url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://jet.com/product/A-Sports-Oriented-Approach-to-Introductory-Statistics/2359a3baaed7426885b11a5a970ab6fb | 1,529,508,366,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863650.42/warc/CC-MAIN-20180620143814-20180620163814-00450.warc.gz | 639,244,373 | 83,094 | Free Shipping Over \$35 & Free Returns Details
### Description
No one can deny the obvious link between statistics and sports, and many a sports fan can reel off a lengthy list of statistics about a favorite team or player. The anthology A Sports-Oriented Approach to Introductory Statistics cleverly combines the study of statistics with the high interest area of sports to create a text that delivers an engaging and effective introduction to statistical principles. The first several chapters of the book are the statistical and mathematical offensive linemen, often overlooked and under-appreciated, but necessary for gaining a firm understanding of theory and statistics. This prepares students for the principles covered in subsequent chapters. These later chapters are the skill players of the text, and through them students learn things like how to predict points in the NHL, determine if there is a home field advantage in football, and compare MLB teams to find out if statistically any one team has been the worst over the past ten years. Each example in the book uses real data from the sports world. The central limit theorem is explained by examining salary data from major league baseball. Understanding probability distribution of a discrete random variable is illustrated through the number of goals scored in the World Cup. Regression analysis is explored through evaluating team performance. Each chapter of the text includes examples, accompanied by complete, step-by-step solutions. All problems require students to produce extended, well-thought-out answers using the target principles. Symbols and formulas used within chapters are organized at the end of the chapters for easy reference. This unique approach increases student interest, and delivers important content in a relatable, likeable format. A Sports-Oriented Approach to Introductory Statistics is written for courses in elementary, undergraduate statistics, and is an ideal supplement to standard required texts. The text can also be used as a supplemental text in many math and business courses grounded in statistics. Andrew Wiesner earned his Ph.D. at the University of Pittsburgh. Currently he is a Lecturer of Statistics at Pennsylvania State University. Dr. Wiesner has been teaching statistics for more than ten years. He successfully combined his passion for education with a life-long love of sports, and supervised several undergraduate Honors theses in which students analyze sports data.
• ISBN13: 9781621316404
• Pubilcation Year: 2012
• Format: Paperback
• Pages: 00236
Specifications
Format Paperback Publication Date December 1, 2012 Pages 236 Primary Category Sports & Recreation/Baseball - Statistics
# A Sports-Oriented Approach to Introductory Statistics by Andrew Wiesner
Write a Review
Free Shipping over \$35 and Free Returns
\$83.09
\$0.44 off if you opt out of free returns | 557 | 2,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-26 | latest | en | 0.918405 |
https://notesread.com/uniform-circular-motion/ | 1,603,492,264,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107865665.7/warc/CC-MAIN-20201023204939-20201023234939-00153.warc.gz | 456,888,593 | 17,540 | # uniform circular motion
The most simple case of curvilinear movement is the uniform circular motion, in which a point P is moving with constant speed on a trajectory given by a circle of radius R and center O . In this case, the time interval T , used by point P to make a complete turn on the circumference, is called the period of motion. If in the unit of time (1 second), the point P completes f turns of circumference, the period used for each single turn will be equal to 1 / f seconds; the number fit is called the frequency of motion and is measured in revolutions per second (rev / s) (the unit of measurement of frequency in the International System is the hertz, symbol Hz, where 1 Hz = 1 s -1 ). The period and the frequency in a uniform circular motion are linked by the relationship:
The speed in uniform circular motion
Known the period T of a uniform circular motion, its speed can be easily deduced by remembering that, by definition, the point P completes, in a time interval equal to a period, exactly one turn of circumference, thus covering a space equal to 2R . The relationship between space traveled and time spent therefore leads to a constant value given by:
Or, substituting the frequency f for period T :
Once the intensity value has been established, the velocity as a vector quantity is fully defined by assigning the direction of the tangent to the circumference at point P and the direction of motion as direction (see fig. 4.1). The velocity in uniform circular motion can also be expressed in terms of angular velocity, which represents the displacement of the anglefollowing the motion of the point P on the circumference. The relationship between the angular velocityand the velocity of the point P is given by:
Acceleration in uniform circular motion
In uniform circular motion, while the modulus of the velocity vector remains constant, its direction changes continuously, since, as the point moves along the circumference, the position of the tangent to the curve changes continuously. It is therefore possible to express this variation by introducing an acceleration vector. In this case the tangential component of the acceleration is zero, while the centripetal component can be determined. To calculate the acceleration, it may be useful to use a graphic method, illustrated starting from figure 4.2; on a circumference of radius R , the velocity vectors v are displayed1v2v3and v4, relating to the various positions P1, P2, P3and P4of the point P in four instants of time t1, t2, t3and t4. Now imagine to transport all the speed vectors parallel to themselves until their origins coincide in a single point (see fig. 4.2 B); their opposite ends will therefore draw a circumference of radius equal to the modulus v of the speed (this circumference does not coincide with the original trajectory of the motion). The velocity vector of the arrow moves on this new circumference so as to make a full turn in a period T equal to that of the motion P . A whole turn of the new circumference thus equates to a space:
(being the modulus of v equal to the radius).
By applying the usual relationship between space and time in a uniform circular motion, we obtain the velocity of the velocity vector of P , i.e. the value of its acceleration, at :
By substituting its value v = (2R ) / T at speed , we obtain:
The acceleration vector (or velocity of the velocity) will then, again according to what has been said for uniform circular motions, direction tangent to the velocity trajectory curve, i.e. direction perpendicular to the radius of the circumference built with the velocity vectors and, therefore, to the vector speed v . But, returning to the original circumference (see fig. 4.2 A), v has a direction perpendicular to the radius R , and therefore the modulus vector a will have the direction of the perpendicular to the perpendicular to the radius, i.e. the direction of the radius itself.
The direction of a , as shown in figure 4.3, will point to the center of the circumference; for this reason, the acceleration thus constructed is called centripetal. | 922 | 4,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-45 | latest | en | 0.876457 |
https://blog.mbedded.ninja/electronics/circuit-design/what-are-transfer-functions-poles-and-zeroes/ | 1,722,740,168,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00602.warc.gz | 94,370,533 | 44,717 | What Are Transfer Functions, Poles, And Zeroes?
Overview
Transfer functions are a way of describing the frequency and phase response of a LTI (linear time invariant) system. The system can be anything with a measurable input and output, e.g. mechanical spring/mass/dampers, electronic RLC circuits, e.t.c. This page will put an emphasis on electrical transfer functions in the continuous-time domain.
The Laplace Domain
Transfer functions are usually written in the Laplace domain using the variable $s$, where $s$ is equal to:
\begin{align} s = \sigma + j\omega \end{align}
where:
$\sigma$ is the real part which determines the exponential increase/decrease
$j$ is the imaginary number ($j = \sqrt{-1}$), also seen as $i$ in maths ($j$ is used in electronics as to not get $i$ confused with current)
$\omega$ is the angular frequency, in units $rads^{-1}$. Remember that $\omega = 2{\pi}f$
Because we normally only care about what happens to steady-state sinusoidal signals (i.e. transients have all died away), we can simplify the equation by setting $\sigma = 0$, as $\sigma$ encodes the exponentially increasing/decaying components. This is essentially simplifying the general Laplace transform to the simpler Fourier transform, and thus $s$ becomes:
\begin{align} \label{eq:s-eq-j-w} s = j\omega \end{align}
Transfer functions describe the relationship between output and input (typically of voltage, but it doesn’t have to be). The below equation shows this relationship. We will use the symbol $H(s)$ is represent the transfer function (we are referring to a continuous-time system here, you might see $H(z)$ used for a discrete-time system).
\begin{align} H(s) = \frac{v_{out}(s)}{v_{in}(s)} \end{align}
The above equation is great, but is really just one of those “be definition” equations and doesn’t tell you anything about an electrical system (e.g. filter). The real usefulness comes in when you can write $\frac{v_{out}(s)}{v_{in}(s)}$ in terms of the circuit components, typically using KVL/KCL. This lets you write the transfer function with a polynomial on top and bottom:
\begin{align} H(s) = \frac{6s}{s^2 + 2s + 9} \end{align}
This is much more useful! It tells use everything we need to know about the electrical system (more on this below).
Magnitude and Phase
Note that because we are using the complex variable $s$, the transfer function $H(s)$ encodes both magnitude and phase relationships between the output and input. On it’s own, it’s complex in nature and not really useful for deducing anything “measurable”. But, using two clever tricks, you can separately extract the magnitude and phase response equations from $H(s)$. The below diagram shows the value of $H(s)$ at a single frequency $\omega$, and how the magnitude and phase information is encoded in this (more on this below).
Bear in mind that the above plot shows $H(s)$ at a single frequency. This point will move around as the frequency changes. The magnitude response is found by taking the magnitude of $H(s)$, which is $|H(s)|$ as shown in the below equation.
\begin{align} | H(s) | = \left| \frac{v_{out}(s)}{v_{in}(s)} \right| \end{align}
You might see the magnitude (gain) written as $G(\omega)$. This is equivalent to the magnitude of $H(s)$, i.e.:
\begin{align} | H(s) | = G(\omega) \end{align}
A $\omega$ is used with $G$ for the gain rather than $s$ because the process of taking the magnitude of $H(s)$ removes all imaginary components, leaving a function which is just dependent on $\omega$ and not $j\omega$.
To calculate the magnitude of the numerator and denominator, you generally:
1. Substitute in $j\omega$ for $s$ into the polynomials on the top and bottom of the transfer function.
2. Simplify any $j$‘s that are now risen to powers. For example, $j^2 = -1$, $j^3 = -j$, e.t.c.
3. Group all the real components together, and group all the imaginary components together, so it’s in the form $a + jb$.
4. Now that we’ve grouped the real and imaginary components, we can use the rule $| a + jb | = \sqrt{a^2 + b^2}$. This removes the imaginary component from the equation.
5. Simplify as needed.
The phase response $\angle H(s)$ is found by finding the angle of the complex number from the positive x-axis, as shown below:
\begin{align} \angle H(s) &= Arg{\left(\frac{\Im \{H(s)\}}{\Re \{H(s)\}}\right)} \end{align}
where:
$Arg$ is the argument (implemented with $atan2(y, x)$ in many software packages)
$\Im\{H(s)\}$ is the imaginary part of $H(s)$
$\Re\{H(s)\}$ is the real part of $H(s)$
Rather than finding the imaginary and real parts of the entire transfer function to calculate the phase, it can be easier to work out the real/imaginary parts for the numerator/denominator separately:
\begin{align} \angle H(s) = \angle (numerator) - \angle (denominator) \end{align}
$\angle H(s)$ is sometimes written as $\theta(\omega)$. When calculating the phase response it removes the imaginary components of the complex numbers and you’re left with an angle that is a function of just $\omega$.
Poles and Zeroes
A value of $s$ that causes the a transfer function to be 0 is called a zero, and a value of $s$ that causes the transfer function to be infinite is called a pole. Zeroes generally occur when a factor in the numerator is 0 (one notable exception is that a zero can also occur as $s \rightarrow \infty$, if the denominator is of higher order than the numerator), poles generally occur when a factor in the denominator is 0. Poles that have an imaginary component always come in pairs (conjugate pairs).
Intuitively, you can think of zeroes as places in where the system completely blocks a certain frequency (as the numerator goes to 0, so does the entire function). A poles is a frequency where the system has infinite response (at least mathematically, as the denominator goes to 0, the function goes to infinity). Poles in the right-half of the Argand diagram (which have a positive real component) cause the system to diverge towards infinity, and your system will be unstable.
The zeroes are the roots of the numerator polynomial, and the poles are the roots of the denominator polynomial. For this reason they are also referred to generally as roots.
The poles and zeros of a system can tell you much about how the system performs — it can tell you if the system is stable, and how fast it responds. In fact, the poles and zeroes completely characterize the filter, except for the overall gain constant $K$1.
For example, the transfer function below has a pole at the origin and a zero at infinity. This simple transfer function represents an integrator. A constant voltage applied to it will result in an output climbs without any limit. However, at high frequencies, the output is essentially zero as the positive and negative parts of the waveform are averaged out over time.
\begin{align} H(s) = \frac{1}{s} \end{align}
Pole Zero Plots (Argand Diagrams)
Poles and zeroes are plotted in a Argand diagram in what is called a pole-zero plot to give the reader an understanding on how the circuit responds.
• Zeroes contribute +90 of phase and increase the magnitude, above the zero frequency.
• Poles contribute -90 of phase and decrease the magnitude, above the pole frequency.
Poles are normally drawn as X’s on the graph, and zeroes as O’s. Unless you are building an oscillator, poles in the right-hand half of the plane (having a positive real component) are a bad thing, as they represent an instability.
Group Delay
The group delay $D(\omega)$ is defined as the negative of the slope of the phase vs. frequency plot. This can be written mathematically as:
\begin{align} D(\omega) \triangleq -\frac{d}{d\omega} \theta(\omega) \end{align}
where:
$\theta(\omega)$ is the phase response of the filter, also written as $\angle H(s)$
Intuitively, you can think of group delay as the time delay in seconds that a signal takes to pass through a filter as a function of frequency. Group delay has units of seconds. All casual filters (e.g. analogue filters) will have a non-zero, positive group delay. “Flatish” group delay plots in the passband are generally desirable as this means all frequencies will take the same time to pass through the filter, and thus the signal at the output will have minimal distortion (distortion is a result of different frequencies being delayed for different amounts of time).
Group delay for a number of 4-order filter tunings is shown below. The Bessel filter tuning aims to have maximally flat group delay across the pass-band of the signal (at the expense of other metrics we don’t talk about here). You can see this in the plot with the straight green line.
Group delay can be calculated in Python with scipy. scipy provides the scipy.signal.group_delay() function which takes as input the numerator and denominator coefficients of the transfer function and returns the calculated group delay.
The Transfer Function Of A Low-Pass RC Filter
A first-order low-pass RC filter has the transfer function:
\begin{align} H(s) &= \frac{1}{1 + \b{s}RC} \end{align}
Using $Eq.\ \ref{eq:s-eq-j-w}$, we can replace $s$ with $j\omega$ to get:
\begin{align} H(\omega) &= \frac{1}{1 + j\omega RC} \end{align}
This system has a pole at $\omega = -\frac{1}{jRC} = \frac{j}{RC}$ and a zero at $\omega = \infty$.
We can find the magnitude response of this low-pass RC filter by taking the magnitude of $H(f)$, remembering that the magnitude of a complex number is defined as:
\begin{align} |a + jb | = \sqrt{a^2 + b^2} \end{align} \begin{align} | H(\omega) | &= \left| \frac{1}{1 + j\omega RC} \right| \nonumber \\ &= \frac{1}{\sqrt{(1 + j\omega RC)(1 - j\omega RC)}} \nonumber \\ &= \frac{1}{\sqrt{1 - (j\omega RC)^2}} \nonumber \\ &= \frac{1}{\sqrt{1 - (-1)(\omega RC)^2}} \nonumber \\ &= \frac{1}{\sqrt{1 + (\omega RC)^2}} \\ \end{align}
The above equation shows the final result. Notice that by finding the magnitude, the imaginary components are gone! We can plot this on a graph.
We can find the phase response of the low-pass RC filter by using rule in $Eq.\ \ref{eq:xfer-fn-phase}$.
\begin{align} \angle H(j\omega) &= Arg\left(H(j\omega)\right) \nonumber \\ &= Arg\left(\frac{1}{1 + j\omega RC}\right) \nonumber \\ &= Arg(1) - Arg(1 + j\omega RC) \nonumber \\ &= 0 - arctan\left(\frac{j\omega RC}{1}\right) \nonumber \\ &= -arctan\left(j\omega RC\right) \\ \end{align}
Transfer Function Design Tools
Wolfram Alpha
Wolfram Alpha can take a transfer function and calculate many of it’s properties. It recognises the keywords transfer function in front of the numerator/denominator. For example, if you input in it’s search bar (which is a 2nd-order Bessel-tuned filter):
It will spit back at you things like the unit step response, state space representation, zeroes and poles, bode plot, Nyquist plot, Nichols plot, Root locus plot, gain margin and phase margin2. Click here to jump to these results in Wolfram Alpha.
OKAWA Transfer Function Analysis and Design Tool
The OKAWA Electric Design website has a Transfer Function Analysis and Design tool which can calculate many of the same things the Wolfram Alpha tool can3. | 2,796 | 11,131 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 95, "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": 16, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-33 | latest | en | 0.881661 |
https://www.nagwa.com/en/videos/685141270659/ | 1,586,190,362,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371637684.76/warc/CC-MAIN-20200406133533-20200406164033-00191.warc.gz | 1,043,753,249 | 5,392 | # Video: Identifying Triangles
Choose the triangle.
00:49
### Video Transcript
Choose the triangle.
We’re shown five different 2D shapes. We have to choose the one which is a triangle. Triangles have three straight sides and three corners. This is a circle, doesn’t have any corners or any straight sides. This shape has four straight sides, so it can’t be a triangle. These two shapes also have four straight sides. So we know they’re not triangles either. This shape has three straight sides and three corners. It’s a triangle. We chose the triangle. | 124 | 557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-16 | latest | en | 0.886751 |
https://brainly.com/question/104205 | 1,484,572,891,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279176.20/warc/CC-MAIN-20170116095119-00024-ip-10-171-10-70.ec2.internal.warc.gz | 791,216,703 | 8,258 | # In what order would you preform the operations to correctly evaluate the expression 2+(3-4)×9? What is the result
2
by nesss
## Answers
2014-08-29T00:19:32-04:00
Parenthesis first then multiply by 9 and add 2
so you would do 3-4 first which is -1 then multiply that by 9 which is -9 and then add 2
• Brainly User
2014-08-29T00:28:21-04:00
Use PEMDAS:
P: parentheses
E: exponents
M/D: multiplication/division
A/S: addition/subtraction
2 + (3 - 4) × 9
2 + (-1) × 9
2 + (-1 × 9)
2 + -9
2 - 9
-7 | 190 | 503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.844307 |
http://www.chegg.com/homework-help/questions-and-answers/mass-160-kg-stretches-verticalspring-0317-m-spring-isstretched-additional-0140-m-andreleas-q452946 | 1,472,061,480,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982292607.17/warc/CC-MAIN-20160823195812-00095-ip-10-153-172-175.ec2.internal.warc.gz | 368,951,676 | 13,771 | A mass of 1.60 kg stretches a verticalspring 0.317 m. If the spring isstretched an additional 0.140 m andreleased, how long does it take to reach the (new) equilibriumposition again?
1
s | 53 | 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} | 2.546875 | 3 | CC-MAIN-2016-36 | latest | en | 0.730257 |
http://www.edx.org/course/reliability-in-engineering-design | 1,591,474,039,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348519531.94/warc/CC-MAIN-20200606190934-20200606220934-00513.warc.gz | 145,943,727 | 54,961 | • Length:
17 Weeks
• Effort:
7–9 hours per week
• Price:
FREE
Add a Verified Certificate for \$2,250 USD
• Institution
• Subject:
• Level:
Intermediate
• Language:
English
• Video Transcript:
English
• Course Type:
Instructor-led on a course schedule
## Prerequisites
Undergraduate mechanics of materials course. A learner should understand (or be willing to learn) the concepts of shear force, bending moment, and stress.
The course is aimed at providing an engineering view (as opposed to a purely statistical view or a management view) of reliability analysis as well as reliable product design. The goal is to make the student familiar with both the statistical tools as well as the failure physics that enable one to model time to failure of products and to use such models during design phase to ensure reliable product designs.
### What you'll learn
Skip What you'll learn
What You Will Learn:
• Probability rules and conditional probabilities
• Expectation and variance of continuous functions and their manipulation
• Failure rate modeling
• Normal, lognormal, exponential, Weibull, binomial and Poisson distributions
• Reliability, mean time to failure and availability
• Data fitting and reliability estimation
• Multimodal distributions and mixed multiple failure mechanisms
• Reliability block diagrams
• Monte Carlo simulation
• Load-strength interference and probabilistic design
• First-order reliability methods
• Accelerated tests and acceleration factors
• Time to failure modeling for selected failure mechanisms in mechanical and electronic systems
### Syllabus
Skip Syllabus
Week 1:
• Introduction and Overview
• Rules of Probability
Week 2:
• Probability Examples
• Conditional Probability
Week 3:
• Expectations and Variance Definition
• Expectation and Variance of Continuous Functions
• Normal Distribution PDF and CDF
Week 4:
Week 5:
• Material Degradation and Time to Failure Modeling
• Practice Problems for Test 1
Week 6:
• Lognormal Distribution, Reliability, Hazard Rate and MTTF
• Test 1
Week 7:
• Exponential Distribution and Examples of MTTF Estimation
• Weibull Distribution
Week 8:
• Multimodal Distributions and Mixed Multiple Failure Mechanisms
• Goodness of Fit
Week 9:
• Binomial and Poisson Distributions
• Practice Problems for Test 2
Week 10:
• Reliability Block Diagrams
• Test 2
Week 11:
• Monte Carlo Simulation
• Uncertainty in Geometry, Load and Strength
• Covariance and Correlation
Week 12:
• Covariance and Correlation Examples
• First Order Reliability Methods Introduction
Week 13:
• First Order Reliability Methods Examples
• Reliability Review During Design
Week 14:
• Accelerated Testing and Acceleration Factors
• Practice Problems for Test 3
Week 15:
• Time to Failure Models for Mechanical Systems
• Test 3
Ganesh Subbarayan
Professor, School of Mechanical Engineering
Purdue University
### Pursue a Verified Certificate to highlight the knowledge and skills you gain\$2250.00
• #### Official and Verified
Receive an instructor-signed certificate with the institution's logo to verify your achievement and increase your job prospects | 682 | 3,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-24 | latest | en | 0.872751 |
https://math.stackexchange.com/questions/3249395/are-there-useful-algebra-of-sets-laws-regarding-cartesian-products-how-to-manip | 1,580,332,937,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251802249.87/warc/CC-MAIN-20200129194333-20200129223333-00013.warc.gz | 541,717,114 | 32,337 | # Are there useful algebra of sets laws regarding cartesian products? How to manipulate cartesian products algebraically?
The following post : Prove: $(A \times C) \setminus (B \times C) = (A \setminus B) \times C$ made me think of the question I am now asking.
Are there frequently used / well known laws for cartesian products in the context of set algebra.
In case such laws exist, can they be proved without analysing the statements in terms of membeship relation ( I mean without using set theory proper)?
Is it possible to " manipulate" cartesian products algebraically and mechanically in the same way one "manipulates" more ordinary sets using DeMorgan's Law, Idempotency Law or Domination law ( for sets) etc. ?
• Sure there are, but you need to prove that those formulas are also true, which leads to the same type of problem that you're trying to solve. – Michael Burr Jun 3 '19 at 11:05
There indeed do exist many well-known laws for Cartesian products, such as the following:
$$(A \cap B) \times (C \cap D) = (A \times C) \cap (B \times D),$$
$$A \times (B \cap C) = (A \times B) \cap (A \times C)$$ (distributivity of intersection),
$$A \times (B \cup C) = (A \times B) \cup (A \times C)$$ (distributivity of union),
$$A \times (B \backslash C) = (A \times B) \backslash (A \times C)$$ (distributivity of set difference),
and various other laws. For more, see here: https://en.wikipedia.org/wiki/Cartesian_product#Most_common_implementation_(set_theory)
Perhaps the following facts are useful:
$$A=B$$ iff $$I_A=I_B$$ where $$I_E(x)=1$$ or $$0$$ according as $$x\in E$$ or not.
$$I_{A\cap B}=I_AI_B$$
$$I_{E\setminus F} =I_E -I_F$$ if $$F \subset E$$.
$$I_{A\times B} (x,y)=I_A(x)I_B(y)$$.
The kind of laws @auscrypt presents would generally count as relational algebra.
While one could use such rules as a vehicle for presenting mathematical arguments (in the same way that elementary set algebra is used in much ordinary mathematics), this is not very common. Except for very simple arguments, trying to phrase things as algebraic manipulations generally seem to make things more complex than ad hoc element-for-element reasoning, and then there's little point to try.
Relational algebra is, however, widely used for implementation and optimization of database operations. In that settings the manipulations need to be carried out by a computer that is incapable of coming up with ad-hoc reasoning, so it is necessary to have a formal notation that can be manipulated symbolically. | 645 | 2,516 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2020-05 | latest | en | 0.88238 |
https://www.physicsforums.com/threads/theoretical-mechanics.86804/ | 1,670,086,367,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710933.89/warc/CC-MAIN-20221203143925-20221203173925-00789.warc.gz | 997,746,121 | 13,949 | # Theoretical Mechanics
akinoshigure
So I'm stuck on the second part of this problem and really don't know hwere to go from here... let me type it up and show where I got stuck at.
3. The motion of particle of chage q in an electromagnetic field is governed by the Lorentz force (for low velocities v<<c): F=qE + qvxB.
With both constant B=B k and E=Ey j + Ez K show that:
z(t)=z(sub-o)+v(sub-zo)t+qE(sub-z)t^2/2m
vx(t)=Asin(omega-t)+E(sub-y)/B
vx(t)=+Acos(omega-t)
I did F=qE+qVxB=m (dv/dt)
dvx/dt= q/m(vyBz)
dvy/dt= q/m(Ey-vxBo)
dvz/dt= q/m (Ez)
I think I'm suppose to now take a second derivative and find the second order differential equation but I'm not too sure how to approach that.
Homework Helper
akinoshigure said:
So I'm stuck on the second part of this problem and really don't know hwere to go from here... let me type it up and show where I got stuck at.
3. The motion of particle of chage q in an electromagnetic field is governed by the Lorentz force (for low velocities v<<c): F=qE + qvxB.
With both constant B=B k and E=Ey j + Ez K show that:
z(t)=z(sub-o)+v(sub-zo)t+qE(sub-z)t^2/2m
vx(t)=Asin(omega-t)+E(sub-y)/B
vx(t)=+Acos(omega-t)
I did F=qE+qVxB=m (dv/dt)
dvx/dt= q/m(vyBz)
dvy/dt= q/m(Ey-vxBo)
dvz/dt= q/m (Ez)
One slight error here: the strength of the magnetic field is just "B", not
"Bz" or "Bo".
I think I'm suppose to now take a second derivative and find the second order differential equation but I'm not too sure how to approach that.
What you are "supposed" to do is solve those equations. Since the last one (for dvz/dt) does not involve the other two components, you can solve it directly. The other two are "entwined". One method of solving a pair of equations is to differentiate the first (so that you have dvy/dt on the right side) and then replace dvy/dt from the second equation. That will give you one second order differential equation for vx/ | 583 | 1,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-49 | latest | en | 0.917066 |
http://lara.epfl.ch/w/using_automata_to_decide_presburger_arithmetic?rev=1429629985&do=diff | 1,558,692,855,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257601.8/warc/CC-MAIN-20190524084432-20190524110432-00530.warc.gz | 111,049,623 | 6,317 | • English only
# Differences
This shows you the differences between two versions of the page.
using_automata_to_decide_presburger_arithmetic [2009/04/29 10:53]
vkuncak
using_automata_to_decide_presburger_arithmetic [2015/04/21 17:26] (current)
Line 28: Line 28:
v: 1 0 0 0 0 v: 1 0 0 0 0
This string represents the satisfying assignment $\{(x,4),\ (y,3),\ (z,4),\ (v,1)\}$ for the above formula. If we take the valid formula This string represents the satisfying assignment $\{(x,4),\ (y,3),\ (z,4),\ (v,1)\}$ for the above formula. If we take the valid formula
-$+\begin{equation*} \lnot (x = y + v \land z = y + v\ \land\ v=1)\ \lor\ (x = z) \lnot (x = y + v \land z = y + v\ \land\ v=1)\ \lor\ (x = z) -$+\end{equation*}
the corresponding automaton accepts all strings. **(End of example.)** the corresponding automaton accepts all strings. **(End of example.)**
Line 83: Line 83:
Therefore, to check if $F$ is satisfiable, we construct $A(F)$ automaton and check whether the graph of $A(F)$ has a reachable accepting state. Therefore, to check if $F$ is satisfiable, we construct $A(F)$ automaton and check whether the graph of $A(F)$ has a reachable accepting state.
+**Example:** Step-by-step construction of automaton for
+\begin{equation*}
+ \lnot (x = y + v \land z = y + v\ \land\ v=1)\ \lor\ (x = z)
+\end{equation*} | 453 | 1,341 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-22 | latest | en | 0.54998 |
http://www.bankersadda.com/2017/12/quantitative-aptitude-for-rbi-assistant_12.html | 1,527,053,373,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865450.42/warc/CC-MAIN-20180523043959-20180523063959-00369.warc.gz | 349,201,700 | 47,679 | Dear Students,
Quantitative Aptitude Questions for RBI Assistant Mains 2017
Quantitative Aptitude is a very important section you must prepare if you are aiming for a job in Bank or Insurance sector. These two weeks are very important as IBPS Clerk and RBI Assistant Mains are lined up. So, these 15 questions can help you practice three very important topics of Quant Section.
Directions (Q. 1-5): Study the following line graph and answer the questions based on it.
Q1. What is the difference between the number of 10 rupee coins issued by RBI in 2016 and 2017 ?
(a) 50000
(b) 39000
(c) 33000
(d) 28000
(e) 27000
Q2. The total no. of coins issued by RBI in 2014 is approximately what percent of total no. of coins issued by RBI in 2013 ?
(a) 105 %
(b) 104 %
(c) 118 %
(d) 110 %
(e) 115 %
Q3. What is the average numbers of 5 rupee coins issued by RBI over the given period ? (rounded off to nearest integer )
(a) 123667
(b) 118667
(c) 128576
(d) 131223
(e) 120500
Q4. In which of the following years, the increment in 5 rupee coins is maximum than that in previous year ?
(a) 2013
(b) 2014
(c) 2015
(d) 2016
(e) 2017
Q5. The number of 10 rupee coins issued by RBI in 2015 is approximately what per cent of the number of 5 rupee coins issued by RBI in the same year ?
(a) 73%
(b) 61%
(c) 52%
(d) 65%
(e) 64%
Q6. A bus started its journey from Jodhpur and reached Jaipur in 75 minutes at its average speed of 66 kmph. If the average speed of the bus is increased by 2 kmph, how much time will it take to cover the same distance?
(a)64 minutes
(b)65 minutes
(c)73 minutes
(d)60 minutes
(e)79 minutes
Q7. In a test, a candidate secured 456 marks out of maximum marks ‘x’. If the maximum marks ‘x’ had been converted into 600 marks, he would have secured 342 marks. What was the maximum marks of the test?
(a)500
(b)650
(c)600
(d)800
(e)700
Q8. The manufacturer of an article makes a profit of 6%, the wholesale dealer makes a profit of 15%, and the retailer makes a profit of 20%. Find the manufacturing price of the article if the retailer sold it for Rs. 10,971.
(a) Rs. 8000
(b) Rs. 7500
(c) Rs. 7000
(d) Rs. 7950
(e) None of these
Q9. In every 45 minutes a watch gains 6 minutes. After setting the correct time at 6 a.m., what time will the watch show after 8 hours ?
(a)3 : 04 pm
(b)5:00 pm
(c)4 : 56 pm
(d)2 : 45 pm
(e)none of these
Q10. Out of the total number of students in a college 10% are interested in sports. 4/5 of the total number of students are interested in dancing. 6% of the total number of students are interested in singing and the remaining 25 students are not interested in any of the activities. What is the total number of students in the college?
(a)615
(b)620
(c)625
(d)630
(e)635
Directions (11-15): Find which term is wrong in the following number series and that does-not follow the sequence followed by other terms of that series?
Q11. 2, 11, 22, 37, 57, 87
(a) 87
(b) 11
(c) 57
(d) 37
(e) 22
Q12. 1, 2.4, 5.76, 14.824, 33.1776
(a) 2.4
(b) 14.824
(c) 33.1776
(d) 1
(e) 5.76
Q13. 4, 11, 20, 33, 52, 77
(a) 77
(b) 11
(c) 20
(d) 33
(e) 52
Q14. 8, 4, 10, 5, 13.5, 6.25
(a) 4
(b) 13.5
(c) 10
(d) 5
(e) 6.25
Q15. 256, 400, 576, 788, 1024
(a) 400
(b) 576
(c) 788
(d) 1024
(e) 1156 | 1,109 | 3,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-22 | latest | en | 0.925891 |
https://physics.stackexchange.com/questions/227054/applying-kirchoff-voltage-law-to-a-short-circuit | 1,720,818,248,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514452.62/warc/CC-MAIN-20240712192937-20240712222937-00871.warc.gz | 374,902,622 | 39,248 | Applying Kirchoff voltage law to a short circuit
If you consider an ideal wire with no resistance that shorts an ideal battery, the only voltage drop that exists is the emf of the battery, with nothing to balance it.
Obviously in the real world such a scenario is impossible, for the wire will have some resistance, but in this ideal example, is KVL not violated?
• No, it's not violated, you just have to use it in a limit with $R_{wire}\to 0$. This is generally true for all physical phenomena where something becomes infinite. While the limit may not exist, we can still extract valuable information from the asymptotic behavior. Commented Jan 2, 2016 at 0:54
• Voting to close as a duplicate but, in solidarity with the majority of newcomers, I wont make any effort to look up the duplicate. Commented Jan 2, 2016 at 1:37 | 195 | 828 | {"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} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.926633 |
https://www.gamedev.net/forums/topic/625301-problem-in-c-with-negamax-alpha-beta-connect-4/ | 1,500,922,840,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424909.50/warc/CC-MAIN-20170724182233-20170724202233-00054.warc.gz | 793,006,403 | 31,697 | Followers 0
# Problem in C with NegaMax / Alpha-Beta (Connect 4)
## 59 posts in this topic
Hi everyone,
I'm facing a problem with negamax. Think it comes from my eval function but i'm not sure at all.
First, can you tell me if my negamax and his call are ok ?
Because i don't find any solution, even if i read a lot about it these days.
Before showing you the code, i have to tell you that my verification function is based on (x,y) position (correspond to the last coin inserted with X = column and Y = line). Is this a portable method with negamax/alpha-beta or do i have to check the entire grid ?
Here is my negamax function:
[CODE]
int IA_NegaMaxAlphaBeta(int profondeur, int alpha, int beta, int grille[LIGNES][COLONNES], int emplacements[COLONNES], int colonne, int emplacement, int joueur)
{
int x = COLONNES/2, y, i, k = 1;
if (profondeur == 0)
{
return IA_Evaluation(grille, emplacements, colonne, emplacement, joueur);
}
if (check(grille, emplacements, colonne, emplacement, joueur) == 4)
{
if (joueur == 0)
{
return -1000;
}
return 1000;
}
if (GrillePleine(emplacements))
{
return 0;
}
for(i = 0; i < COLONNES && alpha < beta; i++)
{
// EFFICACITE: Balayer les colonnes du milieu vers l'extérieur (alternativement gauche/droite)
x += i*k;
k = -k;
if (emplacements[x] > -1)
{
y = InsererJeton(grille, emplacements, x, joueur);
alpha = max(alpha, -IA_NegaMaxAlphaBeta(profondeur-1, -beta, -alpha, grille, emplacements, x, y, joueur^1));
AnnulerJeton(grille, emplacements, y, x);
}
}
return alpha;
}
[/CODE]
Is there something wrong in it ?
And here comes my negamax call:
[CODE]
int x = COLONNES/2, y, i, k = 1, scoreCoup, scoreMeilleur = -1000000, col = -1;
for(i = 0; i < COLONNES; i++)
{
// EFFICACITE: Balayer les colonnes du milieu vers l'extérieur (alternativement gauche/droite)
x += i*k;
k = -k;
if (emplacements[x] > -1)
{
y = InsererJeton(grille, emplacements, x, joueur);
scoreCoup = IA_NegaMaxAlphaBeta(4, -100000, 100000, grille, emplacements, colonne, emplacement, joueur);
if (scoreCoup > scoreMeilleur)
{
scoreMeilleur = scoreCoup;
col = x;
}
AnnulerJeton(grille, emplacements, y, x);
}
}
[/CODE]
If you want to see my eval function, just ask. I don't show it right now because there are maybe errors in my negamax.
Oh and by the way... Sorry for my english if i made mistakes ! I'm from Belgium [img]http://public.gamedev.net//public/style_emoticons/default/laugh.png[/img]
0
##### Share on other sites
[code] x += i*k;
k = -k;[/code]
That's cute, but probably incorrect. You may want to add (i+1)*k to x instead.
Or even better, you may want to wait until you have a working search before you add any embelishments of this type.
EDIT: Nevermind, I now think that is correct. I still would take it out until you know your search is working. Edited by alvaro
0
##### Share on other sites
I totally agree with your point of view, but my grid (columns) is [0,6] so grid[7] is overflow.
See with my code:
[CODE]
- explore column 3
- explore column 2
- explore column 4
- explore column 1
- explore column 5
- explore column 0
- explore column 6
[/CODE]
But if i add 1 to i like you said before, it's going to be wrong:
[CODE]
- explore column 4
- explore column 2
- explore column 5
- explore column 1
- explore column 6
- explore column 0
- explore column 7
[/CODE]
Any other error except that one ? [img]http://public.gamedev.net//public/style_emoticons/default/sad.png[/img]
0
##### Share on other sites
I can't find any flaws in your code. Perhaps you should try debugging a depth-1 search first and see how that goes.
0
##### Share on other sites
Don't you think that the problem may come from my grid's verification method ?
Because it doesn't scan the entire grid, but just all the directions from the last coin inserted.
I can see there is a big problem with all of it but can't find it [img]http://public.gamedev.net//public/style_emoticons/default/mellow.png[/img]
About my eval function, i know it's not good at all so it's maybe the problem. Anyway, do you want me to show that function ?
0
##### Share on other sites
Almost regardless of what the evaluation function returns, your program should be able to find forced victories and avoid tactical losses. Looking only at the last toekn inserted should work just fine to check if the game was won by the player that just played.
But feel free to post the code and I'll take a look.
0
##### Share on other sites
Also, could you describe what the actual problem is? Perhaps you can show us a position that you think shows the program is not working well.
0
##### Share on other sites
If you don't see any problem in my negamax, it probably comes from the eval function.
Have a look (debug function):
[CODE]
Test column 3
Test column 2
Test column 4
Test column 1
FREE column 1 / SCORE: 1000
BEST column 1 (score 1000)
Test column 5
FREE column 5 / SCORE: 900
Test column 0
FREE column 0 / SCORE: 1000
Test column 6
FREE column 6 / SCORE: 1000
[/CODE]
Seems to be OK except for scores so i guess it comes from eval.
In this example, it's not the case here, but almost every time i've the same score for all the columns... So it cannot work.
Here is my eval crap:
[CODE]
int IA_Evaluation(int grille[LIGNES][COLONNES], int emplacements[COLONNES], int colonne, int emplacement, int joueur)
{
int score = 0, flag = check(grille, emplacements, colonne, emplacement, joueur);
if (flag == 4)
{
if (joueur == 0)
{
return -1000;
}
return 1000;
}
if (flag == 3)
{
if (joueur == 0)
{
return -900;
}
return 900;
}
if (flag == 2)
{
if (joueur == 0)
{
return -600;
}
return 600;
}
if (flag == 1)
{
if (joueur == 0)
{
return -300;
}
return 300;
}
return 0;
}
[/CODE]
0
##### Share on other sites
While it is perfectly OK to use your `check' function to detect the victory condition of the game, it is not OK to use if for evaluation purposes. But I would just start with an evaluation function that returns 0 every time and see if you program can find quick tactical wins in easy cases.
And I would still like to see a position together with a description of what the program does and what you think it should do. It's hard to diagnose a disease without knowing any of the symptoms.
0
##### Share on other sites
You're right. When i posted, i thought it was a problem with my negamax so i didn't continue with deep tests.
I'm going to have a closer look to it and test it just like you said.
I'll let you know as soon as possible (like tomorrow if i can).
Thank you
0
##### Share on other sites
Back sooner as i expected.
My IA_Evaluation function returns 0 like you asked.
But it's still the same problem.
I've more details: all the scores are equal to 0 and when there is a winning/losing move it's equal to 1000 (but the AI doesn't act as expected, she inserts coins in columns order in the loop). She fills column by column [img]http://public.gamedev.net//public/style_emoticons/default/sad.png[/img]
Debug:
[CODE]
Test column 3
FREE column 3 / SCORE: 1000
BEST column 3 (score 1000)
Test column 2
FREE column 2 / SCORE: 1000
Test column 4
FREE column 4 / SCORE: 1000
Test column 1
FREE column 1 / SCORE: 1000
Test column 5
FREE column 5 / SCORE: 1000
Test column 0
FREE column 0 / SCORE: 1000
Test column 6
FREE column 6 / SCORE: 1000
[/CODE]
when i'm at this point:
[img]http://img811.imageshack.us/img811/4597/24018014.png[/img]
And before that, all scores to 0.
0
##### Share on other sites
From that position, every move is a winning move. You didn't encourage your program to win in as few moves as possible. If you replace 1000 with 1000-number_of_moves_played it should do the right thing.
EDIT: Actually, that's wrong. Sorry. I don't have time to look at it now, but I will later. Edited by alvaro
0
##### Share on other sites
Ok no problem, thank you.
By the way, with my grid's verification method, i wouldn't be able to generate a score for coins alignment because it doesn't scan the entire grid, would i ?
But if it doesn't matter and if other parameters can be used (like the number of coins red/yellow), it's going to be ok.
I'd be grateful for a clarification about it.
0
##### Share on other sites
I am back to thinking that there might not be anything wrong with your search. You make a depth-4 search after moving in the middle column and you see that you can win. All the other moves return scores of 1000 as well, but the way alpha-beta works, that only means that their scores are not higher than 1000. Then it plays in the center, which is a winning move.
Do you have another position where you think it's doing the wrong thing? Edited by alvaro
0
##### Share on other sites
For now, i don't [img]http://public.gamedev.net//public/style_emoticons/default/wacko.png[/img]
Because it plays column by column (problem with the evaluation)
0
##### Share on other sites
For the evaluation function, you are going to have to identify where threats exist on the board. By "threat" I mean an empty place on the board that is aligned with three pieces of one player, so it the player were to play there he would win. A strong evaluation function needs to know the difference between threats that happen on even and odd rows (with the top row being even), and how multiple threats combine (e.g., player 1 wins with an odd threat, player 2 wins with an even threat, but an odd threat by player 1 beats any number of even threats by player 2). Read [url="http://homepages.cwi.nl/~tromp/c4.html"]this[/url] or buy [url="http://www.amazon.com/Complete-Book-CONNECT-History-Strategy/dp/1402756216"]the book[/url]. You can also give small bonuses for things like controlling squares in the central column.
You also need to be able to search much much further than depth 4. Here are some things that can help:[list]
[*]You are currently using some simple heuristics to determine what order to use to explore moves in the search (3,4,2,5,1,0,6), but one can do much better than that by using statistics on what moves have produced beta cuts before. You can see how "history heuristic" is used in chess and adapt it for connect 4.
[*]You then need transposition tables, because in this game it's very easy to reach the same position through different paths, and you want to reuse what you learned when you saw this position before. You can also use the hash tables to store the most promising move, to be searched first when you encounter this position again.
[*]Instead of doing a fixed-depth search, you should use iterative deepening, where you search depth 1, depth 2, depth 3... until you run out of time. The early searches are not a waste of time because they fill out the hash tables and make the later searches faster. Iterative deepening also allows for proper time control.
[/list]
That should keep you busy for a while.
0
##### Share on other sites
[quote name='alvaro' timestamp='1337866828' post='4942889']
A strong evaluation function needs to know the difference between threats that happen on even and odd rows (with the top row being even), and how multiple threats combine (e.g., player 1 wins with an odd threat, player 2 wins with an even threat, but an odd threat by player 1 beats any number of even threats by player 2).
[/quote]
Sorry but i don't get it. Do you have an example for that "even/odd" thing ?
0
##### Share on other sites
Should it be something like this ?
Knowing that "nbAlign" contains the max number of coins aligned for "joueur" in "colonne".
[CODE]
int IA_Evaluation(int colonne, int joueur, int nbAlign)
{
int score = 0;
if (nbAlign == 3)
{
score = SCORE_3J;
}
else
{
if (nbAlign == 2)
{
score = SCORE_2J;
}
else
{
if (nbAlign == 1)
{
score = SCORE_1J;
}
}
}
if (colonne == 2 || colonne == 3 || colonne == 4)
{
score += SCORE_COL_MIL;
}
else
{
score += SCORE_COL_AUT;
}
if (joueur == 0)
{
return -score;
}
return score;
}
[/CODE]
Because it's the same problem, almost everytime the same score for all columns.
So it fills column by column like before: 3-2-4-1-5-0-6 [img]http://public.gamedev.net//public/style_emoticons/default/sad.png[/img] Edited by Giustino
0
##### Share on other sites
Now, the whole thing works (i think, right now).
Except this problem:
[CODE]
Test column 3
FREE column 3 / SCORE: 1000
BEST column 3 (score 1000)
Test column 2
FREE column 2 / SCORE: 1000
Test column 4
FREE column 4 / SCORE: 1000
Test column 1
FREE column 1 / SCORE: 1
Test column 5
FREE column 5 / SCORE: 1000
Test column 0
FREE column 0 / SCORE: 1000
Test column 6
FREE column 6 / SCORE: 1000
[/CODE]
And it should be the opposite [img]http://public.gamedev.net//public/style_emoticons/default/sad.png[/img]
Everytime i can beat the AI, the "bad" column has a low score and the others have 1000.
Any idea where it comes from ?
Finally, the code above is a good result but it should take the MIN and not the MAX in that case.
What's the problem ? Edited by Giustino
0
##### Share on other sites
I don't understand your description of the situation where you think there is a problem. You keep posting those blocks of output, but I don't really know what they mean, since the code you posted doesn't print anything.
If alpha-beta has found a move with a score of 1000, if the score of any other move is reported to be 1000 or less it just means that it is not more than 1000. So you should essentially ignore the exact value returned.
[quote name='Giustino' timestamp='1337884435' post='4942970']
Finally, the code above is a good result but it should take the MIN and not the MAX in that case.
What's the problem ?
[/quote]
Ah, I see an error now:
[code]scoreCoup = IA_NegaMaxAlphaBeta(4, -100000, 100000, grille, emplacements, colonne, emplacement, joueur);[/code]
should be
[code]scoreCoup = -IA_NegaMaxAlphaBeta(4, -100000, -scoreMeilleur, grille, emplacements, colonne, emplacement, joueur);[/code]
EDIT: Actually, that is still not right. Let's try again:
[code]scoreCoup = -IA_NegaMaxAlphaBeta(4, -100000, -scoreMeilleur, grille, emplacements, colonne, emplacement, joueur^1);[/code] Edited by alvaro
0
##### Share on other sites
Ok i'll have a look.
For my example, here is a cleaner explanation:
[CODE]
Test column 3
FREE column 3 / SCORE: 1000
BEST column 3 (score 1000)
Test column 2
FREE column 2 / SCORE: 1000
Test column 4
FREE column 4 / SCORE: 1000
Test column 1
FREE column 1 / SCORE: 1000
Test column 5
FREE column 5 / SCORE: 1000
Test column 0
FREE column 0 / SCORE: 3
Test column 6
FREE column 6 / SCORE: 1000
[/CODE]
produced by:
[img]http://img715.imageshack.us/img715/4127/18786785.png[/img]
Can you see the problem ? It always happened in those cases.
This is weird...
Initial call for negamax:
When i changed the code you've shown above, goes wrong.
But like this, it works (except the problem i've told before):
[CODE]
y = InsererJeton(grille, emplacements, x, joueur^1);
scoreCoup = IA_NegaMaxAlphaBeta(4, -1000000000, 1000000000, grille, emplacements, colonne, emplacement, joueur);
[/CODE]
Is it normal that i've to write "!joueur" and not simply "joueur" ? Edited by Giustino
0
##### Share on other sites
I edited my previous post. It looks like your initial call to IA_NegaMaxAlphaBeta was pretty messed up. It should be virtually identical to the recursive call made inside of IA_NegaMaxAlphaBeta, except instead of beta you can use the fixed value 1000000000. Edited by alvaro
0
##### Share on other sites
Think i tried that too, but i'm going to test it and i'll tell you.
Yep i already tested it. makes things worse (let's just say... not better).
Where can be the problem ? [img]http://public.gamedev.net//public/style_emoticons/default/wacko.png[/img] Edited by Giustino
0
##### Share on other sites
I would change the depth from 4 to 1 or even 0 and try to use a debugger to see what's going on.
0
##### Share on other sites
Ok, it works better since i've changed something in my eval function.
But there is still the same problem... When it can loose, it doesn't block the player (because the score of that column is the lower -> OK for the MIN but should be the biggest value to make things right).
Do you understand what i mean ? It's the same thing i told you above, just look at the image and the debug code to help you if it's not clear in your mind.
I really don't know what could be crap in all of that [img]http://public.gamedev.net//public/style_emoticons/default/sad.png[/img]
0
## Create an account
Register a new account | 4,523 | 16,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.371674 |
https://www.enotes.com/homework-help/rationalize-1-x-y-339094 | 1,516,306,759,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00136.warc.gz | 905,940,199 | 9,633 | # Rationalize: 1/³√x + ³√yshow step by step solution to explain the answer
thilina-g | Certified Educator
or if you asked to rationalize `1/(root(3)x+root(3)y)`
We know,
`(x+y) = ((root(3)x)^3+(root(3)y)^3)`
`((root(3)x)^3+(root(3)y)^3) = (root(3)x)+root(3)y)((root(3)x)^2-root(3)xroot(3)y+(root(3)y)^2)`
Therefore,
`1/(root(3)x+root(3)y) = ((root(3)x)^2-root(3)xroot(3)y+(root(3)y)^2)/(x+y)`
This is the rationalized form.
thilina-g | Certified Educator
`1/root(3)x + root(3)y`
`= (1+root(3)(x)xxroot(3)y)/root(3)x`
Multiplying both numerator and denominator by `(root(3)x)^2`
`=((root(3)x)^2(1+root(3)(x)xxroot(3)y))/((root(3)x)^2root(3)x)`
`= ((root(3)x)^2+xroot(3)y)/x`
Therefore,
`1/root(3)x+root(3)y = ((root(3)x)^2+xroot(3)y)/x` | 319 | 751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-05 | latest | en | 0.344577 |
https://riddles.guru/short-riddles/page/4/ | 1,716,934,426,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059160.88/warc/CC-MAIN-20240528220007-20240529010007-00399.warc.gz | 419,449,752 | 10,696 | One is to three as three is to five and five is to four and four is the magic number.
What is the pattern?
What can go up a chimney down but can’t go down a chimney up? | 43 | 170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.982899 |
https://www.badulakebilbao.com/casino-on-the-web-playing-program-good-development-method/ | 1,660,133,810,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571153.86/warc/CC-MAIN-20220810100712-20220810130712-00650.warc.gz | 581,800,672 | 8,805 | If you chat about the casino on the web betting method, you will uncover there are a lot of people who will discourage you. They will say that betting on the web genuinely is not a very good useful resource to make money. But I will say that it is very simple to generate from on the web on line casino game titles, if you know the casino online betting methods. In fact funds management expertise is what most of the gamblers deficiency. 해외배팅사이트 That’s why some are presently bankrupt while some are enjoying an affluent life. Did any person of you listen to about “Optimistic Development Program”, this is a single of the very effectively identified casino on the internet betting approach.
You can say this is a logic that tells you the possibilities of successful four instances in a solitary row. At the beginning or just at the preliminary phase the bet is of one device, the 2nd wager is of three models, the third bet is of two models and the fourth wager is of 6 models. Consequently it is also named the 1-three-two-6 program.
I will illustrate this casino online betting method in detail, to give you a clear understanding. For instance you place your very first guess of \$10. The 2nd bet is meant to be \$30 – when you win the initial guess, your \$10 will get extra up with the \$20 already positioned on the desk. The whole will come to \$thirty.
So the next guess you spot would be of \$thirty. The grand whole before you engage in the third guess will be of \$60 whole (the \$thirty wager positioned by you in the second wager mixed together with the next guess winning already positioned on the desk). From the \$sixty you take away \$forty and the third bet is of \$20.
Your third wager will be of \$twenty and right after successful the third wager you will earn \$40. Now, for the fourth bet you will incorporate \$twenty a lot more to the complete \$forty to make it a \$sixty guess for the forth bet you place.
Successful the fourth guess you will be still left with \$a hundred and twenty. This is the net revenue you make from this on line casino on-line betting program. To carry on the sport you will once again area a bet of \$ten and stick to the “Good Development System” once once more. Right after ending the forth guess, you start off more than again. Additionally, every time you loose a wager, commence yet again with preliminary \$ten bet. | 518 | 2,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.952561 |
http://gmatclub.com/forum/chicago-gsb-admit-weekend-matriculation-thread-43821.html?fl=similar | 1,485,000,832,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281069.89/warc/CC-MAIN-20170116095121-00020-ip-10-171-10-70.ec2.internal.warc.gz | 119,371,775 | 52,545 | Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 21 Jan 2017, 04:13
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
Author Message
TAGS:
### Hide Tags
GMAT Club Legend
Affiliations: HHonors Diamond, BGS Honor Society
Joined: 05 Apr 2006
Posts: 5926
Schools: Chicago (Booth) - Class of 2009
GMAT 1: 730 Q45 V45
Followers: 314
Kudos [?]: 2039 [0], given: 7
### Show Tags
28 Mar 2007, 07:21
Now that we actually have some admits to the GSB... I thought I'd see who was going to come to the Admit Weekend.
GMATCram
cappy
lxa
09app
little.nemo
yb
cmns18
Any of you decided on the GSB? cappy, I know you are debating Kellogg as well. Others?
Senior Manager
Joined: 05 Oct 2005
Posts: 485
Followers: 1
Kudos [?]: 4 [0], given: 0
### Show Tags
28 Mar 2007, 07:49
I'll be there! I'm not sure what the schedule is for the event - April 20? What time?
I'm planning on being there Fri-Sun.
GMAT Club Legend
Affiliations: HHonors Diamond, BGS Honor Society
Joined: 05 Apr 2006
Posts: 5926
Schools: Chicago (Booth) - Class of 2009
GMAT 1: 730 Q45 V45
Followers: 314
Kudos [?]: 2039 [0], given: 7
### Show Tags
28 Mar 2007, 07:58
yb wrote:
I'll be there! I'm not sure what the schedule is for the event - April 20? What time?
I'm planning on being there Fri-Sun.
Come in on Thursday before 7pm if you can - people share drinks the night before. You'll have access to the schedule today as soon as you get your admit portal login.
Intern
Joined: 19 Nov 2006
Posts: 44
Location: Toronto
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
28 Mar 2007, 08:17
Hey rhyme,
I've gone back and forth so many times over the past few weeks. I'm either bipolar or too emotional...
Now that I can get into the GSB admitted portal I hope to make a better comparison of what the experience will be like. The one thing that keeps jumping out at me is the instant finance credibility that I will have with GSB. AS someone with a Social Science undergrad and only sales experience I see great value in that, no matter what career I choose.
On the flipside, I think I might be a better fit at Kellogg as the atmosphere suits my background a bit more. Although, based on all of interactions with GSB and Kellogg I have been more impressed with the GSB people. I am slightly leaning towards GSB at this moment so I might see you at admit weekend.
The debate continuees...
GMAT Club Legend
Affiliations: HHonors Diamond, BGS Honor Society
Joined: 05 Apr 2006
Posts: 5926
Schools: Chicago (Booth) - Class of 2009
GMAT 1: 730 Q45 V45
Followers: 314
Kudos [?]: 2039 [0], given: 7
### Show Tags
28 Mar 2007, 08:41
cappy wrote:
Hey rhyme,
I've gone back and forth so many times over the past few weeks. I'm either bipolar or too emotional...
Now that I can get into the GSB admitted portal I hope to make a better comparison of what the experience will be like. The one thing that keeps jumping out at me is the instant finance credibility that I will have with GSB. AS someone with a Social Science undergrad and only sales experience I see great value in that, no matter what career I choose.
On the flipside, I think I might be a better fit at Kellogg as the atmosphere suits my background a bit more. Although, based on all of interactions with GSB and Kellogg I have been more impressed with the GSB people. I am slightly leaning towards GSB at this moment so I might see you at admit weekend.
The debate continuees...
Well, in all honesty, if you are going to make a well informed decision you should come to the admit weekend and experience things for yourself. Its funny you bring up the "impressed by" thing, because I felt the same way.
Manager
Joined: 08 Aug 2006
Posts: 82
Location: Michigan
Followers: 1
Kudos [?]: 1 [0], given: 0
### Show Tags
28 Mar 2007, 08:55
I think I'll be there. I'm going to Ross the weekend before so I will be able to compare. Right now still leaning Ross because of the advantage in general management/sustainablity....but GSB has rhyme.
Intern
Joined: 14 Sep 2005
Posts: 33
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
28 Mar 2007, 18:24
I wish I could go, but I can't.
Enjoy the event, guys!
Intern
Joined: 19 Nov 2006
Posts: 44
Location: Toronto
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
01 Apr 2007, 13:58
I will be there. Unfortunately, can't get in early enough for Thursday night.
Hey Rhyme - where in the admitted student website did you see all of that information about how many people accepted offers, etc?
Senior Manager
Joined: 05 Oct 2005
Posts: 485
Followers: 1
Kudos [?]: 4 [0], given: 0
### Show Tags
01 Apr 2007, 14:23
cappy wrote:
I will be there. Unfortunately, can't get in early enough for Thursday night.
Hey Rhyme - where in the admitted student website did you see all of that information about how many people accepted offers, etc?
Mine's working - there's a link in my acceptance email.
GMAT Club Legend
Affiliations: HHonors Diamond, BGS Honor Society
Joined: 05 Apr 2006
Posts: 5926
Schools: Chicago (Booth) - Class of 2009
GMAT 1: 730 Q45 V45
Followers: 314
Kudos [?]: 2039 [0], given: 7
### Show Tags
01 Apr 2007, 14:27
cappy wrote:
I will be there. Unfortunately, can't get in early enough for Thursday night.
