similarity measures the angle between two vectors.
We calculate word embedding’s using machine learning methods, but this is often a pre-stage of implementing machine learning algorithms on top. For example, let’s say we have access to the tweets of several thousand Twitter users. Let’s also assume that we know which Twitter users bought the house. To estimate the probability of a new Twitter user buying a home, we can combine Word2Vec with logistic regression.
You can train the word embedding yourself or get a pre-trained (transfer learning) set of word vectors. To download pre-trained word vectors in 157 different languages, look at Fast Text.
## Summary
Studying these methods thoroughly and fully understanding the basics of each can serve as a solid starting point for further study of more advanced algorithms and methods.
There is no best way or one size fits all. Finding the right algorithm is partly just trial and error – even highly experienced data scientists can’t tell whether an algorithm will work without trying it out. But algorithmic selection also depends on the size and type of data you’re working with, the insights you want to derive from the data, and how those insights will be used. | 5,646 | 29,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.878729 |
https://www.askmen.com/sports/bodybuilding_900/974_the-golden-ratio.html | 1,569,170,144,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575596.77/warc/CC-MAIN-20190922160018-20190922182018-00086.warc.gz | 732,489,184 | 13,096 | # The Golden Ratio
## The Science Behind A Perfect Body: The Golden Ratio
Page 1 of 2
In my last article, I laid out what I consider to be five of the best strategies for building the ultimate male physique, which really is a discussion about developing a body that others consider “sexy,” for lack of a better word. It's a really interesting idea: building a sexy body. We all know what makes for one, but we don't always think about why those characteristics are associated with sexiness.
For example, why do women want men to have broad shoulders? The answers to those questions are grounded in hard science. We view certain physical traits as desirable in members of each sex. An X-shaped physique for men implies virility and strength, just as an hourglass body on a woman suggests fertility. We view what is attractive partially through the eyes of people looking for mates because of scientifically proven numbers.
### Sex Appeal: It's In The Numbers
Believe it or not, our definition of what’s attractive is only partially a matter of “taste” and is more a matter of math. Put another way, the immediate internal decision about whether we’re attracted to something or someone is, in large part, a highly mathematical process. That process, like all math, is based on certain formulas — in this case, something we call the golden ratio.
This is a number that shows up all over the human body (for example, the length of the arms and legs, compared to the torso), and it seems to define what proportions look best.
In fact, artists and sculptors have known about the golden ratio for a long time and have used it to create sculptures and artwork of the ideal human figure. To this day, surgeons and dentists use it to restructure the human face.
### The Golden Ration Defined
OK, so what is the golden ratio, exactly? It is, in mathematical terms, a comparison of any two aspects that leads us to proportion them in the ideal way. Algebraically, if you have two numbers, A and B, it has to be such that (A + B) divided by A = A divided by B.
In most cases, this is going to be a comparison result in a ratio of 1:1.618. This appears naturally all over your body. For example, if the length of the hand has the value of 1, then the combined length of hand and forearm has the approximate value of 1.618. Similarly, the proportion of upper arm to hand + forearm is in the same ratio of 1:618.
Measure your lower body and you’ll find the same: If the foot is 1, then the length of the foot + the shin is 1.618.
Looking elsewhere on the body, the face is another great example. In fact, the human face abounds with examples of the golden ratio. The head forms a golden rectangle with the eyes at its midpoint.The mouth and nose are each placed at golden sections of the distance between the eyes and the bottom of the chin.
Now, while all of that is interesting in terms of physiology, what is far more interesting, I think, is the effect is has on psychology. You see, our brains are programmed to look for symmetry and balance everywhere; programmed to be attracted to it and to try to create it. This means that what we consider a good body is really based on what we view as a body that projects certain characteristics and bodily symmetry, and that symmetry is determined by the golden ratio.
How can you use the golden ration to increase your sex appeal? Find out, next... | 740 | 3,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-39 | longest | en | 0.953922 |
https://support.sas.com/documentation/cdl/en/lrdict/64316/HTML/default/a001016947.htm | 1,628,047,438,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00539.warc.gz | 535,811,018 | 12,136 | Functions and CALL Routines
# PROBMC Function
Returns a probability or a quantile from various distributions for multiple comparisons of means.
