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zxj160 February 17, 2012 11:51
velocity cyclic BC-groovyBC
Hi,
1 How to set a fixedvalue of velocity to the cyclic BC?
2 But the pressure is not cyclic, how to set a fixedgradient of pressure?
--> FOAM FATAL IO ERROR:
inconsistent patch and patchField types for
I want to set the following fixedvalue using groovyBC. How to do that?
inlet
{
type cyclic;
type groovyBC;
//type fixedValue
// log-law expression
uniform (0 0 0);
variables "u_f=0.23;z0=0.9;z_f=pos().z-10;vel=u_f*log(z_f/z0);";
valueExpression "vector (vel,0,0)";
}
outlet
{
type cyclic;
type groovyBC;
//type fixedValue
// log-law expression
uniform (0 0 0);
variables "u_f=0.23;z0=0.9;z_f=pos().z-10;vel=u_f*log(z_f/z0);";
valueExpression "vector (vel,0,0)";
}
nimasam February 18, 2012 08:47
i think its impossible, you should write your BC, with existing BC you can not!
zxj160 February 18, 2012 11:12
But I have installed the groovyBC. You mean I should write another BC? I donot know how to do that.
1 How to set a fixedvalue of velocity to the cyclic BC (inlet and outlet)?
2 But the pressure is not cyclic, how to set a fixedgradient of pressure?
nimasam February 18, 2012 12:48
i think you should write new BC
however try this, may be it works
1) use funkySetFields and define your velocity function in whole domain at time 0
2) uses a fan boundary condition with one constant ( constant value is the fixed pressure gradient * pipe length)
gschaider February 19, 2012 19:40
Quote:
Originally Posted by zxj160 (Post 344970) Hi, 1 How to set a fixedvalue of velocity to the cyclic BC? 2 But the pressure is not cyclic, how to set a fixedgradient of pressure? --> FOAM FATAL IO ERROR: inconsistent patch and patchField types for I want to set the following fixedvalue using groovyBC. How to do that? inlet { type cyclic; type groovyBC; //type fixedValue // log-law expression uniform (0 0 0); variables "u_f=0.23;z0=0.9;z_f=pos().z-10;vel=u_f*log(z_f/z0);"; valueExpression "vector (vel,0,0)"; } outlet { type cyclic; type groovyBC; //type fixedValue // log-law expression uniform (0 0 0); variables "u_f=0.23;z0=0.9;z_f=pos().z-10;vel=u_f*log(z_f/z0);"; valueExpression "vector (vel,0,0)"; }
Basically the problem is that a patch defined in the mesh as cyclic will accept no other boundary type than cyclic for fields (no groovyBC, no fixedValue, no ...).
zxj160 July 2, 2012 11:30
Quote:
Originally Posted by gschaider (Post 345218) Basically the problem is that a patch defined in the mesh as cyclic will accept no other boundary type than cyclic for fields (no groovyBC, no fixedValue, no ...).
Hi gschaider,
Could you give me some idea of how to set fixedValue and zeroGradient for cyclic inlet and outlet patches respectively? Is it possible to achieve that?
Best regards,
Jian
All times are GMT -4. The time now is 12:31. | 897 | 3,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | latest | en | 0.796755 |
https://courses.lumenlearning.com/waymakermath4libarts/chapter/putting-it-together-finance/ | 1,722,965,192,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640497907.29/warc/CC-MAIN-20240806161854-20240806191854-00518.warc.gz | 141,977,394 | 18,747 | ## Financing a Refrigerator: Three Scenarios
In the beginning of this module, we presented three options for buying a new refrigerator. In one scenario, you could rent to own, with the following terms $17.99 per week for 2 years, which is 104 weeks. The total cost is: $104\times17.99=1870.96$ Scenario two involved a loan of$1299 from your brother at 20% interest for one full year. To calculate the total amount, use the simple interest formula,
$I=P_0rt$
In this situation, the principle amount is $P_0=1299$, rate is $r=20\%=0.20$, and the time is $t=1$ year. Therefore, the interest due to your brother would be:
$I=1299\times0.20\times1=259.80$
Adding the interest back to the principle, the total cost of the refrigerator would amount to $1558.80. That’s quite a bit less than the$1870.96 that the rent-to-own store would ultimately have received from you. But your brother wants the money in one year, so let’s figure out what the weekly payment would be. Simply divide the total by 52 weeks.
$1558.80\div52=29.98$
This is a higher weekly payment than the rent-to-own store is offering, but if you can afford it, then you’ll save money in the long run.
Finally, let’s explore the third option. This time we use the loans formula,
$P_0=\Large\frac{d\left(1-\left(1+\frac{r}{k}\right)^{-Nk}\right)}{\left(\frac{r}{k}\right)}$
The principle is the same, $P_0=1299$, but the rate is now $r=15\%=0.15$. Because the compounding is monthly, we have $k=12$. Finally, $N=3$ represents the total number of years for the loan. We must solve for $d$.
$1299=\Large\frac{d\left(1-\left(1+\frac{0.15}{12}\right)^{-3\left(12\right)}\right)}{\left(\frac{0.15}{12}\right)}$
$1299=\Large\frac{d\left(1-\left(1.0125\right)^{-36}\right)}{0.0125}$
$1299=\Large\frac{d\left(0.36059\right)}{0.0125}$
$d=1299\times0.0125\div0.36059=45.03$
This calculation gives the monthly payment (since the compounding is monthly), $d=45.03$ If we want to see how this compares against our previous scenarios, we can find an equivalent weekly payment. The best way to do this is to multiply d by 12 and then divide by 52. This gives a weekly payment of about $45.03\times12\div52=10.39$, by far the lowest weekly payment, but what is the total cost?
Finally, to calculate the total cost, multiply the monthly payment by the number of months in 3 years, that is, 36 months.
$45.03\times36=1621.09$
Option three, the line of store credit. This option seemed pretty good at first. However, because of the long loan period and compounding interest, the total cost is actually more than the $1558.80 from scenario two. Let’s compare the details of each scenario shown in the table below. Note, the total interest is found by subtracting the list price of the refrigerator ($1299) from the total paid amount.
Rent to Own Brother’s Offer Store Loan Payments $17.99 per week$29.98 per week $45.03 per month ($10.39 per week) Length of Term 2 years 1 year 3 years Total Paid $1870.96$1558.80 $1621.09 Total Interest$571.96 $259.80$322.09
Which scenario would you choose? | 906 | 3,060 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.867938 |
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Various units of measurement o...
Various units of measurement of area.
Updated On: 27-06-2022
Text Solution
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
study about various units of measurement of area video mein padenge ki area kya hai agar hamare pass koi bhi hai aur Humne use per hi figure Banaya what is region ki measurement ko Ham bolenge area Agar hamare pass square Hai uski Ek site 128 Baki side nibandh CM Hogi is where area jo hoga vah hoga 1 cm square what do you CM square the standard unit is standard unit m square Dam square to m square km square
area Mein Hamesha Kisi bhi unit ke Ham Koi Bhi Ho 12 ka Sahi hota hai ki ismein site-to-site multiply Karte Hain To Jab side to side Hoti Hai Jaise 1 m into one metre is unit we multiply ho jate latest 1 metre per 1 cm square and so on hamare pass airline Hai usmein DM CM mm units a m a chote Hote aur Dam hm kilometre M Se Badi unit se DM ko sentimeter Mein convert kar sakte hain aur is video Mein Aage Karenge kilometre hm aur Dam ko metre Mein converter sabse pahle Ham DM ko sentimeter Mein convert Karenge 1 dm equals to 10 cm right then 1 decimeter to metre into
centimetre into centimetre centimetre square and 1 dm into 1 decimeter is 1 dm to 1 dm square = 210 now Agaram convert Kare Dam ko metre me so one dam into one dam equals to write one dam into one dam is 1 Dam square and 10 metre into 10 metre is hundred metre square then 182 metre = 200 metre light Agar Ham Deccan distance AK ha m m Main To
1 and 2 in a zoo distance Hota Hai distance from 10 to 10 into 10 that is 10 metre which is hundred metre then 1 hectare metre into 1 hectare metre = 200 metre in 201 = 200 200 that is 1000 metre square now 1 km equals to 1000 distance is a chord and I'm smart card 10 into 10 into 10 1000 in 21 km equals to 1000 INR to 1000 M M M M right death 1002 1000
M update 10 ki power 6 ki prapt 60 10th Ki Party ka 1 km square equals to 10 ki power 6 metre square | 583 | 2,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.747193 |
http://www.ipredict.it/methods/CurveAnalysis/ | 1,701,604,658,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00315.warc.gz | 65,822,594 | 2,761 | # Curve Analysis
The methods related to curve analysis assume that the underlying signal is quasi-sinusoidal or quasi-periodic. These methods constitute a kernel of methods that will be further expanded in future releases to analyze quasi-periodic waves and signals.
## Peaks and Bottoms
These methods compute respectively the number of peaks and the number of bottoms in the time-series.
## Curve Similarity
Curve similarity computes an index of similarity that is not the “simple” mean squared difference between the two curves and is computed on the first derivative of the signal. The lower the index the more similar the two curves, the higher the more different they are. The index is always positive.
## Average Period
The method computes the average period of the signal that is supposed to be quasi-sinusoidal. The computation is useful when there are more than three periods in the signal, for less than three periods the average is not statistically significant.
## Average Period Deviation
The simple measure of average period is not enough to analyze quasi-sinusoidal time-series. This method computes the average deviation from the mean period of the signal.
This is an example of a quasi-periodic signal:
Its average period is 119.75 and its average period deviation is 5.875. This nearly 5% variation in the period of the quasi-sinusoid and its average period are very important information that allows us to compute a reliable forecast.
Forecasting Methods Holt Winter’s, Series Decomposition and Wavelet Benchmarks Time Series Forecasting Use of the Moving Average in Time-series Forecasting Forecasting Concepts Denoising Techniques Error Statistics Computational Performance Fast Fourier Transform Moving Averages Kernel Smoothing Active Moving Average Savitsky-Golay Smoothing Fractal Projection Downloading Financial Data from Yahoo Multiple Regression Digital Signal Processing Principal Component Analysis Curve Analysis Options Pricing with Black-Scholes Markowitz Optimal Portfolio Time-series preprocessing | 380 | 2,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.87653 |
https://www.mcqbuddy.com/programming-questions/mcq/14871 | 1,726,000,468,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00143.warc.gz | 807,785,206 | 7,478 | M
Mr. Dubey • 51.17K Points
Coach
# `Q. What is the value of k after the following code fragment?~~~int k = 0;~~~int n = 12~~~while (k < n)~~~{~~~k = k + 1;~~~}`
(A) 0.
(B) 11.
(C) 12.
(D) 10.
###### Explanation by: Monu Rathod
```You might wonder that how this value will be 12. Since k is initialised with 0 and n with 12.
Now while loop checks whether value of k is smaller than n or not. So this condition will be true unless te value of k becomes 11.
Now the value of k is 11 and the condition is true to while loop will be executed and now k will become 12.
So the final value of k will be 12. ```
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https://oeis.org/A263317 | 1,586,515,496,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371893683.94/warc/CC-MAIN-20200410075105-20200410105605-00034.warc.gz | 596,852,270 | 4,312 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A263317 Least prime p > n such that the numbers sigma(k^2)/k^2 (k = 1,...,n) are pairwise incongruent modulo p, where sigma(m) is the sum of the divisors of m. 5
2, 5, 5, 7, 7, 29, 37, 37, 37, 37, 37, 43, 43, 43, 53, 79, 101, 101, 101, 101, 101, 101, 101, 101, 131, 131, 131, 131, 131, 131, 131, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 283, 317, 389, 389, 389, 389, 389, 389, 389, 389, 389 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Conjecture: a(n) exists for any n > 0, and a(n) < n^2 for all n > 2. This implies that all the rational numbers sigma(n^2)/n^2 = Sum_{d|n^2} 1/d (n = 1,2,3,...) are pairwise distinct. We have verified that the numbers sigma(n^2)/n^2 (n = 1..10^5) are indeed pairwise distinct, and noted that sigma(26334^2)/26334^2 - sigma(6^2)/6^2 = 127/36 - 91/36 = 1. We guess that for each k = 2,3,... all the numbers sigma(n^k)/n^k = Sum_{d|n^k} 1/d (n = 1,2,3,...) are pairwise distinct. See also A001157 for a similar conjecture. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..2000 EXAMPLE a(1) = 2 since 2 is the least prime greater than sigma(1^2)/1^2 = 1. a(2) = 5 since sigma(1^2)/1^2 = 1 and sigma(2^2)/2^2 = 7/4 are incongruent modulo the prime 5 > 2, but 1 is congruent to 7/4 modulo the prime 3. MATHEMATICA rMod[m_, n_]:=rMod[m, n]=Mod[Numerator[m]*PowerMod[Denominator[m], -1, n], n, -n/2] f[n_]:=f[n]=DivisorSigma[1, n^2]/n^2 Le[n_, m_]:=Le[m, n]=Length[Union[Table[rMod[f[k], Prime[m]], {k, 1, n}]]] Do[n=1; m=1; Label[aa]; If[m>PrimePi[n]&&Le[n, m]==n, Goto[bb], m=m+1; Goto[aa]]; Label[bb]; Print[n, " ", Prime[m]]; If[n<60, n=n+1; Goto[aa]]] CROSSREFS Cf. A000203, A000290, A001157, A065764. Sequence in context: A233565 A121359 A082087 * A256300 A023850 A175649 Adjacent sequences: A263314 A263315 A263316 * A263318 A263319 A263320 KEYWORD nonn AUTHOR Zhi-Wei Sun, Oct 14 2015 EXTENSIONS Definition corrected by Omar E. Pol, Oct 24 2015 STATUS approved
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Last modified April 10 06:23 EDT 2020. Contains 333392 sequences. (Running on oeis4.) | 971 | 2,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-16 | latest | en | 0.653596 |
http://www.waybuilder.net/sweethaven/ModElec/Basic_Electronics/ee/CircuitEE.asp?iNum=511 | 1,521,706,279,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647782.95/warc/CC-MAIN-20180322073140-20180322093140-00748.warc.gz | 495,031,812 | 1,264 | Given:R = 940 WC = 854.7 nFf = 490 Hz Find:XCZt qt
XC = 380 W
ZT = 1.0 kW
qT = -22.0° | 50 | 86 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | longest | en | 0.269959 |
https://gmatclub.com/forum/if-x-1-and-y-1-is-x-y-1-sqrt-x-y-2-sqrt-72350.html | 1,529,782,902,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865181.83/warc/CC-MAIN-20180623190945-20180623210945-00462.warc.gz | 625,662,673 | 45,361 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# If x>1 and y>1, is X>Y 1. sqrt (x) > y 2. sqrt
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If x>1 and y>1, is X>Y 1. sqrt (x) > y 2. sqrt [#permalink]
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02 Nov 2008, 14:35
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If x>1 and y>1, is X>Y
1. sqrt (x) > y
2. sqrt (y) < x
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02 Nov 2008, 20:22
[quote="bigfernhead"]If x>1 and y>1, is X>Y
1. sqrt (x) > y
2. sqrt (y) 1
and sqrt (x) > y then obviously x > y
however if sqrt y y
eg y=9 x=4
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Re: DS: [#permalink] 02 Nov 2008, 20:22
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# Problem P. 3869. (February 2006)
P. 3869. A sample of ideal gas is taken through the cyclic process shown in the figure. The temperature of the gas at state A is TA=200 K. At states B and C the temperature of the gas is the same.
a) What is the greatest temperature of the gas during the cyclic process?
b) Graph the cyclic process in the T-V diagram.
(4 pont)
Deadline expired on March 13, 2006.
### Statistics:
125 students sent a solution. 4 points: 54 students. 3 points: 30 students. 2 points: 2 students. 1 point: 22 students. 0 point: 8 students. Unfair, not evaluated: 9 solutions.
Problems in Physics of KöMaL, February 2006 | 227 | 763 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.858303 |
https://blog.csdn.net/evolone/article/details/46790203 | 1,521,744,821,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647901.79/warc/CC-MAIN-20180322170754-20180322190754-00048.warc.gz | 542,690,590 | 12,397 | # leetcode-202 Happy Number
217人阅读 评论(0)
Write an algorithm to determine if a number is “happy”.
A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
Example: 19 is a happy number
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1
class Solution {
public:
unordered_set<int> hashset;
bool isHappy(int n) {
hashset.insert(n);
int tmp=0;
while(n > 0)
{
int i = n%10;
n=n/10;
tmp = tmp + i*i;
}
if(1 == tmp)
return true;
else
{
if(hashset.find(tmp) != hashset.end())
return false;
else
isHappy(tmp);
}
}
};
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Author Comment/Response Elio Cecchetto 03/14/99 7:39pm I have the following differential equation to solve in the functions th[z] and phi[z].As first I used the EulerEquations to expand the differential equation in two of them. But then I can't introduce boundary conditions in two different points at the same time. phi[0]==th[0]==0, phi[L]==Pi/2 th[L]==0 where L,k1,k2,k3,Ea and V are all fixed parameters (k1 Cos [th[z]]^2+k3 Sin[th[z]]^2)/2 th'[z]^2+ 1/2 Cos[th[z]]^2(k2 Cos[th[z]]^2+k3 Sin[th[z]]^2)phi'[z]^2 - Ea V^2/(8Pi) Sin[th[z]]^2 == 0 Do anyone have any suggest? Probably the solution is trivial, but I do not get it! Elio URL: ,
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# Mention against each of the following whether regular or diffused reflection will take place when a beam of light strikes. Justify your answer in each case.(a) Polished wooden table (b) Chalk powder(c) Cardboard surface (d) Marble floor with water (e) Mirror (f) Piece of paper
Last updated date: 13th Jun 2024
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Hint Regular Reflection
1. Reflection from a polished surface is called regular reflection.
2. Parallel rays remain parallel after reflection.
Irregular Reflection:
1. Reflection from a rough surface is called diffuse reflection.
2. Parallel rays do not remain parallel after reflection.
Step by step solution
As we know that Regular reflection occurs at perfectly smooth surfaces and diffused reflection occurs at rough surfaces.
(a). Polished wooden table: Regular reflection - A polished surface is an example of a smooth surface. A polished wooden table has a smooth surface. Hence, reflections from the polished table will be regular.
(b). Chalk powder: Diffused reflection - Chalk powder spread on a surface is an example of an irregular surface. It is not smooth. Therefore, diffused reflection will take place from chalk powder.
(c). Cardboard surface: Diffused reflection - Cardboard surface is also an example of an irregular surface. Hence, diffused reflection will take place from a cardboard surface.
(d). Marble floor with water spread over it: Regular reflection - Marble floor with water spread over it is an example of a regular surface. This is because water makes the marble surface smooth. Hence, regular reflection will take place from this surface.
(e). Mirror: Regular reflection - Mirror has a smooth surface. Therefore, it will give a regular reflection.
(f). Piece of paper: Diffused reflection - Although a piece of paper may look smooth, but it has many irregularities on its surface. Due to this reason, it will give a diffused reflection.
Note
Reflection is an instance of reflecting; especially: the return of light or sound waves from a surface.
Regular reflection
Diffuse reflection: | 461 | 2,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-26 | latest | en | 0.87722 |
https://byjus.com/question-answer/state-whether-the-following-statement-is-true-or-false-cfrac-tan-theta-1-cot-theta/ | 1,721,256,265,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00478.warc.gz | 129,449,157 | 33,115 | 1
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Question
# State whether the following statement is true or false.tanθ1−cotθ+cotθ1−tanθ=1+tanθ+cotθ
A
True
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B
False
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Solution
## The correct option is A Truetanθ1−cotθ+cotθ1−tanθ=1+tanθ+cotθL.H.S=tanθ1−cotθ+cotθ1−tanθ =tanθ1−1tanθ+1tanθ1−tanθ =tan2θtanθ−1+1tanθ(1−tanθ) =tan3θtanθ(tanθ−1)−1tanθ(tanθ−1) =tan3θ−1tanθ(tanθ−1) =(tanθ−1)(tan2θ+1+tanθ)tanθ(tanθ−1) =tan2θ+1+tanθtanθ =tanθ+cotθ+1 =1+tanθ+cotθ =R.H.S∴ tanθ1−cotθ+cotθ1−tanθ=1+tanθ+cotθ ----- Hence proved
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label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Oct 3rd, 2015
Let's write down what we know.
There is 1 gallon of red paint, 1 gallon of blue paint, and 1 gallon of yellow paint.
Ann used 3/8 of the red paint, 1/4th of the blue paint, and 1/2 of the yellow.
Margie used 1/2 of the red paint, 5/8 of the blue paint, and 1/8 of the yellow.
Thus, we have have the equations
1=3/8+1/2+ r (r is the left over red paint)
1=1/4+5/8+b (b is the left over blue paint )
1=1/2+1/8+y (y is the left over yellow paint).
Now, we just have to solve for r, b, and y.
Step 1: Solve for r
1=3/8+1/2+r (let's multiply the whole equation by 8 to get rid of the denominators)
8=7+8r (subtract 7 from both sides)
1=8r (divide by 8 from both sides)
r=1/8
Therefore there is 1/8 gallon of red paint left.
Step 2: Solve for b
1=1/4+5/8+b (multiply the whole equation by 8)
8=7+8b (subtract seven from both sides)
1=8b (divide by 8 from both sides)
b=1/8
There there is 1/8 gallon of red paint left.
Step 3: Solve for y
1=1/2+1/8+y (multiply by 8 to the whole equation)
8=5+8y (subtract five from both sides)
3=8y (divide by 8 from both sides)
y=3/8
Therefore there is 3/8 gallons of yellow paint left.
Problem 2: Just add what's left of the yellow and blue paints together
1/8+3/8=4/8=1/2 gallons of paint.
Please let me know if you need any clarification. I'm always happy to answer your questions. Leave a positive review if you are happy!
Sep 15th, 2015
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Oct 3rd, 2015
...
Oct 3rd, 2015
Sep 22nd, 2017
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Since Differentiation is in the news, I thought it was time for us to investigate best practices in special education math classes. And as a counterpoint to that article, here is Grant Wiggins’ response.
Differentiation is widely accepted as being at the forefront of best practices in special education. But, when it comes to best practices in special education mathematics classrooms, not much has been clearly defined. Here’s one take about inclusion from a Huffington Post article at the beginning of the year:
One of these “best-teaching practices” is including visual cues to accompany any words in a worksheet or presentation of information. This idea is best represented in mathematics as mathematical models. The What Works Clearinghouse goes into detail about the importance of incorporating visual models of mathematical concepts into lessons for struggling students. The recommendation also illustrates how to use the concrete-representational-abstract approach in math instruction.
Concrete – Representational – Abstract
I’d like to explore how we can use this kind of instructional modeling during the problem solving process. Can we pose problems in the concrete-representational-abstract model for students with disabilities? The concrete stage would be physically representing the problem with a hands-on activity (which is another special education “best practice” and it will get it’s own blog post), but can we pose problems to students in the visual representation stage?
I attempted it and it got complicated when I told my colleagues our class focus for the trimester would be problem solving.
“Word problems?!?” They asked, “That’ll be hard since most of the students struggle with reading.”
“I didn’t say we were studying word problems.” I answered, “I said, we were going to be studying problem solving.”
“Yeah, right. Word problems.”
“Ok.”
When I think of problem solving these are examples of what comes to mind: Martin Gardner, James Tanton, Fawn Nguyen, and The Math Forum.
Not this
Can we create situations where students are solving problems, but not solving “word problems?” Matthew Peterson, from the MIND Research Institute says we can…
Solutions to mathematical problems can also be presented without words. Many mathematical proofs are accompanied with a visual representation. One of my favorites is the way Archimedes proved the approximate value of pi.
Proofs and the use of visuals have been written about by many general education teachers like Michael Pershan, Justin Aion and Joe Schwartz, just name a few.
