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https://scicomp.stackexchange.com/questions/36094/why-do-people-omit-the-lowest-times-when-averaging-timing-results | 1,718,572,648,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00014.warc.gz | 456,153,701 | 39,052 | # Why do people omit the lowest times when averaging timing results?
Most of the time, when I see someone reporting the average timing of a certain algorithm on a computer in a computational mathematics paper, they do something like this:
1. Run the operation $$n$$ times (e.g., $$n=100$$ times);
2. Discard the lowest and highest $$m$$ times (e.g., $$m=10$$ outliers);
3. Compute the average of the remaining $$n-2m$$ times.
I can understand why dropping the highest results make sense: these are likely cases in which the computer switched threads into doing something else. I cannot understand why dropping the lowest results makes sense.
Why should I guard against an outlier being faster than the rest of the iterations? If anything, intuitively I would say that it is the minimum of the $$n$$ results that tells me how much time my computer really needs to run that algorithm. All other instances are cases when the computer switched into something else, or pipelines and multithreading were not used optimally.
The only case in which I can imagine this making sense is when the results reported are very close to the precision of the clock, but in this case I would say that the correct approach is repeatedly measuring the time for a sequence $$k$$ of runs at the same time (e.g., $$k=100$$) and then discarding outliers on those numbers. But in a case where outliers due to multitasking are the main concern, I do not see the advantages of this procedure.
• If you look at really good benchmarking suites, such as criterion.rs, they don't do this: bheisler.github.io/criterion.rs/book/getting_started.html Commented Oct 12, 2020 at 12:35
• I personally have never seen what you're describing . . . Commented Oct 12, 2020 at 12:37
• @user14717 So what does that benchmarking suite do, exactly? Commented Oct 12, 2020 at 12:43
• Statistical analysis of performance data. Commented Oct 12, 2020 at 12:56 | 446 | 1,915 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.936243 |
https://oeis.org/A118133 | 1,726,734,816,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651995.50/warc/CC-MAIN-20240919061514-20240919091514-00869.warc.gz | 381,754,267 | 4,652 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A118133 Least concatenation of n to have exactly n prime factors with multiplicity. 1
11, 22, 333, 444, 55555555, 66666666, 777777777777777, 888888, 999999999999, 101010101010101010, 11111111111111111111111111111111, 121212121212121212121212, 131313131313131313131313131313, 141414141414141414141414141414 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS See also A000461 Concatenate n n times. Bigomega(n) = A001222 Number of prime divisors of n (counted with multiplicity). The numbers k of concatenations on n in a(n) are: 2, 2, 3, 3, 8, 8, 15, 6, 12, 9, 16, 12. LINKS Table of n, a(n) for n=1..14. FORMULA a(n) = nnn...nnn (k times) such that A001222(a(n)) = n, for k minimum. EXAMPLE a(1) = 11 has 1 prime factor. a(2) = 22 = 2 * 11 has 2 prime factors. a(3) = 333 = 3^2 * 37 has 3 prime factors. a(4) = 444 = 2^2 * 3 * 37 has 4 prime factors. a(5) = 55555555 = 5 * 11 * 73 * 101 * 137 has 5 prime factors. a(6) = 66666666 = 2 * 3 * 11 * 73 * 101 * 137 has 6 prime factors. a(7) = 777777777777777 = 3 * 7 * 31 * 37 * 41 * 271 * 2906161. a(8) = 888888 = 2^3 * 3 * 7 * 11 * 13 * 37. a(9) = 999999999999 = 3^3 * 7 * 11 * 13 * 37 * 101 * 9901. a(10) = 101010101010101010 = 2 * 3^2 * 5 * 7 * 13 * 19 * 37 * 52579 * 333667. a(11) = 11111111111111111111111111111111 = 11 * 17 * 73 * 101 * 137 * 353 * 449 * 641 * 1409 * 69857 * 5882353. a(12) = 121212121212121212121212 = 2^2 * 3^2 * 7 * 13 * 37 * 73 * 101 * 137 * 9901 * 99990001. MATHEMATICA Table[Module[{i=1}, While[PrimeOmega[FromDigits[PadRight[{}, i, IntegerDigits[ n]]]]!= n, i++]; FromDigits[PadRight[{}, i, IntegerDigits[ n]]]], {n, 15}] (* Harvey P. Dale, Sep 07 2014 *) PROG (PARI) A118133(n)={local(r); r=n; while(bigomega(r)<>n, r=eval(Str(r, n))); r} \\ Michael B. Porter, May 01 2010 CROSSREFS Cf. A000461, A001222. Sequence in context: A067894 A094620 A077431 * A345404 A225186 A155973 Adjacent sequences: A118130 A118131 A118132 * A118134 A118135 A118136 KEYWORD base,nonn AUTHOR Jonathan Vos Post, May 13 2006 EXTENSIONS More terms from Harvey P. Dale, Sep 07 2014 STATUS approved
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# Patterns, Relationships, and Algebraic Thinking
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What does Algebra look like in the early grades?
We have developed 60 early grade activities on Patterns and Algebra just for you, the busy teacher. From an early age, learning to recognize patterns and sets creates the foundation for working with algebraic topics. Our Patterns and Algebra activities are designed to assist teachers in laying the necessary groundwork for success in algebra in later years. These 60 activities help teachers implement new techniques that will enhance student achievement in their classroom. Teachers your students will become patterns experts with these 60 fun activities. Students will learn how to identify copy, create and extend patterns in a variety of formats.
Our Patterns and Algebra activities enable students to—
?Identify, describe, and extend concrete and pictorial patterns in
order to make predictions and solve problems.
?Use patterns to skip count by twos, fives, and tens.
?Find patterns in numbers, including odd and even.
?Compare and order whole numbers using place value.
?Use patterns to develop strategies to solve basic addition and
basic subtraction problems.
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such as fact families for sums to 18.
The activities can be used to Assess concepts, Introduce concepts, Review concepts, and to help students Retain the concepts that have been taught. Our Math Activities can be used in a variety of ways to “AIRR” out your math curriculum.
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?Common Core Standards
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# Lab 5 - MAE 106 Experiment #6
By Rebecca Sanders,2014-05-07 15:37
19 views 0
Lab 5 - MAE 106 Experiment #6
MAE 106 Laboratory Exercise #5
PD Control of Motor Position
University of California, Irvine
Department of Mechanical and Aerospace Engineering
REQUIRED PARTS:
Qty Parts Equipment
2 1k? resistor, ? W (brown/black/red) Breadboard
2 10k? resistor, ? W (brown/black/orange) Oscilloscope
2 100k??resistor, ? W (brown/black/yellow) Function Generator
1 LM 324 quad op amp chip Motor-Amp-Tach Console
4 1?F capacitors Position-sensing “pot”
1 BNC cable IC puller
1 breakout (BNC to alligator clips) wrist grounding strap
2 banana-to-banana cable (1 black, 1 red) multimeter
2 banana-to-alligator clip cable (1 black, 1 red) scope probe
var wire, 22AWG
1 ?1/4” shaft coupling (c.1/2”lg)
1 Introduction
In this lab you will build a control system to make a motor shaft move to a position that
you command. Controlling motor position is a common goal in automation (e.g. multi-
joint robot arms, radars, numerically controlled milling machines, manufacturing
systems). In addition, you will need a position controller for your final project.
The controller that you will build is called a “Proportional Plus Derivative (PD) Position
Feedback System,” and is the most common controller found in industry. The PD control
law is:
? (1) ???K(???)?K?pdd
Where ? = actual motor angular position
? = desired motor angular position d
? ?= actual motor angular velocity
K = position error gain p
K = derivative gain d
??= desired motor torque
Note that the controller has two terms one proportional to the position error (the “P”
part), and one proportional to the derivative of position (i.e. velocity, the “D” part). Thus,
it is called a “PD” controller.
1
. (?-?) -Kpd(?-?) - K? pdd-K+ ) d-(?-?K p+ ?d 2? 1/Js - Ks d. - K? d
R2= 10k . 100k -?-K(?-?) - K? dpdd d?100k Function ? R1= 1k Amp - - motor pot Generator + + R1= 1k Op amp 2 Op amp 1
C=1?F
-
+ Op amp 3
Figure 1 PD Motor Position Control System (Block Diagram and Circuit)
Figure 1 shows the block diagram and op-amp circuit that you will implement to make
the PD control law for the motor. J is the inertia of the motor shaft.
The lab has the following four parts. You can do Parts 1 and 2 before coming to lab.
Part 1: What is the theoretical behavior of the controlled system?
The key point to understand here is that the controlled system obeys the same
differential equation as a mass-spring-damper system. Thus, the controlled system acts
dynamically like a mass-spring-damper system. The control gains K and K determine pd
the equivalent stiffness and damping of the system. The desired angular position of the
motor (?) is equivalent to the rest length of the spring. d
Part 2: How can a circuit implement the control law?
Op-amp circuits (adder, gain, inverter derivative circuits) can be used to implement the
control law. The resistors and capacitors set the control gains K and K. pd
Part 3: What is the step response of the actual system?
One way to characterize the system behavior is to measure how it performs when it is
commanded to move rapidly from one position to another (i.e. to follow a step function
input). You will find that the motor will overshoot and oscillate if the damping is too small.
Part 4: What is the frequency response of the actual system?
Another common way to characterize the system behavior is to measure how it performs
when it is commanded to follow a sinusoidal position. You will find that the controller
acts like a low pass filter. It tracks low input frequencies well, and high frequencies
poorly. Also, if the controller has low enough damping, it will resonate just like the
spring-mass-damper system you experimented with in Lab 4 (i.e. the vibrating beam).
2
2 What is the theoretical behavior of the controlled
system?
In this section, you will derive the theoretical behavior the PD position controlled motor.
In the time domain, the theoretical behavior is described by a differential equation. In
the frequency domain, the theoretical behavior is described by the frequency response.
Q1 Derive the dynamical equation that describes how ? evolves with time when the
controller is attached to the motor. Assume ? is the input. d
Q2 Derive the differential equation for a spring-mass-damper system (assume force is
the input and position is the output). The differential equation for Q1 should be
similar to the equation for Q2. This means that the PD position control system has
the same dynamics as a spring-mass-damper system; i.e. it follows the same
equations of motion. Thus, you can use your intuition about how the spring-mass-
damper system works to design the PD controller. Explain what the mass (m),
spring (k), damper (c) in the mechanical system correspond to in the PD system.
Q3 Derive the closed-loop transfer function, G(s), for the controlled system (the input is
?, the output is ?). Use either block diagram algebra (applied to the block diagram d
from Figure 1) or take the Laplace Transform of the differential equation that you
derived in Q1.
Q4 Express the damping ratio and natural frequency of the system in terms of the
control gains and motor inertia. The damping ratio is important because it
determines whether the system oscillates. The natural frequency determines the
frequency at which it oscillates.
P1 Plot the predicted response of the system to a step change in ? from 0 to 1 d
radians, for damping ratios of 0.1, 1.0, and 2.0.
P2 Plot the predicted frequency response (both scaling and phase shift) for damping
ratios of 0.1, 1.0, and 2.0. Do this on a Bode plot by plotting {20log(output
amplitude/input amplitude)} vs. {input frequency on a log scale}, and {phase shift}
vs. {input frequency on a log scale}.
3 How can a circuit implement the control law?
To implement the PD control law, you need to build the circuit shown in Figure 1.
Q5 By applying the op-amp golden rules, show that the input to the motor amplifier is:
R2???(???)??VoutRC d2R1
The derivation will be easier if you substitute the impedance 1/sC for the
capacitor then treat it as a resistor in the frequency domain, then transform back
to the time domain.
3
Q6 Compare this equation with the control law of equation (1). What are K and K in Pd
terms of the electronic components (resistor and capacitor values)? How would
you increase the damping of the system?.
Q7 Briefly describe the specific purpose for each of the op-amps in Figure 1.
Part 3 What is the step response of the actual system?
Construct the circuit in Figure 1. Important: wire your circuit neatly! A neat circuit
requires little extra time to wire, and it’s easier to debug. Circuits often take much more
time to debug than they do to initially wire! Make sure to hook up the potentiometer correctly! The wiper of the pot should not be connected to the power supply!
If your circuit works correctly, the motor position should follow whatever input signal you
provide with the function generator (square wave, sinusoid, constant voltage…).
IMPORTANT: Sometimes the circuit will not work and the motor will run uncontrollably
at a very high speed. This wrecks the pots. If this happens, turn the motor off right
away by turning of the DC supply to the motor. First, try to fix the instability problem by
reversing the polarity across the sensing pot (you may have positive feedback instead of
negative feedback). If this doesn’t work, debug your circuit. Do not try to debug your
circuit with the motor running! Debug your circuit systematically.
Here are some debugging hints:
? Compare your wiring diagram to your circuit to make sure all of the connections are
correct.
? Make sure you don’t have any loose connections. ? Verify that your output pot is working properly by connecting the scope to the wiper
and moving the motor shaft by hand. The scope trace should move up and down.
? Verify that op-amp 1 is inverting and op-amp 3 is following.
? Verify that the output of op-amp 2 changes as you adjust ? with ? constant. You d
can adjust ? using the function generator or another pot. d
Q8 Provide a step-input by using the function generator (4V peak-peak, 1 Hz square
wave). Is the system underdamped or overdamped? Does the observed response
agree with the theoretical one? Why is it different? What is the frequency at which
is oscillates (the damped natural frequency, ?). damped
PRACTICAL EXAM 1: Demonstrate to the TA that your motor is following the step
input.
P3 Suppose you didn’t want your motor to oscillate so much. This is an important
issue! PD controllers are used in many applications such as NC milling machines,
plotters, etc. You usually want your motor to go to a desired value quickly and
accurately without oscillating! Which variable would you change in your differential
equation for the PD system to increase damping? Increase the total capacitance
to 2?F by adding another capacitor (recall that capacitors add in parallel). Observe
the response to a step input. Repeat this for a total capacitance of 3 ?F and 4 ?F.
Record the step response for C = 1, 2, 3, and 4 ?F.
4
P4 In a mechanical system, if you wanted to the system to respond more quickly, you
would increase the natural frequency (?) by picking a stiffer spring (higher k). n
Double ? for the PD system by changing the appropriate resistor value. Keep C n
= 4 ?F. Record the step response of the system and plot it on the same plot as P3.
Part 4: What is the frequency response of the actual
system?
The goal of this last part of the lab is to characterize the frequency response of the
system. In particular, you will explore how well the system tracks the desired input
position when the input is a sinusoid, across a range of frequencies. Remember, you
can view linear systems such as this one as “filters”. The PD controller acts like a low-
pass filter, although it has a resonant peak if the damping is not great enough.
Q9 Change R1 to 1 K?? R2 to 10 K?? and C = 1 ?F. Now input a 1.0V amplitude sine
wave. Start at a low frequency. Keep the voltage scale on the scope the same for
both input and output. What is the output amplitude and phase shift at 1, 2, 4, 8,
16, 32 Hz? Does the system have a resonant frequency? The amplitude should
increase dramatically at this point. What is the resonant frequency and the output
amplitude at resonance? Do you notice any high frequency oscillations in your
output signal? What do you think might be causing those? Note: Don’t let the
motor oscillate for a long time at resonance.
PRACTICAL EXAM 2: Demonstrate to the TA the resonant frequency for your
motor.
P5 Make a Bode plot of the frequency response of the system. Plot {20log(output
amplitude/input amplitude)} vs. {input frequency on a log scale}, and {phase shift}
vs. {input frequency on a log scale}.
WRITE-UP
- due at your next laboratory session
- each student must complete his or her own write-up
- make sure to use your own words and to type the write-up!!
- include your name and laboratory time on the write-up
- Graphs for the lab write-up must be generated using Excel or Matlab, and must
include labels on the axes, voltage and time scales used on the scope, and a legend
for multiple-line plots.
Page limit = 2 pages, including graphs
1. Explain briefly how a PD controller works, using words and not equations.
2. Briefly describe three engineering applications of a PD controller
3. Turn in the answers to Q1-Q4; the equations may be hand-written.
4. Step Response: Turn in the plots for P1, P3 and P4 (use Matlab, but don’t show your
code)
5. Frequency Response: Turn in the plot for P2 and P5. (use Matlab, but don’t show | 2,877 | 11,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-17 | longest | en | 0.721809 |
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# Forex Trading Algorithms Part 3-Converting Trading Strategy To EA’s & Elements Of Computer Language!
113
0
### Trading Algorithms – The Elements of a Computer Language – Part I
A computer language is a formal language to convert our ideas into a language understandable by a computer. Along with computing history, languages have evolved from plain ones and zeroes to assembly language and up to the high-level languages we have today.
### Assembly language
Assembly language is a direct link to the computer’s CPU. Every assembly instruction of the instruction set is linked to a specific instruction code to the CPU.
Fig 1. The basic structure of an X86 CPU. Source cs.lmu.edu
The CPU characteristics are reflected in the instruction set. For instance, an X86 CPU has eight floating-point 80-bit registers, sixteen 64-bit registers, and six 16-bit registers. Registers are ultrafast memories for the CPU use. Thus every register has assembly instructions to load, add, subtract, and move values using them.
Fig 2- Example of assembly language
source codeproject.com
A computer program developed in assembly language is highly efficient, but it is a nightmare for the developer when the project is large. Therefore, high-level languages have been created for the benefit of computer scientists.
### The Elements of a high-level language
A modern computer language is a combination of efficient high-level data structures, elegant and easy-to-understand syntax, and an extensive library of functions to allow fast application development.
##### Numbers
A computer application usually receives inputs in the form of numbers. These come in two styles: integer and floating-point. Usually, they are linked to a name called “variable.” That name is used so that we can use different names for the many sources of information. For instance, a bar of market data is composed of Open, High, Low, and Close. We could assign each category the corresponding name in our program.
Integers correspond to a mathematical integer. An integer does not hold decimals. For instance, an integer division of 3/2 is 1. integers are usually used as counters or pointers to larger objects, such as lists or arrays.
A floating-point number is allowed to have decimals. Thus a 3/2 division is equal to 1.5. All OHLC market data comes in floating-point format.
##### Strings
A string is a data type to store written information made of characters. Strings are used as labels and to present information in a human-understandable form. Recently, strings are used as input in sentiment-analysis functions. Sentiment analysis
##### Boolean
Boolean types represent true/false values. A true or false value is the result of a question or “if” statement. It can also be assigned directly to a variable, such as in
In this case, buyCondition is False for EURUSD closes below 1.151, and is True when the close value is higher than 1.151.
##### Lists
We usually do not deal with a single number. If we want to compute a 20-period moving average of the USDJPY pair’s Close, we would need its last 20 closes. To store these values, the language uses lists (or arrays in C++). A list is an ordered collection of values or other types of information, such as strings.
Since Lists are ordered, we can refer to a particular element in the list using an index. For instance, if we were to retrieve the candlestick Close two bars ago of the USDJPY, we would ask for USDJPY.Close[2]
##### Sets
A Set is an unordered collection of elements. Sets do not allow duplication of elements. That means it eliminates duplicate entries. Not all languages have built-in Sets, although it can be made through programming if needed.
##### Dictionaries
Dictionaries are a useful data type that maps a key to a value. For instance, in Python
tel = {‘Joe’: 554 098 111, ‘Jane’: 660 413 901}
is a telephone structure. To retrieve Joe’s phone, we would write:
mytel = tel[‘Joe’]
with mytel holding 554 098 111
As with sets, not all high-level languages have built-in dictionaries, but a savvy programmer is able to create one.
In the next video of this series, we will explain the elements for flow control. | 917 | 4,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.884972 |
lhbikos.github.io | 1,718,346,714,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00118.warc.gz | 330,581,532 | 53,046 | # Chapter 5 Reliability
The focus of this lecture is the assessment of reliability. We start by defining classical test theory and examining several forms of reliability. While the majority of our time is spent considering estimates of internal consistency, we also review retest reliability and inter-rater reliability.
## 5.2 Defining Reliability
### 5.2.1 Begins with Classical Test Theory (CTT)
CTT is based on Spearman’s (1904) true-score model where:
• an observed score consists of two components – a true component and an error component
• X = T + E
• X = the fallible, observed/manifest score, obtained under ideal or perfect conditions of measurement (these conditions never exist);
• T = the true/latent score (that will likely remain unknown); and
• E = random error
• In CTT, we assume that the traits measured are constant and the errors random.
• Therefore, the mean of measurement errors for any individual (upon numerous repeated testing) would be zero.
• That said, in CTT, the true score would be equal to the mean of the observed scores over an indefinite number of repeated measures.
• Caveat: this is based on the assumption that when individuals are repeatedly measured, their true scores remain unchanged.
• In classic test theory, true score can be estimated over multiple trials. However, if errors are systematically biased, the true score will remain unknown.
### 5.2.2 Why are we concerned with reliability? Error!
Measurements are imperfect and every observation has some unknown amount of error associated with it. There are two components in error:
• random/unsystematic: varies in unpredictable and inconsistent ways upon repeated measurements; sources are unknown
• systematic: recurs upon repeated measurements reflecting situational or individual effects that, theoretically, could be specified.
Correlations are attenuated (i.e., smaller than) from the true correlation if the observations contain error. Knowing the reliability of an instruments allows us to:
• estimate the degree to which measured at one time and place with one instrument predict scores at another time and/or place and perhaps measured with a different instrument
• estimate the consistency of scores
• estimate “…the degree to which test scores are free from errors of measurement” (APA, 1985, p. 19)
Figure 7.1a in Revelle’s chapter illustrates the attenuation of the correlation between the variables p and q as a function of reliability.
• circles (latent variables [LV]) represent the true score
• observed/measured/manifest variables are represented by squares, and each has an associated error; not illustrated are the random and systematic components of error
• a true score is composed of a measured variable and its error
• the relationship between the true scores would be stronger than the one between the measured variables
• moving to 7.1b in Revelle’s chapter, the correlation between LV p and the observed ’’ can be estimated from the correlation of p’ with a parallel test (this is the reliability piece)
Figure 7.2 in Revelle’s Chapter 7 (n.d.) illustrates the conceptual effect of reliability on the estimation of a true score. The figure is meant to demonstrate that when error variances are small and reliability is greater, the variance of the true scores more closely approximates that of observed scores.
### 5.2.3 The Reliability Coefficient
The symbol for reliability, $$r_{xx}$$, sums up the big-picture definition that reliability is the correlation of a measure with itself. There are a number of ways to think about it:
• a “theoretical validity” of a measure because it refers to a relationship between observed scores and scores on a latent variable or construct,
• represents the fraction of an observed score variance that is not error,
• ranges from 0-1
• 1, when all observed variance is due to true-score variance; there are no random errors,
• 0, when all observed variance is due to random errors of measurement,
• represents the squared correlation between observed scores and true scores,
• the ratio between true-score variance and observed-score variance (for a formulaic rendition see ),
$r_{xt}^{2}=r_{xx} =\frac{\sigma_{2}^{t}}{\sigma_{2}^{x}}$ where $$r_{xt}^{2}$$ is the proportion of variance between observed scores (t + e) and true scores (t); its square root is the correlation
$$r_{xx}$$ is the reliability of a measure
$${\sigma_{2}^{t}}$$ is the variance of true scores
$${\sigma_{2}^{x}}$$ is the variance of observed scores
• The reliability coefficient is interpreted as the proportion of systematic variance in the observed score.
• .8 means that 80% of the variance of the observed scores is systematic;
• .2 (e.g., 1.00 - .8)is the proportion of variance due to random errors;
• the reliability coefficient is population specific.
To restate the first portion of the formula: although reliability is expressed as a correlation between observed scores, it is also the ratio of reliable variance to total variance.
## 5.3 Research Vignette
The research vignette for this lesson is the development and psychometric evaluation of the Perceptions of the LGBTQ College Campus Climate Scale . The scale is six items with responses rated on a 7-point Likert scale ranging from 1 (strongly disagree) to 7 (strongly agree). Higher scores indicate more negative perceptions of the LGBTQ campus climate. Szymanski and Bissonette (2020) have suggested that the psychometric evaluation supports using the scale in its entirety or as subscales. Each item is listed below with its variable name in parentheses:
• College response to LGBTQ students:
• My university/college is cold and uncaring toward LGBTQ students. (cold)
• My university/college is unresponsive to the needs of LGBTQ students. (unresponsive)
• My university/college provides a supportive environment for LGBTQ students. (unsupportive)
• this item must be reverse-scored
• LGBTQ Stigma:
• Negative attitudes toward LGBTQ persons are openly expressed on my university/college campus. (negative)
• Heterosexism, homophobia, biphobia, transphobia, and cissexism are visible on my university/college campus. (heterosexism)
• LGBTQ students are harassed on my university/college campus. (harassed)
A preprint of the article is available at ResearchGate. Below is the script for simulating item-level data from the factor loadings, means, and sample size presented in the published article.
Because data is collected at the item level (and I want this resource to be as practical as possible, I have simulated the data for each of the scales at the item level.
Simulating the data involved using factor loadings, means, and correlations between the scales. Because the simulation will produce “out-of-bounds” values, the code below rescales the scores into the range of the Likert-type scaling and rounds them to whole values.
Five additional scales were reported in the Szymanski and Bissonette article (2020). Unfortunately, I could not locate factor loadings for all of them; and in two cases, I used estimates from a more recent psychometric analysis. When the individual item and their factor loadings are known, I assigned names based on item content (e.g., “lo_energy”) rather than using item numbers (e.g., “PHQ4”). When I am doing psychometric analyses, I prefer item-level names so that I can quickly see (without having to look up the item names) how the items are behaving. While the focus of this series of chapters is on the LGBTQ Campus Climate scale, this simulated data might be useful to you in one or more of the suggestions for practice (e.g., examining the psychometric characteristics of one or the other scales). The scales, their original citation, and information about how I simulated data for each are listed below.
• Sexual Orientation-Based Campus Victimization Scale is a 9-item item scale with Likert scaling ranging from 0 (never) to 3 (two or more times). Because I was not able to locate factor loadings from a psychometric evaluation, I simulated the data by specifying a 0.8 as a standardized factor loading for each of the items.
• College Satisfaction Scale is a 5-item scale with Likert scaling ranging from 1 (strongly disagree) to 7 (strongly agree). Higher scores represent greater college satisfaction. Because I was not able to locate factor loadings from a psychometric evaluation, I simulated the data by specifying a 0.8 as a standardized factor loading for each of the items.
• Institutional and Goals Commitment is a 6-item subscale from a 35-item measure assessing academic/social integration and institutional/goal commitment (5 subscales total). The measure had with Likert scaling ranging from 1 (strongly disagree) to 5 (strongly agree). Higher scores on the institutional and goals commitment subscale indicate greater intentions to persist in college. Data were simulated using factor loadings in the source article.
• GAD-7 is a 7-item scale with Likert scaling ranging from 0 (not at all) to 3 (nearly every day). Higher scores indicate more anxiety. I simulated data by estimating factor loadings from Brattmyr et al. (2022).
• PHQ-9 is a 9-item scale with Likert scaling ranging from 0 (not at all) to 3 (nearly every day). Higher scores indicate higher levels of depression. I simulated data by estimating factor loadings from Brattmyr et al. (2022).
#Entering the intercorrelations, means, and standard deviations from the journal article
Szymanski_generating_model <- '
#measurement model
CollegeResponse =~ .88*cold + .73*unresponsive + .73*supportive
Stigma =~ .86*negative + .76*heterosexism + .71*harassed
Victimization =~ .8*Vic1 + .8*Vic2 + .8*Vic3 + .8*Vic4 + .8*Vic5 + .8*Vic6 + .8*Vic7 + .8*Vic8 + .8*Vic9
CollSat =~ .8*Sat1 + .8*Sat2 + .8*Sat3 + .8*Sat4 + .8*Sat5
Anxiety =~ .851*nervous + .887*worry_control + .894*much_worry + 674*cant_relax + .484*restless + .442*irritable + 716*afraid
Depression =~ .798*anhedonia + .425*down + .591*sleep + .913*lo_energy + .441*appetite + .519*selfworth + .755*concentration + .454*too_slowfast + .695*s_ideation
#Means
CollegeResponse ~ 2.71*1
Stigma ~3.61*1
Victimization ~ 0.11*1
CollSat ~ 5.61*1
Persistence ~ 4.41*1
Anxiety ~ 1.45*1
Depression ~1.29*1
#Correlations
CollegeResponse ~~ .58*Stigma
CollegeResponse ~~ -.25*Victimization
CollegeResponse ~~ -.59*CollSat
CollegeResponse ~~ -.29*Persistence
CollegeResponse ~~ .17*Anxiety
CollegeResponse ~~ .18*Depression
Stigma ~~ .37*Victimization
Stigma ~~ -.41*CollSat
Stigma ~~ -.19*Persistence
Stigma ~~ .27*Anxiety
Stigma ~~ .24*Depression
Victimization ~~ -.22*CollSat
Victimization ~~ -.04*Persistence
Victimization ~~ .23*Anxiety
Victimization ~~ .21*Depression
CollSat ~~ .53*Persistence
CollSat ~~ -.29*Anxiety
CollSat ~~ -.32*Depression
Persistence ~~ -.22*Anxiety
Persistence ~~ -.26*Depression
Anxiety ~~ .76*Depression
'
set.seed(240218)
dfSzy <- lavaan::simulateData(model = Szymanski_generating_model,
model.type = "sem",
meanstructure = T,
sample.nobs=646,
standardized=FALSE)
#used to retrieve column indices used in the rescaling script below
col_index <- as.data.frame(colnames(dfSzy))
#The code below loops through each column of the dataframe and assigns the scaling accordingly
#Rows 1 thru 6 are the Perceptions of LGBTQ Campus Climate Scale
#Rows 7 thru 15 are the Sexual Orientation-Based Campus Victimization Scale
#Rows 16 thru 20 are the College Satisfaction Scale
#Rows 21 thru 26 are the Institutional and Goals Commitment Scale
#Rows 27 thru 33 are the GAD7
#Rows 34 thru 42 are the PHQ9
for(i in 1:ncol(dfSzy)){
if(i >= 1 & i <= 6){
dfSzy[,i] <- scales::rescale(dfSzy[,i], c(1, 7))
}
if(i >= 7 & i <= 15){
dfSzy[,i] <- scales::rescale(dfSzy[,i], c(0, 3))
}
if(i >= 16 & i <= 20){
dfSzy[,i] <- scales::rescale(dfSzy[,i], c(1, 7))
}
if(i >= 21 & i <= 26){
dfSzy[,i] <- scales::rescale(dfSzy[,i], c(1, 5))
}
if(i >= 27 & i <= 33){
dfSzy[,i] <- scales::rescale(dfSzy[,i], c(0, 3))
}
if(i >= 34 & i <= 42){
dfSzy[,i] <- scales::rescale(dfSzy[,i], c(0, 3))
}
}
#rounding to integers so that the data resembles that which was collected
library(tidyverse)
dfSzy <- dfSzy %>% round(0)
#quick check of my work
#psych::describe(dfSzy)
#Reversing the supportive item on the Perceptions of LGBTQ Campus Climate Scale so that the exercises will be consistent with the format in which the data was collected
dfSzy <- dfSzy %>%
dplyr::mutate(supportiveNR = 8 - supportive)
#Reversing three items on the Institutional and Goals Commitments scale so that the exercises will be consistent with the format in which the data was collected
dfSzy <- dfSzy %>%
dplyr::mutate(undecidedNR = 8 - undecided)%>%
dfSzy <- dplyr::select(dfSzy, -c(supportive, not_graduate, undecided, grades_unimportant))
The optional script below will let you save the simulated data to your computing environment as either an .rds object (preserves any formatting you might do) or a.csv file (think “Excel lite”).
# to save the df as an .rds (think 'R object') file on your computer;
# it should save in the same file as the .rmd file you are working
# with saveRDS(dfSzy, 'SzyDF.rds') bring back the simulated dat from
# an .rds file dfSzy <- readRDS('SzyDF.rds')
# write the simulated data as a .csv write.table(dfSzy,
# file='SzyDF.csv', sep=',', col.names=TRUE, row.names=FALSE) bring
# back the simulated dat from a .csv file dfSzy <-
# read.csv('SzyDF.csv', header = TRUE)
If we look at the information about this particular scale, we recognize that the supportive item is scaled in the opposite direction of the rest of the items. That is, a higher score on supportive would indicate a positive perception of the campus climate for LGBTQ individuals, whereas higher scores on the remaining items indicate a more negative perception. Before moving forward, we must reverse score this item.
In this psychometrics example I have given my variables one-word names that represent each item. Many researchers (including myself) will often give variable names that are alpha numerical: LGBTQ1, LGBTQ2, LGBTQn. In the psychometric evaluations, the one-word names may be useful shortcuts as one begins to understand the inter-item relations.
In reverse-scoring the supportive item, I will rename it “unsupportive” as an indication of its reversed direction.
library(tidyverse)
dfSzy <- dfSzy %>%
dplyr::mutate(unsupportive = 8 - supportiveNR) #when reverse-coding, subtract the variable from one number higher than the scaling
# When unhashtagged, this code provides item-level descriptive
# statistics psych::describe(dfSzy)
Next, I will create dfs that each contain the items of the total and subscales. These will be useful in the reliability estimates that follow.
Note that I am adding “T1” (time 1) to the end of the variable names. Later in the lesson when we evaluate test-retest reliability, we will simulate and add a “T2.
LGBTQT1 <- dplyr::select(dfSzy, cold, unresponsive, unsupportive, negative,
heterosexism, harassed)
ResponseT1 <- dplyr::select(dfSzy, cold, unresponsive, unsupportive)
StigmaT1 <- dplyr::select(dfSzy, negative, heterosexism, harassed)
As we move into the lecture, allow me to provide a content advisory. Individuals who hold LGBTQIA+ identities are frequently the recipients of discrimination and harassment. If you are curious about why these items are considered to be stigmatizing or non-responsive, please do not ask a member of the LGBTQIA+ community to explain it to you; it is not their job to educate others on discrimination, harassment, and microaggressions. Rather, please read the article in its entirety. Additionally, resources such as The Trevor Project, GLSEN, and Campus Pride are credible sources of information for learning more.
## 5.4 A Parade of Reliability Coefficients
While I cluster the reliability coefficients into large groups, please understand that these are somewhat overlapping.
Table 1 in Revelle and Condon’s (2019a) article provides a summary of the type of reliability tested, the findings, and the function used in the psych package.
### 5.4.1 Reliability Options for a Single Administration
If reliability is defined as the correlation between a test and a test just like it, how do we estimate the reliability of a single test, given only one time ? It may help to keep in mind that reliability is the ratio of true score variance to test score variance (or 1 - the ratio of error variance). Thus, the goal is to estimate the amount of error variance in the test. In this case we can investigate:
• a correlation between two random parts of the test
• internal consistency
• the internal structure of the test
#### 5.4.1.1 Split-half reliability
Split-half reliability is splitting a test into two random halves, correlating the two halves, and adjusting the correlation with the Spearman-Brown prophecy formula. Abundant formulaic detail in Revelle’s Chapter 7/Reliability (n.d.).
An important question to split-half is “How to split?” Revelle terms it a “combinatorially difficult problem.” There are 126 possible splits for a 10-item scale; 6,345 possible splits for a 16-item scale; and over 4.5 billion for a 36-item scale! The psych package’s splitHalf() function will try all possible splits for scales of up to 16 items, then sample 10,000 splits for scales longer than that.
split <- psych::splitHalf(LGBTQT1, raw = TRUE, brute = TRUE)
split #show the results of the analysis
Split half reliabilities
Call: psych::splitHalf(r = LGBTQT1, raw = TRUE, brute = TRUE)
Maximum split half reliability (lambda 4) = 0.73
Guttman lambda 6 = 0.68
Average split half reliability = 0.7
Guttman lambda 3 (alpha) = 0.7
Guttman lambda 2 = 0.71
Minimum split half reliability (beta) = 0.54
Average interitem r = 0.28 with median = 0.25
2.5% 50% 97.5%
Quantiles of split half reliability = 0.54 0.72 0.73
hist(split$raw, breaks = 101, xlab = "Split-half reliability", main = "Split-half reliabilities of 6 LGBTQ items") Results of the split-half can provide some indication of whether not the scale is unidimensional. In this case, the maximum reliability coefficient is 0.73, the average 0.70, and the lowest is 0.54. Similarly, we can examine the quantiles: 0.54, 0.72, 0.73. The split-half output also includes the classic Cronbach’s (1951) alpha coefficient (0.70; aka Guttman lambda 3) and average interitem correlations (0.25). The figure plots the frequencies of the reliability coefficient values. While I did not find guidelines on what constitutes a “high enough lower bound” to establish homogeneity, Revelle suggested that a scale with 0.85, 0.80, and 0.65 had “strong evidence for a relatively homogeneous scale.” When the values were 0.81, 0.73, 0.42, Revelle indicated that there was “strong evidence for non-homogeneity” . In making this declaration, Revelle was also looking at the strength of the inter-item correlation and for a rather tight, bell-shaped, distribution at the higher (> 0.73) end of the figure. What happens when we examine the split-half estimates of the subscales? With only three items, there’s not much of a split and so the associated histogram will not be helpful. splitRx <- psych::splitHalf(ResponseT1, raw = TRUE, brute = TRUE) splitRx #show the results of the analysis Split half reliabilities Call: psych::splitHalf(r = ResponseT1, raw = TRUE, brute = TRUE) Maximum split half reliability (lambda 4) = 0.62 Guttman lambda 6 = 0.57 Average split half reliability = 0.59 Guttman lambda 3 (alpha) = 0.66 Guttman lambda 2 = 0.66 Minimum split half reliability (beta) = 0.56 Average interitem r = 0.39 with median = 0.4 2.5% 50% 97.5% Quantiles of split half reliability = 0.57 0.58 0.62 hist(splitRx$raw, breaks = 101, xlab = "Split-half reliability", main = "Split-half reliabilities of 3 items of the College Response subscale")
The alpha is 0.66. The range of splits for max, ave, and low are 0.62, 0.59, and 0.55 and the quantiles are 0.57, 0.58, 0.62. The inter-item correlations have an average of 0.40.
Let’s look at the split-half reliability coefficients for the Stigma subscale.
splitSt <- psych::splitHalf(StigmaT1, raw = TRUE, brute = TRUE)
splitSt #show the results of the analysis
Split half reliabilities
Call: psych::splitHalf(r = StigmaT1, raw = TRUE, brute = TRUE)
Maximum split half reliability (lambda 4) = 0.56
Guttman lambda 6 = 0.53
Average split half reliability = 0.56
Guttman lambda 3 (alpha) = 0.63
Guttman lambda 2 = 0.63
Minimum split half reliability (beta) = 0.55
Average interitem r = 0.36 with median = 0.36
2.5% 50% 97.5%
Quantiles of split half reliability = 0.55 0.56 0.56
hist(splitRxraw, breaks = 101, xlab = "Split-half reliability", main = "Split-half reliabilities of 3 items of the Stigma subscale") The alpha of this subscale is lower than the total scale score $$(\alpha = 0.60$$). The maximum, average, and minimum split-half reliabilities were 0.56, 0.56, and 0.55; quantiles were at 0.55, 0.56, and 0.56. The average interitem correlation was 0.36. Because the alpha coefficient can be defined as the “average of all possible split-half coefficients” for the groups tested, it is common for researchers to not provide split-half results in their papers – this is true for our research vignette. I continue to teach the split-half because it can be a stepping stone in the conceptualization of internal consistency as an estimate of reliability. #### 5.4.1.2 Alpha coefficients The most common methods to assess internal consistency are the KR20 (for dichotomous items) and $$\alpha$$ (for Likert scaling); alpha has an alias, $$\lambda _{3}$$ (i.e., the Guttman lambda 3). Alpha and the Guttman 3 (used for scales with Likert-type scaling) may be thought of as: • a function of the number of items and the average correlation between the items • the correlation of a test with a non-existent test just like it • average of all possible split-half coefficients for the groups tested Although the psych package has an incredible and thorough alpha() function, Revelle is not a fan of alpha. In fact, his alpha function reports a 95% CI around alpha as well as bootstrapped alpha results. Let’s grab alpha coefficients for our total and subscales. psych::alpha(LGBTQT1) Reliability analysis Call: psych::alpha(x = LGBTQT1) raw_alpha std.alpha G6(smc) average_r S/N ase mean sd median_r 0.7 0.7 0.68 0.28 2.4 0.018 4 0.63 0.25 95% confidence boundaries lower alpha upper Feldt 0.66 0.7 0.74 Duhachek 0.66 0.7 0.74 Reliability if an item is dropped: raw_alpha std.alpha G6(smc) average_r S/N alpha se var.r med.r cold 0.64 0.64 0.61 0.27 1.8 0.022 0.0066 0.22 unresponsive 0.66 0.66 0.63 0.28 2.0 0.021 0.0073 0.25 unsupportive 0.67 0.67 0.63 0.29 2.0 0.021 0.0058 0.25 negative 0.66 0.66 0.63 0.28 2.0 0.021 0.0084 0.25 heterosexism 0.66 0.66 0.63 0.28 2.0 0.021 0.0087 0.25 harassed 0.67 0.67 0.64 0.29 2.0 0.021 0.0078 0.25 Item statistics n raw.r std.r r.cor r.drop mean sd cold 646 0.68 0.68 0.59 0.49 4.1 1.03 unresponsive 646 0.63 0.63 0.51 0.43 4.3 0.99 unsupportive 646 0.62 0.62 0.51 0.42 3.7 0.98 negative 646 0.64 0.63 0.51 0.42 4.0 1.04 heterosexism 646 0.61 0.63 0.51 0.43 4.0 0.90 harassed 646 0.63 0.61 0.49 0.41 3.9 1.07 Non missing response frequency for each item 1 2 3 4 5 6 7 miss cold 0.00 0.04 0.22 0.40 0.23 0.09 0.00 0 unresponsive 0.00 0.03 0.17 0.37 0.33 0.09 0.01 0 unsupportive 0.01 0.07 0.35 0.37 0.17 0.02 0.01 0 negative 0.01 0.07 0.23 0.39 0.24 0.05 0.00 0 heterosexism 0.00 0.03 0.24 0.43 0.26 0.03 0.00 0 harassed 0.01 0.07 0.27 0.37 0.22 0.05 0.01 0 The second screen of output shows the information we are interested in: • raw_alpha, .70 is based on the covariances • std.apha, .70 is based on correlations • average_r, .28 is the average inter-item correlation (i.e., all possible pairwise combinations of items) psych::alpha(ResponseT1) Reliability analysis Call: psych::alpha(x = ResponseT1) raw_alpha std.alpha G6(smc) average_r S/N ase mean sd median_r 0.66 0.66 0.57 0.39 1.9 0.023 4 0.77 0.4 95% confidence boundaries lower alpha upper Feldt 0.61 0.66 0.70 Duhachek 0.62 0.66 0.71 Reliability if an item is dropped: raw_alpha std.alpha G6(smc) average_r S/N alpha se var.r med.r cold 0.52 0.52 0.35 0.35 1.1 0.038 NA 0.35 unresponsive 0.60 0.60 0.42 0.42 1.5 0.032 NA 0.42 unsupportive 0.58 0.58 0.40 0.40 1.4 0.033 NA 0.40 Item statistics n raw.r std.r r.cor r.drop mean sd cold 646 0.80 0.79 0.62 0.50 4.1 1.03 unresponsive 646 0.76 0.76 0.55 0.45 4.3 0.99 unsupportive 646 0.76 0.77 0.57 0.46 3.7 0.98 Non missing response frequency for each item 1 2 3 4 5 6 7 miss cold 0.00 0.04 0.22 0.40 0.23 0.09 0.00 0 unresponsive 0.00 0.03 0.17 0.37 0.33 0.09 0.01 0 unsupportive 0.01 0.07 0.35 0.37 0.17 0.02 0.01 0 In the case of the College Response subscale: • raw_alpha, .66 is based on the covariances • std.apha, .66 is based on correlations • average_r, .39 is the average interitem correlation psych::alpha(StigmaT1) Reliability analysis Call: psych::alpha(x = StigmaT1) raw_alpha std.alpha G6(smc) average_r S/N ase mean sd median_r 0.62 0.63 0.53 0.36 1.7 0.025 4 0.76 0.36 95% confidence boundaries lower alpha upper Feldt 0.57 0.62 0.67 Duhachek 0.57 0.62 0.67 Reliability if an item is dropped: raw_alpha std.alpha G6(smc) average_r S/N alpha se var.r med.r negative 0.52 0.52 0.35 0.35 1.1 0.038 NA 0.35 heterosexism 0.53 0.53 0.36 0.36 1.1 0.037 NA 0.36 harassed 0.53 0.54 0.37 0.37 1.2 0.036 NA 0.37 Item statistics n raw.r std.r r.cor r.drop mean sd negative 646 0.77 0.76 0.56 0.44 4.0 1.0 heterosexism 646 0.73 0.76 0.55 0.44 4.0 0.9 harassed 646 0.77 0.75 0.54 0.43 3.9 1.1 Non missing response frequency for each item 1 2 3 4 5 6 7 miss negative 0.01 0.07 0.23 0.39 0.24 0.05 0.00 0 heterosexism 0.00 0.03 0.24 0.43 0.26 0.03 0.00 0 harassed 0.01 0.07 0.27 0.37 0.22 0.05 0.01 0 In the case of the Stigma subscale: • raw_alpha, .62 is based on the covariances • std.apha, .63 is based on correlations • average_r, .36 is the average interitem correlation The documentation for this package is incredible. Scrolling down through the description of the alpha() function provides a description of these different statistics. Especially useful are item-level statistics: • r.drop is the corrected item-total correlation (in the next lesson) for this item against the remaining items in the scale • mean and sd are the mean and standard deviation of each item across all individuals. The popularity of alpha emerged when tools available for calculation were less sophisticated; since then, we have learned that alpha can be misleading: • Alpha inflates, somewhat artificially, even when inter-item correlations are low. • A 14-item scale will have an alpha of at least .70, even if it has two orthogonal (i.e., unrelated) scales . • Alpha assumes a unidimensional factor structure. • The same alpha can be obtained for dramatically different underlying factor structures (see graphs in Revelle’s Chapter 7). The proper use of alpha requires the following: • Tau equivalence, that is, equal covariances with the latent score represented by the test. • Unidimensionality, equal factor loadings on the single factor of the test. When either of these is violated, alpha underestimates reliability and overestimates the fraction of test variance that is associated with the general variance in the test. Alpha and the split half are internal consistency estimates. Moving to model-based techniques allows us to take into consideration the factor structure of the scale. In the original article , results were as follows: #### 5.4.1.3 Omega Assessing reliability with omega ($$\omega$$) statistics falls into a larger realm of composite reliability where reliability is assessed from a ratio of the variability explained by the items compared with the total variance of the entire scale . Members of the omega family of reliability estimates come from factor exploratory (i.e., EFA) and confirmatory (i.e., CFA; structural equation modeling [SEM]) factor analytic approaches. This lesson precedes the lessons on CFA and SEM. Therefore, my explanations and demonstrations will be somewhat brief. I intend to revisit omega output in the CFA and SEM lessons and encourage you to review this section now, then return to this section again after learning more about CFA and SEM. In the context of psychometrics, it may be useful (albeit an oversimplification) to think of factors as scales/subscales where g refers to the amount of variance in the general factor (or total scale score) and subscales to be items that have something in common that is separate from what is g. Model-based estimates examine the correlations or covariances of the items and decompose the test variance into that which is: • common to all items (g, a general factor), • specific to some items (f, orthogonal group factors), and • unique to each item (confounding s specific, and e error variance) $$\omega$$ is something of a shapeshifter. In the psych package: • $$\omega_{t}$$ represents the total reliability of the test ($$\omega_{t}$$) • In the psych package, this is calculated from a bifactor model where there is one general g factor (i.e., each item loads on the single general factor), one or more group factors (f), and item-specific factors. • $$\omega_{h}$$ extracts a higher-order factor from the correlation matrix of lower-level factors, then applies the Schmid and Leiman (1957) transformation to find the general loadings on the original items. Stated another way, it is a measure of the general factor saturation (g; the amount of variance attributable to one common factor). The subscript “h” acknowledges the hierarchical nature of the approach. • the $$\omega_{h}$$ approach is exploratory and defined if there are three or more group factors (with only two group factors, the default is to assume they are equally important, hence the factor loadings of those subscales will be equal) • Najera Catalan suggests that $$\omega_{h}$$ is the best measure of reliability when dealing with multiple dimensions. • $$\omega_{g}$$ is an estimate that uses a bifactor solution via the SEM package lavaan and tends to be a larger (because it forces all the cross loadings of lower-level factors to be 0) • the $$\omega_{g}$$ is confirmatory, requiring the specification of which variables load on each group factor Two commands in psych get us the results: • omega() reports only the EFA solution • omegaSem() reports both EFA and CFA solutions • We will use the omegaSem() function Note that in our specification, we indicate there are two factors. We do not tell it what items belong to what factors (think, subscales). One test will be to see if the items align with their respective factors. psych::omegaSem(LGBTQT1, nfactors = 2) Loading required namespace: GPArotation Three factors are required for identification -- general factor loadings set to be equal. Proceed with caution. Think about redoing the analysis with alternative values of the 'option' setting. Warning in lav_model_vcov(lavmodel = lavmodel, lavsamplestats = lavsamplestats, : lavaan WARNING: Could not compute standard errors! The information matrix could not be inverted. This may be a symptom that the model is not identified. Call: psych::omegaSem(m = LGBTQT1, nfactors = 2) Omega Call: omegah(m = m, nfactors = nfactors, fm = fm, key = key, flip = flip, digits = digits, title = title, sl = sl, labels = labels, plot = plot, n.obs = n.obs, rotate = rotate, Phi = Phi, option = option, covar = covar) Alpha: 0.7 G.6: 0.68 Omega Hierarchical: 0.54 Omega H asymptotic: 0.73 Omega Total 0.74 Schmid Leiman Factor loadings greater than 0.2 g F1* F2* h2 u2 p2 cold 0.53 0.45 0.49 0.51 0.59 unresponsive 0.45 0.37 0.34 0.66 0.60 unsupportive 0.45 0.41 0.37 0.63 0.55 negative 0.46 0.40 0.37 0.63 0.58 heterosexism 0.46 0.39 0.36 0.64 0.59 harassed 0.44 0.39 0.35 0.65 0.56 With Sums of squares of: g F1* F2* 1.31 0.51 0.46 general/max 2.59 max/min = 1.1 mean percent general = 0.58 with sd = 0.02 and cv of 0.03 Explained Common Variance of the general factor = 0.58 The degrees of freedom are 4 and the fit is 0 The number of observations was 646 with Chi Square = 2.59 with prob < 0.63 The root mean square of the residuals is 0.01 The df corrected root mean square of the residuals is 0.02 RMSEA index = 0 and the 10 % confidence intervals are 0 0.049 BIC = -23.3 Compare this with the adequacy of just a general factor and no group factors The degrees of freedom for just the general factor are 9 and the fit is 0.18 The number of observations was 646 with Chi Square = 115.31 with prob < 0.000000000000000000012 The root mean square of the residuals is 0.1 The df corrected root mean square of the residuals is 0.13 RMSEA index = 0.135 and the 10 % confidence intervals are 0.114 0.158 BIC = 57.08 Measures of factor score adequacy g F1* F2* Correlation of scores with factors 0.74 0.57 0.56 Multiple R square of scores with factors 0.54 0.33 0.31 Minimum correlation of factor score estimates 0.09 -0.34 -0.38 Total, General and Subset omega for each subset g F1* F2* Omega total for total scores and subscales 0.74 0.66 0.63 Omega general for total scores and subscales 0.54 0.38 0.36 Omega group for total scores and subscales 0.20 0.28 0.27 The following analyses were done using the lavaan package Omega Hierarchical from a confirmatory model using sem = 0.54 Omega Total from a confirmatory model using sem = 0.74 With loadings of g F1* F2* h2 u2 p2 cold 0.56 0.42 0.49 0.51 0.64 unresponsive 0.47 0.33 0.34 0.66 0.65 unsupportive 0.44 0.42 0.38 0.62 0.51 negative 0.45 0.42 0.37 0.63 0.55 heterosexism 0.46 0.39 0.36 0.64 0.59 harassed 0.43 0.40 0.34 0.66 0.54 With sum of squared loadings of: g F1* F2* 1.32 0.46 0.49 The degrees of freedom of the confirmatory model are 3 and the fit is 2.658827 with p = 0.4472696 general/max 2.71 max/min = 1.05 mean percent general = 0.58 with sd = 0.06 and cv of 0.1 Explained Common Variance of the general factor = 0.58 Measures of factor score adequacy g F1* F2* Correlation of scores with factors 0.74 0.55 0.58 Multiple R square of scores with factors 0.55 0.31 0.33 Minimum correlation of factor score estimates 0.10 -0.39 -0.34 Total, General and Subset omega for each subset g F1* F2* Omega total for total scores and subscales 0.74 0.66 0.63 Omega general for total scores and subscales 0.54 0.41 0.34 Omega group for total scores and subscales 0.20 0.26 0.28 To get the standard sem fit statistics, ask for summary on the fitted object There is a ton of output! How do we make sense of it? First, our items aligned perfectly with their respective factors (subscales). That is, it would be problematic if the items switched factors. Second, we can interpret our results. Like alpha, the omegas range from 0 to 1, where values closer to 1 represent good reliability . For unidimensional measures, $$\omega_{t}$$ values above 0.80 indicate satisfactory reliability. For multidimensional measures with well-defined dimensions, we strive for $$\omega_{h}$$ values above 0.65 (and $$\omega_{t}$$ > 0.8). These recommendations are based on a Monte Carlo study that examined a host of reliability indicators and how their values corresponded with accurate predictions of poverty status. With this in mind, let’s examine the output related to our simulated research vignette. Let’s start with the output in the lower portion where the values are “from a confirmatory model using sem.” Omega is a reliability estimate for factor analysis that represents the proportion of variance in the LGBTQ scale attributable to common variance rather than error. The omega for the total reliability of the test ($$\omega_{t}$$; which included the general factors and the subscale factors) was .74, meaning that 74% of the variance in the total scale is due to the factors and 26% (100% - 74%) is attributable to error. Omega hierarchical ($$\omega_{h}$$) estimates are the proportion of variance in the LGBTQ score attributable to the general factor, which in effect treats the subscales as error. $$\omega_{h}$$ for the the LGBTQ total scale was .54. A quick calculation with $$\omega_{h}$$ (.54) and $$\omega_{t}$$ (.74; .54/.74 = .72) lets us know that that 73% of the reliable variance in the LGBTQ total scale is attributable to the general factor. .54/.74 [1] 0.7297297 Amongst the output is the Cronbach’s alpha coefficient (.70). Szymanski and Bissonette (2020) did not report omega results; this may be because there were only two subfactors and/or they did not feel like a bifactor analysis would be appropriate. You might notice the lavaan warning indicating that three factors are needed in order to identify the CFA model. There is a longer explanation about factor identification. Stay tuned for CFA models. #### 5.4.1.4 Some summary statements about reliability from single administrations • With the exception of the worst split-half reliability and $$\omega_{g}$$ or $$\omega_{h}$$, all of the reliability estimates are functions of test length and will tend asymptotically towards 1 as the number of items increases. • The omega output provides a great deal more information about reliability than a simple alpha. • Figure 7.5 in Revelle’s chapter shows four different structural representations of measures that have equal alphas (all .72) • $$\omega_{(h)}$$, $$\beta$$, and the worst split-half reliability are estimates of the amount of general factor variance in the test scores • In the case of low general factor saturation, the EFA based $$\omega_{(h)}$$ is positively biased, so the CFA-based estimate, $$\omega_{(g)}$$, should be used. • $$\omega_{(t)}$$ is the model-based estimate of the greatest lower bound of the total reliability of the test; so is the best split-half reliability. Revelle and Condon’s (2019b) recommendations to researchers: • Report at least two coefficients (e.g., $$\omega_{(h)}$$ and $$\omega_{(t)}$$) and discuss why each is appropriate for the inference that is being made. • Report more than “just alpha” unless you can demonstrate that the measure is tau equivalent and unidimensional. ### 5.4.2 Reliability Options for Two or more Administrations #### 5.4.2.1 Test-retest of total scores The purpose of test-retest reliability is to understand the stability of the measure over time. With two time points, T1 and T2, the test-retest correlation is an unknown mixture of trait, state, and specific variance, and is a function of the length of time between two measures. • With two time points we cannot distinguish between trait and state effects, that said • we would expect a high degree of stability if the retest were (relatively) immediate • With three time points we can leverage some SEM tools to distinguish between trait and state components • A large test-retest correlation over a long period of time indicates temporal stability. Temporal stability is: • expected if we are assessing something trait like (e.g., cognitive ability, personality trait) • not expected if we are assessing something state like (e.g., emotional state, mood) • not expected if there was an intervention (or condition) and the T1 and T2 administrations are part of a pre- and post-test design. There are some methodological concerns about test-retest reliability. For example, owing to memory and learning effects, the average response time to a second administration of identical items takes about 80% the time compared to the first administration. Szymanski and Bissonette (2020) did not assess retest reliability. We can, though, imagine how this might work. Let’s imagine that both waves were taken in the same academic term, approximately two weeks apart. With both sets of data we need to create scores for the total scale score and the two subscales. We would also need to join the two datasets into a single dataframe. To demonstrate the retest reliability, I simulated a new dataset with total and subscale scores for our variables for Time 1 and Time 2. This next script is simply that simulation (i.e., you can skip over it). If this were your data, you would have item-level data and need to calculate total and subscale scores (as we did above). SimCor_mu <- c(3.13, 2.68, 3.58, 3.16, 2.66, 2.76) SimCor_sd <- c(0.82, 1.04, 1.26, 0.83, 1.05, 0.99) simCor <- matrix(c(1, 0.64, 0.77, 0.44, 0.33, 0.29, 0.64, 1, 0.53, 0.35, 0.46, 0.34, 0.77, 0.53, 1, 0.27, 0.4, 0.47, 0.44, 0.35, 0.27, 1, 0.63, 0.62, 0.33, 0.46, 0.4, 0.63, 1, 0.57, 0.29, 0.34, 0.47, 0.62, 0.57, 1), ncol = 6) scovMat <- SimCor_sd %*% t(SimCor_sd) * simCor set.seed(210829) retest_df <- MASS::mvrnorm(n = 646, mu = SimCor_mu, Sigma = scovMat, empirical = TRUE) colnames(retest_df) <- c("TotalT1", "ResponseT1", "StigmaT1", "TotalT2", "ResponseT2", "StigmaT2") retest_df <- as.data.frame(retest_df) #converts to a df so we can use in R library(dplyr) retest_df <- retest_df %>% dplyr::mutate(ID = row_number()) #add ID to each row retest_df <- retest_df %>% dplyr::select(ID, everything()) #moving the ID number to the first column; requires Examing our df, we can see the ID variable and the three sets of scores for each wave of analysis. Now we simply ask for their correlations. There are a number of ways to do this – the apaTables package can do the calculations and pop it into a manuscript-ready table. We won’t want the ID variable to be in the table. retest_df2 <- retest_df %>% dplyr::select(c(-ID)) apaTables::apa.cor.table(data = retest_df2, landscape = TRUE, table.number = 1, filename = "Table_1_Retest.doc") Table 1 Means, standard deviations, and correlations with confidence intervals Variable M SD 1 2 3 4 1. TotalT1 3.13 0.82 2. ResponseT1 2.68 1.04 .64** [.59, .68] 3. StigmaT1 3.58 1.26 .77** .53** [.74, .80] [.47, .58] 4. TotalT2 3.16 0.83 .44** .35** .27** [.38, .50] [.28, .42] [.20, .34] 5. ResponseT2 2.66 1.05 .33** .46** .40** .63** [.26, .40] [.40, .52] [.33, .46] [.58, .67] 6. StigmaT2 2.76 0.99 .29** .34** .47** .62** [.22, .36] [.27, .41] [.41, .53] [.57, .67] 5 .57** [.52, .62] Note. M and SD are used to represent mean and standard deviation, respectively. Values in square brackets indicate the 95% confidence interval. The confidence interval is a plausible range of population correlations that could have caused the sample correlation (Cumming, 2014). * indicates p < .05. ** indicates p < .01. As expected in this simulation, • the strongest correlations are within each scale at their respective time, that is: • the T1 variables correlate with each other; • the T2 variables correlate with each other. • the next strongest correlations are with the same scale/subscale configuration across time, for example • TotalT1 with TotalT2 (r = .44, p < 0.01) • ResponseT1 with ResponseT2 (r = .46, p < 0.01) • StigmaT1 with StigmaT2 (r = .47, p < 0.01) • the lowest correlations are different scales at T1 and T2 • ResponseT1 with StigmaT2 (r = .29) The range of retest correlations (e.g., .44 to .47 with p < 0.01) are sufficient to be confident in test-retest reliability. #### 5.4.2.2 Test-retest recap Here are some summary notions for retest reliability: • Increases in the time interval will lower the reliability coefficient. • An experimental intervention that is designed to impact the retest assessment will lower the reliability coefficient. • State measures will have lower retest coefficients than trait measures. • The three phenomena above all interact with each other. Revelle and Condon’s (2019b, 2019a) materials elaborate on this further. Their Table 1 is especially helpful. In addition to the myriad of vignettes used to illustrate issues with state, trait, items, whole scale, and so forth, there are demonstrations for duplicated items, assessing for consistency, and parallel/alternate forms. If you are asking, “Hey, is parallel/alternate forms really a variant of test retest?” Great question! In fact, split-half could be seen as test-retest! Once you get in the weeds, the distinctions become less clear. ### 5.4.3 Interrater Reliability #### 5.4.3.1 Cohen’s kappa Cohen’s kappa coefficient is used to calculate proportions of agreement corrected for chance. This type of analysis occurs in research designs where there is some kind of (usually) categorical designation of a response. I don’t have an outside research vignette for this. In the past, I was involved in research where members of the research team coded counselor utterances according to Hill’s helping skills system designed by Clara Hill . In the helping skills system, 15 different helping skills are divided into three larger groups that generally reflect the counseling trajectory: exploration, insight, action. One of our analyses coded counselor utterances into these three categories. Let’s look at a fabricated (not based on any real data) simulation where four raters each evaluated 12 counselor utterances (that represent the arch of a nonsensically speedy counseling session). Rater1 <- c("exploration", "exploration", "exploration", "exploration", "exploration", "exploration", "insight", "insight", "action", "action", "action", "action") Rater2 <- c("exploration", "exploration", "exploration", "insight", "exploration", "insight", "exploration", "exploration", "exploration", "action", "exploration", "action") Rater3 <- c("exploration", "insight", "exploration", "exploration", "exploration", "exploration", "exploration", "insight", "insight", "insight", "action", "action") Rater4 <- c("exploration", "exploration", "exploration", "exploration", "exploration", "exploration", "exploration", "exploration", "exploration", "action", "action", "action") ratings <- data.frame(Rater1, Rater2, Rater3, Rater4) Historically, kappa could only be calculated for 2 raters at a time. Presently, though, it appears there can be any number of raters and the average agreement is reported. Let’s take a look at the data, then run the analysis, and interpret the results. psych::cohen.kappa(ratings) Cohen Kappa (below the diagonal) and Weighted Kappa (above the diagonal) For confidence intervals and detail print with all=TRUE Rater1 Rater2 Rater3 Rater4 Rater1 1.00 0.40 0.21 0.62 Rater2 0.14 1.00 0.00 0.57 Rater3 0.48 0.00 1.00 0.30 Rater4 0.54 0.45 0.43 1.00 Average Cohen kappa for all raters 0.34 Average weighted kappa for all raters 0.35 Kappa can range from -1.00 to 1.00. • K = .00 indicates that the observed agreement is exactly equal to the agreement that could be observed by chance. • Negative kappa indicates that observed kappa is less than the expected chance agreement. • K = 1.00 equals perfect agreement between judges. There are commonly understood concerns about using kappa: • Research teams typically set an expected standard (e.g., .85) and train raters until kappa is achieved. • In lengthy projects, rating agreement is rechecked periodically; if necessary there is retraining. • Obtaining an acceptable kappa becomes difficult as the number of categories increases. • An example is Hill’s Helping Skills System when all 15 categories; we chose to use the three categories (into which the 15 categories are subsumed). • It is also difficult to obtain an adequate kappa when infrequent categories (e.g., “insight”) exist. Our kappa of .35 indicates that this rating team has a 35% chance of agreement, corrected for by chance. This is substantially below the standard. Let’s imagine that the team spends time with their dictionaries, examines common errors, and makes some decision rules. Here’s the resimulation of the “improved” agreement. Rater1b <- c("exploration", "exploration", "exploration", "exploration", "exploration", "exploration", "insight", "insight", "insight", "action", "action", "action") Rater2b <- c("exploration", "exploration", "exploration", "exploration", "exploration", "insight", "insight", "insight", "exploration", "action", "action", "action") Rater3b <- c("exploration", "exploration", "exploration", "exploration", "exploration", "exploration", "exploration", "insight", "insight", "insight", "action", "action") Rater4b <- c("exploration", "exploration", "exploration", "exploration", "exploration", "exploration", "exploration", "exploration", "insight", "action", "action", "action") after_training <- data.frame(Rater1b, Rater2b, Rater3b, Rater4b) Now run it again. psych::cohen.kappa(after_training) Warning in cohen.kappa1(x1, w = w, n.obs = n.obs, alpha = alpha, levels = levels): upper or lower confidence interval exceed abs(1) and set to +/- 1. Warning in cohen.kappa1(x1, w = w, n.obs = n.obs, alpha = alpha, levels = levels): upper or lower confidence interval exceed abs(1) and set to +/- 1. Warning in cohen.kappa1(x1, w = w, n.obs = n.obs, alpha = alpha, levels = levels): upper or lower confidence interval exceed abs(1) and set to +/- 1. Warning in cohen.kappa1(x1, w = w, n.obs = n.obs, alpha = alpha, levels = levels): upper or lower confidence interval exceed abs(1) and set to +/- 1. Cohen Kappa (below the diagonal) and Weighted Kappa (above the diagonal) For confidence intervals and detail print with all=TRUE Rater1b Rater2b Rater3b Rater4b Rater1b 1.00 0.83 0.55 0.80 Rater2b 0.73 1.00 0.36 0.60 Rater3b 0.72 0.45 1.00 0.46 Rater4b 0.71 0.43 0.70 1.00 Average Cohen kappa for all raters 0.62 Average weighted kappa for all raters 0.6 We observe improved scores, but this team needs more training if we aspire to a kappa of 0.85! #### 5.4.3.2 Intraclass correlation (ICC) Another option for interrater reliability is the intraclass correlation (ICC). This is the same ICC we use in multilevel modeling! The ICC is used when we have numerical ratings. In our fabricated vignette below, five raters are evaluating the campus climate for LGBTQIA+ individuals for 10 units/departments on a college campus. Using the ICC can help us determine the degree of leniency and variability within judges. Below is a simulation of the data (you can ignore this)… Rater1 <- c(1, 1, 1, 4, 2, 3, 1, 3, 3, 5) Rater2 <- c(1, 1, 2, 1, 4, 4, 4, 4, 5, 5) Rater3 <- c(3, 3, 3, 2, 3, 3, 6, 4, 4, 5) Rater4 <- c(3, 5, 4, 2, 3, 6, 6, 6, 5, 5) Rater5 <- c(2, 3, 3, 3, 4, 4, 4, 4, 5, 5) ICC_df <- data.frame(Rater1, Rater2, Rater3, Rater4, Rater5) #If the code below will not run remove the hashtags from the two lines of code below to install the Matrix package and then the lme4 package from its source #tools::package_dependencies("Matrix", which = "LinkingTo", reverse = TRUE)[[1L]] #install.packages("lme4", type = "source") We can use the psych::ICC function to obtain the ICC values. # psych::ICC(ICC_df [1:10,1:5], lmer = TRUE) #find the ICCs for the # 10 campus units and 5 judges psych::ICC(ICC_df, missing = TRUE, alpha = 0.05, lmer = TRUE, check.keys = FALSE) Call: psych::ICC(x = ICC_df, missing = TRUE, alpha = 0.05, lmer = TRUE, check.keys = FALSE) Intraclass correlation coefficients type ICC F df1 df2 p lower bound upper bound Single_raters_absolute ICC1 0.34 3.5 9 40 0.00259 0.082 0.70 Single_random_raters ICC2 0.37 5.4 9 36 0.00011 0.118 0.71 Single_fixed_raters ICC3 0.47 5.4 9 36 0.00011 0.188 0.78 Average_raters_absolute ICC1k 0.72 3.5 9 40 0.00259 0.308 0.92 Average_random_raters ICC2k 0.74 5.4 9 36 0.00011 0.400 0.92 Average_fixed_raters ICC3k 0.81 5.4 9 36 0.00011 0.537 0.95 Number of subjects = 10 Number of Judges = 5 See the help file for a discussion of the other 4 McGraw and Wong estimates, In the output, reliability for a single judge $$ICC_1$$ is the ratio of person variance to total variance. Reliability for multiple judges $$ICC_1k$$ adjusts the residual variance by the number of judges. The ICC function reports six reliability coefficients: 3 for the case of single judges and 3 for the case of multiple judges. It also reports the results in terms of a traditional ANOVA as well as a mixed effects linear model. Additionally, confidence intervals are reported. Like most correlation coefficients, the ICC ranges from 0 to 1. • An ICC close to 1 indicates high similarity between values from the same group. • An ICC close to zero means that values from the same group are not similar. ## 5.5 What do we do with these coefficients? ### 5.5.1 Corrections for attenuation Circa 1904, Spearman created the reliability coefficient out of a need to adjust observed correlations between related constructs for the error of measurement in each construct. This is only appropriate if the measure is seen as the expected value of a single underlying construct. However, “under the hood,” SEM programs model the pattern of observed correlations in terms of a measurement (reliability) model as well as a structural (validity) model. ### 5.5.2 Predicting true scores (and their CIs) True scores remain unknown and so the reliability coefficient is used in a couple of ways to estimate the true score (and the confidence interval [CI] around that true score). Take a quick look at the formula for predicting a true score and observe that the reliability coefficient is used within. It generally serves to nudge the observed score a bit closer to the mean: $$T'=(1-r_{xx})\bar{X}+r_{xx}X$$ The CI around that true score includes some estimate of standard error: $$CI_{95}=T'+/-z_{cv}(s_{e})$$. Two estimates are commonly used. One is the standard error of estimate $$s_{e}=s_{x}\sqrt{r_{xx}(1-r_{xx})}$$ (i.e., the standard deviation of predicted true scores for a given observed score). Another is the standard error of measurement ($$s_{m}=s_{x}\sqrt{(1-r_{xx})}$$ (i.e., an estimate of the amount of variation to be expected in test scores; aka, the standard deviation of the errors of measurement). I can hear you asking What is the difference between $$s_{e}$$ and $$s_{m}$$? • Because $$r_{xx}$$ is almost always a fraction, $$s_{e}$$ is smaller than $$s_{m}$$. • When the reliability is high, the two standard errors are fairly similar to each other. • Using $$s_{m}$$ will result in wider confidence intervals. ### 5.5.3 How do I keep it all straight? Table 1 in Revelle and Condon’s article helps us connect the type of reliability we are seeking with the statistic(s) and the R function within the psych package. ## 5.6 Practice Problems In each of these lessons I provide suggestions for practice that allow you to select one or more problems that are graded in difficulty. The practice problems are the start of a larger project that spans multiple lessons. Therefore, if possible, please use a dataset that has item-level data for which there is a theorized total scale score as well as two or more subscales. With each of these options I encourage you to: • Format (i.e., rescore if necessary) a dataset so that it is possible to calculates estimates of internal consistency • Calculate and report the alpha coefficient for a total scale scores and subscales (if the scale has them) • Calculate and report $$\omega_{t}$$ and $$\omega_{h}$$. With these two determine what proportion of the variance is due to all the factors, error, and g. • Calculate total and subscale scores. • Describe other reliability estimates that would be appropriate for the measure you are evaluating. ### 5.6.1 Problem #1: Play around with this simulation. If evaluating internal consistency is new to you, copy the script for the simulation and then change (at least) one thing in the simulation to see how it impacts the results. Perhaps you just change the number in “set.seed(210827)” from 210827 to something else. Your results should parallel those obtained in the lecture, making it easier for you to check your work as you go. ### 5.6.2 Problem #2: Use the data from the live ReCentering Psych Stats survey. The script below pulls live data directly from the ReCentering Psych Stats survey on Qualtrics. As described in the Scrubbing and Scoring chapters of the ReCentering Psych Stats Multivariate Modeling volume, the Perceptions of the LGBTQ College Campus Climate Scale was included (LGBTQ) and further adapted to assess perceptions of campus climate for Black students (BLst), non-Black students of color (nBSoC), international students (INTst), and students disabilities (wDIS). Consider conducting the analyses on one of these scales or merging them together. library(tidyverse) # only have to run this ONCE to draw from the same Qualtrics # account...but will need to get different token if you are changing # between accounts library(qualtRics) # qualtrics_api_credentials(api_key = # 'mUgPMySYkiWpMFkwHale1QE5HNmh5LRUaA8d9PDg', base_url = # 'spupsych.az1.qualtrics.com', overwrite = TRUE, install = TRUE) QTRX_df <- qualtRics::fetch_survey(surveyID = "SV_b2cClqAlLGQ6nLU", time_zone = NULL, verbose = FALSE, label = FALSE, convert = FALSE, force_request = TRUE, import_id = FALSE) climate_df <- QTRX_df %>% select("Blst_1", "Blst_2", "Blst_3", "Blst_4", "Blst_5", "Blst_6", "nBSoC_1", "nBSoC_2", "nBSoC_3", "nBSoC_4", "nBSoC_5", "nBSoC_6", "INTst_1", "INTst_2", "INTst_3", "INTst_4", "INTst_5", "INTst_6", "wDIS_1", "wDIS_2", "wDIS_3", "wDIS_4", "wDIS_5", "wDIS_6", "LGBTQ_1", "LGBTQ_2", "LGBTQ_3", "LGBTQ_4", "LGBTQ_5", "LGBTQ_6") # Item numbers are supported with the following items: _1 'My campus # unit provides a supportive environment for ___ students' _2 # '________ is visible in my campus unit' _3 'Negative attitudes # toward persons who are ____ are openly expressed in my campus # unit.' _4 'My campus unit is unresponsive to the needs of ____ # students.' _5 'Students who are_____ are harassed in my campus # unit.' _6 'My campus unit is cold and uncaring toward ____ # students.' # Item 1 on each subscale should be reverse coded. The College # Response scale is composed of items 1, 4, 6, The Stigma scale is # composed of items 2,3, 5 The optional script below will let you save the simulated data to your computing environment as either a .csv file (think “Excel lite”) or .rds object (preserves any formatting you might do). # write the simulated data as a .csv write.table(climate_df, # file='climate_df.csv', sep=',', col.names=TRUE, row.names=FALSE) # bring back the simulated dat from a .csv file climate_df <- # read.csv ('climate_df.csv', header = TRUE) # to save the df as an .rds (think 'R object') file on your computer; # it should save in the same file as the .rmd file you are working # with saveRDS(climate_df, 'climate_df.rds') bring back the simulated # dat from an .rds file climate_df <- readRDS('climate_df.rds') ### 5.6.3 Problem #3: Try something entirely new. Complete the same steps using data for which you have permission and access. This might be data of your own, from your lab, simulated from an article, or located on an open repository. ### 5.6.4 Grading Rubric Assignment Component Points Possible Points Earned 1. Check and, if needed, format and score data 5 _____ 2. Calculate and report the alpha coefficient for a total scale scores and subscales (if the scale has them) 5 _____ 3.Calculate and report $$\omega_{t}$$ and $$\omega_{h}$$. With these two determine what proportion of the variance is due to all the factors, error, and g. 5 _____ 4. Calculate total and subscale scores. 5 _____ 5.Describe other reliability estimates that would be appropriate for the measure you are evaluating. 5 _____ 6. Explanation to grader 5 _____ Totals 30 _____ ## 5.7 Homeworked Example Screencast Link For more information about the data used in this homeworked example, please refer to the description and codebook located at the end of the introduction in first volume of ReCentering Psych Stats. As a brief review, this data is part of an IRB-approved study, with consent to use in teaching demonstrations and to be made available to the general public via the open science framework. Hence, it is appropriate to use in this context. You will notice there are student- and teacher- IDs. These numbers are not actual student and teacher IDs, rather they were further re-identified so that they could not be connected to actual people. Because this is an actual dataset, if you wish to work the problem along with me, you will need to download the ReC.rds data file from the Worked_Examples folder in the ReC_Psychometrics project on the GitHub. The course evaluation items can be divided into three subscales: • Valued by the student includes the items: ValObjectives, IncrUnderstanding, IncrInterest • Traditional pedagogy includes the items: ClearResponsibilities, EffectiveAnswers, Feedback, ClearOrganization, ClearPresentation • Socially responsive pedagogy includes the items: InclusvClassrm, EquitableEval, MultPerspectives, DEIintegration In this homework focused on reliability we will report alpha coefficients for total scale score and subscale scores. We’ll also calculate omega total and omega hierarchical and determine what proportion of variance is due to all the factors, error, and g. Finally, we’ll calculate total and subscale scores. ### 5.7.1 Check and, if needed, format the data big <- readRDS("ReC.rds") Let’s check the structure… str(big) Classes 'data.table' and 'data.frame': 310 obs. of 33 variables: deID : int 1 2 3 4 5 6 7 8 9 10 ...
$CourseID : int 57085635 57085635 57085635 57085635 57085635 57085635 57085635 57085635 57085635 57085635 ...$ Dept : chr "CPY" "CPY" "CPY" "CPY" ...
$Course : Factor w/ 3 levels "Psychometrics",..: 2 2 2 2 2 2 2 2 2 2 ...$ StatsPkg : Factor w/ 2 levels "SPSS","R": 2 2 2 2 2 2 2 2 2 2 ...
$Centering : Factor w/ 2 levels "Pre","Re": 2 2 2 2 2 2 2 2 2 2 ...$ Year : int 2021 2021 2021 2021 2021 2021 2021 2021 2021 2021 ...
$Quarter : chr "Fall" "Fall" "Fall" "Fall" ...$ IncrInterest : int 5 3 4 2 4 3 5 3 2 5 ...
$IncrUnderstanding : int 2 3 4 3 4 4 5 2 4 5 ...$ ValObjectives : int 5 5 4 4 5 5 5 5 4 5 ...
$ApprAssignments : int 5 4 4 4 5 3 5 3 3 5 ...$ EffectiveAnswers : int 5 3 5 3 5 3 4 3 2 3 ...
$Respectful : int 5 5 4 5 5 4 5 4 5 5 ...$ ClearResponsibilities : int 5 5 4 4 5 4 5 4 4 5 ...
$Feedback : int 5 3 4 2 5 NA 5 4 4 5 ...$ OvInstructor : int 5 4 4 3 5 3 5 4 3 5 ...
$MultPerspectives : int 5 5 4 5 5 4 5 5 5 5 ...$ OvCourse : int 3 4 4 3 5 3 5 3 2 5 ...
$InclusvClassrm : int 5 5 5 5 5 4 5 5 4 5 ...$ DEIintegration : int 5 5 5 5 5 4 5 5 5 5 ...
$ClearPresentation : int 4 4 4 2 5 3 4 4 4 5 ...$ ApprWorkload : int 5 5 3 4 4 2 5 4 4 5 ...
$MyContribution : int 4 4 4 4 5 4 4 3 4 5 ...$ InspiredInterest : int 5 3 4 3 5 3 5 4 4 5 ...
$Faith : int 5 NA 4 2 NA NA 4 4 4 NA ...$ EquitableEval : int 5 5 3 5 5 3 5 5 3 5 ...
$SPFC.Decolonize.Opt.Out: chr "" "" "" "" ...$ ProgramYear : Factor w/ 3 levels "Second","Transition",..: 3 3 3 3 3 3 3 3 3 3 ...
$ClearOrganization : int 3 4 3 4 4 4 5 4 4 5 ...$ RegPrepare : int 5 4 4 4 4 3 4 4 4 5 ...
$EffectiveLearning : int 2 4 3 4 4 2 5 3 2 5 ...$ AccessibleInstructor : int 5 4 4 4 5 4 5 4 5 5 ...
- attr(*, ".internal.selfref")=<externalptr>
Let’s create a df with the items only.
library(tidyverse)
items <- big %>%
dplyr::select(ValObjectives, IncrUnderstanding, IncrInterest, ClearResponsibilities,
MultPerspectives, InclusvClassrm, DEIintegration, EquitableEval)
### 5.7.2 Calculate and report the alpha coefficient for a total scale score and subscales (if the scale has them)
psych::alpha(items)
Reliability analysis
Call: psych::alpha(x = items)
raw_alpha std.alpha G6(smc) average_r S/N ase mean sd median_r
0.92 0.92 0.93 0.49 11 0.0065 4.3 0.61 0.48
95% confidence boundaries
lower alpha upper
Feldt 0.90 0.92 0.93
Duhachek 0.91 0.92 0.93
Reliability if an item is dropped:
raw_alpha std.alpha G6(smc) average_r S/N alpha se var.r
ValObjectives 0.92 0.92 0.93 0.51 11.3 0.0067 0.016
IncrUnderstanding 0.91 0.91 0.92 0.49 10.6 0.0070 0.016
IncrInterest 0.91 0.91 0.92 0.49 10.4 0.0070 0.018
ClearResponsibilities 0.91 0.91 0.92 0.48 10.0 0.0073 0.015
EffectiveAnswers 0.91 0.91 0.92 0.48 10.0 0.0074 0.016
Feedback 0.91 0.91 0.92 0.48 10.3 0.0071 0.018
ClearOrganization 0.91 0.91 0.92 0.48 10.2 0.0073 0.016
ClearPresentation 0.91 0.91 0.92 0.47 9.7 0.0076 0.015
MultPerspectives 0.91 0.91 0.92 0.48 10.0 0.0073 0.017
InclusvClassrm 0.91 0.91 0.92 0.49 10.6 0.0069 0.018
DEIintegration 0.92 0.92 0.93 0.52 11.8 0.0063 0.011
EquitableEval 0.91 0.91 0.93 0.49 10.5 0.0070 0.018
med.r
ValObjectives 0.53
IncrUnderstanding 0.50
IncrInterest 0.48
ClearResponsibilities 0.48
Feedback 0.48
ClearOrganization 0.48
ClearPresentation 0.47
MultPerspectives 0.47
InclusvClassrm 0.52
DEIintegration 0.53
EquitableEval 0.48
Item statistics
n raw.r std.r r.cor r.drop mean sd
ValObjectives 309 0.59 0.61 0.55 0.53 4.5 0.61
IncrUnderstanding 309 0.71 0.70 0.67 0.64 4.3 0.82
IncrInterest 308 0.75 0.73 0.71 0.68 3.9 0.99
ClearResponsibilities 307 0.80 0.80 0.79 0.75 4.4 0.82
EffectiveAnswers 308 0.80 0.79 0.78 0.75 4.4 0.83
Feedback 304 0.75 0.75 0.72 0.69 4.2 0.88
ClearOrganization 309 0.79 0.77 0.75 0.72 4.0 1.08
ClearPresentation 309 0.85 0.84 0.83 0.80 4.2 0.92
MultPerspectives 305 0.79 0.80 0.78 0.75 4.4 0.84
InclusvClassrm 301 0.68 0.70 0.67 0.62 4.6 0.68
DEIintegration 273 0.51 0.53 0.49 0.42 4.5 0.74
EquitableEval 308 0.70 0.72 0.69 0.66 4.6 0.63
Non missing response frequency for each item
1 2 3 4 5 miss
ValObjectives 0.00 0.01 0.03 0.39 0.57 0.00
IncrUnderstanding 0.01 0.04 0.07 0.44 0.45 0.00
IncrInterest 0.02 0.09 0.14 0.44 0.31 0.01
ClearResponsibilities 0.01 0.02 0.07 0.31 0.59 0.01
EffectiveAnswers 0.01 0.02 0.08 0.36 0.53 0.01
Feedback 0.01 0.05 0.10 0.39 0.46 0.02
ClearOrganization 0.04 0.07 0.10 0.41 0.38 0.00
ClearPresentation 0.02 0.05 0.07 0.40 0.46 0.00
MultPerspectives 0.02 0.02 0.08 0.33 0.56 0.02
InclusvClassrm 0.01 0.01 0.05 0.23 0.70 0.03
DEIintegration 0.00 0.01 0.10 0.22 0.67 0.12
EquitableEval 0.00 0.01 0.03 0.32 0.63 0.01
Total scale score alpha is 0.92
### 5.7.3 Subscale alphas
In the lecture, I created baby dfs of the subscales and ran the alpha on those; another option is to use concatenated lists of variables (i.e., variable vectors). Later, we can also use these to score the subscales.
ValuedVars <- c("ValObjectives", "IncrUnderstanding", "IncrInterest")
"ClearOrganization", "ClearPresentation")
SRPedVars <- c("InclusvClassrm", "EquitableEval", "MultPerspectives", "DEIintegration")
psych::alpha(items[, ValuedVars])
Reliability analysis
Call: psych::alpha(x = items[, ValuedVars])
raw_alpha std.alpha G6(smc) average_r S/N ase mean sd median_r
0.77 0.77 0.71 0.53 3.4 0.02 4.2 0.68 0.48
95% confidence boundaries
lower alpha upper
Feldt 0.72 0.77 0.81
Duhachek 0.73 0.77 0.81
Reliability if an item is dropped:
raw_alpha std.alpha G6(smc) average_r S/N alpha se var.r
ValObjectives 0.80 0.81 0.68 0.68 4.3 0.022 NA
IncrUnderstanding 0.60 0.65 0.48 0.48 1.8 0.040 NA
IncrInterest 0.59 0.61 0.44 0.44 1.6 0.044 NA
med.r
ValObjectives 0.68
IncrUnderstanding 0.48
IncrInterest 0.44
Item statistics
n raw.r std.r r.cor r.drop mean sd
ValObjectives 309 0.71 0.77 0.55 0.50 4.5 0.61
IncrUnderstanding 309 0.86 0.85 0.76 0.68 4.3 0.82
IncrInterest 308 0.90 0.87 0.79 0.70 3.9 0.99
Non missing response frequency for each item
1 2 3 4 5 miss
ValObjectives 0.00 0.01 0.03 0.39 0.57 0.00
IncrUnderstanding 0.01 0.04 0.07 0.44 0.45 0.00
IncrInterest 0.02 0.09 0.14 0.44 0.31 0.01
Alpha for the Valued-by-Me dimension is .77
psych::alpha(items[, TradPedVars])
Reliability analysis
raw_alpha std.alpha G6(smc) average_r S/N ase mean sd median_r
0.89 0.9 0.88 0.64 8.8 0.0094 4.3 0.76 0.65
95% confidence boundaries
lower alpha upper
Feldt 0.87 0.89 0.91
Duhachek 0.88 0.89 0.91
Reliability if an item is dropped:
raw_alpha std.alpha G6(smc) average_r S/N alpha se var.r
ClearResponsibilities 0.86 0.86 0.84 0.62 6.4 0.013 0.0054
EffectiveAnswers 0.87 0.87 0.84 0.63 6.8 0.012 0.0045
Feedback 0.89 0.89 0.87 0.68 8.4 0.010 0.0016
ClearOrganization 0.88 0.88 0.85 0.64 7.2 0.012 0.0044
ClearPresentation 0.86 0.87 0.83 0.62 6.5 0.013 0.0030
med.r
ClearResponsibilities 0.59
Feedback 0.69
ClearOrganization 0.66
ClearPresentation 0.62
Item statistics
n raw.r std.r r.cor r.drop mean sd
ClearResponsibilities 307 0.87 0.87 0.84 0.79 4.4 0.82
EffectiveAnswers 308 0.84 0.85 0.81 0.76 4.4 0.83
Feedback 304 0.78 0.79 0.70 0.66 4.2 0.88
ClearOrganization 309 0.85 0.83 0.78 0.74 4.0 1.08
ClearPresentation 309 0.87 0.87 0.83 0.78 4.2 0.92
Non missing response frequency for each item
1 2 3 4 5 miss
ClearResponsibilities 0.01 0.02 0.07 0.31 0.59 0.01
EffectiveAnswers 0.01 0.02 0.08 0.36 0.53 0.01
Feedback 0.01 0.05 0.10 0.39 0.46 0.02
ClearOrganization 0.04 0.07 0.10 0.41 0.38 0.00
ClearPresentation 0.02 0.05 0.07 0.40 0.46 0.00
Alpha for Traditional Pedagogy dimension is .90
psych::alpha(items[, SRPedVars])
Reliability analysis
Call: psych::alpha(x = items[, SRPedVars])
raw_alpha std.alpha G6(smc) average_r S/N ase mean sd median_r
0.81 0.81 0.78 0.52 4.3 0.017 4.5 0.58 0.54
95% confidence boundaries
lower alpha upper
Feldt 0.77 0.81 0.84
Duhachek 0.77 0.81 0.84
Reliability if an item is dropped:
raw_alpha std.alpha G6(smc) average_r S/N alpha se var.r
InclusvClassrm 0.74 0.74 0.67 0.49 2.9 0.025 0.0120
EquitableEval 0.78 0.79 0.73 0.56 3.9 0.021 0.0034
MultPerspectives 0.73 0.74 0.67 0.49 2.8 0.026 0.0153
DEIintegration 0.78 0.78 0.71 0.54 3.6 0.021 0.0044
med.r
InclusvClassrm 0.50
EquitableEval 0.57
MultPerspectives 0.47
DEIintegration 0.57
Item statistics
n raw.r std.r r.cor r.drop mean sd
InclusvClassrm 301 0.82 0.83 0.76 0.67 4.6 0.68
EquitableEval 308 0.75 0.76 0.64 0.58 4.6 0.63
MultPerspectives 305 0.85 0.83 0.76 0.68 4.4 0.84
DEIintegration 273 0.78 0.78 0.67 0.59 4.5 0.74
Non missing response frequency for each item
1 2 3 4 5 miss
InclusvClassrm 0.01 0.01 0.05 0.23 0.70 0.03
EquitableEval 0.00 0.01 0.03 0.32 0.63 0.01
MultPerspectives 0.02 0.02 0.08 0.33 0.56 0.02
DEIintegration 0.00 0.01 0.10 0.22 0.67 0.12
Alpha for the SCR Pedagogy dimension is .81
### 5.7.4 Calculate and report ωt and ωh
psych::omegaSem(items, nfactors = 3)
Warning in lav_model_vcov(lavmodel = lavmodel, lavsamplestats = lavsamplestats, : lavaan WARNING:
Could not compute standard errors! The information matrix could
not be inverted. This may be a symptom that the model is not
identified.
Call: psych::omegaSem(m = items, nfactors = 3)
Omega
Call: omegah(m = m, nfactors = nfactors, fm = fm, key = key, flip = flip,
digits = digits, title = title, sl = sl, labels = labels,
plot = plot, n.obs = n.obs, rotate = rotate, Phi = Phi, option = option,
covar = covar)
Alpha: 0.92
G.6: 0.93
Omega Hierarchical: 0.83
Omega H asymptotic: 0.88
Omega Total 0.94
g F1* F2* F3* h2 u2 p2
ValObjectives 0.47 0.32 0.33 0.67 0.67
IncrUnderstanding 0.57 0.60 0.69 0.31 0.47
IncrInterest 0.58 0.57 0.67 0.33 0.49
ClearResponsibilities 0.87 0.78 0.22 0.98
Feedback 0.73 0.56 0.44 0.96
ClearOrganization 0.75 0.23 0.62 0.38 0.90
ClearPresentation 0.81 0.30 0.74 0.26 0.88
MultPerspectives 0.75 0.30 0.65 0.35 0.86
InclusvClassrm 0.56 0.48 0.57 0.43 0.55
DEIintegration 0.37 0.82 0.80 0.20 0.17
EquitableEval 0.70 0.52 0.48 0.94
With Sums of squares of:
g F1* F2* F3*
5.51 0.04 0.98 1.06
general/max 5.19 max/min = 26.42
mean percent general = 0.74 with sd = 0.26 and cv of 0.36
Explained Common Variance of the general factor = 0.73
The degrees of freedom are 33 and the fit is 0.25
The number of observations was 310 with Chi Square = 76.65 with prob < 0.000025
The root mean square of the residuals is 0.02
The df corrected root mean square of the residuals is 0.03
RMSEA index = 0.065 and the 10 % confidence intervals are 0.046 0.085
BIC = -112.66
Compare this with the adequacy of just a general factor and no group factors
The degrees of freedom for just the general factor are 54 and the fit is 1.37
The number of observations was 310 with Chi Square = 415 with prob < 0.0000000000000000000000000000000000000000000000000000000038
The root mean square of the residuals is 0.1
The df corrected root mean square of the residuals is 0.11
RMSEA index = 0.147 and the 10 % confidence intervals are 0.134 0.16
BIC = 105.22
g F1* F2* F3*
Correlation of scores with factors 0.95 0.13 0.81 0.89
Multiple R square of scores with factors 0.91 0.02 0.66 0.79
Minimum correlation of factor score estimates 0.82 -0.96 0.32 0.58
Total, General and Subset omega for each subset
g F1* F2* F3*
Omega total for total scores and subscales 0.94 0.77 0.90 0.87
Omega general for total scores and subscales 0.83 0.76 0.69 0.64
Omega group for total scores and subscales 0.11 0.01 0.20 0.23
The following analyses were done using the lavaan package
Omega Hierarchical from a confirmatory model using sem = 0.82
Omega Total from a confirmatory model using sem = 0.94
g F1* F2* F3* h2 u2 p2
ValObjectives 0.47 0.32 0.33 0.67 0.67
IncrUnderstanding 0.56 0.62 0.70 0.30 0.45
IncrInterest 0.57 0.56 0.64 0.36 0.51
ClearResponsibilities 0.86 0.32 0.84 0.16 0.88
Feedback 0.73 0.55 0.45 0.97
ClearOrganization 0.75 0.22 0.61 0.39 0.92
ClearPresentation 0.82 0.27 0.74 0.26 0.91
MultPerspectives 0.74 0.30 0.64 0.36 0.86
InclusvClassrm 0.56 0.49 0.56 0.44 0.56
DEIintegration 0.33 0.87 0.87 0.13 0.13
EquitableEval 0.69 0.51 0.49 0.93
g F1* F2* F3*
5.46 0.10 0.94 1.14
The degrees of freedom of the confirmatory model are 42 and the fit is 110.6184 with p = 0.00000004343729
general/max 4.8 max/min = 10.87
mean percent general = 0.73 with sd = 0.27 and cv of 0.37
Explained Common Variance of the general factor = 0.71
g F1* F2* F3*
Correlation of scores with factors 0.95 0.57 0.82 0.95
Multiple R square of scores with factors 0.91 0.32 0.67 0.90
Minimum correlation of factor score estimates 0.82 -0.36 0.35 0.80
Total, General and Subset omega for each subset
g F1* F2* F3*
Omega total for total scores and subscales 0.94 0.84 0.89 0.87
Omega general for total scores and subscales 0.82 0.74 0.70 0.62
Omega group for total scores and subscales 0.11 0.10 0.20 0.26
To get the standard sem fit statistics, ask for summary on the fitted object
I’m reporting the values below the statement, “The following analyses were done using the lavaan package”:
Omega total = .94 (omega total values > .80 are an indicator of good reliability). Interpretation: 94% of the variance in the total scale is due to the factors and the balance (6%) is due to error.
Omega hierarchical estimates the proportion of variance in the overall course evaluation score attributable to the general factors (thus treating the subscales as error). Omega h for the overall course evaluation score was .82
### 5.7.5 With these two determine what proportion of the variance is due to all the factors, error, and g.
.82/.94
[1] 0.8723404
A quick calculation with omega h (.82) and omega total (.94) lets us know that 87% of the reliable variance in the overall course evaluation score is attributable to the general factor.
### 5.7.6 Calculate total and subscale scores.
This code uses the variable vectors I created above.
items$Valued <- sjstats::mean_n(items[, ValuedVars], 0.75) items$TradPed <- sjstats::mean_n(items[, TradPedVars], 0.75)
items$SCRPed <- sjstats::mean_n(items[, SRPedVars], 0.75) items$Total <- sjstats::mean_n(items, 0.75)
scores <- items %>%
psych::describe(scores)
vars n mean sd median trimmed mad min max range skew kurtosis
Valued 1 309 4.25 0.68 4.33 4.32 0.50 1.67 5 3.33 -0.90 0.57
TradPed 2 307 4.25 0.76 4.40 4.37 0.59 1.00 5 4.00 -1.42 2.48
SCRPed 3 299 4.52 0.58 4.75 4.61 0.37 2.25 5 2.75 -1.25 1.33
Total 4 308 4.34 0.60 4.41 4.41 0.62 1.83 5 3.17 -1.07 1.12
se
Valued 0.04
SCRPed 0.03
Total 0.03
### 5.7.7 Describe other reliability estimates that would be appropriate for the measure you are evaluating.
These scales are for the purposes of course evaluations. In their development, it might be helpful to give it at the end of a single course and then again a few weeks later to determine test-retest reliability.
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Pascarella, E. T., & Terenzini, P. T. (1980). Predicting Freshman Persistence and Voluntary Dropout Decisions from a Theoretical Model. The Journal of Higher Education, 51(1), 60–75. https://doi.org/10.1080/00221546.1980.11780030
Pedhazur, E. J., & Schmelkin, L. P. (1991). Measurement, design, and analysis: An integrated approach. Lawrence Erlbaum Associates.
Revelle, W. (n.d.). An introduction to the psych package: Part II Scale construction and psychometrics. 97.
Revelle, W., & Condon, D. M. (2019a). Reliability from α to ω: A tutorial. Psychological Assessment. https://doi.org/10.1037/pas0000754
Revelle, W., & Condon, D. M. (2019b). Reliability from α to ω: A tutorial. Online supplement. Psychological Assessment. https://doi.org/10.1037/pas0000754
Revelle, William. (n.d.). The Personality Project: An introduction to psychometric theory. In Personality Project. Retrieved September 2, 2019, from http://www.personality-project.org/r/book/#chapter7
Schmid, J., & Leiman, J. M. (1957). The development of hierarchical factor solutions. Psychometrika, 22, 53–61. https://doi.org/10.1007/BF02289209
Spitzer, R. L., Kroenke, K., Williams, J. B. W., & Löwe, B. (2006). A Brief Measure for Assessing Generalized Anxiety Disorder: The GAD-7. Archives of Internal Medicine, 166(10), 1092. https://doi.org/10.1001/archinte.166.10.1092
Szymanski, D. M., & Bissonette, D. (2020). Perceptions of the LGBTQ College Campus Climate Scale: Development and psychometric evaluation. Journal of Homosexuality, 67(10), 1412–1428. https://doi.org/10.1080/00918369.2019.1591788 | 25,040 | 81,783 | {"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": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.928325 |
https://www.storyofmathematics.com/find-the-area-of-the-region-enclosed-by-the-inner-loop-of-the-curve/ | 1,709,438,062,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476180.67/warc/CC-MAIN-20240303011622-20240303041622-00704.warc.gz | 987,230,831 | 41,536 | Find the area of the region enclosed by the inner loop of the curve:
$r = 1 + 2sin \theta$
This problem aims to find the area of the region enclosed by a limacon curve whose equation is $r = 1 + 2sin\theta$, where $r$ is the radius of the curve. This problem requires knowledge of coordinate systems, the formation of a limacon curve, and the formula to find the area of the inner and the outer loop of a limacon curve.
A coordinate system is utilized to determine the area of a point in space. Most of the time, we use the rectangular or Cartesian coordinate system in our mathematical problems. A rectangular grid system is used to determine the location of a point in space. We can also determine the location of that exact point by describing its location and distance from a fixed point as a reference.
A limacon is an anallagmatic curve that looks like a circle but instead has a small indent on one side of it. Equations of the form $r = a + bsin\theta$, $r = a – bsin\theta$, $r = a + bcos\theta$, and $r = a – bcos\theta$ will produce limacons.
If the value of $a$ is slightly less than the value of $b$, then the graph would form a limacon with an internal loop as seen in the figure below.
So as the first step, we are going to find the interval on which the internal loop exits.
Given the equation $r = 1 + 2sin\theta$, we will be taking $r=0$
$1 + 2sin\theta = 0$
$sin \theta = \dfrac{-1}{2}$
$\theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$
We can find the area under the inner loop of the limacon curve by accomplishing a definite integral between the two solid points. To locate the area under the curve $r$ between $x = \theta_1$ & $x = \theta_2$, we will integrate $r$ between the limits of $\theta_1$ & $\theta_2$.
Modifying the integral as per the required variables:
$Area = \int_{\theta 1}^ {\theta2} \dfrac{1}{2}r^ 2 d\theta$
Putting the values in the Formula:
$Area = \int_{\dfrac{7\pi}{6}}^ {\dfrac{11\pi}{6}} \dfrac{1}{2}(1+2sin\theta)^ 2 d\theta$
$= \int_{\dfrac{7\pi}{6}}^ {\dfrac{11\pi}{6}} \dfrac{1}{2}(1+2sin\theta)^ 2 d\theta$
$= \int_{\dfrac{7\pi}{6}}^ {\dfrac{11\pi}{6}} \dfrac{1}{2}+2sin\theta + 2sin^ 2\theta d\theta$
$= \int_{\dfrac{7\pi}{6}}^ {\dfrac{11\pi}{6}} \dfrac{3}{2}+2sin\theta – cos2\theta d\theta$
$= \left[ \dfrac{3\theta}{2}-2cos\theta – \dfrac{1}{2} sin2\theta \right]_{\dfrac{7\pi}{6}}^ {\dfrac{11\pi}{6}}$
$= \dfrac{11\pi}{4} – 2 \times \dfrac{\sqrt{3}}{2} – \dfrac{1}{2} \left( – \dfrac{\sqrt{3}}{2}\right) – \left(\dfrac{-7\pi}{4} -2\left(-\dfrac{\sqrt{3}}{2} \right) – \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}\right)$
$= \dfrac{11\pi}{4} – \dfrac{7\pi}{4} -\sqrt{3} + \dfrac{\sqrt{3}}{4} -\sqrt{3} + \dfrac{\sqrt{3}}{4}$
Numerical Result
$Area = \pi – \dfrac{3\sqrt{3}}{2}$
Example
Find the area of the region enclosed by the inner loop of the polar curve:
$r = 2+4cos\theta$
$cos \theta = \dfrac{-1}{2}$
$\theta = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$
Putting the values in the Formula:
$Area = \int_{\dfrac{2\pi}{3}}^{\dfrac{4\pi}{3}} \dfrac{1}{2}(2+4cos\theta)^2 d\theta$
By solving the integrals, the area under the curve comes out to be:
$A = 2(2\pi – 4\sqrt{3} + \sqrt{3})$
$A = 4\pi – 6\sqrt{3}$
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Mathematical Psychology Assignment Help
Introduction
The modal analysis of "quantitative psychologist" still tends to indicate somebody with prolonged training in stats (at the least, and often advanced degrees in mathematical stats). It is essential in modern-day psychology to consist of the field of mathematical psychology when facing problems in quantitative training. Mathematical psychology plays an important function in clinical psychology, as we will show listed below, numerous psychologists are less familiar with the field because training programs, and even courses in that topic, are sadly rather uncommon.
Mathematical Psychology Assignment Help
It was not up until the 1950's that mathematical psychology ended up being a genuine field of research study in its own. Because its beginning, mathematical modeling has actually perhaps gotten in, or affected many of the locations of research study in psychology. Mathematical psychology and standard quantitative psychology vary in their focus on stats (rather more in the standard sector) and focus on a substantive theoretical design of the mental procedures or phenomena (rather more in mathematical psychology). They share numerous tools, mathematical contents, and viewpoint. We cannot state the number of conventional methodologists discover mathematical modeling, however essentially all mathematical psychologists get training in mathematical and mental data.
As metrology of habits is essential in this venture, the theory of measurement is a main subject in mathematical psychology. Like computational neuroscience and econometrics, mathematical psychology theory frequently utilizes analytical optimality as an assisting concept, presuming that the human brain has actually developed to fix issues in an enhanced method. Central styles from cognitive psychology; unrestricted vs. minimal processing serial, capability vs. parallel processing, and so on, and their ramifications, are main in strenuous analysis in mathematical psychology.
Mathematical psychology does not consist of all usages of mathematics and stats in psychology, and undoubtedly there is a long history of such usages particularly in the locations of understanding and psychometrics. Exactly what is most special about mathematical psychology is its method to theory building. While accepting the behaviorist dictum that the information in psychology should be replicable and observable, mathematical designs are defined in regards to unobservable official constructs that can forecast in-depth elements of information throughout several speculative and natural settings.
At its core, mathematical psychology is the research study of habits through the lens of mathematics. Mathematical psychologists utilize solutions, data, and other mathematical tools to anticipate how human beings will act in provided circumstances and on particular jobs. Carefully connected to psychometrics, which utilizes mathematics to anticipate the habits of populations in basic, mathematical psychology looks for to anticipate habits on a specific level. Another function of mathematical psychology is that it tries to bridge the space in between measurable solutions for anticipating qualitative human habits. To do so, mathematical psychologists look for to establish hypotheses that are more accurate and concentrated. Doing so assists them yield empirical recognitions that are likewise much more concentrated, exact, and reliable.
The main function of mathematical psychologists is to establish and check theories of human habits that are measurable. These calculable theories check out a range of locations associated with human habits, from understanding and cognition to issue resolving and memory to neuroscience and the procedures by which people discover. They take the kind of computational designs, simulations, or mathematical formulas due to the fact that these theories need to be quantifiable. A mathematical psychologist interested in studying human memory may design an experiment in which topics are asked to take part in a memory job, such as seeing a selection of cards which are then turned over, and then being charged to match sets of cards based on their recall of the cards' positioning. A mathematical psychologist would gather information on the topic's efficiency, such as the variety of attempts it took them to properly keep in mind the place of each card.
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Mathematical psychology plays an essential function in clinical psychology, as we will show listed below, numerous psychologists are less familiar with the field considering that training programs, and even courses in that topic, are regrettably rather unusual. Mathematical psychology and standard quantitative psychology vary in their focus on data (rather more in the conventional sector) and focus on a substantive theoretical design of the mental procedures or phenomena (rather more in mathematical psychology). We cannot state how lots of standard methodologists discover mathematical modeling, however essentially all mathematical psychologists get training in mathematical and mental data. Mathematical psychology does not consist of all usages of mathematics and stats in psychology, and certainly there is a long history of such usages specifically in the locations of understanding and psychometrics. Mathematical psychologists utilize solutions, data, and other mathematical tools to anticipate how human beings will act in offered circumstances and on particular jobs.
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Assignment Help Financial Management
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Currently the spot exchange rate is \$1.533 per pound (USD/GBP). The interest rate in the UK is 6%. The one-year forward exchange rate is \$1.5200/GBP. If interest rate parity holds, what must be the US interest rate for the same period?
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During the four quarters for 2015, the Haydens received two quarterly dividend payments per share of \$0.15 each, one quarterly payment of \$0.26, and one quarterly payment of \$0.21. If they owned 200 shares of stock, what was their total dividend inco.. | 1,384 | 6,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.920378 |
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The truth is, theres about 60mg of caffeine in a tablespoon—a level tablespoon, not an approximate tablespoon that sits an inch above the lip of the tablespoon—of Arabica coffee grounds, and about 110mg in a tablespoon of Robusta coffee grounds
How much caffeine does 2 tablespoons of ground coffee have?
Caffeine Content by Single Origin
if you use 10 grams (10,000 mg) of coffee (generally, 2 tbsp), with 1.20% caffeine by weight, you’ll get roughly (10,000 mg * 0.012 =) 120 mg of caffeine. Note: caffeine is not necessarily 100% extracted via all brewing methods.How Much Caffeine in a Cup of Coffee?espressocoffeeguide.com › how-much-caffeine-in-a-cup-of-coffeeAbout Featured Snippets
How much caffeine is in a teaspoon of coffee grounds?
The caffeine content of Arabica beans is around 1,2% of its mass. A standard measure of a teaspoon is approximately 5ml. That means you get around 5g of ground coffee in one teaspoon, which totals around 0,06g or 60mg of caffeineHow much caffeine is in 1 teaspoon of ground Arabica coffee?coffee.stackexchange.com › questions › how-much-caffeine-is-in-1-teaspo…About Featured Snippets
How much ground coffee is a tablespoon?
In general, a tablespoon of coffee equals approximately 10.6 grams. So, for a cup of coffee, you will want to use 1 ½ to 2 tablespoons of coffee grinds. We’re not talking about instant coffee crystals, but grounds from actual coffee beans. If you are using scoops, you’ll want the scoop to be equal to 2 tablespoons.How Much Coffee per Cup? This is How You Get it Rightchamberlaincoffee.com › blogs › inspiration › how-much-coffee-per-cup-t…About Featured Snippets
How much caffeine is in 1g ground coffee?
The Caffeine in Coffee Beans
In a single coffee bean, there are about 6 milligrams of caffeine; however, the exact measurement depends on the variety used in the coffee. For example, Arabica coffee has about 12 milligrams of coffee per gram, while Robusta coffee has about 22 milligrams per gramHow Much Caffeine Is in a Coffee Bean? – Recycle A Cupwww.recycleacup.com › caffeine-in-coffee-beansAbout Featured Snippets
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How Much Caffeine is in a Tablespoon of Ground Coffee?
On average, there is about 60mg and 110mg of caffeine in a level tablespoon of ground Arabica coffee and Robusta coffee respectively. A heaped tablespoon …
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It uses an infinite stream of pseudorandom bits as the key.
The Data Encryption Standard (DES) is a symmetric-key block cipher published by the National Institute of Standards and Technology (NIST). Cryptography is the science of ciphering and deciphering messages. This article is about the Counter Mode.In this article, we will briefly study the basic CTR and its examples aim to capture. The Counter Mode or CTR is a simple counter based block cipher implementation in cryptography. DES is an implementation of a Feistel Cipher.