Category: Probability
## Syntax
PROBMC(distribution, q, prob, df, nparms<, parameters>)
### Arguments
distribution
is a character constant, variable, or expression that identifies the distribution. The following are valid distributions:
Distribution Argument
Analysis of Means
`ANOM`
One-sided Dunnett
`DUNNETT1`
Two-sided Dunnett
`DUNNETT2`
Maximum Modulus
`MAXMOD`
Partitioned Range
`PARTRANGE`
Studentized Range
`RANGE`
Williams
`WILLIAMS`
q
is the quantile from the distribution.
Restriction: Either q or prob can be specified, but not both.
prob
is the left probability from the distribution.
Restriction: Either prob or q can be specified, but not both.
df
is the degrees of freedom.
Note: A missing value is interpreted as an infinite value.
nparms
is the number of treatments.
Note: For DUNNETT1 and DUNNETT2, the control group is not counted.
parameters
is an optional set of nparms parameters that must be specified to handle the case of unequal sample sizes. The meaning of parameters depends on the value of distribution. If parameters is not specified, equal sample sizes are assumed, which is usually the case for a null hypothesis.
The PROBMC function returns the probability or the quantile from various distributions with finite and infinite degrees of freedom for the variance estimate.
The prob argument is the probability that the random variable is less than q. Therefore, p-values can be computed as 1- prob. For example, to compute the critical value for a 5% significance level, set prob= 0.95. The precision of the computed probability is O(10--8) (absolute error); the precision of computed quantile is O(10--5).
Note: The studentized range is not computed for finite degrees of freedom and unequal sample sizes.
Note: Williams' test is computed only for equal sample sizes.
### Formulas and Parameters
The equations listed here define expressions used in equations that relate the probability, prob, and the quantile, q, for different distributions and different situations within each distribution. For these equations, let be the degrees of freedom, df.
### Computing the Analysis of Means
Analysis of Means (ANOM) applies to data that is organized as k (Gaussian) samples, the ith sample being of size ni. Let . The distribution function [1, 2, 3, 4, 5] is the CDF for the maximum absolute of a k-dimensional multivariate vector, with degrees of freedom, and an associated correlation matrix . This equation can be written as
where
where , , and , are the gamma function, the density, and the CDF from the standard normal distribution, respectively.
For , the distribution reduces to:
where
For the balanced case, the distribution reduces to:
where
and
The syntax for this distribution is:
x=probmc('anom', q,p,nu,n,);
where
x is a numeric value with the returned result. q is a numeric value denoting the quantile. p is a numeric value denoting the probability. One of p and q must be missing. nu is a numeric value denoting the degrees of freedom. n is a numeric value denoting the number of samples. alphai, i=1,...,k are optional numeric values denoting the alpha values from the first equation of this distribution (see Computing the Analysis of Means.
### Many-One t-Statistics: Dunnett's One-Sided Test
• This case relates the probability, prob, and the quantile, q, for the unequal case with finite degrees of freedom. The parameters are 1, ..., k, the value of nparms is set to k, and the value of df is set to . The equation follows:
• This case relates the probability, prob, and the quantile, q, for the equal case with finite degrees of freedom. No parameters are passed , the value of nparms is set to k, and the value of df is set to . The equation follows:
• This case relates the probability, prob, and the quantile, q, for the unequal case with infinite degrees of freedom. The parameters are 1, ..., k, the value of nparms is set to k, and the value of df is set to missing. The equation follows:
• This case relates the probability, prob, and the quantile, q, for the equal case with infinite degrees of freedom. No parameters are passed , the value of nparms is set to k, and the value of df is set to missing. The equation follows:
### Many-One t-Statistics: Dunnett's Two-sided Test
• This case relates the probability, prob, and the quantile, q, for the unequal case with finite degrees of freedom. The parameters are 1, ..., k, the value of nparms is set to k, and the value of df is set to . The equation follows:
• This case relates the probability, prob, and the quantile, q, for the equal case with finite degrees of freedom. No parameters are passed, the value of nparms is set to k, and the value of df is set to . The equation follows:
• This case relates the probability, prob, and the quantile, q, for the unequal case with infinite degrees of freedom. The parameters are 1, ..., k, the value of nparms is set to k, and the value of df is set to missing. The equation follows:
• This case relates the probability, prob, and the quantile, q, for the equal case with infinite degrees of freedom. No parameters are passed, the value of nparms is set to k, and the value of df is set to missing. The equation follows:
### Computing the Partitioned Range
RANGE applies to the distribution of the studentized range for n group means. PARTRANGE applies to the distribution of the partitioned studentized range. Let the n groups be partitioned into k subsets of size n1+ ...+ nk= n. Then the partitioned range is the maximum of the studentized ranges in the respective subsets, with the studentization factor being the same in all cases.