Hey Rhyme - where in the admitted student website did you see all of that information about how many people accepted offers, etc?
See the community portal. You have to extrapolate from there, see the 09 class google group for a discussion.
Manager
Joined: 14 Mar 2007
Posts: 60
Followers: 1
Kudos [?]: 0 [0], given: 0
### Show Tags
03 Apr 2007, 05:34
I'll be there. It sounds Holiday Inn isn't honoring the \$99 rate that weekend, so I may have to use hotwire and hope to get lucky again. If anyone knows of a good deal, please post it.
03 Apr 2007, 05:34
Similar topics Replies Last post
Similar
Topics:
7 Admit Weekends? 22 10 Jan 2011, 09:23
****** Chicago Booth Admit Weekend March 5-6 rollcall ****** 2 01 Mar 2010, 17:54
38 Admitted to Chicago Booth Class of 2011 Thread 489 08 Jan 2009, 11:27
**** Chicago Booth Admit Weekend Get-Together **** 0 27 Feb 2009, 07:54
Chicago GSB 5 05 Mar 2007, 09:55
Display posts from previous: Sort by | 1,944 | 6,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.915067 |
http://petrajoly.com/forget/probability-and-its-applications-pdf.php | 1,611,161,464,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703521139.30/warc/CC-MAIN-20210120151257-20210120181257-00730.warc.gz | 77,967,998 | 8,148 | # Applications probability and pdf its
## Introduction To Probability And Its Applications Chegg
Log-Concave Probability and Its Applications. Probability Theory and Applications. Lecture-03-Conditional probability, Independence of events. Lecture-04-Random variables, cumulative density function,, Request PDF on ResearchGate Log-concave probability and its applications In many applications, assumptions about the log-concavity of a probability distribution.
### Probability Models For Dna Sequence Evolution Probability
Introduction to Probability and Its Applications Richard. Why is Chegg Study better than downloaded Introduction to Probability and Its Applications PDF solution manuals? It's easier to figure out tough problems faster using, Stochastic Processes and their Applications publishes papers on the theory and for Mathematical Statistics and Probability. Editor Pack’ PDF; View вЂGuide.
The Law of Large Numbers and its Applications by The probability distribution of a discrete The probability density function for a continuous random Lecture notes contains notes for the topics covered during the course. VIII.4 Probability Plotting : IX. Applications to PRA: An Historical Perspective (PDF
The Law of Large Numbers and its Applications by The probability distribution of a discrete The probability density function for a continuous random Lecture notes contains notes for the topics covered during the course. VIII.4 Probability Plotting : IX. Applications to PRA: An Historical Perspective (PDF
Why is Chegg Study better than downloaded Introduction to Probability and Its Applications PDF solution manuals? It's easier to figure out tough problems faster using Introduction to Probability and Its to Probability and Its Applications possible outcomes probability density function probability distribution
Introduction to Probability Models Tenth Edition Sheldon M. Ross University of Southern California 7 Renewal Theory and Its Applications 421 7.1 Introduction 421 Probability Theory and Applications. Lecture-03-Conditional probability, Independence of events. Lecture-04-Random variables, cumulative density function,
Read Online or Download Introduction to Probability and Its Applications, Third Edition PDF. Best introduction books DOWNLOAD INTRODUCTION TO PROBABILITY AND ITS APPLICATIONS SOLUTIONS introduction to probability and pdf Amazon.com: Introduction to Probability (9780821807491
I N T R O D U C T I O NT OP R O B A B I L I T YA N DI T SA P P L I C A T I O N S This page intentionally le... Statistics and Probability for Engineering Applications from the Probability Density Function as a textbook on statistics and probability for
Title: Probability Models For Dna Sequence Evolution Probability And Its Applications PDF Author: Stephenie Meyer Subject: probability models Keywords Linear Statistical Inference and its Applications. 2nd Edition. Wiley Series in Probability and Statistics
Linear Statistical Inference and its Applications. 2nd Edition. Wiley Series in Probability and Statistics Introduction to Probability and Its to Probability and Its Applications possible outcomes probability density function probability distribution
Unit 14: The Normal Distribution and its Applications What is the probability that monthly food expenditure are less than \$5,500 or greater than \$7,000 ? Download as PDF, TXT or read online Flag for inappropriate content. Student Solutions Manual for Introduction to Probability with Statistical Applications.
Introduction to Probability and Its Applications - Kindle edition by Richard L. Scheaffer, Linda Young. Download it once and read it on your Kindle device, PC, phones Read Online or Download Introduction to Probability and Its Applications, Third Edition PDF. Best introduction books
### Free Probability and its Applications Mathematical
Basics Of Applied Stochastic Processes Probability And Its. Introduction to Probability and Its Applications - Kindle edition by Richard L. Scheaffer, Linda Young. Download it once and read it on your Kindle device, PC, phones, Scarlett Johnson wa-cop.org An Introduction To Probability Theory And Its Applications 2nd Ed An Introduction To Probability Theory And Its Applications 2nd Ed.
Log-concave probability and its applications Request PDF. Title: Free Continuous Time Markov Jump Linear Systems Probability And Its Applications (PDF, ePub, Mobi) Author: Policy Press Subject: Continuous Time Markov Jump, Hand-book on STATISTICAL DISTRIBUTIONS for 2 Probability Density Functions 3 6 Probability content from в€’zto zof Gauss distribution in.
### matrix entries are entered as comma-separated numeric
New PDF release Introduction to Probability and Its. 8581000031741 william feller books amazoncom an introduction to probability theory and its applications vol 2 2nd edition skills-from-classic-films.pdf. https://en.m.wikipedia.org/wiki/Theory_of_Probability_%26_Its_Applications Assignments section contains the exercises Probability And Its Applications To Reliability, Quality Control, And Risk Assessment.
DOWNLOAD PROBABILITY AND ITS APPLICATIONS probability and its applications pdf From the reviews of the first editions: "...Kallenberg's present book would have to DOWNLOAD INTRODUCTION TO PROBABILITY AND ITS APPLICATIONS SOLUTIONS introduction to probability and pdf Amazon.com: Introduction to Probability (9780821807491
Introduction To Probability And Its Applications Scheaffer Solutions Pdf Alan agresti personal home page department of statistics, i was employed by the 2012-06-24В В· INTRODUCTION PROBABILITY & ITS APPLICATIONS BY W. FELLER An Introduction To Probability : Theory And Its Applications Free E-book PDF Download.
The Law of Large Numbers and its Applications by The probability distribution of a discrete The probability density function for a continuous random Read Online or Download Probability Theory and Its Applications in China PDF. Similar probability books
Theory of Probability and Its Applications is a quarterly peer-reviewed scientific journal published by the Society for Industrial and Applied Mathematics. Read Online or Download Introduction to Probability and Its Applications PDF. Best probability & statistics books
Introduction To Probability And Its Applications Scheaffer Solutions Pdf Alan agresti personal home page department of statistics, i was employed by the Why is Chegg Study better than downloaded Introduction to Probability and Its Applications PDF solution manuals? It's easier to figure out tough problems faster using
Title: Probability Models For Dna Sequence Evolution Probability And Its Applications PDF Author: Stephenie Meyer Subject: probability models Keywords Assignments section contains the exercises Probability And Its Applications To Reliability, Quality Control, And Risk Assessment
Title: Free Self Normalized Processes Limit Theory And Statistical Applications Probability And Its Applications (PDF, ePub, Mobi) Author: Allen & Unwin Title: Free Self Normalized Processes Limit Theory And Statistical Applications Probability And Its Applications (PDF, ePub, Mobi) Author: Allen & Unwin
Log-Concave Probability and Its Applications Mark Bagnoli, Ted Bergstrom 1. Introduction. Many interesting propositions in the economics of information are built on From "Fuzzy Logic and Probability Applications: Bridging the Gap" Timothy J. Ross, Jane M. Booker, evolution of any theory and its applications.
THIRD EDITION Introduction to Probability and Its Applications RICHARD L. SCHEAFFER University of Florida LINDA J. YOUNG University of Florida Australia • Brazil Probability Theory and Applications. Lecture-03-Conditional probability, Independence of events. Lecture-04-Random variables, cumulative density function,
Student Solutions Manual for Introduction to Probability. linear statistical inference and its applications. 2nd edition. wiley series in probability and statistics, theory of probability and its applications is a quarterly peer-reviewed scientific journal published by the society for industrial and applied mathematics.).
Introduction To Probability And Its Applications Scheaffer Solutions Pdf Alan agresti personal home page department of statistics, i was employed by the Theory of Probability & Its Applications 17:4, 734-744. Citation PDF (1080 KB)
2012-06-24В В· INTRODUCTION PROBABILITY & ITS APPLICATIONS BY W. FELLER An Introduction To Probability : Theory And Its Applications Free E-book PDF Download. Read Online or Download Introduction to Probability and Its Applications, Third Edition PDF. Best introduction books
Scarlett Johnson wa-cop.org An Introduction To Probability Theory And Its Applications 2nd Ed An Introduction To Probability Theory And Its Applications 2nd Ed I N T R O D U C T I O NT OP R O B A B I L I T YA N DI T SA P P L I C A T I O N S This page intentionally le...
Another significant application of probability theory in everyday life is reliability. — HTML index with links to PostScript files and PDF (first three chapters) Linear Statistical Inference and its Applications. 2nd Edition. Wiley Series in Probability and Statistics
8581000031741 william feller books amazoncom an introduction to probability theory and its applications vol 2 2nd edition skills-from-classic-films.pdf. Probability and Its Applications Published in association with the Applied Probability Trust Editors:J. Gani, C.C. Heyde, P. Jagers, T.G. Kurtz
Title: Free Continuous Time Markov Jump Linear Systems Probability And Its Applications (PDF, ePub, Mobi) Author: Policy Press Subject: Continuous Time Markov Jump Unit 14: The Normal Distribution and its Applications What is the probability that monthly food expenditure are less than \$5,500 or greater than \$7,000 ?
Free modern probability theory and its applications Pdf Books For Download free Introduction to Probability Models Tenth Edition Sheldon M. Ross University of Southern California 7 Renewal Theory and Its Applications 421 7.1 Introduction 421
Introduction to Probability and Its Applications Richard
Introduction to Probability and Its Applications Richard. i n t r o d u c t i o nt op r o b a b i l i t ya n di t sa p p l i c a t i o n s this page intentionally le..., preface these notes grew from an introduction to probability theory taught during the п¬ѓrst and second term of 1994 at caltech. there was a mixed audience of).
Introduction To Probability And Its Applications Chegg
Linear Statistical Inference and its Applications. 2nd. unit 14: the normal distribution and its applications what is the probability that monthly food expenditure are less than \$5,500 or greater than \$7,000 ?, introduction to probability and its to probability and its applications possible outcomes probability density function probability distribution).
Student Solutions Manual for Introduction to Probability
Free Probability and its Applications Mathematical. linear statistical inference and its applications. 2nd edition. wiley series in probability and statistics, introduction to probability models tenth edition sheldon m. ross university of southern california 7 renewal theory and its applications 421 7.1 introduction 421).
matrix entries are entered as comma-separated numeric
Download e-book for iPad Introduction to Probability and. preface these notes grew from an introduction to probability theory taught during the п¬ѓrst and second term of 1994 at caltech. there was a mixed audience of, request pdf on researchgate log-concave probability and its applications in many applications, assumptions about the log-concavity of a probability distribution).
Introduction to Probability and Its Applications Third
Basics Of Applied Stochastic Processes Probability And Its. from "fuzzy logic and probability applications: bridging the gap" timothy j. ross, jane m. booker, evolution of any theory and its applications., free modern probability theory and its applications pdf books for download free).
Preface These notes grew from an introduction to probability theory taught during the п¬Ѓrst and second term of 1994 at Caltech. There was a mixed audience of Assignments section contains the exercises Probability And Its Applications To Reliability, Quality Control, And Risk Assessment
Stochastic Processes and their Applications publishes papers on the theory and for Mathematical Statistics and Probability. Editor Pack’ PDF; View вЂGuide Hand-book on STATISTICAL DISTRIBUTIONS for 2 Probability Density Functions 3 6 Probability content from в€’zto zof Gauss distribution in
Statistics and Probability for Engineering Applications from the Probability Density Function as a textbook on statistics and probability for Introduction to Probability Models Tenth Edition Sheldon M. Ross University of Southern California 7 Renewal Theory and Its Applications 421 7.1 Introduction 421
Title: Free Self Normalized Processes Limit Theory And Statistical Applications Probability And Its Applications (PDF, ePub, Mobi) Author: Allen & Unwin PDF In many applications,assumptions about the log-concavity of a probability distribution allow just enough special structure to yield a workable theory.This paper
Why is Chegg Study better than downloaded Introduction to Probability and Its Applications PDF solution manuals? It's easier to figure out tough problems faster using Stochastic Processes and their Applications publishes papers on the theory and for Mathematical Statistics and Probability. Editor Pack’ PDF; View вЂGuide
DOWNLOAD PROBABILITY AND ITS APPLICATIONS probability and its applications pdf From the reviews of the first editions: "...Kallenberg's present book would have to Title: Probability Models For Dna Sequence Evolution Probability And Its Applications PDF Author: Stephenie Meyer Subject: probability models Keywords | 2,631 | 13,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.846057 |
http://www.pdf-tutorial.com/office/excel/258-excel-analytics-and-programming | 1,534,872,579,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221218391.84/warc/CC-MAIN-20180821171406-20180821191406-00606.warc.gz | 552,678,906 | 5,312 | # Excel Analytics and Programming
Goals of the Workshop
• Learn Excel tools by utilizing them in various cases
• Tools and materials covered here are merely a sample of Excel functionality
• Understand the logic and syntax behind Visual Basic programming, which interacts with the Excel interface
• No programming background required
• Create dynamic algorithms to approach cases
• When data is changed, but retains its original format, the algorithm should be able to automatically handle the transition appropriately
Contents Overview
• Case 1: Multiplication Table
• Case 2: Percentile Calculations
• Tutorial 1: Variables and Arrays
• Case 3: Hello World
• Tutorial 2: Functions
• Tutorial 3: Loops and Decisions
• Case 5: Loop through Data
• Tutorial 4: Recording Macro
• Case 6: Select, Pull, Display
• Tutorial 5: Userform
• Case 7: Subway Data All-Around Analysis
## Course, tutorial Summary Excel Analytics and Programming
This course is intended for a strictly personal use, the file is of format pdf level Advanced , the size of this file is 4.28 MB.
The site also offers courses in Excel 2007, 2010, 2013, 2016, charts, PivotTables and PivotCharts, Formulas, Data and Statistics, Pivot Tables, VLOOKUP, and IF functions and many other tutorials. You have to come and see our Excel. You will find your happiness without problem!
## Excel 2016 - Advanced Excel Tools
This booklet is the companion document to the Excel 2016: Advanced Excel Tools workshop. PDF file by Kennesaw State University. - type of file pdf and size 943.64 KB, tutorial for level Advanced .
## Excel 2013: Advanced Excel Tools
Download free Microsoft Office Excel 2013 Advanced Excel tools, cours tutorial training, a PDF file by Kennesaw State University. - type of file pdf and size 813.73 KB, tutorial for level Advanced .
## Excel How To Use VLOOKUP
Download Microsoft Excel How To Use VLOOKUP Function, free PDF file by timeatlas.com. - type of file pdf and size 673.43 KB, tutorial for level Beginner .
## Excel 2013: Accessibility
In this document, you will learn about the tools available for accessibility. You will also learn how to control the visual appearance of your spreadsheet. - type of file pdf and size 1.51 MB, tutorial for level Advanced .
## Excel 2016 - Accessibility
This document has been developed to provide you with information about accessibility in Microsoft Office Excel 2016. - type of file pdf and size 1.42 MB, tutorial for level Beginner .
## Excel 2013: Introduction
Download free Microsoft Office Introduction to Excel 2013, course tutorial training, a PDF file by Kennesaw State University. - type of file pdf and size 1.41 MB, tutorial for level Beginner . | 616 | 2,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-34 | latest | en | 0.775452 |
https://www.mainmoonwarren.com/beef/how-much-shredded-beef-for-18-people.html | 1,656,887,193,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00555.warc.gz | 899,455,637 | 21,995 | ### How Much Shredded Beef For 18 People?
When preparing a main dish such as steak, roast, chicken, or hog where the meat will be the primary attraction of the meal and will be complemented by a few side dishes, we recommend cooking 3/4 of a pound (12 ounces) of meat for each individual who will be eating the meal.
## How much meat do I need to buy per person?
In general, you should give each person half a pound (225 grams) of boneless meat, but you should give each person three quarters of a pound (340 grams) of bone-in meat.Refer to the table that follows to determine the precise quantity of what you want, or come into our store to place an order over the internet.Have you determined how much you require?You may now place your orders for meats here using our online platform.
## How much beef do I need to cook for 50 people?
You will need roughly 35–40 pounds of cooked beef for 50 people, however the exact amount will depend on how hearty the visitors’ appetites are lol. The normal dish is around 8oz to 12oz (one half to three quarters of a pound) (one half to three quarters of a pound). Also, besides the beef, do you have anything else on the menu?
## How much taco meat do I need to feed 200 people?
According to our calculations, a typical person would require 6.5 ounces (184 grams) of ground beef or shrimp, 5.0 ounces (142 grams) of hamburger meat, or 5.5 ounces (156 grams) of chicken. Let’s figure out exactly how many pounds of taco meat we need in order to serve 200 people:
You might be interested: How To Do Corned Beef?
## What is a good portion size for raw meat?
The amount of raw meat equivalent to 8 ounces (or 227 grams) is an appropriate serving size for any type of protein.When serving meat with only two or three side dishes, use three-quarters of a pound or 12 ounces (340 grams).Keep back one pound for each of the big eaters (athletes, teenagers, and others) The yield is defined as the difference in weight between the raw and cooked versions of an item.
## How much meat do I need for 18 adults?
When the Meat Is the Main Feature of the Meal: When preparing a dish like steak, roast, chicken, or pork, in which the meat is the primary focus of the meal and is accompanied by a few side dishes, we recommend preparing approximately half a pound (eight ounces) of meat for each person.For those with larger appetites and those who enjoy having leftovers, we recommend preparing up to three quarters of a pound of meat per person.
## How many pounds of beef do I need to feed 20 people?
What quantity of meat is required to serve 20 people? Given that there are 16 ounces in a pound of beef, about 10 pounds of meat would be necessary to satisfy the needs of 20 persons. When hosting a dinner party or a barbecue, the Food Network suggests providing between 6 and 8 ounces of meat to each adult attendee, depending on the size of the person.
## How many people will 3lb of beef feed?
How many ounces of steak do I need for five grownups?
Boneless Meat Number of People Bone in meat
2lb / 0.91kg 4 – 5 3lb / 1.36kg
3lb / 1.36kg 6 – 7 4lb / 1.82kg
4lb / 1.82kg 8 – 9 6lb / 2.73kg
5lb / 2.27kg 10 – 11 7lb / 3.18kg
## How much meat do I need for 20 guests?
What weight of meat should I order for each individual?
Boneless Meat Number of People Bone in meat
8lb / 3.64kg 16 – 17 12lb / 5.44kg
9lb / 4.08kg 18 – 19 14lb / 6.35kg
10lb / 4.54kg 20 – 21 16lb / 7.25kg
11lb / 4.98kg 22 – 23 18lb / 8.16kg
## How do you calculate meat per person?
How to Determine How Much Meat to Purchase for Each Individual
1. If you intend to offer the meat as an appetizer to your guests, multiply the number of people you wish to serve by 0.2
2. If meat will be served as part of the main course, but there will also be other filling components, multiply the number of guests by 0.5 to 0.33
You might be interested: FAQ: How To Cook Beef Tips In A Crockpot?
## How much ground beef do you need for 20 hamburgers?
How many pounds of hamburger do I need to cook for a party of twenty people? The United States Department of Agriculture (USDA) considers one serving to be three ounces of ground beef. When all of this is taken into account, 3.75 kilograms (or pounds) of ground beef would be more than enough to feed 20 people.
## How do you calculate food portions for a large group?
When Having Guests Over, How Much Food Should Be Prepared?
1. Two to three servings of an appetizer or snack have to be offered to each individual, or even more in the event that they are the only foods available for consumption
2. 3 ounces of dip or salsa (about a third of a cup for each individual)
3. One cup of soup (or a smaller amount if the soup is thicker, such as chowder)
4. 3 ounces of salad, which is approximately 1 cup
## How many does a 2 pound chuck roast serve?
One pound of boneless beef chuck pot roasts will provide three servings of cooked and trimmed meat weighing three ounces each. Each pound of beef chuck pot roasts with the bone in will provide two to two and a half servings of meat that is cooked and cut to be three ounces each.
## How big of a roast do I need to feed 12 people?
Servings Bone-In Roast Boneless Roast
5–6 adults 6 lb. (3 bones) 5 lb.
6–7 adults 7 lb. (3-4 bones) 6 lb.
8–10 adults 10 lb. (5 bones) 8 lb.
10–12 adults 14 lb. (7 bones) 11 lb.
## How many pounds of meat do you need per person?
Sizes of Meat Portions on Average In general, most chefs will recommend serving each individual 8 ounces of beef (or half a pound).For around 10 to 12 individuals, you will need approximately 5 to 5 and a half pounds of food.You can set aside a quarter of a pound for each type of meat that you will be serving in the dinner if you will be serving more than one dish.For ten to twelve people, you will need two and a half pounds of each type of meat.
## How many sandwiches can you make with a pound of roast beef?
How Many Sandwiches Can You Make With One Pound of Roast Beef? If you slice your deli meat in the sandwich form, you may get up to six sandwiches out of only one pound of deli meat. This is a decent rule of thumb.
You might be interested: How Many Calories In Ball Park All Beef Hot Dogs?
## How much beef do I need for 10?
Allow around 375g of beef per person, which means that you need a joint that is approximately 1.5 kilograms for four people, 2.25 kilograms for six people, 3 kilograms for eight people, and 3.75 kilograms for 10 people.
## How much meat do you need for a taco party?
As a general guideline, one taco shell should contain around four ounces of meat fillings for the typical individual. This will come out to around eight tablespoons, which is equal to roughly a half cup for each taco. It is also good to think about how substantial of an appetite your visitor could have in order to determine how much ground beef you will use for the dish.
## How much prime rib do I need for 15 adults?
How Big of a Prime Rib Roast Should I Purchase? The question ″How much prime rib roast should I buy per person?″ is one of the most popular that people ask. A whole prime rib roast consists of seven (7) ribs, weighs close to 15 to 18 pounds, and is adequate for feeding a group of at least 14 individuals (depending on how big of eaters they are).
## How much beef do I need for 12 burgers?
12 people – 68 oz. (4 lbs.)
## How do you calculate food portions for a large group?
When Having Guests Over, How Much Food Should Be Prepared?
1. Two to three servings of an appetizer or snack have to be offered to each individual, or even more in the event that they are the only foods available for consumption
2. 3 ounces of dip or salsa (about a third of a cup for each individual)
3. One cup of soup (or a smaller amount if the soup is thicker, such as chowder)
4. 3 ounces of salad, which is approximately 1 cup
## How many does 1kg of beef serve?
There is no one definitive answer to this issue because every cut of meat is different; nonetheless, on average, one kilogram of beef is sufficient for around four persons, depending on the ratio of adults to children.This is merely meant to serve as a guide, and we strongly suggest that you estimate more than you actually need rather than the opposite, as it is possible that you won’t have enough.
## How much beef do I need for 10?
Allow around 375g of beef per person, which means that you need a joint that is approximately 1.5 kilograms for four people, 2.25 kilograms for six people, 3 kilograms for eight people, and 3.75 kilograms for 10 people. | 2,143 | 8,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-27 | latest | en | 0.945264 |
https://slidesplayer.com/slide/11563462/ | 1,532,131,369,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592001.81/warc/CC-MAIN-20180720232914-20180721012914-00138.warc.gz | 763,157,853 | 68,756 | # Zetasizer Nano 系列: 培训课程
## Presentation on theme: "Zetasizer Nano 系列: 培训课程"— Presentation transcript:
Zetasizer Nano 系列: 培训课程
Who Are Malvern Instruments?
Zetasizer Nano 能够测量什么参数？
Zetasizer Nano系列: 粒子尺寸 Zeta电位 分子量 胶体颗粒，乳液，高分子溶液…… 高灵敏度 高浓度 维护简单
Contents 动态光散射（第一天） 静态光散射及分子量的测定（第二天） Zeta 电位 测量原理（第二天） 测量原理
Nano系列的优化测量位置 由相关曲线得到粒径信息 – 预算法则 样品要求 样品制备 数据解释 静态光散射及分子量的测定（第二天） Zeta 电位 测量原理（第二天）
Zetasizer Nano是如何测试粒子的粒径的?
Nano 的光学构造
Intensity Time g2 1 = 0 Time 1 Time g2 Intensity = 1 Time Time 1 2 g2 Intensity 在起始时刻，光强乘以其自身，相关性能最好，定义为一 随着时间变化，光强的相关性越来越差，最后在无限长时刻，相关性为0 = 2 Time Time Intensity Time 1 2 3 g2 =
This slide shows a schematic of a correlogram and illustrates the type of information that can be gained. Firstly, the time when the correlation starts to significantly decay indicates the mean size of the sample. Secondly, the gradient of the decay indicates the polydispersity of the sample. The steeper the gradient more monodisperse the sample is. Conversely, the more extended the decay becomes, the greater the sample polydispersity. Thirdly, the baseline of the correlation function gives information about the presence of large particles and or aggregates in the sample. The extrapolation of the data to zero time results in an intercept on the Y axis. This is the signal to noise ratio of the measurement. In the example shown in this slide, the intercept value is approximately zero point seven six. That is, the signal to noise ratio is zero point seven six for this measurement. As discussed earlier, a perfect signal would give an intercept value of 1. This is not possible during a real measurement as there is inherently a certain amount of noise present which reduces the intercept value obtained. The various reasons for the reduction in the intercept will discussed in more detail in future modules. 基线是否归零告诉我们是否有灰尘的存在
Correlate Apply Algorithm Time (s) Intensity (kcps) Small Particles Correlate Apply Algorithm Time (s) Intensity (kcps) Large Particles
STOKES-EINSTEIN EQUATION
D 为扩散系数 d(h) 为流体力学直径 kB 为波尔兹曼常数 T 为绝对温度 h 为粘度 这里 我们通过。。。。。方程将粒子的运动速度和尺寸联系起来 2
STOKES-EINSTEIN EQUATION
2c/b2 为分布系数
z-均 直径 z-均直径(ZD)的定义: 累积距法得到的粒子平均尺寸对应于不同尺寸粒子散射光强的贡献 这里“平均”的概念特指用于光散射试验中
General Purpose Multiple Narrow Modes 这两种算法的差异在于所得到的分布曲线的平滑程度 general purpose 算法适用于大部分分布状况未知的样品 multiple narrow mode 算法适用于分布状况不连续的样品
Zetasizer Nano 软件中的尺寸分布分析
Zetasizer Nano 软件中的尺寸分布分析
Here is an example of an intensity size distribution obtained from a dynamic light scattering measurement. The table shows the 70 size classes which are logarithmically spaced between a lower limit of zero point four nanometres and an upper limit of 10 microns on the X axis. The Y axis consists of the relative percentage of light scattered by particles in each of the size classes. The plot at the top right of the slide is the graphical representation of this data as a histogram. By default, the distribution is displayed as a frequency curve in the Nano software. An example is shown in the bottom right of the slide. There are various size distributions available in the Zetasizer Nano software and we will now look at each one in more detail.
Peak 1 Peak 2 Mean (nm) % Intensity 231 86.3 65.8 13.7 Volume 232 50.3 61.8 49.7 Number 184 2.6 58.2 97.4 To illustrate the relationship between intensity, volume and number size distributions, here is an example of a mixture of different sized polystyrene latex standards. These latex standards were mixed in equal volumes. This record is contained in the example results file which is provided with the Zetasizer Nano software. The z-average diameter obtained was one hundred and sixty eight nanometres with a polydispersity index value of zero point two one five. If we first look at the intensity size distribution, we find that the main peak has a mean diameter of two hundred and thirty one nanometres. This peak constitutes around 86 percent of the distribution. This is because the larger particles in the sample are scattering more light compared to the smaller population When the distribution is viewed in volume, the percentage of each peaks is approximately 50%, which corresponds with the ratio at which the 2 latex standards were mixed at. When the result is viewed in number, the vast majority of the distribution is contained in the smaller sized peak. So if this sample was viewed under an electron microscope, the smaller particles would be dominant. That is, on a number basis, the sample appears to consist mainly of the smaller population of particles. The intensity distribution however, detects the presence of larger particles which may be missed by a number based technique.
As we have discussed in a previous slide, DLS tends to overestimate the width of the peaks in the distribution and this will become even more significant upon transformation into volume or number. In addition, we have seen that a small amount of large particles present in the sample will dominate the intensity size distribution obtained from a DLS measurement. Therefore, it is recommended to use the intensity size distribution for reporting the size of each mode in the distribution but to the use volume or number data for reporting the relative amounts of each particle family in the sample. This recommendation can be highlighted with the example shown here. The intensity and volume particle size distributions shown in this slide were derived from a 2 to 1 volume mixture of 60 and 220nm polystyrene latex standards. The intensity particle size distribution shows a bimodal with peak means of 59 and 220 nanometres respectively. The relative percentages of each mode based upon the intensity data are 21 and 79 percent respectively with the larger peak being due to the larger sized population. However, conversion to volume gives relative percentages of 67 and 33 percent respectively. Therefore, the recommended way of reporting this result is to use the peak mean diameters from the intensity size distribution together with the relative percentage values obtained from the volume distribution. Peak DI (nm) % Int % Wt 1 59 21 67 2 220 79 33
Nano 系列中的优化测量位置
Nano 的光路系统
NIBS: 可调整检测位置 小粒子/ 稀溶液 高浓度溶液 减小散射光体积 较大的散射提体积 降低多次散射影响 检测器 激光 凸透镜 样品池
Zetasizer Nano自动水衰减器有11个光学衰减镜片涵盖100% 到 % 的透射率 透射率指到达样品的激光的强度占光源激光强度的百分比 在测量粒子大小的过程中，自动衰减器会自动调节透射光的强度，直到检测器检测到的光强小于 500kcps
0.01 4 99.97 0.03 5 99.9 0.1 6 99.7 0.3 7 99 8 97 9 90 10 70 30 11 100
International Standard ISO 13321 (1996)
When multiple scattering is insignificant the size will be independent of concentration. The plot shown in this slide illustrates the effect of sample concentration on the mean diameter obtained for both a 90 degree and backscatter instrument. In both instruments, the size obtained is independent of concentration when the sample concentration is low. However, as the concentration of the sample is increased, the presence of multiple scattering will begin to influence the size obtained. In a backscatter instrument such as the Nano S, the concentration over which the sample can be measured correctly is greatly extended compared to the 90 degree system due to the reduction of multiple scattering effects. Instruments which have backscatter detection extend the concentration over which samples can be measured before seeing the effect of multiple scattering. Nano S90 样品浓度
Only limited by the sample material interaction (gelation, aggregation) 10nm to 100nm 1mg/ml 0.1mg/ml 0.1% w/v 5% w/v (assuming a density of 1gcm-3) 100nm to 1μm 0.01mg/ml 0.01% w/v 1% w/v (assuming a density of 1gcm-3) > 1μm An important factor in determining the maximum and minimum concentrations the sample can be measured at is the size of the particles. This table is an approximate guide for obtaining results which are independent of concentration for samples with a density near to 1 gramme per cubic centimetre. If such concentrations cannot be selected easily, it is recommended that various concentrations of the sample should be measured in order to determine if concentration dependent effects such as particle-particle interactions or multiple scattering, are present.
z-均直径重复性 多次z-均直径的测试结果误差应在1%-2%之内 z-均直径增长意味着: z-均直径下降意味着: 粒子聚集
Contents 动态光散射 静态光散射及分子量的测定 测量原理 应用实例 Zeta 电位
Mixtures NO
dn/dc = 折光指数对浓度的增量 Rg = 均方旋转半径 = 检测角度 IA = 绝对光强 (I样品 – I溶剂) no = 溶剂 IT = 标准物光强 (toluene) nT = 标准物 (toluene) 折光指数 RT = 标准物瑞利比 (toluene) 者是静态光散射检测方程。 对于小粒子，同场小于激光波长的1/10 p(o)约等于一，也就是说散射光强不再有角度依赖性
1/截距 = 14.6KDa 斜率 = x 10-4
Contents 动态光散射 静态光散射及分子量的测定 Zeta 电位 测量原理 样品制备 样品测试 测试中的选择 数据解释
Zeta电位 Zeta Potential 电泳光散射 Electrophoretic Light Scattering (ELS) 激光多普勒电泳 Laser Doppler Electrophoresis (LDE) 理论概述
Zeta电位(Zeta Potential)
Slipping plane Zeta电位同时依赖于粒子表面和分散剂的化学性质 对于静电力稳定的分散体系，通常是Zeta电位越高，体系越稳定 体系稳定与否通常以Zeta电位是否大于 30mV为标准 Particle with negative surface charge 影响Zeta电位的因素 Stern layer 对于一个带点粒子，通常对电位由三个定义。由于粒子带点，在粒子的表面的电势称为表面zeta电势。在粒子表面有一层抗衡离子紧密地和表面结合在一起，这个边缘的电势称为stern电势，在外层还有一些离子和带电粒子松散的结合在一起，但是这层离子区别于溶剂中自由运动的离子,这一层电势称作zeta电势 Diffuse layer 影响Zeta电位的因素有： pH变化, 电导率 (浓度，盐的类型) 组成成分浓度的变化 (如高分子，表面活性剂) -100 - { Surface potential Stern potential mV Zeta potential Distance from particle surface
（ELECTROPHORETIC MOBILITY）
Henrys 方程 F(ka) 非极性溶剂 极性溶剂 Huckel 近似 F(ka) = 1.0 Smoluchowski 近似
Nano的光学构造 参考光不通过样品池 衰减镜片调整入射光的光强 这样仪器可以检测很宽浓度范围内的样品
M3 测试技术 高频电场转换1000Hz能够准确地测量zeta电位的平均值，但是分辨率较低
-50 mV -110 低频电场的转换可以给出更好的分辨率但是受到电渗的影响 This is the measurement from fast filed reversal By combining the fast mode and slow mode measurement, we get better resolution and also the speed of the electroosmotic flow 通过M3测试技术，结合高频和低频电场的转换，我们既得到准确地平均zeta电位，又得到了较高的分辨率
FFR（高频电场转换） SFR（低频电场转换） 通过相分析和复利叶变换得到电位的分布 通过相分析得到平均电位 (Ep)
PALS only to obtain mean (Ep) – no distribution or width
Zeta电位（Zeta Potential）
zeta电位试验中的样品测试 Zeta电位测试中对样品的要求不像尺寸测试对样品的要求那样严格，样品可以看上去不很透明
zeta电位试验中的样品测试（SiO2 浆料）
Zeta电位和pH 自动滴定确定等电点 Zeta Potential (mV) pH 等电点 （IEP） 9 2 3 4 5 6 7 8 1
10 11 12 13 14 pH Zeta Potential (mV) +30 -30 等电点 （IEP） 自动滴定确定等电点
TiO2 的pH滴定 zeta size Aggregation at Iso-electric point leads to a size increase.
zeta电位试验中的样品测试 在稀释溶液的过程中应该保持粒子表面的性质 你有稀释溶液吗？ 过滤或者离心一些原溶液，取上清夜作为稀释溶液
Zeta电位的测试中，激光必须穿过样品，因为散射光在向前的角度被检测到 如果样品的浓度太高，被检测到的粒子散射的激光强度会有很大的衰减 为了抵消这些影响，仪器中衰减镜片的位置将被调整到一个比较高的指数，也就是比较高的透射率
Zeta电位（Zeta Potential）
Phase Plot Report Voltage/Current Report Frequency Report 参数-PARAMETERS Conductivity Attenuator
Zeta质量报告 Zeta质量报告对任何一个实验结果进行六项测试
Zeta 电位分布范围限制 默认的Zeta 电位分布范围限制在 +150 到 –150mV
Zeta 电位分布范围限制 有几种原因可能导致这个问题的出现 样品的zeta电位分布很宽 较高的导电率导致施加的电压自动降低
Zeta 电位分布范围限制 Mean zeta = +63mV Wall zeta = +61mV Voltage = 150V
Zeta 电位分布范围限制 Voltage = 100V
Thank you ! 更多信息 或中文网站 www.malvern.com.cn 或email至 Info@malvern.co.uk
Similar presentations | 3,655 | 10,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-30 | latest | en | 0.615835 |
https://www.jiskha.com/display.cgi?id=1317524591 | 1,516,284,844,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887414.4/warc/CC-MAIN-20180118131245-20180118151245-00277.warc.gz | 915,211,628 | 3,541 | # Physics
posted by .
During a baseball game, a batter hits a pop-up to a fielder 83 m away.
The acceleration of gravity is 9.8 m/s^2.
If the ball remains in the air for 6.5 s, how high does it rise?
• Physics -
)A batter hits a pop-up to a fielder 83 m away. If the ball remains in the air for 6.5 s, how high does it rise?
Rise time = fall time = 3.25sec.
Vf = 0 = Vo - 9.8(3.25) or Vo = 31.85m/s
d = Vo^2(sin(2µ))/g
----31.85^2(sin(2µ))/9.8 or µ = 26.65º
Then, height h = Vo^2(sin^(x)/2g or
h = 31.85^2(sin(26.65))/2(9.8) = 10.41m
• Physics -
dafuq
## Similar Questions
1. ### physics.
During a baseball game, a batter hits a pop-up to a fielder 94 m away. The acceleration of grvity is 9.8 m/s^2. If the ball remains in the air for 5.5 seconds, how high does it rise?
2. ### Physics
During a baseball game, a batter hits a pop-up to a fielder 82 m away. The acceleration of gravity is 9.8 m/s2 . If the ball remains in the air for 6.9 s, how high does it rise?
3. ### physics
During a baseball game, a batter hits a pop- up to a fielder 78 m away. The acceleration of gravity is 9.8 m/s2 . If the ball remains in the air for 5.8 s, how high does it rise?
4. ### Physics
During a baseball game, a batter hits a pop up to a fielder 65m away. The acceleration of gravity is 9.8m/s^2. If the ball remains in the air for 5.7s, how high does it rise?
5. ### Physics
During a baseball game, a batter hits a pop-up to a fielder 78 m away. The acceleration of gravity is 9.8 m/s^2. If the ball remains in the air for 5.7 s, how high does it rise?
6. ### physics
During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.59 s, how high does it rise?
7. ### physics
During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.65 s, how high does it rise?
8. ### physics
During a baseball game, a batter hits a pop-up to a fielder 92m away. The acceleration is 9.8 m/s^2. If the ball remains in the air for 5.8 seconds, how high does it rise?
9. ### Physics/Science
During a baseball game, a batter hits a pop- up to a fielder 72 m away. The acceleration of gravity is 9.8 m/s^2. If the ball remains in the air for 5 s, how high does it rise?
10. ### physics
During a baseball game, a batter hits a popup to a fielder 80 m away. The acceleration of gravity is 9.8 m/s2 . If the ball remains in the air for 5.5 s, how high does it rise?
More Similar Questions | 778 | 2,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-05 | latest | en | 0.87412 |
http://www.woollythoughts.com/afghans/pane.html | 1,713,640,354,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817674.12/warc/CC-MAIN-20240420184033-20240420214033-00894.warc.gz | 61,154,270 | 11,003 | Counting Pane
Click to read general information about afghans
Other places to visit
Order Form
Mental Blocks
The World of Illusion Knitting
for this afghan
on the order form
COUNTING PANE
A representation of multiplication tables.
The squares across the top row represent the numbers 1 -10, the second row 11 - 20, and so on. The colours show which numbers, from 1 - 10, divide into the numbers represented by the squares. Each colour shows a different multiplication table.
The original Counting Pane was bought for the Mathematics Collection of the Science Museum (London)
Counting Pane
KNITTING INFORMATION
Ten colours are needed to represent the numbers 1 to 10. These colours must be distinctly different and easily identifiable by a child, to avoid confusion.
RELATED DESIGNS
NUMBER AFGHANS
Double Base
Mere Bagatelle
Fibo-optic
Fibrenacci
Some Square Over The Rainbow
CONSTRUCTION INFORMATION
Knitted in ten strips which are stitched together.
It is often difficult to know where inspiration comes from. Counting Pane was a sudden inspiration in our very early days of representing Maths in knitting. Perhaps it was suggested by the squares, to be found in many classrooms, shaded to show multiples of numbers. Counting Pane shows all those shadings simultaneously. It remains one of our favourites and has probably had more attention over the years than any of the others. Does this make it the most important? Maybe.
It was the first time we used the technique of knitting in strips rather than working with diagonal squares and specific angles. The individual squares were separated by a stripe of the background colour and one column was separated from the next by a similar stripe.
Counting Pane has 100 squares in 10 rows. The first row represents the numbers 1 to 10, the second is 11 to 20, and so on. The colours across the top represent factors. Blue stands for 1, yellow stands for 2, red is 3, green is 4, raspberry is 5, brown is 6, pink is 7, pale blue is 8, orange is 9, and purple is 10.
If a square contains yellow it means the number divides by 2, if it has red then the number divides by 3. If a square divides by 2 and 3 it must divide by 6 so also contains brown.
No factors above 10 are included. All squares which remain plain blue are prime numbers (Except Square 1 - See below.) In the top row the squares with only blue and one other colour are also prime numbers.
The main problem at the designing stage was working out what size the squares must be to fit the different numbers of stripes needed. We decided to use squares with twenty-four ridges of garter stitch as 24 will divide by 1, 2, 3, 4, and 6. We had to cheat a little where we needed 5 or 7 stripes as these numbers don’t divide evenly into 24.
We have made three versions of Counting Pane. The first was bought by the Science Museum. It is slightly different from the others because we tried to resolve the problem of 1 looking like a prime number when technically it isn’t. For versions two and three we decided this was an anomaly we would have to live with and introduce as a discussion point when appropriate. Versions 2 and 3 have other slight cosmetic differences.
At the most basic level this afghan offers a very graphic way for children to understand the difference between odd and even and to realise that this is really the same as knowing whether a number divides by 2. Yellow only appears in half of the columns. These columns contain only even numbers.
Numbers which divide by 5 are also spotted quickly as raspberry is noticeably only in the 5’s and 10’s columns. Putting these two ideas together there is only one column with yellow and raspberry so this must be the column of 10’s.
The average person would consider that they have a reasonable understanding of the numbers 1 - 100 yet the colours in this piece throw up surprising ideas that may not have been considered before. Why is there that ‘backward-sloping’ line of orange? Why is there always a green with light blue?
The original hung in my Mathematics classroom and many pupils became addicted to finding patterns and working out the reasons for them being there. One member of staff - a language support teacher - also became addicted and would frequently rush into the room, saying ‘I’ve just come to check on something.’ He would study Counting Pane for a few minutes then disappear again.
Counting Pane has formed the basis of many exhibitions and workshops. One memorable occasion was at the annual North-East Maths Fair held at Houghton-le-Spring, probably in 1997. We had six or seven afghans on display, some with suggestions for tasks to be attempted, others simply for talking about. Part way through the day a girl, aged 13, wandered in alone and was instantly drawn to Counting Pane - perhaps she was merely attracted by the stunning colours as we later discovered she had been brought to the event by a teacher and didn’t really have much interest in being there.
During the afternoon she returned with a succession of friends and towards the end of the day reappeared yet again with her Maths teacher. She didn’t ask him any questions but proceeded to lecture him with her new found confidence in her own ability. He was astounded.
This served to reinforce some of the views we have long held about teaching Maths. These include -
• not everyone sees things in the same way
• some people have a much more visual approach than others
• Maths does not have to be about writing things down
• pupils may have a very distorted view of their own capabilities
• some pupils need unconventional triggers to begin the thought process
We like to think Counting Pane may have had a lasting effect on that girl - and her teacher.
Click here to read a magazine article about mathematical knitting and crochet, including Curve of Pursuit.
This version was made by Margaret O’Mara
in Tunisian crochet.
This version was made by Pam
Click to see
larger picture
This version was made by Lorraine Ehrlinger.
She named it Prime Squares Afghan. | 1,310 | 6,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.951422 |
https://www.kaysonseducation.co.in/questions/p-span-sty_3835 | 1,670,431,942,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00231.warc.gz | 885,008,935 | 25,297 | Give the speed-time graph of the motion of a car. What is the ratio of the distance travelled by the during the last two seconds to the total distance travelled in seven seconds? : Kaysons Education
# Give The Speed-time Graph Of The Motion Of A Car. What Is The Ratio Of The Distance Travelled By The During The Last Two Seconds To The Total Distance Travelled In Seven Seconds?
#### Video lectures
Access over 500+ hours of video lectures 24*7, covering complete syllabus for JEE preparation.
#### Online Support
Practice over 30000+ questions starting from basic level to JEE advance level.
#### National Mock Tests
Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.
#### Organized Learning
Proper planning to complete syllabus is the key to get a decent rank in JEE.
#### Test Series/Daily assignments
Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.
## Question
### Solution
Correct option is
None fo these
Distance travelled in the last two seconds = area of triangle BDC The distance travelled in seven seconds = area of triangle OAE + area of rectangle ABDE + area of triangle BDC
= 90 m.
#### SIMILAR QUESTIONS
Q1
A freely falling body, falling from a tower of height h covers a distanceh/2 in the last second of its motion. The height of the tower is (take g = 10 ms–2) nearly
Q2
A cyclist starts from the centre O of a circular track of radius r = 1 km, reaches the edge P of the track and then cycles along circumference and stops at point Q. if the total time take is 10 min, what is the average velocity of the cyclist?
Q3
A stone falls from rest. The distance covered by the stone in the last second of its motion equals the distance covered by it during the first three seconds of its motion. How long does the stone take to reach the ground?
Q4
Two balls of masses m1 and m2 are thrown vertically upward with the speed u. If air resistance is neglected, they will pass through their point of projection in the downward direction with a speed
Q5
A train covers half of its journey with a speed of 20 ms–1 and the other half with a speed of 30 ms–1. The average speed of the train during the whole journey is
Q6
The displacement y (in metres) of a body varies with time t (in seconds) as
How long does the body take to come to rest?
Q7
A body goes from A to B with a velocity of 40 kmh–1 and returns from Bto A with a velocity of 60 kmh–1. What is the average velocity of the body during the whole journey?
Q8
The area under the velocity-time graph between any two instants t = t1and t = t2 gives the distance covered in time
Q9
A particle is given a displacement of 4 m in the x-y plane. If the x-component of the displacement vector is 2 m, the y-component will be | 691 | 2,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-49 | longest | en | 0.841018 |
http://blogs.msdn.com/b/rezanour/archive/tags/matrices/ | 1,419,689,446,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447552182.102/warc/CC-MAIN-20141224185912-00084-ip-10-231-17-201.ec2.internal.warc.gz | 12,257,767 | 6,410 | # Reza Nourai's Game Development Ramblings
Random discussions about game development. Particularly math, physics, collision detection/response, and performance.
# Browse by Tags
Tagged Content List
• #### Blog Post: Math Primer Series: Rotation Representations and Quaternions
We saw the power and usefulness of transforms in the previous primer posts, and we’ll be using them quite a bit in our discussions of physics and graphics. One of the first tasks we’ll need to do is determine how to represent our transformations in code. We’re going to be creating, manipulating, and...
• #### Blog Post: Math Primer Series: Matrices III: Affine Transformations and Matrices
We’ve seen matrices used as representations of a basis, and we’ve seen them used as linear transformations, but what else can we do with them? So far, we’ve only been able to use matrices to transform vectors from one vector space to another. What about points and vectors in an affine space? There is...
• #### Blog Post: Math Primer Series: Matrices II: Linear Transformations
In the last installment of the math primer series, we looked at the basics of matrices. Today, we’ll take a look at using matrices for linear transformations, which are one of the most common uses of matrices in games. But, before we dive into linear transformations, let’s take a quick look at what a...
• #### Blog Post: Math Primer Series: Intro, Notation, and References
This is the first part of a multi-part primer and reference for the math we’ll be using extensively throughout this blog. If you’ve been around game or graphics programming for any length of time, you’re probably at least somewhat familiar with many of these topics. However, you might...
Page 1 of 1 (4 items) | 370 | 1,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | longest | en | 0.864061 |
https://www.reddit.com/r/Damnthatsinteresting/comments/wf1o5r/wrapping_hay_bales_the_cheap_way/ | 1,679,791,744,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945376.29/warc/CC-MAIN-20230325222822-20230326012822-00361.warc.gz | 1,083,728,073 | 80,535 | ×
Dismiss this pinned window
[–] 5033 points5034 points (129 children)
One down, 347 to go.
[–] 1467 points1468 points (92 children)
Quick do the math. How many sit ups is that total
[–] 1441 points1442 points (82 children)
Took 13 in the video, presumably one before the camera turned on, but let’s say 15 to keep the math easy:
15 * 348 bales = 3480 + 1740 = 5,220 sit-ups
[–] 779 points780 points (45 children)
Assuming that the center of a person's upper body travels about one foot per sit up, that's a mile of sit-ups.
[–] 655 points656 points (34 children)
They are going to have the abs of a fucking God
[–] 297 points298 points (28 children)
Over 17.5 hours if the length of this video is a guide for the remaining bales
[–] 198 points199 points (21 children)
Abs carved from whatever Thor’s hammer was made of jesus
[–] 377 points378 points (13 children)
Thor”s hammer was not in fact made out of Jesus it’s a common misunderstanding. His body was bread, not hammers.
[–] 87 points88 points (0 children)
Shut up and take my gold damn you. That made me laugh until i almost peed myself
[–] 15 points16 points (0 children)
He also in fact, was gluten free. Purest of pure.
[–] 8 points9 points (0 children)
Yeah, Jesus being a hammer is ridiculous and very unrealistic, he's obviously turned into bread,
[–] 19 points20 points (3 children)
the hammer is made out of ‘Uru,’ a fictional metal from Thor’s native realm of Asgard.”
[–][🍰] 21 points22 points (1 child)
Forged in the heart of a dying star by Peter Dinklage.
[–] 5 points6 points (0 children)
I laughed way too hard at this
[–] 9 points10 points (0 children)
Magically infused silver most likely. Abs of magic and silver.
[–] 17 points18 points (4 children)
384 bales of hay at \$50 per bale = \$19 200
Assuming they shot the farmer and are just stealing the hay, input cost = \$0
\$19 200 for 17.5 hours of work for 3 people = \$365 per person per hour
Not bad for an honest day's chorin'
[–] 5 points6 points (0 children)
Please Stop inventing new measurement methods for the americans.
How many sit ups die you do ? Half a Mile!
[–] 100 points101 points (5 children)
[–][🍰] 13 points14 points (4 children)
[–] 11 points12 points (2 children)
[–] 51 points52 points (1 child)
At least 3
[–] 6 points7 points (0 children)
Lots
[–] 71 points72 points (5 children)
In case anyone’s wondering, this really is cheaper! I measured proportions from the pixels. Assuming that the man is 1.7 meters tall (this is the average height of a man), he’s about 10.1 feet from the hay bale on average. If he’s driving around it about 15 times on average per hay bale, that’s 960 ft of driving per bale.
The average bale of hay costs around \$7 to wrap. This man is driving 0.182 miles for every hay bale, spending about 93 cents per bale in gas money (depending on where he lives). The amount of plastic he’s using costs approximately \$2.43 per bale (according to google). This means that he’s spending about \$3.56 per bale which is only half of the average wrapping price!
The cost of the calories to fuel all of those sit-ups though… that may be a different story.
[–] 6 points7 points (0 children)
Math master 5000
[–] 4 points5 points (0 children)
Plus labor.
[–] 58 points59 points (0 children)
\$84 in gas though :)
[–] 52 points53 points (1 child)
How the fuck did you come up with the exact same number I was gonna use for the exact same comment.
I wasn't gonna spell the word "one" though. I guess you're not me from the future.
[–] 13 points14 points (4 children)
“Why are there so many microplastics in farm runoff?”
[–] 9 points10 points (3 children)
No kidding. Use natural twine and put in the barn. Hate this. Save plastics for when they are the only good option, like wire insulation.
[–] 4210 points4211 points (94 children)
Ab workout too
[–] 321 points322 points (3 children)
Ag workout too
[–] 18 points19 points (0 children)
ugh. workout….
[–] 26 points27 points (0 children)
Some days you find things on the internet that make you happy
[–] 5327 points5328 points (224 children)
Ah kids. Cheap labor.
[–][deleted] 2166 points2167 points (94 children)
"We birthed you because someone needs to repair the fence"
[–] 831 points832 points (79 children)
My mom has a similar one “I had kids so I didn’t have to do chores anymore”
[–] 280 points281 points (21 children)
I have two kids and I seem to do more chores now 😂
[–] 81 points82 points (0 children)
Thats been my experience also.
[–] 73 points74 points (13 children)
That's likely because you don't see them as clone serfs. It makes sense though as kids will be able to better take care of you if their head isn't crushed while wrapping hay.
[–][🍰] 271 points272 points (32 children)
My mum said she had kids for the free slave labour - works too, she hasn’t made her own coffee in 20 years.
[–] 201 points202 points (20 children)
That's an expensive coffeemaker.
[–] 130 points131 points (17 children)
That has baggage and EMOTIONAL DAMAGE
[–] 29 points30 points (2 children)
I guess I’ll just live with his voice and expression in my head forever.
[–] 6 points7 points (0 children)
MENTAL ANGUISH
[–] 49 points50 points (12 children)
Or dedication and deep loving connection. My kids are 16, 18, 21 & 24 they've always loved me to bits. I have fresh ground coffee waiting on my Keurig every morning. That goes both ways though. The more love you put in, the more you'll get out.
[–] 21 points22 points (2 children)
Have you considered requesting a little more love and asking for hand ground beans from a burr grinder and a pour-over or Aeropress?
[–] 5 points6 points (1 child)
If my son started washing the grounds out of the french press for me, I would be in heaven. That’s the worst part for me.
[–][🍰] 9 points10 points (1 child)
I wish it was that, but alas, she’s not been a great parent. I was her coparent from the age of 9, her carer (she’s disabled and bedridden), friend, confidante, problem solver, therapist, scapegoat, emotional punching bag, etc; somehow “daughter” never seemed to be relevant. I honestly don’t know what she’s going to do when my sibling moves out of home, she’s unable to care for herself or her house and relies on her children in entirety to sustain her. But getting a professional carer is too hard and scary, “that’s not something I can deal with right now” repeated ad nauseam about anything she may have to be responsible for.
I haven’t spoken to her in four months and can’t think of a single reason to resume contact, the last few months have been the best my mental health has ever been. Very happy you have a great relationship with your kids, it always warms my heart to see it - give them big hugs for me so I can live vicariously through y’all and your functional relationships haha
[–] 5 points6 points (0 children)
Oh my goodness 💜 big hugs to you! I have a great relationship with my kids because my father was emotionally unavailable and uninvolved like many boomer men. I had no idea how much he was truly detached until my mom died when I was 29. She was an alcoholic my entire childhood. For the last decade I've seen him about once a year for a few hours. He never reaches out first, never talks to my kids, never is a grandparent at all. But, it's whatever. No one and no event in my past is going to make my kids or myself unhappy now. Took me a while to get here but I've learned if you don't expect anything from anyone, they'll never disappoint you.
[–] 39 points40 points (3 children)
Kid: What is my purpose?
Mom: You make coffee.
Kid:Oh...<dejected look>
[–] 4 points5 points (0 children)
I think this is the funniest thing to come out of my comment lol thank you
[–] 15 points16 points (0 children)
When asked why she never bought a dish washer, my mother simply replied 'I have four children.'
[–] 6 points7 points (3 children)
My mom woke me up for years just to tell me to go make coffee lol
[–] 59 points60 points (10 children)
I had kids so I'd have someone to play video games with. Mission accomplished. However, my desire for a family band has so far been thwarted.
[–] 20 points21 points (0 children)
Hear me out. Karaoke night!
[–] 11 points12 points (5 children)
I'm so excited to play video games with my kids! So far they're just 4 and 2, but I already have the 4 year old practicing shooting and movement in the Apex Legends firing range!!
[–] 14 points15 points (0 children)
In 2-3 years the will go from you teaching them how to play video games to them schooling you.
Please enjoy beating them as much as possible before it is too late.
[–] 4 points5 points (1 child)
The wii came out just around the time they were born and that was a nice place to start. The motion tracking swinging a sword or run in place. It was nice to wear them out too.
[–] 9 points10 points (0 children)
[–] 4 points5 points (0 children)
My 6 year old is better than I am at most video games (of the two of us, my husband is the gamer but I'm down to play casually when I can), he has yet to fully realise this and still asks me for help sometimes, which I love! My 23 month old likes the Wii so it's only a matter of time.
[–] 12 points13 points (1 child)
Buy an old rock band game. You'll still be playing video games but it really gives you that family band feel.
[–] 18 points19 points (2 children)
I mean I get it and I like it, but I'm going to wait for the Tesla bot.
[–] 7 points8 points (0 children)
Blech, fixing fence sucks but it's a helluva lot better than bailing hay for picky horses who get respiratory infections from round bales.
[–] 6 points7 points (0 children)
When my grandfather was eight his parents dumped him on their farm and told him if he didn't work he wouldn't eat. He was not a kind grandfather, and it has affected my father in many ways that he himself likely doesn't even realize, which has in-turn affected me.
That said, what's happening in this gif definitely doesn't have the feel of something like that. It looks more like a family chore with everyone sharing in the responsibility.
[–] 4 points5 points (2 children)
That was my great grandparents logic lol, had 18 kids. 18 kids=18 unpaid farm hands
[–] 155 points156 points (5 children)
The Hyundai factory in Alabama agrees.
[–] 55 points56 points (6 children)
I think one is an adult, maybe the mother?
[–] 21 points22 points (4 children)
I heard a Norfolk accent, so likely both the mother AND the sister...