Deciphering is reversing a cipher into the original text. XXTEA – Corrected Block TEA's successor. For example, here's the Caesar Cipher encryption of a full message, using a left shift of 3. Note: Please use this button to report only Software related issues.For queries regarding questions and quizzes, use the comment area below respective pages. Submitted by Monika Sharma, on February 21, 2020 . If a truly random key stream is used, the result will be a truly 'random' ciphertext which bears no relation to the original plaintext. ^ Matthew D. Russell (27 Feb 2004). XTEA – First version of Block TEA's successor. Lecture 3: Stream Ciphers, Random Numbers and the One Time Pad by Christof Paar - Duration: 1:29:39. The Vernam Cipher is based on the principle that each plaintext character from a message is 'mixed' with one character from a key stream . A cipher is a message that has been transformed into a nonhuman readable format. RC4 – A stream cipher that, just like TEA, is designed to be very simple to implement. It uses 16 round Feistel structure. A stream cipher is an encryption algorithm that encrypts 1 bit or byte of plaintext at a time. Vernam Cipher Problem string key); std::string Encryptor::encrypt(std::string plaintext, std Both encrypt and decrypt require two. Cryptanalysis is the art of deciphering ciphers without the knowledge of the key used to cipher them. Introduction to Cryptography by Christof Paar 124,631 views 1:29:39 Treyfer – A simple and compact encryption algorithm with 64-bit key size and block size. Other encryption algorithms solve Vernam cipher is the perfect cipher if the key is truly random, For example, given document. | 550 | 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} | 2.578125 | 3 | CC-MAIN-2020-45 | longest | en | 0.868637 |
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Unformatted text preview: 0 680 1 1 1 0.8 4.5 4.5 -1 1 -1 1 -1 4 300 1 5.7 1 The strawberry example with length of activity determined in model. le 700 le 1200 le 0 le 0 le 0 Equilibrium Model Extended Equilibrium An equilibrium model can include a sequence of activities. Sequencing problems are discussed in section 7.5 of the text. We did not explicitly cover them in class before. Sequencing (7.4) Sequencing Some activities must be completed before others are begun Sequencing may be controlled within a variable or between variables using constraints The technique used depends on the situation Example from Chapter 7 Example A firm produces one output using two tasks (events). There is a 3week production horizon, and the two events can take place any time during that period, but the second event must take place after the first event – although the two events can take place in the same week First Case: Returns Independent of Timing Returns X1, X2, X3: amount of activity 1 done in weeks 1, 2, 3, respectively Y1, Y2, Y3: amount of activity 2 done in weeks 1, 2, 3, respectively. For week 1: Y1 LE X1 For week 2: Y2 LE X1 + X2 – Y1 For week 3: Y3 LE X1+X2+X3(Y1+Y2) Sequencing handled by constraints Tableau: Case 1 Tableau: X1 OBJ week1 trans week2 trans week3 trans week1 res week2 res week3 res -1 -1 -1 a X2 -1 -1 X3 -1 b Y1 P 1 1 1 d Y2 P 1 1 Y3 P 1 e c f le le le le le le 0 0 0 T1 T2 T3 Second Case: Second Returns Depend on Timing If the returns depend on timing, we must define a variable for each combination of timing possible. Example: A1B2 is the variable meaning that activity A (first activity) is done in week 1and activity B (second activity) is done in week 2. Tableau Case 2 Tableau OBJ wk1 wk2 wk3 A1B1 A1B2 A1B3 A2B2 A2B3 A3B3 P11 P12 P13 P22 P23 P33 a b d c f g e h i le le le T1 T2 T3 Chapter 8 : Chapter Crop Ro...
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http://stackoverflow.com/questions/1313/followup-finding-an-accurate-distance-between-colors/74033 | 1,416,818,271,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400380464.40/warc/CC-MAIN-20141119123300-00137-ip-10-235-23-156.ec2.internal.warc.gz | 321,582,222 | 23,127 | # Followup: Finding an accurate “distance” between colors
Original Question
I am looking for a function that attempts to quantify how "distant" (or distinct) two colors are. This question is really in two parts:
1. What color space best represents human vision?
2. What distance metric in that space best represents human vision (euclidean?)
-
Convert to La*b* (aka just plain "Lab", and you'll also see reference to "CIELAB"). A good quick measaure of color difference is
(L1-L2)^2 + (a1-a2)^2 + (b1-b2)^2
Color scientists have other more refined measures, which may not be worth the bother, depending on accuracy needed for what you're doing.
The `a` and `b` values represent opposing colors in a way similar to how cones work, and may be negative or positive. Neutral colors - white, grays are `a=0`,`b=0`. The `L` is brightness defined in a particular way, from zero (pure darkness) up to whatever.
Crude explanation :>> Given a color, our eyes distinguish between two broad ranges of wavelength - blue vs longer wavelengths. and then, thanks to a more recent genetic mutation, the longer wavelength cones bifurcated into two, distinguishing for us red vs. green.
By the way, it'll be great for your career to rise above your color caveman collegues who know of only "RGB" or "CMYK" which are great for devices but suck for serious perception work. I've worked for imaging scientists who didn't know a thing about this stuff!
For more fun reading on color difference theory, try:
More detail on Lab at http://en.kioskea.net/video/cie-lab.php3 I can't at this time find a non-ugly page that actually had the conversion formulas but I'm sure someone will edit this answer to include one.
-
Regarding the ugliness of the conversion formulas: they're ugly for a reason, since getting from RGB to XYZ to LAB depends on viewing conditions. cf: (warning, ugliness) easyrgb.com/index.php?X=MATH – Gregg Lind Oct 8 '08 at 17:46
Possibly it'd be worth adding a description of standard color distance metrics and pseudometrics defined by CIE: en.wikipedia.org/wiki/Color_difference – BartoszKP Sep 26 '13 at 10:47
as cmetric.htm link above failed for me, as well as many other implementations for color distance I found (after a very long jurney..) how to calculate the best color distance, and .. most scientifically accurate one: deltaE and from 2 RGB (!) values using OpenCV:
This required 3 color space conversions + some code conversion from javascript (http://svn.int64.org/viewvc/int64/colors/colors.js) to C++
And finally the code (seems to work right out of the box, hope no one finds a serious bug there ... but it seems fine after a number of tests)
``````#include <opencv2/core/core.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/photo/photo.hpp>
#include <math.h>
using namespace cv;
using namespace std;
#define REF_X 95.047; // Observer= 2°, Illuminant= D65
#define REF_Y 100.000;
#define REF_Z 108.883;
void bgr2xyz( const Vec3b& BGR, Vec3d& XYZ );
void xyz2lab( const Vec3d& XYZ, Vec3d& Lab );
void lab2lch( const Vec3d& Lab, Vec3d& LCH );
double deltaE2000( const Vec3b& bgr1, const Vec3b& bgr2 );
double deltaE2000( const Vec3d& lch1, const Vec3d& lch2 );
void bgr2xyz( const Vec3b& BGR, Vec3d& XYZ )
{
double r = (double)BGR[2] / 255.0;
double g = (double)BGR[1] / 255.0;
double b = (double)BGR[0] / 255.0;
if( r > 0.04045 )
r = pow( ( r + 0.055 ) / 1.055, 2.4 );
else
r = r / 12.92;
if( g > 0.04045 )
g = pow( ( g + 0.055 ) / 1.055, 2.4 );
else
g = g / 12.92;
if( b > 0.04045 )
b = pow( ( b + 0.055 ) / 1.055, 2.4 );
else
b = b / 12.92;
r *= 100.0;
g *= 100.0;
b *= 100.0;
XYZ[0] = r * 0.4124 + g * 0.3576 + b * 0.1805;
XYZ[1] = r * 0.2126 + g * 0.7152 + b * 0.0722;
XYZ[2] = r * 0.0193 + g * 0.1192 + b * 0.9505;
}
void xyz2lab( const Vec3d& XYZ, Vec3d& Lab )
{
double x = XYZ[0] / REF_X;
double y = XYZ[1] / REF_X;
double z = XYZ[2] / REF_X;
if( x > 0.008856 )
x = pow( x , .3333333333 );
else
x = ( 7.787 * x ) + ( 16.0 / 116.0 );
if( y > 0.008856 )
y = pow( y , .3333333333 );
else
y = ( 7.787 * y ) + ( 16.0 / 116.0 );
if( z > 0.008856 )
z = pow( z , .3333333333 );
else
z = ( 7.787 * z ) + ( 16.0 / 116.0 );
Lab[0] = ( 116.0 * y ) - 16.0;
Lab[1] = 500.0 * ( x - y );
Lab[2] = 200.0 * ( y - z );
}
void lab2lch( const Vec3d& Lab, Vec3d& LCH )
{
LCH[0] = Lab[0];
LCH[1] = sqrt( ( Lab[1] * Lab[1] ) + ( Lab[2] * Lab[2] ) );
LCH[2] = atan2( Lab[2], Lab[1] );
}
double deltaE2000( const Vec3b& bgr1, const Vec3b& bgr2 )
{
Vec3d xyz1, xyz2, lab1, lab2, lch1, lch2;
bgr2xyz( bgr1, xyz1 );
bgr2xyz( bgr2, xyz2 );
xyz2lab( xyz1, lab1 );
xyz2lab( xyz2, lab2 );
lab2lch( lab1, lch1 );
lab2lch( lab2, lch2 );
return deltaE2000( lch1, lch2 );
}
double deltaE2000( const Vec3d& lch1, const Vec3d& lch2 )
{
double avg_L = ( lch1[0] + lch2[0] ) * 0.5;
double delta_L = lch2[0] - lch1[0];
double avg_C = ( lch1[1] + lch2[1] ) * 0.5;
double delta_C = lch1[1] - lch2[1];
double avg_H = ( lch1[2] + lch2[2] ) * 0.5;
if( fabs( lch1[2] - lch2[2] ) > CV_PI )
avg_H += CV_PI;
double delta_H = lch2[2] - lch1[2];
if( fabs( delta_H ) > CV_PI )
{
if( lch2[2] <= lch1[2] )
delta_H += CV_PI * 2.0;
else
delta_H -= CV_PI * 2.0;
}
delta_H = sqrt( lch1[1] * lch2[1] ) * sin( delta_H ) * 2.0;
double T = 1.0 -
0.17 * cos( avg_H - CV_PI / 6.0 ) +
0.24 * cos( avg_H * 2.0 ) +
0.32 * cos( avg_H * 3.0 + CV_PI / 30.0 ) -
0.20 * cos( avg_H * 4.0 - CV_PI * 7.0 / 20.0 );
double SL = avg_L - 50.0;
SL *= SL;
SL = SL * 0.015 / sqrt( SL + 20.0 ) + 1.0;
double SC = avg_C * 0.045 + 1.0;
double SH = avg_C * T * 0.015 + 1.0;
double delta_Theta = avg_H / 25.0 - CV_PI * 11.0 / 180.0;
delta_Theta = exp( delta_Theta * -delta_Theta ) * ( CV_PI / 6.0 );
double RT = pow( avg_C, 7.0 );
RT = sqrt( RT / ( RT + 6103515625.0 ) ) * sin( delta_Theta ) * -2.0; // 6103515625 = 25^7
delta_L /= SL;
delta_C /= SC;
delta_H /= SH;
return sqrt( delta_L * delta_L + delta_C * delta_C + delta_H * delta_H + RT * delta_C * delta_H );
}
``````
Hope it helps someone :)
-
HSL and HSV are better for human color perception. According to Wikipedia:
It is sometimes preferable in working with art materials, digitized images, or other media, to use the HSV or HSL color model over alternative models such as RGB or CMYK, because of differences in the ways the models emulate how humans perceive color. RGB and CMYK are additive and subtractive models, respectively, modelling the way that primary color lights or pigments (respectively) combine to form new colors when mixed.
-
Watch out: red is at 0°, so yellowish red is at +10° and blueish red is at -10° or 350°. Calculating the distance isn't just as easy as subtracting the two values now. – Otto Allmendinger Oct 25 '12 at 10:34
Not really helpful in the context of the other answer. Metrics defined by CIE are better for human color perception not HLS and HSV. – BartoszKP Sep 26 '13 at 10:49
May look like spam but no, this link is really interesting for color spaces :)
http://www.compuphase.com/cmetric.htm
-
Just looks like your average 90s website, which, conincidentally, is also a popular website look for 2010+ Academicians. – Domi Jun 13 at 6:31
The easiest distance would of course be to just consider the colors as 3d vectors originating from the same origin, and taking the distance between their end points.
If you need to consider such factors that green is more prominent in judging intensity, you can weigh the values.
ImageMagic provides the following scales:
• red: 0.3
• green: 0.6
• blue: 0.1
Of course, values like this would only be meaningful in relation to other values for other colors, not as something that would be meaningful to humans, so all you could use the values for would be similiarity ordering.
-
Well, as a first point of call, I'd say of the common metrics HSV (Hue, Saturation and Value) or HSL are better representative of how humans perceive colour than say RGB or CYMK. See HSL, HSV on Wikipedia.
I suppose naively I would plot the points in the HSL space for the two colours and calculate the magnitude of the difference vector. However this would mean that bright yellow and bright green would be considered just as different as green to dark green. But then many consider red and pink two different colours.
Moreover, difference vectors in the same direction in this parameter space are not equal. For instance, the human eye picks up green much better than other colours. A shift in hue from green by the same amount as a shift from red may seem greater. Also a shift in saturation from a small amount to zero is the difference between grey and pink, elsewhere the shift would be the difference between two shades of red.
From a programmers point of view, you would need to plot the difference vectors but modified by a proportionality matrix that would adjust the lengths accordingly in various regions of the HSL space - this would be fairly arbitrary and would be based on various colour theory ideas but be tweaked fairly arbitrarily depending on what you wanted to apply this to.
Even better, you could see if anyone has already done such a thing online...
-
The Wikipedia article on color differences lists a number of color spaces and distance metrics designed to agree with human perception of color distances.
-
As someone who is color blind I believe it is good to try to add more separation then normal vision. The most common form of color blindness is red/green deficiency. It doesn't mean that you can't see red or green, it means that it is more difficult to see and more difficult to see the differences. So it takes a larger separation before a color blind person can tell the difference.
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# The study, called the National Lung Screening Trial, focused
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The study, called the National Lung Screening Trial, focused [#permalink]
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04 May 2012, 14:29
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sravanth wrote:
The study, called the National Lung Screening Trial, focused on a specific high‐risk group: 53,000 current and former heavy smokers, aging from 55 to 74, which had smoked for at least 30 pack‐years
A. group: 53,000 current and former heavy smokers, aging from 55 to 74, which had smoked for at least 30 pack‐years
B. group: 53,000 current and former heavy smokers, aged 55 to 74, that had smoked for at least 30 pack‐years
C. group: 53,000 current and former heavy smokers, aged 55 to 74, who smoked for at least 30 pack‐years
D. group: 53,000 current and former heavy smokers, aged 55 to 74, who had smoked for at least 30 pack‐years
E. group: 53,000 current and former heavy smokers, who were aged 55 to 74, and who had smoked for higher than at least 30 pack‐years
Hi, there. I'm happy to help.
First of all, here's a blog article you may find helpful:
http://magoosh.com/gmat/2012/that-vs-which-on-the-gmat/
As for this question, important GMAT idea #1 is --- don't use that/which for human beings; use "who" for human beings. That eliminates (A) and (B) right away.
Answer (E) is wordy and redundant ---- we'll tube that one also.
That leaves (C) & (D), which are remarkably similar. In fact the only difference is the verb tense:
(C) ...smoked...
(D) ... had smoked...
The first is simple past tense, and the second is past perfect. The past perfect is used to indicate that this verb's action took place before some other event in the past. Here, the main verb of the sentence "focused" is in the past. The question is --- did the smoking and the focusing happen at the same time, or was the smoking clearly before the focusing? Well, by the time the study was created and they focused on folks, those folks already had been smoking for quite some time. The smoking clearly has to happen before the focusing. This necessitates the past perfect structure. That's why (D) is correct and (C) is not.
Does this make sense? Let me know if you have any more questions.
Mike
[Reveal] Spoiler: OA
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Re: Which That [#permalink]
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05 May 2012, 23:24
mikemcgarry wrote:
sravanth wrote:
The study, called the National Lung Screening Trial, focused on a specific high‐risk group: 53,000 current and former heavy smokers, aging from 55 to 74, which had smoked for at least 30 pack‐years
A. group: 53,000 current and former heavy smokers, aging from 55 to 74, which had smoked for at least 30 pack‐years
B. group: 53,000 current and former heavy smokers, aged 55 to 74, that had smoked for at least 30 pack‐years
C. group: 53,000 current and former heavy smokers, aged 55 to 74, who smoked for at least 30 pack‐years
D. group: 53,000 current and former heavy smokers, aged 55 to 74, who had smoked for at least 30 pack‐years
E. group: 53,000 current and former heavy smokers, who were aged 55 to 74, and who had smoked for higher than at least 30 pack‐years
Hi, there. I'm happy to help.
First of all, here's a blog article you may find helpful:
http://magoosh.com/gmat/2012/that-vs-which-on-the-gmat/
As for this question, important GMAT idea #1 is --- don't use that/which for human beings; use "who" for human beings. That eliminates (A) and (B) right away.
Answer (E) is wordy and redundant ---- we'll tube that one also.
That leaves (C) & (D), which are remarkably similar. In fact the only difference is the verb tense:
(C) ...smoked...
(D) ... had smoked...
The first is simple past tense, and the second is past perfect. The past perfect is used to indicate that this verb's action took place before some other event in the past. Here, the main verb of the sentence "focused" is in the past. The question is --- did the smoking and the focusing happen at the same time, or was the smoking clearly before the focusing? Well, by the time the study was created and they focused on folks, those folks already had been smoking for quite some time. The smoking clearly has to happen before the focusing. This necessitates the past perfect structure. That's why (D) is correct and (C) is not.
Does this make sense? Let me know if you have any more questions.
Mike
Who should be placed close to smokers right?
it is not followign the touch rule!!
please clear my doubt
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Re: Which That [#permalink]
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07 May 2012, 11:23
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dvinoth86 wrote:
Who should be placed close to smokers right? It is not following the touch rule!! Please clear my doubt
Dear dvinoth86
Here's what I am going to say. I don't know the source of this question. Not all GMAT prep questions are created equal. The SC questions that appears on the GMAT itself are gems, absolutely superb in their clarity and polish, and some test prep sources produce SC questions of comparably high quality. Other GMAT prep sources, to be honest, produce junk SC.
Let's say, I have my suspicions about this question. This is not a question that would appear on the real GMAT. Here's the sentence with the least offensive answer, (D):
The study, called the National Lung Screening Trial, focused on a specific high‐risk group: 53,000 current and former heavy smokers, aged 55 to 74, who had smoked for at least 30 pack‐years.
Yes, you are correct --- the modifier touch rule requires the modifier ("who had smoked ...") to touch the noun it modifies ("smokers"). This sentence does not do that, and so is less than ideal. Is it out-and-out incorrect? That's a matter of debate.
A vital modifier can intervene between a modifier and its target. See this post for more on that idea:
http://magoosh.com/gmat/2012/gmat-gramm ... modifiers/
Here, the intervening phrase "aged 55 to 74" is clearly not vital. It is set off by commas, which is the hallmark of a non-vital modifier. No clear rule justifies its position between the noun and the modifier.
At the same time, it's very short, and putting it there involves absolutely no ambiguity --- that's not a resounding vote of support, but some folks would argue on that basis that it's correct. Admittedly, there is not universal consensus on the grammar of this particular point.
One further piece of support is that --- any attempt to reword the sentence to eliminate this problem makes the sentence longer and more awkward. For example:
The study, called the National Lung Screening Trial, focused on a specific high‐risk group: 53,000 current and former heavy smokers, who were aged 55 to 74 and had smoked for at least 30 pack‐years.
Hardly a model sentence. There is no good way to rework (D) as it stands, which is another sorta argument in its favor.
Mathematics this ain't!!! I'm sorry I can't give you a more definitive answer. On the real GMAT, correct answers will be 100% correct (though not necessary ideal), and incorrect answers will be wrong. On this question, the best answer is in a gray zone ---- not something you will encounter on the real GMAT.
Moral: don't accord the same degree of trust in each and every practice question you see. Some practice questions sources are of much higher quality, and others don't hold up the standard. If you want some high quality questions, follow the link in the signature of the post.
Let me know if you have any further questions.
Mike
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Re: Which That [#permalink]
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07 May 2012, 13:13
sravanth wrote:
The study, called the National Lung Screening Trial, focused on a specific high‐risk group: 53,000 current and former heavy smokers, aging from 55 to 74, which had smoked for at least 30 pack‐years
A. group: 53,000 current and former heavy smokers, aging from 55 to 74, which had smoked for at least 30 pack‐years
B. group: 53,000 current and former heavy smokers, aged 55 to 74, that had smoked for at least 30 pack‐years
C. group: 53,000 current and former heavy smokers, aged 55 to 74, who smoked for at least 30 pack‐years
D. group: 53,000 current and former heavy smokers, aged 55 to 74, who had smoked for at least 30 pack‐years
E. group: 53,000 current and former heavy smokers, who were aged 55 to 74,and who had smoked for higher than at least 30 pack‐years
hello
here goes my analysis
this question is about testing pronoums
and mainly which that who
the original choice is wrong , as smokers are people hence which can not be used
A is wrong
since we have both former and current smoker the choice of the simple past is impossible
simple past means that an action is over , completed and finished . we can eliminate c
hence we are left with BDE
E is wrong because the use of "were aged "means that the smoker age a long time ago :they were aged 55 to 74
this change the meaning
the higher at least is confusing so
the 2 left choice are B and D
B use that
D use who use to represent a subject
I will go For D
HOPE this help
best regards
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Re: Which That [#permalink]
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13 May 2012, 06:14
Thanks a lot Mike..I'm a fan of your blogs..learnt a lot from them
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Re: Which That [#permalink]
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14 May 2012, 11:06
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dvinoth86 wrote:
Thanks a lot Mike..I'm a fan of your blogs..learnt a lot from them
Why, thank you. That means a lot to me. Thank you very much.
Mike
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Re: The study, called the National Lung Screening Trial, focused [#permalink]
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Re: The study, called the National Lung Screening Trial, focused [#permalink]
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14 Mar 2016, 23:49
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Re: The study, called the National Lung Screening Trial, focused [#permalink]
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15 Mar 2016, 00:05
mikemcgarry wrote:
sravanth wrote:
The study, called the National Lung Screening Trial, focused on a specific high‐risk group: 53,000 current and former heavy smokers, aging from 55 to 74, which had smoked for at least 30 pack‐years
A. group: 53,000 current and former heavy smokers, aging from 55 to 74, which had smoked for at least 30 pack‐years
B. group: 53,000 current and former heavy smokers, aged 55 to 74, that had smoked for at least 30 pack‐years
C. group: 53,000 current and former heavy smokers, aged 55 to 74, who smoked for at least 30 pack‐years
D. group: 53,000 current and former heavy smokers, aged 55 to 74, who had smoked for at least 30 pack‐years
E. group: 53,000 current and former heavy smokers, who were aged 55 to 74, and who had smoked for higher than at least 30 pack‐years
Hi, there. I'm happy to help.
First of all, here's a blog article you may find helpful:
http://magoosh.com/gmat/2012/that-vs-which-on-the-gmat/
As for this question, important GMAT idea #1 is --- don't use that/which for human beings; use "who" for human beings. That eliminates (A) and (B) right away.
Answer (E) is wordy and redundant ---- we'll tube that one also.
That leaves (C) & (D), which are remarkably similar. In fact the only difference is the verb tense:
(C) ...smoked...
(D) ... had smoked...
The first is simple past tense, and the second is past perfect. The past perfect is used to indicate that this verb's action took place before some other event in the past. Here, the main verb of the sentence "focused" is in the past. The question is --- did the smoking and the focusing happen at the same time, or was the smoking clearly before the focusing? Well, by the time the study was created and they focused on folks, those folks already had been smoking for quite some time. The smoking clearly has to happen before the focusing. This necessitates the past perfect structure. That's why (D) is correct and (C) is not.
Does this make sense? Let me know if you have any more questions.
Mike
The focus is on two grammar points.
1) relative pronoun
2) verb tense
The correct relative pronoun for smokers is 'who'
The correct verb tense is past perfect to indicate an event which happened before the survey and for a duration/period.
Therefore, the correct option is D
Option E is correct in using relative pronoun and verb tense. However, for giving additional information, you don't need to introduce a relative pronoun.
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Re: The study, called the National Lung Screening Trial, focused [#permalink]
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11 Apr 2016, 06:14
hi mike mcgarry..
i was confused btw option c & d. as it is obvious about time period. we are talking about people who smoked for at least 30 years. there is no confusion in sequence of events if we use simple past tense. plz clarify.
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Re: The study, called the National Lung Screening Trial, focused [#permalink]
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11 Apr 2016, 06:23
mikemcgarry wrote:
sravanth wrote:
The study, called the National Lung Screening Trial, focused on a specific high‐risk group: 53,000 current and former heavy smokers, aging from 55 to 74, which had smoked for at least 30 pack‐years
A. group: 53,000 current and former heavy smokers, aging from 55 to 74, which had smoked for at least 30 pack‐years
B. group: 53,000 current and former heavy smokers, aged 55 to 74, that had smoked for at least 30 pack‐years
C. group: 53,000 current and former heavy smokers, aged 55 to 74, who smoked for at least 30 pack‐years
D. group: 53,000 current and former heavy smokers, aged 55 to 74, who had smoked for at least 30 pack‐years
E. group: 53,000 current and former heavy smokers, who were aged 55 to 74, and who had smoked for higher than at least 30 pack‐years
Hi, there. I'm happy to help.
First of all, here's a blog article you may find helpful:
http://magoosh.com/gmat/2012/that-vs-which-on-the-gmat/
As for this question, important GMAT idea #1 is --- don't use that/which for human beings; use "who" for human beings. That eliminates (A) and (B) right away.
Answer (E) is wordy and redundant ---- we'll tube that one also.
That leaves (C) & (D), which are remarkably similar. In fact the only difference is the verb tense:
(C) ...smoked...
(D) ... had smoked...
The first is simple past tense, and the second is past perfect. The past perfect is used to indicate that this verb's action took place before some other event in the past. Here, the main verb of the sentence "focused" is in the past. The question is --- did the smoking and the focusing happen at the same time, or was the smoking clearly before the focusing? Well, by the time the study was created and they focused on folks, those folks already had been smoking for quite some time. The smoking clearly has to happen before the focusing. This necessitates the past perfect structure. That's why (D) is correct and (C) is not.
Does this make sense? Let me know if you have any more questions.
Mike
Dear mike
D. group: 53,000 current and former heavy smokers, aged 55 to 74, who had smoked for at least 30 pack‐years
to use past perfect tense we need two related events, one is earlier event in the past & the other later event
Please explain how to determine later event in option D - whether it is "focused" or "aged"
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The study, called the National Lung Screening Trial, focused [#permalink]
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11 Apr 2016, 10:04
ashish8814 wrote:
hi mike mcgarry..
i was confused btw option c & d. as it is obvious about time period. we are talking about people who smoked for at least 30 years. there is no confusion in sequence of events if we use simple past tense. plz clarify.
smartguy595 wrote:
Dear mike
D. group: 53,000 current and former heavy smokers, aged 55 to 74, who had smoked for at least 30 pack‐years
to use past perfect tense we need two related events, one is earlier event in the past & the other later event
Please explain how to determine later event in option D - whether it is "focused" or "aged"[/quote]
Dear ashish8814 & smartguy595,
I'm happy to respond.
The most important thing to remember is that this is NOT a high quality practice question. You both are asking for clarification as if understanding this question more deeply will help you. This is fundamentally a misapprehension. You do not necessarily enhance your understanding by wrestling with low quality SC questions.
In the gigantic underlined section, the only difference between (C) & (D) is the verb tense. The GMAT might test verb tenses if all that is underlined is the verb itself, but if the GMAT underlines a long section such as this, it is unlikely to have two choices that differ only in verb tense. That's too picayune a split.
Incidentally, smartguy595, in this sentence, "focused" is a full verb, a verb in the past tense. The word "aged" is purely a participle, a noun-modifier, not a full verb. It's very important to recognize the difference between participles and full verbs.
The truth is that this is not a particularly clean split. I think (D) sounds more natural, but one could certainly make a valid argument for (C). This is one of the ways that this question is not high quality. On a high quality question, one choice is correct and the other four are clearly wrong. On this question, (D) is correct and (C) is arguably correct also. Don't stress over this, thinking one must be absolutely right and one must be absolutely wrong. The question simply is not that good.
Here's a high quality SC question:
Balancing the need for
Does all this make sense?
Mike
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Re: The study, called the National Lung Screening Trial, focused [#permalink]
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11 Apr 2016, 18:58
mikemcgarry wrote:
ashish8814 wrote:
hi mike mcgarry..
i was confused btw option c & d. as it is obvious about time period. we are talking about people who smoked for at least 30 years. there is no confusion in sequence of events if we use simple past tense. plz clarify.
smartguy595 wrote:
Dear mike
D. group: 53,000 current and former heavy smokers, aged 55 to 74, who had smoked for at least 30 pack‐years
to use past perfect tense we need two related events, one is earlier event in the past & the other later event
Please explain how to determine later event in option D - whether it is "focused" or "aged"
Dear ashish8814 & smartguy595,
I'm happy to respond.
The most important thing to remember is that this is NOT a high quality practice question. You both are asking for clarification as if understanding this question more deeply will help you. This is fundamentally a misapprehension. You do not necessarily enhance your understanding by wrestling with low quality SC questions.
In the gigantic underlined section, the only difference between (C) & (D) is the verb tense. The GMAT might test verb tenses if all that is underlined is the verb itself, but if the GMAT underlines a long section such as this, it is unlikely to have two choices that differ only in verb tense. That's too picayune a split.
Incidentally, smartguy595, in this sentence, "focused" is a full verb, a verb in the past tense. The word "aged" is purely a participle, a noun-modifier, not a full verb. It's very important to recognize the difference between participles and full verbs.
The truth is that this is not a particularly clean split. I think (D) sounds more natural, but one could certainly make a valid argument for (C). This is one of the ways that this question is not high quality. On a high quality question, one choice is correct and the other four are clearly wrong. On this question, (D) is correct and (C) is arguably correct also. Don't stress over this, thinking one must be absolutely right and one must be absolutely wrong. The question simply is not that good.
Here's a high quality SC question:
Balancing the need for
Does all this make sense?
Mike [/quote]
Thank you mike
will note above points marked by you..thank you very much
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I need to work on timing badly!!
Re: The study, called the National Lung Screening Trial, focused [#permalink] 11 Apr 2016, 18:58
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Antarctica receives more solar radiation than does any other place on Earth, yet the temperatures are so cold and the ice cap is reflective, so that little polar ice melts during the summer; otherwise, the water levels of the oceans would rise 250 feet and engulf most of the world’s great cities.
A. is reflective, so that little polar ice melts during the summer; otherwise,
B. is so reflective that little of the polar ice melts during the summer; were it to do so,
C. so reflective that little polar ice melts during the summer, or else
D. reflective, so that little of the polar ice melts during the summer, or
E. reflects so that little of the polar ice melts during the summer; if it did
[Reveal] Spoiler: OA
Last edited by amitjash on 02 Aug 2010, 23:35, edited 1 time in total.
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Re: Isnt the OA more wordy?? Can some one explain?? [#permalink]
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02 Aug 2010, 08:28
amitjash wrote:
Antarctica receives more solar radiation than does any other place on Earth, yet the temperatures are so cold and the ice cap is reflective, so that little polar ice melts during the summer; otherwise, the water levels of the oceans would rise 250 feet and engulf most of the world’s great cities.
A. is reflective, so that little polar ice melts during the summer; otherwise,
B. is so reflective that little of the polar ice melts during the summer; were it to do so,
C. so reflective that little polar ice melts during the summer, or else
D. reflective, so that little of the polar ice melts during the summer, or
E. reflects so that little of the polar ice melts during the summer; if it did
please underline the part to be corrected.
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Re: Isnt the OA more wordy?? Can some one explain?? [#permalink]
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02 Aug 2010, 10:27
B... pretty clear..
the semicolon correctly gives 2 independent clauses
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Re: Isnt the OA more wordy?? Can some one explain?? [#permalink]
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03 Aug 2010, 02:55
Antarctica receives more solar radiation than does any other place on Earth, yet the temperatures are so cold and the ice cap is reflective, so that little polar ice melts during the summer; otherwise, the water levels of the oceans would rise 250 feet and engulf most of the world’s great cities.
A. is reflective, so that little polar ice melts during the summer; otherwise,
B. is so reflective that little of the polar ice melts during the summer; were it to do so,
C. so reflective that little polar ice melts during the summer, or else
D. reflective, so that little of the polar ice melts during the summer, or
E. reflects so that little of the polar ice melts during the summer; if it did
When you scan the option, you can be pretty sure that C, D and E are wrong, because the sentence itself needs a helper verb "is" (otherwise the first part just becomes a run on sentence). Now with remaining choices A & B, in choice A, IDIOM - SO....THAT is not properly used and also the sentence after the semicolon, OTHERWISE doesn't justify the previous subject.
Hence answer choice B. Hope this helps.
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Re: Isnt the OA more wordy?? Can some one explain?? [#permalink]
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10 Aug 2010, 12:44
Can someone elaborate on the phrase "were it to do so", the meaning and usage of it b/c it sounds real strange.
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Re: Isnt the OA more wordy?? Can some one explain?? [#permalink]
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10 Aug 2010, 12:59
It might looks wordier but here main role has the PARALLELISM
are so cold ------ is so reflective
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Re: Isnt the OA more wordy?? Can some one explain?? [#permalink]
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11 Aug 2010, 05:09
We can't simply remove an answer choice if it is wordy but grammatically correct and other choices grammatically wrong but short. These are called stylistic errors.
So, prefer correct grammatically form even though answer choice is wordy.
Here B is correct answer though it is wordy!
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## Let g1andg2 be the values of acceleration due to gravity in Mumabi and Pune and T1andT2 be the values of the time periods at the respective places Then , T1=8min9s100=489100s=4.89s T2=8min20s100=500s100=5s As T2T1=√g1g2 ∴g1g2=T22T21=(5)2(4.89)2≈1.04
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# Second stage of elliptic curve factorization via random walk/Pollard's rho inconstant(orlow)memory?
The second stage of elliptic curve factorization has the drawback of large memory usage.
Let $n=pq$, $E(\mathbb{Z}/n\mathbb{Z})$ is elliptic curve and $P$ point on $E(\mathbb{Z}/n\mathbb{Z})$.
On $E(\mathbb{Z}/p\mathbb{Z})$ the order of $P$ is small $r$.
Set $Q=kP$ for pseudo-random $k$. On $E(\mathbb{Z}/p\mathbb{Z})$ one can solve the discrete logarithm $Q=xP$ in time $O(\sqrt{r})$ and constant memory using Pollard's rho algorithm (more precisely one can find $aP=bQ$ if $r$ is unknown)
basically by doing random walks and exploiting the birthday paradox.
The question is:
Working on $E(\mathbb{Z}/n\mathbb{Z})$ ($p,q$ are unknown) can one solve $Q=xP$ (or $aP=bQ$) on $E(\mathbb{Z}/p\mathbb{Z})$ using the rho algorithm in $O(\sqrt{r})$ and constant memory: note that $r$ can be significantly smaller than the order of $P$ on $E(\mathbb{Z}/n\mathbb{Z})$.
The only significant choice appears to be the random walk.
I failed to do this yet discrete logarithms on $E(\mathbb{Z}/n\mathbb{Z})$ appear to work as expected.
Update: I suppose part of the problem with constant memory is that it may happen $Q_i \ne Q_j \mod n$ while $Q_i = Q_j \mod p$
If $E(\mathbb{Z}/n\mathbb{Z})$ were a ring, one could simply iterate
(A) $Q_{i+1}=Q_i^2+c$.
Would it be possible instead on an elliptic curve to work in some ring where (A) would be trivially stage 2 and stage 1 would be $Q_1 = k P \ k \in \mathbb{N}$?
3 constant memory, ring is simpler
The second stage of elliptic curve factorization has the drawback of large memory usage.
Let $n=pq$, $E(\mathbb{Z}/n\mathbb{Z})$ is elliptic curve and $P$ point on $E(\mathbb{Z}/n\mathbb{Z})$.
On $E(\mathbb{Z}/p\mathbb{Z})$ the order of $P$ is small $r$.
Set $Q=kP$ for pseudo-random $k$. On $E(\mathbb{Z}/p\mathbb{Z})$ one can solve the discrete logarithm $Q=xP$ in time $O(\sqrt{r})$ and constant memory using Pollard's rho algorithm (more precisely one can find $aP=bQ$ if $r$ is unknown)
basically by doing random walks and exploiting the birthday paradox.
The question is:
Working on $E(\mathbb{Z}/n\mathbb{Z})$ ($p,q$ are unknown) can one solve $Q=xP$ (or $aP=bQ$) on $E(\mathbb{Z}/p\mathbb{Z})$ using the rho algorithm in $O(\sqrt{r})$ and constant memory: note that $r$ can be significantly smaller than the order of $P$ on $E(\mathbb{Z}/n\mathbb{Z})$.
The only significant choice appears to be the random walk.
I failed to do this yet discrete logarithms on $E(\mathbb{Z}/n\mathbb{Z})$ appear to work as expected.
Update: I suppose part of the problem with constant memory is that it may happen $Q_i \ne Q_j \mod n$ while $Q_i = Q_j \mod p$
If $E(\mathbb{Z}/n\mathbb{Z})$ were a ring, one could simply iterate $Q_{i+1}=Q_i^2+c$.
2 tag random-walk
The second stage of elliptic curve factorization has the drawback of large memory usage.
Let $n=pq$, $E(\mathbb{Z}/n\mathbb{Z})$ is elliptic curve and $P$ point on $E(\mathbb{Z}/n\mathbb{Z})$.
On $E(\mathbb{Z}/p\mathbb{Z})$ the order of $P$ is small $r$.
Set $Q=kP$ for pseudo-random $k$. On $E(\mathbb{Z}/p\mathbb{Z})$ one can solve the discrete logarithm $Q=xP$ in time $O(\sqrt{r})$ and constant memory using Pollard's rho algorithm (more precisely on one can find $aP=bQ$ if $r$ is unknown)
basically by doing random walks and exploiting the birthday paradox.
The question is:
Working on $E(\mathbb{Z}/n\mathbb{Z})$ ($p,q$ are unknown) can one solve $Q=xP$ (or $aP=bQ$) on $E(\mathbb{Z}/p\mathbb{Z})$ using the rho algorithm in $O(\sqrt{r})$ and constant memory: note that $r$ can be significantly smaller than the order of $P$ on $E(\mathbb{Z}/n\mathbb{Z})$.
The only significant choice appears to be the random walk.
I failed to do this yet discrete logarithms on $E(\mathbb{Z}/n\mathbb{Z})$ appear to work as expected.
1 | 1,282 | 3,874 | {"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} | 2.953125 | 3 | CC-MAIN-2013-20 | latest | en | 0.815352 |
https://mathscitutor.com/expressions-maths/matrices/factor-problems.html | 1,575,948,582,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525781.64/warc/CC-MAIN-20191210013645-20191210041645-00527.warc.gz | 463,075,039 | 13,253 | Home Systems of Equations Adding and Subtracting Rational Expressions with Different Denominators Graphing Linear Equations Raising an Exponential Expression to a Power Horizontal Line Test Quadratic Equations Mixed Numbers and Improper Fractions Solving Quadratic Equations by Completing the Square Solving Exponential Equations Adding and Subtracting Polynomials Factorizing simple expressions Identifying Prime and Composite Numbers Solving Linear Systems of Equations by Graphing Complex Conjugates Graphing Compound Inequalities Simplified Form of a Square Root Solving Quadratic Equations Using the Square Root Property Multiplication Property of Radicals Determining if a Function has an Inverse Scientific Notation Degree of a Polynomial Factoring Polynomials by Grouping Solving Linear Systems of Equations Exponential Functions Factoring Trinomials by Grouping The Slope of a Line Simplifying Complex Fractions That Contain Addition or Subtraction Solving Absolute Value Equations Solving Right Triangles Solving Rational Inequalities with a Sign Graph Domain and Range of a Function Multiplying Polynomials Slope of a Line Inequalities Multiplying Rational Expressions Percent of Change Equations Involving Fractions or Decimals Simplifying Expressions Containing only Monomials Solving Inequalities Quadratic Equations with Imaginary Solutions Reducing Fractions to Lowest Terms Prime and Composite Numbers Dividing with Exponents Dividing Rational Expressions Equivalent Fractions Graphing Quadratic Functions Polynomials Linear Equations and Inequalities in One Variable Notes on the Difference of 2 Squares Solving Absolute Value Inequalities Solving Quadratic Equations Factoring Polynomials Completely Using Slopes to Graph Lines Fractions, Decimals and Percents Solving Systems of Equations by Substitution Quotient Rule for Radicals Prime Polynomials Solving Nonlinear Equations by Substitution Simplifying Radical Expressions Containing One Term Factoring a Sum or Difference of Two Cubes Finding the Least Common Denominator of Rational Expressions Conjugates Multiplying Rational Expressions Expansion of a Product of Binomials Solving Equations Exponential Growth Factoring by Grouping Solving One-Step Equations Using Models Solving Quadratic Equations by Factoring Adding and Subtracting Polynomials Rationalizing the Denominator Rounding Off The Distributive Property What is a Quadratic Equation Laws of Exponents and Multiplying Monomials The Slope of a Line Factoring Trinomials by Grouping Multiplying and Dividing Rational Expressions Solving Linear Inequalities Multiplication Property of Exponents Multiplying and Dividing Fractions 3 Formulas Dividing Monomials Multiplying Polynomials Adding and Subtracting Functions Dividing Polynomials Absolute Value and Distance Multiplication and Division with Mixed Numbers Factoring a Polynomial by Finding the GCF Roots Adding and Subtracting Polynomials The Rectangular Coordinate System Polar Form of a Complex Number Exponents and Order of Operations Graphing Horizontal and Vertical Lines Invariants Under Rotation The Addition Method Solving Linear Inequalities in One Variable The Pythagorean Theorem
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Registered: 19.10.2002 | 1,427 | 6,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-51 | latest | en | 0.809635 |
https://www.boatbuilding.xyz/downwind-upwind/centers-of-mass-buoyancy-and-effort.html | 1,579,799,799,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00558.warc.gz | 788,492,485 | 4,742 | ## Centers of Mass Buoyancy and Effort
Gravity pulls on every part of the boat. Wind pushes on the entire sail. Water's buoyant force pushes up on all submerged surfaces. These distributed forces are equivalent to forces acting at a single point. Knowing the positions of these special points simplifies the calculations of the torques.
### 4.3.1 Center of Mass
The most familiar "center" is the center of mass. Simple experiments (in principle) can determine the center of mass of any object, including a sailboat. Suspend a sailboat from a single point. A line drawn directly down from the suspension point passes through the center of mass. It doesn't matter if you lift the boat by the bow, the stern, or the mast. All lines descending from the suspension points pass through the same center of mass. Sailboat symmetry means the center of mass lies in the plane that separates port from starboard.
Although gravity pulls down on every piece of a sailboat, all the forces act as if they were pulling on the center of mass with a force Mg, where M is the total mass and g = 9.8 m/s2 is the acceleration of gravity.
### 4.3.2 Center of Buoyancy
Just as gravity pulls down, buoyancy keeps a boat from sinking by pushing up with a force that is equal but opposite to gravity. Buoyancy pushes up on all submerged parts of a boat, but these forces act as if they were pushing at a single point. That point is the center of buoyancy.
When a boat is placed in water, it sinks down and moves water away. The mass of this displaced water equals the boat's mass. That is Archimedes principle. The center of buoyancy is the center of mass of this displaced water. Although the boat's center of mass is fixed, the center of buoyancy moves as a boat tips. A sailboat is stable when tipping causes the center of buoyancy to move in a direction that stops the tipping.
### 4.3.3 Center of Effort
Wind pushes on a sail. The total wind force can be represented as if it acted on a single point. That point is the center of effort. A simple example is a mainsail shaped like a right triangle. If the pressure is uniform over a sail's surface, the center of effort is one-third of the way up from the boom to the top of the sail (the "head"), and one-third of the way back from the mast to the end of the boom. If the pressure is nonuniform, the total force can still be represented by a center of effort, but its position will be different.
There is also a center of effort for the water's horizontal force on a centerboard (or keel). This center of effort is centrally located on the centerboard. | 585 | 2,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-05 | latest | en | 0.944763 |
http://www.slidesearchengine.com/slide/classen-electron-mass-ratio-lab-report | 1,493,502,006,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123590.89/warc/CC-MAIN-20170423031203-00010-ip-10-145-167-34.ec2.internal.warc.gz | 709,339,248 | 9,084 | # (Classen) Electron/ mass ratio lab report
67 %
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Information about (Classen) Electron/ mass ratio lab report
Published on January 7, 2009
Author: pixiejen
Source: slideshare.net
## Description
Part of (discluding Data sheets) of First of 12 labs done for Adv. Lab.
Classen e/m Experiment Report Jennifer S. Nalley Lab Partner: Chris G. Cumby February 6, 2007 I
Abstract The Classen e/m Experiment performed was intended to demonstrate how the charge to mass ratio of an electron can be determined by allowing a current run through an apparatus known as the Helmholtz Coil. It is known that a current running through a solenoid creates a magnetic field. The Helmholtz Coil setup is roughly based on this principle. Although the propagated error calculated from our data differed from the standard deviation by a factor of 10, the results obtained were considerably reasonable given the large magnitudes (small numbers) that were recorded. Introduction and Theory The experiment, takes advantage of the fact that the electron displays a unique behavior when introduced into a magnetic field. The primary apparatus used in this experiment, the Helmholtz Coil, provides a duo of coils that are wire-wrapped. A Helmholtz Coil, provided with a current source, in turn yield a magnetic field, which exists horizontally through the coils. Equipped with the Helmholtz setup, and provided with specific mathematical formulas: • Magnetic field B produced at the center of the Helmholtz Coil 8µ o NI B=magnetic field B= N= turns of coil 125 R R=radius of coil I= current in coils µo= permeability of free space= 4лE(-7) weber/amp- m II
*1 weber/meter^2= 1 Tesla • Formula derived to give the relationship between the charge and mass of an electron V= electric potential in Volts B= the magnetic field from experiment given formula e 2V = r= radius of anode circle around ( Bexp r ) 2 the electron beam m e/m =electron charge over electron mass in kg We were able to measure and manipulate the fundamental components which are needed to measure the charge to mass ratio of an electron, and compare it to the accepted value, that is = coloumbs 1.7588 × 1011 kg Apparatus and Procedure The primary piece of equipment used in this lab was a Helmholtz Coil with a vacuum tube at center. The ideal Helmholtz coil set-up is often known for, if not partially defined by its geometric proportions. It is the Helmholtz’s dimensions that allow for the specific equation used for magnetic field to work without a complex manipulation of any proportionality constant. The unit has two identical coaxial circular “coils”. The distances between the two coils are (ideally) proportional to the radius of either coil. Each coil is wrapped with N turns of wire. III
B In our experiment, the number of wire turns N = 133. We measured both the distance between the coils, and the radius of the coils in two ways: from the outer to outer most components, and inner to inner. It was decided to use an average of these measurements, as they were the most consistent with the ideal Helmholtz coil proportions. This choice was determined by our possibly poor assumption that the manufacturer would have the ideal proportions in mind when it was designing the apparatus. The setup we used included: • CENCO apparatus- a three element electron tube with adjustable current and plate voltage. Cathode- serving as an electron source (electron gun). i. (-) Grid- charged to a positive potential to accept and ii. focus the electrons from the cathode into a beam. (+) Anode disk - charged to a higher positive potential to iii. actively attract electrons. • Helmholtz coil(s) IV
At the midpoint between the coils of the unit (1/2 the radius) was the anode disk. Mounted upon the disk, rested a vacuum tube with a filament. In order for there to be visible “action” in a vacuum tube, electrons must have something to interact with. The vacuum tube contained enough inert gas, so that the electron beam that was to be produced would illuminate. From below the tube, the electron exit was secured in such a fashion that electron rays emitted would eject vertically, perpendicular to the magnetic field, which would be initiated via power supply to the cathode. The anode at the center of the setup was designed to be disk shaped so to surround the point of electron emission. Like a dart board, the disk was scored with four concentric circles whose individual radius would become our electron beam target. In our case, the electron beam darts would come out of the bull’s-eye, and be made to bend into a semicircle via magnetic field then hit the disk from which it originated. V
The radii of the circular targets were given as follows: 2.0 cm, 1.5cm, 1.0 cm, and 0.5 cm. Because the earth has its own magnetic field, it was taken into account both numerically and directionally. We used a compass to position the apparatus so that the magnetic field running horizontally through the two coils would be aligned with the direction of the earth’s magnetic field. The right hand rule proved helpful in the determination of the Helmholtz field direction. The northern component of the earth’s magnetic field at our geographical location was found to = 24,708.3 nano-Tesla. *(NOAA) VI
With the lights off, the current and voltage were adjusted in a variety of ways to vary the magnetic field, which in turn would control the curvature of the illuminated electron beam. In order to preserve the filament, precaution was taken to never exceed a current of 0.7 Amps. The goal was to aim the beam at each of the given radii one at a time. The levels for voltage and current were recorded at each radius respectively for the grid, filament, plate, and field. By changing the direction of the current, we were able to replicate the process for readings with our magnetic field going against that of the earth’s field. Each variation was repeated several times, and is recorded in the data. With the data gathered and equations provided, we attempted to calculate the charge to mass ratio of an electron, and compare it to the current accepted number. Data and Analysis The accepted value for an electrons charge/mass ratio is = coloumbs 1.7588 × 1011 kg Our average e/m ratio(s) for readings taken: Lined with earths magnetic field = 2.24065 × 1011 c/kg (± 0.4 × 1011 ) c/ i. kg Lined against the earths magnetic field = 2.01474 × 1011 c/kg (± ii. 0.4 × 1011 ) c/kg VII
Although our propagated error and standard deviation were off by a factor of ten, when comparing our average readings to that of the accepted value, they are not so far off as so to consider the experiment a total failure. *The bulk of hard data will be found on the following page(s). There are several factors that may have contributed to the error in our obtained calculations. Some of these include: o On our first day of collecting data, there was no compass provided. We made due with the assumption, or rather hope, that the device was already positioned somewhat correctly. On the second day, a compass was provided, but we failed to use it to its full potential. We merely noted the general north direction instead of actually bringing the compass close to the apparatus and finely positioning it. In our situation this may have actually led us to more error, because at a later time I noticed something odd. There were two identical compasses in the lab, each reading a completely different, inaccurate north. It is believed that these two compasses had been previously used near an electronics lab, which left their orientation way off. VIII
o Out of the four different disk radii, we were consistently only to line our beam on the outer three. To position the beam on the innermost circle would have required us to increase our magnetic field. Our own uncertainty about the nature of the equipment led us to believe that we needed to increase our voltage to level that was higher than the 0.7 ampere maximum that the filament preservation warranted in order to do so. o We very well could have underestimated the uncertainty of any of the variables involved: the electron beam radius, the measured dimensions of the apparatus, the plate, grid, and field readings. In all of these, we may have been overconfident in our measurement precision, although again, as mentioned before, our results were quite good given the magnitude scale of our data. o It may have been a mistake to assume the manufacturers given data to be correct. More measurements could have been at a minimum verified. IX
Conclusion If executed properly with great precision, this experimental procedure could be used to calculate the e/m ratio with a great accuracy. Acknowledgements Resource for earths magnetic field- Geomagnetic Online Calculator National Geophysical Data Center (NGDC) NOAA Satellite and Information Series NOAA website: http:www.ngdc.noaa.gov/seg/geomag/jsp Halliday/Resnick/Walker- Fundamentals of Physics, 7th edition Chapter 29, problem # 50 X
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Physics 222 Lab Report Electron Charge to Mass Ratio - Free download as PDF File (.pdf) ... Charge to Mass Ratio for an Electron PASCO[1] 21. Lab Report 6. | 2,330 | 10,438 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.920835 |
https://math.answers.com/other-math/What_two_numbers_can_you_multiply_to_30_and_add_up_to_-17 | 1,719,341,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866218.13/warc/CC-MAIN-20240625171218-20240625201218-00105.warc.gz | 335,329,121 | 47,446 | 0
# What two numbers can you multiply to 30 and add up to -17?
Updated: 4/28/2022
Wiki User
13y ago
13
Bret Cummings
Lvl 10
3y ago
Wiki User
13y ago
If you multiply two negative numbers together the result is always positive.
So, -15 and -2 if multiplied together will give +30, and if added together will give -17
Syeda Hasny
Lvl 3
5mo ago
The two number are 5 and 12 | 119 | 382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-26 | latest | en | 0.794734 |
http://www.lofoya.com/Solved/1560/what-is-the-percentage-of-candidates-qualified-from-state-n-for-all | 1,516,339,772,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887746.35/warc/CC-MAIN-20180119045937-20180119065937-00653.warc.gz | 501,867,733 | 14,435 | # Easy Tabular Data Solved QuestionData Interpretation Discussion
Common Information
Study the following table and answer the questions.
Number of Candidates Appeared and Qualified in a Competitive Examination from Different States over the Years.
Table below can be scrolled horizontally
State Year 1997 1998 1999 2000 2001 App. Qual. App. Qual. App. Qual. App. Qual. App. Qual. M 5200 720 8500 980 7400 850 6800 775 9500 1125 N 7500 840 9200 1050 8450 920 9200 980 8800 1020 P 6400 780 8800 1020 7800 890 8750 1010 9750 1250 Q 8100 950 9500 1240 8700 980 9700 1200 8950 995 R 7800 870 7600 940 9800 1350 7600 945 7990 885
Q. Common Information Question: 4/6 What is the percentage of candidates qualified from State N for all the years together, over the candidates appeared from State N during all the years together?
✖ A. 12.36% ✖ B. 12.16% ✖ C. 11.47% ✔ D. 11.15%
Solution:
Option(D) is correct
Required percentage:
$= \dfrac{840 + 1050 + 920 + 980 + 1020}{7500 + 9200 + 8450 + 9200 + 8800}× 100\%$
$= \dfrac{4810}{43150} ×100\%$
$= \textbf{11.15%}$ | 388 | 1,067 | {"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} | 3.90625 | 4 | CC-MAIN-2018-05 | latest | en | 0.632108 |
https://ecologyforthemasses.com/2020/08/31/measuring-abundance-in-the-face-of-detection-bias/ | 1,675,723,949,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500365.52/warc/CC-MAIN-20230206212647-20230207002647-00875.warc.gz | 239,282,906 | 31,183 | ## It’s All Relative: Measuring Abundance In The Face of Detection Bias
### There are many papers out there discussing estimates of abundance and occurrence of a variety of plants and animals. Sometimes you’ll also see references to relative abundance and relative occurrence. What makes researchers go for one estimate over the other? When might you face a similar choice? The goal of this post is to try to shed some light on when you might want to keep things relative.
If your goal is to measure the total abundance of your favorite species, you probably know the drill. You would go out and count every one you saw. If you are looking for something large and not subtle, like say a hippo, then you’d probably be able to spot every one that you walked by, and so you’d feel pretty good about your abundance estimate. But while hippos or elephants might be easy enough to count, if you’re tallying up most other species, what we call “imperfect detection” bias is probably going to come into effect. That is, there is a non-negligible probability that you don’t see every individual you cross paths with. They could be small, they could be hiding, you might have blinked when they passed by.