The syntax for this distribution is:
x=probmc('partrange', q,p,nu,k,n1,...,nk);
where
x is a numeric value with the returned result (either the probability or the quantile). q is a numeric value denoting the quantile. p is a numeric value denoting the probability. One of p and q must be missing. nu is a numeric value denoting the degrees of freedom. k is a numeric value denoting the number of groups. ni i=1,...,k are optional numeric values denoting the n values from the equation in this distribution (see Computing the Partitioned Range.
### The Studentized Range
Note: The studentized range is not computed for finite degrees of freedom and unequal sample sizes.
• This case relates the probability, prob, and the quantile, q, for the equal case with finite degrees of freedom. No parameters are passed, the value of nparms is set to k, and the value of df is set to . The equation follows:
• This case relates the probability, prob, and the quantile, q, for the unequal case with infinite degrees of freedom. The parameters are 1, ..., k, the value of nparms is set to k, and the value of df is set to missing. The equation follows:
• This case relates the probability, prob, and the quantile, q, for the equal case with infinite degrees of freedom. No parameters are passed, the value of nparms is set to k, and the value of df is set to missing. The equation follows:
### The Studentized Maximum Modulus
• This case relates the probability, prob, and the quantile, q, for the unequal case with finite degrees of freedom. The parameters are 1, ..., k, the value of nparms is set to k, and the value of df is set to . The equation follows:
• This case relates the probability, prob, and the quantile, q, for the equal case with finite degrees of freedom. No parameters are passed, the value of nparms is set to k, and the value of df is set to . The equation follows:
• This case relates the probability, prob, and the quantile, q, for the unequal case with infinite degrees of freedom. The parameters are 1, ..., k, the value of nparms is set to k, and the value of df is set to missing. The equation follows:
• This case relates the probability, prob, and the quantile, q, for the equal case with infinite degrees of freedom. No parameters are passed, the value of nparms is set to k, and the value of df is set to missing. The equation follows:
### Williams' Test
PROBMC computes the probabilities or quantiles from the distribution defined in Williams (1971, 1972) (See References). It arises when you compare the dose treatment means with a control mean to determine the lowest effective dose of treatment.
Note: Williams' Test is computed only for equal sample sizes.
Let X1, X2, ..., Xk be identical independent N(0,1) random variables. Let Yk denote their average given by
It is required to compute the distribution of
where
Yk is as defined previously Z is an N(0,1) independent random variable S is such that ½S2 is a 2 variable with degrees of freedom.
As described in Williams (1971) (See References), the full computation is extremely lengthy and is carried out in three stages.
1. Compute the distribution of Yk. It is the fundamental (expensive) part of this operation and it can be used to find both the density and the probability of Yk. Let Ui be defined as
You can write a recursive expression for the probability of Yk > d, with d being any real number.
To compute this probability, start from an N(0,1) density function
and recursively compute the convolution
From this sequential convolution, it is possible to compute all the elements of the recursive equation for , shown previously.
2. Compute the distribution of Yk - Z. This computation involves another convolution to compute the probability
3. Compute the distribution of (Yk - Z)/S. This computation involves another convolution to compute the probability
The third stage is not needed when = . Due to the complexity of the operations, this lengthy algorithm is replaced by a much faster one when k 15 for both finite and infinite degrees of freedom . For k 16, the lengthy computation is carried out. It is extremely expensive and very slow due to the complexity of the algorithm.
The MEANS statement in the GLM Procedure of SAS/STAT Software computes the following tests:
• Dunnett's one-sided test
• Dunnett's two-sided test
• Studentized Range
### Example 1: Computing Probabilities by Using PROBMC
This example shows how to compute probabilities.
