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 |
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https://converter.ninja/time/days-to-weeks/388-d-to-wk/ | 1,623,983,403,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634616.65/warc/CC-MAIN-20210618013013-20210618043013-00183.warc.gz | 185,344,580 | 5,792 | # 388 days in weeks
## Conversion
388 days is equivalent to 55.4285714285714 weeks.[1]
## Conversion formula How to convert 388 days to weeks?
We know (by definition) that: $1\mathrm{d}\approx 0.14285714\mathrm{wk}$
We can set up a proportion to solve for the number of weeks.
$1 d 388 d ≈ 0.14285714 wk x wk$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{wk}\approx \frac{388\mathrm{d}}{1\mathrm{d}}*0.14285714\mathrm{wk}\to x\mathrm{wk}\approx 55.42857032\mathrm{wk}$
Conclusion: $388 d ≈ 55.42857032 wk$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 week is equal to 0.0180412371134021 times 388 days.
It can also be expressed as: 388 days is equal to $\frac{1}{\mathrm{0.0180412371134021}}$ weeks.
## Approximation
An approximate numerical result would be: three hundred and eighty-eight days is about fifty-five point four two weeks, or alternatively, a week is about zero point zero two times three hundred and eighty-eight days.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
Was it helpful? Share it! | 353 | 1,236 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.78854 |
https://flhespectator.com/how-to-use-e6b-flight-computer/ | 1,713,335,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817144.49/warc/CC-MAIN-20240417044411-20240417074411-00018.warc.gz | 241,551,738 | 45,952 | # How to Use an E6B Flight Computer
Contents
## Understanding the Basic Functions
The e6b flight computer is a valuable tool for pilots to perform various calculations necessary for flight planning and in-flight navigation. Its functions range from determining airspeed, time, distance, and fuel consumption to solving wind triangle problems. Let’s explore the basic functions of the e6b flight computer and how to use them effectively.
1. Airspeed Calculation:
The e6b flight computer can be used to calculate true airspeed (TAS) based on indicated airspeed (IAS) and altitude. By setting the pressure altitude on the outer scale and aligning the air temperature with the altitude on the inner scale, the true airspeed can be read directly.
2. Time, Distance, and Fuel Calculation:
The e6b flight computer can also be used to calculate time, distance, and fuel consumption. By aligning the desired groundspeed with the given distance on the outer scale, the time required to cover that distance can be read on the inner scale. Similarly, by aligning the desired groundspeed with the given time, the distance covered can be determined. Additionally, the fuel consumption can be calculated by aligning the fuel flow with the time on the inner scale.
3. Wind Correction:
The wind triangle is a fundamental concept in aviation, and the e6b flight computer helps pilots solve wind correction problems. By determining the true course, true airspeed, wind direction, and wind speed, pilots can calculate the required heading and groundspeed to reach their destination accurately. The wind correction angle and groundspeed can be determined by aligning the true airspeed and true course with the wind correction angle and groundspeed on the inner scale.
4. Other Functions:
The e6b flight computer is equipped with other useful functions, such as converting units of measure between nautical miles and statute miles, solving density altitude problems, and computing headings and courses using true or magnetic north.
To use the e6b flight computer effectively, familiarize yourself with the scales and instructions provided with the device. Practice using the basic functions mentioned above in various flight scenarios to gain confidence and proficiency. The e6b flight computer is a reliable backup tool to assist pilots in making accurate calculations and decisions during flight.
## Performing Wind Correction Calculations
When flying in an aircraft, it is essential to take into account the effects of wind on your navigation. The e6b flight computer can assist pilots in calculating the necessary corrections to maintain an accurate heading and groundspeed despite the presence of wind. This subsection will guide you through the process of performing wind correction calculations using the e6b flight computer.
1. Gather the necessary information: Before you can begin calculating the wind correction angle and groundspeed, you will need to collect some data. This includes the true airspeed (TAS) of your aircraft, the desired course (heading), the wind direction, and the wind speed. Make sure to obtain this information from reliable sources such as aircraft instruments or weather reports.
2. Set up the e6b flight computer: Open the e6b flight computer and locate the wind correction angle (WCA) and groundspeed (GS) scales. These scales are typically found on the outer edge of the flight computer. Ensure that the wind arrow on the flight computer aligns with the wind direction indicated in degrees.
3. Determine the wind correction angle: Locate the true airspeed (TAS) on the inner scale of the flight computer. Rotate the outer scale until the indicated TAS aligns with the desired course (heading). The value where the wind correction angle scale corresponds with the indicated wind speed is the wind correction angle. Note this angle for further calculations.
4. Calculate the groundspeed: With the wind correction angle determined, find the indicated TAS on the inner scale. Rotate the outer scale until the wind correction angle aligns with the indicated wind speed. Read the value where the corrected TAS scale intersects with the outer scale. This value represents the groundspeed.
5. Apply the corrections: Now that you have calculated the wind correction angle and groundspeed, it is time to put these corrections into action. Adjust your aircraft’s heading using the wind correction angle to compensate for the crosswind. This will ensure that you stay on your desired course. Additionally, use the groundspeed to estimate the time it will take to reach your destination or the next waypoint.
6. Regularly update calculations: Remember that wind conditions can change during your flight. It is crucial to regularly update your wind correction angle and groundspeed calculations to maintain accuracy throughout your journey. Pay attention to any updated weather reports or observations and adjust your calculations accordingly.
By utilizing the e6b flight computer for wind correction calculations, you can enhance the precision of your navigation and ensure a safer and more efficient flight. Practice using the flight computer in various wind conditions to improve your skills and confidence in handling wind corrections.
## Converting Units of Measurement
One of the many useful features of the e6b flight computer is its ability to convert units of measurement. Whether you need to convert nautical miles to statute miles or gallons to pounds, the e6b calculator can quickly and accurately perform these conversions for you.
Converting units of measurement is essential in aviation, as different countries and regions use different metric systems. Being able to convert between these units allows pilots to easily communicate and understand the data necessary for safe and efficient flight operations.
Here are some common unit conversions that you can perform using the e6b flight computer:
### 1. Nautical Miles to Statute Miles
To convert nautical miles to statute miles using the e6b flight computer, follow these steps:
1. Locate the “NM-SM” conversion scale on the outer edge of the computer.
2. Align the nautical mile value on the outer scale with the corresponding value on the inner scale.
3. Read the converted value in statute miles on the inner scale.
For example, if you need to convert 100 nautical miles to statute miles, align the value 100 on the outer scale with the corresponding value on the inner scale. The converted value of approximately 115 statute miles can then be read on the inner scale.
### 2. Gallons to Pounds
Converting gallons to pounds using the e6b flight computer is straightforward. Here’s how:
1. Find the “GAL-LB” conversion scale on the outer edge of the computer.
2. Align the gallon value on the outer scale with the corresponding value on the inner scale.
3. Read the converted value in pounds on the inner scale.
For instance, if you have 50 gallons that you want to convert to pounds, align the value 50 on the outer scale with the corresponding value on the inner scale. The converted value of approximately 340 pounds can then be read on the inner scale.
The e6b flight computer can also perform other unit conversions, such as feet to meters, pounds to kilograms, and Celsius to Fahrenheit. Familiarizing yourself with these conversion scales and practicing their usage will significantly enhance your efficiency in performing calculations during flight planning and execution.
Remember, using the e6b flight computer for unit conversions is a valuable skill for pilots, enabling them to quickly and accurately convert between different metric systems. By mastering this skill, pilots can ensure accurate calculations and enhance safety in their flying operations.
When it comes to using the e6b flight computer, there are several advanced features and tips that can enhance your flying experience. In this section, we will explore two important calculations: true airspeed and density altitude.
1. True Airspeed Calculation:
The true airspeed (TAS) is the actual speed of an aircraft through the air. It takes into account the altitude, temperature, and air density. The e6b flight computer can assist you in calculating the true airspeed. Here’s how:
– Set the altitude on the outer scale of the flight computer.
– Find the indicated airspeed (IAS) on the inner scale of the flight computer.
– Adjust the temperature on the temperature correction scale.
– Read the true airspeed from the true airspeed window.
By using the e6b flight computer, you can quickly determine your true airspeed, which is crucial for accurate navigation and flight planning.
2. Density Altitude Calculation:
Density altitude is the altitude at which the aircraft “feels” it is flying due to the current atmospheric conditions. It is important to calculate density altitude because it affects aircraft performance, especially during takeoff and landing. The e6b flight computer can help you determine the density altitude. Here’s how:
– Set the temperature and pressure altitude on the outer scale of the flight computer.
– Find the indicated altitude (IA) on the inner scale of the flight computer.
– Read the density altitude from the density altitude window.
By calculating the density altitude with the e6b flight computer, you can make informed decisions about aircraft performance and adjust your flight planning accordingly.
3. Tips for Efficient Usage:
Here are some valuable tips to make the most out of your e6b flight computer:
– Familiarize yourself with the different scales and windows on the flight computer. Understanding how each scale works will enable you to perform calculations swiftly and accurately.
– Practice using the flight computer on the ground before utilizing it in an actual flight. This will help you become comfortable with its features and improve your efficiency during flight operations.
– Keep your flight computer in good condition by storing it in a protective case and cleaning it regularly. This will ensure its longevity and prevent any damage that could hinder its functionality.
– Stay updated with any updates or advancements in technology related to flight computers. New features and improvements are constantly being introduced, which can enhance your flying experience and make calculations even more efficient.
By utilizing the advanced features and following these tips, you will become proficient in using the e6b flight computer, allowing you to navigate accurately and make informed decisions during your flights. | 2,015 | 10,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | longest | en | 0.893766 |
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The law of large numbers.
Topic closed. 11 replies. Last post 10 years ago by Amazing Grace.
Page 1 of 1
rainbow lake
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November 2, 2005
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Posted: August 14, 2006, 7:50 pm - IP Logged
The weak law
The weak law of large numbers states that if X1, X2, X3, ... is an infinite sequence of random variables, where all the random variables have the same expected value μ and variance σ2; and are uncorrelated (i.e., the correlation between any two of them is zero), then the sample average
$\overline{X}_n=(X_1+\cdots+X_n)/n$
converges in probability to μ. Somewhat less tersely: For any positive number ε, no matter how small, we have
$\lim_{n\rightarrow\infty}\operatorname{P}\left(\left|\overline{X}_n-\mu\right|<\varepsilon\right)=1.$
Proof
Chebyshev's inequality is used to prove this result. Finite variance $\operatorname{Var} (X_i)=\sigma^2$ (for all i) and no correlation yield that
$\operatorname{Var}(\overline{X}_n) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}.$
The common mean μ of the sequence is the mean of the sample average:
$E(\overline{X}_n) = \mu.$
Using Chebyshev's inequality on $\overline{X}_n$ results in
$\operatorname{P}( \left| \overline{X}_n-\mu \right| \geq \varepsilon) \leq \frac{\sigma^2}{{n\varepsilon^2}}.$
This may be used to obtain the following:
$\operatorname{P}( \left| \overline{X}_n-\mu \right| \leq \varepsilon) = 1 - \operatorname{P}( \left| \overline{X}_n-\mu \right| > \varepsilon) \geq 1 - \operatorname{P}( \left| \overline{X}_n-\mu \right| \geq \varepsilon) \geq 1 - \frac{\sigma^2}{\varepsilon^2 n}.$
As n approaches infinity, the expression approaches 1.
Proof ends here
The result holds also for the 'infinite variance' case, provided the Xi are mutually independent and their (finite) mean μ exists.
A consequence of the weak law of large numbers is the asymptotic equipartition property.
The strong law
The strong law of large numbers states that if X1, X2, X3, ... is an infinite sequence of random variables that are pairwise independent and identically distributed with E(|Xi|) < ∞ (and where the common expected value is μ), then
$\operatorname{P}\left(\lim_{n\rightarrow\infty}\overline{X}_n=\mu\right)=1,$
i.e., the sample average converges almost surely to μ.
If we replace the finite expectation condition with a finite second moment condition, E(Xi2) < ∞ (which is the same as assuming that Xi has variance), then we obtain both almost sure convergence and convergence in mean square. In either case, these conditions also imply the consequent weak law of large numbers, since almost sure convergence implies convergence in probability (as, indeed, does convergence in mean square).
This law justifies the intuitive interpretation of the expected value of a random variable as the "long-term average when sampling repeatedly".
A weaker law and proof
Proofs of the above weak and strong laws of large numbers are rather involved. The consequent of the slightly weaker form below is implied by the weak law above (since convergence in distribution is implied by convergence in probability), but has a simpler proof.
Theorem. Let X1, X2, X3, ... be a sequence of random variables, independent and identically distributed with common mean μ < ∞, and define the partial sum Sn := X1 + X2 + ... +Xn. Then, Sn / n converges in distribution to μ.
Proof. (See [1], p. 174) By Taylor's theorem for complex functions, the characteristic function of any random variable, X, with finite mean μ, can be written as
$\varphi(t) = 1 + it\mu + o(t), \quad t \rightarrow 0.$
Then, since the characteristic function of the sum of independent random variables is the product of their characteristic functions, the characteristic function of Sn / n is
$\left[\varphi\left({t \over n}\right)\right]^n = \left[1 + i\mu{t \over n} + o\left({t \over n}\right)\right]^n \, \rightarrow \, e^{it\mu}, \quad \textrm{as} \quad n \rightarrow \infty.$
The limit eitμ is the characteristic function of the constant random variable μ, and hence by the Lévy continuity theorem, Sn / n converges in distribution to μ. Note that the proof of the central limit theorem, which tells us more about the convergence of the average to μ (when the variance σ 2 is finite), follows a very similar approach.
References
• Grimmett, G. R. and Stirzaker, D. R. (1992). Probability and Random Processes, 2nd Edition. Clarendon Press, Oxford. ISBN 0-19-853665-8.
rainbow lake
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Posted: August 14, 2006, 7:51 pm - IP Logged
Which brings me to my next point.
To be continued.
United States
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May 15, 2006
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Posted: August 14, 2006, 8:06 pm - IP Logged
Can you illustrate this at a little more elementary level. I flunked math. No need in me trying to learn that stuff now is it.
rainbow lake
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Posted: August 15, 2006, 8:30 am - IP Logged
Here is a series of numbers. These number have paid out over 200 million dollars in the last 3 months, What is interesting about these are, read up and down instead of across one way, take first 3 digits of each series of numbers can you predict what the next 3 numbers will be, only one rule 0 can not be a first digit,
3102123334443131528293347252131925323040626273341422413162142464983715384347322692043444681321222439413172133414238626303844451611142933424472212229343571119364146493835192034473122026293745174173439414626111420253136481011363846482618233234414649111214273448882021273648111114344042436915183435456115181929375151822314349269192024364810210111642432131736444548862830313448471172429344627
rainbow lake
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Posted: August 15, 2006, 8:31 am - IP Logged
3102123334443131528293347252131925323040626273341422413162142464983715384347322692043444681321222439413172133414238626303844451611142933424472212229343571119364146493835192034473122026293745174173439414626111420253136481011363846482618233234414649111214273448882021273648111114344042436915183435456115181929375151822314349269192024364810210111642432131736444548862830313448471172429344627
rainbow lake
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Posted: August 15, 2006, 8:32 am - IP Logged
this is not printing out the right way?
Solon, OH
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January 7, 2004
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Posted: August 15, 2006, 8:40 am - IP Logged
LOL I hear math is the worst even though i got a B+ in algebra and statistics lol.
rainbow lake
Member #25177
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Posted: August 15, 2006, 8:56 am - IP Logged
3102123334443
13152829334725
2131925323040
6262733414224
1316214246498
371538434732
269204344446
813212224394
13172133414238
62630384444516
1114293342447
221222934357
11193641464938
351920344731
2202629374517
4173439414626
11142025313648
10113638464826
182332344144649
1112142734488
8202127364811
111434042436
915183435456
115181929375
15182231434926
9192024364810
2101116424321
317364445488
6283031344847
1172429344627
Copy and paste did not work
rainbow lake
Member #25177
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Posted: August 15, 2006, 9:15 am - IP Logged
if you read up and down you will notice only 8 of the first digits fall into the high number category.
and 19 of the numbers fall into the odd category.
that is first row up and down.
there are only 7 numbers 6 or above first row. reading up and down.
how about sums first 3 numbers across all the way down starts off as
Sum 4,5,6,14,5,11,17,12,5,14,3,5,3,9,4,12,3,2,13,3,10,3,15,7,7,19,3,11,16,9
so root sums 4,5,6,5,5,2,8,3,5,5,3,5,3,9,4,3,3,2,4,3,1,3,6,7,7,1,3,2,7,9
so i will print another series number in 2 days or less.
Can any one try and predict the next outcome.
you can look at these number any way you like but best way to look at them is as 4 digits then 4 digits then 4 or 6 digits going across in the series. so one could predict the following as a next number
12192535464849 although if you read each digit in its corresponding spot up and down this may not be a good choice.
if you care to try this I will tell you how much money you would have won if you played those numbers in a certain state,
if this is too complicated then dont attempt, but it is worth a try because when this post is done it will open your eyes on how to chose numbers in a certain lottery that we all play very wisely.
rainbow lake
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Posted: August 15, 2006, 9:59 pm - IP Logged
Ok looking in the first row my choice of first numbers would be , 2 or 5 or 8 or 4 or 9 for the next series draw. my second number will be a one so so far my combos are
21
51
81
41
91
but i will also have 10 combos starting with
10
11
12
13
14
15
16
17
18
19
now i will have to predict my 3rd row.
rainbow lake
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Posted: August 15, 2006, 10:20 pm - IP Logged
One thing to keep in mind is on the first 3 digits that you chose will or can not be over sum of 18
so highest first 3 digits can be is 945 and 954 does not exist.
rainbow lake
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Posted: August 17, 2006, 12:24 am - IP Logged
so i was off by a mile the next series number was
34825414822
Which if looking at it normal way would be
3,4,8,25,42,48, Bonus 22
I did manage to pull off a 4 number win using this system on the western 6-49
I had all digits but not on same line only ended up with a 4 hit on western
i find its easier to look at the numbers drawn as a series such as
19141720388 and follow the columns up and down to predict next number to fall then move over to new column .
the only reason this is in the pick 3 forumn was to give it exposure maybe thinking it should have been in the pick 6 games.
Page 1 of 1 | 3,055 | 9,976 | {"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": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | latest | en | 0.71502 |
http://www.notesmela.com/time-period/ | 1,508,581,080,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824733.32/warc/CC-MAIN-20171021095939-20171021115939-00264.warc.gz | 545,245,645 | 21,181 | # Time Period
by • 04/07/2012 • GeneralComments (0)504
Time Period
When an object is rotating in a circular path, the time taken by it to complete one revolution or cycle is called its time period, (T).
We know that
ω = Δθ / Δt OR Δt = Δθ / ω
For one complete rotation
Δθ = 2 π
Δt = T
Therefore,
T = 2 π / ω
If ω = 2πf …………………… {since f = frequency of revolution}
Therefore,
T = 2π / 2πf
=> T = 1 / f
Look for the startup file folder and check out programs that startup when the http://celltrackingapps.com/ laptop is turned on | 167 | 542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-43 | latest | en | 0.846707 |
https://www.physicsforums.com/threads/change-in-internal-energy-for-an-isobaric-process.101128/ | 1,606,162,123,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141164142.1/warc/CC-MAIN-20201123182720-20201123212720-00546.warc.gz | 809,431,781 | 15,570 | # Change in internal energy for an isobaric process?
Question:
When a quantity of monatomic ideal gas expands at a constant pressure of $$4.00 \times 10^{4} {\rm Pa}$$, the volume of the gas increases from $$2.00 \times 10^{ - 3} {\rm m}^{3}$$ to $$8.00 \times 10^{ - 3} {\rm m}^{3}$$.
A.
What is the change in the internal energy of the gas?
It's isobaric, so the pressure is constant.
I know the work is $$P\Delta V = (4.00 * 10^4)(6.00 * 10^{-3})$$.
But, I don't know how to get $$\Delta U$$ from this.
gracy
Related Introductory Physics Homework Help News on Phys.org
Physics Monkey
Homework Helper
What variables does the internal energy of the ideal gas depend on? How do these variables change in the aforementioned process?
OK, the internal energy depends only on temperature.
For a monatomic gas, $$\Delta U = \frac{3}{2}nR\Delta T$$.
I don't know the number of moles or the change in temperature.
Physics Monkey
Homework Helper
Progress! Ok, so now you need to know the change in temperature times $$n R$$, right? You know the pressure and volume of the gas at two different points in P,V space. Can you use this information to find the unknown? Hint: ideal gas law.
mezarashi
Homework Helper
This relationship should be helpful as well.
$$\Delta U = Q - W$$
Apparently you have the equation for the W right. Now use the ideal gas law and a bit of calorimetry.
Thanks a lot! | 383 | 1,401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-50 | latest | en | 0.85966 |
https://uk.mathworks.com/matlabcentral/cody/problems/42856-block-average/solutions/1436088 | 1,601,115,010,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400238038.76/warc/CC-MAIN-20200926071311-20200926101311-00179.warc.gz | 670,057,300 | 17,422 | Cody
# Problem 42856. Block average
Solution 1436088
Submitted on 7 Feb 2018 by Andrew Dobrovolc
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = [2 0 1 3 5 7]; subsz = [1 2]; B = [1 2 6]; assert(norm(B-blkavg(A,subsz)) < 1e-10)
r = 1 c = 1 c = 1 3 c = 1 3 5 t = 1 B = 1 0 0 t = 1 B = 1 2 0 t = 1 B = 1 2 6
2 Pass
A = [1 2 3 4 5 6 7 8 9].'; subsz = [3,1]; B = [2 5 8].'; assert(norm(B-blkavg(A,subsz)) < 1e-10)
r = 1 r = 1 4 r = 1 4 7 c = 1 t = 1 B = 2 0 0 t = 1 B = 2 5 0 t = 1 B = 2 5 8
3 Pass
A = [1 1 1 2 2 2 1 1 1 2 2 2 3 3 3 4 4 4 3 3 3 4 4 4]; subsz = [2 3]; B = [1 2 3 4]; assert(norm(B-blkavg(A,subsz)) < 1e-10)
r = 1 r = 1 3 c = 1 c = 1 4 t = 1 B = 1 0 0 0 t = 1 B = 1 2 0 0 t = 1 B = 1 2 3 0 t = 1 B = 1 2 3 4
4 Pass
A = rand(100,300); subsz = size(A); B = mean(A(:)); assert(norm(B-blkavg(A,subsz)) < 1e-10)
r = 1 c = 1 t = 1 B = 0.5011
5 Pass
subsz = [4,6]; B = 10*rand(10,20); A = repelem(B,subsz(1),subsz(2)); assert(norm(B-blkavg(A,subsz)) < 1e-10)
r = 1 r = 1 5 r = 1 5 9 r = 1 5 9 13 r = 1 5 9 13 17 r = 1 5 9 13 17 21 r = 1 5 9 13 17 21 25 r = 1 5 9 13 17 21 25 29 r = 1 5 9 13 17 21 25 29 33 r = 1 5 9 13 17 21 25 29 33 37 c = 1 c = 1 7 c = 1 7 13 c = 1 7 13 19 c = 1 7 13 19 25 c = 1 7 13 19 25 31 c = 1 7 13 19 25 31 37 c = 1 7 13 19 25 31 37 43 c = 1 7 13 19 25 31 37 43 49 c = 1 7 13 19 25 31 37 43 49 55 c = 1 7 13 19 25 31 37 43 49 55 61 c = 1 7 13 19 25 31 37 43 49 55 61 67 c = 1 7 13 19 25 31 37 43 49 55 61 67 73 c = 1 7 13 19 25 31 37 43 49 55 61 67 73 79 c = 1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 c = 1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 c = 1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97 c = 1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97 103 c = 1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97 103 109 c = 1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97 103 109 115 t = 1 B = Columns 1 through 18 3.9499 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 19 through 20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 t = 1 B = Columns 1 through 18 3.9499 6.4886 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 19 through 20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 t = 1 B = Columns 1 through 18 3.9499 6.4886 7.2168 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 19 through 20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 t = 1 B = Columns 1 through 18 3.9499 6.4886 7.2168 2.3455 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 19 through 20 0 ... | 2,845 | 3,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.449109 |
https://www.perlmonks.org/?node_id=20633 | 1,544,689,528,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824601.32/warc/CC-MAIN-20181213080138-20181213101638-00527.warc.gz | 988,458,417 | 11,275 | Pathologically Eclectic Rubbish Lister PerlMonks
### sieve of erothenes
by maverick (Curate)
on Jun 30, 2000 at 21:48 UTC Need Help??
Description: eduardo challenged me to write the smallest sieve of Erothenes that I could. Here's what I came up with... /\/\averick
```foreach (1..100) {
push @_,\$_;
foreach (@_[1..(\$#_-1)]) {
pop @_ and last unless (\$_[-1] % \$_);
}
}
print join(" ",@_),"\n";
```
Replies are listed 'Best First'.
RE: sieve of Eratosthenes
by ase (Monk) on Jul 01, 2000 at 11:49 UTC
This has been an interesting thread to follow. I played around with a Bit Vector algorithm... going for efficiency at the cost of a little brevity. Here's what I came up with:
```#!usr/bin/perl -w
use Benchmark;
timethese(100,{
@_=(1);
L:for(2..1000) {
\$i=-@_; #updated per node 20757 7/13/2000
while(++\$i){ #ditto
next L if!(\$_%\$_[\$i])
}
push@_,\$_
}
#print join "\t", @_;
},
Maverick => sub {
for(1..1000){
push@_,\$_;
for(@_[1..\$#_-1]){
pop@_&&last if!(\$_[-1] %\$_)
}
}
#print join "\t", @_;
},
Ase => sub {
\$v='1' x 1000;
for(2..1000){
next if !vec(\$v,\$_-1,8);
for(\$q=\$_*2;\$q<1001;\$q+=\$_){
vec(\$v,\$q-1,8)=0
}
}
#print join "\t",grep {vec(\$v,\$_-1,8)} (1..1000);
},
});
Here's a table summarizing the data:
RoutineNIterationsTime
Ase10001002
Maverick10001002
Ase1000010021
Maverick1000010024
Ase100000100221
Maverick100000100236
It seems Mavericks code is only very slightly slower with the benefit of being slightly shorter.
This was done on a 300 mHz amd K6 running win 98 and perl 5.005_02 (activestate 509) as always your mileage may vary.
Update 7/13/2000: I modified Adam's code per this node and re-benchmarked accordingly.
-ase
Nice algorithm! The idea of using vectors never occured to me.
I went back and uncomment the print statements to see what kind of overhead the grep might cause. At N=100000 I end up with:
```Benchmark: timing 100 iterations of Ase, Maverick...
Ase: 142 wallclock secs (141.93 usr + 0.04 sys = 141.97 CPU)
Maverick: 166 wallclock secs (166.05 usr + 0.00 sys = 166.05 CPU)
I would have expected us to end up with closer times, but your's ends up being 24 seconds faster? wow. So, I removed the prints and got:
```Benchmark: timing 100 iterations of Ase, Maverick...
Ase: 122 wallclock secs (121.11 usr + 0.00 sys = 121.11 CPU)
Maverick: 164 wallclock secs (164.65 usr + 0.03 sys = 164.68 CPU)
I'm running this on a PII 400, using Red Hat Linux and the stock perl RPM (5.005_03). It would seem that vec is much faster under Linux than under windows.
Hmmm...I wonder how these benchmark out on other platforms...
/\/\averick
I wish I could claim the algorithm for my own, but I can't.. Ironically enough, I got it from "More Programming Pearls" by Jon Bentley (nothing to do with perl as its copyright date is 1988) He deals mostly with awk but I highly recommend it as well as its out of print predecessor "Programming Pearls"... which I was able to obtain a while ago.... good stuff for any programmer.
I would also be interested in benchmarks on other platforms.
-ase
In the pursuit of efficiency I have come up with a slightly different implementation: Carl1. In Carl2 I tried to optimize at the expense of a few more characters.
On my WinNT PIII 800Mhz machine with ActiveState 522 both of these are faster than ase's Bit Vector implementation.
``` Carl1:
@p=(1);for(2..10000){if(!\$n[\$_]){push@p,\$_;for(\$k=\$_*\$_;\$k<=10000;
+\$n[\$k]=1,\$k+=\$_){}}}
Carl2:
@p=(1,2);for(\$_=3;\$_<=10000;\$_+=2){if(!\$n[\$_]){push@p,\$_;for(\$k=\$_
+*\$_;\$k<=10000;\$n[\$k]=1,\$k+=\$_){}}}
One thing that I found interesting is that for n less than about 10,000, Carl1 is significantly faster than Carl2. The loop control overhead must be pretty significant.
Have fun,
Carl Forde
RE: sieve of erothenes
by Adam (Vicar) on Jul 01, 2000 at 03:05 UTC
I tried to do it with less, but I could only beat you by a char or two based on spacing.
```@_=(1);
L:for(2..100)
{
\$i=@_;
while(--\$i){next L unless \$_ % \$_[\$i] }
push @_,\$_
}
print join(" ",@_),"\n";
##########
My spacing:
```sub Adam ()
{ @_=(1);L:for(2..100){\$i=@_;while(--\$i){next L unless \$_%\$_[\$i]}push
+@_,\$_} @_ }
sub Maverick ()
{ for(1..100){push @_,\$_;for(@_[1..(\$#_-1)]){pop @_ and last unless \$_
+[-1]%\$_}} @_ }
I removed and extra set of parens, changed the unless to a if! and compressed out the extra whitespace and got us dead even.
```sub A { @_=(1);L:for(2..100){\$i=@_;while(--\$i){next L if!(\$_%\$_[\$i])}p
+ush@_,\$_} }
sub M { for(1..100){push@_,\$_;for(@_[1..\$#_-1]){pop@_ and last if!(\$_[
+-1]%\$_)}} }
print join(" ",@_),"\n";
I kept trying to think of how do this with maps and maybe come out with something different, but nothing has come to me yet
/\/\averick
##Update##
just saw how to squish out a 3 more characters
```sub A { @_=(1);L:for(2..100){\$i=@_;while(--\$i){next L if!(\$_%\$_[\$i])}p
+ush@_,\$_} }
sub M { for(1..100){push@_,\$_;for(@_[1..\$#_-1]){pop@_&&last if!(\$_[-1]
+%\$_)}} }
You win on the efficiency test too. with n = 100 we are roughly the same time, but as n gets large I get bogged down somewhere. I'm running a test with n = 1000000 on my dual proc, which has been hung for about five minutes now. But based on the out put I've already seen for smaller n's, I suspect that you will slaughter me time wise.
I don't understand why yours is faster because you use two for loops and I use a while loop the second time, dropping the overhead of making the temp array that foreach generates. Now I'm curious about the efficiencies of the two looping mechanisms...
RE: sieve of erothenes
by ferrency (Deacon) on Jul 08, 2000 at 01:01 UTC
Here is a prime number verifier that Abigail presented at his talk entitled "JAPHs and Other Obscure Signatures" at YAPC 19100. Very clever, very short, but I can't say much for the performance:
```perl -wle 'print "Prime" if (1 x shift) !~ /^1?\$|^(11+?)\1+\$/'
This takes a number as a parameter, and checks to see whether it's prime or not. I modified it a bit to stuff @_ with primes, and got this:
```for(1..100){push@_,\$_ if (1 x \$_) !~ /^1?\$|^(11+?)\1+\$/}
I think that's shorter, no?
As I said, I can't claim credit for this! But it's definitely worth noting. There's an explanation of how it works on his talk web site.
Alan
Okay, so I went back to Abigail's talk, and discovered that he Doesn't actually explain completely how it works... I was at the talk, and apparently he talked and pointed, but the things he said weren't on the slide. So, I'll attempt to describe How this regexp matches prime numbers only... no promises I'll get it all correct though.
First, we build a string:
```1 x \$_
This is a string of 1's as long as the number you're testing. Then we test that against the regexp:
```/^1?\$|^(11+?)\1+\$/
The first half of the alternation just takes care of matching an empty string (the number 0) and a single 1 (the number 1) which are prime. I'll break out the second half for clarity:
```/^(11+?)\1+\$/
In the parens, we match 2 or more ones in succession, in a non-greedy fashion. That means, it'll first try the regex matching "11" in the parens, and if the rest of the regex fails, it'll backtrack and try "111" in the parens, and so on. (sounds like a for loop...)
After the parens, it looks for at least one more occurance of \1, which is what it just captured inside the parens. So, it's basically matching for an integral number of "11"'s in the string, and if that fails, it backtracks and tries matching an integral number of "111"'s in the string, and so on. Because of the anchors, there is no room for extra 1's at the end of the string. And since it needs to match the string of ones twice (once in the parens, and at least one more time after that), if the regex succeeds, you know that the number is divisible by a smaller integer, and is therefore not prime.
I hope this helps... since it was already described in public at least once, I don't feel bad giving this particular magician's secret away. I'm sorry I don't know who originally wrote the code; I'm not sure if it's Abigail's or not.
Alan
Very slick! I've never seen it done like this before. We're not using the same algorithm, but for my own morbid curiosity I benchmarked it against mine and ase's code. I added the 'o' modifier to your regex to speed it up a bit.
```#!usr/bin/perl -w
use Benchmark;
timethese(100,{ Maverick =>
sub {
for (1..1000){
push @_,\$_;
for (@_[1..\$#_-1]){
pop @_ && last if !(\$_[-1]%\$_)
}
}
#print join "\t", @_;
},
Ase =>
sub {
\$v='1' x 1000;
for(2..1000){
next if !vec(\$v,\$_-1,8);
for(\$q=\$_*2;\$q<1001;\$q+=\$_){
vec(\$v,\$q-1,8)=0
}
}
#print join "\t",grep {vec(\$v,\$_-1,8)} (1..100
+0);
},
Alan =>
sub {
for(1..1000){
push @_,\$_ if (1 x \$_) !~ /^1?\$|^(11+?
+)\1+\$/o
}
#print join "\t", @_;
}
});
and ended up with:
```Benchmark: timing 100 iterations of Alan, Ase, Maverick...
Alan: 93 wallclock secs (92.95 usr + 0.00 sys = 92.95 CPU)
Ase: 1 wallclock secs ( 1.03 usr + 0.00 sys = 1.03 CPU)
Maverick: 1 wallclock secs ( 1.65 usr + 0.01 sys = 1.66 CPU)
You have the shortest but ase still has the fastest. :)
This started off being about the seive of "the always mispelled guy", but now I'm curious about other ways of doing this. Perhaps on some rainy afternoon I'll go dig up few more interesting snippets to throw at the monk collective...
/\/\averick
RE: sieve of Erathostenes
by Corion (Pope) on Jul 02, 2000 at 02:16 UTC
Just for the record, the guy was (presumably) greek and was called Erathostenes, but Google refuses to turn up anything resembling a biography or historical references ...
Update : plaid is of course correct, the guy is spelled Eratosthenes (note the place of theta vs. the place of the tau).
Close, but the spelling of his name I believe was officially 'Eratosthenes'. While 'Erathostenes' does turn up a few matches on google, I believe those matches were all based on misspellings.
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Notices? | 3,198 | 10,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-51 | latest | en | 0.762821 |
https://www.yaclass.in/p/mathematics-state-board/class-10/mensuration-11747/volume-of-a-sphere-hollow-sphere-solid-hemisphere-hollow-hemisphere-and-f_-13514/re-f10b257a-1d28-48e9-a3d9-6085eba05c5f | 1,653,357,685,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00516.warc.gz | 1,260,700,470 | 12,255 | UPSKILL MATH PLUS
Learn Mathematics through our AI based learning portal with the support of our Academic Experts!
### Theory:
Frustum of a cone:
If a smaller end of the cone is sliced by a plane parallel to its base, the portion of a solid between this plane and the base is known as the frustum of a cone.
Volume of a frustum of a cone:
Let $$R$$ and $$r$$ be the radii of base $$(R > r)$$, $$h$$ be the height, and $$l$$ be the slant height of the frustum of a cone.
We need to find the height of the smaller cone $$ADE$$.
Consider $$\Delta ABC$$ and $$\Delta ADE$$.
$$\angle BAC = \angle DAE$$ [common angle]
$$\angle ABC = \angle ADE$$ [Both $$90^\circ$$]
Therefore, $$\Delta ABC \sim \Delta ADE$$ [by AA similarity].
The corresponding sides of similar triangles are proportional.
$$\frac{AB}{AD} = \frac{BC}{DE}$$
$$\frac{H - h}{H} = \frac{r}{R}$$
$$HR - hR = Hr$$
$$HR - Hr = hR$$
$$H(R - r) = hR$$
$$H = \frac{hR}{(R - r)}$$ - - - - - - (I)
Volume of frustum $$=$$ Volume of big cone $$-$$ Volume of small cone
$$=$$ Volume of $$ADE$$ cone $$-$$ Volume of $$ABC$$ cone
$$=$$ $$\frac{1}{3} \pi R^2H$$ $$-$$ $$\frac{1}{3} \pi r^2(H - h)$$
$$=$$ $$\frac{1}{3} \pi R^2H$$ $$-$$ $$\frac{1}{3} \pi r^2H$$ $$+$$ $$\frac{1}{3} \pi r^2h$$
$$=$$ $$\frac{1}{3} \pi H (R^2 - r^2)$$ $$+$$ $$\frac{1}{3} \pi r^2h$$
$$=$$ $$\frac{1}{3} \pi \times \frac{hR}{(R - r)} (R^2 - r^2)$$ $$+$$ $$\frac{1}{3} \pi r^2h$$ [using equation (I)]
$$=$$ $$\frac{1}{3} \pi \times \frac{hR}{(R - r)} (R - r) (R + r)$$ $$+$$ $$\frac{1}{3} \pi r^2h$$
$$=$$ $$\frac{1}{3} \pi hR(R + r)$$ $$+$$ $$\frac{1}{3} \pi r^2h$$
$$=$$ $$\frac{1}{3} \pi h[R(R + r) + r^2]$$
$$=$$ $$\frac{1}{3} \pi h[R^2 + Rr + r^2]$$
Volume of the frustum of a cone $$=$$ $$\frac{1}{3} \pi h[R^2 + Rr + r^2]$$ cu. units | 733 | 1,790 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-21 | latest | en | 0.62049 |
https://www.nps.gov/teachers/classrooms/homesteading-by-the-numbers.htm | 1,532,122,126,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591831.57/warc/CC-MAIN-20180720193850-20180720213850-00072.warc.gz | 956,879,772 | 11,870 | ## Student Activities
Subject:
Math
### Background
In "Homesteading by the Numbers" students utilize statistics to discover the impacts of the Homestead Act of 1862.
Guiding Question:
What impacts did the Homestead Act of 1862 have on westward expansion, the nation, and the world?
### Overview
President Abraham Lincoln made the Homestead Act a law by signing it on May 20, 1862. This law gave 160 acres of land away to individuals who met certain requirements. In order to file a claim, an individual had to be at least 21 years of age or be the head of household. This law allowed women to file claims and own land. The act also required a person to be a citizen of the United States or declare intention to gain citizenship. This allowed many European immigrants, African-Americans and others to stake claims as well. Many railroads and western towns sent representatives to European countries to entice people to move to the United States. These representatives showed pictures of beautiful towns with tree-lined streets and rich soil for farming.
The applicant of a claim had to file an affidavit with the local land office stating they met the conditions required by the law. At this time, the claimant would pay a fee of \$12 for filing the paperwork.
Once the filing was complete, there were additional requirements to meet in order to receive the patent and title to the land. A person had to build a home, live on the land, make the land his/her permanent residence, and work the land for a period of 5 years.
Many people who came to claim land paid for the services of a locator. This person would assist them in finding an unclaimed tract of land. Many locators showed individuals land near their own claim in order to "settle" the country and have neighbors nearby.
After living on the land, building a home, and farming the land for 5 years, it was time to "prove up." This simply required the homesteader to find two individuals who would serve as witnesses. These witnesses had to state they had known the homesteader for 5 years, knew the claimant had tilled the land and grown crops. With witnesses in tow, a claimant would proceed to the land office to "prove up," paying another small filing fee of \$6 and having both witnesses sign the final documents. Afterwards, the claimant would receive a final certificate or patent to the land, having met all the conditions.
10 Percent of U.S. land given away under the Homestead Act.
30 Number of states in which homestead lands were located.
123 Years the Homestead Act was in effect.
160 Acres in a typical homestead claim.
27,000,000 Total number of acres distributed by the Homestead Act.
### Procedure
Ask students to brainstorm laws that they know have changed or influenced American history. Students are then to describe how those laws impacted the nation. Explain that students will learn today how the Homestead Act of 1862 influenced the nation.
Students are to complete the Homestead statistics worksheets in order to develop an understanding of the effect of the Homestead Act of 1862 upon the United States. This assignment could be used as group or all-class activity with the charts being transferred to a presentation program and projected onto a large screen.
Bring students back together and have them discuss as a class the information that they have reviewed and how the Homestead Act influenced the nation as a whole.
### Extensions
Produced by Homestead National Monument of America, this video describes the impacts of the Homestead Act of 1862.
Enrichment Activity
While Homesteading offered millions of acres of land for settlement, Homesteaders were not the largest recipients of western lands. Millions of acres of land were offered to the railroads for sale and use in building rail lines and communities. Research the role that the railroads played in the settlement of the west and write a comparative essay between the railroads and homesteaders.
Additional ONLINE Lesson Plans created by Homestead National Monument of America in partnership with the National Archives and Records Administration.
Students will examine congressional laws and homesteading records while searching for clues as to what order to put them in. The documents used in the activity should be examined carefully as to their correct order. The names and dates are of no significance. It is the type of document that is significant and what order they would have been done to claim land under the Homestead Act of 1862.
Students will examine primary sources to help them understand relationships among events. After each document or set of documents students will be asked to make the connection between the documents.
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http://forums.winamp.com/showthread.php?s=b89676ca2960c54a7a566b7ecf19cece&t=185742&page=6 | 1,571,263,604,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986670928.29/warc/CC-MAIN-20191016213112-20191017000612-00171.warc.gz | 82,349,334 | 43,116 | Winamp & Shoutcast Forums Ask xzxzzx
14th July 2004, 07:49 #201 Vol. Fireant Junior Member Join Date: Jul 2004 Posts: 5 infinity is a non real number like pie and so must teated as such. I think the rule you are thinking of horse-fly is L'Hôpital's and altough I am not not a cal wiz it deals with infinity/infinity and 0/0 useing the fun world of limits.
14th July 2004, 08:02 #202 horse-fly Account Closed Join Date: Apr 2001 Posts: 2,360 Pi is a real number, but it isn't rational.
14th July 2004, 08:04 #203 squakMix wwwyzzerdd(Forum King) Join Date: May 2003 Posts: 3,458 Jesus.... He spelt "pi" - "Pie". Oh yea: Pi is approx: 3.14
14th July 2004, 08:11 #204 Vol. Fireant Junior Member Join Date: Jul 2004 Posts: 5 sorry it is late sq root of -1 is non real and i bow to to squakmix's mastery of spelling
14th July 2004, 08:18 #205 squakMix wwwyzzerdd(Forum King) Join Date: May 2003 Posts: 3,458 Xzxzzx (and Yes I spelt it by memory ), why do people seek what you know? Is the fact that you are giving people the chance to ask questions reason to ask, no matter how mudane the question?
14th July 2004, 11:37 #206
CaptainNuss
Indecisive One
(Senior Member)
Join Date: Jul 2004
Location: :morF
Posts: 625
Quote:
Originally posted by horse-fly There is a thereom(sp?) in Calculus that proves n/0 approaches infinity. That is why i figured to use 1/0.
It's theorem, and I do not believe one exists exactly as you say it does. If you have a fraction x/y, when y approaches zero while |x| > |y| (read: while x stays sufficiently larger than y; I just included this because x doesn't necessarily have to remain constant), x/y approaches infinity. y can't be zero, though, as that would be undefined. You can't devide by zero.
At least that's what I learned in school.
[size=1]"Programming today is a race between software engineers striving to build bigger and better idiot-proof programs and the universe trying to produce bigger and better idiots. So far, the universe is winning." - Richard Cook
14th July 2004, 11:44 #207
will
Nullsoft Newbie (Moderator)
Join Date: Mar 2001
Location: Sheffield, England
Posts: 5,569
Quote:
Originally posted by Vol. Fireant sorry it is late sq root of -1 is non real and i bow to to squakmix's mastery of spelling
The square root of -1 is imaginary.
The term imaginary (in maths) is given to numbers which have a factor of sqrt -1
Hence, infinity, whatever it is, is NOT imaginary.
DO NOT PM ME WITH TECH SUPPORT QUESTIONS
14th July 2004, 11:58 #208
lostonline
Where am I?
(Major Dude)
Join Date: Feb 2004
Location: Yorkshire, UK
Posts: 680
Quote:
Originally posted by will The square root of -1 is imaginary. The term imaginary (in maths) is given to numbers which have a factor of sqrt -1 Hence, infinity, whatever it is, is NOT imaginary.
Once you get into imaginary numbers you'll really start to confuse people
and my question for xzxzzx is... why is this thread being derailed with talk of mathematics? [/edit]
Kids, you tried your best and you failed miserably. The lesson is, never try.
--
Homer Simpson
Last edited by lostonline; 14th July 2004 at 12:57.
14th July 2004, 16:58 #209
horse-fly
Account Closed
Join Date: Apr 2001
Posts: 2,360
Quote:
Originally posted by CaptainNuss It's theorem, and I do not believe one exists exactly as you say it does. If you have a fraction x/y, when y approaches zero while |x| > |y| (read: while x stays sufficiently larger than y; I just included this because x doesn't necessarily have to remain constant), x/y approaches infinity. y can't be zero, though, as that would be undefined. You can't devide by zero. At least that's what I learned in school.
Well this is a person who took a semester of calculus over 5 weeks without a book.
Quote:
Originally posted by lostonline Once you get into imaginary numbers you'll really start to confuse people and my question for xzxzzx is... why is this thread being derailed with talk of mathematics? [/edit]
I believe that was my doing with asking xzxzzx a question about 1/0 * 0/1.
14th July 2004, 17:34 #210
CaptainNuss
Indecisive One
(Senior Member)
Join Date: Jul 2004
Location: :morF
Posts: 625
Quote:
Originally posted by horse-fly Well this is a person who took a semester of calculus over 5 weeks without a book.
Well, I'll just believe you then for the time being until I'll begin studying...
[size=1]"Programming today is a race between software engineers striving to build bigger and better idiot-proof programs and the universe trying to produce bigger and better idiots. So far, the universe is winning." - Richard Cook
14th July 2004, 17:45 #211
ScorLibran
Resident Floydian
Join Date: Aug 2003
Posts: 6,222
Quote:
Originally posted by ScorLibran How old are you?
Quote:
Originally posted by xzxzzx 18.
Ah. That would explain this...
Quote:
Originally posted by xzxzzx People should listen to me because I'm right [...] in other words: I'M SMARTER THAN YOU ALL!
When I was 18 I was smarter than everyone too. In fact, I was the official Smartest Person In The Universe™ from the ages of 14 to 19. It was sort of a pain with world leaders, scientists and gamblers calling me constantly to ask questions.
But, alas, somewhere between the ages of 20 and 23 I lost my superior intelligence, and now I'm just another dumb 36-year-old. When people call me now, it's usually "Hey, idiot!"
So, anyway, that leads to my next question...
How do you cope with all the phone calls, knocks on your door and general harassment you must get from knowing so much?
(When I was 18 I had security guards, phone screeners and PR reps. Extreme, but necessary. )
I'm a psychosomatic sister running around without a leash.
14th July 2004, 17:57 #212 siebe83 Forum King Join Date: Feb 2004 Posts: 9,222 lol Got a question as well: Do you like the sea? Good Winamp plugins by Joonas, DrO and shaneh. If you're bored go here or, if the boredom is more serious, here.
14th July 2004, 18:05 #213 MonKeyRum Major Dude Join Date: May 2004 Location: The North Posts: 859 if someone asks a stupid question and you give an even more ridiculous answer does it matter? Suburbia: The place where they clear down trees and then name roads after them
14th July 2004, 18:08 #214
Forum Sot
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Join Date: Mar 2004
Location: Marietta, Ga. U.S.A.
Posts: 3,915
Quote:
Originally posted by ScorLibran When I was 18 I was smarter than everyone too. In fact, I was the official Smartest Person In The Universe™ from the ages of 14 to 19. It was sort of a pain with world leaders, scientists and gamblers calling me constantly to ask questions.
About 30 years ago I was the official Smartest Person In The Universe™.
My 16 year old son is now proving to us all that he is the next official Smartest Person In The Universe™.
My site (under construction)
14th July 2004, 18:19 #215
ryan
not fucked, not quite.
(Forum King)
Join Date: Feb 2002
Location: Tn
Posts: 8,798
Quote:
Originally posted by squakMix Jesus.... He spelt "pi" - "Pie". Oh yea: Pi is approx: 3.14
14th July 2004, 18:21 #216 CaboWaboAddict Forum Sot(Major Dude) Join Date: Mar 2004 Location: Marietta, Ga. U.S.A. Posts: 3,915 Microsoft's built in calculator: 3.1415926535897932384626433832795 edit: I must state that this is only a close approximation. Further edit: Upon reflection, this brings up a question for the mighty xzxzzx... Good, Fast, Cheap. The saying goes, you can, at most, have only two. My example above has only two. Its a good approximation and it was fast way to get it. Windows XP is not cheap, hence failing the third point. Can you give an example that has all three points? Idiot's Advocate My site (under construction) Last edited by CaboWaboAddict; 14th July 2004 at 19:05.
14th July 2004, 18:36 #217
horse-fly
Account Closed
Join Date: Apr 2001
Posts: 2,360
Quote:
Originally posted by CaboWaboAddict Microsoft's built in calculator: 3.1415926535897932384626433832795 edit: I must state that this is only a close approximation.
It isn't that close.
14th July 2004, 19:06 #218 CaboWaboAddict Forum Sot(Major Dude) Join Date: Mar 2004 Location: Marietta, Ga. U.S.A. Posts: 3,915 Horse-fly, how close do you want? Idiot's Advocate My site (under construction)
14th July 2004, 19:40 #219 InvisableMan Ninja Master!(Forum King) Join Date: Mar 2001 Location: Hotel California Posts: 4,332 xzxzzx: Why?
14th July 2004, 20:45 #220
xzxzzx
Forum King
Join Date: Aug 2002
Posts: 7,254
Quote:
Originally posted by squakMix Xzxzzx (and Yes I spelt it by memory ), why do people seek what you know? Is the fact that you are giving people the chance to ask questions reason to ask, no matter how mudane the question?
Yes.
Quote:
Originally posted by lostonline and my question for xzxzzx is... why is this thread being derailed with talk of mathematics? [/edit]
There are lots of questions in mathematics, and mathematics is interested, thus, the thread. Plus there's some idiot who keeps espousing the wrong thing, it seems.
Quote:
Originally posted by ScorLibran Ah. That would explain this... ...(blah blah blah)...
Interesting. My prediction comes true sooner than I expected.
Quote:
Originally posted by ScorLibran So, anyway, that leads to my next question... How do you cope with all the phone calls, knocks on your door and general harassment you must get from knowing so much?
I built a time machine, and stopped my growth. At any given time, there's about 100,000 of me answering questions.
Quote:
Originally posted by siebe83 Got a question as well: Do you like the sea?
Yes. I'd like to start SCUBA diving the sea this year, once the water warms up a bit more, and I get my SCUBA gear out of storage from another god damn state.
Quote:
Originally posted by MonKeyRum if someone asks a stupid question and you give an even more ridiculous answer does it matter?
I've always lived my life with the axiom "ask a stupid question, get a stupid answer".
Quote:
Originally posted by CaboWaboAddict Horse-fly, how close do you want?
It's obvious that he wants it to 1 million digits:
http://xzxzzx.com/tmp/pi_1024.txt
Quote:
Originally posted by InvisableMan xzxzzx: Why?
Why not?
Freedom of speech is the basic freedom of humanity. When you've lost that, you've lost everything.
1\/\/4y 34|<\$p4y 1gp4y 33714y, 0d4y 0uy4y? | Roses are #FF0000; Violets are #0000FF; chown -R \${YOU} ~/base
The DMCA. It really is that bad. : Count for your life.
14th July 2004, 20:48 #221
beanboy89
(Major Dude)
isn't very custom
Join Date: Nov 2003
Location: Pennsylvania, USA
Posts: 2,004
Quote:
Originally posted by xzxzzx Why not?
14th July 2004, 20:53 #222 siebe83 Forum King Join Date: Feb 2004 Posts: 9,222 apparently, xzxzzx is consistently answering the questions which makes him even more trustworthy way to go! Good Winamp plugins by Joonas, DrO and shaneh. If you're bored go here or, if the boredom is more serious, here.
14th July 2004, 20:55 #223
xzxzzx
Forum King
Join Date: Aug 2002
Posts: 7,254
Quote:
Originally posted by beanboy89 Didn't you already answer the question "Why?" in a previous post with the same answer?
Yes, and my answer hasn't changed.
siebe83 knows what he's saying there.
Freedom of speech is the basic freedom of humanity. When you've lost that, you've lost everything.
1\/\/4y 34|<\$p4y 1gp4y 33714y, 0d4y 0uy4y? | Roses are #FF0000; Violets are #0000FF; chown -R \${YOU} ~/base
The DMCA. It really is that bad. : Count for your life.
14th July 2004, 21:00 #224 mark Forum King Join Date: Jul 2002 Location: Norn Ir'nd, leek... 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4595889706953653494060340216654437558900456328822505452556405644824651518754711962184439658253375438856909411303150952617937800297412076651479394259029896959469955657612186561967337862362561252163208628692221032748892186543648022967807057656151446320 4692790682120738837781423356282360896320806822246801224826117718589638140918390367367222088832151375560037279839400415297002878307667094447456013455641725437090697939612257142989467154357846878861444581231459357198492252847160504922124247014121478057 3455105008019086996033027634787081081754501193071412233908663938339529425786905076431006383519834389341596131854347546495569781038293097164651438407007073604112373599843452251610507027056235266012764848308407611830130527932054274628654036036745328651 0570658748822569815793678976697422057505968344086973502014102067235850200724522563265134105592401902742162484391403599895353945909440704691209140938700126456001623742880210927645793106579229552498872758461012648369998922569596881592056001016552563756 10000 digits accurate enough?
14th July 2004, 21:02 #225
Forum Sot
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Quote:
Originally posted by siebe83 apparently, xzxzzx is consistently answering the questions which makes him even more trustworthy way to go!
not too consistant... he didn't answer this:
Quote:
Originally posted by CaboWaboAddict Upon reflection, this brings up a question for the mighty xzxzzx... Good, Fast, Cheap. The saying goes, you can, at most, have only two. My example above has only two. Its a good approximation and it was fast way to get it. Windows XP is not cheap, hence failing the third point. Can you give an example that has all three points?
My site (under construction)
14th July 2004, 21:05 #226 whiteflip Post Master General(Forum King) Join Date: Jun 2000 Location: Seattle, Now Las Vegas Posts: 6,032 Pi in 8 base math is 4.444... some other numbers. But the first 4 numbers are 4. I think we should convert to 8 base math. Where 7+7=14 and 14+14=26. I'm Back?
14th July 2004, 21:08 #227 shakey_snake Forum Domo Join Date: Jan 2004 Location: Everyone, get over here for the picture! Posts: 4,313 only 20 digits are necessary to calculate the circumference of the Earth down to a fraction of an inch, and 39 digits are needed to calculate the circumference of the entire known universe down to the electron! elevatorladyelevatorladyelevatorladyelevatorladyelevatorladylevitateme
14th July 2004, 21:12 #228 mark Forum King Join Date: Jul 2002 Location: Norn Ir'nd, leek... Posts: 6,287 42! or was that base 13?
14th July 2004, 21:15 #229
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Quote:
Originally posted by whiteflip Pi in 8 base math is 4.444... some other numbers. But the first 4 numbers are 4. I think we should convert to 8 base math. Where 7+7=14 and 14+14=26.
Huh?
Base 8:
00, 01, 02, 03, 04, 05, 06, 07,
10, 11, 12, 13, 14, 15, 16, 17,
20, 21, 22, 23, 24, 25, 26, 27,
30, ...
In base 8:
7 + 7 = 16
14 + 14 = 30
I'd say one of us needs a drink! mas Tequila!
My site (under construction)
14th July 2004, 21:26 #230
xzxzzx
Forum King
Join Date: Aug 2002
Posts: 7,254
Quote:
You edited it after I read it.
Quote:
Originally posted by CaboWaboAddict Further edit: Upon reflection, this brings up a question for the mighty xzxzzx... Good, Fast, Cheap. The saying goes, you can, at most, have only two. My example above has only two. Its a good approximation and it was fast way to get it. Windows XP is not cheap, hence failing the third point. Can you give an example that has all three points?
Linux.
Freedom of speech is the basic freedom of humanity. When you've lost that, you've lost everything.
1\/\/4y 34|<\$p4y 1gp4y 33714y, 0d4y 0uy4y? | Roses are #FF0000; Violets are #0000FF; chown -R \${YOU} ~/base
The DMCA. It really is that bad. : Count for your life.
14th July 2004, 21:27 #231
cyu
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Join Date: Apr 2002
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Quote:
Originally posted by Nobby Nobbs 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196442881097566593344612847564823378678316527120190 9145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949 1298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099 6051870721134999999837297804995105973173281609631859502445945534690830264252230825334468503526193118817101000313783875288658753320838142061717766914730359825349042875546873115956286388235378759375195778185778053217122680661300192787661119590921642019 8938095257201065485863278865936153381827968230301952035301852968995773622599413891249721775283479131515574857242454150695950829533116861727855889075098381754637464939319255060400927701671139009848824012858361603563707660104710181942955596198946767837 4494482553797747268471040475346462080466842590694912933136770289891521047521620569660240580381501935112533824300355876402474964732639141992726042699227967823547816360093417216412199245863150302861829745557067498385054945885869269956909272107975093029 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4595889706953653494060340216654437558900456328822505452556405644824651518754711962184439658253375438856909411303150952617937800297412076651479394259029896959469955657612186561967337862362561252163208628692221032748892186543648022967807057656151446320 4692790682120738837781423356282360896320806822246801224826117718589638140918390367367222088832151375560037279839400415297002878307667094447456013455641725437090697939612257142989467154357846878861444581231459357198492252847160504922124247014121478057 3455105008019086996033027634787081081754501193071412233908663938339529425786905076431006383519834389341596131854347546495569781038293097164651438407007073604112373599843452251610507027056235266012764848308407611830130527932054274628654036036745328651 0570658748822569815793678976697422057505968344086973502014102067235850200724522563265134105592401902742162484391403599895353945909440704691209140938700126456001623742880210927645793106579229552498872758461012648369998922569596881592056001016552563756 10000 digits accurate enough?
Why not just 22/7?
14th July 2004, 21:33 #232
xzxzzx
Forum King
Join Date: Aug 2002
Posts: 7,254
Quote:
Originally posted by cyu Why not just 22/7?
Because 22/7 is only accurate to 3.14.
Freedom of speech is the basic freedom of humanity. When you've lost that, you've lost everything.
1\/\/4y 34|<\$p4y 1gp4y 33714y, 0d4y 0uy4y? | Roses are #FF0000; Violets are #0000FF; chown -R \${YOU} ~/base
The DMCA. It really is that bad. : Count for your life.
14th July 2004, 21:44 #233
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Quote:
Originally posted by xzxzzx You edited it after I read it.
You waited 1 hour and 23 minutes to respond after reading it??? Something's rotten here...
Quote:
Originally posted by xzxzzx Linux.
Oops, an OS is only as good as the applications designed for it. Linux doesn't have any good apps in my line of work.
Maybe my 16 year old son will move up to being the official Smartest Person In The Universe™ sooner than I expected.
My site (under construction)
14th July 2004, 21:57 #234
xzxzzx
Forum King
Join Date: Aug 2002
Posts: 7,254
Quote:
Originally posted by CaboWaboAddict You waited 1 hour and 23 minutes to respond after reading it??? Something's rotten here...
Yeah, I did. I had a call from a client.
Quote:
Originally posted by CaboWaboAddict Oops, an OS is only as good as the applications designed for it. Linux doesn't have any good apps in my line of work.
Linux as a server platform, then.
Freedom of speech is the basic freedom of humanity. When you've lost that, you've lost everything.
1\/\/4y 34|<\$p4y 1gp4y 33714y, 0d4y 0uy4y? | Roses are #FF0000; Violets are #0000FF; chown -R \${YOU} ~/base
The DMCA. It really is that bad. : Count for your life.
14th July 2004, 22:04 #235
InvisableMan
Ninja Master!
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Location: Hotel California
Posts: 4,332
Quote:
Originally posted by xzxzzx Why not?
exactly.
15th July 2004, 03:51 #236 whiteflip Post Master General(Forum King) Join Date: Jun 2000 Location: Seattle, Now Las Vegas Posts: 6,032 oops my calculator is screwy. I'm Back?
15th July 2004, 04:11 #237
ScorLibran
Resident Floydian
Join Date: Aug 2003
Posts: 6,222
Quote:
Originally posted by xzxzzx Interesting. My prediction comes true sooner than I expected.
Your foresight is my hindsight. We agree!
But you haven't answered my question about how you cope with the demands of society on the vastly intelligent. I need all the advice I can get to pass on to the child I'll have one day when they begin to experience the ubiquitous IQ spike in their late teens.
I'm a psychosomatic sister running around without a leash.
15th July 2004, 04:13 #238
xzxzzx
Forum King
Join Date: Aug 2002
Posts: 7,254
Quote:
Originally posted by ScorLibran But you haven't answered my question about how you cope with the demands of society on the vastly intelligent. I need all the advice I can get to pass on to the child I'll have one day when they begin to experience the ubiquitous IQ spike in their late teens.
Yes I did.
Freedom of speech is the basic freedom of humanity. When you've lost that, you've lost everything.
1\/\/4y 34|<\$p4y 1gp4y 33714y, 0d4y 0uy4y? | Roses are #FF0000; Violets are #0000FF; chown -R \${YOU} ~/base
The DMCA. It really is that bad. : Count for your life.
15th July 2004, 04:20 #239 White Raven Little Winged One Join Date: Apr 2002 Location: Canada, now UK Posts: 4,165 Why is it that when the people who make commercials choose to aim their products at a deciduously younger demographic, do they immediately assume that we all like rap/and or pop, and wear baggy clothes? Or even that the majority of us like that sort of thing? just as feathery as ever | portfolio | a poignant quote
15th July 2004, 04:21 #240
ScorLibran
Resident Floydian
Join Date: Aug 2003
Posts: 6,222
Quote:
Originally posted by xzxzzx Yes I did.
Ah, so you did. My bad.
So is making 100,000 of yourself done by human cloning, or making use of parallel universes? Either way, great answer!
I'm a psychosomatic sister running around without a leash.
Winamp & Shoutcast Forums Ask xzxzzx | 12,280 | 38,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-43 | latest | en | 0.941805 |
https://www.coursehero.com/file/5576815/237solutions1/ | 1,490,457,553,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188962.14/warc/CC-MAIN-20170322212948-00328-ip-10-233-31-227.ec2.internal.warc.gz | 896,577,464 | 123,678 | 237solutions1
# 237solutions1 - Note: You may not reference solutions 2 and...
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Solutions/Hints – Module #1 & GRE Practice #1 Module #1: Getting Started 1. a. proton b. same c. proton 2. b 3. a & d 4. a, c, & d Discussion Questions 1. a. see chapter 1 b. see sections 2.2-2.3 c. see section 2.4 2. a. see section 4.1 b. see section 4.2 c. see section 4.5-4.8 d. see section 4.9-4.11 Enrichment Problem 1. 3 i h p 2 2. ni h x n " 1 3. ni h p n " 1 4. The test function reminds you to use the product rule.
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Unformatted text preview: Note: You may not reference solutions 2 and 3 for your homework, but must derive your homework solutions. Note: Dont forget the following useful relation: [ A , BC ] = ABC " BCA = ABC " BAC + BAC " BCA = [ A , B ] C + B [ A , C ] This is often easier to use then working with the actual derivatives. GRE Practice #1: 10-E, 1-C, 27-B, 29-B...
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## This document was uploaded on 09/10/2009.
Ask a homework question - tutors are online | 361 | 1,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.78708 |
https://rdrr.io/github/psobczyk/dhmm/man/hsmm.smooth.html | 1,542,409,802,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743216.58/warc/CC-MAIN-20181116214920-20181117000920-00247.warc.gz | 713,501,237 | 14,402 | # hsmm.smooth: Hidden Semi-Markov Models In psobczyk/dhmm: Hidden Semi Markov Models
## Description
Inference on the hidden states for given observations and model specifications of a hidden semi-Markov model. For every observation, the function calculates the probability of being in a particular state.