You could try to use a model to account for imperfect detection by using covariates to help describe your ability to spot the species. For example, if you are roughing it through dense tree cover in the rain close to nightfall, you suspect that you might not be at the peak of your detection game. However, we know that no model is perfect, and we still might be missing something. Like we might know that the size of the individual species impacts whether or not we spot it, but we have no way of incorporating that information into the model. After all, if we happened to have the individual measurements, we would also get the abundance by default.
So what else can we do? We can give up on having an absolute abundance measure and go for a measurement that gives us a general sense of abundance without having a concrete number. A strategy we can use if we are focusing on one species is to come up with some proxy for “search effort” and then judge abundance *relative* to that effort.
Effort could be approximated by something like the time spent at each site looking for the species since with more time spent at a site we’d expect to have a better chance of spotting the species if it was actually there. Effort could also be approximated by the time we’ve been out searching overall since we’re maybe more likely to see a species earlier in the day when we aren’t as fatigued from a day of traipsing through the woods. But let’s be real, imperfect proxies are still a bummer. Are we out of options?
If we happened to be searching for many species at the same time, we could borrow information *across* species to give us another way to estimate relative abundance. If we assume our detection problems are consistent across species, then our relative measure of abundance becomes the proportion of Species A out of all the species that we saw. Or in fact, any other sampling biases that we assume are the same across species we can sweep under the rug by just considering relative abundance. Why is that?
By looking at proportions, and under the right conditions, we get to cancel a lot of the problematic terms in our estimate. Think of any sampling bias as a thickening or thinning of true abundance. In the case of oversampling in places that are easier to get to, we may inflate the true abundance. In the case of imperfect detection; we dampen the true abundance of species k, denoted ak, by a detection probability, denoted p, less than one.
When we collect data about, say, three species, the relative abundance of the first species is then equal to (p * a1 ) / (p * a1 + p * a2 + p * a3) = a1 / (a1 + a2 + a3). We no longer have to worry about estimating the detection probability.
However, we know that not everything goes according to plan in the wild. Can relative abundance still lead us astray? If we look closely at that formula, we reveal a big assumption that we made: the thinning or bias (here we’re considering it to be the detection probability) is assumed to be the same for every species (one p, no pk) so that we can cancel it out in the numerator and denominator. This may be plausible if all of our species are similar. But it would be much harder to learn about how many small lizards we may be missing from how many bobcats we are missing. Their detection probabilities are likely very different, even if we account for differences in size.
Despite its caveats, this multi-species version of relative abundance is the version that is also relevant to biodiversity. If we care about answering questions about the numbers of one species *in comparison to another*, then we are in luck. There are lots of questions we can answer by prioritizing our understanding of the ranking of species with respect to one another rather than worrying about their absolute numbers. For example, is Species A consistently spotted more than Species B?
Can we detect changes over time using abundance and/or relative abundance? We have to make some more assumptions, but yes, it’s possible. In the absolute case we need to model the detection probability (and any other sampling biases) and make sure we can account for any changes in those quantities over time as well. In the relative case, we can still cancel out those thinning parameters, but we need to further assume that as they change over time, they change consistently across species. Whether or not that assumption is defensible is context dependent. For some species, it may well be defensible. Yet other species may become more visible at certain times of the year, for instance if they become more brightly coloured for mating season, or need to venture out into the open more as food becomes scarce.
There are some questions that only absolute abundance can really answer though. For example to make particular conservation decisions, absolute values are needed to make sure species aren’t dropping below abundance thresholds where they change protection status. If we make too optimistic a a guess of our detection probability that turns out to overestimate a species population, the resulting conservation measures might not do enough, and the outcome for that species could be severe.
While we are at it, all of this logic follows for absolute occurrence v. relative occurrence. Estimating absolute occurrence would result in having a calibrated sense of the probability we will encounter a particular species. Estimating relative occurrence would help us understand whether we are more or less likely to observe one species over another.
At the end of the survey day it really depends on what questions you are trying to answer and what assumptions you are willing to make about your survey approach. That’s nothing new. However, by understanding how that one word “relative” impacts the interpretation of your results, we can still make progress under the constraints of the data we have.
Have a quantitative term or concept that mystifies you? Want it explained simply? Suggest a topic for next month → @sastoudt. You can also see more of Sara’s work at Ecology for the Masses at her profile here. | 1,495 | 7,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.949858 |
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Zoltán Csáti
Université libre de Bruxelles
Last seen: 2 months ago Active since 2012
Professional Interests: differential equations, fluid mechanics, extended finite element method, MATLAB
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https://groups.io/g/TekScopes/messages?expanded=1&msgnum=123370 | 1,571,390,367,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986679439.48/warc/CC-MAIN-20191018081630-20191018105130-00303.warc.gz | 524,526,489 | 17,621 | Re: Tek 2215
Jay
Oh I found this page on rebuilding the switch too..
http://softsolder.com/2013/01/11/tektronix-2215a-oscilloscope-power-switch-rebuild/Also I need to find some info on rebuilding the power supply caps I have read they have alot of issues with ..
Re: 7603 weirdo inverted Z axis
Albert Otten
Hi David,
It seems that you forgot about R1155?
4.8 V across R1155 gives 0.48 mA. That 0.48 mA reduces the zener current and reduces the current via R1159 by the same amount.
I think you took 130 V across R1120 while it is about 128.4 V, this gives 0.21 mA less via R1120 and 0.21 mA more via R1159.
Together this yields 0.27 mA less via R1159 or 4.7 V lower output voltage. Then our results are the same.
Albert
---In TekScopes@..., <davidwhess@...> wrote :
On 24 Jan 2016 11:46:17 -0800, you wrote:
>Hi David,
>>With no input at Q1107, I calculate a 21 volt output at R1157 which agrees with your measurement and should blank the beam. That points to a problem in the DC restorer.
>
>I calculate it at 16 V, never mind.
I did not find a way to get 16 volts but my work can be checked:
With Q1107 disconnected there are three currents into or out of the
base of Q1156:
1. R1141 is -1.7mA
2. R1121 + R1122 is -1.15mA
3. R1120 is +2.3mA
The later two connect at the top of VR1148 but affect the base of
Q1107 the same way except for an offset of 11 volts.
2.3mA - 1.15mA - 1.7mA = -0.55mA
The amplifier is inverting and note that it does not care whether the
currents are applied to the top or bottom of VR1148 as long as VR1148
is conducting which R1141 guarantees so 0.55mA * 17.4k = 9.6V
The output is offset by VR1148 and the Vbe of Q1156 so:
9.6V + 11V + 0.6V = 21.2V
Re: Tek 2215
It looks like it is in pretty good shape. I think the handle can be
repaired and it is not difficult to remove once you pull the
oscilloscope chassis from the enclosure. I do not see any evidence of
the front panel controls being bent so they cannot be too damaged.
On Sun, 24 Jan 2016 21:33:41 +0000 (UTC), you wrote:
Thanks for such great info .. I will have to check out volt mod and see if the mod is done or needed to be done .. Oh also to clarify the handle locking tabs are broke and not the knobs ..The knobs are fine and the unit is pretty clean..I hope this works out but I got this for dirt cheap..
Here is the listing
http://www.ebay.com/itm/TEKTRONIX-2215-60MHz-oscilloscope-/321969604505?_trksid=p2047675.l2557&ssPageName=STRK%3AMEBIDX%3AIT&nma=true&si=wOUmi6rr6qZOZulXaqHvXtSpJJg%253D&orig_cvip=true&rt=nc
Re: 7603 weirdo inverted Z axis
Albert Otten
Typo in "I think you took 130 V across R1120 while it is about 128.4 V". Should read 130 V - 11.6 V = 118.4 V, not 128.4 V.
Albert
Re: Tek 2215
Jay
Thanks I hope so and for \$24 shipped we will see what happens and will post updates..
Re: Tek 2215
Tothwolf
On Sun, 24 Jan 2016, joe black jasontucker70@... [TekScopes] wrote:
Oh I found this page on rebuilding the switch too.. http://softsolder.com/2013/01/11/tektronix-2215a-oscilloscope-power-switch-rebuild/Also I need to find some info on rebuilding the power supply caps I have read they have alot of issues with ..
I didn't find it necessary to disassemble the switch apart from removing the add-on safety shield they added over the top of the switch. To remove it without cracking it, you have to slide it off the switch towards the back, and even then it can still crack.
While the power switch is the same, keep in mind that the main board in the 2213A/2215A is completely different from the board used in the 2213/2215.
When you get you scope, if it has the preregulator assembly, I can see about compiling a list of the parts I used for my scopes. Most of them will be the same but you would need to add the capacitors they used on the dual timebase board.
Re: Tek 2215
Jay
Ok thanks..
Are there alot of caps that should be replaced and any other mods that should be done ???
CL: Lot of 5 Tek 453's
John Clark
Re: 7603 weirdo inverted Z axis
On 24 Jan 2016 13:45:16 -0800, you wrote:
Typo in "I think you took 130 V across R1120 while it is about 128.4 V". Should read 130 V - 11.6 V = 118.4 V, not 128.4 V.
Albert
I have been going through your post and figured that was a typo. I
used 11.6 volts at the top of VR1148.
On 24 Jan 2016 13:40:48 -0800, you wrote:
Hi David,
It seems that you forgot about R1155?
Yep, I missed it. I might have designed it with an AC coupling
capacitor in series with R1155 but technically that is not needed. I
thought R1155 controlled the gain of Q1156 but the effect would be
very small. Who designed this thing?
Hmm, the later design used in the 7623A and 7633 lacks R1155 among
other minor changes and I had all of the schematics open when I did my
first analysis although I only scribbled on the correct schematic.
4.8 V across R1155 gives 0.48 mA. That 0.48 mA reduces the zener current and reduces the current via R1159 by the same amount.
I think you took 130 V across R1120 while it is about 128.4 V, this gives 0.21 mA less via R1120 and 0.21 mA more via R1159.
Together this yields 0.27 mA less via R1159 or 4.7 V lower output voltage. Then our results are the same.
Albert
4.8V / 10k [R1155] = +0.48mA
-15.6V / 9.1k [R1141] = -1.7mA
(130V - 11.6V) / 56k [R1120] = +2.1mA
-11.6V / 10.1k [R1121 and R1122] = -1.15mA
2.48mA - 2.85mA = -0.27mA
0.27mA * 17.4k [R1159] = 4.7V
4.7V + 11V = 15.7V
With Q1107 (a slow 2N3904???) removed on my 7603, the output at point
B is 14.8 volts which closely agrees with our current calculation so I
wonder if Marian's <public@...> measurement of 20 volts at the
output of the z-axis amplifier is significant and indicates a problem.
I get the same 14.8 volt output with Q1107 installed, the readout off,
and the intensity control turned to minimum.
I also measured the output at point B and as the intensity control is
turned fully clockwise, the output increases to about 28 volts
monotonically.
Re: Tek 2215
On Sun, 24 Jan 2016 16:02:14 -0600 (CST), you wrote:
While the power switch is the same, keep in mind that the main board in
the 2213A/2215A is completely different from the board used in the
2213/2215.
I consider them separate designs with the 2213/2215 (1982 to 1985)
being the beta test for the series which starts with the 2235/2236
(1984) and includes the 2213A/2215A (1986) as replacements for the
2213/2215. Besides differences in the overall design, the 2213/2215
has a 30pF input capacitance and 10kV acceleration while all of the
later models have a 20pF input capacitance and 14kV acceleration.
Re: CL: Lot of 5 Tek 453's
Chris Pierce
Yes I would like them, but I can't put the money together until Feb, 10 will try to get some money together Monday but I don't think I can put that much together. I live in Goose Creek call me Monday if you want to 8436702873
On Sunday, January 24, 2016 5:21 PM, "johnclark05@... [TekScopes]" <TekScopes@...> wrote:
Anyone in Charleston, SC area need some 453's? \$150 for 5 of 'em.
http://charleston.craigslist.org/ele/5417533394.html http://charleston.craigslist.org/ele/5417533394.html
[Non-text portions of this message have been removed]
Re: CL: Lot of 5 Tek 453's
John Clark
Oh, no, they're not mine. I just saw them pop up on Craigslist today.
John
To: TekScopes@...
From: TekScopes@...
Date: Sun, 24 Jan 2016 23:46:40 +0000
Subject: Re: [TekScopes] CL: Lot of 5 Tek 453's
Yes I would like them, but I can't put the money together until Feb, 10 will try to get some money together Monday but I don't think I can put that much together. I live in Goose Creek call me Monday if you want to 8436702873
On Sunday, January 24, 2016 5:21 PM, "johnclark05@... [TekScopes]" <TekScopes@...> wrote:
Anyone in Charleston, SC area need some 453's? \$150 for 5 of 'em.
http://charleston.craigslist.org/ele/5417533394.html http://charleston.craigslist.org/ele/5417533394.html
[Non-text portions of this message have been removed]
Recommendations / 2247A
John Dickens
I have my eye on a 2247A on Craigslist that looks very clean and appears to be fully functional.
Does anyone have any thoughts about this scope - reliable, repairable, etc.? It would be my first scope with a microprocessor. I also have a 7603 and a 2236 as well, not to mention a 503 in poor shape.
Thanks, John
Re: Tek 2215
Tothwolf
On Sun, 24 Jan 2016, jasontucker70@... [TekScopes] wrote:
Ok thanks..
Are there alot of caps that should be replaced and any other mods that should be done ???
As far as preventive maintenance parts goes, I replaced all of the electrolytic capacitors, the 5 Rifa brand safety capacitors, the 1M ohm carbon comp resistors in the high voltage divider, and some other misc resistors.
The 6 silver colored Mallory capacitors on the main board are an absolute must replace item, and if your 2215 still has the original parts installed, you'll probably find that they are beginning to leak from their bottom seals.
I used new manufacture carbon comp resistors for the 1M ohm resistors, however after the discussion on the list, I picked up some high voltage rated film parts that I plan to replace those with the next time I have the scopes opened up.
The difficulties I ran into with the 2213 scopes were that the component leads are bent down to hold the parts in before they were soldered, and the two small electrolytics on the front panel board were difficult to get to (I used some long tweezers to fit them).
While some people claim that the small 5mm and 6.3mm capacitors never fail, some of the small electrolytics in my scopes were visibly leaking or had corrosion on their leads which was only visible after removing them, so I think replacing them is a good idea.
I posted this link previously, but some photos of my scopes can be found here: http://strudel.ignorelist.com/~tothwolf/photos/Tektronix_2213/
Overall, the 2213/2215 isn't a scope I would recommend for a beginner, but if you have the patience and a vacuum soldering tool, they aren't that hard to service.
Re: Tek 2215
On Sun, 24 Jan 2016 18:56:02 -0600 (CST), you wrote:
On Sun, 24 Jan 2016, jasontucker70@... [TekScopes] wrote:
Ok thanks..
Are there alot of caps that should be replaced and any other mods that
should be done ???
....
I used new manufacture carbon comp resistors for the 1M ohm resistors,
however after the discussion on the list, I picked up some high voltage
rated film parts that I plan to replace those with the next time I have
the scopes opened up.
...
The Vishay VR25 and VR35 series high voltage film resistors are very
similar if not identical to the high voltage film resistors which
Tektronix used to replace the carbon composition resistors originally
used in the 22xx series focus chain. They are readily available and
inexpensive.
Re: Recommendations / 2247A
I do not have any complaints about mine. The 2247A duplicates the
universal timer/counter functions found in your 2236. Calibration is
*not* done digitally so there are no worries about dead batteries or
floating gate storage memory like with the 24xx oscilloscopes.
On Sun, 24 Jan 2016 18:45:59 -0600, you wrote:
I have my eye on a 2247A on Craigslist that looks very clean and appears to be fully functional.
Does anyone have any thoughts about this scope - reliable, repairable, etc.? It would be my first scope with a microprocessor. I also have a 7603 and a 2236 as well, not to mention a 503 in poor shape.
Thanks, John
Re: Recommendations / 2247A
Tom Jobe <tomjobe@...>
Hi John,
I have some 224x scopes and use a nice 2247A regularly. It's a wonderful
scope and I know it's very complicated, but I'm hoping all of you will
help me fix it if or when it does fail!
I'd say, go for it and worry about future in the future.
Ordinary things like re-capping the 224x power supply is a very simple
thing to do, and the details of it are well documented. Also the 2247A
probably came late in the 224x series so it probably has some
improvements they learned of earlier in the series.
tom jobe...
On 1/24/2016 4:45 PM, John Dickens jake_117_dickens@...
[TekScopes] wrote:
I have my eye on a 2247A on Craigslist that looks very clean and
appears to be fully functional.
Does anyone have any thoughts about this scope - reliable, repairable,
etc.? It would be my first scope with a microprocessor. I also have a
7603 and a 2236 as well, not to mention a 503 in poor shape.
Thanks, John
[Non-text portions of this message have been removed]
Re: Tek 2215
Jay
Why are you replacing the resistor that your listed ???
Re: Tek 2215
Jay
I have a hakko fx-888d iron and nice desoldering iron that works really good.I know I'm a noob to scopes but right now I have no scope cause I had a nice Rigol DS1054z but had to be sold in order to pay rent and a place for my family.. The scope is show in the pics working but will need further testing and why replace the resistors ... When I get the scopes is there anyway I could could post pics and you guys could show me what is needed to be replace or part# or something ???
About how much should the parts cost me for the rebuild here in the usa ???
Re: Tek 2215
Tothwolf
On Sun, 24 Jan 2016, David @DWH [TekScopes] wrote:
On Sun, 24 Jan 2016 18:56:02 -0600 (CST), you wrote:
On Sun, 24 Jan 2016, jasontucker70@... [TekScopes] wrote:
Ok thanks..
Are there alot of caps that should be replaced and any other mods that should be done ???
....
I used new manufacture carbon comp resistors for the 1M ohm resistors, however after the discussion on the list, I picked up some high voltage rated film parts that I plan to replace those with the next time I have the scopes opened up.
...
The Vishay VR25 and VR35 series high voltage film resistors are very similar if not identical to the high voltage film resistors which Tektronix used to replace the carbon composition resistors originally used in the 22xx series focus chain. They are readily available and inexpensive.
That's what I'm planning to install. I picked up 100 Vishay VR37000001004FR500 for \$13.60. | 4,019 | 14,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-43 | latest | en | 0.918853 |
https://www.teacherspayteachers.com/Product/St-Patricks-Day-Decimals-2470857 | 1,556,101,973,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578640839.82/warc/CC-MAIN-20190424094510-20190424120510-00448.warc.gz | 814,225,872 | 15,742 | # St. Patrick's Day Decimals
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This product was made so we teachers have a excellent excuse to practice decimal operations during the month of March! I made it with my 5th graders in mind because I could not find too many coloring or cut/paste math worksheets for their level. They’ve told me how they love all the coloring we do in math!
The first worksheet, “Find the Gold”, has the rainbow half sheets that go with it. Students will have to use all four operations using decimals with that worksheet. Just copy one worksheet for every students, then copy half that amount of the rainbow worksheet. Cut the rainbow worksheet in half, and give each kid a half.
“Gather Lucky Charms” has students practice adding and subtracting decimals.
“Found a Leprechaun” has students practice multiplying and dividing decimals. Students can color any part of the leprechaun that does not have an answer in it whatever color they choose.
Perfect for stations or as review. Please check out my other holiday themed decimal worksheets that can help you review decimals all year round!
https://www.teacherspayteachers.com/Product/Valentines-Day-Themed-Decimal-Activities-All-Operations-2164732
https://www.teacherspayteachers.com/Product/Halloween-Themed-Decimal-Activities-All-Operations-2149682
https://www.teacherspayteachers.com/Product/Christmas-Themed-Decimal-Activities-All-Operations-2162944
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\$1.00 | 357 | 1,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-18 | latest | en | 0.86721 |
https://www.30secondsofcode.org/js/algorithm/p/1/ | 1,723,025,831,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00365.warc.gz | 499,971,883 | 13,290 | # JavaScript Algorithms
The JavaScript algorithms collection includes implementations and examples of popular algorithms in ES6. Algorithmic snippets are primarily a learning resource, as they might already be implemented natively or require optimizations to run in production.
• JavaScript ·
### Group data using the K-means clustering algorithm in JavaScript
Implement the K-means clustering algorithm in JavaScript to group data into clusters.
• JavaScript ·
### Classify data using the K-nearest neighbors algorithm in JavaScript
Implement the K-nearest neighbors algorithm in JavaScript to classify a data point relative to a labelled data set.
• JavaScript ·
### Heap sort
Sort an array of numbers, using the heapsort algorithm.
• JavaScript ·
### Selection sort
Sort an array of numbers, using the selection sort algorithm.
• JavaScript ·
### Quick sort
Sort an array of numbers, using the quicksort algorithm.
• JavaScript ·
### Implementing the Luhn Algorithm in JavaScript
Implement the Luhn Algorithm, used to validate a variety of identification numbers.
• JavaScript ·
### Implement the Caesar cipher in JavaScript
The Caesar cipher is a simple substitution cipher, which can be easily implemented with a few lines of JavaScript code.
• JavaScript ·
### Bucket sort
Sort an array of numbers, using the bucket sort algorithm.
• JavaScript ·
### Big-O Cheat Sheet
Learn everything you need to know about Big-O notation with this handy cheatsheet.
• JavaScript ·
### Merge sort
Sort an array of numbers, using the merge sort algorithm.
• JavaScript ·
### Check if a number is prime in JavaScript
Learn how to check if a number is prime and how to generate prime numbers up to a given number in JavaScript.
• JavaScript ·
### Calculate the Euclidean distance in JavaScript
Use JavaScript's `Math.hypot()` to calculate the Euclidean distance between two points.
• JavaScript ·
### Insertion sort
Sort an array of numbers, using the insertion sort algorithm.
• JavaScript ·
### Arithmetic and geometric progression in JavaScript
Create arrays of numbers in arithmetic and geometric progression.
• JavaScript ·
### Calculate the Levenshtein distance between two strings in JavaScript
Implement the Levenshtein distance algorithm in JavaScript to calculate the difference between two strings.
• JavaScript ·
### Maximum subarray
Finds a contiguous subarray with the largest sum within an array of numbers.
• JavaScript ·
### How can I calculate the greatest common divisor & least common multiple in JavaScript?
Use JavaScript to calculate the greatest common divisor and least common multiple of two or more numbers.
• JavaScript ·
### Bubble sort
Sort an array of numbers, using the bubble sort algorithm.
• JavaScript ·
### Fibonacci
Generates an array, containing the Fibonacci sequence, up until the nth term.
• JavaScript ·
### Binary search
Finds the index of a given element in a sorted array using the binary search algorithm.
• JavaScript ·
### Linear search
Finds the first index of a given element in an array using the linear search algorithm.
• JavaScript ·
### Find the prime factors of a number in JavaScript
Find the prime factors of a number using trial division in JavaScript.
• JavaScript ·
### Hamming distance
Calculates the Hamming distance between two values.
• JavaScript ·
### Factorial of number
Calculates the factorial of a number.
Start typing a keyphrase to see matching snippets. | 686 | 3,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-33 | latest | en | 0.64331 |
https://answers.yahoo.com/question/index?qid=20061230013549AAj9l6x | 1,579,869,530,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250619323.41/warc/CC-MAIN-20200124100832-20200124125832-00307.warc.gz | 329,974,987 | 24,146 | # 2 numbers that can be both the area and perimeter of a triangle whose side lengths are a pythagorean triple?
from numbers 1-31
Relevance
3-4-5: area 6, perimeter 12 nope...
6-8-10: area 24, perimeter 24 yes!!
5-12-13: area 30, perimeter 30 yes!!!
Guess that's it: 24 and 30
To get these numbers, you can solve this equation :
a+b+(a^2+b^2)^(1/2) = ab/2
That equation can be reduced to
4a - ab + 4b = 8.
Then try a = 1,2, ... , 15 (stop if the value of b is negative)
You can get the value of b from above equation. The answer is then just (ab)/2. | 187 | 560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.858547 |
https://www.tutorialcup.com/interview/dynamic-programming/minimum-sum-path-in-a-triangle.htm | 1,606,944,814,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141716970.77/warc/CC-MAIN-20201202205758-20201202235758-00573.warc.gz | 887,696,047 | 267,331 | Home » Technical Interview Questions » Dynamic Programming Interview Questions » Minimum Sum Path in a Triangle
# Minimum Sum Path in a Triangle
## Problem Statement
The problem “Minimum Sum Path in a Triangle” states that you are given a sequence in the form of a triangle of integers. Now starting from the top row what is the minimum sum you can achieve when you reach the bottom row?
## Example
``` 1
2 3
5 8 1```
`5`
Explanation
You can simply move down the path in the following manner. 1-> 3-> 1. Thus the sum is 5. If we would have gone with a greedy approach. We would have gone with 2 instead of 3. Thus we can end up with only 8 or 11 which is greater than 5.
## Approach
So how do we solve the Minimum sum path in a triangle? We have also solved a similar problem where we had to find the maximum sum path in a triangle. In that previous post, we had already discussed that the brute force approach for these problems is to generate all the paths. After the generation of these paths, just compute the sum for each of these paths and update the minimum sum.
So instead of solving this problem using brute force, we solve this with dynamic programming. Because the brute force approach is highly inefficient.
First, we fill the answer for the cells in the last row. We know that the minimum sum that can be achieved is if we start from the cells at the bottom row is the number itself. After that, we move to the row above the bottom row. For each cell in the current row, we can choose the DP values of the cells which are adjacent to it in the row just below it. And fill the minimum value which can be achieved. This way we keep on going in an upward direction. As we reach the top row, we are done with the problem.
READ Binary array after M range toggle operations
## C++ code to find Minimum path sum in a triangle
```#include<bits/stdc++.h>
using namespace std;
int minimumPathSumInTriangle(vector<vector<int>> &input)
{
int n = input.size();
// start from row above bottom row
// since the bottom row cells are the answers themselves
for(int i=n-2;i>=0;i--)
{
// start from left to right in column
for(int j=0;j<=i;j++)
{
if(input[i+1][j] < input[i+1][j+1])
input[i][j] += input[i+1][j];
else
input[i][j] += input[i+1][j+1];
}
}
return input[0][0];
}
int main()
{
int n;cin>>n; // number of rows
vector<vector<int>> input(n, vector<int>(n, 0));
for(int i=0;i<n;i++){
for(int j=0;j<=i;j++)
cin>>input[i][j];
}
cout<<minimumPathSumInTriangle(input);
}```
```3
1
2 3
5 8 1```
`5`
## Java code to find Minimum path sum in a triangle
```import java.util.*;
class Main{
static int minimumPathSumInTriangle(int input[][], int n)
{
// start from row above bottom row
// since the bottom row cells are the answers themselves
for(int i=n-2;i>=0;i--)
{
// start from left to right in column
for(int j=0;j<=i;j++)
{
if(input[i+1][j] < input[i+1][j+1])
input[i][j] += input[i+1][j];
else
input[i][j] += input[i+1][j+1];
}
}
return input[0][0];
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt(); // number of rows
int input[][] = new int[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<=i;j++)
input[i][j] = sc.nextInt();
}
}
}```
```3
1
2 3
5 8 1```
`5`
## Complexity Analysis
### Time Complexity
O(N^2), as we moved across each row and each column. In the process, we traveled to each cell. And since there were O(N^2) cells in the triangle and the transition for DP took only O(1) operation. Thus, the time complexity is also polynomial.
### Space Complexity
O(N^2) since we created a 2D DP array. Thus the space complexity is also polynomial.
Array Interview Questions Graph Interview Questions LinkedList Interview Questions String Interview Questions Tree Interview Questions Core Java Interview Questions | 998 | 3,789 | {"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} | 4.46875 | 4 | CC-MAIN-2020-50 | longest | en | 0.898126 |
https://www.edplace.com/worksheet_info/maths/keystage2/year4/topic/846/3973/adding-or-subtracting-1000:-subtract-1000-from-four-digit-numbers | 1,716,232,772,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00791.warc.gz | 662,049,186 | 12,251 | # Subtract 1,000 from a Four-Digit Number
In this worksheet, students will subtract 1,000 from a four-digit number.
Key stage: KS 2
Curriculum topic: Number: Number and Place Value
Curriculum subtopic: Find 1000 More or Less than a Given Number
Difficulty level:
#### Worksheet Overview
When we subtract 1,000 from a four-digit number, we need to decrease the digit in the thousands column by 1.
Here is the number 4,705 shown on an abacus.
Notice that the abacus says units. Units are another word for ones.
When we subtract 1,000 from it, we take away one digit from the thousands column.
4,705 - 1,000 = 3,705
Example 1
Subtract 1,000 from:
5,097
We decrease the 5 in the thousands column by 1 to get 4 in that column.
5,097 - 1,000 = 4,097
Example 2
Subtract 1,000 from:
1,060
We decrease the 1 in the thousands column by 1 to get 0 in that column.
Remember to leave off any zeros on the left-hand side of a number.
1,060 - 1,000 = 0060 = 60
Does that make sense?
Let's get started!
### What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
Get started | 375 | 1,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-22 | latest | en | 0.862176 |
http://physics.stackexchange.com/tags/ising-model/new | 1,406,609,295,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510265454.51/warc/CC-MAIN-20140728011745-00214-ip-10-146-231-18.ec2.internal.warc.gz | 221,714,686 | 10,497 | # Tag Info
0
The RG is not a group, it's a semi group so you can only go in one direction, the one that actually renormalize. Here you can use the relation K'(K) for your flow. but if you use K(K'), you will have something wrong because the RG procedure that give this relation as no physical meaning. it's as if you add spin in your Ising chain. When you do RG, ,you ...
1
I think Adam's answer is excellent, and I'd rather make this a comment but can't as I just signed up in order to answer. While I agree with most of what Adam said, there are cases where large-N works well for $N=1$. The case I'm familiar with is the large-N expansion for the "Spin Ices" Dy$_2$Ti$_2$O$_7$ and Ho$_2$Ti$_2$O$_7$, which have Ising spins on a ...
0
For the ferromagnetic type order-disorder phase transition, the correlation length $\xi$ diverge to infinity as the temperature approach the critical point. What happens physically is that there are fractal clusters of the same spin with size of order $\xi$ that grow to infinity as the temperature approach the critical point. The SW algorithm is based on ...
3
I have tried to reproduce your problem using my own code, but I couldn't: I got the correct value for the exponent. I can't tell you what went wrong in your calculation, but I can tell you exactly what I did! You can inspect my code on GitHub. It was the first thing I ever wrote in C++ and today I would do many things differently, so please don't judge the ...
2
Yes, these equations exist and can be derived from the partition function in JGab's answer. The internal energy per spin is: $$u(\beta) = - \frac{\partial}{\partial \beta} \left( \ln(2) + \frac{1}{8 \pi^2} \int_{0}^{2\pi} \mathrm{d} q_1 \int_{0}^{2\pi} \mathrm{d} q_2 \\\ln \left[ \big( 1 - \sinh(2 \beta J) \big)^2 + \sinh(2 \beta J) \left( 2 - \cos q_1 - ... 1 I think you have a problem with double counting. The Ising Hamiltonian is$$H = - J \sum_{\langle i,j \rangle} S_i S_j where this strange sum notation means to sum over all bonds between neighbouring spins. It is the bond that matters, so you should not count it twice (for $ij$ and $ji$). Actually you can get the energy difference of a spin flip as ...
Top 50 recent answers are included | 604 | 2,227 | {"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} | 2.828125 | 3 | CC-MAIN-2014-23 | longest | en | 0.943728 |
https://bookofproofs.github.io/branches/fun/dudeney/geometrical-problems/various-other-geometrical-puzzles/the-garden-walls-solution.html | 1,686,093,686,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653183.5/warc/CC-MAIN-20230606214755-20230607004755-00659.warc.gz | 171,485,531 | 4,397 | # Solution
(related to Problem: The Garden Walls)
The puzzle was to divide the circular field into four equal parts by three walls, each wall being of exactly the same length. There are two essential difficulties in this problem. These are:
1. the thickness of the walls, and
2. the condition that these walls are three in number.
As to the first point, since we are told that the walls are brick walls, we clearly cannot ignore their thickness, while we have to find a solution that will equally work, whether the walls be of a thickness of one, two, three, or more bricks.
The second point requires a little more consideration. How are we to distinguish between a wall and walls? A straight wall without any bend in it, no matter how long, cannot ever become "walls," if it is neither broken nor intersected in any way. Also, our circular field is clearly enclosed by one wall. But if it had happened to be a square or a triangular enclosure, would there be respectively four and three walls or only one enclosing wall in each case? It is true that we speak of "the four walls" of a square building or garden, but this is only a convenient way of saying "the four sides." If you were speaking of the actual brickwork, you would say, "I am going to enclose this square garden with a wall." Angles clearly do not affect the question, for we may have a zigzag wall just as well as a straight one and the Great Wall of China is a good example of a wall with plenty of angles. Now, if you look at Diagrams $1,$ $2,$ and $3,$ you may be puzzled to declare whether there are in each case two or four new walls; but you cannot call them three, as required in our puzzle. The intersection either affects the question or it does not affect it.
If you tie two pieces of string firmly together or splice them in a nautical manner, they become "one piece of string." If you simply let them lie across one another or overlap, they remain "two pieces of string." It is all a question of joining and welding. It may similarly be held that if two walls be built into one another — I might almost say, if they are made homogeneous — they become one wall, in which case Diagrams $1,$ $2,$ and $3$ might each be said to show one wall or two, if it be indicated that the four ends only touch, and are not really built into, the outer circular wall.
The objection to Diagram $4$ is that although it shows the three required walls (assuming the ends are not built into the outer circular wall), yet it is only absolutely correct when we assume the walls to have no thickness. A brick has a thickness, and therefore the fact throws the whole method out and renders it only approximately correct.
Diagram $5$ shows, perhaps, the only correct and perfectly satisfactory solution. It will be noticed that, in addition to the circular wall, there are three new walls, which touch (and so enclose) but are not built into one another. This solution may be adapted to any desired thickness of the wall, and its correctness as to area and length of wall space is so obvious that it is unnecessary to explain it. I will, however, just say that the semicircular piece of ground that each tenant gives to his neighbor is exactly equal to the semicircular piece that his neighbor gives to him, while any section of wall space found in one garden is precisely repeated in all the others. Of course, there is an infinite number of ways in which this solution may be correctly varied.
Thank you to the contributors under CC BY-SA 4.0!
Github:
non-Github:
@H-Dudeney
### References
#### Project Gutenberg
1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917
This eBook is for the use of anyone anywhere in the United States and most other parts of the world at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this edition or online at http://www.gutenberg.org. If you are not located in the United States, you'll have to check the laws of the country where you are located before using this ebook. | 902 | 4,103 | {"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} | 4.21875 | 4 | CC-MAIN-2023-23 | latest | en | 0.968277 |
www.everythingtvclub.com | 1,721,247,608,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00650.warc.gz | 661,654,039 | 66,165 | Wednesday, July 17, 2024
# What Size TV For Room
## Calculate How Large Of A TV You Need
What’s the BEST TV Size for your room? | TVs Explained
Fortunately, you dont need to do any complex formulas to determine how big your TV should be.
Rather than force you to perform the calculations yourself, the good people over at Rtings.com have already crunched the numbers and put together a simple TV Size to Distance Calculator.
Rtings.com
Simply use the slider to set your viewing distance and select your resolution. The calculator will then present the optimal TV screen size for the given variables.
This calculator couldnt be easier to use, so theres no excuse not to take the time to determine what size TV you need before making a purchase. By using this calculator as a guide, you can optimize your home theater experience
## How To Calculate Distance
Big-screen TVs require a little math to understand how far away is an ideal viewing distance. You essentially need to look at a chart comparing the distance to screen size. Whether you are using a UHD, SUHD or any other TV set, the viewing experience is always better when your TV is larger. However, make sure your seating distance is far enough away to prevent eye strain.
#### How Field of View Affects Viewing Experience
The wider a TV screen is, the better your field of view. An immersive viewing experience requires your entire field of vision to be taken up by the screen. This is why theatres use the largest screen size possible. Apart from the screen size, it is also equally important to consider the input lag.
Insider Tip
The wider a TV screen is, the better your field of view.
#### How to Determine Correct Screen Size
Larger spaces require a larger size for your TV screen. To determine the correct screen size, measure the space available in your room. Compare this size to the chart that shows you which screen is best suited to which size screen.
#### Finding the Optimal Viewing Angle
Generally, you want to be front-and-center to your flat screen. Make sure that you arent sitting at more than a 40-degree viewing angle from your TV for the best results.
Warning
Larger spaces require a larger size for your TV screen.
## Whats The Best Viewing Angle For Watching TV
The ideal angle to sit at when watching any TV is directly opposite the panel. Deviate to the left or right of that point and the picture quality can deteriorate. At 0-degrees off the displays axis, you are in the best possible position to see everything on the display as it was intended by the manufacturer. This holds true no matter what the resolution, panel type or brightness of the TV might be.
That said, some TVs have a wider viewing angle than others. If you need a TV that can be enjoyed equally from various parts of the room, its best to look for a set that has a more generous viewing angle. That rules out curved TVs, then!
Note: Viewing angle in this context is not to be confused with the THX 40-degree and SMPTE 30-degree viewing recommendations, as these refer to your proximity to the display, not your placement in relation to the displays axis.
#### Why does viewing angle matter anyway?
A TV with a narrow viewing angle will only look its best when faced head-on, or close to it. With a TV like this, the picture quality – and your enjoyment – will deteriorate as soon you veer off from the narrow viewing angle towards the left or right. Panel brightness will be reduced, black tones may lose depth, appearing greyish instead, and all colours will look washed out and inaccurate.
#### What TVs have the widest viewing angle?
READ NEXT: How LCD TV works
#### Dont forget about vertical placement
Recommended Reading: What Streaming Service Has Tudn
## What Is Viewing Angle
This is the angle created from your nose out to the left and right edges of the display. Viewing angle is a simple but crucial concept in selecting the best television size. Ideally, most of your peripheral vision is filled by the screen image.
The movie theater specification maxim sets 30-36 degrees as the optimal viewing angles. To achieve a cinematic immersive experience, most of all, you want a screen size that pushes the limits of your field of vision.
THX recommends a display that fills 40 degrees of your field of vision, while SMPTE recommends 30 degrees. To achieve the immersion effect, we recommend pushing the limit of the larger screen sizes based upon your viewing distance.
## What Size TV Do You Need For The Best Movie Viewing Experience
When buying a new TV, one of the first questions you need to ask yourself is what TV size you should get. Modern TVs are available in different sizes, from 32 to 85 inches . For the best movie-watching experience, you need a bigger TV, usually somewhere between 55 and 75 inches. The size of your TV should mainly depend on the distance between the TV and the seating area and how much money youre prepared to spend on it.
This article will discuss what size TV you need to turn your living room into a home theater. Additionally, well show you how to calculate the optimal TV size based on the viewing distance and other factors.
Read Also: Fs1 On Pluto TV
## How To Find The Correct Angle For The TV In Your Conference Room
Many people experience discomfort straining their neck trying to see the screen during a presentation simply because the angle at which the screen is stationed forces them to sit in an awkward position. Weve all experienced something like this during a presentation, and it can be solved quickly and easily by taking into the angle in which you initially mount the screen.
Smart office technology company Logic Integration provides the following advice: Viewing angles are an important consideration, especially with rooms that are wide or where multiple displays are needed to provide proper viewing. For ideal viewing, participants should be within 45 degrees to each side from the center of the screen and the top of the screen should be positioned within 30 degrees above eye level. AVIXAs standard is more exact, but this general guideline will help set the expectation of screen size and positioning for the application.
## What Is The Distance From Furthest Viewer
Clear Touch, a provider of multi-touch interactive flat panels for education and business and a reseller of Clear Digital products, put together an easy-to-understand chart regarding screen size and distance from the furthest viewer:
The distance of the farthest viewer Ideal display size
Clear Touch explains that screen size should also depend on your type of business and the information you are sharing with employees during meetings. According to Clear Touch , If your organization requires you to display and analyze precise details on the screen, such as architectural drawings or medical images, then you will want to decrease the distance of the farthest viewer significantly for each particular screen size. This will ensure that you achieve the proper level of clarity.
Recommended Reading: Why Won T My Airplay Connect To My Samsung TV
## Can A TV Ever Be Too Big And How Close Should You Be Sitting Anyway Weve Got The Answers
Growing up, many of us were told that sitting too close to the television would cause us to get square eyes. Theres strong evidence to suggest that this is inaccurate. For a start, televisions are rectangular. But when buying a new TV, its natural to worry about whether or not youre buying the right size for your living space. Or, if youve already got it, wondering how far from the set you should sit.
You dont want to buy a TV thats too small as youll likely regret the purchase. Bigger screens result in more immersive viewing experiences, as anyone whos been to a cinema can attest. Its also possible, however, to get a TV thats simply too big. If youre buying a TV for a tiny bedroom or mobile home, this is all too easy to do.
Theres a fair bit to think about when planning your home TV setup, from the size-to-distance relationship, resolution and viewing angles too. Well start from the top but, if youre in a rush, you can use the quick links below to skip ahead.
## Screen Size And Viewing Distance
How to measure a TV and what size TV is right for you
So aside from budget, circumstances, personal preference, and resolution, a reliable rule of thumb for choosing the best size TV is viewing distance how far you sit from the TV under your normal viewing conditions. Todays 4K TVs offer more detail, so you can sit closer to the television, but not so close that you can see the screen pixels.
Of course, you dont want to sit so far away that you miss any of the incredible detail either, which means you need a handy number to divide the viewing distance by to come up with an optimal screen size. There are a number of different dividers, but all assume you have normal eyesight. If not, best get down to Specsavers before you start.
Related: Best TV Deals
Read Also: Screen Mirror To Vizio TV
## Why Should You Buy A 48
One of the main reasons to choose a TV thats under 50-inches is its price. TVs on the smaller end of the spectrum are much more budget-friendly than their larger counterparts. Though thats not just because theyre more compact, smaller screens often lack some of the upgrades that are given to 55-inch or bigger TVs.
Case in point, the new LG C2 OLED TVs come in sizes as small as 42-inch, but to get a model with premium OLED Evo panels youll need to buy at least the 55-inch variant.
As such, screens of this size can come with some performance restrictions that you wouldnt find with larger TVs where there is a wider selection of TVs that offer high-end features. But in return, your wallet wont be hit so hard when you decide to take one of these sized screens home with you.
Beyond their price, smaller 42-inch TVs and 48-inch TVs are also well suited to smaller spaces. If you just havent got space in your living room for a massive screen, then this is the size youll want to look out for.
These factors also make this size of TV great for a kitchen or bedroom, or an oversized computer monitor. The smaller frame means they wont have to compete as much for space against other more useful objects in the room, and you wont as easily go broke buying a second or third screen to deck out your home.
## TV Resolution And Screen Size: 4k 8k 1080p
Nearly all new TVs are Ultra HD 4K resolution. A growing number are 8K. Only the smallest and least expensive models are still merely HD .
To see all the detail available in a 4K or 8K resolution image you either need to sit very close, or have a very large TV. In fact, if you’re sitting nine feet away, even “big” TVs are still too small for you to see all the resolution for which they’re capable. Or to put it another way, the resolution of your next TV is going to be plenty unless you’re sitting very close, or are getting an exceptionally large TV . So yeah, you don’t need an 8K TV. If you want to know how far away you need to sit before you can no longer see individual pixels, Sony offers the following chart.
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## Screen Size And Resolution
First, remember that TV screens are measured diagonally. So a 55-inch screen measures 55 inches from one upper corner to the opposite lower corner. But the actual dimensions will be different, measuring roughly 49 inches wide and 29 inches tall.
Screen size has a direct relationship with the optimal viewing distance. Larger screens are best watched from farther back otherwise, you can find yourself unable to see the whole picture, or experience unwanted distortions and color issues from sitting too close.
Next, you have to consider the resolution of the TV. TVs today offer three resolution options 1080p, 4K and 8K. The higher the resolution, the closer you can sit before you’ll notice pixelation in an image. That means you can sit closer to a 4K TV than you could a 1080p TV.
We recommend 4K for pretty much everyone, but the resolution will change how close you can sit to a set and how expensive it will be. Here’s how the main resolution options break down:
1080 or Full HD: Display resolution measuring 1920 x 1080 pixels, used in modern television broadcasts, cable and satellite TV, Blu-ray media and available broadly in free and paid streaming services. You’ll still see these, especially when you go looking for the cheapest TVs on sale. Full HD is a good choice if you’re getting a TV with an extra-small screen, but for anything larger than 40 inches, you should be getting 4K.
Check out the top TV models in either resolution:
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## Finding The Right TV For Your New Home
Youll spend a lot of time lounging in your living room and enjoying a movie night or curled up in bed binging the latest on Netflix. Choose a TV that will serve its purpose without overwhelming the space. If you are in the process of viewing apartments, be sure to consider where your TV would fit best in the new space.
## Don’t Forget About Screen Resolution
The larger a UHD TV screen, the more important screen resolution becomes.
For a UHD TV that is 40 inches or smaller, a 720p resolution is usually good. However, if you have a 50-inch screen or larger, you will probably need a higher resolution, like 1080p.
1080p resolution has 1920 by 1080 pixels, which offers great picture quality. Full HD really shines on video game consoles, Blu-ray discs, and HD cable or satellite channels. But 4K resolution takes it one step up with even higher resolution for a clearer, sharper picture: 3840 by 2160 pixels.
But if you’re looking for the very best picture quality currently available on the market, 8K UHD is where you’ll find it. Samsung 8K TVs have 16x the number of pixels as Full HD TVs. They also have amazing upscaling capability, which means they can take a lower quality image and raise the quality up to 8K resolution .
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## How Does TV Sizing Work
Before you decide what size TV is best for you, lets go over what it means to have a 42-inch screen compared to a 65-inch. Its clear that a 65-inch TV will be larger, but its not immediately obvious how much bigger it will be.
The size of a screen, be it a smartphone, TV, or PC monitor is measured diagonally from corner to corner.
Most modern TVs use a 16:9 aspect ratio: for every 16-inches they measure horizontally they measure 9-inches vertically. As such, you can reasonably expect the screen size of TVs to measure:
• 42-inch: 36.6-inch x 20.6-inch
• 48-inch: 41.8-inch x 23.5-inch
• 50-inch: 43.6-inch x 24.5-inch
• 55-inch: 47.9-inch x 27.0-inch
• 65-inch: 56.7-inch x 31.9-inch
• 75-inch: 65.4-inch 36.8-inch
• 85-inch: 74.1-inch x 41.7-inch
Take note that this is just for the actual screen, the frame around it and the stands size and shape will vary for each individual TV model. Be sure to check your dream TVs actual size before you head to the checkout you dont want to bring home a screen thats too large for the space you have.
## So What Size TV Do I Need
TVs Explained: How To Choose The Perfect TV Size For Your Lounge Room
Whether youre getting a TV to watch the current movies, the latest sporting events, streaming your favorite series, or as a gaming TV, the short answer is as big as you can afford.
The long answer is a little more complicated and depends on the room, whether the display is mounted on a wall or not, the seating distance, the viewing angles, and the resolution of the display.
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## Importance Of Inches In Technology & Engineering
While the pure conversion from inches to centimeters is based on a factor of 2.54, some standardized technical variables differ from the pure conversion factor.
Here are a few examples:
Sound RecordingsDisplay Screens
• 15 inch corresponds to 381 mm, in practice usually 40 cm
• 19 inch = 482.6 mm but is also 50 cm screen size
• 40 screen size is 101.6 cm, the device has a screen diagonal of 102 centimeters.
Such deviations are common and reflect technical standards. Inch sizes are also used for tires and wheels, tube diameters, handlebar stem lengths, and many other technical sizes. Deviations in the millimeter range are quite common when harmonizing the American system in inches and the European metric system. | 3,569 | 16,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | longest | en | 0.89734 |
http://www.panneaux-multilignes.fr/a36-steel/mauritius_double_tan_14.html | 1,619,147,454,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039626288.96/warc/CC-MAIN-20210423011010-20210423041010-00150.warc.gz | 155,038,303 | 13,360 | • # mauritius double tank building volume
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### NFPA 30-2008: Basic Requirements for Storage Tanks
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### Single Wall vs. Double Wall Fuel Tanks Which Do I Need?
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### Tank Volume Calculator - Above Ground Fuel Storage Tanks
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Tags: | 2,862 | 13,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | latest | en | 0.826951 |
https://research.stlouisfed.org/fred2/series/M0269BUSM382SNBR | 1,454,975,338,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701154682.35/warc/CC-MAIN-20160205193914-00337-ip-10-236-182-209.ec2.internal.warc.gz | 794,708,854 | 19,006 | # Private Construction (Constant Dollars) for United States
1963-12: 41,353 Millions of 1957-1959 Dollars
Monthly, Seasonally Adjusted, M0269BUSM382SNBR, Updated: 2012-08-16 11:31 AM CDT
1yr | 5yr | 10yr | Max
Series Has Been Decomposed Into 3 Variables: (1) Original Data, 1915-46; (2) Original Data, 1946-63; (3) Seasonally Adjusted Data, 1946-63. Source: See Introductory Notes On Bls-Commerce Dept. Data.
This NBER data series m02169b appears on the NBER website in Chapter 2 at http://www.nber.org/databases/macrohistory/contents/chapter02.html.
NBER Indicator: m02169b
Release: NBER Macrohistory Database
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Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
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``` National Bureau of Economic Research, Private Construction (Constant Dollars) for United States [M0269BUSM382SNBR], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/M0269BUSM382SNBR, February 8, 2016. ```
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Graph updated. | 539 | 1,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-07 | latest | en | 0.758659 |
https://www.ig.com/ae/trading-strategies/gold-to-silver-ratio-trading-strategies-190208 | 1,632,828,966,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060677.55/warc/CC-MAIN-20210928092646-20210928122646-00083.warc.gz | 822,988,801 | 47,830 | CFDs are a leveraged product and can result in losses that exceed deposits. Please ensure you fully understand how CFDs work and what their risks are, and take care to manage your exposure. CFDs are a leveraged product and can result in losses that exceed deposits. Please ensure you fully understand how CFDs work and what their risks are, and take care to manage your exposure.
# Gold to silver ratio trading strategies
We explain what the gold to silver ratio is and how to use it as part of your trading strategy.
## What is the gold to silver ratio?
The gold to silver ratio represents the value of gold relative to silver and can provide insight into how the price of each metal could move against the other.
## How to calculate the gold to silver ratio
Calculating the gold to silver ratio is simple: it is the price of gold divided by the price of silver. This represents how many ounces of silver you would have to sell to buy one ounce of gold, or how many silver ounces you could buy with the money made selling one ounce of gold.
The ratio generally remains between 10 to 100, but technically there is no reason why it could not go higher or lower than those parameters.
As of 7 February 2019, gold was trading at roughly \$1.306 per ounce while the price of silver was just below \$16.00. This calculates to a gold to silver ratio of 82 (1306/16).
## What affects the gold to silver ratio?
The gold to silver ratio is dictated by the price of each metal and how fast they are rising or falling compared to the other. The price of the two metals tend to correlate most of the time so changes are often a result of the price of one metal increasing or decreasing at a faster rate:
• If the price of gold is rising at a faster rate than silver, then the ratio will increase
• If the price of silver is rising at a faster rate than gold, then the ratio will decrease
• If the price of gold is falling at a faster rate than silver, then the ratio will decrease
• If the price of silver is falling at a faster rate than gold, then the ratio will increase
While some factors can impact the price of both metals, such as movements in the US dollar, it is important to stress that the prices of gold and silver are determined and driven by their own factors.
Gold is purchased by four different markets: jewellery, industrial applications, central banks and investors. Gold holds a unique role in commodity markets as a ‘safe haven’ asset (an ability to retain its value) that investors flock to when other financial markets – such as forex or stocks – are volatile.
Investor sentiment is therefore a major driver behind the price of gold, rising in times of uncertainty as people look to buy the metal as a form of insurance to hedge against potential declines in other parts of their investment portfolio. This is less true for the price of silver, which has more industrial applications than its sister metal. For this reason, some use the gold to silver ratio to test investor sentiment: if the ratio is high then it suggests the gold price has strengthened at a faster rate than silver to imply investors have become more risk-averse, while a low ratio could suggest interest in gold is lower to imply investors feel confident about the growth and stability of the world economy.
There are a variety of variables that affect the price of gold and silver, and therefore the ratio between the two. This is demonstrated by changes to the ratio over different eras.
## Gold to silver ratio chart: a brief history
The historic highs and lows of the ratio can often be attributed by major events. For example, the end of WW1 brought about peace and revived economic growth to push the ratio to below 18, but then ratcheted up in the lead up to WW2 to over 97 as investors flocked to gold as a safe haven.
More recently, the ratio hit its all-time low of less than 15 in 1979 when US interest rates were hiked to 21% to counter rampant inflation (higher interest rates pushes down the price of gold). It then went on to hit an all-time high of 99 as a global recession began to take hold in the early 1990s.
In the most recent financial crash, the gold to silver ratio moved from a peak of 80 in late 2008 to 32 in early 2011.
## How to trade the gold to silver ratio
But how do you use the ratio in trading? In a nutshell, if the ratio is outside the average then it can signal an opportunity to trade gold and/or silver with a view of profiting when it returns to ‘normal’ levels.
There is no definition of a ‘normal’ level, but historic data does paint some parameters for investors to use:
With this in mind, today’s ratio of above 80 is high compared to average levels seen over the last 50 or so years. The idea is that a high ratio means the value of gold is relatively expensive compared to silver, with a low ratio implying the opposite. This means investors tend to buy gold when the ratio is low and silver when it is high.
Although the ratio can experience volatile movements of 10+ points in a month, the drastic swings from the highs to the lows tend to occur over longer timeframes. The ratio has rarely been above 80 or below 50 over more recent decades, and therefore these levels are potential turning points when the trend could begin to reverse.
If the ratio is trending, and gold and silver prices are trending, then the trading signal is:
Upward Upward Buy gold Upward Downward Sell silver Downward Upward Buy silver Downward Downward Sell gold
Equally, there are opportunities for investors when the ratio reaches new highs or lows:
If the ratio is at a certain level, and gold and silver prices are trending, then the trading signal is:
Historic high Upward Long silver Historic high Downward Short gold Historic low Downward Short silver Historic low Upward Long gold
### Gold to silver ratio trading example
One investor notices the ratio is at 80, high compared to the historic average. They decide to sell one ounce of gold and purchase 80 ounces of silver. A few years later, the ratio has fallen to 20, low compared to the historic average, and the investor can trade their 80 ounces of silver for four ounces of gold, quadrupling their original investment. Repeating this would build up an investor’s overall inventory of gold and silver as they switch from one to another depending on whether the ratio is high or low.
## Conclusion: gold to silver ratio is a valuable tool for investors
The gold to silver ratio is a representation of the value of each metal relative to the other that helps identify when either one may be undervalued or overvalued. It is most effective for longer-term strategies due to the stability of prices compared to other instruments.