```data probs;
array par{5};
par{1}=.5;
par{2}=.51;
par{3}=.55;
par{4}=.45;
par{5}=.2;
df=40;
q=1;
do test="dunnett1","dunnett2", "maxmod";
prob=probmc(test, q, ., df, 5, of par1-par5);
put test \$10. df q e18.13 prob e18.13;
end;
run; ```
SAS writes the following results to the log:
```DUNNETT1 40 1.00000000000E+00 4.82992196083E-01
DUNNETT2 40 1.00000000000E+00 1.64023105316E-01
MAXMOD 40 1.00000000000E+00 8.02784203408E-01```
### Example 2: Computing the Analysis of Means
```data _null_;
q1=probmc('anom',.,0.9,.,20); put q1=;
q2=probmc('anom',.,0.9,20,5,0.1,0.1,0.1,0.1,0.1); put q2=;
q3=probmc('anom',.,0.9,20,5,0.5,0.5,0.5,0.5,0.5); put q3=;
q4=probmc('anom',.,0.9,20,5,0.1,0.2,0.3,0.4,0.5); put q4=;
run;```
SAS writes the following output to the log:
```q1=2.7892895753
q2=2.4549773558
q3=2.4549773558
q4=2.4532130238```
### Example 3: Comparing Means
This example shows how to compare group means to find where the significant differences lie. The data for this example is taken from a paper by Duncan (1955) (See References) and can also be found in Hochberg and Tamhane (1987) (See References).
The following values are the group means:
49.6 71.2 67.6 61.5 71.3 58.1 61
For this data, the mean square error is s2 = 79.64 (s = 8.924) with = 30.
```data duncan;
array tr{7}\$;
array mu{7};
n=7;
do i=1 to n;
input tr{i} \$1. mu{i};
end;
input df s alpha;
prob= 1-alpha;
/* compute the interval */
x = probmc("RANGE", ., prob, df, 7);
w = x * s / sqrt(6);
/* compare the means */
do i = 1 to n;
do j = i + 1 to n;
dmean = abs(mu{i} - mu{j});
if dmean >= w then do;
put tr{i} tr{j} dmean;
end;
end;
end;
datalines;
A 49.6
B 71.2
C 67.6
D 61.5
E 71.3
F 58.1
G 61.0
30 8.924 .05
;
run;```
SAS writes the following output to the log:
```A B 21.6
A C 18
A E 21.7```
### Example 4: Computing the Partitioned Range
```data _null_;
q1=probmc('partrange',.,0.9,.,4,3,4,5,6); put q1=;
q2=probmc('partrange',.,0.9,12,4,3,4,5,6); put q2=;
run;```
SAS writes the following output to the log:
```q1=4.1022395729
q2=4.788862411```
### Example 5: Computing Confidence Intervals
This example shows how to compute 95% one-sided and two-sided confidence intervals of Dunnett's test. This example and the data come from Dunnett (1955) (See References) and can also be found in Hochberg and Tamhane (1987) (See References). The data are blood count measurements on three groups of animals. As shown in the following table, the third group serves as the control, while the first two groups were treated with different drugs. The numbers of animals in these three groups are unequal.
Treatment Group: Drug A Drug B Control
9.76 12.80 7.40
8.80 9.68 8.50
7.68 12.16 7.20
9.36 9.20 8.24
10.55 9.84
8.32
Group Mean 8.90 10.88 8.25
n 4 5 6
The mean square error s2 = 1.3805 (s = 1.175) with = 12.