(It's an ongoing joke that Norfolk is to the UK what Alabama is to the US)
[–] 119 points120 points (40 children)
An accident waiting to happen….
[–] 75 points76 points (37 children)
This could go wrong so many ways! The tractor driving over the kid's heads being the worst.
[–] 44 points45 points (1 child)
Always have a spare on hand.
[–]Interested 56 points57 points (0 children)
They're usually squishy enough to not damage the tractor much on impact, so I don't think a spare tractor is really necessary here.
[–] 13 points14 points (0 children)
That's why there's a tall one and a short one. The taller one is the guide bar.
[–] 17 points18 points (5 children)
I swear some redditors are so unphysical they see someone doing anything and think they're one millisecond away from decapitating themselves
[–] 6 points7 points (0 children)
Know how expensive a kid is lol. Unless it’s someone else’s kid it’s gonna be pricey
[–][🍰] 2364 points2365 points (61 children)
I was waiting for one of the kids to get stuck in the wrapping...
[–] 574 points575 points (28 children)
"Yes sir, the kid comes free with the hey"
[–] 117 points118 points (18 children)
Hay now
[–]Interested 47 points48 points (16 children)
You’re a rockstar
[–] 32 points33 points (15 children)
Get the show on
[–] 19 points20 points (4 children)
Stop
[–] 11 points12 points (0 children)
Go
[–] 11 points12 points (7 children)
Go playyyyy
[–] 8 points9 points (6 children)
Hay now
[–] 9 points10 points (5 children)
You're an All Star
[–] 8 points9 points (4 children)
Get the show on
[–] 7 points8 points (3 children)
Get laid!
[–] 6 points7 points (0 children)
Wrap hay
[–] 51 points52 points (0 children)
I thought this was /r/whatcouldgowrong until the end
[–][deleted] 10 points11 points (0 children)
I was waiting for the bail to roll on them. But they did a great job.
[–] 48 points49 points (12 children)
Or worse gotten their head run over by the tractor.
[–] 4 points5 points (0 children)
That’s what I was waiting for. This looks like something out of Rescue 911
[–] 9 points10 points (0 children)
They were stuck in the wrapping way before the started that farming art piece
[–] 407 points408 points (4 children)
killer core workout
[–] 667 points668 points (24 children)
Are gay bales insanely heavy?
EDIT: I’m gonna leave it.
[–] 422 points423 points (2 children)
I don't know about the gay bales, but I've heard that Christian Bales can fluctuate a lot in weight depending on the movie.
[–] 34 points35 points (1 child)
Alexa, Subscribe to Bales facts
[–] 31 points32 points (0 children)
bro u have to.
[–] 16 points17 points (0 children)
You've got some serious gay bales coming in here with a question like that
[–] 14 points15 points (0 children)
A 4x5 round silage bale should be about 1000lbs, there's a Lotta wiggle room there though. Depends on a number of things.
[–] 11 points12 points (5 children)
Its probably a "straw" bale. Straw is a loose term for a variety dried grasses or stalks. They are generally not for food, but animal bedding.
One of the easy ways to identify straw is the golden or yellow color, as hay/feed is usually green.
Another big difference is the weight, and straw is much, much lighter. That bale probably weighs 250 to 350 pounds vs around 1000 for hay (alfalfa).
Shockingly even the 1000 lb. Big boys will roll much easier than you would expect. At least every few years someone gets killed by one in my parts cuz they come smashing down a hill....
[–] 6 points7 points (0 children)
Yes
[–] 1334 points1335 points (55 children)
As fascinating as that was to watch, being a life long hard working farm girl, I would like to know why the adult who was jumping around was not helping push also. My guess this was just classic 'farm fun'.😁
[–][deleted] 671 points672 points (21 children)
Parents: "Do you want to go to the boring, tiresome shopping mall..."
"...OR DO YOU WANNA DO SOME SUPER COOL BALE WRAPPING?"
[–] 89 points90 points (9 children)
Having grown up on a farm I have little use for shopping but all the desire in the world for running a tractor in just about any capacity
[–] 68 points69 points (8 children)
I have worked in agriculture my whole life but the best job I ever had was sitting in a large tractor cultivating fields in spring. The sounds, smell, and ability to look back and see what I had accomplished...sublime.
[–] 15 points16 points (5 children)
Running the chopper in the fall is my favorite thing. Listening to the howl of the chopper and the smell of fresh corn silage. Ahhh. The best
[–] 9 points10 points (0 children)
The sweet smell of corn silage remains one of my favourite scents. Much less so on the other end of the cow, but the smell of corn silage coming out of the bunk silo or (when I was much younger) falling down the shaft from the old cement silo holds a special place in my brain.
[–] 28 points29 points (0 children)
Hahaha, sounds familiar!
[–] 47 points48 points (5 children)
And teaching 'you can do hard things' and 'team strength' life long lessons.
[–] 55 points56 points (2 children)
My bet is that he was doing other task as well and just got a free moment
[–] 99 points100 points (1 child)
Nope, sorry this is not the reddit way. Based on those 10.7 seconds of screen time you need to diagnose his entire personality, all of his previous actions off screen, tell a story about someone like him who made you feel irritated, predict his future, and make yourself and everyone reading your comment feel superior to him.
[–] 48 points49 points (12 children)
As someone who's never worked on a farm, could that guy have even done anything useful? Looks to me like the two who are there are more than strong enough to roll it already and adding a third person in the middle would get in the way more than help.
[–] 20 points21 points (5 children)
I’d bet they switch up on who drives the tractor.
When I was younger that’s how we took breaks when bringing in tobacco or hay. The guy on the tractor was resting and still doing something. Then you switch out every so often so everyone gets breaks and the process never stops for long. While switching out drivers everyone gets a drink, and sometimes changes positions.
[–] 10 points11 points (4 children)
That's how we did it, too. Hauling in 300 bales in 90-100 degree heat with NC's insane humidity was brutal. We had to move the hay trailer every few bales, so we'd take turns moving the truck as soon as we were tall enough to reach the pedals. Gave us a couple minutes in the shade and a chance to drink some water. And before putting the hay up in the even hotter barn, we'd jump in the pool fully clothed so the time spent in the barn wouldn't be so unbearable.
[–] 42 points43 points (1 child)
Agreed, but I think this was just farm fun aka wholesome hold my beer or let's see if this will work.
[–] 7 points8 points (0 children)
Also on any form of scale this makes no sense. Even wrapping bales for haylage in long tubes with the machines for it takes hours. This almost definitely is for fun.
[–] 5 points6 points (1 child)
You don’t need the tractor. That dude with all the energy could have just run around the hay bale with the plastic and gotten done even faster.
[–] 5 points6 points (0 children)
My guess would be that it used to be his job but now little sisters are big enough to help. I did they same dance the first time my son mowed the lawn for me.
[–][deleted] 250 points251 points (6 children)
He use to have three daughters……..
[–] 107 points108 points (0 children)
an hour ago. Last month it was six.
[–] 8 points9 points (0 children)
I had 2 different friends growing up that had brothers die in tractor accidents. A lot more common than you'd think.
Don't play around tractors and stay the fuck away from the PTO was our rule.
[–] 141 points142 points (0 children)
Girls gonna have dope legs and abs for the summer
[–] 39 points40 points (1 child)
Ok. Who else watched all 3 minutes of that?
[–] 8 points9 points (0 children)
i did lol, it was like i was in a trance
[–] 104 points105 points (4 children)
[–] 668 points669 points (159 children)
Is it really the cheap way? Seems like they’re using way more wrap than necessary.
[–] 234 points235 points (33 children)
Right, the plastic isn't just there for looks. This plastic is meant to be airtight and tight to cause anaerobic fermentation. The fermentation of hay produces silage. Silage is also used as fodder for animals and should have a higher yield of nutrients and forage quality.
We used this with hay bales to feed a large herd of goats. They definitely prefered silage.
[–][🍰] 77 points78 points (8 children)
Also there's sour silage and sweet silage which depend on the technique, grain types, and available wild/internal bacteria. Rye is a really good grain for silage. Overall I like to think of silage like dry and healthy farmhouse beer for cows!
[–] 61 points62 points (7 children)
I like to think of it as cow kraut
[–]Interested 50 points51 points (5 children)
cowerkraut
[–] 26 points27 points (4 children)
Like Hitler in his bunker
[–] 4 points5 points (0 children)
[–] 21 points22 points (11 children)
What did they do before plastic was invented?
[–] 56 points57 points (8 children)
They'd put it in a silo, and keep it airtight.
And way before that, they would bury it in a pit.
[–] 42 points43 points (3 children)
We still bury it in a pit. Then we cover the pit with very heavy plastic, so that we can reuse the plastic year after year. Same silage, way less plastic waste. The problem is, doing it our way means two guys have to do three hours of work outside, rather than just sitting in a tractor. But we do it in October, which is a great time of year to work outside. Slowly peeling back the plastic cover as you feed out of the pit takes about a half hour a week through feeding season, and admittedly, there are times in December and January where it’s not as much fun to work outside.
[–] 7 points8 points (2 children)
What did they do before silos were invented?
[–] 9 points10 points (0 children)
A lot of fires in hay lofts, which may or may not have killed the entire family living in the attached farmhouse.
[–] 7 points8 points (0 children)
Giant piles that occasionally caught fire.
[–] 6 points7 points (0 children)
This is definitely not being wrapped for silage. It would be far more green.
Hay is also wrapped once dried in rainy environs to keep moisture out of there's not indoor storage. Likely what this is since the hay is clearly dry.
[–] 312 points313 points (52 children)
Check out a video where they are using a normal machine to do the same, they use TONS of plastic, way more than this guy.
Edit: spelling
[–] 30 points31 points (6 children)
It seems like the plastic used to wrap it in that video is like, 2x the price of the hay it is wrapping
[–] 17 points18 points (1 child)
Hay can get pretty expensive and is quite dependent on weather. Currently in the Great Plains you'd probably see \$150-200 per ton. A round bale can be between 500-1200 pounds depending on size/density/etc. So each bale can run at least \$45-50 if not more.
[–] 4 points5 points (0 children)
That seems cheap considering all you have Todo to get that bale. For only \$50 a piece.
[–] 64 points65 points (39 children)
But they didn’t have to buy \$1 million machine to do it
[–] 108 points109 points (18 children)
That’s what I’m saying. They don’t have to buy the expensive machine AND they use less plastic compared to the machine.
[–] 11 points12 points (0 children)
small farmers don't buy the machine anyway. They rent it for a day
[–] 30 points31 points (14 children)
Bale wrappers don't cost a million dollars, you can buy them used for under \$10k, less than a couple minimum wage summer salaries (or hospital bill, etc)
Labour is way more costly than machines you can sell when the work is finished.
https://www.tractorhouse.com/listings/for-sale/bale-wrappers-hay-and-forage-equipment/1238
[–] 12 points13 points (5 children)
Labour* is way more costly than machines you can sell when the work is finished.
[–] 148 points149 points (2 children)
I'm guessing the plastic wrap is nothing compared to random ones opening up unexpectedly
[–] 86 points87 points (3 children)
By the cheaper way I'm assuming they mean without a bale wrapping machine. The plastic wrap is pennies by comparison to the bale itself.
[–] 11 points12 points (5 children)
It could be a \$50 bale with \$7 of wrap, but an automated wrapping machine operator will charge \$15
[–]Interested 12 points13 points (1 child)
Everything about this makes me think that this is a small-scale operation. Possibly a bit more on the "fun hobby farming" side than "this is our whole family's livelihood" kind of setup.
[–] 18 points19 points (3 children)
[–][deleted] 20 points21 points (22 children)
if they dont wrap it enough, the hay is going to ignite when it gets wet. this is the only solution if you dont have enough space in a barn to store the hay / straw, i hate the use of plastic here too but what are they gonna do otherwise
[–] 23 points24 points (0 children)
Seems like, but you'd be wrong. Any farm wanting to be economical and make as much money as possible will naturally use as little as needed, so these videos show you the exact amount needed. Not enough and damp, rot and insects will get in, ruining the bale, wasting the resources needed to produce it, water, energy etc and being far worse for the environment than just using the amount someone who "feels" is too much.
[–] 107 points108 points (7 children)
It's all fun and games until somebody gets their head run over.
[–] 11 points12 points (2 children)
I was going to comment this exact sentence lol
[–] 19 points20 points (1 child)
I've worked on farms and around tracked vehicles in the Army. Dangerous things to play with. That had to be the most unsafe, ludicrously stupid thing I've ever seen.
[–] 13 points14 points (0 children)
Totally. It’s just nuts, a slight mis-move and either a front or back wheel goes over their head, or they get wrapped into the bale. You don’t fuck about with farm machinery. I’ve seen the consequences of accidents with kids on farms far too often.
[–] 56 points57 points (3 children)
Don’t really know if I would be confortable with my kids head so close to a tractor tire.
[–] 53 points54 points (4 children)
All I can think about is one of them not ducking in time and being wrapped up. Become one with the bale, don't be scared.
[–] 174 points175 points (31 children)
this could go incredibly wrong.
[–][deleted] 23 points24 points (1 child)
I’ve watched too many r/idiotsincars to trust any sort of machinery/human interactions
[–] 6 points7 points (0 children)
Yeah, I’m sure he’s being very careful, but It’s one momentary lapse in concentration away from a double filicide
[–] 45 points46 points (7 children)
Aren't these bales heavy as fuck? Like "crush your pelvis and paralyze you" heavy as fuck?
[–] 93 points94 points (30 children)
What a hilarious video. I hope they don’t have a lot of bales to wrap coz it’s gonna take aaaages and they’ll use heeeeaps of plastic wrap! On the plus side, those girls will be getting some mean abs doing that lol
[–] 71 points72 points (21 children)
Bales are often double wrapped by the wrapping machine, so this is going to take ages but use less plastic than the machine, (piles and piles of plastic waste from fodder is a problem for farmers.
Edit: but… it’s better for the bales to be double wrapped, and you can still decide to do a single wrap with a machine and it’s takes 30-60 second.
[–] 13 points14 points (19 children)
Thanks for that, I didn’t realise the balers double wrapped the bales. So this is the more efficient way, just not time-wise :-).
Without sounding like too much of a greenie, it would be good to reduce the amount of plastic being used for hay bales worldwide (or using a more biodegradable product). Edit: sorry, I sounded completely like a greenie :(
[–] 21 points22 points (5 children)
Simply put if any air gets into the bale at all the bale with rot, the wrap cannot be biodegradable at the moment because it’s purpose is to seal the biodegradable bale for many years. I’m from a farm, plastic waste is a arguing point at home, not because of it’s use, but because it’s so hard to dispose of. The larger pieces are 95% collected and recycled but dozens of small pieces are created with the unwrapping (ie cutting open with a sharpish knife) or each bale and they end up buried deep in the soil. Hopefully someone comes up with a better solution.
[–] 19 points20 points (0 children)
Health and safety has entered the chat
[–] 21 points22 points (0 children)
And here we are drinking our beverages from paper straws
[–] 9 points10 points (1 child)
Don't forget to duck!
[–] 8 points9 points (0 children)
Never thought about it but I’m thinking farm girls have some insane abs!
[–] 4 points5 points (0 children)
It amazes me how many people really believe the word is spelled “hey” instead of “hay”… 😂
[–] 43 points44 points (16 children)
This looks immensely stupid and dangerous
[–] 12 points13 points (0 children)
maybe not cheap , gasoline price is up | 7,500 | 28,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-14 | latest | en | 0.902592 |
http://oeis.org/A142955 | 1,606,229,744,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176864.5/warc/CC-MAIN-20201124140942-20201124170942-00291.warc.gz | 70,225,986 | 4,462 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Please make a donation to keep the OEIS running. We are now in our 56th year. In the past year we added 10000 new sequences and reached almost 9000 citations (which often say "discovered thanks to the OEIS"). Other ways to donate
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A142955 Primes of the form 3*x^2+4*x*y-5*y^2 (as well as of the form 3*x^2+10*x*y+2*y^2). 1
2, 3, 19, 31, 59, 67, 71, 79, 103, 107, 127, 151, 167, 179, 211, 223, 227, 307, 331, 379, 383, 431, 439, 487, 523, 547, 563, 599, 607, 659, 683, 743, 751, 787, 811, 827, 839, 863, 887, 907, 911, 971, 983, 991, 1019, 1039, 1063, 1091, 1123, 1171, 1231, 1283 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Discriminant = 76. Class = 2. Binary quadratic forms a*x^2+b*x*y+c*y^2 have discriminant d=b^2-4ac and gcd(a,b,c)=1. REFERENCES Z. I. Borevich and I. R. Shafarevich, Number Theory. D. B. Zagier, Zetafunktionen und quadratische Koerper. LINKS EXAMPLE a(4)=31 because we can write 31=3*3^2+4*3*2-5*2^2 (or 31=3*1^2+10*1*2+2*2^2). CROSSREFS Cf. A142956 (d=76). A038872 (d=5). A038873 (d=8). A068228, A141123 (d=12). A038883 (d=13). A038889 (d=17). A141111, A141112 (d=65). Sequence in context: A265799 A058912 A040145 * A213896 A088790 A283186 Adjacent sequences: A142952 A142953 A142954 * A142956 A142957 A142958 KEYWORD nonn AUTHOR Laura Caballero Fernandez, Lourdes Calvo Moguer, Maria Josefa Cano Marquez, Oscar Jesus Falcon Ganfornina and Sergio Garrido Morales (laucabfer(AT)alum.us.es), Jul 14 2008 EXTENSIONS More terms from Colin Barker, Apr 05 2015 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified November 24 07:53 EST 2020. Contains 338607 sequences. (Running on oeis4.) | 748 | 2,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-50 | latest | en | 0.620889 |
http://mathhelpforum.com/advanced-algebra/63919-equivalence-class-mod-problem.html | 1,481,227,104,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542648.35/warc/CC-MAIN-20161202170902-00481-ip-10-31-129-80.ec2.internal.warc.gz | 173,207,615 | 10,655 | # Thread: Equivalence class mod problem
1. ## Equivalence class mod problem
Let $G = \frac { \mathbb {Q} } { \mathbb {Z} }$ under +, so the elements are the equivalence classes $\hat {r} = \{ s \in \mathbb {Q} : s-r \in \mathbb {Z} \}$. Write $r \equiv s \ (mod \ 1 )$ if $r - s \in \mathbb {Z}$.
Prove that if $r = \frac {a}{b}$ with $a,b \in \mathbb {Z}$, $gcd (a,b) = 1$ and $b > 0$, then $\hat {r}$ has order b and $< \hat {r} > = < \hat { \frac {1}{b} } >$
Proof so far.
So the operation here is addition, $\frac {a}{b} (b) = 1$, so would that prove the order is b?
$< \frac {1}{b} > = \{ \frac {1}{b}, \frac {2}{b}, \frac {3}{b}, ..., \frac {b}{b}=1 \}$, which is the same as $< r >$
Is this correct? Thanks.
2. Originally Posted by tttcomrader
Let $G = \frac { \mathbb {Q} } { \mathbb {Z} }$ under +, so the elements are the equivalence classes $\hat {r} = \{ s \in \mathbb {Q} : s-r \in \mathbb {Z} \}$. Write $r \equiv s \ (mod \ 1 )$ if $r - s \in \mathbb {Z}$.
Prove that if $r = \frac {a}{b}$ with $a,b \in \mathbb {Z}$, $gcd (a,b) = 1$ and $b > 0$, then $\hat {r}$ has order b and $< \hat {r} > = < \hat { \frac {1}{b} } >$
We will use $[ r ]$ instead of $\hat r$.
To show that $\left< [\tfrac{a}{b}] \right> = \left< [\tfrac{1}{b}] \right>$ we will show $\left< [\tfrac{a}{b}] \right> \subseteq \left< [\tfrac{1}{b}] \right> \text{ and }\left< [\tfrac{1}{b}] \right>\subseteq \left< [\tfrac{a}{b}] \right>$.
The first inclusion follows because $a( \tfrac{1}{b} ) = \tfrac{a}{b}$ and therefore $[ \tfrac{a}{b} ] \in \left< [ \tfrac{1}{b} ] \right>$
For the second inclusion pick $c\in \mathbb{Z}$ so that $ac\equiv 1(\bmod b)$, this is possible because $\gcd(a,b)=1$.
Then it follows that, $c(\tfrac{a}{b}) = \tfrac{ac}{b} \equiv \tfrac{1}{b}$ and therefore $[\tfrac{1}{b}] \in \left< [ \tfrac{a}{b} ] \right>$.
To show that the order of $[\tfrac{1}{b}]$ is $b$ we need to find the smallest $k\in \mathbb{Z}^+$ such that $k[\tfrac{1}{b}] = [1] \implies [\tfrac{k}{b}] = [1]$.
Say that $0 < k < b$ with $d=\gcd(k,b)$ then $[\tfrac{k}{b}] = [ \tfrac{1}{b/d} ] \not = [1]$.
Therefore the order is at least $b$. But if $k=b$ we get $b[ \tfrac{1}{b}] = [\tfrac{b}{b}]=[1]$.
Thus, the order is $b$. | 904 | 2,218 | {"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": 45, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.658223 |
milujusteak.cz | 1,632,465,326,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057504.60/warc/CC-MAIN-20210924050055-20210924080055-00206.warc.gz | 445,899,951 | 7,147 | how many kilograms of steel bar in cubic meter-Yantai Rhyther Mining Machinery Co.,Ltd
# how many kilograms of steel bar in cubic meter
Steel volume to weight conversion - Aqua-Calc 1 cubic meter of steel weighs 7 900 kilograms kg 1 cubic foot of steel weighs 493.18089 pounds lbs Steel weighs 7.9 gram per cubic centimeter or 7 900 kilogram per cubic meter i.e. density of steel is equal to 7 900 kg/m& 179;. In Imperial or US customary
Reinforcing Deformed Steel Bar: Size and Weight Comparison The deformed Reinforcing Steel Bar is supplied in length of 9m or 12 m as common sizes. The steel bar diameter applied is different and accordingly the weight varies. We are listing theoretical unit weight list of commonly used wire gauge below for your reference.
Steel Round Bar Weight Calculator SS Round Bar Calculator Steel Round Bar Weight Calculator Free Online Carbon Steel Round Bar and Cast Iron Round Bar Weight Calculator Online Weight Calculator of Stainless Steel Titanium Alloy Steel Inconel Alloy 718 Alloy A286 Hastelloy Copper Nickel Duplex etc.
Kilograms per cubic meter to kilograms per cubic … How to convert kilograms per cubic meter to kilograms per cubic millimeter kg/m& 179; to kg/mm& 179; :ρ kg/mm& 179; = 1.0 & 215; 10-9 & 215; ρ kg/m& 179; How many kilograms per cubic millimeter in a kilogram per cubic meter: If ρ kg/m& 179; = 1 thenρ kg/mm& 179; = 1.0 & 215; 10-9 & 215; 1 = 1.0 & 215; 10-9 kg/mm& 179;
For one meter cube of concrete how much kg steel is … its depend on dia of bar used in concrete for example the weight of steel of one meter rod having six mm dia is two hundrerd and twenty grams and ten mm dia is six hundrerd and twenty grams etc you can find weight of different bars having different diameters by this formula = d*d /one hundrerd sixty two
How to Calculate the Unit Weight of Steel Bars - A Civil … Unit Weight of Steel Bar When Length is in Meter If you put 1-meter length for each diameter of steel bars in the formula then you’ll get the unit weight. Let’s see. W = D& 178;L/162 Unit weight of 10mm& 248; bar = 10& 178; x 1/162 = 0.617 kg/m 12mm& 248; bar = 12& 178; x 1/162 = 0.888 kg/m
How many kilograms of steel.bar in cubic meter How many kilograms of steel.bar in cubic meter Products As a leading global manufacturer of crushing grinding and mining equipments we offer advanced reasonable solutions for any size-reduction requirements including How many kilograms of steel.bar in cubic meter quarry aggregate and different kinds of minerals.
Conversion: Convert kg to m3 The cubic meters is an SI accepted metric system unit of volume equal to 1000 cubic decimetre dm3 or liters 1000000 cubic centimetres cm3 or ml . A cubic meters occupies a volume of 1& 215;1& 215;1 metres. How many cubic meter are in a kilograms? The ratio
Steel Tools - Conversion Tables All steel tools conversion tables and metal weight calculators in one place. Calculates steel weights hardness torque force energy. cubic meter cu m 0.00378541 fluid ounce fl oz milliliters ml 29.57353 fluid ounce fl oz cubic meter cu m 0.00002957
how many steel is required for 1 cubic meter of concrete Calculate how many kilograms kg - kilo of concrete are in 1 cubic meter 1 m3 . Specific unit weight of concrete - amount properties converter for conversion factor exchange from 1 cubic meter m3 equals = 2406.53 kilograms kg - kilo exactly for the masonry material type.
weight of rebar steel per cubic meter How many kg of steel reinforcement in one cubic meter concrete. It all depends on the type and sizes of the steel used. It also depends on the load that the concrete has to take at certain points. 7865kg is the weight of a cubic meter of steel. Read More
Metric ton per cubic meter t/m3 Conversion - Density … slug per cubic foot kilogram per liter pound per cubic inch ounce per cubic inch ounce per Imperial gallon kilogram per cubic meter gram per milliliter pound per Imperial gallon 1 0.5153788 0.0186192 0.2979079 82.6454286 515.3788171 0.5153788 5.1653393 1
average weight of steel in one cubic meter of concrete A typical concrete mix weighs 150 lbs per cubic foot 4050 lbs per cubic yard or 2400 kg per cubic meter. The weight of concrete is determined by its density which can vary based on the amount of aggregate water and air in the mix. Determining the Density of
Bar/ kilogram/cubic meter Conversion Table - Chemical - … Chemical - Henry's Law Conversion table and factors: bar/ kilogram/cubic meter Quantity Reference Unit is equal to Conversion Factor Unit 1 bar/ kilogram/cubic meter = 986.92326671601 atmosphere/ gram/cubic centimeter
Cubic Meters ⇄ Kilograms Conversion - NinjaUnits How many kilograms are in one cubic meter of water? 1 kg wt. = 0.001 m 3 / cu m Definition of cubic meters of water provided by WikiPedia The cubic meter in American English or cubic metre in British English is the derived unit of volume. Its symbol is m 3 5/5 1
How much reinforcement steel in 1 cubic meter of concrete How much reinforcement steel in 1 cubic meter of concrete Products As a leading global manufacturer of crushing grinding and mining equipments we offer advanced reasonable solutions for any size-reduction requirements including How much reinforcement steel in 1 cubic meter of concrete quarry aggregate and different kinds of minerals.
Cubic Meters ⇄ Kilograms Conversion - NinjaUnits How many kilograms are in one cubic meter of water? 1 kg wt. = 0.001 m 3 / cu m Definition of cubic meters of water provided by WikiPedia The cubic meter in American English or cubic metre in British English is the derived unit of volume. Its symbol is m 3
how much does 1 cubic meter of average steel weight in … 20/12/2007& 0183;& 32;The Chemical Engineers' Handbook Perry and Chilton lists 98 different steel alloys there are undoubtedly many more ranging in specific gravity from 7.0 Durichlor 51 to 9.24 Chlorimet 3 . A cubic meter of the lighter would weigh 7000 kg while the same volume of …
Convert cubic meter to kilo gram - Conversion of … Do a quick conversion: 1 cubic meters = 1000 kilograms using the online calculator for metric conversions. Check the chart for more details. More information from the unit converter How many cubic meter in 1 kilo gram? The answer is 0.001. We assume you are
How many kg of steel in 1 cubic meter? - Answers Depending on the type of steel you are referring to between 7450 - 8000 kilograms per cubic metre.
How many kg of steel in 1 cubic meter? - Quora Unit weight of steel is 7850kg/M3 If you are talking on Weight of steel per Cum of concrete means it depends on different type of structural elements. For example For a Footing minimum steel requirement as per thumb rule is 0.6kg/Sqm SimilarlyHow much steel is required for 1 cubic meter of RCC?What is quantity of steel required in 1 meter cube of concrete?How many kilogram of metal is equal to 1 cubic meter?How many kg is in 1 meter of 6mm steel?查看更多结果
How to Calculate the Weight of Steel Bar? - Online … Steel Bar Weight Calculation By Calculation method These are some examples of site level calculation which are not accurate but best to judge the approximate. For MS Flat Bar For 1-meter length of Mild Steel bar of size 40 mm width and 20 mm breadth
How much does 1 cubic meter of average steel weight in … steel weight to be 7850 kg per cubic meter About Bayt.com Bayt.com is the leading job site in the Middle East and North Africa connecting job seekers with employers looking to hire.
many steel bar kilograme in concrete 1 squear meter ? - … how many kgs. of steel reinforcement per cubic meter of … Estimate steel reinforced concrete. How many kilograms of … and the Steel … of a 1-meter length of the bar. … round/square portion where there … & 187;More detailed
how must steel will use for 1 cubic meter.formula of steel … What is the lap length for tension members if you have 16 dia steel bar and a concrete fc=30.9 and steel grade60 fy=414 mpa 4 Answers L and T Suppose you planning to paint newly fabri ed workshop building made of structural steel of 10 tons total mass.
How To Calculate Weight Of Steel Bar Learning … 在必应上点击查看5:399/12/2016& 0183;& 32;How To Calculate Weight Of Steel Bar In this video I will teach you about the calculation of the Mild steel bar in a simple way. Official Website in English www.learningtechnologyofficia L and T - Learning Technology
Weight of common engineering metals in kilograms per … Find the weight of a steel plate 2m x 1.5m x 10mm thick. 1 Volume in cubic meters is 2 x 1.5 x 0.01 = 0.03 m 3 2 Volume x density 0.03 x 7850 = 235.5 kg Home Products & 187; Engineering Steels Stainless non ferrous cast iron Tool Steels Mild Steels
Convert kilo gram to cubic metre - Conversion of … How many kilo gram in 1 cubic metre? The answer is 1000. We assume you are converting between kilogram water and cubic metre. You can view more details on each measurement unit: kilo gram or cubic metre The SI derived unit for volume is the cubic meter.
How many kg of steel reinforcement in one cubic meter … It all depends on the type and sizes of the steel used. It also depends on the load that the concrete has to take at certain points. 7865kg is the weight of a cubic meter of steel.
Weight of steel bars per meter - Weight of steel bars … Weight of steel bars per meter are: 6 mm weighs 0.22 kg/meter 10 mm weighs 0.62 kg/meter 12 mm weighs 0.89 kg/meter 16 mm weighs 1.58 kg per meter D is in millimeter. L is the total length of steel bars of which weight is to be calculated. For example: if we
Metal Weight Calculator - steel weight calculator … Metal weight calculator online - free steel weight calculator. Has pre-entered densities for dozens of commonly-used metals and metal alloys like steel aluminum nickel iron copper cadmium gold silver etc. Calculate the weight of a steel beam bar tube
weight of rebar per cubic meter of concrete How many kg of steel reinforcement in one cubic meter Oct 27 2009& 183; It also depends on the load that the concrete has to take at certain points. 7865kg is the weight of a cubic meter of steel. per cubic meter of concrete rebar for the concrete.
Before Article: flow sheet for coke calcining system
Next Article: bernoulli backflush filter for high viscosity liquid
Home»Crusher Machine>how many kilograms of steel bar in cubic meter | 2,505 | 10,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-39 | latest | en | 0.747414 |
https://training.incf.org/search?f%5B0%5D=topics%3A22&f%5B1%5D=topics%3A34&f%5B2%5D=topics%3A61&f%5B3%5D=topics%3A67&%3Bf%5B1%5D=topics%3A64 | 1,566,757,315,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330786.8/warc/CC-MAIN-20190825173827-20190825195827-00196.warc.gz | 667,973,819 | 13,334 | Lesson type
Lesson title:
The probability of a hypothesis, given data.
Difficulty level: Beginner
Duration: 7:57
Speaker: : Barton Poulson
Lesson title:
Why math is useful in data science.
Difficulty level: Beginner
Duration: 1:35
Speaker: : Barton Poulson
Lesson title:
Why statistics are useful for data science.
Difficulty level: Beginner
Duration: 4:01
Speaker: : Barton Poulson
Lesson title:
Statistics is exploring data.
Difficulty level: Beginner
Duration: 2:23
Speaker: : Barton Poulson
Lesson title:
Graphical data exploration
Difficulty level: Beginner
Duration: 8:01
Speaker: : Barton Poulson
Lesson title:
Numerical data exploration
Difficulty level: Beginner
Duration: 5:05
Speaker: : Barton Poulson
Lesson title:
Simple description of statistical data.
Difficulty level: Beginner
Duration: 10:16
Speaker: : Barton Poulson
Lesson title:
Basics of hypothesis testing.
Difficulty level: Beginner
Duration: 06:04
Speaker: : Barton Poulson
In this lecture, the speaker demonstrates Neurokernel's module interfacing feature by using it to integrate independently developed models of olfactory and vision LPUs based upon experimentally obtained connectivity information.
Difficulty level: Intermediate
Duration: 29:56
Speaker: : Aurel A. Lazar
Lesson title:
This lecture covers an introduction to neuroinformatics and its subfields, the content of the short course and future neuroinformatics applications.
Difficulty level: Beginner
Duration: 34:27
Lesson title:
Introduction to the types of glial cells, homeostasis (influence of cerebral blood flow and influence on neurons), insulation and protection of axons (myelin sheath; nodes of Ranvier), microglia and reactions of the CNS to injury.
Difficulty level: Beginner
Duration: 40:32
Lesson title:
This lecture focuses on how the immune system can target and attack the nervous system to produce autoimmune responses that may result in diseases such as multiple sclerosis, neuromyelitis and lupus cerebritis manifested by motor, sensory, and cognitive impairments. Despite the fact that the brain is an immune-privileged site, autoreactive lymphocytes producing proinflammatory cytokines can cause active brain inflammation, leading to myelin and axonal loss.
Difficulty level: Beginner
Duration: 37:36
Speaker: : Anat Achiron | 516 | 2,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-35 | latest | en | 0.771205 |
https://www.jobilize.com/online/course/0-6-solution-of-the-partial-differential-equations-by-openstax?qcr=www.quizover.com&page=7 | 1,620,492,951,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988882.94/warc/CC-MAIN-20210508151721-20210508181721-00535.warc.gz | 884,871,200 | 21,654 | # 0.6 Solution of the partial differential equations (Page 8/13)
Page 8 / 13
$\begin{array}{c}\frac{{\partial }^{2}\varphi }{\partial {x}^{2}}+\frac{{\partial }^{2}\varphi }{\partial {y}^{2}}=0\hfill \\ \frac{{\partial }^{2}\psi }{\partial {x}^{2}}+\frac{{\partial }^{2}\psi }{\partial {y}^{2}}=0\hfill \end{array}$
The Cauchy-Riemann conditions also imply that the gradient of the velocity potential and the stream function are orthogonal.
$\left(\nabla \varphi \right)•\left(\nabla \psi \right)=\frac{\partial \varphi }{\partial x}\phantom{\rule{0.166667em}{0ex}}\frac{\partial \psi }{\partial x}+\frac{\partial \varphi }{\partial y}\phantom{\rule{0.166667em}{0ex}}\frac{\partial \psi }{\partial y}=0$
If the gradients are orthogonal then the equipotential lines and the streamlines are also orthogonal, with the exception of stagnation points where the velocity is zero.
Since the derivative
$\frac{dw}{dz}=\begin{array}{c}lim\\ \left|\delta z\to 0\right|\end{array}\phantom{\rule{0.277778em}{0ex}}\frac{\delta \phantom{\rule{0.166667em}{0ex}}w}{\delta \phantom{\rule{0.166667em}{0ex}}z}$
is independent of the direction of the differential $\delta z$ in the $\left(x,y\right)$ -plane, we may imagine the limit to be taken with $\delta z$ remaining parallel to the x-axis $\left(\delta z=\delta x\right)$ giving
$\frac{dw}{dz}=\frac{\partial \varphi }{\partial x}+i\phantom{\rule{0.166667em}{0ex}}\frac{\partial \psi }{\partial x}={v}_{x}-i\phantom{\rule{0.166667em}{0ex}}{v}_{y}$
Now choosing z to be parallel to the y-axis $\left(\delta z=i\delta y\right)$ ,
$\frac{dw}{dz}=\frac{1}{i}\phantom{\rule{0.166667em}{0ex}}\frac{\partial \varphi }{\partial y}+\frac{\partial \psi }{\partial y}=-i\phantom{\rule{0.166667em}{0ex}}{v}_{y}+{v}_{x}={v}_{x}-i\phantom{\rule{0.166667em}{0ex}}{v}_{y}$
These equations are a restatement that an analytical function has a derivative defined in the complex plane. Moreover, we see that the real part of ${w}^{\text{'}}\left(z\right)$ is equal to ${v}_{x}$ and the imaginary part of ${w}^{\text{'}}\left(z\right)$ is equal to $-{v}_{y}$ . If v is written for the magnitude of $\mathbf{v}$ and for the angle between the direction of $\mathbf{v}$ and the $x$ -axis, the expression for $dw/dz$ becomes
$\begin{array}{c}\frac{dw}{dz}={v}_{x}-i\phantom{\rule{0.166667em}{0ex}}{v}_{y}=v\phantom{\rule{0.166667em}{0ex}}{e}^{-i\theta }\hfill \\ \mathrm{or}\hfill \\ {v}_{x}=\mathrm{real}\left[\frac{dw}{dz}\right]\hfill \\ {v}_{y}=-\mathrm{imaginary}\left[\frac{dw}{dz}\right]\hfill \end{array}$
Flow Fields. The simplest flow field that we can imagine is just a constant translation, $w=\left(U-iV\right)z$ where $U$ and $V$ are real constants. The components of the velocity vector can be determined from the differential.
$\begin{array}{c}w\left(z\right)=\left(U-iV\right)z=\left(U-iV\right)\left(x+i\phantom{\rule{0.166667em}{0ex}}y\right)=Ux+Vy+i\left(-Vx+Uy\right)=\varphi +i\phantom{\rule{0.166667em}{0ex}}\psi \hfill \\ \frac{dw}{dz}=\left(U-iV\right)={v}_{x}-i\phantom{\rule{0.166667em}{0ex}}{v}_{y}\hfill \\ {v}_{x}=U,\phantom{\rule{1.em}{0ex}}{v}_{y}=V\hfill \\ \varphi =Ux+Vy,\phantom{\rule{1.em}{0ex}}\psi =-Vx+Uy\hfill \end{array}$
Another simple function that is analytical with the exception at the origin is
$\begin{array}{ccc}\hfill w\left(z\right)& =& A\phantom{\rule{0.166667em}{0ex}}{z}^{n}=A\phantom{\rule{0.166667em}{0ex}}{r}^{n}{e}^{i\phantom{\rule{0.166667em}{0ex}}n\phantom{\rule{0.166667em}{0ex}}\theta }\hfill \\ & =& A\phantom{\rule{0.166667em}{0ex}}{r}^{n}cos\phantom{\rule{0.166667em}{0ex}}n\theta +i\phantom{\rule{0.166667em}{0ex}}A\phantom{\rule{0.166667em}{0ex}}{r}^{n}sin\phantom{\rule{0.166667em}{0ex}}n\theta \hfill \\ & =& \varphi +i\phantom{\rule{0.166667em}{0ex}}\psi \hfill \\ & & \mathrm{thus}\hfill \\ \hfill \varphi & =& A\phantom{\rule{0.166667em}{0ex}}{r}^{n}cos\phantom{\rule{0.166667em}{0ex}}n\theta \hfill \\ \hfill \psi & =& A\phantom{\rule{0.166667em}{0ex}}{r}^{n}sin\phantom{\rule{0.166667em}{0ex}}n\theta \hfill \\ \hfill \frac{dw}{dz}& =& n\phantom{\rule{0.166667em}{0ex}}A\phantom{\rule{0.166667em}{0ex}}{z}^{n-1}={v}_{x}-i\phantom{\rule{0.166667em}{0ex}}{v}_{y}\hfill \end{array}$
where $A$ and $n$ are real constants. The boundary condition at stationary solid surfaces for irrotational flow is that the normal component of velocity is zero or the surface coincides with a streamline. The expression above for the stream function is zero for all $r$ when $\theta =0$ and when $\theta =\pi /n$ . Thus these equations describe the flow between these boundaries are illustrated below.
Earlier we discussed the Green's function solution of a line source in two dimensions. The same solution can be found in the complex domain. A function that is analytical everywhere except the singularity at ${z}_{o}$ is the function for a line source of strength $m$ .
$\begin{array}{c}w\left(z\right)=\frac{m}{2\pi }ln\left(z-{z}_{o}\right),\phantom{\rule{1.em}{0ex}}\mathrm{line}\phantom{\rule{0.277778em}{0ex}}\mathrm{source}\hfill \\ \frac{\mathrm{dw}}{\mathrm{dz}}=\frac{m}{2\pi }\frac{1}{\left(z-{z}_{o}\right)}={v}_{x}-i\phantom{\rule{0.166667em}{0ex}}{v}_{y}\hfill \end{array}$
This results can be generalized to multiple line sources or sinks by superposition of solutions. A special case is that of a source and sink of the same magnitude.
$\begin{array}{c}w\left(z\right)=\sum _{i}\frac{{m}_{i}}{2\pi }ln\left(z-{z}_{i}\right),\phantom{\rule{1.em}{0ex}}\mathrm{multiple}\phantom{\rule{0.277778em}{0ex}}\mathrm{line}\phantom{\rule{0.277778em}{0ex}}\mathrm{sources}\hfill \\ w\left(z\right)=\frac{m}{2\pi }ln\left(\frac{z-{z}_{o}}{z+{z}_{o}}\right),\phantom{\rule{1.em}{0ex}}\mathrm{source}-\mathrm{sink}\phantom{\rule{0.277778em}{0ex}}\mathrm{pair}\hfill \end{array}$
The above flow fields can be viewed with the MATLAB code corner.m, linesource.m, and multiple.m in the complex subdirectory.
## Assignment 7.5
Line Source Solution For ${z}_{o}$ at the origin, derive expressions for the flow potential, stream function, components of velocity, and magnitude of velocity for the solution to the line source in terms of $r$ and $\theta$ . Plot the flow potentials and stream functions. Compute and plot the flow potentials and stream function for the superposition of multiple line sources corresponding to the zero flux boundary conditions at $y=+1$ and $-1$ of the earlier assignment.
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
Got questions? Join the online conversation and get instant answers! | 2,937 | 9,434 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 39, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.598322 |
https://www.scribd.com/document/38703898/PF-Assignments | 1,566,807,286,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027331228.13/warc/CC-MAIN-20190826064622-20190826090622-00260.warc.gz | 977,839,990 | 56,833 | You are on page 1of 7
# Education and Research Department
June 2004
## Document No. Authorized By Ver. Signature / Date
Revision
ER/CORP/CRS/LA06/004 Dr. M P Ravindra Ver. 1.1
COMPANY CONFIDENTIAL
Infosys Document Revision History
## Document Revision History
Ver.
Date Author(s) Reviewer(s) Description
Revision
0.0a June 2004 Heena Mehta Sheetal Kale, Draft Version
Dinesh Anantwar,
Dharini V
Dinesh Anantwar,
Dharini V
Contents
## Assignment Day 1............................................................................................................... 1
Assignment Day 2............................................................................................................... 2
Assignment Day 3............................................................................................................... 3
Assignment Day 4............................................................................................................... 4
## ER/CORP/CRS/LA06/004 Version No. 1.1 iii
Infosys Assignment for Programming Fundamentals
Assignment Day 1
1. Write a program to find whether the number entered by the user is a prime number or
not. Extend this program to list all the prime numbers between two given numbers.
2. Do the following for the user-entered number of students.
Find the average marks for a student of his marks in 3 subjects. Print whether he
passed or failed. A student will fail if his average is less than 50. Use for loop
3. Do the following for an unknown number of students. (User will explicitly indicate
when to terminate). Find the average marks for a student of his marks in 3 subjects.
Print whether he passed or failed. A student will fail if his average is less than 50
Use While loop.
4. Write a program, that accepts a integer from the user and print the integer with
reverse digits. For eg: rev(1234) = 4321.
5. The librarian in a library wants an application that will calculate the due date for a
book given the issue date. The no. of days in which the book is due can be decided by
the librarian at the time of issuing a book. For e.g. If the librarian enters the current
date as 14-01-99 and the no of days in which the book is due as 15, then your
program should calculate the due date and give the output as 29-01-99.
6. Find the sum of the digits of a given number.
7. Given a number, determine its absolute value.
8. Given three numbers, determine whether they can form the sides of a triangle.
9. Given a number, determine whether it is a valid year and if so, whether it is a leap
year.
10. Given two numbers, determine whether they are valid years, and if so, list all leap
years between the two years (both included).
## ER/CORP/CRS/LA06/004 Version No. 1.1 1
Infosys Assignment for Programming Fundamentals
Assignment Day 2
For the following assignments, write down the prototypes for the functions used before
writing the functions.
1. Write a program to find the nearest smaller prime number for a given integer; use a
function to decide whether a number is prime or not.
2. Write a program to convert a number in decimal system to binary system. Hint: use
recursion.
3. Write a program that takes a positive integer as input and outputs the Fibonacci
sequence up to that number.
4. Given a function with the following prototype
int date_diff (int d1, int m1, int d2, int m2, int yr)
Where d1 and m1 are the day and month of the first date and d2 & m2 are the day and
month for second date and yr is the year to which both the dates belong. The function
returns the difference between the two dates that fall in different year. Write a
program to find the difference between two dates that do not fall in the same year.
5. Write a program to print whether the number entered is a prime/odd. Use functions
6. Write a program that accepts input of a number of seconds, validates it and outputs
the equivalent number of hours, minutes and seconds.
7. Write a program that can either add or multiply two fractions. The two fractions and
the operation to be performed are taken as input and the result is displayed as output.
8. Write a recursive function to compute the factorial to a given number. Use the
function to write a program which will generate a table of factorials of numbers
ranging from 1 to m where m is the number entered by the user.
9. Write a program which to print the multiplication table from 1 to m for n where m, n
is the values entered by the user.
10. Develop a calculator program which will use functions to add / subtract / divide /
multiply 2 numbers. Code the functions as different source files and link them.
## ER/CORP/CRS/LA06/004 Version No. 1.1 2
Infosys Assignment for Programming Fundamentals
Assignment Day 3
Note: In all the below problems, use and define as many as functions as possible.
1. Write a function, which checks whether one string is a sub-string of another string.
2. Write a program that accepts a sentence and returns the sentence with all the extra
spaces trimmed off. (In a sentence, words need to be separated by only one space; if
any two words are separated by more than one space, remove extra spaces)
3. Write a program, which checks for duplicate string in an array of strings.
4. Write functions to insert and delete a string from an array of strings. Write a program
that displays a menu to the user
a) Insert String
b) Delete Strings
c) Exit
Depending on the user choice the program will call functions that will insert / delete a
string from an array of strings
5. Write a program that will accept a string and character to search. The program will
call a function, which will search for the occurrence position of the character in the
string and return its position. Function should return –1 if the character is not found in
the input string.
6. Write a function, which prints a given number in words.
7. Write an program which will set the array element a[i] to 1 if i is prime, and to 0 if i is
not prime. Assume the array size to be 10000.
8. Write a program to count the number of vowels in a given string.
9. Write a program to obtain the transpose of a 4*4 array. The transpose is obtained by
exchanging the elements of each row with the elements of the corresponding column
10. Write a program which allow to perform any of the following operations on two 3*3
arrays
Multiply Arrays
Subtract Arrays
## ER/CORP/CRS/LA06/004 Version No. 1.1 3
Infosys Assignment for Programming Fundamentals
Assignment Day 4
## 1. Write a function for linear search.
2. Write a function for binary search.
3. Write a function to find whether a string exists in an array of strings. This function
should do a partial search for the string.
4. Write a program to sort 3*3 array elements (row wise)
5. Write a program that accepts up to 10 words and outputs them in dictionary order.
(Hint: sort an array of numbers that represent the words; don’t directly sort the
words.) Do NOT use the sort functions provided in the standard library. Here is a
typical user session:
Enter words: Damn Cold Bothers All Animals
Sorted words: All Animals Bothers Cold Damn
More? (Y/N) N
6. Write a program to count number of words in a given text file.
7. Write a program to check print the highest scorer.
Student details are stored in file in the format as below: -
StudNo 4 chars
StudName 20 chars
Marks 2 digits
8. Write a program to update the marks of a given student Accept the student number
and new marks from the user. Use the same student file of Q7.
9. Write a program to print student details from the student file
10. Write a program to print records randomly. Accept a record number from the user and
print the record. Use the same student file of Q7. | 1,791 | 7,695 | {"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} | 2.96875 | 3 | CC-MAIN-2019-35 | latest | en | 0.785239 |
https://www.coursehero.com/tutors-problems/Physics/9245111-For-a-100-%25CE%25BCC-charge-into-a-region-with-electric-field-E-600x10-3/ | 1,531,760,220,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00457.warc.gz | 887,890,874 | 22,702 | View the step-by-step solution to:
# For a 10.0 C charge into a region with electric field E = 6.00x10 3 N/C i find change in electric potential energy when: a) the charge is moved 8.
For a 10.0 μC charge into a region with electric field E = 6.00x103 N/C i find change in electric potential energy when: a) the charge is moved 8.00 cm in the positive x-direction, b) the charge is moved 8.00 cm in the negative x-direction, c) the charge is moved 8.00 cm in the positive y-direction.
Recall that W = - ΔU then, the change in electric potential energy is ΔU = - W ΔU = - F E . Δ r ΔU = - q E . Δ r hence... View the full answer
### Why Join Course Hero?
Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.
### -
Educational Resources
• ### -
Study Documents
Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.
Browse Documents | 261 | 1,012 | {"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} | 2.859375 | 3 | CC-MAIN-2018-30 | longest | en | 0.917457 |
https://www.koofers.com/files/quiz-ggg1gwdhzy/ | 1,568,587,020,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572436.52/warc/CC-MAIN-20190915215643-20190916001643-00474.warc.gz | 905,686,189 | 12,508 | # Quiz for MTH 121 - Calc with Analytic Geom I with Bannon at Essex County College (ESSEX)
## Quiz Information
Material Type: Quiz 7 Professor: Bannon Class: MTH 121 - Calc with Analytic Geom I Subject: Mathematics University: Essex County College Term: Summer I 2005 Keywords: Final AnswersPoints Extra CreditYou Have...Extra CreditMathematics
## Sample Document Text
MTH 121 - Summer - 2005 Essex County College - Division of Mathematics Quiz # 71 - Printed June 5, 2005 Name: Signature: Show all work clearly and in order, and box your final answers . Justify your answers alge- braically whenever possible. You have 30 minutes to take this 22 point quiz (12 points extra credit that will be added to your overall quiz score). When you do use your calculator, sketch all relevant graphs and write down all relevant mathematics. 1. Given: f (x) = x 2 + 7x + 3 x2 = 1 + 7 x + 3 x2 fprime (x) = ?7x + 6x3 fprimeprime (x) = 14x + 18x4 Figure 1: Graph of f (x) on the interval [?20, 20]. Answer the following questions. (a) x-intercept(s): 2 points (b) y-intercept(s): 2 points 1 (c) vertical asymptote(s): 2 points (d) horizontal asymptote(s): 2 points (e) domain: 2 points (f) range: 2 points (g) local maximum(s): 2 points (h) local minimum(s): 2 points (i) global maximum(s): 2 points (j) global minimum(s): 2 points (k) point(s) of inflection: 2 points 1Thi... | 401 | 1,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-39 | latest | en | 0.780301 |
http://blog.crossjoin.co.uk/2014/03/20/pretty-patterns-with-power-query-and-power-map/ | 1,455,323,011,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701165484.60/warc/CC-MAIN-20160205193925-00039-ip-10-236-182-209.ec2.internal.warc.gz | 22,127,665 | 19,786 | # Pretty Patterns With Power Query And Power Map
Here’s something with no practical use whatsoever. Today, after I finished writing the first draft of the chapter on M of my upcoming Power Query book, I got thinking about how Power View and Power Map get all the attention because of all the eye-catching demos you can create with them. And then I thought – why bother spending time finding real data for these demos when you can generate artificial data in Power Query to create patterns? So I got to work…
As you probably know, you can create animated charts in Power Map so long as you have date-based data. I therefore created a function in Power Query to draw a circle as a series of points on a graph where each point is associated with a date; I also added data for height and colour for each point. Here’s the function definition:
`let`
` //declare function to draw a circle`
` CircleFunction = (CircleRadius as number, StartDate as date, Reverse as logical) =>`
`let`
` //set the radius`
` radius = CircleRadius,`
` //create a list of numbers from 0 to 359`
` anglelist = List.Numbers(0, 359, 1),`
` //function to convert degrees to radians`
` radians = (a) => (a * 2 * Number.PI)/360,`
` //create a list of 360 dates starting from the start date`
` unordereddatelist = List.Dates(StartDate, 360,#duration(1,0,0,0)),`
` //reverse the list of dates if the Reverse parameter is True`
` datelist = if Reverse then List.Reverse(unordereddatelist) else unordereddatelist,`
` //generate the list of points on the graph, one for each angle and date`
` positionlist = List.Transform(anglelist, each `
` {_, datelist{_}, Number.Cos(radians(_)) * radius, `
` Number.Sin(radians(_)) * radius, Date.Month(datelist{_}), `
` Number.Abs(Number.Cos(radians(_)))*10}),`
` //convert the list of points to a table`
` outputtable = Table.FromRows(positionlist, {"Angle", "Date", "x", "y", "Colour", "Size"}),`
` //set data types`
` ChangedType = Table.TransformColumnTypes(outputtable,`
` {{"Angle", type number}, {"Date", type date}, {"x", type number}, `
` {"y", type number}, {"Colour", type number}, {"Size", type number}})`
`in`
` ChangedType`
`in`
` CircleFunction`
I then created another Power Query query to call this function 30 times to create 30 circles with different radiuses:
`let`
` //generate a list of numbers from 0 to 29`
` circlelist = {0..29},`
` //generate a list of 30 dates starting on 1 January 2014`
` datelist = List.Dates(#date(2014,1,1), 30,#duration(1,0,0,0)),`
` //call the Circle() function 30 times`
` tablelist = List.Transform(circlelist, each Circle(_+5, datelist{_}, Number.Mod(_,2)=0)),`
` //combine the resulting tables into a single table`
` positionlist = Table.Combine(tablelist)`
`in`
` positionlist`
And here’s the result of the query plotted on a map using Power Map:
Pretty, isn’t it? You can download the workbook with the Power Query query and the Power Map tour here.
## One thought on “Pretty Patterns With Power Query And Power Map”
1. […] Pretty Patterns With Power Query And Power Map (Chris Webb) […] | 819 | 3,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2016-07 | latest | en | 0.788478 |
https://community.cartalk.com/t/peak-acceleration/103631?page=2 | 1,716,275,601,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058385.38/warc/CC-MAIN-20240521060250-20240521090250-00786.warc.gz | 155,658,265 | 9,556 | # "Peak" acceleration
It looks to me like you might have been better off just leaving it in third and letting the engine rev past the horsepower peak to avoid that last 0.59 second shift time, if the engine can safely rev that high.
That pro stock racing transmission would make a lousy road racing transmission because it’s not designed to downshift, in fact, if you lift your foot off the gas the ramped engagement dogs would immediately cam out of engagement kicking the transmission into neutral, possibly with shifting linkage damage.
The whole operating principle is that the shift fork engages the next higher gear and then the dog clutch in the lower gear cams out of engagement, like the starting crank of an antique farm tractor or Model T car.
I had a 3.55 final drive in it. A 4.10 works better, tops out 4th through the traps. I couldn’t wind out 3rd that far. I was already past the power peak and pushing the RPM a bit for a cast crank (silly rules again!) motor. I was just trying to get a little data and see if I could create dyno curves in various gears. Data was a little noisy for that. It was still fun though…
So, you agree that the fastest way down the 1/4 mile is with a CVT that keeps the engine at max power? If not, please cite references.
;-]
I looked up the power and torque of a 2014 Corvette.
Power peak was 402.48 HP @ 5936 RPM, which comes out to 356 ft-lb of torque.
Torque peak was 400.02 ft-lb @ 4839 RPM which comes out to 368.6 HP.
Let’s assume we use this engine in a car with a CVT.
Driver A programs the CVT to hold the torque peak 4839 rpm.
Driver B programs the CVT to hold the power peak 5936 rpm.
At 10 mph, the rear axle is turning 144 rpm. Driver A’s transmission is set to a 33.6:1 reduction ratio, driver B’s transmission is set to a 41.2:1 reduction ratio.
400.02 ft lb X 33.6 = 13,441 ft lb of torque at the rear wheels.
356 ft lb X 41.2 =14,667 ft lb of torque at the rear wheels.
At 30 mph, the rear axle is turning 432 rpm.
Driver A’s transmission is set to a 11.2:1 gear reduction ratio.
Driver B’s transmission is set to a 13.7:1 gear reduction ratio.
400.02 ft lb X 11.2 = 4480 ft lb at the rear wheels.
356 ft lb X 13.7 = 4891 ft lb at the rear wheels.
It seems to me that driver B will win this drag race.
No matter what speed you choose, the CVT that is holding the power peak rpm is delivering more torque to the rear wheels than the transmission that is holding the torque peak. It’s lower gearing at that road speed more than compensates for the engine’s lower torque.
After careful consideration and @B.L.E 's terrific math example, yes I now agree! Thanks to all who participated in the discussion!