But can my students investigate mathematical situations and stories without learning to decode word problems? Is decoding word problems something that they “need” to be learning? Dan Meyer says that they need to be perplexed
I enjoy perplexing my students and I think much more valuably they enjoy it also. Most of the time my students are pandered to or underestimated, so if I can provide one moment of mathematical perplexity I feel successful. Visual problems are one way that I can facilitate my students’ mathematical perplexity without the anxiety of also testing their reading abilities as well.
Here are some of the visual problems we worked on:
What do you notice? What do you wonder?
What do you notice? What do you wonder?
How much?
(I know the prices are in pounds, but for early elementary this is easily explained as British dollars and solved 1:1, or late elementary and early middle schoolers can investigate exchange rates.)
First Row: What do you notice? What do you wonder?
Second Row: How many sticks?
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4 - Data Explorer
I run a pet relocation company and we have many clients who hold their pet files with us in case of a future move. We need a formula that red flags when a vaccination expiry date is imminent (e.g. in 14 days, in 28 days, etc). Really struggling with how we can incorporate this into airtable.
1 Solution
Accepted Solutions
You’re welcome Kristy :slightly_smiling_face:
There are several ways to do so.
I personally would use the Flag in the formula. So it becomes something like:
`IF(DateTime_Diff({Vaccination Exp Date}, Today(), 'days')<=14,"⚠️ 14 Days or Less Till Vaccination Expiry", IF(DateTime_Diff({Vaccination Exp Date}, Today(), 'days')<=28,"28 Days or Less Till Vaccination Expiry")`
This works for 14 days and 28 days, if you want to add more warnings, you can add more IF statements.
Feel free to add emojis to your formula from here
Your screenshot was not attached btw.
BR,
Mo
4 Replies 4
13 - Mars
Welcome to Airtable Community! :slightly_smiling_face:
You can create a formula field that is something like
`DateTime_Diff({Vaccination Exp Date}, Today(), 'days')`
This will give you the number of days till vaccination expires. You can then create views (or add it in the formula if you want) that filters when this number is 28 or less, 14 or less, and so on.
Hope that helps. If it does, please mark this as a Solution so others can see it.
BR,
Mo
4 - Data Explorer
Hi
Thank you Mohamed. That sounds exactly what we need. How does the flag work? Do we need an additional field for a flag to show, or does it highlight the record. If the latter, how does it highlight it?
(Attachment image001.tiff is missing)
18 - Pluto
You can use nested `IF` statements with `DATETIME_DIFF` in a formula field:
``````IF(
DATETIME_DIFF({vaccineExpiryDate), TODAY(), 'days') < 0,
"vaccine overdue",
IF(
DATETIME_DIFF({vaccineExpiryDate), TODAY(), 'days') < 14,
"vaccine due very soon",
IF(
DATETIME_DIFF({vaccineExpiryDate), TODAY(), 'days') < 28,
"vaccine due soon"
)
)
)
``````
If you want the record to be associated with a color (and you have a pro subscription) you can use record coloring.
You’re welcome Kristy :slightly_smiling_face:
There are several ways to do so.
I personally would use the Flag in the formula. So it becomes something like:
`IF(DateTime_Diff({Vaccination Exp Date}, Today(), 'days')<=14,"⚠️ 14 Days or Less Till Vaccination Expiry", IF(DateTime_Diff({Vaccination Exp Date}, Today(), 'days')<=28,"28 Days or Less Till Vaccination Expiry")`
This works for 14 days and 28 days, if you want to add more warnings, you can add more IF statements.
Feel free to add emojis to your formula from here
Your screenshot was not attached btw.
BR,
Mo | 724 | 2,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.838524 |
http://courses.ncssm.edu/apb11o/labs/L120/L120A_equilibrium.htm | 1,498,578,097,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321458.47/warc/CC-MAIN-20170627152510-20170627172510-00214.warc.gz | 81,301,068 | 3,069 | Submitting assignments Schedules Course Info Ch. Reviews Guides Problems Labs Videos AP Info Announcements Contact Shortcuts WebAssign Canvas Equations Lab FAQ Software IWP
L120A. Equilibrium of Forces Challenge, Part A Complete Part A before the designated WebEx session. You'll do Part B during the WebEx session.
Part A. Introduction
As you read through this introductory material, there will be some questions to answer. Write your answers in WebAssign L120A.
If several forces act on an object that is either at rest or moving at constant velocity, then the forces must add to zero. That is, there is 0 net force. This is simply another way of stating Newton's 1st Law. If you know the magnitudes and directions of the forces, then you can add them as vectors to show that the result is zero.
Suppose three strings are tied together in a knot and then the strings are pulled in 3 different directions as shown in the photo below. If the knot is held motionless, then the strings are in equilibrium, and the forces should add to zero. We can show this to be true by using either graphical or component addition of the force vectors.
1. Let's represent the three forces by vectors A, B, and C as shown in the figure below.
1. On the PencilPad in WebAssign, redraw the three force vectors in a head-to-tail arrangement. Specifically, move the tail of vector B to the head of vector A and the tail of vector C to the head of vector B. Label the vectors.
2. In the text window below the PencilPad, write in a sentence how you can tell from the head-to-tail diagram whether the forces are in equilibrium. | 357 | 1,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-26 | longest | en | 0.911853 |
https://whatisconvert.com/788-feet-second-in-knots | 1,679,453,757,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943749.68/warc/CC-MAIN-20230322020215-20230322050215-00282.warc.gz | 718,419,229 | 7,379 | ## Convert 788 Feet/Second to Knots
To calculate 788 Feet/Second to the corresponding value in Knots, multiply the quantity in Feet/Second by 0.59248380129641 (conversion factor). In this case we should multiply 788 Feet/Second by 0.59248380129641 to get the equivalent result in Knots:
788 Feet/Second x 0.59248380129641 = 466.87723542157 Knots
788 Feet/Second is equivalent to 466.87723542157 Knots.
## How to convert from Feet/Second to Knots
The conversion factor from Feet/Second to Knots is 0.59248380129641. To find out how many Feet/Second in Knots, multiply by the conversion factor or use the Velocity converter above. Seven hundred eighty-eight Feet/Second is equivalent to four hundred sixty-six point eight seven seven Knots.
## Definition of Foot/Second
The foot per second (plural feet per second) is a unit of both speed (scalar) and velocity (vector quantity, which includes direction). It expresses the distance in feet (ft) traveled or displaced, divided by the time in seconds (s, or sec). The corresponding unit in the International System of Units (SI) is the metre per second. Abbreviations include ft/s, ft/sec and fps, and the rarely used scientific notation ft s−1.
## Definition of Knot
The knot is a unit of speed equal to one nautical mile (1.852 km) per hour, approximately 1.151 mph. The ISO Standard symbol for the knot is kn. The same symbol is preferred by the IEEE; kt is also common. The knot is a non-SI unit that is "accepted for use with the SI". Worldwide, the knot is used in meteorology, and in maritime and air navigation—for example, a vessel travelling at 1 knot along a meridian travels approximately one minute of geographic latitude in one hour. Etymologically, the term derives from counting the number of knots in the line that unspooled from the reel of a chip log in a specific time.
## Using the Feet/Second to Knots converter you can get answers to questions like the following:
• How many Knots are in 788 Feet/Second?
• 788 Feet/Second is equal to how many Knots?
• How to convert 788 Feet/Second to Knots?
• How many is 788 Feet/Second in Knots?
• What is 788 Feet/Second in Knots?
• How much is 788 Feet/Second in Knots?
• How many kt are in 788 ft/s?
• 788 ft/s is equal to how many kt?
• How to convert 788 ft/s to kt?
• How many is 788 ft/s in kt?
• What is 788 ft/s in kt?
• How much is 788 ft/s in kt? | 609 | 2,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-14 | latest | en | 0.89055 |
https://millenia.cars.aps.anl.gov/pipermail/ifeffit/2005-July/001893.html | 1,652,711,667,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510138.6/warc/CC-MAIN-20220516140911-20220516170911-00595.warc.gz | 465,130,461 | 2,706 | # [Ifeffit] rbkg
Matt Newville newville at cars.uchicago.edu
Mon Jul 11 22:36:47 CDT 2005
```Silvio,
How Autobk / spline() works:
First, the definition of a spline: A spline is a flexible
function that is made up from a series of polynomial segments.
Each polynomial is 4th order (ie, a 'cubic polynomial' a + b*x +
c*x*x + d*x*x*x). The polynomial segments are joined together
such that the value are first two derivatives are continuous
across the joint (also known as a 'knot'). With this
definition, and a few constraints about the endpoints, a
flexible, smooth function can be defined simply defining the y
values at the knots, and the flexibility of the function is
determined by the number and placement of the knots.
Autobk / spline() uses such a spline with Nbkg knots evenly
spaced in k-space to approximate mu0(E), where
Nbkg = 1 + (2*Delta K * Rbkg )/ pi .
where Delta_K is the k-range of the data (or to be considered,
as using kmin=~0.5 is common and sometimes you specify kmax).
In ifeffit, Nbkg is truncated, not rounded.
That gives the *number* of variables, quantifying the
flexibility of mu0(E).. The values of the mu0(E) spline at the
Nbkg knots are then optimized in a fitting procedure. The
criteria for "best values" is that
chi(R) = FT{chi(k)=[mu(k)-mu0(k)]/step}
be "as small as possible" between 0 and Rbkg.
[When using a standard spectra, 'as small as possible' is
replaced with 'as close to the the standard as possible'].
The fit procedure goes something like this:
guess values of mu0(E_i) for i = 1, N_bkg knots
subtract this background
FT this result
generate sum of squares chi(R) between 0 and R_bkg
refine values of mu0(E_i) and start over
So, Rbkg has two purposes:
a) set the flexibility of mu0(E)
b) set the highest R component to consider in the optimization.
In signal processing, Rbkg is commonly called a Nyquist
frequency. You can also think of it as the frequencey of a
high-pass filter It isn't perfectly sharp, so there can be
changes in spectral content around R=Rbkg with small tweaks to
Rbkg.
I highly recommend playing wtih Rbkg with some decent model
compound data (where you know the first shell species). Try
setting Rbkg=0.2, 1.0, 2.0, and 5.0. With Athena, you can
nicely clone a group and plot all the results together in E, k,
and R-spaces. You'll see that you're removing low frequency
components, that Rbkg really is (roughly) the cutoff, and why
you don't want to use Rbkg=5.0!
Hope that clears up most questions,
--Matt
``` | 703 | 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} | 3.28125 | 3 | CC-MAIN-2022-21 | latest | en | 0.915075 |
https://www.aol.com/article/2011/11/12/do-these-energy-companies-pass-buffetts-test/20104982/ | 1,498,580,890,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321458.47/warc/CC-MAIN-20170627152510-20170627172510-00515.warc.gz | 832,611,242 | 32,298 | # Do These Energy Companies Pass Buffett's Test?
We'd all like to invest like the legendary Warren Buffett, turning thousands into millions or more. Buffett analyzes companies by calculating return on invested capital (ROIC) to help determine whether a company has an economic moat -- the ability to earn returns on its money above that money's cost.
ROIC is perhaps the most important metric in value investing. By determining a company's ROIC, you can see how well it's using the cash you entrust to it and whether it's actually creating value for you. Simply, ROIC divides a company's operating profit by how much investment it took to get that profit. The formula:
ROIC = Net operating profit after taxes / Invested capital
You can get further detail on the nuances of the formula.
This one-size-fits-all calculation cuts out many of the legal accounting tricks, such as excessive debt, that managers use to boost earnings numbers, and it provides you with an apples-to-apples way to evaluate businesses, even across industries. The higher the ROIC, the more efficiently the company uses capital.
Ultimately, we're looking for companies that can invest their money at rates that are higher than the cost of capital, which for most businesses is between 8% and 12%. We prefer to see ROIC above 12% at a minimum, along with a history of increasing returns, or at least steady returns, which indicate some durability to the company's economic moat.
Let's look at Hess (NYS: HES) and three of its industry peers, to see how efficiently they use cash. Here are the ROIC figures for each company over a few periods.
Company
TTM
1 Year Ago
3 Years Ago
5 Years Ago
Hess8.6%10.3%16.1%12%
Eni (NYS: E) 6.5%7.2%17.3%17.9%
BP (NYS: BP) 9.5%(5%)*14.8%12.7%
Source: S&P Capital IQ.
*Because BP did not report an effective tax rate, we used its 35% rate for three years ago.
** Because APC did not report an effective tax rate, we used a 35% effective tax rate.
Hess' returns on invested capital are lower than they were five years ago and have consistently declined over the past three years. The other companies have also seen declines in their ROIC over the same time period. This correlation shouldn't be so surprising, given the commodity nature of this business.
Businesses with consistently high ROIC show that they're efficiently using capital. They also have the ability to treat shareholders well, because they can then use their extra cash to pay out dividends to us, buy back shares, or further invest in their franchise. Warren Buffett has long loved healthy and growing dividends -- and you should, too.
So for more successful investments, dig a little deeper than the earnings headlines to find the company's ROIC. If you'd like, you can add these companies to your Watchlist:
At the time this article was published Jim Royal, Ph.D., owns no shares of any company mentioned here. Try any of our Foolish newsletter servicesfree for 30 days. We Fools don't all hold the same opinions, but we all believe thatconsidering a diverse range of insightsmakes us better investors. The Motley Fool has adisclosure policy. | 710 | 3,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-26 | latest | en | 0.957249 |
https://wirings-diagram.com/lionel-train-wiring-diagram/lionel-train-zw-transformers-wiring-diagram-wiring-diagram-lionel-train-wiring-diagram/ | 1,627,755,723,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154099.21/warc/CC-MAIN-20210731172305-20210731202305-00669.warc.gz | 606,670,419 | 26,497 | # Lionel Train Zw Transformers Wiring Diagram | Wiring Diagram – Lionel Train Wiring Diagram
Lionel Train Zw Transformers Wiring Diagram | Wiring Diagram – Lionel Train Wiring Diagram
Lionel Train Wiring Diagram – lionel train engine wiring diagram, lionel train wiring diagram, Every electric structure is made up of various unique parts. Each part should be placed and connected with other parts in specific manner. If not, the structure will not function as it should be. So as to be certain that the electric circuit is built correctly, Lionel Train Wiring Diagram is needed. How does this diagram help with circuit construction?
The diagram provides visual representation of an electric structure. However, the diagram is a simplified version of the arrangement. This makes the procedure for assembling circuit easier. This diagram provides advice of circuit components as well as their own placements.
## Components of Lionel Train Wiring Diagram and A Few Tips
There are two things that will be present in any Lionel Train Wiring Diagram. The first component is emblem that indicate electrical component in the circuit. A circuit is generally composed by many components. The other thing which you will locate a circuit diagram could be traces. Lines in the diagram show exactly how each element connects to one another.
The rankings of circuit’s parts are comparative, not accurate. The arrangement is also not plausible, unlike wiring schematics. Diagram only reveals where to put component in a spot relative to other components within the circuit. Even though it is simplified, diagram is a fantastic basis for everyone to build their own circuit.
One thing that you must learn before studying a circuit diagram is your symbols. Every symbol that’s shown on the diagram shows specific circuit component. The most common elements are capacitor, resistor, and battery. Additionally, there are other components such as ground, switch, motor, and inductor. It all depends on circuit that is being built.
As stated previous, the traces at a Lionel Train Wiring Diagram represents wires. Occasionally, the wires will cross. But, it does not mean link between the wires. Injunction of 2 wires is usually indicated by black dot on the intersection of two lines. There’ll be primary lines that are represented by L1, L2, L3, and so on. Colors can also be used to differentiate cables.
Ordinarily, there are two chief kinds of circuit links. The primary one is known as string connection. It is the easier type of connection because circuit’s elements are placed within a specified line. Because of the electric current in each and every component is similar while voltage of the circuit is total of voltage in each component.
Parallel relationship is more complex compared to series one. Unlike in string connection, the voltage of every component is similar. It’s because the element is directly linked to power resource. This circuit consists of branches that are passed by distinct electric current amounts. The current joins together when the branches match.
There are several things that an engineer needs to look closely at when drawing wirings diagram. To start with, the symbols used in the diagram ought to be precise. It must represent the specific element necessary to build an intended circuit. After the symbol is wrong or unclear, the circuit will not function as it is supposed to.
It is also highly suggested that engineer brings favorable supply and damaging source symbols for better interpretation. Usually positive supply symbol (+) is located above the line. Meanwhile, the negative supply emblem is set under it. The current flows from the left to right.
In addition to this, diagram drawer is suggested to limit the number of line crossing. The line and component placement ought to be designed to decrease it. But if it is inevitable, use universal symbol to indicate whether there is a junction or if the lines aren’t really connected.
As you can see drawing and interpreting Lionel Train Wiring Diagram may be complicated undertaking on itself. The advice and suggestions which have been elaborated above should be a fantastic kick start, however. Lionel Train Wiring Diagram
Lionel Train Zw Transformers Wiring Diagram | Wiring Diagram – Lionel Train Wiring Diagram Uploaded by Hadir on Friday, February 15th, 2019 in category Wiring Diagram.
Here we have another image Lionel Train Wiring Diagram 38 | Wiring Diagram – Lionel Train Wiring Diagram featured under Lionel Train Zw Transformers Wiring Diagram | Wiring Diagram – Lionel Train Wiring Diagram. We hope you enjoyed it and if you want to download the pictures in high quality, simply right click the image and choose "Save As". Thanks for reading Lionel Train Zw Transformers Wiring Diagram | Wiring Diagram – Lionel Train Wiring Diagram. | 910 | 4,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.91691 |
http://spcchartsonline.com/index.php/online-spc-control-charts/spc-control-charts/basic-statistics-and-tables/ | 1,675,241,418,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00209.warc.gz | 41,270,049 | 11,492 | # Basic Statistics
### Mean
The mean of a set of numbers is the arithmetic average of that set of numbers, as given by the formula below:
$$\Large{\left(\frac{1}{N}\right)\sum_{i=1}^N X_i = \frac{(X_1 + X_2 + \cdots + X_N)}{N}}$$
Example for a set of eight numbers:
{6,3,12,15,4,9,4,8}
the mean value is (6 +3 +12 + 15 + 4 + 9 + 4 + 8)/8 = (61)/8 = 7 5/8
Unlike the median calculation below, it makes no difference if the number of data points is an even number or an odd number.
### Median
The median of a set of numbers is found by ordering the set in numeric order (sorting it by increasing value), and in the case an odd number of data points, choosing the center value of the ordered list. In the case of an even number of data points, the middle two data points would be averaged together to produce the median.
Example for an odd numbered set:
{6,3,12,4,9,4,8}
would produce the ordered set (3,4,4,6,8,9,12}, and the center point of that ordered set is element [3], which as the value 6. So the median value is 6.
Example for an even numbered set:
{6,3,12,15,4,9,4,8}
would produce the ordered set (3,4,4,6,8,9,12,15}, and the center point of that ordered set falls between element [3] and element [4], which are the values 6 and 8 respectively. So the median value is (6+8)/2 = 7. Note that the median value does not actually have to be in the original set of numbers.
### Standard Deviation
The standard deviation of a set of numbers is a measure of the dispersion of the data about the mean of the data. There are two similar formulas used for the calculation of the standard deviation. Which one is used depends on whether or not the set of data values is the total population of data values represented by the process being measured. The one which uses the total population of data values uses the formula:
$$\Large {S = \sqrt{\frac{\sum_{i=1}^N X_i – \bar X}{N} }}$$
But, in SPC Charts, the data being processed does not represent the total population of values. Instead the data represents a sampled subset of possible values. In this case a different formula is used, referred to as the sample standard deviation. It also commonly referred to as the unbiased standard deviation for a sample population.
$$\Large {S = \sqrt{\frac{\sum_{i=1}^N X_i – \bar X}{N-1} }}$$
Note that the difference between the two formulas is the (N) versus (N-1) in the denominator of the fraction.
Every calculation for a standard deviation in this software uses the sample standard deviation formula. | 674 | 2,516 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.71875 | 5 | CC-MAIN-2023-06 | longest | en | 0.885939 |
https://www.hsslive.co.in/2021/10/1-gaj-in-marla-bihar.html | 1,713,631,976,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00673.warc.gz | 751,371,245 | 40,706 | # HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board
## 1 Gaj in Marla in Bihar
In this article, you will learn how to convert 1 Gaj in Marla in Bihar. Gaj is one of the most commonly used measurement units used in almost every state of India. People in Indian states refer to gaj for measurement of land. Here in this page you will learn how to calculate 1 Gaj in Marla unit.
## How to Convert from 1 Gaj to Marla?
You can easily convert gaj to Marla or the reverse with a simple method. You can multiply the figure in gaj by 0.03274 to determine the Marla value. This is all you need to do for undertaking the conversion procedure of Gaj to Marla in Bihar.
## Relationship between Gaj and Marla
It is not difficult to work out the actual relationship between gaj to Marla in most cases. You should know the proper calculations existing between gaj to Marla before you venture to conversion procedures. 1 Gaj to Marla calculations will be 0.03274 Marla.
Once you know these calculations, it will not be hard for you to calculate gaj in an Marla. One gaj can be worked out to various other units as well.
## Formula for Converting Gaj to Marla
The formula to convert Gaj to Marla is the following-
1 Gaj = 0.03274 Marla
Conversely, you can also use this formula-
Gaj = Marla * 0.03274
This is all you need to know for undertaking the conversion procedure on your part. You can also use online calculators or conversion tools for this purpose.
### Other Plot Areas in Marla in Bihar
• 1 Gaj in Marla in Bihar = 0.03274
• 2 Gaj in Marla in Bihar = 0.06548
• 3 Gaj in Marla in Bihar = 0.09822
• 4 Gaj in Marla in Bihar = 0.13096
• 5 Gaj in Marla in Bihar = 0.1637
• 6 Gaj in Marla in Bihar = 0.19644
• 7 Gaj in Marla in Bihar = 0.22918
• 8 Gaj in Marla in Bihar = 0.26192
• 9 Gaj in Marla in Bihar = 0.29466
• 10 Gaj in Marla in Bihar = 0.3274
• 15 Gaj in Marla in Bihar = 0.4911
• 20 Gaj in Marla in Bihar = 0.6548
• 25 Gaj in Marla in Bihar = 0.8185
• 30 Gaj in Marla in Bihar = 0.9822
• 35 Gaj in Marla in Bihar = 1.1459
• 40 Gaj in Marla in Bihar = 1.3096
• 45 Gaj in Marla in Bihar = 1.4733
• 50 Gaj in Marla in Bihar = 1.637
• 55 Gaj in Marla in Bihar = 1.8007
• 60 Gaj in Marla in Bihar = 1.9644
• 65 Gaj in Marla in Bihar = 2.1281
• 70 Gaj in Marla in Bihar = 2.2918
• 75 Gaj in Marla in Bihar = 2.4555
• 80 Gaj in Marla in Bihar = 2.6192
• 85 Gaj in Marla in Bihar = 2.7829
• 90 Gaj in Marla in Bihar = 2.9466
• 95 Gaj in Marla in Bihar = 3.1103
• 100 Gaj in Marla in Bihar = 3.274
## FAQs About 1 Gaj in Marla in Bihar
#### Q: Ek Gaj main Kitne Marla in Bihar Hote hain?
Ek Gaj main 0.03274 Marla in Bihar hote hain.