```data a;
array drug{3}\$;
array count{3};
array mu{3};
array lambda{2};
array delta{2};
array left{2};
array right{2};
/* input the table */
do i = 1 to 3;
input drug{i} count{i} mu{i};
end;
/* input the alpha level, */
/* the degrees of freedom, */
/* and the mean square error */
input alpha df s;
/* from the sample size, */
/* compute the lambdas */
do i = 1 to 2;
lambda{i} = sqrt(count{i}/
(count{i} + count{3}));
end;
/* run the one-sided Dunnett's test */
test="dunnett1";
x = probmc(test, ., 1 - alpha, df,
2, of lambda1-lambda2);
do i = 1 to 2;
delta{i} = x * s *
sqrt(1/count{i} + 1/count{3});
left{i} = mu{i} - mu{3} - delta{i};
end;
put test \$10. x left{1} left{2};
/* run the two-sided Dunnett's test */
test="dunnett2";
x = probmc(test, ., 1 - alpha, df,
2, of lambda1-lambda2);
do i=1 to 2;
delta{i} = x * s *
sqrt(1/count{i} + 1/count{3});
left{i} = mu{i} - mu{3} - delta{i};
right{i} = mu{i} - mu{3} + delta{i};
end;
put test \$10. left{1} right{1};
put test \$10. left{2} right{2};
datalines;
A 4 8.90
B 5 10.88
C 6 8.25
0.05 12 1.175
;
run;```
SAS writes the following output to the log:
```DUNNETT1 2.1210786586 -0.958751705 1.1208571303
DUNNETT2 -1.256411895 2.5564118953
DUNNETT2 0.8416271203 4.4183728797```
### Example 6: Computing Williams' Test
In the following example, a substance has been tested at seven levels in a randomized block design of eight blocks. The observed treatment means are as follows:
Treatment Mean
X0 10.4
X1 9.9
X2 10.0
X3 10.6
X4 11.4
X5 11.9
X6 11.7
The mean square, with (7 - 1)(8 - 1) = 42 degrees of freedom, is s2 = 1.16.
Determine the maximum likelihood estimates Mi through the averaging process.
• Because X0 > X1, form X0,1 = (X0 + X1)/2 = 10.15.
• Because X0,1 > X2, form X0,1,2 = (X0 + X1 + X2)/3 = (2X0,1 + X2)/3 = 10.1.
• X0,1,2 < X3 < X4 < X5
• Because X5 > X6, form X5,6 = (X5 + X6)/2 = 11.8.
Now the order restriction is satisfied.
The maximum likelihood estimates under the alternative hypothesis are:
M0 = M1 = M2 = X0,1,2 = 10.1 M3 = X3 = 10.6 M4 = X4 = 11.4 M5 = M6 = X5,6 = 11.8
Now compute , and the probability that corresponds to k = 6, = 42, and t = 2.60 is .9924467341, which shows strong evidence that there is a response to the substance. You can also compute the quantiles for the upper 5% and 1% tails, as shown in the following table.
SAS Statements Results
`prob=probmc("williams",2.6,.,42,6);`
`0.99244673`
`quant5=probmc("williams",.,.95,42,6);`
`1.80654052`
`quant1=probmc("williams",.,.99,42,6);`
`2.49087829`
Functions:
Guirguis, G. H. and R. D. Tobias. 2004. "On the computation of the distribution for the analysis of means." Communications in Statistics: Simulation and Computation 33: 861-887. Nelson, P. R. 1981. "Numerical evaluation of an equicorrelated multivariate non-central t distribution." Communications in Statistics: Part B - Simulation and Computation 10: 41-50. Nelson, P. R. 1982. "Exact critical points for the analysis of means." Communications in Statistics: Part A - Theory and Methods 11: 699-709. Nelson, P. R. 1982a. "An Approximation for the Complex Normal Probability Integral." BIT 22(1): 94-100. Nelson, P. R. 1988. "Application of the analysis of means." Proceedings of the SAS Users Group International Conference 13: 225-230. Nelson, P. R. 1991. "Numerical evaluation of multivariate normal integrals with correlations ." The Frontiers of Statistical Scientific Theory and Industrial Applications 2: 97-114. Nelson, P. R. 1993. "Additional Uses for the Analysis of Means and Extended Tables of Critical Values." Technometrics 35: 61-71.
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# mba 522 The At Least Problem - Understanding Two Terms At...
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Understanding Two Terms: At Least vs. No More Than Students often mix up these terms when answering probability questions! Here are two non-threatening English examples. Just use the same interpretation when working with numbers in a statistics class. At Least A student says: This chapter is very difficult. I think I need “at least” 7 hours of study time before I fully understand it. Analysis: At least 7 hours implies that anything equal to or greater than 7 hours will be sufficient. On the other hand if he spends less than 7 hours, he will not be successful. Hours ≥ 7 will lead to success. hours < 7 will not lead to success. No More Than I am at a store with my daughter who is now on an allowance and can’t use my hard-earned dollars very freely when she is shopping. She picks up a skirt and it does not have a tag, so she must look for a manager to ask the price. She tells me that she really likes it, but she would pay “ no more than” \$10 for it.