I had to stew on this a couple days to put it into words. B. L. E. summed it up real well. In a given gear, the max acceleration will be at the engine’s torque peak. At a given vehicle speed, a gear that puts the engine at its horsepower peak will provide the maximum thrust. In most car engines these days the torque peak is at a fairly high RPM, usually between 4,500 and 5,000. In B. L. E’s example above, the torque peak and the horsepower peak are only about 1,100 RPM apart…Not a whole lot when you’re over 4,500, So I’d have to guess that, given that the horsepower peak is way up there near 6,000, shifting at or near redline is gonna give you the fastest acceleration. | 851 | 3,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.95181 |
https://www.teachoo.com/13886/596/Example-2-ii/category/Examples/ | 1,723,686,062,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00444.warc.gz | 786,683,399 | 21,646 | Examples
Chapter 4 Class 11 Complex Numbers
Serial order wise
### Transcript
Example 2 Express the following in the form of a + bi (ii) ( ) (2 ) ( 1/8 )^3 ( ) (2 ) ( 1/8 )^3 = ( )(2 ) ( 1/8)^3 ^3 = 1 2 ( 1/(8 8 8)) ( 3) = 2 1/(8 8 8) ( ^(3+1+1) ) = 1 1/(4 8 8) ( 5) = 1/256 ( 4) = 1/256 ( 2)2 = 1/256 ( 1)^2 = 1/256 (1) = 1/256 = 0 + 1/256 | 181 | 342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-33 | latest | en | 0.585941 |
https://joningram.org/questions/Algebra/675483 | 1,701,236,352,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100056.38/warc/CC-MAIN-20231129041834-20231129071834-00840.warc.gz | 395,774,368 | 5,187 | # Find the Inverse f(x)=x/8-7
Find the Inverse f(x)=x/8-7
Replace with .
Interchange the variables.
Solve for .
Rewrite the equation as .
Add to both sides of the equation.
Multiply both sides of the equation by .
Simplify both sides of the equation.
Cancel the common factor of .
Cancel the common factor.
Rewrite the expression.
Simplify .
Apply the distributive property.
Multiply by .
Solve for and replace with .
Replace the with to show the final answer.
Set up the composite result function.
Evaluate by substituting in the value of into .
Simplify each term.
Apply the distributive property.
Cancel the common factor of .
Cancel the common factor.
Rewrite the expression.
Multiply by .
Combine the opposite terms in .
Since , is the inverse of .
Do you know how to Find the Inverse f(x)=x/8-7? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.
### Name
Name one billion three hundred thirteen million six hundred sixty-two thousand six hundred eleven
### Interesting facts
• 1313662611 has 16 divisors, whose sum is 1787235840
• The reverse of 1313662611 is 1162663131
• Previous prime number is 719
### Basic properties
• Is Prime? no
• Number parity odd
• Number length 10
• Sum of Digits 30
• Digital Root 3
### Name
Name two hundred seventy-two million nine hundred twenty-eight thousand five hundred four
### Interesting facts
• 272928504 has 64 divisors, whose sum is 1235730816
• The reverse of 272928504 is 405829272
• Previous prime number is 163
### Basic properties
• Is Prime? no
• Number parity even
• Number length 9
• Sum of Digits 39
• Digital Root 3
### Name
Name one billion ninety-eight million four hundred twenty-seven thousand four hundred four
### Interesting facts
• 1098427404 has 32 divisors, whose sum is 3317403600
• The reverse of 1098427404 is 4047248901
• Previous prime number is 149
### Basic properties
• Is Prime? no
• Number parity even
• Number length 10
• Sum of Digits 39
• Digital Root 3 | 523 | 2,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.786013 |
http://mathoverflow.net/questions/133568/smooth-map-to-the-stack-of-g-bundles/134149 | 1,469,582,312,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825125.30/warc/CC-MAIN-20160723071025-00302-ip-10-185-27-174.ec2.internal.warc.gz | 102,375,693 | 15,295 | # Smooth map to the stack of G-bundles
Let $G$ a semisimple group and $B$ a Borel subgroup. We denote by $Bun_{G}$ the stack of G-bundles.
Is it true that a certain open subset $Bun_{B,r}$ maps smoothly to $Bun_{G}$?
My question comes from Lemma 14 .2.1 from http://arxiv.org/pdf/math/0611323.pdf
but I'm not sure to understand well.
-
You need to formulate the question more precisely - in this way it is obvious (any map between reduced algebraic stacks locally of finite type is generically smooth). – Alexander Braverman Jun 13 '13 at 0:36
I fixed the typo in your title, which you can also do if you notice them in future. – David Roberts Jun 13 '13 at 1:02
If $H$ is a subgroup of $G$, the map from the stack of $H$-bundles to the stack of $G$-bundles is a fiber bundle with fiber $G/H$. – Angelo Jun 13 '13 at 4:17
I don't understand the purpose of the lemma 14.2.1 then, why restricting to an open subset if everything is already smooth with connected fibers? – prochet Jun 13 '13 at 6:11
Angelo: what you wrote is wrong (it is only true for bundles over a point). – Alexander Braverman Jun 13 '13 at 15:57
The short answer is the following: for a $G$-torsor $E_G$ over $C$ and for the associated projective scheme $E_{G,B} := E_G/B$, a lift $E_B$ of $E_G$ to a $B$-torsor over $C$ is the same thing as a section $\sigma:C\to E_{G,B}$ of the projection $\pi:E_{G,B}\to C$. Via infinitesimal deformation theory of the Hilbert scheme, this section is unobstructed if $H^1(C,\sigma^*(\Omega_\pi)^\vee)$ is zero. Finally, $\sigma^*(\Omega_\pi)^\vee$ turns out to be $E_B \times^B \text{Lie}(U^{-})$. In fact, since $E_B$ has a further reduction of structure group to a maximal torus $T$, i.e., $E_B$ equals $B \times^T E_T$ for a $T$-torsor $E_T$, the bundle $E_B\times^B \text{Lie}(U^{-})$ turns out to equal $E_T\times^T \text{Lie}(U^{-})$, which splits as a direct sum of invertible sheaves on $C$ (because every representation of $T$ is a direct sum of characters). Thus $E^T\times^T \text{Lie}(U^{-})$ has vanishing $h^1$ if and only if each of these summands has vanishing $h^1$. That is precisely the condition imposed by Gaitsgory and Nadler to define the open subset $\text{Bun}_{B,r}$ inside $\text{Bun}_B$. | 700 | 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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.864751 |
https://www.shaalaa.com/question-bank-solutions/pythagoras-theorem-radius-circle-inscribed-triangle_1838 | 1,513,628,187,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948623785.97/warc/CC-MAIN-20171218200208-20171218222208-00494.warc.gz | 805,999,283 | 10,656 | # Solution - Pythagoras Theorem
Account
Register
Share
Books Shortlist
ConceptPythagoras Theorem
#### Question
In a right triangle ABC, right-angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is
(A) 4
(B) 3
(C) 2
(D) 1
#### Solution
You need to to view the solution
Is there an error in this question or solution?
#### Reference Material
Solution for concept: Pythagoras Theorem. For the course CBSE
S | 133 | 458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.802882 |
https://www.daniweb.com/programming/software-development/threads/235459/problems-with-my-if-loop | 1,713,193,696,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817002.2/warc/CC-MAIN-20240415142720-20240415172720-00601.warc.gz | 693,284,066 | 15,276 | Hi guys:
The following code is supposed to the following:
1 - Receive a random number and divide the number by 1000
2 - Then it needs to check if the number is greater than the series number and the cut point and convert the divided random number to either 1 or -1.
3 - Then do the sum of the (-1 and 1) and this will be my new series number.
4 - Then it will check again.
There is an initial condition that the sum[0] = 1. The cut points I need to use are either 7.5, 5 or 2.5.
My problem is that the code only returns a bunch of -1. I am certain that my problem is in my if ...else if statements. Can anyone point me into the right direction Please don't hesitate to ask any questions, I take criticism pretty well. Here is the code.
G
``````/Program name = random.cpp
#include <cstdlib>
#include <iostream>
using namespace std;
int number_generator();
int main()
{
int series;
double average, times, cut_point;
double series_number[1000001];
double time_series_average[100];
double sum[100];
cout << "How long you want your series: ";
cin >> series;
cout << "How many times do you want to run the series: ";
cin >> times;
cout << "What is your cutting point: ";
cin >> cut_point;
for (int j=0; j<times; j++)
{
for (int i=0; i<series; i++)
{
sum[0] = 1;
double number_to_use = rand() % 9999;
double actual_random_integer = number_to_use/1000;
//cout << "number to use = " << number_to_use <<
//cout << i << "= " << actual_random_integer << endl;
if (actual_random_integer > cut_point && sum[j] > 0)
{
series_number[i] = 1;
}
else if (actual_random_integer > cut_point && sum[j] < 0)
{
series_number[i] = -1;
}
else if (actual_random_integer < cut_point && sum[j] > 0)
{
series_number[i] = 1;
}
else if (actual_random_integer < cut_point && sum[j] < 0)
{
series_number[i] = -1;
}
//cout << "Series nummber [" << i << "]" << series_number[i] << endl;
sum[j]+=series_number[i];
//cout << "sum = " << sum[j] << endl;
}
time_series_average[j] = sum[j]/series;
cout << "(" << j << ") " << time_series_average[j] << endl;
//delete [] time_series_average;
}
//cout << sum << endl;
//average = sum/series;
//cout << "AVERAGE = " << average << endl;
}``````
I'm not sure than sum[] array is initialized except for index 0. also, there is no call to srand, so I imagine you'd get the same results every time. You need to initialize the sum[] array before you use it.
Be a part of the DaniWeb community
We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge. | 697 | 2,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-18 | latest | en | 0.728417 |
http://people.sc.fsu.edu/~jburkardt/m_src/hermite_product_display_test/hermite_product_display_test.html | 1,561,065,335,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999273.79/warc/CC-MAIN-20190620210153-20190620232153-00526.warc.gz | 137,436,747 | 2,439 | # HERMITE_PRODUCT_DISPLAY Display 2D Cartesian Products of Hermite Polynomials
HERMITE_PRODUCT_DISPLAY, a MATLAB program which displays an image of a function created by the Cartesian product of two Hermite polynomials, such as f(x,y) = h(3,x) * h(1,y).
There are five types of Hermite polynomial available. Perhaps the best behaved are "Hen(n,x)" and "Hf(n,x)", which don't blow up within the plotting interval as fast as the other functions do.
The physicist's Hermite polynomial H(n,x) can be defined by:
H(n,x) = (-1)^n exp(x^2/2) * d^n/dx^n ( exp(-x^2/2) )
The normalized physicist's Hermite polynomial Hn(n,x) is scaled so that
Integral ( -oo < X < +oo ) exp ( - X^2 ) * Hn(M,X) Hn(N,X) dX = delta ( N, M )
The probabilist's Hermite polynomial He(n,x) is related to H(n,x) by:
He(n,x) = H(n,x/sqrt(2)) / sqrt ( 2^n )
The normalized probabilist's Hermite polynomial Hen(n,x) is scaled so that
Integral ( -oo < X < +oo ) exp ( - 0.5*X^2 ) * Hen(M,X) Hen(N,X) dX = delta ( N, M )
The Hermite function Hf(n,x) is related to H(n,x) by:
Hf(n,x) = H(n,x) * exp(-x^2/2) / sqrt ( 2^n * n! * sqrt ( pi ) )
and is scaled so that:
Integral ( -oo < X < +oo ) Hf(M,X) Hf(N,X) dX = delta ( N, M )
### Usage:
hermite_product_display ( 'name', i, j )
where
• 'name' is 'h', 'hn', 'he', 'hen', or 'hf';
• i is the index of the polynomial in the X direction;
• j is the index of the polynomial in the Y direction.
### Languages:
HERMITE_PRODUCT_DISPLAY is available in a MATLAB version.
### Related Data and Programs:
FEM_BASIS_Q4_DISPLAY, a MATLAB program which displays a basis function associated with a linear quadrilateral ("Q4") mesh.
FEM_BASIS_T3_DISPLAY, a MATLAB program which displays a basis function associated with a 3-node triangle "T3" mesh.
FEM_BASIS_T4_DISPLAY, a MATLAB program which displays a basis function associated with a 4-node triangle "T4" mesh.
FEM_BASIS_T6_DISPLAY, a MATLAB program which displays a basis function associated with a 6-node triangle "T6" mesh.
HERMITE_POLYNOMIAL, a MATLAB library which evaluates the physicist's Hermite polynomial, the probabilist's Hermite polynomial, the Hermite function, and related functions.
hermite_product_display, a MATLAB program which displays an image of a function created by the Cartesian product of two Hermite polynomials, such as f(x,y) = h(3,x) * h(1,y).
HERMITE_PRODUCT_POLYNOMIAL, a MATLAB library which defines Hermite product polynomials, creating a multivariate polynomial as the product of univariate Hermite polynomials.
POLYGONAL_SURFACE_DISPLAY, a MATLAB program which displays a surface in 3D described as a set of polygons;
### Examples and Tests:
Here we plot all the possible products of orders 0 through 5, using the "hen" polynomial: | 766 | 2,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-26 | latest | en | 0.833049 |
https://i.reddit.com/r/space/comments/q7mksu/shatner_in_space/hgla0cf/.compact | 1,670,015,334,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00623.warc.gz | 339,509,277 | 8,396 | # space
-8 points-7 points 1 year ago
You know exatly what they meant by being in orbit, i'm just poking fun at your ridiculous pendency. Enjoy :)
[deleted] 5 points6 points 1 year ago
I do know exactly what they mean by being in orbit. And my comment still stands. Using orbit is just as arbitrary as anything else. Why are you like this lmao
-6 points-5 points 1 year ago
I didn’t orbit anything". You thought the up and down argument was making a point, it wasnt, as traveling that distance you would have orbited the planet to some degree due to its rotation.
But thats ok, not the end of the world, there seems to be planty of discussion around the level at which space starts to keep you engaged, i'd just refrain from using poor analogies.
2 points3 points 1 year ago
To be in “orbit” requires the ship to be in constant freefall. Meaning that the ship won’t fall back to earth on its own. Orbit isn’t simply moving around the earth.
If you like space and like video games, you should try playing Kerbal Space Program. That game is better at teaching the basics of orbital mechanics than almost anything else.
[deleted] 0 points1 point 1 year ago
Are you really under the impression that the rotation of the Earth has anything to do with making an orbit an orbit? The rotation of the planet has nothing to do with an orbit. An orbit is around the center of mass of the planet, not in relation to a specific point on the surface.
Also, you retain your horizontal velocity when going up, it’s not like you just lose all that velocity at some point and the surface rotates away under you. This is why countries try to launch rockets as close to the equator as possible. Launch near the equator and you get about 1,000 mph of “free” horizontal velocity. Launch near the poles where the surface is not rotating as fast and you have to make up for that loss in speed.
It’s like if you toss a ball in the air while going 60mph down the road in a car. The ball goes straight up (relative to you) and straight back down, it doesn’t go flying out the back at 60mph. Mythbusters did this. Check it out.
If you would consider my example as meeting this definition of being in space, then you’d have to consider Bezos and Shatner as having gone to space too. Your logic is wrong, as explained above, but even if it wasn’t, it would apply to the Blue Origin flights too.
It’s ok though. Orbital mechanics is hard, not everyone can grasp these concepts. Keep trying and you’ll get there!
Btw, OP said their definition was reaching orbital velocity and necessary altitude to stay there. If I go straight up 1,000 miles and fall back down, I never reached orbital velocity. You don’t even know what you’re arguing here.
1 point2 points 1 year ago
I stand corrected, went away and did a bit more research on the definition of an orbit. Youre original comments were clearly not as pithy as i had thought they were. My bad.
Everyday is a school day :)
-1 points0 points 1 year ago
That is quite the combination of arrogance and complete nonsense.
0 points1 point 1 year ago
The irony of someone who knows nothing about what they are talking about calling someone else pedantic. Dunning-Kruger at it's finest. | 743 | 3,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-49 | latest | en | 0.963301 |
https://mynotesadda.com/2019/12/26/maths-tricks-in-hindi-pdf-3/ | 1,582,900,492,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147234.52/warc/CC-MAIN-20200228135132-20200228165132-00158.warc.gz | 472,255,846 | 28,404 | ### Maths Tricks In Hindi PDF
SSC CGL, BANK, RAILWAYS, RRB NTPC, LIC AAO, etc. Exams are starting after a few months. In those exams, a lot of questions are coming from Maths Tricks In Hindi PDF, so Maths Tricks In Hindi PDF is important in all exams. In Our Mynotesadda Website providing you an Important PDF of Maths Tricks In Hindi PDF which is helpful for Students who preparing for all such competitive exams. Common questions are placed in Maths Tricks In Hindi PDF which has been put together in most examinations, you can download this PDF Notes very simply by clicking on the Download Button at the bottom.
MyNotesAdda.com is an online Educational Platform, where you can download free PDF for UPSC, SSC CGL, BANK, RAILWAYS, RRB NTPC, LIC AAO, and many other exams.
There are around 20-25 questions in each Government Exams related to GK Questions in Hindi and you can solve 18-20 questions out of them very easily by reading these Notes of Maths Tricks In Hindi PDF. The complete PDF of Maths Tricks In Hindi PDF is attached below for your reference, which you can download by clicking at the Download Button.
MyNotesAdda.com will update many more new pdfs and study materials and exam updates, keep Visiting and share our post, So more people will get this.
### Topics Includes In Maths Tricks In Hindi PDF
• NUMBER SET
• ALGEBRA
• GEOMETRY
• TRIGONOMETRY
• MATRICS AND DETERMINATION V
• VECTOR
• ANALYTIC GEOMETRY
• DIFFERENTIAL CALCULAS
• INTEGRAL CALCULAS
• DIFFERENTIAL EQUATION
• SERIES PROBABILITY
#### Maths Tricks In Hindi PDF RELATED TOPICS
Some Important Questions Related To Maths Tricks In Hindi PDF
1. 1043 को किसी संख्या से भाग देने पर भागफल 11 तथा शेषफल 20 प्राप्त होता है, भाजक ज्ञात कीजिए ?
• (A) 95
• (B) 93
• (C) 73
• (D) 97
B
2. किसी संख्या को 195 से भाग देने पर 47 शेष बचते हैं, इस संख्या को 15 से भाग देने पर शेष क्या बचेगा ?
• (A) 5
• (B) 4
• (C) 3
• (D) 2
D
3. 8765 * 974 – 8765 * 874 = ?
• (A) 870500
• (B) 876500
• (C) 870000
• (D) 877700
B
4. 1994 * 1994 = ?
• (A) 3776036
• (B) 3976036
• (C) 3976037
• (D) 3976000
B
5. 883 * 883 – 117 * 117 = ?
• (A) 767000
• (B) 766006
• (C) 766000
• (D) 866000
C
6. एक पुस्तक का क्रयमूल्य ₹110 तथा विक्रयमूल्य ₹123.20 है। इसे बेचने पर पुस्तक विक्रेता को कितने प्रतिशत लाभ होगा ?
• (A) 11%
• (B) 12%
• (C) 13%
• (D) 14%
7. एक कुर्सी को ₹873 में बेचने से व्रिकेता को 10% हानि होती है, कुर्सी का क्रय-मूल्य कितना है ?
• (A) ₹970
• (B) ₹900
• (C) ₹950
• (D) ₹980
8. यदि किसी वस्तु को 5% हानि की अपेक्षा 10% लाभ पर बेचा जाता तो विक्रेता को ₹75 अधिक मिलते। वस्तु का क्रयमूल्य कितना है ?
• (A) ₹550
• (B) ₹500
• (C) ₹530
• (D) ₹520
9. रमेश ने एक गाय ₹8580 में बेचकर 4% लाभ कमाया, उसने यह गाय कितने रूपये में खरीदी ?
• (A) ₹8270
• (B) ₹8250
• (C) ₹8650
• (D) ₹9250
10. एक फल विक्रेता ने ₹15 के 6 की दर से केले खरीद कर ₹12 के 4 की दर से बेच दिए, उसका लाभ अथवा हानि प्रतिशत ज्ञात कीजिये।
• (A) 17%
• (B) 19%
• (C) 20%
• (D) 22%
General Science Notes | 1,179 | 2,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2020-10 | latest | en | 0.563047 |
https://www.hrpub.org/journals/article_info.php?aid=11418 | 1,716,567,048,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00782.warc.gz | 704,642,436 | 3,777 | Journals Information
Mathematics and Statistics Vol. 9(5), pp. 825 - 834
DOI: 10.13189/ms.2021.090523
Reprint (PDF) (865Kb)
Numerical Solution of Ostrovsky Equation over Variable Topography Passes through Critical Point Using Pseudospectral Method
Nik Nur Amiza Nik Ismail , Azwani Alias *, Fatimah Noor Harun
Faculty of Ocean Engineering Technology and Informatics, Universiti Malaysia Terengganu, 21030 Kuala Nerus, Terengganu, Malaysia
ABSTRACT
Internal solitary waves have been documented in several parts of the world. This paper intends to look at the effects of the variable topography and rotation on the evolution of the internal waves of depression. Here, the wave is considered to be propagating in a two-layer fluid system with the background topography is assumed to be rapidly and slowly varying. Therefore, the appropriate mathematical model to describe this situation is the variable-coefficient Ostrovsky equation. In particular, the study is interested in the transition of the internal solitary wave of depression when there is a polarity change under the influence of background rotation. The numerical results using the Pseudospectral method show that, over time, the internal solitary wave of elevation transforms into the internal solitary wave of depression as it propagates down a decreasing slope and changes its polarity. However, if the background rotation is considered, the internal solitary waves decompose and form a wave packet and its envelope amplitude decreases slowly due to the decreasing bottom surface. The numerical solutions show that the combination effect of variable topography and rotation when passing through the critical point affected the features and speed of the travelling solitary waves.
KEYWORDS
Solitary Wave, Nonlinear Equation, Ostrovsky Equation, Variable Topography, Background Rotation, Polarity Change, Pseudospectral Method
Cite This Paper in IEEE or APA Citation Styles
(a). IEEE Format:
[1] Nik Nur Amiza Nik Ismail , Azwani Alias , Fatimah Noor Harun , "Numerical Solution of Ostrovsky Equation over Variable Topography Passes through Critical Point Using Pseudospectral Method," Mathematics and Statistics, Vol. 9, No. 5, pp. 825 - 834, 2021. DOI: 10.13189/ms.2021.090523.
(b). APA Format:
Nik Nur Amiza Nik Ismail , Azwani Alias , Fatimah Noor Harun (2021). Numerical Solution of Ostrovsky Equation over Variable Topography Passes through Critical Point Using Pseudospectral Method. Mathematics and Statistics, 9(5), 825 - 834. DOI: 10.13189/ms.2021.090523. | 577 | 2,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.838223 |
http://directlenders.ml/giga/matrices-homework-help-866.php | 1,544,427,316,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823318.33/warc/CC-MAIN-20181210055518-20181210081018-00276.warc.gz | 78,804,940 | 5,765 | Skip Nav
# Operations On Matrices
## How it works:
❶Watch this video lesson to learn how easy it is to perform row operations on a matrix.
Further, the elements should be so arranged that each of them is capable of being subscripted by its i th row and j th column to locate its position in the matrix. Thus, if an element say, 15 is subscripted as 15 1. In a disorderly arrangement of numbers like the above ones a number cannot be subscripted by its i th row and j th column. A matrix always consists of certain rows and columns in which all its elements are arranged.
The number of such rows and columns may be one or more and there may or may not be equality between the number of rows and the number of columns.
But a columns or a row must be complete with some elements. Thus a group of rows and columns not completed with all its elements as follows will not amount to a matrix:. A group of orderly arranged numbers or symbols to be called a matrix must be enclosed by some brackets viz. Thus, a groups of following numbers and symbols not encompassed by any bracket will not constitute a matrix. The magnitude of the order of a matrix refers to the number of rows and column with which a matrix is constituted.
Every matrix must be denominated properly for making a reference to it in the course of computational works. Conventionally, all the matrices are denominated or named by some letters of upper case viz. The answer goes in position 2, 1.
Finally, we multiply 2nd row of the first matrix and the 2st column of the second matrix. The answer goes in position 2, 2. I designed this web site and wrote all the lessons, formulas and calculators. If you want to contact me, probably have some question write me using the contact form or email me on.
Math Calculators, Lessons and Formulas It is time to solve your math problem. Simplifying Adding and Subtracting Multiplying and Dividing. Simplifying Multiplying and Dividing Adding and Subtracting. Introduction Exponential Equations Logarithmic Functions. Arithmetic Progressions Geometric Progressions. Substitution Integration by Parts Integrals with Trig. Area Volume Arc Length. Line in 3D Planes. Definitions Addition and Multiplication Gauss-Jordan elimination.
Introduction to Determinants Applications of Determinants. Random Quote The human mind has never invented a labor-saving machine equal to algebra. Random Quote A mathematician is a device for turning coffee into theorems. More help with radical expressions at mathportal. Arithmetic operations with matrices - online calculator.
## Main Topics
Operation with Matrices in Linear Algebra. Addition and Multiplication.
### Privacy FAQs
Matrices homework help and essay writing with chat with custom writing service The result is what order should a research paper be in a privilege that can be used simultaneously. Mind - bending pictures to invoke familiar experiences in critical scenarios. | 601 | 2,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-51 | latest | en | 0.927688 |
http://www.jiskha.com/search/index.cgi?query=How+many+grams+of+SO3+are+produced+when+20.0+g+FeS2+reacts+with+16.0+g+O2+according+to+this+balanced+equation%3F | 1,498,200,780,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320023.23/warc/CC-MAIN-20170623063716-20170623083716-00146.warc.gz | 582,435,813 | 10,841 | # How many grams of SO3 are produced when 20.0 g FeS2 reacts with 16.0 g O2 according to this balanced equation?
46,625 results
Chemistry
SO3 can be produced in the following two-step process: FeS2 + O2 ---> Fe2O3 + SO2 S02 + O2 ----> SO3 Assuming that all the FeS2 reacts, how many grams of SO3 are produced when 20.0 g of the FeS2 reacts with 16.0 g of O2?
Chemistry
How many grams of SO3 are produced when 20 g FeS2 reacts with 16g O2 according to the balanced equation? 4FeS2(s)+15 O2(g)>>>2Fe2O3+8SO3(g)
Chem
f 147 grams of FeS2 is allowed to react with 88 grams of O2 according to the following unbalanced equation, how many grams of Fe2O3 are produced? FeS2 + O2 → Fe2O3 + SO2
chemistry
1. How many grams of P4 will react with 35.0 grams of H2O according to the balanced equation P4+5O2+6H2O=4H3PO4? 2. If you have 50 grams of P4, 40 grams of H2O, and an unlimited supply of O2, which reactant (according to the above balanced equation) is the limiting reactant? ...
Chem
how many grams SO2 are produced when 20(g) FeS2 reacts with 16(g) O2? 4FeS2(g)+15O2(g)=2Fe2O2(s)+8SO2(g)
Chemistry
1. Iron oxide and carbon monoxide react according to the following equation . Identify which reactant is limiting and which is excess if 43.5 grams of Fe2O3 reacts with 43.5 grams of CO Fe2O3 + 3CO2 --> 3CO2 + 2Fe 2. Calculate the percent yield of Carbon dioxide if 20.0 ...
chemistry
Gaseous arsine reacts according to the balanced equation. If a total of -3663 KJ of heat is produced in the reaction as written, how many grams of AsH3 are consumed? (Enthalpy= -1867.2) 2 AsH3(g) + 3 O2(g) --> As2O3(s) + 3 H2O(l)
chemistry
Gaseous carbon monoxide reacts according to the balanced equation. If a total of 1040 kJ of heat is generated in the reaction as written, how many grams of CO2 are produced? enthalpy=-566kJ 2 CO(g) + O2(g) → 2 CO2(g)
Chemistry
Determine the mass of calcium hydroxide produced when calcium carbide (CaC2) reacts with 0.64 grams of water according to the following balanced chemical equation: CaC2 + 2H2O -> Ni(OH)2 + NaNO3
chemistry
Zinc metal reacts with hydrochloric acid (HCl) according to the equation: Zn(s)+HCl(aq)=ZnCl2(aq)=H2(g) How many grams of H2 are produced if 100 grams of zinc reacts?
Chemistry
1. How many moles of ammonia NH3 are produced when 0.45 moles of nitrogen N2 reacts? N2 + 3H2 yields 2NH3 2. How many mililiters of water will be produced when 2.05 g of hydrogen reacts? The density of water is 1.00 g/mL ? 2H2 + O2 yields 2H2O 3. What is the limiting reactant ...
Chem Help!
I really need help for my homework on how to write the balanced equation for the following reactions. 1) Sodium reacts with iron(lll)oxide to produce sodium oxide and iron. Type of reaction: Balanced Equation: 2) Hydrogen bromide forms from hydrogen and bromine. Type of ...
Chemistry
Many important metals occur as sulfide, arsenide, and antimonide minerals, especially in the Sudbury mineral complex. The first step in processing these ores involves “roasting” the ore in air to produce the metal or metal oxide, along with nonmetal oxides that can be ...
chemistry
Element Symbol Atomic Mass --------------------------------------- Bromine Br 79.904 Calcium Ca 40.078 Carbon C 12.011 Chlorine Cl 35.4527 Cobalt Co 58.93320 Copper Cu 63.546 Flourine F 18.9984032 Hydrogen H 1.00794 Iodine I 126.904 Iron Fe 55.847 Lead Pb 207.2 Magnesium Mg 24...
AP Chemistry
What mass in grams of sodium hydroxide is produced if 20.0 g of sodium metal reacts with excess water according to the chemical equation? 2Na(s)+2H20(l)=2NaOH(aq)+Hs(g) How many moles of Ag can be produced if 350. g of Cu are reacted with excess AgNO3 according to the equation...
Chemistry
I don't understand this question? What is the maximum amount in grams of SO3 that can be produced by the reaction 1.0 g of S with 1.0 g of O2 EQUATION: 2 S(s) +3 O2(g)--->2 SO3(g) The choices are: A) 3.8 B) 2.5 C) 0.27 D) 1.7 Shouldn't it be 2.0, or am I forgetting to do ...
Chemistry
If the percent yeild of the following equation is 55%, how many grams of H2SO4 are produced when 4.88 x 10^24 molecules of SO3 are combined with excess water? SO3(g)+H2O(l)->H2SO4(aq). what would be the formula to solve this equation
Chemistry
pyrite, FeS2 reactswhen it is heated strongly in the air according to the equation 4 FeS2(s) + 11 O2 rightarrow 2 Fe2O3(s) + 8 SO2(g) What volume of sulfir dioxide ( measured at room temprature and pressure) would be produced by heating 100g of pyrite in excess of air Molar ...
chemistry
Zinc reacts with hydrochloric acid according to the following equation: Zn + 2HCl --> ZnCl2 + H2 If 3.63 grams of zinc react, how many grams of zinc chloride is produced?
Chemistry
(1)Aluminum reacts with oxygen to produce aluminum oxide according to the following reaction: 4Al(s) + 3O2(g) → 2Al2O3(s) Calculate the moles of Al2O3 produced when the reaction is performed with 31.06 g of each reactant. (2)Nitrogen dioxide reacts with water to produce ...
chemistry
1)The reaction of methane and oxygen yields carbon dioxide gas and water. The unbalanced reaction is as follows: CH4(g) + O2(g)CO2(g) + H2O(g) A. If 0.718 grams of methane are reacted, how many grams of water vapor are produced? B. If 1.621 grams CO2 are released, how many ...
chemistry
Propane, C3H8, reacts with oxygen according to the following balance equation. If 44 grams of propane react completely with sufficient oxygen, how many grams of water are produced?
Chemistry 130
Can someone please help me with these questions? 1 (7). Rank the following in order of increasing radius based on their positions on the periodic table: S, S-2, Cl a. S, S-2, Cl b. S-2, Cl, S c. Cl, S, S-2 d. Cl, S-2, S e. S, Cl, S-2 6 (16). Methane burns in oxygen according ...
chemistry
If 146 g of Pb(NO3)2 reacts stoichiometrically according to the balanced equation, how many grams of solid PbI2 are produced? Pb(NO3)2(aq) + 2NaI(aq) ¡æ PbI2(s) + 2NaNO3(aq) Molar Mass (g/mol) Pb(NO3)2 331.21 PbI2 461.00 Molar Volume (L) 22.4 at STP Gas Constant (L.atm.mol-1...
Chemistry
Hydrogen gas,H2 , reacts with nitrogen gas,N2 , to form ammonia gas,NH3 , according to the equation 3H2+N2-2NH3 How many grams ofNH3 can be produced from 2.90mol ofN2 ? How many grams ofH2 are needed to produce 13.75g of NH3 ? How many molecules (not moles) ofNH3 are produced ...
chem
how would i figure out this question? assume that 5.60 L of H2 at STP reacts with CuO according to the equation: CuO+H2 yields Cu+H2O a. how many moles of H2 react? b. how many moles of Cu are produced? c. how many grams of Cu are produced? i have no clue what im doing! can ...
CHEMISTRY
If 4.32 g of solid C reacts stoichiometrically according to the balanced equation, how many grams of solid TiO2 are required? 3TiO2(s) + 4C(s) + 6Cl2(g) → 3TiCl4(l) + 2CO2(g) + 2CO(g)
Chemistry
How many grams of HCl are produced according to the following equation, H2 + Cl2 = 2 HCl, when 4.0 g of hydrogen reacts completely?
chemistry
How many grams of HCl are produced according to the following equation, H2 + Cl2 = 2 HCl, when 4.0 g of hydrogen reacts completely?
chemistry
Liquid quinoline reacts according to the balanced equation. When 1200 g of C9H7N is reacted, calculate the sign and magnitude (kJ) of the heat produced. 4 C9H7N(l) + 43 O2(g) → 2 N2(g) + 36 CO2(g) + 14 H2O(l) .
Chemistry
Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation 3H2(g)+N2(g)→2NH3(g) PArt 1: How many grams of NH3 can be produced from 3.26 mol of N2 and excess H2. Part 2: How many grams of H2 are needed to produce 13.17 g of NH3? part 3...
Science
What is the mass (in grams) of water that is produced when 100.0 grams of nitrogen monoxide is reacted with 75.0 grams of hydrogen according tho the following balanced eguation? 2NO + 5H2 = 2NH3 + 2H2O
Chemistry
Assume that 7.2 L of I2 are produced at STP according to the following balanced equation: 2KI+Cl - 2KCl+ I2 How many grams of KI (MOLAR MASS=166.002G/MOL) reacted
Chemistry
Assume that 7.2 L of I2 are produced at STP according to the following balanced equation: 2KI+Cl - 2KCl+ I2 How many grams of KI (MOLAR MASS=166.002G/MOL) reacted
Chemistry
Assume that 7.2 L of I2 are produced at STP according to the following balanced equation: 2KI+Cl - 2KCl+ I2 How many grams of KI (MOLAR MASS=166.002G/MOL) reacted
chemistry
Assume that 7.2 L of I2 are produced at STP according to the following balanced equation: 2KI+Cl - 2KCl+ I2 How many grams of KI (MOLAR MASS=166.002G/MOL) reacted
chemistry.
What amount (in moles) of FeS2(s) are required to produced 64g of SO2(g) according to the following equation? 4FeS2(s) + 110(g)_________2FeO3(s)__8SO2(g)...remember the first long bar is an arrow.
Chemistry
what mass in grams of sodium hydroxide is produced if 20.0g of sodium metal reacts with excess water according ro the chemical equation, Na(s)+H2O4->naoh(aq)+h2(g)
chemistry
Given the balanced equation: 2 Al + Fe2O3 → Al2O3 + 2 Fe how many grams of Al2O3 will be produced if 68.1 g Al reacts with 112 g Fe2O3?
chemistry
Calcium reacts with water to yield calcium hydroxide and water according to the following equation: Ca + 2H2O ==> Ca(OH)2 + H2 . How many moles of hydrogen will be produced if 3115.8 grams of calcium hydroxide is produced?
chemistry
Suppose 50.0 grams of magnesium reacts with 71.0 grams of hydrochlorie acid according to the foloowing equation: Mg(s) + 2HCl(aq) -> MgCl2(s) + H2(g) 1) What is the maximum mass of magesium chloride that can be produced? Maximum mass of magnesium chloride 2) which is the ...
Chemistry
Hydrogen gas, , reacts with nitrogen gas, , to form ammonia gas, , according to the equation 3H2+N2-2NH3 How many grams of can be produced from 2.90 of ?
Chemistry
Acetylene gas, C2H2, reacts with oxygen according to the following equation. If 46.0 g of acetylene reacts with 100.0 g of oxygen, what is the theoretical yield of heat produced from the reaction? ___ C2H2(g) + ___ O2(g) → ___ CO2(g) + ___ H2O(g); balanced reaction ...
Chemistry
Did I do these right? Can someone let me know if I did them correctly? I had someone help me with the problems but know I want to make sure their the right answers? 1. Methane burns in oxygen according to the following equation: CH4 + 2O2 ----> CO2 + 2H2O What is the mass ...
stoichiometry
Ammonia gas reacts with oxygen gas according to the following equation: 4NH3 + 5O2----4NO + 6H2O a. How many moles of oxygen gas are needed to react with 23 moles of ammonia? (29 mole) b. How may grams of NO are produced when 25 moles of oxygen gas react with an excess of ...
chem
For this problem, will someone please check my answer? How many grams of oxygen are produced in the decomposition of 5.00g of potassium chlorate? KClO3(s)--->KCl(s)+O2(g) [the equation may not be balanced] My answer is 1.96 g of O2. The balanced equation is 2KClO3 ==> ...
Chemistry
If 18.80 grams of hydrochloric acid are reacted with a large amount of aluminum metal, what volume of hydrogen gas would be produced at STP? (Hint: start by writing a balanced equation for the reaction.) 3HCl + Al -> 3H + AlCl3 18.80g moles= grams/mw 18.80/109.5 0....
Chemistry
For the balanced equation shown below, if 40.1 grams of C2H3O2Br were reacted with 8.01 grams of O2, how many grams of H2O would be produced? 2C2H3O2Br+O2->4CO+2H2O+2HBr Thank you!
Chemistry
For the balanced equation shown below, if 40.1 grams of C2H3O2Br were reacted with 8.01 grams of O2, how many grams of H2O would be produced? 2C2H3O2Br+O2->4CO+2H2O+2HBr Thank you!
Chemistry
How many grams of pure H2SO4 can be obtained from 250 grams of iron ore if the ore is 82% FeS2? The reactions involved are given below: 4FeS2 + 11 O2 --> 2 Fe2O3 + 8 SO2 (95% efficient) 2SO2 + O2 --> 2SO3 (90% efficient) SO3 + H2O --> H2SO4 (90% efficient)
Chemistry
I blanked how to solve this: How many grams of carbon dioxide are produced when 2.5 grams of sodium bicarbonate reacts with excess citric acid according to: 3NaHCO3+H3C6H5O7 -> Na3C6H5O7+3CO2+3H2O Could you include the steps, please? Thanks.
Chemistry
Hydrogen gas, , reacts with nitrogen gas, , to form ammonia gas, , according to the equation 3H2+N2-2NH3 How many grams of are needed to produce 13.75g of NH3 ? How many molecules (not moles) of are produced from 8.26×10−4g of H2 ?
Chemistry
If 0.870 mol of liquid Br2 and 720 mL of 0.958 M aqueous NaI are reacted stoichiometrically according to the balanced equation, how many grams of solid I2 are produced? 2NaI(aq) + Br2(l) → 2NaBr(aq) + I2(s) help please. :)
chemistry
If the initial pressure of SO3(g) is 4.775 atm, calculate the % of SO3(g) left over after the reaction reaches equilibrium according to the balanced equation. The value of Kp at 1000 K is 24.40. The initial pressure of the reaction products is 0 atm. 2SO3(g) = 2SO2(g)+O2(g) ...
Chemistry
How many grams of silver nitrate must react to give 67.5 grams of Ag according to the balanced equation. Cu(s) + 2 AgNO3(aq) --> Cu(NO3)2(aq) + 2 Ag(s)
Chemistry
Copper metal reacts with concentrated nitric acid according to the following balanced equation: 3Cu(s) + 8HNO3(aq) ¨ 3Cu(NO3)2(s) + 2NO(g) + 4H2O(l) Calculate the mass in grams of the excess reagent remaining after the complete reaction of 4.63 g of Cu with 9.26 g of HNO3. ...
Chemistry
Iodine, I2, reacts with aqueous thiosulfate ion in neutral solution according to the balanced equation I2(aq)+2S2O2−3(aq)→S4O2−6(aq)+2I−(aq). How many grams of I2 are present in a solution if 33.00 mL of 0.165 M Na2S2O3 solution is needed to titrate the...
Chemistry
Based on the following chemical equation answer the question below show all your calculation The balanced chemical reaction of iodine with hydrogen sulfide h2s(g)+i2(s)-----2HI(g) + S(s) (a) Based on the above chemically balanced equation, how many grams of HI should be ...
chemistry
iron 3 oxide reacts with carbon monoxide according to the following equation: fe2O3+3Co-->2Fe+3Co2 a) how many moles of ions are produced when one mole of iron 3 oxide reacts b) calculate the volume of carbondioxide produced when one mole of iron 3 oxidereacts
chemistry
Based on the following chemical equation, answer the questions below. Show all your calculation The balanced chemical reaction of iodine with hydrogen sulfide is as follows: H2S(g)+I2(s)-----2HI(g) + S(s) (a) Based on the above chemically balanced equation, how many grams of ...
chemistry
For the balanced equation shown below, how many grams of O2 reacted, if 98.0 grams of HCl are produced? 2C6H4Cl2+7O2=>12CO+2H2O+4HCl
CHEMISTRY
2NaBrO3 reacts to 2NaBr + 3O2 How many grams of NaBrO3 are required to produce 19 grams of O3? How many liters of 02 are produced from 4 moles of NaBrO3? How many atoms of NaBr are produced when 1 mole of O2 is produced?
science
How many grams of hydrochloric acid are required to react completely with 4.30g of zinc? How many molecules of gas will be produced? 1. Write the balanced chemical equation. Zn + 2HCl ==> ZnCl2 + H2 2. Convert grams Zn to mols. mols = grams/atomic mass Zn. 3. Using the ...
Chemistry
2Mg + O2 ¨ 2MgO How many grams of MgO are produced when 79.5 grams of O2 react completely with Mg according to this equation?
Chemistry
1) Consider the formation of sulfur trioxide according to the reaction: 2 SO2(g) + O2(g) -> 2 SO3(g) delta H = -198 How much heat is evolved in the formation of 750 grams of SO3? I am not sure how to start this problem.
chemistry
480 g of FeS2 is combined with 12.000mol of molecular oxygen. balanced chemical reaction is 4FeS2 + 11O2 ---> 2Fe2O3 + 8SO2 a. Which cheical will run out first? b. What mass of Fe2O3is produced?
Chemistry
according to the balanced chemical equation 2SO2<--- 2 is a subscript...so anyway it's 2SO2+ O2 ----> 2SO3 How many liters of oxygen would be needed to produce 0.5 moles of sulfur trioxide, SO3?
If 0.836 g of solid Al reacts stoichiometrically according to the balanced equation in a reaction solution with a total volume of 1160 mL, what mass (g) of gaseous H2 is produced? 6 HClO4(aq) + 2 Al(s) → 3 H2(g) + 2 Al(ClO4)3(aq) 0.836g/molecular weight Al=mole of Al ...
Chemistry
Some drain cleaners contain a mixture of sodium hydroxide and aluminum which reacts with water as follows: 2Al(s)+2 NaOH(aq)+6H2O(l)->2Na[Al(OH)4](aq)+3H2(g) What mass of hydrogen gas can be produced when 5g Al reacts with sufficient NaOH and H2O) according to the above ...
Chemistry
How many grams of zinc must react to give 13.0 grams of ZnO according to the balanced equation. 2 Zn(s) + O2(aq) --> 2 ZnO(s)
Chemistry
Question 2 Marks: 2 How many grams of oxygen gas must react to give 174.3 grams of ZnO according to the balanced equation. 2 Zn(s) + O2(aq) --> 2 ZnO(s)
Chemistry
Question 2 Marks: 2 How many grams of oxygen gas must react to give 174.3 grams of ZnO according to the balanced equation. 2 Zn(s) + O2(aq) --> 2 ZnO(s)
Chemistry
Question 2 Marks: 2 How many grams of oxygen gas must react to give 174.3 grams of ZnO according to the balanced equation. 2 Zn(s) + O2(aq) --> 2 ZnO(s)
Chemistry
a)How many grams of NH3 can be produced from 2.84mol of N2. b)How many grams of H2 are needed to produce 14.86g of NH3? balanced equation: 3H2+N2->2HN3
chemistry
For the balanced equation shown below, how many grams of H2SO4 reacted, if 50.9 grams of H2 are produced? Mg+H2SO4=>MgSO4+H2
chemistry
What mass of carbon dioxide will be produced when 0.250 mol of propane(C3H8) reacts with 25.0 g of oxygen? Write a balanced equation first.
chemistry
If 0.0490 mol of solid Al reacts stoichiometrically according to the balanced equation with 990 mL of aqueous Cl-, what molarity (M) of Cl- is required? 3 CuCl2(aq) + 2 Al(s) → 3 Cu + 2AlCl3(aq)
chemistry Emergency TEST
1.How many moles of sodium oxide are produced when3.9 moles of sodium combine with excess oxygen. 2. Sodium chloride decomposes into elemental sodium and elemental chlorine.How many grams of chlorine will be produced if 39.4 grams of sodium chloride decomposes 3.If 6.7 moles ...
chemistry
What mass in grams of sodium hydroxide is produced if 20.0 g of sodium metal reacts with excess water according to the chemical equation 2Na(s)+2H20(l)=2NaOH(aq)+H2(g)? Please show me step by step how to do this because I have 5 other questions like this one on a worksheet and...
chemistry
7. Write the balanced equation for the equation for the reaction of magnesium with HCl. What is the mole ratio between magnesium and magnesium chloride will be produced? If 2.5g of magnesium react, how many moles of magnesium chloride will be produced? How many grams of ...
chemistry
If 120 mL of 2.8333 M aqueous HCl reacts stoichiometrically according to the balanced equation, how many milliliters of 4.43 M aqueous FeCl3 are produced? Fe2S3(s) + 6HCl(aq) ¡æ 3H2S(g) + 2FeCl3(aq) Molar Mass (g/mol) HCl 36.461 FeCl3 162.21 Molar Volume (L) 22.4 at STP Gas ...
Chemistry
In the formation of acid rain, sulfur dioxide reacts with oxygen and water in the air to form sulfuric acid. Write the balanced chemical equation for the reaction. 2 SO2 + 1 O2 + 2 H2O --> 2 H2SO4 If 5.45 g SO2 react with excess oxygen and water, how many grams of H2SO4 are...
science
2 H2 + O2 -> 2 H2O 2 grams + 32 grams -> _ _ _ _ _ grams how would the mass of water formed in the reaction compare to the mass of oxygen and hydrogen that reacts? include types of change illustrated in the balanced equation. explain the conversation of matter and ...
science
2 H2 + O2 -> 2 H2O 2 grams + 32 grams -> _ _ _ _ _ grams how would the mass of water formed in the reaction compare to the mass of oxygen and hydrogen that reacts? include types of change illustrated in the balanced equation. explain the conversation of matter and ...
science
2 H2 + O2 -> 2 H2O 2 grams + 32 grams -> _ _ _ _ _ grams how would the mass of water formed in the reaction compare to the mass of oxygen and hydrogen that reacts? include types of change illustrated in the balanced equation. explain the conversation of matter and ...
science
2 H2 + O2 -> 2 H2O 2 grams + 32 grams -> _ _ _ _ _ grams how would the mass of water formed in the reaction compare to the mass of oxygen and hydrogen that reacts? include types of change illustrated in the balanced equation. explain the conversation of matter and ...
Chemistry
Calcium carbonate reacts w/stomach acid according to the following chemical equation. CaCO3+2HCl(aq)-> CaCl2(aq)+H2O(l)+CO2(g) A. Balance the eqaution B. Tums is one commercially sold antacid that contains CACO3. If Tums is added to 20.0ml of 0.400 M HCl, how many grams of ...
chemistry
Stoichiometrics: a 36 grams sample of calcium hydroxide is allowed to react with 40.5 grams of phosphoric acid according to the following reaction. 3Ca(OH)2 + 2H3PO4 = Ca3(PO4)2 + 6H2O how many grams of calcium phosphate are produced? how many moles of Calcium phosphate are ...
science need help
2 H2 + O2 -> 2 H2O 2 grams + 32 grams -> _ _ _ _ _ grams how would the mass of water formed in the reaction compare to the mass of oxygen and hydrogen that reacts? include types of change illustrated in the balanced equation. explain the conversation of matter and ...
chemistry
Copper reacts with sulfuric acid according to the following equation: 2H3SO4 + Cu-> CuSO4 + 2H2O + SO2 How many grams of sulfur dioxide are created by this reaction if 14.2 g of copper reacts with 18 g of sulfuric acid?
chemistry
Copper reacts with sulfuric acid according to the following equation: 2H3SO4 + Cu-> CuSO4 + 2H2O + SO2 How many grams of sulfur dioxide are created by this reaction if 14.2 g of copper reacts with 18 g of sulfuric acid?
chemistry
Copper reacts with sulfuric acid according to the following equation: 2H3SO4 + Cu-> CuSO4 + 2H2O + SO2 How many grams of sulfur dioxide are created by this reaction if 14.2 g of copper reacts with 18 g of sulfuric acid?
chemistry
a 36 grams sample of calcium hydroxide is allowed to react with 40.5 grams of phosphoric acid according to the following reaction. how many grams of calcium phosphate are produced? how many moles of calcium phosphate are produced? how many moles of excess reagent remain? if 43...
Chemistry
1) Consider the formation of sulfur trioxide according to the reaction: 2 SO2(g) + O2(g) -> 2 SO3(g) delta H = -198 How much heat is evolved in the formation of 750 grams of SO3? So I did 193 X (750/2*80.0642) = 904. The answer choices given are 10.6, 928, 1.86 X 10^3, 7.43...
Chemistry
given this reaction: 2CO(g) + O2(g)---2CO2(g) what volume of CO2 will be produced when 8.45 liter of CO reacts with enough oxygen? I thought the answer would be 8.45 liters of CO2 because there was a ratio in the balanced equation of 2CO yielding 2CO2, so I thought the ration ...
chemistry
When cane sugar reacts with oxygen in living systems, carbon dioxide and water are produced. What weight of carbon dioxide can be produced from the reaction of 15.0 grams of cane sugar with 15.0 grams of oxygen? (Hint: balance the equation) ___C12H22O11(s) + ___O2(g) ...
chemistry
Calculate the volume of H2 produced at 300 K and 98.65 kPa when 48 g of Mg reacts with excess HCl according to the following equation: Mg(s) 2HCl(aq) £ MgCl(aq) H2(g)
Chemistry
2 SO2 (g)+O2 (g)→2 SO3↑ How many grams of SO3(g) are formed if 8.45 grams of oxygen are used?
Chemistry
Given the balanced equation N2(g)+3H2(g)==>2NH3 Calculate: a)the volume H2 that reacts with 12L of N2 b)the volume of NH3 produced Iron 4L ofN2 c)the volume of N2 and H2 to produced 60L of NH3 Assume that all volume measurements are made under identical conditions
science
1. How many molecules of ammonia, NH3, are produced from 12.5 grams of hydrogen in the following balanced equation?
1. Pages:
2. 1
3. 2
4. 3
5. 4
6. 5
7. 6
8. 7
9. 8
10. 9
11. 10
12. 11
13. 12
14. 13
15. 14
16. 15
17. Next>>
Post a New Question | 7,211 | 23,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-26 | latest | en | 0.900114 |
https://bettingiscool.com/2016/12/27/risk-vs-uncertainty/?amp=1 | 1,701,400,122,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100264.9/warc/CC-MAIN-20231201021234-20231201051234-00246.warc.gz | 152,016,252 | 16,893 | # Risk vs Uncertainty
In sports betting it’s sometimes hard to judge where you stand on your way to profits. Is it a matter of good fortune (or bad fortune for that matter) or are you genuinely holding an edge over the bookmaker(s)?
To answer this question a proper understanding of risk and uncertainty helps.
Risk, as first articulated by the economist Frank H. Knight in 1921, is something that you can put a price on. Say that you’ll win a poker hand unless your opponent draws to an inside straight: the chances of that happening are exactly 1 chance in 11. This is risk. It is not pleasant when you take a bad beat in poker, but at least you know the odds of it and can account for it ahead of time. In the long run, you’ll make a profit from your opponent making desperate draws with insufficient odds.
Uncertainty, on the other hand, is risk that is hard to measure. You might have some vague awareness of the demons lurking out there. You might even be acutely concerned about them. But you have no real idea how many of them there are or when they might strike. Your back-of-the-envelope estimate might be off by a factor of 100 or by a factor of 1,000; there is no good way to know. This is uncertainty. Risk greases the wheels of a free-market economy; uncertainty grinds them to a halt.
Translating this into sports betting would simply mean that you need to make every effort to put a (fair) price on every bet you make. If you do this you will know your expectation, you will know your RISK. Put that in contrast to a tipster service you follow. Your tipster is delivering hundreds if not thousands of bets, but you will never be quite sure what the expected value of each bet really is – is it +5%, is it -2%? You don’t know your RISK. You might assume that he (your tipster) knows his stuff due to his ‘brilliant’ track record, but what if he loses his edge? You will not realise this until it’s too late and you have already blown a big part of your bankroll. What you are up against with most tipsters is UNCERTAINTY rather than RISK. This is not what you want as an investor, really! | 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} | 2.546875 | 3 | CC-MAIN-2023-50 | latest | en | 0.96248 |
https://softkeys.uk/blogs/blog/how-to-convert-seconds-to-minutes-in-excel | 1,718,359,167,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00212.warc.gz | 482,545,133 | 57,178 | Blog
# How to Convert Seconds to Minutes in Excel?
Have you ever needed to work out how many minutes are in a certain number of seconds? It’s a tricky calculation to make if you don’t have a calculator handy, but luckily you can easily convert seconds to minutes in Excel. In this article, we’ll show you how to quickly and easily convert seconds to minutes in Excel, making sure you always have the right answer at your fingertips.
How to Convert Seconds to Minutes in Excel?
• Open the Excel document containing the seconds you want to convert.
• Select the cell with the number of seconds.
• Select the “Format Cells” option from the pop-up menu.
• In the “Number” tab, select “Time” from the category list.
• Choose the “m:ss” option from the Type list.
• Select “OK” to apply the changes.
• The selected cell will now display the converted minutes.
## How to Utilize Excel to Convert Seconds to Minutes
Excel is a powerful tool to help users manipulate data, and one of the most commonly used functions is to convert seconds to minutes, or vice versa. To do this, users can employ Excel’s built-in formulas or functions to easily and quickly convert between units of time. In this article, we will explain the steps involved in converting seconds to minutes in Excel.
### Using Excel Formulas to Convert Seconds to Minutes
The most straightforward way to convert seconds to minutes in Excel is to use the built-in formulas. To do this, users will first need to enter the number of seconds into a cell in the worksheet. Then, in the cell where the converted minutes are to be displayed, enter the formula: =SUM(A1/60). This formula will take the value in cell A1 and convert it to minutes.
Users can also use the formula =A1/60 to display the converted minutes in the same cell. However, this formula will only display the result and not sum the seconds, as the previous formula does.
### Using Excel Functions to Convert Seconds to Minutes
Another way to convert seconds to minutes in Excel is to use the built-in functions. Excel has several functions that can be used to convert seconds to minutes, including the TIMEVALUE, MINUTE, and MOD functions.
The TIMEVALUE function is used to convert a text representation of a time into a decimal number that represents the same time. For example, to convert the text “1:30” to a decimal number, the formula =TIMEVALUE(“1:30”) can be used.
The MINUTE function can be used to convert a decimal number that represents a time into the number of minutes. For example, to convert the decimal number 0.5 (which represents 30 minutes) to the number of minutes, the formula =MINUTE(0.5) can be used.
The MOD function can be used to calculate the remainder of a division operation. This can be used to calculate the number of remaining seconds when a number of minutes is divided by 60. For example, to calculate the number of remaining seconds after dividing 30 minutes by 60, the formula =MOD(30,60) can be used.
### Using Excel to Automatically Convert Seconds to Minutes
Excel also provides users with the ability to automatically convert seconds to minutes using the built-in Autofill feature. To do this, users will need to enter the number of seconds into a cell in the worksheet. Then, select the cell and drag the Autofill handle down to the desired number of rows. As the rows are filled, the seconds will automatically be converted to minutes.
### Using Excel to Convert Minutes to Seconds
Finally, users can also use Excel to convert minutes to seconds. To do this, users can use the TIMEVALUE, MINUTE, and MOD functions, as explained above. However, these functions must be used in reverse. For example, to convert 30 minutes to seconds, the formula =TIMEVALUE(MINUTE(0.5)) can be used.
### Conclusion
In conclusion, Excel provides users with a variety of built-in formulas and functions that can be used to easily and quickly convert between units of time. By following the steps outlined in this article, users can quickly convert seconds to minutes in Excel.
## Few Frequently Asked Questions
### Question 1: What is a Second?
A second is a unit of time equal to 1/60th of a minute. It is the smallest unit of time in the International System of Units (SI). It is often used to measure short lengths of time, such as the time between two events or the amount of time that has passed since an event occurred.
### Question 2: What is a Minute?
A minute is a unit of time equal to 60 seconds. It is the most commonly used unit of time and is often used to measure longer lengths of time, such as the amount of time between two events or the amount of time that has passed since an event occurred.
### Question 3: How to Convert Seconds to Minutes in Excel?
To convert seconds to minutes in Excel, use the formula =A1/60, where A1 is the cell containing the number of seconds. This formula will return the number of minutes in the cell.
### Question 4: What is the Formula for Seconds to Minutes?
The formula for converting seconds to minutes is =A1/60, where A1 is the cell containing the number of seconds. This formula will return the number of minutes in the cell.
### Question 5: What is the Excel Function for Seconds to Minutes?
The Excel function for converting seconds to minutes is the DIVIDE function. To use the function, enter the formula =DIVIDE(A1,60), where A1 is the cell containing the number of seconds. This formula will return the number of minutes in the cell.
### Question 6: How to Format the Result of Seconds to Minutes in Excel?
To format the result of seconds to minutes in Excel, select the cell containing the result, right-click, and select Format Cells. Under the Number tab, select the Custom category and enter the format code “mm:ss”. This will format the result to show the number of minutes and seconds.