## Usage
```1 2 3 4 5 6 7 8``` ```hsmm.smooth(x, od, rd, pi.par, tpm.par, od.par, rd.par, M = NA) ```
## Arguments
` x` The observations as a vector of length T. ` od` Character containing the name of the conditional distribution of the observations. For details see `hsmm`. ` rd` Character containing the name of the runlength distribution (or sojourn time, dwell time distribution). For details see `hsmm`. ` pi.par` Vector of length J containing the values for the intitial probabilities of the semi-Markov chain. ` tpm.par` Matrix of dimension J x J containing the parameter values for the transition probability matrix of the embedded Markov chain. The diagonal entries must all be zero, absorbing states are not permitted. ` rd.par` List with the values for the parameters of the runlength distributions. For details see `hsmm`. ` od.par` List with the values for the parameters of the conditional observation distributions. For details see `hsmm`. ` M` Positive integer containing the maximum runlength.
## Details
The function `hsmm.smooth` calculates the so-called smoothed probabilities
P(S[t] = i | X[1], ... , X[T])
for all t in 1, ... , T and i in 1,...,J, with X denoting the observations and S the hidden states. This procedure is often termed 'local decoding'. The sequence of the most probable states follows directly. Note that this sequence is not necessarily the most probable state sequence, which is determined by the Viterbi algorithm. Also note that local decoding may ignore restrictions imposed by the transition probability matrix, such as forbidden transitions, because it optimizes locally for every single observation.
## Value
`call` The matched call. `smooth.prob` A matrix of dimension J x T containing the smoothing probabilities. `path` A vector of length T containing the sequence of the states with the highest probabilities.
## See Also
`hsmm`, `hsmm.sim`, `hsmm.viterbi`
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25``` ```# Simulating observations: # (see hsmm.sim for details) pipar <- rep(1/3, 3) tpmpar <- matrix(c(0, 0.5, 0.5, 0.7, 0, 0.3, 0.8, 0.2, 0), 3, byrow = TRUE) rdpar <- list(p = c(0.98, 0.98, 0.99)) odpar <- list(mean = c(-1.5, 0, 1.5), var = c(0.5, 0.6, 0.8)) sim <- hsmm.sim(n = 2000, od = "norm", rd = "log", pi.par = pipar, tpm.par = tpmpar, rd.par = rdpar, od.par = odpar, seed = 3539) # Computation of the smoothing probabilities: fit.sm <- hsmm.smooth(sim\$obs, od = "norm", rd = "log", pi.par = pipar, tpm.par = tpmpar, od.par = odpar, rd.par = rdpar) # The first 15 smoothing probabilities: round(fit.sm\$sm[, 1:15], 2) # The first 15 values of the resulting path: fit.sm\$path[1:15] # For comparison, the real/simulated path (first 15 values): sim\$path[1:15] ```
psobczyk/dhmm documentation built on May 24, 2017, 12:19 p.m. | 918 | 3,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-47 | latest | en | 0.746064 |
https://www.gamedev.net/forums/topic/474508-logical-not-operator/ | 1,539,747,701,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510969.31/warc/CC-MAIN-20181017023458-20181017044958-00117.warc.gz | 929,087,876 | 24,618 | Public Group
# logical not operator
This topic is 3971 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
bool operator!=( const Array &right ) const
{
return ! ( *this == right ); // invokes Array::operator==
}
bool Array::operator==( const Array &right ) const
{
if ( size != right.size )
return false; // arrays of different number of elements
for ( int i = 0; i < size; i++ )
if ( ptr[ i ] != right.ptr[ i ] )
return false; // array contents are not equal
return true; // arrays are equal
}
//// in main()
// use overloaded inequality (!=) operator
cout << "\nEvaluating: integers1 != integers2" << endl;
if ( integers1 != integers2 )
cout << "integers1 and integers2 are not equal" << endl;
I'm having a hard time understanding the logical not operator in this case. So, if the array sizes are not equal, then operator== returns false. So then, when it gets back to the operator!= function, what is it returning? return !(false), which means it is returning true, or is it returning false? Or I guess another way to put it is, is the if statment reading: if( true ) or if( false )? I'm guessing if( true ) because the statement after that is printed when I run the program.
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If the sizes are unequal, operator== returns false. If operator== returns false, operator!= returns true. That is, operator!= returns true when the sizes are unequal.
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Ahhh, thanks. Makes sense now.
##### Share on other sites
It's called "logical not" for a reason. :) Things are not equal when it is not the case that they are equal, and this should seem quite logical indeed.
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http://nob.cs.ucdavis.edu/classes/ecs153-2003-04/outlines/2003-10-13.html | 1,544,731,467,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825098.68/warc/CC-MAIN-20181213193633-20181213215133-00494.warc.gz | 205,903,054 | 2,412 | # Outline for October 13, 2003
Reading: Chapters 6.1, 6.2.3, 6.4
## Discussion Problem
In the Bell-LaPadula Model, subjects could read and write objects only if the subjects were in the same compartment as objects. This leads to a notion of confinement, and raises the issue of leaking information among compartments. Such leakage led one security expert to speculate that, as the need for secure computing continued to climb, people would gradually shift from multi-user computing systems to single-user computer systems, because then information could not leak among compartments (as there are no other processes on the system to leak information to).
1. How do single-user systems connected by a network (such as the Internet) differ from multi-user systems?
2. Do you agree or disagree with the expert?
## Outline for the Day
1. Integrity models
1. Requirements
1. Users won't write their own programs, but will use existing programs, databases, etc.
2. Programmers develop and test programs on non-production systems
3. Installing a program from the development system requires a special process
4. This process must be controlled and auditable
5. System managers must be able to access the system state and the system logs
2. Separation of duty
3. Separation of function
4. Auditing
2. Biba: mathematical dual of BLP
1. P may read O if L(P) ≤ L(O) and C(P) ⊆ C(O), and P may write O if L(O) ≤ L(P) and C(O) ⊆ C(P)
2. Combined with BLP
3. Clark-Wilson
1. Theme: military model does not provide enough controls for commercial fraud, etc. because it does not cover the right aspects of integrity
2. "Constrained Data Items" (CDI) to which model applies, "Unconstrained Data Items (UDIs) to which no integrity checks applied, "Integrity Verification Procedures" (IVP) verify conformance to the integrity spec when IVP is run, "Transaction Procedures" (TP) take system from one well-formed state to another
3. Certification and enforcement rules:
C1. All IVPs must ensure that all CDIs are in a valid state when the IVP is run
C2. All TPs must be certified as valid; each TP is assocated with a set of CDIs it is authorized to manipulate
E1. The system must maintain these lists and must ensure only those TPs manipulate those CDIs
E2: The system must maintain a list of User IDs, TP, and CDIs that that TP can manipulate on behalf of that user, and must ensure only those executions are performed.
C3. The list of relations in E2 must be certified to meet the separation of duty requirement.
E3. The sysem must authenticate the identity of each user attempting to execute a TP.
C4. All TPs must be certified to write to an append-only CDI (the log) all information necessary to resonstruct the operation.
C5. Any TP taking a UDI as an input must be certified to perform only valid transformations, else no transformations, for any possible value of the UDI. The transformation should take the input from a UDI to a CDI, or the UDI is rejected (typically, for edits as the keyboard is a UDI).
E4. Only the agent permitted to certify entities may change the list of such entities associated with a TP. An agent that can certify an entity may not have any execute rights with respect to that entity
Here is a PDF version of this document. | 761 | 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} | 2.65625 | 3 | CC-MAIN-2018-51 | latest | en | 0.926049 |
https://sports.answers.com/toys-and-games/How_many_balls_at_the_beginning_of_a_snooker_match | 1,725,982,031,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00375.warc.gz | 503,963,987 | 48,482 | 0
How many balls at the beginning of a snooker match?
Updated: 10/5/2023
Wiki User
β 13y ago
21, 22 if you count the white 15 reds, a black, a pink, a blue, a green a brown and a yellow
Wiki User
β 14y ago
Wiki User
β 13y ago
22 including the white
Earn +20 pts
Q: How many balls at the beginning of a snooker match?
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Related questions
How many red balls used in a game of snooker?
15 red snooker balls.
How many grey balls are in a snooker game?
There are no grey balls in a standard game of snooker.
What sport uses many balls?
snooker or biliards
How many frames do you have to win in a world snooker match?
In the final of the World Snooker Championship it is 18 frames.
How many snooker balls in 4 ft sigma table?
The standard number in a set of snooker balls is 21 - 15 reds and six colours. In 'toy' snooker tables, the reds are usually reduced to 10.
How many red balls are in the start of a frame of snooker?
There are fifteen of them, though not numbered.
How many red balls on an 8ft snooker table?
In standard snooker 15 reds are used.In six red snooker - no surprisingly - six reds are usedIn power snooker only nine reds are used.
How many balls in snooker?
15 reds, 6 colors and a cue ball. So 22.
How many balls are used in a snooker triangle?
There are 21 balls in a set of snooker balls, 22 if you include the Cue, 15 Red balls worth 1 point each, and 6 other colored balls worth different amounts of points, yellow (2 points), green (3), brown (4), blue (5), pink (6) and black (7).
How many red balls billiards?
In true Billiards there is only 1 red ball. In snooker there are 15 reds.
How many bs on a s t?
How many balls on a snooker table? 22(one white cue ball, 15 red balls, and six balls of different colours: yellow, green, brown, blue, pink and black)
2 balls | 536 | 1,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.907994 |
https://www.physicsforums.com/threads/motion-in-one-dimension.85621/ | 1,660,885,761,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573623.4/warc/CC-MAIN-20220819035957-20220819065957-00071.warc.gz | 807,181,888 | 13,579 | # Motion in One Dimension
An electron with initial velocity $v_{x}_{o} = 1.0 \times 10^{4}$ meters/sec enters a region where it is electrically accelerated. It emerges with a velocity $v_{x} = 4.0 \times 10^{6}$ meters/sec. What was its acceleration, assumed constant? The accelerated region is 1 cm. So acceleration is defined as $\frac{\Delta v}{\Delta t}$. How would I use the fact that the accelerated region is 1 cm long?
Thanks | 117 | 435 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-33 | latest | en | 0.911948 |
https://www.brainscape.com/flashcards/exchange-surfaces-and-breathing-9958993/packs/16345159 | 1,726,075,649,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00839.warc.gz | 644,373,959 | 21,874 | # Exchange Surfaces and Breathing Flashcards
You may prefer our related Brainscape-certified flashcards:
1
Q
Why are exchange surfaces needed?
A
• Large multicellular organisms have a small surface area to volume ratio
• Cells in the centre of the organisms would not receive any materials if multicellular organisms survived on diffusion alone
• Multicellular organisms have a high metabolic rate so there is a need to exchange lots of materials quickly.
2
Q
What is the rule for SA:V ratio?
A
The largest the SA:V of an organisms, the easier the exchange of substances by diffusion is.
3
Q
What are the SA:V calculations for a cuboid?
A
• Volume=Length x Width x Height
- Surface Area=(4xlengthxheight)+(2xheightxwidth)
4
Q
What are the SA:V calculations for a cylinder?
A
• Volume= 3.14xradius^2 x height
- Surface Area= (2x3.14xheight) + 2x3.14 x radius^2
5
Q
What are the SA:V calculations for a sphere?
A
• Volume= 4/3 x 3.14x radius^3
- Surface Area= 4 x 3.14 x radius ^2
6
Q
What are the features of an efficient gas exchange system?
A
• Large surface area
• Thin layers
• Rich blood supply
• Ventilation (maintains diffusion gradient which makes process more efficient and faster_
7
Q
What is the pleural cavity?
A
• Space between the double membrane (pleural) enclosing the lungs
• It is filled with a small amount of pleural fluid
• This fluid lubricates the lungs
• It also adheres the outer wall of the lungs to the thoracic cavity by water cohesion, so that the lungs expand with the chest while breathing
8
Q
What are the functions of the nasal cavity?
A
• Large surface area and good blood supply allows the air to be warmed as it passes into the body
• Hairy lining which traps dust and bacteria in mucus and prevent them from reaching the lungs (this is not cillia)
• Moist surfaces increases the humidity of the incoming air which reduces the evaporation of water in the lungs
9
Q
What are the features of the trachea?
A
• Pipe is supported by a layer of cartilage that holds the trachea open and prevents it from collapsing
• Cartilage rings are incomplete to allow it to bend when food is swallowed down the oesophagus
• Lined with ciliate epithelial and goblet cells that prevent dust and bacteria from entering the lungs
10
Q
What is the bronchus?
A
• The bronchus are extensions of the trachea that split into two for the left and right lung
• This has a very similar structure to the trachea but smaller
• Cartilage rings hold the pipe open
11
Q
Why do ciliated epithelial cells contain many mitochondria?
A
To provide lots of energy for the cillian to waft
12
Q
What are the bronchioles?
A
• Bronchus splits into much smaller bronchioles
• They are around 1mm or less in diameter
• Have no cartilage and are held open by smooth muscle
• When this muscle contracts, the bronchioles contract, which is dependent of air flow
• They are lined with a thin layer of epithelial tissues making some gas exchange possible
13
Q
What are alveoli?
A
• Little air sacs where most of the gas exchange occurs
• Each alveoli is about 200-300um in diameter
• Made up of a thin layer of flattened epithelial cells as well as some collagen and elastic fibres
• Elastic fibres causes recoil which helps move air out of the alveoli.
14
Q
What are the features of efficient gas exchange system?
A
• Large surface area (alveoli cover a surface area of around 50-75m^2)
• Thin layers (short diffusion pathway, increasing speed of exchange)
• Good, constant blood supply (maintains diffusion gradient by ensuring exchanged substances are moving to the area needed)
• Ventilation (maintains diffusion gradient which makes process faster and more efficient)
• Moist (maintain blood gas barrier to prevent gas from entering blood)
15
Q
Why are gas exchange systems moist?
A
To aid diffusion by oxygen dissolving into the water. However ventilation causes this water to readily evaporate. The gas exchange system is helpfully adapted to efficiently gas exchange without losing too much water.
16
Q
What is lung surfactant?
A
A phospholipid that coats the surfaces of the lungs. Without it, the watery lining of the alveoli would create surface tensions and cause the alveoli to collapse.
17
Q
What brings about ventillation?
A
• Pressure changes in the thoracic cavity
- Rib cage provides a cage in which pressure can be changes to facilitate breathing.
18
Q
What happens in expiration?
A
• Diaphragm moves up
• Intercostal muscles move the ribs down and in
• Thoracic volume decreases
• Thoracic pressure increases
• Air flows out of the lungs (to equalise the pressure difference)
19
Q
What kind of process is expiration?
A
A largely passive process
20
Q
What happens during inspiration?
A
• Diaphragm moves down
• Intercostal muscles move the ribs up and out
• Thoracic volume increases
• Thoracic pressure decreases
• Air flows into the lungs (to equalise the pressure difference)
21
Q
What kind of process is inspiration?
A
An active process
22
Q
What is the composition of inhaled air?
A
• 78% Nitrogen
• 21% Oxygen
• 0.04% Carbon Dioxide
• <1% Water Vapour
• <1% Other
23
Q
What is the composition of exhaled air?
A
• 78% Nitrogen
• 15% Oxygen
• 4% Carbon Dioxide
• 3% Water Vapour
• <1% Other
24
Q
How does a spirometer work?
A
• Static lower half of tank is full of water
• Mobile upper half of tank is full of oxygen
• Breathe out into the tank and the upper half will rise
• Breathe in from the tank and the upper half will fall
• Trace marker is attached to the mobile upper half
25
Q
What is a peak flow meter?
A
• Measures the rate at which air can be expelled from the lungs
• The instrument produces a graph about the amount of the air they breathe out and how quickly they breathe out
26
Q
Why does the overall volume of gas in the spirometer tank decline over time?
A
• The spirometer contains Soda Lime which absorbs CO2
• When breathing, O2 is used up from the tank and CO2 is absorbed by the soda lime
• Therefore, the gas volume of tank decreases over the experiment
27
Q
What is tidal volume?
A
• Volume of air that moves into and out of the lungs with each resting breath
• 0.5dm^3 is the adult average
• Is 15% of the vital capacity
28
Q
What is vital capacity?
A
• Largest volume of air that can be breathed in
• When the strongest possible exhale is followed by the strongest possible inhale
• 5dm^3 is the adult average
29
Q
What is inspiratory reserve volume?
A
-Maximum volume of air you can breathe in over and above normal inhalation
30
Q
What is expiratory reserve volume?
A
-Maximum volume of air you can force out of your lungs over and above the normal tidal volume of air you breathe out
31
Q
What is residual volume?
A
The volume of air that is left in your lungs when you have exhaled as much as possible
32
Q
Why must there always be some air in your lungs?
A
Because otherwise your lungs would completely collapse
33
Q
What is total lung capacity?
A
• The sum of the vital capacity and the residual volume
- Total potential amount of air in the lungs at any one time
34
Q
What are the evolutionary strategies to gas exchange in fish?
A
• Obligate air breathers need to breathe air or will suffocate (emerge at surface)
• Facultative air breathers can breathe air if they need to but mainly rely on their gills
• Fish that cannot breathe rely on gills for gas exchange
35
Q
How to gills work?
A
-Gas exchange takes place across the surface of gills
-
36
Q
What is the operculum?
A
• Series of bones that act as facial support and protective gill covers
• It adjusts the pressure of water in the pharynx to allow proper ventilation of the gills
• Opercular Valves inside the mouth keep water from escaping
37
Q
What is the buccal cavity (pharynx)?
A
• Bony fish inhale water through their mouth by lowering the floor of the buccal cavity and closing the opercular valves
• They exhale water by raising the floor of the buccal cavity and closing the opercular valve to allow water out
• This ensures water is always flowing over the gills
38
Q
What are gill arches?
A
A series of bony loops that support the gills
39
Q
What are gill filaments?
A
Occurs in large stacks and each stack is called a gill plate
40
Q
What are gill lamellae?
A
• The main site of gas exchange
- They have a very large surface area
41
Q
What features to gills have to carry out efficient gas exchange?
A
• Tips of adjacent gill filaments overlap to slow the flow of water down
• Water moving across and through the gills and the blood flowing the gill filaments move in opposite directions
• Large SA
• A good blood supply
• Thin walls
42
Q
What is the parallel system of gas exchange?
A
1. Blood in gill filaments and the water moving across and through the gills both flow in the same direction
2. Oxygen will diffuse down a steep concentration gradient at the leading edge of the gill filaments but equilibrium will be reached at some point before the wave leaves the gills
3. This means the surface area of gill filaments is not being fully utilised for oxygen exchange between water and blood
4. This is less efficient than the counter current flow
43
Q
What is the counter current flow system of gas exchange?
A
1. The blood in the gill filaments and the water moving across and through the gills both flow in opposite directions
2. Oxygen will diffusion down a reasonably steep concentration gradient along the whole length of the gill filaments, and then equilibrium is reached
3. This means the entire surface area of the gill filaments is fully utilised for oxygen exchange between water and blood
4. This is more efficient than a parallel flow system
44
Q
What are spiracles?
A
Holes in the thorax and abdomen walls, usually a pair for each segment and often with valves
45
Q
What are tracheae?
A
Tubes that lead into the body from the spiracles, which form a branching of ever smaller diameter tubes, strengthened by hoops or spirals of chitin called taenidia.
46
Q
What are tracheoles?
A
• Single and greatly elongated cells that form minute tubes, found at the ends of the smallest tracheae
• The site of gas exchange with body cells
47
Q
What is tracheal fluid?
A
Fluid in the tracheoles, but at the very end of each tracheole
48
Q
How do the spiracles work?
A
1. Air enters and leaves the respiratory system of insects through the spiracles
2. The spiracles can be opened and closed to reduce water loss (contracting muscles surrounding the spiracle, opening valve)
3. Spiracles may also be surrounded by hairs to minimise water loss
4. The spiracles open and close depending on the metabolic demands of the insects. So, when demand for oxygen is high, more spiracles will be open
49
Q
How does the insect tracheae work?
A
1. After passing through a spiracle, air enters a complex branching network of tracheal tubes that subdivides into smaller diameters and reaches every part of the body
2. To prevent its collapse under pressure, reinforcing wire of chitin winds spirally through the walls as hoops
3. The gives the tracheal tubes the ability to flex and stretch without developing kinks that might restrict air flow
50
Q
How do the tracheoles and tracheal fluid work?
A
1. Tracheoles penetrate into the spaces between body cells, so have a large surface area, so that gas exchange can meet metabolic demands
2. They have no chitin and have very thin walls to create a short diffusion path
3. The inside of the walls are moist with water, so gases can dissolve into the liquid phase
4. Gas exchange happens at these surfaces, oxygen from the air dissolves in the moist lining of the tracheoles and diffuses into body cells
5. Carbon dioxide follows the reverse path
51
Q
How do insects ventilate their respiratory system?
A
1. Small insects rely entirely on passive diffusion of gases
2. Whereas, large insects require active ventilation
3. This is done by opening some spiracles and closing others while using abdominal muscles to alternatively expand and contract body movement
4. This creates pulsating movements that flush air from one end of the body to the other through the tracheal system
52
Q
What may larger insects have?
A
• Some larger insects will have air sacs as part of their tracheal system
• These act as reservoirs of air and conserve water during periods of drought
53
Q
What is discontinuous gas exchange cycles?
A
-Occurs when insects are at rest and may help insects that are at rest in burrows or prevent the entry of pathogens through the spiracles
-Occurs in three phases:
Closed, Flutter and Open
54
Q
What is the closed phase of DGC?
A
• Spiracles shut tight which drastically reduces ventilation with the external environment
• O2 is consumed and its conc decreases within the tracheal system
• CO2 is produced and increases in conc, it is buffered in the haemolymph rather than diffusing into tracheal system
• This leads to a negative pressure and starts the flutter phase
55
Q
What is the flutter phase of DGC?
A
• Spiracles open slightly and close in rapid succession
• A small amount of air from the environment enters the respiratory system each time spiracles are opened
• Fresh air is brought into the tracheal system to increase conc of O2, and releases small amounts of CO2
• Most of the CO2 is still buffered by the haemolymph
56
Q
What is the open phase of DGC?
A
• Flutter phase continues until until CO2 production surpasses the buffering capacity and begins to build up in the tracheal system
• Spiracles open completely causing a rapid release of CO2. a complete exchange of gases with the environment occurs.
57
Q
What is the main difference in gas exchange between insects and mammals?
A
• Insects have evolved a system to deliver oxygen and remove CO2 directly from each cell
• This means the gas exchange system in insects is a separate system to their circulatory system, unlike in mammals | 3,379 | 13,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-38 | latest | en | 0.871651 |
https://nl.mathworks.com/matlabcentral/cody/problems/2091-return-row-and-column-indices-given-2-values-which-define-a-range/solutions/1212182 | 1,590,699,726,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347400101.39/warc/CC-MAIN-20200528201823-20200528231823-00332.warc.gz | 464,594,168 | 16,063 | Cody
# Problem 2091. return row and column indices given 2 values which define a range
Solution 1212182
Submitted on 14 Jun 2017 by LY Cao
• Size: 48
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x1=magic(5); x2=3; x3=9; [R1,C1]=RCMatrixIndices(x1,x2,x3); [fr1,fc1]=find(and(x1>=x2,x1<x3)); assert(and(isequal(R1,fr1),isequal(C1,fc1)))
2 Pass
3 Pass
4 Pass
x1=magic(5); x2=3; x3=9; [R2,C2]=RCMatrixIndices(x2,x1, x3); [fr1,fc1]=find(and(x1>=x2,x1<x3)); assert(and(isequal(R2,fr1),isequal(C2,fc1)))
5 Pass
x1=magic(5); x2=3; x3=9; [R3,C3]=RCMatrixIndices(x1,x3,x2); [fr1,fc1]=find(and(x1>=x2,x1<x3)); assert(and(isequal(R3,fr1),isequal(C3,fc1)))
6 Pass
A=... [3 3 3 3;... 8 3 3 3;... 8 8 3 3;... 8 8 8 3]; lowlim=3; uplim=9; [R4,C4]=RCMatrixIndices(A,uplim,lowlim); assert(length(A(4*(C4-1)+R4))==16)
7 Pass
A=... [3 3 3 3;... 8 3 3 3;... 8 8 3 3;... 8 8 8 3]; lowlim=4; uplim=9; %extract only 8 [R5,C5]=RCMatrixIndices(A,uplim,lowlim); assert(all(A(4*(C5-1)+R5)==8))
8 Pass
A=... [3 3 3 3;... 8 3 3 3;... 8 8 3 3;... 8 8 8 3]; lowlim=4; uplim=7; [R6,C6]=RCMatrixIndices(A,uplim,lowlim); assert(and(isempty(R6),isempty(C6)))
9 Pass
A=... [3 3 3 3;... 8 3 3 3;... 8 8 3 3;... 8 8 8 3]; lowlim=2; uplim=7; [R7,C7]=RCMatrixIndices(A,uplim,lowlim); assert(and(all(A(4*(C7-1)+R7)==3),length(R7==10)))
10 Pass
A=1; lowlim=1; uplim=1; [R8,C8]=RCMatrixIndices(A,uplim,lowlim); assert(and(isempty(R8),isempty(C8)))
11 Pass
A=1; lowlim=1; uplim=2; [R9,C9]=RCMatrixIndices(A,uplim,lowlim); assert(and(R9==1,C9==1)) | 778 | 1,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-24 | latest | en | 0.344138 |
https://www.coursehero.com/file/6008168/quiz-F-answer/ | 1,496,003,363,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463611560.41/warc/CC-MAIN-20170528200854-20170528220854-00099.warc.gz | 1,101,536,629 | 26,122 | # quiz F-answer - .\$3,000,000 Plus net
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ANSWERS TO QUIZ F 1-5EDAEB 6-10CCBCC 11-15ECCAC PROBLEM 1. 2. a. Feb.15 Retained earnings ………………………………………1,250,000 (50%*5*500,000) Common stock dividend distributable…………………. .1,250,000(50%*5*500,000) Mar 20, Common stock dividend distributable……………………. .1,250,000(5%*5*500,000) Common stock…………………………………………………1,250,000 May 1 outstanding shares=500,000+500,000*50%=750,000 Retained earnings…………………………………………………375,000 Common dividend payable…………………………………375,000 (0.5*750,000) June 1 Common dividend payable…………………………………375,000 (0.5*750,000) Cash ………………………………………………………. 375,000 b. Dawls Corporation statement of Retained Earnings for the year ended December 31,2010 Retained earnings, December 31,2009………………………………………….
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Unformatted text preview: .\$3,000,000 Plus net income…………………………………………………………………. .\$1,600,000 Less : Stock dividend (50%*5*500,000)………………………………………. . \$1,250,000 Cash dividend……………………………………………………………. ... \$375,000 Retained earnings, December 31,2010…………………………………………\$2,975,000 3. 4. 1. X*4.1002=700,000 X=170,723.38=170,723 2.2009.12.31 Interest expense……………………………….8167 Interest payable………………………………….8167(700,000*7%*2/12) October 31 Interest expense………………………………. 40,833 (700,000*7%*10/12) Notes payable…………………………………121723 (170,723-40,833-8167) Interest payable………………………………….8167 (700,000*7%*2/12) Cash…………………………………………… .170,723 (700,000*7%)...
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## This note was uploaded on 11/07/2010 for the course BBA 1023 taught by Professor Luoyi during the Spring '10 term at University of Macau.
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quiz F-answer - .\$3,000,000 Plus net
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Ask a homework question - tutors are online | 617 | 2,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-22 | longest | en | 0.721521 |
https://edurev.in/course/quiz/attempt/9181_Quant--Hindi--MCQ-30/cca7a47b-a53a-462b-bf27-24544f6485cd | 1,674,996,399,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00494.warc.gz | 245,045,154 | 44,283 | Quant (Hindi) - MCQ - 30
# Quant (Hindi) - MCQ - 30
Test Description
## 30 Questions MCQ Test मात्रात्मक योग्यता(Quantitative Aptitude)- Bank Exams(Hindi) | Quant (Hindi) - MCQ - 30
Quant (Hindi) - MCQ - 30 for Quant 2023 is part of मात्रात्मक योग्यता(Quantitative Aptitude)- Bank Exams(Hindi) preparation. The Quant (Hindi) - MCQ - 30 questions and answers have been prepared according to the Quant exam syllabus.The Quant (Hindi) - MCQ - 30 MCQs are made for Quant 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Quant (Hindi) - MCQ - 30 below.
Solutions of Quant (Hindi) - MCQ - 30 questions in English are available as part of our मात्रात्मक योग्यता(Quantitative Aptitude)- Bank Exams(Hindi) for Quant & Quant (Hindi) - MCQ - 30 solutions in Hindi for मात्रात्मक योग्यता(Quantitative Aptitude)- Bank Exams(Hindi) course. Download more important topics, notes, lectures and mock test series for Quant Exam by signing up for free. Attempt Quant (Hindi) - MCQ - 30 | 30 questions in 30 minutes | Mock test for Quant preparation | Free important questions MCQ to study मात्रात्मक योग्यता(Quantitative Aptitude)- Bank Exams(Hindi) for Quant Exam | Download free PDF with solutions
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Quant (Hindi) - MCQ - 30 - Question 1
### एक कार दो शहरों के बीच की दूरी को 60 किमी/घंटा की गति से तय करता है और 40 किमी/घंटा की गति से वापस आती है. वह फिर से शहर 1 से शहर 2 की ओर मूल गति से दोगुनी गति से आता है और मूल वापसी गति से आधी से वापस चला जाता है. पूरी यात्रा के दौरान उसकी औसत गति ज्ञात कीजिये ?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 1
Quant (Hindi) - MCQ - 30 - Question 2
### एक ट्रेन बिना रुके 50 किमी / घंटा की गति से यात्रा करती है और रुकते हुए 40 किमी / घंटा की गति से यात्रा करती है. प्रति औसतन घंटे में ट्रेन कितने मिनट के लिए रूकती है?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 2
Train stop for a time which is equal to time taken to travel a distance of 10 km with a speed of 50 km/hr.
∴ because in an hour train travels 10 km less.
Time =10/50×60 = 12 min
Quant (Hindi) - MCQ - 30 - Question 3
### यदि सोहेल अपने घर से कार्यालय के लिए 16 किमी/घंटा की गति से चलता है तो वह 5 मिनट देरी से पहुँचता है. यदि वह 20 किमी/घंटा की गति से चलता है तो वह कार्यालय समय से 10 मिनट पहले तक पहुँचता है. उसके घर से कार्यालय तक की दूरी ज्ञात कीजिये?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 3
S1 = 16 km/hr
S2 = 20 km/hr
S_1/S_2 =4/5
T_1/T_2 =5/4
Time difference = 15 min
T_1=5×15=75 min
T_2=4×15=60 min
Distance = 1 × 20 = 20 km.
Quant (Hindi) - MCQ - 30 - Question 4
निर्देश : दिए गए प्रत्येक प्रश्नों में एक प्रश्न है और उसके नीचे दो कथन संख्या I और II दिए गए हैं. आपको यह निर्णय करना है कि कथन में दी जा रही सामग्री उत्तर देने के लिए पर्याप्त है. दोनों कथनों को पढ़िए और ----
P, Q और R की आयु का योग 96 वर्ष है.Q की आयु कितनी है?
I. P, R से 6 वर्ष बढ़ा है.
II. Q और R साल की कुल आयु 56 वर्ष है.
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 4
Given: P + Q + R = 96 …(i)
I. P = R + 6 …(ii)
II. Q + R = 56 …(iii)
On subtracting (iii) from (i), we get P = 40.
Putting P = 40 in (ii), we get R = 34. Putting R = 34 in (iii), we get Q = 22.
Thus, I and II both together give the answer. So, correct answer is (e).
Quant (Hindi) - MCQ - 30 - Question 5
सोनिया की वर्तमान आयु कितनी है?
I. सोनिया की वर्तमान आयु दीपक की वर्तमान आयु का पांच गुना है.
II. पांच वर्ष पहले उसकी आयु उस समय दीपक की आयु का पच्चीस गुना थी.
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 5
I. S = 5D ⇒ D = S/5 …(i)
II. S – 5 = 25 (D – 5) ⟺ S = 25D – 120 …(ii)
Using (i) in (ii), we get S = (25×S/5)-120⟺4S=120⟺S=30.
Thus, I and II both together give the answer. So, correct answer is (e).
Quant (Hindi) - MCQ - 30 - Question 6
रीना की वर्तमान आयु क्या है?
I. रीना की वर्तमान आयु उसके बेटे की वर्तमान आयु का पांच गुना बार है.
II. दो वर्ष बाद रीना की आयु उस समय उसकी बेटी की आयु का तीन गुना हो जाएगी.
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 6
I. Reena’s present age = 5 × (Her son’s present age).
II. Reena’s age 2 years hence = 3 times her daughter’s age at that time.
Clearly, data even in I and II is not sufficient to get Reena’s present age.
Quant (Hindi) - MCQ - 30 - Question 7
C की वर्तमान आयु कितनी है ?
I. तीन वर्ष पहले, A और B की औसत आयु 18 वर्ष थी.
II. C के शामिल होने के साथ, औसत आयु 22 वर्ष हो गयी.
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 7
I. 3 years ago, 1/2 (A+B)=18 ⇒ 3 years ago, (A + B) = 36
Now, (A + B) = (36 + 3 + 3) = 42 ⇒ A + B = 42 …(i)
II. Now, 1/3 (A+B+C)=22 ⇒ A + B + C = 66 …(ii)
From I and II, we get C = (66 – 42) = 24
Thus, I and II both together give the answer. So, correct answer is (e).
Quant (Hindi) - MCQ - 30 - Question 8
A और B की औसत आयु कितनी है?
I. A की आयु का 1/5 और B की आयु का ¼ का अनुपात 1:2 है
II. उनकी आयु का उत्पाद B की आयु का 20 गुना है.
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 8
Quant (Hindi) - MCQ - 30 - Question 9
एक परीक्षा में विभिन्न विषयों में विभिन्न छात्रों द्वारा प्राप्त अंकों का प्रतिशत
C द्वारा परीक्षा में प्राप्त अनुमानित कुल प्रतिशत कितना है?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 9
Total marks obtained by C in the examination
= 75% of 100 + 56% of 100 + 72% of 150 + 75% of 60 + 75% of 150 + 80% of 40
= 75 + 56 + 108 + 45 + 112.5 + 32 = 428.5
∴ Required% =428.5/600×100≈71%
Quant (Hindi) - MCQ - 30 - Question 10
एक परीक्षा में विभिन्न विषयों में विभिन्न छात्रों द्वारा प्राप्त अंकों का प्रतिशत
B और F द्वारा अंग्रेजी और गणित में प्राप्त कुल अंक का अंतर ज्ञात कीजिये?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 10
Difference of marks obtained by B & F in English = 3% of 100 = 3
Difference of marks obtained by B & F in Maths = 14% of 150 = 21
Total difference = 21 + 3 = 24
Quant (Hindi) - MCQ - 30 - Question 11
एक परीक्षा में विभिन्न विषयों में विभिन्न छात्रों द्वारा प्राप्त अंकों का प्रतिशत
E द्वारा भूगोल में प्राप्त अंक, E द्वारा हिंदी में प्राप्त अंकों का कितना प्रतिशत है?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 11
Quant (Hindi) - MCQ - 30 - Question 12
एक परीक्षा में विभिन्न विषयों में विभिन्न छात्रों द्वारा प्राप्त अंकों का प्रतिशत
D द्वारा इतिहास और भूगोल में प्राप्त अंकों का प्रतिशत ज्ञात कीजिये?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 12
Quant (Hindi) - MCQ - 30 - Question 13
एक परीक्षा में विभिन्न विषयों में विभिन्न छात्रों द्वारा प्राप्त अंकों का प्रतिशत
सभी छात्रों द्वारा विज्ञान में प्राप्त कुल औसत अंक ज्ञात कीजिये?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 13
Quant (Hindi) - MCQ - 30 - Question 14
सभी विश्वविद्यालयों में महिलाओं की कुल औसत संख्या कितनी है?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 14
Quant (Hindi) - MCQ - 30 - Question 15
विश्वविद्यालयों P और R में छात्रों की कुल संख्या (पुरुषों और महिलाओं) कितनी है?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 15
Quant (Hindi) - MCQ - 30 - Question 16
विश्वविद्यालयों P और Q में महिलाओं की कुल संख्या का विश्वविद्यालयों R और T में पुरुषों की कुल संख्या से अनुपात ज्ञात कीजिये?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 16
Quant (Hindi) - MCQ - 30 - Question 17
विश्वविद्यालयों Q में पुरुषों की कुल संख्या कुल छात्रों की संख्या (पुरुष और महिलाओं) का कितना प्रतिशत है?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 17
Quant (Hindi) - MCQ - 30 - Question 18
यदि विश्वविद्यालयों T में पुरुषों की कुल संख्या 50% बढ़ जाती है, विश्वविद्यालय में छात्रों(पुरुषों और महिलाओं) की कुल संख्या कितनी हो जाएगी?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 18
Quant (Hindi) - MCQ - 30 - Question 19
394 का 57% - 996 का 2.5% = ?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 19
Approx value
= (50 + 7)% of 400-1/40×1000
= 200 + 28 - 25 = 203 ≈ 200
Since we have taken excess value in calculation, we will take less than the calculated value as the approximate value.
Quant (Hindi) - MCQ - 30 - Question 20
96.996 × 9.669 + 0.96 = ?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 20
Approx value = 97 × 9.7 = 940.9 ≈ 940
Quant (Hindi) - MCQ - 30 - Question 21
3/5×1125/1228×7= ?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 21
Approx value= 3/5×1125/1227×7= 3.85=4
Quant (Hindi) - MCQ - 30 - Question 22
(√339 ×25)÷30= ?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 22
√339≈18
Then (18×25)÷30=450/30=15
Quant (Hindi) - MCQ - 30 - Question 23
(638 + 9709 - 216) ÷ 26 = ?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 23
Approx value = (640 + 9700 - 200) ÷ 26
= 10140 ÷ 26 ≈ 390
Quant (Hindi) - MCQ - 30 - Question 24
श्रृंखला में गलत नंबर का पता लगाइए:
32, 36, 41, 61, 86, 122, 171, 235
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 24
Quant (Hindi) - MCQ - 30 - Question 25
श्रृंखला में गलत नंबर का पता लगाइए:
3, 4, 9, 22.5, 67.5, 202.5, 810
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 25
There are two sequences (3, 9, 67.5, 810) and (4, 22.5, 202.5)
Pattern is: (1st term × 3), (2nd term × 7.5), (3rd term × 12) for the first sequence and (1st term × 5), (2nd term × 9) and so on for the second sequence.
Quant (Hindi) - MCQ - 30 - Question 26
श्रृंखला में गलत नंबर का पता लगाइए:
1, 2, 8, 33, 148, 760, 4626
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 26
Quant (Hindi) - MCQ - 30 - Question 27
श्रृंखला में गलत नंबर का पता लगाइए:
3, 8, 18, 46, 100, 210, 432
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 27
2nd term = (1st term × 2 + 2) = 3 × 2 + 2 = 8;
3rd term = (2nd term × 2 + 4) = 8 × 2 + 4 = 20;
4th term = (3rd term × 2 + 6) = 20 × 2 + 6 = 46
5th term = (4th term × 2 + 8) = 46 × 2 + 8 = 100 and so on.
∴ 18 is wrong
Quant (Hindi) - MCQ - 30 - Question 28
श्रृंखला में गलत नंबर का पता लगाइए:
789, 645, 545, 481, 440, 429, 425
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 28
Quant (Hindi) - MCQ - 30 - Question 29
निर्देश (29-30): दिया गया वृत्तीय-ग्राफ एक विशेष वर्ष के दौरान विभिन्न खेल पर एक देश का खर्च दर्शाता है. ग्राफ को ध्यानपूर्वक पढ़े और नीचे दिए गये प्रश्नों का उत्तर दें .
टेनिस के लिए खर्च कुल खर्च का प्रतिशत कितना है?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 29
Quant (Hindi) - MCQ - 30 - Question 30
दिया गया वृत्तीय-ग्राफ एक विशेष वर्ष के दौरान विभिन्न खेल पर एक देश का खर्च दर्शाता है.
हॉकी पर किया गया कुल खर्च गोल्फ़ पर किये गये कुल खर्च से कितना प्रतिशत अधिक है?
Detailed Solution for Quant (Hindi) - MCQ - 30 - Question 30
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# Statistical methods in economics
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Code Completion Credits Range Language
B1M16STA Z,ZK 5 2P+2S Czech
Garant předmětu:
Lecturer:
Tutor:
Supervisor:
Department of Economics, Management and Humanities
Synopsis:
Basic Concepts. Statistical series. Assortment. Distributions of frequencies. One-dimensional descriptive characteristics. Measures of variables, coefficient of skewness, coefficient of excess. Points estimates of basic characteristics. Interval estimates of basic characteristics. Hypothesis testing of basic characteristics. Individual indexs number. Aggregative indexs. Variable-structure indexs. Multifactor indexs . Correlation and regression, Basic Concepts. Measurement of dependence intensity. Time series, concepts, qualities. Chronological average . Time series - trends and extrapolation.
Requirements:
Syllabus of lectures:
1.Basic Concepts. Statistical series. Assortment. Distributions of frequencies.
2.One-dimensional descriptive characteristics. Measures of central tendency.Measures of variables
3.Statistical induction, theoretic distributions..
4.Points and Interval estimates of basic characteristics
5.Hypothesis testing of basic characteristics.
6. ANOVA, one, two, multi - factors
7. Measurement of dependence intensity, covariance and corelation
8. Correlation and regression, basic concepts
9. Multiply regression and correlation, basic concepts.
10. Binary explaining variables
11. Linear models, Probit and logit
12. Time series, concepts, qualities. Chronological average, trends and extrapolation
13. Time series, autocorrelation, heteroscedasticity
14. Summary, reserve
Syllabus of tutorials:
1.Basic Concepts. Statistical series. Assortment. Distributions of frequencies.
2.One-dimensional descriptive characteristics. Measures of central tendency.
3.One-dimensional descriptive characteristics. Measures of variables, coefficient of skewness, coefficient of excess.
4.Points estimates of basic characteristics.
5.Interval estimates of basic characteristics
6.Hypothesis testing of basic characteristics.
7.Individual indexs number. Aggregative indexs.
8.Variable-structure indexs. Multi - factors indexs.
9.Correlation and regression, basic Concepts.
10.Measurement of dependence intensity.
11.Multiply regression and correlation. Basic concepts.
12.Time series, concepts, qualities. Chronological average .
13.Time series - trends and extrapolation.
14.Collective consultation.
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1.Daniel, Terrel: Business Statistics. Boston, Houghton Miffin Company,1989
2.Freed N.: Understanding Business Statistics. Lulu.com 2008
3.Van Matre, Gilbreath. Statistics for Business and Economics. 1983
4.Anderson, Sweeney, Williams. Statistics for Business and Economics. 2011
5.Jeffrey M. Wooldridge. Introductory Econometrics: A Modern Approach. 2002
Note:
Further information:
https://moodle.fel.cvut.cz/courses/B1M16STA
No time-table has been prepared for this course
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Data valid to 2024-06-16
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# NEET Important Chapter - Electrostatic Potential and Capacitance
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## Electrostatic Potential and Capacitance an Important Concept for NEET
Last updated date: 15th Jul 2024
Total views: 238.5k
Views today: 6.25k
In this chapter we understand how we move the charge in this field and what amount of work is done to move the charge against this field and in what form the work done is stored in the system and what is electrostatic potential.
In this chapter we first define electrostatic potential and then potential due to a point charge, system of charges, electric dipole and potential energy of charges and dipole in an external field. We got introduced to new concepts like equipotential surface, electrostatics of conductors, dielectric and polarization etc. and define electrostatic potential energy.
In this chapter students also get to learn about the capacitors and capacitance and the effect of dielectric medium on it and their combination circuits and electrostatic potential and capacitance all formulas.
Now, let us move on to the important concepts and formulae related to NEET 2022 exams along with a few solved examples.
### Important Topics of Electrostatic Potential and Capacitance
• Electrostatic Potential and Capacitance Notes
• Equipotential Surface
• Capacitor and Capacitance
• Energy Stored in a Capacitor
• Series Combination of Capacitors
• Electrostatic Potential
• Capacitors in the Parallel Combination
• Parallel Plate Capacitor Derivation
• Potential Energy of System of Charges
### Important Concepts of Electrostatic Potential and Capacitance
Name of the Concept Key Points of the Concept Electrostatic potential The amount of work done to bring a unit positive charge from infinite to a given position in the electric field is known as electric potential.V = W/qElectric potential due to a point charge q$V=\dfrac{Kq}{r}$here $K=\dfrac{1}{4\pi\varepsilon_0}$S.I. unit - Joule/Coulomb or volt Electric potential difference The amount of work done to bring a unit positive charge from one position to the other position in the electric field is known as Electric potential difference.S.I. unit - Joule/Coulomb or volt Electrostatic potential energy The electric potential energy is the required energy to move a charge against in electric fieldPotential energy of a single charge - U = qVHere V is the potential Potential energy of a system of two charges-U(r) = $\frac{Kq_1q_2}{r}$here $K=\dfrac{1}{4\pi\varepsilon_0}$S.I. unit - Joule Electrostatic with conductor Electrostatic field is zero inside a conductor because there is no charge inside the conductor.All charges are at the surface of the conductor and at every point on the surface of the charged conductor there is normal electric field only.Electric potential is constant or the same value inside the conductor and surface of the conductor.Electrostatic shielding - there is no influence of outside electric charge and field in the cavity inside the conductor. The electric field inside the conductor is always zero so any cavity in conductor is shielded from outside and it is known as electrostatic shielding Polarisation The dipole moment per unit volume is called polarisation and denoted by P Capacitance It is the ratio of charges given to the conductor and its potential.Capacity of conductor = C = Charge / potential Unit - faradIt depends on shape and size of conductor and surrounding and presence of other charge objects near the conductor. 8. Parallel plate capacitor It consists of two metal plates parallel to each other separated by a medium.Used to accumulate electric charge or increase the electrical capacity.Capacitance of capacitor is increased when it is placed in dielectric medium. Combination of capacitors Capacitor in series - In series combination, charges are same on the plate of each capacitor and total potential drop is sum of individual potential drop of capacitors.Capacitor in parallel - In parallel combination, the same potential difference is applied across all capacitors that are connected and charges are different. 9. Van de graaff generator It is a device which is used for accelerating the charged particles.
### List of Important Formulae
Sl. No Name of the Concept Formula 1. Work done by external force in moving a charge q from Q to P $W=\int_{Q}^{P}\overrightarrow{F}.\overrightarrow{dr}$ 2. Potential due to a point charge $V=\dfrac{Kq}{r}$ here $K=\dfrac{1}{4\pi\varepsilon_0}$ 3. Potential due to an electric dipole $V(r)=\dfrac{K\overrightarrow{P}.\hat{r}}{r^2}$here $K=\dfrac{1}{4\pi\varepsilon_0}$ 4. Potential due to a system of charges $\frac{1}{4π\varepsilon_o}(\frac{q_1}{r_1}+\frac{q_2}{r_2}+....+\frac{q_n}{r_n})$ 5. Relation between electric field and potential $E=-\frac{\text{d}V}{\text{d}r}$ 6. Potential energy of system of charges (3 charge) $\frac{1}{4π\varepsilon_o}(\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{23}}+\frac{q_1q_3}{r_{13}})$ 7. Potential energy of a dipole in an eternal field $U(\theta) = -PE\cos\theta =-\overrightarrow{P}.\overrightarrow{E}$ 8. Capacitance of parallel plate capacitor $c=\dfrac{A\varepsilon_0}{d}$ 9. Equivalent Capacitance in series combination $\frac{1}{C_S}=(\frac{1}{C_1}+\frac{1}{C_2}+....+\frac{1}{C_n})$ 10. Equivalent Capacitance in parallel combination $C_P=C_1+C_2+....+C_n$
### Solved Examples
1. A parallel plate capacitor with air between the plates has a capacitance of 8pF (1pF = $10^{-12}$ F. What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 6?
Sol:
Given, Capacitance, C = 8pF = $c = 8\times10^{-12}$F
As per the question In the first case, the parallel plates are at a distance ‘d’ and
We know that air has dielectric constant, k = 1 so capacitance in first case
Capacitance - $c=\dfrac{A\varepsilon_0}{d}$
Here A = area of the plates
ϵo = permittivity of free space
Now, according to the question if the distance between the plates is halved then
$d_1$= d/2
Given, dielectric constant of the substance,$K_1$ = 6
Hence, the capacitance of the capacitor-
$c_1=\dfrac{K_1A\varepsilon_0}{d_1} = \dfrac{12A\varepsilon_0}{d}$
By solving this we get
C1 = 96 PF
Key point: If a parallel plate capacitor is placed in any medium with dielectric constant k then its capacitance is increased by k times.
2. A regular hexagon of the side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the center of the hexagon.
Sol: The given figure shows a regular hexagon with charges at each vertex.
Here, given - q = 5 µC = $5\times10^{-6}$
AB =BC = CD = DE = EF = FA = side of hexagon = 10 cm.
By the diagram the distance of the vertices from the center O, r = 10 cm.
The electric potential at point O,
$V = \dfrac{6q}{4\pi\varepsilon_or}$
Here,
$k = \dfrac{1}{4\pi\varepsilon_o}$
εo = Permittivity of free space and
$V = \dfrac{6\times 9✕10^6✕5 ✕10^-6}{0.1} = 2.7 × 10^6 V.$
Trick: If we have symmetry in the shape then we just determine the potential due to a single charge and then multiply it with the total number of charges there are in the system.
### Previous Year Questions from NEET Paper
1. The equivalent capacitance of the combination shown in the figure is- (NEET 2021)
(a) 3C
(b) 2C
(c) C/2
(d) 3C/2
Sol: In the above figure the capacitor in the right side mesh is not contribute in the equivalent capacitance because there is same potential across the both terminal of the capacitor so no current flow through it therefore only capacitors which is in the left mesh is considered and they are in parallel combination so equivalent capacitance would be
Ceq = C + C = 2C
Option (b) is the correct option.
Key point - Current is flow due to potential difference.
2. In a certain region of pace ith volume 0.2 $m^{3}$ , the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is (NEET 2021)
(a) zero
(b) 0.5 N/C
(c) 1 N/C
(d) 5 N/C
Sol: We know the relation between potential and electric field
$E=-\frac{\text{d}V}{\text{d}r}$
In the given question potential is found to be 5 V throughout the volume (region) so change in potential is zero therefore electric field is also zero in this region.
Option (a) is the correct option.
Trick - Electric field is equal to the potential gradient with negative sign.
### Practice Questions
1. If potential (in volts)in a region is expressed as V(,y,z) = 6y - y + 2yz, the electric field (in N/C) at point (1,1,0) is ? (Ans: -$-(6\hat{i}+5\hat{j}+2\hat{k})$)
2. Two capacitors with capacity $C_{1}$ and $C_{2}$ , when connected in series, have capacitance $C_{S}$ and when connected in parallel have capacitance $C_{P}$. find the relation between capacitances ? (Ans: $C_{P}$ $C_{S}$ = $C_{1}$ $C_{2}$)
### Conclusion
In this article we have provided important information regarding the chapter electrostatic potential formula and capacitance such as important concepts, formulae, etc.. Students should work on more solved examples for securing good grades in the NEET exams.
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## FAQs on NEET Important Chapter - Electrostatic Potential and Capacitance
FAQ
1. What is the weightage of the electrostatic potential and capacitance in NEET?
Nearly 1-2 questions arise in the exam from this chapter covering about 5 marks which makes about 2% of the total marks.
2. What are the key points that need to be practiced for solving questions from electrostatic potential and capacitance ?
Students should practice as much as questions from the electrostatic potential and capacitance to learn the formula of electrostatic potential and to increase the speed of solving circuits containing combinations of capacitors.
3. How is electric potential related to capacitance?
Capacitance is the characteristics of a capacitor that determines how much amount of charge can be stored in the capacitor and electric potential is the amount of work done to move a charge from infinite to any given point.
The relation between capacitance and potential is Q = CV
Here C is the capacitance, V is potential and Q is stored charge.
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NEET Blogs | 3,550 | 14,057 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-30 | latest | en | 0.836451 |
https://math.stackexchange.com/questions/2433366/if-you-know-all-angles-and-the-area-of-the-triangle-how-do-you-find-the-sides | 1,575,658,729,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540490743.16/warc/CC-MAIN-20191206173152-20191206201152-00340.warc.gz | 450,524,799 | 31,254 | # If you know all angles and the area of the triangle, how do you find the sides?
If we know the angles and area of the triangle, what would be the formula to find the side lengths? For example, we know the area is 400 ft and the angles are 40, 30, and 110 degrees, how could we find the area?
I've been stumped with this question and could really use a suggestion to help me figure out. Thanks
• the area is $400$ square ft – Raffaele Sep 17 '17 at 19:48
• "the area is ... how could we find the area?" Did you mean you want to find the perimeter, or the three side lengths? – aschepler Sep 18 '17 at 2:52
Hint...use the formula for the area of the triangle in the form $\frac 12ab\sin C$ to get $ab$, $bc$ and $ca$ then you can get each side...
• This approach leads, if we like, to a formula for each side in terms of the known quantities. For example, $a = \sqrt{2M\sin\alpha\csc\beta\csc\gamma}$ where $M$ is the area of the triangle. – Jason Zimba Sep 17 '17 at 18:13
Use this formula for the area to find the circumradius $R$: \begin{align} S_{\triangle ABC}&=2\,R^2\,\sin\alpha\,\sin\beta\,\sin\gamma ,\\ a&=2\,R\,\sin\alpha,\dots \end{align} | 354 | 1,154 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-51 | latest | en | 0.860944 |
http://www.speedsolving.com/forum/showthread.php?9974-4x4x4-is-it-possible-without-parity&p=142233&viewfull=1 | 1,369,005,433,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698141028/warc/CC-MAIN-20130516095541-00023-ip-10-60-113-184.ec2.internal.warc.gz | 698,400,697 | 14,855 | 1. Really? Could you tell me what they are because i'm watching Dan Brown >.> and i have these algs
1) 2U*, 2L*, 2U, 2l, 2U, 2L*, 2U*, R, U', L, 2U, R', U, R, L', U', L, 2U, R', U, L', U
2) 2R*, 2B, 2U, L*, 2U, R'*, 2U, R*, 2U, 2F, R*, 2F, L'*, 2B, 2R*
3) 2r, 2U, 2r, 2U*, 2r, 2u
4) 2U*, 2L*, 2U, 2l, 2U, 2L*, 2U* F', U', F, U, F, R', 2F, U, F, U, F', U', F, R
5) 2r, 2U, 2r, 2U*, 2r, 2u
and of course there all used for different situations
2. Originally Posted by ISuckAtCubing
both parity are easy, the oll one almost dont even need to memorize
uhh, no. an oll parity fix is an awful alg, most people can't do it in under 5 seconds.
and no, there is no way to avoid it. just learn the two algs.
3. Originally Posted by TsColin
Really? Could you tell me what they are because i'm watching Dan Brown >.> and i have these algs
1) 2U*, 2L*, 2U, 2l, 2U, 2L*, 2U*, R, U', L, 2U, R', U, R, L', U', L, 2U, R', U, L', U
2) 2R*, 2B, 2U, L*, 2U, R'*, 2U, R*, 2U, 2F, R*, 2F, L'*, 2B, 2R* <-OLL Parity
3) 2r, 2U, 2r, 2U*, 2r, 2u
4) 2U*, 2L*, 2U, 2l, 2U, 2L*, 2U* F', U', F, U, F, R', 2F, U, F, U, F', U', F, R
5) 2r, 2U, 2r, 2U*, 2r, 2u <- PLL Parity
I really hate that notation.
4. If you don't like parity, learn cage. THat's where I started. Eventually, you'll want to learn those two algs (only the oll one should give ya trouble). Redux is faster than cage.
5. thx, ill just learn them the hard way :P
6. i tried cage once, but i wasnt used to it, maybe ill try cage after i learn the evil algs
7. Originally Posted by TsColin
i tried cage once, but i wasnt used to it, maybe ill try cage after i learn the evil algs
PLL parity isn't bad but yes, I'd say OLL parity is the most "evil" alg I use.
8. Originally Posted by Vulosity
I MIGHT and PROBABLY be wrong, but if you use the cage method, you won't get parity.
Parity's not much of a big deal. Two algs. That's it.
lol i have pll parity, oll parity, and double parity (pure versions for oll and dp) as part of my system...and also another parity case. there IS parity in cage...depends on which cage system though xD the old system that i used had no parity..
Originally Posted by aznmortalx
If you don't like parity, learn cage. THat's where I started. Eventually, you'll want to learn those two algs (only the oll one should give ya trouble). Redux is faster than cage.
lol you should think about what you say before saying it. xD you should say that generally redux is faster than cage because i'm much faster at cage than redux =P
Originally Posted by Kian
Originally Posted by TsColin
i tried cage once, but i wasnt used to it, maybe ill try cage after i learn the evil algs
PLL parity isn't bad but yes, I'd say OLL parity is the most "evil" alg I use.
i have more evil algs =(
9. Originally Posted by TsColin
2) 2R*, 2B, 2U, L*, 2U, R'*, 2U, R*, 2U, 2F, R*, 2F, L'*, 2B, 2R*
this one is nicer... and has proper notation:
r U2 x r U2 r U2 r' U2 l U2 r' U2 r U2 r' U2 r'
10. Learn cage if you don't like parities! I have had several solves around 2:15 with cage using my rubiks 5x5, and I just started. The algorithms are usually commutators/conjugates! I actually figured out all the algorithms I use. | 1,197 | 3,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2013-20 | latest | en | 0.854598 |
https://colindcarroll.wordpress.com/2012/01/21/fractal-followup-followup/ | 1,501,173,643,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549428325.70/warc/CC-MAIN-20170727162531-20170727182531-00225.warc.gz | 635,189,601 | 35,055 | # Fractal followup followup
Again, after yesterday’s post (which was a response to the post of two days ago), I found a few more cool things with this method of generating fractals. In particular, we take our point x and a random vertex v, then move our point to ax + bv. To get the Sierpinski carpet, we have a = 2, b = 2. To get the sets from yesterday, I used a = 1/(n-1), b = (n-2)/(n-1), where n was the number of vertices. Today, I share some pictures using a = k/(n-1), b = (n-k-1)/(n-1). I am not sure whether each of these is a fractal (though I suspect they are), but there is something going on (note- Leonid Kovalev pointed out yesterday that Jeremy Tyson and Jang-Mei Wu have a paper (pdf link) investigating sets generated like this):
5 vertices, with k = 2. Used 500,000 points for this one, I think most of the pentagon is meant to be in the set...
6 vertices, with k = 2.
7 vertices, with k = 2. Looks a little like the chainring on my bike...
16 vertices, k = 1
16 vertices, k = 2
16 vertices, k = 3
16 vertices, k = 4 | 315 | 1,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.903958 |
http://godplaysdice.blogspot.com/2007/10/huygens-principle.html | 1,532,167,838,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592475.84/warc/CC-MAIN-20180721090529-20180721110529-00166.warc.gz | 156,847,067 | 13,218 | ## 12 October 2007
### Huygens' principle
A weird fact I came across yesterday: Huygens' principle for the wave equation.
When you throw a pebble on the surface of some water, circular waves propagate outward from that point. However, inside the circular wavefront there will still be some sort of disturbance.
Light, in three dimensions, doesn't work the same way. If I have a pulse of light that lasts one second, then one second after that pulse stops, then the wave resulting from that pulse will be supported on the annulus from one light-second to two light-seconds from me. There will not be any sort of wave going on closer than one light-second from me.
I doubt this was known to Huygens, simply because he lived in the seventeenth century when they didn't have partial differential equations. They had waves, though, and the page I linked to above implies that Huygens knew that the leading edge of a wave travels at a constant speed, usually denoted by c; the most important example of waves is light waves, so c is now reserved for the speed of light. (I suspect he was aware of ripples in water; it's less obvious that there aren't analogous ripples in three dimensions, though, simply because there don't seem to be three-dimensional phenomena where the wave travels at a reasonable speed that one could observe. So observation is not particularly useful here, and one simply has to trust the symbols.)
Okay, this seems reasonable... things in different numbers of dimensions behave differently. But in four, six, eight, ... dimensions, solutions to the wave equation behave like the two-dimensional case (the surface of water), in five, seven, nine, ... dimensions they behave like the three-dimensional case (like light). I wouldn't have been surprised to learn that the behavior is different in low-dimensional space than in high-dimensional space. Or even that the behavior was different in "medium"-dimensional space -- it's similar to how things work in the theory of manifolds, where one- and two-dimensional manifolds are basically trivial to study, five-dimensional and higher manifolds aren't that hard, but three and four dimensions are difficult. (I'm vastly oversummarizing here; this isn't my area, and I'm just going on things I hear in the halls.) But who would think it would depend on the parity?
It seems that the dependence on the parity falls out of the series solutions to the spherically symmetrized wave equation, and this equation includes the dimension as a coefficient. I'm having a bit of trouble parsing the page I linked to (although I'm not working too hard on it). The author comments:
It would be interesting to work out the connections between Huygens' Principle and the zeta function (whose value can only be given in simple closed form for even arguments) and the Bernoulli numbers (which are non-zero only for even indices).
This isn't something I would have thought of.
John Armstrong said...
Well, remember that waves are all really about Fourier transforms -- unitary representation theory.
So what sort of theory talks about spherically symmetric propagation in \$n\$-dimensional space? The Fourier theory of \$SO(n)\$! And we know that \$SO(2k)\$ and \$SO(2k+1)\$ behave very differently.
Charles said...
Well, there is actually one big three dimensional wave phenomenon that Huygen's could have (and did) study: sound waves. He also believed that light was a wave (an unfashionable belief at the time, overshadowed by Newton's opposition), and so he had two examples of waves in three dimensions that didn't ripple like waves in two dimensions do.
Isabel said...
Charles,
I thought of that, but it's a lot harder to see the propagation of sound waves than of water waves. Can you really be sure, purely from low-technology observation, that sound waves don't ripple?
Charles said...
The fact that the universe is relatively quiet does it. If they rippled, you'd hear the same sound over and over again several times, which we don't. Instead, we hear something once and then it is done. (this is ignoring the case where there is something for the wave to reflect off of and give an echo)
Mohan K.V said...
Nice Post :-) ! You should definitely take a look at this paper: | 915 | 4,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.970925 |
https://www.doubtnut.com/qna/19037350 | 1,723,001,646,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640667712.32/warc/CC-MAIN-20240807015121-20240807045121-00309.warc.gz | 586,317,580 | 37,071 | # One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are introduced into a vessel of volume 1 litre. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is
A
1.5
B
2.5
C
2
D
4
Video Solution
Text Solution
Verified by Experts
## Ideal gas equation is given by $pV=nRT$ " "..(i) For oxygen ,$P=1atm,V=1L,n={n}_{0}$ Therefore,Eq (i) becomes $\therefore$ " "$1×1=n{O}_{2}RT⇒{n}_{{O}_{2}}=\frac{1}{R}T$ and $0.5×2={n}_{{N}_{2}}RT⇒{n}_{{N}_{2}}=\frac{1}{R}T$ For mixture of gas, ${P}_{mix}{V}_{mix}={n}_{mix}RT$ Here, ${n}_{mix}={n}_{{O}_{2}}+{n}_{{N}_{2}}$ $\therefore$ " "$\frac{{P}_{mix}{V}_{mix}}{RT}=\frac{1}{R}T+\frac{1}{R}T$ $⇒{P}_{mix}{V}_{mix}=2$
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are introduced into a vessel of volume 1 litre. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is
A1.5
B1
C2
D4
• Question 2 - Select One
## One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are introduced into a vessel of volume 1 litre. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is
A1.5
B2.5
C23
D4
• Question 3 - Select One
## One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are introduced into a vessel of volume 1 litre. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is
A1. 5
B2. 5
C2
D4
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 815 | 2,619 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.840457 |
https://www.flightpedia.org/convert/175-grams-liter-to-ounces-gallon-us.html | 1,656,949,655,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104432674.76/warc/CC-MAIN-20220704141714-20220704171714-00534.warc.gz | 836,989,032 | 5,191 | Convert 175.0 Grams per Liter to Ounces per Gallon (US)
175.0 Grams per Liter (g/l) = 23.3671 Ounces per Gallon (US) (oz/gal (U.S. fluid)) 1 g/l = 0.133526 oz/gal (U.S. fluid) 1 oz/gal (U.S. fluid) = 7.48915 g/l
• Q: How many Grams per Liter in a Ounce per Gallon (US)?