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http://www.theory.caltech.edu/people/patricia/exp3a.html | 1,369,167,898,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700497024/warc/CC-MAIN-20130516103457-00008-ip-10-60-113-184.ec2.internal.warc.gz | 740,612,408 | 1,803 | ## Why no gravitational force in 3 dimensions?
Electromagnetic radiation is well-described by a wave equation where the basic mode of radiation is called the dipole, as pictured in the figure to the left. A dipole wave only needs one space direction to oscillate in, plus one direction to travel in, in addition to time. Therefore electromagnetic radiation can be described mathematically in 2+1 spacetime dimensions.
If we solve the Einstein equation and look for solutions that give gravitational radiation solutions, we find that the lowest wave mode of oscillation for gravitation radiation is the quadrupole, as pictured in the figure to the right.
But a quadrupole wave needs two space directions to oscillate in, and if we only have two space dimensions in our spacetime, then the wave still needs one more direction to travel in. So the lowest dimension spacetime where gravitational radiation is possible according to General Relativity is 3+1 dimensions. And that happens to be the number of spacetime dimensions we measure in our world.
The implications of this are that curvature in 2+1 spacetime dimensions can only exist locally in regions where matter is present. I.e that means that for the example of a single point mass, the spacetime everywhere around the mass will be flat according to the Einstein equations.
But curvature can still be measured in a spacetime with 2+1 dimensions with a point mass. An observer will feel no gravitational force from the mass itself, because force has to be transmitted causally and that means by a wave equation, and we don't have that here. But an observer who travels a closed path around the point mass can measure the total curvature located at the point mass, and we'll show that in detail in the next section.
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### 1 Answer |Add Yours
Top Answer
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
The pyramid connects a square base to an apex.
Since you did not specified the type of pyramide whose TSA and volume are found out, we'll recall the formulas for triangular pyramid and the square pyramid.
If the pyramid is triangular, the base is a triangle and the three faces are also triangles
The total surface area of a pyramid is given by the addition of the area of the base and the lateral area of the pyramid.
TSA = Base area + Perimeter*Side Length/2
The volume of triangular pyramid is:
V = Base area*Height/3
If the pyramid has a square base, the TSA is:
TSA = Base area + Perimeter*Slant Length/2
The formula of the volume remains the same.
Therefore, the formula that gives TSA for a triangular pyramid is different from the onr that gives TSA for a square pyramid, but the formula of the volume is the same for both types.
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## Details
Forces
Glossary for Forces and Motion
41
Science
Not Applicable
02/23/2005
Term
Acceleration
Definition
Any change in the velocity of an object. If the object moves faster, or slower, or changes directions, it has accelerated. Acceleration is the change in velocity per second.
Term
Acclerometer
Definition
A device for detecting and measuring acceleration; it indicates zero when it is stationary or moving at constant velocity.
Term
Archimedes
Definition
A greek mathematician , physicist, and inventor. Archimde
Term
Archimedes
Definition
A greek mathematician , physicist, and inventor. Archimde
Term
Aristotle
Definition
A Greek philosopher, pupil of Plato, and tutor of Alexander the Great. Aritotle saw the importance of careful observation and detailed classification, which are parts of the scientific method.
Term
Balance
Definition
An instrument used to compare masses, traditionally with tow equal pans. Known masses are palced in one pan to balance an object in the other pan.
Term
Buoyancy
Definition
The apparent loss of weight of a body partly or completely immersed in a fluid, due to the wieght of fluid displaced by the object.
Term
Centripetal
Definition
Directed towards the center, for example the force requeired to keep an obeject moving in a circular path or orbit.
Term
Deceleration
Definition
Slwoing down, braking, or negatice acceleration.
Term
Density
Definition
The mass of an object divided by its volume. The amount of matter (mass) in a unit volume of a material: for example, the density of gold is 19.3 grams per cubic centimeter.
Term
Dependent Variable
Definition
The quanity plotted on the vertical axis of a graph, which depends on the value of the (independent) variable on the horizontal axis.
Term
equilibrium
Definition
A situation in which the resultant force on a obejct is zero. The object is at rest or moving at constant velocity in a straight line.