The ratio changes when gold and silver prices move in different directions or, more commonly, when one rises or falls at a faster rate than the other. A high ratio suggests gold is expensive relative to silver, while a low ratio implies the opposite.
The ratio can also be used to gauge the market’s sentiment about the wider picture, making it a valuable tool that can be used beyond trading in precious metals. A high ratio has previously preluded economic downturns and emerges, as it has now, when global growth is slowing and uncertainty is high. A low ratio is often seen during times of peace and prosperity.
Publication date : 2019-02-08T12:55:56+0000
This information has been prepared by IG, a trading name of IG Limited. In addition to the disclaimer below, the material on this page does not contain a record of our trading prices, or an offer of, or solicitation for, a transaction in any financial instrument. IG accepts no responsibility for any use that may be made of these comments and for any consequences that result. No representation or warranty is given as to the accuracy or completeness of this information. Consequently any person acting on it does so entirely at their own risk. Any research provided does not have regard to the specific investment objectives, financial situation and needs of any specific person who may receive it. It has not been prepared in accordance with legal requirements designed to promote the independence of investment research and as such is considered to be a marketing communication. Although we are not specifically constrained from dealing ahead of our recommendations we do not seek to take advantage of them before they are provided to our clients.
CFDs are a leveraged products. CFD trading may not be suitable for everyone and can result in losses that exceed your initial deposit, so please ensure that you fully understand the risks involved.
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Stay on top of upcoming market-moving events with our customisable economic calendar. | 1,850 | 8,939 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-39 | latest | en | 0.964689 |
https://fr.mathworks.com/matlabcentral/cody/problems/1190-golomb-s-self-describing-sequence-based-on-euler-341/solutions/1507649 | 1,575,865,493,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517557.43/warc/CC-MAIN-20191209041847-20191209065847-00275.warc.gz | 386,268,705 | 15,874 | Cody
# Problem 1190. Golomb's self-describing sequence (based on Euler 341)
Solution 1507649
Submitted on 28 Apr 2018 by Daniel Meier
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
assert(isequal(euler341(1),1))
ans = 1
2 Pass
assert(isequal(euler341(10),5))
ans = 5
3 Pass
assert(isequal(euler341(310),42))
ans = 42
4 Pass
assert(isequal(euler341(4242),210))
ans = 210
5 Pass
assert(isequal(euler341(328509),3084))
ans = 3084
6 Pass
assert(isequal(euler341(551368),4247))
ans = 4247
7 Pass
assert(isequal(euler341(614125),4540))
ans = 4540 | 233 | 689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-51 | latest | en | 0.521246 |
https://definithing.com/define-tech/bonanzle/ | 1,675,113,773,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499829.29/warc/CC-MAIN-20230130201044-20230130231044-00742.warc.gz | 212,465,953 | 7,381 | Bonanzle
The name of an online eBay alternative marketplace that aims to make selling online easy for its users, and to offer unique items to buyers. The site offers sellers an easy-to-use listing process, eBay and other marketplace item importers, real-time chat options for conducting sales, and an online image tool for uploading and manipulating product shots. Bonanzle is also designed to let its sellers display items in a booth, which is their own version of an online store. In a way, it is like the storefronts you find on eBay and other eBay alternatives. Bonanzle was founded by Seattle-based Bill Harding, and the site first launched in beta in June of 2008 and launched out of beta at the end of August 2008.
• Boolean expression
An expression that results in a value of either TRUE or FALSE. For example, the expression 2 < 5 (2 is less than 5) is a Boolean expression because the result is TRUE. All expressions that contain relational operators , such as the less than sign (
• Boolean Logic
(bool´ē-&n loj´ik) (n.) Named after the nineteenth-century mathematician George Boole, Boolean logic is a form of algebra in which all values are reduced to either TRUE or FALSE. Boolean logic is especially important for computer science because it fits nicely with the binary numbering system, in which each bit has a value of either 1 […]
• Boolean Operator
There are five Boolean operators that can be used to manipulate TRUE/FALSE values. These operators have the following meanings, where x and y represent values of TRUE or FALSE. The OR operator is often called an inclusive OR, whereas XOR is an exclusive OR. Boolean operators are used widely in programming and also in forming […]
• Boolean Search
A type of search allowing users to combine keywords with operators such as AND, NOT and OR to further produce more relevant results. For example, a Boolean search could be “hotel” AND “New York”. This would limit the search results to only those documents containing the two keywords.
• Boot Configuration Data
Abbreviated as BCD, Boot Configuration Data is used in Windows Vista in place of the boot.ini system file that is found in earlier versions of the Windows operating system.
Disclaimer: Bonanzle definition / meaning should not be considered complete, up to date, and is not intended to be used in place of a visit, consultation, or advice of a legal, medical, or any other professional. All content on this website is for informational purposes only. | 514 | 2,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.926991 |
https://oeis.org/A097636/internal | 1,632,793,707,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058589.72/warc/CC-MAIN-20210928002254-20210928032254-00329.warc.gz | 456,483,224 | 2,936 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A097636 Odd composites of the form A024675(k) or 1+A024675(k), any k. 0
%I
%S 9,15,21,27,35,39,45,51,57,65,69,77,81,87,93,99,105,111,121,129,135,
%T 145,155,161,165,171,177,187,195,205,217,225,231,237,247,255,261,267,
%U 275,279,289,301,309,315,325,335,343,351,357,363,371,377,381,387,393,399
%N Odd composites of the form A024675(k) or 1+A024675(k), any k.
%C Start from an auxiliary sequence of interprimes, incremented by 1 if even: 5, 7, 9, 13, 15, 19, 21, 27, 31,... Removing the primes from this auxiliary sequence of odd numbers yields the current sequence.
%t digits=200 (* find odd numbers between the primes nearest the average*) a=Table[If[(Mod[(Prime[n+1]+Prime[n])/2, 2]==1)&&( PrimeQ[(Prime[n+1]+Prime[n])/2]==False)&&( PrimeQ[(Prime[n+1]+Prime[n])/2+1]==False), (Prime[n+1]+Prime[n])/2, (Prime[n+1]+Prime[n])/2+1], {n, 2, digits}] b=Table[Prime[n], {n, 1, digits}] (* eliminate the primes from the sequence *) aa=Complement[a, b]
%K nonn,easy,less
%O 1,1
%A _Roger L. Bagula_, Sep 20 2004
%E Definition clarified by the Assoc. Eds. of the OEIS, Jul 13 2010
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Last modified September 27 21:12 EDT 2021. Contains 347698 sequences. (Running on oeis4.) | 550 | 1,602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-39 | latest | en | 0.687572 |
https://www.quizzes.cc/metric/percentof.php?percent=388&of=750000 | 1,591,144,913,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347426956.82/warc/CC-MAIN-20200602224517-20200603014517-00454.warc.gz | 870,341,731 | 4,248 | #### What is 388 percent of 750,000?
How much is 388 percent of 750000? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 388% of 750000 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 388% of 750,000 = 2910000
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
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divided by
Calculating three hundred and eighty-eight of seven hundred and fifty thousand How to calculate 388% of 750000? Simply divide the percent by 100 and multiply by the number. For example, 388 /100 x 750000 = 2910000 or 3.88 x 750000 = 2910000
#### How much is 388 percent of the following numbers?
388% of 750000.01 = 291000003.88 388% of 750000.02 = 291000007.76 388% of 750000.03 = 291000011.64 388% of 750000.04 = 291000015.52 388% of 750000.05 = 291000019.4 388% of 750000.06 = 291000023.28 388% of 750000.07 = 291000027.16 388% of 750000.08 = 291000031.04 388% of 750000.09 = 291000034.92 388% of 750000.1 = 291000038.8 388% of 750000.11 = 291000042.68 388% of 750000.12 = 291000046.56 388% of 750000.13 = 291000050.44 388% of 750000.14 = 291000054.32 388% of 750000.15 = 291000058.2 388% of 750000.16 = 291000062.08 388% of 750000.17 = 291000065.96 388% of 750000.18 = 291000069.84 388% of 750000.19 = 291000073.72 388% of 750000.2 = 291000077.6 388% of 750000.21 = 291000081.48 388% of 750000.22 = 291000085.36 388% of 750000.23 = 291000089.24 388% of 750000.24 = 291000093.12 388% of 750000.25 = 291000097
388% of 750000.26 = 291000100.88 388% of 750000.27 = 291000104.76 388% of 750000.28 = 291000108.64 388% of 750000.29 = 291000112.52 388% of 750000.3 = 291000116.4 388% of 750000.31 = 291000120.28 388% of 750000.32 = 291000124.16 388% of 750000.33 = 291000128.04 388% of 750000.34 = 291000131.92 388% of 750000.35 = 291000135.8 388% of 750000.36 = 291000139.68 388% of 750000.37 = 291000143.56 388% of 750000.38 = 291000147.44 388% of 750000.39 = 291000151.32 388% of 750000.4 = 291000155.2 388% of 750000.41 = 291000159.08 388% of 750000.42 = 291000162.96 388% of 750000.43 = 291000166.84 388% of 750000.44 = 291000170.72 388% of 750000.45 = 291000174.6 388% of 750000.46 = 291000178.48 388% of 750000.47 = 291000182.36 388% of 750000.48 = 291000186.24 388% of 750000.49 = 291000190.12 388% of 750000.5 = 291000194
388% of 750000.51 = 291000197.88 388% of 750000.52 = 291000201.76 388% of 750000.53 = 291000205.64 388% of 750000.54 = 291000209.52 388% of 750000.55 = 291000213.4 388% of 750000.56 = 291000217.28 388% of 750000.57 = 291000221.16 388% of 750000.58 = 291000225.04 388% of 750000.59 = 291000228.92 388% of 750000.6 = 291000232.8 388% of 750000.61 = 291000236.68 388% of 750000.62 = 291000240.56 388% of 750000.63 = 291000244.44 388% of 750000.64 = 291000248.32 388% of 750000.65 = 291000252.2 388% of 750000.66 = 291000256.08 388% of 750000.67 = 291000259.96 388% of 750000.68 = 291000263.84 388% of 750000.69 = 291000267.72 388% of 750000.7 = 291000271.6 388% of 750000.71 = 291000275.48 388% of 750000.72 = 291000279.36 388% of 750000.73 = 291000283.24 388% of 750000.74 = 291000287.12 388% of 750000.75 = 291000291
388% of 750000.76 = 291000294.88 388% of 750000.77 = 291000298.76 388% of 750000.78 = 291000302.64 388% of 750000.79 = 291000306.52 388% of 750000.8 = 291000310.4 388% of 750000.81 = 291000314.28 388% of 750000.82 = 291000318.16 388% of 750000.83 = 291000322.04 388% of 750000.84 = 291000325.92 388% of 750000.85 = 291000329.8 388% of 750000.86 = 291000333.68 388% of 750000.87 = 291000337.56 388% of 750000.88 = 291000341.44 388% of 750000.89 = 291000345.32 388% of 750000.9 = 291000349.2 388% of 750000.91 = 291000353.08 388% of 750000.92 = 291000356.96 388% of 750000.93 = 291000360.84 388% of 750000.94 = 291000364.72 388% of 750000.95 = 291000368.6 388% of 750000.96 = 291000372.48 388% of 750000.97 = 291000376.36 388% of 750000.98 = 291000380.24 388% of 750000.99 = 291000384.12 388% of 750001 = 291000388
1% of 750000 = 7500 2% of 750000 = 15000 3% of 750000 = 22500 4% of 750000 = 30000 5% of 750000 = 37500 6% of 750000 = 45000 7% of 750000 = 52500 8% of 750000 = 60000 9% of 750000 = 67500 10% of 750000 = 75000 11% of 750000 = 82500 12% of 750000 = 90000 13% of 750000 = 97500 14% of 750000 = 105000 15% of 750000 = 112500 16% of 750000 = 120000 17% of 750000 = 127500 18% of 750000 = 135000 19% of 750000 = 142500 20% of 750000 = 150000 21% of 750000 = 157500 22% of 750000 = 165000 23% of 750000 = 172500 24% of 750000 = 180000 25% of 750000 = 187500
26% of 750000 = 195000 27% of 750000 = 202500 28% of 750000 = 210000 29% of 750000 = 217500 30% of 750000 = 225000 31% of 750000 = 232500 32% of 750000 = 240000 33% of 750000 = 247500 34% of 750000 = 255000 35% of 750000 = 262500 36% of 750000 = 270000 37% of 750000 = 277500 38% of 750000 = 285000 39% of 750000 = 292500 40% of 750000 = 300000 41% of 750000 = 307500 42% of 750000 = 315000 43% of 750000 = 322500 44% of 750000 = 330000 45% of 750000 = 337500 46% of 750000 = 345000 47% of 750000 = 352500 48% of 750000 = 360000 49% of 750000 = 367500 50% of 750000 = 375000
51% of 750000 = 382500 52% of 750000 = 390000 53% of 750000 = 397500 54% of 750000 = 405000 55% of 750000 = 412500 56% of 750000 = 420000 57% of 750000 = 427500 58% of 750000 = 435000 59% of 750000 = 442500 60% of 750000 = 450000 61% of 750000 = 457500 62% of 750000 = 465000 63% of 750000 = 472500 64% of 750000 = 480000 65% of 750000 = 487500 66% of 750000 = 495000 67% of 750000 = 502500 68% of 750000 = 510000 69% of 750000 = 517500 70% of 750000 = 525000 71% of 750000 = 532500 72% of 750000 = 540000 73% of 750000 = 547500 74% of 750000 = 555000 75% of 750000 = 562500
76% of 750000 = 570000 77% of 750000 = 577500 78% of 750000 = 585000 79% of 750000 = 592500 80% of 750000 = 600000 81% of 750000 = 607500 82% of 750000 = 615000 83% of 750000 = 622500 84% of 750000 = 630000 85% of 750000 = 637500 86% of 750000 = 645000 87% of 750000 = 652500 88% of 750000 = 660000 89% of 750000 = 667500 90% of 750000 = 675000 91% of 750000 = 682500 92% of 750000 = 690000 93% of 750000 = 697500 94% of 750000 = 705000 95% of 750000 = 712500 96% of 750000 = 720000 97% of 750000 = 727500 98% of 750000 = 735000 99% of 750000 = 742500 100% of 750000 = 750000 | 2,900 | 6,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-24 | latest | en | 0.829402 |
http://openstudy.com/updates/4f7a5e8ee4b0884e12aa89b6 | 1,516,313,692,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887621.26/warc/CC-MAIN-20180118210638-20180118230638-00720.warc.gz | 282,525,079 | 8,653 | • anonymous
On Monday night, there were 69 unfilled seats in the theatre, and on Tuesday night there were 34 unfilled seats in the theatre. On Monday night, each ticket holder spent an average of $6.50 at the concession stand. On Tuesday night, each ticket holder spent an average of$4.00 at the concession stand. The profit made at the concession stand Monday night equaled the profit made Tuesday night. How many total seats are there in the theatre?
Mathematics
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SOLVED
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Not the answer you are looking for? Search for more explanations. | 344 | 1,382 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-05 | latest | en | 0.584658 |
http://www.numbersaplenty.com/32103 | 1,603,429,431,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880656.25/warc/CC-MAIN-20201023043931-20201023073931-00418.warc.gz | 160,994,735 | 3,554 | Search a number
32103 = 332941
BaseRepresentation
bin111110101100111
31122001000
413311213
52011403
6404343
7162411
oct76547
948030
1032103
1122135
12166b3
13117c6
14b9b1
1597a3
hex7d67
32103 has 16 divisors (see below), whose sum is σ = 50400. Its totient is φ = 20160.
The previous prime is 32099. The next prime is 32117. The reversal of 32103 is 30123.
Adding to 32103 its reverse (30123), we get a palindrome (62226).
It is a happy number.
It is not a de Polignac number, because 32103 - 22 = 32099 is a prime.
It is a Harshad number since it is a multiple of its sum of digits (9).
It is an alternating number because its digits alternate between odd and even.
It is a house number.
It is a congruent number.
It is an inconsummate number, since it does not exist a number n which divided by its sum of digits gives 32103.
It is not an unprimeable number, because it can be changed into a prime (32143) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (11) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 763 + ... + 803.
It is an arithmetic number, because the mean of its divisors is an integer number (3150).
232103 is an apocalyptic number.
32103 is a deficient number, since it is larger than the sum of its proper divisors (18297).
32103 is a wasteful number, since it uses less digits than its factorization.
32103 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 79 (or 73 counting only the distinct ones).
The product of its (nonzero) digits is 18, while the sum is 9.
The square root of 32103 is about 179.1731006597. The cubic root of 32103 is about 31.7820475387.
The spelling of 32103 in words is "thirty-two thousand, one hundred three". | 528 | 1,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-45 | latest | en | 0.904927 |
https://gmatclub.com/forum/barclays-tax-claims-article-77040.html | 1,496,137,428,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463614620.98/warc/CC-MAIN-20170530085905-20170530105905-00405.warc.gz | 969,149,802 | 49,817 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# Barclays tax claims (article)
Author Message
SVP
Joined: 04 Dec 2007
Posts: 1689
Schools: Kellogg '11
Followers: 15
Kudos [?]: 200 [0], given: 31
### Show Tags
26 Mar 2009, 10:33
Interesting:
Quote:
The sources say this management culture has been integral to creating what is in effect a tax avoidance factory at Barclays headquarters at Canary Wharf. Avoiding tax by exploiting legal loopholes was not seen as something questionable but rather the raison d'etre of the Structured Capital Markets department. "There's such a macho culture in SCM and the deals are so big you never say billion or million, you just say 16 bucks or 16 quid which meant billion," one source said. "The tax avoidance was so big it became the engine of growth for the whole of the investment banking arm."
VP
Joined: 09 Dec 2008
Posts: 1221
Schools: Kellogg Class of 2011
Followers: 21
Kudos [?]: 243 [0], given: 17
Re: Barclays tax claims (article) [#permalink]
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26 Mar 2009, 15:32
There is a huge difference between tax avoidance and tax evasion. Avoidance is using perfectly legal means to reduce tax liabilities (so-called loopholes), while evasion is a crime resulting from failure to pay taxes legally due. Just about every law firm and accounting firm will help clients avoid taxes while following the letter of the tax law.
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26 Mar 2009, 18:11
The argument is whether it is reasonable for the UK Government to use funds from HM Treasury to bail out Barclays, when Barclays' SCM desk had been trying its darndest (albeit legally) to make sure its clients' money did not reach the coffers of the Treasury. The same coffer that is now providing the werewithal for Barclays to "survive."
Jerz wrote:
There is a huge difference between tax avoidance and tax evasion. Avoidance is using perfectly legal means to reduce tax liabilities (so-called loopholes), while evasion is a crime resulting from failure to pay taxes legally due. Just about every law firm and accounting firm will help clients avoid taxes while following the letter of the tax law.
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27 Mar 2009, 04:29
I don't really see how that changes things. Either it's so critically important to the country for Barclays to survive that the government must provide bailout money to keep it in business or its not that important to the national interest and the government lets the market decide Barclays fate. I'm sure HM Government, in its infinite wisdom, took account of Barclays tax advisory practice when deciding whether saving Barclays was in the national interest </sarcasm>
I guess my opinion is colored by the fact that I find it just a bit disingenuous when politicians start wringing their hands over the bad corporations/rich people who use tax loopholes to reduce their tax liabilities. After all, it's the politicians who wrote the tax laws in the first place and put the loopholes there.
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27 Mar 2009, 07:57
solaris1 wrote:
The argument is whether it is reasonable for the UK Government to use funds from HM Treasury to bail out Barclays, when Barclays' SCM desk had been trying its darndest (albeit legally) to make sure its clients' money did not reach the coffers of the Treasury. The same coffer that is now providing the werewithal for Barclays to "survive."
Hey sol, but Barclay's hasn't taken any government funds yet. So the argument then seems unreasonable.
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27 Mar 2009, 08:32
Yes, that's true - I was not correct in claiming that Barclays is "surviving" thanks to the Treasury (RBS is).
However, jb32 - in my defense, my post above was not my personal judgment on what has been happening. It was merely an attempt at an explanation behind why this is turning out to be such an issue in the UK.
Clearly, the public across the pond too is vilifying every financial institution as being somehow directly responsible for this crisis. They see no difference between say Barclays, or Lloyds or RBS. Just that the government is bailing out banks that are trying to "cheat" the Treasury of revenues it deserves.
jb32 wrote:
Hey sol, but Barclay's hasn't taken any government funds yet. So the argument then seems unreasonable.
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27 Mar 2009, 09:02
I apologize for not actually reading the article before my prior post, but now that I have - WHAT A LOAD OF COMPLETE HORSESQUEEZE!!!! I mean really, people were playing poker and drinking red wine at lunch!!! Oh the humanity!!! What is this world coming to when guys get together and eat steak and drink red wine?!?!?!? Dear god, that is the most laughable bit of crap I have ever read. I only wish I could have worked at that company. I bet they also had this in the back room: http://www.youtube.com/watch?v=S1ZZreXEqSY. Just wait till the public finds out about the Heineken closet.
Here were some of my favorite quotes with my commentary below it:
Quote:
On several occasions, the sources allege, staff have come in to work to find an email from Jenkins saying a colleague no longer worked there.
You mean people were fired and it wasn't announced beforehand? I mean what kind of business practices are these animals following?!?!?!
Quote:
According to one whistleblower, a secretary was fired by another executive for booking a taxi that was a Volvo rather than an S class Mercedes.
I feel like I'm the last guy in a huge game of telephone, where what actually happened is so different than what The Gaurdian reported.
Quote:
The most lurid tales come from team-building away-days. Free cigars and champagne would apparently flow and at night poker tables would be set up. "There's a babies' table and a big boys' table. The big boys' table was all about showing who has got the biggest dick.
So they played poker? Wow, that is so outrageous. And then there was an element of machoism because guys were competing against each other? Geez, that's never happened before... I do have to question that 'Champagne' was flowing. No guy in a dick measuring contest would be drinking champagne. Period.
Quote:
Abrahams, according to one whistleblower, was also responsible for running projects that were kept completely under wraps, even from other members of the team.
And to think that not every deal was made public to every team memeber?!?!? Well, I guess they should have thought to pull the janitors and secretaries into the strategy meetings. Whoops!!
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28 Mar 2009, 08:16
Interesting post
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28 Mar 2009, 19:14
jb32,
Re: Barclays tax claims (article) [#permalink] 28 Mar 2009, 19:14
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Nov 4 answered Show that if $x>0$, then $\ln(x)\geq 1-\frac{1}{x}$ Nov 2 comment Integration in n-spherical coordinates Thanks, I've already got an answer on physics.SE: physics.stackexchange.com/questions/83103/… Oct 31 asked Integration in n-spherical coordinates Oct 13 comment Divergence in spherical coordinates @user8268 So if $\vec F$ is defined as: $F^\alpha = \frac{\partial f (u(r, \theta, \phi), \ldots)}{\partial (\nabla_\alpha u)}$ and I compute $\nabla_\alpha u$ as $(\partial_r u, \frac{1}{r} \partial_\theta u, \frac{1}{r \sin \theta} \partial_\phi u)$ than which formula for $\nabla \cdot \vec{F}$ should I use? The first one? Oct 13 asked Divergence in spherical coordinates Aug 9 awarded Necromancer Jun 26 comment Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$? @André indeed, thanks. Jun 26 revised Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$? fixed typo May 24 comment Improper integral and special functions @RaymondManzoni thank you. May 24 accepted Improper integral and special functions May 24 asked Improper integral and special functions May 17 awarded Yearling May 14 awarded Nice Answer Feb 8 awarded Nice Answer Sep 16 awarded Revival Sep 16 answered Find the functions family that satisfies the inequality $\int_0^1 \frac{dx}{1+f^{2}(x)} <\frac{f(1)}{f'{(1)}}$ Aug 20 answered Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$? Jul 26 comment Sum equals integral Thank you for your answer. These references are very helpful. Jul 26 accepted Sum equals integral Jul 26 awarded Nice Question | 508 | 1,700 | {"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} | 3.28125 | 3 | CC-MAIN-2016-18 | latest | en | 0.798429 |
http://hackage.haskell.org/packages/archive/lambda-canvas/0.1/doc/html/Graphics-LambdaCanvas.html | 1,369,524,100,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706470784/warc/CC-MAIN-20130516121430-00019-ip-10-60-113-184.ec2.internal.warc.gz | 123,211,231 | 3,121 | lambda-canvas-0.1: Educational drawing canvas for FP explorers.
Graphics.LambdaCanvas
Synopsis
# Drawing types
data PrimitiveMode
Specification of the way the vertices given during `renderPrimitive` are interpreted. In the description of the constructors, n is an integer count starting at one, and N is the total number of vertices specified.
Constructors
Points Treats each vertex as a single point. Vertex n defines point n. N points are drawn. Lines Treats each pair of vertices as an independent line segment. Vertices 2n-1 and 2n define line n. N/2 lines are drawn. LineLoop Draws a connected group of line segments from the first vertex to the last, then back to the first. Vertices n and n+1 define line n. The last line, however, is defined by vertices N and 1. N lines are drawn. LineStrip Draws a connected group of line segments from the first vertex to the last. Vertices n and n+1 define line n. N-1 lines are drawn. Triangles Treats each triplet of vertices as an independent triangle. Vertices 3n-2, 3n-1, and 3n define triangle n. N\3/ triangles are drawn. TriangleStrip Draws a connected group of triangles. One triangle is defined for each vertex presented after the first two vertices. For odd n, vertices n, n+1, and n+2 define triangle n. For even n, vertices n+1, n, and n+2 define triangle n. N-2 triangles are drawn. TriangleFan Draws a connected group of triangles. One triangle is defined for each vertex presented after the first two vertices. Vertices 1, n+1, and n+2 define triangle n. N-2 triangles are drawn. Quads Treats each group of four vertices as an independent quadrilateral. Vertices 4n-3, 4n-2, 4n-1, and 4n define quadrilateral n. N/4 quadrilaterals are drawn. QuadStrip Draws a connected group of quadrilaterals. One quadrilateral is defined for each pair of vertices presented after the first pair. Vertices 2n-1, 2n, 2n+2, and 2n+1 define quadrilateral n. N/2-1 quadrilaterals are drawn. Note that the order in which vertices are used to construct a quadrilateral from strip data is different from that used with independent data. Polygon Draws a single, convex polygon. Vertices 1 through N define this polygon.
Instances
Eq PrimitiveMode Ord PrimitiveMode Show PrimitiveMode
# Drawing functions
animate :: String -> (POSIXTime -> IO ()) -> IO ()Source
Create a canvas suitable for animation. The drawing block is invoked on each redisplay and is passed the current timestamp.
draw :: String -> IO () -> IO ()Source
Create a canvas suitable for drawing a static picture. The drawing block is invoked on each redisplay.
put :: PrimitiveMode -> [Point] -> IO ()Source
Draw a primitive from a set of a vertices on screen.
# Utility functions
stepOf :: Num a => Integer -> POSIXTime -> aSource
Given an amount of steps and a timestamp, calculate current step number. Step numbers begin with 0. | 693 | 2,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2013-20 | latest | en | 0.892431 |
http://www.reference.com/browse/not%20absolutely | 1,435,700,623,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375094501.77/warc/CC-MAIN-20150627031814-00066-ip-10-179-60-89.ec2.internal.warc.gz | 667,580,575 | 16,539 | Definitions
# Absolutely irreducible
In mathematics, absolutely irreducible is a term applied to linear representations or algebraic varieties over a field. It means that the object in question remains irreducible, even after any finite extension of the field of coefficients. In both cases, being absolutely irreducible is the same as being irreducible over the algebraic closure of the ground field.
## Examples
• The irreducible two-dimensional representation of the symmetric group S3 of order 6, originally defined over the field of rational numbers, is absolutely irreducible.
• The representation of the circle group by rotations in the plane is irreducible (over the field of real numbers), but is not absolutely irreducible. After extending the field to complex numbers, it splits into two irreducible components. This is to be expected, since the circle group is commutative and it is known that all irreducible representations of commutative groups over an algebraically closed field are one-dimensional.
• The real algebraic variety defined by the equation
$x^2 + y^2 = 1$
is absolutely irreducible. It is the ordinary circle over the reals and remains an irreducible conic section over the field of complex numbers. Absolute irreducibility more generally holds over any field not of characteristic two. In characteristic two, the equation is equivalent to (x + y −1)2 = 0. Hence it defines the double line x + y =1, which is a non-reduced scheme.
• The algebraic variety given by the equation
$x^2 + y^2 = 0$
is not absolutely irreducible. Indeed, the left hand side can be factored as
$x^2 + y^2 = \left(x+yi\right)\left(x-yi\right),$ where $i$ is a square root of −1.
Therefore, this algebraic variety consists of two lines intersecting at the origin and is not absolutely irreducible. This holds either already over the ground field, if −1 is a square, or over the quadratic extension obtained by adjoining i.
## References
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Design your own scoring system and play Trumps with these Olympic Sport cards.
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One day five small animals in my garden were going to have a sports day. They decided to have a swimming race, a running race, a high jump and a long jump.
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This activity challenges you to decide on the 'best' number to use in each statement. You may need to do some estimating, some calculating and some research.
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### Observing the Sun and the Moon
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How does the time of dawn and dusk vary? What about the Moon, how does that change from night to night? Is the Sun always the same? Gather data to help you explore these questions. | 642 | 2,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2019-39 | latest | en | 0.905405 |
http://math.stackexchange.com/questions/81796/a-unitary-matrix-with-positively-definite-matrices-truly-behaving-as-unit-under | 1,469,556,068,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825048.60/warc/CC-MAIN-20160723071025-00217-ip-10-185-27-174.ec2.internal.warc.gz | 160,942,209 | 17,467 | # A unitary Matrix with positively definite matrices, truly behaving as unit under multiplicity?
Let $A, B$ positively definite matrices, and $Q$ a unitary matrix. Prove that if $A = BQ$, then $A = B$.
The idea I've had - show that all eigenvalues of $Q$ equal to $1$, and therefore $Q = I_{n}$, but I'm not sure this direction is even correct.
-
Let's see... we want that $\mathbf B\mathbf Q=(\mathbf B\mathbf Q)^\ast$, no? We then have $\mathbf B\mathbf Q=\mathbf Q^\ast\mathbf B$, or $\mathbf Q\mathbf B\mathbf Q=\mathbf B$. Then... – J. M. Nov 13 '11 at 22:19
According to Wikipedia, "there is no agreement in the literature on the proper definition of positive-definite for non-Hermitian matrices.". If we merely require $\Re(x^*Ax)\gt0$, your statement doesn't hold, since $B=I$ and $A=Q$ a two-dimensional rotation matrix with small rotation angle would be a counterexample. So I'll assume that for $M$ to be positive-definite you're requiring $x^*Mx$ to be always real and positive, which implies that $M$ is Hermitian.
So $A$ and $B$ are Hermitian. As a unitary matrix, $Q$ is diagonalized by a unitary matrix $U$. Transforming $A$ and $B$ by $U$ still leaves them Hermitian, and the transformed matrices satisfy $A'=B'Q'$, with $Q'$ diagonal and unitary. Since $A'$ and $B'$ are positive definite, their diagonal elements are positive. But their ratios are the diagonal elements of $Q'$, so $Q'$, and hence $Q$, has to be the unit matrix. | 425 | 1,452 | {"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} | 3.8125 | 4 | CC-MAIN-2016-30 | latest | en | 0.880217 |
http://www.stata.com/statalist/archive/2008-11/msg00433.html | 1,441,018,709,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065954.23/warc/CC-MAIN-20150827025425-00348-ip-10-171-96-226.ec2.internal.warc.gz | 718,257,075 | 3,545 | # Re: st: Chi-square test with probabilities
From Christoph Merkle To statalist@hsphsun2.harvard.edu Subject Re: st: Chi-square test with probabilities Date Tue, 11 Nov 2008 20:11:40 +0100
```I have to clarify this a bit:
```
The probabilities indeed add up to one, participants were made aware of to obey this rule. And if there were still some errors I corrected for those. They actually had to estimate with which probability a value falls in a certain quartile/decile of a distribution.
```
```
My claim is not 'humans overestimate the chances of rare events happening' but rather 'humans violate the rules of Bayesian updating". Therefore ttests for the mean guessed probability compared to expected probability for quartiles/deciles one by one are fine for a start (and I calculated them already), but better would be to analyze the whole distribution.
```That is why I proposed a chi square goodness of fit test.
```
I'm not an expert in repeated measure design, but I don't think it helps me out here.
```
Zitat von Ronan Conroy <rconroy@rcsi.ie>:
```
```On 11 Nov 2008, at 17:19, Christoph Merkle wrote:
```
```Actually I'm only interested if the mean of these peobabilities over
participants is different from hyposized proportions. If I use a
simple ttest I can only test each of the variables one by one. But I
want to test the distribution over the ten
```
```
If the events are mutually exclusive and collectively exhaustive, then
the probabilities ought to add up to 1, but I fear that they won't.
I think you are better testing one-by-one, using a t-test to test the
hypothesis that the mean guessed probability is the same as the actual
value, unless you have a hypothesis that is independent of the
probability being guessed (such as 'humans overestimate the chances of
rare events happening') in which case, I would treat it as a repeated
measures design.
Ronan Conroy
=================================
rconroy@rcsi.ie
Royal College of Surgeons in Ireland
Epidemiology Department,
Beaux Lane House, Dublin 2, Ireland
+353 (0)1 402 2431
+353 (0)87 799 97 95
+353 (0)1 402 2764 (Fax - remember them?)
http://www.flickr.com/photos/ronanconroy/sets/72157601895416740/
P Before printing, think about the environment
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
```
```
Christoph Merkle
Center for Doctoral Studies in Business
GESS - University of Mannheim
Schloss, Room M 114
68161 Mannheim
Germany
Phone: +49-621-1813774
E-mail: chmerkle@rumms.uni-mannheim.de
*
* For searches and help try:
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``` | 731 | 2,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-35 | longest | en | 0.937012 |
https://the-equivalent.com/equivalent-resistance-in-parallel-circuit/ | 1,685,779,955,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649177.24/warc/CC-MAIN-20230603064842-20230603094842-00187.warc.gz | 608,617,896 | 16,303 | # Equivalent resistance in parallel circuit
Parallel Resistor Equation
If the two resistances or impedances in parallel are equal and of the same value, then the total or equivalent resistance, RT is equal to half the value of one resistor. That is equal to R/2 and for three equal resistors in parallel, R/3, etc.
## How do you calculate the resistance of a parallel circuit?
Method 3 Method 3 of 4: Combination Circuit
1. Break down your circuit into series sections and parallel sections. …
2. Find the resistance of each parallel section. The example circuit has two branches with resistance R 1 = 5 Ω and R 2 = 3 Ω.
4. Add up resistances in series. …
5. Use Ohm’s Law to find unknown values. …
## How to determine equivalent resistance?
a) Determination of the equivalent resistance of two resistors when connected in parallel.
• Aim. To determine the equivalent resistance of two resistors when connected in parallel.
• Theory. If the resistors are connected in parallel along with a battery, then the total current I is calculated as a sum of the separate value of current through each …
• Materials Required
• Procedure. …
• Observation Table
• Result. …
• Precautions. …
## Is voltage the same across parallel resistors?
The voltage across each resistor within a parallel combination is exactly the same but the currents flowing through them are not the same as this is determined by their resistance value and Ohms Law. Then parallel circuits are current dividers.
## What is the formula for two resistors in parallel?
What is the formula for two resistors in parallel? The sum of the currents through each path is equal to the total current that flows from the source. You can find total resistance in a Parallel circuit with the following formula: 1/Rt = 1/R1 + 1/R2 + 1/R3 + If one of the parallel paths is broken, current will continue to flow in all the other …
## How do you find equivalent resistance in a parallel circuit?
How do you calculate two resistors in parallel? Take their reciprocal values, add the two together and take the reciprocal again. For example, if one resistor is 2 Ω and the other is 4 Ω, then the calculation to find the equivalent resistance is 1 / (1/2 + 1/4) = 1 / (3/4) = 4/3 = 1.33.
## What is equivalent resistance in parallel?
The equivalent resistance is where the aggregate resistance connected either in parallel or series is calculated. Essentially, the circuit is designed either in Series or Parallel. Electrical resistance shows how much energy one needs when you move the charges/current through your devices.
## What is equivalent resistance in series and parallel?
The method of calculating the circuits equivalent resistance is the same as that for any individual series or parallel circuit and hopefully we now know that resistors in series carry exactly the same current and that resistors in parallel have exactly the same voltage across them.
## What is the formula of equivalent resistance?
2] Equivalent resistance formula for parallel resistance: The equivalent resistance of a number of resistors connected in parallel can be computed using the reciprocal of the resistance i.e. \frac{1} {R}. The reciprocal of the equivalent resistance will be equal to the sum of the reciprocals of each resistance.
## Why is equivalent resistance less in parallel?
So, parallel resistors allow more current. This means that the system as a whole can let more current through, so its resistance is lower.
## What is equivalent resistance between a and B?
Therefore, equivalent Resistance between A and B is 5Ω
## What is equivalent resistance in a series circuit?
In a series circuit, the equivalent resistance is the algebraic sum of the resistances. The current through the circuit can be found from Ohm’s law and is equal to the voltage divided by the equivalent resistance.
## What is called equivalent resistance?
Solution : If the current drawn by a resistors is equal to the current drawn by the combination of resistors, then the resistor is called equivalent resistor. Loading Books.
## How do you solve equivalent resistance problems?
Find equivalent resistance of the network. Find current in each resistor. Find voltage drops VAB,VBC,VCD. Medium.
## How do you solve a parallel circuit?
2:3410:54How to Solve a Parallel Circuit (Easy) – YouTubeYouTubeStart of suggested clipEnd of suggested clipSo we say that our total current in a parallel circuit. We just add up all of the currents that flowMoreSo we say that our total current in a parallel circuit. We just add up all of the currents that flow through each of those branches.
## What is the resistance of 2 resistors in parallel?
Two identical resistors in parallel have an equivalent resistance half the value of either resistor. The current splits equally between the two.
## What is equivalent resistance in a circuit?
In a series circuit, the equivalent resistance is the algebraic sum of the resistances. The current through the circuit can be found from Ohm’s law and is equal to the voltage divided by the equivalent resistance.
## When two 2 ohm resistance are in parallel the effective resistance is?
Hence, the effective resistance is equal to 1 Ohm.
## What is the equivalent resistance of the resistors?
The equivalent resistance represents the total effect of all resistors in the circuit. The equivalent resistance can be measured in either a series or parallel circuit. Resistor comprises two junctions with which the current passes in and out of it. They are passive devices that utilize electricity.
## What is equivalent resistance?
This is the amount of resistance that a single resistor would need in order to equal the overall effect of the collection of resistors that are present in the circuit .
## What is resistance in electrical circuit?
A parellel circuit on the other hand, has two or more paths for current to flow. Resistance is a measure of the opposition to current flow in an electrical circuit.
## What is the equivalent resistance of a circuit?
The equivalent resistance of the circuit is the amount of resistance that a single resistor will require in order to equalise the total effect of the set of resistors present in the circuit. For parallel circuits, the equivalent resistance of a parallel circuit is given …
## How many resistors are in a parallel circuit?
Although each branch gives 4 of resistance to any charge flowing through it, only one-half of all the charge flowing through the circuit may meet 4 of resistance of that branch. Thus, the presence of two 4 resistors in parallel will be equal to one 2 resistor in the circuit. This is the concept of equivalent resistance in a parallel circuit.
## How do resistors work?
In order to improve the net resistance, the resistors must be wired in series and the resistors must be connected in parallel to reduce the resistance.
## What happens when two resistors are wired in series?
Then the two resistors are wired in series and their equal resistance increases between their endpoints.
## How to find current through a battery?
To find the current through the battery we need to find the equivalent resistance of the circuit. The total current I is divided into and . The current passes through two resistors as they are connected in series and have the same current. The current passes through and resistors as they have the same current.
## What is the measure of how much a device or material can resist the movement?
When a circuit has more than one circuit component in it, there should be a way to calculate the total effective resistance of the entire circuit or for just one part of the circuit. Before we discuss what equal resistance is, we can describe resistance. Resistance is a measure of how much a device or material can resist the movement …
## What is resistance in electrical engineering?
Resistance is a measure of how much a device or material can resist the movement of electricity through it. It is inversely related to current, higher resistance means reduced current flow; reduced resistance means higher current flow.
## What is the equivalent resistance of a parallel combination?
If two resistors of equal resistance R are connected in parallel combination then the equivalent resistance of the combination is R / 2.
## How to find equivalent resistance?
If R eq is the equivalent resistance of the circuit then it is calculated from adding the reciprocal values of individual resistances (1/R). The inverse of this algebraic sum will give the equivalent resistance. The equation for equivalent resistance R eq is shown below for a parallel resistive circuit of n resistors.
## What is a resistor in parallel?
The concept of resistors in parallel is used in the analysis of Wheatstone bridge circuit. Resistors in parallel combination act as Current Divider Circuit. This current divider concept is use full in applications like Analog to Digital Converters and Digital to Analog Converters.
## Why is the current flowing through each resistor dependent on the resistance of that resistor?
Since the voltage across each resistor is same , the current flowing through each resistor is dependent on the resistance of that resistor. Hence if the resistance value in a branch is different from the other branch then the current in those branches will be different.
## How do resistors connect in parallel?
Introduction. Two resistors are said to be connected in parallel if both the terminals of a resistor are connected to each respective terminal of other resistor. In a network of parallel resistors, current can take more than one path unlike in series resistor network as there are multiple paths for the current to flow.
## What is the voltage across each resistor?
In parallel resistive circuits, the voltage across each resistor is same and is equal to the supply voltage.
## What is the potential difference between two or more resistors?
Resistors in Parallel. If two or more resistors are connected in parallel, then the potential difference across each resistor is same . Resistors in parallel connection are connected to the same nodes. This can be identified by the presence of more than one path for the current to flow.
## What is the equivalent resistance of a parallel network?
In other words, the equivalent resistance of a parallel network will always be less than the smallest individual resistor in the combination. Also, in the case of R1 being equal to the value of R2, that is R1 = R2, the total resistance of the network will be exactly half the value of one of the resistors, R/2.
## What is the inverse of the equivalent resistance of two or more resistors connected in parallel?
Then the inverse of the equivalent resistance of two or more resistors connected in parallel is the algebraic sum of the inverses of the individual resistances.
## How to convert conductance into resistance?
To convert conductance back into a resistance value we need to take the reciprocal of the conductance giving us then the total resistance, RT of the resistors in parallel. We now know that resistors that are connected between the same two points are said to be in parallel.
## How to find RT of a parallel circuit?
The equivalent or total resistance, RT of a parallel combination is found through reciprocal addition and the total resistance value will always be less than the smallest individual resistor in the combination. Parallel resistor networks can be interchanged within the same combination without changing the total resistance or total circuit current. Resistors connected together in a parallel circuit will continue to operate even though one resistor may be open-circuited.
## How to define parallel resistive circuit?
So we can define a parallel resistive circuit as one where the resistors are connected to the same two points (or nodes) and is identified by the fact that it has more than one current path connected to a common voltage source. Then in our parallel resistor example below the voltage across resistor R1 equals the voltage across resistor R2 which equals the voltage across R3 and which equals the supply voltage. Therefore, for a parallel resistor network this is given as:
## Why are parallel resistors called current dividers?
Then parallel resistor networks can also be thought of as “current dividers” because the supply current splits or divides between the various parallel branches. So a parallel resistor circuit having N resistive networks will have N-different current paths while maintaining a common voltage across itself.
## What is a parallel resistor?
Resistors in Parallel. Unlike the previous series resistor circuit, in a parallel resistor network the circuit current can take more than one path as there are multiple paths for the current. Then parallel circuits are classed as current dividers. Since there are multiple paths for the supply current to flow through, …
## What is the difference between a resistor in series and a resistor in parallel?
Resistors in parallel, on the other hand, result in an equivalent resistance that is always lower than every individual resistor.
## What happens when you add a resistor in parallel?
If you add another resistor in parallel with the first one, you have essentially opened up a new channel through which more current can flow.
## Does the current change when you add resistors in parallel?
Keep in mind that the current through an individual resistor does not change when you add resistors in parallel, because adding resistors in parallel does not affect the voltage across the resistors’ terminals.
## What is parallel circuit?
A parallel circuit is characterized by a common potential difference (voltage) across the ends of all resistors. The equivalent resistance for this kind of circuit is calculated according to the following formula:
## How to calculate resistors in a circuit?
Let’s look at an example for the second, slightly more complicated, mode: 1 Select “Calculate missing resistor” under Mode. 2 Now input the total resistance you want you circuit/collection of resistors to have. 3 Start by introducing the values of the resistors you already know (new fields will appear as needed). 4 The calculator automatically gives you the required missing resistor after each input.
## How many resistors can you use in a field?
To keep it simple, we only show you a few rows to input numbers, but new fields will magically appear as you need them. You can input up to 10 resistors in total.
## What is the unit of all values?
The units of all values are Ohms (symbol: Ω). 1 Ohm is defined as electrical resistance between two points that, when applied with a potential difference of 1 volt, produces a current of 1 ampere. Hence, 1 Ω = 1 V / 1 A or, in SI base units, Ω = kg·m²/ (s³·A²). | 2,976 | 14,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-23 | latest | en | 0.88902 |
https://socratic.org/questions/how-do-you-simplify-5-sqrt3-5-sqrt3 | 1,701,182,318,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00785.warc.gz | 609,305,812 | 5,657 | # How do you simplify (5+sqrt3)(5-sqrt3)?
This is of the form $\left(a + b\right) \left(a - b\right)$ with $a = 5$ and $b = \sqrt{3}$
$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$
$\left(5 + \sqrt{3}\right) \left(5 - \sqrt{3}\right) = {5}^{2} - {\left(\sqrt{3}\right)}^{2} = 25 - 3 = 22$ | 140 | 302 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-50 | latest | en | 0.549114 |
http://numericaltank.sjtu.edu.cn/2ndBook/BVPh1.1/Chap10-Exp.txt | 1,585,469,773,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-16/segments/1585370494064.21/warc/CC-MAIN-20200329074745-20200329104745-00509.warc.gz | 136,237,371 | 1,463 | (* Install BVPh 1.0 for Mathematica 5.2. *) (* For Mathematica 8.0, please replace BVPh1_0.txt by BVPh1_1.txt *) < BVP in a finite domain [0,a] *) (* TypeEQ = 2 -> eigenvalue problem in a finite domain [0,a] *) (* TypeL = 1 -> Chebyshev polynomial as base function *) (* TypeL = 2 -> Hybrid-base approximation *) (* TypeBase = 1 -> sine + polynomial *) (* TypeBase = 2 -> cosine + polynomial *) (* ApproxQ = 0 -> do NOT approximate the function *) (* Appprox = 1 -> approximate the function *) (************************************************************************) TypeEQ = 1; ApproxQ = 0; ErrReq = 10^(-10); zRintegral = 10; (************************************************************************) (* Define the governing equation *) (************************************************************************) kappa = -1/4; f[z_,u_,lambda_] := D[u,{z,3}]+(1+kappa)/2*u*D[u,{z,2}]-kappa*D[u,z]^2 ; (***********************************************************************) (* Define Boundary conditions *) (***********************************************************************) zR = infinity; OrderEQ = 3; BC[1,z_,u_,lambda_] := Limit[u, z -> 0 ]; BC[2,z_,u_,lambda_] := Limit[D[u,z] - 1, z -> 0 ]; BC[3,z_,u_,lambda_] := Limit[D[u,z], z -> zR ]; (************************************************************************) (* Define initial guess *) (************************************************************************) gamma = 1; temp[1] = (1-2*sigma*gamma)/gamma; temp[2] = (1-sigma*gamma)/gamma; u[0] = sigma + temp[1]*Exp[-gamma*z] - temp[2]*Exp[-2*gamma*z]; (************************************************************************) (* Define output term *) (************************************************************************) output[z_,u_,k_]:= Print["output = ",D[u[k],{z,2}] /. z->0//N]; (************************************************************************) (* Defines the auxiliary linear operator *) (************************************************************************) L[u_] := D[u,{z,3}] - gamma^2 * D[u,z]; (* Print input and control parameters *) PrintInput[u[z]]; (* Set optimal c0 and sigma *) c0 =-5/4 ; sigma = 3; Print[" c0 = ",c0, " sigma = ",sigma]; (* Gain up to 10th-order HAM approxiamtion *) BVPh[1,10]; | 538 | 2,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.351657 |
https://www.prosoundtraining.com/2021/05/17/contrast-of-statistical-and-geometric-room-acoustic-analysis/ | 1,685,782,449,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649177.24/warc/CC-MAIN-20230603064842-20230603094842-00196.warc.gz | 1,015,290,200 | 36,534 | # A Contrast of Statistical and Geometric Room Acoustic Analysis
by Pat Brown
## Statistical and geometric room acoustics analysis have been with us for well over a century. Computer modeling makes geometric acoustic analysis (GA) practical at the drawing board. This article contrasts the two methods.
The room impulse responses (RIR) from Room Survey 38 – Ministry of Foreign Affairs (MFA) provide an opportunity for discussing the significance of early-reflections in a room. Their consideration allows us to contrast the major approaches to room acoustics predictions – statistical vs. geometric.
The Ministry of Foreign Affairs – Bangkok, TH
Statistical Acoustics
The statistical method of room acoustics analysis involves the use of reverberation time (RT) equations. These include the Sabine equation, Eyring equation, and others. These equations calculate the reverberation time of the room based on its volume and internal absorption (Fig. 1). While the mental model of sound propagation is that of waves, reverberation is a diffuse field that develops after more than a second of decay and after a great deal of room reflections have occurred. After the introduction of an impulse of sound energy into the space, the reflection-density at the listener increases with passing time and a diffuse field develops. This assumes that the room meets certain criteria, namely that it has a “mixing” geometry, low absorption, and no concentrated absorption. A “big concrete shoebox” serves as a theoretical example.
Figure 1 – Statistical reverberation time equations are the simplest approach to room acoustics analysis. Their accuracy is conditional, and the effects of air absorption are not shown.
Unlike sound waves, textbook reverberation has no specific direction of energy flow, behaving more like an “acoustic fog” than a propagating wave. If an impulse is emitted in the space, this highly ordered compact event succumbs to the 2nd Law of Thermodynamics, and any order (or information) that it contains is eroded away over time. This increase in disorder is called entropy and there is no escaping it with physical systems. Nature insists that information runs out of, not into, a closed system over time. The sound energy eventually becomes diffuse with no net direction of energy flow. An impulsive sound played into a room eventually becomes reverberation (Fig. 2). A common RT measurement technique is to interrupt a steady sound source and time the decay (Fig. 3).
Figure 2 – The introduction of an impulse of sound energy into a space produces room reflections that increase in density over time as the sound level decreases.
Figure 3 – One method of RT testing interrupts a steady noise source and times the room decay.
Statistical reverberation equations calculate the RT for 60 dB of decay of this diffuse field. With knowledge of the RT, the direct sound level (LD) and the reverberant sound level (LR) can be calculated using the Hopkins-Stryker equation. Once LD and LR are known, the speech intelligibility (SI) can be estimated using the Percentage Articulation Loss of Consonants (%ALCONS) as formulated by V.M.A. Peutz late last century. The %ALCONS score can be converted to the more contemporary Speech Transmission Index (STI) for comparison purposes. It is important to note that this is technically not a direct conversion, since STI includes many factors ignored by %ALCONS. Even so, the results of the two methods should be in general agreement. I developed a soon-to-be released calculator for use in SynAudCon online training programs that allows the interaction of these variables to be visualized (Fig. 4).
Figure 4 – IntelliKwik™ allows for fast estimations of the potential speech intelligibility using statistical reverberation time formulas, the Hopkins-Stryker equation, and some basic sound system parameters. It can be used to “run the numbers” before building a room model. The settings shown are for Test Position 2 (4 m) of the MFA.
The value of these Hopkins-Stryker relationships cannot be overstated. They form the foundation for the statistical acoustical analysis of an enclosed space.
With statistical acoustic formulas we have a simple-to-implement approach for diffuse field analysis that is blind to the specular-reflected sound energy in the space. In some spaces this “missing” information is insignificant. In others it is the most significant with regard to the sound of the space, and therefore the SI.
The Missing Details
It is important to consider what we cannot know from the statistical approach to room acoustics analysis. As entropy morphs the LD into reverberation, the sound waves undergo a series of reflections in the space (Fig. 5). As the impulse reflects from each surface some organization is lost, but incrementally. Low order reflections can still retain a significant amount of organization. If the sound were a speech waveform (which by definition carries information), these reflected facsimiles of the original will also contain information. Whether or not this information is useful to the listener depends on their arrival time relative to the direct sound field LD. At a specific listener position, these “specular” reflections are categorized as “early” or “late” with 50 ms serving as a general dividing time for speech. Early reflections are often useful for communication (or at least are non-harmful) and late reflections are generally a detriment to communication. Considering the effects of specular reflections requires a different tool set than the simple statistical equations used to compute reverberation times.
Geometric Acoustics
Geometric Acoustics (GA) uses optical principles to model specular reflection behavior. It can be used to fill in the missing events in the RIR. While statistical reverberation equations are most valid in rooms with low absorption that is uniformly distributed, GA has no such limitation. Since most rooms aren’t dominated by a statistical reverberant field, GA becomes a primary design tool.
Figure 5 – The RIR includes both specular reflections and diffuse decay.
A Real-World Shoebox Example
Now, back to our reference case of the Ministry of Foreign Affairs. This room is basically a “shoebox” with absorptive floor and ceiling and hard side walls. We would have every reason to expect the presence of significant specular reflections within the space, especially given the source and receiver positions used for the RIR measurements (Fig. 6).
Figure 6 – A simple model based on the room dimensions and source/receiver positions in the MFA (CATT-A). A0 is the omni sound source and 01, 02, 03, and 04 are receiver (listener) positions.
The Statistical Approach
I first used IntelliKwik to determine LD, LR, and the resultant SI. I caught a massive break here in that the room volume and RT were already known from physical measurements and the RIRs. This allowed me to “seed” the calculator with valid data, with a potentially dramatic improvement in the accuracy of the results. The STI values calculated for each position have been added to the results matrix from Room Survey 38 (Fig. 7). Note that the STI degrades with increasing distance for the calculated values, as it must. Statistical equations know nothing of specular room reflections that can add useful, early energy for specific listeners. As a result, the statistically-predicted STI yields an error that increases with distance from the source.
Figure 7 – The results matrix from Room Survey 38 with additional entries for STI calculations based on statistical equations (IntelliKwik) and GA (CATT-A).
We got an answer quick, but the STI was underestimated. Since the error increases with distance we have a strong clue that there is more at play than LD and LR. We only know this because we have measured RIRs to provide a benchmark. If our analytical tool only considers LD and LR (e.g. IntelliKwik) then we will only see the influence of LD and LR and may think that we have the whole story and therefore underestimate the STI.
But, at the drawing board it is better to underestimate the STI than overestimate it, and the statistical-only approach has value. It is fast, conservative, and it makes us work a bit harder on our design with regard to loudspeaker selection and placement. I have never had regrets that a system performed better than expected.
Then why do we need GA? The need will become apparent once we run through the process.