#### 1 Gaj main kitna Marla in Bihar Hota Hain?
1 Gaj main 0.03274 Marla in Bihar hote hain
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Copyright © HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board About | Contact | Privacy Policy | 964 | 2,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-18 | latest | en | 0.866451 |
https://tioj.ck.tp.edu.tw/problems/1850 | 1,582,424,922,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145742.20/warc/CC-MAIN-20200223001555-20200223031555-00035.warc.gz | 604,346,138 | 4,188 | 75.0% (3/4)
58.3% (14/24)
Sample Input 1
2 2
Sample Input 2
3 3
# Sample Output
Sample Output 1
1001
Sample Output 2
020110120210221112122200010
# Hints
$13000|65000$, 所以他真的沒大腦了。
# Problem Source
De Bruijn Sequence
No. Testdata Range Score
1 0 16
2 1 12
3 2 12
4 3 12
5 4 12
6 5 12
7 6 12
8 7 12
# Testdata and Limits
No. Time Limit (ms) Memory Limit (KiB) Output Limit (KiB) Subtasks
0 3000 131072 262144 1
1 3000 131072 262144 2
2 3000 131072 262144 3
3 3000 131072 262144 4
4 3000 131072 262144 5
5 3000 131072 262144 6
6 3000 131072 262144 7
7 3000 131072 262144 8 | 280 | 579 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-10 | latest | en | 0.292314 |
https://socratic.org/questions/how-do-you-solve-12-5-x-72 | 1,575,972,809,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527205.81/warc/CC-MAIN-20191210095118-20191210123118-00118.warc.gz | 538,344,878 | 5,769 | # How do you solve 12( 5- x ) = 72?
Apr 13, 2018
$x = - 1$
$60 - 12 x = 72$
$- 12 x = 72 - 60$
$- 12 x = 12$
$x = - 1$ | 69 | 120 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-51 | latest | en | 0.582573 |
http://mathhelpforum.com/calculus/141674-volumes-rotate-about-y-axis.html | 1,529,717,586,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864848.47/warc/CC-MAIN-20180623000334-20180623020334-00606.warc.gz | 204,072,876 | 9,661 | 1. ## volumes rotate about y axis
The region of the curve $\displaystyle y = \frac{e^x}{x}$ bounded by x = 1 and x = 2 and y = 0, is rotated about the y axis. Find the volume of the solid.
i drew the diagram and it doesn't look doable. Would I have to consider volumes by slicing?
thanks
2. It certainly is "doable". The region in the xy-plane is a "quadrilateral with one curved side, vertices at (1, 0), (2, 0), $\displaystyle (2, e^4/2)$ and (1, e). There would be a problem with doing it as "washers" since there would be a change of formula at (1, e).
If you "slice" it parallel to the y-axis, rotating around the y-axis, each "slice" would be a cylinder, of thickness dx, and height the distance from y= 0 to $\displaystyle y= e^x/x$ which is $\displaystyle e^x/x$.
Since this is rotated around the x-axis, the radius of each cylinder is x and the area is $\displaystyle \pi r^2h= \pi x^2(e^x/x)= \pi xe^x$. The volume of each thin cylinder is $\displaystyle \pi x e^x dx$ and the entire volume is given by the integral
$\displaystyle \pi \int_1^2 x e^x dx$. | 321 | 1,070 | {"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} | 3.5625 | 4 | CC-MAIN-2018-26 | latest | en | 0.926471 |
http://www.physicsforums.com/showthread.php?t=303950 | 1,386,315,882,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163050081/warc/CC-MAIN-20131204131730-00002-ip-10-33-133-15.ec2.internal.warc.gz | 491,739,500 | 10,552 | by KFC
Tags: ising, model
P: 369 I am learning the 1D ising model (spin 1/2), without external field and considering the nearest site interaction, the hamiltonian for 1D chain is simple $$H = -J\sum_i S_iS_{i+1}$$ Since each spin can only take either +1 or -1, we can write the transition matrix as $$T = \left( \begin{matrix} e^{K} & e^{-K} \\ e^{-K} & e^{K} \end{matrix} \right)$$ where $$K=\beta J$$ Now I try to learn 2D case, I read some book on it but seems quite complicated, so I started with the simplest case (no external field, only nearest interaction, rectangular lattics with only 2 rows and N columns and perodic boundary condition). The hamiltonian is $$H = -J\sum_{i=1}^{2}\sum_{j=1}^N S_{ij}S_{i, j+1}$$ right? What I am really consfuing is how to find the transition matrix? Now each site has four nearest neighbor (of course, to avoid double counting, we only need to count two one at a time, let's say we count the one next to and below the current site) and each spin can take 2 values, so what's the dimension of the transition matrix? and what does $$T_{ij}$$ means?
P: 369 Here is my idea. Since we have to sum over all possible sites, so consider the site (1, 1), and (2, 1) (i.e. the current site and the site below it), the energy is $$-J\left[\left(S_{11}S_{12}+S_{11}S_{21}\right) + \left(S_{21}S_{22} + S_{21}S{11}\right)\right]$$ All other sites has the same case as this specific site (1, 1), and in this case, we only need to consider four spin $$S_{11}, S_{12}, S_{21}, S_{22}$$. Note that each of them can take two values so there are totally 16 possible values. The transition matrix would be 4x4, right?
Sci Advisor P: 5,396 Your Hamiltonian is incorrect- it's missing a term like S_ij S_i+1,j .The 2-D ising model is significantly more complicated than the 1-D. There's a brief treatment in Landau and Lifgarbagez, vol. 5 (pp 498-506) and Chaikin and Lubensky "Principles of condensed matter physics" has additional detail along with the O_n model.
P: 369
Quote by Andy Resnick Your Hamiltonian is incorrect- it's missing a term like S_ij S_i+1,j .The 2-D ising model is significantly more complicated than the 1-D. There's a brief treatment in Landau and Lifgarbagez, vol. 5 (pp 498-506) and Chaikin and Lubensky "Principles of condensed matter physics" has additional detail along with the O_n model.
Thanks. I will check
P: 369
Quote by Andy Resnick Your Hamiltonian is incorrect- it's missing a term like S_ij S_i+1,j .The 2-D ising model is significantly more complicated than the 1-D. There's a brief treatment in Landau and Lifgarbagez, vol. 5 (pp 498-506) and Chaikin and Lubensky "Principles of condensed matter physics" has additional detail along with the O_n model.
I checked it, now the Hamiltonian is modified as
$$H = -J\sum_{i=1}^{2}\sum_{j=1}^N S_{ij}S_{i, j+1} + S_{ij}S_{i+1,j}$$
So, the for site (i, j), the energy could take 16 possible values. For
$$S_{ij}, S_{i+1, j}, S_{i, j+1}, S_{i+1, j+1} \Longrightarrow E$$
++++: 4
+++-: 2
++-+: 2
++--: 0
+-++: 2
+-+-: 0
+--+:-2
+---: -2
-+++: 2
-+-+: 0
-+--: -2
--++: 0
--+-: -2
---+: -2
----: 4
So could I just use this values to setup the transition matrix?
P: 981 The exact solution of the 2D Ising model was done by Onsager. See here for a brief outline: http://www.nyu.edu/classes/tuckerman..._26/node2.html
P: 369
Quote by genneth The exact solution of the 2D Ising model was done by Onsager. See here for a brief outline: http://www.nyu.edu/classes/tuckerman..._26/node2.html
Thanks. I read that before, but I found this quite confusing.
The above outline is about n x n lattices, now let's apply his result to 2 x n, we get the matrix element
$$T_{jk} = \exp\left[\beta J\left(\sigma_{1j}\sigma_{1k} + \sigma_{1j}\sigma_{2j} + \sigma_{2j}\sigma_{2k} + \sigma_{2j}\sigma_{1j}\right)\right]$$
For the last term, we apply the periodic boundary condition.
My doubt is $$\sigma_{ij}$$ is the eigenvalue and could be -1 or +1, so each terms in above expression couble be either +1 or -1, how come do we get a certain value for specific matrix element $$T_{jk}$$ ?
P: 2,141 Solvng the 2D Ising model is almost trivial by transforming it to a close packed dimer model. The dimer model can be solved using very simple combinatorial techiques, you don't need complicated transfer matrix techniques. See e.g. here: http://arxiv.org/abs/cond-mat/0212363
P: 369
Quote by Count Iblis Solvng the 2D Ising model is almost trivial by transforming it to a close packed dimer model. The dimer model can be solved using very simple combinatorial techiques, you don't need complicated transfer matrix techniques. See e.g. here: http://arxiv.org/abs/cond-mat/0212363
Thank you very much. I will read that later.
I am learning this problem because there is one chapter about transfer matrix in my text and this method is useful in some other place. So I want to learn it by studying 2D ising model as an example.
P: 2,141
Quote by KFC Thank you very much. I will read that later. I am learning this problem because there is one chapter about transfer matrix in my text and this method is useful in some other place. So I want to learn it by studying 2D ising model as an example.
Then you should read this book:
http://tpsrv.anu.edu.au/Members/baxter/book
You can download it free of charge. The transfer matrix technique that most textbooks explain for solving the Ising model is of no use for most other models. The Ising model is a so-called "free fermion model", the transfer matrix can then be diagonalized using a Bogoliubov transformation. This won't work for the vast class of integrable models.
So, if you want to learn about solving models, you should learn about the Bethe Ansatz, the Yang-Baxter equation etc. etc. This is explained in detail in the book by Baxter.
P: 527 Using the 2D Ising model to learn about the Transfer matrix is probably not the best approach. First, the Transfer Matrices are somewhat obscure (although certainly doable), and second, it doesn't even get you half way there in solving the model (i.e. you still need perform some other steps as well - steps that are quite specific to the Ising model). But just to throw in another book, I know that this one solves the Ising model using Transfer matrices: http://www.amazon.com/Equilibrium-St.../dp/9810216424 (Chapter 5.1)
P: 369 ok. Thank you for all your help.
P: 8 Is the Onsager Solution extendable to 3D ? is it do-able in principle ?
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## How many times can you subtract 6 from 60 riddle?
Thus, you cannot subtract 6 again from 42. 7 times, because 42 divided by 6 is 7. After you subtract 6, 7 times from 42, then there is nothing left to subtract from.
Infinite : It could be infinite because we can subtract 4 from 44 any number of times and in each attempt the number 44will remain same. 2. Eleven : If we subtract 4 from 44 it would result 40. Then subtracting 4 from 40 results 36.
How many times can you subtract the number 4 from 40?
So, you can subtract 5 from 35 seven times.
It is once, because after you subtract 6 from 42 once, you are left with 36. Thus, you cannot subtract 6 again from 42. 7 times, because 42 divided by 6 is 7.
How many times can you take away 9 from 27?
Thus, you cannot subtract 6 again from 60. 10 times, because 60 divided by 6 is 10.
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Q: 30% off 75
A: 30% off of 75 is 52.5. 75 - (30% of 75); = 75 - (0.3 * 75); = 75 - 22.5; = 52.5
Question
Asked 9/7/2011 8:21:51 AM
Updated 1/4/2015 12:24:30 PM
Rating
3
30% off of 75 is 52.5.
75 - (30% of 75);
= 75 - (0.3 * 75);
= 75 - 22.5;
= 52.5
Added 1/4/2015 12:24:30 PM
This answer has been confirmed as correct and helpful.
Confirmed by yumdrea [1/4/2015 12:36:27 PM]
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https://www.hackmath.net/en/calculator/fraction?input=5+3%2F4 | 1,631,848,424,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780054023.35/warc/CC-MAIN-20210917024943-20210917054943-00495.warc.gz | 811,431,548 | 10,580 | # Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 53/4 = 23/4 = 5 3/4 = 5.75
Spelled result in words is twenty-three quarters (or five and three quarters).
### How do you solve fractions step by step?
1. Conversion a mixed number 5 3/4 to a improper fraction: 5 3/4 = 5 3/4 = 5 · 4 + 3/4 = 20 + 3/4 = 23/4
To find new numerator:
a) Multiply the whole number 5 by the denominator 4. Whole number 5 equally 5 * 4/4 = 20/4
b) Add the answer from previous step 20 to the numerator 3. New numerator is 20 + 3 = 23
c) Write a previous answer (new numerator 23) over the denominator 4.
Five and three quarters is twenty-three quarters
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Dividends
The three friends divided the win by the invested money. Karlos got three-eighths, John 320 permille, and the rest got Martin. Who got the most and which the least?
• Buing
Brother got to buy 240 CZK and could buy for 1/8 what he wanted. Could he pay the rest of the purchase for 200 CZK?
• Simplest form of a fraction
Which one of the following fraction after reducing in simplest form is not equal to 3/2? a) 15/20 b) 12/8 c) 27/18 d) 6/4
• Torque
Torque and Mari each multiplied 1/8 inch times 5/8 inch. Tartaric 5/8 squares point inches. And Marie got 5/64 squared thought inches tall. Which student found a corrupt area?
• Ten fractions
Write ten fractions between 1/3 and 2/3
• Paper collecting
At the paper collecting contest gathered Franta 2/9 ton, Karel 1/4 ton, and Patrick 19/36 tons of paper. Who has gathered the most and the least?
• Comparing and sorting
Arrange in descending order this fractions: 2/7, 7/10 & 1/2
• Stones in aquarium
In an aquarium with a length 2 m; width 1.5 m and a depth of 2.5 m is a water level up to three-quarters of the depth. Can we place stones with a volume of 2 m3 into the aquarium without water being poured out?
• Equivalent fractions
Are these two fractions equivalent -4/9 and 11/15?
• Turtles 2
A box turtle hibernates in the sand at 11 5/8. A spotted turtle hibernates at 11 16/25 feet. Which turtle is deeper? Write answer as number 1 or 2.
• Giraffes to monkeys
The ratio of the number of giraffes to the number of monkeys in a zoo is 2 to 5. Which statement about the giraffes and monkeys could be true? A. For every 10 monkeys in the zoo, there are 4 giraffes. B. For every giraffe in the zoo, there are 3 monkeys.
• Arrange
Arrange the following in descending order: 0.32, 2on5, 27%, 1 on 3
• Evaluate mixed expressions
Which of the following is equal to 4 and 2 over 3 divided by 3 and 1 over 2? A. 4 and 2 over 3 times 3 and 2 over 1 B. 14 over 3 times 2 over 7 C. 14 over 3 times 7 over 2 D. 42 over 3 times 2 over 31 | 1,620 | 5,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | longest | en | 0.839552 |
http://dailydoseofexcel.com/archives/2011/08/10/overlapping-dates/ | 1,718,816,060,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861828.24/warc/CC-MAIN-20240619154358-20240619184358-00351.warc.gz | 7,690,910 | 14,765 | # Overlapping Dates
Given a Bill Date and a Cycle Days, Kimberly wants to determine how many days in the cycle fall in Winter. For our purposes, Winter is defined as November 1st to April 30th. I put the season start dates in D1 and D2 and use this formula to find the difference.
=MIN(C5,B5-IF(B5>\$D\$2,DATE(YEAR(B5),MONTH(\$D\$2),DAY(\$D\$2)),IF(B5>\$D\$1,DATE(YEAR(B5),MONTH(\$D\$1),DAY(\$D\$1)),DATE(YEAR(B5)-1,MONTH(\$D\$2),DAY(\$D\$2)))))
It’s not as onerous as it might seem at first. Let’s start with the case of Bill Date after November 1, but before January 1. If we were restricted to that time period, the formula would be
=MIN(C5,B5-DATE(YEAR(B5),11,1))
Take the date in question and subtract the start of Winter. Take the smaller of that or the cycle time. Once you have that basic formula, it’s easy to expand it. There are three time frames that I care about: pre-May 1, post-Nov 1, and the time in between. If it’s after November 1, I use the formula above. If it’s after May 1, I use the formula above but use May 1 as the date. If it’s before May 1, I use November again, but the year prior. If the Cycle Days spans both dates, well, there’s trouble, so let’s call that a bug that we don’t care to fix.
You might think those days are one off. They match some sample data I was provided, so I assume it’s good to go. If you want the actual number of days from 1-Nov-2011 to 16-Nov-2011, you need to add 1 to the above formula.
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## One thought on “Overlapping Dates”
1. Reuvain says:
Dick, It seems that the formula is not working for row 7. It is returning the number of cycle days NOT in winter instead of the number of cycle days that ARE in winter.
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# Calculation of banking required reserves
Suppose the banking system's nonborrowed reserves total \$48.3 billion, with total legal reserves standing at \$51.2 billion. What must borrowed reserves be? This morning the Federal Reserve decided to undertake the sale of \$500 million in government securities through open market operations. What will the new level of nonborrowed reserves? If interest rates do not change, what will be the new level of total reserves? What must you assume to make this calculation? If interest rates do change, which way are they likely to move?
#### Solution Preview
Nonborrowed reserves consist of total reserves (member bank deposits in Federal Reserve Banks, plus vault cash) less funds borrowed (borrowed reserves) at the Federal Reserve Discount Window. So borrowed reserves are total reserves minus nonborrowed reserves, or 51.2 - 48.3 = \$2.9 billion. Open market operations allow the Federal Reserve to adjust directly the supply of nonborrowed reserves in the banking system. To increase reserves, the Federal Reserve ...
#### Solution Summary
Calculation of banking required reserves and relationship between the Federal funds rate and banking reserves.
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# What is this fraction lowest term 21 over 27?
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7/21 = 1/3
### What is 21 over 49 in lowest term?
21 over 49 in lowest terms is 3/7 | 315 | 983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-38 | latest | en | 0.923045 |
https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-4-number-representation-and-calculation-chapter-4-test-page-246/18 | 1,576,416,107,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541308149.76/warc/CC-MAIN-20191215122056-20191215150056-00504.warc.gz | 721,724,434 | 12,325 | ## Thinking Mathematically (6th Edition)
${{250}_{\text{six}}}$.
To multiply numerals of same bases other than base ten,multiply unit digits first according to base 10 then convert them to their respective base. Then continue the same process for other digits.Solve the provided numerals as follows: \begin{align} & {{4}_{\text{six}}}\,\times {{3}_{\text{six}}}\,={{12}_{\text{ten}}} \\ & \,={{\left( 2\times 6 \right)}_{{}}}+\left( 0\times 1 \right) \\ & ={{20}_{\text{six}}} \end{align} Base ten product of $4\times 3=12$which is larger than base six. So, can be written as $2$ times six and $0$times one. \begin{align} & \underline{\begin{align} & \overset{2}{\mathop{5}}\,{{4}_{\text{six}}} \\ & \times {{3}_{\text{six}}} \end{align}} \\ & \text{ }0 \\ \end{align} Now, \begin{align} & {{5}_{\text{six}}}\,\times {{3}_{\text{six}}}\,+{{2}_{\text{six}}}\,={{17}_{\text{ten}}} \\ & \,={{\left( 2\times 6 \right)}_{{}}}+\left( 5\times 1 \right) \\ & ={{25}_{\text{six}}} \end{align} Base ten product of $5\times 3+2=17$which is larger than base six.So, can be written as $2$ times six and $5$ times one. \begin{align} & \underline{\begin{align} & \overset{2}{\mathop{5}}\,{{4}_{\text{six}}} \\ & \times {{3}_{\text{six}}} \end{align}} \\ & 25{{0}_{\text{six}}} \end{align} Hence, the result is${{250}_{\text{six}}}$. | 474 | 1,318 | {"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": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2019-51 | latest | en | 0.542867 |
https://www.trenchlesspedia.com/definition/4522/rock-failure-criterion | 1,696,326,478,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00125.warc.gz | 1,116,309,087 | 28,229 | Advertisement
# Rock Failure Criterion
Published: December 7, 2019 | Last updated: July 5, 2023
## What Does Rock Failure Criterion Mean?
The rock failure criterion is crucial in predicting the failure of rocks or rock masses. The criteria includes methods of predicting how and when a rock will fail when acted upon by external forces.
During the drilling of a wellbore, the use of optimal mud weight, location of casing seats, and drilling parameters that minimize swab and surge are important parameters that ensure that the newly created borehole does not collapse in on itself.
To calculate the precise parameters, engineers turn to the rock failure criterion for adequate calculations. Several failure criteria have been developed over the past years such as the Mogi-Coulomb criterion and the Mohr-Coulomb criterion.
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## Trenchlesspedia Explains Rock Failure Criterion
Mogi-Coulomb rock failure criterion is used in wellbore stability analysis understanding rock strength under true triaxial condition. It is found that the intermediate principal stress influences the ultimate strength of a rock.
When ignoring the intermediate stress and utilize solely the Mohr-Coulomb criterion greatly underestimate the overall rock strength. Conversely, some believe the Mogi-Coulomb criterion greatly overestimates overall rock strength.
Mohr-Coulomb rock failure criterion is the most commonly used shear failure criteria proposed for rocks. In the Mohr-Coulomb criterion, only the maximum (σ1) and minimum (σ3) principal stresses are considered and it is assumed that the intermediate stress (σ2) has no influence on the strength of a rock.
This criterion ignores the strengthening effect of the intermediate stress σ2 and hence is considered to be too conservative in estimating critical mud weight essential to maintain the stability of the wellbore.
As rock failure is one of the uncontrollable instability factors that often lead to the collapse of a wellbore, the use of the Mohr-Coulomb failure criterion is frequently employed to understand the risks of collapse.
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Horizontal Directional Drilling | 459 | 2,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-40 | latest | en | 0.868002 |
https://thejoboverflow.com/p/p949/ | 1,669,986,637,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710902.80/warc/CC-MAIN-20221202114800-20221202144800-00724.warc.gz | 599,683,630 | 5,139 | Question: BNY Mellon, Recently Asked Questions in IIT-G, November, 2022
0
Entering edit mode
# Question 1
Merge Elements
0
Entering edit mode
Question 1
• Observation 1 - There can be n final positions at max, all that matters is if a position is possible or not, if a position corresponding to an element is possible, then that element will never move from its place.
• Observation 2 - If our final position corresponds to an element `aj`, the all `ai <= aj` will for sure move to this position trivially. `if a < = b implies a < = 2*b`
• Observation 3 - If the position corresponding to an element is possible, then all the positions corresponding to values greater than that element will also be possible. Now all we need to find is the smallest element corresponding to which the mentioned operation in the question is possible.
Solution
We will sort the array `arr` and take prefix sum. It is obvious that for the last element the answer is always possible (from observation 2). We will have a pointer `pt` from the second last element and check if `2*prefix[pt] >= arr[pt+1]` then move the pointer one step lower to the next smaller element, i.e `pt--`. If we can reach the element at pt, then we can reach the next element as well, which is understood from the condition we check before moving the pointer. We check `2*prefix[pt]` because we know from observation 2 that all smaller elements can reach here trivially and then value will sum up till that position.
In the end we will get the number of elements for which we can reach a single element, hence all position corresponding to this element will be the positions which can be reached, and the final answer would hence be the number of elements from which we can use the mentioned operation to reach a single element.
pseudo code:
``````for(int i = n-2; i>= 0; i--)
{
if(2*prefix[i] >= arr[i+1])
continue;
else
{
ans = n - i - 1;
}
}
`````` | 472 | 1,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-49 | latest | en | 0.83886 |
https://mathexamination.com/lab/matrix-consimilarity.php | 1,621,063,629,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991378.48/warc/CC-MAIN-20210515070344-20210515100344-00040.warc.gz | 403,209,105 | 8,600 | ## Take My Matrix Consimilarity Lab
As pointed out above, I utilized to write an easy as well as uncomplicated mathematics lab with only Matrix Consimilarity However, the easier you make your lab, the less complicated it ends up being to obtain stuck at the end of it, after that at the beginning. This can be really irritating, and all this can take place to you due to the fact that you are using Matrix Consimilarity and/or Modular Equations improperly.
With Modular Equations, you are currently utilizing the wrong equation when you obtain stuck at the start, otherwise, after that you are most likely in a dead end, as well as there is no feasible way out. This will just worsen as the trouble ends up being a lot more intricate, yet then there is the concern of exactly how to wage the trouble. There is no other way to effectively set about solving this sort of mathematics trouble without being able to promptly see what is taking place.