In conclusion, converting seconds to minutes in Excel is a simple process. With a few clicks of your mouse, you can quickly and easily convert seconds to minutes. This tutorial has provided step-by-step instructions to help you get the job done quickly and accurately. So now that you know how to convert seconds to minutes in Excel, you can take advantage of the powerful features that Excel has to offer.
Related Articles | 1,341 | 6,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-26 | latest | en | 0.861982 |
https://www.queryhome.com/puzzle/37192/three-numbers-ratio-and-their-average-what-the-largest-number | 1,601,073,428,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400228998.45/warc/CC-MAIN-20200925213517-20200926003517-00020.warc.gz | 992,888,732 | 31,685 | # Three numbers are in the ratio 3 : 6 : 9 and their average is 66. What is the largest number?
40 views
Three numbers are in the ratio 3 : 6 : 9 and their average is 66. What is the largest number?
posted Aug 17
99
Numbers are 3x, 6x and 9x
Their average is
(3x+6x+9x)/3=66
6x=66
x= 11
The largest number is
9x= 9*11= 99
the largest number is 99
Solution,
x + 2x + 3x / 3 = 66, then x= 33
33 x 3 = 99 | 162 | 405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-40 | longest | en | 0.918228 |
https://math.stackexchange.com/questions/4374858/is-the-the-transformation-of-the-transposed-matrix-equal-to-transpose-of-the-tra/4374878 | 1,713,895,841,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818732.46/warc/CC-MAIN-20240423162023-20240423192023-00422.warc.gz | 330,140,501 | 34,277 | # Is the the transformation of the transposed matrix equal to transpose of the transformed matrix?
Let $$A$$ be a square matrix expressed in a basis $$\{\mathbf e_i\}$$ and let $$C$$ be the transition matrix from $$\{\mathbf e_i\}$$ to a new basis $$\{\tilde{\mathbf e}_i\}$$, so $$\widetilde {A}=DAC$$, with $$D=C^{-1}$$. This question has come up for me: is the transpose of a transformed matrix equal to the transformation of the transposed matrix? That is,
$$\widetilde {A}^T\stackrel{?}{=}\widetilde {A^T} \tag{1}$$
• If you want $(1)$ to hold for all $A$, then $CC^T$ must be a scalar multiple of the identity matrix. If one only finds that $(1)$ holds for some $A$, I don't think we can say anything about $C$ apart from the fact that it is nonsingular. Consider $A=0$ for instance. Feb 5, 2022 at 18:30
$$\widetilde {A}^T=(DAC)^T=C^TA^TD^T$$ $$\widetilde {A^T}=DA^TC$$
And we see that $$\widetilde {A}^T=\widetilde {A^T}$$ if
$$D^{-1}C^TA^TD^TC^{-1}=A^T$$
$$CC^TA^TD^TD=A^T$$
So this only happens in the case of orthogonal transformations so that $$CC^T=I$$ and $$D^TD=I$$. | 373 | 1,085 | {"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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-18 | latest | en | 0.773375 |
https://www.coursehero.com/file/6690484/IA35/ | 1,490,731,729,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189884.21/warc/CC-MAIN-20170322212949-00415-ip-10-233-31-227.ec2.internal.warc.gz | 874,755,220 | 66,275 | # IA3.5 - Score:_________________/12 3. (3 pt) Every summer,...
This preview shows pages 1–2. Sign up to view the full content.
Math 64 Name: ___________________ J. Frewing Date: ____________________ Score:_________________/12 Inclass Activity 3.5 1. (3 pt) In a right triangle, the measure of the smallest angle is 26 o . What is the measure of the middle angle? Legend: Picture: Formula: Equation: Sentence : 2. (3pt) Kay wants to bend a 106 inch wire into a rectangle so that the width is 24 inches. What is the length of this rectangle? Legend: Picture: Formula: Equation: Sentence :
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Math 64 Name: ___________________ J. Frewing Date: ____________________
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: Score:_________________/12 3. (3 pt) Every summer, Andy hikes a mountain trail in southwestern Colorado. The route he takes is 72 miles long. If he can average 18 miles per day, how many days does it take him to hike this trail? Legend: Picture: Formula: Equation: Sentence : 4. (3 pt) A developer has built a house on a rectangular piece of land that has an area of 1,200 square yards. If the width of this land is 25 yards, what is its length? Legend: Picture: Formula: Equation: Sentence :...
View Full Document
## This note was uploaded on 01/13/2012 for the course MATH 64 taught by Professor Janetfrewing during the Spring '10 term at Riverside Community College.
### Page1 / 2
IA3.5 - Score:_________________/12 3. (3 pt) Every summer,...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 437 | 1,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-13 | longest | en | 0.858645 |
https://nrich.maths.org/public/leg.php?code=114&cl=1&cldcmpid=8296 | 1,511,581,666,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809392.94/warc/CC-MAIN-20171125032456-20171125052456-00252.warc.gz | 666,346,647 | 5,830 | # Search by Topic
#### Resources tagged with 2D representations of 3D shapes similar to How Do You See It?:
Filter by: Content type:
Stage:
Challenge level:
### There are 11 results
Broad Topics > 3D Geometry, Shape and Space > 2D representations of 3D shapes
### A City of Towers
##### Stage: 1 Challenge Level:
In this town, houses are built with one room for each person. There are some families of seven people living in the town. In how many different ways can they build their houses?
### New House
##### Stage: 2 Challenge Level:
In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
### Pupils' Recording or Pupils Recording
##### Stage: 1, 2 and 3
This article, written for teachers, looks at the different kinds of recordings encountered in Primary Mathematics lessons and the importance of not jumping to conclusions!
### Building Blocks
##### Stage: 2 Challenge Level:
Here are some pictures of 3D shapes made from cubes. Can you make these shapes yourself?
### The Development of Spatial and Geometric Thinking: Co-ordinating Space in Drawings
##### Stage: 1
This second article in the series refers to research about levels of development of spatial thinking and the possible influence of instruction.
### Next Size Up
##### Stage: 2 Challenge Level:
The challenge for you is to make a string of six (or more!) graded cubes.
### Three Cubed
##### Stage: 2 Challenge Level:
Can you make a 3x3 cube with these shapes made from small cubes?
### The Third Dimension
##### Stage: 1 and 2 Challenge Level:
Here are four cubes joined together. How many other arrangements of four cubes can you find? Can you draw them on dotty paper?
### Thinking 3D
##### Stage: 2 and 3
How can we as teachers begin to introduce 3D ideas to young children? Where do they start? How can we lay the foundations for a later enthusiasm for working in three dimensions?
### The Development of Spatial and Geometric Thinking: 5 to 18
##### Stage: 1, 2, 3 and 4
This is the first article in a series which aim to provide some insight into the way spatial thinking develops in children, and draw on a range of reported research. The focus of this article is the. . . .
### Air Nets
##### Stage: 2, 3, 4 and 5 Challenge Level:
Can you visualise whether these nets fold up into 3D shapes? Watch the videos each time to see if you were correct. | 586 | 2,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2017-47 | latest | en | 0.916415 |
https://www.coursehero.com/file/5586550/SampleExam1/ | 1,519,299,309,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814101.27/warc/CC-MAIN-20180222101209-20180222121209-00791.warc.gz | 847,598,715 | 28,333 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
SampleExam1
# SampleExam1 - MEEN 221 — Fall 2007 Exam 1 Oct 11 2007 Dr...
This preview shows pages 1–7. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: MEEN 221 — Fall 2007 Exam 1 Oct. 11, 2007 Dr. Gao Section: 506 Name: Student Number: "Aggies do not lie, cheat, or steal, nor do they tolerate those who do. ” — Aggie Code of Honor By my signature below I pledge that my conduct on this exam is consistent in every way with the Aggie Code of Honor: Signature: This exam consists of Five (5) problems that are egually weighted. Some problems have multiple parts. Be sure to carefully read and properly analyze each question that is asked. Do not jump to unfounded conclusions, but also do not overlook or oversimplify problems either. 4. Be sure to show all work, including sketches, Free Body Diagrams, and calculations, and organize your solution procedure as clearly and systematically as possible. 5. Work problems in the space provided on the exam sheets. Clearly indicate continuation of the problem if extra sheets are used. Work efficiently and neatly. Clearly indicate final answers by enclosing in a "box" or place answer in box if box is grovided. Include any and all appropriate units. 9N? T‘?’ Problem 1: _ [20 Problem 2: I 20 Problem 3: __ I 20 Problem 4: _ I20 Problem 5: — I20 TOTAL: Hugs—2: 3, H fie com..an Hie \$59.8 A 85 m :95 3:»: mem mum m8 flooQoamnowzw 033mg man: 9m” 9m 89:35 moHoo mombm 62,28: 99: gm m Bmmswcaa om n: 52 man mm E893 macaw 55 Au. Uaflowambm 9w EH 9 En Summon 5 03% AG 85 wn 85 9a Bmwm S om go: @508. 2on n5“ NEQSSW N.“ o: \$m Essa. Ewan v55 35203 5 9a cox 303mg c302. 5652: t N Am: 635\$ wofiam 5% BE En mcwwo: m 98% Emr 33 Ba .86 ~55 \$303? 3N mcw 2:8 mm @585. ‘25 Summon 5 9a Em: 2:6 33 mm c EA. >350 \$5 29%: om So SW8 man 90 62am 8.: ca 55039 Ema :5 330m 5 9m mg 2:8 wO mum wb «Sm Ea 38w SwamBEmm cw :5 can 5%. Emoa v55 mbmémam 5 9a cox 633mg. 355—2: “5 Ana com—:3 ‘25 wwwoo mrcam Wm Radium W03 03: <38: w: mmqoswfi 93 25 9.3m A: man a3 8 83.3.2 :5 3305. 135 H2 .5 mm 5098a m: m wow: 8 S. o. o v 85 gm 9 drama o», moo Sm m: 85N 9m “\$3030: + N 8 £525. 136 HQ .3 Wm 5033 a Alwoxfi +mo\~u 3 gm :3 m 955» mm<g cw 90 mega \$33 .3 H Iwoo mu m + So 3 a + moo a» w. \$5 5 R“ N oooawsmfio mwmflmB mm mm @525. 2:: 9o 3me 8 \$5 00 o». 30 E58 9539 \$93 mm 9m 8a: 8082: 93 mm 833m 8 388 5a mrsaa 935m: :5 00 ~52: 3 9m x 83m“ \$ :5 R mxmmu m5; 3 So N Same 3 gm: Wm Em Bmmiamo ow 5089: om Emma H.088 mwofi m =3 mxfimbamsm W03 Em CG 8 So 8% cm 90 85 126 8w 3,90 8m mm 5889 92133.“ P Asks; Ewan v58. 852on m: :5 cox 32an @052. 5632: n A»? A; 633\$ 3 Q wowamv Eng :6 3:30: 0». \$5 £3me :50 m: So mmfiow A5 Q @0589 35¢ 90 38:0: om :5 % 803330 om \$5 025.05 om :6 magma m8? 380 v55 «Sm/29¢ E Em Emmwm Eofiaaa c052. AII|V . .. mo BB €2u|lw Evan n 3632: t Aw Cc 13:an 508333 :5 2-80amst om 90 8333 ow 90 £938 m8» @625 6302. Ewen v55 “59,88 w: :5 Swan going c335 mo BB 3622: t m E: com—:3 m3 9a @850 \$.52? 38356 3 En 38¢ 5 5% mod E 9o roaNosS— man 338: 85605053 0m 90 namofios mega m: 9.0:: A. 0:85 9m.” m: @038 So 3509 Ram. Ewao v55 36\$on 3 9m cox Had/Anon \$302. Semk‘irmfa: an: k I n 33C R r .fifl m V Wham Ankh We. 2 Circa!” W ...
View Full Document
{[ snackBarMessage ]}
### Page1 / 7
SampleExam1 - MEEN 221 — Fall 2007 Exam 1 Oct 11 2007 Dr...
This preview shows document pages 1 - 7. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 1,543 | 3,860 | {"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} | 3 | 3 | CC-MAIN-2018-09 | latest | en | 0.826823 |
http://brainden.com/forum/topic/1561--/ | 1,511,529,235,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808133.70/warc/CC-MAIN-20171124123222-20171124143222-00238.warc.gz | 47,427,647 | 14,258 | BrainDen.com - Brain Teasers
• 0
## Question
Being the math crazed maniac I was in school, one of my teachers challenged me to prove that 1=2.
Ok, so he said I would fail if I could not prove it....
3 months later I came upwith the solution....
Where is the Flaw?
A=B
A^2 = AB
A^2 - B^2 = AB - B^2
(A+B)(A-B) = B(A-B)
A+B = B
2B = B
2=1
What I did each step: (sorry forum not conducive to spaces)
A=B - Given
A^2 = AB - Multiplicative Property of Equality (A)
A^2 - B^2 = AB - B^2 - Additive Property of Equality (-B^2)
(A+B)(A-B) = B(A-B) - Evaluation
A+B = B - Reduction
2B = B - Substitution and Addition (since first statement say a=b)
2=1 - Reduction (by B)
The "^" means to the power of.
A-B = 0
## 3 answers to this question
• 0
A=B
A^2 = AB
A^2 - B^2 = AB - B^2
(A+B)(A-B) = B(A-B)
A+B = B
2B = B
2=1
If A=B ; A-B=0
##### Share on other sites
• 0
Being the math crazed maniac I was in school, one of my teachers challenged me to prove that 1=2.
Ok, so he said I would fail if I could not prove it....
3 months later I came upwith the solution....
Where is the Flaw?
A=B
A^2 = AB
A^2 - B^2 = AB - B^2
(A+B)(A-B) = B(A-B)
A+B = B
2B = B
2=1
What I did each step: (sorry forum not conducive to spaces)
A=B - Given
A^2 = AB - Multiplicative Property of Equality (A)
A^2 - B^2 = AB - B^2 - Additive Property of Equality (-B^2)
(A+B)(A-B) = B(A-B) - Evaluation
A+B = B - Reduction
2B = B - Substitution and Addition (since first statement say a=b)
2=1 - Reduction (by B)
The "^" means to the power of.
A-B = 0
You are dividing by zero.
(A+B)(A-B) = B(A-B)
You then divide by A-B. And if A=B, then A-B is zero.
##### Share on other sites
• 0
What I did each step: (sorry forum not conducive to spaces)
A=B - Given
A^2 = AB - Multiplicative Property of Equality (A)
A^2 - B^2 = AB - B^2 - Additive Property of Equality (-B^2)
(A+B)(A-B) = B(A-B) - Evaluation
A+B = B - Reduction
2B = B - Substitution and Addition (since first statement say a=b)
2=1 - Reduction (by B)
The "^" means to the power of.
spaces can be kept in code tags ... like this ...
```A=B - Given
A^2 = AB - Multiplicative Property of Equality (A)
A^2 - B^2 = AB - B^2 - Additive Property of Equality (-B^2)
(A+B)(A-B) = B(A-B) - Evaluation
A+B = B - Reduction
2B = B - Substitution and Addition (since first statement say a=b)
2=1 - Reduction (by B)[/code]
"power of" can be easily used as "super-script" ... eg. A[sup]2[/sup] = AB
both code and super-script are buttons when you write posts
and I almost forgot ... already posted in a few variations ... check this
## Create an account or sign in to comment
You need to be a member in order to leave a comment
## Create an account
Sign up for a new account in our community. It's easy!
Register a new account | 913 | 2,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2017-47 | latest | en | 0.892667 |
https://www.investoo.com/trade-bearish-bullish-pennants/ | 1,571,711,030,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987795403.76/warc/CC-MAIN-20191022004128-20191022031628-00064.warc.gz | 940,723,181 | 14,088 | # How to Trade Bearish and Bullish Pennants
Video Transcript:
Hello traders. Welcome to the third module of the advanced technical analysis course, chart patterns. In this lesson we are going to teach you how to trade the bullish and the bearish pennant.
Remember that the bullish and the bearish pennant are continuation patterns that are found in the middle of directional moves. They look like triangles, but they are much smaller and the consolidation area inside of them is much choppier. Let’s go through the lesson and see how we can profit from these continuation patterns.
Let’s start with the entries, the stop loss’s levels, the target levels, and how to trade these continuation patterns. It is quite simple to trade these continuation patterns, and actually it’s a lot like flags. The difference lies in the stop loss area and how to calculate the targets. Let’s go through the lesson.
Of course you need to wait for the breakout of the pennant for you to be able to trade it. Pennants are continuation patterns that are found in the middle of directional moves. They look like triangles, but they are smaller and the area of consolidation is choppier. The way to trade it is to wait for the breakout and trade in the direction of the move. Because they are continuation patterns, you will be looking to go long if you are in an up move and you find a pennant, or go short if you are in a down move and you find a pennant.
As we already told you, they look like triangles, but the price action inside of them can be very choppy. This is because after a directional move, traders will take small profits or take all of their positions out, which will make prices drop, and then traders will try to get in on the long position. Also, traders will try to push price down. That creates a consolidation period inside of this triangle-like formation. When it breaks out on the direction of the move you will be trading a long position in an up move on a pennant breakout.
Now for the stop losses. The stop losses should go below the pennant support of the previous low when in an up move, and in a down move, the stop loss should go below the pennant’s resistance or the previous high. Here’s an example. As you can see, we are in an up move, and we are trapped inside a pennant.
We already know that this is a continuation pattern, so we are looking for a breakout of this trend line. When we have the breakout of this trend line, we get a signal to go long, and our stop loss levels should go below the pennant’s support and below the previous low, because if price goes against us and breaks this ascending support and this low, it will mean that we are in a reversal and our long idea is no longer valid.
The targets are calculated like we calculate a flag breakout. When calculating a pennant target, you should measure the pull. The pull is the previous directional move before the consolidation period, and when you have the measurement in pips you should extrapolate it to the breakout zone. Here’s an example of how to calculate the target on a pennant breakout.
As you can see here, we are in an up move. Right before we got into this pennant, we calculate from below until the pennant’s start, and we extrapolate this measurement to the breakout zone, so we can get our target. When we do have the breakout, we have our stop loss levels and our target levels. If our stop loss levels get hit, it would mean that our trade idea was not valid, and if our targets get hit it would mean that we have made a good trade.
In either case the risk to reward ratio on a pennant is greater than one to two most of the time, so if you are looking to trade continuation patterns, you should be trading a very profitable strategy.
Let’s go through a trade analysis and see a live pennant in play. In this case we are going to go with the short position on a bearish pennant. You can see that here we are in a down move, and then we have a consolidation period. As you can see, it’s super choppy inside this pennant, but in any case we are making lower highs and higher lows, which means that we are inside of a pennant.
What we look for is a breakout to the downside, so we can go short and profit from the momentum of the breakout. You can see here that this candle gave us a fake out. This is why you need to have strong moves to the downside of strong breakouts for you to be able to profit from them, because if you trade false breakouts like this one, you will be trapped inside the range for a while, and you will be in a non directional trade, which will mean that your trade will not be making money and, most certainly, will be losing money.
The first thing we are going to do is calculate the targets of the pennant breakout. As we said before, we go for the high until the pennant starts. Here we have 100 pips. Then we extrapolate it from the breakout area, and we get our targets, which are 100 pips below the pennant. Then we just need to wait for the breakout, and in this case we have a very successful trade with a 2.2 to 1 risk to reward ratio.
As you can see, we have reversal patterns and we have continuation patterns. If you focus on trading these barters and actually being patient enough for them to complete and give you the signal to enter the market, you will be a profitable trader.
WordPress database error: [Table 'wp_investooprod.wp_icl_string_packages' doesn't exist]
`SELECT CONCAT(kind_slug, '-', name) FROM wp_icl_string_packages /* From [www.investoo.com/trade-bearish-bullish-pennants/] in [/nas/content/live/investooprod/wp-content/plugins/wpml-string-translation/classes/package/class-domains.php:34] */` | 1,249 | 5,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-43 | longest | en | 0.940039 |
http://lists.electorama.com/htdig.cgi/election-methods-electorama.com/2003-January/074648.html | 1,558,716,680,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257699.34/warc/CC-MAIN-20190524164533-20190524190533-00326.warc.gz | 122,648,312 | 2,249 | # [EM] 01/21/03 - Re: Wasted votes and quotas:
Donald Davison donald at mich.com
Tue Jan 21 03:11:25 PST 2003
01/21/03 - Re: Wasted votes and quotas:
Greetings Doug and list members,
Doug, you wrote: "The definition of "wasted" that I had in mind when I was
writing my
piece on quotas was the votes that remained with the runner-up after
all seats in an election had been filled." (Date: Sat, 18 Jan 2003)
Donald here, I understood perfectly what you were saying, you were clear.
Doug: "My reason for asking about the Meek counting method is that it
seemed to provide a method by which you could start with a Hare quota and
then reduce it using Meek ending up with all elected candidates elected
with an equal quota that utilised the maximum of
Donald: You are correct, but a very good way to avoid the conflict of Hare
vs Droop is to not have any quota. If no quota, then no surplus votes. If
no surplus votes, then no fractional transfer of any surplus votes. If no
fractional transfers, then no secondary fractional transfers of fractional
`paper'.
Wow! Look at all the math we will be avoiding.
This is possible with your plan because you will be using the new
elimination rule. If you think about it, you should come to the
realization that all the votes above average will be transferred
automatically by the new elimination rule anyway. Having a quota would be
moot.
If we were to do this, STV becomes moot.
The method is a variant of Bottoms Up aka Alternative Vote for Multi-Seat
elections. I call this variant `Davison-Bottoms Up'.
---- | 398 | 1,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-22 | longest | en | 0.962165 |
http://geoscien.neigae.ac.cn/EN/abstract/article/1000-0690/17634 | 1,726,621,914,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00508.warc.gz | 12,229,875 | 19,409 | 2012 , Vol. 32 >Issue 10: 1236 - 1240
Orginal Article
A New Algorithm of Vector Date Compression Based On the Tolerance of Area Error in GIS
• MI Xue-jun ,
• SHENG Guang-ming ,
• ZHANG Jing ,
• BAI Huan-xin ,
• HOU Wei
Expand
• Institute of Remote Sensing Applications Southern Yangtze,Shanghai 200436,China
Received date: 2012-03-22
Request revised date: 2012-07-10
Online published: 2012-10-20
Copyright
Abstract
：In GIS, vector data is the most commonly used data structure. Data compression is an issue in vector data processing and applications. In this paper,several commonly used algorithms of vector data compression are analyzed and a new efficient algorithm is proposed to resolve the problems of the classic algorithms. The Douglas-Peucker algorithm and the vertical distance tolerance algorithm are commonly used algorithms in vector data compression.The Douglas-Peucker algorithm have the advantage that it has invariance in translation and rotation, but at the same time the result has a big area error and there is a contradiction between the compression ratio and retention of feature points for the curvature change. The advantages of the vertical distance tolerance algorithm is fast, but the area error and the characteristics of the retention curve space are very poor. In this paper,a new algorithm is proposed which improved vertical distance tolerance algorithm and resolved the shortcomings of the Douglas-Peucker algorithm and the vertical distance tolerance algorithm.The basic idea of the new algorithm is based on the vertical distance tolerance algorithm which increase an area error tolerance by adopting the method of straight line fitting to approximate the axis of the polyline in order to resolve the problem of the area error and declination of segment in space. An experiment is included, which verified the new algorithm is efficient by the example of dealing with the boundary contour vector of Chongming Island, Shanghai. In the experiment ,the new algorithm has only 1 km2 error, but the classic algorithms has 6 km2 of the error. The most advantage of the new algorithms is that the area error can be controlled in a specified range. The experiments show that comparing with the two classic algorithms,the new algorithm has no substantial advantage in the compression ratio,but greatly improved the performance of two targets, area error and declination of segment in space. It proved that the new algorithm is efficient in the data procession which has high requirement in area accuracy and spatial characteristics such as the use of land resources.
Cite this article
MI Xue-jun , SHENG Guang-ming , ZHANG Jing , BAI Huan-xin , HOU Wei . A New Algorithm of Vector Date Compression Based On the Tolerance of Area Error in GIS[J]. SCIENTIA GEOGRAPHICA SINICA, 2012 , 32(10) : 1236 -1240 .
2 面积偏差控制下的线段动态拟合矢量数据压缩法
2.1 基本思想与算法实现
s=SACD+SEDF+SFGH+SHIB/|AB| （1）
2.2 案例分析
Fig. 5 Spatial overlay of among the original data , the classic algorithm and the author′s algorithm results in drawing of partial enlargement
Table 1 Area error comparison between the classic algorithm and the author’s
2、3号节点间 11.97 明显偏移 5.0 无明显偏移
4、5号节点间 7.06 明显偏移 5.0 无明显偏移
5、6号节点间 12.34 明显偏移 5.0 无明显偏移
6、7号节点间 5.87 明显偏移 5.0 无明显偏移
7、8号节点间 0.18 无明显偏移 0.18 无明显偏移
8、9号节点间 4.5 无明显偏移 4.5 无明显偏移
10、11号节点间 3.45 无明显偏移 3.45 无明显偏移
12、13号节点间 10.06 明显偏移 5.0 无明显偏移
15、16号节点间 17.64 明显偏移 5.0 无明显偏移
3 结 论
The authors have declared that no competing interests exist.
[1] 黄杏元. 地理信息系统概论[M].高等教育出版社,1989.
[2] 刘兆礼. 遥感影像屏幕数字化高效方法研究[J].地理科学,1999,19(3):250~253.
[3] 赵斌. 导航地理数据生产系统及其关键技术研究[D].郑州:中国人民解放军信息工程大学,2007.
[4] 黄培之. 具有预测功能的曲线矢量数据压缩方法[J].测绘学报,1995,24(4):316~319.
[5] 杨得志,王杰臣,闾国年.矢量数据压缩的Douglas-Peucker算法的实现与改进[J].测绘通报,2002,(7):18~19.
[6] 刘晓红,李树军.矢量数据压缩的角度分段道格拉斯算法研究[J].四川测绘,2005,28(2):51~52.
[7] 杨云,孙群,朱长青.曲线数据压缩的总体最小二乘算法[J].西安电子科技大学学报,2008,35(5):946~950.
[8] 陈飞翔,周治武,张建兵.基于动态规划算法的矢量数据压缩改进算法[J].计算机应用,2008,28(1):168~170.
[9] Jaafar J.Line Generalization: Least Square with Double Tolerance[C].Third International Conference on Management Information Systems Incorporating GIS & Remote Sensing. Southampton: Wessex Institute of Technology, 2002:135-144.
[10] Hershberger J, Snoeyink J.An O (nlogn) implementation of the Douglas-Peucker algorithm for line simplification[C].Proceedings of the Tenth Annual Symposium on Computational Geometry,1994,(6):383-384.
[11] Ramer U.An Iterative Procedure for the polygonal Approximation of Plane Curves[J]. Computer Graphics and Image Processing,1972,(1):244-256 .
[12] 杨海军,邵全琴.GIS 空间分析技术在地理数据处理中的应用研究[J].地球信息科学,2007,9(5):70~75.
[13] 彭认灿,董箭,郑义东,等. 垂距法与道格拉斯-普克法删除冗余顶点效率的比较[J].测绘通报,2010,(03): 66~67.
[14] 陈春,王野乔,薄立群,等.地理信息系统中矢量数据快速求交及其应用[J].地理科学,1990,10(2): 134~141.
Options
Outlines
/
〈 〉 | 1,571 | 4,757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.889875 |
http://math.stackexchange.com/questions/822422/what-is-mathbbq-sin2-pi-5-mathbbq-when-i-in-mathbbq-xi | 1,406,360,187,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997895170.20/warc/CC-MAIN-20140722025815-00020-ip-10-33-131-23.ec2.internal.warc.gz | 233,715,719 | 15,716 | # What is $[\mathbb{Q}(\sin(2 \pi / 5)) : \mathbb{Q}]$ when $i \in \mathbb{Q}(\xi)$ and when $i \not\in \mathbb{Q}(\xi)$? [closed]
I know how to find $[\mathbb{Q}(\cos( 2\pi / n )) : \mathbb{Q}]$ but for this I am lost! I have been working a looong time on this problem, any help would be greatly appreciated.
-
## closed as unclear what you're asking by Derek Holt, Najib Idrissi, user1729, Claude Leibovici, Davide GiraudoJun 6 at 9:29
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.
What is $\xi$ in this case? Also is $i$ the imaginary number? – DanZimm Jun 6 at 2:06
A related question with some discussion. – Jyrki Lahtonen Jun 6 at 6:26
It would be helpful if you stated the question in entirety in the body of your post...(Also, what do you mean by $[P:R]$? Index?) – user1729 Jun 6 at 8:27
I'm unsure what you mean by
when $i \in \mathbb{Q}(\xi)$ and when $i \not\in \mathbb{Q}(\xi)$
but I believe I can provide some help to finding $[\mathbb{Q}(\sin(2 \pi / 5)) : \mathbb{Q}]$.
Hint:
Take different powers of $\sin(2 \pi / 5)$ and create a linear combination of them with arbitrary constants in $\mathbb{Q}$. See if you can find constants so that the linear combination equals $0$. If so this is your minimal polynomial for $\sin(2 \pi /5)$
For example one way to find $[\mathbb{Q}(\sqrt{2}+\sqrt{3}) : \mathbb{Q}]$ is by doing the following:
$$(\sqrt{2} + \sqrt{3})^1 + a_0 = 0 \text{ has no solutions for } a_i \in \mathbb{Q} \\ (\sqrt{2} + \sqrt{3})^2 + a_1(\sqrt{2} + \sqrt{3})^1 + a_0 = 0 \text{ has no solutions for } a_i \in \mathbb{Q} \\ (\sqrt{2} + \sqrt{3})^3 + a_2(\sqrt{2} + \sqrt{3})^2 + a_1(\sqrt{2} + \sqrt{3})^1 + a_0 = 0 \text{ has no solutions with } a_i \in \mathbb{Q} \\ (\sqrt{2} + \sqrt{3})^4 + a_3(\sqrt{2} + \sqrt{3})^3 + a_2(\sqrt{2} + \sqrt{3})^2 + a_1(\sqrt{2} + \sqrt{3})^1 + a_0 = 0 \\\text{ has no solutions with } a_i \in \mathbb{Q}$$ since if we notice $$(\sqrt{2} + \sqrt{3})^2 = 5 + 2\sqrt{2}\sqrt{3} \\ (\sqrt{2} + \sqrt{3})^3 = (\sqrt{2} + \sqrt{3})(5 + 2\sqrt{2}\sqrt{3}) = 5\sqrt{2} + 5\sqrt{3} + 4 \sqrt{3} + 6\sqrt{2} = 11\sqrt{2} + 9\sqrt{3} \\ (\sqrt{2} + \sqrt{3})^4 = (\sqrt{2} + \sqrt{3})(11\sqrt{2} + 9\sqrt{3}) = 49 + 20 \sqrt{2}\sqrt{3}$$ so then we need to solve $$49 + 20\sqrt{2}\sqrt{3} + 11a_3\sqrt{2} + 9a_3\sqrt{3} + 5a_2 + 2a_2\sqrt{2}\sqrt{3} + a_1\sqrt{2} + a_1\sqrt{3} + a_0 = 0$$ with $a_i \in \mathbb{Q}$. This gives us the following equations: $$49 + 5a_2 + a_0 = 0 \\ 20 + 2a_2 = 0 \\ 11 a_3 + a_1 = 0 \\ 9a_3 + a_1 = 0$$ so we get that $a_1 = a_3 = 0, a_2 = -10$ and $$49 + 5(-10) + a_0 = 0 \implies a_0 = 1$$ so that the minimal polynomial of $\sqrt{2} + \sqrt{3}$ is $p(x) = x^4 - 10x^2 + 1 = 0$ thus $$[\mathbb{Q}(\sqrt{2} + \sqrt{3}) : \mathbb{Q}] = 4$$ Do you see how to apply this process to your polynomial?
- | 1,248 | 3,083 | {"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} | 3.546875 | 4 | CC-MAIN-2014-23 | longest | en | 0.825271 |
https://www.geocities.ws/gstumpff/fog/fog.html | 1,627,524,762,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153814.37/warc/CC-MAIN-20210729011903-20210729041903-00003.warc.gz | 794,681,533 | 8,784 | Glenn’s Treatise on a Preferred Color, or the Lack Thereof, of Fog Lights
For years, I was one of the many people who succumbed to the myth that fog lights are supposed to be yellow—all because I knew that Earth’s atmosphere scatters blue light better than that of longer wavelengths, thus explaining the blue sky and why the sun appears reddish at sunset. Never mind the fact that I was aware that fog droplets are a lot larger than air molecules and scatter differently. Never mind the fact that I’ve long had a “computational tool” with which to actually answer the question. I just made a “natural” generalization. Well, a couple of years back, when I first saw cars with those really pretty bluish-purple fog lights, I decided to actually use what I knew and answer the question. In case anyone else cares, and because I’m really bored evenings, I’ve documented my results here. (The conclusions here are not new. See, for example,
http://www.visualexpert.com/Resources/weather.html
or
First, the idea behind a preferred color is that it would give better visibility or penetration into the fog. “Penetration” relates to something known more technically as “extinction.” The process by which a beam of light becomes dimmer as it propagates through some medium is called “extinction.” (The medium extinguishes the light.) The less the extinction, the greater the “penetration.” The amount of extinction is quantified by an extinction coefficient. Call it “ALPHA”. If a beam of light has intensity I0 at one point, it will have intensity
I = I0 * EXP(-ALPHA * L)
after travelling a distance L from that point. (“EXP(X)” refers to raising the natural number “e” to the X power. I will always be less than I0.) Note that ALPHA generally depends on the wavelength. (It should be clear that wavelength is the physical parameter that correlates with what we call “color.” Visible light has wavelengths from .4 (blue/violet) to .7 (red) um. (“um” means micrometer (or micron). The “u” is supposed to be the Greek letter mu, but I’m doing this without using “fancy word processor features.”) So, the fundamental idea here is that fog lights should be of a color that makes ALPHA significantly smaller than at other colors. For the case in which the wavelengths involved are all much larger than the particles causing the scattering, ALPHA decreases smoothly as wavelength increases. This is shown in the plot below. This plot shows the extinction coefficient (ALPHA) as a function of wavelength for water drops of various sizes or various distributions of sizes. Consider the curve labelled “.005 um”, and use the scale on the right of the plot if you actually want to read values from this curve. This curve gives ALPHA for the case in which there are 200 drops per cm^3, all of .005 um diameter. This diameter is much smaller than any of the wavelengths our eyes are sensitive to and ALPHA does indeed decrease smoothly as wavelength increases. (Atmospheric molecules, which are of course even smaller, would show a curve that decreases in the same basic way.) The question arises that, if the “.005 um” curve was in fact descriptive of fog, why not use red light instead of yellow? After all, ALPHA is even smaller at .7 um than at .575 um (roughly where “yellow” is in the spectrum). All I’ve ever been able to come up with is that the human eye is generally more sensitive to yellow light than red and so it doesn’t make much sense to optimize lighting with wavelengths we don’t detect all that well.
The ultimate problem with using the “.005 um” curve as a guide to designing fog lights is that fog is composed primarily of droplets much larger than .005 um. Fog in fact contains a distribution of droplet sizes. The exact sizes involved, and the maximum diameters encountered, depend on the particular fog. Particles with sizes near or larger than the wavelength scatter light much differently than do very small particles. Small particles tend to scatter unpolarized light (e.g., the kind of light emitted by headlights and fog lights) more or less equally in all directions. Larger particles don’t scatter light so uniformly. (They tend to scatter light more into the forward hemisphere, i.e., in the general direction that the light was already heading. Strangely, the physics of what causes this actually decreases the beam intensity.) The following diagram shows the unpolarized scattering pattern of a very small water drop and that of a typical fog droplet. There simply is very little or no similarity between the two patterns. An air molecule scatters visible light (and longer wavelengths) with the same basic pattern as that of the small water drop in the diagram. (The patterns are relative. No implication is intended that a small water drop actually scatters as much into the forward direction as a fog drop or that the backscattering from a fog drop is actually less than that from a very small water drop. The backscattering from a fog drop is just much less than forward scattering.) This diagram should make it clear that whatever benefits yellow fog lights might have, they have absolutely nothing to do with what happens with sunlight and air molecules in the atmosphere.
The following plot shows two different models of the size distribution of fog droplets. (Size distributions that consist of non-constant particle sizes are called “polydispersions.” A distribution of constant-sized particles would be a “monodispersion.”) Do not get hung up on the difference between “advection fogs” and “radiation fogs.” You may have enough trouble figuring out how to interpret the curves, especially if you think about those strange units on the vertical scale. “per cm^3 per um”, what’s that? Don’t get hung up on that either. The really important point is that, as mentioned before, fog is composed of a whole bunch of different sized droplets. The fog examples shown here are most dense for drop radii of 10 or 2 um. But if you really want to understand those units on the vertical scale, okay, here goes how to interpret those curves. The curves represent density functions. But they don’t just give the drop density as a function of drop radius. There’s a double meaning behind the term “density function” here. They give the drop density per increment of drop size (radius). Let “dr” be a small interval of radius and “r” symbolize the radius. Symbolize either curve as the mathematical function “n(r)”. The way you would interpret a curve in this figure is to say that, for any given value of r, there are n(r)dr drops per cm^3 with radii between r and r + dr. More technically, dr is an “infinitesimal” quantity. The number of drops in a cubic centimeter with radii between r1 and r2 is given by integrating n(r)dr from r1 to r2. The total number of drops per cm^3 irrespective of size is given by integrating n(r)dr over all values of r (theoretically, r1 = 0 and r2 = infinity). For both curves, that all-encompassing integral gives N = 200 drops / cm^3 for the total density (the same value used to generate the “.005 um” extinction coefficient curve in the first figure).
Okay, with that digression out of the way, we can continue. Before considering all drop sizes at once, just consider a distribution of drops of diameter 20 um (where the advection fog curve maximizes). The curve labelled “20 um” in the first figure gives the extinction coefficient for this large drop size (with 200 drops / cm^2). As with all curves in that figure other than the “.005 um” one, use the vertical scale on the left. That curve in no way decreases smoothly as wavelength increases. It oscillates somewhat and very gradually increases with increasing wavelength (but not enough to make blue fog lights a particularly good idea). There is no clear preferred color for fog lights indicated there. The radiation fog drop size distribution maximizes at a diameter of 4 um. Even with this much smaller size the “4 um” curve in the extinction coefficient plot doesn’t show any particular preferred color for fog lights. The extinction coefficient is again virtually independent of wavelength.
Just to counter any arguments that this lack of a preferred color is being caused by assuming a monodispersion and that considering the full distribution of drop sizes might show something different, the initial extinction coefficient plot also shows ALPHA for the two polydispersions of the second plot (showing the size distribution functions). The oscillations of the “20 um” plot are greatly reduced in the “heavy advection fog” curve. Both it and the “moderate radiation fog” curve again show essentially no dependence on wavelength. Just to show the extinction behavior’s dependence with wavelength perhaps more explicitly, the following figure plots 100 * I / I0 (the percentage transmittance) from the above equation as a function of wavelength, for L = 10 m, using the extinction coefficient functions for the two polydispersive fog curves of the first plot. Because of the greatly expanded vertical scales, the slight oscillatory nature of the curves is shown. (Note that this same basic oscillatory behavior would’ve been seen in the first plot if a different vertical scale had been used.) In the figure below, again note that the lefthand vertical axis is used for one curve and the righthand vertical scale is used for the other. Again, looking at the average behavior with wavelength and ignoring the insignificant oscillations (remember that greatly expanded scales are used), if there is any systematic trend in this figure, it favors fog light colors corresponding to short wavelengths (blue or violet), not longer ones (yellow or red).
It might be noted that if some particular fog was composed of drops small enough for the phenomenon described by the “.005 um” extinction coefficient curve to apply (it’s called “Rayleigh scattering,” by the way), from the actual values of the extinction coefficients, it should be apparent that the extinction is so low even for blue light, that there would be no need to worry about the optimum color to use! As a final note on the alleged advantage of yellow light in regard to penetrability, if there was any significant preference for the shorter wavelengths to be scattered out of the beam, white headlights on oncoming cars would appear reddish (not, in fact, yellow), just as the setting sun sometimes appears reddish. I’ve never observed such an effect. (All I see is white.) | 2,274 | 10,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.929873 |
https://engineeringstatics.org/Chapter_05-2d-rigid-body-equilibrium.html | 1,721,567,577,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00006.warc.gz | 189,218,465 | 19,439 | Engineering Statics: Open and Interactive
Section5.42D Rigid Body Equilibrium
Two-dimensional rigid bodies have three degrees of freedom, so they only require three independent equilibrium equations to solve. The six scalar equations of (5.3.3) can easily be reduced to three by eliminating the equations which refer to the unused $$z$$ dimension. For objects in the $$xy$$ plane there are no forces acting in the $$z$$ direction to create moments about the $$x$$ or $$y$$ axes, so the reduced set of three equations is
\begin{align*} \{1\} \amp = \begin{cases} \sum F_x \amp =0 \\ \sum F_y \amp =0 \\ \sum M_A \amp =0 \end{cases} \end{align*}
where the subscript $$z$$ has been replaced with a letter to indicate an arbitrary moment center in the $$xy$$ plane instead of a perpendicular $$z$$ axis.
This is not the only possible set of equilibrium equations. Either force equation can be replaced with a linearly independent moment equation about a point of your choosing
1
Labels $$A\text{,}$$ $$B$$ and $$C$$ in these equations are representative. They don’t have to correspond to points $$A\text{,}$$ $$B$$ and $$C$$ on your problem.
, so the other possible sets are
\begin{align*} \{2\} \amp = \begin{cases} \sum F_x \amp =0 \\ \sum M_B \amp =0 \\ \sum M_A \amp =0 \end{cases} \amp \{3\} \amp = \begin{cases} \sum M_C \amp =0 \\ \sum F_y \amp =0 \\ \sum M_A \amp =0 \end{cases} \amp \{4\} \amp = \begin{cases} \sum M_C \amp =0 \\ \sum M_B \amp =0 \\ \sum M_A \amp =0 \end{cases} \end{align*}
For set four, moment centers $$A\text{,}$$ $$B\text{,}$$ and $$C$$ must form a triangle to ensure the three equations are linearly independent.
You have a lot of flexibility when solving rigid-body equilibrium problems. In addition to choosing which set of equations to use, you are also free to rotate the coordinate system to any orientation you like, pick different points for moment centers, and solve the equations in any order or simultaneously.
This freedom raises several questions. Which equation set should you choose? Is one choice ‘better’ than another? Why bother rotating coordinate systems? How do you select moment centers? Students want to know “how to solve the problem,” when in reality there are many ways to do it.
The actual task is to choose an efficient approach and carry it out. An efficient solution is one which avoids mathematical complications and makes the problem easy to solve. Complications include unpleasant geometries, unnecessary algebra, and particularly simultaneous equations, which are algebra intensive and error prone.
So how do you set up an efficient approach? First, stop, think, and look for opportunities to make the solution more efficient. Here are some recommendations.
1. Equation set one is usually a good choice and should be considered first.
2. Inspect your free-body diagram and identify the unknown values in the problem. These may be magnitudes, directions, angles or dimensions.
3. Align the coordinate system with at least one unknown force.
4. Take moments about the point where the lines of action of two unknown forces intersect, which eliminates them from the equation.
5. Solve equations with one unknown first.
Example5.4.1.Pin and Roller.
The L-shaped body is supported by a roller at $$B$$ and a frictionless pin at $$A\text{.}$$ The body supports a $$\lb{250}$$ vertical force at $$C$$ and a $$\ftlb{500}$$ couple-moment at $$D\text{.}$$ Determine the reactions at $$A$$ and $$B\text{.}$$
This problem will be solved three different ways to demonstrate the advantages and disadvantages of different approaches.
Solution 1.
Solutions always start with a free-body diagram, showing all forces and moments acting on the object. Here, the known loads $$C = \lb{250}$$ (down) and $$D= \ftlb{500}$$ (CCW) are red, and the unknown reactions $$A_x\text{,}$$ $$A_y$$ and $$B$$ are blue.
The force at $$B$$ is drawn along its known line-of-action perpendicular to the roller surface, and drawn pointing up and right because that will oppose the rotation of the frame about A caused by load C and moment D. The force at $$A$$ is represented by unknown components $$A_x$$ and $$A_y\text{.}$$ The sense of these components is unknown, so we have arbitrarily assigned the arrowheads pointing left and up.
We have chosen the standard coordinate system with positive $$x$$ to the right and positive $$y$$ pointing up, and resolved force $$A$$ into components in the $$x$$ and $$y$$ directions.
The magnitude of force $$B$$ is unknown but its direction is known, so the $$x$$ and $$y$$ components of B can be expressed as
\begin{align*} B_x \amp= B \sin \ang{60} \amp B_y \amp = B \cos \ang{60} \text{.} \end{align*}
We choose to solve equation set $$\{A\}\text{,}$$ and choose to take moments about point $$A,$$ because unknowns $$A_x$$ and $$A_y$$ intersect there. Substituting the variables into the equation and solving for the unknowns gives
\begin{align*} \sum F_x \amp = 0 \\ B_x - A_x \amp = 0\\ A_x \amp = B\ \sin \ang{60} \amp \amp (1)\\ \\ \sum F_y \amp = 0\\ B_y - C + A_y \amp= 0\\ A_y \amp= C - B\ \cos \ang{60}\amp \amp (2)\\ \\ \sum M_A \amp = 0 \\ -B_x( 3) -B_y(7)+C(4) + D \amp = 0\\ 3 B \cos \ang{60} + 7 B \sin \ang{60} \amp = 4 C + D \\ B (3 \sin \ang{60} + 7 \cos \ang{60})\amp = 4 C + D \\ B \amp = \frac{4 C + D}{6.098} \amp \amp (3) \end{align*}
Of these three equations only the third can be evaluated immediately, because we know $$C$$ and $$D\text{.}$$ In equations $$(1)$$ and $$(2)$$ unknowns $$A_x$$ and $$A_y$$ can’t be found until $$B$$ is known. Inserting the known values into $$(3)$$ and solving for $$B$$ gives
\begin{align*} B \amp = \frac{4 (250) + 500 }{6.098} \\ \amp = \frac{\ftlb{1500}}{\ft{6.098}} \\ \amp = \lb{246.0} \end{align*}
Now with the magnitude of $$B$$ known, $$A_x$$ and $$A_y$$ can be found with $$(1)$$ and $$(2)\text{.}$$
\begin{align*} A_x \amp = B \sin \ang{60} \\ \amp = 246.0 \sin \ang{60} \\ \amp = \lb{213.0}\\ \\ A_y \amp= C - B \cos \ang{60}\\ \amp= 250 - 246.0 \cos \ang{60}\\ \amp= \lb{127.0} \end{align*}
The positive signs on these values indicate that the directions assumed on the free-body diagram were correct.
The magnitude and direction of force $$\vec{A}$$ can be found from the scalar components $$A_x$$ and $$A_y$$ using a rectangular to polar conversion.
\begin{gather*} A = \sqrt{A_x^2 + A_y^2} = \lb{248.0}\\ \\ \theta = \tan^{-1} \left | \frac{ A_y}{A_x} \right | = \ang{30.8} \end{gather*}
The final values for $$\vec{A}$$ and $$\vec{B}\text{,}$$ with angles measured counter-clockwise from the positive $$x$$ axis are
\begin{equation*} \vec{A} = \lb{248.0}\ \measuredangle\ \ang{149.2}\text{,} \end{equation*}
\begin{equation*} \vec{B} = \lb{246.0}\ \measuredangle \ 30°\text{.} \end{equation*}
This solution demonstrates a fairly standard approach appropriate for many statics problems which should be considered whenever the free-body diagram contains a frictionless pin. Start by taking moments there.
Solution 2.
In this solution, we have rotated the coordinate system $$\ang{30}$$ to align it with force $$\vec{B}$$ and also chosen the components of force $$\vec{A}$$ to align with the new coordinate system.
There is no particular advantage to this approach over the first one, but with two unknown forces aligned with the $$x'$$ direction, $$A_{y'}$$ can be found directly after breaking force $$C$$ into components.
\begin{align*} \sum F_{x'} \amp = 0 \\ B - C_{x'} + A_{x'} \amp = 0\\ A_{x'} \amp = -B + C \sin \ang{30} \amp \amp (1)\\ \\ \sum F_{y'} \amp = 0\\ -C_{y'} + A_{y'} \amp= 0\\ A_{y'} \amp= C \cos \ang{30} \amp \amp (2)\\ \\ \sum M_A \amp = 0 \\ -B_x( 3) -B_y(7)+C(4) + D \amp = 0\\ 3 B \cos \ang{60} + 7 B \sin \ang{60} \amp = 4 C + D \\ B (3 \cos \ang{60} + 7 \sin \ang{60})\amp = 4 C + D \\ B \amp = \frac{4 C + D}{7.56} \amp \amp (3) \end{align*}
Solving equation (2) yields
\begin{equation*} A_{y'} = \lb{216.5}\text{.} \end{equation*}
Solving equation (3) yields the same result as previously
\begin{equation*} B = \lb{246.0}\text{.} \end{equation*}
Substituting $$B$$ and $$C$$ into equation (1) yields
\begin{align*} A_{x'} \amp = -B + C \sin \ang{30} \\ \amp = - 246.0 + 250 \sin \ang{30} \\ \amp = -\lb{121.0} \end{align*}
The negative sign on this result indicates that our assumed direction for $$A_{x'}$$ was incorrect, and that force actually points $$\ang{180}$$ to the assumed direction.
Resolving the $$A_{x'}$$ and $$A_{y'}$$ gives the magnitude and direction of force $$\vec{A}\text{.}$$
\begin{gather*} A = \sqrt{A_{x'}^2 + A_{y'}^2} = \lb{248.0}\\ \\ \theta = \tan^{-1} \left | \frac{ A_y}{A_x} \right | = \ang{60.8}\\ \alpha = \ang{180} - (\theta - \ang{30}) = \ang{149.2} \end{gather*}
Again, the final values for $$\vec{A}$$ and $$\vec{B}\text{,}$$ with angles measured counter-clockwise from the positive $$x$$ axis are
\begin{equation*} \vec{A} = \lb{248.0}\ \measuredangle\ \ang{149.2}\text{,} \end{equation*}
\begin{equation*} \vec{B} = \lb{246.0}\ \measuredangle \ 30° \end{equation*}
This approach was slightly more difficult than solution one because of the additional trigonometry involved to find components in the rotated coordinate system.
Solution 3.
For this solution, we will use the same free-body diagram as solution one, but will use three moment equations, about points $$B\text{,}$$ $$C$$ and $$D\text{.}$$
\begin{align*} \sum M_B \amp = 0 \\ - A_x(3) + A_y(7) - C(3) + D \amp = 0 \\ - 3 A_x + 7 A_y \amp = 250 \amp \amp (1)\\ \\ \sum M_C \amp = 0\\ -A_x(3) + A_y(4) - B_y(3) + D \amp= 0\\ -3A_x + 4 A_y - 3 B \cos \ang{60} \amp= -D \\ 3A_x - 4 A_y + 1.5 B \amp = 500 \amp \amp (2)\\ \\ \sum M_D \amp = 0 \\ - A_x(1.5) -B_x( 1.5 ) -B_y(7) + C(4) + D \amp = 0\\ 1.5 A_x + 1.5 B \sin \ang{60} + 7 B \cos \ang{60} \amp = 4 C + D \\ 1.5 A_x + 4.799 B \amp = 1500 \amp \amp (3) \end{align*}
This set of three equations and three unknowns can be solved with some algebra.
\begin{align*} 3 A_y + 1.5 B \amp = 750 \amp \amp (4) \end{align*}
Dividing equation (2) by 2 and subtracting it from (3) gives
\begin{align*} 2 A_y + 4.049 B = 1250 \amp \amp (5) \end{align*}
Multiplying (4) by 2/3 and subtracting from (5) eliminates $$A_y$$ and gives
\begin{gather*} 3.049 B = 750 \end{gather*}
\begin{gather*} B = \lb{246.0}\text{,} \end{gather*}
the same result as before.
Substituting $$B$$ into (3) gives $$A_x = \lb{213.0}\text{,}$$ and substituting this into (1) gives $$A_y = \lb{127.0}\text{,}$$ again the same result as before.
An alternate approach is to set these three equations up for a matrix solution and use technology to do the algebra, as done here with Sage.
\begin{equation*} \begin{bmatrix} -3 \amp 7 \amp 0 \\ 3 \amp -4 \amp 1.5 \\ 1.5 \amp 0 \amp 4.799 \end{bmatrix} \begin{bmatrix} A_x \\ A_y \\ B \end{bmatrix} = \begin{bmatrix} 250 \\ 500 \\ 1500 \end{bmatrix} \end{equation*}
A = Matrix([[-3,7,0],[3,-4, 1.5],[1.5,0,4.799]])
B = vector([250, 500, 1500])
x = A.solve_right(B)
x
This is a good example of an inefficient solution because of all the algebra involved. The issue here was the poor choice of $$B\text{,}$$ $$C$$ and $$D$$ as moment centers. Whenever possible you should take moments about a point where the line of action of two unknowns intersect as was done in solution one. This gives a moment equation which can be solved immediately for the third unknown. | 3,645 | 11,288 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-30 | latest | en | 0.844675 |
https://avt-mig.ru/problem-solving-with-linear-functions-key-12252.html | 1,624,470,526,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00150.warc.gz | 123,409,607 | 9,386 | # Problem Solving With Linear Functions Key
Tags: Essays To Write For CollegeWriting Lab Reports And Scientific PapersExamples Of Research Paper FormatWriting Ability Early Childhood ThesisHow To Write Essays For College ApplicationsSmall Manufacturing Business Plan
In this section, we will explore examples of linear function models.How can they calculate how much they will charge for an evening of babysitting?Emily is a college student who plans to spend a summer in Seattle.The rate of change is constant, so we can start with the linear model $$M(t)=mt b$$.Then we can substitute the intercept and slope provided.To find the x-intercept, we set the output to zero, and solve for the input.$\begin 0&=−400t 3500 \ t&=\dfrac \ &=8.75 \end$ The x-intercept is 8.75 weeks.The amount of money she has remaining while on vacation depends on how long she stays.We can use this information to define our variables, including units. The problem should list the Y- intercept, a starting amount of something and a slope, or a rate of change. You can tell that you need to create a linear equation by the information the problem gives you.
## Comments Problem Solving With Linear Functions Key
• ###### Problem Solving with Linear Graphs CK-12 Foundation
Feb 24, 2012. Learn how to use graphs to solve real-life linear problems. Step 1 Understand the problem and underline or highlight key information. Use these rates to draw a graph of distance in miles as a function of time in minutes.…
• ###### Modeling with Linear Functions - Mathematics LibreTexts
When modeling and solving a problem, identify the variables and. Linear models may be built by identifying or calculating the.…
• ###### Write linear functions to solve word problems Algebra 1. - IXL
Improve your math knowledge with free questions in "Write linear functions to solve word problems" and thousands of other math skills.…
• ###### Linear function word problems — Basic example video.
Watch Sal work through a basic Linear functions word problem. Solving linear equations and linear inequalities — Harder example · Interpreting linear.…
• ###### Problem Solving with Linear Graphs - Algebra Socratic
The best videos and questions to learn about Problem Solving with Linear Graphs. It makes sense to define a function that represents a dependency of the.…
• ###### Writing linear functions word problems Algebra practice.
Find a linear function that represents a real-world relationship that is given verbally. Functions are written using function notation.… | 520 | 2,536 | {"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} | 3.703125 | 4 | CC-MAIN-2021-25 | latest | en | 0.847546 |
https://cboard.cprogramming.com/c-programming/86735-question-about-double-float.html | 1,495,751,313,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608617.6/warc/CC-MAIN-20170525214603-20170525234603-00401.warc.gz | 736,784,356 | 12,057 | 1. ## question about double and float
Hi fellow programmers,
I am new to programming and working on learning C. I have am confused about the following. I tried to program celsius to fahrenheit converter. If i use float for celsius and fahrenheit, it works. but if i use those two variables as double then it only shows 32 as an answer. Is there something i neeed to know about doubles?? Here is the program.
Code:
```#include <stdio.h>
main(){
double celsius;
double fahrenheit;
printf("Enter temperature in Celsius: ");
scanf("%f", &celsius);
fahrenheit = 32 + celsius * 9.0 / 5.0;
printf("The temperature in Fahrenheit is %.2f" , fahrenheit);
fflush(stdin);
getchar();
}```
Thanks guys!
2. %lf for double in scanf
3. It worked! Thanks Bro! | 191 | 746 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-22 | longest | en | 0.749156 |
https://learn.podium.school/math/fractions-mutiplication-division/ | 1,725,938,727,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00487.warc.gz | 339,925,934 | 57,254 | # Everything You Need To Know About Multiplication and Division of Fractions
The Difference between Arithmetic and simple mathematics is confusing to people including kids. Arithmetic is a branch of mathematics that only deals with the study of numbers. Maths includes everything. The basic or the fundamentals of Mathematics are addition, subtraction, multiplication, and division.
Many cases talk about their experiences of kids loving addition and subtraction. But, when it came to multiplication it did not favor them much. Further in this article, we will put our focus on multiplication and division of fractions.
Before moving ahead with the topic the kids must know what is multiplication? Multiplication is denoted by the symbol ‘x’ or by an asterisk. In simple, the multiplication of whole numbers can be said as repeated addition. We could say that the multiplication of two numbers is the same as adding many copies to them.
The simple theory of multiplication is as follows:-
3×4=12, when broken down it, will look 4+4+4=12. This is the basics of multiplication.
## Properties of Multiplication
There are various properties of multiplication but, only three of them are considered important. These three properties are highly used in major parts of the world. The three main properties of multiplication are as follows:-
• Commutative multiplication
• Associative multiplication
• Identity multiplication
These are the three main properties of multiplication.
### Commutative multiplication
This property of multiplication says that just changing the order of the factor does not change the product. Here is an example:
4×3=3×4
Multiplying both sides would give the same result. The first paragraph contains an example for this.