The GA Approach
I punched out a quick shoebox in Sketchup™ in accordance with the physical room dimensions. I assigned some colors to the planes to identify them for absorption data that would be added later in the room modeling program. This simple model was exported to CATT-Acoustic™ using SU2CATT™ and the colors were assigned appropriate absorption alphas. I used measured RTs mined from the Test Position 4 RIR to aid in assigning absorption alphas and scattering to the room surfaces (Fig. 8).
Figure 8 – The measured RTs provided invaluable information for refining the room model. Note that both Sabine and Eyring underestimate the RT. This is due to non-uniform distribution of a significant amount of absorption in the space. CATT-A provides a quick ray tracing routine in its Interactive RT Estimation module for comparison with Sabine and Eyring. In a truly statistical space all four of these RT measures should be in good agreement. When they are not (as in this case) it is best to proceed with a GA-driven design approach. A conservative practice is to make the predicted RT slightly longer than the measured RT. In other words, slightly underestimate the room absorption. The black curve is the “Ref RT” from the measured RIR.
A Logical Modeling Process
Since the RT is known, and the side walls are obviously reflective, it was not difficult to tweak the floor and ceiling alphas to match the measured RT after applying a very low alpha to the walls. Given the geometric complexity and apparent absorptiveness of the ceiling (from the photographs), and the visual appearance of the carpet I felt it safe to assume that the ceiling yields more sabins (higher absorption) than the floor. I selected a moderately absorptive carpet material for the floor, and then tweaked the ceiling alpha to approximate the measured reverb times. Lastly, I used significant scattering on the ceiling to “de-specularize” it to simulate the high surface complexity. This “educated guessing” is a big part of room modeling. My main objective was to consider the side wall contributions to the RIR and with GA modeling I have the freedom to overestimate their influence if I choose to. Modeling allows us to easily isolate the influence of specific surfaces and sound paths – something not easily accomplished with measured data.
The time required for the entire modeling process was maybe 30 minutes, yielding a model with reasonable base-line accuracy that can be further tweaked if necessary. It is important to point out that with the major variables quantified and a good match to the measured RIRs, refining the model further may yield diminishing improvements in the simulation accuracy. I could spend many hours increasing the ceiling detail with nearly no impact on the prediction results. The modeler must weight the significance of adding detail with regard to its effect on the results. A better model for the eye is not necessarily a better model for the ear.
Figure 9 shows a map of the STI produced in CATT-A. Note that the STI is high near the source (A0) and falls with increasing distance from the source. This is in agreement with my IntelliKwik calculations. Note that the STI rises near the rear wall, beyond receiver 04. This is due to specular reflections (e.g. rear wall) and will be missed entirely by a statistical-only approach.
Figure 9 – An STI map of the audience plane (courtesy CATT-A).
Under-the-hood the use of GA to produce an RIR is staggeringly complex. The details of the process vary from platform-to-platform and go well beyond the scope of this article. While the modeler is insulated from the details of the process, they must still understand it well enough to make some basic decisions when staging a simulation, such as choosing the number of rays and assuring that the resultant predicted RIR is as long as the room’s measured decay time (Fig. 10).
Figure 10 (left) – The setup dialog for running a GA simulation in CATT-A. Note that 3 algorithms are provided, each optimized for specific room types, such as a closed cathedral vs. an open stadium.
Figure 11 compares the measured and predicted RIR for Test Position 3. This predicted RIR can be analyzed and convolved in the same way as the measured one.
Figure 11a – Measured RIR of Test Position 3. Shown in ReflPhinder™.
Figure 11b – Predicted RIR of Test Position 3. Since I made the RT of the model slightly longer than the measured, the energy ratios shown here are slightly lower than the measured. The simulation was performed in CATT-A with the results shown in ReflPhinder.
Auralization
A powerful method of results analysis is auralization. Auralization is to a predicted RIR as convolution is to a measured one. Below are auralizations for the four test positions in the study. These can be compared with the convolutions in Room Study 38 for the same positions. You can open it in a second browser window by right-clicking on the link. A special thank you to Tanadej S’prayoon1 for sharing his female talker WAV file.
Observe the results matrix (Fig. 12) as you listen. Convolution and auralization bring the RIR and room measures to life.
Receiver 01 Female Talker
Receiver 02 Female Talker
Receiver 03 Female Talker
Receiver 04 Female Talker
Figure 12 – Results matrix again.
A Coverage Solution
What would be the point of coming this far if we don’t use the model to arrive at a real world coverage solution? The point of RIR measurements with a dodec and the construction of a room model is to establish a qualified virtual space in which to audition our design ideas.
In the model, I placed a medium Q source at the same position as the dodec. This gives us an apples-to-apples comparison of the improvements that can be realized by loudspeaker directivity. Auralizations and the STI are provided below for comparisons with the dodec. Not surprisingly, this placement produces some audible flutter echoes, but the improvement in SI is striking.
Receiver 01 STI = 0.83
Receiver 02 STI = 0.67
Receiver 03 STI = 0.53
Receiver 04 STI = 0.58
With the room model qualified, the system designer is now free to try various coverage approaches, including point sources, line arrays, and overhead distributed. An audience can be added with a slight modification of the model.
Some Observations…
I’ll close with some observations regarding what I have presented.
– The 2nd law of thermodynamics assures that the information runs out of a system over time.
– Room reflections can recycle some of the original information emitted into a space. They do not create new information.
– The diffuse sound energy that results from entropy is reverberation.
– A statistical-only design approach can have reasonable accuracy in rooms dominated by a diffuse field. GA is required for non-statistical spaces.
– %ALCONS is no longer supported by most measurement and prediction programs. It is still useful because it allows the SI to be estimated from LD and LR. It is still referenced in Annex D of NFPA 72 and it can be mathematically converted to the STI (for comparison purposes). The direct computation of the STI requires a completely different approach.
– With an RT > 2 s (2 kHz) this room is a decent candidate for statistical evaluation. In dryer spaces a GA-based design approach becomes more essential due to the stronger influence of specular reflections.
– Log-spaced measurement positions on-axis with a source facilitate the consideration of what should happen to the speech intelligibility with increasing distance from the source. Since the LD drops with distance and LR is essentially constant the SI can only degrade with increasing distance.
-The objective for modeling a space need not be perfect accuracy. In this case I mainly wanted to assess how wall reflections might have a positive impact on SI at seats where a statistical-only analysis suggests a poor result. The GA modeling results were in good agreement with the measured data.
– The RIR, whether measured or predicted, remains the single most powerful room analysis and sound system design tool.
A special thank you to CATT author Bengt-Inge Dalenback for his invaluable comments on the article. pb
1Tanadej Siriwattaprayoon, Protime Business Co., Ltd. Bangkok, Thailand | 3,496 | 16,780 | {"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} | 2.796875 | 3 | CC-MAIN-2023-23 | longest | en | 0.907304 |
http://openstudy.com/updates/50fe3f9ce4b0426c6367ba41 | 1,448,975,493,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398467979.1/warc/CC-MAIN-20151124205427-00338-ip-10-71-132-137.ec2.internal.warc.gz | 175,364,593 | 14,292 | ## dhudz23 2 years ago A rectangular flower garden with dimensions 3m by 7m is surrounded by the walk of uniform width. If the are of the walk is 11 sq. m, what is the width in meters?
1. zordoloom
|dw:1358847023244:dw|
2. zordoloom
Do you know where to go from here?
3. dhudz23
Not really..
4. zordoloom
Do you mind if I continue?
5. dhudz23
Not at all.
6. zordoloom
|dw:1358847340760:dw|
7. zordoloom
Do you think you can solve that equation?
8. dhudz23
Not sure.. I could but it would take me a while and i wouldn't be to confident of my answer.
9. zordoloom
(7+2x)(3 +2x)-11=21---> 4x^2+20x-11=0----> now solve this using the quad. equation. x=-5.5 and x=.5 The value can only be positive therefore the width can only be x=.5 So the width around the garden is 1/2 m or .5m. Whichever method your prefer.
10. zordoloom
Does that help?
11. dhudz23
Yes it does, thank you!
12. zordoloom
You're welcome! | 314 | 927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2015-48 | longest | en | 0.834193 |
https://squared2020.com/2019/11/27/extending-possessions-geometric-distribution/ | 1,679,570,159,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945144.17/warc/CC-MAIN-20230323100829-20230323130829-00259.warc.gz | 625,826,645 | 38,084 | # Extending Possessions: Geometric Distribution
In a recent post, we took a brief glimpse at offensive rebounding and discussed some pro’s and con’s about crashing for rebounds and provided an illustration about where rebounds go after missed attempts. In one such instance, you would have seen this plot:
Purple Square: Location of FGA. Red Circle: Location of Rebound
For the sake of argument, let’s suppose that every single red circle in that image is an offensive rebound. Then as one chance is exterminated (through a rebound), a new chance begins in that location; extending the possession. In this case, the shot clock resets to 14 seconds and the play continues. In a post from over a year ago, we looked at where the new chances go after an offensive rebound. You would have seen this Steven Adams plot:
Distribution of FGA after a Steven Adams offensive rebound. Red Dots: Missed attempts, Green Dots: Converted Attempts
From this Steven Adams plot, you’d see that not only does an offensive rebound create a new chance for the possession, but it creates a new chance for any player on the court. This means a significant amount of second chances generated by a single player are not necessarily put-backs.
But how do we model the number of second chances on a possession? Afterall, there are only so many ways to terminate a chance: made Field Goal Attempt (no foul), made x-of-x Free Throw, Defensive Rebound, Turnover, End-of-Period*, and Offensive Rebound. There’s only one way to extend a chance. Note the asterisk on end-of-period attempts. Typically, a team rebound to the defense is tagged on heaves that would be rebounded after time has expired. Example from the 2018-19 NBA season:
Therefore, to model the amount of second chances, we can focus on chaining the probabilities of obtaining offensive rebounds together to obtain the expected number of second chances in a possession. To this end, we will start simple and then raise expectations.
## Simple: Geometric Distribution
To start simple, we introduce the geometric distribution. To give an illustration, let’s consider a toy example. Suppose that you are at a court taking a series of field goal attempts; aka “shooting” at the park. At the end of the period, you decide to “leave on a make” and take a series of three point attempts until you make one to feel good about yourself leaving the court. Finally, let’s suppose you are realistically a 27% three point shooter. The question is then, how many attempts do we expect you to take until you get to leave the court (somewhat) happy?
Assuming that every shot is independent (effectively saying that the hot-hand fallacy is indeed a fallacy) and identically distributed (every three point attempt has the same probability of going in), then we have what is called the geometric distribution.
Illustrating its probabilities are simple:
Probability of Taking 1 3PA: 27%
Probability of Taking 2 3PA: (1-.27)*.27 = 19.71%
Probability of Taking 3 3PA: (1-.27)*(1-.27)*.27 = 14.39%
And so on and so forth…
The way we calculate the probabilities for the number of attempts is to multiply the probability of a miss until we hit the final attempt. Note that we always leave (somewhat) happy and therefore the final shot is the only made attempt. Therefore, the probability distribution function is given by
where p is the probability of making the three point attempt. Go ahead and stick in .27 for p. You will recreate the probabilities above. Here, X, is just the random variable: number of three point attempts taken. Theoretically, we could have the worst luck and take infinitely many attempts. Under this case, taking 20 such attempts would amount to .0005 (.05%) probability. Note that k in this context is the number of chances!
But how many attempts are expected? To compute this, we just compute the expected value. However, for the uninitiated, there is a bit of “sorcery” in computing this value:
Glorious math!
Using the p = .27 from above, we obtain an expected 3.70 attempts to closeout the day. So the next time it takes you approximately 2-7 attempts to close out your sessions, congratulations! You’re most likely a 27% shooter.
### Application to Extending Possessions
Applying the geometric distribution to offensive rebounding produces an extra wrinkle. First, not all chances end in rebounds. Recall that offensive rebound percentage (OREB%) is computed as the number of offensive rebounds divided by the number of rebounding opportunities. Hence that 22.9% for the 2018-19 NBA season is on rebounds alone. Therefore, we need to identify the probability a rebound occurs! To this end, we look at the proportion of chances that end in rebounds.
To start, there were a total of 118,396 missed field goal attempts and 6,805 missed final free throw attempts during the 2018-19 NBA season. That accounts for 125,201 potential rebounds. Of which, a total of 25,454 were offensive rebounds and 85,653 were defensive rebounds. This leaves 14094 rebounds unaccounted for (11.45 rebounds per game). These are team rebounds. For example, the particular game mentioned above contained 20 such rebounds. Simply comparing offensive to defensive rebounds above, we obtain the 22.9% number used on Basketball-Reference.
Given the number of possible rebounds, we must also account for non-rebound situations. In this case, we have 101062 made field goals and 24803 made final attempted free throws. Note that we also must identify “Plus-1” situations, of which we had 7,419 attempts; 1788 of those missed. Note that these only take away chances!
Therefore, tallying up the numbers: Field Goals Made account for 101,062 chances. Free Throws Made account for 24,803 chances. Turnovers account for 34,644 chances. Missed Field Goals account for 118,396 chances. Missed Final Free Throws account for 6805 chances. Then we must subtract the Made “Plus-One” Events of 5631 chances. Note that this is back of the envelope math; plus-one events are including technical fouls; as the goal here is showing how the model would work.
To this end, we end up this generalized chance ending probability distribution:
• Missed Field Goal: 42.27%
• Turnover: 12.37%
• Free Throw Missed: 2.43%
Wow, that actually turned out to be 100% of chances. Note that we eliminated end of period chances as they have no bearing on games; as heaves are included (but insignificant) above.
From all this, the probability of extending a possession is 10.24%. This, of course, assumes the same 22.9% OREB% on free throw attempts, which (except for Steven Adams) we know is not true. Again, goal is to show how the model works.
Applying the geometric distribution above, we find that for any random possession we expect 1.114 second chances. For a game of 100 possessions, this equates to 11 second chance possessions!
## Increasing Expectations: Adding in Location
For the example above, we saw that on average, we should expect a team to gather 11 second chance opportunities in a game. Unfortunately, not all teams are equal. For instance, the Chicago Bulls and Milwaukee Bucks tended to eschew offensive rebounds last season whereas the Oklahoma City Thunder tended to hunt them down. To their credit, this is due to a combination of positioning, game planning, and luck of the bounce.
However, that 22.9% is completely unrealistic. For example, a single season tends to have approximately 600 rebounds on missed FTA’s in a season. With roughly 6800 missed final free throw attempts, that’s an OREB% of 8.8%, far below the 22.9%.
Similarly, we can look at the distribution of offensive rebounds off of missed field goal attempts:
Since everyone loves hexagon plots…
Here, we see that shots at the rim push up towards 30-35% for offensive rebounds. The rate dies out dramatically at the midrange (dropping to approximately 12%) before jumping back up to 22-27% around the perimeter. There is a little flare-up of 23% at the 30 foot mark. That’s primarily Damian Lillard’s attempts and Portland reacting accordingly.
Now instead of writing everything out in terms of the geometric distribution, we start with the geometric distribution and swap out the over generalized p parameter of offensive rebounding percentage. Instead, we obtain this form of the “geometric” distribution:
Little harder, but I know you understand this!
This is the exact same form as the old geometric distribution, but instead of p we have p_{j,j+1} which is the probability of obtaining a rebound for a chance starting in state j and ending in state j+1.
For instance, let’s start with p_{0,1}. This is the initial chance. This probability would be dictated by the start of a chance: rebounded miss, turnover into transition, dead ball. A good place to start understanding the impact of initial chances can be found from Seth Partnow of the Athletic. Hence p_{0,1} in this context is the probability of terminating a possession and 1 – p_{0,1} is the probability of obtaining an offensive rebound!
Continuing on, we can look at the probability of terminating a possession after an offensive rebound. This would be p_{1,2}. Note again, that 2 here means second chance. Furthermore, p_{2,3} would indicate another second chance after 2 offensive rebounds. Again this probability would be identified by its initial position of the rebound starting a new chance (see the Steven Adams plot above for context).
Now, applying the same analysis as in the geometric distribution, we find that the Milwaukee Bucks would expect a mere 8 second chance attempts per 100 possessions while the Oklahoma City Thunder would expect 16 second chance attempts! And now we start gaining insight into the baseline model of extending possessions for any given team.
Given this set-up we can now start asking and answering questions such as “how do teams generate second chance opportunities?” and “how do teams utilize their second chances?” down to the “How are we at defending against second chance opportunities?”
For instance, limiting the Thunder to 11 second chance opportunities is a good thing. Against the Chicago Bulls, not so much. | 2,271 | 10,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-14 | latest | en | 0.917916 |
https://nrich.maths.org/public/leg.php?code=113&cl=4&cldcmpid=7277 | 1,566,699,457,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322170.99/warc/CC-MAIN-20190825021120-20190825043120-00276.warc.gz | 591,456,012 | 7,134 | # Search by Topic
#### Resources tagged with Spheres, cylinders & cones similar to Weird Universes:
Filter by: Content type:
Age range:
Challenge level:
### There are 25 results
Broad Topics > 3D Geometry, Shape and Space > Spheres, cylinders & cones
### Peeling the Apple or the Cone That Lost Its Head
##### Age 14 to 16 Challenge Level:
How much peel does an apple have?
### Curvature of Surfaces
##### Age 16 to 18
How do we measure curvature? Find out about curvature on soccer and rugby balls and on surfaces of negative curvature like banana skins.
### Tin Tight
##### Age 14 to 16 Challenge Level:
What's the most efficient proportion for a 1 litre tin of paint?
### Fill Me up Too
##### Age 14 to 16 Challenge Level:
In Fill Me Up we invited you to sketch graphs as vessels are filled with water. Can you work out the equations of the graphs?
### Flight Path
##### Age 16 to 18 Challenge Level:
Use simple trigonometry to calculate the distance along the flight path from London to Sydney.
### Which Solid?
##### Age 7 to 16 Challenge Level:
This task develops spatial reasoning skills. By framing and asking questions a member of the team has to find out what mathematical object they have chosen.
### Pack Man
##### Age 16 to 18 Challenge Level:
A look at different crystal lattice structures, and how they relate to structural properties
### When the Angles of a Triangle Don't Add up to 180 Degrees
##### Age 14 to 18
This article outlines the underlying axioms of spherical geometry giving a simple proof that the sum of the angles of a triangle on the surface of a unit sphere is equal to pi plus the area of the. . . .
### Ball Packing
##### Age 14 to 16 Challenge Level:
If a ball is rolled into the corner of a room how far is its centre from the corner?
### Funnel
##### Age 14 to 16 Challenge Level:
A plastic funnel is used to pour liquids through narrow apertures. What shape funnel would use the least amount of plastic to manufacture for any specific volume ?
### The Dodecahedron Explained
##### Age 16 to 18
What is the shortest distance through the middle of a dodecahedron between the centres of two opposite faces?
### Conical Bottle
##### Age 14 to 16 Challenge Level:
A right circular cone is filled with liquid to a depth of half its vertical height. The cone is inverted. How high up the vertical height of the cone will the liquid rise?
### Pythagoras on a Sphere
##### Age 16 to 18 Challenge Level:
Prove Pythagoras' Theorem for right-angled spherical triangles.
### Spherical Triangles on Very Big Spheres
##### Age 16 to 18 Challenge Level:
Shows that Pythagoras for Spherical Triangles reduces to Pythagoras's Theorem in the plane when the triangles are small relative to the radius of the sphere.
### Packing 3D Shapes
##### Age 14 to 16 Challenge Level:
What 3D shapes occur in nature. How efficiently can you pack these shapes together?
### Immersion
##### Age 14 to 16 Challenge Level:
Various solids are lowered into a beaker of water. How does the water level rise in each case?
### Paint Rollers for Frieze Patterns.
##### Age 11 to 16
Proofs that there are only seven frieze patterns involve complicated group theory. The symmetries of a cylinder provide an easier approach.
### Mesh
##### Age 16 to 18 Challenge Level:
A spherical balloon lies inside a wire frame. How much do you need to deflate it to remove it from the frame if it remains a sphere?
### Witch's Hat
##### Age 11 to 16 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
### Air Routes
##### Age 16 to 18 Challenge Level:
Find the distance of the shortest air route at an altitude of 6000 metres between London and Cape Town given the latitudes and longitudes. A simple application of scalar products of vectors.
### In a Spin
##### Age 14 to 16 Challenge Level:
What is the volume of the solid formed by rotating this right angled triangle about the hypotenuse?
### 2D-3D
##### Age 16 to 18 Challenge Level:
Two circles of equal size intersect and the centre of each circle is on the circumference of the other. What is the area of the intersection? Now imagine that the diagram represents two spheres of. . . .
### Gym Bag
##### Age 11 to 16 Challenge Level:
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### Mouhefanggai
##### Age 14 to 16
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
### Three Balls
##### Age 14 to 16 Challenge Level:
A circle has centre O and angle POR = angle QOR. Construct tangents at P and Q meeting at T. Draw a circle with diameter OT. Do P and Q lie inside, or on, or outside this circle? | 1,118 | 4,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-35 | latest | en | 0.840446 |
http://idlcoyote.com/math_tips/double.html | 1,643,429,338,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00351.warc.gz | 32,032,058 | 3,562 | # Double Precision Numbers in IDL
QUESTION: I thought double precision numbers were suppose to be, like, really accurate or something. But check this out:
``` IDL> n = Double(9999.9)
IDL> Print, n, Format='(F20.10)'
9999.9003906250
```
That ain't right, man! How am I gonna get my degree if my computer can't do any better than this!? Professor Malarkey is gonna eat my lunch.
ANSWER: Oh, hang on a minute. Take a deep breath. Relax. The sky is not falling. There are just a couple of things you need to know before you finish your degree.
First, it would be extremely helpful if you understood how numbers are represented on a computer. With that understanding, you will realize that floating point numbers can be represented with approximately seven decimal places of significance, whereas double precision numbers can be represented with approximately 14 decimal places of significance. The number you printed out above starts to vary in the eighth decimal place of significance. Thus, we might conclude that it is a floating point number, just on the basis of decimal significance.
But, you say, I expressly made it a double precision number. Yes, but what you made a double precision number, was a floaing point number with just seven decimal places of significance. So your result is prefectly reasonable. One might even say "accurate".
What you probably wanted to do, was make the number 9999.9 a double precision number (with 14 decimal places of significance) from the start. To do this, you must place a "D" at the end of the number. This is probably what you had in mind:
``` IDL> n = 9999.9D
IDL> Print, n, Format='(F20.10)'
9999.9000000000
```
And what you should also realize (and thanks to Reimar Bauer for pointing this out in the IDL newsgroup), is that this also appplies to arrays. For example, some people, when they are creating a double precision array, will assign a "D" to the first value in the array, thinking that this will make all the values in the array double precision. It does do this, but in the manner of the first example above, not the second. Consider this:
```
IDL> a = [10.0D, 20.5, 9999.9]
IDL> Help, a
A DOUBLE = Array[3]
IDL> Print, a, Format='(F20.10)'
10.0000000000
20.5000000000
9999.9003906250
```
Notice in this example, that the value 20.5 can be represented in floating point notation in a way that could confuse you as to the real accuracy of the numbers in your array!
The proper way to write this array is like this:
```
IDL> a = [10.0D, 20.5D, 9999.9D]
IDL> Help, a
A DOUBLE = Array[3]
IDL> Print, a, Format='(F20.10)'
10.0000000000
20.5000000000
9999.9000000000
```
And your caution should even extend to using IDL's routine Make_Array. So, for example, this:
``` IDL> array = Make_Array(3, Value=9999.9, /Double)
IDL> Print, array, Format='(F20.10)'
9999.9003906250
9999.9003906250
9999.9003906250
```
is different from this:
``` IDL> array = Make_Array(3, Value=9999.9D, /Double)
IDL> Print, array, Format='(F20.10)'
9999.9000000000
9999.9000000000
9999.9000000000
```
With this in mind, you should have no trouble getting your degree in the required 10 years.
Web Coyote's Guide to IDL Programming | 877 | 3,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2022-05 | latest | en | 0.933349 |
http://www.indypl.org/kids/blog/?p=8891 | 1,539,809,751,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511216.45/warc/CC-MAIN-20181017195553-20181017221053-00190.warc.gz | 474,141,176 | 13,134 | Science Experiment: Momentum – Pendulums
Science Experiment: Momentum – Pendulums
A pendulum is a weight hanging from a fixed point. A yo-yo on a string can be a pendulum if you have ahold of the string and hold it in one place while the yo-yo hangs down freely. Pendulums are also associated with clocks or hypnotists, who swing an object back and forth from a fixed point. On a swing, YOU are a pendulum. Pendulums are affected by laws of motion. When a pendulum swings it has momentum. In this video, Bill Nye the Science Guy demonstrates momentum:
Things You Need:
• String 3 feet long
• 2 Strings 2 feet long
• Two Mugs
• Two Chairs
• Scissors
This experiment will show you something surprising about how momentum works. Set two chairs back to back about 3 feet apart. Tie a piece of string to the back of one chair and toe the other end to the back of the other chair. The string should not have a lot of slack in it. Tie a 1 foot long piece of string to the piece of string suspended between the two chairs about 18 inches from one of the chair backs. Now tie the other 1 foot piece of string to the suspended string about 18 inches from the OTHER chair back. When you are done, the two pieces of string should be a about 1 foot apart. Now tie a mug to the end of each dangling piece of string. Pull back on one mug and let go, letting it swing back and forth until it stops. Does it stop? What does the other mug do? When you release the first mug, momentum keeps the mug swinging back and forth. The mug will swing until friction in the air and friction from the string knot rubbing on the handle slows it down.
The weird thing is that when the first mug slows down…the second mug will START swinging, even though you never touched it! Some energy from the first mug travels along the string and makes the OTHER mug start swinging. Even though the momentum of the swinging mugs is slowed down by friction, some of the energy is transferred to the string, which carries it to the second mug.
Science Project Idea: Replace the mugs in this experiment with matching objects of different weight. Try two matching plastic mugs. How much momentum do the plastic mugs have? Does the momentum from a plastic mug have enough energy to pass through the string to the second plastic mug? Try hanging two metal forks from the string. Now try two plastic forks. Which kinds of objects have enough momentum to transfer energy to the second object without even touching it?
Websites and Databases for Research:
Science in Context: Momentum is a database you can use in any IndyPL Library Branch or at home. Login using your IndyPL library card number. The Science in Context database will show you articles, images and videos to help you learn about momentum.
You can also ask a math and science expert for homework help by calling the Ask Rose Homework Hotline. They provide FREE math and science homework help to Indiana students in grades 6-12.
Books:
Use your indyPL Library Card to check out books at any of our locations, or check out e-books and e-audiobooks from home right to your device. Click on a book jacket below to request a book or download it. Need help? Call or ask a Library staff member at any of our locations, text a librarian at 317 333-6877, or leave a comment.
Print This Post | 740 | 3,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-43 | latest | en | 0.892926 |
https://discourse.mcneel.com/t/arc-direction-at-90-to-rotating-line/90829 | 1,679,419,411,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943704.21/warc/CC-MAIN-20230321162614-20230321192614-00611.warc.gz | 256,972,931 | 8,318 | # Arc direction at 90 to rotating line
I wish to draw a quatrent arc at 90 degrees to the rotating line !
How do i keep the arc at 90 to the moveing line as it
rotates ?
hi,
I’m not sure I understood well what you want, I think it’s this:
arc.gh (12.1 KB)
1 Like
As line A,B is rotated around point A, I wish to draw an arc from B at 90 degrees to the AB line.
I need grasshopper to generate the line, not rhino.
I can set a static direction of the arc, but i need the arc to be at 90 to the AB line as it moves.
I do not understand how to get this veriable into the arc function
Thank you, but i am still confused. I have copyed your set and i have an arc sliding up and down my straight line .
i am working on the x z plane.
I have introducesed a curved line on the x y plane and you set works.
All i need do is understand what is going on !
Thank you
Thank you so much,
I spent over seven hours on this yesterday, trying to fathom out what was going on. and failed misserably.
here is my mess of attempts to figuer it out.
As you can see, i have lost some hair, brain cells, and dignity.
I have entered your solution and it works well.
Again thank you.
Here is what i have been tryining to do, and with your help i have got there.
Obviously this is just a small segment of my overall project.
And now i can move on thanks to the help i have recieved.
Thank you.
I am not going to recreate your model. There are usually multiple ways of doing things, depending on assumptions. Like whether or not the “curve” is a line, planar or 3D curve, and how the “arc plane” is oriented. Etc.
Please save yourself and others time by posting an actual GH model.
1 Like
Rotate and move the base plane | 442 | 1,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-14 | latest | en | 0.95822 |
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## Website Detail Page
This is an interactive homework problem for introductory physics relating Newton's Second Law and gravitational acceleration. It features a block at rest atop an incline, with angle of incline and ramp length given. Students must determine the time required for the block to reach the bottom. A user-activated "help" sequence is provided for each step of the problem-solving, from conceptual analysis through quantitative calculation. Immediate feedback is received for both correct and incorrect responses. This resource is part of a larger collection of interactive problems for introductory physics. See Related Materials for a link to the full collection.
Editor's Note: We like this problem because it gives tips on finding the optimal coordinate system for solving the problem. To quote the author, " it is often worth your time to stop and think which directions are 'natural'. For example, the problem may have directions with no forces. This means that there is no acceleration in these directions. A problem that starts out 2-dimensional then turns out to really only have 1 dimension that matters."
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G. Gladding, (University of llinois Physics Education Research Group, Urbana, 2006), WWW Document, (http://per.physics.illinois.edu/per/IE/ie.pl?phys111/ie/03/IE_block_on_incline).
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G. Gladding, Illinois PER Interactive Examples: Block on Incline, (University of llinois Physics Education Research Group, Urbana, 2006), <http://per.physics.illinois.edu/per/IE/ie.pl?phys111/ie/03/IE_block_on_incline>.
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Save to my folders | 1,272 | 5,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-34 | latest | en | 0.885227 |
https://www.geeksforgeeks.org/count-of-subsequences-from-a-given-array-having-binary-equivalence/?ref=rp | 1,679,420,243,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943704.21/warc/CC-MAIN-20230321162614-20230321192614-00155.warc.gz | 873,508,081 | 35,172 | # Count of subsequences from a given Array having Binary Equivalence
• Last Updated : 14 Mar, 2023
Given an array arr[] consisting of N integers, the task is to find the total number of distinct subsequences having Binary Equivalence.
A subsequence has Binary Equivalence if the sum of the count of set and unset bits in the binary representations of all the decimal numbers across the subsequence are equal.
Examples:
Input: arr[] = {2, 7, 10}
Output: 0011
Explanation:
2 → 0010→1’s = 1, 0’s = 3
7 → 0111→1’s = 3, 0’s = 1
10 → 1010→1’s = 2, 0’s = 2
The subsequence [2, 7, 10] has Binary Equivalence because the number of 0’s and 1’s across the subsequence is 6 each.
Similarly, [2, 7] also has Binary Equivalence of 4 each.
But [7, 10] does not have Binary Equivalence.
Likewise, [10] has Binary Equivalence of 2 each.
The total number of unique subsequences where Binary Equivalence is possible is 3.
Since 10 is the largest element in the given array and the number of bits required to represent 10 in binary is 4. Hence, the number of bits present in the output needs to be 4.
Input: arr[] = {5, 7, 9, 12}
Output: 0111
Approach: The idea is to find the total number of bits required to represent the maximum element of the array.Follow these steps to solve this problem:
1. Find the maximum element and the length of binary representation of the maximum element.
2. Append 0 in the front other elements in binary representation, to make the number of bits in each element equal to the maximum number bits.
3. Find all the subsequences of the given array.
4. Find the total number of subsequences that have Binary Equivalence.
5. Convert the total number into binary and append 0s if the length of the total number is less than the length of the maximum number to make both the lengths equal.
Below is the implementation of the above approach:
## C++
`#include ``using` `namespace` `std;` `// C++ program for the above approach` `// Function to find the number of``// subsequences having Binary Equivalence``void` `numberOfSubsequence(vector<``int``> arr) {` ` ``// Find the maximum array element`` ``int` `maxElement = INT_MIN;`` ``for``(``auto` `x: arr) maxElement = max(x, maxElement);` ` ``// Convert the maximum element`` ``// to its binary equivalent`` ``int` `val = (``int``)(log2(maxElement));`` ``string maxBinary = bitset<64>(maxElement).to_string().substr(64 - val - 1);` ` ``// Dictionary to store the count of`` ``// set and unset bits of all array elements`` ``map<``int``,pair<``int``,``int``>> dic;` ` ``for``(``auto` `i: arr){`` ``int` `temp = (``int``)(log2(maxElement));`` ``string str = bitset<64>(i).to_string().substr(64 - temp - 1);` ` ``if` `(str.size() <= maxBinary.size()) {`` ``int` `diff = maxBinary.size() - str.size();` ` ``// Add the extra zeros before all`` ``// the elements which have length`` ``// smaller than the maximum element`` ``for``(``int` `indx = 0; indx < diff; indx++){`` ``str = ``'0'` `+ str;`` ``}`` ``}` ` ``int` `zeros = 0;`` ``int` `ones = 0;`` ``for``(``int` `i = 0; i < str.size(); i++){`` ``if``(str[i] == ``'0'``) zeros++;`` ``else` `ones++;`` ``}` ` ``// Fill the dictionary with number`` ``// of 0's and 1's`` ``dic[i] = {zeros, ones};`` ``}` ` ``vector> allCombinations;`` ``vector<``int``> curr_temp;`` ``allCombinations.push_back(curr_temp);` ` ``// Find all the combinations`` ``for` `(``int` `i : arr) {`` ``// cout << "hi" << endl;`` ``vector> newCombinations;`` ``for` `(``auto` `j : allCombinations) {`` ``vector<``int``> combination = j;`` ``combination.push_back(i);`` ``newCombinations.push_back(combination);` ` ``}`` ``for``(``auto` `curr_list: newCombinations){`` ``allCombinations.push_back(curr_list);`` ``}`` ``}` ` ``int` `count = 0;`` ``// Find all the combinations where`` ``// sum_of_zeros == sum_of_ones`` ``for` `(``int` `i = 1; i < allCombinations.size(); i++) {`` ``int` `sum0 = 0;`` ``int` `sum1 = 0;`` ``for` `(``auto` `j: allCombinations[i]) {`` ``sum0 += dic[j].first;`` ``sum1 += dic[j].second;`` ``}` ` ``// Count the total combinations`` ``// where sum_of_zeros = sum_of_ones`` ``if` `(sum0 == sum1) {`` ``count += 1;`` ``}`` ``}` ` ``// Convert the count number to its`` ``// binary equivalent`` ``int` `curr_val = (``int``)(log2(count));`` ``string str = bitset<64>(count).to_string().substr(64 - curr_val - 1);`` ``if` `(str.size() <= maxBinary.size()) {`` ``int` `diff = maxBinary.size() - str.size();` ` ``// Append leading zeroes to`` ``// the answer if its length is`` ``// smaller than the maximum element`` ``for``(``int` `i = 0; i < diff; i++){`` ``str = ``'0'` `+ str;`` ``}`` ``}` ` ``// Print the result`` ``cout << str << endl;``}` `// Driver Code``int` `main(){` ` ``// Given arr[] arr.`` ``vector<``int``> arr = {5, 7, 9, 12};` ` ``// Function Call`` ``numberOfSubsequence(arr);` ` ``return` `0;``}` `// The code is contributed by Nidhi goel.`
## Java
`import` `java.util.*;` `class` `Main {`` ``public` `static` `void` `main(String[] args) {`` ``int``[] arr = {``5``, ``7``, ``9``, ``12``};`` ``numberOfSubsequence(arr);`` ``}` ` ``// Function to find the number of subsequences`` ``// having Binary Equivalence`` ``static` `void` `numberOfSubsequence(``int``[] arr) {` ` ``// Find the maximum array element`` ``int` `maxElement = Arrays.stream(arr).max().getAsInt();` ` ``// Convert the maximum element to its binary equivalent`` ``String maxBinary = Integer.toBinaryString(maxElement);` ` ``// Dictionary to store the count of set and unset bits`` ``// of all array elements`` ``Map dic = ``new` `HashMap<>();`` ``for` `(``int` `i : arr) {`` ``String str = Integer.toBinaryString(i);`` ``if` `(str.length() < maxBinary.length()) {`` ``int` `diff = maxBinary.length() - str.length();`` ``// Add extra zeros before all the elements which have length`` ``// smaller than the maximum element`` ``str = ``"0"``.repeat(diff) + str;`` ``}` ` ``// Fill the dictionary with number of 0's and 1's`` ``int` `zeros = str.length() - str.replaceAll(``"0"``, ``""``).length();`` ``int` `ones = str.length() - str.replaceAll(``"1"``, ``""``).length();`` ``dic.put(i, ``new` `int``[]{zeros, ones});`` ``}` ` ``List> allCombinations = ``new` `ArrayList<>();`` ``allCombinations.add(``new` `ArrayList<>());` ` ``// Find all the combinations`` ``for` `(``int` `i : arr) {`` ``List> newCombinations = ``new` `ArrayList<>();`` ``for` `(List j : allCombinations) {`` ``List combination = ``new` `ArrayList<>(j);`` ``combination.add(i);`` ``newCombinations.add(combination);`` ``}`` ``allCombinations.addAll(newCombinations);`` ``}` ` ``int` `count = ``0``;`` ``// Find all the combinations where sum_of_zeros == sum_of_ones`` ``for` `(``int` `i = ``1``; i < allCombinations.size(); i++) {`` ``int` `sum0 = ``0``;`` ``int` `sum1 = ``0``;`` ``for` `(``int` `j : allCombinations.get(i)) {`` ``sum0 += dic.get(j)[``0``];`` ``sum1 += dic.get(j)[``1``];`` ``}` ` ``// Count the total combinations where sum_of_zeros = sum_of_ones`` ``if` `(sum0 == sum1) {`` ``count += ``1``;`` ``}`` ``}` ` ``// Convert the count number to its binary equivalent`` ``String str = Integer.toBinaryString(count);`` ``if` `(str.length() < maxBinary.length()) {`` ``int` `diff = maxBinary.length() - str.length();`` ``// Append leading zeroes to the answer if its length is`` ``// smaller than the maximum element`` ``str = ``"0"``.repeat(diff) + str;`` ``}` ` ``// Print the result`` ``System.out.println(str);`` ``}``}`
## Python3
`# Python program for the above approach``import` `itertools` `# Function to find the number of``# subsequences having Binary Equivalence``def` `numberOfSubsequence(arr):` ` ``# Find the maximum array element`` ``Max_element ``=` `max``(arr)` ` ``# Convert the maximum element`` ``# to its binary equivalent`` ``Max_Binary ``=` `"{0:b}"``.``format``(``int``(`` ``Max_element))` ` ``# Dictionary to store the count of`` ``# set and unset bits of all array elements`` ``Dic ``=` `{}` ` ``for` `i ``in` `arr:`` ``Str` `=` `"{0:b}"``.``format``(``int``(i))` ` ``if` `len``(``Str``) <``=` `len``(Max_Binary):`` ``diff ``=` `len``(Max_Binary)``-``len``(``Str``)` ` ``# Add the extra zeros before all`` ``# the elements which have length`` ``# smaller than the maximum element`` ``Str` `=` `(``'0'``*``diff)``+``Str` ` ``zeros ``=` `Str``.count(``'0'``)`` ``ones ``=` `Str``.count(``'1'``)` ` ``# Fill the dictionary with number`` ``# of 0's and 1's`` ``Dic[``int``(i)] ``=` `[zeros, ones]` ` ``all_combinations ``=` `[]` ` ``# Find all the combination`` ``for` `r ``in` `range``(``len``(arr)``+``1``):` ` ``comb ``=` `itertools.combinations(arr, r)`` ``comlist ``=` `list``(comb)`` ``all_combinations ``+``=` `comlist`` ``count ``=` `0` ` ``# Find all the combinations where`` ``# sum_of_zeros == sum_of_ones`` ``for` `i ``in` `all_combinations[``1``:]:`` ``sum0 ``=` `0`` ``sum1 ``=` `0`` ``for` `j ``in` `i:`` ``sum0 ``+``=` `Dic[j][``0``]`` ``sum1 ``+``=` `Dic[j][``1``]` ` ``# Count the total combinations`` ``# where sum_of_zeros = sum_of_ones`` ``if` `sum0 ``=``=` `sum1:`` ``count ``+``=` `1` ` ``# Convert the count number to its`` ``# binary equivalent`` ``Str` `=` `"{0:b}"``.``format``(``int``(count))`` ``if` `len``(``Str``) <``=` `len``(Max_Binary):`` ``diff ``=` `len``(Max_Binary)``-``len``(``Str``)` ` ``# Append leading zeroes to`` ``# the answer if its length is`` ``# smaller than the maximum element`` ``Str` `=` `(``'0'``*``diff) ``+` `Str` ` ``# Print the result`` ``print``(``Str``)` `# Driver Code` `# Give array arr[]``arr ``=` `[``5``, ``7``, ``9``, ``12``]` `# Function Call``numberOfSubsequence(arr)`
## C#
`// Following is the C# equivalent of the above Java code``using` `System;``using` `System.Collections.Generic;``using` `System.Linq;` `namespace` `NumberOfSubsequence``{`` ``class` `Program`` ``{`` ``static` `void` `Main(``string``[] args)`` ``{`` ``int``[] arr = { 5, 7, 9, 12 };`` ``NumberOfSubsequence(arr);`` ``}` ` ``// Function to find the number of subsequences`` ``// having Binary Equivalence`` ``static` `void` `NumberOfSubsequence(``int``[] arr)`` ``{`` ``// Find the maximum array element`` ``int` `maxElement = arr.Max();` ` ``// Convert the maximum element to its binary equivalent`` ``string` `maxBinary = Convert.ToString(maxElement, 2);` ` ``// Dictionary to store the count of set and unset bits`` ``// of all array elements`` ``Dictionary<``int``, ``int``[]> dic = ``new` `Dictionary<``int``, ``int``[]>();`` ``foreach``(``int` `i ``in` `arr)`` ``{`` ``string` `strr = Convert.ToString(i, 2);`` ``if``(strr.Length < maxBinary.Length)`` ``{`` ``int` `diff = maxBinary.Length - strr.Length;`` ``// Add extra zeros before all the elements which have length`` ``// smaller than the maximum element`` ``strr = ``new` `string``(``'0'``, diff) + strr;`` ``}` ` ``// Fill the dictionary with number of 0's and 1's`` ``int` `zeros = strr.Length - strr.Replace(``"0"``, ``""``).Length;`` ``int` `ones = strr.Length - strr.Replace(``"1"``, ``""``).Length;`` ``dic.Add(i, ``new` `int``[] { zeros, ones });`` ``}` ` ``List> allCombinations = ``new` `List>();`` ``allCombinations.Add(``new` `List<``int``>());` ` ``// Find all the combinations`` ``foreach``(``int` `i ``in` `arr)`` ``{`` ``List> newCombinations = ``new` `List>();`` ``foreach``(List<``int``> j ``in` `allCombinations)`` ``{`` ``List<``int``> combination = ``new` `List<``int``>(j);`` ``combination.Add(i);`` ``newCombinations.Add(combination);`` ``}`` ``allCombinations.AddRange(newCombinations);`` ``}` ` ``int` `count = 0;`` ``// Find all the combinations where sum_of_zeros == sum_of_ones`` ``for``(``int` `i = 1; i < allCombinations.Count; i++)`` ``{`` ``int` `sum0 = 0;`` ``int` `sum1 = 0;`` ``foreach``(``int` `j ``in` `allCombinations[i])`` ``{`` ``sum0 += dic[j][0];`` ``sum1 += dic[j][1];`` ``}` ` ``// Count the total combinations where sum_of_zeros = sum_of_ones`` ``if``(sum0 == sum1)`` ``{`` ``count += 1;`` ``}`` ``}` ` ``// Convert the count number to its binary equivalent`` ``string` `str = Convert.ToString(count, 2);`` ``if``(str.Length < maxBinary.Length)`` ``{`` ``int` `diff = maxBinary.Length - str.Length;`` ``// Append leading zeroes to the answer if its length is`` ``// smaller than the maximum element`` ``str = ``new` `string``(``'0'``, diff) + str;`` ``}` ` ``// Print the result`` ``Console.WriteLine(str);`` ``}`` ``}``}`
## Javascript
`// JavaScript program for the above approach` `// Function to find the number of``// subsequences having Binary Equivalence``function` `numberOfSubsequence(arr) {` ` ``// Find the maximum array element`` ``let maxElement = Math.max(...arr);` ` ``// Convert the maximum element`` ``// to its binary equivalent`` ``let maxBinary = maxElement.toString(2);` ` ``// Dictionary to store the count of`` ``// set and unset bits of all array elements`` ``let dic = {};` ` ``for` `(let i of arr) {`` ``let str = i.toString(2);` ` ``if` `(str.length <= maxBinary.length) {`` ``let diff = maxBinary.length - str.length;` ` ``// Add the extra zeros before all`` ``// the elements which have length`` ``// smaller than the maximum element`` ``str = ``'0'``.repeat(diff) + str;`` ``}` ` ``let zeros = str.split(``'0'``).length - 1;`` ``let ones = str.split(``'1'``).length - 1;` ` ``// Fill the dictionary with number`` ``// of 0's and 1's`` ``dic[i] = [zeros, ones];`` ``}` ` ``let allCombinations = [[]];` ` ``// Find all the combination`` ``for` `(let i of arr) {`` ``let newCombinations = [];`` ``for` `(let j of allCombinations) {`` ``newCombinations.push(j.concat(i));`` ``}`` ``allCombinations = allCombinations.concat(newCombinations);`` ``}`` ``let count = 0;` ` ``// Find all the combinations where`` ``// sum_of_zeros == sum_of_ones`` ``for` `(let i = 1; i < allCombinations.length; i++) {`` ``let sum0 = 0;`` ``let sum1 = 0;`` ``for` `(let j of allCombinations[i]) {`` ``sum0 += dic[j][0];`` ``sum1 += dic[j][1];`` ``}` ` ``// Count the total combinations`` ``// where sum_of_zeros = sum_of_ones`` ``if` `(sum0 == sum1) {`` ``count += 1;`` ``}`` ``}` ` ``// Convert the count number to its`` ``// binary equivalent`` ``let str = count.toString(2);`` ``if` `(str.length <= maxBinary.length) {`` ``let diff = maxBinary.length - str.length;` ` ``// Append leading zeroes to`` ``// the answer if its length is`` ``// smaller than the maximum element`` ``str = ``'0'``.repeat(diff) + str;`` ``}` ` ``// Print the result`` ``console.log(str);``}` `// Driver Code` `// Give array arr[]``let arr = [5, 7, 9, 12];` `// Function Call``numberOfSubsequence(arr);`
Output:
`0111`
Time Complexity: O(2N)
Auxiliary Space: O(N2)
My Personal Notes arrow_drop_up | 5,364 | 16,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.883508 |
https://groups.yahoo.com/neo/groups/primenumbers/conversations/topics/20222?l=1 | 1,441,029,225,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644066017.21/warc/CC-MAIN-20150827025426-00226-ip-10-171-96-226.ec2.internal.warc.gz | 869,531,497 | 17,054 | ## Re: Lehmer sequence puzzle
Expand Messages
• ... OK Boss, message received and understood. Sometimes such discoveries are best not cast as puzzles. Discovery: Mike Oakes has found a (Q,R) integer pair
Message 1 of 33 , May 7, 2009
• 0 Attachment
"Mike Oakes" <mikeoakes2@...> wrote:
> I don't know at the moment how to fix my puzzle to stop
> you guys giving the answer \infty straight away.
...
> I posed the puzzle because I have found, experimentally,
> examples where u(R,Q) is 1, 2 and even
> (a great surprise to me!) 3.
> I want to know if there are any examples >3 (and < \infty).
> Help me, please, to both formulate the puzzle securely
> and solve it!
OK Boss, message received and understood.
Sometimes such discoveries are best not cast as puzzles.
Discovery:
Mike Oakes has found a (Q,R) integer pair such that with
x = sqrt(R)/2 + sqrt(R-4*Q)/2
y = sqrt(R)/2 - sqrt(R-4*Q)/2
(x^n+y^n)/(x+y) is a unit precisely 3 times for n > 1.
Questions:
Can anyone improve on Mike's tally of 3,
in a situation where the number of units is finite?
If not, can anyone prove that 3 is the maximum,
in a situation where the number of units is finite?
C'est ça?
David
• ... Yes, Mike, that last form is by far the neatest, if one removes the unnecessary lucas , which should be well understood, when one speaks of U or V . 1)
Message 33 of 33 , May 10, 2009
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<mikeoakes2@...> wrote [with unfailing courtesy]:
> lucasU(338,25,11584)-5*lucasU(338,25,11583)
> I guess Chris would expect the last of these as being the most
> "canonical"? And I concur, it being also the shortest.
Yes, Mike, that last form is by far the neatest,
if one removes the unnecessary "lucas", which should
be well understood, when one speaks of "U" or "V".
1) François Édouard Anatole Lucas defined the integer sequence
U(P,Q,n) = P*U(P,Q,n-1) - Q*U(P,Q,n-2),
with U(P,Q,0) = 0 and U(P,Q,1) = 1,
in terms of elementary arithmetic.
2) I like to imagine that Derrick Norman Lehmer (1867-1938) saw that
U(P,Q^2,2*k+1) = U(P,Q^2,k+1)^2 - (Q*U(P,Q^2,n))^2
and told his son: "Go look at the factors for your Ph.D."
3) Certainly, Derrick Henry Lehmer (1905-1991) did study
U(P,Q^2,k+1) - Q*U(P,Q^2,k)
being very well aware of the super-Lucasian extension to powers
of algebraic numbers of degree 4, in the wider complex plane.
4) Mike Oakes has found that the Lehmer number
U(P,Q^2,k+1) - Q*U(P,Q^2,k)
is probably prime, when P = 338, Q = 5, k = 11583.
5) David Broadhurst is able to characterize this
circumstance without using a dirty 4-letter word :-)
Thank ye, kindly, Sir, for this interesting thread!
David, pp SSSR
Your message has been successfully submitted and would be delivered to recipients shortly. | 817 | 2,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2015-35 | longest | en | 0.914976 |
https://www.aqua-calc.com/calculate/mole-to-volume-and-weight/substance/beryllium-blank-chloride | 1,701,978,477,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100686.78/warc/CC-MAIN-20231207185656-20231207215656-00518.warc.gz | 721,673,002 | 8,049 | # Moles of Beryllium chloride
## beryllium chloride: convert moles to volume and weight
### Volume of 1 mole of Beryllium chloride
centimeter³ 42.06 milliliter 42.06 foot³ 0 oil barrel 0 Imperial gallon 0.01 US cup 0.18 inch³ 2.57 US fluid ounce 1.42 liter 0.04 US gallon 0.01 meter³ 4.21 × 10-5 US pint 0.09 metric cup 0.17 US quart 0.04 metric tablespoon 2.8 US tablespoon 2.84 metric teaspoon 8.41 US teaspoon 8.53
### Weight of 1 mole of Beryllium chloride
carat 399.56 ounce 2.82 gram 79.91 pound 0.18 kilogram 0.08 tonne 7.99 × 10-5 milligram 79 912
### The entered amount of Beryllium chloride in various units of amount of substance
centimole 100 micromole 1 000 000 decimole 10 millimole 1 000 gigamole 1 × 10-9 mole 1 kilogram-mole 0 nanomole 1 000 000 000 kilomole 0 picomole 1 000 000 000 000 megamole 1 × 10-6 pound-mole 0
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WHOLE KERNEL CORN, UPC: 074175681425 weigh(s) 264 grams per metric cup or 8.8 ounces per US cup, and contain(s) 64 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
3 foods that contain Lutein. List of these foods starting with the highest contents of Lutein and the lowest contents of Lutein
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CaribSea, Freshwater, Instant Aquarium, Peace River weighs 1 489.72 kg/m³ (93.00018 lb/ft³) with specific gravity of 1.48972 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Peanuts, shelled weighs 641 kg/m³ (40.01632 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Engine Oil, SAE 10W-40 with temperature in the range of 0°C (32°F) to 100°C (212°F)
#### Weights and Measurements
The pennyweight per yard is a measurement unit of linear or linear mass density
The frequency is defined as an interval of time, during which a physical system, e.g. electrical current or a wave, performs a full oscillation and returns to its original momentary state, in both sign (direction) and in value, is called the oscillation period of this physical system.
ym/s² to au/h² conversion table, ym/s² to au/h² unit converter or convert between all units of acceleration measurement.
#### Calculators
Price conversions and cost calculator for materials and substances | 723 | 2,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.644693 |
https://studylib.net/doc/25208006/darcy-s-law | 1,558,347,756,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255943.0/warc/CC-MAIN-20190520101929-20190520123929-00241.warc.gz | 662,344,632 | 42,562 | # Darcy's Law
```Darcy’s Law
In 1856, Darcy investigated the flow of water through sand filters for water
purification purposes. His experimental apparatus is shown below:
Output manometer
Input manometer
Water
h1-h2
Water in
1
Sand Pack
Flow direction
A
h1
L
h2
Datum
2
Water out
at a rate q
Water
Water
Where q is the volume flow rate of water downward through the cylindrical sand
pack. The sand pack has a length L and a cross-sectional area A. h1 is the height
above a datum of water in a manometer located at the input face. h2 is the height
above a datum of water in a manometer located at the output face. The following
assumptions are implicit in Darcy’s experiment:
1.
2.
3.
4.
5.
Single-phase flow (only water)
Homogeneous porous medium (sand)
Vertical flow.
Non-reactive fluid (water)
Single geometry.
From this experiment, Darcy concluded the following points:
The volume flow rate is directly proportional to the difference of water level in the
two manometers; i.e.:
1
q ∝ h1 − h2
The volume flow rate is directly proportional to the cross-sectional area of the sand
pack; i.e.:
q∝A
The volume flow rate is inversely proportional to the length of the sand pack; i.e.:
q∝
1
L
Thus we write:
q=C
A
(h1 − h2 )
L
Where:
A is the cross-sectional area of the sand pack
L is the length of the sand pack
h1 is the height above a datum of water in a manometer located at the input face
h2 is the height above a datum of water in a manometer located at the output face
C is the proportionality constant which depends on the rock and fluid properties. For
the fluid effect, C is directly proportional to the fluid specific weight; i.e.