It is clear that Matrix Consimilarity and Modular Equations are difficult to find out, and it does take practice to develop your own feeling of instinct. But when you wish to fix a math issue, you need to make use of a tool, as well as the tools for discovering are made use of when you are stuck, and they are not made use of when you make the incorrect step. This is where lab Assist Solution comes in.
For example, what is wrong with the inquiry is incorrect concepts, such as getting a partial worth when you do not have enough functioning parts to finish the whole work. There is a great factor that this was wrong, and also it refers logic, not intuition. Reasoning enables you to adhere to a detailed treatment that makes sense, as well as when you make an incorrect step, you are normally compelled to either try to go forward and also deal with the error, or attempt to go backward and do an in reverse step.
Another instance is when the student does not recognize a step of a procedure. These are both rational failings, and there is no way around them. Even when you are embeded a location that does not allow you to make any type of sort of action, such as a triangular, it is still essential to understand why you are stuck, so that you can make a far better step as well as go from the action you are stuck at to the following area.
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Allow's speak about the very first instance, which associates with the Matrix Consimilarity mathematics lab. This is not also complicated, so let's initial review how to begin. Take the adhering to process of connecting a part to a panel to be made use of as a body. This would certainly require three measurements, and also would be something you would require to connect as part of the panel.
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Nonetheless, instead of keeping in mind the added measurement, you can use what is called a "mental shortcut" to help you remember that added measurement. As you make your first step, visualize on your own taking the measurement as well as attaching it to the component you intend to attach to, and then see how that makes you really feel when you duplicate the process.
Visualisation is a very powerful method, as well as is something that you should not skip over. Visualize what it would certainly seem like to actually affix the component and also have the ability to go from there, without the dimension.
Now, allow's consider the 2nd example. Allow's take the exact same procedure as in the past, now the pupil needs to bear in mind that they are mosting likely to return one action. If you tell them that they have to return one step, but then you eliminate the suggestion of needing to move back one step, after that they will not know exactly how to proceed with the issue, they will not know where to seek that action, and also the process will certainly be a mess.
Rather, utilize a mental faster way like the mental representation to emotionally reveal them that they are mosting likely to return one step. and also place them in a position where they can move on from there. without needing to think about the missing out on a step.
## Hire Someone To Do Your Matrix Consimilarity Lab
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## Pay Someone To Take My Matrix Consimilarity Lab
Matrix Consimilarity is used in a lot of schools. Some instructors, nevertheless, do not use it extremely effectively or utilize it inaccurately. This can have an unfavorable impact on the student's understanding.
So, when appointing projects, use a great Matrix Consimilarity aid service to aid you with each lab. These services give a variety of valuable services, consisting of:
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There are many ways that you can educate on your own to work well with the course as well as achieve success. Utilize an appropriate Matrix Consimilarity assistance solution to guide you as well as obtain the work done. Matrix Consimilarity is just one of the hardest courses to learn however it can be quickly mastered with the right assistance. | 2,119 | 10,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-21 | latest | en | 0.974749 |
https://worldbuilding.stackexchange.com/questions/59461/could-modern-structural-engineer-be-useful-to-medieval-master-mason/59522 | 1,582,008,823,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143635.54/warc/CC-MAIN-20200218055414-20200218085414-00437.warc.gz | 636,267,280 | 33,258 | # Could modern structural engineer be useful to medieval Master Mason?
I'm creating a world opposite of Les Visiteurs, where modern age structural engineer accidentally time travels to 13th century. When guards of the local lord arrest him on suspicion of being a spy, due to his weird clothing and the way he talks, he claims that he's mason from abroad looking for work. The lord is just starting a new castle and needs skilled workers, so he hires him but feels deeply suspicious of him, so he tells the Master Mason to keep an eye on the newcomer.
Would the Master Mason have any use of modern structural engineer knowledge or their level of technology is too low for any of it to be practical?
• @klingedion The question asks about a structural engineer in Medieval times not a master mason in modern times. – Bellerophon Oct 24 '16 at 21:21
• @Bellerophon Whoops, I got too excited about trying locate Les Visiteurs on Netflix and forgot to actually read the question. – kingledion Oct 24 '16 at 22:19
• Do you know anything about structural engineering? – JDługosz Oct 25 '16 at 0:53
Your structural engineer will have to have a very scientific mind - undoubtedly he has many skills which will be extremely useful, but he has to be able to convince his new colleagues of his theories.
Material strength, beam theory and formulae for the elasticity of materials had not yet been invented; these would allow for much more accurate calculations during construction. This would equate to great savings in terms of material cost; instead of throwing up all the superstructure possible to ensure a building is strong enough, they can now calculate the required strength and use only what is necessary, with a little factor of safety included.
Calculus had not yet been invented. This is one of an engineer's most important mathematical tools today, used to determine the forces in complex configurations of structural elements, shear strength of soil foundations, and lateral earth pressure and slope stability - all very useful for a castle which may be built atop a mound. Calculus is also useful for friction calculations.
Modern materials such as concrete could have been invented much sooner, as long as your engineer knows the correct properties and composition required.
But all of this relies on his ability to prove these mad ideas to his superior - whether through models and examples, or mathematical proofs. | 497 | 2,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.960012 |
https://www.teachoo.com/2675/1799/Example-22---Straight-lines-work-as-plane-mirror-for-a-point/category/Chapter-10-Class-11th-Straight-Lines/ | 1,548,122,070,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583823140.78/warc/CC-MAIN-20190122013923-20190122035923-00040.warc.gz | 928,979,374 | 12,737 | 1. Class 11
2. Important Question for exams Class 11
3. Chapter 10 Class 11 Straight Lines
Transcript
Example 22 Assuming that straight lines work as the plane mirror for a point, find the image of the point (1, 2) in the line x 3y + 4 = 0 . Let line AB be x 3y + 4 = 0 & point P be (1, 2) Let Q (h, k) be the image of point P (1, 2) in line AB Since line AB is mirror Point P & Q are at equal distance from line AB, i.e. PR = QR, i.e. R is the mid point of PQ Image is formed perpendicular to mirror i.e. line PQ is perpendicular to line AB Since R is the mid point of PQ We know that mid point of a line joining (x1, y1) & (x2, y2) = (( _1+ _2)/2, ( _1+ _2)/2) Mid point of PQ joining (1, 2) & (h, k) is = ((1 + )/2, (2 + )/2) Coordinate of point R = ((1 + )/2, (2 + )/2) Since point R lies on the line AB It will satisfy the equation of line AB Putting x = ( + 1)/2 & y = ( + 2)/2 in equation of AB x 3y + 4 = 0 (( + 1)/2) 3(( + 2)/2) + 4 = 0 ( + 1 3( + 2) + 4 2)/2 = 0 h + 1 3k 6 + 8 = 0 h 3k + 1 6 + 8 = 0 h 3k + 3 = 0 h 3k = 3 Also, PQ is perpendicular to AB We know that If two lines are perpendicular then product of their slope is equal to 1 Slope of AB Slope of PQ = 1 Slope of PQ = ( 1)/( ) Finding slope of AB Equation of line AB is x 3y + 4 = 0 x + 4 = 3y 3y = x + 4 y = ( + 4)/3 y = (1/3)x + 4/3 The above equation is of the form y = mx + c where m = Slope of line So, Slope of line AB = 1/3 Now, Slope of line PQ = ( 1)/( ) = ( 1)/(1/3) = 3 Now, Line PQ is formed by joining points P(1, 2) & Q(h, k) Slope of PQ = ( _2 _1)/( _2 _1 ) 3 = ( 2)/( 1) 3(h 1) = k 2 3h + 3 = k 2 3h k = 2 3 3h k = 5 (3h + k) = 5 3h + k = 5 Now, our equations are h 3k = 3 (1) & 3h + k = 5 (2) From (1) h 3k = 3 h = 3k 3 Putting value of h in (2) 3h + k = 5 3(3k 3) + k = 5 9k 9 + k = 5 9k + k = 5 + 9 10k = 14 k = 14/10 k = 7/5 Putting k = 7/5 in (1) 3h + k = 5 3h + 7/5 = 5 3h = 5 7/5 3h = (5(5) 7)/5 3h = (25 7)/5 3h = 18/5 h = 18/(5 3) h = 6/5 Hence Q = (6/5, 7/5) Hence, image is (6/5, 7/5)
Chapter 10 Class 11 Straight Lines
Class 11
Important Question for exams Class 11 | 914 | 2,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2019-04 | latest | en | 0.9042 |
https://thinkinator.com/2013/12/13/book-recommendation-longitudinal-structural-equation-modeling/?replytocom=91 | 1,656,145,716,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034877.9/warc/CC-MAIN-20220625065404-20220625095404-00503.warc.gz | 638,459,822 | 29,328 | Book recommendation: Longitudinal Structural Equation Modeling
Longitudinal Structural Equation Modeling, Todd D. Little, Guilford Press 2013.
Let me start by saying that this is one of the best textbooks I’ve ever read. It was written as if the author was our mentor, and I really get the feeling that he’s sharing his wisdom with us rather than trying to be pedagogically correct. The book is full of insights on how he thinks about building and applying SEMs, and the lessons he’s learned the hard way.
It’s a little risky to endorse a highly-targeted book like this: it’s dedicated to longitudinal SEMs that don’t include categorical variables. But the discussion and advice that it has is so valuable that I’d recommend it to anyone who is interested in SEMs.
For example, his discussion on scale setting — you have to nail down something regarding a latent variable (“construct”) in order to provide a defined scale — and in addition to the way that most SEM software does it (fixing one of the loading factors to 1), there are two other ways to do it, both of which provide several important advantages. (The book’s website provides example Lisrel, Mplus, and R lavaan code illustrating select chapters, including the one that illustrates scaling.)
As another example, he talks about phantom constructs (latent variables), which is an expert’s trick for modifying your model to change coefficients into more interpretable forms. As one example, you can convert the SEM’s native covariance between latent variables into correlations, post-estimation, by simple algebra. But you couldn’t take two of these correlations (from different times in a longitudinal study, say) and use the model to do a chi-squared test of the significance of the difference between them. Using phantom constructs, with certain constraints, the model can yield correlations directly, so you could properly test their significance.
I could go on, but the bottom line is the book is well-written, entertaining, enlightening, well-illustrated, insightful, and it covers many areas of basic SEM as well as the specifics of the more complicated longitudinal SEM. It’s \$60 for a 380-page hardback, and worth every penny.
One thought on “Book recommendation: Longitudinal Structural Equation Modeling”
1. Aleksandr Blekh
Wayne, thank you for sharing this! Best regards, Alex. | 490 | 2,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.944563 |
http://math.stackexchange.com/questions/797/whats-so-natural-about-the-base-of-natural-logarithms/800 | 1,469,724,146,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828286.80/warc/CC-MAIN-20160723071028-00206-ip-10-185-27-174.ec2.internal.warc.gz | 158,897,890 | 22,666 | # What's so “natural” about the base of natural logarithms?
There are so many available bases. Why is the strange number $e$ preferred over all else?
Of course one could integrate $\frac{1}x$ and see this. But is there more to the story?
-
$\frac11>\ln\frac21>\frac12>\ln\frac32>\frac13>\ln\frac43>\dots$ This infinite system of inequalities is not true for other bases. – Akiva Weinberger Dec 5 '14 at 5:14
$1+\frac12+\frac13+\dotsb+\frac1n\approx\ln n$, for large $n$. – Akiva Weinberger Dec 5 '14 at 5:16
$\ln(1+x)\approx x$ when $x$ is small. For example, $\ln(1.00001)=0.000009999950000\dots$ This isn't really true for other bases. – Akiva Weinberger Dec 5 '14 at 5:18
$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\dotsb$ when $-1<x\le1$. In particular, $1-\frac12+\frac13-\frac14+\dotsb=\ln2$. The other bases don't have nice series like that. – Akiva Weinberger Dec 5 '14 at 5:20
In a sense, $\ln(-1)=\pi i$, but this probably looks meaningless to you. – Akiva Weinberger Dec 5 '14 at 5:23
If you know some linear algebra, then here is an abstract reason: $e^x$ is the unique eigenvector of eigenvalue $1$ of the derivative $D$ acting on, say, the space of smooth functions on $\mathbb{R}$. Why is this important? The study of solutions of linear differential equations with constant coefficients is equivalent to the study of nullspaces of operators which are polynomials in $D$, e.g. operators of the form $\sum a_k D^k$. Any such operator automatically commutes with $D$, so this nullspace splits up into eigenspaces of $D$. That's why solutions to linear differential equations with constant coefficients can be expressed as sums of complex exponentials. The choice of $e$ makes it particularly easy to see what the eigenvalue is: the eigenvalue of the eigenvector $e^{\lambda x}$ is $\lambda$.
-
I would think that very few function are eigenvectors of the Derivative operator. In fact, I'd be curious to know if there are any others besides the exponentials and even some sums of sinusoidals/hyperbolics. – crasic Nov 9 '10 at 8:36
@crasic: it's just the exponentials. However, functions of the form x^k e^{lambda x} are generalized eigenvectors of the derivative operator, so these are the full set of functions you need to solve homogeneous constant-coefficient linear ODEs. – Qiaochu Yuan Nov 9 '10 at 9:07
This is a nice explanation of the relation between D and exponentials, but I don't see how it answers the original question, i.e. why is eigenvalue 1 more important than any other eigenvalue? Except for the fact that 1 is a nice number (being a multiplicative identity and all), I don't see any rational argument why e should be preferred over any other base. – Marek Nov 9 '10 at 13:26
@Marek: e^x is the trivial representation, and the trivial representation is always special (for example it is the identity object in the monoidal category of representations). – Qiaochu Yuan Nov 9 '10 at 15:21
@Marek: I'm not talking about any action of R. I'm talking about certain subrepresentations of the representation of C[X] on C^{\infty}(R) where X acts as the derivative. The trivial representation is the one where X acts trivially (that is, is sent to the identity operator), and it occurs with multiplicity 1 and is spanned by e^x. – Qiaochu Yuan Nov 10 '10 at 0:42
Differentiation and integration is precisely why it is considered natural, but not just because $$\displaystyle\int \frac{1}{x} dx=\ln x$$
$e^x$ has the two following nice properties
$$\frac{d}{dx} e^x=e^x$$
$$\int e^x dx=e^x+c$$
If we looked at $a^x$ instead, we would get:
$$\frac {d} {dx} a^x= \frac{d}{dx} e^{x\ln(a)}=\ln(a) \cdot a^x$$
$$\int a^x dx= \int e^{x\ln(a)} dx=\frac{a^x}{\ln(a)}+c$$
So $e$ is vital to the integration and differentiation of exponentials.
-
The wikipedia article on e tells a bit of the story.
One example is an account that starts with 1.00 and pays 100% interest per year. If the interest is credited once, at the end of the year, the value is 2.00; but if the interest is computed and added twice in the year, the 1 is multiplied by 1.5 twice, yielding 1.00×1.5² = $2.25. Compounding quarterly yields 1.00×1.254 = 2.4414…, and compounding monthly yields 1.00×(1.0833…)12 = 2.613035…. Bernoulli noticed that this sequence approaches a limit (the force of interest) for more and smaller compounding intervals. Compounding weekly yields 2.692597…, while compounding daily yields 2.714567…, just two cents more. Using n as the number of compounding intervals, with interest of 100%/n in each interval, the limit for large n is the number that came to be known as e; with continuous compounding, the account value will reach 2.7182818…. More generally, an account that starts at$1, and yields (1+R) dollars at simple interest, will yield eR dollars with continuous compounding.
Additionally, it is the base of the exponential function y = k^x, finding a specific value for k where d/dx k^x = k^x. That is, the rate of change of the exponential curve at any point is equal to the y value of the curve at that point.
-
The Wikipedia article about e lists many properties of the constant that make it naturally occurring.
I think the biggest reason it is natural when it comes to exponentiation/logarithms is that it is the only number that satisfies
$$\frac{d}{dt} e^t =e^t$$ while every other number satisfies
$$\frac{d}{dt} a^t = c \cdot a^t$$ where $c$ is some constant, different than 1. This makes it "normalized" in a sense.
-
$c = ln(a)$, isn't it? – mikhailcazi Jul 26 '13 at 14:02
If you consider all exponential equations $a^x$, they all have $y$-intercept $(0,1)$. If you wanted to specify an archetypal exponential equation to refer to as you work through Calculus, a natural choice would be to choose the one whose tangent line at $(0,1)$ has slope 1. The equation $e^x$ is the unique exponential equation with that property.
-
I'm surprised I never answered this; maybe I was deterred by the fact that several other answers are here.
One short answer is this: An exponential function $y=a^x$ grows at a rate proportional to its present size, but only when the base is $e$ does it grow at a rate equal to its present size. In other words $$\frac{d}{dx} a^x = \left(\text{constant}\cdot a^x \right),$$ but only when $a=e$ is the "constant" equal to $1$.
The number $a=2$ is too small for this to happen. To see that consider $$\frac{d}{dx} 2^x = \lim_{h\to0} \frac{2^{x+h}-2^x}{h} = \lim_{h\to0}2^x\frac{2^h-1}{h}$$ This last limit is equal to $\displaystyle 2^x \lim_{h\to0}\frac{2^h-1}{h}$. That step can be done because $2^x$ is a "constant", but in this instance, "constant" means "not depending on $h$". Then observe that $\displaystyle\lim_{h\to0}\frac{2^h-1}{h}$ is a "constant", where "constant" now means "not depending on $x$".
So $$\frac{d}{dx} 2^x = \left(\text{constant}\cdot 2^x\right).$$ But what number is this "constant"? Notice that as $x$ increases from $0$ to $1$, $2^x$ increases from $1$ to $2$, so the average slope on that interval is $\dfrac{2-1}{1-0}=1$. Since the curve gets steeper as $x$ increases, it's not yet that steep at $x=0$. Its slope at $x=0$ is $\left.\dfrac{d}{dx}2^x\right|_{x=0}=\left(\text{constant}\cdot2^0\right)$, so that "constant" must be less than $1$.
A similar argument shows that if $4$ is used as the base, the "constant" is more than $1$. This is done by using the interval from $-1/2$ to $0$ instead of the interval from $0$ to $1$.
So $2$ is to small, and $4$ is too big, to be the natural base. $e$ must be somewhere between $2$ and $4$. In a similar way one can show that $3$ is to big, but that's where the previously simple arithmetic gets messy. Use the interval from $-1/6$ to $0$ for that.
-
A few months ago when I was learning LaTeX, I wrote a quick page to test its math rendering. I never though it'd help anyone, but here it is.
Logarithmic Growth Model
The first sentence is key.
The logarithmic growth model is defined by its proportionality of rate of growth to population size.
-
But the fact is that you could have used log to any other base. – user218 Jul 27 '10 at 21:02
The proportion N'(t) ~ N(t) can be written as an example "The growth rate of a bacteria colony is proportional to the population of the bacteria." This makes sense because an increase in number of bacteria will directly increase its growth. Thus, a function that is its own derivative is needed. If you were to guess and check each base, you would find that the number e is the only solution. This section should answer your original question if you know what you're looking for. – Vortico Jul 27 '10 at 21:14
If the condition is just “The growth rate is proportional to the population” as you say, then any exponential function $a^x$ satisfies this, not just $e^x$. – Peter LeFanu Lumsdaine Dec 2 '12 at 8:52
## protected by Marvis Dec 23 '12 at 7:48
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 2,608 | 9,090 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2016-30 | latest | en | 0.85354 |
https://www.numberempire.com/880000 | 1,596,511,982,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735851.15/warc/CC-MAIN-20200804014340-20200804044340-00384.warc.gz | 718,126,826 | 7,314 | Home | Menu | Get Involved | Contact webmaster
# Number 880000
eight hundred eighty thousand
### Properties of the number 880000
Factorization 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5 * 5 * 5 * 5 * 11 Divisors 1, 2, 4, 5, 8, 10, 11, 16, 20, 22, 25, 32, 40, 44, 50, 55, 64, 80, 88, 100, 110, 125, 128, 160, 176, 200, 220, 250, 275, 320, 352, 400, 440, 500, 550, 625, 640, 704, 800, 880, 1000, 1100, 1250, 1375, 1408, 1600, 1760, 2000, 2200, 2500, 2750, 3200, 3520, 4000, 4400, 5000, 5500, 6875, 7040, 8000, 8800, 10000, 11000, 13750, 16000, 17600, 20000, 22000, 27500, 35200, 40000, 44000, 55000, 80000, 88000, 110000, 176000, 220000, 440000, 880000 Count of divisors 80 Sum of divisors 2389860 Previous integer 879999 Next integer 880001 Is prime? NO Previous prime 879979 Next prime 880001 880000th prime 13506011 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 11010110110110000000 Octal 3266600 Duodecimal 365314 Hexadecimal d6d80 Square 774400000000 Square root 938.08315196469 Natural logarithm 13.687677186454 Decimal logarithm 5.9444826721502 Sine 0.8093091421774 Cosine -0.5873829350501 Tangent -1.3778220201586
Number 880000 is pronounced eight hundred eighty thousand. Number 880000 is a composite number. Factors of 880000 are 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5 * 5 * 5 * 5 * 11. Number 880000 has 80 divisors: 1, 2, 4, 5, 8, 10, 11, 16, 20, 22, 25, 32, 40, 44, 50, 55, 64, 80, 88, 100, 110, 125, 128, 160, 176, 200, 220, 250, 275, 320, 352, 400, 440, 500, 550, 625, 640, 704, 800, 880, 1000, 1100, 1250, 1375, 1408, 1600, 1760, 2000, 2200, 2500, 2750, 3200, 3520, 4000, 4400, 5000, 5500, 6875, 7040, 8000, 8800, 10000, 11000, 13750, 16000, 17600, 20000, 22000, 27500, 35200, 40000, 44000, 55000, 80000, 88000, 110000, 176000, 220000, 440000, 880000. Sum of the divisors is 2389860. Number 880000 is not a Fibonacci number. It is not a Bell number. Number 880000 is not a Catalan number. Number 880000 is not a regular number (Hamming number). It is a not factorial of any number. Number 880000 is an abundant number and therefore is not a perfect number. Binary numeral for number 880000 is 11010110110110000000. Octal numeral is 3266600. Duodecimal value is 365314. Hexadecimal representation is d6d80. Square of the number 880000 is 774400000000. Square root of the number 880000 is 938.08315196469. Natural logarithm of 880000 is 13.687677186454 Decimal logarithm of the number 880000 is 5.9444826721502 Sine of 880000 is 0.8093091421774. Cosine of the number 880000 is -0.5873829350501. Tangent of the number 880000 is -1.3778220201586 | 1,126 | 2,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.441097 |
https://cboard.cprogramming.com/cplusplus-programming/59915-copy-function-changing-value-printable-thread.html | 1,493,194,503,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121216.64/warc/CC-MAIN-20170423031201-00265-ip-10-145-167-34.ec2.internal.warc.gz | 781,151,546 | 3,737 | # Copy Function changing value?
• 12-20-2004
GMHummerH1
Copy Function changing value?
Ok the result of my copy function is weird. For people going to look at it. I am using an array of records where everything stored is in the same array. So if i put in 3 and 2 as one set then 4 and 7 as one set has
location: 0 | .info 3 | .next NUL
location: 1 | .info 2 | .next 0
location: 2 | .info 4 | .next NUL
location: 3 | .info 7 | .next 2
If i were to print out the original list BEFORE i exit the copy function it prints it fine and everything is correct.
EDITED FROM ORIGINAL POST:
It is also weird that I pass the original list in as a constant so it SHOULD not be allowed to changed
END EDITED FROM ORIGINAL POST:
However if I print it in my driver it prints ONLY the first term wrong
Exampe:
Enter a polynomial function in the form of: "2x^2 + 3x^2" ==> 4x^5 - 3 + 3x^5
The original fucntion in copy function: 4x^5 - 3 + 3x^5
(now in driver:)
The original function is: 1075464688 - 3 + 3x^5
The term with the highest highest degree is: 4x5
and the degree is: 5
The derivative is: 20x^40 + 15x^4
I have no clue why this is happening. I traced my program, when I print the location where the index is printing in the driver the index is correct and is 0. So it is not pointing to the wrong index. FOr some reason my .info changes when I try and print it inside my driver but in my copy function it is not changed.