Term
Equilibrium
Definition
A situation in which the resultant force on a object is zero. The obejct is at terst or moving at constant velocity in a straight line.
Term
Fluid
Definition
Any liquid, gas, or powdered material, which flows, and lacks a derinite shape.
Term
Force
Definition
When applied to an object, a force tends to make the object move, change the motion of the obejct, or change its shape. Forces are measured in newtons and are vectors.
Term
Fricton
Definition
The force that resists sliding motion between two surfaces in contact. Vehicles moving trhough air experience friciton, commonly called air resistnace or drag.
Term
Gravity
Definition
The floce of attraction between objects, which increases as the mass of the objects increases but decreases as the separation of the objects increases.
Term
Independent Variable
Definition
The quantity plotted on the horizontal axis of a graph; for example, time is a common independent variable.
Term
Inertia
Definition
The tendency of an object to remain at rest (stationary) or to continue moving in a straight line at constant speed. Inertia increases as the mass of the object increases.
Term
kilogram
Definition
The SI unit of mass; 1000 grams; approximately 2.2 pounds. Written as "kilogram" symbol "kg"
Term
mass
Definition
is the amount of substnace in an object, is fixed, and measured in grams and kilograms,. The mass of any object is independent of the location of the object.
Term
meter
Definition
The SI unit of length; equal to 39.37 inches (US) Written as "meter", symbol"m".
Term
momentum
Definition
A moving object has mass and a velocity in a certian direction. Momentum can be calculated by multiplying mass times velocity. To change the momentum of an object, a force must act on it. Momentum is a vector because its direction is as important as its magnitude.
Term
Newton
Definition
The SI unit of force, written as "newton" symbol "N". This is the force that makes a one kilgram mass accelerate at one meter per second per second.
Term
Newton, Isaac
Definition
(1642-1727) English mathematician and physicist who studied motion, gravity, optics, and the orbits of the planets. His laws of motion an theory of gravitation were set out in a famous book, the Principia.
Term
Newton's 1st Law
Definition
An object remains at rest or continues to to moce in a straight line at constant velocity, unless it is acted on by an external force.
Term
Newton's 2nd Law
Definition
The acceleration of an object is proportional to the applied force and takes place inthe direction of the force. The accerleration is inversely proportional to the mass of the object: for example heavier objects accerlerate more slowly.
Term
Newton's 3rd Law
Definition
Actiona nd reaction are equal and opposite. The weight of an object presses on the floor, the floor provides a reaction force on the object equal to the object's weight.
Term
Orbit
Definition
The path of a planet aroudn the Sun, or of a moon around a planet, or of a satellite around the Earth. The path may be circular or elliptical. The orbit is maintained by the force of gravity and the speed of the body in its orbit.
Term
Reaction
Definition
The equal and opposite force that arises when a force is applied to anobject or system. The weight of a building is supported by the reaction of the ground.
Term
Resultant
Definition
The single force that is equivalent to the sum of the separate forces acting on an object.
Term
SI
Definition
Systeme International d'Unites; the international system of scientific untis of measurement.
Term
Speed
Definition
The distance a moving object covers in unit time; for example five centimeters per second, 20 miles per hour; the rate of movement.
Term
Spring Scale
Definition
A device containing a spring for measuring forces. The size of the force is indicated by the length the spring stretches.
Term
Terminal Velocity
Definition
The maximun velocity of an object falling through air or water. Gravity accerlerates the obejct until air or water resistnace prevents it accerlerating any more. The velocity is then constant.
Term
Upthrust
Definition
The buoymant force experienced by objects in any fluid. The upward force provided by the fluid.
Term
Vector
Definition
A quantity such as velocity or gorce, which must be specified by its size and direction.
Term
Velocity
Definition
The speed and direction of an object; a vector describing the motion of an object.
Term
Venn Diagram
Definition
A method for respresenting groups of similar objects (sets) using circules. Relationships between sets are expressed by overlapping the circles.
Term
Weight
Definition
The force of gravity on any object. Weight is dependent on the location of the object and for the force of gravity acting on the object.
Term
Weightlessness
Definition
A condition experienced by astronaunts orbiting the Earth. Gravity and the astronauts' speed keep them and the spacecraft in orbit.
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