### Associative multiplication
These properties of multiplication say that changing the group of factors does not change the product. In fact, the product for three or more numbers will remain the same. But, there are certain terms and conditions for this. Here is an example:-
(2×3) x 4= 2 x (3×4)
Solve “=” first. Solving the first part will result in the following:
(2×3) x 4
=6×4
=24
Now, we will move towards the right-hand side of the problem. Follow the same process as in the first part.
2 x (3×4)
=2 x 12
= 24
Moreover, we can see that both sides equal 24 as the final answer. In fact, we did not even multiply them by the same number. Multiply first half by 2 and 3, and 3 and 4 in the right side.
### Identity multiplication
This is the simplest property of multiplication. In fact, this property of multiplication says that the product of 1 or any number is that number. Moreover, any number multiplied by 1 will be the original number. Here is an example for you:-
8×1=8
Additionally, it does not matter if 1 comes before or after the consequence will be the same. Here is another example for the same:-
1×8=8
Three highly used and common properties of multiplication are –
## What Are The 3 Steps To Multiplying Fractions?
In mathematics, fractions are part of the Arithmetic branch. A fraction consists of a number that expresses a quotient. Moreover, this quotient includes a numerator which is dividing the denominator.
Proper fractions are those in which the numerator is less than the denominator. Improper fractions are those in which the numerator is greater than the denominator. Mixed factors are sums of whole numbers and proper fractions.
You can also add, subtract, or divide a fraction and multiply it. To multiply fractions, one can do it in three simple steps.
### Steps To multiplying fractions:-
1. Simply multiply the top numbers. They are numerators.
2. Multiply the bottom ones after numerators. These lower ones are denominators.
3. If necessary, simplify the fraction.
There is a simple example for multiplication fraction below:-
1/2 x 2/5
First, Multiply the top numbers or numerators.
1/2 x 2/5 = 1×2 =2 (numerator answer)
Second, Multiply bottom numbers or denominators.
1/2 x 2/5 = 1x 2/2×5 = 2/10
Thirdly, if the final result can be simplified into short-form then just simplify it if possible.
2/10= 1/5
The simplification is used with various techniques for the kids to understand better. There is the pizza method, pen, and paper method, the rhyme method, and many more.
## Multiplying Fractions with Whole Numbers
Fractions and multiplications can be done with various numbers and types. The same way fraction with whole numbers becomes slightly different but, easy. In fact, it is one of the easiest ways to solve a whole number fraction.
The example for a whole number with a fraction is given below:-
5 x 2/3 here 5 will be counted as 5/1
2/3 x 5/1
Firstly, again we have to follow the first step to multiply the numerators.
2 x 5 will be the multiplication for the numerator and 3 x 1 will be for the denominator.
So, the final answer to them would be 10/3.
In fact, the same can be in another way where we do not take any denominator under a whole number. But, that might be slightly confusing for the kids to understand fractions at this age. Moreover, there is multiplication with mixed fractions as well. Overall, there might be various types but, the steps for them remain the same.
## How Do You Multiply Fractions With Different Denominators?
Multiplying fractions with numerators is quite easy but when it comes to denominators that become quite tough. This is especially for the kMultiplying fractions with numerators is easy, but denominators are quite tough. This is especially true for kids who are in the fourth-seventh grade. We know that in every fraction, there is a top number and a bottom number to deal with.
The numerator in the fraction tells us how many units we have of a whole. On the other hand, the denominators tell us how many units make up the whole. For example, if we take 2/3, 2 here is the numerator and 3 is the denominator.
We can see that there are two units as a whole, but when it comes to a fraction that is not the case. Firstly, to multiply fractions the basic way has been discussed above. Additionally, fractions will be given on both sides and they will need to be multiplied. In fact, the multiplication sequence will be numerator x numerator and denominator x denominator.
### Steps to Multiply Fractions with Different Denominators
Unlike denominators is also very easy to multiply. One can easily do a simple multiplication with unlike denominators. The steps to do so are quite the same as multiplication of like fractions. Below is an example of the multiplication of fractions with unlike denominators.
Example: Multiplication fraction of 4/12 x 16/24
There are two different methods to solve the above-mentioned problem. The first one is given below:
1. Multiply the numerators to make it simple, 4 x 16= 64
2. Follow the same procedure to multiply the denominators, 12 x 24= 288
3. The final answer that we get here by solving the fraction is 64/288. Moreover, this number can be reduced into a much simpler form. Therefore, we will then get 2/9 which is the final answer.
### Alternate Method
Interestingly, the same example with the same numbers can be solved by another simple method. Moreover, in this method, we will simplify the fractions among themselves. After doing this, we will be multiplying the numerators then, the denominators will be multiplied.
Example: Multiplication fraction of 4/12 x 16/24
Step 1. Simplify the fractions among themselves without multiplication. So, the fraction can be reduced to 1/3 x 2/3. This is the first and simple step to reduce and simplify the fraction.plication. So, the fraction can be now reduced to 1/3 x 2/3. This is the first and simple step to reduce and simplify the fraction.
Step 2. Simplify the numerator. 1 x 2= 2
Step 3. The denominators would need to be simplified. In fact, the denominator cannot be simplified prior to the numerators. Unfortunately, doing so will just cause a mess to the fraction and the result will be wrong. The denominators are, 3 x 3 = 9.
Step 4. Therefore, the final answer by solving the fraction we get is 2/9.
## Fractions with Mixed Numbers Multiplication
Mixed fractions are quite different to solve as compared to other variants. Moreover, mixed fractions consist of a whole number and a proper fraction. In fact, the fraction needs to be converted with the whole number by multiplication. 23/4 is a mixed fraction, where 2 is a whole number and ¾ is a proper fraction.
Firstly, to multiply the mixed fraction, we need to change the mixed fraction into a simple fraction. Now, for example, if the mixed fraction is 22/3, we can change that into 8/3. An example is given below for a better understanding.ge the mixed fraction into a simple fraction. Now, for example, if the mixed fraction is 22/3, we can change that into 8/3. An example is given below for a better understanding.
Example: Fraction multiplication of 22/3 and 31/4
1. The first step in this mixed fraction will be to convert it into a simple fraction. The whole number 2 will be multiplied by the denominator 3 which will result in 6. Moreover, after this, result 6 will need to be added by the numerator 2 that is 6+2=8. So, the answer to the first problem will be 8/3 x 13/4.
2. Now, the numerators of the improper fractions will be multiplied followed by denominators. The final result after that will come as 104/12.
3. Now, simply convert the fraction to a much simpler form by dividing the denominator with the numerator. Here, that is possible and the answer will be 26/3.
4. Interestingly, the final answer can again be converted back to a mixed fraction. So, doing that the final result will be 82/3.
This is how multiplication with mixed fractions is done. Additionally, there are other ways and techniques to do it but, this is the best and simple way.
## Improper Fractions of Multiplication
We have learned two types of fractions and how to multiply with them. Moreover, even fractions with different denominators are very easy to multiply. But, multiplying improper fractions can be a little tricky. This is where the fractions need to be simplified and again bring the result back to mixed fractions.
Moreover, when there are two improper fractions to multiply, we frequently end up with an improper fraction. Let’s take an example with two improper fraction multiplication.
Example: 3/2 x 7/5
Step 1: Firstly, the numerators will need to be multiplied and followed by the denominators. So, (3 x 7)/ (2 x 5) = 21/10
Step 2: Interestingly, the result of solving the above question leads to an improper fraction. In fact, this improper fraction cannot be reduced into a much simpler form.
Step 3: Therefore, the final answer to the above question is 21/10 which can be changed into a mixed fraction. The result will then be, 21/10.
Improper fractions can be tricky sometimes but, if the base knowledge is correct then that might not happen. Additionally, we have discussed all the forms of fractions with multiplications. Above were the basic terms for conducting multiplications with fractions.
## Dividing Fractions
The Division is one of the important operations under the four mathematical operations. In fact, division more or less works quite the same as subtraction. Moreover, the primary aim for division is to split the large groups into equal smaller groups.
In fact, the division is a primary arithmetic operation where various numbers are combined and divided. Now, these numbers are combined in such a way that it forms a new number. The same way division is used very often when it comes to fractions.
## Division in Fractions
The base formula of division remains the same but, slightly changes when done in fractions. In fact, dividing two fractions is the same as multiplying the first by reciprocal and the second by fraction. Moreover, the first step of dividing fractions is just to find the reciprocal of the second fraction.
The next simple step to follow is to multiply the two numerators followed by the denominators. Finally, one can simplify the fraction if needed, or else the answer will remain as it is.
The example for fraction division is given below:
5/8 ÷ 15/16
1: Firstly, we will substitute the value of the numerators then followed by the denominators.
2: The result after substituting will become: 5/8 ÷ 15/16 = 5/8 x 16/15 = 2/3.
3: Now, if we simplify the above answer then the final answer would turn to 5/8 ÷ 15/16 = 2/3.
This is the basic concept for conducting fractions using division operations in mathematics. Now, we will be talking about how to simplify fractions with whole numbers.
## Division with Whole Number Fractions
The Division of fractions is quite different when they are compared to multiplication. Now, division with whole numbers is quite the same process as multiplication. Firstly, we need to multiply the denominator here of the fraction with the whole number.
In fact, the first step with the whole number will be the same as with multiplication. Still, let us take an example for the following:
2/3 ÷ 4 = 2/3 x1/4
= 1/6
Now, the third step after this step would just be to simplify the result. Therefore, for the above answer, we get 1/6 as the final answer.
## Dividing Fractions with Decimals
Before moving any forward we must know what is decimal. Decimal is a subpart of algebra which is another branch in Mathematics. It can be defined as a number whose whole number part and the fractional part are separated by a point. Now, this separation part of the number is called decimal.
Moreover, the dot that we put in between the numbers is called the decimal point. In fact, the digits following after the dot showcases the value smaller than one.
Now, decimal numbers are a fraction to base 10. In most cases, we can represent the decimal in the fractional form and then divide them. There are two simple steps to divide fractions with decimals and they are given below:
• Firstly, convert the given decimal to a fraction to make it look easier.
• Secondly, and lastly divide both the fractions using the simple method.
Now, if we take the example of, 4/5 ÷ 0.5. Here, we can see 0.5 as the decimal that needs to be divided in the fraction. Interestingly, the 0.5 here can be converted to 5/10 or 1/2. Moreover, now the division in fraction can be done very easily.
So, the simplified question would now be 4/5 by 1/2. Further simplifying the problem would turn into 4/5 ÷ 1/2 = 4/5 x 2/1 = 8/5. This is how decimals can be changed into a fraction and then divided with the other numbers.
## Two Ways of Dividing Fractions
There are three-four ways of dividing fractions but, we will talk about the highly used ones. In fact, the first method of dividing fractions is given above. The next two methods are given below:-https://learn.podium.school/downloads/division-with-unit-fractions-fractions-3/
### Method 1: Cross-Multiplication
Step 1: This method of dividing the fraction is quite simple. Firstly, it consists of multiplying the numerator of the first fraction with the denominator of the second. This will get you a result that needs to be written down in the resulting fraction’s numerator.
Step 2: Secondly, we will then multiply the denominator of the first fraction with the numerator of the second. Again, we will need to write the answer in the resulting fraction’s denominator.
Step 3: Thirdly, after we get an answer for both sides just simplify it if possible.
Now, let us take an example for such a case.
Example: 3/4: 6/10
The first step here would be to multiply the first fraction 3 with the denominator of the second 10. Now, doing that will give us the following fraction: 3 x 10 = 30. This answer will be written in the resulting fraction’s numerator.
Secondly, we have to multiply the denominator of the first fraction 4 with the numerator of the second 6. Now, doing that will result in 4 x 6 = 24. This answer will be written in the resulting fraction’s denominator.
Thirdly, the last step will be to simplify the fraction. Since both the numbers are divisible by 6, so we can simply divide the numerator and denominator by 6. Now, doing that will result in 30 ÷ 6 = 5 and 24 ÷ 6 =.
Moreover, the final result or the answer to the question would be 5/4.
### Method 2: Inverting and Multiplying
This is another great way to solve fractions with division. In fact, we could say that it is a cross-multiplication process but, with a slight change. Let us look at the steps to divide in fractions using this method.
Step 1: The second fraction of the question must be inverted. In simple, you just need to swap the numerator for the denominator.
Step 2: Secondly, you just need to simplify the numerator with any denominator given in the question.
Step 3: Thirdly, and lastly, the interesting part is to multiply them across. This will then get you a different result and if possible just simplify that.
An example for such a method is given below:-
Example: 12/6: 6/4
1: As we have mentioned earlier that we need to invert the second fraction in the question. So, 6/4 will be 4/6.
2: Secondly, the numerators in the question will need to be simplified with the denominators. So, the numerators are:
12 = 2 x 2 x 3
4 =2 x 2
Denominators are:
5 = 5
6 = 2 x 3
Now, we can just simplify the numbers if they come out in common or divisible by any number. Doing this process will make the division method quite easy.
• What are the basic rules for multiplying fractions?
Answer- Moreover, there are two simple rules when it comes to multiplying fractions. The first rule is to multiply numerators, and then the denominators. Now, the second rule is to simplify the obtained fraction and get the final answer.
• Why there is a need for multiplication in fraction?
Answer- A fraction is multiplied because it can break into smaller parts to make it simpler. In fact, these smaller parts can be chosen.
• What is multiplication for kids?
Answer- Multiplication is nothing but taking one number and then adding it together with a number of times.
## Conclusion
After some of the intriguing concept accompanied with wonderful tips and tricks, you might be curious much to explore more. Right? So, for more information, do not forget to visit our Blog site. | 4,199 | 18,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2024-38 | latest | en | 0.937698 |
http://www.sevenforums.com/general-discussion/138157-need-help-couple-homework-questions.html | 1,485,053,831,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281331.15/warc/CC-MAIN-20170116095121-00320-ip-10-171-10-70.ec2.internal.warc.gz | 683,281,263 | 14,631 | Join Forum | Login | Today's Posts | Tutorials | Windows 10 Forum | Windows 8 Forum
Welcome to Windows 7 Forums. Our forum is dedicated to helping you find support and solutions for any problems regarding your Windows 7 PC be it Dell, HP, Acer, Asus or a custom build. We also provide an extensive Windows 7 tutorial section that covers a wide range of tips and tricks.
# Windows 7: Need help with a couple of homework questions??
16 Jan 2011 #1 skers5xs windows 7 2 posts Lincoln, NE Need help with a couple of homework questions?? I am working with a computer that has a total storage space of 1TB...of this total space I have used 78GB. 1. Assuming that I should not use more than 85% of the total available space, how much total storage should I use? 2. If I use an average of 500MB of storage each week, how many weeks before I run out of storage space? How do I figure out these questions? so #1 do I do 1TB x .85? #2 do I do 1TB divided by 500? Please help....thanks My System Specs
.
16 Jan 2011 #2 Punkster Windows 7 Ultimate SP1 (64-bit) 2,310 posts Valencia, VE. 1TB HDD: ~931,32GB or ~953671,68MB Used Space on HDD: ~80GB or ~81920MB Available Space on HDD: ~871751,68MB 85% of Available Space on HDD: ~740988,93MB ~500MB per Week ~1481 Weeks or ~28 years until 85% of the HDD fills up (with the previous data) Of course this is theoretical. Real world is another thing but that'll give you an idea. This is only an approach, you need to find out how many GBs do you get left after formatting. then substract the used GBs to that number, then multiply the available space by 0.85 (which is the percentage you want to use), then divide by the number of MB per week you want and then you have the number of weeks, divide weeks by 52 to get how many years and that's it My System Specs
16 Jan 2011 #3 xxxdannyxxx Windows 7 Home Premium x64 SP1 3,278 posts England There are 1024 gig in a TB(binary) therefore 85% of that is 870.4 gig (1024/100*85) There are 1000 MB in a gig (non binary) so roughly thats a gig used every 2 weeks so 870gig *2=1740 weeks or 33.4 years. Your ok for space at the moment My System Specs
.
16 Jan 2011 #4 skers5xs windows 7 2 posts Lincoln, NE Thanks a bunch guys! Much appreciated. My System Specs
16 Jan 2011 #5 Kari Windows 10 Pro x64 EN-GB 17,819 posts A Finnish ex-pat in Leipzig, Germany Hi Skers5xs, welcome to the Seven Forums. You got the idea, I see. Question 1 is simple: 1 TB = 1024 GB, so 0.85 * 1024 = 870.4. You should use maximum of 870 GB of that hard disk. The second question: if it is a direct follow up of question 1, then divide 870 GB (1 GB = 1024 MB so 870 GB = 890,880 MB) with 500 MB > 890,880 / 500 = 1,781.76, so you'd need 1,781.76 weeks to fill that space (about 34 years). If the second question means how long to fill the complete 1 TB, then it is 1,048,676 MB divided with 500 ( 1 TB = 1024 GB = 1,048,676 MB), so 1,048,676 / 500 = 2,097.152. Filling 1 TB hd takes in this case 2,097.152 weeks (about 40 years) Kari My System Specs
16 Jan 2011 #6 Cato Windows 10 Home 275 posts Home of the Missouri Mule. And if you're my age, you have more HD space than years left. I have 3TB total, so I better get busy if I want to fill them up. My System Specs | 956 | 3,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-04 | latest | en | 0.897954 |
http://questionbankcollections.blogspot.com/2011/09/571101-statistics-for-management-anna.html | 1,511,265,978,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806353.62/warc/CC-MAIN-20171121113222-20171121133222-00148.warc.gz | 238,862,868 | 14,331 | Search This Blog
571101 STATISTICS FOR MANAGEMENT ANNA UNIVERSITY MBA QUESTION PAPER JAN 2011 REGULATION 2010
Sunday, September 11, 2011 ·
M.B.A. DEGREE EXAMINATION, JANUARY 2011.
First Semester
571101 — STATISTICS FOR MANAGEMENT
(Regulation 2010)
Time : Three hours Maximum : 100 marks
Statistical tables are permitted.
PART A — (10 × 2 = 20 marks)
1. Define probability by the way of mathematical approach.
2. Write the probability mass function of Poisson distribution
3. Write the standard error of sampling distributions of mean and proportion.
4. What is the use of central limit theorem?
5. What do you mean by Type I and Type II errors?
6. Write the use of sign test.
7. Explain the term “Run” with an example.
8. KW test is the non-parametric ANOVA – Why?
9. What do you interpret if the r = 0 and r = –1?
10. What do you mean by error variation?
PART B — (5 × 16 = 80 marks)
11. (a) A disciplinary committee is formed from the staff of XYZ company which
has three departments Marketing. Finance and Production of size 10, 5
and 20 members respectively. All departments have two female staff
each. A department is selected at random and from which two members
are selected for the committee,
(i) What is the probability that both the team members are female? (6)
(ii) If the committee formed with female only, then find the Bayesian
probability that the committee has come from Marketing, Finance
and Production departments respectively. (10)
Or
(b) (i) A sales representative can convert a customer as potential buyer
with the probability of 70%. If he is able to meet the 10 customers
in day, find the probability of converting (9)
(1) atleast one customer
(2) not even a single customer
(3) exactly one customer.
(ii) N = 1000; n = 5; p = 50% then find P(X = 2) by binomial distribution.
(7)
12. (a) Explain Non-random sampling methods in detail stating the merits and
demerits.
Or
(b) (i) Explain the properties of good (point) estimator. (8)
(ii) What do you mean by interval estimation? Give examples. (8)
13. (a) (i) Explain the procedure for testing the two sample proportion with
population proportion comparison. (8)
(ii) IQ test result of randomly selected five employees in an
organization is given below. Test whether minimum requirement of
average IQ level 87 is maintained in that company or not. (8)
Employee Code : 234 232 121 343 111
IQ test : 85 95 90 93 87
Or
(b) Test whether the association of income level and interest on buying a
new model car is significant or spurious from a study conducted from
2000 members randomly selected from an area.
Income group Buying a new model car
Interested Not interested
Low income (below Rs. 10000) 200 200
Medium income (Rs. 10000-50000) 400 600
High income (above Rs. 50000) 200 400
14. (a) (i) Distinguish Nonparametric methods over parametric methods. (8)
(ii) Explain the KW test procedure with appropriate examples. (8)
Or
(b) Time of service by two cashiers in a period in a Bank are given below :
Customer number : SB 12309 CuA/c 32 SB 453 SB 0093 CuA/c 21 SB 123
Cashier A (in min) : 12 23 4 5 16 17
Customer number : SB 30909 CuA/c 12 SB 678 SB 0093 SB A/c 121
Cashier B (in min) : 2 3 10 8 12
Test whether the service time varies significantly between the operators
using Mann-Whitney U test.
15. (a) The following table presents the results of a survey of 8 randomly
selected families :
Annual Income (in 000 Rs.) : 8 12 9 24 13 37 10 16
Per cent allocation for investment : 36 25 33 15 28 19 20 22
Find Karl Pearson’s correlation and Spearman’s rank correlation
methods for the above data. (10 + 6)
Or
(b) (i) Explain various methods of trend analysis for financial time series
data. (8)
(ii) An electronics and appliance store sells three different brands of
DVD players. The three brands sold by the store are Brand A,
Brand B, and Brand C. The unit prices for the years 2000 and 2010.
with the volume of sales (units sold) for 2000, are given below.
Unit price \$ Unit sold
DVD Player 2000 (P0) 2010 (Pt) 2000
Brand A 700 900 200
Brand B 500 600 300
Brand C 300 420 500
(1) Compute an unweighted aggregate price index for DVD
players with 2000 as the base period. (4)
(2) Compute a weighted aggregate price index (Laspeyre’s index)
for 2010 with 2000 as the base period. (4) | 1,178 | 4,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-47 | longest | en | 0.891321 |
https://vustudents.ning.com/group/mth641-functional-analysis | 1,639,010,329,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363641.20/warc/CC-MAIN-20211209000407-20211209030407-00092.warc.gz | 653,819,887 | 18,096 | www.vustudents.ning.com
We non-commercial site working hard since 2009 to facilitate learning Read More. We can't keep up without your support. Donate.
# MTH641 Functional Analysis
## Information
MTH641 Functional Analysis.
Download / Upload Video Lectures, Handouts, Helping Materials, Assignment Solution, Online Quizzes, GDB, Past Papers, Solved Papers & more….
Members: 23
Latest Activity: 9 hours ago
## Discussion Forum
### MTH641 Functional Analysis Assignment No.01 Fall 2021 Solution / Discussion
Started by + M.Tariq Malik. Last reply by + M.Tariq Malik 9 hours ago.
### MTH641 Functional Analysis Assignment No 02 Fall 2020 Solution / Discussion Due Date:15-02-2021
Started by + M.Tariq Malik. Last reply by + M.Tariq Malik Feb 11.
### MTH641 Functional Analysis GDB Fall 2020 Solution / Discussion
Started by + M.Tariq Malik. Last reply by Asad waqas Jan 27.
### MTH641 Grand Quiz + Mid Term Quiz Fall 2020 Prepartion Material Due Date: 28-12-2020
Started by + M.Tariq Malik. Last reply by + M.Tariq Malik Dec 28, 2020.
### MTH641 Functional Analysis Online Quiz No 01 Fall 2020 Solution / Discussion
Started by + M.Tariq Malik. Last reply by + M.Tariq Malik Nov 30, 2020.
### MTH641 - Functional Analysis Handouts, PPTs Slides, Recommend Books, FAQs, Glossary & More Helping Material
Started by + M.Tariq Malik Nov 8, 2020.
### MTH641 Current Final Term Papers Spring 2020 & Solved MCQs, Short Notes, Solved Past Papers, Solved Online Quizzes, E-Books, FAQs, Short Questions Answers & More
Started by + M.Tariq Malik. Last reply by + M.Tariq Malik Sep 16.
### MTH641 Functional Analysis Assignment Spring 2020 Solution & Discussion
Started by + M.Tariq Malik. Last reply by + M.Tariq Malik Aug 21, 2020.
### MTH641 Current Final Term Papers Fall 2019 (15 to 26 February 2020) & All Solved Past Papers, Solved MCQs & Helping Material
Started by + M.Tariq Malik. Last reply by + M.Tariq Malik Feb 21, 2020.
### MTH641 Functional Analysis Assignment No 01 Fall 2019 Solution & Discussion
Started by + M.Tariq Malik. Last reply by + M.Tariq Malik Nov 28, 2019.
### MTH641 Final Term Papers Spring 2019 (24 August ~ 04 September 2019) & All Solved Past Papers, Solved MCQs & Helping Material
Started by + M.Tariq Malik. Last reply by + M.Tariq Malik Aug 30, 2019.
### MTH641 Assignment No 02 Spring 2019 Solution & Discussion Due Date: 25-07-2019
Started by + M.Tariq Malik. Last reply by Rao Waqar Jul 23, 2019.
### MTH641 Solution
Started by Mudassar Iqbal Jun 3, 2019.
### Untitled
Started by chaudry umer Jun 2, 2019.
### MTH641- Assignment#01 Due date 03-Jun-2019
Started by Rao Waqar. Last reply by + M.Tariq Malik Jun 3, 2019.
## Comment Wall
Comment
Comment by Waheed Ahmad on May 31, 2019 at 12:10am
mth641 ka solution file send kry agr kisi k pas hy tho
1
2
3
4
5
## Latest Activity
Attas updated their profile
4 hours ago
Ahmed Salman, Abdul Wahab, mubashar and 4 more joined Virtual University of Pakistan
5 hours ago
Abdul Wahab joined + M.Tariq Malik's group
### MGT503 Principles of Management
5 hours ago
mubashar, AbdulRabecca and Wasim joined + M.Tariq Malik's group
### PAD603 Governance, Democracy and Society
5 hours ago
Abdul Wahab joined + M.Tariq Malik's group | 952 | 3,244 | {"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} | 2.75 | 3 | CC-MAIN-2021-49 | latest | en | 0.714186 |
https://rdrr.io/cran/plot3logit/man/effect.html | 1,623,828,136,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487622234.42/warc/CC-MAIN-20210616063154-20210616093154-00567.warc.gz | 402,762,461 | 8,992 | # effect: Draw a change in the probability distribution on an existing... In plot3logit: Ternary Plots for Trinomial Regression Models
## Description
Given the first two probabilities of a trinomial distribution before and after a change, `effect()` adds an arrow to an existing ternary plot. If the probability distribution does not change, a point (instead of an arrow) is added to the plot.
## Usage
`1` ```effect(x, y, ..., length = 0.05) ```
## Arguments
`x, y` `numeric` vectors of the first two probabilities. If the probability distribution is unchanged, `x` and `y` should have length one. `...` other graphical parameters such as `xpd` and the line characteristics `lend`, `ljoin` and `lmitre`. See `graphics::par()`. `length` length of the edges of the arrow head (in inches).
## Warning
Only when `effect()` is passed to `Ternary::AddToTernary()` as the first argument, arrows and points are drawn consistently with ternary coordinate system, otherwise `effect` draws both arrows and points according to a Cartesian coorinate system centered on (0,0.5,0.5).
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15``` ```TernaryPlot() # Arrow plot3logit:::effect(c(0.1, 0.2), c(0.3, 0.2)) # Point plot3logit:::effect(0.3, 0.2) # Compare the two coordinate systems plot3logit:::effect(0.1, 0.2, pch = 19, col = 'red') AddToTernary(plot3logit:::effect, list(c(0.1, 0.2, 0.7)), pch = 19, col = 'blue') # The origin of Cartesian coordinate systems plot3logit:::effect(0, 0, pch = 19, col = 'red') AddToTernary(plot3logit:::effect, list(c(0, 0.5, 0.5)), pch = '+', col = 'white') ```
plot3logit documentation built on Feb. 26, 2021, 5:06 p.m. | 522 | 1,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-25 | latest | en | 0.720711 |
https://www.engineersedge.com/engineering-forum/search.php?s=5c5736974163f48fc00eadaec04d6738&searchid=1430439 | 1,563,858,678,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528869.90/warc/CC-MAIN-20190723043719-20190723065719-00326.warc.gz | 665,327,128 | 11,719 | Search:
Type: Posts; User: Hudson
Page 1 of 6 1
1. Thread: how do they do it ?
by Hudson
Replies
2
Views
240
Are they multi-million dollar dies? Years ago,...
Are they multi-million dollar dies? Years ago, when the earth was thought to be cooling, auto makers used a free machining, low melting alloy called Kirksite for fender dies.
More recently,...
2. Thread: How to Interpret Geometric Tolerance and Dimensional Tolerance
by Hudson
Replies
4
Views
1,580
You have a .002 cylindricity and a roundness call...
You have a .002 cylindricity and a roundness call out. That seems like double dimensioning to me.
Could you just call out a .005 cylindricity and a surface finish on the basic shaft diameter and...
3. Thread: How much does the flow of a pipe get reduced if undersized couplings are used?
by Hudson
Replies
2
Views
645
Look harder for 1" couplings. They should be...
Look harder for 1" couplings. They should be less than a dollar for the PVC type.
4. Thread: how to calculate the steel weight?
by Hudson
Replies
4
Views
1,781
.283 lbs/cubic inch
.283 lbs/cubic inch
by Hudson
Replies
1
Views
473
If you know the distance from one pivot to...
If you know the distance from one pivot to another on the links, it is a simple triangle problem:
With angle theta and the distance from the two base pivot points and the length of the link...
6. Thread: backup power for old sailboat
by Hudson
Replies
6
Views
712
Take another look at the Parker catalog (fine...
Take another look at the Parker catalog (fine folks they are) and see if you haven’t misread the volume as gallons when the actual figure is in liters (57).
57L is more like 15 gallons or 2...
7. Thread: How to manufacture this spiral part?
by Hudson
Replies
2
Views
359
Much simpler to mold is a vortex spiral.
Much simpler to mold is a vortex spiral.
8. Thread: backup power for old sailboat
by Hudson
Replies
6
Views
712
Regarding hydraulics, you cannot compress a...
Regarding hydraulics, you cannot compress a liquid so your accumulator piston must work against some sort of spring like compressed air or steel. So the real energy storage is in the spring.
...
9. Thread: Case hardening steel processes
by Hudson
Replies
10
Views
771
by Hudson
Replies
3
Views
751
You might look at a stepper motor and ball screw...
You might look at a stepper motor and ball screw configuration that would be positioned under the table to provide linear force.
Look for a 400 mm travel length linear stepper motor on Amazon. ...
by Hudson
Replies
3
Views
751
Most mechanism to accomplish this motion will...
Most mechanism to accomplish this motion will require a gear reduction. Since we don't know the motor speed, friction etc. it is not possible to give you a direct answer.
You want to move 300...
12. Thread: Help needed to design a bed on a winch/pulley system!
by Hudson
Replies
5
Views
870
Look at some existing designs for folding or...
Look at some existing designs for folding or Pullman berths. Some are ceiling mount - some are wall mount. Also consider getting double duty out of another piece of furniture like a table or a...
13. Thread: Tubing size for cantilevered aluminum beam
by Hudson
Replies
5
Views
571
If you recall, the deflection equation used a...
If you recall, the deflection equation used a length cubed term. The difference between 42 cubed and 36 cubed is considerable.
We are now looking for a tube with a section moment of...
14. Thread: Tubing size for cantilevered aluminum beam
by Hudson
Replies
5
Views
571
Increasing the allowable deflection by 2X reduces...
Increasing the allowable deflection by 2X reduces the moment of inertia required by ½.
A 3” square tube with an 1/8” wall has a section moment of inertia of approximately 1.98 inches^4. A...
15. Thread: Tubing size for cantilevered aluminum beam
by Hudson
Replies
5
Views
571
We don’t know quite enough to design a davit for...
We don’t know quite enough to design a davit for you.
What we can guess is that that 200 lbf end load can get a lot higher under sea if the dinghy fills with storm water and starts to...
16. Thread: Help with Proper Torque Spec
by Hudson
Replies
6
Views
907
117 N/mm2 is a stress level equivalent to 117 MPa...
117 N/mm2 is a stress level equivalent to 117 MPa or approximately 17,000 psi.
This rather low stress for a bolt. This might be based on the unusual thermal requirements of a fireproof...
by Hudson
Replies
6
Views
1,137
Are you attaching this gusset by welding?
Are you attaching this gusset by welding?
by Hudson
Replies
8
Views
592
19. Thread: Self locking aluminum extruded U channels
by Hudson
Replies
7
Views
414
Why not use rectangular tubing?
Why not use rectangular tubing?
20. Thread: Can anyone help me identify which material is better?
by Hudson
Replies
4
Views
392
Also think about heat treating. Most steel gears...
Also think about heat treating. Most steel gears are carburized.
21. Thread: Hydraulic Total Cost of Ownership
by Hudson
Replies
2
Views
612
Ask the sales force of the high priced hydraulic...
Ask the sales force of the high priced hydraulic fluid to justify the cost. If they don't have good numbers, move on to your next project.
22. Thread: Controlling low air pressure
by Hudson
Replies
2
Views
431
You might use two valves in series. The first...
You might use two valves in series. The first would open a 2psi and the second close at 3.5 psi. You could also combine an electrical with a mechanical valve to handle separate conditions.
23. Thread: Deck Joist Bending Calculations
by Hudson
Replies
3
Views
413
I attach a segment from "Building Construction...
I attach a segment from "Building Construction Illustrated" by Francis D. K. Ching (A favorite of mine on the subject) which suggests that cantilevers of 24" or less might not need engineering...
24. Thread: Deck Joist Bending Calculations
by Hudson
Replies
3
Views
413
Your efforts to calculate the forces and...
Your efforts to calculate the forces and deflection are commendable however, your should look into any local codes to see if they have a 'span table' that they use for structures that are not...
25. Thread: Three (3) Point Suspension System for Measuring Flatness
by Hudson
Replies
3
Views
738
You have not shown us the part that you are...
You have not shown us the part that you are checking. Can you make a fixture that has two supports with an indicator between them so that you can place the fixture directly upon the part in the...
Results 1 to 25 of 147
Page 1 of 6 1 | 1,627 | 6,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-30 | latest | en | 0.87562 |
http://lexic.us/definition-of/decile | 1,513,374,710,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948579567.73/warc/CC-MAIN-20171215211734-20171215233734-00073.warc.gz | 166,960,215 | 8,330 | ### Definition of Decile
1. Noun. (statistics) any of nine points that divided a distribution of ranked scores into equal intervals where each interval contains one-tenth of the scores.
Category relationships: Statistics
### Definition of Decile
1. Noun. (statistics) Any of the values in a series that divides the distribution of individuals in that series into ten groups of equal frequency. ¹
2. Noun. Any one of the ten subsets or groups so divided. ¹
¹ Source: wiktionary.com
### Definition of Decile
1. a statistical interval [n -S]
### Medical Definition of Decile
1. An aspect or position of two planets, when they are distant from each other a tenth part of the zodiac, or 36 deg . Origin: F. Decil, fr. L. Decem ten cf. It. Decile. Source: Websters Dictionary (01 Mar 1998)
### Decile Pictures
Click the following link to bring up a new window with an automated collection of images related to the term: Decile Images
### Lexicographical Neighbors of Decile
deciduous plantdeciduous teethdeciduous toothdeciduousnessdeciduousnessesdecievedecigradedecigradesdecigramdecigramme decigrammesdecigramsdecikataldecikatalsdecile (current term)decilesdeciliterdecilitersdecilitredecilitres decilliondecillionsdecillionthdecillionthsdecimaldecimal arithmeticdecimal digitdecimal dozendecimal fraction
### Literary usage of Decile
Below you will find example usage of this term as found in modern and/or classical literature:
1. Gold, Prices, and Wages Under the Greenback Standard by Wesley Clair Mitchell (1908)
"Lowest 1st decile 2d decile 3d decile 4th decile Median 6th decile 7th decile 8th decile January April ..."
2. Fact Finding Report: Commission on the Future of Worker-Management Relations by John T. Dunlop (1994)
"Measures of the gap between the earnings of workers in the highest decile of earnings and those in the lowest decile show that the US earnings distribution ..."
3. A Lifelong Passion: Nicholas and Alexandra: Their Own Story by Andrei Maylunas (2005)
"... cancer at several sites, especially cancers of the stomach and lung among men. The relative risk of lung cancer between the highest and lowest decile ..."
4. Rice Market Liberalization and Poverty in Viet Nam by Nicholas Minot, Francesco Goletti (2000)
"Figure 6—Rice consumption by expenditure decile Kilograms/person/year 200 175 150 125 100 1 2 3 4 5 6 7 8 9 Expenditure decile 10 Source: Data from the Viet ..."
5. La Niäna and Its Impacts: Facts and Speculation by Michael H. Glantz (2002)
"3-15 Regions with rainfall below decile 3 during November-January of warm ENSO events initiated in 1957, 1972. 1976. 1986. and 1991. ..."
6. Rural Growth Linkages: Household Expenditure Patterns in Malaysia and Nigeria by P. B. R. Hazell, Ailsa Röell (1983)
"Marginal budget shares by per capita expenditure decile in Muda, 1972/73 13. Marginal budget shares by per capita expenditure decile in Gusau, 1976/77 14. ..." | 758 | 2,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | longest | en | 0.772495 |
http://www.flow3d.com/resources/news_10/assessing-mesh-resolution-for-boundary-layer-accuracy.html | 1,398,305,032,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223204388.12/warc/CC-MAIN-20140423032004-00304-ip-10-147-4-33.ec2.internal.warc.gz | 572,012,253 | 8,210 | Home | Contact Us | Users Site SEARCH:
# Assessing Mesh Resolution for Boundary Layer Accuracy
## How does mesh cell size affect my flow simulation near an obstacle?
When simulating flow around and over solid obstacles like river beds, submerged obstacles, or duct walls, FLOW-3D calculates the near-wall velocity based on the flow being laminar or turbulent. From the near-wall velocity, other values, such as shear velocity, wall shear stress, and velocity farther from the wall are computed. The size of the mesh near obstacles is important in order to get physical results for these variables. This article describes a simplified way of assessing the best size for cells near the wall, and some ways to implement this in practice.
## How should I set friction and roughness?
The friction coefficient (FRCOF globally, or OFRCOF for individual components) should be negative (the default value is -1) so the surface is treated as a no-slip surface. If the flow is laminar everywhere in the simulation, the surface roughness of the obstacle component (ROUGH) should be set to zero. If the flow is turbulent or transitional, ROUGH should be the effective grain size corresponding to uniformly spaced elements. This may be derived from empirical formulae (Manning's n or Strickler's k, for example) if elements are not uniformly sized or spaced. Having selected appropriate friction and roughness coefficients for our components, we turn to mesh resolution issues.
## Mesh cell resolution in the boundary layer
The first cell adjacent to an obstacle surface is where the logarithmic or laminar wall velocity profile is applied. The cells along the surface are either normal to the surface, if the surface is on a gridline, or they contain the wall surface. If the flow is laminar, direct differentiation of the velocity profile will be performed, so the cell average will always be correct. In this case, the velocity profile is better resolved as the mesh is refined. The optimal cell size depends only on the required profile accuracy and the tolerated computational time, both of which increase as the cells get smaller.
If a turbulence model is active, the first cell near the obstacle always gets its velocity according to a logarithmic profile corresponding to the log-law region shown in Figure 1. The first cells along the wall should be sized so that they include the viscous sub-layer and end well within the log-law region of the boundary layer. If the first cell outer edge falls in the viscous sub-layer or extends into the outer or free-stream region, then the actual near-wall velocity and shear stress deviate from the log-law calculation (Figure 1). Finding the correct cell size is a matter of estimating the height of the boundary layer regions normal to the solid surface.A helpful value for this is the non-dimensional normal distance from the wall y+, sometimes referred to as the viscous length.
## Finding an appropriate cell size
In the equations below (illustrated in Figure 1), uτ is the shear velocity, τw is the shear stress on the solid, y is the normal distance from the solid, ρf is the fluid density, and μf is the fluid dynamic (molecular) viscosity.
In order to estimate y+, the shear stress τw must be estimated manually, and the interested reader is referred to the hydraulics literature for help with this. In general, y+ (as a function of the cell size) should be greater than 30 (where the inner layer transitions smoothly into the log-law region) and less than a value that depends on the Reynolds number of the flow and the thickness of the boundary layer (generally 100 to 500 is a reasonable upper limit). When manual estimation of τw is impossible, multiple simulations can be used to iterate toward a ‘best fit’, where the observed value (shear stress or velocity) levels off as shown in Figure 2.
## Tips on implementing boundary-layer cell size
Implementation of a ‘best’ cell size is fairly simple. If the component surface is angled so that it matches the direction of the gridlines (as shown in Figure 2), then fixed points should be used at the surface itself and at the appropriate distance from the surface (so that the first cell distance y meets the y+ criteria just explained). If the obstacle surfaces are not parallel to the mesh gridlines, use nested mesh blocks so that the cells nearest the surfaces are of an appropriate size.
The approximations used to calculate parameters at solid surfaces assume that flow is fully developed, and caution should be exercised when interpreting the results for undeveloped flows.
^ back to top | 940 | 4,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2014-15 | latest | en | 0.890577 |
https://www.teacherspayteachers.com/Store/All-Things-Algebra/Type-of-Resource/Graphic-Organizers?ref=filter/resourcetype/top | 1,537,518,928,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156901.47/warc/CC-MAIN-20180921072647-20180921093047-00345.warc.gz | 862,368,503 | 30,277 | # All Things Algebra
(21,057)
United States - Virginia
4.0
You Selected:
Types
CUSTOM CATEGORIES
Subject
Prices
Top Resource Types
• Graphic Organizers
My Products
sort by:
Rating
view:
Linear Equations Line Match Activity: This activity includes 32 cards in which students write a linear equation given either standard form, the graph of the line, a point and slope, or two points. Every four cards will represent the same equation.
Subjects:
Types:
\$3.00
398 ratings
4.0
Pre-Algebra Review (For Algebra 1 Students) Flip Book #prealgreview This flip book was created to use with Algebra 1 students as a review/refresher of the following Pre-Algebra topics: • Properties (commutative, associative, identify, inverse,
Subjects:
Types:
\$4.00
220 ratings
4.0
Use this flip book to review systems of equations, including: types of solutions (one solution, no solution, and infinite solution), methods for solving systems of equations (graphing, substitution, and elimination), and applications. There are 18
Subjects:
Types:
CCSS:
\$3.00
212 ratings
4.0
Algebra 1 Review (For Geometry and Algebra 2 Students) Flip Book This flip book was created to use with Geometry or Algebra 2 students as a review/refresher of the following Algebra 1 topics: • Multi-Step Equations • Exponent Rules (simplifying
Subjects:
Types:
\$4.00
211 ratings
4.0
This flip book provides a compact way for students to review and practice transformations. The topics are divided into the following 5 tabs:• Reflections• Translations• Rotations• Dilations • Identifying Transformations and Writing RulesReflections
Subjects:
Types:
CCSS:
\$3.00
186 ratings
4.0
Exponent Rules Review BookThis book reviews the following exponent rules: product rule, quotient rule, and power rule, negative exponent rule, and zero exponent rule. Adding and subtracting monomials (combining like terms) is also included.In
Subjects:
Types:
\$3.00
155 ratings
4.0
Exponent Rules Graphic Organizer This is graphic organizer reviews the following exponent rules (or laws): product rule, power rule, quotient rule, negative exponent rule, zero exponent rule, and adding/subtracting (like terms only). It is a great
Subjects:
Types:
FREE
148 ratings
4.0
Circles Review (Arcs, Angles, Special Segments) Flip Book Students can use this flip book to review concepts taught during the circles unit, including: -Identifying Parts of Circles: Center, Chord, Diameter, Radius, Central Angle, Inscribed
Subjects:
Types:
\$4.00
147 ratings
4.0
Congruent Triangles (SSS, SAS, ASA, AAS, and HL) Flip Book This flip book will help students organize methods of proving triangles congruent. This includes Side-Side-Side (SSS), Side-Angle-Side (SAS), Angle-Side-Angle (ASA), Angle-Angle-Side
Subjects:
Types:
\$3.00
135 ratings
4.0
Factoring Polynomials Review BookThis book reviews factoring polynomials using the following methods: GCF (greatest common factor), difference of squares, trinomials were a = 1, trinomials where a > 1, and grouping of four terms. A section with
Subjects:
Types:
\$3.00
135 ratings
4.0
Parent Functions and Transformations Reference Book This reference book was created to use as a review of transformations and the following function families: linear, absolute value, quadratic, cubic, square root, cube root, exponential,
Subjects:
Types:
\$4.00
131 ratings
4.0
Trigonometry Flip Book This trigonometry flip book will help students review how to find missing sides and angle measures in right triangles using the sine, cosine, and tangent ratios. Organized practice, mixed practice, and applications with
Subjects:
Types:
\$3.00
125 ratings
4.0
Quadrilaterals Flip Book This flip book reviews properties of quadrilaterals, including parallelograms, rectangles, rhombi, squares, trapezoids, and isosceles trapezoids. Practice problems also included in which students apply the properties.
Subjects:
Types:
\$3.00
115 ratings
4.0
Parent Function Posters for Algebra 2 This is a set of nine parent function posters to display in your classroom. Parent functions included: linear, absolute value, quadratic, cubic, square root, cube root, reciprocal, exponential, and
Subjects:
Types:
\$3.00
115 ratings
4.0
Volume and Surface Area Flip Book Students can use this flip book to review finding the volume and surface area of prisms, cylinders, pyramids, cones, spheres, and hemispheres. Some problems require the Pythagorean Theorem to find a missing
Subjects:
Types:
\$3.00
110 ratings
4.0
Factoring Polynomials Graphic Organizer I use this graphic organizer to review all types of factoring just before I play my Factoring "Spin to Win" game. It is a good visual to help students with factoring polynomials. It covers the
Subjects:
Types:
FREE
96 ratings
4.0
Subjects:
Types:
\$3.00
96 ratings
4.0
Quadratic Equations: Flip Book This flip book is intended as a review of graphing quadratic equations, along with solving quadratic equations by factoring, square roots, completing the square, and the quadratic formula. Each topic includes a short
Subjects:
Types:
\$3.00
96 ratings
4.0
Graphing Polynomial Functions Flip Book This flip book was created to be used as a stations activity to provide extra practice with graphing polynomial functions and identifying the following key characteristics: Turning Points (Relative Minimum
Subjects:
Types:
\$3.00
87 ratings
4.0
Direct and Inverse Variation Cut and Paste Activity The first page is a side-by-side reference sheet for students to review the direct and inverse variation concepts. This helps reinforce the following: -How to check for direct/inverse variation
Subjects:
Types:
\$2.00
79 ratings
4.0
showing 1-20 of 43
### Ratings
Digital Items
4.0
Overall Quality:
4.0
Accuracy:
4.0
Practicality:
4.0
Thoroughness:
4.0
Creativity:
4.0
Clarity:
4.0
Total:
91,292 total vote(s)
TEACHING EXPERIENCE
I have taught Math 8, Algebra, Honors Algebra, and Geometry here in Virginia!
MY TEACHING STYLE
I am a fan of structured notes that are easy to for students to read, followed by lots of fun activities to keep them engaged and practice their new skills! I am ALWAYS looking for new ideas that present concepts to my students in a fun and exciting way. I believe that the appearance of what we give our students matters greatly. In a day where we are competing with the newest gadgets, we need to grab their attention and interests.
HONORS/AWARDS/SHINING TEACHER MOMENT
MY OWN EDUCATIONAL HISTORY
I graduated from the State University of New York at Brockport in 2006 with a degree in Mathematics, Computer Science, and Education.
Originally from Orchard Park, NY, I relocated to Virginia with my husband, and our two young boys. I have taught Pre-Algebra, Algebra I, and Geometry to 8th graders and LOVE IT! Not only is teaching my passion, but also creating materials that students can relate to, find more appealing, and make learning more fun. | 1,654 | 6,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-39 | latest | en | 0.880617 |
https://learn.k20center.ou.edu/lesson/572 | 1,686,101,917,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653501.53/warc/CC-MAIN-20230607010703-20230607040703-00688.warc.gz | 397,213,952 | 7,733 | # My Teacher's Record-Breaking Cup
## Partial Quotients Division Method
K20 Center, Christine Cox | Published: September 22nd, 2020 by Oklahoma Young Scholars/Javits
• Subject Mathematics
• Course Elementary Mathematics
• Time Frame 1-2 class period(s)
• Duration 120 minutes
### Summary
Students will use a fun problem to discover and practice the partial quotients process for multiple-digit division.
### Essential Question(s)
How can a large number be divided?
### Snapshot
Engage
Students engage in a division problem dealing with their teacher's favorite cup.
Explore
Students use manipulatives to construct different ways to divide a multi-digit number.
Explain
Students first explain how and why they divided the multi-digit number, and then the teacher formalizes new knowledge by connecting to the method of partial quotients for division.
Extend
Students use the Commit and Toss strategy to practice writing and solving division problems using the partial quotients method.
Evaluate
Students individually solve another division problem using the partial quotient method.
### Materials
• The teacher's favorite water bottle, coffee cup, or other beverage container
• 1000, 100, 10, and 1 blocks
• Paper/pencils
• Whiteboard/markers for practice
### Engage
Go to slide 3. Display a reusable cup or mug and share your LOVE for your favorite beverage (coffee, water, soda, etc). Explain that you refill this cup five (5) times a every day.
Go to slide 4. Students will use a Think, Pair, Share activity and collaborate to find the solutions for each of the following questions:
• How many refills would I have in one school week (Monday through Friday only)?
• How many refills would I have in one month?
• About how many refills would I have in a year?
### Explore
Go to slide 5. Explain to students that you have tracked the number of times you have refilled your cup, but not the number of days. Tell them, "I have refilled my cup 5,235 times, and I refill it 5 times each day."
Go to slide 6. Pass out copies of the Strategy Harvest handout from the attachments. Ask students to use the Strategy Harvest strategy and figure out how many days the cup has been used.
Allow students to think about how to represent the number 5,235 and also attempt to solve the problem. Tell them that it might be helpful to think about how we represent large numbers first. Students might first rewrite the number using expanded notation (5000+200+30+5) or base 10 blocks or rewrite it into friendly numbers like 5000+200+35.
Go to slide 7. As students begin to finish the problem, instruct them to find others who are done and compare notes about what strategies each used to solve the problem. Students should record details about others' strategies on their Strategy Harvest handout.
### Explain
Go to slide 8. As a class, have students share strategies that they have used to solve the problem. Build upon student responses and be ready to connect their strategies to the partial quotients strategy.
Go to slide 9. Walk through the partial quotients method by talking about how students divided the quotient to start. You can point out how different groups approached the problems for examples.
• When you divide the quotient, do you start with the smallest pieces or the biggest pieces first? The thousands or the ones?
• Model how to “show the work.”
Go to slide 10. Work through this problem as a class:
759 ? 3
Go to slide 11. Have students work in pairs to solve the problems below. After each problem, have the students pair up with another pair and review the other pair's strategies and answers.
1,926 ? 6
3,380 ? 4
### Extend
Go to slide 12. Follow the prompts on slides 12-17 to engage in a version of the Commit and Toss strategy that will help students practice estimating and evaluating their work.
1. (Slide 12) Write a division problem on the paper. Toss and pick up a new paper.
2. (Slide 13) Use estimation to find a reasonable answer for the written problem. Discuss, toss, and pick up a new paper.
3. (Slide 14) Solve the division problem on the paper. Discuss, toss, and pick up a new paper.
4. (Slide 15) Evaluate/provide feedback for the solution.
5. (Slide 16) Write a “real-world” situation to match the problem. Remember that key words such as “per,” “each,” and “evenly between” often cue us that a problem is a division problem.
6. (Slide 17) Share problems with the class as time permits.
### Evaluate
Go to slide 18. Pass out copies of the Exit Ticket handout and have students answer the questions below.
Mrs. Kennedy is preparing a STEM activity where students will build a bridge out of plastic drinking straws. She has 642 straws that need to be shared between the 6 groups of students. Estimate how many straws each group gets.
Next, calculate the exact number of straws each group gets. Remember to show your work. | 1,096 | 4,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-23 | longest | en | 0.894113 |
https://api-project-1022638073839.appspot.com/questions/what-was-the-pre-sale-price-of-a-tv-that-is-35-off-and-has-a-new-price-of-195 | 1,591,294,733,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347445880.79/warc/CC-MAIN-20200604161214-20200604191214-00279.warc.gz | 242,135,909 | 6,141 | # What was the pre-sale price of a TV that is 35% off and has a new price of $195? ##### 1 Answer Nov 13, 2014 Let $x$be the pre-sale price of the TV. Since35% off means that 100-35=65% of the pre-sale price, we have $\frac{65}{100} x = 195$by multiplying by $100$, $\implies 65 x = 19500$by dividing by 65, $\implies x = 300$Hence, the pre-sale price of the TV was$300.
I hope that this was helpful. | 135 | 401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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} | 4.03125 | 4 | CC-MAIN-2020-24 | latest | en | 0.970001 |
https://www.sawaal.com/calendar-questions-and-answers/the-last-day-of-a-century-cannot-be_3151 | 1,702,219,879,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102469.83/warc/CC-MAIN-20231210123756-20231210153756-00200.warc.gz | 1,070,241,379 | 14,018 | 904
Q:
# The last day of a century cannot be
A) Monday B) Wednesday C) Tuesday D) Friday
Explanation:
Q:
Lata decides to go for a walk every alternate day. In a particular month she starts on a Friday. On which of the following days will she be mina for a walk the following week?
A) Wednesday, Friday, Sunday B) Monday, Wednesday, Friday C) Thursday, Saturday, Sunday D) Tuesday, Thursday and Saturday
Explanation:
16 2546
Q:
Afroze was born on the $2nd$of February 2015, while Avash was born 555 days later. On which date was Avash born?
A) 11thAugust 2016 B) 9thAugust 2016 C) 10thAugust 2016 D) 8thAugust 2016
Explanation:
16 3999
Q:
Rohit remembers that his birthday is before 12th December and after 6th December. His wife remembers that Rohit’s birthday is after 7th December. His son remembers
that Rohit’s birthday is before 10th December. On which date of December is Rohit’s birthday?
A) 8 B) 9 C) 10 D) 8 or 9
Explanation:
6 1668
Q:
What day of the week was 29 June 2010 ?
A) Sunday B) Monday C) Wednesday D) Tuesday
Explanation:
9 5698
Q:
Which year will have the same calendar as that of 2005?
A) 2009 B) 2010 C) 2011 D) 2008
Explanation:
7 6820
Q:
If 19th May 2012 was a Saturday, what day was 1st April 2014?
A) Friday B) Tuesday C) Monday D) Thursday
Explanation:
19 3273
Q:
What day would it be on 1st March 2020?
A) Monday B) Saturday C) Friday D) Sunday
Explanation:
17 3605
Q:
5th February 2018 is a Monday. 8th April 2019 will be a
A) Wednesday B) Tuesday C) Monday D) Sunday | 480 | 1,542 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-50 | latest | en | 0.91454 |
https://proofwiki.org/wiki/Leigh.Samphier/Sandbox/Definition:Circuit_Axioms_(Matroid)/Definition_3 | 1,618,357,403,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038075074.29/warc/CC-MAIN-20210413213655-20210414003655-00429.warc.gz | 583,744,683 | 10,095 | # Leigh.Samphier/Sandbox/Definition:Circuit Axioms (Matroid)/Definition 3
## Definition
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$.
$\mathscr C$ is said to satisfy the circuit axioms if and only if:
$(C1)$ $:$ $\displaystyle \O \notin \mathscr C$ $(C2)$ $:$ $\displaystyle \forall C_1, C_2 \in \mathscr C:$ $\displaystyle C_1 \neq C_2 \implies C_1 \not \subseteq C_2$ $(C3'')$ $:$ $\displaystyle \forall X \subseteq S \land \forall x \in S:$ $\displaystyle \paren{\forall C \in \mathscr C : C \not \subseteq X} \implies \paren{\exists \text{ at most one } C \in \mathscr C : C \subseteq X \cup \set x}$ | 221 | 644 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-17 | longest | en | 0.393496 |
https://newbedev.com/how-to-get-the-mse-of-the-node-in-the-decisiontreeregressor-of-scikit-learn | 1,680,270,759,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00062.warc.gz | 480,309,507 | 6,939 | # How to get the MSE of the node in the DecisionTreeRegressor of scikit-learn?
Nice question. You need tree_reg.tree_.impurity.
tree_reg = tree.DecisionTreeRegressor(max_depth=2)
tree_reg.fit(X_train, y_train)
extracted_MSEs = tree_reg.tree_.impurity # The Hidden magic is HERE
for idx, MSE in enumerate(tree_reg.tree_.impurity):
print("Node {} has MSE {}".format(idx,MSE))
Node 0 has MSE 86.873403833
Node 1 has MSE 40.3211827171
Node 2 has MSE 25.6934820064
Node 3 has MSE 19.0053469592
Node 4 has MSE 74.6839429717
Node 5 has MSE 38.3057346817
Node 6 has MSE 39.6709615385
## Long answer using the boston dataset with visual output:
import pandas as pd
import numpy as np
from sklearn import ensemble, model_selection, metrics, datasets, tree
import graphviz
X_train, X_test, y_train, y_test = model_selection.train_test_split(
pd.DataFrame(house_prices.data, columns=house_prices.feature_names),
pd.Series(house_prices.target, name="med_price"),
test_size=0.20, random_state=42)
tree_reg = tree.DecisionTreeRegressor(max_depth=2)
tree_reg.fit(X_train, y_train)
extracted_MSEs = tree_reg.tree_.impurity # YOU NEED THIS
print(extracted_MSEs)
#[86.87340383 40.32118272 25.69348201 19.00534696 74.68394297 38.30573468 39.67096154]
# Compare visually
dot_data = tree.export_graphviz(tree_reg, out_file=None, feature_names=X_train.columns)
graph = graphviz.Source(dot_data)
#this will create an boston.pdf file with the rule path
graph.render("boston")
Compare MSE values with visual Output: | 433 | 1,505 | {"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} | 2.515625 | 3 | CC-MAIN-2023-14 | latest | en | 0.682094 |
http://www.ehow.co.uk/how_8083131_calculate-pipe-size.html | 1,532,078,126,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591575.49/warc/CC-MAIN-20180720080634-20180720100634-00580.warc.gz | 441,130,994 | 11,726 | DISCOVER
# How to Calculate a Pipe Size
Updated March 23, 2017
Wider pipes transmit fluid at higher rates, with all other factors remaining constant. The wider pipe contains a larger volume of water in any given length of pipe. A required volumetric flow rate therefore corresponds with needed pipe width. Other factors also affect volumetric flow, and you must take them into account when calculating the pipe's width from the flow rate. These factors include the length of the pipe and the viscosity of the fluid.