I know this question may be hard for you guys to answer but if I can recieve ANY help i be grateful.
GetIndex() returns the location of the first item (basically where NUL is
and
GetListData(): Returns the location of the last item
so location 1 for the original list and location 2 for the new
GetNode(location2, storage);
finds the next free node in storage and puts it value into location2
Code:
void Polynomial::operator=(const Polynomial old)
{
int location = old.GetIndex();
int location2 = 999; // weird
Terms toCopy;
for (int i = location; i <= old.GetListData(); i++)
{
// makes a node, index=locatio2
GetNode(location2, storage);
if (i == old.GetIndex())
SetIndex(location2);
toCopy = storage.nodes[i].info;
storage.nodes[location2].info = toCopy;
storage.nodes[location2].next = listData;
listData = location2;
}
}
• 12-20-2004
[Ren]
My guess is Terms does not copy right. May I see code for Terms?
• 12-20-2004 | 672 | 2,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.865443 |
https://www.jmp.com/support/help/en/16.2/jmp/ems-traditional-model-fit-reports.shtml | 1,696,147,673,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510810.46/warc/CC-MAIN-20231001073649-20231001103649-00521.warc.gz | 895,795,298 | 4,149 | Publication date: 11/10/2021
## EMS (Traditional) Model Fit Reports
Caution: The use of EMS is not recommended. REML is the recommended method.
When EMS is selected as the fitting method, four new reports are displayed. The Effect Tests report is not shown, as tests for both fixed and random effects are conducted in the Tests wrt Random Effects report.
### Expected Mean Squares
The expected mean square for a model effect is a linear combination of variance components and fixed effect values, including the residual error variance. This table gives the coefficients that define each model effect’s expected mean square. The rows of the matrix correspond to the effects, listed on the left. The columns correspond to the variance components, identified across the top. Each expected mean square includes the residual variance with a coefficient of one. This information is given beneath the table.
Figure 3.63 shows the Expected Mean Squares report for the Investment Castings.jmp sample data table. Run the Model - EMS script and then run the model.
Figure 3.63 Expected Mean Squares Report
As indicated by the table, the expected mean square for Casting[Temperature] is
### Variance Component Estimates
Estimates of the variance components are obtained by equating the expected mean squares to the corresponding observed mean squares and solving. The Variance Component Estimates report gives the estimated variance components.
Component
Lists the random effects.
Var Comp Est
Gives the estimate of the variance component.
Percent of Total
Gives the ratio of the variance component to the sum of the variance components.
CV
Gives the coefficient of variation for the variance component. It is 100 times the square root of the variance component, divided by the mean response.
Note: Appears only if you right-click in the report and select Columns > CV.
### Test Denominator Synthesis
For each effect to be tested, an F statistic is constructed. The denominator for this statistic is the mean square whose expectation is that of the numerator mean square under the null hypothesis. This denominator is constructed, or synthesized, from variance components and values associated with fixed effects.
Source
Shows the effect to be tested.
MS Den
Gives the estimated mean square for the denominator of the F test.
DF Den
Gives the degrees of freedom for the synthesized denominator. These are constructed using Satterthwaite’s method (Satterthwaite 1946).
Denom MS Synthesis
Gives the variance components used in the denominator synthesis. The residual error variance is always part of this synthesis.
### Tests wrt Random Effects
Tests for fixed and random effects are presented in this report.
Source
Lists the effects to be tested. These include fixed and random effects.
SS
Gives the sum of squares for the effect.
MS Num
Gives the numerator mean square.
DF Num
Gives the numerator degrees of freedom.
F Ratio
Gives the F ratio for the test. It is the ratio of the numerator mean square to the denominator mean square. The denominator mean square can be obtained from the Test Denominator Synthesis report.
Prob > F
Gives the p-value for the effect test.
Caution: Standard errors for least squares means and denominators for contrast F tests use the synthesized denominator. In certain situations, such as tests involving crossed effects compared at common levels, these tests might not be appropriate. Custom tests are conducted using residual error, and leverage plots are constructed using the residual error, so these also might not be appropriate.
### EMS Profiler
When you use the EMS method and select Factor Profiling > Profiler, the profiler gives predictions and conditional mean confidence intervals based on the fixed-effects model. These values are not based on the predicted values for the random effects. | 766 | 3,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-40 | latest | en | 0.878134 |
https://kubuntu-art.org/how-to-measure-golf-club-length-driver-references/ | 1,653,328,285,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00186.warc.gz | 403,506,660 | 16,687 | ## How To Measure Golf Club Length Driver References
How To Measure Golf Club Length Driver References. As you can tell from the above steps, measuring the length of a golf club and driver shaft is really simple. For this, you need a ruler of 48 inches or greater.
Place the club in the playing position so that the sole of the club / driver is touching the ground. You can easily measure the driver length in golf with measuring tape. Use this chart to see the average length of each individual type of golf club.
### Kids Golf Is One Of The Largest Producers Of Junior Golf Clubs, And They Offer An Online Fitting Center To Help You Make The Right Purchase For Your Junior Golfer.
Measure from the end of the grip to the end of the sole. Simply hold your golf club at 45 degrees to the ground or as in your playing position. Golf club length is an important factor to consider when.
### The Length Is Defined As The Distance From The Point Of The Intersection Between The Two Planes To The End Of The Grip.
Lay the club flat so that the toe of the club faces directly up. How to measure golf club length driver. This will give you you an accurate and usga standard for measuring your golf driver.
### The Ruler Must Be Kept Such That The Golf Club Rests On It.
Measure the distance from the edge of the grip cap to the heel. It involves placing the golf club on the ground and using your 48 inch ruler to accurately read how long the club or driver is. Here are the lengths of a standard mens’ golf club set.
### Measure To Butt Of Grip.
Golf club lengths for junior golfers. As you can tell from the above steps, measuring the length of a golf club and driver shaft is really simple. What is the standard length of a golf driver?
### How To Measure The Length Of Your Driver Shaft:
You can easily measure the driver length in golf with measuring tape. Position a 48” golf club ruler behind the golf club making sure the ruler tip touches the ground by the heel of the club. Use a tape measure to find the length in inches between the end of the grip and the farthest point that the sole contacts the horizontal surface. | 455 | 2,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-21 | longest | en | 0.903525 |
https://www.jiskha.com/display.cgi?id=1373923708 | 1,511,185,723,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806066.5/warc/CC-MAIN-20171120130647-20171120150647-00282.warc.gz | 782,899,487 | 4,279 | # alg2 check?
posted by .
solve for n:6(5n-1)=2(15n-3)
I got 30n-6=30n-6 but I think I was suppose to keep going but don't know how. Do I divide or multiply
• alg2 check? -
At this point, since the two sides of the equation have identical values, the equation is true for ALL values of n.
There is no value of n which makes the equation not true.
If you want to keep going, just do what you usually do. Add 6 to both sides:
30n=30n
subtract 30n from both sides:
0=0
Not much else to do now.
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A 30 N fishing pole is 2.00 m long and has its center of gravity .350 m from the heavy end. A fisherman holds the end of the pole in his left hand as he lifts a 100 N fish. If his right hand is .800 m from the heavy end, how much force …
6. ### physics
A metal rod of length 90cm has a disc of radius 24cm fixed rigidly at its centre,and the assembly is pivoted from its centre. Two forces each of magnitude 30N,are applied normal to the rod at each end so as to produce a turning effect.A …
7. ### physics
A metal rod of length 90cm has a disc of radius 24cm fixed rigidly at its centre,and the assembly is pivoted from its centre. Two forces each of magnitude 30N,are applied normal to the rod at each end so as to produce a turning effect.A …
8. ### Science
check answer please. An object is pulled with 3 forces: 1 at 20n to the right another at 40n to the right and the third at 30n to the left. The net force is 30n to right
9. ### alg2 check?
solve: 2x+2(x+3)=2(x+7) what I got was x=4 but I really think I calculated wrong. Can you show me the step by step process please?
10. ### physics
. A 90cm uniform lever has a load of 30N suspended at 15cm from one of its ends. If the fulcrum is at the centre of gravity, calculate the force that must be applied at its other end to keep it horizontal
More Similar Questions | 765 | 2,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-47 | latest | en | 0.928996 |
mathematicspractice.com | 1,475,021,587,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661289.41/warc/CC-MAIN-20160924173741-00132-ip-10-143-35-109.ec2.internal.warc.gz | 183,929,605 | 3,467 | ## What are prime numbers?
A Prime Number is any numbers that can not be divided by any number other than itself and the number one. For example, 11 is a prime number because there are no other numbers other than 1 and 11 that will divide 11
Still dont get it? Take the number 12, 12 can be divided by 1 and 12 but it can also be divided by the numbers 2, 6, 3 or 4. Therefore 12 is not a Prime Number but 11 is!
See an example of working with prime numbers!
# Prime numbers and fractions made easy!
Special design of the book and game make learning prime numbers fun and easy to learn today!
The math game interactively challenges you to "factor down" a number into it’s prime numbers. Lessons start out easy and get progressively challenging,
helping you to build on your knowledge.
The math book on CD contains a clear explanation of what prime numbers are, and then reinforces each lesson using fractions. The reason why fractions are used when teaching prime numbers is to help you to think of larger numbers as combinations of smaller ones, smaller ones that become prime numbers.
As you get better at learning and underdstanding prime numbers working with fractions then becomes easier and easier!
Looking for help with prime numbers and fractions?
## How do prime numbers help when working with fractions?
Again, consider the number 12. Twelve easily divided by 2, 6, 3, 4 (and its most basic factors 1 or 12). Prime numbers help when working with fractions because every number can be factored down to prime numbers.
See an example on how prime numbers help with reducing fractions! | 352 | 1,604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2016-40 | longest | en | 0.933211 |
https://educationexpert.net/physics/2276336.html | 1,620,739,078,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989614.9/warc/CC-MAIN-20210511122905-20210511152905-00200.warc.gz | 245,473,326 | 7,189 | 26 February, 08:59
# The radius RH of a black hole, also known as the event horizon, marks the location where not even light itself can escape from the black hole. That is, no information about the interior of the black hole may escape to any observer located outside of the black hole. According to general relativity, RH = (2GM / c^2), where M is the mass of the black hole and c is the speed of light. you want to study a black hole by getting near it ith a radial distance of 50 RH. However, you don't want the difference in gravitational acceleration between your head and your feet to exceed 10 m/s^2.a) As a multiple of the Sun's mass, approximate what is the limit to the mass of the black hole you can tolerate at the given radial distance.b) Is the limit an upper limit (you can tolerate smaller masses) or a lower limit (you can tolerate larger masses) ?
+1
1. 26 February, 10:48
0
Solution:
a) We know acceleration due to gravity, g = GM/r²
Differential change, dg/dr = - 2GM/r³
Here, r = 50*Rh = 50*2GM/c² = 100GM/c ²
My height, h=dr = 1.7 m
Difference in gravitational acceleration between my head and my feet, dg = - 10 m/s²
or, dg/dr = - 10/1.7 = - 2GM / (100GM/c²) ³
or, 5.9*100³*G²*M² = 2c⁶
or, M = 0.59*c³ / (1000G) = 2.39*1032 kg = [ (2.39*1032) / (1.99*1030) ]Ms = 120*Ms
Mass of black hole which we can tolerate at the given distance is 120 time the mass of Sun.
b) This limit an upper limit, we can tolerate smaller masses only. | 440 | 1,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.909476 |
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# If the total enrollment in institutions of higher education
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If the total enrollment in institutions of higher education [#permalink] 16 May 2017, 01:56
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If the total enrollment in institutions of higher education in 1972 was 5,000,000, approximately how many students were enrolled in private 4-year institutions in 1995?
A) 1,000,000
B) 1,100,000
C) 1,250,000
D) 1,500,000
E) 1,650,000
[Reveal] Spoiler: OA
_________________
Display posts from previous: Sort by | 340 | 1,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.914589 |
https://docs.unrealengine.com/5.0/en-US/normal-calculation-methods-with-the-proxy-geometry-tool-in-unreal-engine/ | 1,685,670,140,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648245.63/warc/CC-MAIN-20230602003804-20230602033804-00003.warc.gz | 259,190,049 | 5,180 | # Normal Calculation Methods
## Adjusting the Normal Calculation Method for the simplified geometry.
The Proxy Geometry tool will allow you to specify which method should be used when computing the Vertex Normals of a given Static Mesh. In the following How-To we will take a look at how you can change the Vertex Normal calculation method and the effect that will have on your generated Static Meshes.
## Steps
In the following section, we will take a look at how you can adjust the method that is used to calculate the Normals that are used for a Static Mesh.
1. First, locate a Static Mesh or group of Static Meshes that you want to generate new geometry for and select the Mesh or Meshes inside the Viewport.
2. Next, open the Merge Actor tool by going to Window > Developer Tools > Merge Actors.
3. In the Merge Actor tool, click the second icon to access the Proxy Geometry tool and then expand the Proxy Settings.
4. Set the Screen Size value to 50 and make sure that the Normal Calculation Method is set to Angle Weighted.
By setting the Screen Size to a value of 50 we are telling the Proxy Geometry tool to reduce the amount of triangles used in the selected Static Mesh.
5. Next, click the Merge Actors button and input a name and location in the Content Browser for the newly created Static Mesh. Then click the Save button to start the merging process.
6. Once that has completed, double-click the Static Mesh to open it in the Static Mesh Editor, where you can see the results.
## End Results
Getting the best results is going to take some time and iteration as each object could require a different Normal Calculation Method to achieve the desired results. The results can also be very subtle depending on the type of object that you are using. The following image comparison shows how the Static Mesh used in this example looks when the Normal Calculation Method is set to Angle Weighted, Area Weighted and Equal Weighted.
The following image slider shows the results that each of the three Normal Calculation Methods can achieve. First you will see Angle Weighted then Area Weighted and finally Equal Weighted.
Help shape the future of Unreal Engine documentation! Tell us how we're doing so we can serve you better. | 484 | 2,249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-23 | latest | en | 0.84049 |
https://www.matlabhelp.com/using-global-variables/ | 1,601,414,187,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402088830.87/warc/CC-MAIN-20200929190110-20200929220110-00658.warc.gz | 931,673,236 | 13,513 | Select Page
Using Global Variables
Engineers often need to estimate the pressures and volumes of a gas in a container. ideal gas law provides one way of making the estimate. The law is . RT I p = V More accurate estimates can be made with the van der Waals equation: RT a p= V -b V2
where the term b is a correction for the volume of the molecules, and the term a/y2 is correction for molecular attractions. The gas constant is R, the absolute temperature is T and the gas specific volume is Y. The value of R is the same in both equations and for gases; it is R = 0.08206 L atmosphere/mol K. The values of a and b depend on the type of gas. For example, for Chlorine (Ch), a = 6.49 and b = 0.0562 in these units. Write two user-defined functions, one to compute the pressure using the ideal g law, and one using the van der Waals equation. Develop a solution without using global variables; and one using global variables .
Solution
We can write the following function for the ideal gas law:
function P = ideal_l(T,Vhat,Rl
P = R*T. Mat; .
Note that we should use array division because the user might call the function using Vha t as a vector. However, we need not use array multiplication between R and T because R will always be a scalar. This function is independent of the units used for T, Y, and R, as long as·they are consistent with each other. The difficulty with this function is that we must always enter the value of R when we call the function. To avoid this, we can rewri e’the function as follows:
function P = ideal_2(T,Vhatl % This-requires liter, atmosphere,
R = 0.08206;
P = R*T.Vhat;
and mole units!
Now the value of R is “hard-wired” into the function. However, the danger with this approach is that the value of R depends on the units being used for Y and T. For example, in SI metric units, R = 8.314 J=mol So if the user enters values for Vhat and T in SI units, the computed value of P will be incorrect. Thus it is a good idea to put a comment to that effect within the function.However, it is not necessary to read the file in order call the function, and users will not see the comment unless they read the file! A solution to this problem is to declare R a global variable. Then the function file would be: function P = ideal_3(T,Vhat global R
P = R*T Vhat;
With this approach, you must then declare R global wherever you call the function, either from the base workspace or from another function. For example, the following session computes the gas pressures in gases having V = 20, at two temperatures: T = 300 K and
330 K, respectively. ,
»global R
»R = 0.08206;
»ideal_l([300,330) ,20)
ans =
1. 2309 1.3540
The pressures. are approximately 1.23 and 1.35 atmospheres. With this approach, you must always enter the value of R, but this might force you to think about what units are being used for vhat , T, and P. If you are sure that you will always use the same units, then the simplest approach is to hard-wire the value of R into the function, as was done with ideal_2. In the van der Waals equation, the constants a and b depend on the particular gas. A
function r that is independent of the u~t.s used and the particular gas is function = v der waals_l(T,Vhat,R,a,b)
P = R*T./(Vhab)-a./Vhat.A2; The difficulty with this function is that you must always enter the values of R, a, and b. If you use the equation to analyze chlorine only, then you can hard-wire the constants into the function along with the value of R, as follows, but if you do so be sure you always use the proper ,units in the function’s arguments! function P = v der waals_2 (T, Vhat) % For chlorine only, and liter, atmosphere, and mole units!
R 0.08206;
a = 6.49;b = 0.0562;
P = R*T (Vhat-b) a Vhat.A2;
Note that the variable P is local to all these functions, and is available only if you assign it a value when calling the function. For example, consider the following session. > der waals_3(300,20)
ans
1.3416
»P
?? Undefined function or variable ‘P’.
»P = der waals_3(300,20)
P
1.3416
. In addition to using the global command to make values accessible within functions, we can also use it to make global variables that would otherwise be local. For .example, suppose we want to make the chlorine value of and b available to other functions. Then we could modify the der waals_2 function by adding the statement global
a b. Then any other function declaring a and b global would have access to those
values. | 1,173 | 4,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.841203 |
https://bestincentivetours.com/best-time-to-travel/how-far-can-a-snail-travel-in-a-year.html | 1,653,253,841,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662546071.13/warc/CC-MAIN-20220522190453-20220522220453-00457.warc.gz | 184,302,114 | 18,376 | # How Far Can A Snail Travel In A Year?
What is the most distance a snail can go in a year? It was discovered that garden snails could only travel up to 30,660 feet in 365 days, which equates to around 5.8 miles in a year. An aquatic snail, on the other hand, may move up to 66 miles in a single year. The Roman snail, which travels at a speed of around 150 inches per hour, might cover up to 21 miles in a year.
## How do you track snail movement?
The snails were tracked over a 72-hour period using a sample size of 450 snails, with the speed, distance, and places of their travels recorded. Researchers put LED lights on top of some of the snails’ shells, while others were covered with ultraviolet paint, in order to follow the snails’ movements.
## How much energy do snails use?
According to the findings of the study, a snail’s energy expenditure for slime production alone might account for up to 30% of its total energy expenditure. The study was commissioned as a resource for dog owners whose pets are at danger from the lungworm Angiostrongylus vasorum, a potentially lethal parasite carried by slugs and snails that can be fatal if not treated promptly.
## Why do snails move in convoy?
Scientists at Exeter University revealed that they even travel in convoys, piggybacking on the slime of other snails in order to conserve energy. They studied the routines of 450 garden snails, photographing their motions with LED lights, UV paints, and time-lapse photography to better understand their behavior.
## How far will a snail travel?
What is the maximum distance a snail can go in 5 minutes? Garden snails can move at a speed of around 0.65 inches per minute on average. The snail will traverse around 3.25 inches in 5 minutes, which is a significant distance.
We recommend reading: Question: How To Szoom On The Journey Map Mod?
## How far do snails travel in an hour?
Snails can travel at a reasonably rapid one metre per hour, researchers have discovered
## How far can a snail travel overnight?
Researchers discovered that snails may move up to 25 metres in a 24-hour period and that they seek for areas of refuge, such as tall grass, trees, or things, such as dog’s toys, that have been left in the yard overnight. According to the findings of the study, a snail’s energy expenditure for slime production alone might account for up to 30% of its total energy expenditure.
## How far does a slug travel in a day?
Every day, these gluttonous creatures consume several times their own body weight in plant material, and they will travel as far as 40 feet to obtain their nourishment. Slug and snail damage is most common during periods of moist, temperate weather, such as in the spring and late fall.
## What’s the lifespan of a snail?
8. The Life Span of a Snail. In captivity, snail life lengths might vary, but it is feasible for your snail to live into his or her adolescent years. Most wild snails live for two to five years on average, however some bigger species may survive for as long as fifteen years in extreme cases.
## What is the worlds fastest snail?
The typical snail’s pace is around 0.03 miles per hour, but in 1995, a snail named Archie reached nearly twice that speed, achieving a world record time of 2 minutes and 20 seconds as a result of his efforts.
## How long would it take a snail to cross a football field?
543 minutes, or nearly 9 hours and 30 minutes.
We recommend reading: What Waves Does Light Travel In?
## Do snails sleep?
Snails, in contrast to humans, do not observe the distinction between night and day. Snails typically sleep for 13 to 15 hours at a time, alternating between periods of rest and sleep. After then, they get a sudden burst of energy that lasts for the next 30 hours, allowing them to do all of their snail duties!
## What do snails eat?
Snails and slugs have evolved to consume a wide variety of foods; they are herbivorous, carnivorous, omnivorous, and detritivorous in their diets (eating decaying waste from plants and other animals). A variety of species, both specialized and generalist, prey on worms, plant material (including decaying plant material), animal feces, fungus, and other snails.
## Can snails swim?
1. Do Snails Have the Ability to Swim?
2. Sea snails are the only snails that are actually capable of swimming.
3. In order to travel from one location to another, land snails and freshwater snails utilize their ″feet.″ Even while certain creatures are capable of wriggling their bodies on the water’s surface in order to effectively push themselves through the water, they are not technically swimming.
## Can you take a snail on a plane?
As a result, I’ve learnt that transporting garden snails across provinces by plane is entirely acceptable, but they must be handled with the same care as any other animal. To put it another way, you cannot bring them in your carry-on; they must instead be checked.
## How far does 12 gauge slug travel?
A slug’s accuracy decreases as the distance between them increases; out to 100 yards, it loses around five inches and has a maximum range of about 400 yards. Centerfire rounds fired from rifles, on the other hand, may travel for kilometers. A shotgun slug’s effectiveness is best suited for close-range applications.
We recommend reading: FAQ: Fahrenheit 451 And How It Relates To Montags Journey?
## How fast does a slug move in mph?
Snails and slugs move at speeds ranging from slow (0.013 m/s) to very slow (0.0028 m/s). Snails and slugs have a range of speeds that range from slow to extremely slow. Approximately the speed of a snail.
Bibliographic Entry Result (w/surrounding text) Standardized Result
The World Almanac and Book of Facts 1999. New Jersey: Primedia, 1998: 572. ‘Garden snail, 0.03 mph’ 0.013 m/s
## How far can a slug travel NRA?
A little shot can travel 200–350 yards with a good trajectory. A powerful shot may travel more than 600 yards. Slugs are capable of traveling more than 800 yards. | 1,400 | 5,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-21 | latest | en | 0.972853 |
https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.18/share/doc/Macaulay2/Graphs/html/_top__Sort.html | 1,652,877,048,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2022-21/segments/1652662522270.37/warc/CC-MAIN-20220518115411-20220518145411-00028.warc.gz | 313,598,453 | 3,028 | topSort -- outputs a hashtable containing original digraph, new digraph with vertices topologically sorted and a map from vertices of original digraph to new digraph.