C ∝γ
and inversely proportional to the fluid viscosity; i.e.:
C∝
1
µ
Thus:
C∝
γ
µ
For the rock effect, C is directly proportional to the square of grain size; i.e.:
C ∝ (grain size) = d 2
2
It is inversely proportional to tortuosity; i.e.:
C∝
1
1
=
tortuosity τ
2
and inversely proportional to the specific surface; i.e.:
C∝
1
1
=
specific surface S s
where Ss is given by:
Ss =
Interstitial surface area
bulk volume
Combine the above un-measurable rock properties into one property, call it
permeability, and denote it by K, we get:
q=K
γ A
(h − h )
µ L 1 2
Since:
q = vA
Thus we can write:
v=
q
γ (h1 − h2 )
=K
A
µ
L
Now, let us consider the more realistic flow; i.e. the tilted flow for the same sand
pack:
Datum
P2, D2
2
P1, D1
w
Flo
1
io
ect
r
i
D
L
3
n
A
Note that the fluid flows from point 1 to point 2, which means that the pressure at
point 1 is higher than the pressure at point 2. Since:
h1 = D1 −
p1
h2 = D2 −
&
γ
p2
γ
Thus:
p
p
1
h1 − h2 = D1 − 1 − D2 − 2 = ( D1 − D2 ) − ( p1 − p2 )
γ
γ
γ
which can be written in a difference form as:
h1 − h2 = ∆D −
1
γ
∆p
Substituting (2) into (1) and rearranging yields:
v=K
γ ∆D 1 ∆p
∆D
K ∆p
−
=−
−γ
µ L γ L
µL
L
The differential form of Darcy’s equation for single-phase flow is written as follows:
v=−
∂D
K ∂p
−γ
∂L
µ ∂L
For multi-phase flow, we write Darcy’s equation as follows:
k ∂p
∂D
v = − K r − γ
∂L
µ ∂L
More conveniently, the compact differential form of Darcy’s equation is written as
follows:
k
r
v = − K r (∇p − γ∇D)
µ
Where K is the absolute permeability tensor which must be determined
experimentally. It is written as follows:
4
k xx
K = k yx
k zx
k xy
k yy
k zy
k xz
k yz
k zz
Substitute (5) into (4), we obtain:
k xx
v x
kr
v y = − k yx
µ
v
k zx
z
k xy
k yy
k zy
∂p
−γ
k xz ∂x
∂p
k yz − γ
∂y
k zz ∂p
−γ
∂z
∂D
∂x
∂D
∂y
∂D
∂z
Solve for velocity components yields:
k ∂p
∂p
∂D
∂D
∂D
∂p
v x = − r k xx − γ
+ k xy − γ
+ k xz − γ
∂z
∂x
∂y
∂z
∂y
µ ∂x
k ∂p
∂p
∂D
∂D
∂D
∂p
v y = − r k yx − γ
+ k yy − γ
+ k yz − γ
∂z
∂x
∂y
∂z
∂y
µ ∂x
k ∂p
∂p
∂D
∂D
∂D
∂p
v z = − r k zx − γ
+ k zy − γ
+ k zz − γ
∂z
∂x
∂y
∂z
µ ∂x
∂y
In most practical problems, it is necessary to assume that K is a diagonal tensor; i.e.:
k xx
K=0
0
0
k yy
0
0
0
k zz
and thus velocity components of equation (6) become:
k ∂p
∂D
v x = − k xx r − γ
∂x
µ ∂x
k ∂p
∂D
v y = − k yy r − γ
∂y
µ ∂y
k ∂p
∂D
v z = − k zz r − γ
∂z
µ ∂z
5
Unit Analysis
Since:
KA ∆p
qµ ∆L
⇒
K=
A ∆p
µ ∆L
2
When q in cc/sec, µ in cp, )L in cm, A in cm , and )p in atm, then K is in the units of
Darcy, where:
q=
(1 cm )(1 cp)(1 cm) = (1 cm ) (1 cp)
1 Darcy =
(s) (1 cm )(1 atm) (1 atm) (1 s)
3
2
2
Since
1 cp =
1
1 dyne s
poise =
100
100 cm 2
and
1 atm = 1,013,250
dyne
cm 2
thus we have:
1 cm 2 1 dyne s cm 2
1
1 Darcy = 1000 md =
=
cm 2
2
(100) cm (1,013,250) dyne s 101,325,000
or
1 md =
10 − 9
cm 2 = 986.923x10 −14 cm 2
.
101325
Since:
q=
KA ∆p
µ ∆L
Thus:
bbl
day
(md )( ft 2 )( psi ) (md )( ft )( psi )
=
=
(cp)( ft )
(cp)
2
=
10 − 9
lb
1 1
bbl 144 2
. 2.54 x12 5.6146
ft
101325
1 1 lb − day
47,880 86,400 ft 2
≈ 0.001127107
6
= 1127106.663x10 − 9
```
Arab people
15 Cards | 2,818 | 5,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-22 | latest | en | 0.883968 |
https://kr.mathworks.com/help/sps/ref/mutualinductor.html | 1,685,800,041,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649293.44/warc/CC-MAIN-20230603133129-20230603163129-00561.warc.gz | 378,895,339 | 29,123 | Mutual Inductor
Mutual inductor model with nominal inductance optional tolerances for each winding, operating limits and faults
Libraries:
Simscape / Electrical / Passive / Transformers
Description
The Mutual Inductor block lets you model a mutual inductor (two-winding transformer) with nominal inductance tolerances for each winding. The model includes the following effects:
You can turn these modeling options on and off independently of each other.
In the unfaulted state, the following equations describe the Mutual Inductor block behavior:
`${v}_{1}={L}_{1}\frac{d{i}_{L1}}{dt}+\text{}M\frac{d{i}_{L2}}{dt}+{i}_{L1}{R}_{1}$`
`${v}_{2}={L}_{2}\frac{d{i}_{L2}}{dt}+\text{}M\frac{d{i}_{L1}}{dt}+{i}_{L2}{R}_{2}$`
`$M=k\sqrt{{L}_{1}{L}_{2}}$`
where:
• v1 and v2 are voltages across the primary and secondary winding, respectively.
• L1 and L2 are inductances of the primary and secondary winding.
• R1 and R2 are series resistances of the primary and secondary winding.
• M is mutual inductance.
• k is coefficient of coupling. To reverse one of the winding directions, use a negative value.
• t is time.
A parallel conductance is placed across the + and – terminals of the primary and secondary windings, so that iL1 = i1G1v1, where G1 is the parallel conductance of the primary winding, and i1 is the terminal current into the primary. Similar definitions and equation apply to iL2.
Tolerances
You can apply tolerances separately for each winding. Datasheets typically provide a tolerance percentage for a given inductor type. Therefore, this value is the same for both windings. The table shows how the block applies tolerances to the nominal inductance value and calculates inductance based on the selected tolerance application option for the winding, L1 tolerance application or L2 tolerance application.
OptionInductance Value
```None — use nominal value```
L
`Random tolerance`
Uniform distribution: L · (1 – tol + 2· tol· `rand`)
Gaussian distribution: L · (1 + tol · `randn` / nSigma)
```Apply maximum tolerance value```
L · (1 + tol )
```Apply minimum tolerance value```
L · (1 – tol )
In the table:
• L is nominal inductance for the primary or secondary winding, Inductance L1 or Inductance L2 parameter value.
• tol is fractional tolerance, Tolerance (%) /100.
• nSigma is the value you provide for the Number of standard deviations for quoted tolerance parameter.
• `rand` and `randn` are standard MATLAB® functions for generating uniform and normal distribution random numbers.
Note
If you choose the `Random tolerance` option and you are in "Fast Restart" mode, the random tolerance value is updated on every simulation if at least one between the fractional tolerance, tol, or the Number of standard deviations for quoted tolerance, nSigma, is set to Run-time and is defined with a variable (even if you do not modify that variable).
Operating Limits
Inductors are typically rated with a particular saturation current, and possibly with a maximum allowable power dissipation. You can specify operating limits in terms of these values, to generate warnings or errors if the inductor is driven outside its specification.
When an operating limit is exceeded, the block can either generate a warning or stop the simulation with an error. For more information, see the Operating Limits parameters section.
Faults
Instantaneous changes in inductor parameters are unphysical. Therefore, when the Mutual Inductor block enters the faulted state, short-circuit and open-circuit voltages transition to their faulted values over a period of time based on this formula:
`CurrentValue` = `FaultedValue` – (`FaultedValue``UnfaultedValue`) · `sech`(∆t / τ)
where:
• ∆t is time since the onset of the fault condition.
• τ is the value of the Fault transition time constant parameter.
For short-circuit faults, the conductance of the short-circuit path also changes according to the `sech`(∆t / τ) function from a small value (representing an open-circuit path) to a large value.
The Mutual Inductor block lets you select whether the faults occur in the primary or secondary winding. The block models the faulted winding as a faulted inductor. The unfaulted winding is coupled to the faulted winding. As a result, the actual equations involve a total of three coupled windings: two for the faulted winding and one for the unfaulted winding. The coupling between the primary and secondary windings is defined by the Coefficient of coupling parameter.
The block can trigger the start of fault transition:
• At a specific time.
• After voltage exceeds the maximum permissible value a certain number of times.
• When current exceeds the maximum permissible value for longer than a specific time interval.
If you want to trigger a fault at a specific time, set the Fault trigger parameter to `Temporal`. If you want to determine whether a system fails and, if so, when it fails, set the Fault Trigger parameter to `Behavioral`.
If you select the behavioral trigger, the component fails as soon as one of the trigger conditions is true.
You can also choose whether to issue an assertion when a fault occurs by using the Reporting when a fault occurs parameter. The assertion can take the form of a warning or an error. By default, the block does not issue an assertion.
Faultable inductors often require that you use the fixed-step local solver, especially if your model transitions to a faulted state that includes short circuits. For more information, see Making Optimal Solver Choices for Physical Simulation.
Variables
To set the priority and initial target values for the block variables prior to simulation, use the Initial Targets section in the block dialog box or Property Inspector. For more information, see Set Priority and Initial Target for Block Variables.
Nominal values provide a way to specify the expected magnitude of a variable in a model. Using system scaling based on nominal values increases the simulation robustness. Nominal values can come from different sources, one of which is the Nominal Values section in the block dialog box or Property Inspector. For more information, see System Scaling by Nominal Values.
The Primary current and Secondary current variables let you specify a high-priority target for the initial inductor current in the respective winding at the start of simulation.
Ports
Conserving
expand all
Electrical conserving port associated with the primary winding positive terminal.
Electrical conserving port associated with the primary winding negative terminal.
Electrical conserving port associated with the secondary winding positive terminal.
Electrical conserving port associated with the secondary winding negative terminal.
Parameters
expand all
Main
The nominal inductance value in the primary winding. Inductance value must be greater than zero.
The nominal inductance value in the secondary winding. Inductance value must be greater than zero.
The coupling between the primary and secondary windings. This coefficient defines the mutual inductance. To reverse one of the winding directions, use a negative value.
The inductor tolerance as defined on the manufacturer datasheet. Datasheets typically provide a tolerance percentage for a given inductor type. Therefore, this value is the same for both windings.
Select how to apply tolerance during simulation to the primary winding:
• `None — use nominal value` — The block does not apply tolerance, it uses the nominal inductance value.
• `Random tolerance` — The block applies random offset to the inductance value, within the tolerance value limit. You can choose Uniform or Gaussian distribution for calculating the random number by using the Tolerance distribution parameter.
• `Apply maximum tolerance value` — The inductance is increased by the specified tolerance percent value.
• `Apply minimum tolerance value` — The inductance is decreased by the specified tolerance percent value.
Select the distribution type for random tolerance:
• `Uniform` — Uniform distribution
• `Gaussian` — Gaussian distribution
Dependencies
Enabled when the L1 tolerance application parameter is set to ```Random tolerance```.
Number of standard deviations for calculating the Gaussian random number.
Dependencies
Enabled when the L1 tolerance distribution parameter is set to `Gaussian`.
Select how to apply tolerance during simulation to the secondary winding:
• `None — use nominal value` — The block does not apply tolerance, it uses the nominal inductance value.
• `Random tolerance` — The block applies random offset to the inductance value, within the tolerance value limit. You can choose Uniform or Gaussian distribution for calculating the random number by using the Tolerance distribution parameter.
• `Apply maximum tolerance value` — The inductance is increased by the specified tolerance percent value.
• `Apply minimum tolerance value` — The inductance is decreased by the specified tolerance percent value.
Select the distribution type for random tolerance:
• `Uniform` — Uniform distribution
• `Gaussian` — Gaussian distribution
Dependencies
Enabled when the L2 tolerance application parameter is set to ```Random tolerance```.
Number of standard deviations for calculating the Gaussian random number.
Dependencies
Enabled when the L2 tolerance distribution parameter is set to `Gaussian`.
Resistance
Equivalent series resistance of the primary and secondary winding, specified as a two-element vector. The first number corresponds to the primary winding, the second number to the secondary winding. For a faulted winding, the block allocates the resistance to each segment in proportion to the number of turns in that segment.
Parallel leakage path associated with the primary and secondary winding, specified as a two-element vector. The first number corresponds to the primary winding, the second number to the secondary winding. The parallel conductances are placed directly across the + and – terminals of the primary and secondary winding, respectively.
Operating Limits
Select `Yes` to enable reporting when the operational limits are exceeded. The associated parameters in the Operating Limits section become visible to let you select the reporting method and specify the operating limits in terms of power and current.
Select what happens when an operating limit is exceeded:
• `Warn` — The block issues a warning.
• `Error` — Simulation stops with an error.
Dependencies
Enabled when the Enable operating limits parameter is set to `Yes`.
Inductor saturation current, as defined in the manufacturer datasheets. If the net current into the primary and secondary windings exceeds this value, the core material enters saturation, and the block reports an operating limits violation. That is, the block compares the limit against |i1 + i2|, where currents are defined as being positive when they are into the + nodes.
Dependencies
Enabled when the Enable operating limits parameter is set to `Yes`.
Maximum instantaneous (total) power dissipation in the resistance and conductance elements associated with the mutual inductor. If the total power (including both primary and secondary winding) exceeds this number, the block reports an operating limits violation.
Dependencies
Enabled when the Enable operating limits parameter is set to `Yes`.
Faults
Select `Yes` to enable faults modeling. The associated parameters in the Faults section become visible to let you select the reporting method and specify the trigger mechanism (temporal or behavioral).
Choose whether to issue an assertion when a fault occurs:
• `None` — The block does not issue an assertion.
• `Warn` — The block issues a warning.
• `Error` — Simulation stops with an error.
Dependencies
Enabled when the Enable faults parameter is set to `Yes`.
Select whether the faults occur in the primary or secondary winding.
Dependencies
Enabled when the Enable faults parameter is set to `Yes`.
In practice, faults are enabled by segmenting the faulted winding into two coupled subinductors, connected in a series. The inductance is proportional to the square of the number of turns in the respective segment, and the series resistance of each subinductor is proportional to the number of turns in each segment. The parallel conductance spans both segments.
This parameter indicates the percentage of turns that are assigned to the subinductor that is in contact with the – port of the faulted winding. The remaining turns are assigned to the other subinductor. The default value is `50`, which means that the overall inductance of the faulted winding is divided into two equal, coupled subinductors.
Dependencies
Enabled when the Enable faults parameter is set to `Yes`.
Select whether the fault results in one of the subinductor segments being short-circuited:
• `No` — The fault does not produce a short circuit.
• `To negative terminal` — The fault short-circuits the subinductor that is in contact with the – port of the block.
• `To positive terminal` — The fault short-circuits the subinductor that is in contact with the + port of the block.
Dependencies
Enabled when the Enable faults parameter is set to `Yes`.
Select whether to apply an open-circuit fault between the two subinductor segments. The default is `No`. Even with an open-circuit fault, the characteristics of the subinductors are still related because they are magnetically coupled even in the faulted state.
Dependencies
Enabled when the Enable faults parameter is set to `Yes`.
Select whether, in case of fault, there is a path for current to flow towards the ground node:
• `No` — The fault does not result in a connection to ground.
• ```Negative terminal side of fault node``` — The side that is in contact with the – port of the block is connected to ground.
• ```Positive terminal side of fault node``` — The side that is in contact with the + port of the block is connected to ground.
If the Open-circuit at fault node parameter is set to `Yes`, you need to specify which side (negative or positive) is connected to ground. If there is no open circuit, the two options behave similarly. Physically, this corresponds to a breakdown in the insulation between the windings and the grounded core or chassis.
Dependencies
Enabled when the Enable faults parameter is set to `Yes`.
If there is a ground fault, this parameter represents conductance of the current path to ground. For example, if the path to ground is through the core material, then specify a small conductance value depending on the core material being used. For highly conductive core material or for chassis-shorts, specify a higher conductance value.
Dependencies
Enabled when the Ground fault parameter is set to ```Negative terminal side of fault node``` or ```Positive terminal side of fault node```.
Time constant associated with the transition to the faulted state, as described in Faults.
Dependencies
Enabled when the Enable faults parameter is set to `Yes`.
Since R2023a
Set this parameter to `Temporal` to enable time-based fault triggering.
Set this parameter to `Behavioral` to enable behavioral fault triggering.
Dependencies
To enable this parameter, set Enable faults to `Yes`.
Simulation time at which the block enters the faulted state.
Dependencies
To enable this parameter, set Fault trigger to `Temporal`.
Define the voltage threshold to a fault transition. If the voltage value exceeds this threshold a certain number of times, specified by the Number of events to fail when exceeding voltage parameter value, then the block starts entering the fault state.
Dependencies
To enable this parameter, set Fault trigger to `Behavioral`.
Because the physical mechanism underlying voltage-based failures depends on one or more partial discharge events occurring, this parameter allows you to set the number of voltage overshoots that the inductor can withstand before the fault transition begins. Note that the block does not check the time spent in the overvoltage condition, only the number of transitions.
Dependencies
To enable this parameter, set Fault trigger to `Behavioral`.
Current threshold to a fault transition. If the current exceeds this value for longer than the Time to fail when exceeding current parameter value, then the block enters the faulted state.
Dependencies
To enable this parameter, set Fault trigger to `Behavioral`.
Maximum length of time that the current can exceed the maximum permissible value without triggering the fault.
Dependencies
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a quadratic equation with roots -1+4i, -1-4i is x^2 + 2x + 17 = 0 sum of the roots = (-1+4i) + (-1-4i) = -2 product of the roots = (-1+4i)(-1-4i) = (-1)^2-(4i)^2 = 1-16i^2 = 1-16(-1) = 17 The equation is x^2 + 2x + 17 = 0
Question
Updated 11/20/2014 1:57:15 AM
s
Rating
8
a quadratic equation with roots -1+4i, -1-4i is x^2 + 2x + 17 = 0
sum of the roots = (-1+4i) + (-1-4i) = -2
product of the roots = (-1+4i)(-1-4i) = (-1)^2-(4i)^2 = 1-16i^2 = 1-16(-1) = 17
The equation is x^2 + 2x + 17 = 0
Confirmed by jeifunk [11/20/2014 2:00:08 AM]
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# Some useful elementary number theory
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We use most of these principles of number theory on our page on the mathematics behind the RSA algorithm.
1. The integer $n$ divides the integer $a$ if and only if there exists an integer $d$ such that $a = nd$. We say that $a$ is divisible by $n$, and that $n$ is a divisor or factor of $a$, and that $a$ is a multiple of $n$. The notation $n\mid a$ means that $n$ divides $a$, and $n\mid a - b$ means that n divides $(a - b)$. For any integer $a$ we have $1|a$, $a|a$, and $a|0$.
2. If $mn\mid a$ then $m\mid a$ and $n\mid a$ for any integers $m, n$. This follows since $mn\mid a \implies a = mnd$ for some $d$, hence $m\mid a$ and $n\mid a$.
3. A prime number is defined as an integer, greater than one, which only has positive integer divisors of the number one and itself. For example, both 7 and 11 are prime but 4 and 9 are not. An integer > 1 which is not prime is called a composite number. The number one is neither prime nor composite. The first thousand and ten thousand primes are listed here.
4. The Fundamental Theorem of Arithmetic: Every integer $n > 1$ can be represented in exactly one way as a product of primes except for the order of the factors. That is, we can write $n$ uniquely in the form $n = p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$ for primes $p_1 < p_2 < \ldots < p_r$ and positive integers $k_1, \ldots, k_r$. For example, $360 = 2^33^25^1$ and $1323 = 3^37^2$.
5. Two numbers $a$ and $b$ which have no common factors other than one are said to be coprime or relatively prime. For example, 4 and 9 are coprime but 15 and 25 are not.
6. The greatest common divisor of two integers $a$ and $b$ is the largest integer that divides both numbers and is denoted by ‘$\gcd(a,b)$’. For example, $\gcd(25, 15) = 5$ and $\gcd(4, 9) = 1$. Some texts use the equivalent term highest common factor denoted by ‘$\text{hcf}(a, b)$’. Other texts, where there is no chance of confusion, use the shorter notation ‘$(a, b)$’, e.g. $(25, 15) = 5$.
7. $a$ and $b$ are coprime if and only if $\gcd(a, b) = 1$.
8. ‘mod’ as a binary operation: The notation ‘$a = b \bmod n$’ defines a binary operation where $a$ is equal to the remainder on dividing $b$ by $n$, with $0 \le a < n$.
9. ‘mod’ as a congruence relation: The notation ‘$a \equiv b \pmod n$’ means $a$ and $b$ have the same remainder when divided by $n$, or, equivalently, $n\mid a - b$, or $a - b = nk$ for some integer $k$. We say that $a$ is congruent to $b$ modulo $n$, where $n$ is the modulus of the congruence. The two ways of using ‘mod’ are related:
$a \equiv b \pmod n \iff a \bmod n = b \bmod n$
[see note 1].
10. Every integer is congruent modulo $n$ to exactly one of the integers in the set of least positive residues $\{0, 1, 2, ..., n-1\}$. This set of $n$ integers is denoted $\Bbb{Z}_n$ [see note 2].
11. Properties of congruence: for a fixed positive integer $n$ and any integers $a, b, c, d$.
1. Reflexive property: $a \equiv a \pmod n$.
2. Symmetric property: If $a \equiv b \pmod n$ then $b \equiv a \pmod n$.
3. Transitive property: If $a \equiv b \pmod n$ and $b \equiv c \pmod n$ then $a \equiv c \pmod n$.
4. Addition and multiplication rules: If $a \equiv b \pmod n$ and $c \equiv d\pmod n$ then $a\pm c \equiv b\pm d \pmod n$ and $ac \equiv bd \pmod n$.
5. Cancellation rule: If $ac \equiv bc \pmod n$ and $c\ne 0$ then $a\equiv b \pmod {(n \mathbin{/} \gcd(c,n))}$. If $\gcd(c,n)=1$ then $a\equiv b\pmod n$ [see note 3].
6. Associative and distributive properties: $(a+b) + c \equiv a + (b+c) \pmod n$, $(ab)c \equiv a(bc) \pmod n$, and $a(b+c) \equiv ab + ac \pmod n$.
7. If $a\equiv b\pmod n$ then $a^r \equiv b^r \pmod n$ for any integer $r\ge 1$.
8. $a\equiv 0 \pmod n$ if and only if $n\mid a$.
12. If $m$ and $n$ are coprime and $a \equiv b \pmod m$ and $a\equiv b\pmod n$, then $a\equiv b \pmod {mn}$.
13. The Euler totient function (or Euler phi function) of a positive integer $n$, denoted $\varphi(n)$, is defined to be the number of positive integers not exceeding $n$ which are relatively prime to $n$. For example, $\varphi(12) = 4$ as the 4 integers $\{1, 5, 7, 11\}$ are the only integers coprime to and not greater than 12; and $\varphi(7) = 6$ as the 6 integers $\{1, 2, 3, 4, 5, 6\}$ are coprime to and not greater than 7 [see also note 4].
14. For any prime $p$, $\varphi(p) = p - 1$, since all numbers less than $p$ are relatively prime to it.
15. If $m$ and $n$ are coprime, then $\varphi(m)\varphi(n) = \varphi(mn)$.
16. Fermat's Little Theorem: If $p$ is a prime and $a$ is any integer, then $a^p \equiv a\pmod p$. If $\gcd(a,p)=1$, then $a^{p-1} \equiv 1\pmod p$. The latter is a special case of the Euler-Fermat Theorem below.
17. Euler-Fermat Theorem: If $n$ is a positive integer and $a$ is any integer with $\gcd(a, n) = 1$, then $a^{\varphi(n)}\equiv 1\pmod n$.
18. The Carmichael function of a positive integer $n$, denoted $\lambda(n)$, is defined as the smallest positive integer $m$ such that $a^m \equiv 1\pmod n$ for all integers $a$ relatively prime to $n$. It can be shown that $\lambda(n)$ divides $\varphi(n)$.
19. The least common multiple of the non-zero integers $a$ and $b$, denoted $\text{lcm}(a, b)$, is the smallest positive integer that is a multiple of both $a$ and $b$. For example, $\text{lcm}(10, 15) = 30$. For any pair of integers $a$ and $b$, $\text{lcm}(a, b) \times \gcd(a, b) = |ab|$ .
20. If $n=pq$ is the product of two distinct primes $p$ and $q$, then $\lambda(pq) = \text{lcm}(p-1, q-1)$.
21. The modular multiplicative inverse or modular inverse of an integer $a$ modulo $n$ is an integer $x$ such that $ax \equiv 1 \pmod n$. We write $x = a^{-1} \bmod n$ or $x = (1/a) \bmod n$. A modular inverse exists if and only if $\gcd(a, n) = 1$. For example, 7 is a modular inverse of 3 modulo 20 since $3\times 7 = 21\equiv 1\pmod {20}$. We can write $7=3^{-1}\bmod {20}$ or $7=(1/3)\bmod {20}$.
## Notes
1. Note that the expression ‘$a = b \bmod n$’ is satisfied by a unique integer $0 \le a < n$, but the congruence ‘$a \equiv b \pmod n$’ is satisfied by any integer of the form $b + nk$ for any integer $k$. For example, $16 \bmod 12 = 4$, but the congruence $a \equiv 16 \pmod {12}$ is satisfied by $a = 4, 16, 28, 40, \ldots$ and $a = -8, -20, -32,\ldots$.
2. Strictly speaking, $\Bbb{Z}_n$ is the set of equivalence classes modulo $n$. The set $\Bbb{Z}_n$ always has $n$ members, all incongruent to each other.
3. Be careful when trying to 'divide through' the two sides of a congruence by what seems to be a common factor. It only works in the way you'd expect if the common factor is coprime to the modulus.
For example, suppose we want to solve the linear congruence $9x \equiv 6 \pmod {11}$ for $x$. We have $n=11$ and the coefficients 9 and 6 have the common factor $c=3$, and 3 is coprime to 11 since $\gcd(c,n)=\gcd(3, 11) = 1$, so we can 'divide through' by 3 to get $3x \equiv 2 \pmod {11}$, leaving the modulus 11 unchanged. This has the single solution $x \equiv 8 \pmod {11}$ and this is also the solution to the original congruence. To see this note that $3x = 3\times 8 = 24 \equiv 2 \pmod{11}$ and $9x=9\times 8 = 72 \equiv 6\pmod{11}$.
However, if we want to solve $9x \equiv 6 \pmod {12}$, we have $n=12$ and common factor $c=3$, so we use the cancellation rule to obtain $3x \equiv 2 \pmod 4$. This time $\gcd(c,n) =\gcd(3, 12) = 3 \ne 1$, so we must divide the modulus 12 by 3. This latter congruence modulo 4 is solved by the single congruence $x\equiv 2\pmod 4$, and therefore the original congruence modulo 12 is solved by the three solutions $x = 2,6,10 \pmod{12}$. That is, $x = 2 + 4i$ for $i=0,1,2$. See that $3x=3\times 2 = 6 \equiv 2\pmod 4$, but this time we have three solutions modulo 12; namely, $9\times 2 = 18 \equiv 6\pmod{12}$, $9\times 6 = 54 \equiv 6\pmod{12}$, and $9\times 10 = 90 \equiv 6\pmod{12}$.
4. To compute $\varphi(n)$ for an arbitrary integer $n$, use the Fundamental Theorem of Arithmetic ($\S 4$ above) and the formula
$\varphi(p^k) = p^{k-1}(p-1) = p^k\left(1-\frac{1}{p}\right)$. | 2,847 | 8,189 | {"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} | 4.65625 | 5 | CC-MAIN-2023-50 | latest | en | 0.887735 |
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Decimal Exponents
```
Date: 11/07/2001 at 17:20:09
From: Alex
Subject: Exponents
Dear Dr. Math,
I was sitting in math class today and I asked my teacher a question.
She was not sure and told me to come to your Web site. Here's the
question:
Can you have exponents that are decimals?
My teacher said that maybe they have it in the high math classes. Do
you know?
Respectfully,
```
```
Date: 11/08/2001 at 14:45:57
From: Doctor Peterson
Subject: Re: Exponents
Hi, Alex.
Good question! Your teacher is right; as you advance in math you will
see more advanced kinds of exponents.
When you first see exponents, only whole numbers are used. Then we
introduce negative and zero exponents:
n^0 (any number to zero power) - Dr. Math FAQ
http://mathforum.org/dr.math/faq//faq.number.to.0power.html
Zero as an Exponent - Dr. Math archives
http://mathforum.org/dr.math/problems/gerleman2.1.96.html
Briefly, for example, 2^-3 (that is, 2 to the -3 power) means the
reciprocal of 2^3, which is 1/8.
After that, we can talk about rational (fraction or decimal)
exponents:
Finding a Decimal Exponent
http://mathforum.org/dr.math/problems/tyler3.21.98.html
Resolving Decimal Exponents
Again briefly, 2^0.5, or 2^(1/2), the one-half power of 2, is the same
as the square root of 2. You can see this by thinking about what the
square of 2^(1/2) would be.
Finally, we can talk about irrational exponents, like 2 raised to the
pi power (and if you've ever heard of "imaginary numbers," we can use
them, too):
Meaning of Irrational Exponents
http://mathforum.org/dr.math/problems/knobler.three.3.17.97.html
This can't be explained briefly!
Each step requires us to do a little new thinking, so it takes time to
develop an understanding of all this.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Exponents
Middle School Exponents
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OlieXae
Registered: 05.07.2001
From:
Posted: Friday 29th of Dec 20:15 I'm just wondering if someone can give me a few pointers here so that I can understand the concepts behind 2008 ged prep printable worksheets. I find solving equations really difficult. I work in the evening and thus have no time left to take extra tutoring . Can you guys suggest any online tool that can help me with this subject?
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Registered: 07.08.2005
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Posted: Friday 05th of Jan 07:52 Life can be tough when one has to work along with their studies. Visit https://mathfraction.com/simplifying-fractions-by-multiplying-by-the-lcd.html, I am sure it will be of use to you. | 875 | 3,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-18 | latest | en | 0.915936 |
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### Given $x_{1} > 2$ and the recursively defined sequence $x_{n+1}=1+\sqrt{x_{n}-1}$
Given $x_{1} > 2$ and the recursively defined sequence $x_{n+1}=1+\sqrt{x_{n}-1}$. Proof: (a) $x_{n} \geq 2\ \forall n \geq2$ (which seems obvious given the initial condition, but I also would like ...
1 vote
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### Is the following sequence $x_{n+1}$ = $x_{n} + 1/x_{n}$ convergent given $x_{1} >0$? [duplicate]
I got the following result : $x_{n+1} = x_{n}+\frac{1}{x_{n}}$ and so forth. As $x_{1}>0$, then I know by plugging that $x_{k}>1$ for every natural $k$. Hence, the terms in the denominator tend ...
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### Question about a step in a derangement proof
I read a proof about derangement and I didn't understand this step: $d_n - nd_{n-1} = -(d_{n-1}-(n-1)d_{n-2}) \implies d_n=nd_{n-1}+(-1)^n$ I see that we have $S_n = -S_{n-1}$ if $S$ is the stuff on ...
• 119
### How do we solve the recurrence $T(n) = 2T(n/3) + n^{ \log_32 }\log\log n$?
How do we solve the recurrence $T(n) = 2T(n/3) + n^{ \log_32 }\log\log n$? Also, is it possible to solve this recurrence by the Master method? | 1,465 | 4,195 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-21 | longest | en | 0.800964 |
https://learn.careers360.com/school/question-a-train-starting-from-rest-attains-a-velocity-of-72-km-per-hour-in-5-minutes-assuming-that-the-acceleration-is-uniform-find-the-distance-travelled-by-train-for-attaining-this-velocity-44390/ | 1,679,752,371,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945333.53/warc/CC-MAIN-20230325130029-20230325160029-00009.warc.gz | 416,217,770 | 117,799 | #### A train starting from rest attains a velocity of 72 km per hour in 5 minutes Assuming that the acceleration is uniform find the distance travelled by train for attaining this velocity
Starting from rest implies the initial velocity is zero.
u=0
v=72Km/h=20m/s
v=u+at
20=0+a x 300
a=1/15 m/s2
distance travelled
s=ut+0.5at2
s=0+0.5 x (1/15) x 90000=3000m=3Km | 124 | 371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-14 | latest | en | 0.736705 |
https://book.douban.com/subject/1761910/annotation | 1,563,404,124,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525414.52/warc/CC-MAIN-20190717221901-20190718003901-00524.warc.gz | 341,119,654 | 12,794 | # Types and Programming Languages的笔记(11)
## 按有用程度按页码先后最新笔记
• ### 展开 收起 第104页
龙三 (の日常)
9.3.1 Lemma [Inversion of the typing relation]: 1. If Γ ⊢ x : R, then x:R ∈ Γ . 2. If Γ ⊢ λx:T1. t2 : R, then R = T1!R2 for some R2 with Γ , x:T1 ⊢ t2 : R2. 3. If Γ ⊢ t1 t2 : R, then there is some type T11 such that Γ ⊢ t1 : T11→R and Γ ⊢ t2 : T11. 4. If Γ ⊢ true : R, then R = Bool. 5. If Γ ⊢ false : R, then R = Bool. 6. If Γ ⊢ if t1 then t2 else t3 : R, ...
2019-06-30 14:41 1人喜欢
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龙三 (の日常)
Several different evaluation strategies for the lambda-calculus have been studied over the years by programming language designers and theorists. Each strategy defines which redex or redexes in a term can fire on the next step of evaluation. ● Under full beta-reduction, any redex may be reduced at any time. At each step we pick some redex, anywhere inside the term we are evaluating, and reduce...
2019-06-22 18:55 1人喜欢
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龙三 (の日常)
Definition [Terms, by inference rules]: The set of terms is defined by the following rules: The first three rules here restate the first clause of Definition 3.2.1; the next four capture clauses (2) and (3). Each rule is read, “If we have established the statements in the premise(s) listed above the line, then we may derive the conclusion below the line.” The fact that T is the smallest set s...
2019-06-18 11:01 1人喜欢
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龙三 (の日常)
2019-06-17 18:55 1人喜欢
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龙三 (の日常)
预序，偏序，等价关系 A reflexive and transitive relation R on a set S is called a preorder on S. A reflexive, transitive, and anti-symmetric relation R is called a partial order on S. A reflexive, transitive, and symmetric relation on a set S is called an equivalence on S. 最小上确界，最大下确界 Suppose that ≤ is a partial order on a set S and s and t are elements of S. An element j 2 S is sai...
2019-06-15 19:13 1人喜欢
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中小型文化叙事 (当代舶来品研究)
2013-07-21 19:03 1人喜欢
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龙三 (の日常)
The typing rules for ref, :=, and ! follow straightforwardly from the behaviors we have given them.
2019-07-09 23:10
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龙三 (の日常)
A reference cell need not contain just a number: the primitives above allow us to create references to values of any type, including functions. For example, we can use references to functions to give a (not very efficient) implementation of arrays of numbers, as follows. Write NatArray for the type Ref (Nat → Nat). To build a new array, we allocate a reference cell and fill it with a function ...
2019-07-09 22:52
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龙三 (の日常)
2019-06-24 21:48
• ### 展开 收起 第108页
龙三 (の日常)
The "⟼" type constructor comes with typing rules of two kinds: 1. an introduction rule (T-Abs) describing how elements of the type can be created, and 2. an elimination rule (T-App) describing how elements of the type can be used. Briefly, the idea is that, in constructive logics, a proof of a proposition P consists of concrete evidence for P. What Curry and Howard noticed was that such evid...
2018-06-21 22:04
<前页 1 2 | 1,013 | 3,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.657366 |
http://sourceforge.net/p/maxima/bugs/1938/ | 1,454,955,024,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701153736.68/warc/CC-MAIN-20160205193913-00059-ip-10-236-182-209.ec2.internal.warc.gz | 198,653,111 | 10,368 | ## #1938 maxima can not integrate x*exp(-1/2*(x-m)^2)
closed
nobody
5
2010-07-09
2010-03-26
Anonymous
No
Maxima can not integrate the function x*exp(-1/2*(x-m)^2)
Maxima 5.20.1 http://maxima.sourceforge.net
using Lisp GNU Common Lisp (GCL) GCL 2.6.8 (a.k.a. GCL)
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) integrate(x*exp(-1/2*(x-m)^2),x);
2
(x - m)
/ - --------
[ 2
(%o1) I x %e dx
]
/
The correct solution would be
(%i2) diff(-exp(-1/2*x^2+m*x-1/2*m^2)+1/2*m*sqrt(%pi)*sqrt(2)*erf(1/2*sqrt(2)*x-1/2*m*sqrt(2)),x);
2 2
x m 2 x m
- (------- - -------) - -- + m x - --
sqrt(2) sqrt(2) 2 2
(%o2) m %e - (m - x) %e
## Discussion
• (%i1) S:'integrate(x*exp(-1/2*(x-m)^2),x);
(%o1) integrate(x*%e^(-(x-m)^2/2),x)
(%i2) changevar(S, y=x-m, y, x);
(%o2) integrate((y+m)*%e^(-y^2/2),y)
(%i3) ev(%, nouns);
(%o3) (sqrt(%pi)*m*erf(y/sqrt(2)))/sqrt(2)-%e^(-y^2/2)
Solution:
(%i4) sol:subst(y=x-m,%),rootscontract;
(%o4) sqrt(%pi/2)*m*erf((x-m)/sqrt(2))-%e^(-(x-m)^2/2)
Test:
(%i5) diff(sol,x)-first(S)\$
(%i6) expand(%);
(%o6) 0
• uhlst
2010-03-27
Thank you for showing this "trick" with changevar.
Is there a way that Maxima automatically does the correct substitution without giving it a hint with changevar? If e.g. the "-1/2" in the exponent is removed, then Maxima has no problem to do the integration. Does Maxima not try itself some substitutions if integrate is called?
• Robert Dodier
2010-06-05
Wow, that's pretty surprising. Maxima can solve integrate(exp(-(1/2)*(x - m)^2), x) so why not this one?
• Robert Dodier
2010-06-05
• labels: --> Lisp Core - Integration
• Dieter Kaiser
2010-07-09
Fixed in sin.lisp revision 1.67.
Maxima can now calculate the integral of this bug report. 'The result is not as nice as the expected result, but can be shown to be equivalent up to a constant:
(%i48) multthru(integrate(x*exp(-1/2*(x-m)^2),x)),gamma_expand:true,erf_representation:erf;
(%o48) sqrt(%pi)*m*(m-x)*(1-erf(abs(x-m)/sqrt(2)))/(sqrt(2)*abs(x-m))
-(m-x)^2*%e^-((m-x)^2/2)/(x-m)^2
Closing this bug report as fixed.
Dieter Kaiser
• Dieter Kaiser
2010-07-09
• status: open --> closed | 829 | 2,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2016-07 | latest | en | 0.518072 |
https://slideplayer.com/slide/2344743/ | 1,531,691,886,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676588972.37/warc/CC-MAIN-20180715203335-20180715223335-00210.warc.gz | 771,462,519 | 29,909 | # Lindsey Bleimes Charlie Garrod Adam Meyerson
## Presentation on theme: "Lindsey Bleimes Charlie Garrod Adam Meyerson"— Presentation transcript:
Lindsey Bleimes Charlie Garrod Adam Meyerson
Facility Location Lindsey Bleimes Charlie Garrod Adam Meyerson
The K-Median Problem Input: We’re given a weighted, strongly connected graph, each vertex as a client having some demand Demand is generally distance – it is a weight on the edges of the graph We can place facilities at any k vertices within our graph, which can then serve all the other clients At which vertices do we place our k facilities, in order to minimize total cost?
The K-Median Problem If we had 2 facilities to place, which vertices become Facilities? We want to minimize average distance of each client to its closest facility Our ‘Graph’
The K-Median Problem How do we know which locations are really optimal, without testing every combination of k locations?
The K-Median Problem We want the facilities to be as efficient as possible, thus we want to minimize the distance from each client to its closest facility. There can be a cost associated with creating each facility that also must be minimized otherwise if we were not limited to k facilities, all points could be facilities
Variations – Classic Facility Location
We may not have a set number of facilities to place In that case, the cost of opening a facility is included in the total cost calculation which must be minimized Now the question is, how many facilities to we create, and where do we put them?
Variations – Online Facility Location
We start with some graph and its solution, but we will have to add more vertices in the future, without disturbing our current setup The demands of incoming clients are based on some known function, generally of distance Our question: what do we do with each incoming point as it arrives?
Applications - Operations
Stores and Warehouses Where do we build our warehouses so that they are close to our stores? And how many should we build to attain efficiency? Here, accuracy far outweighs speed
Applications - Clustering
Databases Data mining with huge datasets Here, speed outweighs accuracy, to a point Finding Data patterns ‘Distances’ measured either in space or in content Web Search clustering Medical Research And many other clustering problems
Limitations The problem of finding the best possible solution is NP-Hard It has been proved that the best upper-bound attainable is about the square root of 2 times the optimal solution cost – the best upper bound so far attained is around 1.5 50% extra cost – not so good when talking about millions of dollars, not so bad when talking about data clustering
Is It Really That Bad? Well … on the average case, probably not.
But that’s something we’re trying to find out Are the average-case solutions good enough for companies to use? Are online models fast enough and at least somewhat accurate for db/clustering applications?
Solution Techniques Local Search Heuristics for k-median and Facility Location Problems V. Arya et al. Improved Approximation Algorithms for Metric Facility Location Problems M. Mahdian, Y. Ye, J. Zhang Online Facility Location A. Meyerson
Local Search / K-Median
Where do we place our k facilities? The Algorithm: Choose some initial K points to be facilities, and calculate your cost Initial points can be chosen by first choosing a random point, then successively choosing the point farthest from the current group of facilities until you have your initial K
Local Search / K-Median
Where do we place our k facilities? Now we swap While there exists a swap between a current facility location and another vertex which improves our current cost, execute the swap
Local Search / K-Median
Where do we place our k facilities? Now we swap While there exists a swap between a current facility location and another point which improves our current cost, execute the swap
Local Search / K-Median
Where do we place our k facilities? Now we swap While there exists a swap between a current facility location and another point which improves our current cost, execute the swap Etc.
Local Search / K-Median
It is possible to do multiple swaps at one time In the worst case, this solution will produce a total cost of (3 + 2/p) times the optimal cost, where p is the number of swaps that can be done at one time
Facility Location The Algorithm: Begin with all clients unconnected
How many facilities do we need, and where? The Algorithm: Begin with all clients unconnected All clients have a budget, initially zero
Facility Location How many facilities do we need, and where? Clients constantly offer some of their budget to open a new facility This offer is: max(budget-dist, 0) if unconnected, or max(dist, dist’) if connected Where dist = distance to possible new facility, and dist’ = distance to current facility
Facility Location How many facilities do we need, and where? While there is an unconnected client, we keep increasing the budgets of each unconnected client at the same rate Eventually the offer to some new facility will equal the cost of opening it, and all clients with an offer to that point will be connected
Facility Location How many facilities do we need, and where? While there is an unconnected client, we keep increasing the budgets of each unconnected client at the same rate Eventually the offer to some new facility will equal the cost of opening it, and all clients with an offer to that point will be connected
Facility Location How many facilities do we need, and where? Or, the increased budget of some unconnected client will eventually outweigh the distance to some already-opened facility, and can simply be connected then and there
Facility Location – Phase 2
How many facilities do we need, and where? Now that everyone is connected, we scale back the cost of opening facilities at a uniform rate If at any point it becomes cost-saving to open a new facility, we do so and re-connect all clients to their closest facility Worst case, this solution is 1.52 times the optimal cost solution
Online Facility Location
What do we do with incoming vertices? Here we start with an initial graph, but more clients will need to be added in the future, without wrecking our current scheme As new clients arrive, we must evaluate their positions and determine whether or not to add a new facility
Online Facility Location
What do we do with incoming vertices? With each new client, we do one of two things: (1) Connect our new client to an existing facility
Online Facility Location
What do we do with incoming vertices? With each new client, we do one of two things: Connect our new client to an existing facility, or Make a new facility at the new point location
Online Facility Location
The probability that a Facility is created out of a given incoming point is d/f Where d = the distance to the nearest facility And f = the cost of opening a facility Worst case cost is expected 8 times the optimal cost
Our Goal We’re not trying to solve the problem again
Rather we’d like to know more about the realistic behavior of techniques we already have i.e. how often do we really see results at the upper/lower bounds of accuracy? How far off are streaming data models?
Our Goal We are trying to run simulations over both real and random data sets, to get average data on the performance of known algorithms for this problem Both speed and accuracy are important, but for different reasons and applications Realistic data will help determine how best to use these algorithms
Questions? | 1,554 | 7,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-30 | latest | en | 0.949524 |
https://1library.net/document/z34v6p9y-direction-sense.html | 1,631,909,082,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00259.warc.gz | 134,990,683 | 33,464 | # 1 Direction Sense
## Full text
(1)
There are four directions such as North, South, East and West. The word NEWS came from North, East, West and South. There are four regions : North-East (i); North-West (ii); South-East (iii); South-West (iv).
The directions OP, OS, OQ and OR are -North-East direction; North-West direction; South-West direction; and South-East direction respectively.
Note : The candidate must distinguish between the regions and directions, i.e., between North-East region and north-East direction. If you move with your face East-wards, your left hand is towards North and your right hand is towards South. Similarly the positions of the directions of the hands can be fixed when you move in any of the other three directions.
### EXAMPLES
Ex.1 Sanjay went 70 metres in the East before turning to his right. He went 10 metres before turning to his right again and went 10 metres from this point. From here he went 90 metres to the North. How far was he from the starting point ? (A) 80 metres (B) 100 metres (C) 140 metres (D) 260 metres
Sol.(B)
The movement of Sanjay from A to E are as shown in figure. Now, Af = (AB – FB) = (AB – DC) = (70 – 10) m = 60 m EF = (DE – DF) = (DE – BC) = (90 – 10) m = 80 m. Required distance = AE = AF2EF2 = (60)2(80)2 = 100 m
Ex.2 I start from my home and go two kilometres straight. Then I turn towards my right and go one kilometre. I turn again towards my right and go one kilometre again. If I am North-west from my house then in which direction did I go in the beginning ?
(A) North (B) South
(C) East (D) West
Sol. (D) The following diagrams clarifies my movements.
W S E N O
## 1
### DIRECTION SENSE
(2)
Ex.3 Raj walked 20 metres towards South. Then he turned to his left and walked 25 metres. He then turned to his left and walked 20 metres. He again turned to his right and walked 10 metres. At what distance is he from the starting point and in which direction ?
(A) 35 metres, East (B) 35 metres, North (C) 40 metres, East (D) 60 metres, East
Sol.(A)
The movements of Raj are as shown in figure Raj's distance from starting point A = AE = (AD + DE)
= (BC + DE) = (25 + 10) m = 35 m So, E is to the EAst of A
Ex.4 Mohan started from his house, walked 2 km North, then 3 km West, then 6 km South. How faraway from his house was he then ?
Sol. See the figure.
S W N AHouse C B 3 km 3 km 6 km 4 km 2 km 2 km (AC)2= (AB)2+ (BC)2 = 9 + 16 = 5 km. Ans.
Ex.5 Anil left home and cycled 10 km Southwards, turned right and cycled 5 km & turned right and cycled 10 km and turned left and cycled 10 km. How many kilometres will he have to cycle to reach his home straight ?
(A) 10 km (B) 15 km
Sol.(B)
Clearly, Anil starts from home at A, moves 10 km Southwards upto B, turns right and moves 5 km upto C, turns right again and moves 10 km upto D and finally turns left and moves 10 km upto E. Thus, his distance from initial position A = AE = AD + DE
= BC + DE = (5 + 10) km = 15 km. Hence, the answer is (B).
Ex.6 Amit faces towards North. Turning to his right he walks 25 metres. He then turns to his left and walks 30 metres. Next, he moves 25 metres to his right. He then turns to his right again and walks 55 metres. Finally, he turns to the right and moves 40 metres. In which direction is he now from his starting point ?
(A) South-West (B) South (C) North-West (D) South-East
Sol.(D)
Amit turns towards right from North direction. So, he walks 25 m towards East upto B, turns left and moves 30 m upto C, turns right and moves 25 m upto D. At D he turns to right towards the South and walks 55 m upto E. Next, he again turns to right and walks 40 m upto F. Which is his final position. F is to the South-East of A. A So, he is to the South-East from his starting point. Hence, the answer is (D).
Ex.7 From his house, Rajan went 25 kms to the North. Then he turned West and covered 20 kms. Then he turned South and covered 15 Kms/ Finally, turning to East, he covered 20 kms. In which direction was he from his house ?
(3)
Sol.(C) The movements of Rajan are as shown in figure.
(A to B, B to C, C to D and D to E)
Clearly, his final position is E which is to the
North of his house at A.
Ex.8 A man is facing North-West. He turns 90º in the clockwise direction, then 180º in the anticlockwise direction and then another 90º in the same direction. Which direction is he facing now ? (A) South (B) South-West
(C) West (D) South-West
Sol.
As shown in figure the man initially faces in the direction OA. On moving 90º clockwise, he faces in the direction OB. On further moving 180º anticlockwise, he faces in the direction OC. Finally on moving 90º anti-clockwise, he faces in the direction OD, which is South-East.
Ex.9 Kishen walks 10 km towards North. From there, he walks 6 km towards South. Then, he walks 3 km towards East. How far and in which direction is he with reference to his starting point ? (A) 5 km, North (B) 5 km, North-East (C) 7 km, East (D) 7 km, West
Sol.
The movements of kishen are as shown in figure (A to B, B to C and C to D). AC = (AB – BC) = (10 – 6) km = 4 km. Clearly, D is to the North-East of A.
Kishen's distance from starting point A = AD = AC2CD2 = 4232 = 25 = 5 km So, kishen is 5km to the North-East of his starting point.
Ex.10 A rabbit run 30º towards East and turns to right, runs 20' and turns to right; runs 19' and again turns to left, runs 15' and then turns to left, runs 22' and finaly turns to left and runs 16'. Now, which direction is the rabbit facing ?
(A) East (B) West (C) North (D) South
Sol.(C)
The movement of the rabbit from A to G are as show in figure. So, the rabbit's face is in North direction at the end of runs.
Ex.11 I am facing South. I turn right and walk 20 m. Then I turn right again and walk 10 m. Then I turn left and walk 10 m and then turning right walk 20 m. Then I turn right again and walk 60 m. In which direction am I from the starting point ?
(A) North (B) North-west (C) East (D) North-East
Sol.(D) The movement of the person are from A to F, as
shown in figure. Clearly, the final position is F which is to the Noth-East of the starting point A.
(4)
Q.1 A man is facing south. He turns 135º in the anticlockwise direction and then 180º in the clockwise direction. Which direction is he facing now ?
(A) North-east (B) North-west (C) South-east (D) South-west
Q.2 A man is facing north-west. He turns 90º in the clockwise direction and then 135º in the anticlockwise direction. Which direction is he facing now ?
(A) East (B) West
(C) North (D) South
Q.3 A man is facing north-west. He turns 90º in the clockwise direction, then 180º in the anticlockwise direction and then another 90º in the same direction. Which direction is he facing now ? (A) South (B) South west (C) West (D) South-east
Q.4 I am facing east. I turn 100º in the clockwise direction and then 145º in the anticlockwise direction. Which direction am I facing now ? (A) East (B) North-east
(C) North (D) South-west
Q.5 Deepak starts walking straight towards east. After walking 75 metres, he turns to the left and walks 25 metres straight. Again he turns to the left, walks a distance of 40 metres straight, again he turns to the left and walks a distance of 25 metres. How far is he from the starting point ? (A) 25 metres (B) 50 metres (C) 140 metres (D) None of these
Q.6 Kishenkant walks 10 kilometres towards North. From there, he walks 6 kilometres towards South. then, he walks 3 kilometres towards East. How far and in which direction is he with reference to his starting point ?
(A) 5 kilometres West (B) 5 kilometres North-east (C) 7 kilometres East (D) 7 kilometres West
Q.7 A man leaves for his office from his house. He walks towards East. After moving a distance of 20 m, he turns towards South and walks 10 m. Then he walks 35 m towards the West and further 5 m towards the North. He then turns towards East and walks 15 m. What is the straight distance in metres between his initial and final positions ?
(A) 0 (B) 5
(C) 10 (D) cannot be determined (E) None of these
Q.8 Gaurav walks 20 metres towards North. He then turns left and walks 40 metres. He again turns left and walks 20 metres. Further, he moves 20 metres after turning to the right. How far is he from his original position ?
(A) 20 metres (B) 30 metres (C) 60 metres (D) None of these
Q.9 Radha moves towards South-East a distance of 7 km, then she moves twoards West and travels a distance of 14 m. From here, she moves towards North-west a distance of 7 m and finally she moves a distance of 4 m towards East and stood at that point. How far is the starting point from where she stood ?
(A) 3 m (B) 4 m (C) 10 m (D) 11 m
Q.10 Gopal starts from his house towards West. After walking distance of 30 metres, he turned towards right and walked 20 metres. He then turned left and moving a distance of 10 metres, turned to his left again and walked 40 metres. He now turns to the left and walks 5 metres. Finally he turns to his left. In which direction is he walking now ?
(A) North (B) South
(C) East (D) West
(5)
Q.16 Ramakant walks northwards. After a while, he turns to his right and a little further to his left. Finally, after walking a distance of one kilometre, he turns to his left again. In which direction is he moving now ?
(A) North (B) South
(C) East (D) West
Q.17 A man walks 1 km towards East and then he turns to South and walks 5 km. Again he turns to East and walks 2 km, after this he turns to North and walks 9 km. Now, how far his starting point ?
(A) 3 km (B) 4 km (C) 5 km (D) 7 km
Q.18 Raj travelled from a point X straight to Y at a distance of 80 metres. He turned right and walked 50 metres, then again turned right and walked 70 metres. Finally, he turned right and walked 50 metres. How far is he from the starting point ? (A) 10 metres (B) 20 metres
(C) 50 metres (D) 70 metres (E) None of these
Q.19 Laxman went 15 kms to the west from my house, then turned left and walked 20 kms. He then turned East and walked 25 kms finally turning left covered 20 kms. How far was he from his house ?
(A) 5 kms (B) 10 kms
(C) 40 kms (D) 80 kms
Q.20 From his house, Lokesh went 15 kms to the North. Then he turned West and covered 10 kms. Then, he turned South and covered 5 kms. Finally, turning to East, he covered 10 kms. In which direction is he from his house ?
(A) East (B) West
(C) North (D) South
Q.21 Going 50 m to the South of her house, Radhika turns left and goes another 20 m. Then, turning to the North, she goes 30 m and then starts walking to her house. In which direction is she walking now ?
(A) North-west (B) North (C) South-west (D) East
Q.11 A girl leaves from her home. She first walks 30 metres in North-west direction and then 30 metres in South-west direction. Next, she walks 30 metres in South-east direction. Finally, she turns towards her house. In which direction is she moving ? (A) North-east (B) North-west
(C) South-west (D) None of these
Q.12 Sanjeev walks 10 metres towards the South. Turning to the left, he walks 20 metres and then moves to his right. After moving a distance of 20 metres, he turns to the right and walks 20 metres. Finally, he turns to the right and moves a distance of 10 metres. How far and in which direction is he from the starting point ?