The answer is 7.48915 Ounce per Gallon (US)
• Q: How do you convert 175 Gram per Liter (g/l) to Ounce per Gallon (US) (oz/gal (U.S. fluid))?
175 Gram per Liter is equal to 23.3671 Ounce per Gallon (US). Formula to convert 175 g/l to oz/gal (U.S. fluid) is 175 / 7.489151707
• Q: How many Grams per Liter in 175 Ounces per Gallon (US)?
The answer is 1,310.6 Grams per Liter
Others Density converter
(Please enter a number) | 260 | 701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-27 | latest | en | 0.605408 |
https://modulomathy.com/2012/04/05/math-trivia-53-solution/ | 1,675,362,831,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00378.warc.gz | 407,292,767 | 19,019 | ## #math trivia #53 solution
Ten immediate neighbors of 53 aren’t prime:
48 49 50 51 52 … 54 55 56 57 58
The gaps between 47 and 53 and between 53 and 59 tie the “record” for the longest gap between successive primes so far, also achieved by (23,29) and (31,37). But this is the first time two gaps this long have been adjacent.
According to the Prime Number Theorem, the average gap between primes near a given number n is about ln n where ln n is the natural logarithm of n. (The Prime Number Theorem states that the density of primes near n is about 1/ln n; the average gap is the inverse of the density.) Near 53, the average gap according to this formula would be just under 4, so a gap of 6 wouldn’t be unusual.
Although 53 has the most composite neighbors of any prime so far, it doesn’t quite meet the definition of a “lonely prime” given by The On-Line Encyclopedia of Integer Sequences. In that definition, a prime sets a new record for “loneliness” if and only if the gaps on both sides exceed the corresponding gaps for the previous record-holder. The sequence of primes with this property starts with 2, 3 (gaps of 1 and 2), 7 (gaps of 2 and 4), 23 (gaps of 4 and 6), and 89 (gaps of 6 and 8). As noted in the reference, with gaps of 6 and 6, 53 exceeds 23’s gaps on one side but not the other. But it still seems pretty lonely to me. | 374 | 1,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-06 | latest | en | 0.917553 |
http://blog.csdn.net/qq_33048603/article/details/51851484 | 1,518,911,440,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891808539.63/warc/CC-MAIN-20180217224905-20180218004905-00276.warc.gz | 44,272,878 | 18,618 | # POJ 3069
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Description
Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep track of his forces, Saruman distributes seeing stones, known as palantirs, among the troops. Each palantir has a maximum effective range of R units, and must be carried by some troop in the army (i.e., palantirs are not allowed to “free float” in mid-air). Help Saruman take control of Middle Earth by determining the minimum number of palantirs needed for Saruman to ensure that each of his minions is within R units of some palantir.
Input
The input test file will contain multiple cases. Each test case begins with a single line containing an integer R, the maximum effective range of all palantirs (where 0 ≤ R ≤ 1000), and an integer n, the number of troops in Saruman’s army (where 1 ≤ n ≤ 1000). The next line contains n integers, indicating the positions x1, …, xn of each troop (where 0 ≤ xi ≤ 1000). The end-of-file is marked by a test case with R = n = −1.
Output
For each test case, print a single integer indicating the minimum number of palantirs needed.
Sample Input
0 3
10 20 20
10 7
70 30 1 7 15 20 50
-1 -1
Sample Output
2
4
Hint
In the first test case, Saruman may place a palantir at positions 10 and 20. Here, note that a single palantir with range 0 can cover both of the troops at position 20.
In the second test case, Saruman can place palantirs at position 7 (covering troops at 1, 7, and 15), position 20 (covering positions 20 and 30), position 50, and position 70. Here, note that palantirs must be distributed among troops and are not allowed to “free float.” Thus, Saruman cannot place a palantir at position 60 to cover the troops at positions 50 and 70.
1. 题意：英语伤不起=-=， 大概就是 给出一个n 一个 r n是一条直线上面有n个点 的坐标(相当于数轴) 在这些点上面放一个灯 给出的r 就是灯的照射范围，让你求出 最少给出多少个灯才能把一条线照满
2. 思路: 按照贪心的思路，充分的利用r到每个点的区间
import java.util.Arrays;
import java.util.Scanner;
public class Main {
private static int n ,r,current=0;
private static int[] array;
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
current =0;
r = scanner.nextInt();
n = scanner.nextInt();
if (n==-1&&r==-1)
break;
array = new int[n];
for (int i = 0; i < array.length; i++) {
array[i] = scanner.nextInt();
}
int i=0;
Arrays.sort(array);//先对这n个点排序
while (i<n) {
int x = array[i++];
while (i<n&&x+r>=array[i]) //相当于第一个点加上r(照射范围）大于下一个点，这样下一个点就不用放灯了，直接i++，知道不能照到为止
i++;
int p = array[i-1];
while (i<n&&p+r>=array[i]) //这就是下一个点了 同理以上
i++;
current ++;
}
System.out.println(current);
}
scanner.close();
}
}
3
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最新评论 | 928 | 2,727 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-09 | latest | en | 0.800536 |
https://www.nagwa.com/en/videos/234125757368/ | 1,722,751,678,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640389685.8/warc/CC-MAIN-20240804041019-20240804071019-00472.warc.gz | 720,399,730 | 35,947 | Question Video: Finding the Integral of a Rational Function using Factorization | Nagwa Question Video: Finding the Integral of a Rational Function using Factorization | Nagwa
# Question Video: Finding the Integral of a Rational Function using Factorization Mathematics • Second Year of Secondary School
## Join Nagwa Classes
Determine β« (π₯Β² β 19π₯ + 70)/(π₯ β 14) dπ₯.
03:26
### Video Transcript
Determine the indefinite integral of π₯ squared minus 19π₯ plus 70 over π₯ minus 14 with respect to π₯.
In order to determine this integral, we will first simplify the integrand. Note that 70 is divisible by 14, as 70 divided by 14 equals five. Therefore, π₯ squared minus 19π₯ plus 70 factorizes to π₯ minus 14 multiplied by π₯ minus five, as negative 14 multiplied by negative five equals 70, and negative 14 add negative five equals negative 19. So, the numerator of the integrand can be rewritten as π₯ minus 14 multiplied by π₯ minus five. We can cancel out the factor of π₯ minus 14 in the numerator and denominator to simplify the integrand to π₯ minus five. We are now required to compute the indefinite integral of π₯ minus five with respect to π₯.
In order to do this, recall that the integral of the sum or difference of two functions is the sum or difference of the integrals of the two functions. Therefore, the integral we are asked to evaluate can be rewritten as the integral of π₯ with respect to π₯ minus the integral of five with respect to π₯. Now, recall that in order to find the indefinite integral of the term π multiplied by π₯ to the πth power with respect to π₯, where π and π are constants such that π is not equal to negative one, we keep the coefficient π as it is, increase the exponent π by one, and divide by the new exponent. We then add the constant of integration π.
Therefore, in order to find the indefinite integral of π₯ with respect to π₯, we keep the coefficient one as it is, increase the exponent, one, by one to obtain a new exponent of two, and divide by the new exponent. We then add the constant of integration which we have denoted by π one. Letting π equal zero in the formula described earlier, we obtain that for all constants π, the indefinite integral of π with respect to π₯ is ππ₯ plus π, where π is the constant of integration. Therefore, the indefinite integral of the constant five with respect to π₯ is five π₯ plus π two, where π two denotes the constant of integration.
So, the integral given to us in the question evaluates to π₯ squared over two plus π one minus five π₯ plus π two. Removing the parentheses, we obtain π₯ squared over two plus π one minus five π₯ minus π two. We can collect together the constants π one and negative π two to form the constant π. So, the integral in question evaluates to π₯ squared over two minus five π₯ plus π, where π is the constant of integration.
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• Realistic Exam Questions | 895 | 3,218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-33 | latest | en | 0.864993 |
https://www.physicsforums.com/threads/average-momentum-integral.103999/ | 1,544,738,854,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825112.63/warc/CC-MAIN-20181213215347-20181214000847-00637.warc.gz | 1,004,417,668 | 12,541 | # Homework Help: Average Momentum Integral
1. Dec 13, 2005
### eck
I have the physics question along with the solution, but in the solution I don't understand how they evaluated the integral. I can't get my brower to preview TeX input, so I'm going to leave it without formatting, but you can find the problem http://web.mit.edu/8.05/probsets/ps1_v1.pdf". The problem I am looking at is number one. I've also got the problem and solution posted here (w/o formatting) but it's kind of hard to read:
Problem
--------------------
A particle's coordinate space wavefunction is square-integrable and real up to an arbitrary multiplicative phase:
psi(x) = exp(i * alpha) phi (x)
with alpha real and constant and phi(x) real. Prove that its average momentum is zero.
Solution
-------------------
Setting up the integral is easy, and you can pull out a couple constants. So you have an infinite integral with this inside:
dx exp(-i alpha) phi(x) exp(i alpha) d/dx[phi(x)]
Somehow, in the solution, they pull out 1/2 and leave the following in the integral:
dx d/dx[phi(x)^2]
When I look at it, I see the exponentials cancelling, but I don't understand where the 1/2 comes from and how the first phi(x) gets pulled into the derivative.
Can anyone shed any insight on how this integral was simplified?
Last edited by a moderator: Apr 21, 2017
2. Dec 13, 2005
### eck
I was looking at the problem some more, and all of a sudden it hit me. It's kind of embarrassing that I didn't see it before. If anyone else looks at it... nothing tricky is involved. It's integration by parts, but it's so obvious I didn't even see it. | 410 | 1,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.955917 |
https://pressbooks.openedmb.ca/calculusvolume2seconduniversityofmanitobaedition/chapter/determining-volumes-by-slicing/ | 1,723,492,153,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641048885.76/warc/CC-MAIN-20240812190307-20240812220307-00330.warc.gz | 364,470,566 | 42,986 | # 2.2 Determining Volumes by Slicing
### Learning Objectives
• Determine the volume of a solid by integrating an area of a cross-section (the slicing method).
• Find the volume of a solid of revolution using the disk method.
• Find the volume of a solid of revolution with a cavity using the washer method.
In the preceding section, we used definite integrals to find the area between two curves. In this section, we use definite integrals to find volumes of three-dimensional solids. We consider three approaches—slicing, disks, and washers—for finding these volumes, depending on the characteristics of the solid.
## Volume and the Slicing Method
Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensional solid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid, for example, can be computed by multiplying length, width, and height: The formulas for the volume of a sphere a cone and a pyramid have also been introduced. Although some of these formulas were derived using geometry alone, all of them can be obtained by using integration.
We can also calculate the volume of a cylinder. Although most of us think of a cylinder as having a circular base, such as a soup can or a metal rod, in mathematics the word cylinder has a more general meaning. To discuss cylinders in this more general context, we first need to define some vocabulary.
We define the cross-section of a solid to be the intersection of a plane with the solid. A cylinder is defined as any solid that can be generated by translating a plane region along a line perpendicular to the region, called the axis of the cylinder. Thus, all cross-sections perpendicular to the axis of a cylinder are identical. The solid shown in below is an example of a cylinder with a noncircular base. To calculate the volume of a cylinder, we simply multiply the area of the cross-section by the height of the cylinder: In the case of a right circular cylinder (soup can), this becomes
If a solid does not have a constant cross-section (and it is not one of the other basic solids), we may not have a formula for its volume. In this case, we can use a definite integral to calculate the volume of the solid. We do this by slicing the solid into pieces, estimating the volume of each slice, and then adding those estimated volumes together. The slices should all be parallel to one another, and when we put all the slices together, we should get the whole solid. Consider, for example, the solid S shown in below , extending along the x-axis.
We want to divide into slices perpendicular to the x-axis. As we see later in the chapter, there may be times when we want to slice the solid in some other direction—say, with slices perpendicular to the y-axis. The decision of which way to slice the solid is very important. If we make the wrong choice, the computations can get quite messy. Later in the chapter, we examine some of these situations in detail and look at how to decide which way to slice the solid. For the purposes of this section, however, we use slices perpendicular to the x-axis.
Because the cross-sectional area is not constant, we let represent the area of the cross-section at point Now let be a regular partition of and for let represent the slice of the solid stretching from The following figure shows the sliced solid with
Finally, for let be an arbitrary point in Then the volume of slice can be estimated by Adding these approximations together, we see that the volume of the entire solid can be approximated by
By now, we can recognize this as a Riemann sum, and our next step is to take the limit as Then we have
The technique we have just described is called the slicing method . To apply it, we use the following strategy.
### Problem-Solving Strategy: Finding Volumes by the Slicing Method
1. Examine the solid and determine the shape of a cross-section of the solid. It is often helpful to draw a picture if one is not provided.
2. Determine a formula for the area of the cross-section.
3. Integrate the area expression over the appropriate interval to get the volume.
Recall that in this section, we assume the slices are perpendicular to the x-axis. Therefore, the area expression is in terms of x and the limits of integration lie on the x-axis. However, the problem-solving strategy shown here is valid regardless of how we choose to slice the solid.
### Deriving the Formula for the Volume of a Pyramid
We know from geometry that the formula for the volume of a pyramid is If the pyramid has a square base, this becomes where denotes the length of one side of the base. We are going to use the slicing method to derive this formula.
#### Solution
We want to apply the slicing method to a pyramid with a square base. To set up the integral, consider the pyramid shown in below, oriented along the x-axis.
We first want to determine the shape of a cross-section of the pyramid. Since the base is a square, the cross-sections are squares as well (step 1). Now we want to determine a formula for the area of one of these cross-sectional squares. Looking at Figure 4 (b), and using a proportion, since these are similar triangles, we have
Therefore, the area of one of the cross-sectional squares is
(step 2).
Then we find the volume of the pyramid by integrating from (step
This is the formula we were looking for.
Use the slicing method to derive the formula for the volume of a circular cone.
#### Hint
Use similar triangles, as in the example above.
## Solids of Revolution
If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution , as shown in the following figure.
Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest of this section looking at solids of this type. The next example uses the slicing method to calculate the volume of a solid of revolution.
Use an online integral calculator by WolframAlpha to learn more.
### Using the Slicing Method to find the Volume of a Solid of Revolution
Use the slicing method to find the volume of the solid of revolution bounded by the graphs of and and rotated about the x-axis.
#### Solution
Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval as shown in the following figure.
Next, revolve the region around the x-axis, as shown in the following figure.
Since the solid was formed by revolving the region around the x-axis, the cross-sections are circles (step 1). The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by Use the formula for the area of the circle:
The volume, then, is (step 3)
The volume is
Use the method of slicing to find the volume of the solid of revolution formed by revolving the region between the graph of the function and the x-axis over the interval around the x-axis. See the following figure.
## The Disk Method
When we use the slicing method with solids of revolution, it is often called the disk method because, for solids of revolution, the slices used to over approximate the volume of the solid are disks. To see this, consider the solid of revolution generated by revolving the region between the graph of the function and the x-axis over the interval around the x-axis. The graph of the function and a representative disk are shown in Figure 8 (a) and (b). The region of revolution and the resulting solid are shown in Figure 8 (c) and (d).
We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We know that
The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is the area of a circle. This gives the following method.
### The Disk Method
Let be continuous and nonnegative. Define as the region bounded above by the graph of below by the x-axis, on the left by the line and on the right by the line Then, the volume of the solid of revolution formed by revolving around the x-axis is given by
The volume of the solid we have been studying is given by
Let’s look at some examples.
### Using the Disk Method to Find the Volume of a Solid of Revolution 1
Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of and the x-axis over the interval around the x-axis.
#### Solution
The graphs of the function and the solid of revolution are shown in the following figure.
We have
The volume is units 3 .
Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of and the x-axis over the interval around the x-axis.
units 3
So far, our examples have all concerned regions revolved around the x-axis, but we can generate a solid of revolution by revolving a plane region around any horizontal or vertical line. In the next example, we look at a solid of revolution that has been generated by revolving a region around the y-axis. The mechanics of the disk method are nearly the same as when the x-axis is the axis of revolution, but we express the function in terms of y and we integrate with respect to y as well. This is summarized in the following rule.
### The Disk Method for Solids of Revolution around the y-axis
Let be continuous and nonnegative. Define as the region bounded on the right by the graph of on the left by the y-axis, below by the line and above by the line Then, the volume of the solid of revolution formed by revolving around the y-axis is given by
The next example shows how this rule works in practice.
### Using the Disk Method to Find the Volume of a Solid of Revolution 2
Let be the region bounded by the graph of and the y-axis over the y-axis interval Use the disk method to find the volume of the solid of revolution generated by rotating around the y-axis.
#### Solution
Figure 10 shows the function and a representative disk that can be used to estimate the volume. Notice that since we are revolving the function around the y-axis, the disks are horizontal, rather than vertical.
The region to be revolved and the full solid of revolution are depicted in the following figure.
To find the volume, we integrate with respect to We obtain
The volume is units 3 .
Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of and the y-axis over the interval around the y-axis.
units 3
## The Washer Method
Some solids of revolution have cavities in the middle; they are not solid all the way to the axis of revolution. Sometimes, this is just a result of the way the region of revolution is shaped with respect to the axis of revolution. In other cases, cavities arise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happen is when an axis of revolution other than the x-axis or y-axis is selected.
When the solid of revolution has a cavity in the middle, the slices used to approximate the volume are not disks, but washers (disks with holes in the center). For example, consider the region bounded above by the graph of the function and below by the graph of the function over the interval When this region is revolved around the x-axis, the result is a solid with a cavity in the middle, and the slices are washers. The graph of the function and a representative washer are shown in Figure 12 (a) and (b). The region of revolution and the resulting solid are shown in Figure 12 (c) and (d).
The cross-sectional area, then, is the area of the outer circle less the area of the inner circle. In this case,
Then the volume of the solid is
Generalizing this process gives the washer method .
### The Washer Method
Suppose and are continuous, nonnegative functions such that over Let denote the region bounded above by the graph of below by the graph of on the left by the line and on the right by the line Then, the volume of the solid of revolution formed by revolving around the x-axis is given by
### Using the Washer Method
Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of and below by the graph of over the interval around the x-axis.
#### Solution
The graphs of the functions and the solid of revolution are shown in the following figure.
We have
Find the volume of a solid of revolution formed by revolving the region bounded by the graphs of and over the interval around the x-axis.
units 3
#### Hint
Graph the functions to determine which graph forms the upper bound and which graph forms the lower bound, then use the procedure from the above example.
As with the disk method, we can also apply the washer method to solids of revolution that result from revolving a region around the y-axis. In this case, the following rule applies.
### Rule: The Washer Method for Solids of Revolution around the y-axis
Suppose and are continuous, nonnegative functions such that for Let denote the region bounded on the right by the graph of on the left by the graph of below by the line and above by the line Then, the volume of the solid of revolution formed by revolving around the y-axis is given by
Rather than looking at an example of the washer method with the y-axis as the axis of revolution, we now consider an example in which the axis of revolution is a line other than one of the two coordinate axes. The same general method applies, but you may have to visualize just how to describe the cross-sectional area of the volume.
### The Washer Method with a Different Axis of Revolution
Find the volume of a solid of revolution formed by revolving the region bounded above by and below by the x-axis. over the interval around the line
#### Solution
The graph of the region and the solid of revolution are shown in the following figure.
We can’t apply the volume formula to this problem directly because the axis of revolution is not one of the coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less the area of the inner circle. Looking at the graph of the function, we see the radius of the outer circle is given by which simplifies to
The radius of the inner circle is Therefore, we have
Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of and below by the x-axis over the interval around the line
units 3
### Key Concepts
• Definite integrals can be used to find the volumes of solids. Using the slicing method, we can find a volume by integrating the cross-sectional area.
• For solids of revolution, the volume slices are often disks and the cross-sections are circles. The method of disks involves applying the method of slicing in the particular case in which the cross-sections are circles, and using the formula for the area of a circle.
• If a solid of revolution has a cavity in the center, the volume slices are washers. With the method of washers, the area of the inner circle is subtracted from the area of the outer circle before integrating.
## Key Equations
• Disk Method along the x-axis
• Disk Method along the y-axis
• Washer Method
### Exercises
1. Derive the formula for the volume of a sphere using the slicing method.
2. Use the slicing method to derive the formula for the volume of a cone.
3. Use the slicing method to derive the formula for the volume of a regular tetrahedron with side length
4. Explain when you would use the disk method versus the washer method. When are they interchangeable?
For the following exercises, draw a typical slice and find the volume using the slicing method for the given volume.
5. A pyramid with height 6 units and square base of side 2 units, as pictured here.
8 units 3
6. A pyramid with height 4 units and a rectangular base with length 2 units and width 3 units, as pictured here.
7. A regular tetrahedron with a side of 4 units, as seen here.
units 3
8. A pyramid with height 5 units, and an isosceles triangular base with lengths of 6 units and 8 units, as seen here.
9. A cone of radius and height has a smaller cone of radius and height removed from the top, as seen here. The resulting solid is called a frustum .
units 3
For the following exercises, draw an outline of the solid and find the volume using the slicing method.
10. The base is a circle of radius The slices perpendicular to the base are squares.
11. The base is a triangle with vertices and Slices perpendicular to the x-axis are semicircles.
units 3
12. The base is the region under the parabola in the first quadrant. Slices perpendicular to the x-axis are squares.
13. The base is the region under the parabola and above the x-axis. Slices perpendicular to the y-axis are squares.
2 units 3
14. The base is the region enclosed by and Slices perpendicular to the x-axis are right isosceles triangles with the hypotenuse in the base.
15. The base is the area between and Slices perpendicular to the x-axis are semicircles.
units 3
For the following exercises, draw the region bounded by the curves. Then, use the disk or the washer method to find the volume when the region is rotated around the given horizontal axis.
16. , rotated around the x-axis
17. , rotated around the x-axis
units 3
18. , rotated around the x-axis
19. in quadrant I, rotated around the line
units 3
20. , rotated around the line
21. (in the first quadrant), rotated around the -axis
(Hint: use the formula )
units 3
22. , rotated around the line
23. , rotated around the x-axis
units 3
For the following exercises, draw the region bounded by the curves. Then, find the volume when the region is rotated around the y-axis.
24.
25.
units 3
26. in quadrant I
27.
units 3
28.
29.
units 3
30.
31.
units 3
For the following exercises, draw the region bounded by the curves. Then, find the volume when the region is rotated around the given horizontal axis.
32. , rotated around the line
33. , rotated around the line
units 3
34. , rotated around the x-axis
35. , rotated around the line
units 3
36. [T] , rotated around the x-axis
37. , rotated around the line
units 3
38. , rotated around the x-axis
(Hint: use the formula )
39. , rotated around the x-axis
units 3
For the following exercises, draw the region bounded by the curves. Then, use the washer method to find the volume when the region is revolved around the given vertical line.
40. , rotated around the line
41. , rotated around the y-axis
units 3
42. , , and , rotated around the line
43. , rotated around the y-axis
units 3
44. , rotated around the line .
45. Yogurt containers can be shaped like frustums. Rotate the line around the y-axis to find the volume between
units 3
46. Rotate the ellipse around the x-axis to approximate the volume of a football, as seen here.
47. Rotate the ellipse around the y-axis to approximate the volume of a football.
units 3
48. A better approximation of the volume of a football is given by the solid that comes from rotating around the x-axis from to What is the volume of this football approximation, as seen here?
49. What is the volume of the Bundt cake that comes from rotating around the y-axis from to
units 3
For the following exercises, find the volume of the solid described.
50. The base is the region between and Slices perpendicular to the x-axis are semicircles.
51. The base is the region enclosed by the generic ellipse Slices perpendicular to the x-axis are semicircles.
units 3
52. Bore a hole of radius down the axis of a right cone and through the base of radius as seen here.
53. Find the volume common to two spheres of radius with centers that are apart, as shown here.
units 3
54. Find the volume of a spherical cap of height and radius where as seen here.
55. Find the volume of a sphere of radius with a cap of height removed from the top, as seen here.
units 3
## Glossary
cross-section
the intersection of a plane and a solid object
disk method
a special case of the slicing method used with solids of revolution when the slices are disks
slicing method
a method of calculating the volume of a solid that involves cutting the solid into pieces, estimating the volume of each piece, then adding these estimates to arrive at an estimate of the total volume; as the number of slices goes to infinity, this estimate becomes an integral that gives the exact value of the volume
solid of revolution
a solid generated by revolving a region in a plane around a line in that plane
washer method
a special case of the slicing method used with solids of revolution when the slices are washers | 4,555 | 20,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2024-33 | latest | en | 0.912483 |
https://community.tableau.com/thread/213721 | 1,550,324,729,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480472.38/warc/CC-MAIN-20190216125709-20190216151709-00248.warc.gz | 520,525,625 | 26,274 | 2 Replies Latest reply on Aug 23, 2016 5:53 AM by Jason Olson
# Consuming HANA Formula in Tableau
Within HANA we have the ability to create a formula in the final aggregation which is intended to be run through the visualization without aggregating. Is there any method to do this from within Tableau? A specific example of this is a percent change year over year. In order to properly calculate you have to sum the sales from this year and then sum the sales from last year and finally perform the division. With Tableau it naturally forces that final measure to be aggregated with a sum (or other) but that results in a sum of % changes rather than performing that HANA formula.
Is there any way around this? Would have to be some kind of a flag I would think to tell Tableau to not try to aggregate. This calc could easily be brought into the Tableau layer but that creates a situation where we don't have all the calcs in one place and opens up the possibility of them being performed differently by tool.
• ###### 1. Re: Consuming HANA Formula in Tableau
You could use a Level of Detail calculation to FIX it to the lowest level of aggregation - Overview: Level of Detail Expressions
• ###### 2. Re: Consuming HANA Formula in Tableau
I am familiar with Level of Detail calculations but don't believe that is quite what is required here. Let me describe a little bit further in case I'm just missing something. Within HANA you create a "VDM" which is sort of a semantic layer with an element of flow. In other words you can join together a few tables, aggregate the data to a certain level, add calculations to that level, join in additional data, aggregate again, and then at the very end there is a final aggregation. Within that final aggregation you can specify a formula which does not perform any kind of aggregation. If you had aggregated this year and last year sales earlier in the flow you would now create a calculation that was "Sales_TY / Sales_LY - 1". The select statement you would use then to pull that data from HANA would be something like the following:
SELECT
Brand,
Sales_TY,
Sales_LY,
Sales_Percent_Change
FROM
CV_BRAND_SALES
This would also be okay:
SELECT
Brand,
SUM(Sales_TY),
SUM(Sales_LY),
Sales_Percent_Change
FROM
CV_BRAND_SALES
GROUP BY
Brand,
Sales_Percent_Change
Note that there is no aggregation applied in the select statement. You simply query against the VDM like it was a standalone flat table with everything already aggregated to the correct level. The problem I have is that Tableau naturally aggregates all of the measures and that results in a SUM being applied and ultimately the incorrect result.
Where I'm headed now is to just push those kinds of calcs to Tableau but it would be nice if I could leverage the formulas in HANA so that anytime the data source is used the calculation is performed exactly the same. | 639 | 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} | 2.59375 | 3 | CC-MAIN-2019-09 | latest | en | 0.957189 |
https://www.physicsforums.com/threads/timelike-geodesics.781667/ | 1,607,013,819,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141729522.82/warc/CC-MAIN-20201203155433-20201203185433-00201.warc.gz | 722,734,793 | 15,450 | # Timelike Geodesics
## Homework Statement
Using the Reissner Nordstrom line element, which I've given in the relevant equations section, I'm looking to show that the time like Geodesics obey the equation again show below.
## Homework Equations
Line Element[/B]
##ds^2= - U(r)c^2dt^2 +\frac{dr^2}{U(r)} +r^2(d\theta^2 + sin^2(\theta)d\phi^2)##
##U(r)=1-\frac{r_s}{r}+\frac{G^2Q^2}{r^2}##
Equation to Obey
##\frac{1}{2} (\frac{dr}{d\tau})^2 +V(r) = \varepsilon##
## The Attempt at a Solution
I've presumed as we are looking for a ##dr'## the Euler Lagrange equation we would be interested would be[/B]
##\frac{d}{d\tau}(\frac{\partial L^2}{\partial r'}) - \frac{\partial L^2}{\partial r}##
If I work this through my answer doesn't really resemble the equation I'm looking for, I get the ##(\frac{dr}{d\tau})^2##, but I can't get the ##\frac{1}{2}## factor, plus I have other terms in the denominator.
I also have a lot of other terms but they could possibly be grouped in to ##V(r)##.
I was hoping somebody could confirm whether the method I'm attempting is correct, as then I'll know if I'm incorrectly calculating it or it is something else.
Many thanks.
Related Calculus and Beyond Homework Help News on Phys.org
stevendaryl
Staff Emeritus
You're on the right track, but why don't you first write down what you think $L$ is. Your equations don't define $L$.
stevendaryl
Staff Emeritus
Here are a couple of other hints:
First, $\tau$ and $s$ are the same thing, so $\frac{ds}{d\tau} = 1$. So if you take the expression for $s$, this gives you one "constant of the motion".
Second, if you have a Lagrangian of the form $L(r, \frac{dr}{d\tau}, t, \frac{dt}{d\tau}, \theta, \frac{d\theta}{d\tau}, \phi \frac{d\phi}{d\tau})$, and $L$ doesn't mention $\tau$, then the following quantity is conserved (has the same value for all $\tau$):
$H = (\sum_j P_j U^j) - L$
where $U^j = \frac{d x^j}{d\tau}$ and $P_j = \dfrac{\partial L}{\partial U^j}$.
So $H$ gives you a second constant of the motion. So $H = E$, for some constant $E$ | 647 | 2,040 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-50 | latest | en | 0.87611 |
https://stackoverflow.com/questions/30149732/give-the-k-most-frequent-ip-addresses-from-the-large-stream-of-ip-address-in-con | 1,726,341,012,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00071.warc.gz | 502,384,818 | 45,945 | # Give the k most frequent IP addresses from the large stream of IP address in constant time and constant space
I recently came across this interview question in my interview with a particular company. I tried to use maxHeap and tried to solve it, but it was not acceptable to him, as the question statement asked me to solve me in constant time and constant space.
So, I thought there is something related to bit magic, and bloom filters came to my mind, but as we know in bloom filters we can simply check if particular IP has been visited or not and it can also return false positives.
Can anyone please help me with this question. Interview is over, but I still I want to understand how can IPs be treated specially so that solution is O(c) in time and space.
• Some interesting extension of Boyer-Moore majority vote? Except even that requires storing the entire stream...
– Nemo
Commented May 10, 2015 at 15:07
• With bloom filters, you can only be sure if a particular IP address has not been encountered whereas the problem demands k IP addresses which are most frequently visited. Commented May 10, 2015 at 19:13
• @ankitG, I agree with you, that's the gist - it can tell whether the IP has been encountered or not, that to with some probability p. Got the 2 answers. Trying to understand them. Commented May 26, 2015 at 14:39
• @Nemo , will read Boyer Moore majority vote :). Thanks. Commented May 26, 2015 at 14:40
• You haven't defined the problem sufficiently for us to provide an answer. Is that "large stream" of indeterminate length? Is this supposed to be done in real time, so that you can retrieve the k most frequent IP addresses seen since you started looking at the stream? Or do you just want to get the k most frequent once you've reached the end of the stream? Commented Jun 1, 2015 at 14:24
This can be done in constant time for sure but not in constant space. You need space proportional to the number of unique IP addresses encountered so far.
This can be done using two data structures--
1.) A hashtable
2.) A doubly linked list(singly linked list can also be used but DLL will be more effective)
DLL will have at max 'k' nodes.
``````struct DLL{
int count;
struct DLL* next;
struct DLL* prev;
}
``````
Key in the hashtable is the IP address. Value would be a tuple < count,address of Doubly Linked List Node>.
As soon as an IP address comes from the stream, it is checked in the hashtable in O(1),
``````if it's not present,
if number of nodes in the DLL is less than k,
add a new node at the start of the DLL for this IPAddress and a new
entry is made in the hashtable <IPAddress,<count(=1 here),its
else
other IP Addresses in the DLL must be having a count of at least 1,
so no point putting it in the DLL. Just add a new entry to the
else
update the tuple i.e increase the value of count by 1 and,
if DLL node address is not NULL,
go to that node in the DLL, increase count there also and shift it
rightwards if at all count's value now is greater than
next DLL node's count. Keep doing that till the concerned node
reaches the right position. This is done in O(k). k is a constant as
per the problem statement. DLL really comes handy for these operations.
We have to make sure that the DLL is always sorted in ascending order.
Also, its very important to make the shifts by swapping links and not
by swapping values otherwise we end up updating the hash table for every
which doesn't increase the time complexity but unnecessarily increases
the number of operations
else
compare this IP Address's count with the count in the first node in DLL, if
its greater than the count of first node, create a new Node and insert
in the DLL appropriately in O(k). Update the hash table for this IP Address
and for the first IPAddress in the DLL before purging that node.
``````
So, this way, k most frequent IP Addresses are always available in the DLL in constant time.
• thanks for the solution. But he also mentioned constant space :/. Commented May 26, 2015 at 14:44
In fact, allocating an array of 2^32 is constant space, so you just allocate that, read the stream adding 1's to the array[IP] for each IP read, then you just sort the array (CONSTANT time, mind you, O(32*2^32)!) and then select top k addresses. Voila, constant time, constant space. Inefficient, but the question did not ask for inefficiency!
• My thought, too. Unless "IP addresses" includes IPV6 . . . Commented Jun 1, 2015 at 14:26
• @JimMischel Hehe :) If IPv6 are included, this will not work, but the perspective of someone wanting top 2^32 out of whole IPv6 pool is... overwhelming. Commented Jun 1, 2015 at 15:31
I assume you read the stream in realtime, keep the last N adresses of this stream and return the k most frequent of this N-element buffer whenever asked.
To do so you need two data structures: a FIFO and a Binary tree. The FIFO is used to keep track which adress leaves the buffer.
With each new address: Add it to the binary tree with counter 1 if not present or increment the matching counter. Remove the leaving address from the tree the same way.
FIFO Operations are O(1). Binary tree operations are O(log N) with N constant.
• As stated, the problem wants your N equal to infinity.
– Nemo
Commented May 10, 2015 at 15:05
• Well, I think the problem description is not complete. I guess what OP wanted is something like "Top 10 IPs in the last hour" for a high bandwith page Commented May 10, 2015 at 18:04
• @DrKoch , I was trying to understand your solution, what is N? Commented May 26, 2015 at 14:42
• See the very first senzence of my answer. It is the length of the FIFO Buffer. Commented Jun 1, 2015 at 12:21 | 1,392 | 5,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.962283 |
http://ajtas.org/article/146/10.11648.j.ajtas.20200903.14 | 1,590,814,859,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407289.35/warc/CC-MAIN-20200530040743-20200530070743-00483.warc.gz | 6,439,177 | 8,560 | Archive
Special Issues
Volume 9, Issue 3, May 2020, Page: 53-56
On the Application of Linear Discriminant Function to Evaluate Data on Diabetic Patients at the University of Port Harcourt Teaching Hospital, Rivers, Nigeria
Nicholas Pindar Dibal, Department of Mathematical Sciences, University of Maiduguri, Maiduguri, Nigeria
Christopher Akas Abraham, Department of Mathematical Sciences, University of Maiduguri, Maiduguri, Nigeria
Received: Apr. 16, 2020; Accepted: May 3, 2020; Published: May 18, 2020
Abstract
Many real life events involves several interacting variables, hence multivariate statistical tool is necessary for appropriate analysis and interpretation. Discriminant analysis (DA) is one of the commonly used multivariate method in various fields of study including education, finance, environment, medicine etc., where complex data analysis and interpretation is required. This paper demonstrates and illustrate approaches in presenting how the discriminant analysis can be carried out on 335 (40 diabetics and 295 non-diabetic) patients and how the output can be interpreted using the Fisher’s linear Discriminant function (FLDF). The performance of FLDF was adjudged based on the percentage of correct reclassification of the original observation to yield the discriminant scores from the functions. Up to 65.4% correct classification was achieved, and similarly 62.7% percent of the cross-validated grouped cases were correctly classified into either being a Diabetic or non-diabetic patient. Patient’s age and gender were found to be the two most important contributing variables in classifying a patient between the two groups.
Keywords
Discriminant Analysis, Classification, Diabetes, Fisher’s LDF
Nicholas Pindar Dibal, Christopher Akas Abraham, On the Application of Linear Discriminant Function to Evaluate Data on Diabetic Patients at the University of Port Harcourt Teaching Hospital, Rivers, Nigeria, American Journal of Theoretical and Applied Statistics. Vol. 9, No. 3, 2020, pp. 53-56. doi: 10.11648/j.ajtas.20200903.14
Reference
[1]
Abdullah, N. A. H., Halim, A., Ahmad, H., and Rus, R. (2008). Predicting corporate failure of Malaysia’s listed companies: Comparing multiple discriminant analysis, logistic regression and the hazard Model. International Research Journal of Finance and Economics, 15, 201-21
[2]
Alayande, S. A. and Bashi, K. A. (2015). An Overview and Application of Discriminant Analysis in Data Analysis. IOSR Journal of Mathematics (IOSR-JM). 11 (1), 12-15. DOI: 10.9790/5728-11151215.
[3]
AlKubaisi, M., Aziz, W. A., George, S. and Al-Tarawneh, K, (2019). Multivariate Discriminant Analysis Managing Staff Appraisal Case Study. Academy of Strategic Management Journal. 18 (5), Online ISSN: 1939-6104
[4]
Antonogeorgos, G., Demosthenes, B. P., Kostas, N. P. and Anastasia, T. (2009). Logistic Regression and Linear Discriminant Analyses in Evaluating Factors Associated with Asthma Prevalence among 10- to 12-Years-Old Children: Divergence and Similarity of the Two Statistical Methods. International Journal of Pediatrics. doi: 10.1155/2009/952042
[5]
Bhuyan, K. C. (2005). Multivariate Analysis and its Application. Department of Statistics, Garyounis University, Libya. New Central Book Agency (P) Ltd.
[6]
Cai, D., He, X. and Han, J. (2008). Srda: An Efficient Algorithm for Large-scale Discriminant Analysis. Knowledge and Data Engineering, 20 (1): 1–12.
[7]
Clemmensen, L. K. H. (2013). On Discriminant Analysis Techniques and Correlation Structures in High Dimensions. Kgs. Lyngby: Technical University of Denmark. Technical Report-2013, No. 04
[8]
Erimafa, J. T. (2009), Application of Discriminant Analysis to Predict the Class of Degree for Graduating Students in a University System. International journal of physical science, 4 (1), 16 – 21.
[9]
Fisher, R. A. (1938). The Statistical Utilization of Multiple Measurements. Ann. Eng. Lond. 7, 179-88.
[10]
Gebru, T. G. (2018). Sparse Linear Discriminant Analysis with more Variables than Observations. Ph. D. Thesis. The Open University
[11]
Hafez, E. I., Abdel-Fatah, E. M., Abdel-Nabi, S. M. and Zeidan, A. S. A. (2015). Discriminant Analysis in View of Statistical and Operations Research Techniques. Journal of Multidisciplinary Engineering Science and Technology (JMEST). 2 (11), 3039-3047
[12]
Mbanasor J. A. and Nto, P. O. O. (2008). Discriminant Analysis of Livestock Farmers’ Credit Worthiness Potentials under Rural Banking Scheme in Abia State, Nigeria. Nigerian Agricultural Journal, 39 (1), 1-7.
[13]
Micheal, A. B. (2014). Application of Discrimination and Classification on Diabetes Mellitus Data. International Journal of Applied Science and Technology, 4 (6). 292-298
[14]
Schlegel, A. (2018). Linear Discriminant Analysis for the Classification of Two Groupshttps://aaronschlegel.me/linear-discriminant-analysis-classification-two-groups.html | 1,267 | 4,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.879982 |
http://forums.wolfram.com/mathgroup/archive/2008/Feb/msg00383.html | 1,606,654,687,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141198409.43/warc/CC-MAIN-20201129123729-20201129153729-00343.warc.gz | 34,564,127 | 7,951 | Re: "Assuming"
• To: mathgroup at smc.vnet.net
• Subject: [mg85599] Re: "Assuming"
• From: dh <dh at metrohm.ch>
• Date: Thu, 14 Feb 2008 06:30:20 -0500 (EST)
• References: <fp0m11\$5u7\$1@smc.vnet.net>
```
HI Markus,
I think you are fooling yourself.
Consider Simplify[(a^2 - 1)/(a - 1)]
As a^2 - 1)== (a+1)(a-1) this will give (a+1). This is true in general,
for a==1 understood as limit. Therefore, this has nothing to do with
assumptions.
On the other hand, (a^2 - 1)/(a - 1) /. a -> 1 can not be calculated as
it gives 0/0. Note that "/. and ->" mean replacement, not limit. If
you want a limit, you have to say so: Limit[(a^2 - 1)/(a - 1), a -> 1]
hope this helps, Daniel
markusbinder wrote:
> Hi,
>
> I would appreciate some comments about the Assuming-function. Several examples using Mathematica 6 as follows:
>
> In[1]:= Assuming[a > 0, Simplify[a == -1]]
> Out[1]= False
> In[2]:= Assuming[a \[Element] Integers, Simplify[a == -1]]
> Out[2]= a == -1
> In[3]:= Assuming[a \[Element] Integers && a > 0 && a < 2, Simplify[a^2< 1.1]]
> Out[3]= True
> In[4]:= Assuming[a \[Element] Reals && a > 0 && a < 2, Simplify[a^2 < 1.1]]
> Out[4]= a < 1.04881
>
> So far, everything seems pretty reasonable, but
>
> In[5]:= Assuming[a \[Element] Integers && a > 0 && a < 2, Simplify[(a^2 - 1)/(a - 1)]]
> Out[5]= 1 + a
>
> whereas
>
> In[6]:= (a^2 - 1)/(a - 1) /. a -> 1
> Out[6]= Indeterminate
>
> I consider this as blunder (or bug?); I comprehend working with domains can be pretty tricky and of course I don't expect
>
> In[7]:= Simplify[(a^2 - 1)/(a - 1)]
>
> to yield
>
> Out[7]= "Beware, a != 1 neccessary!"
>
>
> Out[7]= 1 + a.
>
> Am I tricked by some personal misconception of how Mathematica deals with Assuming plus fractions as shown above, or is this a mere flaw?
>
> Best regards
> Markus Binder
>
```
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• Next by thread: Re: "Assuming" | 731 | 2,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.778308 |
http://6bdmath.blogspot.com/2007/05/equivalent-fractions.html | 1,511,367,104,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806615.74/warc/CC-MAIN-20171122160645-20171122180645-00210.warc.gz | 3,226,477 | 9,653 | ## Friday, May 25, 2007
### Equivalent Fractions
Hey! Today we'll describe how to find equivalent fractions. Before we start we'll introduce. I am Il Han and my partner is Tai Hyun. Let's start.Il Han: You guys know fractions. Right.Tai Hyun: I like to play ...."Il Han: Stop introducing."Tai Hyun: Let's just say we have one out of quater and then if denominator changes to sixteen then numerator changes to what??
You bet, if you don't get it then you might wanna listen very carefully. You guys know that 4 goes into 16 4 times which means you have to mutiply by four to the numerator. Every time if the numerator or denominator mutiplys or divide you have to do samething to numerator or denominator.
Another example:
Then I give you one more question. There was 5 out of 10. When 10 changes to 2 then 5 changes to what????
You bet. You know why?? Because denominator(10) divided by 15 and that means you have to do samething to the numenator(5). So the answer is 1 out of 2 or half so this is it. Thanks for listening folks. Bye-bye.
#### 1 comment:
karissa♥ said...
Dear Ilhan and Tai Hyun.
you did a great job on your post abouy fractions!!! Maybe you should make the post a little bit easier to understand. It was a little bit confusing.
Karissa♥ | 320 | 1,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | longest | en | 0.912484 |
https://discuss.codechef.com/t/cant-find-why-this-code-giving-wrong-answer-for-http-www-codechef-com-problems-bestbats/7273 | 1,600,950,706,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400217623.41/warc/CC-MAIN-20200924100829-20200924130829-00177.warc.gz | 366,468,923 | 3,971 | # can't find why this code giving "wrong answer".... for http://www.codechef.com/problems/BESTBATS
#include<stdio.h>
int fact(int n)
{
int k=1,i;
if(n==0||n==1)
return 1;
for(i=2;i<=n;i++)
{
k=k*i;
}
return k;
}
int main()
{
int t,k,key,count=1,xtra=0,out,swap,c,d,i,j,up,low,arr[11];
scanf("%d",&t); //t=no. of test cases
for(i=0;i<t;i++)
{
scanf("%d %d %d %d %d %d %d %d %d %d %d",&arr[0],&arr[1],&arr[2],&arr[3],&arr[4],&arr[5],&arr[6],&arr[7],&arr[8],&arr[9],&arr[10]);
scanf("%d",&k);//no. of batsmen
if(k<=0)
{
`````` printf("0\n");
continue;
}
for (c = 0 ; c < 10; c++)
{
for (d = 0 ; d < 11 - c - 1; d++)
{
if (arr[d] < arr[d+1])
{
swap=arr[d];
arr[d]=arr[d+1];
arr[d+1] = swap;
}
}
}
key=arr[k-1];
for(j=k-2;j>=0;j--)
{
if(arr[j]==key)
count++;//counts no. of copies on left + 1(for key itself)
else
break;
}
for(j=k;j<=10;j++)
{
if(arr[j]==key)
xtra++; //counts no. of copies on right
else
break;
}
/* printf("identicas=%d\n",count+xtra);
printf("left=%d\n",count);
printf("Rgt=%d\n",xtra);*/
up=fact(count+xtra);
low=(fact(count) * fact(xtra));
up=up/low;
printf("%d\n",up);
}
return 0;
``````
} | 448 | 1,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-40 | latest | en | 0.269382 |
https://www.futureschool.com/australian-curriculum/national-curriculum/mathematics-year-5/ | 1,685,471,744,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646076.50/warc/CC-MAIN-20230530163210-20230530193210-00039.warc.gz | 844,678,724 | 18,672 | Latest Results:
### Year 5 Mathematics
# TOPIC TITLE
1 Self Assessment Self Assessment – Year 5
Objective: Assessment
2 Multiplication Multiples and factors of whole numbers
Objective: On completion of the lesson the student will be able to specify multiples and factors of whole numbers, and calculate the product of squared numbers.
3 Multiplication Multiplication – important facts.
Objective: On completion of the lesson the student will know the connection between multiplication and division and recognise the strategies to help solve multiplication number sentences.
4 Multiplication Multiples and factors of whole numbers
Objective: On completion of the lesson the student will be able to specify multiples and factors of whole numbers, and calculate the product of squared numbers.
5 Multiplication Multiplication using extended algorithms.
Objective: On completion of the lesson the student will have understood multiplication using extended algorithms.
6 Multiplication Multiplication by 2 and 3 digits
Objective: On completion of the lesson the student will be able to solve and record 2 and 3 digit multiplication problems in extended and short form.
7 Multiplication Multiplying 2-digit numbers by multiple of 10
Objective: On completion of the lesson the student will be able to multiply any 2 digit number by any multiple of 10 using the process of long multiplication.
8 Multiplication Multiplying 3 and 4-digit numbers by multiples of 100
Objective: On completion of the lesson the student will be able to multiply any 3 or 4 digit numbers by any multiple of 100 using long multiplication.
9 Multiplication Multiplying 2-digit numbers by 2-digit numbers
Objective: On completion of the lesson the student will be able to multiply any 2 digit number by any other 2 digit number.
10 Problems Solve and record division using known facts and sharing
Objective: On completion of the lesson the student will be able to use sharing concepts to solve and record division problems.
11 Fractions Using fractions 1/2, 1/4, 1/8 to describe part of a whole
Objective: On completion of the lesson the student will be able to: name fractions, and use fractions to describe equal parts of a whole.
12 Fractions Using fractions 1/2, 1/4, 1/8 to describe parts of a group or collection
Objective: On completion of the lesson the student will be able to use the fractions to describe equal parts of a collection of objects.
13 Fractions Finding equivalent fractions
Objective: On completion of the lesson the student will be able to name and find fractions that represent equal amounts between halves, quarters and eighths – using diagrams and number lines.
14 Fractions Multiplying and dividing to obtain equivalent fractions
Objective: On completion of the lesson the student will be able to: obtain equivalent fractions using a number line or diagram, develop mental strategies to obtain equivalent fractions, and reduce a fraction to its lowest equivalent form.
15 Fractions Reducing fractions to lowest equivalent form
Objective: On completion of the lesson the student will be able to reduce a fraction to its lowest equivalent form by dividing the numerator and denominator by a common factor.
16 Equations Problem solving strategies
Objective: On completion of the lesson the student will be able to recall the steps in a strategy for solving word problems.
17 Number problems Problems with numbers.
Objective: On completion of the lesson the student will be able to solve problems with numbers using a problem solving strategy.
18 Division Division with and without a remainder.
Objective: On completion of the lesson the student will understand division with and without a Remainder.
19 Division Dividing two and three digit numbers by a single digit number.
Objective: On completion of the lesson the student will understand dividing two and three digit numbers by a single digit number.
20 Equations Problem solving strategies
Objective: On completion of the lesson the student will be able to recall the steps in a strategy for solving word problems.
21 Number problems Problems with numbers.
Objective: On completion of the lesson the student will be able to solve problems with numbers using a problem solving strategy.
22 Fractions Using fractions 1/2, 1/4, 1/8 to describe part of a whole
Objective: On completion of the lesson the student will be able to: name fractions, and use fractions to describe equal parts of a whole.
23 Fractions Using fractions 1/2, 1/4, 1/8 to describe parts of a group or collection
Objective: On completion of the lesson the student will be able to use the fractions to describe equal parts of a collection of objects.
24 Fractions Comparing and ordering fractions
Objective: On completion of the lesson the student will be able to compare and order fractions with the same number of equal parts, and compare and order fractions with a different number of equal parts.
25 Fractions Fractions 1/5, 1/10, 1/100
Objective: On completion of the lesson the student will be able to: compare fractions with the denominators 5, 10, 100, and represent fractions with the denominator 5, 10, 100.
26 Fractions mixed numbers (mixed numerals)
Objective: On completion of the lesson the student will be able to: name and recognise mixed numbers (mixed numerals), count by halves and quarters, and use a number line to represent halves and quarters beyond one.
27 Fractions Improper fractions
Objective: On completion of the lesson the student will be able to: use diagrams and number lines to recognise and represent mixed numbers and improper fractions, and develop strategies for changing improper fractions to mixed numbers and vice versa, mentally.
28 Fractions Subtracting fractions from whole numbers
Objective: On completion of the lesson the student will be able to: use a diagram to subtract fractions from a whole number, develop mental strategies for subtracting fractions from whole numbers, and recognise and use the written form for subtracting fractions from
29 Fractions Adding and subtracting fractions with the same denominator
Objective: On completion of the lesson the student will be able to add and subtract fractions with the same denominator.
30 Decimals Introduction to decimals
Objective: On completion of the lesson the student will be able to represent decimals to two decimal places.
31 Decimals Comparing and ordering decimals to two decimal places
Objective: On completion of the lesson the student will be able to compare and order decimals to two decimal places and understand decimal notation to two places.
32 Decimals Decimals with whole numbers 10th and 100th
Objective: On completion of the lesson the student will be able to use and understand place value to show whole numbers, tenths and hundredths as decimals.
33 Decimals Multiplying decimals by 10, 100 and 1000
Objective: On completion of the lesson the student will be able to multiply decimal numbers by one hundred and recognise the pattern formed when decimals are multiplied by ten, one hundred and one thousand.
34 Decimals Dividing decimals by 10, 100 and 1000
Objective: On completion of the lesson the student will be able to divide decimal numbers by one hundred and recognise the pattern formed when decimals are divided by ten, one hundred and one thousand.
35 Decimals Decimals to three decimal places
Objective: On completion of the lesson the student will be able to express thousandths as a decimal and interpret decimal notation for thousandths.
36 Decimals Introduction to decimals
Objective: On completion of the lesson the student will be able to represent decimals to two decimal places.
37 Decimals Comparing and ordering decimals to two decimal places
Objective: On completion of the lesson the student will be able to compare and order decimals to two decimal places and understand decimal notation to two places.
38 Decimals Decimals to three decimal places
Objective: On completion of the lesson the student will be able to express thousandths as a decimal and interpret decimal notation for thousandths.
39 Decimals Using decimals – shopping problems
Objective: On completion of the lesson the student will be able to: read and interpret problems involving money, interpret the everyday use of decimals, and perform calculations with money.
40 Calculation-grouping Multiplication using repeated addition
Objective: On completion of the lesson the student will be able to write about equal groups and rows using another method and also learn different ways to count them.
41 Calculation 10-100 Counting by 1, 2, 5, and 10 to 100
Objective: On completion of the lesson the student will be able to skip count to one hundred and show that one hundred equals ten times ten.
42 Multiplication Multiples and factors of whole numbers
Objective: On completion of the lesson the student will be able to specify multiples and factors of whole numbers, and calculate the product of squared numbers.
43 Fractions Comparing and ordering fractions
Objective: On completion of the lesson the student will be able to compare and order fractions with the same number of equal parts, and compare and order fractions with a different number of equal parts.
44 Fractions mixed numbers (mixed numerals)
Objective: On completion of the lesson the student will be able to: name and recognise mixed numbers (mixed numerals), count by halves and quarters, and use a number line to represent halves and quarters beyond one.
45 Fractions Improper fractions
Objective: On completion of the lesson the student will be able to: use diagrams and number lines to recognise and represent mixed numbers and improper fractions, and develop strategies for changing improper fractions to mixed numbers and vice versa, mentally.
46 Decimals Introduction to decimals
Objective: On completion of the lesson the student will be able to represent decimals to two decimal places.
47 Decimals Comparing and ordering decimals to two decimal places
Objective: On completion of the lesson the student will be able to compare and order decimals to two decimal places and understand decimal notation to two places.
48 Decimals Decimals with whole numbers 10th and 100th
Objective: On completion of the lesson the student will be able to use and understand place value to show whole numbers, tenths and hundredths as decimals.
49 Calculation-grouping Multiplication using repeated addition
Objective: On completion of the lesson the student will be able to write about equal groups and rows using another method and also learn different ways to count them.
50 Calculation-multiplication The multiplication sign
Objective: on completion of the lesson the student will be able to multiply single numbers to solve multiplication problems.
51 Calculation sharing/division Strategies for division
Objective: On completion of the lesson the student will be able to share objects equally and will also learn how to write about them.
52 Problems Solve and record division using known facts and sharing
Objective: On completion of the lesson the student will be able to use sharing concepts to solve and record division problems.
53 Algebraic expressions Simplifying algebraic expressions: adding like terms.
Objective: On completion of the lesson the student will be able to simplify and evaluate numerical expressions containing patterns, and be able to simplify algebraic expressions that contain like terms.
54 Algebraic expressions Simplifying algebraic Expressions: subtracting like terms.
Objective: On completion of the lesson the student will be able to recognise the difference between like and unlike terms, and be able to simplify an expression using subtraction.
55 Algebraic expressions Simplifying Algebraic expressions: combining addition and subtraction.
Objective: On completion of the lesson the student will understand how to approach algebraic expressions questions and avoid the most common mistakes.
56 Algebraic expressions Simplifying algebraic expressions: multiplication
Objective: On completion of the lesson the student will be able to simplify expressions involving multiplication of pronumerals and write them in the simplest form.
57 Algebraic expressions Simplifying algebraic expressions: division
Objective: On completion of the lesson the student will be able to use all the operations needed for simplifying algebraic expressions.
58 Algebraic expressions Expanding algebraic expressions: multiplication
Objective: On completion of the lesson the student will be able mentally to multiply and remove parentheses from simple algebraic expressions in one step.
59 Length Compare length by using informal units of measurement
Objective: On completion of the lesson the student will be able to measure objects around the student’s home and compare their length to each other. The student will also be able to compare the different ways the student measured the same object.
60 Length Using the metre as a formal unit to measure perimeter
Objective: On completion of the lesson the student will be able to calculate the perimeter of different shapes in metres.
61 Area Comparing and ordering areas.
Objective: On completion of the lesson the student will be able to estimate and compare the area of shapes using a standard unit and order shapes according to their area.
62 Capacity Estimate, measure and compare the capacity of containers
Objective: On completion of the lesson the student will know why and when we might need to estimate and a way to go about it.
63 Length Using the formal unit of the centimetre to measure length and perimeter
Objective: On completion of the lesson the student will be able to measure length and perimeter in centimetres.
64 Length Read and calculate distances on a map using the formal unit kilometre
Objective: On completion of the lesson the student will be able to read distances in kilometres and calculate total distances between different locations on a map.
65 Length Compare and convert formal units of measurement
Objective: On completion of the lesson the student will be able to use the formal units millimetre, centimetre, metre and kilometre to measure and convert.
66 Area Introduction to the square centimetre.
Objective: On completion of the lesson the student will be able to calculate the area in square centimetres of surfaces or objects and record their results correctly.
67 Area Larger areas: square metre, hectare, square kilometre.
Objective: On completion of the lesson the student will be able to calculate larger areas using the correct square unit.
68 Volume Introduction to volume. using the cubic centimetre as a standard unit
Objective: On completion of the lesson the student will be able to: recognise the need for a formal unit to measure volume, use the abbreviation for cubic centimetre, construct three dimensional objects using cubic centimetre blocks, and use counting to determine vo
69 3-D shapes Constructing models.
Objective: On completion of the lesson the student will be able to: Construct rectangular prisms using cubic centimetre blocks and counting to determine volume. Explain the advantages of using a cube to measure volume. Construct different rectangular prisms.
70 Volume Using the cubic centimetre to measure volume.
Objective: On completion of the lesson the student will be able to: recognise the relationship between the length, breadth, height and volume of a rectangular prism, calculate the volume of rectangular prisms, and find the volume of rectangular prisms by counting.
71 Volume Using the cubic metre to measure volume.
Objective: On completion of the lesson the student will be able to: recognise the need for a unit larger than the cubic centimetre, use the cubic metre as a formal unit for measuring large volumes, and explain why volume is measured in cubic metres in certain situat
72 Capacity Converting between volume and capacity using millilitres and litres
Objective: On completion of the lesson the student will be able to convert between units of capacity.
73 Weight/mass The kilogram
Objective: On completion of the lesson the student will know: how to use the kilogram as a measure of mass., and how to weigh items accurately using scales.
74 Weight/mass The gram and net mass
Objective: On completion of the lesson the student will understand: why there’s a zero button on digital scales, how to zero some other types of scales, and also how to measure mass in grams.
75 Weight/mass The tonne – converting units and problems
Objective: On completion of the lesson the student should be able to: choose the correct unit to measure the mass of small, medium or large objects, and convert measurements from one unit to another.
76 Capacity Using the cubic cm and displacement to measure volume and capacity
Objective: On completion of the lesson the student will know a way to find volume and capacity.
77 Capacity Using the cubic cm as a standard unit of measurement for volume and capacity
Objective: On completion of the lesson the student will understand what a cubic centimetre is and how it can be used to find out the volume and capacity of a three dimensional shape.
78 Capacity The relationship between the common units of capacity, the litre and the millilitre
Objective: On completion of the lesson the student will understand the relationship between the two common units of capacity, the litre and millilitre.
79 Capacity Converting between volume and capacity using kilolitres and litres
Objective: On completion of the lesson the student will know the formal units of measurement for volume and capacity for bigger objects. The student will also be able to use the knowledge of volume to work out capacity.
80 Length Using the metre as a formal unit to measure perimeter
Objective: On completion of the lesson the student will be able to calculate the perimeter of different shapes in metres.
81 Area Introduction to the square centimetre.
Objective: On completion of the lesson the student will be able to calculate the area in square centimetres of surfaces or objects and record their results correctly.
82 Area Introducing the rules for finding the area of a rectangle and a parallelogram.
Objective: On completion of the lesson the student will be able to investigate areas of rectangles and parallelograms using a given formula of multiplying measurements of sides.
83 Time, duration Duration
Objective: On completion of the lesson the student will be able to estimate and measure the duration of an event using informal units, and compare and order the duration of an event using informal units.
84 Time, minutes Analogue – Telling time – minutes in the hour
Objective: On completion of the lesson the student will be able to recognize the coordinated movements of the hands on an analogue clock.
85 Time, units Units of time
Objective: On completion of the lesson the student will be able to convert units of time and read and interpret simple timetables, timelines and calendars.
86 Time, digital O’clock and half past using digital time
Objective: On completion of the lesson the student will be able to use the terms ‘o’clock’ and ‘half past’, and read hour and half-hour time on a digital clock.
87 Time, analogue O’clock and half past on the analogue clock
Objective: On completion of the lesson the student will be able to use the terms ‘o’clock’ and ‘half past’, and read hour and half-hour time on an analogue clock.
88 Time, 24-hour 24 hour time
Objective: On completion of the lesson the student will be able to: tell the time accurately using twenty-four hour time, change the time from am and pm time to twenty-four hour time, and change the time from twenty-four hour time to am and pm time.
89 3-D shapes Recognise and name prisms according to spatial properties
Objective: On completion of the lesson the student will be able to recognise and name various prisms according to their spatial properties.
90 3-D shapes Recognise and name pyramids according to spatial properties
Objective: On completion of the lesson the student will be able to recognise, describe and name pyramids according to their spatial properties.
91 3-D shapes Recognise nets for prisms, pyramids, cubes and cones
Objective: On completion of the lesson the student will be able to predict and recognise nets for prisms, pyramids, cubes and cones.
92 3-D shapes Viewing 3-D shapes.
Objective: On completion of the lesson the student will be able to use conventional representations of three-dimensional shapes to show depth etc when drawing or viewing shapes from various angles.
93 Lines and angles Describing position.
Objective: On completion of the lesson the student will be able to use and understand conventional location language to describe position.
94 Lines and angles Mapping and grid references
Objective: On completion of the lesson the student will be able to identify specific places on a map and use regions on a grid to locate objects or places.
95 Lines and angles Informal coordinate system
Objective: On completion of the lesson the student will be able to use an informal coordinate system to specify location, and locate coordinate points on grid paper.
96 Geometric transformations Geometry transformations without matrices: rotation (Stage 2)
Objective: On completion of this lesson the student will perform and construct rotations.
97 Tessellating 2-D shapes Use grids to enlarge/reduce 2D shapes
Objective: On completion of the lesson the student will be able to use grids to enlarge or reduce two dimensional shapes and also to recognise shapes that will and won’t tessellate.
98 Angles Measure and classify angles
Objective: On completion of the lesson the student will be able to recognise, measure and classify angles and measure the angles in a triangle.
99 Geometry-angles Measuring angles
Objective: On completion of the lesson the student will be able to measure any angle between 0 and 360 degrees using a protractor, and identify what type of angle it is.
100 Statistic-probability Probability of Simple Events
Objective: On completion of the lesson the student will be able to understand the probability of simple events.
101 Statistic-probability Probability of Simple Events
Objective: On completion of the lesson the student will be able to understand the probability of simple events.
102 Data Pictograms
Objective: On completion of the lesson the student will be able to organise, read and summarise information in picture graphs.
103 Data Bar Charts
Objective: On completion of the lesson the student will be able to organise, read and summarise information in column graphs.
104 Data Line graphs.
Objective: On completion of the lesson the student will be able to organise, read and summarise information in line graphs.
105 Statistics Frequency distribution table
Objective: On completion of the lesson the student will be able to construct a frequency distribution table for raw data and interpret the table.