Synopsis
• Usage:
topSort(D)
topSort(D,S)
• Inputs:
• D, an instance of the type Digraph,
• S, ,
• Outputs:
• ,
Description
This method outputs a HashTable with keys digraph, map and newDigraph, where digraph is the original digraph, map is the relation between old ordering and the new ordering of vertices and newDigraph is the Digraph with topologically sorted vertices. This method needs the input to be directed adyclic graph (DAG). S provides the preference given to the vertices in order to break ties and provide unique topological sorting to the DAG.
Permissible values of S are: "random", "max", "min", "degree".
S = "random" randomly sort the vertices of graph which have same precedence at any instance of the algorithm to break the ties.
S = "min" sort the vertices according to their indices (from low to high) to break the ties.
S = "max" sort the vertices according to their indices (from high to low) to break the ties.
S = "degree" sort the vertices according to their degree (from low to high) to break the ties.
i1 : G = digraph{{5,2},{5,0},{4,0},{4,1},{2,3},{3,1}} o1 = Digraph{0 => {} } 1 => {} 2 => {3} 3 => {1} 4 => {0, 1} 5 => {2, 0} o1 : Digraph i2 : H = topSort G o2 = SortedDigraph{digraph => Digraph{0 => {} } } 1 => {} 2 => {3} 3 => {1} 4 => {0, 1} 5 => {2, 0} map => HashTable{0 => 4} 1 => 6 2 => 3 3 => 5 4 => 2 5 => 1 newDigraph => Digraph{1 => {3, 4}} 2 => {4, 6} 3 => {5} 4 => {} 5 => {6} 6 => {} o2 : SortedDigraph i3 : H#digraph o3 = Digraph{0 => {} } 1 => {} 2 => {3} 3 => {1} 4 => {0, 1} 5 => {2, 0} o3 : Digraph i4 : H#map o4 = HashTable{0 => 4} 1 => 6 2 => 3 3 => 5 4 => 2 5 => 1 o4 : HashTable i5 : topSort(G,"min") o5 = SortedDigraph{digraph => Digraph{0 => {} } } 1 => {} 2 => {3} 3 => {1} 4 => {0, 1} 5 => {2, 0} map => HashTable{0 => 3} 1 => 6 2 => 4 3 => 5 4 => 1 5 => 2 newDigraph => Digraph{1 => {3, 6}} 2 => {3, 4} 3 => {} 4 => {5} 5 => {6} 6 => {} o5 : SortedDigraph i6 : topSort(G,"max") o6 = SortedDigraph{digraph => Digraph{0 => {} } } 1 => {} 2 => {3} 3 => {1} 4 => {0, 1} 5 => {2, 0} map => HashTable{0 => 6} 1 => 5 2 => 3 3 => 4 4 => 2 5 => 1 newDigraph => Digraph{1 => {3, 6}} 2 => {5, 6} 3 => {4} 4 => {5} 5 => {} 6 => {} o6 : SortedDigraph i7 : topSort(G,"random") o7 = SortedDigraph{digraph => Digraph{0 => {} } } 1 => {} 2 => {3} 3 => {1} 4 => {0, 1} 5 => {2, 0} map => HashTable{0 => 5} 1 => 6 2 => 2 3 => 4 4 => 3 5 => 1 newDigraph => Digraph{1 => {2, 5}} 2 => {4} 3 => {5, 6} 4 => {6} 5 => {} 6 => {} o7 : SortedDigraph i8 : topSort(G,"degree") o8 = SortedDigraph{digraph => Digraph{0 => {} } } 1 => {} 2 => {3} 3 => {1} 4 => {0, 1} 5 => {2, 0} map => HashTable{0 => 3} 1 => 6 2 => 4 3 => 5 4 => 1 5 => 2 newDigraph => Digraph{1 => {3, 6}} 2 => {3, 4} 3 => {} 4 => {5} 5 => {6} 6 => {} o8 : SortedDigraph
Ways to use topSort :
• "topSort(Digraph)"
• "topSort(Digraph,String)"
For the programmer
The object topSort is . | 1,224 | 3,065 | {"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} | 2.96875 | 3 | CC-MAIN-2022-21 | latest | en | 0.816108 |
https://www.coursehero.com/file/163621/MATH311ansHW3/ | 1,493,572,892,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125719.13/warc/CC-MAIN-20170423031205-00218-ip-10-145-167-34.ec2.internal.warc.gz | 888,328,715 | 48,235 | MATH311ansHW3
# MATH311ansHW3 - Math 311, Homework 3 partial answers and...
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Math 311, Homework 3 partial answers and solutions 2.2D.2. Answer: u v w x y = s 1 - 1 1 0 0 0 + s 2 - 1 0 1 0 0 + s 3 1 0 0 1 0 + s 4 1 0 0 0 1 + 1 0 0 0 0 . 2.2D.4. Partial solution: a possible reduced form of this matrix (there are many) is 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 . Each column contains a leading entry. 2.2D.10. Partial solution: the given system is equivalent to 1 0 2 0 1 - 3 0 0 0 x y z = 1 1 0 . Note that the solution space does not contain 0 . 2.2D.11. Row reduction gives 1 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 x y z w = 0 0 0 0 . Thus the solution space of this system is the plane
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## This note was uploaded on 04/29/2008 for the course MATH 311 taught by Professor Anshelvich during the Spring '08 term at Texas A&M.
### Page1 / 3
MATH311ansHW3 - Math 311, Homework 3 partial answers and...
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Ask a homework question - tutors are online | 462 | 1,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.8497 |
https://www.fdotstokes.com/2022/09/22/what-is-300-ppi-in-mb/ | 1,713,496,392,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00169.warc.gz | 686,959,828 | 13,085 | # What is 300 PPI in MB?
## What is 300 PPI in MB?
7.2 MB
The required digital file size for an 8” x 10” print at 300 ppi would be 7.2 MB.
## What is 300 PPI MP?
We could also look at it the other way – if the request is for a photo that will print to 300 ppi on 8″ x 10″ paper, we can multiply 300 ppi by those dimensions and we’ll get 3000 x 2400 pixels or 7.2 Mp.
How do I create a 300 PPI image?
Open your image in Preview. Go to Tools > Adjust size… In the Resolution box you’ll see the DPI of your image. If it’s different than 300, uncheck the “Resample image” box and enter your desired DPI (300).
How many pixels are in 1 MB?
The number of pixels in one megabyte depends on the color mode of the picture. 8-bit (256 color) picture, there are 1048576, or 1024 X 1024 pixels in one megabyte. 16-bit (65536 colors) picture, one megabyte contains 524288 (1024 X 512) pixels.
### How many pixels is 0.3 MP?
307,200 pixels
1-1 of 1 Answer 3MP translates into 307,200 pixels. You can think of a pixel as 1 dot on a computer screen, so this camera can take pictures and display them using 307,200 pixels.
### How do I convert MP to PPI?
A good quality picture for a photo album is printed at 300 dpi. To calculate the resolution in megapixels, multiply the number of pixels of length and width and divide them by mega (1 million)….Result.
Pixel total Pixel [ Mpx]
dpi / ppi dpi
Is 300 pixels per inch the same as 300 dpi?
So, when you ask yourself, “what is 300 DPI in Pixels Per Inch per image,” the answer is 300 because 300 DPI in an image means there are 300 pixels per inch in your web design image. Finding the number of pixels you need to get 300 DPI for your image is simple.
What is the difference between 72 dpi vs 300 dpi?
The difference between 72 dpi and 300 dpi is the amount of dots for every square inch of the image.The more pixels an image contains the sharper the image will print. If you try to print an image with a 72 dpi it’ll cause it to print blurry.
#### How do I know if my photo is 300 dpi?
– Open your photo. – Click Tools > Adjust size. – Make sure to uncheck the Resample Image box. – Type in a higher width and height. … – Your resolution should be exactly 300 pixels per inch or more if you’re printing a picture into a larger photo size.
#### What does 300 pixels per inch mean?
PPI, or pixels per inch, is the amount of pixels that get printed (horizontally and vertically) in a 1 inch line. Therefore, a square inch of print at 300 PPI will consist of 90,000 pixels. The higher the PPI, the clearer your image will be because the pixels will be denser. | 701 | 2,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-18 | latest | en | 0.872458 |
http://www.r-bloggers.com/search/ANova/page/2/ | 1,386,686,128,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164020864/warc/CC-MAIN-20131204133340-00020-ip-10-33-133-15.ec2.internal.warc.gz | 618,331,249 | 20,397 | # 226 search results for "ANova"
## How Do You Write Your Model Definitions?
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## Science at the speed of ligth
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## “Statistical Models with R” Course – Milano, October 24-25, 2013
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## informative hypotheses (book review)
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## Mixed models; Random Coefficients, part 2
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## Mixed models; Random Coefficients, part 1
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## Latent Variable Analysis with R: Getting Setup with lavaan
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Getting Started with Structural Equation Modeling Part 1Getting Started with Structural Equation Modeling: Part 1 Introduction For the analyst familiar with linear regression fitting structural equation models can at first feel strange. In the R environment, fitting structural equation models involves learning new modeling syntax, new plotting...
## The joy and martyrdom of trying to be a Bayesian
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## More REML exercise
August 25, 2013
By
Last week I tried exercise 1 of the SAS(R) proc mixed with R libraries lme4 and MCMCglm. So this week I aimed for exercise 2 but ended up redoing exercise 1 with nlme.Exercise 2 results gave me problems with library lme4 and latter parts of the ex... | 806 | 3,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2013-48 | longest | en | 0.940711 |
http://quizlet.com/4505842/hw-3-flash-cards/ | 1,427,570,659,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131297689.58/warc/CC-MAIN-20150323172137-00175-ip-10-168-14-71.ec2.internal.warc.gz | 224,099,603 | 18,892 | # HW #3
### 20 terms by watchamx
#### Study only
Flashcards Flashcards
Scatter Scatter
Scatter Scatter
## Create a new folder
### Refer to the above diagram. A decrease in supply is depicted by a:
shift from S2 to S1.
### Refer to the above diagram. An increase in quantity supplied is depicted by a:
move from point y to point x.
P = 4 + 1/3Q.
### An improvement in production technology will:
shift the supply curve to the right.
### If producers must obtain higher prices than previously to produce various levels of output, the following has occurred
a decrease in supply.
### Assume a drought in the Great Plains reduces the supply of wheat. Noting that wheat is a basic ingredient in the production of bread and that potatoes are a consumer substitute for bread, we would expect the price of wheat to
rise, the supply of bread to decrease, and the demand for potatoes to increase.
### An increase in the excise tax on cigarettes raises the price of cigarettes by shifting the:
supply curve for cigarettes leftward.
\$2
### Refer to the above data. If the price in this market was \$4:
farmers would not be able to sell all their wheat.
### Refer to the above data. If price was initially \$4 and free to fluctuate, we would expect:
the quantity of wheat supplied to decline as a result of the subsequent price change.
\$1.60.
### At the equilibrium price:
there are no pressures on price to either rise or fall.
### The rationing function of prices refers to the:
capacity of a competitive market to equate the quantity demanded and the quantity supplied
### Assume in a competitive market that price is initially below the equilibrium level. We can predict that price will:
increase, quantity demanded will decrease, and quantity supplied will increase.
### In which of the following instances will the effect on equilibrium price be dependent on the magnitude of the shifts in supply and demand?
demand rises and supply rises.
### Refer to the above diagram, in which S1 and D1 represent the original supply and demand curves and S2 and D2 the new curves. In this market:
demand has increased and equilibrium price has decreased.
### Refer to the above diagram, in which S1 and D1 represent the original supply and demand curves and S2 and
an increase in demand has been more than offset by an increase in supply.
### Data from the registrar's office at Gigantic State University indicate that over the past twenty years tuition and enrollment have both increased. From this information we can conclude that
the supply of education provided by GSU has also increased over the twenty-year period.
### A price floor means that:
government is imposing a minimum legal price that is typically above the equilibrium price.
### An effective ceiling price will:
result in a product shortage
Example: | 593 | 2,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-14 | latest | en | 0.917472 |
https://leetcode.ca/2021-07-19-2058-Find-the-Minimum-and-Maximum-Number-of-Nodes-Between-Critical-Points/ | 1,708,780,333,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474533.12/warc/CC-MAIN-20240224112548-20240224142548-00198.warc.gz | 358,956,638 | 8,894 | # 2058. Find the Minimum and Maximum Number of Nodes Between Critical Points
## Description
A critical point in a linked list is defined as either a local maxima or a local minima.
A node is a local maxima if the current node has a value strictly greater than the previous node and the next node.
A node is a local minima if the current node has a value strictly smaller than the previous node and the next node.
Note that a node can only be a local maxima/minima if there exists both a previous node and a next node.
Given a linked list head, return an array of length 2 containing [minDistance, maxDistance] where minDistance is the minimum distance between any two distinct critical points and maxDistance is the maximum distance between any two distinct critical points. If there are fewer than two critical points, return [-1, -1].
Example 1:
Input: head = [3,1]
Output: [-1,-1]
Explanation: There are no critical points in [3,1].
Example 2:
Input: head = [5,3,1,2,5,1,2]
Output: [1,3]
Explanation: There are three critical points:
- [5,3,1,2,5,1,2]: The third node is a local minima because 1 is less than 3 and 2.
- [5,3,1,2,5,1,2]: The fifth node is a local maxima because 5 is greater than 2 and 1.
- [5,3,1,2,5,1,2]: The sixth node is a local minima because 1 is less than 5 and 2.
The minimum distance is between the fifth and the sixth node. minDistance = 6 - 5 = 1.
The maximum distance is between the third and the sixth node. maxDistance = 6 - 3 = 3.
Example 3:
Input: head = [1,3,2,2,3,2,2,2,7]
Output: [3,3]
Explanation: There are two critical points:
- [1,3,2,2,3,2,2,2,7]: The second node is a local maxima because 3 is greater than 1 and 2.
- [1,3,2,2,3,2,2,2,7]: The fifth node is a local maxima because 3 is greater than 2 and 2.
Both the minimum and maximum distances are between the second and the fifth node.
Thus, minDistance and maxDistance is 5 - 2 = 3.
Note that the last node is not considered a local maxima because it does not have a next node.
Constraints:
• The number of nodes in the list is in the range [2, 105].
• 1 <= Node.val <= 105
## Solutions
• /**
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
int first = 0, last = 0;
int i = 1;
int[] ans = new int[] {Integer.MAX_VALUE, Integer.MIN_VALUE};
while (curr.next != null) {
if (curr.val < Math.min(prev.val, curr.next.val)
|| curr.val > Math.max(prev.val, curr.next.val)) {
if (last == 0) {
first = i;
last = i;
} else {
ans[0] = Math.min(ans[0], i - last);
ans[1] = i - first;
last = i;
}
}
++i;
prev = curr;
curr = curr.next;
}
return first == last ? new int[] {-1, -1} : ans;
}
}
• /**
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
int first = 0, last = 0;
int i = 1;
vector<int> ans(2, INT_MAX);
while (curr->next) {
if (curr->val < min(prev->val, curr->next->val) || curr->val > max(prev->val, curr->next->val)) {
if (last == 0)
first = i;
else {
ans[0] = min(ans[0], i - last);
ans[1] = i - first;
}
last = i;
}
++i;
prev = curr;
curr = curr->next;
}
if (first == last) return {-1, -1};
return ans;
}
};
• # Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def nodesBetweenCriticalPoints(self, head: Optional[ListNode]) -> List[int]:
first = last = None
i = 1
ans = [inf, -inf]
while curr.next:
if curr.val < min(prev.val, curr.next.val) or curr.val > max(
prev.val, curr.next.val
):
if last is None:
first = last = i
else:
ans[0] = min(ans[0], i - last)
ans[1] = i - first
last = i
i += 1
prev, curr = curr, curr.next
return ans if first != last else [-1, -1]
• /**
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
first, last := 0, 0
i := 1
ans := []int{math.MaxInt32, 0}
for curr.Next != nil {
if curr.Val < min(prev.Val, curr.Next.Val) || curr.Val > max(prev.Val, curr.Next.Val) {
if last == 0 {
first, last = i, i
} else {
ans[0] = min(ans[0], i-last)
ans[1] = i - first
last = i
}
}
i++
prev, curr = curr, curr.Next
}
if first == last {
return []int{-1, -1}
}
return ans
}
• /**
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function nodesBetweenCriticalPoints(head: ListNode | null): number[] {
let idx = 1;
let nums = [];
if (pre < val && val > post) {
nums.push(idx);
}
if (pre > val && val < post) {
nums.push(idx);
}
pre = val;
idx++;
}
let n = nums.length;
if (n < 2) return [-1, -1];
let min = Infinity;
for (let i = 1; i < n; i++) {
min = Math.min(nums[i] - nums[i - 1], min);
}
return [min, nums[n - 1] - nums[0]];
} | 1,581 | 5,063 | {"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} | 3.859375 | 4 | CC-MAIN-2024-10 | longest | en | 0.870593 |
https://www.neuron.yale.edu/phpBB/viewtopic.php?f=12&t=1687&p=5947 | 1,582,726,483,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146342.41/warc/CC-MAIN-20200226115522-20200226145522-00435.warc.gz | 800,795,494 | 8,891 | Measurements of Synchrony?
Moderator: wwlytton
Bill Connelly
Posts: 86
Joined: Thu May 22, 2008 11:54 pm
Location: Australian National University
Measurements of Synchrony?
Hi,
So with Teds help, I've got a lovely interneuron network making 40Hz oscillations. Now I'm planning on doing cruel things to it, and seeing what that does to the quality of the oscillations. I'm only used to the slice arena, where you'd use a spike triggered averaged from cell 1 to cell 2 to measure how strong the oscillation is.
However, in a large scale network, there must be a better way to measure synchrony. Does anyone know of one?
--EDIT--
This is one I came up with. It varies from -1 for a completely random network, to 1, for a completely synchronous (all cells fire within 1ms of eachother) network. I'd love to know if people thought this was a robust metric or not. It would require some slight modification if you were dealing with networks of hetrogenous cells, but otherwise it seems quite good.
Code: Select all
``````// Assume you have a network of cells ncells long
// You the data for when cell i fires it jth action potential in the 2D array AP_cell_num[i][j]
// You have the number of action potentials cell j fires during the simulation in the vector forcellAPnum.x(j)
// For ever action potential, you compare it to the temporally nearest action potential in every other cell
// If the action potential you compare to is within 1ms you add 1 to the system score
// if the compared action potential is not within 1ms you subtract 1 from the system score
// (you exclude the 1st and last action potential from each cell because you can not know whether the other action potentials are the nearest
// without an action potential before and after each event to compare it to)
// You divide the system score by the number of spikes in the network (-2 because you loose the first and last events)
// Finally you devide that by (ncells*ncells-ncells) as I believe this is the number of possible comparisons (you never compare a cell to itself).
systemScore=0
for n=0, ncells-1 {
for i=0, forcellAPnum.x(n)-1 {
for m=0, ncells-1 {
for j=0, forcellAPnum.x(m)-1 {
if (n!=m) {
matchingAP=AP_cell_num[n][i]
previousAP=AP_cell_num[m][j-1]
thisAP=AP_cell_num[m][j]
nextAP=AP_cell_num[m][j+1]
lagtopreviousAP=abs(matchingAP-previousAP)
lagtothisAP=abs(matchingAP-thisAP)
lagtonextAP=abs(matchingAP-nextAP)
if(j-1>=0 && nextAP!=0) { //ignore the first and last events
if(lagtopreviousAP>lagtothisAP && lagtonextAP>lagtothisAP) { //if the event in the other cell is the closest to this one
if(matchingAP-thisAP>-1 && matchingAP-thisAP<=0) { //is it within 1msec of the matchingAP
systemScore+=1
} else {
systemScore-=1
}
j = forcellAPnum.x(m)-1 // once the nearest AP is compared, skip to next AP
}
}
}
}
}
}
}
APpercells=0
for i = 0, ncells-1 {
APpercells += forcellAPnum.x(i)
}
APpercells/=ncells
print "Average action potentials per cell = ", APpercells
print "Synch score = ", (systemScore/(APpercells-2))/(ncells*ncells-ncells)
``````
wwlytton
Posts: 65
Joined: Wed May 18, 2005 10:37 pm
Contact:
Re: Measurements of Synchrony?
Paul Tiesinga has a large number of sync measures and other ways of interpreting population spike patterns
eg
mean ISI for neuron n is tauN, CV is (stdev of tauNi)/<tauN>
associate a particular interval statistic with the midpoint of the ISI
The interspike interval of the combined set of network spikes is tau_v=t_{v+1} - t_v
Note that these interspike intervals are between different neurons. The coefficient
of variation is CV_P = sqrt{<tau_v^2>_v - <tau_v>_v^2/<tau_v>_v} <>_v denotes ave
over all intervals
Rapid temporal modulation of synchrony by competition in cortical interneuron
networks PHE Tiesinga & TJ Sejnwoski
also see
Discovering Spike Patterns in Neuronal Responses
Jean-Marc Fellous, Paul H. E. Tiesinga, Peter J. Thomas, and Terrence J. Sejnowski
The Journal of Neuroscience, 2004, 24(12):2989-3001
J Neurophysiol 91: 194-205, 2004.
Influence of Ionic Conductances on Spike Timing Reliability of Cortical
Neurons for Suprathreshold Rhythmic Inputs
Susanne Schreiber,1,5 Jean-Marc Fellous,2 Paul Tiesinga,1,3 and Terrence J. Sejnowski1,2,4
Methods for finding and validating neural spike patterns
Neurocomputing, 69:1362-1365, 2006
J. Vincent Toups and Paul H.E. Tiesinga
Tiesinga PH, Jose JV. (2000) Spiking statistics in noisy hippocampal interneurons
Neurocomputing 26-27 299-304.
Tiesinga PH, Jose JV. (2000) Synchronous clusters in a noisy inhibitory neural network.
Journal Computational Neuroscience 9 49-65
Bill Connelly
Posts: 86
Joined: Thu May 22, 2008 11:54 pm
Location: Australian National University
Re: Measurements of Synchrony?
Currently I'm trying to work on measure proposed by Dr. David Golomb
http://www.scholarpedia.org/article/Syn ... l_networks
However it doesn't seem to be working.
Assuming I've got a vector storing the average membrane potential of the cells in the network in averageCell
Also, I've got a vector storing the membrane potential of every cell n in the network in cellV[n]
Then the synchrony score proposed in that article should be
Code: Select all
``````AvVar = averageCell.sumsq()/averageCell.size() - averageCell.mean()*averageCell.mean()
for n=0, n<ncells-1 {
cellVar.x(n)=cellV[n].sumsq()/cellV[n].size() - cellV[n].mean()*cellV[n].mean()
}
syncScore = AvVar/cellVar.mean()
``````
Right? Golomb says it should vary betwen 0 and 1, but my network gets 13. Any idea on what I'm doing wrong?
ted
Posts: 5686
Joined: Wed May 18, 2005 4:50 pm
Location: Yale University School of Medicine
Contact:
Re: Measurements of Synchrony?
The first term on the RHS of your formula for AvVar should take dt into account. Ditto for the formula for cellVar.
Bill Connelly
Posts: 86
Joined: Thu May 22, 2008 11:54 pm
Location: Australian National University
Re: Measurements of Synchrony?
Actually I found the problem. Its the fact that dt was getting reset by neuron halfway through the script, so arrays were different lengths. | 1,695 | 6,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.931099 |
https://rdrr.io/cran/nvctr/man/R2xyz.html | 1,708,697,916,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474412.46/warc/CC-MAIN-20240223121413-20240223151413-00529.warc.gz | 495,509,657 | 7,789 | # R2xyz: Find the three rotation angles about new axes in the xyz... In nvctr: The n-vector Approach to Geographical Position Calculations using an Ellipsoidal Model of Earth
## Description
The angles (called Euler angles or Tait–Bryan angles) are defined by the following procedure of successive rotations: Given two arbitrary coordinate frames A and B, consider a temporary frame T that initially coincides with A. In order to make T align with B, we first rotate T an angle x about its x-axis (common axis for both A and T). Secondly, T is rotated an angle y about the NEW y-axis of T. Finally, T is rotated an angle z about its NEWEST z-axis. The final orientation of T now coincides with the orientation of B. The signs of the angles are given by the directions of the axes and the right hand rule.