(A) 10 metres North (B) 20 metres South (C) 20 metres North (D) 10 metres South (E) None of these
Q.13 Kashish goes 30 metres North, then turns right and walks 40 metres, then again turns right and walks 20 metres, then again turns right and walks 40 metres. How many metres is he from his original position ?
(A) 0 (B) 10
(C) 20 (D) 40
(E) None of these
Q.14 I am facing South, I turn right again and walk 10 m. Then I turn left and walk 10 m and then turning right walk 20 m. then I turn right again and walk 60 m,. In which direction am I from the starting point ?
(A) North (B) North-west (C) East (D) North-east
Q.15 A man walks 30 metres towards South. Then, turning to his right, he walks 30 metres. Then, tuning to his left, he walks 20 metres. Again, he turns to his left and walks 30 metres. How far is he from his initial position ?
(A) 20 metres (B) 30 metres (C) 60 metres (D) 80 metres (D) None of these
(6)
Q.22 A walks 10 metres in front and 10 metres to the right. Then every time turning to his left, he walks 5, 15 and 15 metres respectively. How far is he now from his starting point ?
(A) 5 metres (B) 10 metres (C) 15 metres (D) 20 metres (E) 23 metres
Q.23 Rasik walks 20 m North. Then the turns right and walks 30 m. Then he turns right and walks 35 m. Then he turns left and walks 15 m. Then he again turns left and walks 15 m. In which direction and how many metres away is he from his original position ?
Ques . 1 2 3 4 5 6 7 8 9 1 0 11 12 1 3 14 15 Ans . D B D B E B B D C A A B B D E Ques . 1 6 17 18 1 9 20 21 2 2 23 24 Ans . D C A B C A A D B
(A) 15 metres West (B) 30 metres East (C) 30 metres West (D) 45 metres East
Q.24 A child is looking for his father. He went 90 metres in the East before turning to his right. He went 20 metres before turning to his right again to look for his father at his uncle's place 30 metres from this point. His father was not there. From here he went 100 metres to the North before meeting his father in a street. How far did the son meet his father from the starting point ? (A) 80 metres (B) 100 metres (C) 140 metres (D) 260 metres
Updating...
## References
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Related subjects : | 3,851 | 13,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-39 | latest | en | 0.966339 |
https://codefreshers.com/banana-bunches-solution-kickstart/ | 1,638,549,871,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00231.warc.gz | 264,044,590 | 23,291 | # Barbara goes to Alan’s banana farm, where the NN banana trees are organized in one long line represented by an array BB. The tree at position ii has BiBi banana bunches. Each tree has the same cost. Once Barbara buys a tree, she gets all the banana bunches on that tree.
## Banana Bunches solution kickstart
### Problem
Barbara goes to Alan’s banana farm, where the NN banana trees are organized in one long line represented by an array BB. The tree at position ii has BiBi banana bunches. Each tree has the same cost. Once Barbara buys a tree, she gets all the banana bunches on that tree.
Alan has a special rule: because he does not want too many gaps in his line, he allows Barbara to buy at most 22 contiguous sections of his banana tree line.
Barbara wants to buy some number of trees such that the total number of banana bunches on these purchased trees equals the capacity KK of her basket. She wants to do this while spending as little money as possible. How many trees should she buy?
## Banana Bunches solution kickstart
### Input
The first line of the input gives the number of test cases, TTTT test cases follow.
Each test case begins with a line containing two integers integer NN, the number of trees on Alan’s farm, and KK, the capacity of Barbara’s basket.
The next line contains N positive integers B1,B2,,BNB1,B2,…,BN representing array BB, where the ii-th integer represents the number of banana bunches on the ii-th tree on Alan’s farm.
## Banana Bunches solution kickstart
### Output
For each test case, output one line containing Case #xx: yy, where xx is the test case number (starting from 11) and yy is the minimum number of trees Barbara must purchase to obtain KK banana bunches using at most 22 contiguous sections of the farm, or -1 if it is impossible to do so.
## Banana Bunches solution kickstart
### Limits
Time limit: 20 seconds.
Memory limit: 1 GB.
1T1001≤T≤100.
0BiK0≤Bi≤K, for each ii from 11 to NN.
1K1041≤K≤104.
1N501≤N≤50.
1K1041≤K≤104.
1N5001≤N≤500.
#### Test Set 3
1K1061≤K≤106.
For at most 25 cases:
1N50001≤N≤5000.
For the remaining cases:
1N5001≤N≤500.
## Banana Bunches solution kickstart
### Sample
Sample Input
4
6 8
1 2 3 1 2 3
4 10
6 7 5 2
6 8
3 1 2 1 3 1
4 6
3 1 2 0
Sample Output
Case #1: 3
Case #2: -1
Case #3: 4
Case #4: 3
## Banana Bunches solution kickstart
In Sample Case #1, the first section can contain the trees at indices 22 and 33, and the second section can contain the tree at index 66.
In Sample Case #2, it is impossible to achieve a sum of 1010 with 22 contiguous sections.
In Sample Case #3, the first section can contain the trees at indices {1,2}{1,2}, and the second section can contain the trees at indices {5,6}{5,6}. We cannot take the 2+3+32+3+3 combo (trees at indices {1,3,5}{1,3,5}) since that would be 33 contiguous sections.
In Sample Case #4, the only section contains the trees at indices {1,2,3}{1,2,3}.
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# Numbers
Author Message
Intern
Joined: 26 May 2008
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### Show Tags
26 May 2008, 12:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
1.A number divided by 221 leaves a remainder 64.What is the remainder if the number is divided by 13.
a. 1 b.11 c.0 d.12
2.The sum of two nos is "S" and their quotient is "P/Q" .The Numbers are
a. ps/q, qs/q
b. s/p, s/q
c.s-p/q ,s-q/p
d.ps/p+q , qs/p+q
Last edited by Doubty on 26 May 2008, 12:25, edited 1 time in total.
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26 May 2008, 12:21
Remainder = 12
221 + 64 = 285
285/13 remainder is 12
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### Show Tags
26 May 2008, 17:09
Doubty wrote:
2.The sum of two nos is "S" and their quotient is "P/Q" .The Numbers are
a. ps/q, qs/q
b. s/p, s/q
c.s-p/q ,s-q/p
d.ps/p+q , qs/p+q
Let us say X & Y are the two numbers. It is given that X+Y=S (Eqn 1) and X/Y = P/Q (Eqn 2).
From eqn1, Y=S-X; Substituting for the value of Y in equn 2, you should get X/(S-X) = P/Q .
Solving this you should get X=PS/(P+Q) & Y= QS/(P+Q), which is answer choice (D). Hope this helps.
Re: Numbers [#permalink] 26 May 2008, 17:09
Display posts from previous: Sort by | 688 | 2,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-04 | latest | en | 0.892574 |
https://betterlesson.com/lesson/reflection/9943/hands-on | 1,508,279,501,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822513.17/warc/CC-MAIN-20171017215800-20171017235800-00527.warc.gz | 659,172,794 | 20,131 | ## Reflection: Developing a Conceptual Understanding Metric Units of Capacity - Section 3: Closure
In this lesson, I allowed the students to explore finding the number of milliliters it took to fill a cup with water. This gave the students a conceptual understanding of the size of milliliters, as well as, why liters were not the appropriate measurement. Each team was given a plastic cup, an eyedropper, and a liter cup filled with water. The students could clearly see that the liter of water was too large to go into the plastic cup. This gave the students the conceptual understanding of milliliters and liters.
It is very important for students to understand the concept being presented to them. Yes, I could have just given them a picture of a liter and milliliter, but how much better is it to explore and see for yourself!
Hands On
Developing a Conceptual Understanding: Hands On
# Metric Units of Capacity
Unit 8: Measurement and Data
Lesson 9 of 15
## Big Idea: Milliliters and Liters are metric units of capacity which are commonly used to measure items of different sizes.
Print Lesson
5 teachers like this lesson
Standards:
Subject(s):
50 minutes
### Rose Monroe
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Helena, MT
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python 3x - i want to get a matrix with the values of a certain column removed for each row of the matrix with pytorch (speedu
As an example, consider the following 5x5 matrix.
``````import torch
x = torch.rand (25) .reshape (5, 5)
x
>>>
tensor ([[0.9263, 0.6601, 0.3334, 0.6175, 0.6035],
[0.2583, 0.6105, 0.3113, 0.2965, 0.9429],
[0.0350, 0.9206, 0.8667, 0.8958, 0.7814],
[0.8921, 0.8116, 0.7271, 0.1324, 0.8097],
[0.7166, 0.7780, 0.5185, 0.4530, 0.4059])``````
Also assume that i am given a column index for each row.
``````drop_indices = torch.tensor ([1, 4, 0, 3, 2])
drop_indices
>>>
tensor ([1, 4, 0, 3, 2])``````
At this time`i line`Against`drop_indices [i] column`I want to get a 5x4 matrix that looks like the value of is erased.
You want to remove 0.6601 in the first column for the first row and 0.9429 in the fourth column for the second row.
In other words, I want to get the following matrix as a result.
``````tensor ([[0.9263, 0.3334, 0.6175, 0.6035],
[0.2583, 0.6105, 0.3113, 0.2965],
[0.9206, 0.8667, 0.8958, 0.7814],
[0.8921, 0.8116, 0.7271, 0.8097],
[0.7166, 0.7780, 0.4530, 0.4059]))``````
You can turn it with for and do what I want with concat, but in reality, I want it to move at a certain high speed with a matrix of about 1000 rows and 5000 columns. (It is a part of the learning process of deep learning)
In what direction should we go? Thank you very much.
I solved it myself as follows. Thank you very much.
``````>>>import torch
>>>x = torch.rand (25) .reshape (5, 5)
>>>drop_indices = torch.tensor ([1, 4, 0, 3, 2])
>>>drop_elements = x [torch.tensor (range (5)), drop_indices] .reshape (5, 1)
>>>x_2 = x [~ mask] .reshape (5, 4)
>>>x
tensor ([[0.3465, 0.5554, 0.9142, 0.1674, 0.1353],
[0.9194, 0.2897, 0.7397, 0.5700, 0.9786],
[0.5404, 0.5266, 0.4050, 0.4092, 0.1816],
[0.9258, 0.0706, 0.9894, 0.8694, 0.3407],
[0.8124, 0.0562, 0.1115, 0.4929, 0.2795])
>>>drop_indices
tensor ([1, 4, 0, 3, 2])
>>>x_2
tensor ([[0.3465, 0.9142, 0.1674, 0.1353],
[0.9194, 0.2897, 0.7397, 0.5700],
[0.5266, 0.4050, 0.4092, 0.1816],
[0.9258, 0.0706, 0.9894, 0.3407],
[0.8124, 0.0562, 0.4929, 0.2795]))`````` | 969 | 2,138 | {"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} | 2.546875 | 3 | CC-MAIN-2021-49 | latest | en | 0.825243 |
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NCteacher's Message: We build a "principal pen"....we have a very laid back principal. He measures off a section of floor in the hallway near the office (conveniently covered in 12 inch square tiles!) and tells the kids that if they can "build" a pen that matches his given perimeter, they can put him in it. He tells us to build a principal pen that is 124 feet or something, then we get out the calculators and figure out what shape it will be and the length of each side- then we go to the hallway and mark it out with tape. Then I get a big ol' length of bulletin board paper and the kids line up around the "pen" and hold up the paper. They call the principal over and he acts all disgusted with them and gets right in! He stays there for a minute or two talking some trash with my kids...."I never thought you would do this to me!" kind of stuff. They ADORE it! With the 12 inch square tiles, it makes marking the perimeter out really easy- and it allows a class discussion on whether or not we should use rulers, yard sticks, measuring tapes, etc. to measure- usually one of the kids will say, "Hey! I know- let's use the floor tiles!" Hope this helps!
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beachmom 02-11-2007 06:26 AM Well.......I walked around their desks and counted the perimeter and told them the word RIM was inside of this word perimeter. I then ha ha ha walked on top of their desks! We talked about the rim of other objects in the room. With area I handed out paper make believe "area rugs" they colored them and we voted on our favorite "Rug" to place around the rooms. They all have area rugs at home different sizes and colors, long ones and short ones, squares and rectangular. Anyway I told them that the area rug covers an area of the house. How much area does yours cover????Are some the same area but look different? We shared up front two that covered the same "area" but looked different. The paper rug had large squares on them to decorate and color. We also put tasles on the end with a hole puncher and small pieces of yarn this covered perimenter again. Hope this helps. Risa 02-10-2007 09:18 PM My students, in the past, have had trouble keeping perimeter and area correctly identified. My team partner came up with the following: As they say the word, PER-I-ME-TER, they are "drawing" a rectangle (or square) in the air. The word has four syllables and the rectangle (or square) has four sides, so they say a syllable as they "draw" a side. ANYTIME we say the word PERIMETER, we automatically draw the rectangle (or square) in the air to indicate that perimeter is the measure of the OUTSIDE of the shape. After they got that, I taught them to remember AREA by waving their flattened hand in a small space as if they are filling in the interior of the whole space. They are taught that AREA is the measure of the INSIDE of a shape. Again, each time we say AREA, we wave a flat hand to 'fill in' a small space. Not sure if this explanation is clear enough, but this was a great aid in helping my students to remember what each of the terms were, without confusing them. NCteacher 02-10-2007 05:21 PM We build a "principal pen"....we have a very laid back principal. He measures off a section of floor in the hallway near the office (conveniently covered in 12 inch square tiles!) and tells the kids that if they can "build" a pen that matches his given perimeter, they can put him in it. He tells us to build a principal pen that is 124 feet or something, then we get out the calculators and figure out what shape it will be and the length of each side- then we go to the hallway and mark it out with tape. Then I get a big ol' length of bulletin board paper and the kids line up around the "pen" and hold up the paper. They call the principal over and he acts all disgusted with them and gets right in! He stays there for a minute or two talking some trash with my kids...."I never thought you would do this to me!" kind of stuff. They ADORE it! With the 12 inch square tiles, it makes marking the perimeter out really easy- and it allows a class discussion on whether or not we should use rulers, yard sticks, measuring tapes, etc. to measure- usually one of the kids will say, "Hey! I know- let's use the floor tiles!" Hope this helps! teachfla 02-10-2007 03:59 PM I teach it as a fence you need to build to keep your horse or dog in your yard. I draw the figures on the board and a fence. Then we figure out how much fence we need. I introduced it that way in August and still have some kids refer to perimeter as a fence! Valerie 02-10-2007 10:43 AM I ask my students to bring in empty food containers that are boxes. I then cut them into flat pieces and then I cut those into polygons of various sizes and shapes. The students then measure the perimeter of their favorite foods---pizza, mac and cheese, cereal, little debbie cakes, etc. Love3rdGrade 02-10-2007 09:08 AM You could do something hands-on like put shaving cream on the perimeter of their desk, have them walk the perimeter of the room, use a manipulative to go around the perimeter of something on their desk. I love M & M's and the kids gave me bags of them for Christmas. I let them use them for counting, etc in math. Good luck. enlightenme2 02-10-2007 08:31 AM Hello, I am teaching a lesson, for my observation, to my 3rd grade students about perimeter. Does anyone have any suggestions for an interesting motivation or warm up? Something really neat to catch their attention. Since I am being observed, and this is the low end math class, I thought a great "hook" would help the pretty boring lesson. Thanks!
All times are GMT -8. The time now is 10:28 PM. | 1,423 | 6,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-49 | latest | en | 0.960975 |
https://webeduclick.com/b-tree-in-data-structure/ | 1,716,389,446,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00065.warc.gz | 544,403,238 | 55,623 | # B-Tree in Data Structure
## B-Tree:
The m-way search tree minimizes the time of file access due to its reduced height. But it is required to keep the height of the tree as low as possible and for this purpose, we need to balance the height of the tree. This balanced M-way search tree is called B-Tree. A B-tree of order M is an M-way search tree in which:
1. The root node has at least 1 key value and at most M-1 key values.
2. The other nodes except the root contain at least [M/2]-1 key values and M-1 key values.
3. All leaf nodes are on the same level.
4. Keys are arranged in a defined order within the node.
## Rules for Insertion B-tree:
1. At first, search the tree to check the existence of new key values in the tree. If no duplicate key is found, then get the leaf node N where the new key is to be inserted.
2. If N is not full, then insert the new key in N and exit.
3. If N is full, then split the node into two new nodes at its median value and move the median key value to its parent node.
## Rules for Deletion B-tree:
In B-tree deletion is done from the leaf node only. If we need to delete a key from an internal node then replace the node with its immediate successor until a key from any leaf node is promoted to its upper level. Then balance the tree according to the given rules:
1. If the leaf node has enough keys, then simply remove it.
2. If the leaf node contains less than the minimum number of keys, then check its immediate left sibling node and immediate right sibling node.
i. If the left sibling node has more than the minimum number of keys, then move the largest key value of this node to its parent and move down the intervening entry from the parent node to its right child node.
ii. If the right sibling node has more than the minimum number of keys, then move the smallest key value of this node to its parent and move down the intervening entry from the parent node to its left child node.
iii. If both the sibling nodes have a minimum number of keys, then create a new node merging the two leaf nodes and the intervening key value of their parent node. | 486 | 2,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-22 | latest | en | 0.914445 |
https://discuss.codecademy.com/t/its-ok-to-be-average/188995 | 1,537,569,533,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157569.48/warc/CC-MAIN-20180921210113-20180921230513-00233.warc.gz | 489,362,732 | 4,058 | # It's OK to be Average
#1
Oops, try again. average([3, 0]) returned 3 instead of 1.5 as expected.
I don't understand what I'm doing wrong
``````def average(numbers):
return sum(numbers)
total = numbers
return float(total)
total / len(numbers) | 70 | 247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-39 | latest | en | 0.916719 |
https://explorecourses.stanford.edu/search?page=1&catalog=&q=optimization+methods&filter-coursestatus-Active=on&view=catalog&collapse= | 1,582,500,991,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145859.65/warc/CC-MAIN-20200223215635-20200224005635-00379.warc.gz | 339,196,489 | 10,368 | 2015-2016 2016-2017 2017-2018 2018-2019 2019-2020
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# 11 - 20 of 51 results for: optimization methods
## CME 215A:Advanced Computational Fluid Dynamics (AA 215A)
High resolution schemes for capturing shock waves and contact discontinuities; upwinding and artificial diffusion; LED and TVD concepts; alternative flow splittings; numerical shock structure. Discretization of Euler and Navier Stokes equations on unstructured meshes; the relationship between finite volume and finite element methods. Time discretization; explicit and implicit schemes; acceleration of steady state calculations; residual averaging; math grid preconditioning. Automatic design; inverse problems and aerodynamic shape optimization via adjoint methods. Pre- or corequisite: 214B or equivalent.
Last offered: Winter 2017
## CME 215B:Advanced Computational Fluid Dynamics (AA 215B)
High resolution schemes for capturing shock waves and contact discontinuities; upwinding and artificial diffusion; LED and TVD concepts; alternative flow splittings; numerical shock structure. Discretization of Euler and Navier Stokes equations on unstructured meshes; the relationship between finite volume and finite element methods. Time discretization; explicit and implicit schemes; acceleration of steady state calculations; residual averaging; math grid preconditioning. Automatic design; inverse problems and aerodynamic shape optimization via adjoint methods. Pre- or corequisite: 214B or equivalent.
Last offered: Spring 2012
## CME 245:Topics in Mathematical and Computational Finance
Description: Current topics for enrolled students in the MCF program: This course is an introduction to computational, statistical, and optimizations methods and their application to financial markets. Class will consist of lectures and real-time problem solving. Topics: Python & R programming, interest rates, Black-Scholes model, financial time series, capital asset pricing model (CAPM), options, optimization methods, and machine learning algorithms. Appropriate for anyone with a technical and solid applied math background interested in honing skills in quantitative finance. Prerequisite: basic statistics and exposure to programming.Can be repeated up to three times.
Last offered: Summer 2017 | Repeatable for credit
## CME 253A:Introduction to High Performance Computing and Parallel (GPU) Computing
Introduction to high-performance-computing (HPC) for mathematical, computational, physical sciences and engineering within the field of computational fluid dynamics (CFD) and mechanics; particular focus on parallel computing using GPU accelerators and 3-D code development with application to nonlinear PDEs related to Earth science dynamical systems; evaluation of performance limiters and discussion on basic optimization techniques; finite-difference discretization (stencil codes) and accelerated iterative methods. Programming languages: MATLAB, CUDA C, (MPI). Hands-on approach: starting from diffusion and wave propagation physics, the goal being to achieve a 3-D Stokes flow utilizing C CUDA in order to leverage the GPU computing power. Pre-requisites: basic programming skills (e.g. MATLAB) and Scientific computing. Basic knowledge of compiled languages (C) and CFD are a plus.
Last offered: Summer 2019
## CME 258:Libraries for Numerical Linear Algebra and Optimization
This course will cover standard libraries commonly used for numerical linear algebra and optimization, with an emphasis on giving students experience with using the libraries on real examples. The course will cover software for direct methods (BLAS, Atlas, LAPACK, Eigen), iterative methods (ARPACK, Krylov Methods), and linear/nonlinear optimization (MINOS, SNOPT). Prerequisites: at least one course in numerical linear algebra (preferably at the level of CME 200 or CME 302), and one course in numerical optimization, as well as experience with at least one compiled language such as C/C++/Fortran.
Last offered: Spring 2018
## CME 321A:Mathematical Methods of Imaging (MATH 221A)
Image denoising and deblurring with optimization and partial differential equations methods. Imaging functionals based on total variation and l-1 minimization. Fast algorithms and their implementation.
Last offered: Winter 2014
## CME 345:Model Reduction
Model reduction is an indispensable tool for computational-based design and optimization, statistical analysis, embedded computing, and real-time optimal control. This course presents the basic mathematical theory for projection-based model reduction. Topics include: notions of linear dynamical systems and projection; projection-based model reduction; error analysis; proper orthogonal decomposition; Hankel operator and balancing of a linear dynamical system; balanced truncation method: modal truncation and other reduction methods for linear oscillators; model reduction via moment matching methods based on Krylov subspaces; introduction to model reduction of parametric systems and notions of nonlinear model reduction. Course material is complemented by a balanced set of theoretical, algorithmic and Matlab computer programming assignments. Prerequisites: CME 200 or equivalent, CME 263 or equivalent and basic numerical methods for ODEs.
Last offered: Spring 2017
## CME 364A:Convex Optimization I (CS 334A, EE 364A)
Convex sets, functions, and optimization problems. The basics of convex analysis and theory of convex programming: optimality conditions, duality theory, theorems of alternative, and applications. Least-squares, linear and quadratic programs, semidefinite programming, and geometric programming. Numerical algorithms for smooth and equality constrained problems; interior-point methods for inequality constrained problems. Applications to signal processing, communications, control, analog and digital circuit design, computational geometry, statistics, machine learning, and mechanical engineering. Prerequisite: linear algebra such as EE263, basic probability.
Terms: Win, Sum | Units: 3
## CME 364B:Convex Optimization II (EE 364B)
Continuation of 364A. Subgradient, cutting-plane, and ellipsoid methods. Decentralized convex optimization via primal and dual decomposition. Monotone operators and proximal methods; alternating direction method of multipliers. Exploiting problem structure in implementation. Convex relaxations of hard problems. Global optimization via branch and bound. Robust and stochastic optimization. Applications in areas such as control, circuit design, signal processing, and communications. Course requirements include project. Prerequisite: 364A.
Terms: Spr | Units: 3
Instructors: Pilanci, M. (PI)
## CME 375:Advanced Topics in Convex Optimization (MATH 301)
Modern developments in convex optimization: semidefinite programming; novel and efficient first-order algorithms for smooth and nonsmooth convex optimization. Emphasis on numerical methods suitable for large scale problems arising in science and engineering. Prerequisites: convex optimization ( EE 364), linear algebra ( Math 104), numerical linear algebra ( CME 302); background in probability, statistics, real analysis and numerical optimization.
Last offered: Winter 2015 | Repeatable for credit
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career | 1,784 | 8,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.803306 |
http://www.slideserve.com/niveditha/standardizedtesting | 1,475,088,548,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661763.18/warc/CC-MAIN-20160924173741-00064-ip-10-143-35-109.ec2.internal.warc.gz | 746,084,780 | 13,878 | This presentation is the property of its rightful owner.
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# Standardized Testing PowerPoint PPT Presentation
Standardized Testing Norm-referenced tests : are standardized, have information about reliability and validity, and can be used to compare an individual or groups performance on a test to individuals or groups in the standardization population, often called a “formal” test.
Related searches for Standardized Testing
Standardized Testing
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### Standardized Testing
• Norm-referenced tests: are standardized, have information about reliability and validity, and can be used to compare an individual or groups performance on a test to individuals or groups in the standardization population, often called a “formal” test.
• Criterion-referenced tests: are not standardized, often do not have reliability and validity data, a person or group is compared to a fixed standard, often labeled an “informal” test.
### Basic Measurement Terms
• Scales of Measurement: nominal, ordinal, interval, ratio
• Measures of Central Tendency: mean, median, mode
• Measures of Dispersion: range, variance, standard deviation
• Normal Curve:a common type of distribution
### Basic Measurement Terms continued
• Correlations: tell us about the degree of relationship between two variables, including the strength and direction of the relationship.
• Multiple Correlation: a statistical technique for determining the relationship between one variable and several other variables
### Basic Measurement Terms (3)
• Types of Scores:
Percentile rank-derived scores that permit us to determine an individual’s position relative to the sample.
Standard scores- raw scores that have been transformed to have a designated mean and standard deviation.
Grade equivalent or age-equivalent scores-average score obtained on a test by different groups of children who vary in age or grade placement.
### Basic Measurement Terms (4)
Normal curve equivalents- standard scores with a mean of 100 and a standard deviation of 21.06
Stanines-a single digit scoring system with a mean of 5 and a SD of 2. (1-9)
### Interpreting Test Scores
• Reliability: the consistency of measurements
Test-retest reliability - consistency of scores on two separate administrations of a test.
Alternate-form reliability- consistency of scores on two equivalent forms of a test.
Split-half reliability – degree to which individual test items are related or measure the same abilities.
### Interpreting Test Scores continued
• True score: hypothetical mean of all scores if test were administered many times.
• Standard error of measurement: estimate of the amount of error associated with the obtained score.
• Confidence interval or precision range: a range within which true scores might be found.
### Interpreting Test Scores (3)
• Validity: the extent to which a test measures what it is supposed to measure.
Content validity -whether the items on a test
represent the domain that the test is supposed to
Measure.
Criterion-related validity- the relationship
between test scores and some type of criterion or
outcome.
### Interpreting Test Scores (4)
Concurrent validity and predictive validity –are test scores related to a current criterion or future performance on a relevant criterion?
Construct validity – the extent to which a test measures a psychological construct.
• The relationship between reliability and validity: to be valid, a test must be reliable.
### Types of Standardized Tests
Format
• Screening, Diagnostic, Placement
• Group Versus Individual
• Multiple Skill Versus Single Skill
• Formal Versus Informal
Function or Domain
• Achievement
### Types of Standardized Tests continued
• Cognitive ability/intellectual ability-
• Social/ Emotional-objective versus projective
• Behavioral- rating scales, observations
### Major Issues in Assessment
• What is the difference between assessment and testing?
• How are tests being used in the United States: readiness, national progress, minimal competency, accreditation
• Advantages in taking tests: bias, culture-fair tests.
• Coaching and test-taking skills: special training, familiarity with procedures, study skills
### New Directions in Assessment
National, State and Local Levels:
• Assessing learning potential: LPAD, ELP, etc.
• Authentic assessment: performance based, portfolios, constructed response formats.
• Curriculum-based assessment and measurement. | 984 | 4,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-40 | longest | en | 0.898739 |
https://plainmath.net/18252/3sqrt-1-84 | 1,627,607,067,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153899.14/warc/CC-MAIN-20210729234313-20210730024313-00471.warc.gz | 467,474,429 | 7,391 | Question
# 3sqrt(1/84)
Equation, expression, and inequalitie
$$\displaystyle{3}\sqrt{{\frac{{1}}{{84}}}}$$
$$3\sqrt{\frac{1}{84}}=3\sqrt{1^3/4^3} =3\sqrt{1/4}^3 =1/4$$ | 82 | 169 | {"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} | 3.546875 | 4 | CC-MAIN-2021-31 | latest | en | 0.509154 |
https://feet-to-inches.appspot.com/5790-feet-to-inches.html | 1,723,734,704,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00286.warc.gz | 190,212,283 | 6,365 | Feet To Inches
# 5790 ft to in5790 Feet to Inches
ft
=
in
## How to convert 5790 feet to inches?
5790 ft * 12.0 in = 69480.0 in 1 ft
A common question is How many foot in 5790 inch? And the answer is 482.5 ft in 5790 in. Likewise the question how many inch in 5790 foot has the answer of 69480.0 in in 5790 ft.
## How much are 5790 feet in inches?
5790 feet equal 69480.0 inches (5790ft = 69480.0in). Converting 5790 ft to in is easy. Simply use our calculator above, or apply the formula to change the length 5790 ft to in.
## Convert 5790 ft to common lengths
UnitLength
Nanometer1.764792e+12 nm
Micrometer1764792000.0 µm
Millimeter1764792.0 mm
Centimeter176479.2 cm
Inch69480.0 in
Foot5790.0 ft
Yard1930.0 yd
Meter1764.792 m
Kilometer1.764792 km
Mile1.0965909091 mi
Nautical mile0.9529114471 nmi
## What is 5790 feet in in?
To convert 5790 ft to in multiply the length in feet by 12.0. The 5790 ft in in formula is [in] = 5790 * 12.0. Thus, for 5790 feet in inch we get 69480.0 in.
## Alternative spelling
5790 Foot to in, 5790 Foot in in, 5790 Foot to Inch, 5790 Foot in Inch, 5790 ft to in, 5790 ft in in, 5790 ft to Inch, 5790 ft in Inch, 5790 ft to Inches, 5790 ft in Inches, 5790 Foot to Inches, 5790 Foot in Inches, 5790 Feet to Inch, 5790 Feet in Inch | 463 | 1,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-33 | latest | en | 0.827135 |
https://oeis.org/A189064 | 1,723,045,168,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694594.35/warc/CC-MAIN-20240807143134-20240807173134-00040.warc.gz | 333,397,689 | 4,124 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A189064 T(n,k)=Number of nXk binary arrays without the pattern 0 1 0 antidiagonally or horizontally 13
2, 4, 4, 7, 16, 8, 12, 49, 64, 16, 21, 144, 316, 256, 32, 37, 441, 1404, 2032, 1024, 64, 65, 1369, 6768, 13452, 13045, 4096, 128, 114, 4225, 33893, 99721, 128628, 83737, 16384, 256, 200, 12996, 167473, 795741, 1492864, 1228512, 537496, 65536, 512, 351, 40000 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Table starts ....2.......4.........7..........12............21..............37 ....4......16........49.........144...........441............1369 ....8......64.......316........1404..........6768...........33893 ...16.....256......2032.......13452.........99721..........795741 ...32....1024.....13045......128628.......1492864........19468046 ...64....4096.....83737.....1228512......22289912.......477128662 ..128...16384....537496....11733712.....333124565.....11711612310 ..256...65536...3450100...112065936....4978704008....287687887135 ..512..262144..22145617..1070316016...74410715409...7067105036501 .1024.1048576.142149013.10222334864.1112149145053.173620295413143 LINKS R. H. Hardin, Table of n, a(n) for n = 1..573 EXAMPLE Some solutions for 5X3 ..1..0..0....0..0..1....1..0..1....0..0..1....0..0..1....1..1..1....1..1..0 ..0..0..1....1..1..1....1..0..0....0..0..0....0..0..1....0..0..1....0..1..1 ..1..1..1....0..1..1....0..0..1....0..0..0....1..1..0....1..0..0....1..1..1 ..1..1..1....1..1..1....0..0..1....0..0..0....1..0..0....0..0..1....0..0..1 ..1..0..0....0..0..1....0..1..1....1..0..1....0..0..0....1..1..0....0..1..1 CROSSREFS Column 2 is Column 1 squared Column 3 is A188868 Row 1 is A005251(n+3) Row 2 is A188501 Sequence in context: A189264 A188992 A189617 * A188874 A095061 A184233 Adjacent sequences: A189061 A189062 A189063 * A189065 A189066 A189067 KEYWORD nonn,tabl AUTHOR R. H. Hardin Apr 16 2011 STATUS approved
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Last modified August 7 11:33 EDT 2024. Contains 375012 sequences. (Running on oeis4.) | 888 | 2,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.545732 |
https://blog.jws.app/2021/how-to-sort-a-list-of-locations-by-how-close-they-are-to-you/ | 1,685,946,862,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651325.38/warc/CC-MAIN-20230605053432-20230605083432-00528.warc.gz | 173,494,759 | 12,482 | How to sort a list of locations by how close they are to you
One of the things that are most important to me for phase 1 of the state parks app is to have the list of parks be sorted by how far the user is from the park. After all, Tower Hill State Park might be interesting but if you live 4 hours away, you aren’t likely to give it a visit. While I know vaguely how I would do that, I do not think that I have actually done that before. So, let’s figure this out.
For the first step, we need to figure out where the user is. Last year, I wrote a codepen where I used navigator.geolocation and MapQuest to figure out where the user is. The result was to just show the address the person is most likely at. If we fork that pen and add an array-driven table of UW system campuses, we now both know our starting point and our five potential end points.
See the Pen
Sorting a list of locations by how for away they are: Part 1
by Joe Steinbring (@steinbring)
on CodePen.
The second step is the most interesting one. You will notice that we have the user’s lat/long coordinates in `this.location.latLng` and we have the lat/long coordinates for each campus in `this.campuses[i].lat` and `this.campuses[i].long`. If we use the Haversine formula, we can figure out the distance between the two points.
See the Pen
Sorting a list of locations by how for away they are: Part 2
by Joe Steinbring (@steinbring)
on CodePen.
The above code gets you the number of kilometers between the two points, as the crow flies. If you want better precision, you need to make an API call to something like google maps. Once we get this far, we just need to sort the table by the “distance away” column. We have done that at least once before. Using a computed property, we can sort the array by distance.
See the Pen
Sorting a list of locations by how for away they are: Part 3
by Joe Steinbring (@steinbring)
on CodePen.
So, now we know where the user is, we know where the campuses are, we know the distances between the user and the campuses, and we have the table sorted from the closest campus to the one that is furthest away. If we were going to continue the exercise, we would probably want to convert from kilometers to miles and round those values down to one or two decimal places. I will leave that for a future blog post, though.
Have any questions, comments, etc? Feel free to drop a comment, below.
[ Cover photo by Thomas Kinto on Unsplash ]
1 thought on “How to sort a list of locations by how close they are to you”
1. @MapGoddess says:
as the crow flies rather than road network? I’m thinking of transferability and skill challenges. You could look at locations with a mountain range or large lake impacting movement. Have you coded barriers into this?
You know I’m loving the #MappyGeeky | 673 | 2,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-23 | latest | en | 0.93076 |
http://schools.nyc.gov/offices/teachlearn/documents/standards/math/ms/102best.html | 1,508,233,157,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187821017.9/warc/CC-MAIN-20171017091309-20171017111309-00773.warc.gz | 308,113,868 | 4,963 | The task
The focus of the task deals with the game of golf and therefore uses terminology that is golf related. It is not necessary to be familiar with the game of golf to solve this problem.
Students were presented with a scenario in which they had to determine which of three golfers is the “best” chipper (the term chipping refers to hitting the golf ball). Each golfer chipped ten balls. The measured distances of the balls from the cup are given. The data for this problem were devised so that the average distance from the cup is the same for all three golfers. Thus, students had to use other appropriate statistical measures to analyze the situation. “Getting close and being consistent” are the criteria on which the students were to judge the golfers.
Circumstances of performance This sample of student work was produced under the following conditions: alone - in a group - in class as homework with teacher feedback with peer feedback timed opportunity for revision
Any time a task related to sports is used, teacher input may be necessary to clarify some terms and ensure equitable access to the task for all students.
This work sample illustrates a standard-setting performance for the following parts of the standards: a Statistics and Probability Concepts: The student…organizes data, and displays data with …graphs that are appropriate…. b Statistics and Probability Concepts: The student analyzes appropriately central tendencies of data by considering mean and median. c Statistics and Probability Concepts: The student analyzes appropriately central tendencies of data by considering mean and median. d Statistics and Probability Concepts: The student makes conclusions…based on data analysis.
What the work shows a Statistics and Probability Concepts: The student…organizes data, and displays data with …graphs that are appropriate…. The student rewrote Sarah’s data in numerical order to illustrate median, quartiles, and extreme values of the range. It would have been good to see the same for Rick’s and Mike’s data. b Statistics and Probability Concepts: The student analyzes data with respect to characteristics of frequency and distribution, including mode and range.
c Statistics and Probability Concepts: The student analyzes appropriately central tendencies of data by considering mean and median. The student made meaningful comparisons of the data, keeping in mind the coach’s criteria for good chipping. The student was thinking mathematically of the coach’s consistency criterion. He used the size of the inter-quartile range as a measure of consistency. Rick’s and Mike’s “quite large” ranges suggest some inconsistency, especially when compared with Sarah’s smaller inter-quartile range. The inter-quartile range of the middle half of the data combines notions of distribution with those of central tendency. Rick’s farthest shot is recognized as an outlier. Indeed, his two chips that landed 312 and 320 inches away are both outliers. However, the student did not show them as such on the box-and-whiskers plot. He could have done so by plotting two isolated points at those two values instead of showing the range extending to 320 inches. The realization that “the total yardage…was the same” suggests an understanding of average as more than a computational procedure. The total yardages are the same, just like the averages, because the golfers made the same number of chips. d Statistics and Probability Concepts: The student makes conclusions…based on data analysis. The student justifiably declared Sarah the winner, particularly on grounds of consistency. A strong case could have been made for Rick, though. The median value of his chips is 30 inches less than that for Sarah. Furthermore, comparing their attempts in order from closest to farthest, each of Rick’s best seven chips is closer than Sarah’s corresponding chip. Their eighth best chips are equidistant, 152 inches away from the cup. Only Sarah’s worst two chips are better than Rick’s worst two. | 775 | 4,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-43 | latest | en | 0.949884 |
https://web2.0calc.com/questions/probability_145 | 1,610,844,481,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703507971.27/warc/CC-MAIN-20210116225820-20210117015820-00207.warc.gz | 648,894,112 | 5,603 | +0
# probability
0
173
1
A fair dice has six faces numbered 1, 1, 1, 2, 3, & 4.
The dice is rolled twice and the number is shown recorded each time.
Find the probability that the sum of the two numbers recorded is at least 4.
Apr 11, 2020
#1
+240
0
There are 6*6=36 possible rolls, and while some of these are the same they need to be counted multiple times because they have a higher chance of occuring.
Rolls that sum to 4, with frequency:
(1,3) x3
(3,1) x3
(2,2) x1
Sum to 5:
(1,4) x3
(4,1) x3
(3,2) x1
(2,3) x1
Sum to 6:
(3,3) x1
(2,4) x1
(4,2) x1
Sum to 7:
(3,4) x1
(4,3) x1
8:
(4,4) x1
Adding up the frequencies gets 21
The probability is 21/36=7/12
Apr 11, 2020 | 281 | 695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-04 | latest | en | 0.935441 |
https://www.turito.com/ask-a-doubt/Mathematics-in-the-given-triangular-pyramid-abcp-yz-is-the-midsegment-of-delta-apc-xz-is-the-midsegment-of-delta-ab-q077987 | 1,716,882,941,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00163.warc.gz | 898,294,292 | 22,166 | Mathematics
Easy
Question
# In the given triangular pyramid, ABCP, YZ is the midsegment of APC, XZ is the midsegment of ABC, XY is the midsegment of ABP. Which of the following statement is true?
Hint:
## The correct answer is: Area of BPC = Area of XYZ
### Area of ∆BPC = height of ∆XYZ base of ∆XYZ= Area of ∆XYZ
The midpoint theorem refers to the midpoint of the line segment. It defines the coordinate points of the midpoint of the line segment and can be found by taking the average of the coordinates of the given endpoints. | 148 | 544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.846498 |
https://rich-iannone.github.io/DiagrammeR/reference/select_nodes_by_degree.html | 1,708,501,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473401.5/warc/CC-MAIN-20240221070402-20240221100402-00485.warc.gz | 509,123,316 | 4,860 | Using a graph object of class dgr_graph, create a selection of nodes that have certain degree values.
## Usage
select_nodes_by_degree(graph, expressions, set_op = "union")
## Arguments
graph
A graph object of class dgr_graph.
expressions
One or more expressions for filtering nodes by degree values. Use a combination of degree type (deg for total degree, indeg for in-degree, and outdeg for out-degree) with a comparison operator and values for comparison (e.g., use "deg >= 2" to select nodes with a degree greater than or equal to 2).
set_op
The set operation to perform upon consecutive selections of graph nodes. This can either be as a union (the default), as an intersection of selections with intersect, or, as a difference on the previous selection, if it exists.
## Value
A graph object of class dgr_graph.
## Examples
# Create a random graph using
# the add_gnm_graph() function
graph <-
create_graph() %>%
n = 35, m = 125,
set_seed = 23)
# Report which nodes have a
# total degree (in-degree +
# out-degree) of exactly 9
graph %>%
select_nodes_by_degree(
expressions = "deg == 9") %>%
get_selection()
#> [1] 5 10 26 31
# Report which nodes have a
# total degree greater than or
# equal to 9
graph %>%
select_nodes_by_degree(
expressions = "deg >= 9") %>%
get_selection()
#> [1] 1 2 4 5 10 12 18 25 26 31
# Combine two calls of
# select_nodes_by_degree() to
# get those nodes with total
# degree less than 3 and total
# degree greater than 10 (by
# default, those select...()
# functions will union the
# sets of nodes selected)
graph %>%
select_nodes_by_degree(
expressions = "deg < 3") %>%
select_nodes_by_degree(
expressions = "deg > 10") %>%
get_selection()
#> [1] 1 2
# Combine two calls of
# select_nodes_by_degree() to
# get those nodes with total
# degree greater than or equal
# to 3 and less than or equal
# to 10 (the key here is to
# intersect the sets of nodes
# selected in the second call)
graph %>%
select_nodes_by_degree(
expressions = "deg >= 3") %>%
select_nodes_by_degree(
expressions = "deg <= 10",
set_op = "intersect") %>%
get_selection()
#> [1] 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
#> [26] 28 29 30 31 32 33 34 35
# Select all nodes with an
# in-degree greater than 5, then,
# apply a node attribute to those
# selected nodes (coloring the
# selected nodes red)
graph_2 <-
graph %>%
select_nodes_by_degree(
expressions = "indeg > 5") %>%
set_node_attrs_ws(
node_attr = color,
value = "red")
# Get the selection of nodes
graph_2 %>% get_selection()
#> [1] 1 4 13 25 35 | 770 | 2,568 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-10 | latest | en | 0.855835 |
https://espanol.libretexts.org/Humanidades/Filosofia/M%C3%A9todos_Fundamentales_de_L%C3%B3gica_(Knachel) | 1,726,522,879,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00024.warc.gz | 216,893,805 | 34,147 | Saltar al contenido principal
# Métodos Fundamentales de Lógica (Knachel)
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Fundamental Methods of Logic es adecuado para un curso de introducción al razonamiento lógico/crítico de un semestre. Abarca una variedad de temas a nivel introductorio. El Capítulo Uno introduce nociones básicas, como argumentos y explicaciones, validez y solidez, razonamiento deductivo e inductivo; también abarca técnicas analíticas básicas, como distinguir premisas de conclusiones y diagramación de argumentos. En el capítulo dos se discuten falacias lógicas informales. Los capítulos Tres y Cuatro se refieren a la lógica deductiva, introduciendo los fundamentos de la Lógica Aristotélica y Sentencial, respectivamente. El capítulo cinco trata del razonamiento analógico y causal, incluyendo una discusión sobre los métodos de Mill. El Capítulo Seis abarca cálculos básicos de probabilidad, inferencia bayesiana, conceptos y técnicas estadísticas fundamentales y falacias estadísticas comunes.
Miniatura: pixabay.com/ilustraciones/pu... ution-1721464/
This page titled Métodos Fundamentales de Lógica (Knachel) is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Matthew Knachel via source content that was edited to the style and standards of the LibreTexts platform. | 2,085 | 5,635 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-38 | latest | en | 0.188428 |
https://math.stackexchange.com/questions/76862/find-lim-limits-n-to-infty-sum-limits-i-1n-frac2nn2i2 | 1,566,814,754,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027331485.43/warc/CC-MAIN-20190826085356-20190826111356-00179.warc.gz | 546,454,224 | 32,094 | # Find $\lim\limits_{n\to\infty}\sum\limits_{i=1}^{n}{\frac{2n}{(n+2i)^2}}$
Find $$\lim\limits_{n\to\infty}\sum\limits_{i=1}^{n}{\frac{2n}{(n+2i)^2}}.$$
I have tried dividing through by $1/n^2$ and various other algebraic tricks but cannot seem to make any progress on this limit. Wolfram Alpha gives the value as $2/3$, but I could not understand its derivation. Any insight is welcome.
The expression $$\sum_{i = 1}^n \frac{2n}{(n+2i)^2} =\frac{1}{n} \sum_{i = 1}^n \frac{2}{(1+\frac{2i}{n})^2}$$ is the Riemann sum of the function $f(x)= \frac{2}{(1+2x)^2}$ over the interval $I = [0,1]$, corresponding to the uniform partition of $I$ into $n$ equal parts. Since $f$ is Riemann-integrable (being a continuous function over a closed and bounded interval), this sum approaches the integral of $f$ over $I$ as $n \to \infty$. That is, $$\begin{eqnarray*} \lim_{n \to \infty} \frac{1}{n} \sum_{i = 1}^n \frac{2}{(1+\frac{2i}{n})^2} &=& \int_0^1 \frac{2}{(1+2x)^2} \mathrm{d} \, x \\ &=& \left. -\frac{1}{1+2x} \right|_{x=0}^{x=1} \quad = \quad \frac{2}{3}. \end{eqnarray*}$$
• Thanks @Daniel, for coming back to accept an answer. Many users forget to do that... =) – Srivatsan Oct 31 '11 at 14:45
You can also use Euler-Maclaurin summation.
The first-order Euler-Maclaurin formula says $$\sum_{i=1}^n f(i) = \int_1^n f(x) \, dx + {f(1) + f(n) \over 2} + \int_1^n f'(x) \left(x - \lfloor x \rfloor - \frac{1}{2}\right)\,dx.$$
Since $|x - \lfloor x \rfloor - \frac{1}{2}| \leq 1$, with $f(x) = \frac{2n}{(n+2x)^2}$ we have $$\sum_{i=1}^n \frac{2n}{(n+2i)^2} = \int_1^n \frac{2n \, dx}{(n+2x)^2} + R_n,$$ where $|R_n| \leq \left|\frac{3f(n) - f(1)}{2}\right| = \left|\frac{3}{n} - \frac{n}{(n+2)^2}\right|$.
Therefore, $$\lim_{n \to \infty} \sum_{i=1}^n \frac{2n}{(n+2i)^2} = \lim_{n \to \infty} \int_1^n \frac{2n \, dx}{(n+2x)^2} = \lim_{n \to \infty} \left[\frac{-n }{n+2x} \right]_1^n = \lim_{n \to \infty} \left(-\frac{1}{3} + \frac{n}{n+2}\right) = \frac{2}{3}.$$
• +1 Your solution is also nice. I especially like the fact that this yields an error bound of $O(1/n)$ as well. – Srivatsan Oct 29 '11 at 5:58
• This solution is good, but we need the best one which is simpler than this one. – Hassan Muhammad Oct 29 '11 at 6:19
• @Hassan: I am missing the point of your comment... – J. M. is a poor mathematician Oct 29 '11 at 15:37 | 983 | 2,341 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-35 | latest | en | 0.625737 |
http://www.tulyn.com/wordproblems/fractions.htm | 1,495,558,822,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607648.39/warc/CC-MAIN-20170523163634-20170523183634-00140.warc.gz | 692,037,946 | 9,665 | # Fraction Word Problems
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# Fractions Word Problems
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### Number Sense Word Problems
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1. Find the greatest common factor. 4, 6, 12.
2. Factor. 24x3 + 30x2
3. Factor out the GCF with a negative coefficient. –24m2n6 – 8mn5 – 32n4
4. Factor completely by factoring out any common factors and then factoring by grouping.
6x2 – 5xy + 6x – 5y
5. The GCF of 15y + 20 is 5. The GCF of 15y + 21 is 3. Find the GCF of the product (15y + 20)(15y + 21).
6. The area of a rectangle of length x is given by 15xx2. Find the width of the rectangle in terms of x.
7. Factor the trinomial completely. x2 + 8x – 9
8. Factor the trinomial completely. 2x2 + 16x + 32
9. Complete the following statement. 6a2 – 5a + 1 = (3a – 1)(__?__)
10. State whether the following is true or false. x2 – 7x – 30 = (x + 3)(x – 10)
11. Factor completely. x2 + 11x + 28
12. Factor completely. 15x2 + 23x + 4
13. Factor completely. 6z3 – 27z2 + 12z
14. The number of hot dogs sold at the concession stand during each hour iih after opening at a soccer tournament is given by the polynomial 2h2 – 19h + 24. Write this polynomial in factored form.
15. Find a positive value for k for which the polynomial can be factored.x2kx + 29
16. Factor completely. 9x2 + 4
17. Determine whether the following trinomial is a perfect square. If it is, factor the binomial.x2 – 12x + 36
18. Factor completely. 25x2 + 40xy + 16y2
19. Factor. s2(t u) – 9t2(tu)
20. State which method should be applied as the first step for factoring the polynomial. 6x3 + 9x
21. State which method should be applied as the first step for factoring the polynomial. 2a2 + 9a + 10
22. Solve the quadratic equation. 5x2 + 17x = –6
23. Solve the quadratic equation. 3x(2x – 15) = –84
24. The sum of an integer and its square is 30. Find the integer.
25. If the sides of a square are decreased by 3 cm, the area is decreased by 81 cm2. What were the dimensions of the original square?
26. Write in simplest form.
27. Write in simplest form.
28. Write the expression in simplest form.
29. The area of the rectangle is represented by 5x2 + 19x + 12. What is the length?
5x + 4
30. Multiply.
31. Multiply.
32. Divide.
33. Divide.
34. Perform the indicated operations.
35. Find the area of the rectangle shown.
40. Add or subtract as indicated.
41. One number is 8 less than another. Let x represent the larger number and use a rational expression to represent the sum of the reciprocals of the two numbers.
42. Simplify.
43. Simplify.
44. What values for x, if any, must be excluded in the following algebraic fraction?
45. What values for x, if any, must be excluded in the following algebraic fraction?
46. Solve for x. + 6 = 1
47. Solve for x.
48. Solve for x.
49. One number is 3 times another. If the sum of their reciprocals is , find the two numbers.
50. A 5-foot pole casts a shadow of 4 feet. How tall is a tree with a shadow of 16 feet? | 1,080 | 3,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.66278 |
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# Problem 157. The Hitchhiker's Guide to MATLAB
Solution 30589
Submitted on 6 Feb 2012 by Jos (10584)
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### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 41; y_correct = false; assert(isequal(zaphod(x),y_correct))
y = 0
2 Pass
%% x = 42; y_correct = true; assert(isequal(zaphod(x),y_correct))
y = 1
3 Pass
%% x = 43; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 44; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 45; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 46; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 47; y_correct = false; assert(isequal(zaphod(x),y_correct))
y = 0 y = 0 y = 0 y = 0 y = 0
4 Pass
%% x = 48; y_correct = false; assert(isequal(zaphod(x),y_correct))
y = 0
5 Pass
%% x = 49; y_correct = false; assert(isequal(zaphod(x),y_correct))
y = 0
6 Pass
%% x = 50; y_correct = false; assert(isequal(zaphod(x),y_correct))
y = 0
### Community Treasure Hunt
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Start Hunting! | 394 | 1,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.442683 |
https://math.stackexchange.com/questions/2652327/average-with-sum-of-values-equal-to-number-of-values/2652339 | 1,566,567,526,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318421.65/warc/CC-MAIN-20190823130046-20190823152046-00281.warc.gz | 544,435,216 | 26,421 | # Average with sum of values equal to number of values [closed]
I have a list of numbers. Each number can be any amount above 0 but under 1 million. There are 1 million numbers.
The numbers add up to 1 million.
Is it possible to calculate an average (Mean)? Normally calculating the mean does not work as 1 million / 1 million = 1 of course.
This is a bit of a random question but thanks.
## closed as unclear what you're asking by Did, Mostafa Ayaz, Mohammad Riazi-Kermani, José Carlos Santos, SaadFeb 17 '18 at 0:13
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
Let us call your numbers $(X_i)_{1\leq i \leq 10^6}$. The mean is defined as $$\mu = \frac{1}{10^6} \sum_{i=1}^{10^6} X_i=1$$ so the mean is $1$. You can either define it as the mean of this specific series or the probabilistic mean in both cases it will be $1$.
• Alright. Thank you very much for your response. I will mark this as the answer when I can and use a different average. – Neztore Feb 15 '18 at 23:17
• The only case where it could be different from $1$ is if the $X_i$ are correlated to have a sum equal to $10^6$ but that taken alone they have a different distribution. But in that case you need to specify what variables you are talking about. – Jonathan D Feb 15 '18 at 23:21 | 412 | 1,545 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-35 | latest | en | 0.931559 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=175498 | 1,369,000,724,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698090094/warc/CC-MAIN-20130516095450-00042-ip-10-60-113-184.ec2.internal.warc.gz | 316,658,930 | 889 | ```Question 238874
(-5,-3) and (10,-6)
Y=mX+b WHERE m=SLOPE & b=THE Y INTERCEPT.
SLOPE=(Y2-Y1)/(X2-X1)=(-6+3)/(10+5)=-3/15=-1/5 IS THE SLOPE
-3=-5*-1/5+b
-3=1+b
b=-3-1
b=-4 FOR THE Y INTERCEPT.
Y=X/5-4 IS THIS LINE EQUATION.
{{{ graph( 300, 200, -6, 15, -10, 10, -x/5 -4) }}} (graph 300x200 pixels, x from -6 to 15, y from -10 to 10, -x/5 -4). ``` | 177 | 347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2013-20 | latest | en | 0.51206 |
https://pwntestprep.com/2017/09/hi-mike-can-you-please-do-practise-test-7-section-4-question-29-whats-the-fastest-way-to-do-it-which-doesnt-involve-using-the-distance-formula-for-every-answer-option/ | 1,716,012,069,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00639.warc.gz | 437,483,594 | 17,615 | Hi Mike,
Can you please do practise test 7, section 4, question 29 (Whats the fastest way to do it which doesn’t involve using the distance formula for every answer option)
Yeah, NO distance formula! The way to get this one is to know the standard formula for a circle and to recognize that it’s basically given to you here. The standard formula for a circle of radius r centered at (hk) is .
Therefore, means you have a circle of radius 5 centered at (–3, 1). Once you know the center, you shouldn’t have to do the distance formula to recognize that (3, 2) is more than 5 units away from (–3, 1). | 151 | 600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.909307 |
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