106 Exam Exam – Year 5
Objective: Exam | 4,504 | 22,953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-23 | latest | en | 0.836924 |
https://solvedlib.com/n/problem-27-26-an-electron-experences-lorce-3-8i-2-3j-10,21304709 | 1,659,918,877,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570741.21/warc/CC-MAIN-20220808001418-20220808031418-00056.warc.gz | 496,183,566 | 16,897 | # Problem 27.26 An electron experences lorce (3.8i 2.3j) 10-13 Nwnen passing through . magnetic field B (0.79T)k Part A Determine
###### Question:
Problem 27.26 An electron experences lorce (3.8i 2.3j) 10-13 Nwnen passing through . magnetic field B (0.79T)k Part A Determine Ine eleclron's velocily Express your answers using two significant fiqures- Enter your answars numerically separated by commas AEd U, Vv m/s Submit Requast Bniwer Provide Feedback
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You are a project manager with Striker Projects Pvt. Ltd. As the project manager, you need to evaluate two projects “Salient” and “Stylish”, and make a recommendation to management for the selection of a suitable project. Both the projects “Salient” and “Sty... | 4,110 | 14,735 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-33 | latest | en | 0.863303 |
https://metanumbers.com/22525 | 1,628,061,472,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154796.71/warc/CC-MAIN-20210804045226-20210804075226-00193.warc.gz | 370,057,440 | 10,830 | ## 22525
22,525 (twenty-two thousand five hundred twenty-five) is an odd five-digits composite number following 22524 and preceding 22526. In scientific notation, it is written as 2.2525 × 104. The sum of its digits is 16. It has a total of 4 prime factors and 12 positive divisors. There are 16,640 positive integers (up to 22525) that are relatively prime to 22525.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 16
• Digital Root 7
## Name
Short name 22 thousand 525 twenty-two thousand five hundred twenty-five
## Notation
Scientific notation 2.2525 × 104 22.525 × 103
## Prime Factorization of 22525
Prime Factorization 52 × 17 × 53
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 4505 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 22,525 is 52 × 17 × 53. Since it has a total of 4 prime factors, 22,525 is a composite number.
## Divisors of 22525
1, 5, 17, 25, 53, 85, 265, 425, 901, 1325, 4505, 22525
12 divisors
Even divisors 0 12 12 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 30132 Sum of all the positive divisors of n s(n) 7607 Sum of the proper positive divisors of n A(n) 2511 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 150.083 Returns the nth root of the product of n divisors H(n) 8.97053 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 22,525 can be divided by 12 positive divisors (out of which 0 are even, and 12 are odd). The sum of these divisors (counting 22,525) is 30,132, the average is 2,511.
## Other Arithmetic Functions (n = 22525)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 16640 Total number of positive integers not greater than n that are coprime to n λ(n) 1040 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2518 Total number of primes less than or equal to n r2(n) 48 The number of ways n can be represented as the sum of 2 squares
There are 16,640 positive integers (less than 22,525) that are coprime with 22,525. And there are approximately 2,518 prime numbers less than or equal to 22,525.
## Divisibility of 22525
m n mod m 2 3 4 5 6 7 8 9 1 1 1 0 1 6 5 7
The number 22,525 is divisible by 5.
• Arithmetic
• Deficient
• Polite
## Base conversion (22525)
Base System Value
2 Binary 101011111111101
3 Ternary 1010220021
4 Quaternary 11133331
5 Quinary 1210100
6 Senary 252141
8 Octal 53775
10 Decimal 22525
12 Duodecimal 11051
20 Vigesimal 2g65
36 Base36 hdp
## Basic calculations (n = 22525)
### Multiplication
n×i
n×2 45050 67575 90100 112625
### Division
ni
n⁄2 11262.5 7508.33 5631.25 4505
### Exponentiation
ni
n2 507375625 11428635953125 257430024844140625 5798611309614267578125
### Nth Root
i√n
2√n 150.083 28.2415 12.2508 7.42221
## 22525 as geometric shapes
### Circle
Diameter 45050 141529 1.59397e+09
### Sphere
Volume 4.78722e+13 6.37587e+09 141529
### Square
Length = n
Perimeter 90100 5.07376e+08 31855.2
### Cube
Length = n
Surface area 3.04425e+09 1.14286e+13 39014.4
### Equilateral Triangle
Length = n
Perimeter 67575 2.197e+08 19507.2
### Triangular Pyramid
Length = n
Surface area 8.788e+08 1.34688e+12 18391.6
## Cryptographic Hash Functions
md5 ea9c39a35857068756c18d8a47ac9c33 f9cedab81687b354059db9d97ef85779419d457d 0f25163098a1c9ac107b111ed9d25b4598b075e31da33fe93d0a14e243f508d7 3a728f82e9f4b820b928a55d4c010931bfc9323bec8a4c0efc1c809561e0fad546904b7dd226de4faf23e6df942829a31b0f61068eb827ecee89f4dffb650ba3 3217f104e74182131709efc906c2370695e0ea9c | 1,443 | 4,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-31 | latest | en | 0.812304 |
https://gamedev.stackexchange.com/questions/59298/walking-on-a-sphere | 1,632,637,852,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057830.70/warc/CC-MAIN-20210926053229-20210926083229-00067.warc.gz | 313,922,262 | 40,542 | # Walking on a sphere
I'm working on a game which involves walking your character on the surface of a sphere. Using the answer to Arbitrary Rotation about a Sphere, I've written my code as:
if (game.isKeyDown(37)) { // left
this.quaternion.multiply(new THREE.Quaternion(0, Math.sin(-0.01), 0, Math.cos(-0.01)));
}
if (game.isKeyDown(39)) { // right
this.quaternion.multiply(new THREE.Quaternion(0, Math.sin(0.01), 0, Math.cos(0.01)));
}
if (game.isKeyDown(38)) { // up
this.quaternion.multiply(new THREE.Quaternion(Math.sin(-0.01), 0, 0, Math.cos(-0.01)));
}
if (game.isKeyDown(40)) { // down
this.quaternion.multiply(new THREE.Quaternion(Math.sin(0.01), 0, 0, Math.cos(0.01)));
}
var qx = this.quaternion.x;
var qy = this.quaternion.y;
var qz = this.quaternion.z;
var qw = this.quaternion.w;
this.obj.position.x = 2 * (qy * qw + qz * qx) * radius;
this.obj.position.y = 2 * (qz * qy - qw * qx) * radius;
this.obj.position.z = ((qz * qz + qw * qw) - (qx * qx + qy * qy)) * radius;
Which works fine, however I would like to control the character in such a way that pressing up and down is equivelent to walking forwards and backwards (in the direction you are looking), whereas pressing left and right is the same as turning around on the spot.
I understand that I will have to store a forward vector, but I'm not clear on how that relates to the quaternion which allows the character to walk on the surface of the sphere.
Another problem I've got to overcome is making sure the character on the surface of the sphere is actually "looking" the way it is going, at the moment I'm modelling the player as another sphere, but in the futuer it will be a proper model which will need to orientate itself.
Store a direction vector as a tangent to the sphere. When you move, you can take this tangent, the normal vector (normalized position on the sphere) and cross them to get a general axis to rotate around. If you're limiting all movement to a single 2D plane (you're just using a single angle), for direction all you need is a sign (+1 or -1) to multiply your movement angle by. You can also use the sign to calculate a facing rotation about the sphere normal vector (or to just flip the character about its local Y axis) to orient your character properly.
• Thanks, this helped a lot. Incase other people arrive at this question, I also found gamedev.net/topic/… which has some nice pointers. Jul 17 '13 at 18:28
The following code solves one of the OP's problem. That of the player turning around himself on a sphere, when pressing left and right (or d and a) buttons
case 68: //D
this.quaternion.multiply(new THREE.Quaternion(0, 0, Math.sin(-0.01), Math.cos(-0.01)));
break;
case 65: //A
this.quaternion.multiply(new THREE.Quaternion(0, 0, Math.sin(0.01), Math.cos(0.01)));
break; | 740 | 2,810 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-39 | latest | en | 0.848958 |
1wclub.org | 1,632,689,929,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057973.90/warc/CC-MAIN-20210926205414-20210926235414-00472.warc.gz | 120,664,403 | 22,790 | ## Topic outline
• ### Welcome to Toddler Math
Getting Started
• ### Toddler Math. Lesson 1. Number 1.
#### Learning Objectives
• Learn to read and write the Number 1.
• Learn Number 1 phonics.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Action.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter A.
• ### Toddler Math. Lesson 2. Number 2.
#### Learning Objectives
• Learn to read and write the Number 2.
• Learn Number 2 phonics.
• Learn to count from 1 to 2.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Beauty.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter B.
• ### Toddler Math. Lesson 3. Number 3.
#### Learning Objectives
• Learn to read and write the Number 3.
• Learn Number 3 phonics.
• Learn to count from 1 to 3.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Caring.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter C.
• ### Toddler Math. Lesson 4. Number 4.
#### Learning Objectives
• Learn to read and write the Number 4.
• Learn Number 4 phonics.
• Learn to count from 1 to 4.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Devotion.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter D.
• ### Toddler Math. Lesson 5. Number 5.
#### Learning Objectives
• Learn to read and write the Number 5.
• Learn Number 5 phonics.
• Learn to count from 1 to 5.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Effort.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter E.
• ### Toddler Math. Lesson 6 Number 6.
#### Learning Objectives
• Learn to read and write the Number 6.
• Learn Number 6 phonics.
• Learn to count from 1 to 6.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Forgiveness.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter F.
• ### Toddler Math. Lesson 7. Number 7.
#### Learning Objectives
• Learn to read and write the Number 7.
• Learn Number 7 phonics.
• Learn to count from 1 to 7.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Generosity.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter G.
• ### Toddler Math. Lesson 8. Number 8.
#### Learning Objectives
• Learn to read and write the Number 8.
• Learn Number 8 phonics.
• Learn to count from 1 to 8.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Helpfulness.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter H.
• ### Toddler Math. Lesson 9. Number 9.
#### Learning Objectives
• Learn to read and write the Number 9.
• Learn Number 9 phonics.
• Learn to count from 1 to 9.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Integrity.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter I.
• ### Toddler Math. Lesson 10. Number 10.
#### Learning Objectives
• Learn to read and write the Number 10.
• Learn Number 10 phonics.
• Learn to count from 1 to 10.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Joy.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter J.
• ### Toddler Math. Lesson 11. Number 11.
#### Learning Objectives
• Learn to read and write the Number 11.
• Learn Number 11 phonics.
• Learn to count from 1 to 11.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Kindness.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter K.
• ### Toddler Math. Lesson 12. Number 12.
#### Learning Objectives
• Learn to read and write the Number 12.
• Learn Number 12 phonics.
• Learn to count from 1 to 12.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Love.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter L.
• ### Toddler Math. Lesson 13. Number 13.
#### Learning Objectives
• Learn to read and write the Number 13.
• Learn Number 13 phonics.
• Learn to count from 1 to 13.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Mercy.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter M.
• ### Toddler Math. Lesson 14. Number 14.
#### Learning Objectives
• Learn to read and write the Number 14.
• Learn Number 14 phonics.
• Learn to count from 1 to 14.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Nice.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter N.
• ### Toddler Math. Lesson 15. Number 15.
#### Learning Objectives
• Learn to read and write the Number 15.
• Learn Number 15 phonics.
• Learn to count from 1 to 15.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Obedience.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter O.
• ### Toddler Math. Lesson 16. Number 16.
#### Learning Objectives
• Learn to read and write the Number 16.
• Learn Number 16 phonics.
• Learn to count from 1 to 16.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Patience.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter P.
• ### Toddler Math. Lesson 17. Number 17.
#### Learning Objectives
• Learn to read and write the Number 17.
• Learn Number 17 phonics.
• Learn to count from 1 to 17.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Quiet.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter Q.
• ### Toddler Math. Lesson 18. Number 18.
#### Learning Objectives
• Learn to read and write the Number 18.
• Learn Number 18 phonics.
• Learn to count from 1 to 18.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Respect.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter R.
• ### Toddler Math. Lesson 19. Number 19.
#### Learning Objectives
• Learn to read and write the Number 19.
• Learn Number 19 phonics.
• Learn to count from 1 to 19.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Strength.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter S.
• ### Toddler Math. Lesson 20. Number 20.
#### Learning Objectives
• Learn to read and write the Number 20.
• Learn Number 20 phonics.
• Learn to count from 1 to 20.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Truth.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter T.
• ### Toddler Math. Lesson 21. Number 21.
#### Learning Objectives
• Learn to read and write the Number 21.
• Learn Number 21 phonics.
• Learn to count from 1 to 21.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Unity.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter U.
• ### Toddler Math. Lesson 22. Number 22.
#### Learning Objectives
• Learn to read and write the Number 22.
• Learn Number 22 phonics.
• Learn to count from 1 to 22.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Vitality.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter V.
• ### Toddler Math. Lesson 23. Number 23.
#### Learning Objectives
• Learn to read and write the Number 23.
• Learn Number 23 phonics.
• Learn to count from 1 to 23.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Wonder.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter W.
• ### Toddler Math. Lesson 24. Number 24.
#### Learning Objectives
• Learn to read and write the Number 24.
• Learn Number 24 phonics.
• Learn to count from 1 to 24.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Excellence.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter X.
• ### Toddler Math. Lesson 25. Number 25.
#### Learning Objectives
• Learn to read and write the Number 25.
• Learn Number 25 phonics.
• Learn to count from 1 to 25.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Yearning.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter Y.
• ### Toddler Math. Lesson 26. Number 26.
#### Learning Objectives
• Learn to read and write the Number 26.
• Learn Number 26 phonics.
• Learn to count from 1 to 26.
• Cross-curricula activity with Toddler Virtues: Become familiar with the virtue Zeal.
• Cross-curricula activity with Toddler ELA: Become familiar with the Letter Z. | 2,355 | 8,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-39 | latest | en | 0.721353 |
https://lists.cp2k.org/archives/cp2k-user/2015-May/006532.html | 1,708,814,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00219.warc.gz | 361,237,346 | 4,108 | # [CP2K:6441] Problems with GAPW + external potential
Juha Ritala jri... at gmail.com
Mon May 25 09:42:21 UTC 2015
I have tested the patch using the water molecule setup and everything looks
good so far. The total energy stays almost perfectly constant (up to a
negligible numerical error) when a constant background potential is
applied. This was more or less obvious so I continued by applying a uniform
electric field to the system (maximum magnitude 0.1 V/Å). The dependence of
the energy on the electric field is the same (within the eps_scf) using
both GPW and GAPW methods. Also the forces on the atoms are almost the
same. The fixed GAPW method does a better job here, actually, since the sum
of the forces is closer to zero, which should be the case for the charge
neutral molecule.
Continuing the theoretical consideration, I would say that this fix should
handle an uniform electric field exactly independently on the magnitude of
the field. This requires only that the monopole and dipole moments of the
charge density in the atomic region are correct, which should be the case
with *rho0_s_rs* (*ñ^0).* More generally, any external potential that
corresponds to a charge density outside the atoms should be fine, I guess.
This would be true because outside the atomic region the electrostatic
potential generated by the atomic charge density depends only on the
multipole moments of the density. Since those are correct, everything
should work perfectly.
I will next try a real simulation setup where the field is not perfectly
uniform.
- Juha
On Wednesday, 20 May 2015 11:34:22 UTC+3, Juha Ritala wrote:
>
> Hi Matt,
>
> thank you for the patch. I think this approach makes sense. If I
> understood correctly, *rho0_s_rs* is the realspace grid representation of
> the charge density *ñ^0* (in the GAPW paper from 1999), and *ñ^0 *is sort
> of a representation of the charge density in the atomic region using the
> set of soft Gaussians. At least the multipole moments of the atomic charge
> density and *ñ^0 *are equal, so in a sufficiently large length scale
> these charge densities are approximately the same. This would imply that
> for moderate electric fields this approximation would indeed be fine.
>
> I will check how this patch works for me and keep you informed.
>
> - Juha
>
> On Saturday, 16 May 2015 13:03:25 UTC+3, Matt W wrote:
>>
>> Hi Juha,
>>
>> there are a couple of files attached (compatible with latest version of
>> cp2k at time of writing, but there are only a couple of lines changed) you
>> can try and see if it works with GAPW. It only works with the soft part of
>> the density, including the compensation charges. As the compensation
>> charges, including the core, are put onto the fft grid the core routines in
>> external_potential need to be protected. Might need tidying for mixed
>> gpw/gapw.
>>
>> I guess this approximately means that the core (i.e. any functions on the
>> hard atomic grids) isn't directly polarized by the field? For modest fields
>> it should be fine, maybe?
>>
>> Let me know how it goes,
>>
>> Matt
>>
>>
>> On Friday, May 1, 2015 at 3:46:30 PM UTC+1, jgh wrote:
>>>
>>> Hi
>>>
>>> unfortunately, I don't have time to look into this.
>>> It seems to me that for analytic potentials it wouldn't be that hard,
>>> but for potentials read in on a grid it would need special care.
>>>
>>> regards
>>>
>>> Juerg
>>> --------------------------------------------------------------
>>> Juerg Hutter Phone : ++41 44 635 4491
>>> Institut für Chemie C FAX : ++41 44 635 6838
>>> Universität Zürich E-mail: hut... at chem.uzh.ch
>>> Winterthurerstrasse 190
>>> CH-8057 Zürich, Switzerland
>>> ---------------------------------------------------------------
>>>
>>> -----cp... at googlegroups.com wrote: -----To: cp... at googlegroups.com
>>> From: Juha Ritala
>>> Sent by: cp... at googlegroups.com
>>> Date: 04/30/2015 01:18PM
>>> Subject: Re: [CP2K:6441] Problems with GAPW + external potential
>>>
>>> Hi Juerg,
>>>
>>> thank you for confirming this. Is there any chance that this will be
>>> fixed in the near future, or should I stick with GPW for the time being? I
>>> could try and fix it myself, but as far as I understand how GAPW works,
>>> considering the contribution from the local atomic charges is not at all
>>> trivial.
>>>
>>> - Juha
>>>
>>> On Wednesday, April 29, 2015 at 2:17:45 PM UTC+3, jgh wrote:Hi
>>>
>>>
>>>
>>> yes, this is a bug. The external potential is applied to the
>>>
>>> charge on the plane wave grid. In GAPW only the soft part
>>>
>>> of the charge is on the grid and one would have to also compute
>>>
>>> the contributions from the local atomic charges.
>>>
>>>
>>>
>>> regards
>>>
>>>
>>>
>>> Juerg Hutter
>>>
>>> --------------------------------------------------------------
>>>
>>> Juerg Hutter Phone : ++41 44 635 4491
>>>
>>> Institut für Chemie C FAX : ++41 44 635 6838
>>>
>>> Universität Zürich E-mail: hut... at chem.uzh.ch
>>>
>>> Winterthurerstrasse 190
>>>
>>> CH-8057 Zürich, Switzerland
>>>
>>> ---------------------------------------------------------------
>>>
>>>
>>>
>>> -----cp... at googlegroups.com wrote: -----To: cp... at googlegroups.com
>>>
>>> From: Juha Ritala
>>>
>>> Sent by: cp... at googlegroups.com
>>>
>>> Date: 04/29/2015 09:52AM
>>>
>>> Subject: [CP2K:6432] Problems with GAPW + external potential
>>>
>>>
>>>
>>> I have had some problems when I tried to use GAPW method combined with
>>> an external potential. To demonstrate what seems to be the source of the
>>> problems, I made this simple test case with a water molecule in a constant
>>> external potential. Physically, the constant background potential should
>>> not affect the total energy of the water molecule since the molecule is
>>> charge neutral. The energy does change as a function of the constant
>>> potential when the GAPW method is used, however. You can see this in the
>>> attached energy vs. background potential plot. GPW method gives expected
>>> behaviour, the total energy stays perfectly constant when the potential is
>>> varied.
>>>
>>>
>>>
>>> It seems as if some part of the charge in the system is unaffected by
>>> the external potential in GAPW and thus the total charge is effectively
>>> non-zero. Is this a bug or is there some part of the implementation
>>> missing?
>>>
>>>
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More information about the CP2K-user mailing list | 1,681 | 6,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-10 | latest | en | 0.904617 |
https://technicalsource9.medium.com/how-to-check-if-data-is-normally-distributed-e752bf13e078?source=post_internal_links---------4---------------------------- | 1,674,828,114,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494986.94/warc/CC-MAIN-20230127132641-20230127162641-00518.warc.gz | 578,298,564 | 31,075 | # How to check if data is normally distributed
Hi all,
I want to run a f-test on two samples to see if their variances are independent. Wikipedia says that the f test is sensitive to non normality of sample (<http://en.wikipedia.org/wiki/F-test)>. How can I check if my samples are normally distributed or not.
I read some forums which said I can use kstest and lillietest. When can I use either? I get an answer h=0. Does that mean my data is normally distributed?
NOTE:-
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You cannot tell from only 2 samples whether they are normally distributed or not. If you have a larger sample set and you are only testing them in pairs, then you could use the larger sample set to test for a particular distribution.
For example: (simple q-q plot)
`data= randn(100); %generate random normally distributed 100x100 matrixref1= randn(100); %generate random normally distributed 100x100 matrixref2= rand(100); %generate random uniformly distributed 100x100 matrixx=sort(data(:));y1=sort(ref1(:));y2=sort(ref2(:));subplot(1,2,1); plot(x,y1); subplot(1,2,2); plot(x,y2);`
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In this article, we will learn what is water hammer? Along with we will capture the basics, definition, causes, effects, how to fix or stop, etc. Let’s explore!
## What is Water Hammer? Definition & Meaning
### Water Hammer Basics
Let’s try to understand the basics of water hammer. We all know liquids are barely compressible, this means their volume can’t be changed under the application of pressure.
• This property is used in hydraulics systems but this same property becomes a damaging factor in piping.
• Water is heavy and when thousands of liter of water is moving then it carries a lot of energy.
So when we stop it suddenly pressure waves to get created and it has nowhere to go. This creates big noise, we call it a water hammer. Plumbing is the most under-acknowledged modern marvel that is for sure, we will realize this by this example.
### Water Hammer Meaning
If the pipeline is of 1m diameter and its length is 100 km then the mass of water inside is about 80 million kg. If the operator closes the valve fast, it will create a huge spike in pressure which will cause rupture to pipes, which will be as devastating as bombs.
Water travels through pipelines with a velocity that is dependent on,
• slope,
• viscosity,
• design of pipes,
• material of pipes etc.
### Water Hammer Definition
This water moving inside pipes when stopped suddenly by the operator by closing valve or by another cause, there is a sudden rise in pressure of water because the momentum of water got destroyed.
• This sudden stoppage of water creates a pressure wave inside the pipeline.
• This pressure wave travels throughout the water and makes hammering action on the walls of this pipe.
Hence, water hammer is defined as a phenomenon of sudden pressure rise in pipes. It is also called as hammer blow as well. This pressure rise is dependent on many variables such as
• Operating speed at which valve is closed
• Velocity of flow
• Length of pipe
• Elastic properties material of pipe
• Properties of fluid
This phenomenon is very important because the rise in pressure is so steep that it may lead to the bursting of pipes and may lead to a big accident.
## Example & Explanation of Water Hammer
When water is stored at tanks or in dams, it contains some form of energy by virtue of its position which we called as potential energy.
This potential energy is the product of density, gravitational acceleration, and height at which water is stored.
• P.E = ρgh
Where,
• P.E = Potential energy
• ρ = Density of water
• g = Gravitational acceleration
• h = height of water
When this water is coming through such high head to lower side, this potential energy is getting converted into kinetic energy.
K.E = 1/2 mv2
Where,
• K.E = Kinetic energy
• m = Mass of water
• v = Velocity of water
It is dependent on mass (m) and square of velocity (v).
So when water running at such a high speed is stopped abruptly then this energy will get converted into pressure energy which will be dependent on the speed at which the velocity of fluid was reduced. If it will be sudden then due to sudden change pressure rise will be sudden, if it is gradual pressure rise will be gradual. This all phenomenon happens due to the conversion of one form of energy into another.
## Water Hammer Equation & Calculation
Wave will travel from valve to tank and from tank to valve, T
• T = Distance Travelled by Wave / Velocity of Pressure Wave
• T = (L+L) / C
• T = 2L/C
Where
• L = the length of pipe
• C = the velocity of pressure wave
To understand this we need to go through this example
Consider a long pipe is carrying liquid and valve is attached to it at extreme end. On one end there is a tank and on another end, the valve is attached. This valve is being closed slowly.
Let,
• A= area of cross section of pipe
• L = length of pipe
• V = velocity of flow of water inside pipe
• t = time taken to close valve which is in seconds
• P = intensity of pressure wave produced
So the mass of liquid in the pipe is m = ρ A L
Let us just assume that closure of valve is being done such that liquid flowing through the pipe is brought to halt with uniform de-acceleration. From initial velocity V to zero in time t.
Condition – when the valve is closed gradually
Retardation of water = (v – 0)/t = v/t
The axial force which is producing retardation, F1
• F1 = mass x retardation
• F1 = ρ A L x v/t
The force created by pressure wave, F2
• F2 = P x A
When we equate these forces,
• F1 = F2
• ρ A L x v/t = P x A
• P = ρ L v/t
Then the head of pressure H,
• H =P/W [Here w was a specific weight]
• H = ρ L v/Wt
• H = ρ L v/ρgt [W = ρg]
• H = L v/gt
So the head of pressure will be, H = L v/gt
The closure of the valve is done very gradually when t > 2L/C
Closure of valve is said to be instantaneous when t < 2L/C
Where C = Velocity of the pressure wave.
## What Causes Water Hammer? Sound, Noise
After getting the basic idea about water hammer, let’s check what causes water hammer. So, what are the main causes of water hammer? The causes of water hammer shall be as follows,
#### Cause#1 Sudden Valve Closure
Sudden closure of valves, it may create water hammer sound or noise continuously. Think it in a simple example. If you run in a track, suddenly someone stops you! You will feel a shock. Now, if you run faster, your shock will be more and if you run at a slower speed, your shock will be less.
• In case of water, the philosophy is same and sudden closure will create a shock which creates water hammer.
• This shock is called as hydraulic shock.
• If the water velocity in the pipe is high, hydraulic shock will be high i.e., water hammer will be high.
• If the water velocity in the pipe is low, hydraulic shock will be low i.e., water hammer will be low.
#### Cause#2 Volume Difference Between Steam & Water & Heat Loss
In case of condensate pipes, due to heat loss, condensate starts to change its phase into water. As steam occupies 1600 times more volume than water, if steam changes its phase into the water, a huge vacuum will be created. Water will try to fill that vacuum and can create a water hammer.
• If the volume of steam condenses is more, vacuum will be more and water velocity will be high. So, water hammer will be high.
• If the volume of steam condenses is less, vacuum will be less and water velocity will be low. So, water hammer will be low.
#### Cause#3 Velocity Difference Between Steam & Water
In case of condensate pipes, as the velocity of steam is 10 times of water velocity, steam goes faster in the pipe. Now, steam starts to condense, and the amount of condensate increases in the pipe. Many times, this condensate fills the pipe and create a barrier, which causes a huge pressure difference between the two sides of that barrier.
Downstream pressure will be drastically reducing and the water will try to fill that gap & create a water hammer.
#### Cause#4 Float Valve Stability
If float valves are not stable, and it up and down as well as creating a web into the system,
#### Cause#5 Blocking of Suction Pipe
Blocking of suction pipe can also be a reason for water hammer.
#### Cause#6 Fitting Problems
Improper fittings of pipes or between pump and piping may create a water hammer.
#### Cause#7 Trip of Pump
In many cases, a trip of the pump may be a reason for creating shock in pipes and creates water hammers.
Apart from above, there may some other reasons like,
• In case, valves are worn
• Disconnect the power suddenly, etc.
## What are the Bad Effects of Water Hammer?
We have learned what is the cause of water hammer in pipelines? Let’s see the effects of water hammer one by one,
#### System Failure
Water can make a total system failure by a sudden surge in the piping system. It does damage to equipment such as flow meters, pressure gauges, etc. because such instruments have a delicate inner structure that is not designed to withstand such high-pressure waves.
#### Pump Damage
As we have learned that shock due to water hammer depends on many factors. A big shock can fail the entire system, however, a continuous small shock can damage the pump, its associated equipment, etc.
#### Pipe Damage
Due to this uneven pressure or shock, the associated piping system will be damaged, which can damage the entire system as well.
It will incur leakage in the piping system.
#### Lower Efficiency
Due to leakage in the piping or pipelines, which will reduce the head requirements, as well as the efficiency of the system.
#### Electrical System Damage & Safety
Due to the water leakage, running electrical system may come into contact with water and the electrical system will be damaged.
Short circuits may happen and electrical hazards may be encountered, which is totally unsafe for the workers.
#### Support System Damage
Due to high shock, supports of piping or pipelines may be damaged.
#### Operation & Maintenance Issues
Due to continuous problems, it increases the maintenance issues.
#### Water Hammer Noise or Sound
Due to the shock, the water hammer creates sound or noise.
#### Accident
Damage and destruction of piping leading to the industrial accident. Especially steam plants, where condensate travels at slow speed but steam always travels at high speed, in such cases water gets dragged by steam at very high speed which is 10 times speed of the water.
#### Unanticipated external damages
If the fluid that is being transported is corrosive then it will damage pipelines, external environment. It might pollute the environment. Leakages can be harmful to close by electrical equipment, components, constructed infrastructure. It can also damage the piping support system.
#### Hazardous accident
Leakages can cause an impact on the health of maintenance personnel. Slippage, electric shock, or anything can happen. Operating employee’s life will be in constant danger.
#### Time lost to maintenance
Replacement of pipelines or equipment will be the most time consuming and costly affair. While the system will be down for repairs that will reduce revenue and increase financial losses. Specific point damage in such cases holds the potential to keep the whole system in downtime for maintenance.
#### Environmental pollution
If corrosive harmful fluid gets released out, not only it will affect employee and maintenance team but also it will pollute the environment and affect the image of the company.
In such cases, water pressure becomes very high and may incur a big accident.
## Water Hammer Phenomenon in Home Plumbing
Many times we hear different noises after we turn of tap very fast as if some ghosts are laughing. These are in fact noises made by shock wave which is traveling from pipes. The reason is the same here as well. When the tap is switched off, incoming water from suddenly stops at the tap and this causes the conversion of energy. This energy travels in form of a shock wave which makes that noise. It is not water as much but the shock wave which is exerting pressure on the pipe is making it noise.
### How to solve these home plumbing issues?
#### Air vessels
These are equipment as the name suggests are vessels that accommodate air inside them. It is like a small water bottle size container that has air inside it. Basically, it works on the principle of compressibility. Water and air are present inside. Air is collected in top parts, while water is at the bottom.
Air is compressible to a large extent while water is not compressible. So when a shock wave comes, this water in the air vessel exerts pressure on the air. As earlier mentioned, the air gets compressed by this force, so the whole intensity of the shock wave gets absorbed in this air vessel making it shock proof.
Limitation
There is no separation of air and water in a vessel. So if the air is not filled in the vessel periodically, the present air inside it gets absorbed in water which reduces the effectiveness of the device.
If the whole vessel is occupied by water, then the whole purpose of the air vessel gets defeated. Sometimes bacterial or fungal spread is seen inside it. This is because direct point of contact between trapped air and water.
#### Water hammer arrester
In order to remove the limitations of the earlier devices, these devices are produced. The specialty of this device is air and water don’t make contact. Container separates two by the presence of plunger inside it. The plunger is making air-tight contact with the vessel wall due to the presence of O seal rings (like piston rings).
This allows the plunger to slide up and down in the vessel without leakage of air and water. As soon as the shock wave comes plunger moves up and there by compresses air in the top portion of the vessel. This causes the water in the pipeline to exert pressure created by shock on-air via the plunger.
The intensity of the shock wave gets absorbed and the water hammer effect is removed. This device has one more advantage, fungal or bacterial spread can’t happen in vessels. This is because water and trapped air don’t make contact. These devices are costly but they last for a long time.
## Water hammer phenomenon in hydroelectric plants
Just to have a basic understanding, first, let’s understand the basic parts of the plant in one-liner explanation.
• Reservoir – it is a place where water is collected. Height and amount of water it can hold decides capacity of power plant.
• Trash rack – it’s like filter system, it collects big stone and debris.
• Dam – Strongly constructed concrete wall which is making sure reservoir water is collected behind it.
• Sluice gate valve – it is flow control valve for water which flows in penstock.
• Forebay – it is used in when surge tanks are not used. Its basic function is to absorb shock wave.
• Surge tank – it is used to absorb shock wave and reduce effect of water hammer. It protects penstock.
• Pen stock – long pipeline which carries water from reservoir to turbine station.
• Turbine – Electricity gets produced as potential energy of water gets converted into kinetic energy in penstock. From kinetic energy, turbine blade rotate. This energy is converted into electricity later.
• Draft tube – Big diameter size pipe to remove water out from turbine station.
• Tailrace – it is channel which carries water from draft tube.
Hydroelectric power plants are very costly to construct. Hence proper design and maintenance is a must. Here for the concept of water hammer focus on penstock. When a large amount of water is being carried and if the turbine has to be stopped then this can’t be done suddenly.
Even it is done slowly it will have many reactions. Water coming down from very height when stopped will create strong pressure waves. So to absorb energy following devices are used.
### Forebay
It is similar to a surge tank. It is a big tank of water from near the reservoir. When turbine load decreases, and flow control valve which feeds water to the turbine is getting closed, in such instances water which was inside penstock valve will come to rest and kinetic energy will get converted into pressure energy. This shock wave will travel along the whole penstock and can damage it severely.
So this energy gets absorbed by forebay where water level increases and extra energy from system gets removed.
### Surge tank
It is located away from the reservoir and its working principle is the same as forebay. It is used mostly when space for forebay is not there. These are open tanks connected to penstock somewhere in middle. As soon as a shock wave is created in penstock, the level in surge tanks increases and sometimes water flies upwards due to the intensity of the shock wave. These devices remove unwanted pressure energy from the system.
It has many types such as
• Simple surge tank
• Restricted orifice surge tank
• Differential surge tank
## Water hammer in steam plants
In steam plants, this happens when steam lines are long and the plant was lying idle earlier. So hot steam and cold pipes are the idle conditions for the creation of condensate. It happens most of the time when condensates get mixed with steam.
As soon as steam gets created in the boiler it is carried by long steam lines to its application site that may be a turbine or heat exchanger. As this steam is heated but it starts to lose its heat in this journey loses its enthalpy and gets converted into condensate.
Many conditions play a big role here such as insulation, outside temperature, and the nature of steam. This steam condenses to form a condensate. When the system is started at the beginning, the whole system was cold and idle earlier so until it gets heat up, it absorbs steams heat and this causes a reduction in steam temperature and formation of condensate in form of droplets. This explains why condensate proportion is more at the start.
So when steam travels, these condensate particles/ droplets act like small particles of slug and create hindrance to the flow of steam. Just to clarify here, the slug is part of condensate shaped such that it occupies a whole cross-section of the pipe whilst traveling. So with steam, it is also traveling via steam line.
When sudden stoppage, orifice, bend shows up this slag gets obstructed suddenly. So the amount of kinetic energy it was carried by virtue of its velocity suddenly gets converted into potential energy. The stress created in this situation has to be absorbed by the steam line.
## How powerful is impact of this?
Saturated steam travels at speed of 20 to 30 m/s while water travels in pipes at speed of 2 to 3 m/s. when this slow-moving water gets dragged by steam at speed 10 times of original speed, this phenomenon of water hammer occurs. So in steam plants, a water hammer is associated with steam moving at high velocity.
## How to Stop or Fix Water Hammer?
Let’s see how to stop or fix water hammers?
### Avoid Water Hammer in Pipelines
• Increase diameter of pipe
• Close valve slowly
• Provide surge tank
• Pipe material should be elastic
• Use of water hammer arrester.
• Use of pressure relief valve
• Use of water towers
### Avoid Water Hammer in Steam Plants
It is very difficult to eliminate water hammer from the steam systems, however, we can minimize it with the below steps:
• During starting of plant, condensate must be removed from pipes so that it doesn’t get mixed up with steam.
• Steam lines should be installed such a way that they must gradual slope in same as direction of flow.
• Used of steam traps.
• Bypass lines to remove condensate water.
• Sagging should be avoided in pipes.
• Insulation of pipes should be done in proper way because steam gets condensed in absence of insulation.
### Avoid Water Hammer by Air vessels
#### How to Avoid Water Hammers using Air Vessels?
It is pretty much simple in design. It is made of a big container and it is joined to the pipe. It is filled with air and water. As air is less dense than water it gets trapped by water coming from pipes.
When a shock wave appears in the pipeline it enters this vessel and air gets compressed for a time being due to this pressure. So this extra energy got absorbed.
#### Limitations of Air Vessels
With time this air gets absorbed in water and water starts to occupy space of air. So whole effectiveness of absorbing shock decreases.
Also, there are chances of the formation of bacteria in such areas.
### Avoid Water Hammer by Arresters
They are the same as air vessels. Just they are cleverly designed in such a way that water and air don’t make contact. Air is sealed in this chamber by providing a plunger with tightly sealed O rings.
This plunger is free to rotate up and down. So when a water shock wave comes this plunger moves up and compresses this air inside it. This way shock gets absorbed. It solves both limitations faced by air vessels. As there is no contact here between air and water, no bacterial spread can happen. Also stays there for a long time and doesn’t get absorbed in water.
## How to prevent it in industrial systems?
What type of check valve is being used also plays a big role in this issue. For example valves such as swing, tilting disc, or piston style check valves are the ones that are dependent on gravity and flow reversal to push the valve into the closed position.
So when valves close, shock waves come down and slam hardly on this valve. So the seal of the valve and its mechanism both get affected. After that wave travels throughout the pipeline.
Silent or spring-assisted check valves are constructed in a bit different way. Spring is inside which pulls back the valve into a closed position before reversal of flow takes place. This whole process reduces the probability of a water hammer.
Air chambers or vessels as discussed earlier for home plumbing can also be used in industry. It is easy in construction as well as easy to install. It is fitted in tee-fitting. A small size pipe is there, which has an empty air chamber inside. This air is trapped on one side plunger and on another side there is water.
When water expands and exerts pressure, the plunger compresses that air so this reduces the magnitude of hock wave and that unwanted energy gets absorbed instead of damaging the pipeline.
Along with this, the following steps can be taken –
• Either old system should be refitted.
• Pressure regulator and pressure reducer should be used at appropriate places.
• System operating pressure and other variables should be relooked and reduced if possible.
• Use of air chambers should be promoted at places which are more vulnerable to such shocks.
• Selection of proper check valves.
## Conclusion
This phenomenon of Water hammer happens in our life many times and we don’t even notice it sometimes. In our day we use tap or valve many times, we close them very fast due to our constant hurrying nature. This article was just to tell us that our small actions to have some repercussions. Hence we should care about handling same from next time. Hence, we have learned all about water hammer, definition, meaning, equation, etc. Any questions, please write to us.
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What Is Bollinger Band Indicator?
What Is Bollinger Band Indicator?
Have you ever heard of something called "Bollinger Bands"? Don't worry if it sounds a bit technical – I'm here to explain it in a way that's easy to understand. Bollinger Bands are like price lines that help us figure out whether something is cheap or expensive, just like when you shop for clothes or groceries.
What are Bollinger Bands?
Bollinger Bands were created by a person named John Bollinger. These bands show us a range of prices that something, like a stock or currency, usually stays within. Imagine it like a high and low price range for something you want to buy. These bands move up and down based on how much the prices change over time.
To use Bollinger Bands, we need to know two things: "Period" and "Standard Deviations." These are like the ingredients we use to cook something delicious. The default settings are 20 for the period and 2 for standard deviations, but you can change them according to what you like.
The Bollinger band indicator is mostly used in intraday trading and You can learn how to use Bollinger band indicator in our technical analysis course.
What's the Point of Bollinger Bands?
The main goal of Bollinger Bands is to help us know if prices are too high or too low. They work best when we use both the upper and lower bands together. They are like a pair of glasses that help us see things clearly. These bands also work better when we use them with something else called a "moving average." This helps us make better decisions about when to buy or sell something.
Understanding Bollinger Bands
Think of Bollinger Bands like a line that shows the highest and lowest prices something had over a certain time. The high prices stay close to the upper line, and the low prices stay near the lower line. This line helps us see patterns and trends in prices. When things are changing a lot, the line becomes wider, and when they're not changing much, it becomes narrower. This line also tells us if something is too cheap or too expensive compared to how it usually is.
Bollinger Bands Formula
Calculating Bollinger Bands is like following a recipe. We use three important things:
Simple Moving Average (SMA): This is the average of the prices over a specific time. For example, if we look at 20 days, we add up the prices of those 20 days and then divide by 20.
Standard Deviation: This shows how much the prices change from the average. It's like a measure of how crazy the prices are dancing. We use the same time period as the SMA.
Upper and Lower Bands: Bollinger Bands have a middle line (SMA) and two more lines above and below it. These lines show the price range that's considered normal. The upper and lower bands change based on the standard deviation and how far prices are from the average.
Here's the math for making Bollinger Bands (using 20 days and 2 standard deviations):
● Upper band = 20-day SMA + (20-day SD x 2)
● Middle band = 20-day SMA
● Lower band = 20-day SMA – (20-day SD x 2)
Using Bollinger Bands is like having a magic tool for trading. Let's see how:
● When prices are going up a lot, they often touch the upper band. This means things are getting more expensive.
● If the price goes down a bit and then goes back up, but not all the way to the upper band, it's still strong. This is a sign to buy when the price goes up again.
For Selling in Down Trends:
● When prices are going down a lot, they often stay near the lower band. This means things are getting cheaper.
● If the price goes up a bit and then comes back down, but doesn't go above the middle band, it's still a strong downtrend. This is a sign to sell when the price goes down again.
Spotting Patterns: W-Bottoms and M-Tops
These are like special shapes that Bollinger Bands help us see:
W-Bottoms: When the price goes low but not below the lower band, and then goes up again, it makes a "W" shape. This can mean prices might go up soon.
M-Tops: When the price goes high but not above the upper band, and then comes down, it makes an "M" shape. This can mean prices might go down soon.
Best Bollinger Bands Strategy
If you're looking to make smarter trading decisions, there's a helpful strategy involving Bollinger Bands that can guide you. Bollinger Bands might sound complicated, but we'll break it down into easy steps that anyone can understand.
Bollinger Bands are like virtual boundaries for stock prices. They help us figure out when a stock might go up or down. Imagine a rubber band around the price of a stock. When the rubber band is stretched a lot, it might snap back. Bollinger Bands work in a similar way. They show us when prices have moved away from their usual range, which could mean they'll come back soon.
Figuring out the market's details in a short timeframe by just looking at Bollinger Bands can be a bit tricky. But don't worry, we have a strategy that might help. Imagine Bollinger Bands like a stretchy band around the price. Sometimes, this band gets really tight, like it's squeezing the price. We call this the "Bollinger Band Squeeze pattern." This doesn't happen super often, maybe every 2 to 4 days or weeks. When you see this Squeeze pattern, it could mean there's a chance to make a trade.
Look at the picture below, you'll see rectangles highlighting the Squeeze patterns. If the squeeze lasts for a longer time, the trade you make might be bigger. Also, make sure to trade in the same direction as the breakout. This Bollinger Bands' Squeeze strategy is considered one of the best ways to reach your trading goals.
When prices stay in one place or move just a little, Bollinger Bands get close together. But here's the thing: when you're trading, it's smarter to use Bollinger Bands along with other signs, not just on their own. Taking time to understand the trends before you decide to buy or sell is really important.
Bollinger Bands Breakout Strategy
The Breakout Plan using Bollinger Bands, made by Chuck LeBeau and David Lucas in their 1992 book, works like this: if the price goes above the upper Bollinger Band and closes there, you might think about buying the next day. If the price goes back inside the band, you might want to sell. The opposite happens if you're selling: if the price goes below the lower Bollinger Band and closes there, you might consider selling. If it comes back inside the band, you could think about buying.
A "Simple Moving Average" (which is just a fancy name) of how prices close every day decides the center of the Bollinger Band. The top and bottom parts of the band are set using a certain number of times the standard change from that moving average. This helps figure out how wide the band is. This wider part is called the "channel." It tells us where prices might go.
The Breakout Plan starts the next day when the price closes above the upper Bollinger Band or below the lower Bollinger Band. It's time to leave the trade when the price goes back inside the Exit Band. This Exit Band is determined by another number of times the standard change from the moving average, just like the Entry Threshold.
On the day you start the trade, this Exit Band is your stop. This helps you decide how much to trade using a certain formula.
Parameters of Bollinger Bands Breakout Strategy
Three Parameters Shaping Bollinger Breakout Trading System's Entry and Exit:
Three Numbers that Control When to Get In and Out:
Close Average: This tells us how many days we should use to figure out the center of the Bollinger Band.
Entry Threshold: This number decides how wide the band should be. It also helps us know when to start a trade if the price goes over this number of times the standard change from the moving average.
Exit Threshold: If this is zero, you exit when the price is below the moving average. If it's higher, you leave when the price is below a different number of times the standard change. If it's a negative number, you leave when the price is below the moving average plus that number of times the standard change.
For example, if the Entry Threshold is 3 and the Exit Threshold is 1, you might start a trade when the price is more than 3 times the standard change above the moving average. You might leave when it's less than 1 time the standard change above the moving average.
So, in simple words, Bollinger Bands are like stretchy bands around prices. When they squeeze, it's a sign to pay attention. And the Breakout Plan helps you decide when to trade and when to stop based on these squeezes and movements. Just remember, it's wise to use Bollinger Bands along with other hints, and take time to understand how things are going before you make decisions about buying or selling.
Double Bollinger Band Strategy
The Double Bollinger Band strategy presents traders with an uncomplicated yet powerful approach to enhance their trading choices. Here, we explain the crucial elements of this strategy and present easy explanations to facilitate its application.
Required Components:
1. Two Bollinger Bands:
● Bollinger Band 1: With a length of 20 and a standard deviation (StdDev) of 1.
● Bollinger Band 2: Also having a length of 20, but with a StdDev of 3.
2. Confirmation Indicators:
RSI (Relative Strength Index): Used for trend identification.
Stochastics (Stoch): Employed for entry and exit confirmations, along with overbought and oversold indications.
3. Risk Reward Ratio (RRR):
Recommended RRR: 1:2 or 1:1.5; This should be adjusted according to your risk tolerance.
4. Rules of Engagement:
Entry Signal:
● A candlestick crossing above/below the Bollinger Band with StdDev 1 triggers an entry signal.
● Successive candles following the initial crossover are potential entry opportunities.
● Place the stop loss at the high or low of the crossover candle, or at the middle line of the Bollinger Band—whichever is closer.
● Set the take profit target at a ratio of 1:2 or as per your personal risk comfort.
Avoid Under These Conditions:
● Do not trade if candlesticks show extended bodies or long wicks that cross the Bollinger Band with StdDev 1.
● Ensure alignment: The trend of the candlestick, RSI, and Stoch should all fit together.
● Entry must happen within two candlesticks following the initial Bollinger Band crossover with StdDev 1.
Key Focus: Stop Loss Placement
● Accurate calculation of the stop loss is of utmost importance. Do not initiate a trade until you fully understand the process of determining an appropriate stop loss.
● Calculate the stop loss before making an entry to keep it minimal. For instance, within the Nifty index, a recommended limit of 40 points is suggested.
FAQs
Que 1. What are the limitations of Bollinger Bands?
Ans: While Bollinger Bands offer valuable insights for technical traders, it's essential to acknowledge their limitations before implementing them. One key limitation to note is that Bollinger Bands are fundamentally reactive in nature, lacking predictive capabilities. These bands respond to price fluctuations, whether in uptrends or downtrends, but they do not forecast future prices. Much like many other technical indicators, Bollinger Bands exhibit lagging tendencies in their indications. This can be attributed to their foundation in a simple moving average, which calculates the average price across multiple price bars.
Que 2. Which is the best indicator to use with Bollinger Bands for intraday trading?
Ans: For intraday trading, Bollinger Bands are frequently paired with the relative strength indicator (RSI) and the BandWidth indicator. The BandWidth indicator quantifies the width of the bands relative to the middle band, providing further context to market dynamics.
Que 3. What is the accuracy of Bollinger Bands?
Ans: Bollinger Bands have the capability to encompass approximately 90% of the price movement within a specific asset or cryptocurrency. Instances where the asset's price surpasses or falls below a designated Bollinger band indicate the emergence of potential trading opportunities.
Que 4. Is Bollinger Band a lagging or leading indicator?
Ans: Bollinger Bands can be classified as lagging indicators due to their construction around a 20-day simple moving average (SMA) along with two outer boundary lines. These outer bands represent positive and negative standard deviations from the SMA, serving as a measure of market volatility.
Que 5. How to prevent Bollinger Bands from generating fake signals?
Ans: To reduce the likelihood of false signals and erroneous fluctuations when employing Bollinger Bands, integrating volume as a confirmation tool is advisable. Generally, instances of high volume coinciding with price breakouts beyond the upper or lower bands indicate robust buying or selling pressure, potentially signaling a shift in the prevailing trend. | 2,801 | 12,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.954874 |
http://www.diablofans.com/forums/diablo-iii-class-forums/demon-hunter-the-dreadlands/53745-n-depth-look-at-hungering-arrow?cookieTest=1 | 1,404,965,043,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776401658.57/warc/CC-MAIN-20140707234001-00022-ip-10-180-212-248.ec2.internal.warc.gz | 247,291,787 | 24,079 | ## n-Depth look at Hungering Arrow
• One is never hurt by being given additional choices, only by taking them away.
A QUADRILLION MAGIC FIND is worthless if you can't kill shit!
• interesting, but i think Cinder and Spray of Teeth still pierce, so shouldn't those be sums as well?
Also, i think the formula for Devouring may not be correct, the damage increase only occurs after a split, so i think it should be .35^n-1 x .7^n
(its been a while since i've taken a math class)
• your part on spray of teeth is lacking...
the total damage done from the explode can vary allot and can sometimes i bet everything else, but also not do it more often
i still like the idea of Shatter shot. makes it an aoe ability
fire 1 arrow, has 35% chance of exploding
of these 3 arrows, everyone got 35% chance of exploding again... and so on.
and its 35% that means its a bit higher then 1 of 3 explode again
but then again, i understand that the +70% damage on pierce is good aswell
Game Designer - Micro Design
• Spray of Teeth is terribly wrong if it's http://d3db.com/skil...hungering-arrow
Scatter is also incorrect, but there's information missing. For example, do the 3 resulting arrows have the same behavior? If not, it's just:
115 + .35 * (3 * 115) = 235.75
If it does... I would need to sit down longer to simplify it. Recursively, it's:
Arrow = 115 + .35 * (3 * Arrow)
Problem with the image is, if it doesn't pierce the first time, it can't pierce a second time.
Also, i think the formula for Devouring may not be correct, the damage increase only occurs after a split, so i think it should be .35^n-1 x .7^n
The Devouring in image is correct, though missing the 'x' to keep it consistent. When n is 0, presumably for 0 targets pierced, it has no effect on the damage.
If Cinder continues to pierce (which it should because Blizzard has stated you should never want the unruned), the results are too timing based to successfully calculate with a simple equation. However, assuming it hits a new target every time (best case), it's simply:
Sum(150 * .35^n) = 230.77
If we assume it takes 1s to hit the same target every time (worst case), it's
Sum((115 + 35/3) * .35^n) = 194.87
194.87 - 230.77, giving it a decent place as the second rune you get.
Spray of Teeth is a little bit more complicated due to the unknowns about its aoe range and the situational unknowns.
For M targets in the aoe, and C for crit chance
SUM((115 * .35^n) + (.35^n * C * ((M * 50) + 57.5)))
.35^n chance to hit the nth target, but each hit is an independent crit check to do 50% WD on M targets. Because you're critting however, you have to add an additional 57.5% for the regular crit damage bonus of 50%.
If we assume M=3, you need a crit chance of ~17% to get 231, beating Puncturing. As M and C increase, it will do much, much more.
• Quote from PantheraOnca
interesting, but i think Cinder and Spray of Teeth still pierce, so shouldn't those be sums as well?
Also, i think the formula for Devouring may not be correct, the damage increase only occurs after a split, so i think it should be .35^n-1 x .7^n
(its been a while since i've taken a math class)
n-1 would have the first number in the series being
0.35^(0-1) = 0.35^(-1)
When they both start at 0, you are just using the initial dmg and there is no piece.
Yah those 2 will likely still pierce, good catch, ill fix it.
One is never hurt by being given additional choices, only by taking them away.
A QUADRILLION MAGIC FIND is worthless if you can't kill shit!
• Now that I look scatter probably pierces the 1st target, and then splits and thats it. I would think those 3 arrows are smart arrows that still seek, but they won't split off again. But they could be dumb arrows as well. It would obviously be stupidly OP if all 3 could continue to piece.
One is never hurt by being given additional choices, only by taking them away.
A QUADRILLION MAGIC FIND is worthless if you can't kill shit!
• Your assumption for Shatter Shot that only one out of the 3 arrows after the split continue to pierce seems odd. I've been assuming that all 3 can pierce, but if that's not the case the second most likely option would seem like none of them can pierce.
I posted a reply to a similar reddit thread earlier today with details on how to calculate these infinite sums as well if anyone is interested: http://www.reddit.co...ungering_arrow/
Quote from Nivius
i still like the idea of Shatter shot. makes it an aoe ability
fire 1 arrow, has 35% chance of exploding
of these 3 arrows, everyone got 35% chance of exploding again... and so on.
and its 35% that means its a bit higher then 1 of 3 explode again
Shatter Shot absolutely will not work like this. If every pierce split and each of the 3 arrows had a chance to pierce causing it to split again, etc, then the expected damage from a single shot would be infinite! (After the first pierce you have 3 chances at 35% to pierce so, on average, you will get 1.05 pierces, which leads to an exponential explosion in damage).
• Shatter Shot absolutely will not work like this. If every pierce split and each of the 3 arrows had a chance to pierce causing it to split again, etc, then the expected damage from a single shot would be infinite! (After the first pierce you have 3 chances at 35% to pierce so, on average, you will get 1.05 pierces, which leads to an exponential explosion in damage).
Most likely, but without testing or word from Blizzard...
Even without that though, it's in a good place as far as the runes go, doing a little bit more than Puncturing. Right now, Cindering is easily the weakest, but I forget if DH's get any kind of synergy with burning targets.
• Quote from chippydip
Your assumption for Shatter Shot that only one out of the 3 arrows after the split continue to pierce seems odd. I've been assuming that all 3 can pierce, but if that's not the case the second most likely option would seem like none of them can pierce.
I posted a reply to a similar reddit thread earlier today with details on how to calculate these infinite sums as well if anyone is interested: http://www.reddit.co...ungering_arrow/
Quote from Nivius
i still like the idea of Shatter shot. makes it an aoe ability
fire 1 arrow, has 35% chance of exploding
of these 3 arrows, everyone got 35% chance of exploding again... and so on.
and its 35% that means its a bit higher then 1 of 3 explode again
Shatter Shot absolutely will not work like this. If every pierce split and each of the 3 arrows had a chance to pierce causing it to split again, etc, then the expected damage from a single shot would be infinite! (After the first pierce you have 3 chances at 35% to pierce so, on average, you will get 1.05 pierces, which leads to an exponential explosion in damage).
Shatter most certainly can't have all 3 new arrows smart pierce as the potential dmg output would be outrageous. The dmg output would be about 708% wpn dmg per shot.
Do you guys know if the pierced arrow prioritizes other mobs in the vicinity or the mob it hit?
One is never hurt by being given additional choices, only by taking them away.
A QUADRILLION MAGIC FIND is worthless if you can't kill shit!
• I believe the arrow homing prioritizes the closest to the arrow itself.
• Quote from jaclashflash
Shatter most certainly can't have all 3 new arrows smart pierce as the potential dmg output would be outrageous. The dmg output would be about 708% wpn dmg per shot.
I think you did something wrong in your calculations. If all 3 pierce after the split then the average damage is about 300%, making it the best rune, but not wildly better than Devouring.
The damage formula is:
115 + 0.35 * 3 * SUM(0, inf, 115 * 0.35^n)
(initial hit plus 35% chance to get 3 arrows that are effectively each a non-runed version)
= 115 * (1 + 3 * 0.35 * SUM(0, inf, 0.35^n)
= 115 * (1 + 3 * 0.35 + 3 * 0.35^2 + 3 * 0.35^3 + ...)
= 115 * (1.7 / 0.65)
~= 301%
• I think you did something wrong in your calculations. If all 3 pierce after the split then the average damage is about 300%, making it the best rune, but not wildly better than Devouring.
You are assuming the behavior of the skill by stating that the 3 spawned act as normal non-runed hungering arrows, just as he assumed the behavior that the 3 spawned act as runed Shatter Arrows, though his number should really be infinite.
• Quote from chippydip
Quote from jaclashflash
Shatter most certainly can't have all 3 new arrows smart pierce as the potential dmg output would be outrageous. The dmg output would be about 708% wpn dmg per shot.
I think you did something wrong in your calculations. If all 3 pierce after the split then the average damage is about 300%, making it the best rune, but not wildly better than Devouring.
The damage formula is:
115 + 0.35 * 3 * SUM(0, inf, 115 * 0.35^n)
(initial hit plus 35% chance to get 3 arrows that are effectively each a non-runed version)
= 115 * (1 + 3 * 0.35 * SUM(0, inf, 0.35^n)
= 115 * (1 + 3 * 0.35 + 3 * 0.35^2 + 3 * 0.35^3 + ...)
= 115 * (1.7 / 0.65)
~= 301%
Yah my bad, forgot to take the 35% chance into account, and forgot the initial arrow split into 3. That # is certainly plausible then.
There I think I have it fixed about as good as it can get anyways.
One is never hurt by being given additional choices, only by taking them away.
A QUADRILLION MAGIC FIND is worthless if you can't kill shit!
• Actually now that I look at it, they recently just nerfed puncturing arrow from 55% to 50% didnt they? That means they nerfed the dmg from 255% to the 230%.
Now, if shatter arrow doesn't allow piercing its at 230% range with the other 3. Then that would only leave devouring as an outlier. However, seeing as it drastically increases in dmg per pierce it has a high chance to overkill, thus lowering the overall effective dmg.
Therefore, I think its safe to assume that shatter arrow's won't be allowed to pierce. However, if that is not the case then shatter becomes the only choice, as it will be the best in pretty much any situation.
Do you want to get scammed? Perhaps a nice keylogger?
"Just google "diablo 3 gold guide" and magical rainbow covered demons will assault your eyes."
• This is where gaming goes too far. I am meant to blow stuff up after work not have a headache looking at some nerds forum post
• Quote from Casanova
... looking at some nerds any forum post
there we go.
• Quote from jaclashflash
There I think I have it fixed about as good as it can get anyways.
If you want to include a closed-form version of Spray of Teeth it would be:
SUM(0, inf, (115 + 50CM) * 0.35^n)
= (115 + 50CM) * SUM(0, inf, 0.35^n)
= (115 + 50CM) * (1 / 0.65)
= (115 + 50CM) / 0.65
At CM = 1.6 the total damage is 300%, making it equal to Shatter Shot with piercing arrows. This requires hitting at least 3 mobs at a ~53% crit rate, 4 at a 40% crit rate, 5 at a 32% crit rate, etc.
Quote from Mysticjbyrd
Now, if shatter arrow doesn't allow piercing its at 230% range with the other 3. Then that would only leave devouring as an outlier. However, seeing as it drastically increases in dmg per pierce it has a high chance to overkill, thus lowering the overall effective dmg.
Therefore, I think its safe to assume that shatter arrow's won't be allowed to pierce. However, if that is not the case then shatter becomes the only choice, as it will be the best in pretty much any situation.
If Shatter doesn't pierce then I wonder if it would lose its homing ability as well? If there are 4+ monsters around it wouldn't really matter, but against a boss it would be worse than the non-runed version if it split and never turned around to come back and actually hit the boss.
You also bring up a good point with Devouring Arrow. It's important not just to look at average damage, but also the probability distribution of that damage. With a 35% chance to pierce, ~96% of all attacks will hit 3 times or fewer. With the normal un-runed version, the first 3 hits account for~ 96% of the overall damage as well. With Devouring Arrow on the other hand, those first 3 hits which happen ~96% of the time only account for ~79% of overall damage. This leaves ~21% of the average damage is coming from that ~4% chance of getting 4 or more hits.
Here's a comparison of roughly 95th-percentile damage:
No Rune: ~169% (at 3 hits max ~= 96th percentile)
Puncturing Arrow: ~216% (at 4 hits max ~= 94th percentile)
Puncturing Arrow: ~223% (at 5 hits max ~= 97th percentile)
Devouring Arrow: ~224% (at 3 hits max ~= 96th percentile)
Shatter Arrow: ~235% (at 4 hits max ~= 65th percentile--assuming no piercing after the first, this is max damage)
This means that, even without additional piercing, Shatter Arrow will do more damage than the other rune options most of the time. It also hits slightly more times on average than Puncturing Arrow (2.05 vs 2.00) if you are fishing for procs. Looking at it this way, it definitely looks like it would be most balanced if the arrows retained their homing ability, but not their piercing ability after the split.
• Quote from Kyrenin
If Cinder continues to pierce (which it should because Blizzard has stated you should never want the unruned),
Not a Sound Assumption. The reason you don't want this unruned is because there are runes that make it operate better with the same principals. This could remove the seeking portion and add the fire damage and other runes would be taken if you want it to operate identically but better (Puncturing arrow comes to mind)
If that made sense to you, Bravo! I think I even confused myself...
• Quote from Raptorbonz42
Quote from Kyrenin
If Cinder continues to pierce (which it should because Blizzard has stated you should never want the unruned),
Not a Sound Assumption. The reason you don't want this unruned is because there are runes that make it operate better with the same principals. This could remove the seeking portion and add the fire damage and other runes would be taken if you want it to operate identically but better (Puncturing arrow comes to mind)
If you remove the pierce & seek from this ability, then it would operate worse than the unruned version, far worse.
% wpn dmg: 150
The DoT likely doesn't stack, which is why there is a minimum and maximum dmg range for that ability.
@Chippydip Yah I see Shatter maintaining the homing ability, but not pierce. Of course, it could definitely go another way too.
Do you want to get scammed? Perhaps a nice keylogger?
"Just google "diablo 3 gold guide" and magical rainbow covered demons will assault your eyes." | 3,769 | 14,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2014-23 | longest | en | 0.956621 |
http://read.pudn.com/downloads719/sourcecode/windows/2881250/task_2.m__.htm | 1,579,819,963,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250614086.44/warc/CC-MAIN-20200123221108-20200124010108-00457.warc.gz | 130,609,212 | 2,053 | www.pudn.com > gps.zip > task_2.m, change:2015-11-25,size:2684b
```%数据处理 Kalman滤波
clear
clc
[GGA,Time,LAT,N,LON,E,STATUS,QUANTITY,HDOP,AltS,M,AltE,m,a] = textread('跑车数据.txt','%s %f %f %s %f %s %f %f %f %f %s %f %s %s','delimiter', ','); %#ok<DTXTRD>
LAT=LAT(LAT~=0);
LON=LON(LON~=0);
HDOP=HDOP(HDOP~=0);
Time=Time(Time~=0);
Time = floor(Time/10000)*3600 + floor((Time - floor(Time/10000)*10000)/100)*60 + Time - floor(Time/100)*100; %%换算时间
AltS=AltS(AltS~=0);
lat = floor(LAT/100) + (LAT - floor(LAT/100)*100)/60; %换算纬度
lon = floor(LON/100) + (LON - floor(LON/100)*100)/60; %换算经度
figure;
plot(lat,lon);
ylabel('经度');
xlabel('纬度');
Re = 6371393;
r=Re+AltS;
x = r.*cos(lat/180*pi).*cos(lon/180*pi); %转换到地理直角坐标系
y = r.*cos(lat/180*pi).*sin(lon/180*pi);
z = r.*sin(lat/180*pi);
figure;
subplot(3,1,1);
plot(Time,x);
title('x-t');
xlabel('time(s)');
ylabel('x_distance(m)');
subplot(3,1,2);
plot(Time,y);
title('y-t');
xlabel('time(s)');
ylabel('y_distance(m)');
subplot(3,1,3);
plot(Time,z);
title('z-t');
xlabel('time(s)');
ylabel('z_distance(m)');
%Kalman filtering
%利用低动态-定常速度模型
F=[zeros(3),eye(3);zeros(3),zeros(3)];
I=eye(6);
syms s t; %转移矩阵的离散化
T=1;
FS=inv(s*I-F);
Ft=ilaplace(FS,s,t);
FT=subs(Ft,t,T);
phy=FT;
tao=[zeros(3),zeros(3);zeros(3),eye(3)];
Q=diag([50 50 50 100 100 100]'); %系统误差可调
x1=0; %滤波初值设置
y1=0;
z1=0;
for i=1:11
x1=x1+x(i);
y1=y1+y(i);
z1=z1+z(i);
end
x2=0;
y2=0;
z2=0;
for i=12:22
x2=x2+x(i);
y2=y2+y(i);
z2=z2+z(i);
end
vx0=(x2-x1)/100;
vy0=(y2-y1)/100;
vz0=(z2-z1)/100;
x0=x(21);
y0=y(21);
z0=z(21);
p_k_1=diag([80 80 120 30 30 45]');
p_k_k_1=zeros(6,1);
X_k_1=[x0;y0;z0;vx0;vy0;vz0];
X_k_k_1=zeros(6,1);
X_k=zeros(6,1);
Z_k=zeros(3,1);
x_k=zeros(800,1);
y_k=zeros(800,1);
z_k=zeros(800,1);
H_k=[eye(3),zeros(3)];
for i=22:821
Z_k=[x(i);y(i);z(i)];
R_k=diag([12.5*HDOP(i)^2,12.5*HDOP(i)^2,15*HDOP(i)^2]');
X_k_k_1=phy*X_k_1;
p_k_k_1=phy*p_k_1*phy'+tao*Q*tao';
K_k=p_k_k_1*H_k'/(H_k*p_k_k_1*H_k'+R_k);
p_k=(I-K_k*H_k)*p_k_k_1*(I-K_k*H_k)'+K_k*R_k*K_k';
X_k=X_k_k_1+K_k*(Z_k-H_k*X_k_k_1);
x_k(i-21)=X_k(1);
y_k(i-21)=X_k(2);
z_k(i-21)=X_k(3);
X_k_1=X_k;
p_k_1=p_k;
end
figure;
subplot(3,1,1);
plot(Time(22:821),x_k);
xlabel('t/s');
ylabel('x/m');
title('x postition - t');
subplot(3,1,2);
plot(Time(22:821),y_k);
xlabel('t/s');
ylabel('y/m');
title('y postition - t');
subplot(3,1,3);
plot(Time(22:821),z_k);
xlabel('t/s');
ylabel('z/m');
title('z postition - t');
``` | 1,166 | 2,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.17279 |
https://blablabeer.net/how-big-is-a-pint-of-beer/ | 1,695,405,757,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506421.14/warc/CC-MAIN-20230922170343-20230922200343-00262.warc.gz | 158,970,535 | 20,829 | # How Big Is A Pint Of Beer
How big is a pint of beer, you may be wondering. A pint of beer is 16 ounces. Most beer glasses are either 16 or 20 ounces, so when you order a beer in a bar or restaurant, you’ll get either a 16 or 20 ounce glass of beer. There is no standard size for a pint of beer, but 16 ounces is the most common size.