## Usage
`1` ```R2xyz(R_AB) ```
## Arguments
`R_AB` a 3x3 rotation matrix (direction cosine matrix) such that the relation between a vector v decomposed in A and B is given by: v_A = R_AB * v_B
## References
Kenneth Gade A Nonsingular Horizontal Position Representation. The Journal of Navigation, Volume 63, Issue 03, pp 395-417, July 2010.
`xyz2R`, `R2zyx` and `zyx2R`.
```1 2 3 4 5 6``` ``` R_AB <- matrix( c( 0.9980212 , 0.05230407, -0.0348995 , -0.05293623, 0.99844556, -0.01744177, 0.03393297, 0.01925471, 0.99923861), nrow = 3, ncol = 3, byrow = TRUE) R2xyz(R_AB) ``` | 408 | 1,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-10 | latest | en | 0.809874 |
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Telephonic:
Round 1
Round 2
1. Find the number of islands (all 8 corners)
2. Dont remember
Round 3
1.word frequency of a stream of words (no code required, explain trie method)
Round 4
1. Next larger element (use stacks)
Time and space complexity must for all questions
Start from brute force and go to optimized solution.
All edge cases are to be covered in code?
Think out loud all the time (so that even when questions are easy they will catch your thinking with greater importance)
by Expert (19,470 points)
0 like 0 dislike
Hi, Recently i was interviewed for Amazon SDE-1 Position in Hyderabad.There a telephonic round followed by 4 F2F rounds.
Telephonic Round:
1. Inserting an element into a BST
2. A array is increasing and then decreasing find the point where it stops increasing.
F2F Round 1:
1. Replace all the elements in the array with its next highest element to its right
Expected O(n) Solution.
2. Given a binary tree and a value k. A path is called heavy path if the sum of the elements in the path (path from root to leaf) > k remove all the paths from the tree which are not heavy i.e., tree should contain only heavy paths.
F2F Round 2:
1. Given a array find all the triplets which satisfy the triangle preoperty(sum of 2 sides should be greater than third side)
Sol: sort then o(n^2 log(n)) using binary search.
2. Given a dependency where for example process p1,p2,p3
p1:{p2,p3}
p2:{p3}
p3:{}
This means p1 starts once p2 and p3 are complete
p2 starts p3 is complete
p3 can start as it does not have any dependence.
Figure out strategy to find the order of execution of processes.
Ans:Topological sorting.
F2F Round 3:
1. Design a stack with push pop and find min operations in o(1) time.
Ans:can be done using 2 Stacks
2. Given an input string and a dictionary of words, find out if the input string can be segmented into a space-separated sequence of dictionary words.
Solution https://www.geeksforgeeks.org/dynamic-programming-set-32-word-break-problem/
F2F Round 4:
Discussion of projects and current work experience.
Diameter of a binary tree in o(n).
First devised o(n^2) then optimized to o(n)
by Expert (19,470 points)
0 like 0 dislike
Hi, Recently I was interviewed for the SDE-I for Amazon, Chennai.
Round 1(Online Round):
e.g 1–>2–>3
3–>4
Output: 1–>5–>7
I cleared the online coding round and I was asked to come down for the face to face discussions.
Round 2(F2F):
1. Find if the given binary tree is height balanced.
Round 3(F2F):
1. Find if the given tree is the subtree of the big tree.
2. Given a sorted matrix (row-wise and column wise) , find kth smallest element.
3. Given an array. Pop min element in O(1) time.
4. Implement increment operator for the array.
Round 4(F2F):
1. Print the level-order traversal from bottom to up in a given binary tree.
2. What is process and thread?
3. What is copy constructor?
4. Design parking lot.
Round 5(F2F) (Hiring manager round):
1. Questions on achievements, challenges faced, area of improvement.,etc.
2. Given a file with millions of URLs. Some URLs repeating , some unique. Find the first unique URL.
Coding was required in each and every round. Think loud. Tell the interviewers what you think. Give them different approaches. Be confident about your solution.
by Expert (19,470 points)
0 like 0 dislike
Hi, Recently i was interviewed for Amazon SDE-1 Position in Hyderabad.There a telephonic round followed by 4 F2F rounds.
2. Given a array find whether it is majority array or not.
Complexity: O(n) Space complexity O(1)
F2F Round 1:
1. Given a dependency where for java packages p1,p2,p3
p1:{p2,p3}
p2:{p3}
p3:{}
This means p1 can be compiled when compilation of p2 and p3 done
p2 can compile when p3 is compiled
p3 can start as it does not have any dependence.
Figure out strategy to find the order of compilation of processes.
Ans:Topological sorting
2. Discussion on project
Asked all concept related to my final year project.
3. Current project in company
Challenges faced at company
2F Round 2:
1. Given a binary tree print its side view from left from bottom to top and right side view as up to downward .
Eg. For image 1 output will be 3, 1, 2, 5, 6, 7, 8
and for Example 2( image 2) output should be C, B, A, D, F, I, L
3. What is virtual memory?
4. What is paging in OS?
2F Round 3:
1. Why are you looking for change in 6 months of experience .
2. Discussion on how to send notification to friends when a user post on social network like facebook
The discussion was on when to load friend list of user and he was expecting we should load friend list when user login.
2F Round 4:
1. Discussion on current project challenges faced
Situation when you have to take decision in absence of your team lead/Manager
2. Why are you looking for change in 6 months of experience
3. Given a special binary tree structure given in image 3 A node have 3 node pointers left,right and nextTwo adjacent node share there left and right as shown in treeGive tree whose node are set with next pointer as null.
you have to set the next pointer as shown in the figure .
If any node is null then the next of its prev node should be next of that eg. a -> b -> c (a’s next is b ,b’s next is c) if b is null then a’s next should be c.
4. What is virtual memory?
5. What is segmentation fault?
Note :- They expect fully working code in all round.
Tips:
1. Think loud they always support you
2. Ask for hint if not getting.
by Expert (19,470 points)
0 like 0 dislike
Recently I am interviewed for Amazon SDE-1 position for Bangalore. There are 3 F2F rounds followed by a telephonic round.
As it was a drive, they asked everyone to write code for these problems
2. Longest Palindrome in a String
1st F2F:
1. Why do you want to leave your current company?
2. Why Amazon?
3. Find median in a stream
I told him min heap & max heap method, then he asked me who can you do it using trees.
4. There is a mxn matrix which contains only 1 & 0’s. You have to print the unique rows. I solved it using tries, then he asked why cannot I use hash map and asked to write the code using tries.
2nd F2F (Managerial Round):
1. Why Amazon?
2. Why leaving your current company so early?
3. Areas of Improvement , strengths & weakness
4. Given a prefix expression , convert into prefix tree and extended the qtsn for infix expression, time complexities etc.
5. Given a dictionary , and we have to query for anagrams for the word. extended the qstn to while typing the word we have to provide the autotype. Gave him a solun using tries.
Telephonic Round (Bar Riser):
1. Why Amazon? | 1,738 | 6,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.857268 |
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Dissertation Statistics Help
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https://whatsupwhimsy.com/table-algebra-32 | 1,674,833,496,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494986.94/warc/CC-MAIN-20230127132641-20230127162641-00490.warc.gz | 632,997,322 | 17,931 | # Table algebra
Free math lessons and math homework help from basic math to algebra, geometry and beyond. Students, teachers, parents, and everyone can find solutions to their math problems instantly.
## Complete a function table from an equation (Algebra 1 practice)
So what about a Table Of Values? Since, as we just wrote, every linear equation is a relationship of x and y values, we can create a table of values for any line. These are just the x and y values
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## Recognizing functions from table (video)
A table algebra is (equivalent to) the Core Structure with R ⊆ C such that 1 C ∈ R, and S = R ∩ R ≥ 0. Definition 1.12. A table algebra is the Core Structure with R = C, S = R ≥ 0, B
## Algebra Tables
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## Table algebras
Calculus: Integral with adjustable bounds. example. Calculus: Fundamental Theorem of Calculus
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Need help with math homework? Our math homework helper is here to help you with any math problem, big or small. | 542 | 2,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-06 | latest | en | 0.925334 |
https://math.stackexchange.com/questions/2930250/proving-soundness-property-of-a-hilbert-system | 1,571,853,737,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00448.warc.gz | 577,396,430 | 34,554 | # Proving soundness property of a Hilbert system
Now that I have a better understanding of soundness, I'd like to try this again.
My goal is to prove that the classical Hilbert system has the soundness property:
$$\Gamma \vdash \varphi \implies \Gamma \models \varphi$$
For a set of wffs $$\Gamma$$ and wff $$\varphi$$.
This soundness property being "If $$\varphi$$ is provable from $$\Gamma$$, then $$\varphi$$ is also true under every interpretation where $$\Gamma$$ is satisfied (i.e. when all its propositions are true)."
We can induct on the length of the proof, which we denote as a sequence of wffs $$\varphi_1, \varphi_2, \varphi_3, ..., \varphi_n = \varphi$$.
We start with the case of $$n=1$$, where we only have a one-line proof $$\varphi_1$$. There are two cases:
1. $$\varphi$$ is an axiom of the Hilbert system. We can write out the truth tables and show that for every interpretation, the axiom is true.
Axiom I:
$$\begin{array}{|c|c|ccc|} \hline (A & \to & ( B & \to & A )) \\ \hline F & T & T & F & F \\ F & T & F & T & F \\ T & T & T & T & T \\ T & T & F & T & T \\ \hline \end{array}$$
Axiom II:
$$\begin{array}{|ccccc|c|ccccccc|} \hline ((A & \to & (B & \to & C)) & \to & ((A & \to & B) & \to & (A & \to & C))) \\ \hline F & T & F & T & F & T & F & T & F & T & F & T & F \\ F & T & F & T & T & T & F & T & F & T & F & T & T \\ F & T & T & F & F & T & F & T & T & T & F & T & F \\ F & T & T & T & T & T & F & T & T & T & F & T & T \\ T & T & F & T & F & T & T & F & F & T & T & F & F \\ T & T & F & T & T & T & T & F & F & T & T & T & T \\ T & F & T & F & F & T & T & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$$
Axiom III:
$$\begin{array}{|ccc|c|ccccc|} \hline ((A & \to & B) & \to & (\neg & B & \to & \neg & A)) \\ \hline F & T & F & T & T & F & T & T & F \\ F & T & T & T & F & T & T & T & F \\ T & F & F & T & T & F & F & F & T \\ T & T & T & T & F & T & T & F & T \\ \hline \end{array}$$
1. $$\varphi$$ is a element of $$\Gamma$$. Since we only care about the situation where $$\Gamma$$ is satisfied, all elements of $$\Gamma$$ will be true.
Moving onto the case of $$n > 1$$, our inductive hypothesis is that $$\Gamma \models \varphi_k$$ holds for all $$1 \leq k < n$$ for all interpretations that satisfy $$\Gamma$$. It is possible that $$\varphi_n$$ is an axiom or an element of $$\Gamma$$, which are cases we've already covered. But since $$n > 1$$, we now look at a new possible case where $$\varphi_n$$ can be the result of modus ponens proven from two earlier wffs $$\varphi_i$$ and $$\varphi_j = \varphi_i \to \varphi_n$$ with indices $$i, j < n$$. By inductive hypothesis we know $$\Gamma \models \varphi_i$$ and $$\Gamma \models \varphi_j$$, i.e. $$\varphi_i$$ and $$\varphi_j$$ are both true in every interpretation where $$\Gamma$$ is satisfied.
Using truth tables:
$$\begin{array}{|c|ccc||c|} \hline \varphi_i & \varphi_i & \to & \varphi_n & \varphi_n\\ \hline F & F & T & F & F \\ F & F & T & T & T \\ T & T & F & F & F \\ T & T & T & T & T \\ \hline \end{array}$$
We see in the last case where $$\varphi_i$$ is true and when $$\varphi_i \to \varphi_n$$ is true, $$\varphi_n$$ is true as well. Thus modus ponens is sound, and we have covered all cases. This closes the inductive step.
Now we can conclude that if $$\varphi$$ is provable from $$\Gamma$$, then $$\varphi$$ is true in all interpretations where $$\Gamma$$ is satisfied.
Have I proven that the Hilbert system has the soundness property?
• Quite perfect... "By inductive hypothesis we know $\varphi_i,\varphi_j$ are both true." NO; we know that they are logical cons of $\Gamma$, i.e. that $\Gamma \vDash \varphi_i$ and $\Gamma \vDash \varphi_j$. – Mauro ALLEGRANZA Sep 25 '18 at 14:19
• Thus, because modus ponens is sound, we conclude that: in every int where all formulas of $\Gamma$ are true, also $\varphi_n$ must be true. – Mauro ALLEGRANZA Sep 25 '18 at 14:21
• Updated with truth tables. – user525966 Sep 25 '18 at 14:25
• @MauroALLEGRANZA Is that not saying the same thing? Or I guess if $\Gamma \models \varphi_i$ then is it incorrect to say $\varphi_i$ is true, period -- or would it have been better to say "true under every satisfiable interpretation of $\Gamma$" – user525966 Sep 25 '18 at 14:26
• Perfect......... – Mauro ALLEGRANZA Sep 25 '18 at 14:40
• I don't follow. Isn't something like $A \to (B \to A)$ already presuming that $A$ and $B$ are wffs? And what do you mean by sound instantiation? That seems like a very different thing. What do you mean by "full form"? What substitution rule? – user525966 Sep 25 '18 at 20:07
• When you substitute AFAIK you use parentheses, so it's like substituting each instance of $A$ with $(A \to (B \to A))$, so you'd get $(A \to (B \to A)) \to (B \to (A \to (B \to A)))$ – user525966 Sep 25 '18 at 20:12 | 1,666 | 4,838 | {"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": 42, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-43 | latest | en | 0.679055 |
http://linfographik.com/how-to-calculate-steady-state-error-from-graph/ | 1,529,271,863,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859817.15/warc/CC-MAIN-20180617213237-20180617233237-00079.warc.gz | 195,689,114 | 5,743 | # How To Calculate Steady State Error From Graph
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Mar 30, 2011. steady state error command. Learn more about steady state error Control System Toolbox.
When you retire, you’ll no longer receive a steady. contend with state taxes on top of federal taxes, unless you live in one of the states that doesn’t impose these.
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We can calculate the steady-state error for this system from. at steady state we do have. Feel free to zoom in on different areas of the graph to observe.
How to find steady-error value. How to find steady-error value from the response graph? is there any command to find the steady state error from the response graph?
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. How to find Steady-State Error. So I already know how to find steady-state by going to the graph and. I trying to calculate the steady-state error.
Feb 8, 2010. In the context of control engineering feedback loops, these slides describe how to find the steady-state error between a target and the system.
May 2, 2011. hi all, I was wondering that when i plot a graph from Simuink, is there any way of determining the response parameters, like Steady-state error,
Many VR insiders cite research firm Gartner’s hype cycle graph as evidence that everything’s fine. Growth will be slow and steady, not explosive, which Schell.
Error. ▫ Steady-state error. Final value theorem. (Suppose CL system is stable!!!). Kp, Kv, Ka : ability to reduce steady-state error. 6. Characteristic equation.
CTM: Steady State Error – Steady-state error can be calculated from the open or closed-loop transfer function for unity feedback.
This syntax is useful when you know that the expected steady-state system. The response has settled when the error |y(t) – yfinal| becomes smaller than a. The upper threshold RT(2) is also used to calculate SettlingMin and SettlingMax.
New Gallup global income data estimate the median household income across 131 countries at \$9,733, and median per-capita income at \$2,920. Incomes in the 10.
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We can also find out the steady state rates. Have you ever wondered how a pocket calculator works out the square root of a number?. graph, the x axis.
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The simulation program operates on the theoretical model basis and returns the result as a graph. avoid any error in the calculation process. Therefore, a partial loss of results is inevitable, but this error can be negligible. Steady state. | 861 | 3,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-26 | latest | en | 0.888841 |
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# Length of arrays in matrices
• #1 ✎ 3 ENIGMA I don't have a lot of experience with smilebasic and I am trying to find the length of a single array within a 2x2 matrix, or thr length of the row. Technically, becaude the matrix im using is a square, i could just use the length of the matrix, but I want the program to work on non-square matrices as well. I tried len(mat[0]) assuming it would work but it didnt. Everything ive seen usually used a variable to set up the matrix deminsions and used that same variable again later, but I reading in the array, so that won't work. I'm at a loss for what to do which really sucks because I'm pretty sure my idea for my program will work. So, does anybody know a way to find the length of a row in a 2d array? Ot any work arounds? Posted
• #2 ✎ 86 Simeon Scholar Received for knowing a great deal about programming topics Achievements Amazing Page Hidden Achievements Drawing I like to draw! Hobbies Here is the code: ```DEF SIZE A OUT X,Y VAR T=A[1,0],B[0],L=LEN(A) A[1,0]=1<<31 COPY B,A FOR Y=0 TO L-1 IF B[Y]==A[1,0] THEN BREAK NEXT A[1,0]=T X=L/Y END``` Usage:```DIM MYARRAY[12,8] SIZE MYARRAY OUT X,Y ?"DIM MYARRAY[";X;",";Y;"]"``` How it works: When you have a two dimensional array, copying it to a new array will convert it into a one dimensional array. So we can change the value in [1,0] and see what index it gets mapped to. That will give us the size of one dimension, we can easily calculate the other. Note: The smallest array that this works on is [2,1]. The time complexity of this algorithm is O(X*Y+X), copying the whole array is linear, then we have to loop through Y once more, to get O(X*(Y+1)). I believe this is the fastest algorithm to do this. I'd love to see if someone can do better. Posted Edited by Simeon
• #3 ✎ 90 HackTheWorlds Second Year My account is over 2 years old Website Summer 2016 Contest Participant I participated in the SmileBASIC Source Summer 2016 Contest! Programming Contest Reading I like to read books! Hobbies Non-square arrays are not possible in SmileBASIC, and there is no built in way to determine the dimensions of an array. However I would recommend checking out these posts: Getting the Dimension Lengths from 2D Arrays Changing dimensions of an array Posted Edited by HackTheWorlds
• #4 ✎ 90 HackTheWorlds Second Year My account is over 2 years old Website Summer 2016 Contest Participant I participated in the SmileBASIC Source Summer 2016 Contest! Programming Contest Reading I like to read books! Hobbies On a side note, welcome to the website! :) Posted
• #5 ✎ 3 ENIGMA I don't really know a lot about this language including a bunch of syntax stuff so really i just dont know whats going on with that. I started to understand a bit but I dont fully get it. You'll basically have to explain it like i dont know the lanuage at all. Posted
• #6 ✎ 86 Simeon Scholar Received for knowing a great deal about programming topics Achievements Amazing Page Hidden Achievements Drawing I like to draw! Hobbies Well, okay, although it's a fun programming challenge, you have no reason to calculate the dimensions of a two dimensional array since you'll need those dimensions to initialize it in the first place. ```VAR W=8,H=12 VAR MYARRAY[H,W] ?"Width: ";W ?"Height: ";H``` Typing a "?" is the same thing as typing "PRINT", it's just easier to use Inside a print statement, we can use ";" to concatenate variables. Also note that VAR and DIM are the same. Posted Edited by Simeon
• #7 ✎ 86 Simeon Scholar Received for knowing a great deal about programming topics Achievements Amazing Page Hidden Achievements Drawing I like to draw! Hobbies ``` 'Make a function named SIZE that takes in 'a variable and outputs two variables X and Y DEF SIZE A OUT X,Y 'T is a temporary backup of A[1,0] 'B is an empty array 'L is width*height VAR T=A[1,0],B[0],L=LEN(A) 'Set A[1,0] to -2147483648 (bitshifting) A[1,0]=1<<31 'Copy the contents of A into B 'setting B's size to width*height COPY B,A 'Y will loop from 0 to width*height-1 FOR Y=0 TO L-1 'Check if we reach the -2147483648 inside B 'Once we do, break out of the loop 'Then Y will be the index of -2147483648 IF B[Y]==A[1,0] THEN BREAK NEXT 'Put back the value inside of A[1,0] A[1,0]=T 'width=width*height/height X=L/Y END ``` Posted Edited by Simeon
• #8 ✎ 3 ENIGMA I am seeing now that i could have used the variables i used to instantiate the 2d array, since i did actually need those numbees to instantiate it. I was thinking I guess that i just was given an array with unknown values but the only way to instatiate the array it seems is to make it blank and add everything after. And i now get that i was incorrect in assuming len(mat) still gave number of rows because i guess its not an array of arrays so when this finds the size it returns the size of both dimensions. I then see T is a temporary variable that for the first element of the second row because youre going to change that elements value, though I dont know why 2 other things seperated by commas are next to it exactly. I think i otherwise get it though Posted
• #9 ✎ 284 spaceturtles Video Games I like to play video games! Hobbies Avatar Block I didn't change my avatar for 30 days. Website Intermediate Programmer I can make programs, but I still have trouble here and there. Programming Strength Which two other things separated by commas on which line? Posted
• #10 ✎ 95 niconii Video Games I like to play video games! Hobbies Expert Programmer Programming no longer gives me any trouble. Come to me for help, if you like! Programming Strength Drawing I like to draw! Hobbies `VAR T=A[1,0],B[0],L=LEN(A)` is the same as ```VAR T=A[1,0] VAR B[0] VAR L=LEN(A)``` Posted | 1,508 | 5,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-43 | longest | en | 0.913034 |
https://www.jiskha.com/display.cgi?id=1194681373 | 1,516,298,923,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887535.40/warc/CC-MAIN-20180118171050-20180118191050-00626.warc.gz | 939,711,005 | 3,846 | # math
posted by .
I have a question in
(( distance between two points ))
show that A(4,2), B(-2,-2) and C(2,-8)are the vertices of an isoceles right triangle.
please in detales .. thank u
• math -
The lengths of the sides are
AB: sqrt(6^2 + 4^2) = sqrt 52 (side c)
BC: sqrt (4^2 + 6^2) = sqrt 52 (side a)
AC: sqrt (2^2 + 10^2) = sqrt 104 (side b)
This proves that two sides are equal and that it is a right triangle since
c^2 + a^2 = b^2.
## Similar Questions
1. ### isosceles triangle
An isosceles triangle is a triangle that has two equal sides. Find AB, BC and AC. Determine if the triangle is isoseles or not based on the lengths of the three sides. can someone show me how to solve?
2. ### math
hi i will ask u in math :) in (( distance between two points )) * three vertices of rectanglr WXYZ are W (4,3), X (-1,3), and Y (-1,-1), what sre the coordinates of Z ?
3. ### math
i have a question in math (( distance between two points )) A quardilateral has vertices P(3,5), Q(-4,3), R(-3,-2), and S(5,-4). find the lengths of the diagonals,to the nearest tenth . please in detales .. thank u
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i have a question in math (( distance between two points )) the vertices of a right triangle are S(-2,-2), T(10,-2), and R(4,5). find the area of the triangle. please in detales .. thank u
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have a question in math (( distance between two points )) show that C(-5,-1)is the midpoint of the line segment joining A(-2,5) and B(-8,-7) please in detales .. thank u
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have a question in math (( distance between two points )) show that C(-5,-1)is the midpoint of the line segment joining A(-2,5) and B(-8,-7) please in detales .. thank u
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have a question in math (( distance between two points )) the coordinate of the diameter of a circle are (6,4) and (-2,0) find the length of the radius of the crcle. please in detales .. thank u
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9. ### Math
1. the points given are two of the 3 vertices of a right triangle. Which could be the third vertex of the triangle given 2 points are (-4,-5) & (3,2) ? | 676 | 2,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-05 | latest | en | 0.873774 |
https://www.physicsforums.com/threads/newtons-second-law-for-a-system-of-particles.446368/ | 1,627,470,963,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153709.26/warc/CC-MAIN-20210728092200-20210728122200-00448.warc.gz | 987,304,766 | 14,043 | # Newtons Second Law For A System Of Particles
## Homework Statement
In figure (a), a 5.4 kg dog stands on a 16 kg flatboat at distance D = 6.1 m from the shore. It walks 2.1 m along the boat toward shore and then stops. Assuming no friction between the boat and the water, find how far the dog is then from the shore.