Multiply the fluid's viscosity by 8. Water, for instance, has a viscosity of 0.01 Poise: 0.01 x 8 = 0.08.
Multiply the distance over which the fluid must travel, which forms the pipe's length. With a distance, for instance, of 300 centimetres: 0.08 x 300 = 24.
Multiply the flow rate, measured in cubic centimetres per second, by this answer. If you must move, for instance, 400,000 cubic centimetres each second: 400,000 x 24 = 9,600,000.
Divide the answer by the pressure drop across the pipe. With a pressure drop, for instance, of 250 dynes per cubic centimetre: 9,600,000 / 250 = 38,400.
Divide the answer by pi: 38,400 / 3.142 = 12,221.5.
Raise your answer to the power of 0.25: 12,221.5 ^ 0.25 = approximately 10.5 centimetres. You need a pipe with a 10.5 centimetre radius. | 357 | 1,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-30 | latest | en | 0.881616 |
https://questioncove.com/updates/4df8e6260b8b370c28bdd422 | 1,725,848,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00570.warc.gz | 451,004,391 | 5,675 | Mathematics 42 Online
OpenStudy (anonymous):
-x^2-3x+9
OpenStudy (anonymous):
whats the prob here helpme? do you wanna solve for x?
OpenStudy (anonymous):
find the vertex
OpenStudy (anonymous):
ah ok, did you try first graphing this? you can see it if you graph it....say y=-x^2-3x+9
OpenStudy (amistre64):
id give it a shot, but apparently my negatives signs are off today ;)
OpenStudy (anonymous):
OpenStudy (anonymous):
idk what i keep doing wrong
OpenStudy (amistre64):
-x^2-3x+9 the mirror image over the x axis is: x^2+3x -9; so whatever the y value for this is at the vertex, change its sign
OpenStudy (amistre64):
(-3/2,-(y)) 9 9(2) 9(4) 9-18 -36 -45 --- - ----- - ---- = ---------- = ---- ; opposite that maybe 4 2(2) 1(4) 4 4
OpenStudy (anonymous):
A plot is attached.
OpenStudy (amistre64):
(-3/2 , 45,4) ??
Latest Questions
Alexis1415: What are some pills I could take for heart and liver pains?
2 hours ago 0 Replies 0 Medals
Bones: Matching pfps @alexis1415
4 hours ago 37 Replies 3 Medals
Kaydi: Why won't my computer let me zoom in
9 hours ago 0 Replies 0 Medals
Jesslovesjess: My cramps are slowly killing me
1 hour ago 24 Replies 3 Medals
brianagatica14: Look at my view from my alley....
7 hours ago 7 Replies 3 Medals
Coco233: what do I do if I'm still inlove but he isn't ?
5 hours ago 1 Reply 0 Medals | 432 | 1,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.876618 |
https://quantumcomputing.stackexchange.com/questions/6917/how-to-amplify-a-specific-part-of-the-quantum-state/6925 | 1,656,693,693,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00189.warc.gz | 526,742,735 | 68,911 | # How to amplify a specific part of the quantum state
I understand that, according to amplitude amplification, I can amplify states according to a partition over the state space. However, suppose I want to amplify or de-amplify a specific portion of the state after putting it into a superposition, (e.g. I have $$\frac{1}{2\sqrt(2)}\sum_{x=000}^{111}\left|x\right>$$ and I want to amplify/de-amplify $$\left|000\right>$$). Is this possible without amplitude amplification?
• I'm not clear why you don't want to use amplitude amplification, which does exactly what you are asking for
– glS
Aug 1, 2019 at 11:22
• @glS The reason I don't want to use amplitude amplification is that I want to de-amplify one specific, constant state and I want to avoid the $O\left(\sqrt{N}\right)$ time required to do that completely. Aug 2, 2019 at 4:52
If you know, in advance, that the state you want to deamplify is specifically $$|000\rangle$$, there are a couple of strategies that you could follow.
For example, introduce an ancilla and perform the multi-controlled not, targeting the ancilla, where it is controlled off every qubit in the original state being in the $$|0\rangle$$ state. So, you'd be doing $$\frac{1}{\sqrt{8}}\sum_x|x\rangle|0\rangle\mapsto \frac{1}{\sqrt{8}}\left(|000\rangle|1\rangle+\sum_{x\neq 000}|x\rangle|0\rangle\right).$$
If you want to completely get rid of the $$|000\rangle$$ term, just measure your ancilla in the standard (0/1) basis. If you get the 0 answer, you've succeeded (here, this happens with probability 7/8). If you get the 1 answer, you've failed. You produce your state again and repeat until success.
If all you want to do is decrease the amplitude of $$|000\rangle$$ rather than completely remove it, you can correspondingly increase your probability of success. A simple strategy is to apply the POVM/filtering operation $$\left(\begin{array}{cc} 1 & 0 \\ 0 & \alpha \end{array}\right)$$ for $$\alpha<1$$. Afterwards, you'd repeat your multi-controlled-not operation to disentangle the ancilla.
These strategies are perhaps conceptually simpler to understand than amplitude amplification, and work well if the amplitude you're trying to de-amplify is small enough. However, if the amplitude you're trying to amplify is too small, you won't do any better than the scaling resulting from amplitude amplification, and you'd be much better off seeing if you can change the circuit that's producing your state so that you don't have so much of the $$|000\rangle$$ component in the first place. For instance, in your example, instead of applying $$H^{\otimes 3}$$, you'd find a slightly more complicated circuit such as where $$U|0\rangle=(\sqrt{3}|0\rangle+2|1\rangle)/\sqrt{7}$$ and $$V|0\rangle=(|0\rangle+\sqrt{2}|1\rangle)/\sqrt{3}$$.
• I don't see how you can do POVM-filtering. That seems like the best option to me but I don't see how you can build a circuit to do that. Can you please provide more information about this? Aug 6, 2019 at 1:46
• @Woody1193 Do you mean for the POVM? It's not unitary. It only works with a certain probability of success, but is heralded, meaning that if it works, you know it's worked. Aug 23, 2019 at 5:34
• Sorry, I had to look up more information on it to really understand what it was doing. Thanks for the reply Sep 5, 2019 at 2:14
• If I had a state $\frac{1}{\sqrt{2}} \left( \left|000\right> + \left|101\right> \right)$ and I used POVM to measure the third qubit with a bias toward $\left|1\right>$, would that make the probability of measuring $\left|101\right>$ increase above 50% or would it not change? Sorry to keep bothering you about this Sep 19, 2019 at 9:56
There is a method called inversion about the mean, which is used in Grover's search algorithm (and many more algortihms). In Grover's search algorithm, the preferred outcome (e.g. what you are searching for) is a certain measurement outcome. To increase the possibility of this outcome (compared to other outcomes), the probability of this outcome (e.g. its amplitude) is increased through this 'inversion about the mean' method.
Essentially what it does, is flipping (through phase-inversion) a certain outcome's phase (in your case the $$|000\rangle$$ state) about the mean of all the states. Then this outcome is further apart from the mean - updating all the states in some way that bring them close to the mean then can increase the amplitude of this outcome.
A nice introduction can be found here, but it doesn't have any graphical explanation. That can be found here instead.
• isn't this essentially the amplitude amplification algorithm?
– glS
Aug 1, 2019 at 11:21
• Welp, yes, sorry for that. My bad, I didn't really check the link, this renders this answer obsolete!
– JSdJ
Aug 1, 2019 at 13:08
The probability amplitudes of a quantum system comprising of several qubits can be changed (amplified / de-amplified) with the application of suitable Rotation Gate. However, the sum of probabilities of collapsing to all possible states will remain unity, so if some states are getting amplified then others will get de-amplified. Is there any practical utility of such a gate or is it a hypothetical question?
• Thanks for the answer. The reason I want to do this is that I have an algorithm which produces a number of states and the state $\left|000\right>$, which is a junk state. However, due to the nature of the algorithm, this state is the most probable. So, I want to de-amplify this state while amplifying the others. Aug 2, 2019 at 4:57
• OK, though naively Rotation Gate appears to help in a single qubit case but in multi qubit system, it may not be very helpful and linearity of quantum mechanics may pose a challenge in designing such a gate, i.e. you may not be able to beat performance of Amplitude Amplification technique, though again I have not come across any proofs in this regard...so All the Best.. Aug 2, 2019 at 5:13 | 1,542 | 5,908 | {"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-27 | longest | en | 0.894276 |
https://www.physicsforums.com/threads/derivation-of-momentum-expectancy.549208/ | 1,695,391,089,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506420.84/warc/CC-MAIN-20230922134342-20230922164342-00098.warc.gz | 1,050,007,651 | 14,175 | # Derivation of momentum expectancy
• A_B
#### A_B
Hi,
I'm working through Griffiths' Introduction to QM, In the derivation for the expectation value of momentum, he uses that
$$\left( x \left. \left( \Psi^* \frac{\partial\Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x}\Psi \right) \right) \right|_{-\infty}^{+\infty} = 0$$
Why is this so? It's easy to see that this is zero if the x weren't there, but I can't figure out why the limits are zero with the x in.
Thanks,
A_B
Last edited:
since you said that it's easy to see that this is zeero if the x weren't there, wouldn't that mean by your reasoning that everything inside the brackets would equal zero.
I'm sorry, the x is included in the evaluation, I edited the first post to make that clear, thanks. | 215 | 771 | {"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} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.946127 |
https://www.howmany.wiki/wv/How-many--tbsp--of--sliced+banana--in--1--pound | 1,571,186,123,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660829.5/warc/CC-MAIN-20191015231925-20191016015425-00425.warc.gz | 895,755,744 | 19,490 | ## HowMany.wiki
Please get in touch with us if you:
1. Have any suggestions
2. Have any questions
3. Have found an error/bug
4. Anything else ...
# How many tbsp of sliced banana in 1 pound?
β
How many US tablespoons of sliced banana in 1 pound? How much is 1 pound of sliced banana in tbsp?
1 pound of sliced banana equals 32 ( ~ 32 1/4) US tablespoons(*)
### Inputs?Notes: the results in this calculator are rounded (by default) to 2 significant figures. The conversion factors are approximate once it is intended for recipes measurements. This is not rocket sciece ☺.
From: To: gram kilogram pound ounce cup, US milliliter (ml) ounce, US, Fluid cup cup, Imperial cup, Canada cup, metric liter ounce, UK, Fluid tablespoon, US tablespoon, UK tablespoon, metric teaspoon, UK teaspoon, US teaspoon, metric ?Please, select the weight (mass) unit, then choose the volume unit to which you want to convert. Weight (mass): Ingredient: ?Please, fill a value for volume you want to convert in the left box, choose an ingredient by typing its name, then click on the button 'Calculate!'. Significant Figures: 2 3 4 5
### Results
1 pound of sliced banana equals 32 ( ~ 32 1/4) US tablespoons. (*)
(*) or precisely 32.256 US tablespoons. All values are approximate.
## Pound to US tablespoon Conversion Chart - Sliced banana
Note: Fractions are rounded to the nearest 8th fraction. Values are rounded to 3 significant figures.
pounds to US tablespoons of Sliced banana
1 pound = 32.3 ( 32 1/4 ) US tablespoons
2 pounds = 64.5 ( 64 1/2 ) US tablespoons
4 pounds = 129 ( 129 ) US tablespoons
5 pounds = 161 ( 161 1/4 ) US tablespoons
8 pounds = 258 ( 258 ) US tablespoons
1/16 pound = 2.02 ( 2 ) US tablespoons
1/8 pound = 4.03 ( 4 ) US tablespoons
1/4 pound = 8.06 ( 8 1/8 ) US tablespoons
1/3 pound = 10.8 ( 10 3/4 ) US tablespoons
1/2 pound = 16.1 ( 16 1/8 ) US tablespoons
2/3 pound = 21.5 ( 21 1/2 ) US tablespoons
3/4 pound = 24.2 ( 24 1/4 ) US tablespoons
1 1/16 pounds = 34.3 ( 34 1/4 ) US tablespoons
1 1/8 pounds = 36.3 ( 36 1/4 ) US tablespoons
1 1/4 pounds = 40.3 ( 40 1/3 ) US tablespoons
1 1/3 pounds = 43 ( 43 ) US tablespoons
1 1/2 pounds = 48.4 ( 48 1/3 ) US tablespoons
1 2/3 pounds = 53.8 ( 53 3/4 ) US tablespoons
1 3/4 pounds = 56.4 ( 56 1/2 ) US tablespoons
2 1/16 pounds = 66.5 ( 66 1/2 ) US tablespoons
2 1/8 pounds = 68.5 ( 68 1/2 ) US tablespoons
2 1/4 pounds = 72.6 ( 72 1/2 ) US tablespoons
2 1/3 pounds = 75.3 ( 75 1/4 ) US tablespoons | 806 | 2,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-43 | latest | en | 0.76194 |
https://www.airmilescalculator.com/distance/psp-to-scc/ | 1,606,211,174,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176049.8/warc/CC-MAIN-20201124082900-20201124112900-00579.warc.gz | 571,898,659 | 100,549 | Distance between Palm Springs, CA (PSP) and Deadhorse, AK (SCC)
Flight distance from Palm Springs to Deadhorse (Palm Springs International Airport – Deadhorse Airport) is 2788 miles / 4486 kilometers / 2423 nautical miles. Estimated flight time is 5 hours 46 minutes.
Driving distance from Palm Springs (PSP) to Deadhorse (SCC) is 4077 miles / 6562 kilometers and travel time by car is about 74 hours 57 minutes.
Map of flight path and driving directions from Palm Springs to Deadhorse.
Shortest flight path between Palm Springs International Airport (PSP) and Deadhorse Airport (SCC).
How far is Deadhorse from Palm Springs?
There are several ways to calculate distances between Palm Springs and Deadhorse. Here are two common methods:
Vincenty's formula (applied above)
• 2787.776 miles
• 4486.491 kilometers
• 2422.511 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 2784.902 miles
• 4481.865 kilometers
• 2420.013 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
Airport information
A Palm Springs International Airport
City: Palm Springs, CA
Country: United States
IATA Code: PSP
ICAO Code: KPSP
Coordinates: 33°49′46″N, 116°30′25″W
Country: United States
IATA Code: SCC
ICAO Code: PASC
Coordinates: 70°11′40″N, 148°27′53″W
Time difference and current local times
The time difference between Palm Springs and Deadhorse is 1 hour. Deadhorse is 1 hour behind Palm Springs.
PST
AKST
Carbon dioxide emissions
Estimated CO2 emissions per passenger is 309 kg (681 pounds).
Frequent Flyer Miles Calculator
Palm Springs (PSP) → Deadhorse (SCC).
Distance:
2788
Elite level bonus:
0
Booking class bonus:
0
In total
Total frequent flyer miles:
2788
Round trip? | 494 | 1,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.793517 |
https://socratic.org/questions/how-do-you-simplify-ln-1-2#244171 | 1,726,518,226,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00121.warc.gz | 496,719,899 | 5,916 | # How do you simplify ln 1/2?
Mar 24, 2016
I found: $- \ln \left(2\right) = - 0.69315$ when the original question stated $\ln \left(\frac{1}{2}\right)$...!
#### Explanation:
I would use a property of the logs where you have:
$\log x - \log y = \log \left(\frac{x}{y}\right)$
To write:
$\ln \left(\frac{1}{2}\right) = \ln \left(1\right) - \ln \left(2\right) = 0 - \ln \left(2\right) = - \ln \left(2\right) = - 0.69315$
Mar 24, 2016
$0$
#### Explanation:
Assuming you meant $\ln \frac{1}{2}$ and not $\ln \left(\frac{1}{2}\right)$:
$\ln 1 = 0$, so
$\ln \frac{1}{2} = \frac{0}{2} = 0$ | 240 | 591 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.600711 |
https://www.thestudentroom.co.uk/showthread.php?t=7475029 | 1,718,832,483,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861832.94/warc/CC-MAIN-20240619185738-20240619215738-00206.warc.gz | 911,140,303 | 45,465 | # A level physics mechanics question
Why are we equating the moments of T and F if they are both pulling clockwise?
(edited 1 month ago)
Original post by gcsestudent56
Why are we equating the moments of T and F if they are both pulling clockwise?
Its a lever so the aim is to gain a mechanical advantage by increasing the force. F is the force on the lever and T is the force on the ship.. An equal but opposite force to T would be being applied to the lever (force pair).
(edited 1 month ago)
Original post by mqb2766
Its a lever (not in equilibrium) so the aim is to gain a mechanical advantage by increasing the force. Also, F is the force on the lever, but T is the force on the ship. An equal but opposite force would be being applied to the lever.
im ngl i still don't get what any of that means
Original post by Gcsestudent56
im ngl i still don't get what any of that means
If you consider the lever, then the two forces being applied are F and -T. -T is anticlockwise.
Original post by mqb2766
If you consider the lever, then the two forces being applied are F and -T. -T is anticlockwise.
oh wait so if the lever is being pulled down by force F, there is going to be tension downwards as the boat weight is pulling it down. So that why we consider the tension to be anti clockwise? Might be waffling, sorry im awful at mechanics
Original post by mqb2766
Its a lever so the aim is to gain a mechanical advantage by increasing the force. F is the force on the lever and T is the force on the ship.. An equal but opposite force to T would be being applied to the lever (force pair).
Also I dont get the last sentence.
Original post by Gcsestudent56
oh wait so if the lever is being pulled down by force F, there is going to be tension downwards as the boat weight is pulling it down. So that why we consider the tension to be anti clockwise? Might be waffling, sorry im awful at mechanics
The lever is pulling the boat up (tension upwards) and the boat is pulling the lever downwards (tension downwards). In the diagram T is upwards so when you think about the lever, the relevant force is -T and its acting anticlockwise (on the lever).
Original post by mqb2766
The lever is pulling the boat up (tension upwards) and the boat is pulling the lever downwards (tension downwards). In the diagram T is upwards so when you think about the lever, the relevant force is -T and its acting anticlockwise (on the lever).
I get it now thanks. I really need to work on mechanics. Is there a specific youtuber that helped you learn mechanics? Cuz I really need help
Original post by Gcsestudent56
I get it now thanks. I really need to work on mechanics. Is there a specific youtuber that helped you learn mechanics? Cuz I really need help
Sorry, dont know. A fair bit of it is to try and follow the guidelines and make sure you understand the terms. So sketch force diagrams, make sure you only put on forces on the specific object, ... | 698 | 2,939 | {"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} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.951589 |
https://www.sqlservercentral.com/Forums/Topic462584-143-1.aspx | 1,516,278,733,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887253.36/warc/CC-MAIN-20180118111417-20180118131417-00701.warc.gz | 976,254,277 | 34,773 | ## What about MONTH and YEAR functions?
Author Message ruben ruvalcaba Valued Member Group: General Forum Members Points: 55 Visits: 104 Hi, First of all thank you, it's a good point to be considered. In fact, I use functions on the WHERE clause very often, specially when filtering dates ... how do you optimize a query like this: SELECT SUM(Amount) FROM myEntries WHERE YEAR(myEntries.myDate) = @Year AND MONTH(myEntries.myDate) <= @MonthAny clue? thank you regards Michael Valentine Jones SSC-Dedicated Group: General Forum Members Points: 35752 Visits: 11933 Do queries like this as a selection against a range in this form:where date >= @StartOfRange and date < @EndOf RangeIn the following code, it will be the same as this:where date >= '2008-01-01' and date < '2008-03-01'`declare @year intdeclare @month intselect @year = 2008, @month = 2select *from MyTablewhere -- Date on or after start of year MyTable.MyDate >= dateadd(month,(12*@Year)-22801+1,0) and -- Date before start of next month MyTable.MyDate < dateadd(month,(12*@Year)-22801+@Month+1,0)` Jack Corbett SSC Guru Group: General Forum Members Points: 103208 Visits: 15045 If this is how you will most often be querying your data you should have month and year columns in table with an index. Or, write your queries with explicit date ranges. If you want your users to be able to enter month and year paramters then convert them to dates after entry. So year=2008 and month=3 would become. Startdate=1/1/2008 & endate=3/31/2008. Jack Corbett Consultant Straight Path Solutions Dont let the good be the enemy of the best. -- Paul FlemingAt best you can say that one job may be more secure than another, but total job security is an illusion. -- Rod at workCheck out these links on how to get faster and more accurate answers: Forum Etiquette: How to post data/code on a forum to get the best helpNeed an Answer? Actually, No ... You Need a QuestionHow to Post Performance ProblemsCrosstabs and Pivots or How to turn rows into columns Part 1Crosstabs and Pivots or How to turn rows into columns Part 2 Lynn Pettis SSC Guru Group: General Forum Members Points: 223230 Visits: 40398 here is a slight variation to what Michael Valentine Jones provided:`declare @year smallint, @month tinyintset @year = 2007set @month = 2select sum(Amount)from dbo.myEntrieswhere myEntries.myDate >= dateadd(yy,(@year - 1900), 0) and myEntries.myDate < dateadd(mm, @month, dateadd(yy,(@year - 1900), 0))` Lynn PettisFor better assistance in answering your questions, click hereFor tips to get better help with Performance Problems, click hereFor Running Totals and its variations, click here or when working with partitioned tablesFor more about Tally Tables, click hereFor more about Cross Tabs and Pivots, click here and hereManaging Transaction LogsSQL Musings from the Desert Fountain Valley SQL (My Mirror Blog) Jack Corbett SSC Guru Group: General Forum Members Points: 103208 Visits: 15045 The thing about functions in where clauses is that, if on a column, you have to run every row through the function while using them on a parameter or constant the optimizer can run it once. This is why Michael's and Lynn's solutions are more effecient. Jack Corbett Consultant Straight Path Solutions Dont let the good be the enemy of the best. -- Paul FlemingAt best you can say that one job may be more secure than another, but total job security is an illusion. -- Rod at workCheck out these links on how to get faster and more accurate answers: Forum Etiquette: How to post data/code on a forum to get the best helpNeed an Answer? Actually, No ... You Need a QuestionHow to Post Performance ProblemsCrosstabs and Pivots or How to turn rows into columns Part 1Crosstabs and Pivots or How to turn rows into columns Part 2 Jeff Moden SSC Guru Group: General Forum Members Points: 505817 Visits: 44255 For too many reasons to list in this short amount of space, I strongly recommend that you never process dates by individual Year and/or Month components. Some of the others have already suggested how to handle things... always treat dates as a range... yep... even if the date is for one measely day. WHERE clauses should always follow the general format of...` WHERE somedatecol >= @StartDate AND somedatecol < @EndDate+1`That is assuming, of course, that you are working with whole dates that either have no literally expressed time component (defaults to midnight) or has a time component of precisely 00:00:00.000. Other considerations will need be made if @StartDate or @EndDate have a non-midnight time component. Doesn't matter if "somedatecol" does or not and that's the beauty of the method shown above. And, it'll allow for very high performance Index SEEKs if the correct indexes are available.You will also find those that suggest that you use one of the following...` WHERE somedatecol BETWEEN @StartDate AND DATEADD(ms,-3,@EndDate+1) WHERE somedatecol BETWEEN @StartDate AND @EndDate+'23:59:59.997`Treat them just like street drugs... just say "NO". Do they work? Yes, today they do... when 2008 comes out, it will "depend". --Jeff ModenRBAR is pronounced ree-bar and is a Modenism for Row-By-Agonizing-Row.First step towards the paradigm shift of writing Set Based code: Stop thinking about what you want to do to a row... think, instead, of what you want to do to a column.If you think its expensive to hire a professional to do the job, wait until you hire an amateur. -- Red Adair Helpful Links:How to post code problemsHow to post performance problemsForum FAQs | 1,344 | 5,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.800284 |
http://stackoverflow.com/questions/2428404/tricky-interview-question-for-c | 1,398,006,207,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00119-ip-10-147-4-33.ec2.internal.warc.gz | 223,348,835 | 18,329 | # tricky interview question for C++
Given the code below, how would you create/implement SR.h so that it produces the correct output WITHOUT any asterisks in your solution?
I got bummed by this question. I would like to know some of the different approaches that people use for this problem.
``````#include <cstdio>
#include "SR.h"
int main()
{
int j = 5;
int a[] = {10, 15};
{
SR x(j), y(a[0]), z(a[1]);
j = a[0];
a[0] = a[1];
a[1] = j;
printf("j = %d, a = {%d, %d}\n", j, a[0], a[1]);
}
printf("j = %d, a = {%d, %d}\n", j, a[0], a[1]);
}
``````
output
``````j = 10, a = {15, 10}
j = 5, a = {10, 15}
``````
Second one:
``````#include <cstdio>
#include "SR.h"
int main()
{
int sum = 0;
for (int i = 1; i < 100; i++) {
SR ii(i);
while (i--)
sum += i;
}
printf("sum = %d\n", sum);
}
//The output is "sum = 161700".
``````
-
Why shouldn't this be a real question? Am I missing something? – sbi Mar 11 '10 at 20:27
@sbi Interview questions, particularly artificial ones, usually are not "real". Not that I downvoted or close voted this one myself. – anon Mar 11 '10 at 20:29
This is indeed an EXCELLENT question, both for general knowledge and interview purposes. – Matthieu N. Mar 11 '10 at 21:38
@Kirill et al.: Given that three separate people came up with the correct answer within about twenty minutes of the question being posted, I'm confident that it could "be reasonably answered in its current form" :P . – Bill Mar 11 '10 at 22:15
@Neil: Call me conservative, but to me a question seems "real" when it comes in the form of a grammatical question (i.e., something you can put a question mark after), poses a problem, and, in theory, allows allows you to write up something which answers it. Whether it is "artificial" (whatever that means - wouldn't almost all homework questions be artificial?) seems subjective to me, the criteria I gave seems objective. – sbi Mar 12 '10 at 6:14
SR is acting as a captured-variable-restorer. When it goes out of scope it restores some value that it previously captured.
The constructor will do two things: capture a reference, and capture the value of that reference. The destructor will restore the original value to that reference.
``````class SR
{
public:
SR(int& var) : capture(var), value(var) {}
~SR() { capture = value; }
private:
int& capture;
int value;
};
``````
Edit: Just a guess, but I assume SR is supposed to stand for ScopeRestorer?
-
My thought was SR = SaveRestore. – Mark Ransom Mar 11 '10 at 21:26
@Mark: SaveRestore seems a lot more likely, thanks! – Bill Mar 11 '10 at 22:17
I don't have time to write code but, you need to use references &int in constructor. And you would need to restore original values to references in the destructor. When SR goes out of scope it needs to restore original values that were passed in during construction.
-
+1 Cool and Crisp! – bragboy Mar 11 '10 at 20:31
For the first one:
``````class SR
{
int &ref;
int orig;
public:
SR(int& r)
:ref(r), orig(r)
{
}
~SR()
{
ref = orig;
}
};
``````
For the second snippet, should it be the same SR or some other SR?
-
This solution works for both, so I assume it's meant to satisfy both `main` functions. – Bill Mar 11 '10 at 20:52
yes...it's meant for both – aherlambang Mar 11 '10 at 21:04
Your code wont compile until you place a colon after public – Matthieu N. Mar 11 '10 at 21:31
Sorry, typo. Fixed. – Seva Alekseyev Mar 11 '10 at 22:06
``````#define printf myprintf
void myprintf(int, int, int, int) {
printf("j = 10, a = {15, 10}\nj = 5, a = {10, 15}");
exit(0);
}
void myprintf(int, int) {
printf("sum = 161700");
exit(0);
}
``````
Or, in other words, I think the concept of the scope restorer macro is really cool, but I don't like the way the question was worded :)
-
haha, funny! This could actually be a smart ass answer for this problem. – Grim May 31 '11 at 9:50
Since nobody posted a solution to #2:
``````#define _SR_H_
int count = 0;
class SR {
private:
int& ref;
public:
SR(int& val) : ref(val) {
count++;
}
~SR() {
if (count == (161700 + 1)) {
ref = 100;
} else {
ref = 1;
}
}
};
#endif
``````
I know this solution is bit ugly and runs the for loop 161700 times to add the numbers. This would work for any number, but I am not sure why 161700 was chosen. It doesn't factorize nicely either.
- | 1,284 | 4,295 | {"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} | 2.515625 | 3 | CC-MAIN-2014-15 | latest | en | 0.908439 |
http://stackoverflow.com/questions/676824/how-to-calculate-the-difference-between-two-dates-using-php | 1,469,293,847,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823133.4/warc/CC-MAIN-20160723071023-00036-ip-10-185-27-174.ec2.internal.warc.gz | 231,186,423 | 43,110 | Dismiss
Announcing Stack Overflow Documentation
We started with Q&A. Technical documentation is next, and we need your help.
Whether you're a beginner or an experienced developer, you can contribute.
# How to calculate the difference between two dates using PHP?
I have two dates of the form:
``````Start Date: 2007-03-24
End Date: 2009-06-26
``````
Now I need to find the difference between these two in the following form:
``````2 years, 3 months and 2 days
``````
How can I do this in PHP?
-
2 years 94 days. Calculating the months, taking into account leap years, would be problematic. How accurate does this need to be? – dbasnett Mar 24 '09 at 12:35
I dont think this is a necessary question given that there is plenty of Q&A about calculating the difference between dates. Just look at the related section to the right. – Gordon Nov 30 '12 at 15:38
Though accepted answer fulfill your requirements but, if you are using PHP 5.3 or greater than please check the following answer on this question stackoverflow.com/a/3923228/631652 – Parixit Oct 16 '13 at 7:13
possible duplicate of How do I calculate relative time? – Cole Johnson Jun 26 '14 at 5:00
For PHP < 5.3 otherwise see jurka's answer below
You can use strtotime() to convert two dates to unix time and then calculate the number of seconds between them. From this it's rather easy to calculate different time periods.
``````\$date1 = "2007-03-24";
\$date2 = "2009-06-26";
\$diff = abs(strtotime(\$date2) - strtotime(\$date1));
\$years = floor(\$diff / (365*60*60*24));
\$months = floor((\$diff - \$years * 365*60*60*24) / (30*60*60*24));
\$days = floor((\$diff - \$years * 365*60*60*24 - \$months*30*60*60*24)/ (60*60*24));
printf("%d years, %d months, %d days\n", \$years, \$months, \$days);
``````
Edit: Obviously the preferred way of doing this is like described by jurka below. My code is generally only recommended if you don't have PHP 5.3 or better.
Several people in the comments have pointed out that the code above is only an approximation. I still believe that for most purposes that's fine, since the usage of a range is more to provide a sense of how much time has passed or remains rather than to provide precision - if you want to do that, just output the date.
Despite all that, I've decided to address the complaints. If you truly need an exact range but haven't got access to PHP 5.3, use the code below (it should work in PHP 4 as well). This is a direct port of the code that PHP uses internally to calculate ranges, with the exception that it doesn't take daylight savings time into account. That means that it's off by an hour at most, but except for that it should be correct.
``````<?php
/**
* Calculate differences between two dates with precise semantics. Based on PHPs DateTime::diff()
* implementation by Derick Rethans. Ported to PHP by Emil H, 2011-05-02. No rights reserved.
*
* See here for original code:
* http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/tm2unixtime.c?revision=302890&view=markup
* http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/interval.c?revision=298973&view=markup
*/
function _date_range_limit(\$start, \$end, \$adj, \$a, \$b, \$result)
{
if (\$result[\$a] < \$start) {
\$result[\$b] -= intval((\$start - \$result[\$a] - 1) / \$adj) + 1;
\$result[\$a] += \$adj * intval((\$start - \$result[\$a] - 1) / \$adj + 1);
}
if (\$result[\$a] >= \$end) {
}
return \$result;
}
function _date_range_limit_days(\$base, \$result)
{
\$days_in_month_leap = array(31, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
\$days_in_month = array(31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
_date_range_limit(1, 13, 12, "m", "y", &\$base);
\$year = \$base["y"];
\$month = \$base["m"];
if (!\$result["invert"]) {
while (\$result["d"] < 0) {
\$month--;
if (\$month < 1) {
\$month += 12;
\$year--;
}
\$leapyear = \$year % 400 == 0 || (\$year % 100 != 0 && \$year % 4 == 0);
\$days = \$leapyear ? \$days_in_month_leap[\$month] : \$days_in_month[\$month];
\$result["d"] += \$days;
\$result["m"]--;
}
} else {
while (\$result["d"] < 0) {
\$leapyear = \$year % 400 == 0 || (\$year % 100 != 0 && \$year % 4 == 0);
\$days = \$leapyear ? \$days_in_month_leap[\$month] : \$days_in_month[\$month];
\$result["d"] += \$days;
\$result["m"]--;
\$month++;
if (\$month > 12) {
\$month -= 12;
\$year++;
}
}
}
return \$result;
}
function _date_normalize(\$base, \$result)
{
\$result = _date_range_limit(0, 60, 60, "s", "i", \$result);
\$result = _date_range_limit(0, 60, 60, "i", "h", \$result);
\$result = _date_range_limit(0, 24, 24, "h", "d", \$result);
\$result = _date_range_limit(0, 12, 12, "m", "y", \$result);
\$result = _date_range_limit_days(&\$base, &\$result);
\$result = _date_range_limit(0, 12, 12, "m", "y", \$result);
return \$result;
}
/**
* Accepts two unix timestamps.
*/
function _date_diff(\$one, \$two)
{
\$invert = false;
if (\$one > \$two) {
list(\$one, \$two) = array(\$two, \$one);
\$invert = true;
}
\$key = array("y", "m", "d", "h", "i", "s");
\$a = array_combine(\$key, array_map("intval", explode(" ", date("Y m d H i s", \$one))));
\$b = array_combine(\$key, array_map("intval", explode(" ", date("Y m d H i s", \$two))));
\$result = array();
\$result["y"] = \$b["y"] - \$a["y"];
\$result["m"] = \$b["m"] - \$a["m"];
\$result["d"] = \$b["d"] - \$a["d"];
\$result["h"] = \$b["h"] - \$a["h"];
\$result["i"] = \$b["i"] - \$a["i"];
\$result["s"] = \$b["s"] - \$a["s"];
\$result["invert"] = \$invert ? 1 : 0;
\$result["days"] = intval(abs((\$one - \$two)/86400));
if (\$invert) {
_date_normalize(&\$a, &\$result);
} else {
_date_normalize(&\$b, &\$result);
}
return \$result;
}
\$date = "1986-11-10 19:37:22";
print_r(_date_diff(strtotime(\$date), time()));
print_r(_date_diff(time(), strtotime(\$date)));
``````
-
If you're using the DateTime class you can go for \$date->format('U') to get the unix timestamp. – Jon Cram Aug 7 '09 at 13:26
It's not true if you have to deal with summer/winter time. In this particular case when you adjust summer/winter time, one day equals 23 or 25 hours. – Arno Dec 21 '09 at 15:54
Well, the same argument could be made for leap years. It doesn't take that into account either. Still, I'm not convinced that you even want to take that into account since we're discussing a range here. The semantics for a range are somewhat different than for an absolute date. – Emil H Dec 21 '09 at 20:35
This function is incorrect. It's good for an approximation, but incorrect for exact ranges. For one, it assumes there are 30 days in a month, which is to say it will have the same difference of days between February 1st and March 1st as it will for July 1st to August 1st (regardless of leap year). – enobrev Apr 11 '11 at 19:14
In PHP, reference variables are in the function signature, not the call. Move all your `&` to the signatures. – Paul Tarjan Mar 19 '13 at 9:32
I suggest to use DateTime and DateInterval objects.
``````\$date1 = new DateTime("2007-03-24");
\$date2 = new DateTime("2009-06-26");
\$interval = \$date1->diff(\$date2);
echo "difference " . \$interval->y . " years, " . \$interval->m." months, ".\$interval->d." days ";
// shows the total amount of days (not divided into years, months and days like above)
echo "difference " . \$interval->days . " days ";
``````
From the manual:
As of PHP 5.2.2, DateTime objects can be compared using comparison operators.
``````\$date1 = new DateTime("now");
\$date2 = new DateTime("tomorrow");
var_dump(\$date1 == \$date2); // bool(false)
var_dump(\$date1 < \$date2); // bool(true)
var_dump(\$date1 > \$date2); // bool(false)
``````
-
+1 This OOP approach is much more succinct than the functional floor() approach. – systemovich Dec 14 '10 at 13:56
note that DateTime->diff() is php 5.3+ – cerberos Apr 1 '11 at 8:58
+1 DateTime handles leap years and time-zones properly and there's a good book for the shelf: phparch.com/books/… – hakre Aug 7 '11 at 12:03
note that PHP 5.3 is out since 2009 – feeela Feb 9 '12 at 13:32
Is there a method that gives the total number of seconds between the two DateTimes? (without adding up the components, that is) – potatoe Feb 19 '12 at 3:52
The best course of action is using PHP's `DateTime` (and `DateInterval`) objects. Each date is encapsulated in a `DateTime` object, and then a difference between the two can be made:
``````\$first_date = new DateTime("2012-11-30 17:03:30");
\$second_date = new DateTime("2012-12-21 00:00:00");
``````
The `DateTime` object will accept any format `strtotime()` would. If an even more specific date format is needed, `DateTime::createFromFormat()` can be used to create the `DateTime` object.
After both objects were instantiated, you substract one from the other with `DateTime::diff()`.
``````\$difference = \$first_date->diff(\$second_date);
``````
`\$difference` now holds a `DateInterval` object with the difference information. A `var_dump()` looks like this:
``````object(DateInterval)
public 'y' => int 0
public 'm' => int 0
public 'd' => int 20
public 'h' => int 6
public 'i' => int 56
public 's' => int 30
public 'invert' => int 0
public 'days' => int 20
``````
To format the `DateInterval` object, we'll need check each value and exclude it if it's 0:
``````/**
* Format an interval to show all existing components.
* If the interval doesn't have a time component (years, months, etc)
* That component won't be displayed.
*
* @param DateInterval \$interval The interval
*
* @return string Formatted interval string.
*/
function format_interval(DateInterval \$interval) {
\$result = "";
if (\$interval->y) { \$result .= \$interval->format("%y years "); }
if (\$interval->m) { \$result .= \$interval->format("%m months "); }
if (\$interval->d) { \$result .= \$interval->format("%d days "); }
if (\$interval->h) { \$result .= \$interval->format("%h hours "); }
if (\$interval->i) { \$result .= \$interval->format("%i minutes "); }
if (\$interval->s) { \$result .= \$interval->format("%s seconds "); }
return \$result;
}
``````
All that's left now is to call our function on the `\$difference` `DateInterval` object:
``````echo format_interval(\$difference);
``````
And we get the correct result:
20 days 6 hours 56 minutes 30 seconds
The complete code used to achieve the goal:
``````/**
* Format an interval to show all existing components.
* If the interval doesn't have a time component (years, months, etc)
* That component won't be displayed.
*
* @param DateInterval \$interval The interval
*
* @return string Formatted interval string.
*/
function format_interval(DateInterval \$interval) {
\$result = "";
if (\$interval->y) { \$result .= \$interval->format("%y years "); }
if (\$interval->m) { \$result .= \$interval->format("%m months "); }
if (\$interval->d) { \$result .= \$interval->format("%d days "); }
if (\$interval->h) { \$result .= \$interval->format("%h hours "); }
if (\$interval->i) { \$result .= \$interval->format("%i minutes "); }
if (\$interval->s) { \$result .= \$interval->format("%s seconds "); }
return \$result;
}
\$first_date = new DateTime("2012-11-30 17:03:30");
\$second_date = new DateTime("2012-12-21 00:00:00");
\$difference = \$first_date->diff(\$second_date);
echo format_interval(\$difference);
``````
-
DateTime() not working on my php server – Sagar G. Jul 22 '13 at 16:15
`DateTime()` is not a function, it's an object, and it's there since PHP 5.2. Make sure that your server supports it. – Madara Uchiha Jul 22 '13 at 16:21
@SecondRikudo DateTime::Diff need PHP 5.3.0 – PhoneixS Jul 7 '14 at 9:21
View Hours and Minuts and Seconds..
``````\$date1 = "2008-11-01 22:45:00";
\$date2 = "2009-12-04 13:44:01";
\$diff = abs(strtotime(\$date2) - strtotime(\$date1));
\$years = floor(\$diff / (365*60*60*24));
\$months = floor((\$diff - \$years * 365*60*60*24) / (30*60*60*24));
\$days = floor((\$diff - \$years * 365*60*60*24 - \$months*30*60*60*24)/ (60*60*24));
\$hours = floor((\$diff - \$years * 365*60*60*24 - \$months*30*60*60*24 - \$days*60*60*24)/ (60*60));
\$minuts = floor((\$diff - \$years * 365*60*60*24 - \$months*30*60*60*24 - \$days*60*60*24 - \$hours*60*60)/ 60);
\$seconds = floor((\$diff - \$years * 365*60*60*24 - \$months*30*60*60*24 - \$days*60*60*24 - \$hours*60*60 - \$minuts*60));
printf("%d years, %d months, %d days, %d hours, %d minuts\n, %d seconds\n", \$years, \$months, \$days, \$hours, \$minuts, \$seconds);
``````
-
Probably this will not give the accurate result. – Dolphin Aug 5 '11 at 8:24
And is a terrible solution unless you're forced to use a terribly outdated version of PHP ... – rdlowrey Mar 24 '12 at 17:50
Not so DRY. For instance, 60*60*24 is repeated 15 times. Long live copy-paste reuse! – Peter Mortensen Apr 8 '14 at 23:07
What about leap years? A year is not 365 days on average. – Peter Mortensen Apr 8 '14 at 23:10
This code assumes a month is 30 days on average. Even assuming 365 days for a year, an average month is 365 / 12 = 30.42 days (approx.). – Peter Mortensen Apr 8 '14 at 23:16
Take a look at the following link. This is the best answer I've found so far.. :)
``````function dateDiff (\$d1, \$d2) {
// Return the number of days between the two dates:
return round(abs(strtotime(\$d1) - strtotime(\$d2))/86400);
} // end function dateDiff
``````
It doesn't matter which date is earlier or later when you pass in the date parameters. The function uses the PHP ABS() absolute value to always return a postive number as the number of days between the two dates.
Keep in mind that the number of days between the two dates is NOT inclusive of both dates. So if you are looking for the number of days represented by all the dates between and including the dates entered, you will need to add one (1) to the result of this function.
For example, the difference (as returned by the above function) between 2013-02-09 and 2013-02-14 is 5. But the number of days or dates represented by the date range 2013-02-09 - 2013-02-14 is 6.
http://www.bizinfosys.com/php/date-difference.html
-
Summarize the answer here, then you can provide a link to more detailed information. – Anders Abel Aug 6 '11 at 11:20
What about DST? – toon81 Feb 6 '13 at 10:41
@Anders Abel: I've added the quoted text from provided link.. Hope its clear now! – casper123 Feb 14 '13 at 19:17
@casper123 what about Daylight savings time? – Shackrock Mar 8 '13 at 20:56
The question asked for the difference as the number of years, months and days, not the total number of days. – Peter Mortensen Apr 8 '14 at 23:41
I voted for jurka's answer as that's my favorite, but I have a pre-php.5.3 version...
I found myself working on a similar problem - which is how I got to this question in the first place - but just needed a difference in hours. But my function solved this one pretty nicely as well and I don't have anywhere in my own library to keep it where it won't get lost and forgotten, so... hope this is useful to someone.
``````/**
*
* @param DateTime \$oDate1
* @param DateTime \$oDate2
* @return array
*/
function date_diff_array(DateTime \$oDate1, DateTime \$oDate2) {
\$aIntervals = array(
'year' => 0,
'month' => 0,
'week' => 0,
'day' => 0,
'hour' => 0,
'minute' => 0,
'second' => 0,
);
foreach(\$aIntervals as \$sInterval => &\$iInterval) {
while(\$oDate1 <= \$oDate2){
\$oDate1->modify('+1 ' . \$sInterval);
if (\$oDate1 > \$oDate2) {
\$oDate1->modify('-1 ' . \$sInterval);
break;
} else {
\$iInterval++;
}
}
}
return \$aIntervals;
}
``````
And the test:
``````\$oDate = new DateTime();
\$oDate->modify('+111402189 seconds');
var_dump(\$oDate);
var_dump(date_diff_array(new DateTime(), \$oDate));
``````
And the result:
``````object(DateTime)[2]
public 'date' => string '2014-04-29 18:52:51' (length=19)
public 'timezone_type' => int 3
public 'timezone' => string 'America/New_York' (length=16)
array
'year' => int 3
'month' => int 6
'week' => int 1
'day' => int 4
'hour' => int 9
'minute' => int 3
'second' => int 8
``````
I got the original idea from here, which I modified for my uses (and I hope my modification will show on that page as well).
You can very easily remove intervals you don't want (say "week") by removing them from the `\$aIntervals` array, or maybe adding an `\$aExclude` parameter, or just filter them out when you output the string.
-
Thanks for this. It was a very helpful base from which to work when tackling a similar problem. – Walf Sep 27 '11 at 5:30
Unfortunately this doesn't return the same thing as DateInterval because of year/month overflows. – Stephen Harris Aug 18 '12 at 10:56
@StephenHarris: I haven't tested this, but by reading the code I'm pretty confident it should return the same result - provided that you delete the `week` index in `\$aIntervals` (since `DateDiff` never uses that). – Alix Axel Nov 1 '12 at 19:07
I don't know if you are using a PHP framework or not, but a lot of PHP frameworks have date/time libraries and helpers to help keep you from reinventing the wheel.
For example CodeIgniter has the `timespan()` function. Simply input two Unix timestamps and it will automatically generate a result like this:
``````1 Year, 10 Months, 2 Weeks, 5 Days, 10 Hours, 16 Minutes
``````
http://codeigniter.com/user_guide/helpers/date_helper.html
-
``````<?php
\$today = strtotime("2011-02-03 00:00:00");
\$myBirthDate = strtotime("1964-10-30 00:00:00");
printf("Days since my birthday: ", (\$today - \$myBirthDate)/60/60/24);
?>
``````
-
The question asked for the difference as the number of years, months and days. This outputs the difference as the total number of days. – Peter Mortensen Apr 8 '14 at 23:40
i suggest to use this code to get difference between date.
``````\$date1 = new DateTime(\$date_1);
\$date2 = new DateTime(\$date_2);
\$interval = \$date1->diff(\$date2);
echo "difference " . \$interval->y . " years, " . \$interval->m." months, ".\$interval->d." days ";
// shows the total amount of days (not divided into years, months and days like above)
echo "difference " . \$interval->days . " days ";
``````
-
This is almost a complete copy of Jurka's answer... – Spencer May Feb 26 at 18:34
You can use the
``````getdate()
``````
function which returns an array containing all elements of the date/time supplied:
``````\$diff = abs(\$endDate - \$startDate);
\$my_t=getdate(\$diff);
print("\$my_t[year] years, \$my_t[month] months and \$my_t[mday] days");
``````
If your start and end dates are in string format then use
``````\$startDate = strtotime(\$startDateStr);
\$endDate = strtotime(\$endDateStr);
``````
before the above code
-
doesn't seem to work. I get a date at the begining of the timestamp era. – Sirber Jul 26 '10 at 17:34
It is important to understand that you need to do a `\$my_t["year"] -= 1970` to get the correct number of years. You also need to subtract your hour difference from GMT to get the hours right. You need to subtract 1 from month and date as well. – Salman A Feb 21 '12 at 7:55
``````// If you just want to see the year difference then use this function.
// Using the logic I've created you may also create month and day difference
// which I did not provide here so you may have the efforts to use your brain.
// :)
\$date1='2009-01-01';
\$date2='2010-01-01';
echo getYearDifference (\$date1,\$date2);
function getYearDifference(\$date1=strtotime(\$date1),\$date2=strtotime(\$date2)){
\$year = 0;
while(\$date2 > \$date1 = strtotime('+1 year', \$date1)){
++\$year;
}
return \$year;
}
``````
-
Does "strtotime('+1 year', \$date1)" take leap years into account? – Peter Mortensen Apr 8 '14 at 23:38
I have some simple logic for that:
``````<?php
per_days_diff('2011-12-12','2011-12-29')
function per_days_diff(\$start_date, \$end_date) {
\$per_days = 0;
\$noOfWeek = 0;
\$noOfWeekEnd = 0;
\$highSeason=array("7", "8");
\$current_date = strtotime(\$start_date);
\$current_date += (24 * 3600);
\$end_date = strtotime(\$end_date);
\$seassion = (in_array(date('m', \$current_date), \$highSeason))?"2":"1";
\$noOfdays = array('');
while (\$current_date <= \$end_date) {
if (\$current_date <= \$end_date) {
\$date = date('N', \$current_date);
array_push(\$noOfdays,\$date);
\$current_date = strtotime('+1 day', \$current_date);
}
}
\$finalDays = array_shift(\$noOfdays);
//print_r(\$noOfdays);
\$weekFirst = array("week"=>array(),"weekEnd"=>array());
for(\$i = 0; \$i < count(\$noOfdays); \$i++)
{
if (\$noOfdays[\$i] == 1)
{
//echo "This is week";
//echo "<br/>";
if(\$noOfdays[\$i+6]==7)
{
\$noOfWeek++;
\$i=\$i+6;
}
else
{
\$per_days++;
}
//array_push(\$weekFirst["week"],\$day);
}
else if(\$noOfdays[\$i]==5)
{
//echo "This is weekend";
//echo "<br/>";
if(\$noOfdays[\$i+2] ==7)
{
\$noOfWeekEnd++;
\$i = \$i+2;
}
else
{
\$per_days++;
}
//echo "After weekend value:- ".\$i;
//echo "<br/>";
}
else
{
\$per_days++;
}
}
/*echo \$noOfWeek;
echo "<br/>";
echo \$noOfWeekEnd;
echo "<br/>";
print_r(\$per_days);
echo "<br/>";
print_r(\$weekFirst);
*/
\$duration = array("weeks"=>\$noOfWeek, "weekends"=>\$noOfWeekEnd, "perDay"=>\$per_days, "seassion"=>\$seassion);
return \$duration;
?>
``````
-
There seems to be something missing at the end of the sample code (an ending brace and "?>" ?). – Peter Mortensen Apr 8 '14 at 23:52
This is my function. Required PHP >= 5.3.4. It use DateTime class. Very fast, quick and can do the difference between two dates or even the so called "time since".
``````if(function_exists('grk_Datetime_Since') === FALSE){
function grk_Datetime_Since(\$From, \$To='', \$Prefix='', \$Suffix=' ago', \$Words=array()){
# Est-ce qu'on calcul jusqu'à un moment précis ? Probablement pas, on utilise maintenant
if(empty(\$To) === TRUE){
\$To = time();
}
# On va s'assurer que \$From est numérique
if(is_int(\$From) === FALSE){
\$From = strtotime(\$From);
};
# On va s'assurer que \$To est numérique
if(is_int(\$To) === FALSE){
\$To = strtotime(\$To);
}
# On a une erreur ?
if(\$From === FALSE OR \$From === -1 OR \$To === FALSE OR \$To === -1){
return FALSE;
}
# On va créer deux objets de date
\$From = new DateTime(@date('Y-m-d H:i:s', \$From), new DateTimeZone('GMT'));
\$To = new DateTime(@date('Y-m-d H:i:s', \$To), new DateTimeZone('GMT'));
# On va calculer la différence entre \$From et \$To
if((\$Diff = \$From->diff(\$To)) === FALSE){
return FALSE;
}
# On va merger le tableau des noms (par défaut, anglais)
\$Words = array_merge(array(
'year' => 'year',
'years' => 'years',
'month' => 'month',
'months' => 'months',
'week' => 'week',
'weeks' => 'weeks',
'day' => 'day',
'days' => 'days',
'hour' => 'hour',
'hours' => 'hours',
'minute' => 'minute',
'minutes' => 'minutes',
'second' => 'second',
'seconds' => 'seconds'
), \$Words);
# On va créer la chaîne maintenant
if(\$Diff->y > 1){
\$Text = \$Diff->y.' '.\$Words['years'];
} elseif(\$Diff->y == 1){
\$Text = '1 '.\$Words['year'];
} elseif(\$Diff->m > 1){
\$Text = \$Diff->m.' '.\$Words['months'];
} elseif(\$Diff->m == 1){
\$Text = '1 '.\$Words['month'];
} elseif(\$Diff->d > 7){
\$Text = ceil(\$Diff->d/7).' '.\$Words['weeks'];
} elseif(\$Diff->d == 7){
\$Text = '1 '.\$Words['week'];
} elseif(\$Diff->d > 1){
\$Text = \$Diff->d.' '.\$Words['days'];
} elseif(\$Diff->d == 1){
\$Text = '1 '.\$Words['day'];
} elseif(\$Diff->h > 1){
\$Text = \$Diff->h.' '.\$Words['hours'];
} elseif(\$Diff->h == 1){
\$Text = '1 '.\$Words['hour'];
} elseif(\$Diff->i > 1){
\$Text = \$Diff->i.' '.\$Words['minutes'];
} elseif(\$Diff->i == 1){
\$Text = '1 '.\$Words['minute'];
} elseif(\$Diff->s > 1){
\$Text = \$Diff->s.' '.\$Words['seconds'];
} else {
\$Text = '1 '.\$Words['second'];
}
return \$Prefix.\$Text.\$Suffix;
}
}
``````
-
### Use example :
``````echo time_diff_string('2013-05-01 00:22:35', 'now');
echo time_diff_string('2013-05-01 00:22:35', 'now', true);
``````
### Output :
``````4 months ago
4 months, 2 weeks, 3 days, 1 hour, 49 minutes, 15 seconds ago
``````
### Function :
``````function time_diff_string(\$from, \$to, \$full = false) {
\$from = new DateTime(\$from);
\$to = new DateTime(\$to);
\$diff = \$to->diff(\$from);
\$diff->w = floor(\$diff->d / 7);
\$diff->d -= \$diff->w * 7;
\$string = array(
'y' => 'year',
'm' => 'month',
'w' => 'week',
'd' => 'day',
'h' => 'hour',
'i' => 'minute',
's' => 'second',
);
foreach (\$string as \$k => &\$v) {
if (\$diff->\$k) {
\$v = \$diff->\$k . ' ' . \$v . (\$diff->\$k > 1 ? 's' : '');
} else {
unset(\$string[\$k]);
}
}
if (!\$full) \$string = array_slice(\$string, 0, 1);
return \$string ? implode(', ', \$string) . ' ago' : 'just now';
}
``````
-
if I want to determine if the difference is bigger then 30 minute, what should I do? – Ofir Attia Feb 19 '14 at 10:07
@OfirAttia: you have a bunch of questions like that here on SO, just use search. Simple demo – Glavić Feb 19 '14 at 10:50
I found your article on the following page, which contains a number of references for PHP date time calculations.
Calculate the difference between two Dates (and time) using PHP. The following page provides a range of different methods (7 in total) for performing date / time calculations using PHP, to determine the difference in time (hours, munites), days, months or years between two dates.
-
Very simple:
`````` <?php
\$date1 = date_create("2007-03-24");
echo "Start date: ".\$date1->format("Y-m-d")."<br>";
\$date2 = date_create("2009-06-26");
echo "End date: ".\$date2->format("Y-m-d")."<br>";
\$diff = date_diff(\$date1,\$date2);
echo "Difference between start date and end date: ".\$diff->format("%y years, %m months and %d days")."<br>";
?>
``````
PHP: date_diff - Manual
Note that it's for PHP 5.3.0 or greater.