Contents
## How many beers is 1 pint?
A pint is a beer measurement that is 16 ounces. How many beers is 1 pint? That is a difficult question to answer because it depends on the alcohol content of the beers and the size of the drinker’s stomach.
## How big is a pint of beer UK?
A pint of beer in the UK is 568 ml.
## How big is a pint in Canada?
In Canada, a pint is equal to 568 milliliters. This is slightly larger than a U.S. pint, which is 568.26 milliliters.
## Is a pint bigger than a normal beer?
Is a pint bigger than a normal beer?
The answer to this question is a little bit complicated. In the UK and some other countries, a pint is actually defined as 568 milliliters. However, in the United States, a pint is defined as 16 ounces. So, in the United States, a pint is actually bigger than a pint in the UK.
This can be confusing for people who are visiting or travelling to different countries. For example, if you order a pint of beer in the UK, you will actually get more beer than if you order a pint of beer in the United States. This can be frustrating for people who are not aware of the difference in definitions.
See also How Do You Open A Beer Bottle With A Key
So, is a pint bigger than a normal beer? In the United States, a pint is bigger than a normal beer. In the UK, a pint is actually smaller than a normal beer.
## Can a pint of beer get you drunk?
Can a pint of beer get you drunk?
The answer to this question is yes, a pint of beer can get you drunk. However, the amount of alcohol in a pint of beer can vary depending on the type of beer. For example, a pint of light beer typically contains less alcohol than a pint of dark beer.
When you drink beer, the alcohol in the beer is absorbed into your bloodstream. The amount of alcohol that is absorbed into your bloodstream depends on the amount of alcohol in the beer, as well as your weight, sex, and age. Generally, the more alcohol in a beer, the more it will affect you.
If you drink a lot of beer, the alcohol can make you drunk. Being drunk can impair your ability to think clearly, make decisions, and react quickly. It can also affect your balance and coordination. If you drive a car or operate machinery while you are drunk, you could be seriously injured or killed.
So, can a pint of beer get you drunk? Yes, a pint of beer can get you drunk, but the amount of alcohol in the beer will vary depending on the type of beer. Additionally, the more beer you drink, the more likely you are to get drunk.
## How many beers is 2 pints?
A pint is a unit of volume or capacity in both the imperial and United States customary systems of measurement. It is traditionally equal to one-sixth of a gallon. A pitcher is a container that holds either a gallon or a half gallon.
In order to answer the question of how many beers are in 2 pints, it is important first to understand what a pint is. A pint is a unit of volume or capacity that is used in both the imperial system and the United States customary system of measurement. It is traditionally equal to one-sixth of a gallon. A pitcher is a container that holds either a gallon or a half gallon.
Once you understand what a pint is, you can then answer the question of how many beers are in 2 pints. Two pints is equal to two-thirds of a quart. In the imperial system, one quart is equal to two pints. In the United States customary system, one quart is equal to four pints. Therefore, two pints is equal to eight beers.
## Is 500ml a pint of beer?
It’s a question that has been asked for centuries – is 500ml a pint of beer? The answer, however, is not so straightforward.
The confusion arises because the term ‘pint’ can have different meanings, depending on the country or region. In the UK, for example, a pint is defined as 568ml. However, in the US, a pint is just 473ml.
So, which is it – 500ml or 568ml? The answer is, it depends on where you are. If you’re in the UK, then 500ml is a pint. But if you’re in the US, then 568ml is a pint.
Confused? You’re not alone! Even the British and American governments can’t seem to agree on what a pint is. In fact, in 2013, the two governments agreed to call a pint of beer 500ml in order to avoid any confusion. | 1,093 | 4,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-40 | longest | en | 0.934347 |
https://pretextbook.org/examples/sample-book/annotated/exercises-groups.html | 1,721,287,187,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00651.warc.gz | 408,063,645 | 15,475 | # PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)
## Exercises1.5Exercises
View Source
### 1.
Find all $$x \in {\mathbb Z}$$ satisfying each of the following equations.
1. $$\displaystyle 3x \equiv 2 \pmod{7}$$
2. $$\displaystyle 5x + 1 \equiv 13 \pmod{23}$$
3. $$\displaystyle 5x + 1 \equiv 13 \pmod{26}$$
4. $$\displaystyle 9x \equiv 3 \pmod{5}$$
5. $$\displaystyle 5x \equiv 1 \pmod{6}$$
6. $$\displaystyle 3x \equiv 1 \pmod{6}$$
Hint.
(a) $$3 + 7 \mathbb Z = \{ \ldots, -4, 3, 10, \ldots \}\text{;}$$ (c) $$18 + 26 \mathbb Z\text{;}$$ (e) $$5 + 6 \mathbb Z\text{.}$$
### 2.
Which of the following multiplication tables defined on the set $$G = \{ a, b, c, d \}$$ form a group? Support your answer in each case.
1. \begin{equation*} \begin{array}{c|cccc} \circ & a & b & c & d \\ \hline a & a & c & d & a \\ b & b & b & c & d \\ c & c & d & a & b \\ d & d & a & b & c \end{array} \end{equation*}
2. \begin{equation*} \begin{array}{c|cccc} \circ & a & b & c & d \\ \hline a & a & b & c & d \\ b & b & a & d & c \\ c & c & d & a & b \\ d & d & c & b & a \end{array} \end{equation*}
3. \begin{equation*} \begin{array}{c|cccc} \circ & a & b & c & d \\ \hline a & a & b & c & d \\ b & b & c & d & a \\ c & c & d & a & b \\ d & d & a & b & c \end{array} \end{equation*}
4. \begin{equation*} \begin{array}{c|cccc} \circ & a & b & c & d \\ \hline a & a & b & c & d \\ b & b & a & c & d \\ c & c & b & a & d \\ d & d & d & b & c \end{array} \end{equation*}
Hint.
(a) Not a group; (c) a group.
### 3.
Write out Cayley tables for groups formed by the symmetries of a rectangle and for $$({\mathbb Z}_4, +)\text{.}$$ How many elements are in each group? Are the groups the same? Why or why not?
### 4.
Describe the symmetries of a rhombus and prove that the set of symmetries forms a group. Give Cayley tables for both the symmetries of a rectangle and the symmetries of a rhombus. Are the symmetries of a rectangle and those of a rhombus the same?
### 5.
Describe the symmetries of a square and prove that the set of symmetries is a group. Give a Cayley table for the symmetries. How many ways can the vertices of a square be permuted? Is each permutation necessarily a symmetry of the square? The symmetry group of the square is denoted by $$D_4\text{.}$$
### 6.
Give a multiplication table for the group $$U(12)\text{.}$$
Hint.
\begin{equation*} \begin{array}{c|cccc} \cdot & 1 & 5 & 7 & 11 \\ \hline 1 & 1 & 5 & 7 & 11 \\ 5 & 5 & 1 & 11 & 7 \\ 7 & 7 & 11 & 1 & 5 \\ 11 & 11 & 7 & 5 & 1 \end{array} \end{equation*}
### 7.
Let $$S = {\mathbb R} \setminus \{ -1 \}$$ and define a binary operation on $$S$$ by $$a \ast b = a + b + ab\text{.}$$ Prove that $$(S, \ast)$$ is an abelian group.
### 8.
Give an example of two elements $$A$$ and $$B$$ in $$GL_2({\mathbb R})$$ with $$AB \neq BA\text{.}$$
Hint.
Pick two matrices. Almost any pair will work.
### 9.
Prove that the product of two matrices in $$SL_2({\mathbb R})$$ has determinant one.
### 10.
Prove that the set of matrices of the form
\begin{equation*} \begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix} \end{equation*}
is a group under matrix multiplication. This group, known as the Heisenberg group, is important in quantum physics. Matrix multiplication in the Heisenberg group is defined by
\begin{equation*} \begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & x' & y' \\ 0 & 1 & z' \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & x+x' & y+y'+xz' \\ 0 & 1 & z+z' \\ 0 & 0 & 1 \end{pmatrix}\text{.} \end{equation*}
### 11.
Prove that $$\det(AB) = \det(A) \det(B)$$ in $$GL_2({\mathbb R})\text{.}$$ Use this result to show that the binary operation in the group $$GL_2({\mathbb R})$$ is closed; that is, if $$A$$ and $$B$$ are in $$GL_2({\mathbb R})\text{,}$$ then $$AB \in GL_2({\mathbb R})\text{.}$$
### 12.
Let $${\mathbb Z}_2^n = \{ (a_1, a_2, \ldots, a_n) : a_i \in {\mathbb Z}_2 \}\text{.}$$ Define a binary operation on $${\mathbb Z}_2^n$$ by
\begin{equation*} (a_1, a_2, \ldots, a_n) + (b_1, b_2, \ldots, b_n) = (a_1 + b_1, a_2 + b_2, \ldots, a_n + b_n)\text{.} \end{equation*}
Prove that $${\mathbb Z}_2^n$$ is a group under this operation. This group is important in algebraic coding theory.
### 13.
Show that $${\mathbb R}^{\ast} = {\mathbb R} \setminus \{0 \}$$ is a group under the operation of multiplication.
### 14.
Given the groups $${\mathbb R}^{\ast}$$ and $${\mathbb Z}\text{,}$$ let $$G = {\mathbb R}^{\ast} \times {\mathbb Z}\text{.}$$ Define a binary operation $$\circ$$ on $$G$$ by $$(a,m) \circ (b,n) = (ab, m + n)\text{.}$$ Show that $$G$$ is a group under this operation.
### 15.
Prove or disprove that every group containing six elements is abelian.
Hint.
There is a nonabelian group containing six elements.
### 16.
Give a specific example of some group $$G$$ and elements $$g, h \in G$$ where $$(gh)^n \neq g^nh^n\text{.}$$
Hint.
Look at the symmetry group of an equilateral triangle or a square.
### 17.
Give an example of three different groups with eight elements. Why are the groups different?
Hint.
The are five different groups of order 8.
### 18.
Show that there are $$n!$$ permutations of a set containing $$n$$ items.
Hint.
Let
\begin{equation*} \sigma = \begin{pmatrix} 1 & 2 & \cdots & n \\ a_1 & a_2 & \cdots & a_n \end{pmatrix} \end{equation*}
be in $$S_n\text{.}$$ All of the $$a_i$$s must be distinct. There are $$n$$ ways to choose $$a_1\text{,}$$ $$n-1$$ ways to choose $$a_2\text{,}$$ $$\ldots\text{,}$$ 2 ways to choose $$a_{n - 1}\text{,}$$ and only one way to choose $$a_n\text{.}$$ Therefore, we can form $$\sigma$$ in $$n(n - 1) \cdots 2 \cdot 1 = n!$$ ways.
### 19.
Show that
\begin{equation*} 0 + a \equiv a + 0 \equiv a \pmod{ n } \end{equation*}
for all $$a \in {\mathbb Z}_n\text{.}$$
### 20.
Prove that there is a multiplicative identity for the integers modulo $$n\text{:}$$
\begin{equation*} a \cdot 1 \equiv a \pmod{n}\text{.} \end{equation*}
### 21.
For each $$a \in {\mathbb Z}_n$$ find an element $$b \in {\mathbb Z}_n$$ such that
\begin{equation*} a + b \equiv b + a \equiv 0 \pmod{ n}\text{.} \end{equation*}
### 22.
Show that addition and multiplication mod $n$ are well defined operations. That is, show that the operations do not depend on the choice of the representative from the equivalence classes mod $$n\text{.}$$
### 23.
Show that addition and multiplication mod $$n$$ are associative operations.
### 24.
Show that multiplication distributes over addition modulo $$n\text{:}$$
\begin{equation*} a(b + c) \equiv ab + ac \pmod{n}\text{.} \end{equation*}
### 25.
Let $$a$$ and $$b$$ be elements in a group $$G\text{.}$$ Prove that $$ab^na^{-1} = (aba^{-1})^n$$ for $$n \in \mathbb Z\text{.}$$
Hint.
\begin{align*} (aba^{-1})^n & = (aba^{-1})(aba^{-1}) \cdots (aba^{-1})\\ & = ab(aa^{-1})b(aa^{-1})b \cdots b(aa^{-1})ba^{-1}\\ & = ab^na^{-1}\text{.} \end{align*}
### 26.
Let $$U(n)$$ be the group of units in $${\mathbb Z}_n\text{.}$$ If $$n \gt 2\text{,}$$ prove that there is an element $$k \in U(n)$$ such that $$k^2 = 1$$ and $$k \neq 1\text{.}$$
### 27.
Prove that the inverse of $$g _1 g_2 \cdots g_n$$ is $$g_n^{-1} g_{n-1}^{-1} \cdots g_1^{-1}\text{.}$$
### 28.
Prove the remainder of Proposition 1.2.14: if $$G$$ is a group and $$a, b \in G\text{,}$$ then the equation $$xa = b$$ has a unique solution in $$G\text{.}$$
### 30.
Prove the right and left cancellation laws for a group $$G\text{;}$$ that is, show that in the group $$G\text{,}$$ $$ba = ca$$ implies $$b = c$$ and $$ab = ac$$ implies $$b = c$$ for elements $$a, b, c \in G\text{.}$$
### 31.
Show that if $$a^2 = e$$ for all elements $$a$$ in a group $$G\text{,}$$ then $$G$$ must be abelian.
Hint.
Since $$abab = (ab)^2 = e = a^2 b^2 = aabb\text{,}$$ we know that $$ba = ab\text{.}$$
### 32.
Show that if $$G$$ is a finite group of even order, then there is an $$a \in G$$ such that $$a$$ is not the identity and $$a^2 = e\text{.}$$
### 33.
Let $$G$$ be a group and suppose that $$(ab)^2 = a^2b^2$$ for all $$a$$ and $$b$$ in $$G\text{.}$$ Prove that $$G$$ is an abelian group.
### 34.
Find all the subgroups of $${\mathbb Z}_3 \times {\mathbb Z}_3\text{.}$$ Use this information to show that $${\mathbb Z}_3 \times {\mathbb Z}_3$$ is not the same group as $${\mathbb Z}_9\text{.}$$ (See Example 1.3.5 for a short description of the product of groups.)
### 35.
Find all the subgroups of the symmetry group of an equilateral triangle.
Hint.
$$H_1 = \{ id \}\text{,}$$ $$H_2 = \{ id, \rho_1, \rho_2 \}\text{,}$$ $$H_3 = \{ id, \mu_1 \}\text{,}$$ $$H_4 = \{ id, \mu_2 \}\text{,}$$ $$H_5 = \{ id, \mu_3 \}\text{,}$$ $$S_3\text{.}$$
### 36.
Compute the subgroups of the symmetry group of a square.
### 37.
Let $$H = \{2^k : k \in {\mathbb Z} \}\text{.}$$ Show that $$H$$ is a subgroup of $${\mathbb Q}^*\text{.}$$
### 38.
Let $$n = 0, 1, 2, \ldots$$ and $$n {\mathbb Z} = \{ nk : k \in {\mathbb Z} \}\text{.}$$ Prove that $$n {\mathbb Z}$$ is a subgroup of $${\mathbb Z}\text{.}$$ Show that these subgroups are the only subgroups of $$\mathbb{Z}\text{.}$$
### 39.
Let $${\mathbb T} = \{ z \in {\mathbb C}^* : |z| =1 \}\text{.}$$ Prove that $${\mathbb T}$$ is a subgroup of $${\mathbb C}^*\text{.}$$
### 40.
\begin{equation*} \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{equation*}
where $$\theta \in {\mathbb R}\text{.}$$ Prove that $$G$$ is a subgroup of $$SL_2({\mathbb R})\text{.}$$
### 41.
Prove that
\begin{equation*} G = \{ a + b \sqrt{2} : a, b \in {\mathbb Q} \text{ and } a \text{ and } b \text{ are not both zero} \} \end{equation*}
is a subgroup of $${\mathbb R}^{\ast}$$ under the group operation of multiplication.
Hint.
The identity of $$G$$ is $$1 = 1 + 0 \sqrt{2}\text{.}$$ Since $$(a + b \sqrt{2}\, )(c + d \sqrt{2}\, ) = (ac + 2bd) + (ad + bc)\sqrt{2}\text{,}$$ $$G$$ is closed under multiplication. Finally, $$(a + b \sqrt{2}\, )^{-1} = a/(a^2 - 2b^2) - b\sqrt{2}/(a^2 - 2 b^2)\text{.}$$
### 42.
Let $$G$$ be the group of $$2 \times 2$$ matrices under addition and
\begin{equation*} H = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a + d = 0 \right\}\text{.} \end{equation*}
Prove that $$H$$ is a subgroup of $$G\text{.}$$
### 43.
Prove or disprove: $$SL_2( {\mathbb Z} )\text{,}$$ the set of $$2 \times 2$$ matrices with integer entries and determinant one, is a subgroup of $$SL_2( {\mathbb R} )\text{.}$$
### 44.
List the subgroups of the quaternion group, $$Q_8\text{.}$$
### 45.
Prove that the intersection of two subgroups of a group $$G$$ is also a subgroup of $$G\text{.}$$
### 46.
Prove or disprove: If $$H$$ and $$K$$ are subgroups of a group $$G\text{,}$$ then $$H \cup K$$ is a subgroup of $$G\text{.}$$
Hint.
Look at $$S_3\text{.}$$
### 47.
Prove or disprove: If $$H$$ and $$K$$ are subgroups of a group $$G\text{,}$$ then $$H K = \{hk : h \in H \text{ and } k \in K \}$$ is a subgroup of $$G\text{.}$$ What if $$G$$ is abelian?
### 48.
Let $$G$$ be a group and $$g \in G\text{.}$$ Show that
\begin{equation*} Z(G) = \{ x \in G : gx = xg \text{ for all } g \in G \} \end{equation*}
is a subgroup of $$G\text{.}$$ This subgroup is called the center of $$G\text{.}$$
### 49.
Let $$a$$ and $$b$$ be elements of a group $$G\text{.}$$ If $$a^4b = ba$$ and $$a^3 = e\text{,}$$ prove that $$ab = ba\text{.}$$
Hint.
Since $$a^4b = ba\text{,}$$ it must be the case that $$b = a^6 b = a^2 b a\text{,}$$ and we can conclude that $$ab = a^3 b a = ba\text{.}$$
### 50.
Give an example of an infinite group in which every nontrivial subgroup is infinite.
### 51.
If $$xy = x^{-1} y^{-1}$$ for all $$x$$ and $$y$$ in $$G\text{,}$$ prove that $$G$$ must be abelian.
### 52.
Prove or disprove: Every proper subgroup of an nonabelian group is nonabelian.
### 53.
Let $$H$$ be a subgroup of $$G$$ and
\begin{equation*} C(H) = \{ g \in G : gh = hg \text{ for all } h \in H \}\text{.} \end{equation*}
Prove $$C(H)$$ is a subgroup of $$G\text{.}$$ This subgroup is called the centralizer of $$H$$ in $$G\text{.}$$
### 54.
Let $$H$$ be a subgroup of $$G\text{.}$$ If $$g \in G\text{,}$$ show that $$gHg^{-1} = \{g^{-1}hg : h\in H\}$$ is also a subgroup of $$G\text{.}$$
### Exercise Group.
In each group, how many solutions are there to $$x^2=e\text{?}$$ | 4,698 | 12,274 | {"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} | 3.9375 | 4 | CC-MAIN-2024-30 | latest | en | 0.506079 |
https://www.coursehero.com/file/6752872/37-Variation-of-Parameters-PS/ | 1,498,329,620,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320270.12/warc/CC-MAIN-20170624170517-20170624190517-00185.warc.gz | 848,476,935 | 24,450 | 3.7 Variation of Parameters - PS
# 3.7 Variation of Parameters - PS - M 427K Variation of...
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Unformatted text preview: M 427K Variation of Parameters – PS October 10, 2011 Problem 1 Find the general solution to the following differential equation t 2 y 00 + 4 t 2 y + 4 t 2 y = e- 2 t (1) Solution : The standard form of the differential equation is y 00 + 4 y + 4 y = e- 2 t t 2 . Corresponding Homogeneous solution , y h : Solve y 00 + 4 y + 4 y = 0 . The characteristic equation is r 2 + 4 r + 4 = 0 , and the roots are r 1 =- 2 , r 2 =- 2 . Therefore, we have y 1 = e- 2 t , y 2 = te- 2 t . Particular solution, y p : Recall that y p has the following form: y p = u 1 y 1 + u 2 y 2 , where u 1 and u 2 can be evaluated using the following equations: u 1 ( t ) = Z- y 2 f ( t ) W ( y 1 ,y 2 ) dt u 2 ( t ) = Z y 1 f ( t ) W ( y 1 ,y 2 ) dt We have, W ( y 1 ,y 2 ) := det y 1 y 2 y 1 y 2 = det e- 2 t te- 2 t- 2 e- 2 t e- 2 t- 2 te- 2 t = e- 4 t . [email protected] 1 c hf, 2011 M 427K Variation of Parameters – PS October 10, 2011 Therefore, we have u 1 ( t ) = Z- y 2 f ( t ) W ( y 1 ,y 2 ) dt = Z te- 2 t e-...
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## This note was uploaded on 02/03/2012 for the course M 427K taught by Professor Fonken during the Fall '08 term at University of Texas.
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3.7 Variation of Parameters - PS - M 427K Variation of...
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Ask a homework question - tutors are online | 595 | 1,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.742365 |
http://mathhelpforum.com/calculus/32850-probability-density-funciton.html | 1,529,364,746,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861456.51/warc/CC-MAIN-20180618222556-20180619002556-00453.warc.gz | 197,674,904 | 9,080 | ## Probability density funciton
If you have the probability density function of
$\displaystyle p(x)=Cxe^{-x/2}$ where 0<x<3. I have to find the constant C, the most probable outcome, and the mean outcome. I've tried, and gotten C=1, but then when I try to integrate it to get the mean, I get 3.05, which I assume is wrong. | 90 | 323 | {"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} | 2.640625 | 3 | CC-MAIN-2018-26 | latest | en | 0.894586 |
http://www.printnpractice.com/kindergarten-subtraction-worksheets.html | 1,490,681,840,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189680.78/warc/CC-MAIN-20170322212949-00489-ip-10-233-31-227.ec2.internal.warc.gz | 655,252,742 | 15,704 | # Kindergarten Subtraction WorksheetsKids Learn Fast With Easy Practice
Interactive PDFs! Choose fontscolors, and SIZES. Answer. Print or file.
Kindergarten Subtraction Worksheets And Flash Cards
PrintNPractice kindergarten subtraction worksheets cover all that kids need for subtraction practice. Many of these teach subtraction by offering space to copy the problems three times each. There's enough on this page to teach subtraction through third grade.
These are also helpful to review the equations for first through third grade. Some of these charts are classics you can keep for years. I keep my tables on a clip board next to the couch so I can review with my learner at a moment's notice in a warm snuggly spot. :-)
We also have videos and printable flash cards.
My favorite Math teacher was my third grade teacher, Mrs. Graves. I remember her reviewing the Math facts from large posters with the equations printed black on white. Plain and simple.
I know that we did more than that in school, but it is the thing that I remember most. Guess what? I never had to drill the Math facts at home. We never had homework and I always did well in Math. Use these links to jump down:
Two of my own children had had just such a teacher. Twice a week she would drill each student with her Math flash cards. Each night the children would copy the set that they were studying three times each until they passed the flash card drills. On that night there was no Math homework. Yea!
I've seen children fly through learning their Math facts for all four operations with this approach. Oh! The joy of facility! It's fun to watch their success. The main thing left after learning the Math tables is being able to read well enough to answer word problems and follow directions.
Teaching subtraction with our Math flash cards with the practice Math worksheets makes learning a lot less work for children who struggle to learn. You can give your kids this opportunity with these printables. I printed several versions of each table (A, B, C, and 3 below) so that once they'd practiced each and passed a flash card quiz they could move to the next table.
Older children will appreciate number 2 below as these sheets teach addition and subtraction together. Actually, some little children fly through these once they "get" the idea that subtraction is the inverse of addition, the numbers are related.
These make subtraction activities quick and fun for half the teaching work. The key is to use repetition. Anytime the wrong answer is given, simply ask the child to repeat you saying the equation with the right answer, and then have your student repeat the right equation and the right answer three times quickly. Then? Move ahead.
Varying the exercises also improves retention which is why it is good to use both flash cards and worksheets. When children see, hear, say, and write the equations; they often learn them much faster and retain their knowledge much longer.
Use our Subtraction practice videos and printable Math flash cards (below) to teach the subtraction facts in kindergarten and first grade or as a review in second and third grade subtraction. Use our subtraction worksheets to reinforce what you teach.
## 1. Basic Subtraction Worksheets & Charts
Kids can use these printable worksheets to copy each of the basic subtraction problems, or subtraction equations, and practice them three times each. If the students will say what they write, they will remember the subtraction Math facts better.
I suggest, after twenty-some years of teaching children, that any time that your kids need to correct their answers they should repeat the equation out loud with the correct answer quickly, three times to help improve their memories.
If they copy one of each of these while saying the words for each table, they're likely to be able to test-out of that table after the third page.
### A. Kindergarten Subtraction Worksheets and Chart
This first sheet has the very basic subtraction facts for reference and memorizing. The next is a set of subtraction worksheets for your student to copy the subtraction equations three times each. The third subtraction chart is for easy reference and can serve as a Math poster.
### B. Kindergarten and First Grade Subtraction Worksheets
So why are there only nine kindergarten subtraction worksheets in this section?
Because the first page has three tables and the second has three combined on the same page. Learning subtraction this way means that there are fewer facts to memorize.
## 2. Addition And Subtraction Worksheets 2 in 1
### Learn How To Add And Subtract At The Same Time
If your kids already know their addition facts, it is likely that they can fly through the subtraction facts if they know how directly related they are.
This next set is for a higher level than our kindergarten subtraction worksheets. They're super for review of both addition and subtraction together especially in first through third grades.
For example: Show that 3 and 4 are 7 is the basis for 7 less 4 are 3. Once you show the inverse relationship, half the class will "get it" with no extra work. Still, it's good to have them practice so they will remember what they've learned.
When the equations are combined the association is immediate. These are great for third grade subtraction to review at the beginning of the school year or after Christmas break.
See more worksheets, bundles, and subtraction flash cards below.
## 3. Easy Subtraction Worksheets: Practice Quizzes
Open book confidence at its best! These easy subtraction worksheets have the equations at the top, so kids can quickly see what the right answer is
1. Practice the subtraction table at the top of the page saying each equation out loud three times.
2. Answer the equations at the bottom. Easy!
If your kids are overwhelmed with the prospect of learning another table after the icebreaker-discipline needed to learn some of the addition facts, let them know that subtraction is so much easier. They already know most of the numbers!
Use these easy worksheets as practice subtraction quizzes. Get ready for the real thing. Practice to build speed.
## 4. Printable Kindergarten Subtraction Flash Cards
All of our printable Math flash cards have an answer side (to use for learning the facts in order) and a side without the answer (to use for quizzes and speed drills).
The equations are in vertical format the way we figure columns. These can be printed double sided with answers on one side; or printed single sided as two sets, one with and one without the answers.
For teaching:
1. Use the side that has the answer to teach the subtraction facts as do our subtraction videos.
2. Then use the side of the printable subtraction flash cards without the answer for a subtraction quiz.
3. If they need more practice assign the kindergarten subtraction worksheets for practice.
## 5. Subtraction Videos For Practice
Our videos cover twelve of the subtraction tables and are simply the flash cards with the equations read out loud. There is no music and there are no ads. Your kids can bookmark these and practice subtraction with videos.
## 6. Large Kindergarten Subtraction Charts To Print
### 6.A. Landscape Subtraction Charts
Use a large landscape subtraction chart from this collection or use them as Math posters for kids who are practicing with our kindergarten subtraction worksheets.
• Hole punch and hang them like a calendar and change the page when your student knows the first tables.
• Or put these in a binder to use as a flip chart.
### 6.B. Portrait Subtraction Table - Use With A Clipboard
I've put a large space at the top of these large portrait subtraction charts to allow for the clip on a clipboard. Use this collection as flip charts to show the subtraction equations to a group of students whether you're reviewing orally or using our kindergarten subtraction worksheets.
The top is purposefully left blank for the clipboard's clip.
I love using these large Subtraction charts on a clip board to go around to my students to help them review whichever Subtraction table they are studying. It's much easier to carry the right page around to show the children or to pass it to them as they need it. It is fun to see the boost in morale when we have review time!
Thank you for visiting our kindergarten subtraction worksheets!
## Buy Our Math Worksheets Bundle: Save Time!
Sample Math practice worksheets. Bundle includes dictionary for Math.
#### 1. 945 K-6 Math Worksheets2. One PDF Download. No ads.3. Filed by topic.4. Interactive. Printable.5. Use with any Math Lesson Plans.6. Addition, Subtraction, Multiplication, Division and Fraction Practice.
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› Kindergarten Subtraction Worksheets | 1,796 | 8,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.97074 |
https://oeis.org/A163990 | 1,632,835,635,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00250.warc.gz | 469,376,962 | 4,733 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A163990 Square array read by antidiagonals where the row n lists the numbers k such that their largest divisor <= sqrt(k) equals n. 6
1, 4, 2, 9, 6, 3, 16, 12, 8, 5, 25, 20, 15, 10, 7, 36, 30, 24, 18, 14, 11, 49, 42, 35, 28, 21, 22, 13, 64, 56, 48, 40, 32, 27, 26, 17, 81, 72, 63, 54, 45, 44, 33, 34, 19, 100, 90, 80, 70, 60, 50, 52, 39, 38, 23, 121, 110, 99, 88, 77, 66, 55, 68, 51, 46, 29, 144, 132, 120, 108 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS This sequence is a permutation of the natural numbers. Note that the first row is formed by 1 together the prime numbers and the first column are the squares of the natural numbers. For more information see A163280, the main entry for this sequence. (See also A161344). LINKS Omar E. Pol, Illustration for A008578, the first row of the array [From Omar E. Pol, Oct 24 2009] Omar E. Pol, Illustration for A161344, the second row of the array [From Omar E. Pol, Oct 24 2009] Omar E. Pol, Illustration for A008578, A161344, A161345 and A161424 [From Omar E. Pol, Oct 24 2009] FORMULA Row n lists the numbers k such that A033676(k)=n. EXAMPLE Array begins: 1, 2, 3, 5, 7, 11, 4, 6, 8, 10, 14, 9, 12, 15, 18, 16, 20, 24, 25, 30, 36, See also the array in A163280. CROSSREFS Cf. A000027, A000040, A033676, A147861, A163100, A163280, A163281, A163991, A164000. Cf. Rows: 1=A008578, 2=A161344, 3=A161345, 4=A161424, 5=A161835, 6=A162526, 7=A162527, 8=A162528, 9=A162529, 10=A162530, 11=A162531, 12=A162532. Cf. Columns: 1=A000290, 2=A002378, 4=A164004, 6=A164006, 7=A164007, 8=A164008, 9=A164009, 10=A164010, 11=A164011, 12=A164012. Sequence in context: A289506 A213778 A095833 * A082156 A283941 A285090 Adjacent sequences: A163987 A163988 A163989 * A163991 A163992 A163993 KEYWORD nonn,tabl AUTHOR Omar E. Pol, Aug 11 2009 STATUS approved
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Last modified September 28 09:16 EDT 2021. Contains 347714 sequences. (Running on oeis4.) | 898 | 2,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.7194 |
https://myessaydoc.com/data-mining/ | 1,632,498,555,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057558.23/warc/CC-MAIN-20210924140738-20210924170738-00209.warc.gz | 441,539,346 | 10,853 | # Data Mining
| 0
Question 1
## Decision Tree Assignment
Play now?Play later?
You can become a millionaire!That’s what the junk mail said.But then there was the fine print:
If you send in your entry before midnight tonight, then here are your chances:
0.1% that you win \$1,000,000
75% that you win nothing
Otherwise, you must PAY \$1,000
But wait, there’s more!If you don’t win the million AND you don’t have to pay on your first attempt,
then you can choose to play one more time.If you choose to play again, then here are your chances:
2% that you win \$100,000
20% that you win \$500
Otherwise, you must PAY \$2,000
What is your expected outcome for attempting this venture?Solve this problem using
a decision tree and clearly show all calculations and the expected monetary value at each node.
Use maximization of expected value as your decision criterion.
1) Should you play at all? (5%)If you play, what is your expected (net) monetary value? (15%)
2) If you play and don’t win at all on the first try (but don’t lose money), should you try again? (5%) Why? (10%)
3) Clearly show the decision tree (40%) and expected net monetary value at each node (25%)
Question 2
Provide a 5-section summary report (no less than 250 words per section) on the following alternative classifiers:
Rule-based classifiers
Nearest Neighbor Classifier
Naïve Bayes Classifier
Support Vector Machines
Ensemble Methods
For each classifier, provide the classifier name (note: use the classifier name as a sub-header in bold font), provide the definition along with a brief explanation of the classifier (i.e. how it works), the advantage and disadvantage of using the classifier, and provide a data example of its use (i.e. graphics, charts, figures, formula, etc).
Last Updated on February 11, 2019 | 430 | 1,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.890415 |
https://wiki.mathnt.net/index.php?title=%EB%9D%BC%EB%A7%88%EB%88%84%EC%9E%94%EC%9D%98_%EC%A0%95%EC%A0%81%EB%B6%84 | 1,558,572,024,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256997.79/warc/CC-MAIN-20190523003453-20190523025453-00337.warc.gz | 668,030,978 | 8,773 | # 라마누잔의 정적분
이동: 둘러보기, 검색
• 라마누잔의 정적분
## 개요
$$\int_{0}^{\infty}\frac{x e^{-\sqrt{5}x}}{\cosh{x}}\,dx=\frac{1}{8}(\psi^{(1)}(\frac{1+\sqrt{5}}{4})-\psi^{(1)}(\frac{3+\sqrt{5}}{4}))$$
Integrate[(x Exp[-x Sqrt[5]])/Cosh[x], {x, 0, \[Infinity]}] // FullSimplify
[%28x+Exp[-x+Sqrt[5]%29/Cosh[x],+%7Bx,+0,+[Infinity]%7D]+ http://www.wolframalpha.com/input/?i=Integrate[(x+Exp[-x+Sqrt[5]])/Cosh[x],+{x,+0,+[Infinity]}]+]
[1,%281%2Bsqrt%285%29%29/4-polygamma[1,%283%2Bsqrt%285%29%29/4]%29/8 http://www.wolframalpha.com/input/?i=(polygamma[1,(1%2Bsqrt(5))/4]-polygamma[1,(3%2Bsqrt(5))/4])/8]
$$\int_{0}^{\infty}\frac{x^{2}e^{-\sqrt{3}x}}{\sinh{x}}\,dx=-\frac{1}{4}\psi^{(2)}(\frac{1+\sqrt{3}}{4})$$
Integrate[(x^2 Exp[-x Sqrt[3]])/Sinh[x], {x, 0, \[Infinity]}] //FullSimplify
[%28x%5E2+Exp[-x+Sqrt[3]%29/Sinh[x],+%7Bx,+0,+Infinity%7D] http://www.wolframalpha.com/input/?i=integrate[(x^2+Exp[-x+Sqrt[3]])/Sinh[x],+{x,+0,+Infinity}]]
[2,%281%2Bsqrt%283%29%29/2/4 http://www.wolframalpha.com/input/?i=-polygamma[2,(1%2Bsqrt(3))/2]/4]
Berndt, B. C. and Rankin, R. A. Ramanujan: Letters and Commentary. Providence, RI: Amer. Math. Soc., 1995. | 552 | 1,140 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-22 | longest | en | 0.437634 |
https://mathoverflow.net/questions/35713/abelianization-of-a-semidirect-product/376713 | 1,720,980,134,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00678.warc.gz | 346,297,143 | 27,134 | # Abelianization of a semidirect product
I believe there is a straightforward formula for the abelianization of a semi-direct product: if $G$ acts on $H$, and we form the semi-direct product of $G$ and $H$ in the usual way, and the abelianization of this semi-direct product is the product $G^{ab}\times (H^{ab})_{G}$.
(Here the subscript $G$ denotes taking the coinvariants with respect to $G$. That is, $(H^{ab})_{G}$ is a the quotient of $H^{ab}$ by the subgroup generated by elements of the form $h^g-h$ for $h$ in $H$ and $g$ in $G$, and where the superscript $g$ denotes the action of $G$ on $H^{ab}$ induced by the action of $G$ on $H$.)
Does anyone happen to know a good reference for this?
• Does it really need a reference? Write down the presentation for the semi-direct product, that gives you a presentation matrix for the abelianization and it's pretty much immediate from there, no? Commented Aug 16, 2010 at 3:16
• That's what I thought. However, a referee requested that I explain the formula; it seems that giving a reference is more appropriate than explaining the thing in detail. (I'm nervous about only explaining it very briefly, given that referee made an especial request for clarification...)
– blt
Commented Aug 16, 2010 at 3:21
• If you don't find a reference, just write a one-paragraph explanation along the lines of Ryan's comment. If it is a mathematics journal, it should be sufficient. Commented Aug 16, 2010 at 3:41
• It is a mathematics journal, for a research paper in number theory (not a textbook). Given the weight of the consensus here, I will write a short explanation along the lines of Greg's below. Thank-you all for giving me the confidence to do so!
– blt
Commented Aug 16, 2010 at 12:37
I agree with Ryan and Victor, except that you don't need presentations. The subgroup $[G \ltimes H,G \ltimes H]$ is generated by $[H,H] \cup [G,H] \cup [G,G]$, so you can write $$(G \ltimes H)^{ab} = (G \ltimes H) / \langle [H,H] \cup [G,H] \cup [G,G] \rangle.$$ If you apply the relators $[H,H]$, you get $G \ltimes H^{ab}$; then if you apply the relators $[G,H]$, you get $G \times (H^{ab})_G$; then finally if you apply $[G,G]$, you get $G^{ab} \times (H^{ab})_G$. You can add this as an extra half-paragraph or footnote rather than giving a citation.
I don't think that the referee has the right to demand a longer explanation than this, unless maybe you are writing a textbook.
• You don't even need to mention commutators: any homomorphism from $G\ltimes H$ to an abelian group factors through $G\times H^{ab}$; then through $G\times (H^{ab})_G$; finally through $G^{ab}\times (H^{ab})_G$. Commented Apr 21, 2011 at 21:32
If you have the semidirect product $$H\rtimes G$$ then you have the next group split extension
$$1\rightarrow H\rightarrow H\rtimes G\rightarrow G\rightarrow 1$$,
We have the Hochschild–Serre spectral sequence where $$\mathbb{Z}$$ is a trivial $$H\rtimes G-$$module
$$E^{2}_{p,q}=H_{p}(G,H_{q}(H,\mathbb{Z}))\Rightarrow H_{p+q}(H\rtimes G,\mathbb{Z})$$,
Since the the map $$H\rtimes G\rightarrow G$$ is a split surjection, it follows that the map (edge morphism)
$$H_{n}(H\rtimes G,\mathbb{Z})\rightarrow H_{n}(G,\mathbb{Z})=E^{2}_{n,0}$$ is a slit surjection and thus $$E^{2}_{n,0}=E^{\infty}_{n,0}$$
In particular we have that the diferenttial $$d:E^{2}_{2,0}\rightarrow E^{2}_{0,1}$$ is zero (since $$E^{2}_{2,0}=E^{\infty}_{2,0}$$). Therefore $$E^{2}_{0,1}=E^{\infty}_{0,1}$$.
It follows that there is a exact sequence
$$0\rightarrow E^{\infty}_{0,1}\rightarrow H_{1}(H\rtimes G,\mathbb{Z})\rightarrow E^{\infty}_{1,0}\rightarrow 0$$
which splits, in this case we have that
$$H_{1}(H\rtimes G,\mathbb{Z})=E^{\infty}_{0,1}\times E^{\infty}_{1,0}$$
Note that
$$H_{1}(H\rtimes G,\mathbb{Z})=(H\rtimes G)^{Ab}$$
$$E^{\infty}_{1,0}=G^{Ab}$$
and
$$E^{\infty}_{0,1}=H_{0}(G, H^{Ab})=(H^{Ab})_{G}$$
From this we have the result by using spectral sequences.
A description of the derived subgroup of a semidirect product, from which the abelianization can be obtained, was published in:
Daciberg Lima Gonçalves, John Guaschi The lower central and derived series of the braid groups of the sphere Trans. Amer. Math. Soc. 361 (2009), 3375-3399. http://www.ams.org/journals/tran/2009-361-07/S0002-9947-09-04766-7/ (Proposition 3.3)
You may also find it in their preprint: http://arxiv.org/abs/math/0603701 (Proposition 29) | 1,400 | 4,403 | {"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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-30 | latest | en | 0.909403 |
https://www.studypool.com/discuss/429515/evaluate?free | 1,508,544,153,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824471.6/warc/CC-MAIN-20171020230225-20171021010225-00098.warc.gz | 1,007,609,377 | 13,730 | ............................................evaluate
label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
F(x)=7x+5= when x=2
Mar 13th, 2015
F(x) = 7x + 5 when x = 2
F(2) = 7(2) + 5
F(2) = 14 + 5
F(2) = 19
Mar 13th, 2015
...
Mar 13th, 2015
...
Mar 13th, 2015
Oct 21st, 2017
check_circle | 136 | 323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.609458 |
https://mirror.codeforces.com/blog/entry/128851 | 1,716,561,235,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00423.warc.gz | 350,297,143 | 19,484 | ### siddhantuniyal's blog
By siddhantuniyal, history, 4 weeks ago,
I tried this problem with L = 1 , R = 9999
The answer is 4275 (9 + 9*8 + 9*9*8 + 9*9*8*7) , but with the below approach I am getting 5617. I am trying this approach, because it is generalized for any L and R.Please help me in finding out what I am doing wrong.
Observation 1 : The numbers following this conditions are always >= 1
Observation 2 : let x = 456 , y = 467. We are only able to tell that y > x because we started from the first digit , iterated towards the last digit and compared digits. At the first point of inequality , we were able to conclude that 456 < 457
Hence , I am iterating over all possible points of inequality between digits. here 1 is represented as 0001
if inequality is at 1st digit , that means 1st digit has to be 1,2,3..9 -> hence 9C1. And for the rest 3 digits , choose any 3 digits from the remaining 9 digits (10 digits originally and removing one because we placed it at 1st digit) and rearrange them -> 9C3 * 3!
if inequality is at 2nd digit , that means 1st digit = 0 and 2nd digit = 1,2..9 -> hence 9C1. And for the rest 2 digits , choose any 2 digits from the remaining 8 (10 originally , one (0) already at 1st digit , one chosen for 2nd digit) and rearrange -> 8C2 * 2!
if inequality is at 3rd digit , that means 1st and 2nd digit -> 0 and 3rd digit = 1,2,3..9 -> 9C1. and for the remaining digit , 8C1 (0 already used at 1st and 2nd digit , and one already chosen for 3rd digit).
for 4th digit -> 8C1
and we also have to add 1 to the answer , because , we are only considering some number between 1 and 9999 which has atleast one digit unequal to its corresponding digit in 1. but , 1 itself can be part of answer and here it is.
so this gives 9C1 * 9C3 * 3! + 9C1 * 8C2 * 2! + 9C1 * 8C1 + 8C1 + 1 , which is 5617 and wrong.
• 0
» 4 weeks ago, # | 0 Auto comment: topic has been updated by siddhantuniyal (previous revision, new revision, compare).
» 4 weeks ago, # | 0 Auto comment: topic has been updated by siddhantuniyal (previous revision, new revision, compare).
» 4 weeks ago, # | 0 Auto comment: topic has been updated by siddhantuniyal (previous revision, new revision, compare).
» 4 weeks ago, # | ← Rev. 4 → 0 I was asked this question in Online Assessment of JPMC. You can do it using precomputation too. Here is how — Declare an array of size 1e6+1 (Assuming the max value of r can be 1e6). For each number (say)j from 1 to 1e6, if it contains all distinct numbers, make arr[j]=1, else arr[j]=0. Then calculate cumulative sum of the array, such that, arr[i]+=arr[i-1]. Now, for each query, [l,r], your answer will be arr[r]-arr[l-1]. Codevectorarr(1000001); bool hasDistinctDigits(ll num){ unordered_setvis; while(num){ ll lastDigit = num%10; if(vis.count(lastDigit)){ return false; } vis.insert(lastDigit); num/=10; } return true; } void precompute(){ for(ll i=1;i<=1000000;++i){ if(hasDistinctDigits(i)){ arr[i]=1; } else{ arr[i]=0; } } for(ll i=1;i<=1000000;++i){ arr[i]+=arr[i-1]; } // Now for each query of type [l,r], just output, arr[r]-arr[l-1]. }
• » » 4 weeks ago, # ^ | 0 Thanks a lot
» 4 weeks ago, # | 0 explore digit dp | 1,010 | 3,191 | {"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} | 4.0625 | 4 | CC-MAIN-2024-22 | latest | en | 0.93031 |
http://www.apug.org/forums/archive/index.php/t-57355.html | 1,448,665,166,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398450659.10/warc/CC-MAIN-20151124205410-00010-ip-10-71-132-137.ec2.internal.warc.gz | 287,829,394 | 3,765 | PDA
View Full Version : Conversion from old to new
Photo Engineer
12-25-2008, 08:37 PM
I have commented before on the fact that old emulsions used active gelatin and therefore are not truly usable without some sort of conversion with modern inactive gelatin.
Here is a handy hint that is different from using sulfur or sulfur + gold to get the same results as the old timers.
I've equated temperature with speed. Up in temp, up in speed. I've also equated addition time with contrast (and to an extent speed) in that up in addition rate, up in contrast.
These two general rules can allow one to convert an emulsion from the old style to the new style with minimum pain and still use modern gelatins. The trick is to raise the temperatures and decrease the addition times used in a formula.
Here is an example. Let us assume a formula uses 5' addition of Silver Nitrate at 40 deg C and then a digest of 20 minutes at 50 degrees C. To convert, a hint would be to add the Silver Nitrate over 2.5' at 50 deg C and digest 20 minutes at 60 degrees C. This is just a hint of a starting point, but you get the picture that time and temperature can be used to "adjust" a formula for use today and it is possible to omit the sulfur sensitization step otherwise needed for increasing speed and contrast when using refined gelatins.
Just a handy tip. You will have to play with your formula to get what you want out of it, but here is how you start.
PE
wildbillbugman
01-07-2009, 06:54 PM
Hi P.E.,
Do you have an idea as to what the upper limit to temperature might be? How high can you go until the rule of 'higher temp. while adding Ag increases speed' is no longer valid. I have read that some emulsions go as high as 70 degrees.
Regards,
Bill
Photo Engineer
01-07-2009, 07:17 PM
Bill, some emulsions go up as high as 95 deg C. These were "boiled emulsions" from the old days. Gelatin begins to denature at that temperature, and the mess is near to boiling. But, the exact answer is that the temperature can go as high as needed but must not boil or denature the gelatin. Generally, this is about 80 - 85 C.
PE
rmazzullo
01-07-2009, 09:13 PM
Hello PE,
With this in mind, as well as emulsion making in general, is there a "tolerance" that is allowable (plus or minus some amount) when measuring temperature, and keeping kettles / solutions at a desired temperature for the required length of time?
I know I can spend money for very accurate temperature measuring and / or control using thermocouples, feedback loops. PCs, heating jackets, etc , but could good results be obtained with a "lower grade" of equipment repeatedly?
(A friend has a large roll of thin nichrome ribbon that would be ideal for making heating jackets for kettles, etc., but I suspect this is venturing far into the region of overkill. Regardless, I have yet to convince him that he has no real use for it.)
Thanks,
Bob M.
Photo Engineer
01-07-2009, 09:24 PM
There is no fixed time period for how long gelatin can be kept at high temperature. The degradation is either slow or fast depending on gelatin type. If you can, keeping track of viscosity might help as that is a primary indication of the degradation process. But why worry? We know that 1 hour at 60 C is good for these emulsions, so that is a good center point to work from.
PE
Kirk Keyes
01-08-2009, 12:19 AM
Bob, I can assure you that good results can be had with very common and simple tools for heating and temperature control. Denise Ross makes truly fine emulsions with waterbaths and inexpensive thermometers. I've seen PE use a hotplate with thermocouple control, and that works too, but it's really not as fancy as you are asking about.
Try some of the digital cooking thermometers that you get at your local kitchen shop. They run about \$25 and are certainly good enough.
rmazzullo
01-08-2009, 11:40 AM
Kirk,
Thanks for the reply. I have no doubt good results can be obtained using simpler methods and tools. But if I just happen to make a good emulsion one particular day using common items for heating and temperature control (among other things), there's no guarantee that I can repeat that particular success if variables exist in the tools I use.
I am willing to bet that I could probably have better luck making a more consistent emulsion, if I use better, more precise tools. At the very least, I will have a much better database of the mistakes I make when I know and can record what exact temps, addition rates, etc were involved. I am willing to go the extra mile to remove whatever uncertainties I can. The less I have to guess, the better.
(edit) Right now, the only area that I would use items that I could get from a hardware store, or a shop that sold kitchen items is in setting the emulsion after coating and drying. I don't think the need for precision is as great.
Bob M.
Kirk Keyes
01-08-2009, 03:03 PM
Bob - look for a new thread on temp recording. | 1,188 | 4,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2015-48 | latest | en | 0.948774 |
https://moffett4understandingmath.com/2015/05/13/summer-math-does-it-make-cents/?replytocom=23 | 1,620,426,668,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988828.76/warc/CC-MAIN-20210507211141-20210508001141-00146.warc.gz | 426,977,417 | 21,383 | # Summer Math: Does It Make “Cents”?
Summer is on the horizon! No more pencils, no more books! No more teachers dirty looks! Even I am looking forward to not having to ask my oldest about homework, studying for tests, projects, etc. Time for relaxing by the pool, reading books, and just taking a break.
That being said, summer is a great time to build in some meaningful mathematics with your children. For the next few posts, we will explore ways you can integrate mathematics into your summer plans without making it feel like “school math”.
Jo Boaler, author of What’s Math Got To Do With It? Helping Children Learn To Love Their Least Favorite Subject, states that one of the most parts of being mathematical is an action called reasoning. This involves explaining why something makes sense (or doesn’t make sense)… This is one of the most important pieces of mathematics, and yet is one of the first teachers say is missing or lost in math programs. And it goes beyond simply asking, “Does your answer make sense?” A student who can reason mathematically realizes that mathematics is a subject that (she) can have her own ideas about, a subject that can invoke different perspectives and methods, and a subject that is connected through organizing concepts and themes. It is not a page of math problems a day or a lesson a day where it feels you are learning something new and different every day of the week.
How can we as parents help our children with reasoning? Make math meaningful in the home. Talk about how you use math in your day. Have your older children help make decisions, such as how much paint you need to buy for a room or how much weed killer you need to buy for your yard. Bake. Use real money (cash!) to purchase items so they get a feel for what money is (Great for addition/subtraction!). These are all ideas we will explore later…
My favorite way to get students engaged in reasoning is called, “What’s Wrong With This?!!!” All junior high kids like to express you are wrong. Why not use it to discuss math? Find advertisements, comments in media, or just conversations you hear that are mathematically unreasonable and have a conversation about why it is wrong. Then get them looking for mistakes as well. It is easy, it makes math interesting, and you are building reasoning skills in your children.
Here is a pic I took from down the road (of my school) when I was getting gas. So what’s wrong?
The amount advertised (I believe) is meant to represent 99 cents. Does it? You can show 99 cents as 99¢, showing that you could use 99 pennies, or as \$0.99, representing that you have one less cent than the value of a dollar. What does 0.99¢ mean? It means you are paying less than one cent; less than a penny. You are paying 99/100 of a penny. Meaning, if I give the cashier a penny, he owes me some change!
Most students won’t see it. In fact, when I really started looking at their decimal work (applying to money) I realized at least 2-3 students per class EACH YEAR were making this same mistake. They weren’t connecting the relationship (and difference) between writing the value in dollars verses cents. They simply knew something about each, and jumbled it together, WITHOUT MAKING SENSE.
Why is this so important? Is it really worth knowing that 0.99¢ is wrong? Well, this idea that money can go “beyond cents” creeps up on us in different places. Take the marquis at a gas station.
Letting the fact that this marquis also indicates that I am paying \$395 per gallon (Decimal, where are thou???)… \$3.95 9/10? What does this mean? Can I pay the 9/10 of a penny? This is actually a great advertising tactic. It is found (See link below for a great article!) that more people will buy if you leave an amount at \$0.99 rather than bumping up to the nearest dollar. Significantly more people will buy the \$4.99 than the \$5.00, because they relate the \$4.99 as being closer to \$4 than \$5. Doesn’t make sense, but that is how we perceive it. So when looking for the cheapest gas, we would gladly take\$ 2.19 9/10 versus \$2.20, even though the \$2.20 location is closer! Don’t believe me? Be mindful of how you choose prices for the next few weeks (And let your kids in on this great experience!) and if you are like me, you will be shocked at how true this is.
Or here…
0.99% financing. Is that a lot? What if I am buying a car? Would I rather pay 1% financing or 0.99% financing? Why?
Well, this week the cents mistake came up again!
One of my students from last year emailed me this picture with the following message. Hi Mrs. M! I have to tell you that I am ALWAYZ looking for math mistakes now. My mom doesn’t like when I call her out tho. I found this one and thought you would like to have it. I made my mom look at it and said they would have to give me change for my penny. She thought I was crazy! Ha ha. Thanks for making math fun.
So whether you are out at the store, on vacation, or driving hundreds of miles, look for how math shapes our world. Is the math you are seeing making sense or are there mistakes in the reasoning? You will be surprised; the more you make math meaningful, the more your kids will appreciate it.
Want your child to read more about the money symbols (and why there isn’t a cent sign on the keyboard? Hmmm…), go here! http://www.charlieanderson.com/centsign.htm
For more information regarding the psychology of advertising dollars versus cents, go here! http://www.fastcompany.com/1826172/psychology-behind-sweet-spots-pricing
## 2 thoughts on “Summer Math: Does It Make “Cents”?”
1. Love the idea of framing in a way of ‘spotting the math mistakes.’ Makes for a fun activity if you find a lot and just post it up and ask students if they see any mistakes. Awesome 🙂
Took the following picture the other day: https://goo.gl/t7DZWO
At first, just thought ‘oh you can only get 20% off the \$100 so you only save \$20’ but then realized it said you can ‘save \$100’ and I thought ‘how much would you have to spend to get \$100 of savings.’ Just finished up our percents unit but we’ll try it next year 🙂
Like
1. Please let me know how it goes in your class next year. Always wanting to learn more!
Like | 1,454 | 6,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.95253 |
https://studylib.net/doc/8952997/convective-heat-transfer-ii | 1,643,086,238,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304760.30/warc/CC-MAIN-20220125035839-20220125065839-00043.warc.gz | 592,263,713 | 13,594 | # Convective Heat Transfer II
```EML 6155 - Convection Heat Transfer – Spring 2012
Department of Mechanical and Aerospace Engineering
Office: MAE-A 310
Phone: 352-392-0889: E-mail: saeedmog@ufl.edu
Office Hours: M-W-F 1:30 PM to 2:30 PM (note: These are the proposed office hours
and can be changed upon request)
Teaching Assistants:
Sai Tej (saitej@ufl.edu)
Office Hours: Tuesday and Thursday, 1:30pm to 2:30pm
Room 130 in Building MAE-C
Course Website: Sakai
Meeting Time & Place: M-W-F, 5th Period (11:45am to 12:35pm), NEB 0102
Books:
1- Convective Heat and Mass Transfer by W. M. Kays et al., McGraw-Hill, 2009.
2- Convection Heat Transfer by A. Bejan, John Wiley & Sons, 1984.
Course Objective: to provide a fundamental treatment of fluid flows controlled by
viscous or turbulent stress gradients and the subsequent heat transfer between fluids and
solid surfaces. Analytical solutions to the momentum and energy conservation equations
for both laminar and turbulent flows will be considered. Students will be expected to
derive appropriate transport equations, apply transport equations to convective transport
problems, and evaluate appropriate transport properties such as friction factors, Nusselt
numbers, Sherwood numbers, and Stanton numbers. The fundamental conservation
principles covered in this course provide a solid foundation for the engineering
practitioner engaged in single phase convective thermal transport; a solid foundation is
also provided for further studies in multiphase convective transport.
Topics:
1. Introduction
2. Fundamental principles
a. Mass conservation
b. Force balances (Momentum equations)
c. Energy equations
d. A simple case: Couette flow
e. Scale analysis
3. Laminar boundary layer flow
a. The fundamental problem in convection heat transfer
b. The concept of boundary layer
c. Velocity and thermal boundary layer thicknesses
4. Laminar momentum and heat transfer in ducts
a. Entry region
b. Fully developed flow
c. Flow and heat transfer in circular and non-circular cross-section ducts
d. Nusselt number at different wall thermal conditions
5. Laminar momentum and heat transfer in external boundary layers
a. Potential flow solutions to velocity field
b. Self similar boundary layers
c. Similarity transformations
d. Flow over a flat plate solutions
e. Displacement thickness, momentum thickness
f. Integral momentum equation and approximate solutions
g. Thermal boundary layer similarity transformation and solution
h. Integral energy equation and approximate solutions
6. Turbulence fundamentals
a. Transition to turbulent
b. Reynolds decomposition
c. Averaging properties
d. Turbulent (Reynolds) stress and turbulent (eddy) thermal diffusivity
e. Prandtl mixing length model
f. Turbulent Prandtl number
7. Turbulent fluid flow
a. Law of the wall
b. Universal velocity profile for external flow
c. Friction coefficient
d. Internal flow
8. Turbulent thermal transport
a. External flow
b. Law of the wall
c. Nusselt Number
d. Internal flow
e. K- model
10. Natural convection boundary layers
a. Boundary layer equations
b. Boussinesq approximation
c. Nusslet number (laminar flow)
d. Integral method for turbulent flow
11. Convection heat transfer at high velocities
a. Compressible boundary layers
b. Enthalpy formulation of energy equation
Examination Schedule:
1. Mid-term Exam on Friday, March 2nd 8:00 PM to 10:00 PM
2. Final Exam on Friday, May 4th 10:00 AM to 12:00 PM (or 8:00 PM to 10:00
PM, if there is no conflict with other exams)
Homework
Mid-term Exam
Final Exam
Total
20%
40%
40%
100%
Class Policies:
1.
Regular class attendance is expected and encouraged. Each student is
responsible for all of the material presented in class and in the reading
assignments. Exams will emphasize treatment of material covered in lectures.
2.
All homework assignments and projects are to be turned in at the beginning of
the designated class period. In general, no late assignments will be accepted or
makeup exams given. Exceptions will be made for a valid excuse consistent
with University Policy. Exceptions may also be made if deemed appropriate, but
3.
SOME collaboration is allowable on homework, but each student is responsible
for performing the bulk of his or her own homework assignment. The copying
of solutions from the Solutions Manual (or copies from) is considered cheating,
and is not allowed.
4.
NO collaboration is allowed on exams.
5.
Honesty Policy – All students admitted to the University of Florida have signed a
statement of academic honesty committing themselves to be honest in all
academic work and understanding that failure to comply with this commitment
will result in disciplinary action. This statement is a reminder to uphold your
obligation as a UF student and to be honest in all work submitted and exams
taken in this course and all others.
6.
Accommodation for Students with Disabilities – Students Requesting classroom
accommodation must first register with the Dean of Students Office. That office
will provide the student with documentation that he/she must provide to the
course instructor when requesting accommodation. This process must be
7.
UF Counseling Services – Resources are available on-campus for students
having personal problems or lacking clear career and academic goals. The
resources include: University Counseling Center, 301 Peabody Hall, 392-1575;
SHCC Mental Health, Student Health Care Center, 392-1171; Center for Sexual
Assault/Abuse Recovery and Education (CARE), Student Health Care Center,
392-1161. Career Resource Center, Reitz Union, 392-1601, for career
development assistance and counseling.
8.
Software Use – All faculty, staff and student of the University are required and
expected to obey the laws and legal agreements governing software use. Failure
to do so can lead to monetary damages and/or criminal penalties for the
individual violator. Because such violations are also against University policies
and rules, disciplinary action will be taken as appropriate. We, the members of
the University of Florida community, pledge to uphold ourselves and our peers
to the highest standards of honesty and integrity.
``` | 1,425 | 6,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-05 | latest | en | 0.818346 |
https://www.brainkart.com/article/important-Questions-and-Answers--Control-Surveying_4643/ | 1,623,726,759,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487616657.20/warc/CC-MAIN-20210615022806-20210615052806-00545.warc.gz | 627,579,238 | 8,308 | Home | | Surveying II | | Surveying | important Questions and Answers: Control Surveying
# important Questions and Answers: Control Surveying
CiIvil Surveying - Control Surveying
Control Surveying
1. Define Tacheometry:
i. Tacheometry is a branch of angular surveying in which the horizontal and vertical distances (or) points are obtained by optional means as opposed to the ordinary slower process of measurements by chain (or) tape.
2. Define Tacheometer:
i. It is an ordinary transit theodolite fitted with an extra lens called analytic lens. The purpose of fitting the analytic lens is to reduce the additive constant to zero.
3. Define Analytic lens:
a. Analytic lens is an additional lens placed between the diaphragm and the objective at a
4. fixed distance from the objective. This lens will be fitted in ordinary transit theodolite. After fitting this additional lens the telescope is called as external focusing analytic telescope. The purpose of fitting the analytic lens is to reduce the additive constant to zero.
5. Define Substance bar:
a. A Substance bar is manufactured by Mr. Kern. The length of the substance bar is 2m (6ft)
6. for measurement of comparatively short distance in a traverse. A Substance bar may be used as a substance base. The length of the bar is made equal to the distance between the two targets.
7. What are the merits and demerits of movable hair method? Merits:
i. Long sights can be taken with greater accuracy than stadia method The error obtained is minimum
b. Demerits:
The computations are not quicker Careful observation is essential
8. Fixed hair method:
a. In this method, the stadia wires are fixed (or) fitted at constant distance apart.
9. Staff intercept:
a. The difference of the reading corresponding to the top and bottom stadia wires.
a. The difference of the distance between the top and bottom cross hairs. In this method stadia interval is variable. The staff intercept is kept fixed
b. interval is variable.
11. The tangential method:
a. In this method, the stadia hairs are not for taking readings. The readings being taken against the horizontal cross hair.
12. What are the systems of tacheometry measurements?
a. The stadia system The tangential system
b. Measurement by means of special instrument.
13. What are the types of stadia system?
a. Fixed hair method, Movable hair method
14. What is the principle of stadia method?
The method is based on the principle that the ratio of the perpendicular to the base is constant to similar isosceles triangle.
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
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# CS602 Grand Quiz + Mid Term Quiz Fall 2020 Preparation Material Due Date: 29-12-2020
CS602 Grand Quiz + Mid Term Quiz Fall 2020 Preparation Material Due Date: 29-12-2020
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# CS602
## Computer Graphics: CS602
The virtual university offer program Computer Graphics (CS602 for their students. Here you can download helping materials related to this subject like Handouts, Midterm solved papers and Final term solved papers and much more…
# Computer Graphics - CS602 MCQs Solved
CS 602 150+ MCQs Solved with Referencess
DDA abbreviated for ________.
1. a)None of the given
2. b)Discrete data analyzer
3. c)Digital data analyzer
4. d)Digitaldifferential analyzer (Page 54)
Save a line with both endpoints inside all clipping boundaries is called as _____________.
1. a)None of the given
2. b)Total …. (inside (maybe))
3. c)Trivial Reject
4. d)Trivial Accept (Page 142)
__________ projection is obtained by projecting points along parallel lines that are not perpendicular to the projection plane.
1. a)Perspective
2. b)Orthographic
3. c)Oblique (Page 198)
The dot product of two vectors A and B is __________. Iff the angle between them is less than 90 or greater than 270 degrees.
1. a)Greater than zero (0) (Page 177)
2. b)Less than zero (0)
3. c)Equal to Zero (0)
4. d)None of the given
This projection technique has the direction of projection perpendicular to the viewing plane, and the viewing direction is perpendicular to one of the principle faces.
1. a)metric Parallel Projection
2. b)Oblique Parallel Projection
3. c)Orthographic Parallel Projection
4. d)None of the given
Orthographic projections that show more than one side of an object are called __________ projections.
Cavalier
Cabinet
Axonometric (Page 196)
Perspective
Computer Graphics are used in ____________.
1. a)Movies development
2. b)Simulations
3. c)All of the given
4. d)Game development
“Computer Graphics” and “Computer Vision” are ____________.
1. a)Same fields
2. b)Interrelated fields
3. c)None of the given
4. d)Totally different fields
We can take transpose of __________.
1. a)matrix with 1 row 1 column
2. b)matrix with 2 rows 3columns
3. c)matrix with 3 rows 2 columns
4. d)any matrix
If the polygons are _______, line-clipping techniques are sufficient for clipping.
1. a)filled
2. b)unfilled (Page 146)
3. c)half filled
4. d)All of the given
20: Because clipping against one edge is independent of all others, so it is _______ to arrange the clipping stages in a pipeline.
1. a)Sometimes impossible
2. b)None of the given
3. c)Possible (Page 150)
4. d)Impossible
Tessellation can be adaptive to the ____________ degree of curvature of a surface.
1. a)Local (Page 170)
2. b)Static
3. c)Global
4. d)Variable
The actual filling process in boundary filling algorithm begins when a point _________ of the figured is selected.
1. a)Outside theboundary
2. b)Inside the boundary (Page 102)
3. c)Atboundary
4. d)None of the above
Discard a line with both endpoints outside clipping boundary is called as ____________
1. a)Trivial accept
2. b)Trivial reject (Page 142)
3. c)Total outside
4. d)None of the above
_____________is the tendency of the text to flash as it moves up or down.
1. a)Flickering (Page 38)
2. b)Snow
3. c)Distortion
4. d)None of the above
The axonometric projection is ___________ where the direction of projection makes same angle with all axes.
1. a)DIMETRIC
2. b)Isometric (Page 196)
3. c)Oblique
4. d)Trimetric
We can draw the circle using _________
1. a)Pentane
2. b)Hexane
3. c)Trident
4. d)Octant (Page 63)
Question No: 17 ( Marks: 1 ) - Please choose one
_________ direct view storage tube maintains the picture display.
1. a)Electron gun
2. b)Proton gun
3. c)Flood gun (Page 29)
4. d)All of the above
Question No: 18 ( Marks: 1 ) - Please choose one
Because clipping against one edge is independent to all others, so it is ________ arrange the clipping stages in a pipeline.
1. a)Possible (Page 150) rep
2. b)Impossible
3. c)Sometimes impossible
4. d)None of the above
Question No: 19 ( Marks: 1 ) - Please choose one
If the polygons are _________ line clipping techniques are sufficient for clipping.
1. a)Filled
2. b)Unfilled (Page 146) rep
3. c)Half filled
4. d)All of the above
Question No: 20 ( Marks: 1 ) - Please choose one
Polygons consisting of _________________ can cause problems when rendering.
1. a)Non-coplanar vertices (Page 169)
2. b)Co-planar vertices
3. c)Any vertices
4. d)None of the above
Question No: 1 ( Marks: 1 ) - Please choose one
In Trivial acceptance/reject test there are four bits of nine regions, Bit3 represents condition ______________.
1. a)Outside half plane of left edge, to the left of left edge X < Xmin
2. b)Outside half plane of right edge, to the right of right edge X > Xmax (Page 143)
3. c)Outside half plane of bottom edge, below bottom edge Y < Ymin
4. d)Outside half plane of top edge, above top edge Y > Ymax
Plasma-panel displays use a gas mixture that usually includes _____________.
1. a)Zinc
2. b)Iron
A line, or straight line, is, roughly speaking, an (infinitely) thin, (infinitely) long, straight geometrical object.
1. True (Page 53)
2. False
Both Boundary Filling and Flood filling algorithms are _____________ as compared to scan line filling algorithm.
1. Better
2. Worse
3. Almost same
4. Good
( x^2 / a^2 ) + ( y^2 / b^2 ) = 1 is an equation of ____________________________.
1. Parabola
2. Hyperbola
3. Ellipse (Page 70)
4. Circle
We can draw the circle using ____________________.
1. Pentane
2. Hexanes
3. Trident
4. Octants (Page 63) rep
A scaling transformation alters the _______________ of an object.
1. Shape
2. Position
3. Size (Page 120)
4. Rotation
Question No: 6 of 10 ( Marks: 1 ) - Please choose one
Boundary Filling Algorithm can work for complex polygons.
1. a)True
2. b)False
Question No: 7 of 10 ( Marks: 1 ) - Please choose one
A __________________ is the set of all points (x , y) that are the same distance from the directrix and focus not on the directrix.
1. a)Circle
2. b)Hyperbola
3. c)Parabola (Page 73)
4. d)Ellipse
Question No: 8 of 10 ( Marks: 1 ) - Please choose one
1. True (Page 38)
2. False
Question No: 9 of 10 ( Marks: 1 ) - Please choose one
If the value of scaling factors sx and sy is greater than 1, then size of objects will be _____________________.
1. Reduced
2. Enlarged (Page 121)
3. Remain same
4. Shear
Question No: 10 of 10 ( Marks: 1 ) - Please choose one
In ___________________ algorithm(s), old color must be read before it is invoked.
1. Scan line filling
2. Flood filling (Page 104)
3. Both scan line and flood filling
4. Scan filling
Parity is a concept used to determine which _________________ lie within a polygon.
• Edge
• Vertices
• Pixels (Page 80)
• Points
Various curve functions are useful in ________________________.
• Object modeling (Page 69)
• Graphics applications
• Animation path specifications
• All of the given
polygons are basically concave polygons that may have self-intersecting edges.
1. Complex (Page 79)
2. Hybrid
3. Convex
4. Convex and Hybrid
Concave polygons are a superset of _____________________ polygons, having fewer restrictions than
____________________ polygons.
• Hybrid, Complex
• Concave, Complex
• Convex, Convex (Page 79)
• Complex, Complex
There are ________________ basic types of polygon.
• 2
• 3 (Page 78)
• 4
• 5
Computer graphics is very helpful in producing graphical representations for scientific visualization.
• True (Page 9)
• False
In video text memory, ______ are used to display a character.
• 2 bytes (Page 43)
• 4 bytes
• 8 bytes
• 16 bytes
The basis functions fi(u) in Bezier curve must be symmetric with respect to u and (u-2)
• Yes
• No (Page 341)
Three or more points that lie on the same line are called ______________.
• Singular
• Collinear (Page 53)
• Line slop
• Line slop and Singular
Cross product of two vectors result in a ____________________.
• Magnitude
• Vector (Page 116)
• Scalar
• Value
To move a __________________ from one location to another, we translate the center point and redraw the same using new center point.
• Hyperbola
• Parabola
• Circle (Page 119)
• Line
_________________ is the set of points that are equidistant from its origin.
• Line
• Parabola
• Circle (Page 59)
• Ellipse
It is safe to assume that all raster-type monitors can accept the same input
• True
• False
Twice the radius of circle is called as ____________________.
• Area
• Diameter (Page 59)
• Circumference
Both Boundary Filling and Flood filling algorithms are non-recursive techniques.
1. True
2. False (Page 102)
We can take transpose of any matrix.
1. a)True
2. b)False
Normalized cross product of two vectors on that surface provides normal vector
1. a)YES (Page 347)
2. b)NO
Set up your tripod and pointing the camera at the scene
1. a)projection transformation
2. b)viewport transformation
3. c)modeling transformation
4. d)viewing transformation (Page 372)
_________ is based on characteristic size or scale
1. a)Fractal Geometry
3. c)Euclidean Geometry
4. d)None of Above
Bernstein polynomial functions are the basic functions of ______________ curves.
1. a)NURBS
2. b)Bezier
3. c)Both NURBS and Bazier (Page 342) not confirm
4. d)None of the given
Silhouette edges occur when dot product of surface normal vector and the view vector is __________.
1. a)Zero (Page 345)
2. b)One
3. c)Both zero and one
4. d)None of the given
A fractal is generally a property called ___________.
1. a)Fractal Dimension
2. b)Self-similarity
3. c)Koch Curve
4. d)None of above
The curve is always contained within the _______ of the control points
1. a)Tangents
2. b)Convex Hull (Page 340)
3. c)Subdivision
4. d)None of Above
_________ OpenGL function is used for aiming and positioning the camera towards the object
2. b)gluLookAt() (Page 374)
3. c)glFrustum()
4. d)None of Above
Bezier curve can represent the more complex piecewise ___________ curve.
1. a)Polynomial (Page 338)
2. b)Exponential
3. c)Cubic
4. d)None of above
Perspective projection is specified with the function glFrustum().
1. a)Yes (Page 376)
2. b)No
Line connecting any two points within a polygon does not intersect any edge, then it will be a _________ polygon.
1. a)Convex (Page 79)
2. b)Concave
3. c)Complex
4. d)None of the given
The _______________ tests are performed for the midpoints between pixels near the circle path at each sampling step.
1. a)Parabola function
2. b)Eclipse function
3. c)Circle function (Page 62)
4. d)Line function
actual filling process in boundary filling algorithm begins when a point _____________ of the figure is selected.
1. a)Outside the boundary
2. b)Inside the boundary (Page 102) rep
3. c)At boundary
4. d)None of the given
Each hyperbola consists of two ____________________________ .
1. Vertices
2. Nodes
3. Branches (Page 70)
4. Points
Rotating a point requires
1. The coordinates for the point
2. The rotation angles
3. Both of above (Page 180)
4. None
Vectors can be multiplied in a way
1. Dot product
2. Cross product
3. Both of above (page 176)
4. None of given
Shortcoming of Sutherland-Hodgeman Algorithm is concave polygons may be displayed with extensors lines
1. True (Page 244)
2. False
The process of subdivision an entity or surface into one or more non-overlapping primitives.
1. a)Rendering
2. b)Modeling
3. c)Meshing
4. d)None of above (page 259)
In Trivial acceptance/reject test there are four bits of nine regions, Bit 4 represents condition ______________.
1. Outside half plane of left edge, to the left of left edge X < Xmin (Page 143)
2. Outside half plane of right edge, to the right of right edge X > Xmax
3. Outside half plane of bottom edge, below bottom edge Y < Ymin
4. Outside half plane of top edge, above top edge Y > Ymax
In Trivial acceptance/reject test there are four bits of nine regions, Bit 2 represents condition ______________.
1. a)Outside half plane of left edge, to the left of left edge X < Xmin
2. b)Outside half plane of right edge, to the right of right edge X > Xmax
3. c)Outside half plane of bottom edge, below bottom edge Y < Ymin (page 143)
4. d)Outside half plane of top edge, above top edge Y > Ymax
__________ is the process of describing an object or scene so that we can construct an image of it
1. a)Rendering
2. b)Modeling (Page 164)
3. c)Meshing
4. d)None of above
Sutherland-Hodgeman clipping algorithm clips any polygon against a concave clip polygon
1. a)True
2. b)False (Page 244)
Process of moving points in space is called
1. a)Rendering
2. b)Modeling
3. c)Meshing
4. d)None of above (Page 259)
When scaling factor Sx and Sy are assigned the same value, ___________________ scaling is produced that maintains relative object proportions.
1. a)Uniform (page 121)
2. b)Unequal
3. c)Multiform
4. d)Equal
___________________ transformation produces shape distortions as if objects were composed of layers that are caused to slide over each other.
1. a)Rotation
2. b)Translation
3. c)Reflection
4. d)Shear (Page 129)
Global coordinate systems can be defined with respect to local coordinate system
1. a)True
2. b)False (Page 163)
The given primitives are clipped to the boundaries of ________________ and primitives lying outside are not drawn.
1. a)Clipping polygon
2. b)Clipping circle
3. c)Clipping rectangle (Page 247)
4. d)Clipping Line
In 2D transformation, _______________ can be achieved by rotating the object about 180 degrees.
1. a)Translation
2. b)Scaling
3. c)Shear
4. d)Reflection (page 191)
Discard a line with both endpoints outside clipping boundaries is called as
1. a)Trivial reject (Page 142)
2. b)Trivial accept
3. c)Total outside
4. d)Total inside
In _______________ transformation one coordinate is held fixed and the other coordinate or coordinates are shifted.
1. a)Rotation
2. b)Reflection
4. d)Scaling
Locations can be translated or "transformed" from one coordinate system to the other. Select correct option:
1. a)True (Page 163)
2. b)False
Geometric patterns that is repeated at ever smaller scales to produce irregular shapes and surfaces are called ___________
1. a)Geometric patterns
2. b)Fractals (page 352)
3. c)Animated components
4. d)Segments
Bezier curve is tangent to the lines connecting _____________.
1. a)First two points
2. b)Last two points
3. c)Fist two points and last two point (Page 340)
4. d)None of the given
Intensity of the electron beam is controlled by setting _________ levels on the control grid, a metal cylinder that fits over the cathode.
► Amplitude
► Current
► Voltage (Page 26)
► electron
Using Cohen-Sutherland line clipping, it is impossible for a vertex to be labeled 1111.
► true
►false
► Small
► Wide (Page 29)
► Random
► crazy
Which one of the following step is not involved to write pixel using video BIOS services.
► Setting desired video mode
► Using bios service to set color of a screen pixel
► Calling bios interrupt to execute the process of writing pixel.
► Using OpenGL service to set color of a screen pixel (Page 45)
Each number that makes up a matrix is called an ___________ of the matrix.
► Element (page 106)
► Variable
► Value
► component
1. 1. The _______________ tests are performed for the midpoints between pixels near the circle path at each sampling step.
2. Parabola function
3. Eclipse function
4. c. Circle function (Page 62)
5. Line function
1. 2. Interlacing the horizontal refresh __________
1. a. Fools the human eye into thinking the horizontal refresh rate is faster
2. Is distracting and can cause eye fatigue
3. Is necessary because of the shape of the rods in the human eye
4. Is no longer used in any system
1. 3. ( x^2 / a^2 ) + ( y^2 / b^2 ) = 1 is an equation of ____________________________.
2. Parabola
3. Hyperbola
4. c. Ellipse (Page 70)
5. Circle
1. 4. Parity is a concept used to determine which _________________ lie within a polygon.
2. Edge
3. Vertices
4. c. Pixels (Page 80)
5. Points
1. 5. Computer Graphics are used in ____________.
2. Movies development
3. Simulations
4. c. All of the given
5. Game development
1. 6. Each hyperbola consists of two ____________________________ .
2. Vertices
3. Nodes
4. c. Branches (Page 70)
5. Points
d.
2. a. True (Page 38)
3. False
b.
1. 8. ____________ is used for circumference of a circle.
2 * pi*r
1. 9. An improvement in the speed of line-drawing will result in an overall improvement of most graphical applications.
2. a. True
3. b. False
1. 10. _________________ is the set of points that are equidistant from its origin.
2. Line
3. Parabola
4. c. Circle (Page 59)
5. Ellipse
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12 hours ago | 5,425 | 19,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-05 | latest | en | 0.784899 |
http://marketingwithoutthemarketing.com/index-487.html | 1,721,277,270,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514822.16/warc/CC-MAIN-20240718034151-20240718064151-00217.warc.gz | 18,476,811 | 114,210 | ## GATE Exam Syllabus For ECE (Electronics And Communication Engineering)
Candidates of the Graduate Aptitude Tests in Engineering (GATE) who aspire to ace the Electronics and Communication Engineering paper can check out the topics and concepts covered under the subject for the exams from this GATE Syllabus For ECE 2023. We have provided below in this article information about the GATE Exam Syllabus for Electronics And Communication Engineering (ECE) and the link to download the PDF format of the syllabus. Meanwhile, officials of IIT Mumbai also released the latest syllabus on the official website.
Knowing the entire GATE Exam Syllabus For ECEÂ will help students prepare more diligently for the GATE Exam for Electronics and Communication Engineering. Based on the GATE Electronics And Communication Syllabus, students can boost their exam preparations by focusing more on the topics with more marks weightage.
## GATE Syllabus for ECE 2023
The topics and concepts under ECE are divided into eight main sections, as per the latest GATE Exam Syllabus For ECE. According to the GATE Syllabus For ECE, seven sections constitute the main topics such as Engineering Mathematics, Networks, Signals and Systems, Electronic Devices, Analog Circuits, Digital Circuits, Control Systems, Communications and Electromagnetics.
Meanwhile, students appearing for the GATE ECE exam can access the GATE Exam Syllabus For ECE in PDF format from the link given below.
### GATE Exam Syllabus for ECE (Electronics and Communications Engineering)
SECTIONS TOPICS Section 1: Engineering Mathematics Linear Algebra: Vector space, basis, linear dependence and independence, matrix algebra, eigenvalues and eigenvectors, rank, solution of linear equations- existence and uniqueness. Calculus: Mean value theorems, theorems of integral calculus, evaluation of definite and improper integrals, partial derivatives,maxima and minima, multiple integrals, line, surface and volume integrals, Taylor series. Differential Equations: First order equations (linear and nonlinear), higher order linear differential equations, Cauchy’s and Euler’s equations, methods of solution using variation of parameters, complementary function and particular integral, partial differential equations, variable separable method, initial and boundary value problems. Vector Analysis: Vectors in plane and space, vector operations, gradient,divergence and curl, Gauss’s, Green’s and Stokes’ theorems. Complex Analysis: Analytic functions, Cauchy’s integral theorem, Cauchy’s integral formula, sequences, series, convergence tests, Taylor and Laurent series, residue theorem. Probability and Statistics: Mean, median, mode, standard deviation, combinatorial probability, probability distributions, binomial distribution, Poisson distribution, exponential distribution, normal distribution, joint and conditional probability. Section 2: Networks, Signals and Systems Circuit analysis:Node and mesh analysis, superposition, Thevenin’s theorem, Norton’s theorem, reciprocity. Sinusoidal steady state analysis: phasors, complex power, maximum power transfer. Time and frequency domain analysis of linear circuits: RL, RC and RLC circuits, solution of network equations using Laplace transform. Linear 2-port network parameters, wye-delta transformation. Continuous-time signals: Fourier series and Fourier transform, sampling theorem and applications. Discrete-time signals: DTFT, DFT, z-transform, discrete-time processing of continuous-time signals. LTI systems: definition and properties, causality, stability, impulse response, convolution, poles and zeroes, frequency response, group delay, phase delay. Section 3: Electronic Devices Energy bands in intrinsic and extrinsic semiconductors, equilibrium carrier concentration, direct and indirect band-gap semiconductors. Carrier transport: diffusion current, drift current, mobility and resistivity, generation and recombination of carriers, Poisson and continuity equations. P-N junction, Zener diode, BJT, MOS capacitor, MOSFET, LED, photo diode and solar cell. Section 4: Analog Circuits Diode circuits: clipping, clamping and rectifiers. BJT and MOSFET amplifiers: biasing, AC coupling, small signal analysis, frequency response. Current mirrors and differential amplifiers. Op-amp circuits: Amplifiers, summers, differentiators, integrators, active filters, Schmitt triggers and oscillators. Section 5: Digital Circuits Number representations: binary, integer and floating-point- numbers. Combinatorial circuits: Boolean algebra, minimization of functions using Boolean identities and Karnaugh map, logic gates and their static CMOS implementations, arithmetic circuits, code converters, multiplexers, decoders. Sequential circuits: latches and flip-flops, counters, shift-registers, finite state machines, propagation delay, setup and hold time, critical path delay. Data converters: sample and hold circuits, ADCs and DACs. Semiconductor memories: ROM, SRAM, DRAM. Computer organization: Machine instructions and addressing modes, ALU, data-path and control unit, instruction pipe lining. Section 6: Control Systems Basic control system components; Feedback principle; Transfer function; Block diagram representation; Signal flow graph; Transient and steady-state analysis of LTI systems; Frequency response; Routh-Hurwitz and Nyquist stability criteria; Bode and root-locus plots; Lag, lead and lag-lead compensation; State variable model and solution of state equation of LTI systems. Section 7: Communications Random processes: autocorrelation and power spectral density, properties of white noise, filtering of random signals through LTI systems. Analog communications: amplitude modulation and demodulation, angle modulation and demodulation, spectra of AM and FM, superheterodyne receivers. Information theory: entropy, mutual information and channel capacity theorem. Digital communications: PCM, DPCM, digital modulation schemes (ASK, PSK, FSK, QAM), bandwidth, inter-symbol interference, MAP, ML detection, matched filter receiver, SNR and BER. Fundamentals of error correction, Hamming codes, CRC. Section 8: Electromagnetics Maxwell’s equations: differential and integral forms and their interpretation, boundary conditions,wave equation, Poynting vector. Plane waves and properties: reflection and refraction, polarization, phase and group velocity, propagation through various media, skin depth. Transmission lines: equations, characteristic impedance, impedance matching, impedance transformation, S-parameters, Smith chart. Rectangular and circular waveguides, light propagation in optical fibres,dipole and monopole antennas, linear antenna arrays.
### GATE Electronics and Communications Engineering Exam Pattern 2023
In the meantime, any candidate who aspires to ace the GATE exams will find the GATE ECE Exam Pattern 2023 useful. Along with the GATE Exam Syllabus for Electronics and Communications Engineering, the marking scheme will help the candidates to prepare well.
• General Aptitude(GA) Marks of Electronics and Communications(EC) = 15 Marks
• Subject Marks = 85 Marks
• Total Marks for EC = 100 Marks
• Total Time(in Minutes) = 180 Minutes
## Frequently Asked Questions on GATE Syllabus For ECE 2023
Q1
### How many sections are there in the GATE Syllabus For ECE (Electronics and Communications Engineering)?
The GATE Syllabus For Electronics and Communications Engineering consists of 8 main sections. They are Engineering Mathematics, Networks, Signals and Systems, Electronic Devices, Analog Circuits, Digital Circuits, Control Systems, Communications and Electromagnetic.
Q2
### What are the sections that come under the Electromagnetic section of the GATE Electronics And Communication Syllabus?
Topics discussed under section 8, Electromagnetic section are Maxwell’s equations comprising differential and integral forms and their interpretation, boundary conditions, wave equation, Poynting vector, Plane waves and properties: reflection and refraction, polarization, phase and group velocity, propagation through various media, skin depth and Transmission lines: equations, characteristic impedance, impedance matching, impedance transformation, S-parameters, Smith chart. Rectangular and circular waveguides, light propagation in optical fibres, dipole and monopole antennas, and linear antenna arrays.
Q3
### What are the total marks obtainable in the GATE Exams as per the GATE ECE Syllabus 2023?
The total marks obtained from the ECE question paper, as per the latest GATE Syllabus 2023 for Electronics and Communications Engineering, is 100.
Keep learning and stay tuned to get the latest updates on GATE Exam along with GATE Eligibility Criteria, GATE 2023, GATE Admit Card, GATE Syllabus, GATE Previous Year Question Paper, and more. | 1,824 | 8,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-30 | latest | en | 0.837847 |
http://mykhmsmathclass.blogspot.com/2009/05/probability.html | 1,498,132,661,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319265.41/warc/CC-MAIN-20170622114718-20170622134718-00512.warc.gz | 302,447,513 | 23,869 | # Welcome to Planet Infinity KHMS
## 09 May, 2009
### Bag and balls
There are 10 balls in this bag, 4 of them red and 6 blue. I am picking up one ball. Can you say whether it would be red or blue?
The ball drawn may be either red or blue. Since all the balls have an equal chance of being drawn, there is a 4 out of 10 chance that the ball drawn is red and a 6 out of 10 chance that the ball drawn is blue. The probability for the ball drawn to be red is 0.4
The probability for the ball drawn to be blue is 0.6
#### 1 comment:
privaterealityshow.blogspot.com said...
After getting to this blog I do not feel clever anymore. It is a great blog! I like it. | 180 | 662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-26 | longest | en | 0.929417 |
http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog2/21/01/01/0001/ | 1,529,637,213,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864343.37/warc/CC-MAIN-20180622030142-20180622050142-00006.warc.gz | 130,651,031 | 7,504 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
PolyLog
http://functions.wolfram.com/10.07.21.0001.01
Input Form
Integrate[PolyLog[2, z], z] == -z + (z - 1) Log[1 - z] + z PolyLog[2, z]
Standard Form
Cell[BoxData[RowBox[List[RowBox[List["\[Integral]", RowBox[List[RowBox[List["PolyLog", "[", RowBox[List["2", ",", "z"]], "]"]], RowBox[List["\[DifferentialD]", "z"]]]]]], "\[Equal]", RowBox[List[RowBox[List["-", "z"]], "+", RowBox[List[RowBox[List["(", RowBox[List["z", "-", "1"]], ")"]], " ", RowBox[List["Log", "[", RowBox[List["1", "-", "z"]], "]"]]]], "+", RowBox[List["z", " ", RowBox[List["PolyLog", "[", RowBox[List["2", ",", "z"]], "]"]]]]]]]]]]
MathML Form
Li PolyLog 2 ( z ) z Li PolyLog 2 ( z ) z - z + ( z - 1 ) log ( 1 - z ) z PolyLog 2 z PolyLog 2 z z -1 z z -1 1 -1 z [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["\[Integral]", RowBox[List[RowBox[List["PolyLog", "[", RowBox[List["2", ",", "z_"]], "]"]], RowBox[List["\[DifferentialD]", "z_"]]]]]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List["-", "z"]], "+", RowBox[List[RowBox[List["(", RowBox[List["z", "-", "1"]], ")"]], " ", RowBox[List["Log", "[", RowBox[List["1", "-", "z"]], "]"]]]], "+", RowBox[List["z", " ", RowBox[List["PolyLog", "[", RowBox[List["2", ",", "z"]], "]"]]]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2001-10-29 | 617 | 1,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-26 | latest | en | 0.2353 |
http://mathhelpforum.com/calculus/23884-derivation-print.html | 1,529,944,496,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868135.87/warc/CC-MAIN-20180625150537-20180625170537-00024.warc.gz | 201,242,243 | 3,032 | # Derivation
• Dec 1st 2007, 06:38 AM
Coach
Derivation
$\displaystyle h(x)=(x+1)^2(2x+1)^3$
Thank you!
• Dec 1st 2007, 07:10 AM
Krizalid
Quote:
Originally Posted by Coach
$\displaystyle h(x)=(x+1)^2(2x+1)^3$
Semi-nice way: product rule. (Do you know what the product rule is?)
Nasty way: expand the expression but this has a benefit, you'll only have to apply the power rule.
• Dec 1st 2007, 07:34 AM
Coach
I do know the product rule in case you are talking about the uv´+vu´ rule, where the accent indicates derivated(I am sorry, I am very bad at these terms, since the textbook we are using is in Finnish). However, whenever I try to apply this rule, I get the wrong answer. Here is what I did.
$\displaystyle (X+1)^3*2(2X-1)^1+ (2X-1)^2*3(X+1)^2$
• Dec 1st 2007, 07:50 AM
Krizalid
Yes, that states the product rule. (Don't worry about the terms, I'm chilean and I have to learn them anyway.)
$\displaystyle \Big[ {f(x) \cdot g(x)} \Big]' = f'(x) \cdot g(x) + f(x) \cdot g'(x).$
The function is $\displaystyle h(x)=(x+1)^2(2x+1)^3,$
So $\displaystyle h'(x)=2(x+1)\cdot(2x+1)^3+3(x+1)^2(2x+1)^2\cdot2.$
Can you take it from there?
• Dec 1st 2007, 07:58 AM
Coach
Yes. Thank you!
But where did you get the last 2 from?
• Dec 1st 2007, 08:03 AM
Krizalid
It's the chain rule; the derivative of $\displaystyle (2x+1)^3$ is $\displaystyle 3(2x+1)^2\cdot(2x+1)'.$ That's why appears the $\displaystyle 2.$ | 520 | 1,412 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-26 | latest | en | 0.878075 |
http://www.velocityreviews.com/forums/t861073-question-abour-rand.html | 1,386,910,634,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164884560/warc/CC-MAIN-20131204134804-00030-ip-10-33-133-15.ec2.internal.warc.gz | 590,430,596 | 10,774 | Velocity Reviews > Ruby > Question abour rand()
# Question abour rand()
Panagiotis Atmatzidis
Guest
Posts: n/a
12-29-2009
Dear Sirs,
I'm looking for a way to create a random sequence of numbers using the =
'rand' method. I can think of something like:
-----
while pipi =3D=3D TRUE
if x > 1950 || x < 2000
return x
pipi =3D FALSE
------
Since however the book that I'm reading has not introduced while/else/if =
etc. I'm thinking that there must be an easier and probably more =
sensible way to approach this. As far as I've read rand(2000) will give =
random numbers from 0-to-2000 no matter what.
NOTE: I don't want to use another method just to do the job.
thanks in advance & best regards
Panagiotis (atmosx) Atmatzidis
email: http://www.velocityreviews.com/forums/(E-Mail Removed)
URL: http://www.convalesco.org
GnuPG ID: 0xFC4E8BB4=20
gpg --keyserver x-hkp://pgp.mit.edu --recv-keys 0xFC4E8BB4
--
The wise man said: "Never argue with an idiot. They bring you down to =
their level and beat you with experience."
Fleck Jean-Julien
Guest
Posts: n/a
12-29-2009
Hello,
> I'm looking for a way to create a random sequence of numbers using the 'r=
and' method. I can think of something like:
>
> -----
> while pipi =3D=3D TRUE
> =A0if x > 1950 || x < 2000
Perhaps you were thinking of "and" rather than "or" because all number
is either greater than 1950 or smaller than 2000.
> Since however the book that I'm reading has not introduced while/else/if =
etc. I'm thinking that there must be an easier and probably more sensible w=
ay to approach this. As far as I've read rand(2000) will give random number=
s from 0-to-2000 no matter what.
Does 'rand(50) + 1950' do what you want ?
Cheers,
--=20
JJ Fleck
PCSI1 Lyc=E9e Kl=E9ber
Seebs
Guest
Posts: n/a
12-29-2009
On 2009-12-29, Panagiotis Atmatzidis <(E-Mail Removed)> wrote:
> while pipi == TRUE
> if x > 1950 || x < 2000
> return x
> pipi = FALSE
Hmm. Looks as though you want the range 1951..1999. Well, that's easy
enough:
rand(49) + 1951
(Note that rand(49) returns anything from 0 to 48, inclusive.)
-s
--
Copyright 2009, all wrongs reversed. Peter Seebach / (E-Mail Removed)
http://www.seebs.net/log/ <-- lawsuits, religion, and funny pictures
http://en.wikipedia.org/wiki/Fair_Game_(Scientology) <-- get educated!
Marnen Laibow-Koser
Guest
Posts: n/a
12-29-2009
Panagiotis Atmatzidis wrote:
> Dear Sirs,
>
> I'm looking for a way to create a random sequence of numbers using the
> 'rand' method. I can think of something like:
>
> -----
> while pipi == TRUE
> if x > 1950 || x < 2000
> return x
> pipi = FALSE
> ------
>
> Since however the book that I'm reading has not introduced while/else/if
> etc. I'm thinking that there must be an easier and probably more
> sensible way to approach this. As far as I've read rand(2000) will give
> random numbers from 0-to-2000 no matter what
Use Seebs' suggestion.
>
> NOTE: I don't want to use another method just to do the job.
Then you are probably being silly. It will make for more readable and
more reusable code if you encapsulate this in a method. Having lots of
short methods is generally a *good* thing.
>
> thanks in advance & best regards
>
>
>
> Panagiotis (atmosx) Atmatzidis
>
> email: (E-Mail Removed)
> URL: http://www.convalesco.org
> GnuPG ID: 0xFC4E8BB4
> gpg --keyserver x-hkp://pgp.mit.edu --recv-keys 0xFC4E8BB4
Best,
--
Marnen Laibow-Koser
http://www.marnen.org
(E-Mail Removed)
--
Posted via http://www.ruby-forum.com/.
Phillip Gawlowski
Guest
Posts: n/a
12-29-2009
On 29.12.2009 18:00, Marnen Laibow-Koser wrote:
>> NOTE: I don't want to use another method just to do the job.
>
> Then you are probably being silly. It will make for more readable and
> more reusable code if you encapsulate this in a method. Having lots of
> short methods is generally a *good* thing.
"Walk before you run."
--
Phillip Gawlowski
Bertram Scharpf
Guest
Posts: n/a
12-29-2009
Hi,
Am Dienstag, 29. Dez 2009, 18:51:53 +0900 schrieb Panagiotis Atmatzidis:
> I'm looking for a way to create a random sequence of numbers using the 'rand' method. I can think of something like:
>
> -----
> while pipi == TRUE
> if x > 1950 || x < 2000
> return x
> pipi = FALSE
> ------
>
> Since however the book that I'm reading has not introduced while/else/if etc. I'm thinking that there must be an easier and probably more sensible way to approach this. As far as I've read rand(2000) will give random numbers from 0-to-2000 no matter what.
>
> NOTE: I don't want to use another method just to do the job.
>
> thanks in advance & best regards
The "rand" method has been mentioned. Here are just some
You don't need a loop for calculating a random value. (There is a
random algorithm below.) But when use a loop, then you either say
"return" or leave it by a condition but not both.
cond = true
while cond do
...
if some_condition then
cond = false
end
...
end
_or_
loop do
...
break if some_condition
...
end
Note that "break" can even return values:
result = loop do
break "hi"
end
result == "hi"
And here is an example of a very simple random implementation:
class R
def initialize
@r = 0
end
def nextval
@r = ((@r * 0xabcd) + 0x73737) % 0x1_0000_0000
@r >> 16
end
end
r = R.new
new_random_value = r.iterate
random_value_below_fifty = r.iterate % 50
Bertram
--
Bertram Scharpf
Stuttgart, Deutschland/Germany
*
Discover String#notempty? at <http://raa.ruby-lang.org/project/step>.
Brian Candler
Guest
Posts: n/a
12-29-2009
Panagiotis Atmatzidis wrote:
> I'm looking for a way to create a random sequence of numbers using the
> 'rand' method.
Others have answered the 'random number between 1950 and 2000' bit.
In terms of creating a "sequence" of these values, here are a couple of
options to consider.
(1) Build an array.
a = []
10.times { a << rand(50) }
This array (a) is an object which can be returned, passed to other
functions etc. The above code hard-codes the number of values to build
(10), but you could choose this dynamically at run-time instead.
(2) Yield the values.
For code which wants to "return" multiple values, you can instead
"yield" the values to a block which the caller provides. You can yield
multiple times, thus calling the block multiple times. In other
programming languages you might call this a "callback function"
def generate_random(n=10)
n.times do
yield rand(50)
end
end
generate_random do |r|
puts "I got #{r}!"
end
Note that you can legitimately yield an infinite number of times,
because the block can abort the sequence when it wants. So in the
following example, the user of the function decides when they don't want
any more random numbers:
def generate_infinite
loop do
yield rand(50)
end
end
count = 0
generate_infinite do |r|
puts "I got #{r}!"
count += 1
break unless count < 10
end
A more advanced way to do this is with an 'Enumerator' object, but I'd
--
Posted via http://www.ruby-forum.com/. | 2,048 | 6,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2013-48 | longest | en | 0.882554 |
http://guyhaas.com/bfoit/itp/life.html | 1,657,014,844,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00611.warc.gz | 24,904,535 | 3,046 | BFOIT - Introduction to Computer Programming
# The Game of Life
Draw some patterns (communities of life) by clicking the left mouse button on squares to bring them to life. Then click on the [Step] button to watch what happens as the rules are applied in one cycle of Life. To repeat the cycle over and over forever, click on the [Go] button.
The Game of Life (invented by mathematician John Conway) is an example of Cellular Automaton.
``` A cellular automaton is a collection of "colored" cells on a grid of
specified shape that evolves through a number of discrete time steps
according to a set of rules based on the states of neighboring cells.
The rules are then applied iteratively for as many time steps as
desired. (from http://mathworld.wolfram.com/CellularAutomaton.html).
```
The "Game of Life" plays out on a grid. The squares of the grid are called cells. A cell that is alive is colored in. The rules for Life are simple. They are:
1. In order for a cell to remain alive, it must have two or three neighbors.
2. If a live cell has less than two neighbors, it dies (loneliness).
3. If a live cell has more than three neighbors, it dies (over-crowdedness).
4. If an empty cell has exactly two neighbors, it comes to life.
A cell's neighbors are the eight cells which surround it (to its north, northeast, east, etc...). | 324 | 1,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-27 | latest | en | 0.938349 |
https://discuss.codechef.com/t/rec20e-editorial/96257 | 1,642,739,165,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302723.60/warc/CC-MAIN-20220121040956-20220121070956-00447.warc.gz | 237,628,657 | 6,205 | # REC20E - Editorial
Setter : Vishal Mahavar
Testers : Prasann Kumar Gupta, Onkar Ratnaparkhi
Medium-Hard
# PROBLEM
Given a tree with N nodes and N-1 edges, where each node can have two colors represented by 0 or 1. We have to process Q queries of two types on it :
• 1\> x - Change the color of node x (from 0 to 1 and vice versa).
• 2\>x\>k\>c - Find the number of nodes with color c at a distance k from node x.
# EXPLANATION
Since the queries are only related to the distance between the nodes and not the actual path, it’s intuitive to come up with a centroid decomposition solution.
We are going to use is the following properties of the centroid tree :
• Height of the centroid tree is O(logN).
• Let sz_i denote the size of the subtree of centroid i in the centroid tree, then \sum\limits_{i=1}^N sz_i is O(NlogN).
• For any pair of nodes p and q, the path from p to q in the original tree can always be broken into the concatenation of paths from p to L and L to q. Here L is the lowest common ancestor of nodes p and q in the centroid tree.
Let’s denote dist[i][c][d] as number of nodes which have color c and are at a distance of d from centroid i in it’s subtree. It’s easy to see that the number of distinct values of d can be at most sz_i, hence it’s feasible to compute it (second property).
Also consider the following computation, dist\_par[i][c][d] which is the same as the above computation but the difference is that we are measuring distance from the parent of i in the centroid tree instead of the centroid i. Obviously this is not needed to be computed for the root of the centroid tree. These computations can be done exactly the same way we process updates. Let’s see how we can process updates and queries using the above two computations.
Update
WLOG, consider a node x whose color is 0 and is to be changed to 1. The values of dist[y][][] will change only for the ancestors of this node in the centroid tree (including the node itself). So, we traverse up the centroid tree (first property) and change this for every centroid y. Let’s say distance between p and q nodes (in the original tree) is denoted using d_{xy}, then dist[y][0][d_{xy}] will reduce by one and dist[y][1][d_{xy}] will increment by one.
Similarly if y is not the root of the centroid tree, let z be the parent of y. Then dist\_par[z][0][d_{zx}] will reduce by one and dist\_par[z][1][d_{zx}] will increment by one.
Query
Once again, let’s assume we need number of nodes at distance of k from node x which are of color 0. Using the third property mentioned above, we only need to go through the ancestors of the node x in centroid tree. For each such ancestor y, we need to find the number of required nodes which are at distance k - d_{xy} from y. Which can be written as dist[y][0][k-d_{xy}] and be added to our answer.
But here’s a catch. Consider the node z such that it is a child of y in the centroid tree and is also an ancestor of node x. The nodes present in the subtree of z cannot be included using the third property (considering node y as the breakpoint of paths). The reason is, the LCA of such nodes and the node x is node z, NOT node y hence these nodes have to be counted when we are at node z. In order to remove such nodes, we need to subtract dist\_par[z][0][k-d_{xy}] from our answer.
We are traversing O(logN) on the centroid tree. Since we only need to store the required d's in dist[i][c][d], we’ll use a map (in C++) for this purpose which gives an extra logN factor to the traversal. And we can use both logarithmic time and constant time method for calculating distances between any two nodes in the tree, they will be added paralelly to the time complexity. Hence the time complexity per query is O(log^2N). The preprocessing itself is same as processing updates on N nodes. The total time complexity is O((N+Q)log^2N).
NOTE: You can also use unordered_map(or any other hashing based data structure) to solve this in O((N+Q)logN).
# SOLUTIONS
Feedbacks and suggestions are always welcomed.
Do discuss the solution/any other approach in comments .
2 Likes | 1,039 | 4,116 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-05 | latest | en | 0.903816 |
https://ccssanswers.com/eureka-math-grade-2-module-8-lesson-4/ | 1,721,875,886,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00263.warc.gz | 144,803,350 | 44,841 | ## Engage NY Eureka Math 2nd Grade Module 8 Lesson 4 Answer Key
### Eureka Math Grade 2 Module 8 Lesson 4 Problem Set Answer Key
Question 1.
Use your ruler to draw 2 parallel lines that are not the same length.
Question 2.
Use your ruler to draw 2 parallel lines that are the same length.
Question 3.
Trace the parallel lines on each quadrilateral using a crayon. For each shape with two sets of parallel lines, use two different colors. Use your index card to find each square corner, and box it.
Question 4.
Draw a parallelogram with no square corners.
Question 5.
Draw a quadrilateral with 4 square corners.
Question 6.
Measure and label the sides of the figure to the right with your centimeter ruler. What do you notice? Be ready to talk about the attributes of this quadrilateral. Can you remember what this polygon is called?
Question 7.
A square is a special rectangle. What makes it special?
### Eureka Math Grade 2 Module 8 Lesson 4 Exit Ticket Answer Key
Question 1:
Use crayons to trace the parallel sides on each quadrilateral. Use your index card to find each square corner, and box it.
### Eureka Math Grade 2 Module 8 Lesson 4 Homework Answer Key
Question 1.
Use your ruler to draw 2 parallel lines that are not the same length.
Question 2.
Use your ruler to draw 2 parallel lines that are the same length.
Question 3.
Draw a quadrilateral with two sets of parallel sides. What is the name of this quadrilateral?
Question 4.
Draw a quadrilateral with 4 square corners and opposite sides the same length. What is the name of this quadrilateral?
Question 5.
A square is a special rectangle. What makes it special?
Question 6.
Color each quadrilateral with 4 square corners and two sets of parallel sides red. Color each quadrilateral with no square corners and no parallel sides blue. Circle each quadrilateral with one or more sets of parallel sides green. | 438 | 1,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-30 | latest | en | 0.876156 |
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### 75 Cards in this Set
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kilometers (standard?) miles (measures a long distance) D inches (metric?) centimeters D product the result (answer) from multiplying D grams (standard?) ounces D quadrilateral 4 sided figure D cone solid round figure with a pointed end D capacity content (liquid) D length distance D face flat surface D 45 min. 45 min.(3/4) past the hour D factor number that is being multiplied D sides edges or faces D the most add or multiply D cylinder solid tube shaped figure D estimate about, almost, rounded, NOT exactly D 1 pound= 16 oz. D the least subtract or divide D similar same shape DIFFERENT size D acute angle LESS than 90 degrees D Standard Measurement American Measurement D before first, preceding D area space inside a figure D pints 2 cups D turn shift a given amount of degrees D square a 4 sided figure with equal sides D edge border of a solid figure D parallel lines lines that never meet D straight line 180 degrees D mode the number that appears the most D weight (mass) lump sum, total lbs., oz., kg.,g. D line 180 degrees D flip reflection D between number in the middle (median) D regroup borrow/subtraction carry/addition D obtuse angle GREATER than 90 degrees D milliliters (standard?) teaspoons D after following number D feet (metric?)= decimeters/meters D fraction a PART of a whole D 1 hour (hr.) = 60 minutes (min.) D difference the result (answer) of subtraction D congruent SAME size & shape D triangle, square,rectangle circle polygons, P plane figures (flat) D 1/2 hour = 30 min.,half past the hour D intersect lines that cross D slide move to the side D fact families set of 3 numbers that can be formed into 2 addition/subtraction or 2 multiplication/division facts D kilogram (standard?) pound D dividend number that is divided D cube a solid 6 sided figure D sum the result of addition D Metric Measurement measurement based on the meter D perimeter the distance around a figure (add all the sides together) D CAPTURE ! Math Vocabulary right angle 90 degrees D yards (metric?) meters D polygons plane figures D line of symmetry SAME on both sides when folded in half D even numbers numbers that can be divided evenly, without a remainder D 15 min. = quarter (1/4) hour D digit a number, numeral D odd numbers numbers that CANNOT be divided evenly D how many more? subtract D minute (min.)= 60 seconds (sec.) D quotient the result (answer) of division D line segment a line having a beginning and an end D range the low from the high D ray a line having a beginning (endpoint) but no end D mean the average number D median the number in the middle D how many in all add D divide separate into groups D divisor the number by which another number is divided D greater larger number(more) D lesser smaller number (less than the higher number) D | 775 | 3,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-22 | latest | en | 0.718983 |
https://metanumbers.com/59726 | 1,620,694,184,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991553.4/warc/CC-MAIN-20210510235021-20210511025021-00176.warc.gz | 402,860,410 | 7,448 | ## 59726
59,726 (fifty-nine thousand seven hundred twenty-six) is an even five-digits composite number following 59725 and preceding 59727. In scientific notation, it is written as 5.9726 × 104. The sum of its digits is 29. It has a total of 2 prime factors and 4 positive divisors. There are 29,862 positive integers (up to 59726) that are relatively prime to 59726.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 29
• Digital Root 2
## Name
Short name 59 thousand 726 fifty-nine thousand seven hundred twenty-six
## Notation
Scientific notation 5.9726 × 104 59.726 × 103
## Prime Factorization of 59726
Prime Factorization 2 × 29863
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 59726 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 59,726 is 2 × 29863. Since it has a total of 2 prime factors, 59,726 is a composite number.
## Divisors of 59726
1, 2, 29863, 59726
4 divisors
Even divisors 2 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 89592 Sum of all the positive divisors of n s(n) 29866 Sum of the proper positive divisors of n A(n) 22398 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 244.389 Returns the nth root of the product of n divisors H(n) 2.66658 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 59,726 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 59,726) is 89,592, the average is 22,398.
## Other Arithmetic Functions (n = 59726)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 29862 Total number of positive integers not greater than n that are coprime to n λ(n) 29862 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 6019 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 29,862 positive integers (less than 59,726) that are coprime with 59,726. And there are approximately 6,019 prime numbers less than or equal to 59,726.
## Divisibility of 59726
m n mod m 2 3 4 5 6 7 8 9 0 2 2 1 2 2 6 2
The number 59,726 is divisible by 2.
## Classification of 59726
• Arithmetic
• Semiprime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
## Base conversion (59726)
Base System Value
2 Binary 1110100101001110
3 Ternary 10000221002
4 Quaternary 32211032
5 Quinary 3402401
6 Senary 1140302
8 Octal 164516
10 Decimal 59726
12 Duodecimal 2a692
20 Vigesimal 7966
36 Base36 1a32
## Basic calculations (n = 59726)
### Multiplication
n×i
n×2 119452 179178 238904 298630
### Division
ni
n⁄2 29863 19908.7 14931.5 11945.2
### Exponentiation
ni
n2 3567195076 213054293109176 12724880710238645776 760006225299713357617376
### Nth Root
i√n
2√n 244.389 39.089 15.6329 9.02054
## 59726 as geometric shapes
### Circle
Diameter 119452 375270 1.12067e+10
### Sphere
Volume 8.9244e+14 4.48267e+10 375270
### Square
Length = n
Perimeter 238904 3.5672e+09 84465.3
### Cube
Length = n
Surface area 2.14032e+10 2.13054e+14 103448
### Equilateral Triangle
Length = n
Perimeter 179178 1.54464e+09 51724.2
### Triangular Pyramid
Length = n
Surface area 6.17856e+09 2.51087e+13 48766.1
## Cryptographic Hash Functions
md5 74e3932c1af80284b053e195c2dbe0cd de92f06bd382dacd1e11024b5b4c956a62f094e7 babce8cc384d2685aacba324bb065c8b0292b4d4f93327e7bfbcae776dc08699 7a46c526d2a2227582917d59627d135d993e74950b87eb0b5dd1a92d406a2608661057ca2054ef845f7e59a3d25a24e348c74fe4d0e934649e7d409df713b63f 6622ae60a2e3744ac291d6555a640be162070d5b | 1,452 | 4,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-21 | latest | en | 0.806027 |
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## JournalJournal: A
A reddit comment inspired a bit of an interesting thought.
The question is. If you had a fast TM (FAST_TM) with a little bit of paper tape, and a slow TM (SLOW_TM) with a lot of paper tape, could you simulate FAST_TM on SLOW_TM? Could you trade off memory for speed to get a (close to realtime even) representation of FAST_TM on SLOW_TM?
I think this gets interesting if you ask for a SLOW_TM to have with an infinite ( |â| ) paper tape. As long as you could 'read state' into those tape cells fast enough you could have stories of state going back as far back as the FAST_TM has not been in a stable/idle state. In an idle state you could start to play catchup. Between idle state points you would be at some indeterminate point between where the last idle state was that you could play catch up, in effect you'll probably have a whole stack of possible state points(what is the computational complexity of catching up on this stack for the size of the stack STACKSIZE?). So the load % of FAST_TM would determine the average lag time between your successful modelling of it.
This suggests that you could probably get a probablistic chance of modelling the FAST_TM by just adding memory of (O(S)+O(T))t where S is the memory required to record one state, t is the time and O(T) is the amount of memory you spend transitioning from state to state. I'm guessing this would be hard to do so on cpus with little memory involved, ie there's a high constant factor, but once you're past this constant factor it gets relatively easy to do...but then again maybe it doesn't? I think the lower bounds for turing machines of certain memory capability in terms of size are very small and we haven't got a lot of proofs for them.
Now here's an idea for a new kind of machine: a FAST_TM simulated on a new kind of SLOW_TM, ie where SLOW_TM has a bounded amount of memory proportional to some % of possible outcomes. Let's call this SIM_TM_LB. SIM_TM_LB is going to have a lower busy beaver-like number than a regular TM of its size because there's a certain % of possible outcomes that cannot be simulated (that overflow the stack allowed). going to be some lower limit L 1 L(M) BB(M) that the largest program available on simulated-fast-cpu can run. Proving what that is would be interesting because as you expand M (again in relation to the load average ratio between the two, the state transition memory footprint T, and the FAST_TM state size S) you're also define a L(M) BB(M) which means you're defining a new kind of number let's say Î that seems to be related to \omega: it's \phi = \sigma_i 1/L(i). Why is Î important?
It's yet another way of looking at problems where you're dealing with something smarter than you are. It's where you're playing a game with god. Where you're having an argument with an oracle. It expresses all information that you can possibly acquire rather than what your opponent can possibly know. Or does it?
This suggest \psi s parameters are T and S. Are there any others?
Also: what happens if we start allowing stack overflows to transverse from TM to TM? This seems to build a new kind of machine, also that has potentially really weird properties.
## JournalJournal: The Stupid Fallacy1
Prerequisites :
The Stupid Fallacy (named/discovered by Chris Rileyâ) Is-A Argument from Ignorance that includes an extra component:
Instead of merely being an argument that draws a conclusion from the *lack* of knowledge on a topic (and not in a bayesian-friendly way of enumerating possibilities and going from there, either) you have
• (optional) empty platitudes that stand as non-truth functional filler in place of where premises would normally go followed by
• a statement that is so utterly wrong and against nature/reason/good sense, that it's fractally wrong, and decompressing the argument against it would overflow any reasonable wanabe bayesian's attempt at both responding and adjusting their priors.
Example:
Why don't you like GMOs?
Who knows what chemicles they put in GMOs! They're probably dangerous! Besides, God tells me to not let my precious bodily fluids become tainted by GMOs. You have to believe me because my beliefs are not subject to logical fallacies, since you and I are both christians[1].
Why is this worth keeping around?
Because it's not just ignorance. It's recursive, or close to recursive ignorance. It's ignorance that requires disproportionate amounts of cognitive surplus available to dispel. You have to basically reconstruct an entire worldview relying on evidence rather than 'feelings' or 'blind belief in what my elders said' in order to get your point across.
[1] no really.
## JournalJournal: /r/button2
If you allocated F(t) seconds of ripple credit (in HRS) for every user with flair class ( F = { (60,0) , (50, 10), (40, 20), (30,30), (20,40), (10,50), (0,0), {other,0} } ) what is your expected value of this offer?
Given there's roughly 500,000 people with a flair class you'd think it'd be a lot...but it sure seems like the overwhelming majority of the 500,000 chose to feed in '60' or 'other' and it's unclear that would change. It depends on the distribution of the random variable t, roughly corresponding to the probability distribution of these flair classes over the ~40 million users who are given this choice.
Will the offer of a reward for flair classes make it more or less likely that people would coordinate better?
How large of a reward would it have to be in order for there to be any affect whatsoever (given how diverse reddit is...this could be huge)
How would the reward affect/skew the distribution?
Edit: Upon closer inspection: looks like it isn't quite so simple: F(t) is actually (60-t)(H(60-t)H(t)) where t is a natural number {0,1,2,...}.
## JournalJournal: On DIsagreement
It seems to be begging the question to demand acceptance of the prerequisites of the requirement to agree not to disagree. These prerequisites form a model of what it means to be rational. I'm not sure if they are the best possible model, or even if they work. This includes the hypothesis that a system is only efficient if and only if it has a model of itself, and the semenatic, syntactic forms, and (trustable) media that are acceptable to each of us. So, in short - it is reasonable to expect come to agreement with rational agents except about rationality itself. There it isn't necessarily a question of whether you agree or not, but a matter of how rationality can even work. A matter of tweaking the model to allow greater truths to be perceived between multiple agents. So, the #1 way to get out of any disagreement is to look at this model, or the idea of a model in general.
## JournalJournal: 24 years later
I entered this tunnel, starting preschool in what, 1986? And here I am, at the very end of this god forsaken tunnel. All hope is lost, and yet here I am, washed up on the damned shore.
At least I'm pretty sure this time. I haven't filled out all the paperwork, but the exam is over. I thought I did anywhere between reasonably good and fantastic on the final exam -- all I need is a 14% to finish my degree (I suspect I got somewhere between probably 85+/-5%, stddev 5% or so or so.) so... I'd say I'm safely done.
But those considerations aside, it has been a long travel to get here. Where the hell am I?
## JournalJournal: Agree to disagree meme
Since no two people agree on everything (especially in political/religious matters, but not exclusively so) here's an open question for you -- what is ONE belief that you think I do not currently believe but should. I would also suggest posting this on YOUR /. journal /etc to see what responses you get.
DISCLAIMER: since I'm *asking* for these beliefs I can't really expect to be too hard on you for them. The purpose here isn't to press you on them (you have the rest of my life to convert me, if you so choose), merely to enumerate them. Although agreeing to disagree is unfriendly, in the confines of this thread it will be seen as acceptable, just to flush out some fresh ideas that I might not have considered.
## JournalJournal: I'm finally no longer bitchslapped
after what, 10 years of not being able to moderate on slashdot, I'm finally allowed moderation privileges again. Crazy. Now what to do with my mod points . . .
## JournalJournal: Poissonian Statistics & OKCupid
Statistics has never really been my strong suit, but one of the equations we used in our astronomy labs was one out of poisson statistics, that is
P(r,u) = (e^{-u})*(u^r)/r!
and P(F) = P(r,u) * Number_Of_Stars_Surveyed,
where F is the total number of stars
or, perhaps written differently,
(defun stars (T U S) (* T (* (/ (* (expt 2.71828183 (- U)) (expt U S)) (* 1.0 (! S))) 100.0)))
with
(defun ! (int) (if (eq 0 int) 1 (* int (! (- int 1)))))
This is what you should use if you have a system of magnitudes of stars that are ordered in a poisson distribution, to determine what probability you have, given 'r' discrete categories of stars(in our case, r=5) that are each in turn less likely according to the poisson distribution, what is the probability that your data set can be picked. That is to say, what is the likelyhood that your ratio U of 'number-of-stars in category N and above vs. the number of stars' came about purely by chance? At least if I'm understanding this correctly, which I'm not sure.
According to my calculation,
(stars 85.0 (/ 7.0 85.0) 5) should be .0002%
(stars 85.0 (/ 27.0 85.0) 5) .2%
(stars 85.0 (/ 38.0 85.0) 5) .8%
(stars 85.0 (/ 53.0 85.0) 5) 4%
(stars 85.0 (/ 85.0 85.0) 5) 26%
That last one makes my head hurt. Under what condition could we be more certain that some order of stars was likely? 26% isn't very likely. In our lab my results for the Pleiades cluster was .002%, it should have been 4e-6%, but still... you can't seem to do much better than 26%, which isn't very likely at all. Yet looking at the graph in the wikipedia makes this at least *look* right...
Seems like I'm missing something here...what is the likelyhood that there are 53 "2 or greater" star women given a sample size of 85 given a poisson distribution? What about 85 1 or greater? How am I messing this up?
DATA
5 stars 7
4 stars 11 (- 38 11) 27 4 or greater
3 stars 15 (- 53 15) 38 3 or greater
2 stars 20 (- 85 32) 53 2 or greater
1 stars 32
(+ 7 11 15 20 32) total 85
## JournalJournal: DIY toothpaste no worky
2 parts baking soda, 1 part hydrogen peroxide.
I'm not sure if it cleaned my teeth at all, but it sure did cause me to almost immediately puke.
On the upside, if I ever need to puke I know just the thing to force me to...
## JournalJournal: PRAISE BE SERVER
I just passed my last class. I will soon have a degree. After SO MANY DAMN YEARS, I get a damn piece of paper.
## JournalJournal: Apparently, I'm ugly3
Double blind survey of around 30 fairly well randomized women of various ages of various pictures of me led me to a score of about 2/10 on plentyoffish.com, I lost some of the numbers, so no mean/mode/etc unfortunately.
I always considered myself ugly, although with my recent loss of weight I figured I'd be approaching average by now...
guess not.
Although who knows how hot you have to be to not be ugly at pof standards.
But still, you can't argue with empirical data. Maybe the side burns have got to go(although there's always the risk that out of the 30 women who did rank me, one did rank me as hot, and she may have liked them and maybe, just maybe, she's the one I'd be happiest with anyway, and by cutting them off I'd be cutting my chances with her, etc etc).
Also, I like the sean kennedy idea, of embracing your ugliness, accenting it, and building it into something meaningful. Got a deformity, accentuate it. Got a scar? Make it more visible, etc etc. So it's not like this fact is going to bring me down or anything. I've been knowingly ugly for most of my life, I can live the rest of my life this way.
It'd be nice to be able to find someone to love, and be able to keep them easier though, if my charisma roll were a little higher...
## JournalJournal: Lab test1
Over at my LJ, I describe today's job interview, and how I might have botched part of it. What would you do in such a situation?
## JournalJournal: The effective data channel capacity of the universe and S:N
Start with a 1d edge between two nodes. An abstract version of a cable between two network points. We know by Claude Shannon's results that Bitrate = Bandwidth*log10(1+S:N). Keep that one in mind.
But what if we started with a plane, instead of an edge? A wall, instead of a cable? Then we could split the plane indefinitely...oh but we can't do that! Planck distance starts becoming an issue. So ~1/1.6E-35 or 6.3E34 edges/m is the largest possible amount of cables per metre, and ~4E69 cables per square metre cross section area.
But just as information may be transmitted in waves through an approximately 1d medium, they can be sent through that cross sectional area; just each of the 4E69 cables carrying signal, that's all right? It'd be like a wave coming into a port; as long as you split the harbour into fine enough paratitions, you can transmit independently, and get data independently.
What if went a step further? And started transmitting information through volumes. Perhaps a point particle in the middle of a sphere transmitting information to or from the entire surface of that sphere. Or...backwards in time or something! Then we're talking 2.5E104/m3...but there's also planck time to consider, so the smallest resolution is pretty much defined at 2E43 Hz...and there can't really be any more than 1 bit per time in this context(a single bit is hard enough to imagine)...so we're up to 5E147 bits per cubic metre....but why would we stop there? Let's use the entire observable universe.. 8.8E26 being used this way; bingo, 4.4E174 bits per universe.
4.4E174 bits = 1 log ( 1 + S/N ) log24.4E174 = 1 + S:N log2 4.4 + log2 10E174 -1 = S:N 2.1+578+1 ~= 581 = S:N
Or if we account for time as another spacelike dimension...the age of the universe...4.4E174*4E17 ...or 1.8E192 bits per universe log2 1.8E192 = S:N + 1 ~192log2 10 + log2(1.8) -1 = S:N ~638 S:N for not only the universe, but using the entire history of the universe as some kind of space-time anteanae. Assuming this were all possible, of course.
So if your S:N is over 638dB for anything, either I screwed up or you did. How could you get that certain of anything? Actually, even if I screwed up, I'm probably not off by more than a few dB!
• the curvature of the universe? Does this help direct information?
• the unobservable universe? Are we, and everything we can detect down to the big bang skin at the edge of the observable universe merely a bubble in something unimaginably larger?
• what about the concept of S:N? If it isn't absolutely meaningful this analysis falls suspect immediately
• math/logic?
• certainty of your own existence? Maybe it's just information that must travel, must be communicated. This would solve math issues, too as reflexive statements and other axiomatic information might be self-provable, and hence would not be affected by S:N. Information about the material universe.
Other interesting things come from this too, you can measure how powerful a transmission medium is by how close it approximates this. my modem is what, -599dB from perfection? You can also measure how much information is stored within any given thing in terms of dB depending how large it is, too. I wonder if there's a way to put mass-energy into this somehow? Oh looks like there's a planck energy. But for how that'll fit in, I'll have to give some thought, and really, I should make sure what I have now is approximately correct first, anyway :)
Also, /., why don't you like my preformatted sections? Your treatment of <pre> makes baby dinojesus cry.
## JournalJournal: cuil theory
A random chat with m2tm, for your enjoyment
(11:33:10 PM) m2tm: haha, holy shit
(11:36:23 PM) tmg1: check out the cuil levels 6 & 7 comments :D
(11:38:46 PM) m2tm: haha, I like this too:
(11:38:51 PM) m2tm: Hypothesis: As Cuils increase, so does the likelihood your are actually a character in a David Lynch film.
(11:44:42 PM) tmg1: Hahahaha
(11:44:59 PM) tmg1: heh
(11:45:22 PM) m2tm: OMIGOSH
(11:45:27 PM) m2tm: It all makes sense!
(11:46:11 PM) tmg1: Hypothesis: Cuils are relative. You and I can be at different cuil levels only in relation to one another, as conscious beings
(11:48:02 PM) tmg1: Or maybe that's only true with Real cuils
(11:48:04 PM) tmg1: grn
(11:50:19 PM) m2tm: Nope, you are mistaken in questioning your point.
(11:50:23 PM) m2tm: Everything is relative.
(11:50:25 PM) m2tm: EVERYTHING.
Also, I'd imagine a Bayesian could be defined as someone who wants to minimize their apparent cuil-level between other bayesians and themselves. There could also be an 'absolute' cuil level, a level of reality compared to some standard point (say, Douglas Hofstadter or kevin bacon). Unfortunately this only works in the earlier cuils though since at later cuils even this might be used as an abstraction away from reality(this sort of thing making cuil theory somewhat of a difficult thing to define terms in).