-
I'm using the following function which I wrote, when PHP 5.3 (respectively date_diff()) is not available:
`````` function dateDifference(\$startDate, \$endDate)
{
\$startDate = strtotime(\$startDate);
\$endDate = strtotime(\$endDate);
if (\$startDate === false || \$startDate < 0 || \$endDate === false || \$endDate < 0 || \$startDate > \$endDate)
return false;
\$years = date('Y', \$endDate) - date('Y', \$startDate);
\$endMonth = date('m', \$endDate);
\$startMonth = date('m', \$startDate);
// Calculate months
\$months = \$endMonth - \$startMonth;
if (\$months <= 0) {
\$months += 12;
\$years--;
}
if (\$years < 0)
return false;
// Calculate the days
\$measure = (\$months == 1) ? 'month' : 'months';
\$days = \$endDate - strtotime('+' . \$months . ' ' . \$measure, \$startDate);
\$days = date('z', \$days);
return array(\$years, \$months, \$days);
}
``````
-
`DateInterval` is great but it has a couple of caveats:
1. only for PHP 5.3+ (but that's really not a good excuse anymore)
2. only supports years, months, days, hours, minutes and seconds (no weeks)
3. it calculates the difference with all of the above + days (you can't get the difference in months only)
To overcome that, I coded the following (improved from @enobrev answer):
``````function date_dif(\$since, \$until, \$keys = 'year|month|week|day|hour|minute|second')
{
\$date = array_map('strtotime', array(\$since, \$until));
if ((count(\$date = array_filter(\$date, 'is_int')) == 2) && (sort(\$date) === true))
{
\$result = array_fill_keys(explode('|', \$keys), 0);
foreach (preg_grep('~^(?:year|month)~i', \$result) as \$key => \$value)
{
while (\$date[1] >= strtotime(sprintf('+%u %s', \$value + 1, \$key), \$date[0]))
{
++\$value;
}
\$date[0] = strtotime(sprintf('+%u %s', \$result[\$key] = \$value, \$key), \$date[0]);
}
foreach (preg_grep('~^(?:year|month)~i', \$result, PREG_GREP_INVERT) as \$key => \$value)
{
if ((\$value = intval(abs(\$date[0] - \$date[1]) / strtotime(sprintf('%u %s', 1, \$key), 0))) > 0)
{
\$date[0] = strtotime(sprintf('+%u %s', \$result[\$key] = \$value, \$key), \$date[0]);
}
}
return \$result;
}
return false;
}
``````
It runs two loops; the first one deals with the relative intervals (years and months) via brute-forcing, and the second one computes the additional absolute intervals with simple arithmetic (so it's faster):
``````echo humanize(date_dif('2007-03-24', '2009-07-31', 'second')); // 74300400 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'minute|second')); // 1238400 minutes, 0 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'hour|minute|second')); // 20640 hours, 0 minutes, 0 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|day')); // 2 years, 129 days
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|week')); // 2 years, 18 weeks
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|week|day')); // 2 years, 18 weeks, 3 days
echo humanize(date_dif('2007-03-24', '2009-07-31')); // 2 years, 4 months, 1 week, 0 days, 0 hours, 0 minutes, 0 seconds
function humanize(\$array)
{
\$result = array();
foreach (\$array as \$key => \$value)
{
\$result[\$key] = \$value . ' ' . \$key;
if (\$value != 1)
{
\$result[\$key] .= 's';
}
}
return implode(', ', \$result);
}
``````
-
It does not support directly week because there's no need. 7 days is a week... – David Bélanger Oct 1 '13 at 11:03
Does it work for leap years? – Peter Mortensen Apr 9 '14 at 0:04
@PeterMortensen: It should work, but I make no guarantees. Set your timezone and give it a go. – Alix Axel Apr 9 '14 at 1:01
An easy function
``````function time_difference (\$time_1, \$time_2) {
\$val_1 = new DateTime(\$time_1);
\$val_2 = new DateTime(\$time_2);
\$interval = \$val_1->diff(\$val_2);
\$year = \$interval->y;
\$month = \$interval->m;
\$day = \$interval->d;
\$hour = \$interval->h;
\$minute = \$interval->i;
\$second = \$interval->s;
\$output = '';
if(\$year > 0){
if (\$year > 1){
\$output .= \$year." years ";
} else {
\$output .= \$year." year ";
}
}
if(\$month > 0){
if (\$month > 1){
\$output .= \$month." months ";
} else {
\$output .= \$month." month ";
}
}
if(\$day > 0){
if (\$day > 1){
\$output .= \$day." days ";
} else {
\$output .= \$day." day ";
}
}
if(\$hour > 0){
if (\$hour > 1){
\$output .= \$hour." hours ";
} else {
\$output .= \$hour." hour ";
}
}
if(\$minute > 0){
if (\$minute > 1){
\$output .= \$minute." minutes ";
} else {
\$output .= \$minute." minute ";
}
}
if(\$second > 0){
if (\$second > 1){
\$output .= \$second." seconds";
} else {
\$output .= \$second." second";
}
}
return \$output;
}
``````
use like
`echo time_difference (\$time_1, \$time_2);`
-
You can also use following code to return date diff by round fractions up \$date1 = \$duedate; // assign due date echo \$date2 = date("Y-m-d"); // current date \$ts1 = strtotime(\$date1); \$ts2 = strtotime(\$date2); \$seconds_diff = \$ts1 - \$ts2; echo \$datediff = ceil((\$seconds_diff/3600)/24); // return in days
If you use floor method of php instead of ceil it will return you the round fraction down. Please check the difference here, some times if your staging servers timezone is different then the live site time zone in that case you may get different results so change the conditions accordingly.
-
``````\$date1 = date_create('2007-03-24');
\$date2 = date_create('2009-06-26');
\$interval = date_diff(\$date1, \$date2);
echo "difference : " . \$interval->y . " years, " . \$interval->m." months, ".\$interval->d." days ";
``````
-
This will try to detect whether a timestamp was given or not, and will also return future dates/times as negative values:
``````<?php
function time_diff(\$start, \$end = NULL, \$convert_to_timestamp = FALSE) {
// If \$convert_to_timestamp is not explicitly set to TRUE,
// check to see if it was accidental:
if (\$convert_to_timestamp || !is_numeric(\$start)) {
// If \$convert_to_timestamp is TRUE, convert to timestamp:
\$timestamp_start = strtotime(\$start);
}
else {
// Otherwise, leave it as a timestamp:
\$timestamp_start = \$start;
}
// Same as above, but make sure \$end has actually been overridden with a non-null,
// non-empty, non-numeric value:
if (!is_null(\$end) && (!empty(\$end) && !is_numeric(\$end))) {
\$timestamp_end = strtotime(\$end);
}
else {
// If \$end is NULL or empty and non-numeric value, assume the end time desired
// is the current time (useful for age, etc):
\$timestamp_end = time();
}
// Regardless, set the start and end times to an integer:
\$start_time = (int) \$timestamp_start;
\$end_time = (int) \$timestamp_end;
// Assign these values as the params for \$then and \$now:
\$start_time_var = 'start_time';
\$end_time_var = 'end_time';
// Use this to determine if the output is positive (time passed) or negative (future):
\$pos_neg = 1;
// If the end time is at a later time than the start time, do the opposite:
if (\$end_time <= \$start_time) {
\$start_time_var = 'end_time';
\$end_time_var = 'start_time';
\$pos_neg = -1;
}
// Convert everything to the proper format, and do some math:
\$then = new DateTime(date('Y-m-d H:i:s', \$\$start_time_var));
\$now = new DateTime(date('Y-m-d H:i:s', \$\$end_time_var));
\$years_then = \$then->format('Y');
\$years_now = \$now->format('Y');
\$years = \$years_now - \$years_then;
\$months_then = \$then->format('m');
\$months_now = \$now->format('m');
\$months = \$months_now - \$months_then;
\$days_then = \$then->format('d');
\$days_now = \$now->format('d');
\$days = \$days_now - \$days_then;
\$hours_then = \$then->format('H');
\$hours_now = \$now->format('H');
\$hours = \$hours_now - \$hours_then;
\$minutes_then = \$then->format('i');
\$minutes_now = \$now->format('i');
\$minutes = \$minutes_now - \$minutes_then;
\$seconds_then = \$then->format('s');
\$seconds_now = \$now->format('s');
\$seconds = \$seconds_now - \$seconds_then;
if (\$seconds < 0) {
\$minutes -= 1;
\$seconds += 60;
}
if (\$minutes < 0) {
\$hours -= 1;
\$minutes += 60;
}
if (\$hours < 0) {
\$days -= 1;
\$hours += 24;
}
\$months_last = \$months_now - 1;
if (\$months_now == 1) {
\$years_now -= 1;
\$months_last = 12;
}
// "Thirty days hath September, April, June, and November" ;)
if (\$months_last == 9 || \$months_last == 4 || \$months_last == 6 || \$months_last == 11) {
\$days_last_month = 30;
}
else if (\$months_last == 2) {
// Factor in leap years:
if ((\$years_now % 4) == 0) {
\$days_last_month = 29;
}
else {
\$days_last_month = 28;
}
}
else {
\$days_last_month = 31;
}
if (\$days < 0) {
\$months -= 1;
\$days += \$days_last_month;
}
if (\$months < 0) {
\$years -= 1;
\$months += 12;
}
// Finally, multiply each value by either 1 (in which case it will stay the same),
// or by -1 (in which case it will become negative, for future dates).
// Note: 0 * 1 == 0 * -1 == 0
\$out = new stdClass;
\$out->years = (int) \$years * \$pos_neg;
\$out->months = (int) \$months * \$pos_neg;
\$out->days = (int) \$days * \$pos_neg;
\$out->hours = (int) \$hours * \$pos_neg;
\$out->minutes = (int) \$minutes * \$pos_neg;
\$out->seconds = (int) \$seconds * \$pos_neg;
return \$out;
}
``````
Example usage:
``````<?php
\$birthday = 'June 2, 1971';
\$check_age_for_this_date = 'June 3, 1999 8:53pm';
\$age = time_diff(\$birthday, \$check_age_for_this_date)->years;
print \$age;// 28
``````
Or:
``````<?php
\$christmas_2020 = 'December 25, 2020';
\$countdown = time_diff(\$christmas_2020);
print_r(\$countdown);
``````
-
I had the same problem with PHP 5.2 and solved it with MySQL. Might not be exactly what you're looking for, but this will do the trick and return the number of days:
``````\$datediff_q = \$dbh->prepare("SELECT DATEDIFF(:date2, :date1)");
\$datediff_q->bindValue(':date1', '2007-03-24', PDO::PARAM_STR);
\$datediff_q->bindValue(':date2', '2009-06-26', PDO::PARAM_STR);
\$datediff = (\$datediff_q->execute()) ? \$datediff_q->fetchColumn(0) : false;
``````
-
Since everyone is posting code samples, here is another version.
I wanted a function to display differences from seconds to years (just one unit). For periods over 1 day, I wanted it to rollover at midnight (10am Monday seen from 9am Wednesday is 2 days ago, not 1). And for periods over a month, I wanted the rollover to be on the same day of the month (including for 30/31 day months & leap years).
This is what I came up with:
``````/**
* Returns how long ago something happened in the past, showing it
* as n seconds / minutes / hours / days / weeks / months / years ago.
*
* For periods over a day, it rolls over at midnight (so doesn't depend
* on current time of day), and it correctly accounts for month-lengths
* and leap-years (months and years rollover on current day of month).
*
* \$param string \$timestamp in DateTime format
* \$return string description of interval
*/
function ago(\$timestamp)
{
\$then = date_create(\$timestamp);
// for anything over 1 day, make it rollover on midnight
\$today = date_create('tomorrow'); // ie end of today
\$diff = date_diff(\$then, \$today);
if (\$diff->y > 0) return \$diff->y.' year'.(\$diff->y>1?'s':'').' ago';
if (\$diff->m > 0) return \$diff->m.' month'.(\$diff->m>1?'s':'').' ago';
\$diffW = floor(\$diff->d / 7);
if (\$diffW > 0) return \$diffW.' week'.(\$diffW>1?'s':'').' ago';
if (\$diff->d > 1) return \$diff->d.' day'.(\$diff->d>1?'s':'').' ago';
// for anything less than 1 day, base it off 'now'
\$now = date_create();
\$diff = date_diff(\$then, \$now);
if (\$diff->d > 0) return 'yesterday';
if (\$diff->h > 0) return \$diff->h.' hour'.(\$diff->h>1?'s':'').' ago';
if (\$diff->i > 0) return \$diff->i.' minute'.(\$diff->i>1?'s':'').' ago';
return \$diff->s.' second'.(\$diff->s==1?'':'s').' ago';
}
``````
-
"if" the date is stored in MySQL, I find it easier to do the difference calculation at the database level... Then based on the Day, Hour, Min, Sec output, parse and display results as appropriate...
``````mysql> select firstName, convert_tz(loginDate, '+00:00', '-04:00') as loginDate, TIMESTAMPDIFF(DAY, loginDate, now()) as 'Day', TIMESTAMPDIFF(HOUR, loginDate, now())+4 as 'Hour', TIMESTAMPDIFF(MINUTE, loginDate, now())+(60*4) as 'Min', TIMESTAMPDIFF(SECOND, loginDate, now())+(60*60*4) as 'Sec' from User_ where userId != '10158' AND userId != '10198' group by emailAddress order by loginDate desc;
+-----------+---------------------+------+------+------+--------+
| firstName | loginDate | Day | Hour | Min | Sec |
+-----------+---------------------+------+------+------+--------+
| Peter | 2014-03-30 18:54:40 | 0 | 4 | 244 | 14644 |
| Keith | 2014-03-30 18:54:11 | 0 | 4 | 244 | 14673 |
| Andres | 2014-03-28 09:20:10 | 2 | 61 | 3698 | 221914 |
| Nadeem | 2014-03-26 09:33:43 | 4 | 109 | 6565 | 393901 |
+-----------+---------------------+------+------+------+--------+
4 rows in set (0.00 sec)
``````
-
Some time ago I wrote a `format_date` function as this gives many options on how you want your date:
``````function format_date(\$date, \$type, \$seperator="-")
{
if(\$date)
{
\$day = date("j", strtotime(\$date));
\$month = date("n", strtotime(\$date));
\$year = date("Y", strtotime(\$date));
\$hour = date("H", strtotime(\$date));
\$min = date("i", strtotime(\$date));
\$sec = date("s", strtotime(\$date));
switch(\$type)
{
case 0: \$date = date("Y".\$seperator."m".\$seperator."d",mktime(\$hour, \$min, \$sec, \$month, \$day, \$year)); break;
case 1: \$date = date("D, F j, Y",mktime(\$hour, \$min, \$sec, \$month, \$day, \$year)); break;
case 2: \$date = date("d".\$seperator."m".\$seperator."Y",mktime(\$hour, \$min, \$sec, \$month, \$day, \$year)); break;
case 3: \$date = date("d".\$seperator."M".\$seperator."Y",mktime(\$hour, \$min, \$sec, \$month, \$day, \$year)); break;
case 4: \$date = date("d".\$seperator."M".\$seperator."Y h:i A",mktime(\$hour, \$min, \$sec, \$month, \$day, \$year)); break;
case 5: \$date = date("m".\$seperator."d".\$seperator."Y",mktime(\$hour, \$min, \$sec, \$month, \$day, \$year)); break;
case 6: \$date = date("M",mktime(\$hour, \$min, \$sec, \$month, \$day, \$year)); break;
case 7: \$date = date("Y",mktime(\$hour, \$min, \$sec, \$month, \$day, \$year)); break;
case 8: \$date = date("j",mktime(\$hour, \$min, \$sec, \$month, \$day, \$year)); break;
case 9: \$date = date("n",mktime(\$hour, \$min, \$sec, \$month, \$day, \$year)); break;
case 10:
\$diff = abs(strtotime(\$date) - strtotime(date("Y-m-d h:i:s")));
\$years = floor(\$diff / (365*60*60*24));
\$months = floor((\$diff - \$years * 365*60*60*24) / (30*60*60*24));
\$days = floor((\$diff - \$years * 365*60*60*24 - \$months*30*60*60*24)/ (60*60*24));
\$date = \$years . " years, " . \$months . " months, " . \$days . "days";
}
}
return(\$date);
}
``````
-
This answer is just as wrong as khaldonno's answer. It assumes (case 10) that a year has 365 days (every fourth year has 366 days (except for the 100 year / 400 years rules for the Gregorian calendar)), and that a month has 30 days (it is about approximately 30.42 days in non-leap years). Even with better constants it is only correct on average, not necessarily correct for any two particular dates. – Peter Mortensen Apr 9 '14 at 0:00
you can always use the following function that can return the age in years and months (ie. 1 Year 4 Months)
``````function getAge(\$dob, \$age_at_date)
{
\$d1 = new DateTime(\$dob);
\$d2 = new DateTime(\$age_at_date);
\$age = \$d2->diff(\$d1);
\$years = \$age->y;
\$months = \$age->m;
return \$years.'.'.months;
}
``````
or if you want the age to be calculated at the current date, you can use
``````function getAge(\$dob)
{
\$d1 = new DateTime(\$dob);
\$d2 = new DateTime(date());
\$age = \$d2->diff(\$d1);
\$years = \$age->y;
\$months = \$age->m;
return \$years.'.'.months;
}
``````
-
In for a penny, in for a pound: I have just reviewed several solutions, all providing a complex solution using floor() that then rounds up to a 26 years 12 month and 2 days solution, for what should have been 25 years, 11 months and 20 days!!!!
here is my version of this problem: may not be elegant, may not be well coded, but provides a more closer proximity to a answer if you do not count LEAP years, obviously leap years could be coded into this, but in this case - as someone else said, perhaps you could provide this answer:: I have included all TEST conditions and print_r so that you can see more clearly the construct of the results:: here goes,
// set your input dates/ variables::
``````\$ISOstartDate = "1987-06-22";
\$ISOtodaysDate = "2013-06-22";
``````
// We need to EXPLODE the ISO yyyy-mm-dd format into yyyy mm dd, as follows::
\$yDate[ ] = explode('-', \$ISOstartDate); print_r (\$yDate);
\$zDate[ ] = explode('-', \$ISOtodaysDate); print_r (\$zDate);
``````// Lets Sort of the Years!
// Lets Sort out the difference in YEARS between startDate and todaysDate ::
\$years = \$zDate[0][0] - \$yDate[0][0];
// We need to collaborate if the month = month = 0, is before or after the Years Anniversary ie 11 months 22 days or 0 months 10 days...
if (\$months == 0 and \$zDate[0][1] > \$ydate[0][1]) {
\$years = \$years -1;
}
// TEST result
echo "\nCurrent years => ".\$years;
// Lets Sort out the difference in MONTHS between startDate and todaysDate ::
\$months = \$zDate[0][1] - \$yDate[0][1];
// TEST result
echo "\nCurrent months => ".\$months;
// Now how many DAYS has there been - this assumes that there is NO LEAP years, so the calculation is APPROXIMATE not 100%
// Lets cross reference the startDates Month = how many days are there in each month IF m-m = 0 which is a years anniversary
// We will use a switch to check the number of days between each month so we can calculate days before and after the years anniversary
switch (\$yDate[0][1]){
case 01: \$monthDays = '31'; break; // Jan
case 02: \$monthDays = '28'; break; // Feb
case 03: \$monthDays = '31'; break; // Mar
case 04: \$monthDays = '30'; break; // Apr
case 05: \$monthDays = '31'; break; // May
case 06: \$monthDays = '30'; break; // Jun
case 07: \$monthDays = '31'; break; // Jul
case 08: \$monthDays = '31'; break; // Aug
case 09: \$monthDays = '30'; break; // Sept
case 10: \$monthDays = '31'; break; // Oct
case 11: \$monthDays = '30'; break; // Nov
case 12: \$monthDays = '31'; break; // Dec
};
// TEST return
echo "\nDays in start month ".\$yDate[0][1]." => ".\$monthDays;
// Lets correct the problem with 0 Months - is it 11 months + days, or 0 months +days???
\$days = \$zDate[0][2] - \$yDate[0][2] +\$monthDays;
echo "\nCurrent days => ".\$days."\n";
// Lets now Correct the months to being either 11 or 0 Months, depending upon being + or - the years Anniversary date
// At the same time build in error correction for Anniversary dates not being 1yr 0m 31d... see if (\$days == \$monthDays )
if(\$days < \$monthDays && \$months == 0)
{
\$months = 11; // If Before the years anniversary date
}
else {
\$months = 0; // If After the years anniversary date
\$years = \$years+1; // Add +1 to year
\$days = \$days-\$monthDays; // Need to correct days to how many days after anniversary date
};
// Day correction for Anniversary dates
if (\$days == \$monthDays ) // if todays date = the Anniversary DATE! set days to ZERO
{
\$days = 0; // days set toZERO so 1 years 0 months 0 days
};
echo "\nTherefore, the number of years/ months/ days/ \nbetween start and todays date::\n\n";
printf("%d years, %d months, %d days\n", \$years, \$months, \$days);
``````
the end result is:: 26 years, 0 months, 0 days
That's how long I have been in business for on the 22nd June 2013 - Ouch!
-
## protected by Community♦Aug 5 '11 at 13:28
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 13,957 | 45,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.899933 |
http://blog.strobaek.org/2011/01/29/college-statistics/ | 1,723,554,194,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00550.warc.gz | 3,288,212 | 13,646 | # College Statistics
Michael Feathers (@mfeathers) posted a link on twitter to a nice little statistics assignment.
If you would like to solve it yourselves, stop reading when you get to the mathematics, otherwise it will spoil your fun.
Enjoy 🙂
Currently a college senior, Jeremy has had a secret crush on Emma ever since the third grade. Two weeks ago, fearing that his feelings would forever go unrequited, he broke his silence and sent Emma a letter through Campus Mail, acknowledging his twelve/year secret romance.Now, fourteen agonizing days later, he has yet to receive a response. Hoping against hope, Jeremy and his fragile psyche are clinging to the possibility that someone’s letter was lost in the mail.
Assuming that:
1. Emma (who is actually secretly dating Jeremy’s father) has a 70% change of mailing a response if, in face, she had received the letter and
2. The Campus Post Office has a one in fifty change of losing any particular piece of mail,
what is the probability that Emma never received Jeremy’s confession of the heart?
Let $B$ represent the event that Jeremy did not receive a response; let $A_1$ and $A_2$ denote the events that Emma did and did not, respectively, receive, Jeremy’s letter. The objective is to find $P(A_2 | B)$.
Stop reading now, if you do not wish to see the proof. Not that it is very complicated, mind you, but try to extract that old knowledge from years back when you had statistics yourself.
From what we know about Emma’s behavior and the incompetence of the Campus Post Office,
$\displaystyle{P(A_1) = \frac{49}{50}, P(A_2) = \frac{1}{50}, P(B | A_2) = 1}$
Also,
$\displaystyle{P(B | A1) = P(\mbox{Jeremy receive no response} | \mbox{Emma received Jeremy's letter})}$
$\displaystyle{ = P[\mbox{Emma does not respond} \cup (\mbox{Emma responds} \cap \mbox{Post Office loses letter})]}$
$\displaystyle{ = P(\mbox{Emma does not respond}) + P(\mbox{letter is lost} | \mbox{Emma responds}) \times P(\mbox{Emma responds})}$
$\displaystyle{ = 0.30 + 0.70\frac{1}{50}}$
$\displaystyle{ = 0.314}$
Therefore,
$\displaystyle{ P(A_2 | B) = \frac{1\frac{1}{50}}{0.324\frac{49}{50} + 1\frac{1}{50}}}$
$\displaystyle{ = 0.061}$
Sadly, the magnitude of $P(A_2 | B)$ is not good news for Jeremy. If $P(A_2 | B) = 0.061$, it follows that Emma’s probability of having received the letter but not caring enough to respond was almost $94\%$. “Faint heart ne’er won fair lady,” but Jeremy would probaly be well/advised to direct his romantic intentions elsewhere. | 674 | 2,519 | {"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": 15, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.940556 |
http://forum.allaboutcircuits.com/threads/plot-square-wave-where-duty-cycle-is-a-function.121590/ | 1,477,666,433,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988722653.96/warc/CC-MAIN-20161020183842-00185-ip-10-171-6-4.ec2.internal.warc.gz | 93,831,903 | 16,839 | # Plot square wave where duty cycle is a function
Discussion in 'Math' started by anhnha, Mar 5, 2016.
1. ### anhnha Thread Starter Active Member
Apr 19, 2012
757
44
I have a square wave where duty cycle is a function of time as below. How can I plot the square wave?
d(t) = 0.5 + 0.1cos(t).
I am confused because at each point in time there is a duty cycle but the definition of duty cycle should be in a period.
2. ### Alec_t AAC Fanatic!
Sep 17, 2013
5,594
1,070
What is the pulse period? Unless you know that you can't plot the waveform against an actual time scale.
3. ### anhnha Thread Starter Active Member
Apr 19, 2012
757
44
Let's assume that the frequency is 1kHz or period 1ms.
BTW, I think that there is a contradiction here. The duty cycle can be a function of time but it should be constant in a period.
4. ### wayneh Expert
Sep 9, 2010
11,931
2,862
I've used this approach to throb an LED. I set the duty cycle of a PWM output to follow a sine wave. It produces a very nice effect.
Of course: PWM frequency >> throb frequency
5. ### anhnha Thread Starter Active Member
Apr 19, 2012
757
44
With PWM frequency >> throb frequency, so in a period of PWM signal, duty cycle can be approximated as a constant but actually it is not a constant at all. So I am wondering what the PWM signal will be if duty cycle is not a constant.
Here is what I am confused from the power electronics lecture. How is it possible get a square wave gate drive from that duty cycle d(t)?
eanidal likes this.
6. ### Alec_t AAC Fanatic!
Sep 17, 2013
5,594
1,070
I don't see a contradiction. A period can have only one duty-cycle, but the 'mark' duration can vary from one period to another.
7. ### anhnha Thread Starter Active Member
Apr 19, 2012
757
44
I agree with you about this. However, for the definition of duty cycle function in post #1 or #5, it is a continuous with t and in a switching period the duty cycle is not a constant.
8. ### WBahn Moderator
Mar 31, 2012
17,461
4,701
This is similar to the instantaneous frequency in a frequency modulated system except that there it is a better defined since everything is continuous and so it make more sense to talk about different frequencies at different points within the same period.
Here I can imagine a few things. First, imagine that you have not only the rectangular wave output, but a linear ramp that goes from o to 1 over the course of the period. When that ramp first exceeds your d(t) signal that is when your rectangular wave drops from HI to LO. I could also imagine sampling d(t) at the beginning of each period to ensure that each period has one defined duty cycle.
9. ### wayneh Expert
Sep 9, 2010
11,931
2,862
That depends on the details of the PWM generator. In my case I was changing the duty cycle setting every tenth of a second or so. Since the throb took several seconds, updating every tenth second gave very smooth operation.
The PWM would run at a constant duty cycle until the next update. I have no idea how it dealt with a change that arrived during a cycle. It's irrelevant at a practical level, since there were hundreds of full cycles for every one change cycle.
10. ### anhnha Thread Starter Active Member
Apr 19, 2012
757
44
This works but then the duty cycle is not actually the one defined above. It is sampled of d(t) and is discontinuous.
I see how it works in your case because d(t) is discontinuous here.
11. ### Alec_t AAC Fanatic!
Sep 17, 2013
5,594
1,070
To define the pulse width within any period you have to pick an end-of-pulse moment. Do you regard that as continuous or discontinuous?
12. ### anhnha Thread Starter Active Member
Apr 19, 2012
757
44
Not really sure what you mean here.
Here is what I meant by continuous or discontinuous:
d(t) = 0.5 + 0.1cos(t) for all positive value of t: this is a continuous. At each value of t there is one duty cycle. Even in a switching period, there is infinite duty cycle because there is infinite point in time t within that period.
d(t) = 0.5 + 0.1cos(t) is discontinuous if in each switching frequency there is only one value of at which the function is defined.
13. ### wayneh Expert
Sep 9, 2010
11,931
2,862
Suppose the PWM frequency is nearly the same as that of the sine wave, and suppose your job is running the PWM duty cycle in response to that sine wave. How would you decide what to do? There is no singular answer. Different designers would take different approaches. The question cannot be answered without looking at a specific device.
Using LTspice to model a simple 555-based PWM driver might give you what you want. It wouldn't be hard to simulate what you've described.
14. ### Alec_t AAC Fanatic!
Sep 17, 2013
5,594
1,070
Even LTspice will be 'discontinuous', since it calculates in discrete, albeit brief, time steps. Any practical pulse generator will also differ from the ideal continuous implementation.
15. ### wayneh Expert
Sep 9, 2010
11,931
2,862
I was picturing putting a sine wave onto the control pin of the 555, and watching the PWM. Or something like that. I don't know what the TS is looking for but maybe seeing a simulation would trigger the aha moment. | 1,320 | 5,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | longest | en | 0.913246 |
https://braingenie.ck12.org/skills/103259/learn | 1,550,765,670,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247505838.65/warc/CC-MAIN-20190221152543-20190221174543-00348.warc.gz | 535,010,445 | 1,879 | ### Sample Problem
Mr. Clough earns a monthly income of \$3,980. He spends \$218 on food and three times as much on rent. He also spends \$172 on clothing and half as much on transportation. If he saves the rest of his income, what are his monthly savings? (Don’t include a comma in your answer.)
\$.00
#### Solution
Mr. Clough spends \$218 × 3 = \$654 on rent.
He spends \$172 ÷ 2 = \$86 on transportation.
Therefore, he spends \$218 + \$654 + \$172 + \$86 = \$1,130 every month.
Then his monthly savings are \$3,980 – \$1,130 = \$2,850. | 152 | 545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2019-09 | longest | en | 0.961452 |
https://studylib.net/doc/9923946/stoichiometry | 1,638,954,294,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363465.47/warc/CC-MAIN-20211208083545-20211208113545-00061.warc.gz | 607,763,925 | 11,024 | # Stoichiometry
```Stoichiometry
Stoichiometry
Stoichiometry is the study of the mass
relationships of the reactants and the
products in a chemical reaction.
How much stuff it takes to make a certain
amount of new stuff.
Based on the law of conservation of mass
Mass of the reactants must equal the mass
the products
CH3OH(g) + O2(g) CO2(g) + H2O(g)
Chemical reactions are mole ratios between
the reactants and the products – not a mass
ratio!
Mole ratio – the ratio of moles of any two
substances in a chemical reaction
2CH3OH(l) + 3O2(g) 2CO2(g) + 4H2O(g)
2 mol CH3OH : 2 mol CO2
3 mol O2 : 4 mol H2O
There is not a direct relationship between
masses of individual reactants and products.
Must use the mole ratio to convert!
R1
+
R2
→
P1
+
P2
grams
molar mass
R1
moles R1
grams
molar mass
P2
moles P2
moles R1
(mole ratio)
moles P2
Guidelines for Stoichiometry
1.
2.
3.
4.
Write a balanced equation for the chemical
reaction.
Convert grams A to moles of A using the
molar mass.
Convert moles of A to moles of B using the
mole ratio.
Convert moles of B to grams of B using the
molar mass.
3. Examples
24. Sulfuric acid reacts with sodium hydroxide to
form sodium sulfate and water. How many
grams of Na2SO4 are produced from 25.0 g of
H2SO4?
25. In the reaction from #24, I need 550.0 g Na2SO4.
What mass of each reactant is required?
26. How much oxygen will be required to react with
100.0 g KCl to produce KClO3.
27. What mass of hydrogen peroxide must
decompose to produce 0.77 g of water? The
other product is oxygen gas.
28. Space vehicles use solid lithium hydroxide to
remove exhaled carbon dioxide according to the
equation: LiOH + CO2 Li2CO3 + H2O.
Determine the mass of carbon dioxide removed if
the space vehicle carries 1.00 kg of LiOH.
29. How many grams of carbon dioxide are
produced when 2.50 g of baking soda react with
excess citric acid (H3C6H5O7) to produce
Na3C6H5O7, carbon dioxide, and water.
30. Aspirin (C9H8O4) is produced when salicylic acid
(C7H6O3) reacts with acetic anhydride (C4H6O3)
according to the equation C7H6O3 + C4H6O3
C9H8O4 + HC2H3O2. Determine the mass of
aspirin produced when 150.0 g of salicylic acid
reacts with an excess of acetic anhydride.
``` | 768 | 2,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.83437 |
http://slideplayer.com/slide/3963905/ | 1,511,269,760,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806353.62/warc/CC-MAIN-20171121113222-20171121133222-00589.warc.gz | 269,854,930 | 23,042 | Game Theory Game theory is an attempt to model the way decisions are made in competitive situations. It has obvious applications in economics. But it.
Presentation on theme: "Game Theory Game theory is an attempt to model the way decisions are made in competitive situations. It has obvious applications in economics. But it."— Presentation transcript:
Game Theory
Game theory is an attempt to model the way decisions are made in competitive situations. It has obvious applications in economics. But it has also been applied to a huge range of other areas, including politics, philosophy, biology, and computer science.
A basic starting point is the “two person game”. Two players must decide how they will move, and are rewarded or punished accordingly. There are many variations, such as one player moving first and the other responding, or both moving simultaneously, or taking turns for a sequence of moves, etc. The goal of the theory is to analyze how they should move so as to optimize the outcome. It hopes to provide a “rational basis” for playing the game. Naturally, the players do not have to be “persons”. They can be corporations or other entities, and, for applications to biology, they can be species or even Nature.
Of course, “games” in the real world tend to be very complicated, too much so to be readily subjected to analysis. In a way, it is surprising how much we can learn about real situations from simplified formalized versions. They do not tell us exactly how to proceed in various real situations, but they do provide insight into what is going on. One of the most difficult situations to address, either in theory or in practice, is an opponent who behaves irrationally. Most theory about games assumes that all players behave rationally. It is difficult even to see how to incorporate irrational behaviour into the analysis of a game. But even in this situation, we can learn something from game theory.
A simple kind of game is the two-person “matrix game”. The idea is that each player can choose from a list of strategies, and the outcome of the game is determined by the strategies chosen by both of them. For each possible combination of strategies, one for each player, a payoff is specified for each player. For example, suppose one player can choose either of two strategies, a and b, and the other can choose from a list of three: A, B, C. Then the game is completely specified by a table of the payoffs to each player.
Here is such a table: The idea is that the two numbers in each cell represent the payoffs to the two players for each combination of strategies. ABC a2, -13, 1-1, 3 b0, 3-1, 23, -1 Player #1 Player #2 For example, if Player #1 plays strategy a, and Player #2 plays C, then Player #1 loses 1 point and Player #2 wins 3 points. A table like this is called a “matrix”, and that is why this type of game is called a “matrix game”. -1, 3
The horizontal lines in the matrix are called the “rows”, while the vertical lines are the “columns”. ABC a2, -13, 1-1, 3 b0, 3-1, 23, -1 Player #1 Player #2 With this terminology in mind, Player #1 is sometimes called “Rose”, and Player #2 is “Colin”. Rose Colin
How should the players play? Rose would like to get 3 points, which is possible if she plays a or b. But in either case, it is also possible that she will lose a point. ABC a2, -13, 1-1, 3 b0, 3-1, 23, -1 Player #1 Player #2 Colin could get 3 from either A or C, but in either case might lose 1. Rose Colin A “safe” strategy for Colin is B: he will get either 1 or 2.
If Rose observes or predicts that Colin will play it safe with B, then she should play a. Unfortunately for Colin, this limits his winnings. ABC a2, -13, 1-1, 3 b0, 3-1, 23, -1 Player #1 Player #2 If he realizes that she is going to play a, then he should play C. Rose Colin And so it goes. If either player’s intentions are predictable, then the opponent can take advantage of this knowledge.
Example: Stag Hunt This game is used to explore questions of cooperation among players. The idea is that two wolves working together can catch a stag, whereas each on his own can only catch a hare. staghare stag3, 30, 2 hare2, 02, 2 If they work together, they can catch a stag, and each has the highest possible payoff. But if either refuses to hunt a stag and goes after a hare, the other is better off also going after a hare. Wolf #2 Wolf #1
This is a simple version of something that really happens. If people work together and compromise to reach a consensus, they can often achieve the best results for everybody. staghare stag3, 30, 2 hare2, 02, 2 But as soon as one participant chooses to put his self interest first, then there is an incentive for the others to do the same, for fear of losing out. 3, 3 2, 0 2, 2
You could argue that this is what happens to global efforts to control the environmental crisis. A multilateral agreement like the Kyoto Accord offers a good hope for improving things for everybody. staghare stag3, 30, 2 hare2, 02, 2 But if a few countries refuse to sign, it makes others reluctant to sign. Or, if they have signed, it makes them reluctant to act on their commitments. They fear that the cost of emission reductions will put them at a competitive disadvantage. The end result is that everybody may continue polluting, which is bad for everybody.
This offers an interesting insight into the way we often seem to end up doing something that is not really good for anybody. It is especially likely to happen when there are multiple “players”. staghare stag3, 30, 2 hare2, 02, 2 But even two players can easily find themselves in a stalemate in which neither is prepared to budge, even though both could do better if they changed their behaviour. A classic example is the division of property after a divorce.
A special kind of game is one in which the payoffs are each other’s negatives: In this situation, it is not really necessary to write both payoffs. It is enough to write the payoffs for Rose only. ABC a2, -2-1, 1-3, 3 b0, 0-2, 23, -3 Player #1 Player #2 Rose Colin A game like this is called a “zero-sum game”, because the payoffs in any situation add up to zero. ABC a2-3 b0-23 Rose Colin
Many real-world games are zero-sum games. A typical example involves the division of fixed resources, financial or otherwise. An analysis of how best to achieve this can be complicated and difficult, even for simple examples. ABC a2-3 b0-23 Rose Colin
Download ppt "Game Theory Game theory is an attempt to model the way decisions are made in competitive situations. It has obvious applications in economics. But it."
Similar presentations | 1,568 | 6,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-47 | longest | en | 0.971944 |
http://www.openproblemgarden.org/comment/reply/60052 | 1,718,321,358,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00511.warc.gz | 52,098,127 | 5,986 | Importance: High ✭✭✭
Author(s):
Subject: Probability
Keywords: KPZ equation, central limit theorem
Posted by: Tomas Kojar on: October 3rd, 2020
Conjecture Formulate a central limit theorem for the KPZ universality class.
The KPZ equation is given by
where denotes space-time white noise and is a parameter describing the strength of its "asymmetry". It has been conjectured (see [BPRS93, BG97, Cor 12 ,GJ14,HQ18] for a number of results in this direction) that the KPZ equation has a “universal” character in the sense that any one-dimensional model of surface growth should converge to it provided that it has the following features:
• There is a microscopic smoothing mechanism. Pictorially this means that large valleys are quickly filled.
• The system has microscopic fluctuations with short-range correlations. Pictorially this means that height function change depends only on neighboring heights.
• The system has some “lateral growth” mechanism in the sense that the growth speed depends in a nontrivial way on the slope. The vertical effective growth rate depends non-linearly on local slope.
• At the microscopic scale, the strengths of the growth and fluctuation mechanisms are well separated: either the growth mechanism dominates (intermediate disorder) or the fluctuations dominate (weak asymmetry). Growth is drive by noise which quickly decorrelates in space / time and is not heavy tailed.
Here is a concrete surface growth mathematical model to give a sense of the above features. The random deposition model is one of the simplest (and least realistic) models for a randomly growing one-dimensional interface. Unit blocks fall independently and in parallel from the sky above each site of according to exponentially distributed waiting times. Recall that a random variable X has exponential distribution of rate (or mean ) if . Such random variables are characterized by the memoryless property – conditioned on the event that , still has the exponential distribution of the same rate. Consequently, the random deposition model is Markov – its future evolution only depends on the present state (and not on its history). The ballistic deposition (or sticky block) model was introduced by Vold [V59] in 1959 and, as one expects in real growing interfaces, displays spatial correlation. As before, blocks fall according to iid exponential waiting times, however, now a block will stick to the first edge against which it becomes incident. This creates overhangs and we define the height function h(t, x) as the maximal height above x which is occupied by a box.
## Bibliography
[BG97] L. BERTINI and G. GIACOMIN. Stochastic Burgers and KPZ equations from particle systems. Comm. Math. Phys. 183, no. 3, (1997), 571–607.
[BPRS93] L. BERTINI, E. PRESUTTI, B. RUDIGER ¨ , and E. SAADA. Dynamical fluctuations at the critical point: convergence to a nonlinear stochastic PDE. Teor. Veroyatnost. i Primenen. 38, no. 4, (1993), 689–741
[Cor 12] I. Corwin, The Kardar-Parisi-Zhang equation and universality class, Random Matrices Theory Appl. 1 (2012), 1130001, 76. MR 2930377. Zbl 1247.82040. http://dx.doi.org
[GJ14] P. GONC¸ ALVES and M. JARA. Nonlinear fluctuations of weakly asymmetric interacting particle systems. Arch. Ration. Mech. Anal. 212, no. 2, (2014), 597–644
[HQ18] HAIRER, M. and QUASTEL, J. (2018). A class of growth models rescaling to KPZ. Forum Math. Pi 6 e3
[V59] M. J. Vold. A numerical approach to the problem of sediment volume. J. Colloid Sci., 14:168 (1959).
* indicates original appearance(s) of problem. | 874 | 3,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-26 | latest | en | 0.922027 |
http://fullhomework.com/downloads/expert-answers-53/ | 1,603,662,502,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890028.58/warc/CC-MAIN-20201025212948-20201026002948-00182.warc.gz | 43,449,917 | 10,283 | # Expert Answers
Brief Exercises: BE12-1
BE12-1 Bella Company is considering purchasing new equipment for \$450,000. It is expected that the equipment will produce net annual cash flows of \$50,000 over its 10-year useful life. Annual depreciation will be \$45,000. Compute the cash payback period.
Brief Exercises: BE12-4
BE12-4 Caine Bottling Corporation is considering the purchase of a new bottling machine. The machine would cost \$200,000 and has an estimated useful life of 8 years with zero salvage value. Management estimates that the new bottling machine will provide net annual cash flows of \$34,000. Management also believes that the new bottling machine will save the company money because it is expected to be more reliable than other machines, and thus will reduce downtime. How much would the reduction in downtime have to be worth in order for the project to be acceptable? Assume a discount rate of 9%. (Hint: Calculate the net present value.)
Brief Exercises: BE12-5
BE12-5 Beacon Company is considering two different, mutually exclusive capital expenditure proposals. Project A will cost \$400,000, has an expected useful life of 1 0 years, a salvage value of zero, and is expected to increase net annual cash flows by \$70,000. Project B will cost \$280,000, has an expected useful life of 10 years, a salvage value of zero, and is expected to increase net annual cash flows by \$50,000. A discount rate of 9% is appropriate for both projects. Compute the net present value and profitability index of each project. Which project should be accepted?
Brief Exercises: BE12-6
BE12-6 Quillen Company is performing a post-audit of a project completed one year ago. The initial estimates were that the project would cost \$250,000, would have a useful life of 9 years, zero salvage value, and would result in net annual cash flows of \$46,000 per year. Now that the investment has been in operation for 1 year, revised figures indicate that it actually cost \$260,000, will have a useful life of 11 years, and will produce net annual cash flows of \$39,000 per year. Evaluate the success of the project. Assume a discount rate of 10%.
Exercise: E12-5
E12-5 Eisler Corporation is involved in the business of injection molding of plastics. It is considering the purchase of a new computer-aided design and manufacturing machine for \$430,000. The company believes that with this new machine it will improve productivity and increase quality, resulting in an increase in net annual cash flows of \$101,000 for the next 6 years. Management requires a 10% rate of return on all new investments.
Instructions
Calculate the internal rate of return on this new machine. Should the investment be accepted?
Answer the Following Question:
Discuss the significance of recognizing the time value of money in the long-term impact of the capital budgeting decision | 653 | 2,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-45 | latest | en | 0.935069 |
http://www.kylesconverter.com/mass-flow/kilograms-per-week-to-kilograms-per-month | 1,534,228,554,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221208676.20/warc/CC-MAIN-20180814062251-20180814082251-00251.warc.gz | 680,682,370 | 5,505 | # Convert Kilograms Per Week to Kilograms Per Month
### Kyle's Converter > Mass Flow > Kilograms Per Week > Kilograms Per Week to Kilograms Per Month
Kilograms Per Week (kg/wk) Kilograms Per Month (kg/mo)* Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Kilograms Per Month to Kilograms Per Week
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Kilogram per Week:
Mass flow of kilograms across a threshold per unit time of a week. A week containing 604800 seconds. 1 Kilogram per week = 1/604800 kilograms per second (SI base unit). 1 kg/wk ? 0.000 001 653 439 153 439 kg/s.
1 Kilogram per Month:
Mass flow of kilograms across a threshold per unit time of a month. A 30 day month of 2592000 seconds. 1 Kilogram per month = 1/2592000 kilograms per second (SI base unit). 1 kg/mo ? 0.000 000 385 802 469 136 kg/s.
Link to Your Exact Conversion
Conversions Table
1 Kilograms Per Week to Kilograms Per Month = 4.285770 Kilograms Per Week to Kilograms Per Month = 300
2 Kilograms Per Week to Kilograms Per Month = 8.571480 Kilograms Per Week to Kilograms Per Month = 342.8571
3 Kilograms Per Week to Kilograms Per Month = 12.857190 Kilograms Per Week to Kilograms Per Month = 385.7143
4 Kilograms Per Week to Kilograms Per Month = 17.1429100 Kilograms Per Week to Kilograms Per Month = 428.5714
5 Kilograms Per Week to Kilograms Per Month = 21.4286200 Kilograms Per Week to Kilograms Per Month = 857.1429
6 Kilograms Per Week to Kilograms Per Month = 25.7143300 Kilograms Per Week to Kilograms Per Month = 1285.7143
7 Kilograms Per Week to Kilograms Per Month = 30400 Kilograms Per Week to Kilograms Per Month = 1714.2857
8 Kilograms Per Week to Kilograms Per Month = 34.2857500 Kilograms Per Week to Kilograms Per Month = 2142.8571
9 Kilograms Per Week to Kilograms Per Month = 38.5714600 Kilograms Per Week to Kilograms Per Month = 2571.4286
10 Kilograms Per Week to Kilograms Per Month = 42.8571800 Kilograms Per Week to Kilograms Per Month = 3428.5714
20 Kilograms Per Week to Kilograms Per Month = 85.7143900 Kilograms Per Week to Kilograms Per Month = 3857.1429
30 Kilograms Per Week to Kilograms Per Month = 128.57141,000 Kilograms Per Week to Kilograms Per Month = 4285.7143
40 Kilograms Per Week to Kilograms Per Month = 171.428610,000 Kilograms Per Week to Kilograms Per Month = 42857.1429
50 Kilograms Per Week to Kilograms Per Month = 214.2857100,000 Kilograms Per Week to Kilograms Per Month = 428571.4286
60 Kilograms Per Week to Kilograms Per Month = 257.14291,000,000 Kilograms Per Week to Kilograms Per Month = 4285714.2857 | 782 | 2,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | longest | en | 0.596598 |
https://stacks.math.columbia.edu/tag/0D7R | 1,563,382,456,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525355.54/warc/CC-MAIN-20190717161703-20190717183703-00358.warc.gz | 561,696,883 | 6,229 | Lemma 21.28.2. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D})$ be a morphism of ringed topoi. Consider the full subcategory $D' \subset D(\mathcal{O}_\mathcal {C})$ consisting of objects $K$ such that
$Lf^*Rf_*K \longrightarrow K$
is an isomorphism. Then $D'$ is a saturated triangulated strictly full subcategory of $D(\mathcal{O}_\mathcal {C})$ and the functor $Rf_* : D' \to D(\mathcal{O}_\mathcal {D})$ is fully faithful.
Proof. See Derived Categories, Definition 13.6.1 for the definition of saturated in this setting. See Derived Categories, Lemma 13.4.15 for a discussion of triangulated subcategories. The canonical map of the lemma is the counit of the adjoint pair of functors $(Lf^*, Rf_*)$, see Lemma 21.20.1. Having said this the proof that $D'$ is a saturated triangulated subcategory is omitted; it follows formally from the fact that $Lf^*$ and $Rf_*$ are exact functors of triangulated categories. The final part follows formally from fact that $Lf^*$ and $Rf_*$ are adjoint; compare with Categories, Lemma 4.24.3. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 402 | 1,351 | {"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": 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": 2, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-30 | latest | en | 0.79299 |
https://www.bleedingheartland.com/2020/02/03/iowa-caucus-math-still-complicated-less-important/ | 1,725,781,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00001.warc.gz | 660,054,001 | 30,060 | Iowa caucus math: Still complicated, less important
Part 3 in a series on how the Iowa caucuses work
When Iowa Democrats gather to express support for their favorite presidential candidates tonight, the mathematical formula used to determine who gets how many delegates from each precinct will remain the same as it has been for decades.
But other rule changes will limit opportunities for gamesmanship during the realignment process, and reporting three kinds of results from each precinct could downgrade the importance of the final delegate counts.
WHAT HASN’T CHANGED ABOUT IOWA CAUCUS MATH
To earn any delegates from a precinct, a presidential candidate must be “viable.” In most places, which assign four or more county convention delegates, the threshold for viability will be 15 percent of caucus attendees. (Click here for a full list of county convention delegates and state delegate equivalents per precinct.)
In precincts with three delegates, the viability threshold will be one-sixth of attendees. In those assigning two delegates, candidates will need 25 percent of caucus-goers present to be viable. There is no threshold in precincts allocated just one delegate. Rather, caucus attendees will break into preference groups just one time, after which the entire group will elect the delegate by a majority vote.
Many precincts will have more delegates to assign than there are viable groups. In that case, larger groups will get one or more additional delegates. Each precinct chair will have a booklet explaining the mathematical formulas used for these scenarios. These examples come from the party’s Delegate Selection Plan.
Often two groups of different sizes will receive the same number of delegates, because of rounding up or rounding down. My precinct split its delegates 2-2-2 among three viable candidates in 1988 and 2008, even though the largest group (for Paul Simon and John Edwards, respectively) was substantially bigger than the smallest group (for Bruce Babbitt and Hillary Clinton). Those smaller groups were just large enough to round up to two delegates.
The potential to change a delegate count by moving a few people from one candidate to another during realignment created lots of opportunities for experienced caucus-goers to game the system. I gave some examples in this post from four years ago.
LOCKING IN SUPPORT MEANS LESS MISCHIEF
After the first division into preference groups, caucus-goers will count how many people support each candidate and report back to the chair, who will announce the results. Those supporting candidates who did not hit the viability threshold will have a chance to realign. They can either join one of the viable groups or join others backing non-viable candidates to hit 15 percent if possible, either for one of the contenders or for “uncommitted.” (As history buffs may recall, “uncommitted” gained the most delegates in the 1976 caucuses; Jimmy Carter “won” by finishing second.)
Under a new rule the Iowa Democratic Party adopted for this cycle, “a viable candidate cannot lose support at any point in the caucus process. A candidate’s support can only change if they receive additional supporters in the realignment.”
That means, for example, supporters of Joe Biden and Pete Buttigieg can’t send a few people over to the Amy Klobuchar corner to make her viable and stop Bernie Sanders from getting an extra delegate. Or supporters of Sanders and Elizabeth Warren can’t help Andrew Yang become viable in order to keep Biden from getting an extra delegate.
This change should make realignment go a lot more quickly, since most caucus-goers will likely go for a viable candidate in the first alignment. A relatively small number of people in each precinct will be in play. In my precinct in 2016, the small group of Martin O’Malley supporters spent some time trying to persuade those backing Sanders or Hillary Clinton to help them become viable. That kind of scenario won’t be possible tonight. People in non-viable groups will either make a second choice or go home.
ONE UNUSUAL SCENARIO COULD BE MORE COMMON THIS YEAR
In rare cases, a precinct caucus may have more viable groups than there are county delegates to assign. As mentioned above, the viability threshold is 15 percent in most precincts. Given the large field of well-funded candidates, four of whom are polling at least in the mid-teens, there could be quite a few precincts where five candidates are viable. It’s theoretically possible for six candidates to reach that 15 percent mark. Yet not all precincts elect that many delegates to the county convention.
The problem will crop up more often in smaller precincts. It’s easy to envision more than three candidates attracting one-sixth of the caucus-goers in a precinct with three delegates, or three candidates attracting at least 25 percent support in a precinct with only two delegates to assign.
In that case, there will be a second realignment. Those backing the viable candidate with the smallest group in the room will get to make a third choice: join one of the larger groups or go home.
UPDATE: A version of this scenario played out in Josh Hughes’ Warren County precinct, which had two delegates. Warren got one, and Biden and Klobuchar were tied with eleven caucus-goers, exactly the viability threshold. A coin toss gave the second delegate to Biden.
HOW IMPORTANT ARE THE DELEGATE COUNTS?
The Iowa Democratic Party is encouraging news organizations to focus on delegate counts rather than the raw supporter numbers for each candidate, which will be publicly available for the first time this year. The Associated Press has agreed to consider only delegate counts when declaring the Iowa caucus winners.
But knowing how many Iowans showed up for each candidate is more valuable information, to my mind. The old way of reporting made support for non-viable candidates invisible. I’m interested to know how many people preferred each option, even if not all were able to elect a county delegate.
Moreover, it will take many more caucus-goers to elect a state delegate equivalent in urban or college town precincts and counties than in most smaller counties. So looking only at delegates may not provide a representative picture of Iowans’ preferences.
I’m excited that we will be able to see supporter numbers after realignment as well as before. That will provide clues on which candidates are better positioned to unite Democrats going forward.
In my mind, everyone’s voice should carry equal weight, no matter where they caucus. So if there is a mismatch between tonight’s “popular vote” and the delegate counts, I will consider the winner to be the candidate who inspired the largest number of Iowans to stand in their corner.
Top image: Caucus-goers in Des Moines precinct 61 raise their hands to be counted on February 1, 2016. Photo by Phil Roeder, available on Flickr. | 1,372 | 6,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.927452 |
https://www.jiskha.com/questions/1093022/which-property-is-shown-by-5x-6-7-5x6-5-x7-a-commutative-property-of-addition-b | 1,582,551,660,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145960.92/warc/CC-MAIN-20200224132646-20200224162646-00131.warc.gz | 740,409,680 | 5,022 | # math
Which property is shown by -5x(6+7)=-5x6+(-5)x7?
C. Distributive property
1. 👍 0
2. 👎 0
3. 👁 269
1. Nope. C
time to review
1. 👍 0
2. 👎 0
posted by Steve
## Similar Questions
1. which property is shown by 6 + 9 = 9 + 6 A) Commutative property of addition B) Distributiove property C) Identity property of addition D) Associative property of addition 2. Which symbol will make /-8/ ? 8 true? A) > B) < C) =
asked by lindacampoalegre2 on September 25, 2017
2. ### Math-Ms sue
Need Help with my final 2 questions can someone please help What is the additive inverse of 38? A –38 B. 38 C -(- -39) D. -(-(-(-38))) Which property is shown by –6+0= -6 Commutative property of addition identity property of
asked by Frank on September 5, 2018
I'm really bad at these, can someone help me PLEASE? 1. Name the property shown by the statement: 9x + 0 = 9x a. Commutative Property of Addition b. Commutative Property of Multiplication c. Identity Property of Addition d.
asked by Cheryl on March 1, 2011
4. ### math
Which property is shown by 2+0=2? A. Associative property of addition B. Commutative property of addition C. Distributive property D. Identity property of addition Is the answer B?
asked by gwen on September 8, 2014
5. ### Math
Miranda simplified an expression using the steps below: 1. 43 + (57 + 69) 2. (43+57) +69 3. 100 + 69 4. 169 Which property makes step 1 and step 2 equivalent expressions? A. Associative Property of Addition B. Commutative Property
asked by FNAF Girl on September 5, 2018
6. ### Math
1. 16y + 0 = 16y Associate Property of Addition Zero Property of Multiplication Commutative Property of Addition Identity Property of Addition 2. d • r = r • d Commutative Property of Multiplication Identity Property of
asked by EmberShy on January 10, 2017
7. ### Math
which property is shown by 0+9=9 A.) Distributive Property B.) Commutative Property of Addition C.) Identity Property of Addition D.) Associative Property of Addition I think its A but I'm not really sure
asked by Audrey on August 24, 2018
8. ### Math
Which property is shown by 2 + 5 = 5 + 2? A. Associative Property of Addition B. Commutative Property of Addition C. Distributive Property **** D. Identity Property of Addition
asked by SoAestheticallyPleasing on September 10, 2019
9. ### math
which property is shown by -6+0=-6 -commutative property of addition*** -identity property of addition -distributive property -associative property of addition can someone please tell me if im right
asked by Marta on September 19, 2019
10. ### Algebra
55. Find the range of y = 2x + 1. a. all real numbers b. all positive numbers Not sure.. 67. Name the Property: -2 + (10+1) = (-2 + 10) + 1 a. Commutative Property of Multiplication b. Inverse property of addition c. Commutative
asked by mysterychicken on May 16, 2010
More Similar Questions | 856 | 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} | 3.96875 | 4 | CC-MAIN-2020-10 | longest | en | 0.874032 |
http://www.slideshare.net/michaparamitha/financial-management-30799365 | 1,429,466,112,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246639325.91/warc/CC-MAIN-20150417045719-00016-ip-10-235-10-82.ec2.internal.warc.gz | 785,205,739 | 36,035 | Upcoming SlideShare
×
Thanks for flagging this SlideShare!
Oops! An error has occurred.
×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Standard text messaging rates apply
# Financial Management
279
views
Published on
This slide is for Financial Management assignment …
This slide is for Financial Management assignment
I don't really guarantee the answer
0 Likes
Statistics
Notes
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
• Be the first to like this
Views
Total Views
279
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
12
0
Likes
0
Embeds 0
No embeds
No notes for slide
• Project B hanya 6x pembayaran76.06
• ### Transcript
• 1. Quiz 3 Midwest manufacturing Company is considering two mutually exclusive investments. The projects’ expected cash flow as follows. a) Construct NPV (net present value) if the cost of capital is 10 percent. b) Construct NPV (net present value) if the cost of capital is 17 percent. c) If you were told that each project’s cost is 10 percent, which project should be selected? If the cost of capital were 17 percent, what would proper choice be?
• 2. Answer a) Construct NPV (net present value) if the cost of capital is 10 percent.
• 3. Answer b) Construct NPV (net present value) if the cost of capital is 17 percent.
• 4. Answer c) If you were told that each project’s cost is 10 percent, which project should be selected? If the cost of capital were 17 percent, what would proper choice be? 10% Project A : \$7,918.75 Project B : - \$ 405 We would select Project A because the amount is more than 0 17% Project A : \$2,018.75 Project B : - \$ 405 We would select Project A because the amount is more than 0
• 5. Chapter 11: Capital Budget Risk-Adjusted Discount Rates A method for incorporating the project’s level of risk into the capital-budgeting process, in which the discount rate is adjusted upward to compensate for higher than normal risk or downward to adjust for lower than normal risk. FCF IO K* N = = = = the annual expected free cash flow in time period t. the initial cash outlay. the risk-adjusted discount rate. the project’s expected life.
• 6. Example 1 A toy manufacture is considering the introduction of a line of fishing equipment with an expected life of five years. In the past, this firm has been quite conservative in its investment in new products, sticking primarily to standard toys. In this context, the introduction of a line of fishing equipment is considered an abnormally risky project. Management thinks that the normal required rate of return for the firm of 10 percent is not sufficient. Instead, the minimally acceptable rate of return on this project should be 15 percent. The initial outlay would be \$110,000, and the expected free cash flows from this project are as given below:
• 7. Example 1 The project would have been accepted Project B with a net present value \$3,700
• 8. Example 2 Bennett Company wishes to apply the Risk-Adjusted Discount Rate (RADR) approach to determine whether to implement Project A or B. rate of return on Project A : 14% Project B : 11%
• 9. Example 2 The project would have been accepted Project B with a net present value \$9,802
• 10. Certainty Equivalent vs. Risk-Adjusted Discount Rate Methods Certainty Equivalent Risk-Adjusted Discount Rate Step 1 : Adjust the discount rate upward for risk, or down in the case of less than normal risk. Step 2 : Discount the expected free cash flows back to the present using the risk-adjusted discount rate. Step 3 : Apply the normal decision criteria except in the case of the internal rate of return, where the risk-adjusted discount rate replaces the required rate of return as the hurdle rate
• 11. Certainty Equivalent Example The risk-risk free rate of interest is 6 percent. What is the project’s net present value? | 929 | 3,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2015-18 | latest | en | 0.900643 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.