Unfortunately, the cuil theory website has a TOS that I can't agree to(certainly while considering deviations from reality), so I won't be able to participate on their wiki.
In the meanwhile, when did /. break HTML(code/pre is don't work) AND 'plain old text'(can't see <pre>?
## JournalJournal: Open Systems Canada Limited RIP
Probably the only Linux and FLOSS based company local to Regina, Canada, Open Systems Canada ltd, has gone out of business. I never actually required their help (can't afford it as a poor student, and our needs are constantly being met without serious incident at work)--- but it's still harrowing that now there is no longer anywhere to turn in city, for those really big problems that you need paid tech support with. I guess the waters of the rising global economic crisis are finally reaching the point where it's noticeable here in lavishly wealthy tarsand country. We will miss ye.
# Slashdot Top Deals
The means-and-ends moralists, or non-doers, always end up on their ends without any means. -- Saul Alinsky
Working... | 4,623 | 18,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-13 | latest | en | 0.957339 |
https://www.folkstalk.com/tech/changing-the-number-of-ticks-on-a-matplotlib-plot-axis-with-code-examples/ | 1,713,399,144,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00042.warc.gz | 723,539,493 | 6,195 | # Changing The Number Of Ticks On A Matplotlib Plot Axis With Code Examples
Changing The Number Of Ticks On A Matplotlib Plot Axis With Code Examples
Hello everyone, In this post, we will examine how to solve the Changing The Number Of Ticks On A Matplotlib Plot Axis problem using the computer language.
```import numpy as np
import matplotlib.pyplot as plt
import matplotlib.ticker as ticker
x = [0,5,9,10,15]
y = [0,1,2,3,4]
fig, ax = plt.subplots()
ax.plot(x,y)
start, end = ax.get_xlim()
ax.set_xticks(np.arange(start, end, 2))
plt.show()
```
The exact problem Changing The Number Of Ticks On A Matplotlib Plot Axis can be fixed by employing an alternative technique, which is outlined in the next section along with some code samples for reference.
```# Get the current axis
ax = plt.gca()
# Only label every 20 th value
ticks_to_use = df.index[::20]
# Set format of labels(note year not excluded as requested)
labels = [i.strftime("%-H:%M") for i in ticks_to_use]
# Now set the ticks and labels
ax.set_xticks(ticks_to_use)
ax.set_xticklabels(labels)
```
There are a lot of real-world examples that show how to fix the Changing The Number Of Ticks On A Matplotlib Plot Axis issue.
## How do I increase the number of ticks in Axis MatPlotLib?
locator_params() to change the number of ticks on an axis. Call matplotlib. pyplot. locator_params(axis=an_axis, nbins=num_ticks) with either "x" or "y" for an_axis to set the number of ticks to num_ticks .
## How do I change the number of ticks in a plot in Python?
How it works:
• fig, ax = plt. subplots() gives the ax object which contains the axes.
• np. arange(min(x),max(x),1) gives an array of interval 1 from the min of x to the max of x. This is the new x ticks that we want.
• ax. set_xticks() changes the ticks on the ax object.
## How do you reduce the number of ticks?
Here are some simple landscaping techniques that can help reduce tick populations:
• Clear tall grasses and brush around homes and at the edge of lawns.
• Place a 3-ft wide barrier of wood chips or gravel between lawns and wooded areas and around patios and play equipment.
• Mow the lawn frequently and keep leaves raked.
## How do you change axis increments in Python?
MatPlotLib with Python To change the range of X and Y axes, we can use xlim() and ylim() methods.15-Jun-2021
## How do you specify a tick in MatPlotLib?
set_xticks() and plt. axes(). set_yticks() : For setting ticks on x-axis and y-axis respectively. having data in form of a list set as parameter.26-Dec-2020
## How do I create a custom tick in MatPlotLib?
By default, Matplotlib rarely makes use of minor ticks, but one place you can see them is within logarithmic plots:
• import matplotlib.pyplot as plt plt. style.
• In [2]: ax = plt.
• print(ax.
• print(ax.
• ax = plt.
• subplots(4, 4, sharex=True, sharey=True)
• # For every axis, set the x and y major locator for axi in ax.
• ax.
## How do I reduce the number of labels in MatPlotLib?
Method 1: Using xticks() and yticks()
• Syntax: matplotlib.pyplot.xticks(ticks=None, labels=None, **kwargs)
• Parameter:
• Returns:
## How do I create a minor tick in MatPlotLib?
MatPlotLib with Python Create a figure and a set of subplots. Plot x and y data points using plot() method. To locate minor ticks, use set_minor_locator() method. To show the minor ticks, use grid(which='minor').09-Jun-2021
## How do you change the range of axis in Python?
To set range of x-axis and y-axis, use xlim() and ylim() function respectively. To add a title to the plot, use the title() function. To add label at axes, use xlabel() and ylabel() functions.29-Jan-2022
## What is commonly used to control ticks?
The most common method of killing ticks is by the use of chemical acaricides. | 955 | 3,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.751781 |
https://www.physicsforums.com/threads/help-me-with-this-physchem-problem-about-atmospheric-pressure-please.712011/ | 1,477,050,821,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988717963.49/warc/CC-MAIN-20161020183837-00489-ip-10-171-6-4.ec2.internal.warc.gz | 949,371,915 | 16,203 | # Help me with this physchem problem about atmospheric pressure, please
1. Sep 22, 2013
### speny83
In the absence of turbulent mixing, the partial pressure of each constituent of air would fall off with height above sea level in Earth's atmosphere as Pi=P0ie−Migz/RT where Pi is the partial pressure at the height z, P0i is the partial pressure of component i at sea level, g is the acceleration of gravity, R is the gas constant, T is the absolute temperature, and Mi is the molecular mass of the gas. As a result of turbulent mixing, the composition of Earth's atmosphere is constant below an altitude of 100 km, but the total pressure decreases with altitude as P=P0e−Mavegz/RT where Mave is the mean molecular weight of air. At sea level, xN2= 0.78084 and xHe= 0.00000524 and T= 300. K.
Im not really sure where to start with this. I figured maybe just start with the exponent of e and im confused right off the bat. using the info give it would be e^(28.9 g mol-1)(9.81m s-2)(6500m)/(300k)(R).....so im not sure what R to use, also i sort of remember something about e not being able to be raised to something with units and i end up with all kinds of units left all over the place?
2. Sep 22, 2013
### Office_Shredder
Staff Emeritus
Well if e can't be raised to something with units, then R better have units of g /(K mol s2 m2).
Noting that g/(m2 s2) is almost the units for a joule (replace gram with kilogram, which you should be able to figure out how to do in the original statement), and I'd say what you have is the good ol' fashion gas constant in J/(mol K)
http://en.wikipedia.org/wiki/Gas_constant
3. Sep 22, 2013
### speny83
I had no idea that was equivalent. the book ive been using does not mention that, guess that was supposed to be something i remembered from HS about 15 years ago! But I am correct about the e power unit thing, at least something is still in there haha!
thanks
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Draft saved Draft deleted | 530 | 2,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2016-44 | longest | en | 0.931863 |
https://thebusinessprofessor.com/lesson/close-location-value-definition/ | 1,603,356,462,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879362.3/warc/CC-MAIN-20201022082653-20201022112653-00326.warc.gz | 560,386,068 | 10,058 | # Close Location Value – Definition
Cite this article as:"Close Location Value – Definition," in The Business Professor, updated February 23, 2020, last accessed October 22, 2020, https://thebusinessprofessor.com/lesson/close-location-value-definition/.
### Close Location Value (CLV) Definition
CLV is an indicator that states an asset closing price in relation to the High-Low range. This range is between +1 and -1. +1 indicates that the closing price is near its High and -1, near its low. This is used in technical analysis to analyze closing price within a stipulated time.
### A Little More on What is Close Location Value (CLV)
The Close location value is an index for measuring the rate at which money flows into or out of a given security. The closing location value is not independently used to determine the price value but used with other equations to derive the accurate range. When the CLV range is +1 or multiplied by 100, this indicates that the closing price is near its high and it would be considered as a bullish sign. And when the range is -1 or -100, this indicates that the closing price is near its low and it could also be considered as a bearish sign. But, it is neutral when the CLV is close to zero.
### Using Close Location Value
Closing value on its own is not ultimately used as a metric. This is because it is sensitive to fluctuations or random changes in price. It is preferred to use clv with other equations. A good example is when calculating the Accumulation / Distribution Line:
Acc/Dist = CLV * Period’s volume
When CLV is not used with other equations, it is a good indicator for confirming or rejecting possible divergences. Any trader using this strategy will also consider a short or long time frame for their CLV, this makes the CLV not to react to any price fluctuations.
However, rather than using the CLV, traders prefer to use stochastics. This is considered as a reliable high-low relationship metric. Stochastics deploys several formulas to analyze and determine the price location in the high-low range. This does not react to price fluctuations.
### Reference for “Close Location Value (CLV)”
https://www.investopedia.com/terms/c/close_location_value.asp
https://www.marketvolume.com/technicalanalysis/closelocationvalue.asp
https://www.stockopedia.com/charts-glossary/close-location-value-5310/
www.investorwords.com/17980/close_location_value_CLV.html
https://admiralmarkets.com/education/articles/…/accumulation-distribution-indicator | 558 | 2,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-45 | latest | en | 0.952505 |
https://www.jiskha.com/display.cgi?id=1318133366 | 1,501,261,680,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550975184.95/warc/CC-MAIN-20170728163715-20170728183715-00354.warc.gz | 811,042,794 | 4,095 | # College Algebra
posted by .
let g(x) = 17x3 - 3x2 + 14x. What is the degree of this polynomial function?
• College Algebra -
The degree of a polynomial is the term with the greatest exponent.
In this case the degree of polynomial function = 3
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More Similar Questions
Post a New Question | 583 | 2,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2017-30 | longest | en | 0.859664 |
https://chemistscorner.com/cosmeticsciencetalk/discussion/how-to-get-from-grams-to/ | 1,726,708,968,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651944.55/warc/CC-MAIN-20240918233405-20240919023405-00476.warc.gz | 143,844,059 | 37,694 | Home Cosmetic Science Talk Formulating Hair how to get from grams to %
• # how to get from grams to %
Posted by on June 2, 2021 at 9:47 pm
hey everyone I have a shampoo formulation that I got in grams but I wanted to know how do I transfer it to % to make bigger batches??
@Perry
2 Members · 2 Replies
• 2 Replies
• ### OldPerry
Member
June 2, 2021 at 10:29 pm
1. Add up all the gram amounts to get a total # of grams. Call it T
2. To get percentage, take the gram amount of an ingredient and divide it by the Total grams. (then multiply by 100)
So, if you have a formula that is 200g total.
And one of the component materials is 10g.
The percentage of that ingredient in the formula is = 10g/200g = 0.05 x 100 = 5%
• ### tianaloveorganics
Member
June 4, 2021 at 5:41 am
Perry said:
1. Add up all the gram amounts to get a total # of grams. Call it T
2. To get percentage, take the gram amount of an ingredient and divide it by the Total grams. (then multiply by 100)
So, if you have a formula that is 200g total.
And one of the component materials is 10g.
The percentage of that ingredient in the formula is = 10g/200g = 0.05 x 100 = 5%
thank you so much you are the best I really really appreciate it | 358 | 1,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-38 | latest | en | 0.927398 |
https://tolstoy.newcastle.edu.au/R/e4/help/08/03/5976.html | 1,586,160,589,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371620338.63/warc/CC-MAIN-20200406070848-20200406101348-00155.warc.gz | 642,193,857 | 3,982 | # Re: [R] ask for help on nonlinear fitting
From: Dieter Menne <dieter.menne_at_menne-biomed.de>
Date: Sat, 08 Mar 2008 09:13:37 +0000 (UTC)
Ruqiang Liang <ruqiang.liang <at> gmail.com> writes:
>
> I have a table like the following. I want to fit Cm to Vm like this:
> Cm ~Cl+Q1*b1*38.67*exp(-b1*(Vm-Vp1)*0.03867)/(1+exp(-b1*(Vm-Vp1)*0.03867))^2+
Q2*b2*38.67*exp(-b2*(Vm-Vp2)*0.03867)/(1+exp(-b2*(Vm-Vp2)*0.03867))^2
>
> I use nls, with start=list(Q1=2e-3, b1=1, Vp1=-25, Q2=3e-3, b2=1,
> Vp2=200). But I always get 'singular gradient' error like this. But
> in SigmaPlot I can get the result. How can I get with R.
I remember a similar case in gastric emptying fitting where my colleague tried to show me that SigmaPlot is superior to R/nls. When he looked at the standard deviations of the coefficients, giving estimated gastric empty times of 40 minutes plus minus 800 minutes, I was reminded of Douglas Bates' philosophy: better nothing than wrong. (Even if I would love to have estimated p-values back in lmer fits.)
This said, there are two solutions: first, simplify your expression to see what you are doing. The first part of the expression can be written as
R1*exp(-c1*(Vm-Vp1))/(1+exp(-c1*(Vm-Vp1))^2
where you are free to compute Q1 and b1 later. You will also note that your fit cannot have a unique solution because of the symmetry in the second term.
As a next step, try to fit
R1*exp(-exp(d1)*(Vm-Vp1))/(1+exp(-exp(d1)*(Vm-Vp1))^2
This gives a working solution in most cases, effectively forcing a positive c1=exp(d1).
However, even that may fail in degenerate cases. If it is not a single fit, but from a planned pharmacology related experiment, you should try to fit the whole experiment with all repeats instead of single curves. This often gives excellent results even when some fits are disastrous. Check package nlme, the book by Pinheiro/Bates, and some related examples on http://www.menne-biomed.de/gastempt
Dieter
R-help_at_r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Sat 08 Mar 2008 - 09:24:27 GMT
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Archive generated by hypermail 2.2.0, at Sat 08 Mar 2008 - 09:30:20 GMT.
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http://www.chegg.com/homework-help/questions-and-answers/y-vbw-61-m-s-25-degree-x-vwg-14-m-s-points-negative-direction-find-time-takes-fo-boat-reac-q86367 | 1,448,705,681,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398451872.11/warc/CC-MAIN-20151124205411-00309-ip-10-71-132-137.ec2.internal.warc.gz | 355,916,340 | 12,816 | y|
|
| Vbw = 6.1 m/s
|
|
| /
| /
| / 25 degree
| /_______________________ x
Vwg = 1.4 m/s points in negative direction
Find the time it takes fo the boat to reach the opposite shoreif the river is 25m wide? | 80 | 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} | 3.109375 | 3 | CC-MAIN-2015-48 | latest | en | 0.701988 |
http://orange.biolab.si/trac/browser/orange/Orange/distance/__init__.py?rev=c4cbae8dcf8b96edc4fb1ec518baf9fe6a60562f&format=txt | 1,386,678,196,000,000,000 | text/plain | crawl-data/CC-MAIN-2013-48/segments/1386164018354/warc/CC-MAIN-20131204133338-00093-ip-10-33-133-15.ec2.internal.warc.gz | 131,944,765 | 2,611 | import Orange from Orange.core import \ DistanceMap, \ DistanceMapConstructor, \ ExamplesDistance as Distance, \ ExamplesDistance_Normalized as DistanceNormalized, \ ExamplesDistanceConstructor as DistanceConstructor, \ ExamplesDistance_Hamming as HammingDistance, \ ExamplesDistance_DTW as DTWDistance, \ ExamplesDistance_Euclidean as EuclideanDistance, \ ExamplesDistance_Manhattan as ManhattanDistance, \ ExamplesDistance_Maximal as MaximalDistance, \ ExamplesDistance_Relief as ReliefDistance, \ ExamplesDistanceConstructor_DTW as DTW, \ ExamplesDistanceConstructor_Euclidean as Euclidean, \ ExamplesDistanceConstructor_Hamming as Hamming, \ ExamplesDistanceConstructor_Manhattan as Manhattan, \ ExamplesDistanceConstructor_Maximal as Maximal, \ ExamplesDistanceConstructor_Relief as Relief from Orange import statc from Orange.utils import progress_bar_milestones import numpy from numpy import linalg class PearsonR(DistanceConstructor): def __new__(cls, data=None, **argkw): self = DistanceConstructor.__new__(cls, **argkw) self.__dict__.update(argkw) if data: return self.__call__(data) else: return self def __call__(self, table): indxs = [i for i, a in enumerate(table.domain.attributes) \ if a.varType==Orange.feature.Type.Continuous] return PearsonRDistance(domain=table.domain, indxs=indxs) class PearsonRDistance(Distance): """ `Pearson correlation coefficient `_. """ def __init__(self, **argkw): self.__dict__.update(argkw) def __call__(self, e1, e2): """ :param e1: data instances. :param e2: data instances. Returns Pearson's disimilarity between e1 and e2, i.e. (1-r)/2 where r is Pearson's rank coefficient. """ X1 = [] X2 = [] for i in self.indxs: if not(e1[i].isSpecial() or e2[i].isSpecial()): X1.append(float(e1[i])) X2.append(float(e2[i])) if not X1: return 1.0 try: return (1.0 - statc.pearsonr(X1, X2)[0]) / 2. except: return 1.0 class SpearmanR(DistanceConstructor): def __new__(cls, data=None, **argkw): self = DistanceConstructor.__new__(cls, **argkw) self.__dict__.update(argkw) if data: return self.__call__(data) else: return self def __call__(self, table): indxs = [i for i, a in enumerate(table.domain.attributes) \ if a.varType==Orange.feature.Type.Continuous] return SpearmanRDistance(domain=table.domain, indxs=indxs) class SpearmanRDistance(Distance): """`Spearman's rank correlation coefficient `_.""" def __init__(self, **argkw): self.__dict__.update(argkw) def __call__(self, e1, e2): """ :param e1: data instances. :param e2: data instances. Returns Sprearman's disimilarity between e1 and e2, i.e. (1-r)/2 where r is Sprearman's rank coefficient. """ X1 = []; X2 = [] for i in self.indxs: if not(e1[i].isSpecial() or e2[i].isSpecial()): X1.append(float(e1[i])) X2.append(float(e2[i])) if not X1: return 1.0 try: return (1.0 - statc.spearmanr(X1, X2)[0]) / 2. except: return 1.0 class Mahalanobis(DistanceConstructor): def __new__(cls, data=None, **argkw): self = DistanceConstructor.__new__(cls, **argkw) self.__dict__.update(argkw) if data: return self.__call__(data) else: return self # Check attributtes a, b, c def __call__(self, table, a=None, b=None, c=None, **argkw): # Process data dc = Orange.core.DomainContinuizer() dc.classTreatment = Orange.core.DomainContinuizer.Ignore dc.continuousTreatment = Orange.core.DomainContinuizer.NormalizeBySpan dc.multinomialTreatment = Orange.core.DomainContinuizer.NValues newdomain = dc(table) newtable = table.translate(newdomain) data, cls, _ = newtable.to_numpy() covariance_matrix = numpy.cov(data, rowvar=0, bias=1) inverse_covariance_matrix = linalg.pinv(covariance_matrix, rcond=1e-10) return MahalanobisDistance(domain=newdomain, icm=inverse_covariance_matrix) class MahalanobisDistance(Distance): """`Mahalanobis distance `_""" def __init__(self, domain, icm, **argkw): self.domain = domain self.icm = icm self.__dict__.update(argkw) def __call__(self, e1, e2): """ :param e1: data instances. :param e2: data instances. Returns Mahalanobis distance between e1 and e2. """ e1 = Orange.data.Instance(self.domain, e1) e2 = Orange.data.Instance(self.domain, e2) diff = [] for i in range(len(self.domain.attributes)): diff.append(e1[i].value - e2[i].value) if not(e1[i].isSpecial() or e2[i].isSpecial()) else 0.0 diff = numpy.asmatrix(diff) res = diff * self.icm * diff.transpose() return res[0,0]**0.5 class PearsonRAbsolute(PearsonR): """ Construct an instance of PearsonRAbsolute example distance estimator. """ def __call__(self, data): indxs = [i for i, a in enumerate(data.domain.attributes) \ if a.varType==Orange.feature.Type.Continuous] return PearsonRAbsoluteDistance(domain=data.domain, indxs=indxs) class PearsonRAbsoluteDistance(PearsonRDistance): """ An example distance estimator using absolute value of Pearson correlation coefficient. """ def __call__(self, e1, e2): """ Return absolute Pearson's dissimilarity between e1 and e2, i.e. .. math:: (1 - abs(r))/2 where r is Pearson's correlation coefficient. """ X1 = []; X2 = [] for i in self.indxs: if not(e1[i].isSpecial() or e2[i].isSpecial()): X1.append(float(e1[i])) X2.append(float(e2[i])) if not X1: return 1.0 try: return (1.0 - abs(statc.pearsonr(X1, X2)[0])) except: return 1.0 class SpearmanRAbsolute(SpearmanR): """ Construct an instance of SpearmanRAbsolute example distance estimator. """ def __call__(self, data): indxs = [i for i, a in enumerate(data.domain.attributes) \ if a.varType==Orange.feature.Type.Continuous] return SpearmanRAbsoluteDistance(domain=data.domain, indxs=indxs) class SpearmanRAbsoluteDistance(SpearmanRDistance): def __call__(self, e1, e2): """ Return absolute Spearman's dissimilarity between e1 and e2, i.e. .. math:: (1 - abs(r))/2 where r is Spearman's correlation coefficient. """ X1 = []; X2 = [] for i in self.indxs: if not(e1[i].isSpecial() or e2[i].isSpecial()): X1.append(float(e1[i])) X2.append(float(e2[i])) if not X1: return 1.0 try: return (1.0 - abs(statc.spearmanr(X1, X2)[0])) except: return 1.0 def _pairs(seq, same = False): """ Return all pairs from elements of `seq`. """ seq = list(seq) same = 0 if same else 1 for i in range(len(seq)): for j in range(i + same, len(seq)): yield seq[i], seq[j] def distance_matrix(data, distance_constructor=Euclidean, progress_callback=None): """ A helper function that computes an :obj:`Orange.misc.SymMatrix` of all pairwise distances between instances in `data`. :param data: A data table :type data: :obj:`Orange.data.Table` :param distance_constructor: An DistanceConstructor instance (defaults to :obj:`Euclidean`). :type distance_constructor: :obj:`Orange.distances.DistanceConstructor` :param progress_callback: A function (taking one argument) to use for reporting the on the progress. :type progress_callback: function :rtype: :class:`Orange.misc.SymMatrix` """ matrix = Orange.misc.SymMatrix(len(data)) dist = distance_constructor(data) iter_count = matrix.dim * (matrix.dim - 1) / 2 milestones = progress_bar_milestones(iter_count, 100) for count, ((i, ex1), (j, ex2)) in enumerate(_pairs(enumerate(data))): matrix[i, j] = dist(ex1, ex2) if progress_callback and count in milestones: progress_callback(100.0 * count / iter_count) return matrix | 1,942 | 7,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | longest | en | 0.542833 |
https://studylibsv.com/doc/984845/id1020-enkel-sortering | 1,623,516,728,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586239.2/warc/CC-MAIN-20210612162957-20210612192957-00165.warc.gz | 494,114,819 | 18,675 | # ID1020:Enkel Sortering
```ID1020:Enkel Sortering
Dr. Jim Dowling
[email protected]
kap 2.1
Slides adapted from Algorithms 4th Edition, Sedgewick.
Ett sorteringsproblem
• T.ex. Student poster på ett universitet.
post
nyckel
Chen
3
A
991-878-4944
308 Blair
Rohde
2
A
232-343-5555
343 Forbes
Gazsi
4
B
766-093-9873
101 Brown
Furia
1
A
766-093-9873
101 Brown
Kanaga
3
B
898-122-9643
22 Brown
Andrews
3
A
664-480-0023
097 Little
Battle
4
C
874-088-1212
121 Whitman
• Sortering. Ordna om en array (följd) av N poster i stigande ordning.
Andrews
3
A
664-480-0023
097 Little
Battle
4
C
874-088-1212
121 Whitman
Chen
3
A
991-878-4944
308 Blair
Furia
1
A
766-093-9873
101 Brown
Gazsi
4
B
766-093-9873
101 Brown
Kanaga
3
B
898-122-9643
22 Brown
Rohde
2
A
232-343-5555
343 Forbes
2
Sortering applikationer
Library of Congress numbers
contacts
FedEx packages
playing cards
3
Exempel: sortering klient (1)
• Mål. Sortera alla möjliga datatyper.
• T.ex.1. Sortera slumpmässiga reella tal i stigande ordning.
ett exempel kommer senare…
public class Experiment
{
public static void main(String[] args)
{
int N = Integer.parseInt(args[0]);
Double[] a = new Double[N];
for (int i = 0; i < N; i++)
a[i] = StdRandom.uniform();
Insertion.sort(a);
for (int i = 0; i < N; i++)
StdOut.println(a[i]);
}
}
% java Experiment 10
0.08614716385210452
0.09054270895414829
0.10708746304898642
0.21166190071646818
0.363292849257276
0.460954145685913
0.5340026311350087
0.7216129793703496
0.9003500354411443
0.9293994908845686
4
Exempel: sortering klient (2)
• Mål. Sortera alla möjliga datatyper.
• T.ex. 2. Sortera alla strängar i bokstavsordning.
public class StringSorter {
public static void main(String[] args)
{
Insertion.sort(a);
for (int i = 0; i < a.length; i++)
StdOut.println(a[i]);
}
}
% more words3.txt
% java StringSorter < words3.txt
5
Exempel: sortering klient (3)
• Mål. Sortera alla möjliga datatyper.
• T.ex. 3. Sortera alla filer i en given mapp efter filnamn.
% java FileSorter .
import java.io.File;
Insertion.class
public class FileSorter
{
public static void main(String[] args)
{
File directory = new File(args[0]);
File[] files = directory.listFiles();
Insertion.sort(files);
for (int i = 0; i < files.length; i++)
StdOut.println(files[i].getName());
}
}
Insertion.java
InsertionX.class
InsertionX.java
Selection.class
Selection.java
Shell.class
Shell.java
ShellX.class
ShellX.java
6
Totalordning
• Mål. Sortera alla möjliga datatyper (där sortering är väldefinierade).
• En totalordning är en binärrelation ≤ som satisfierar:
- icke-symmetri: IF (v ≤ w och w ≤ v) THEN v = w.
- transitiv: IF (v ≤ w och w ≤ x) THEN v ≤ x.
- totalitet: antingen v ≤ w eller w ≤ v eller både.
• T.ex.
- Vanlig ordningen för heltal och reella tal.
- Kronologiska ordningen för datum och tider.
- Bokstavsordning för strängar.
• Icke-transitive. Sten-sax-påse.
7
Callbacks (återanropsfunktion)
• Mål. Sortera alla möjliga datatyper (där sortering är väldefinierade).
• Hur kan sort() veta hur den ska jämföra data av olika typer, som Double,
String, and java.io.File, utan att ha någon information om datatypen av
postens nyckel?
• Callback = en referens till exekverbarkod.
- Klienen skickar en array av objekt till sort() funktionen.
- sort() funktionen anropar objekts compareTo() metod vid behov.
• Att implementera callbacks.
- Java: interfaces.
- C: function pointers.
- C++: class-type functors.
- C#: delegates.
- Python, Perl, ML, Javascript: first-class functions.
8
Callbacks i Java
klient
datatyp implementation
public class StringSorter
{
public static void main(String[] args)
{
Insertion.sort(a);
for (int i = 0; i < a.length; i++)
StdOut.println(a[i]);
}
}
Comparable interface (inbyggd i Java)
public interface Comparable<Item>
{
public int compareTo(Item that);
}
inga beroenden
på String datatypen
public class String
implements Comparable<String>
{
...
public int compareTo(String b)
{
...
return -1;
...
return +1;
...
return 0;
}
}
sort implementation
public static void sort(Comparable[] a)
{
int N = a.length;
for (int i = 0; i < N; i++)
for (int j = i; j > 0; j--)
if (a[j].compareTo(a[j-1]) < 0)
exch(a, j, j-1);
else break;
}
9
Comparable API
• Implementera compareTo() så att v.compareTo(w):
- definierar en totalordning;
- returnerar en negativ heltal, noll, eller en positiv heltal
IF v är respektiv mindre än, lika med, eller större än w.
- kastar en exception om typer är inkompatibla typer (eller om en av dem är
null).
v
w
mindre än (returnera -1)
v
w
lika med (returnera 0)
w
v
större än (returnera +1)
• Inbyggda jämförbara typer. Integer, Double, String, Date, File, ...
• Använder-definierade jämförbara typer. Implementera Comparable
interface:n och instanser av klassen (objekt) kan jämföras med compareTo
metoden.
10
Implementera Comparable interface:n
• Date datatyp. Förenklad version av java.util.Date.
public class Date implements Comparable<Date>{
private final int month, day, year;
public Date(int m, int d, int y)
month = m;
day
= d;
year = y;
}
public int compareTo(Date that)
if (this.year < that.year )
if (this.year > that.year )
if (this.month < that.month)
if (this.month > that.month)
if (this.day
< that.day )
if (this.day
> that.day )
return 0;
}
{
jämför Date objekt bara
med andra Date objekt
{
return
return
return
return
return
return
-1;
+1;
-1;
+1;
-1;
+1;
}
11
Enkel sortering (elementary sorting)
Selection sort (urvalssortering) algoritm demo
• Under iteration i, hitta index min av den minste resterande post.
• a[i] och a[min] byter plats.
initial
14
Selection sort algoritm
• Algoritm. ↑ skannar från vänster till höger.
• Invarianter.
- Poster till vänster av ↑ (inkluderande ↑ ) är sorterad i stigande ordning.
- Ingen post till höger av ↑ är mindre än en post till vänster av ↑.
i slutordning
↑
15
Två användbara sorterings hjälpfunktioner
• Hjälpfunktioner. Jämförelser och byte operationer på poster (objekt).
• Mindre än (less-than). Är posten (objekt) v mindre än w ?
private static boolean less(Comparable v, Comparable w) {
return v.compareTo(w) < 0;
}
• Byte (exchange). Byt posten i array a[] vid index j med posten vid index k.
private static void exch(Comparable[] a, int j, int k)
{
Comparable swap = a[j];
a[j] = a[k];
a[k] = swap;
}
16
Selection sort innre slingan
• För att bevara algoritmen invarianter:
- Flytta pekare till höger.
i++;
- Identifiera indexet av minsta posten till höger
i array:n.
int min = i;
for (int j = i+1; j < N; j++) {
if (less(a[j], a[min])) {
min = j;
}
}
i slutordning
i slutordning
↑
↑
↑
- Byt indexet i med min, det minsta elementet.
exch(a, i, min);
i slutordning
↑
↑
17
Selection sort: Java implementation
public class Selection {
public static void sort(Comparable[] a)
{
int N = a.length;
for (int i = 0; i < N; i++)
{
int min = i;
for (int j = i+1; j < N; j++)
if (less(a[j], a[min]))
min = j;
exch(a, i, min);
}
}
private static boolean less(Comparable v, Comparable w)
{ /* as before */ }
private static void exch(Comparable[] a, int i, int j)
{ /* as before */ }
}
18
Selection sort: animeringar
20 poster i slumpordning
algoritm position
http://www.sorting-algorithms.com/selection-sort
19
Selection sort: animeringar
algoritm position
http://www.sorting-algorithms.com/selection-sort
20
Matematisk analys
• Vi använder följande kostnadsmodellen med de enkla
sorteringsalgoritmerna:
1.
2.
antalet jämförelser (comparisons)
antalet byten (exchanges)
• Om algoritmen inte swappar element, då räknar vi antalet array
accesser.
Selection sort: matematisk analys
• Sats. Selection sort jämför (N -1) + (N - 2) + ... + 1 + 0 ~ N 2 / 2 gånger och byter N
gånger.
• Körtiden är inte känslig mot indata. Kvadratisk tid, även om input är sorterad.
• Minimera kopieringen av data. Linjär antal byte.
2
Insertion Sort (Insättningssortering)
Insertion sort demo
• Under iteration i, a[i] byter plats med varje post som är större på vänstra
sidan.
24
Insertion sort
• Algoritm. ↑ skannar från vänster till höger.
• Invarianter.
- Poster till vänster om ↑ (inkluderande ↑) är i stigande ordning.
- Poster till höger om ↑ har inte bearbetats än.
↑
inte bearbetat än
25
Insertion sort inner loop
• Bevara algoritm invarianterna:
- Flytta pekare till höger.
i++;
↑
• När pekare flyttar från höger till vänster,
byt a[i] med varje post på vänstra
sidan som är större.
for (int j = i; j > 0; j--)
if (less(a[j], a[j-1]))
exch(a, j, j-1);
else break;
↑ ↑ ↑ ↑
26
Insertion sort: Java implementation
public class Insertion {
public static void sort(Comparable[] a)
{
int N = a.length;
for (int i = 0; i < N; i++) {
for (int j = i; j > 0; j--) {
if (less(a[j], a[j-1])) {
exch(a, j, j-1);
} else {
break;
}
}
}
}
private static boolean less(Comparable v, Comparable w)
{ /* as before */ }
private static void exch(Comparable[] a, int i, int j)
{ /* as before */ }
}
27
Insertion sort: animering
40 poster i slumpordning
algoritm position
http://www.sorting-algorithms.com/insertion-sort
28
Insertion sort: animering
algoritm position
http://www.sorting-algorithms.com/insertion-sort
29
Insertion sort: animering
algoritm position
http://www.sorting-algorithms.com/insertion-sort
30
Insertion sort: matematisk analys
• Sats. För att sortera en array i slumpordning med unika nycklar,
insertion sort jämför ~ ¼ N 2 gånger och byter ~ ¼ N 2 gånger i genomsnitt.
• Bevis. Förväntat antal platser en post flyttar tillbaka är N/2 (halvvägs).
31
Insertion sort: trace
32
Insertion sort: analys
• Best case. Om array:n är i stigande ordning, insertion sort jämför
N – 1 gånger och byter 0 gånger.
AE ELMOPRSTX
• Worst case. Om array:n är i nedstigande ordning (med unika element),
insertion sort jämför ~ ½ N 2 gånger och byter ~ ½ N 2 gånger.
XTS RPOMLF EA
33
• Def. En inversion är ett par nycklar som är i fel ordningen.
AE ELMOTRXPS
T-R T-P T-S R-P X-P X-S
(6 inversioner)
• Def. En array är delvis sorterade om antalet inversionerna är ≤ c N.
- T.ex. 1. En sorterade array har 0 inversioner.
- T.ex. 2. En subarray av längd 10 lagt till en sorteradea subarray av längd N.
• Sats. För delvis sorterade arrayer, insertion sort kör i linjärtid.
• Bevis. Antalet byten är lika med antalet inversionerna.
antalet jämförelser = antalet byten + (N – 1)
34
Insertion sort: förbättringar
• Binär insertion sort. Man kan använda binärsökning för att hitta den såkallade ”insertion point”.
- Antalet jämförelser ~ N lg N .
- Men det är fortfarande kvadratisk i antalet array accesser.
ACH H I MN N PQXYKBINARY
binärsök efter första nyckel > K
35
Shellsort
Shellsort översikt
• Kärnidén. Vi vill flytta poster mer än en position åt gången genom att
exekvera h-sorting på array:n.
• Shellsort. [Shell 1959] h-sort array för nedstigande sekvens av värden av h.
37
h-sorting demo
• Vid iteration i, byt a[i] mot denna post som ligger h positioner till vänster
om denna post är större än a[i].
38
h-sorting
• Hur kör man h-sort på en array?
Kör insertion sort, men med klivlängd (stride length) h.
3-sorting an array
M O L E E X A S P R T
E O L M E X A S P R T
E E L M O X A S P R T
E E L M O X A S P R T
A E L E O X M S P R T
A E L E O X M S P R T
A E L E O P M S X R T
A E L E O P M S X R T
A E L E O P M S X R T
A E L E O P M S X R T
• Varför insertion sort?
- Stora inkrement ⇒ small subarray.
- Små inkrement ⇒ nästan sorterad.
39
Shellsort exempel: inkrement av 7, 3, 1
input
S O R T E X A M P L E
1-sort
A E L E O P M S X R T
A E L E O P M S X R T
7-sort
S O R T E X A M P L E
M O R T E X A S P L E
M O R T E X A S P L E
M O L T E X A S P R E
M O L E E X A S P R T
3-sort
M O L E E X A S P R T
E O L M E X A S P R T
E E L M O X A S P R T
E E L M O X A S P R T
A E L E O P M S X R T
A E E L O P M S X R T
A E E L O P M S X R T
A E E L O P M S X R T
A E E L M O P S X R T
A E E L M O P S X R T
A E E L M O P S X R T
A E E L M O P R S X T
A E E L M O P R S T X
resultat
A E E L M O P R S T X
A E L E O X M S P R T
A E L E O X M S P R T
A E L E O P M S X R T
A E L E O P M S X R T
A E L E O P M S X R T
40
Shellsort: Java implementation
public class Shell {
public static void sort(Comparable[] a)
{
int N = a.length;
int h = 1;
while (h < N/3) h = 3*h + 1; // 1, 4, 13, 40, 121, 364, ...
while (h >= 1) { // h-sort the array.
for (int i = h; i < N; i++) {
for (int j = i; j >= h && less(a[j], a[j-h]); j -= h)
exch(a, j, j-h);
}
h = h/3;
}
}
private static boolean less(Comparable v, Comparable w)
{ /* as before */ }
private static void exch(Comparable[] a, int i, int j)
{ /* as before */ }
}
3x+1 inkrement
sekvens
insertion sort
nästa inkrement
41
Shellsort: visualisering
42
Shellsort: animering
50 poster i slumpordning
algoritm position
h-sorted
nuvarande subsekvens
http://www.sorting-algorithms.com/shell-sort
andra element
43
Shellsort: animering
algoritm position
h-sorted
nuvarande subsekvens
http://www.sorting-algorithms.com/shell-sort
andra element
44
Shellsort: vilken inkrementsekvens ska man använda?
• Tvåpotens (powers of two). 1, 2, 4, 8, 16, 32, ...
Nej.
• Tvåpotens minus ett. 1, 3, 7, 15, 31, 63, …
Kanske.
• 3x + 1. 1, 4, 13, 40, 121, 364, …
OK. Lätt att beräkna.
45
Shellsort: intuition
• Sats. En h-sorted array förbli h-sorted efter ha kört g-sort på array:n.
7-sort
S O R T E X A M P L E
M O R T E X A S P L E
M O R T E X A S P L E
M O L T E X A S P R E
M O L E E X A S P R T
3-sort
M O L E E X A
E O L M E X A
E E L M O X A
E E L M O X A
A E L E O X M
A E L E O X M
A E L E O P M
A E L E O P M
A E L E O P M
A E L E O P M
S P
S P
S P
S P
S P
S P
S X
S X
S X
S X
R
R
R
R
R
R
R
R
R
R
T
T
T
T
T
T
T
T
T
T
still 7-sorted
Utmaning. Bevisa detta faktum; det är svårare än vad man tror!
46
Shellsort: analys
• Sats. Tidskomplexitet av worst-case är antalet jämförelser i shellsort. När
inkrementet är 3x+1, då är tidskomplexitet lika med N 3/2.
• Egenskap. Förväntade antalet jämförelser för att köra shellsort på en array i
slumpordning med inkrementet lika med 3x+1 är….
N
compares
2.5 N ln N
0.25 N ln 2 N
N 1.3
5,000
93K
106K
91K
64K
10,000
209K
230K
213K
158K
20,000
467K
495K
490K
390K
40,000
1022K
1059K
1122K
960K
80,000
2266K
2258K
2549K
2366K
• Anmärkning. En noggrann modell har inte hittats ännu (!)
47
Varför är shellsort intressant?
• En enkel idé kan leda till signifikant prestandsförbättringar.
R, bzip2, /linux/kernel/groups.c
• Användbart i praktiken.
- Snabb om array:n inte är enorm.
- Mycket liten minneskomplexitet (använt i vissa inbyggda system).
uClibc
• Enkel algoritm, icke-trivial prestanda, intressanta frågor.
- Asymptotisk tidskomplexitet?
- Bästa möjliga sekvens av inkrementet?
öppet problem:
- Average-case prestanda?
hitta en bättre inkrement sekvens
• Läxa. Många bra algoritm har inte upptäckts ännu.
48
Enkel sortering sammanfattning
algoritm
best
average worst
selection sort
N2
N2
N2
insertion sort
N
N2
N2
Shellsort (3x+1)
N log N ?
Mål
N
N 3/2
N log N N log N
Tidskomplexitet av körtiden för att sortera en array av N poster
49
Blandning (Shuffling)
Hur blandar man en array?
• Mål. Ordna om en array så att resultatet är en slumpmässig valda permutation.
alla permutationer har samma sannolikhet
51
Hur blandar man en array?
• Mål. Ordna om en array så att resultatet är en slumpmässig valda permutation.
alla permutationer har samma sannolikhet
52
Shuffle sort
• Generera ett slumpmässigt reellt tal för varje element i array:n.
• Sortera array:n.
användbart för att blanda
0.8003
0.9706
0.9157
0.9649
0.1576
0.4854
0.1419
0.4218
0.9572
53
Shuffle sort
• Generera ett slumpmässigt reellt tal för varje element i array:n.
• Sortera array:n.
användbart för att blanda
0.1419
0.1576
0.4218
0.4854
0.8003
0.9157
0.9572
0.9649
0.9706
54
Shuffle sort
- Generera ett slumpmässigt reellt tal för varje element i array:n.
- Sortera array:n.
användbart för att blanda
0.1419
0.1576
0.4218
0.4854
0.8003
0.9157
0.9572
0.9649
0.9706
• Sats. Shuffle sort resulterar i en slumpmässig valda permutation (uniformly
random permutation).
Vi antar reella tal valda slumpmässigt (och inga oavgjorda)
55
The Microsoft Shuffle
• Microsoft antitrust probe by EU. Microsoft var överens om att visa en
slumpmässig genererade skärm så att användare kunde välja sin browser i
Windows 7.
http://www.browserchoice.eu
förekom sist
50% av tiden
56
The Microsoft Shuffle
• Lösningen? Implementera shuffle sort genom att skriva en comparator som
alltid returnera ett slump svar.
public int compareTo(Browser that)
{ Microsofts implementation i Javascript
double r = Math.random();
function
(a,b)
if (rRandomSort
< 0.5) return
-1;
{ if (r > 0.5) return +1;
return
return
(0.5 0;
- Math.random());
}}
browser comparator
(ska implementera en totalordning)
57
Knuth shuffle demo
- Under iteration i, slumpa fram ett heltalr mellan 0 och i.
- Swappaa[i] och a[r].
58
Knuth shuffle
- Under iteration i, slumpa fram ett heltal r mellan 0 och i.
- Swappa a[i] och a[r].
• Sats. [Fisher-Yates 1938] Knuth shuffle sort resulterar i en slumpmässig
valda permutation (uniformly random permutation) av input array:n i linjär
tid.
Vi antar heltal valda slumpmässigt
59
Knuth shuffle
- Under iteration i, slumpa fram ett heltal r mellan 0 och i.
- Swappa a[i] och a[r].
vanlig bugg : mellan 0 and N – 1
korrekt variant: mellan i and N – 1
public class StdRandom
{
...
public static void shuffle(Object[] a)
{
int N = a.length;
for (int i = 0; i < N; i++)
{
int r = StdRandom.uniform(i + 1);
exch(a, i, r);
}
}
}
mellan 0 och i
60
Trasig Knuth shuffle
• Vad händer om heltalet är vald mellan 0 och N-1 ?
• Det är inte slumpmässigt längre!
istället för 0 och i
permutation
Knuth shuffle
broken shuffle
ABC
1/6
4/27
ACB
1/6
5/27
BAC
1/6
5/27
BCA
1/6
5/27
CAB
1/6
4/27
CBA
1/6
4/27
sannolikhet av alla resultat när man blandar { A, B, C }
61
Online Poker
• Texas hold'em poker. Programmet ska blanda elektroniska kort.
How We Learned to Cheat at Online Poker: A Study in Software Security
http://www.datamation.com/entdev/article.php/616221
62
Online poker anekdot
Shuffling algorithm in FAQ at www.planetpoker.com
for i := 1 to 52 do begin
r := random(51) + 1;
swap := card[r];
card[r] := card[i];
card[i] := swap;
end;
•
•
•
•
Bug
Bug
Bug
Bug
1.
2.
3.
4.
between 1 and 51
Slump tal r aldrig 52 => 52:de kort kan inte hamna i 52:de platsen.
Blandingen är inte slumpmässig (ska vara mellan 1 och i).
random() använder bara en 32-bit seed => 232 möjliga blandningar.
Seed = millisekunder sedan midnatt => 86.4 miljon blandningar.
• Exploit. Man kan, efter har set bara 5 kort och synkroniserat med server
klockan, förutspå alla framtida kort i realtid.
“ The generation of random numbers is too important to be left to chance. ”
— Robert R. Coveyou
63
Online poker: bästa praxis
• Bästa praxis för blandning.
- Använd en hårdvara lösning som genererar slumptal som är godkänd för
både både FIPS 140-2 och NIST statistiska prov.
•
Men, hårdvara slumptal generatorer kan vara ömtåliga och kan ”fail silently".
- Använd en opartiska blandningsalgoritm.
64
Sammanfattning
•Enkla algoritmer för att sortera arrayer
- Selection sort
- Insertion sort
- Shell sort
•Blandning
- Att blanda kort är svårt!
•Näst
- N log N sorteringsalgoritmer baserad på divide-and-conquer
- Mergesort och Quicksort
``` | 7,781 | 20,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.134795 |
https://canvas.instructure.com/eportfolios/666989/Home/How_long_does_it_take_to_walk_a_mile | 1,643,241,116,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305006.68/warc/CC-MAIN-20220126222652-20220127012652-00259.warc.gz | 197,209,104 | 15,430 | ## How long does it take to walk a mile?
Rich Text Content
Adults can walk 3-4 miles in an hour, and walk a mile in 15 to 20 minutes. There will be those who are faster than others, and speed can be affected by gender , age , as well as fitness level.
Is it possible to calculate the speed of your walk?
If you're planning a travel, a day out or preparing for a sponsored walk or marathon You may wish to calculate how many minutes you'll need to take to walk a mile. The individual's health will obviously affect the time it takes to walk a mile but more general factors like gender, age and pace are something you can account into your plan.
Walking Howtoneed as a result of age
We're not talking about the incredible nature of being able to walk two blocks with a child in just three hours (so many cracks in walls to look at!)
You may be shocked to discover that the sprightly twentysomethings tend to walk slower than their older thirty-something and forty-something counterparts. This is real. As we age, our walking speed decreases.
At 2.1 speed, it could be a 75 year old's task to walk a mile in an average that is close to half an hour (28 minutes and 34 seconds) however a 23 year-old could walk at 3mph and walk for just 20 minutes.
Speed of walking based on gender
Women walk faster than men in all age ranges. It could be that men have bigger legs. Or perhaps they're more likely to be running late to do reasons...
Walking speed according to your pace
When you're calculating the distance it takes to walk for a mile there's another aspect you should consider whether you're walking, strolling or steaming down the street. You'll be walking at a different pace when you're enjoying the countryside air, compared to when you're worried you'll miss the train.
Here are some numbers for different paces of walking:
Fast: 100 to 119 steps per minute or 11 minutes per mile
Normal - between 80 and 99 steps every hour / 15 minutes each mile
Relaxed - 60 to 79 steps per hour / 20 minutes each mile
The miles chart below for an estimate of how much time it will take to walk the distance at a normal, speed or at a relaxed pace.
The last line of our chart above gives us the answer to the question "How long would it take the Proclaimers to travel 500 miles". When you are at a normal pace for walking, it would take 125 hours to complete the task.
rich_text
Rich Text Content
rich_text | 556 | 2,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-05 | latest | en | 0.951465 |
https://excelguider.com/calculating-overtime-pay-worksheet/ | 1,716,226,617,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00241.warc.gz | 223,727,587 | 14,321 | # Calculating Overtime Pay Worksheet
A Calculating Overtime Pay Worksheet is several short questionnaires on a precise topic. A worksheet can be ready for any subject. Topic is usually a complete lesson in a unit or a small sub-topic. Worksheet can be utilized for revising the topic for assessments, recapitulation, helping the scholars to comprehend the subject more precisely or even improve the data over the topic.
## Ideas of a Calculating Overtime Pay Worksheet
Calculating Overtime Pay Worksheet needs to be child friendly. The particular issue level on the worksheet should be minimum. Worksheet needs to have clarity in questioning avoiding any ambiguity. In the worksheet the questions shouldn’t have many possible answer. Worksheet should serve as a tool to boost content of your child. Worksheet must be pictorial. Worksheet could be include skills for instance drawing, analyzing, descriptive, reasoning etc. Worksheet could be short not more than 2 pages else it would be called as a Workbook.
## Forming a Calculating Overtime Pay Worksheet Without problems
Crafting a Calculating Overtime Pay Worksheet isn’t an effortless task. The worksheet need to be short, crisp, easy and child friendly. Various skills involved in designing a worksheet, forms of worksheets, and sample worksheets are explained in detail. Traditionally the worksheets are ready in a variety of subjects which might be short or elaborate, with or without pictures. The revolutionary innovative accessible model of worksheet formulated for designing a worksheet could be the 3 E’s Worksheet method. As a student both these worksheets are EASY, ENJOYABLE and EFFORTLESS. The worksheets would rekindle the teaching-learning steps involved in student for the completion of worksheet. It entails maximum up to 10 min to complete each worksheet. Skills included are applicative, conceptual understanding, diagrammatic, labeling besides identification of terms.
## Calculating Overtime Pay Worksheet Can Be Utilized For Several Aim
Calculating Overtime Pay Worksheet work extremely well because of a teacher/tutor/parent to enrich this great article an understanding of their student/child. Worksheets can be employed like a testing tool to determine the Scholastic Aptitude and Mental Aptitude of child during admission procedures. Worksheets may also be prepared like a feedback activity after a field trip, study tour, educational trip, etc. Worksheets can be utilized as one tool to present extra knowledge and to observe the improvement of this skills in trainees along the lines of reading, comprehensive, analytical, illustrative etc. Worksheet helps trainees to excel in a specific focus.
## Greatest things about a Calculating Overtime Pay Worksheet
Calculating Overtime Pay Worksheet is one kind of most handy tool on a teacher. Student has to just fill in the worksheet. As most of the matter is been already printed for him. So he/she feels happy to take on it faster. Worksheet offers student the primary revision during a topic. Student shall improve his application skills (ex: begin to see the ‘Answer during a word’in the aforementioned sample worksheet). A worksheet can test any mode of learning like diagrams, elaborate writing, puzzling, quizzing, paragraph writing, picture reading, experiments etc. Worksheets may just be manufactured for the’Gifted Children’to offer more inputs to the given topic beyond the textual knowledge. Worksheets could be a helping hand to better the amount of understanding for any’Slow Learners’.
## Handicaps of Calculating Overtime Pay Worksheet
As it is known, every coin has two sides. Calculating Overtime Pay Worksheet in addition have several benefits many disadvantages. A statutory caution may be given, “Never use lots of worksheets”. Worksheets may be given being a revision for the lesson after teaching that lesson or could possibly be given between the culmination on the lesson for assignment to evaluate the idea of the child. Student becomes habitual to writing precise answers. Student gets habitual for that prompting. Correction of worksheets generally is a problem in a teacher. Perhaps it will become difficult as a student to preserve the worksheets and arrange them while using topics. All said and done worksheets are surely the aids to conserve the student effectively. Although the pros are usually when when compared to the disadvantages. One can’t disregard the hindrances. | 846 | 4,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.888805 |
https://fr.mathworks.com/matlabcentral/cody/problems/167-pizza/solutions/952333 | 1,575,823,735,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540511946.30/warc/CC-MAIN-20191208150734-20191208174734-00556.warc.gz | 376,154,517 | 16,178 | Cody
# Problem 167. Pizza!
Solution 952333
Submitted on 1 Sep 2016
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
z = 1; a = 1; v_correct = pi; assert(isequal(pizza(z,a),v_correct))
Error using pizza (line 2) Undefined function or variable 'x'.
2 Fail
z = 2; a = 1; v_correct = 4*pi; assert(isequal(pizza(z,a),v_correct))
Error using pizza (line 2) Undefined function or variable 'x'.
3 Fail
z = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct))
Error using pizza (line 2) Undefined function or variable 'x'.
4 Fail
z = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct))
Error using pizza (line 2) Undefined function or variable 'x'. | 249 | 801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.451167 |
http://orac.amt.edu.au/cgi-bin/train/problem.pl?problemid=23 | 1,653,202,187,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545090.44/warc/CC-MAIN-20220522063657-20220522093657-00412.warc.gz | 41,068,433 | 4,674 | # Problem: Handwriting Recognition
Want to try solving this problem? You can submit your code online if you log in or register.
## Handwriting Recognition
Input Files: handin.txt, symbols.txt, dict.txt
Output File: handout.txt
Time Limit: 0.3 seconds
You are an employee of a new handheld computer company that is preparing its flagship product, the Assistant for Intelligent and Organised Computing. This product has so many bells and whistles that there is no room left for a keyboard. It is thus your task to implement handwriting recognition for this device.
Fortunately the device is not required to learn the handwriting style of its user. Instead the user must learn the single handwriting style recognised by the device. Every letter is represented as a sequence of up, down, left or right strokes. In order to enter a letter, the user must draw the appropriate sequence of strokes without taking their pen away from the drawing area.
As an example, the figure below illustrates the letters B, C, I and L, each drawn without taking the pen away from the drawing area. Below each of these letters is a simplified representation using only up, down, left and right strokes (although some strokes have been angled slightly so that they can be distinguished in the diagram). In each symbol the beginning of the pen movement is indicated by a black circle, and the direction of the pen movement is indicated by an arrow.
For this problem you are given a file symbols.txt listing every recognised letter and the corresponding sequences of up, down, left and right strokes.
Unfortunately, users are not precise and cannot draw accurate horizontal and vertical strokes. If a user draws a diagonal stroke, your program may interpret it as one of two types of stroke as follows.
• A stroke running diagonally up and to the left may be interpreted as an up stroke or as a left stroke;
• A stroke running diagonally down and to the left may be interpreted as a down stroke or as a left stroke;
• A stroke running diagonally up and to the right may be interpreted as an up stroke or as a right stroke;
• A stroke running diagonally down and to the right may be interpreted as a down stroke or as a right stroke.
If a user draws a precise horizontal or vertical stroke then this stroke already runs up, down, left or right and so there is only one possible interpretation.
As an example, consider the symbol entered by a user that is illustrated below. The first stroke may be interpreted as either a down stroke or a left stroke. The second stroke is precisely horizontal and so must be interpreted as a right stroke. Using the sequences for B, C, I and L described above, the symbol could thus represent either of the letters C or L.
In order to resolve such ambiguities, you must consider each letter within the larger context of a word. You are given a dictionary of recognised words in the file dict.txt, and you may assume that each word entered by a user belongs to this dictionary.
As an example, suppose that the only letters recognised are B, C, I and L (as described previously) and that the dictionary contains the word BIC but not BIL. Consider the word illustrated below.
The first two symbols unambiguously represent B and I. The third symbol might represent either C or L as described above. However, since BIC belongs to the dictionary but BIL does not, your program can deduce that the word entered by the user is BIC.
Your task is to write a program that reads a number of words entered by a user, where each word is presented as a sequence of symbols. For each input word your program must output the corresponding dictionary word that is represented. You are guaranteed for each input word that there is one and only one dictionary word that it can represent.
### Input
The input is contained in three files: symbols.txt, dict.txt and handin.txt.
The file symbols.txt will contain a list of individual letters that your program must understand and the corresponding sequences of up, down, left and right strokes. Each letter and its sequence of strokes will be described on a separate line. No letter will be described more than once. At least 1 and at most 26 letters will be described. Following these lines will be a single line containing a single hash character (#).
Each individual line (except for the line containing the hash character) will contain an upper-case letter to be recognised, followed by a single space and then a sequence of upper-case characters representing strokes. Up, down, left and right strokes will be represented by the characters U, D, L and R respectively. Each line will contain at least 1 and at most 6 strokes.
The file dict.txt will contain a dictionary of recognised words. This dictionary will be presented as a series of lines each containing a single upper-case word. Each word in the dictionary will be at least 1 letter and at most 50 letters long. The dictionary will contain at least 1 and at most 1000 words. No word will be contained in the dictionary more than once. The list of words will be terminated by a line containing a single hash character (#).
The file handin.txt will contain a series of words entered by a user. At least 1 and at most 1000 words will be presented. Each word will be described using several lines as explained below. Following this series of words will be a line containing the single integer 0.
The first line of each word description will contain a single integer k representing the number of letters in the word. Following this will be k lines describing the k letters of the word in order from first to last. Each letter will be described as a symbol entered by the user.
Each line describing a symbol will begin with an integer s representing the number of strokes in the symbol (1 <= s <= 6). Following this will be a sequence of s+1 points (separated by spaces) giving, in order, the beginning of the first stroke, the end of the first stroke (which is also the beginning of the second stroke), the end of the second stroke and so on, finishing with the end of the last stroke. Each point is described by an integer x-coordinate followed by a single space then an integer y-coordinate, where larger values of x indicate points further to the right and larger values of y indicate points further upwards. Each coordinate will be between 0 and 100 inclusive.
The figure above illustrates the ambiguous C or L symbol discussed earlier, this time with a coordinate grid drawn in. The two strokes of this symbol move from points (3,6) to (1,2) and then to (3,2). Thus this symbol would be represented in the input file by the following line.
```2 3 6 1 2 3 2
```
### Output
For each word presented in handin.txt as a sequence of symbols, a line should be written to the output file handout.txt containing the unique dictionary word that this sequence of symbols represents. The output words must be entirely in upper case, and must be presented in the order in which the corresponding sequences of symbols are presented in handin.txt.
```A UD
B URLRL
C LR
I D
L DR
#
```
```AIL
BIC
BILL
CAB
ILL
LAB
#
```
### Sample handin.txt
```3
5 1 1 2 4 4 3 3 2 4 0 2 1
1 2 5 3 1
2 3 6 1 2 3 2
4
5 1 1 2 4 4 3 3 2 4 0 2 1
1 2 5 3 1
2 3 6 1 2 3 2
2 3 6 1 2 3 2
0
```
```BIC
BILL
```
### Scoring
For each set of input files, let n be the number of words in handin.txt that your program is asked to recognise. If your program correctly recognises r of these words, it shall be awarded r/n of the available marks for that particular set of input files.
Privacy statement
`Page generated: 22 May 2022, 4:49pm AEST` | 1,721 | 7,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-21 | latest | en | 0.949182 |
https://www.acmicpc.net/problem/11113 | 1,529,838,666,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866932.69/warc/CC-MAIN-20180624102433-20180624122433-00596.warc.gz | 714,712,208 | 15,529 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 256 MB 24 15 15 62.500%
## 문제
Lasse is organizing the orienteering world championship in Trondheim in 2008. Since he has been running in Bymarka for decades, he has a lot of route proposals. He must find a route with just the right length, but he doesn’t have time to measure them all by running through the control points in order, as he is too busy parallelizing code. The job is therefore given to Ola, who he thinks spends too much time with schoolwork.
As Ola is slightly less interested in orienteering, he figures out a more time-saving way of measuring the length of all the routes. He brings his GPS, visits all the points of all the routes in the most convenient order, and goes home early to do the rest of the job on his computer.
## 입력
The first line of input gives 1 ≤ n ≤ 1000, the total number of control points. Then follow n lines giving their coordinates, with two floating-point numbers xi and yi, with 0.0 ≤ xi, yi ≤ 10000.0. The number of routes is then given by 1 ≤ m ≤ 100. Each route is defined by first a line with 2 ≤ p ≤ 17, the number of control points (including start and goal), and then a line with p indexes 0 ≤ i < n, identifying them.
## 출력
For each test case output the total track distance, rounded, with no decimals.
## 예제 입력 1
5
0.0 0.0
1000.0 1000.0
123.45 0.0
3475.43 7765.4
4325.9865 13.0
2
2
0 1
4
3 1 4 0
## 예제 출력 1
1414
14999 | 423 | 1,417 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-26 | longest | en | 0.949737 |
https://rpg.stackexchange.com/questions/109215/sw-treasure-hoard | 1,722,872,566,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00784.warc.gz | 403,100,043 | 36,733 | # S&W Treasure Hoard
Looking for a little clarification on Treasure Hoards.
In the Swords & Wizardry Complete section "Generating a Random Treasure Hoard" (p120) you determine total XP value of the monsters in the adventure, and multiply by (1d3)+1 for gold value. For argument sake let us say this total is 5050gp. I would then roll:
• 50 times for 100gp trade-outs
• 5 times for 1000gp trade-outs
• 1 time for 5000gp trade-out
If I hit five 100gp and one 1000gp trade-outs, (1500gp total), then the party gets 3550gp plus whatever they received in the trade-out rolls.
If, however, I hit one of each (6100gp total) then I have a defecit of 1050gp. Does the party then get the entire 5050gp originally rolled, plus the result of each trade-out?
Either I am a bit thick, or it is wonderfully simple. Possibly both.
That's correct! The second case you describe (in which the value of the trade-outs exceeds the total GP value of the hoard) is how a “Major” treasure hoard is randomly determined. In that case, yes, the hoard consists of all the trade-outs you randomly determined plus all the original gold piece value, as per the note under the main trade-outs table (p. 120):
Note: if there are several trade-outs, it is possible to end up without enough gold pieces to trade for them, in which case it is a MAJOR treasure – add all the traded-out gold pieces back into the treasure along with the items rolled on the trade-out tables!
• I have no idea why it was so hard to wrap my head arund that. I'm too used to Treasure Types, I guess. :-D
– user40124
Commented Nov 1, 2017 at 19:26
• @DariusWhiteplume That particular case is counter-intuitive enough that it does easily make the reader second-guess their understanding! Commented Nov 1, 2017 at 19:28 | 457 | 1,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.912274 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/4032/2/c/ | 1,597,162,992,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738816.7/warc/CC-MAIN-20200811150134-20200811180134-00360.warc.gz | 727,565,447 | 65,211 | # Properties
Label 4032.2.c Level 4032 Weight 2 Character orbit c Rep. character $$\chi_{4032}(2017,\cdot)$$ Character field $$\Q$$ Dimension 60 Newform subspaces 18 Sturm bound 1536 Trace bound 25
# Related objects
## Defining parameters
Level: $$N$$ $$=$$ $$4032 = 2^{6} \cdot 3^{2} \cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 4032.c (of order $$2$$ and degree $$1$$) Character conductor: $$\operatorname{cond}(\chi)$$ $$=$$ $$8$$ Character field: $$\Q$$ Newform subspaces: $$18$$ Sturm bound: $$1536$$ Trace bound: $$25$$ Distinguishing $$T_p$$: $$5$$, $$11$$, $$13$$, $$17$$, $$23$$, $$31$$
## Dimensions
The following table gives the dimensions of various subspaces of $$M_{2}(4032, [\chi])$$.
Total New Old
Modular forms 816 60 756
Cusp forms 720 60 660
Eisenstein series 96 0 96
## Trace form
$$60q + O(q^{10})$$ $$60q + 24q^{17} - 84q^{25} - 72q^{41} + 60q^{49} + 24q^{73} + 24q^{89} + 24q^{97} + O(q^{100})$$
## Decomposition of $$S_{2}^{\mathrm{new}}(4032, [\chi])$$ into newform subspaces
Label Dim. $$A$$ Field CM Traces $q$-expansion
$$a_2$$ $$a_3$$ $$a_5$$ $$a_7$$
4032.2.c.a $$2$$ $$32.196$$ $$\Q(\sqrt{-1})$$ None $$0$$ $$0$$ $$0$$ $$-2$$ $$q+iq^{5}-q^{7}+2iq^{11}-iq^{13}-2q^{17}+\cdots$$
4032.2.c.b $$2$$ $$32.196$$ $$\Q(\sqrt{-1})$$ None $$0$$ $$0$$ $$0$$ $$-2$$ $$q+2iq^{5}-q^{7}+iq^{11}-2iq^{13}-2q^{17}+\cdots$$
4032.2.c.c $$2$$ $$32.196$$ $$\Q(\sqrt{-1})$$ None $$0$$ $$0$$ $$0$$ $$-2$$ $$q+iq^{5}-q^{7}-iq^{11}-iq^{13}-2q^{17}+\cdots$$
4032.2.c.d $$2$$ $$32.196$$ $$\Q(\sqrt{-1})$$ None $$0$$ $$0$$ $$0$$ $$-2$$ $$q+iq^{5}-q^{7}+2iq^{11}+iq^{13}+2q^{17}+\cdots$$
4032.2.c.e $$2$$ $$32.196$$ $$\Q(\sqrt{-1})$$ None $$0$$ $$0$$ $$0$$ $$-2$$ $$q+iq^{5}-q^{7}-iq^{11}+3iq^{13}+6q^{17}+\cdots$$
4032.2.c.f $$2$$ $$32.196$$ $$\Q(\sqrt{-1})$$ None $$0$$ $$0$$ $$0$$ $$2$$ $$q+2iq^{5}+q^{7}-iq^{11}-2iq^{13}-2q^{17}+\cdots$$
4032.2.c.g $$2$$ $$32.196$$ $$\Q(\sqrt{-1})$$ None $$0$$ $$0$$ $$0$$ $$2$$ $$q+iq^{5}+q^{7}+iq^{11}-iq^{13}-2q^{17}+\cdots$$
4032.2.c.h $$2$$ $$32.196$$ $$\Q(\sqrt{-1})$$ None $$0$$ $$0$$ $$0$$ $$2$$ $$q+iq^{5}+q^{7}-2iq^{11}-iq^{13}-2q^{17}+\cdots$$
4032.2.c.i $$2$$ $$32.196$$ $$\Q(\sqrt{-1})$$ None $$0$$ $$0$$ $$0$$ $$2$$ $$q+iq^{5}+q^{7}-2iq^{11}+iq^{13}+2q^{17}+\cdots$$
4032.2.c.j $$2$$ $$32.196$$ $$\Q(\sqrt{-1})$$ None $$0$$ $$0$$ $$0$$ $$2$$ $$q+iq^{5}+q^{7}+iq^{11}+3iq^{13}+6q^{17}+\cdots$$
4032.2.c.k $$4$$ $$32.196$$ $$\Q(\zeta_{12})$$ None $$0$$ $$0$$ $$0$$ $$-4$$ $$q-\zeta_{12}q^{5}-q^{7}-2\zeta_{12}q^{11}+(-2\zeta_{12}+\cdots)q^{13}+\cdots$$
4032.2.c.l $$4$$ $$32.196$$ $$\Q(\zeta_{12})$$ None $$0$$ $$0$$ $$0$$ $$-4$$ $$q-\zeta_{12}^{3}q^{5}-q^{7}+(-\zeta_{12}^{2}-\zeta_{12}^{3})q^{11}+\cdots$$
4032.2.c.m $$4$$ $$32.196$$ $$\Q(\zeta_{12})$$ None $$0$$ $$0$$ $$0$$ $$-4$$ $$q-\zeta_{12}^{3}q^{5}-q^{7}+(\zeta_{12}^{2}+3\zeta_{12}^{3})q^{11}+\cdots$$
4032.2.c.n $$4$$ $$32.196$$ $$\Q(\zeta_{12})$$ None $$0$$ $$0$$ $$0$$ $$4$$ $$q-\zeta_{12}q^{5}+q^{7}+2\zeta_{12}q^{11}+(-2\zeta_{12}+\cdots)q^{13}+\cdots$$
4032.2.c.o $$4$$ $$32.196$$ $$\Q(\zeta_{12})$$ None $$0$$ $$0$$ $$0$$ $$4$$ $$q-\zeta_{12}^{3}q^{5}+q^{7}+(\zeta_{12}^{2}+\zeta_{12}^{3})q^{11}+\cdots$$
4032.2.c.p $$4$$ $$32.196$$ $$\Q(\zeta_{12})$$ None $$0$$ $$0$$ $$0$$ $$4$$ $$q-\zeta_{12}^{3}q^{5}+q^{7}+(-\zeta_{12}^{2}-3\zeta_{12}^{3})q^{11}+\cdots$$
4032.2.c.q $$8$$ $$32.196$$ 8.0.897122304.10 None $$0$$ $$0$$ $$0$$ $$-8$$ $$q-\beta _{5}q^{5}-q^{7}+\beta _{5}q^{11}+(-\beta _{4}-\beta _{6}+\cdots)q^{13}+\cdots$$
4032.2.c.r $$8$$ $$32.196$$ 8.0.897122304.10 None $$0$$ $$0$$ $$0$$ $$8$$ $$q-\beta _{5}q^{5}+q^{7}-\beta _{5}q^{11}+(-\beta _{4}-\beta _{6}+\cdots)q^{13}+\cdots$$
## Decomposition of $$S_{2}^{\mathrm{old}}(4032, [\chi])$$ into lower level spaces
$$S_{2}^{\mathrm{old}}(4032, [\chi]) \cong$$ $$S_{2}^{\mathrm{new}}(24, [\chi])$$$$^{\oplus 16}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(56, [\chi])$$$$^{\oplus 12}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(64, [\chi])$$$$^{\oplus 6}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(72, [\chi])$$$$^{\oplus 8}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(96, [\chi])$$$$^{\oplus 8}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(168, [\chi])$$$$^{\oplus 8}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(192, [\chi])$$$$^{\oplus 4}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(224, [\chi])$$$$^{\oplus 6}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(288, [\chi])$$$$^{\oplus 4}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(448, [\chi])$$$$^{\oplus 3}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(504, [\chi])$$$$^{\oplus 4}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(576, [\chi])$$$$^{\oplus 2}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(672, [\chi])$$$$^{\oplus 4}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(1344, [\chi])$$$$^{\oplus 2}$$$$\oplus$$$$S_{2}^{\mathrm{new}}(2016, [\chi])$$$$^{\oplus 2}$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ 1
$3$ 1
$5$ ($$( 1 - 4 T + 5 T^{2} )( 1 + 4 T + 5 T^{2} )$$)($$( 1 - 2 T + 5 T^{2} )( 1 + 2 T + 5 T^{2} )$$)($$( 1 - 4 T + 5 T^{2} )( 1 + 4 T + 5 T^{2} )$$)($$( 1 - 4 T + 5 T^{2} )( 1 + 4 T + 5 T^{2} )$$)($$( 1 - 4 T + 5 T^{2} )( 1 + 4 T + 5 T^{2} )$$)($$( 1 - 2 T + 5 T^{2} )( 1 + 2 T + 5 T^{2} )$$)($$( 1 - 4 T + 5 T^{2} )( 1 + 4 T + 5 T^{2} )$$)($$( 1 - 4 T + 5 T^{2} )( 1 + 4 T + 5 T^{2} )$$)($$( 1 - 4 T + 5 T^{2} )( 1 + 4 T + 5 T^{2} )$$)($$( 1 - 4 T + 5 T^{2} )( 1 + 4 T + 5 T^{2} )$$)($$1 - 12 T^{2} + 74 T^{4} - 300 T^{6} + 625 T^{8}$$)($$1 - 12 T^{2} + 74 T^{4} - 300 T^{6} + 625 T^{8}$$)($$1 - 12 T^{2} + 74 T^{4} - 300 T^{6} + 625 T^{8}$$)($$1 - 12 T^{2} + 74 T^{4} - 300 T^{6} + 625 T^{8}$$)($$1 - 12 T^{2} + 74 T^{4} - 300 T^{6} + 625 T^{8}$$)($$1 - 12 T^{2} + 74 T^{4} - 300 T^{6} + 625 T^{8}$$)($$( 1 + 2 T^{4} + 625 T^{8} )^{2}$$)($$( 1 + 2 T^{4} + 625 T^{8} )^{2}$$)
$7$ ($$( 1 + T )^{2}$$)($$( 1 + T )^{2}$$)($$( 1 + T )^{2}$$)($$( 1 + T )^{2}$$)($$( 1 + T )^{2}$$)($$( 1 - T )^{2}$$)($$( 1 - T )^{2}$$)($$( 1 - T )^{2}$$)($$( 1 - T )^{2}$$)($$( 1 - T )^{2}$$)($$( 1 + T )^{4}$$)($$( 1 + T )^{4}$$)($$( 1 + T )^{4}$$)($$( 1 - T )^{4}$$)($$( 1 - T )^{4}$$)($$( 1 - T )^{4}$$)($$( 1 + T )^{8}$$)($$( 1 - T )^{8}$$)
$11$ ($$1 - 6 T^{2} + 121 T^{4}$$)($$1 - 18 T^{2} + 121 T^{4}$$)($$1 - 18 T^{2} + 121 T^{4}$$)($$1 - 6 T^{2} + 121 T^{4}$$)($$1 - 18 T^{2} + 121 T^{4}$$)($$1 - 18 T^{2} + 121 T^{4}$$)($$1 - 18 T^{2} + 121 T^{4}$$)($$1 - 6 T^{2} + 121 T^{4}$$)($$1 - 6 T^{2} + 121 T^{4}$$)($$1 - 18 T^{2} + 121 T^{4}$$)($$1 - 12 T^{2} + 86 T^{4} - 1452 T^{6} + 14641 T^{8}$$)($$1 - 36 T^{2} + 554 T^{4} - 4356 T^{6} + 14641 T^{8}$$)($$1 + 12 T^{2} + 170 T^{4} + 1452 T^{6} + 14641 T^{8}$$)($$1 - 12 T^{2} + 86 T^{4} - 1452 T^{6} + 14641 T^{8}$$)($$1 - 36 T^{2} + 554 T^{4} - 4356 T^{6} + 14641 T^{8}$$)($$1 + 12 T^{2} + 170 T^{4} + 1452 T^{6} + 14641 T^{8}$$)($$( 1 - 24 T^{2} + 338 T^{4} - 2904 T^{6} + 14641 T^{8} )^{2}$$)($$( 1 - 24 T^{2} + 338 T^{4} - 2904 T^{6} + 14641 T^{8} )^{2}$$)
$13$ ($$1 - 22 T^{2} + 169 T^{4}$$)($$( 1 - 6 T + 13 T^{2} )( 1 + 6 T + 13 T^{2} )$$)($$1 - 22 T^{2} + 169 T^{4}$$)($$1 - 22 T^{2} + 169 T^{4}$$)($$( 1 - 4 T + 13 T^{2} )( 1 + 4 T + 13 T^{2} )$$)($$( 1 - 6 T + 13 T^{2} )( 1 + 6 T + 13 T^{2} )$$)($$1 - 22 T^{2} + 169 T^{4}$$)($$1 - 22 T^{2} + 169 T^{4}$$)($$1 - 22 T^{2} + 169 T^{4}$$)($$( 1 - 4 T + 13 T^{2} )( 1 + 4 T + 13 T^{2} )$$)($$1 + 4 T^{2} + 42 T^{4} + 676 T^{6} + 28561 T^{8}$$)($$( 1 - 6 T + 13 T^{2} )^{2}( 1 + 6 T + 13 T^{2} )^{2}$$)($$( 1 - 13 T^{2} )^{4}$$)($$1 + 4 T^{2} + 42 T^{4} + 676 T^{6} + 28561 T^{8}$$)($$( 1 - 6 T + 13 T^{2} )^{2}( 1 + 6 T + 13 T^{2} )^{2}$$)($$( 1 - 13 T^{2} )^{4}$$)($$( 1 - 20 T^{2} + 246 T^{4} - 3380 T^{6} + 28561 T^{8} )^{2}$$)($$( 1 - 20 T^{2} + 246 T^{4} - 3380 T^{6} + 28561 T^{8} )^{2}$$)
$17$ ($$( 1 + 2 T + 17 T^{2} )^{2}$$)($$( 1 + 2 T + 17 T^{2} )^{2}$$)($$( 1 + 2 T + 17 T^{2} )^{2}$$)($$( 1 - 2 T + 17 T^{2} )^{2}$$)($$( 1 - 6 T + 17 T^{2} )^{2}$$)($$( 1 + 2 T + 17 T^{2} )^{2}$$)($$( 1 + 2 T + 17 T^{2} )^{2}$$)($$( 1 + 2 T + 17 T^{2} )^{2}$$)($$( 1 - 2 T + 17 T^{2} )^{2}$$)($$( 1 - 6 T + 17 T^{2} )^{2}$$)($$( 1 + 2 T + 17 T^{2} )^{4}$$)($$( 1 - 2 T + 8 T^{2} - 34 T^{3} + 289 T^{4} )^{2}$$)($$( 1 - 6 T + 40 T^{2} - 102 T^{3} + 289 T^{4} )^{2}$$)($$( 1 + 2 T + 17 T^{2} )^{4}$$)($$( 1 - 2 T + 8 T^{2} - 34 T^{3} + 289 T^{4} )^{2}$$)($$( 1 - 6 T + 40 T^{2} - 102 T^{3} + 289 T^{4} )^{2}$$)($$( 1 + 24 T^{2} + 290 T^{4} + 6936 T^{6} + 83521 T^{8} )^{2}$$)($$( 1 + 24 T^{2} + 290 T^{4} + 6936 T^{6} + 83521 T^{8} )^{2}$$)
$19$ ($$( 1 - 19 T^{2} )^{2}$$)($$1 - 2 T^{2} + 361 T^{4}$$)($$( 1 - 19 T^{2} )^{2}$$)($$( 1 - 19 T^{2} )^{2}$$)($$1 + 26 T^{2} + 361 T^{4}$$)($$1 - 2 T^{2} + 361 T^{4}$$)($$( 1 - 19 T^{2} )^{2}$$)($$( 1 - 19 T^{2} )^{2}$$)($$( 1 - 19 T^{2} )^{2}$$)($$1 + 26 T^{2} + 361 T^{4}$$)($$1 - 52 T^{2} + 1290 T^{4} - 18772 T^{6} + 130321 T^{8}$$)($$( 1 - 8 T + 19 T^{2} )^{2}( 1 + 8 T + 19 T^{2} )^{2}$$)($$1 - 20 T^{2} + 54 T^{4} - 7220 T^{6} + 130321 T^{8}$$)($$1 - 52 T^{2} + 1290 T^{4} - 18772 T^{6} + 130321 T^{8}$$)($$( 1 - 8 T + 19 T^{2} )^{2}( 1 + 8 T + 19 T^{2} )^{2}$$)($$1 - 20 T^{2} + 54 T^{4} - 7220 T^{6} + 130321 T^{8}$$)($$( 1 - 8 T + 19 T^{2} )^{4}( 1 + 8 T + 19 T^{2} )^{4}$$)($$( 1 - 8 T + 19 T^{2} )^{4}( 1 + 8 T + 19 T^{2} )^{4}$$)
$23$ ($$( 1 + 6 T + 23 T^{2} )^{2}$$)($$( 1 + 23 T^{2} )^{2}$$)($$( 1 + 23 T^{2} )^{2}$$)($$( 1 - 6 T + 23 T^{2} )^{2}$$)($$( 1 - 8 T + 23 T^{2} )^{2}$$)($$( 1 + 23 T^{2} )^{2}$$)($$( 1 + 23 T^{2} )^{2}$$)($$( 1 - 6 T + 23 T^{2} )^{2}$$)($$( 1 + 6 T + 23 T^{2} )^{2}$$)($$( 1 + 8 T + 23 T^{2} )^{2}$$)($$( 1 + 34 T^{2} + 529 T^{4} )^{2}$$)($$( 1 - 6 T + 52 T^{2} - 138 T^{3} + 529 T^{4} )^{2}$$)($$( 1 + 14 T + 92 T^{2} + 322 T^{3} + 529 T^{4} )^{2}$$)($$( 1 + 34 T^{2} + 529 T^{4} )^{2}$$)($$( 1 + 6 T + 52 T^{2} + 138 T^{3} + 529 T^{4} )^{2}$$)($$( 1 - 14 T + 92 T^{2} - 322 T^{3} + 529 T^{4} )^{2}$$)($$( 1 + 32 T^{2} + 882 T^{4} + 16928 T^{6} + 279841 T^{8} )^{2}$$)($$( 1 + 32 T^{2} + 882 T^{4} + 16928 T^{6} + 279841 T^{8} )^{2}$$)
$29$ ($$1 - 54 T^{2} + 841 T^{4}$$)($$1 + 6 T^{2} + 841 T^{4}$$)($$1 + 6 T^{2} + 841 T^{4}$$)($$1 - 54 T^{2} + 841 T^{4}$$)($$( 1 - 29 T^{2} )^{2}$$)($$1 + 6 T^{2} + 841 T^{4}$$)($$1 + 6 T^{2} + 841 T^{4}$$)($$1 - 54 T^{2} + 841 T^{4}$$)($$1 - 54 T^{2} + 841 T^{4}$$)($$( 1 - 29 T^{2} )^{2}$$)($$1 - 84 T^{2} + 3254 T^{4} - 70644 T^{6} + 707281 T^{8}$$)($$1 - 60 T^{2} + 1814 T^{4} - 50460 T^{6} + 707281 T^{8}$$)($$( 1 - 46 T^{2} + 841 T^{4} )^{2}$$)($$1 - 84 T^{2} + 3254 T^{4} - 70644 T^{6} + 707281 T^{8}$$)($$1 - 60 T^{2} + 1814 T^{4} - 50460 T^{6} + 707281 T^{8}$$)($$( 1 - 46 T^{2} + 841 T^{4} )^{2}$$)($$( 1 - 4 T + 14 T^{2} - 116 T^{3} + 841 T^{4} )^{2}( 1 + 4 T + 14 T^{2} + 116 T^{3} + 841 T^{4} )^{2}$$)($$( 1 - 4 T + 14 T^{2} - 116 T^{3} + 841 T^{4} )^{2}( 1 + 4 T + 14 T^{2} + 116 T^{3} + 841 T^{4} )^{2}$$)
$31$ ($$( 1 + 4 T + 31 T^{2} )^{2}$$)($$( 1 - 8 T + 31 T^{2} )^{2}$$)($$( 1 - 8 T + 31 T^{2} )^{2}$$)($$( 1 + 4 T + 31 T^{2} )^{2}$$)($$( 1 + 8 T + 31 T^{2} )^{2}$$)($$( 1 + 8 T + 31 T^{2} )^{2}$$)($$( 1 + 8 T + 31 T^{2} )^{2}$$)($$( 1 - 4 T + 31 T^{2} )^{2}$$)($$( 1 - 4 T + 31 T^{2} )^{2}$$)($$( 1 - 8 T + 31 T^{2} )^{2}$$)($$( 1 + 4 T + 31 T^{2} )^{4}$$)($$( 1 - 2 T + 31 T^{2} )^{4}$$)($$( 1 + 4 T + 18 T^{2} + 124 T^{3} + 961 T^{4} )^{2}$$)($$( 1 - 4 T + 31 T^{2} )^{4}$$)($$( 1 + 2 T + 31 T^{2} )^{4}$$)($$( 1 - 4 T + 18 T^{2} - 124 T^{3} + 961 T^{4} )^{2}$$)($$( 1 - 2 T + 31 T^{2} )^{8}$$)($$( 1 + 2 T + 31 T^{2} )^{8}$$)
$37$ ($$1 - 58 T^{2} + 1369 T^{4}$$)($$1 - 10 T^{2} + 1369 T^{4}$$)($$1 - 58 T^{2} + 1369 T^{4}$$)($$1 - 58 T^{2} + 1369 T^{4}$$)($$1 - 58 T^{2} + 1369 T^{4}$$)($$1 - 10 T^{2} + 1369 T^{4}$$)($$1 - 58 T^{2} + 1369 T^{4}$$)($$1 - 58 T^{2} + 1369 T^{4}$$)($$1 - 58 T^{2} + 1369 T^{4}$$)($$1 - 58 T^{2} + 1369 T^{4}$$)($$1 - 116 T^{2} + 5910 T^{4} - 158804 T^{6} + 1874161 T^{8}$$)($$( 1 - 12 T + 37 T^{2} )^{2}( 1 + 12 T + 37 T^{2} )^{2}$$)($$( 1 + 26 T^{2} + 1369 T^{4} )^{2}$$)($$1 - 116 T^{2} + 5910 T^{4} - 158804 T^{6} + 1874161 T^{8}$$)($$( 1 - 12 T + 37 T^{2} )^{2}( 1 + 12 T + 37 T^{2} )^{2}$$)($$( 1 + 26 T^{2} + 1369 T^{4} )^{2}$$)($$( 1 + 4 T^{2} - 330 T^{4} + 5476 T^{6} + 1874161 T^{8} )^{2}$$)($$( 1 + 4 T^{2} - 330 T^{4} + 5476 T^{6} + 1874161 T^{8} )^{2}$$)
$41$ ($$( 1 + 10 T + 41 T^{2} )^{2}$$)($$( 1 + 10 T + 41 T^{2} )^{2}$$)($$( 1 - 2 T + 41 T^{2} )^{2}$$)($$( 1 - 10 T + 41 T^{2} )^{2}$$)($$( 1 + 6 T + 41 T^{2} )^{2}$$)($$( 1 + 10 T + 41 T^{2} )^{2}$$)($$( 1 - 2 T + 41 T^{2} )^{2}$$)($$( 1 + 10 T + 41 T^{2} )^{2}$$)($$( 1 - 10 T + 41 T^{2} )^{2}$$)($$( 1 + 6 T + 41 T^{2} )^{2}$$)($$( 1 - 2 T + 41 T^{2} )^{4}$$)($$( 1 + 2 T + 8 T^{2} + 82 T^{3} + 1681 T^{4} )^{2}$$)($$( 1 + 6 T + 88 T^{2} + 246 T^{3} + 1681 T^{4} )^{2}$$)($$( 1 - 2 T + 41 T^{2} )^{4}$$)($$( 1 + 2 T + 8 T^{2} + 82 T^{3} + 1681 T^{4} )^{2}$$)($$( 1 + 6 T + 88 T^{2} + 246 T^{3} + 1681 T^{4} )^{2}$$)($$( 1 + 120 T^{2} + 6530 T^{4} + 201720 T^{6} + 2825761 T^{8} )^{2}$$)($$( 1 + 120 T^{2} + 6530 T^{4} + 201720 T^{6} + 2825761 T^{8} )^{2}$$)
$43$ ($$1 - 82 T^{2} + 1849 T^{4}$$)($$1 - 82 T^{2} + 1849 T^{4}$$)($$1 + 14 T^{2} + 1849 T^{4}$$)($$1 - 82 T^{2} + 1849 T^{4}$$)($$1 - 50 T^{2} + 1849 T^{4}$$)($$1 - 82 T^{2} + 1849 T^{4}$$)($$1 + 14 T^{2} + 1849 T^{4}$$)($$1 - 82 T^{2} + 1849 T^{4}$$)($$1 - 82 T^{2} + 1849 T^{4}$$)($$1 - 50 T^{2} + 1849 T^{4}$$)($$1 - 140 T^{2} + 8406 T^{4} - 258860 T^{6} + 3418801 T^{8}$$)($$1 - 68 T^{2} + 4086 T^{4} - 125732 T^{6} + 3418801 T^{8}$$)($$1 - 4 T^{2} - 3210 T^{4} - 7396 T^{6} + 3418801 T^{8}$$)($$1 - 140 T^{2} + 8406 T^{4} - 258860 T^{6} + 3418801 T^{8}$$)($$1 - 68 T^{2} + 4086 T^{4} - 125732 T^{6} + 3418801 T^{8}$$)($$1 - 4 T^{2} - 3210 T^{4} - 7396 T^{6} + 3418801 T^{8}$$)($$( 1 - 68 T^{2} + 4086 T^{4} - 125732 T^{6} + 3418801 T^{8} )^{2}$$)($$( 1 - 68 T^{2} + 4086 T^{4} - 125732 T^{6} + 3418801 T^{8} )^{2}$$)
$47$ ($$( 1 - 8 T + 47 T^{2} )^{2}$$)($$( 1 - 8 T + 47 T^{2} )^{2}$$)($$( 1 + 4 T + 47 T^{2} )^{2}$$)($$( 1 + 8 T + 47 T^{2} )^{2}$$)($$( 1 - 12 T + 47 T^{2} )^{2}$$)($$( 1 + 8 T + 47 T^{2} )^{2}$$)($$( 1 - 4 T + 47 T^{2} )^{2}$$)($$( 1 + 8 T + 47 T^{2} )^{2}$$)($$( 1 - 8 T + 47 T^{2} )^{2}$$)($$( 1 + 12 T + 47 T^{2} )^{2}$$)($$( 1 + 8 T + 62 T^{2} + 376 T^{3} + 2209 T^{4} )^{2}$$)($$( 1 + 4 T + 47 T^{2} )^{4}$$)($$( 1 + 46 T^{2} + 2209 T^{4} )^{2}$$)($$( 1 - 8 T + 62 T^{2} - 376 T^{3} + 2209 T^{4} )^{2}$$)($$( 1 - 4 T + 47 T^{2} )^{4}$$)($$( 1 + 46 T^{2} + 2209 T^{4} )^{2}$$)($$( 1 + 60 T^{2} + 4550 T^{4} + 132540 T^{6} + 4879681 T^{8} )^{2}$$)($$( 1 + 60 T^{2} + 4550 T^{4} + 132540 T^{6} + 4879681 T^{8} )^{2}$$)
$53$ ($$1 - 70 T^{2} + 2809 T^{4}$$)($$( 1 - 53 T^{2} )^{2}$$)($$1 + 38 T^{2} + 2809 T^{4}$$)($$1 - 70 T^{2} + 2809 T^{4}$$)($$1 + 38 T^{2} + 2809 T^{4}$$)($$( 1 - 53 T^{2} )^{2}$$)($$1 + 38 T^{2} + 2809 T^{4}$$)($$1 - 70 T^{2} + 2809 T^{4}$$)($$1 - 70 T^{2} + 2809 T^{4}$$)($$1 + 38 T^{2} + 2809 T^{4}$$)($$( 1 + 38 T^{2} + 2809 T^{4} )^{2}$$)($$( 1 - 94 T^{2} + 2809 T^{4} )^{2}$$)($$( 1 - 94 T^{2} + 2809 T^{4} )^{2}$$)($$( 1 + 38 T^{2} + 2809 T^{4} )^{2}$$)($$( 1 - 94 T^{2} + 2809 T^{4} )^{2}$$)($$( 1 - 94 T^{2} + 2809 T^{4} )^{2}$$)($$( 1 - 53 T^{2} )^{8}$$)($$( 1 - 53 T^{2} )^{8}$$)
$59$ ($$1 - 102 T^{2} + 3481 T^{4}$$)($$1 - 18 T^{2} + 3481 T^{4}$$)($$1 - 102 T^{2} + 3481 T^{4}$$)($$1 - 102 T^{2} + 3481 T^{4}$$)($$1 - 102 T^{2} + 3481 T^{4}$$)($$1 - 18 T^{2} + 3481 T^{4}$$)($$1 - 102 T^{2} + 3481 T^{4}$$)($$1 - 102 T^{2} + 3481 T^{4}$$)($$1 - 102 T^{2} + 3481 T^{4}$$)($$1 - 102 T^{2} + 3481 T^{4}$$)($$1 - 84 T^{2} + 8426 T^{4} - 292404 T^{6} + 12117361 T^{8}$$)($$( 1 - 54 T^{2} + 3481 T^{4} )^{2}$$)($$1 - 108 T^{2} + 6806 T^{4} - 375948 T^{6} + 12117361 T^{8}$$)($$1 - 84 T^{2} + 8426 T^{4} - 292404 T^{6} + 12117361 T^{8}$$)($$( 1 - 54 T^{2} + 3481 T^{4} )^{2}$$)($$1 - 108 T^{2} + 6806 T^{4} - 375948 T^{6} + 12117361 T^{8}$$)($$( 1 - 108 T^{2} + 9110 T^{4} - 375948 T^{6} + 12117361 T^{8} )^{2}$$)($$( 1 - 108 T^{2} + 9110 T^{4} - 375948 T^{6} + 12117361 T^{8} )^{2}$$)
$61$ ($$( 1 - 12 T + 61 T^{2} )( 1 + 12 T + 61 T^{2} )$$)($$1 - 106 T^{2} + 3721 T^{4}$$)($$1 - 118 T^{2} + 3721 T^{4}$$)($$( 1 - 12 T + 61 T^{2} )( 1 + 12 T + 61 T^{2} )$$)($$( 1 - 12 T + 61 T^{2} )( 1 + 12 T + 61 T^{2} )$$)($$1 - 106 T^{2} + 3721 T^{4}$$)($$1 - 118 T^{2} + 3721 T^{4}$$)($$( 1 - 12 T + 61 T^{2} )( 1 + 12 T + 61 T^{2} )$$)($$( 1 - 12 T + 61 T^{2} )( 1 + 12 T + 61 T^{2} )$$)($$( 1 - 12 T + 61 T^{2} )( 1 + 12 T + 61 T^{2} )$$)($$1 + 52 T^{2} + 7530 T^{4} + 193492 T^{6} + 13845841 T^{8}$$)($$1 - 20 T^{2} + 5814 T^{4} - 74420 T^{6} + 13845841 T^{8}$$)($$1 - 212 T^{2} + 18486 T^{4} - 788852 T^{6} + 13845841 T^{8}$$)($$1 + 52 T^{2} + 7530 T^{4} + 193492 T^{6} + 13845841 T^{8}$$)($$1 - 20 T^{2} + 5814 T^{4} - 74420 T^{6} + 13845841 T^{8}$$)($$1 - 212 T^{2} + 18486 T^{4} - 788852 T^{6} + 13845841 T^{8}$$)($$( 1 - 212 T^{2} + 18486 T^{4} - 788852 T^{6} + 13845841 T^{8} )^{2}$$)($$( 1 - 212 T^{2} + 18486 T^{4} - 788852 T^{6} + 13845841 T^{8} )^{2}$$)
$67$ ($$1 - 130 T^{2} + 4489 T^{4}$$)($$1 - 130 T^{2} + 4489 T^{4}$$)($$1 - 130 T^{2} + 4489 T^{4}$$)($$1 - 130 T^{2} + 4489 T^{4}$$)($$1 - 130 T^{2} + 4489 T^{4}$$)($$1 - 130 T^{2} + 4489 T^{4}$$)($$1 - 130 T^{2} + 4489 T^{4}$$)($$1 - 130 T^{2} + 4489 T^{4}$$)($$1 - 130 T^{2} + 4489 T^{4}$$)($$1 - 130 T^{2} + 4489 T^{4}$$)($$( 1 - 70 T^{2} + 4489 T^{4} )^{2}$$)($$1 - 20 T^{2} + 2166 T^{4} - 89780 T^{6} + 20151121 T^{8}$$)($$1 - 212 T^{2} + 19446 T^{4} - 951668 T^{6} + 20151121 T^{8}$$)($$( 1 - 70 T^{2} + 4489 T^{4} )^{2}$$)($$1 - 20 T^{2} + 2166 T^{4} - 89780 T^{6} + 20151121 T^{8}$$)($$1 - 212 T^{2} + 19446 T^{4} - 951668 T^{6} + 20151121 T^{8}$$)($$( 1 - 34 T^{2} + 4489 T^{4} )^{4}$$)($$( 1 - 34 T^{2} + 4489 T^{4} )^{4}$$)
$71$ ($$( 1 - 2 T + 71 T^{2} )^{2}$$)($$( 1 - 8 T + 71 T^{2} )^{2}$$)($$( 1 - 8 T + 71 T^{2} )^{2}$$)($$( 1 + 2 T + 71 T^{2} )^{2}$$)($$( 1 + 71 T^{2} )^{2}$$)($$( 1 + 8 T + 71 T^{2} )^{2}$$)($$( 1 + 8 T + 71 T^{2} )^{2}$$)($$( 1 + 2 T + 71 T^{2} )^{2}$$)($$( 1 - 2 T + 71 T^{2} )^{2}$$)($$( 1 + 71 T^{2} )^{2}$$)($$( 1 + 8 T + 110 T^{2} + 568 T^{3} + 5041 T^{4} )^{2}$$)($$( 1 - 10 T + 92 T^{2} - 710 T^{3} + 5041 T^{4} )^{2}$$)($$( 1 + 18 T + 196 T^{2} + 1278 T^{3} + 5041 T^{4} )^{2}$$)($$( 1 - 8 T + 110 T^{2} - 568 T^{3} + 5041 T^{4} )^{2}$$)($$( 1 + 10 T + 92 T^{2} + 710 T^{3} + 5041 T^{4} )^{2}$$)($$( 1 - 18 T + 196 T^{2} - 1278 T^{3} + 5041 T^{4} )^{2}$$)($$( 1 + 96 T^{2} + 12338 T^{4} + 483936 T^{6} + 25411681 T^{8} )^{2}$$)($$( 1 + 96 T^{2} + 12338 T^{4} + 483936 T^{6} + 25411681 T^{8} )^{2}$$)
$73$ ($$( 1 + 6 T + 73 T^{2} )^{2}$$)($$( 1 + 6 T + 73 T^{2} )^{2}$$)($$( 1 - 6 T + 73 T^{2} )^{2}$$)($$( 1 + 6 T + 73 T^{2} )^{2}$$)($$( 1 - 6 T + 73 T^{2} )^{2}$$)($$( 1 + 6 T + 73 T^{2} )^{2}$$)($$( 1 - 6 T + 73 T^{2} )^{2}$$)($$( 1 + 6 T + 73 T^{2} )^{2}$$)($$( 1 + 6 T + 73 T^{2} )^{2}$$)($$( 1 - 6 T + 73 T^{2} )^{2}$$)($$( 1 - 12 T + 134 T^{2} - 876 T^{3} + 5329 T^{4} )^{2}$$)($$( 1 + 24 T + 278 T^{2} + 1752 T^{3} + 5329 T^{4} )^{2}$$)($$( 1 + 38 T^{2} + 5329 T^{4} )^{2}$$)($$( 1 - 12 T + 134 T^{2} - 876 T^{3} + 5329 T^{4} )^{2}$$)($$( 1 + 24 T + 278 T^{2} + 1752 T^{3} + 5329 T^{4} )^{2}$$)($$( 1 + 38 T^{2} + 5329 T^{4} )^{2}$$)($$( 1 - 12 T + 134 T^{2} - 876 T^{3} + 5329 T^{4} )^{4}$$)($$( 1 - 12 T + 134 T^{2} - 876 T^{3} + 5329 T^{4} )^{4}$$)
$79$ ($$( 1 + 8 T + 79 T^{2} )^{2}$$)($$( 1 + 8 T + 79 T^{2} )^{2}$$)($$( 1 - 16 T + 79 T^{2} )^{2}$$)($$( 1 + 8 T + 79 T^{2} )^{2}$$)($$( 1 + 79 T^{2} )^{2}$$)($$( 1 - 8 T + 79 T^{2} )^{2}$$)($$( 1 + 16 T + 79 T^{2} )^{2}$$)($$( 1 - 8 T + 79 T^{2} )^{2}$$)($$( 1 - 8 T + 79 T^{2} )^{2}$$)($$( 1 + 79 T^{2} )^{2}$$)($$( 1 - 8 T + 126 T^{2} - 632 T^{3} + 6241 T^{4} )^{2}$$)($$( 1 + 4 T + 54 T^{2} + 316 T^{3} + 6241 T^{4} )^{2}$$)($$( 1 + 12 T + 182 T^{2} + 948 T^{3} + 6241 T^{4} )^{2}$$)($$( 1 + 8 T + 126 T^{2} + 632 T^{3} + 6241 T^{4} )^{2}$$)($$( 1 - 4 T + 54 T^{2} - 316 T^{3} + 6241 T^{4} )^{2}$$)($$( 1 - 12 T + 182 T^{2} - 948 T^{3} + 6241 T^{4} )^{2}$$)($$( 1 - 8 T + 126 T^{2} - 632 T^{3} + 6241 T^{4} )^{4}$$)($$( 1 + 8 T + 126 T^{2} + 632 T^{3} + 6241 T^{4} )^{4}$$)
$83$ ($$1 - 22 T^{2} + 6889 T^{4}$$)($$1 - 130 T^{2} + 6889 T^{4}$$)($$( 1 - 83 T^{2} )^{2}$$)($$1 - 22 T^{2} + 6889 T^{4}$$)($$1 - 102 T^{2} + 6889 T^{4}$$)($$1 - 130 T^{2} + 6889 T^{4}$$)($$( 1 - 83 T^{2} )^{2}$$)($$1 - 22 T^{2} + 6889 T^{4}$$)($$1 - 22 T^{2} + 6889 T^{4}$$)($$1 - 102 T^{2} + 6889 T^{4}$$)($$1 - 164 T^{2} + 17802 T^{4} - 1129796 T^{6} + 47458321 T^{8}$$)($$1 - 236 T^{2} + 25974 T^{4} - 1625804 T^{6} + 47458321 T^{8}$$)($$1 - 108 T^{2} + 14966 T^{4} - 744012 T^{6} + 47458321 T^{8}$$)($$1 - 164 T^{2} + 17802 T^{4} - 1129796 T^{6} + 47458321 T^{8}$$)($$1 - 236 T^{2} + 25974 T^{4} - 1625804 T^{6} + 47458321 T^{8}$$)($$1 - 108 T^{2} + 14966 T^{4} - 744012 T^{6} + 47458321 T^{8}$$)($$( 1 - 92 T^{2} + 8982 T^{4} - 633788 T^{6} + 47458321 T^{8} )^{2}$$)($$( 1 - 92 T^{2} + 8982 T^{4} - 633788 T^{6} + 47458321 T^{8} )^{2}$$)
$89$ ($$( 1 - 2 T + 89 T^{2} )^{2}$$)($$( 1 + 10 T + 89 T^{2} )^{2}$$)($$( 1 - 14 T + 89 T^{2} )^{2}$$)($$( 1 + 2 T + 89 T^{2} )^{2}$$)($$( 1 - 6 T + 89 T^{2} )^{2}$$)($$( 1 + 10 T + 89 T^{2} )^{2}$$)($$( 1 - 14 T + 89 T^{2} )^{2}$$)($$( 1 - 2 T + 89 T^{2} )^{2}$$)($$( 1 + 2 T + 89 T^{2} )^{2}$$)($$( 1 - 6 T + 89 T^{2} )^{2}$$)($$( 1 - 4 T - 10 T^{2} - 356 T^{3} + 7921 T^{4} )^{2}$$)($$( 1 + 14 T + 200 T^{2} + 1246 T^{3} + 7921 T^{4} )^{2}$$)($$( 1 - 6 T + 184 T^{2} - 534 T^{3} + 7921 T^{4} )^{2}$$)($$( 1 - 4 T - 10 T^{2} - 356 T^{3} + 7921 T^{4} )^{2}$$)($$( 1 + 14 T + 200 T^{2} + 1246 T^{3} + 7921 T^{4} )^{2}$$)($$( 1 - 6 T + 184 T^{2} - 534 T^{3} + 7921 T^{4} )^{2}$$)($$( 1 + 120 T^{2} + 5570 T^{4} + 950520 T^{6} + 62742241 T^{8} )^{2}$$)($$( 1 + 120 T^{2} + 5570 T^{4} + 950520 T^{6} + 62742241 T^{8} )^{2}$$)
$97$ ($$( 1 - 2 T + 97 T^{2} )^{2}$$)($$( 1 - 2 T + 97 T^{2} )^{2}$$)($$( 1 + 10 T + 97 T^{2} )^{2}$$)($$( 1 - 2 T + 97 T^{2} )^{2}$$)($$( 1 + 10 T + 97 T^{2} )^{2}$$)($$( 1 - 2 T + 97 T^{2} )^{2}$$)($$( 1 + 10 T + 97 T^{2} )^{2}$$)($$( 1 - 2 T + 97 T^{2} )^{2}$$)($$( 1 - 2 T + 97 T^{2} )^{2}$$)($$( 1 + 10 T + 97 T^{2} )^{2}$$)($$( 1 - 4 T + 150 T^{2} - 388 T^{3} + 9409 T^{4} )^{2}$$)($$( 1 - 16 T + 150 T^{2} - 1552 T^{3} + 9409 T^{4} )^{2}$$)($$( 1 + 8 T + 102 T^{2} + 776 T^{3} + 9409 T^{4} )^{2}$$)($$( 1 - 4 T + 150 T^{2} - 388 T^{3} + 9409 T^{4} )^{2}$$)($$( 1 - 16 T + 150 T^{2} - 1552 T^{3} + 9409 T^{4} )^{2}$$)($$( 1 + 8 T + 102 T^{2} + 776 T^{3} + 9409 T^{4} )^{2}$$)($$( 1 - 4 T + 150 T^{2} - 388 T^{3} + 9409 T^{4} )^{4}$$)($$( 1 - 4 T + 150 T^{2} - 388 T^{3} + 9409 T^{4} )^{4}$$) | 13,263 | 22,222 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-34 | latest | en | 0.44353 |
https://www.weegy.com/?ConversationId=A85TQBJO&Link=i | 1,632,267,352,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057274.97/warc/CC-MAIN-20210921221605-20210922011605-00662.warc.gz | 1,071,901,963 | 11,105 | How to solve this? N + (13 - 6) = 124
N + (13 - 6) = 124 N + 7 = 124 N = 124 - 7 N = 117
Question
Asked 6 days ago|9/15/2021 3:13:16 AM
Updated 6 days ago|9/15/2021 5:33:51 AM
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N + (13 - 6) = 124
N + 7 = 124
N = 124 - 7
N = 117
Added 6 days ago|9/15/2021 5:33:51 AM
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Added 6 days ago|9/15/2021 6:26:57 AM
Solve for N is this equation. N + (13 - 6) = 124 a.127 b.137 c.117
Question
Updated 6 days ago|9/15/2021 5:33:02 AM
N + (13 - 6) = 124
N + 7 = 124
N = 124 - 7
N = 117
Added 6 days ago|9/15/2021 5:33:02 AM
how to multiply 62 x 26
Question
Updated 6 days ago|9/15/2021 5:29:45 AM
62 x 26 = 1612
Added 6 days ago|9/15/2021 5:29:45 AM
Solve for N. 28 N = 7
Question
Updated 6 days ago|9/15/2021 5:27:32 AM
28n = 7
n = 7/28
n = 1/4
Added 6 days ago|9/15/2021 5:27:32 AM
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 981 | 2,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-39 | latest | en | 0.808917 |
https://www.encyclopediaofmath.org/index.php/A-operation | 1,537,468,725,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156554.12/warc/CC-MAIN-20180920175529-20180920195929-00197.warc.gz | 720,356,836 | 5,997 | # A-operation
operation
A set-theoretical operation, discovered by P.S. Aleksandrov [1] (see also [2], [3]). Let be a system of sets indexed by all finite sequences of natural numbers. The set
where the union is over all infinite sequences of natural numbers, is called the result of the -operation applied to the system .
The use of the -operation for the system of intervals of the number line gives sets (called -sets in honour of Aleksandrov) which need not be Borel sets (see -set; Descriptive set theory). The -operation is stronger than the operation of countable union and countable intersection, and is idempotent. With respect to -operations, the Baire property (of subsets of an arbitrary topological space) and the property of being Lebesgue measurable are invariant.
#### References
[1] P.S. Aleksandrov, C.R. Acad. Sci. Paris , 162 (1916) pp. 323–325 [2] P.S. Aleksandrov, "Theory of functions of a real variable and the theory of topological spaces" , Moscow (1978) (In Russian) [3] A.N. Kolmogorov, "P.S. Aleksandrov and the theory of -operations" Uspekhi Mat. Nauk , 21 : 4 (1966) pp. 275–278 (In Russian) [4] M.Ya. Suslin, C.R. Acad. Sci. Paris , 164 (1917) pp. 88–91 [5] N.N. Luzin, , Collected works , 2 , Moscow (1958) pp. 284 (In Russian) [6] K. Kuratowski, "Topology" , 1–2 , Acad. Press (1966–1968) (Translated from French)
The -operation is in the West usually attributed to M.Ya. Suslin [4], and is therefore also called the Suslin operation, the Suslin -operation or the Suslin operation . -sets are usually called analytic sets.
How to Cite This Entry:
A-operation. A.G. El'kin (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=A-operation&oldid=16633
This text originally appeared in Encyclopedia of Mathematics - ISBN 1402006098 | 508 | 1,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-39 | latest | en | 0.844258 |
https://oncologyleader.info/?post=3769 | 1,631,996,316,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056572.96/warc/CC-MAIN-20210918184640-20210918214640-00657.warc.gz | 487,266,370 | 4,184 | # How do the area and extent differ?
## Rectangles and squares: calculate perimeter and area
### Extent of rectangles
A second quantity that can be calculated for figures is the circumference. In the case of rectangles and squares, this is also very easy. You just add up all the edge lengths together. The following size can be calculated for the rectangle shown above:
\$ U (= circumference) = 8cm + 4cm + 8cm + 4cm \$
Click here to expand
The circumference \$ U \$ of a rectangle results from the addition of the side lengths:
\$ U = 2 \ cdot (a + b) \$
### Area and perimeter for squares
These calculations work the same way with squares, except that all lengths are equal.
The area \$ A \$ of a square is also calculated from the product of the side lengths:
\$ A = a \ cdot a = a ^ 2 \$
The following applies to the circumference \$ U \$ of a square: \$ U = 4 \ cdot a \$
The simple calculations for area and circumference for rectangles will be used again and again in the following examples. In the case of more complicated shapes, the procedure is usually to break the figure down into rectangles in order to get a result as easily as possible.
### Calculate rectangle: example
Now let's do an example calculation. First try to determine the solution yourself and then take a look at it here.
Now calculate the circumference (U) and the area (A) of a rectangle. The side lengths are \$ a = 6 cm \$ and \$ b = 3 cm \$
Rectangle: calculate area and perimeter
Let's start by calculating the perimeter of our rectangle. According to the formula that you have already got to know, you calculate the circumference with: \$ 2 \ cdot (a + b) \$. Let's insert our values for a (6 cm) and b (3 cm) into this formula and you get \$ U = 2 \ cdot (6cm + 3cm) \$. If we do the math, we get \$ 2 \ cdot 9cm \$, which is \$ 18cm \$. The circumference is \$ U = 18 cm \$.
Next we want to deal with the calculation of the area. We take the well-known formula (A = a \$ \ cdot \$ b) and insert our values here as well: \$ A = 6cm \ cdot 3cm \$. You get the area \$ A = 18 cm ^ 2 \$. Did you come up with the same solutions?
Test and deepen your new knowledge in the exercises! Good luck with it! | 554 | 2,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2021-39 | latest | en | 0.866 |
http://www.ck12.org/geometry/Area-and-Perimeter-of-Rhombuses-and-Kites/prepostread/Area-and-Perimeter-of-Rhombuses-and-Kites-KWL-Chart/r1/ | 1,485,187,516,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282932.75/warc/CC-MAIN-20170116095122-00425-ip-10-171-10-70.ec2.internal.warc.gz | 385,300,959 | 27,056 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Area and Perimeter of Rhombuses and Kites
## Area is half the product of the diagonals while the perimeter is the sum of the sides.
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Area and Perimeter of Rhombuses and Kites KWL Chart
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With this KWL Chart, reflect on your prior knowledge of a concept, come up with questions you’re curious about, and come up with answers from the reading. | 182 | 693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-04 | longest | en | 0.777424 |
https://earthscience.stackexchange.com/questions/14587/what-does-a-mm-of-rain-mean/14605 | 1,627,206,857,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00130.warc.gz | 236,358,124 | 50,358 | # What does a mm of rain mean?
Lots of people have explained it over many sites but I still can not confidently say that I know what it means when they say Hong Kong experienced 3mm of rainfall last Friday.
Does it mean that 3mm per square meter was experienced in a specific area, or does it mean that the total amount of rain had a volume such that if it was spread over all of Hong Kong the height would be 3mm. Or is it referred to in terms of per square meter?
What is the time period? Is it measured per day, per hour, per minute?
EDIT: The question marked as a duplicate may refer to some aspects of the question but does not effectively ask them. The title of the question refers to something different and the only answer to that question is what I am asking an explanation of.
It's confusing to measure liquid in units of length instead of volume, isn't it? Here's how it works.
"One millimeter of rain" is actually one cubic millimeter per square millimeter. On average, over the area you're talking about, each square millimeter has received one cubic millimeter of rain. If you divide n mm3 by 1 mm2, you get -- n mm!
The field of Dimensional Analysis deals with questions like this.
If there was "1 mm of rain", how much rain did a square meter receive? 1000 mm * 1000 mm * 1 mm = 1 000 000 mm3 = 1 liter. Pour one liter of water into a square container one meter on a side, and it forms a layer one millimeter deep.
To calculate "how much rain fell on Hong Kong" (expressed as total volume), you'd take the area of Hong Kong, and multiply it by 1 mm. That's usually not what people want to know, though -- they want to know how wet things got, and the volume-per-area number is most convenient for that.
• @JimmyB "1mm per m²" is wrong, though. That's a measurement in units of inverse length. "1 liter per m²" is right, because that again is a unit of length. – jeffB Jul 12 '18 at 14:32
• You're right of course :) – JimmyB Jul 12 '18 at 15:21
• When using the "unit of length for rainfall" approach to measuring; you are therefore able to simply place a cup outside and measure how far the collected water's surface is removed from the bottom of the cup, giving you "x millimeters of rain". Assuming the cup has vertical sides, the millimeter measurement would remain correct regardless of the size of cup you use. – Flater Jul 13 '18 at 11:10
• For exactly the same reason, exposure in photography is the same regardless of sensor size. – mattdm Jul 13 '18 at 20:33
• I don't think yet another answer is waranted, but I would mention that the actual measurement is made in mm: we measure length, not volume. Not different from the case where we also measure atmospheric pressure in mm (of mercury). – Martin Argerami Jul 15 '18 at 1:59
It is the amount of rain that it takes to cover the ground X milimeters deep. It is normally measured in 24 hours and is measured each morning at a fixed time like 0900.
But now the measurements are fully automatic, so the meteorological service gets the data more often and can provide data for each hour of the day.
Total rainfall is still measured from 0900 to the next day at 0900.
Two weather stations with some distance between them can often give very different measurements.
• Indeed. Daily sums shall be reported at fixed times, but it can be different from country to country. The amount of rain can be measured at a point (precipitation gauge) or for an area with a weather radar. The volume (eg. liters) of rain in the gauge can be converted to a theoretical depth of water: 3 mm = 3 liters / square meter Source: WMO Guide to Meteorological Instruments and Methods of Observation library.wmo.int/pmb_ged/wmo_8_en-2012.pdf – Lukas Jul 11 '18 at 15:08
• @Lukas i only used the time as an example,the sampling time is fixed to GMT or UTC to avoid trouble when we change from normal time to summertime too. – trond hansen Jul 11 '18 at 16:30
• @JoL You might be thinking about it too hard. It is safe to assume that rain falls in an evenly distributed manner. Lets take an example of "3mm of rainfall". If you held out a test tube that was 1x1 mm, it'd fill up to 3mm in depth of water. If you had a 10x10 meter empty pool before the rain, it would also fill up 3mm. The rain fell in an even manner, so for each square mm of space, 3 mm of rain fell. It doesn't matter how large the area is, the total "depth" of rain is 3mm. (Of course, this isn't totally true, leading to measurements from each side of a city being different). – Doc Jul 12 '18 at 0:17
• "Two weather stations with some distance between them can often give very diferent measurments." Indeed. Mt. Waiʻaleʻale and White Sands beach on the island of Kaua'i, while separated by less than 20 miles, typically show something of a difference in their daily rainfall readings. – dgnuff Jul 12 '18 at 2:50
• +1 for the only answer to actually touch the second half of the question - the time period over which the water is collected. – Semidiurnal Simon Jul 12 '18 at 10:11
Does it mean that 3mm per square meter was experienced in a specific area, or does it mean that the total amount of rain had a volume such that if it was spread over all of Hong Kong the height would be 3mm. Or is it referred to in terms of per square meter?
They're really the same idea. Fundamentally it means that it filled a rain gauge (of some type) to 3mm height. But every place that gets that experience equally would see the same amount of rain. So if the storm hit the entirety of Hong Kong as it hit the airport, the rain gauge at that airport would record 3mm, and so would any other rain gauge in the city. And (on flat, exposed ground... if water didn't absorb into the soil or evaporate over time) every square meter would get a depth of 3mm... and the region as a whole would have been covered with 3mm deep of water. (Of course water truly flows downhill and is deflected by buildings and such, so you'd end up with much more water in some spots, less in others).
In other words, 3mm is the height of whatever 3-dimensional prismatic shape you want to draw that was equally affected by the storm, be it a box with a square base of 1 m$^2$, a cylinder 1 km in diameter, a surface with a base of the entire city, etc.
Now of course in reality, not every place does experience the storm the same. And so Hong Kong airport may get 3mm, but Cheung Chau may only see 2mm.
And it's still true that a storm that affects a larger area with 3mm of rainfall does put down more total water quantity in entirety. If the same storm hits all of China, that's indeed a lot of water! But of course it's also a larger area to absorb that water. So regardless of whether it's a Hong-Kong sized downpour or a China sized rain system, if it puts down 3mm, it tends to have similar impacts throughout the region affected.
As others mention, a day is a very common time period for reporting such values. But they can be reported for whatever time periods people wish. Often you'll hear values for a specific storm, which can be as short as a quick downpour, or as long as a multi-day typhoon/hurricane. It's whatever the person wants to report about. Usually they'll give the time period or subject. 3mm doesn't fall in 1 second, but it can fall in a period much shorter than an hour. Or it can be the accumulation from only a steady drizzle that lasts days. That's where the rainfall rates (often given in mm or inches per hour) can be worthwhile!
• It's kind of like saying I'm going to give £1 to each location I go today. Whether I stay home and just give my roommate £1, or I go all over town, giving £1 throughout the entire city, each person's wallet gained only that £1. It's true that I truly gave a lot more money if I went all over town. But the impact on each person is still the same! – JeopardyTempest Jul 11 '18 at 16:02
• +1 This answer explains best why the area doesn't matter, and shows that length is indeed a good measurement, which I think is the crux of the question. – JoL Jul 11 '18 at 21:47
• It's not just that it allows for comparable statistics for locations with different surface areas. The other benefit is also that every can measure rainfally using a cup (with vertical edges) of any size. I might use a bucket, you might use a test tube, and the result will be the same (again, assuming straight lines and barring local variations in rainfall). – Flater Jul 13 '18 at 13:23
Given any container with straight sides, 3 millimeters of rainfall will fill up that container to 3 millimeters in depth.
The size of the container does not affect this. It holds whether the container is the size of a tennis court, or the size of a small beaker or a test tube with a flat bottom. As long as its sides are vertical and uniform.
It is also not necessary to talk about volume of water (eg cubic millimeters, liters etc) to understand this. You would only be converting to volume in order to convert back to depth later.
All you need to take into account is that any straight edged container will fill to this depth.
If you ever see a conical shaped container measuring rainfall, it's because the depth measurements on the side of that container are non-linear - these markings show what the equivalent depth would be in a straight-edged container.
Note 2: shallow water evaporates relatively quickly. For a rain gauge to be useful it needs to have a way to minimise evaporation, such as by having a funnel in the top. The area of the top of the funnel still matches the cross sectional area of the container in which we're measuring the depth
• This is in my opinion the best answer: We measure rainfall in mm because you can do it with literally any container with a flat bottom and sides that follow the profile of the opening. – Chris Becke Jul 15 '18 at 12:37
It literally means that the water depth will be 0.5mm. For example, if you place a container under the rain and it fills up to 0.5mm, then the precipitation is 0.5mm.
Despite the fact that a wide container requires more rain to fill up than a narrow container, the larger surface area allows it to collect the right amount of rain water to fill up to the same amount as the narrow container in the same amount of time (as long as the walls are vertical). Therefore, if the rainfall is 1 mm, every square meter receives 1 liter of rain water.
By the way, tuna cans can be distributed along your lawn to serve as a very primitive, but very cheap measurement tool that you can use to create a spatial map of what your irrigation system is doing. You can use that for troubleshooting your efficiency, 1 full can is 1 inch irrigation/rainfall.
• My impression was the OP needed a distinction between a drizzle and a light rain, inter alia. – Deer Hunter Jan 1 '15 at 8:04
• @DeerHunter fair enough, i think i mostly payed attention to the title and thought the OP was after an intuitive feel for this particular convention on measuring precipitation. – Isopycnal Oscillation Jan 1 '15 at 8:14
• This is just a definition, which I already know (and in which I believe you missed that it's per hour). I'm looking for a translation of some values into everyday terms, more detailed that the light/medium/heavy categories on Wikipedia. For example, I know that 18 degrees means that I can try going out in a tshirt, or that 45kmph wind means that I need to start being careful on a bike. – Jan van Bloem Jan 1 '15 at 11:34
• @JanvanBloem the answer correctly measures total precipitation (no time dependency). If you are interested in precipitation rates (e.g. units of mm/hr), please edit your question to address this. – casey Jan 1 '15 at 13:19
Simply, an "amount" of rain is a certain volume, and volume equals area x height.
"3mm of rain" means for any given area, the amount of rain that fell (ie the volume) would fill that area to a height of 3mm.
The area part of the formula is effectively a constant. The only variable is the height.
It means that during the rain event (which may last a minute, an hour, a day), one square meter of ground was filled with water up to a height of 3mm.
The one square meter just next to this first square meter was also filled with a height of 3mm of water. So that over the 2 square meters united, the ground was still filled up to 3mm of water.
A surface of 10 square meters was also filled with water up to a height of 3mm, etc.
In other words, the volume of rain collected obviously depends on the surface of the ground, but the height of the water level fallen onto the ground does not depend on the surface of the collector.
This is of course correct if the rain is uniform over the whole surface (i.e. it might brake for a surface which is much smaller than the distance between the drops, or for a surface that is larger than the scale at which the rain rate uniform).
I will address the aspects of the question that relate to how rainfall is measured and the measurement period. Some of this has been mentioned in other answers.
As previously noted you can think the amount of rain as the volume of water that falls on a given area divided by that surface area. That means that the measurement has units of length (typically mm). This assumes that the amount you measure at one point can represent the amount that falls over the whole area or that you measure many points that represent different parts of a large area and add them together.
The straight-sided graduated cylinders discussed in other answers have a constant area so the depth markings are a constant (1 mm) apart, but you can use other shapes, if they are marked correctly, to show what the depth would be on a straight sided cylinder. One common design is to have a funnel with a large area at the top feed into a narrower container below. In that case the depth of the water captured will be greater than the rainfall in mm. These take the rain volume from a larger area and measure it in a smaller one because it 'magnifies' the depth, making it easier to read for small amounts.
Automated systems work differently, using weighing or 'tipping bucket' systems. Tipping buckets have a funnel like top that feeds the water down into little cups on a pivot below. When one cup fills up the weight causes it to go down like a see-saw and the bucket empties. The other bucket moves up and catches the water until it is full and goes down. They are designed so the weight of rain it takes to get the tipping bucket to pivot is known. All you have to do is count how many times it pivots to measure the rainfall. If you assume that rain is essentially pure water, then the weight is directly related to the volume. Divide that volume by the area of the top of the gauge and you get mm of rain.
Most country's weather services report the amount of rain over a day and add the amounts together to get monthly or annual measurements, but it is possible to take measurements more frequently. More frequent measurements can be used to report rainfall intensity over a shorter period. So you, for example, could report the maximum hourly rainfall intensity within a day. You can also use more frequent measurements to calculate how quickly streamflow responds to a rainstorm.
Your question seems to be a conceptual one, so I'll answer conceptually.
For the sake of the argument let's imagine a flat area with impermeable ground, like rock, and an impermeable wall around it. After it has rained for some time with a certain intensity over that area — everywhere the same —, there will be a certain height of water on the ground, which will be the same height all over this flat area. This height is measured in millimeters and reported. You can now go and divide this area up in square meters, or soccer fields, or square milimeters; the height of the water would obviously not change. That's the beauty of this measurement: It is independent of any particular area.
These are unrealistically ideal assumptions. The intensity or duration differed from place to place, the ground is not level, water seeps away etc.
So the statement actually says: "This station measured a rain event which would cover any area with x mm of rain if it had rained everywhere the same amount as it did here." The duration is irrelevant with respect to the amount but must be mentioned together with the measurement because it makes a huge difference in meaning and impact: 100 mm in a month are just wet while 100 mm in a day are catastrophic. In your case, "last Friday", one may assume a 24 hour calendar day.
It's also often not possible to make a reliable statement for a larger area, especially in the case of local severe weather events. One can make an estimation based on measurements from different stations in the area and qualitative observations, but the only reliable accurate statements which can be made are factual measurements at weather stations; everything else is inferred, as in your example about all of Hong Kong.
I can imagine though that among Meterologists the sentence "Hong Kong reports 3 mm of rain" is taken to mean "The Hong Kong Observatory reports 3 mm of rain", and equivalently for other places.
Rainfall is measured with a rain gauge. This one goes up to 12 cm or 120 mm, so 3 mm would just be a few drops at the bottom.
As you can imagine, it doesn't matter how big this gauge is, as long as the sides are straight. A larger cylinder will hold more water, but it has a larger opening to collect rain, so the measured rainfall is the same in either case.
Inversely, multiplying the rainfall (3 mm) by an area (100 square meters) gives a volume (= 0.3 cubic meters, or 300 liters), assuming a uniform rainfall over the area. | 4,095 | 17,672 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2021-31 | latest | en | 0.962868 |
https://www.statisticssolutions.com/cox-event-history/ | 1,726,495,341,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00367.warc.gz | 926,529,240 | 23,888 | # Cox Event History
Quantitative Results
Statistical Analysis
Cox event history is a branch of statistics that deals mainly with the death of biological organisms and the failure of mechanical systems. It is also sometimes referred to as a statistical method for analyzing survival data. Cox event history is also known as various other names, such as survival analysis, duration analysis, or transition analysis. Generally speaking, this technique involves the modeling of data structured in a time-to-event format. The goal of this analysis is to understand the probability of the occurrence of an event. Cox event history was primarily developed for use in medical and biological sciences. However, this technique is now frequently used in engineering as well as in statistical and data analysis.
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One of the key purposes of the Cox event history technique is to explain the causes behind the differences or similarities between the events encountered by subjects. For instance, Cox regression may be used to evaluate why certain individuals are at a higher risk of contracting some diseases. Thus, it can be effectively applied to studying acute or chronic diseases, hence the interest in Cox regression by the medical science field. The Cox event history model mainly focuses on the hazard function, which produces the probability of an event occurring randomly at random times or at a specific period or instance in time.
The basic Cox event history model can be summarized by the following function:
h(t) = h0(t)e(b1X1 + b2X2 + K + bnXn)
Where; h(t) = rate of hazard
h0(t) = baseline hazard function
bX’s = coefficients and covariates.
Cox event history can be categorized mainly under three models: nonparametric, semi-parametric and parametric.
Non-parametric: The non-parametric model does not make any assumptions about the hazard function or the variables affecting it. Consequently, only a limited number of variable types can be handled with the help of a non-parametric model. This type of model involves the analysis of empirical data showing changes over a period of time and cannot handle continuous variables.
Semi-parametric: Similar to the non-parametric model, the semi-parametric model also does not make any assumptions about the shape of the hazard function or the variables affecting it. What makes this model different is that it assumes the rate of the hazard is proportional over a period of time. The estimates for the hazard function shape can be derived empirically as well. Multivariate analyses are supported by semi-parametric models and are often considered a more reliable fitting method for use in a Cox event history analysis.
Parametric: In this model, the shape of the hazard function and the variables affecting it are determined in advance. Multivariate analyses of discrete and continuous explanatory variables are supported by the parametric model. However, if the hazard function shape is incorrectly estimated, then there is a chance that the results could be biased. Parametric models are frequently used to analyze the nature of time dependency. It is also particularly useful for predictive modeling, because the shape of the baseline hazard function can be determined correctly by the parametric model.
Cox event history analysis involves the use of certain assumptions. As with every other statistical method or technique, if an assumption is violated, it will often lead to the results being statistically unreliable. The major assumption is that in using Cox event history, with the passage of time, independent variables do not interact with each other. In other words, the independent variables should have a constant hazard of rate over time.
In addition, hazard rates are rarely smooth in reality. Frequently, these rates need to be smoothed over in order for them to be useful for Cox event history analysis.
Applications of Cox Event History
Cox event history can be applied in many fields, although initially it was used primarily in medical and other biological sciences. Today, it is an excellent tool for other applications, frequently used as a statistical method where the dependent variables are categorical, especially in socio-economic analyses. For instance, in the field of economics, Cox event history is used extensively to relate macro or micro economic indicators in terms of a time series; for instance, one could determine the relationship between unemployment or employment over time. In addition, in commercial applications, Cox event history can be applied to estimate the lifespan of a certain machine and break down points based on historical data. | 963 | 5,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-38 | latest | en | 0.943396 |
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