## Homework Equations
Center of mass = (x1m1 + x2m2)/(m1+m2)
## The Attempt at a Solution
Since the center of mass cant change i set the two equal.
x1 (position of dog) = 6.1
x2 (center of canoe) = x
(6.1*5.4 + 16x2)/21.4 =
(5.4*(6.1-2.1) + 16(x2+Δcom))/21.4
which gives the change in the center of mass of the boat is .70875 so the dog only moves 2.1-.70875 and starting 6.1 meters away he is 4.70875 but thats wrong
## Answers and Replies
rl.bhat
Homework Helper
When the dog moves towards the shore, the center of gravity of the boat moves away from the boat.Along with that the dog also moves away from the the boat. If x is the distance moves, the equation becomes
[(6.1- 2.1+x)*5.4 + 16*(x2 + x)])/21.4 = the initial center of mass.
Now solve for x. | 343 | 1,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-31 | latest | en | 0.90206 |
https://www.dummies.com/education/economics/idea-gross-domestic-product/ | 1,571,354,970,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677230.18/warc/CC-MAIN-20191017222820-20191018010320-00549.warc.gz | 866,106,063 | 16,275 | The Idea of Gross Domestic Product - dummies
# The Idea of Gross Domestic Product
As of late 2015, the U.S. GDP is more than \$18 trillion, and projections are that it will grow by 2.7 percent to over \$18.5 trillion before 2016 is over.
In other words, if you add the value of all final sales recorded in the U.S. in one year it comes to more than \$18 trillion. But note the qualifier final: To avoid double counting, economists count only the value of final goods produced, not intermediate goods. So, if a car manufacturer produces a car worth \$20,000 but buys component parts worth \$8,000 to physically make the car, the total addition to GDP is just the final \$20,000.
Sometimes you hear people calling GDP the total value added. They do so because you can think of the contribution of a good or service to GDP as the value that is added to it at each stage of production.
For example, the car manufacturer we mentioned has added value worth \$12,000 (by buying the parts for \$8,000 and then selling the car for \$12,000 more). The same applies to the parts for which component manufacturers may have had to buy raw materials for \$2,000 in order to produce the \$8,000 worth of parts. They have then added value worth \$6,000.
As the table shows, the value of the car (the final good being produced) is \$20,000. This is exactly equal to the total value added by the raw materials, the component manufacturer, and the car manufacturer. If a dealer now sells the car for \$23,000, it implies further added value of \$3,000 in advertising and sales services.
Cost of Inputs (\$) Value of Output (\$) Value Added (\$) Raw materials — 2,000 2,000 Component manufacturers 2,000 8,000 6,000 Car manufacturer 8,000 20,000 12,000 Total value added — — 20,000
## Determining national income — and not consuming it all at once
Someone pays for the value of whatever is produced. So, someone gets paid for the production, too. Those payments are income for whoever receives them. Thus, GDP tells you not just the aggregate production but also everyone’s individual income summed in total. And that should be equal to GDP, the aggregate amount of income of the economy. In other words, the total of everyone’s income should equal the total of all expenditures made to buy domestic goods, and both should equal GDP.
To continue the example from the preceding section, the car manufacturer has managed to turn \$8,000 worth of parts into a \$20,000 car, leaving it with a surplus of \$12,000. Some of this surplus goes to pay the workers in the factory (say \$8,000), leaving a profit of \$4,000. Ultimately, people own firms, so this profit also provides an income to somebody. Thus, the manufacturer’s value added is entirely distributed as income either to its workers or its shareholders.
The same is true for the component manufacturer and the owners of the raw materials: The value added must have been paid to someone as income. Because the total value added equals GDP, so too must total income equal GDP.
Of course, when you have your part of national income or GDP, you have to decide what to do with it. Here you have a number of options:
• You can consume it all: Which would mean spending all your income on consumer goods, such as food, entertainment, and similar goods and services. Another way of thinking about this option is that your income gives you a claim to a certain share of national output. By spending it all now, you’re choosing to use your entire claim on consumption today.
• You can consume some of it and save the rest: This would mean you spend only part of your income on consumer goods. By saving some of it, you give up some of your claim to consumption today and trade it for a claim to consumption in the future. The interest rate tells you the terms of this trade. If you save some of your income by putting it in a savings account or perhaps a bond fund that pays 5 percent, it means that by giving up \$100 worth of consumer goods today, you will establish a claim to \$105 worth of consumer goods one year from now. Thus, economists think of saving as a way of converting consumption today into consumption in the future.
• You can consume more than your share: How? By borrowing from someone who wants to save part of his share. The catch is that you’ll have to pay that person back at some point by giving up some of your own future consumption. This is the same trade as before but in reverse. In this trade, every extra \$100 worth of consumer goods you buy today requires a sacrifice of \$105 worth of consumer goods in a year. Thus, economists think of borrowing as a way of converting income in the future into consumption today.
## Watching a nation’s GDP flow
GDP is the same thing as aggregate annual income and so therefore it must be a flow variable, one that’s measured per unit of time — yearly, in this case.
It’s important to remember then that as a flow variable, GDP “restarts” each year. Roughly \$18 trillion worth of goods and service were produced in 2015. On January 1, 2016, the nation started producing all over again and will probably produce \$18.6 trillion by the end of 2016. The point is that because GDP is a flow variable, it has a rate of time or per year dimension. Over 2015 and 2016, the U.S. has been producing at an average rate of \$18.3 trillion worth of goods and services per year, with some acceleration in this pace over time.
Because GDP is a flow, all the component variables such as household consumer spending, private investment spending, and so on, are also flow variables. A new, hopefully bigger pie or flow will be produced next year because we’ll have more workers, more capital, and better technology. | 1,276 | 5,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.939142 |
https://www.teacherspayteachers.com/Product/Calculus-Derivatives-of-Trig-Functions-20-problems-typed-solutions--3935614 | 1,548,101,804,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583807724.75/warc/CC-MAIN-20190121193154-20190121215154-00634.warc.gz | 940,145,644 | 18,348 | # Calculus Derivatives of Trig Functions ( 20 problems, typed solutions )
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Calculus Differentiation - Derivatives of Trig Functions
This resource contains total of 20 problems. Students will practice differentiation of common and composite trig functions.
The packet has 2 worksheets:
⟐ The first worksheet has the students finding the first derivatives of 10 trig functions using differentiation formulas, the product and quotient rules and the derivatives of the three main trig functions.
⟐ The second worksheet is finding the first derivatives of 10 composite trig functions using the chain rule.
The worksheets can be used as class practice, for an extra practice or enrichment, an assessment or homework assignment. It can be also used as a partner activity like:
⟡ Partner A will solve WS # 1 while Partner B solves WS # 2, then they swap papers and Partner A will solve WS # 2 while Partner B solves WS # 1. Once they have completed the work, they compare their results. If there are different answers to one and the same problem, students have to identify and correct any errors.
Typed solutions are provided.
This activity is included in my Calculus 1 DERIVATIVES Growing BUNDLE of Activities & Worksheet Packets - 22 products included and discounted at 25%
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Calculus BIG Growing BUNDLE - Limits, Derivatives, Integrals, Differential Equations (850 problems) - 44 products included and discounted at 32%!
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Contents
An analytic function called PERCENTILE_CONT was introduced in SQL Server 2012. It is capable of calculating the median within a partitioned set. It calculates the median when we pass 0.5 as an argument and specify the order within that dataset.
## Does SQL have a median function?
In simple terms, it may be thought of as the “middle” value of a data set. There is no MEDIAN function in T-SQL.
## How do you find the median of a column in SQL?
“how to find median of a column sql” Code Answer
1. SET @rowindex := -1;
2. SELECT.
4. FROM.
5. (SELECT @rowindex:=@rowindex + 1 AS rowindex,
## How do you select the median?
The median of a finite list of numbers can be found by arranging all the observations from lowest value to highest value and picking the middle one. If there is an even number of observations, then there is no single middle value; the median is then usually defined to be the mean of the two middle values.”
## How does SQL Server calculate median?
If you are using SQL Server 2000/ 2005 / 2008, you can calculate the median by using 50 percent highest value and 50 percent lowest values. This method is quite simple and can be used to calculate the median value of an odd or even number of rows within a specified dataset.
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## What is median SQL?
The median of an array of numbers is the value of the middle item in the array, assuming the array is sorted. If the array has an even number of items, the median represents the average of the two middle values in the array.
## How do you find the median in MySQL?
We calculate the median of the Distance from the demo table. SET @rowindex := -1; SELECT AVG(d. distance) as Median FROM (SELECT @rowindex:=@rowindex + 1 AS rowindex, demo.
## How does MySQL calculate median?
To find the median value using a MySQL query, you need to write a subquery that returns the column where you want to find the median value with a number index attached to each row. The complete solution to find a median value would be similar to the following query: SET @row_index := -1; SELECT AVG(subq.
## Is median same as average?
Average Median
The average is the arithmetic mean of a set of numbers. The median is a numeric value that separates the higher half of a set from the lower half.
## What is the value of median?
Median: To find the median of a data set, arrange the data values in order. from least to greatest or greatest to least; the median is the data value in the middle; if there is an even number of data values in the set, the median is the mean of the two middle values.
## Is median an aggregate function?
An expression that specifies the set of values from which the median is determined. … The expression must return a value that is a built-in numeric data type, CHAR, or VARCHAR data type. In a Unicode database, the expression can also be a GRAPHIC or VARGRAPHIC data type.
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## How does SQL Server calculate percentile?
PERCENT_RANK() The PERCENT_RANK function in SQL Server calculates the relative rank SQL Percentile of each row. It always returns values greater than 0, and the highest value is 1. It does not count any NULL values.
## How do you find the median of an Access query?
To find the median, you sort your list (ascending or descending, makes no difference), count the number of values in your list, and then divide that number by two. If the list has an odd number of values, the result will be a fraction (n. 5).
Categories BD | 839 | 3,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-05 | latest | en | 0.842621 |
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# A five-student committee must be formed to discuss and plan
Author Message
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A five-student committee must be formed to discuss and plan [#permalink]
### Show Tags
18 May 2007, 05:18
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
A five-student committee must be formed to discuss and plan the development of the school’s sporting program. Knowing that each of these five students must be chosen from a group of 10 students, then the possible number of choices is:
(A) 360
(B) 180
(C) 21600
(D) 252
(E) 210
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18 May 2007, 05:45
10!/(10-5)!*5!=10*9*8*7*6/5*4*3*2=252
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18 May 2007, 12:28
same here ! (D)
C(10,5) = (choosing five items out of ten items) = 252
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### Show Tags
18 May 2007, 13:02
KillerSquirrel wrote:
same here ! (D)
C(10,5) = (choosing five items out of ten items) = 252
ME TOO . 10C5 ? since no order
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23 May 2007, 08:04
D for me too! Yeah!
23 May 2007, 08:04
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$\left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\pi \\ x = \dfrac{\pi }{3} + k\pi \end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)$
$$\begin{array}{l} \sin x.\cos y = \dfrac{1}{2}.\left[ {\sin \left( {x + y} \right) + \sin \left( {x – y} \right)} \right]\\ 2\sqrt 2 \sin \left( {x – \dfrac{\pi }{{12}}} \right).\cos x = 1\\ \Leftrightarrow \sqrt 2 .\left[ {2\sin \left( {x – \dfrac{\pi }{{12}}} \right).\cos x} \right] = 1\\ \Leftrightarrow \sqrt 2 .\left[ {\sin \left( {x – \dfrac{\pi }{{12}} + x} \right) + \sin \left( {x – \dfrac{\pi }{{12}} – x} \right)} \right] = 1\\ \Leftrightarrow \sqrt 2 .\left[ {\sin \left( {2x – \dfrac{\pi }{{12}}} \right) + \sin \left( { – \dfrac{\pi }{{12}}} \right)} \right] = 1\\ \Leftrightarrow \sin \left( {2x – \dfrac{\pi }{{12}}} \right) + \dfrac{{ – \sqrt 6 + \sqrt 2 }}{4} = \dfrac{1}{{\sqrt 2 }}\\ \Leftrightarrow \sin \left( {2x – \dfrac{\pi }{{12}}} \right) = \dfrac{{\sqrt 6 + \sqrt 2 }}{4}\\ \Leftrightarrow \left[ \begin{array}{l} 2x – \dfrac{\pi }{{12}} = \dfrac{{5\pi }}{{12}} + k2\pi \\ 2x – \dfrac{\pi }{{12}} = \dfrac{{7\pi }}{{12}} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\pi \\ x = \dfrac{\pi }{3} + k\pi \end{array} \right.\,\,\,\,\,\left( {k \in Z} \right) \end{array}$$ | 649 | 1,421 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.288141 |
https://www.aqua-calc.com/one-to-all/volume/preset/cubic-fathom/7 | 1,642,568,419,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301263.50/warc/CC-MAIN-20220119033421-20220119063421-00703.warc.gz | 649,186,099 | 7,526 | # Convert cubic fathoms [ftm³] to other units of volume
## fathoms³ [ftm³] volume conversions
7 ftm³ = 0.03 acre-foot ftm³ to ac-ft 7 ftm³ = 62 963 341.21 capsules 0 ftm³ to cap 0 7 ftm³ = 47 572 302.26 capsules 00 ftm³ to cap 00 7 ftm³ = 31 251 877.37 capsules 000 ftm³ to cap 000 7 ftm³ = 42 815 072.02 capsules 00E ftm³ to cap 00E 7 ftm³ = 54 891 117.96 capsules 0E ftm³ to cap 0E 7 ftm³ = 89 198 067 capsules 1 ftm³ to cap 1 7 ftm³ = 2 378 615.11 capsules 10 ftm³ to cap 10 7 ftm³ = 4 281 507.2 capsules 11 ftm³ to cap 11 7 ftm³ = 8 563 014.39 capsules 12 ftm³ to cap 12 7 ftm³ = 5 708 676.27 capsules 12el ftm³ to cap 12el 7 ftm³ = 13 379 709.98 capsules 13 ftm³ to cap 13 7 ftm³ = 118 930 755.3 capsules 2 ftm³ to cap 2 7 ftm³ = 158 574 341.1 capsules 3 ftm³ to cap 3 7 ftm³ = 214 075 360.1 capsules 4 ftm³ to cap 4 7 ftm³ = 329 346 707.9 capsules 5 ftm³ to cap 5 7 ftm³ = 1 783 961.33 capsules 7 ftm³ to cap 7 7 ftm³ = 1 529 109.72 capsules Su7 ftm³ to cap Su7 7 ftm³ = 4.28 × 10+31 cubic angstroms ftm³ to ų 7 ftm³ = 1.28 × 10-32 cubic astronomical unit ftm³ to au³ 7 ftm³ = 42 815 072.02 cubic centimeters ftm³ to cm³ 7 ftm³ = 0.01 cubic chain ftm³ to ch³ 7 ftm³ = 42 815.07 cubic decimeters ftm³ to dm³ 7 ftm³ = 0.04 cubic dekameter ftm³ to dam³ 7 ftm³ = 1 512 cubic feet ftm³ to ft³ 7 ftm³ = 5.26 × 10-6 cubic furlong ftm³ to fur³ 7 ftm³ = 4.28 × 10-5 cubic hectometer ftm³ to hm³ 7 ftm³ = 2 612 736 cubic inches ftm³ to in³ 7 ftm³ = 4.28 × 10-8 cubic kilometer ftm³ to km³ 7 ftm³ = 5.06 × 10-47 cubic light year ftm³ to ly³ 7 ftm³ = 42.82 cubic meters ftm³ to m³ 7 ftm³ = 2.61 × 10+24 cubic microinches ftm³ to µin³ 7 ftm³ = 4.28 × 10+19 cubic micrometers ftm³ to µm³ 7 ftm³ = 4.28 × 10+19 cubic microns ftm³ to µ³ 7 ftm³ = 2.61 × 10+15 cubic mils ftm³ to mil³ 7 ftm³ = 1.03 × 10-8 cubic mile ftm³ to mi³ 7 ftm³ = 42 815 072 048 cubic millimeters ftm³ to mm³ 7 ftm³ = 4.28 × 10+28 cubic nanometers ftm³ to nm³ 7 ftm³ = 6.74 × 10-9 cubic nautical mile ftm³ to nmi³ 7 ftm³ = 1.46 × 10-48 cubic parsec ftm³ to pc³ 7 ftm³ = 4.28 × 10+37 cubic picometers ftm³ to pm³ 7 ftm³ = 2.61 × 10+15 cubic thous ftm³ to thou³ 7 ftm³ = 56 cubic yards ftm³ to yd³ 7 ftm³ = 428 150.72 deciliters ftm³ to dl 7 ftm³ = 4 281.51 dekaliters ftm³ to dal 7 ftm³ = 856 301 439 drops ftm³ to gt 7 ftm³ = 4.28 × 10-5 gigaliter ftm³ to Gl 7 ftm³ = 150 687.92 Imperial cups ftm³ to imperial c 7 ftm³ = 1 506 879.25 Imperial fluid ounces ftm³ to imperial fl.oz 7 ftm³ = 9 418 Imperial gallons ftm³ to UK gal 7 ftm³ = 75 343.96 Imperial pints ftm³ to imperial pt 7 ftm³ = 37 671.98 Imperial quarts ftm³ to UK qt 7 ftm³ = 42.82 kiloliters ftm³ to kl 7 ftm³ = 42 815.07 liters ftm³ to l 7 ftm³ = 0.04 megaliter ftm³ to Ml 7 ftm³ = 171 260.29 metric cups ftm³ to metric c 7 ftm³ = 4 281 507.2 metric dessertspoons ftm³ to metric dstspn 7 ftm³ = 2 854 338.13 metric tablespoons ftm³ to metric tbsp 7 ftm³ = 8 563 014.39 metric teaspoons ftm³ to metric tsp 7 ftm³ = 42 815 072.02 milliliters ftm³ to ml 7 ftm³ = 269.3 oil barrels ftm³ to bbl 7 ftm³ = 4.28 × 10-8 teraliter ftm³ to Tl 7 ftm³ = 180 968.73 US cups ftm³ to US c 7 ftm³ = 5 790 999.27 US dessertspoons ftm³ to US dstspn 7 ftm³ = 1 447 749.82 US fluid ounces ftm³ to fl.oz 7 ftm³ = 11 310.55 US gallons ftm³ to US gal 7 ftm³ = 90 484.36 US pints ftm³ to pt 7 ftm³ = 45 242.18 US quarts ftm³ to US qt 7 ftm³ = 2 895 499.64 US tablespoons ftm³ to US tbsp 7 ftm³ = 8 686 498.94 US teaspoons ftm³ to US tsp
#### Foods, Nutrients and Calories
TOMATO PEPPERS, UPC: 783963003102 contain(s) 20 calories per 100 grams (≈3.53 ounces) [ price ]
4827 foods that contain Aspartic acid. List of these foods starting with the highest contents of Aspartic acid and the lowest contents of Aspartic acid
#### Gravels, Substances and Oils
Substrate, Clay/Laterite weighs 1 019 kg/m³ (63.61409 lb/ft³) with specific gravity of 1.019 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
ZA-27, Zinc aluminum alloy (27% aluminum) weighs 5 000 kg/m³ (312.1398 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-508B, liquid (R508B) with temperature in the range of -106.67°C (-160.006°F) to -6.65°C (20.03°F)
#### Weights and Measurements
A square inch per minute (in²/min) is an imperial or United States customary measurement unit of kinematic viscosity.
Entropy is an extensive state function that describes the disorder in the system.
oz t/US tsp to gr/cm³ conversion table, oz t/US tsp to gr/cm³ unit converter or convert between all units of density measurement.
#### Calculators
Quadratic equation calculator. Find its real and complex roots | 1,974 | 4,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-05 | latest | en | 0.159694 |
https://www.neetprep.com/question/29793-geometrical-isomers-possible-followingCHCHCHCHCHCH------ | 1,563,411,424,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525483.62/warc/CC-MAIN-20190718001934-20190718023934-00397.warc.gz | 746,600,814 | 25,907 | # NEET Questions Solved
How many geometrical isomers are possible of the following?
CH3-CH=CH-CH=CH-CH3
1. 2 2. 3
3. 4 4. 6
(2) In case of symmetrical alkene, no. of geometrical isomers = 2n-1 + 2n/2 -1
=22-1 + 22/2 -1
=2+1 = 3
Difficulty Level:
• 21%
• 33%
• 42%
• 7% | 130 | 301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-30 | latest | en | 0.716074 |
https://utzx.com/units/convert-24-6-cm-to-inches/ | 1,726,610,173,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00739.warc.gz | 543,719,128 | 21,470 | # Convert 24.6 CM to Inches
## One centimeter is how many inches?
Do you wanna to convert 24.6 centimeters to a result in inches? first, you should know how many inches 1 centimeter is equal to.
You may use the cm to inches conversion formula to reverse the conversion.
## The length unit Centimeter
Centimeters or centimetres is the unit for length measurement in metric systems. It is abbreviated by the letter cm . The length unit meter has been defined internationally to the “International System of Units”, but the unit the unit cm is not. But a centimeter is equal to 100 meters. It’s also about 39.37 in.
An Anglo-American length unit for measuring is the inch (its symbol is in).. The symbol is in. In a variety of other European local languages, the word “inch” is similar to or is derived from “thumb”. Because a man’s thumb is approximately an inch in width.
• Electronic components, for example, the dimensions of the PC screen.
• Dimensions of truck and car tires.
## How Do I Convert 24.6 c to inch?
Convert centimeters into inches with the cm converter. It is possible to calculate the number of cm to inches through using this fundamental.
From the above, you have a good understanding for cm into inches. Using this simple formula, you can answer the following related questions:
• What’s the formula to convert inches from 24.6 cm?
• How do I convert cm to inches?
• How to change 24.6 cm to inches?
• What is cm to inch ratio?
• What is 24.6 cm equal to in inches?
cm inches 24.2 cm 9.52754 inches 24.25 cm 9.547225 inches 24.3 cm 9.56691 inches 24.35 cm 9.586595 inches 24.4 cm 9.60628 inches 24.45 cm 9.625965 inches 24.5 cm 9.64565 inches 24.55 cm 9.665335 inches 24.6 cm 9.68502 inches 24.65 cm 9.704705 inches 24.7 cm 9.72439 inches 24.75 cm 9.744075 inches 24.8 cm 9.76376 inches 24.85 cm 9.783445 inches 24.9 cm 9.80313 inches 24.95 cm 9.822815 inches | 524 | 1,891 | {"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} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.835736 |
https://kr.mathworks.com/matlabcentral/cody/problems/44243-ternary-conditional-operator/solutions/3341298 | 1,606,234,744,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176864.5/warc/CC-MAIN-20201124140942-20201124170942-00214.warc.gz | 363,918,780 | 17,098 | Cody
# Problem 44243. Ternary Conditional Operator
Solution 3341298
Submitted on 23 Oct 2020
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
mkdir @function_handle movefile colon1.m @function_handle/colon.m
The function colon is a built-in MATLAB function. Please use a different function name.
2 Fail
assert(isequal((2 > 1) : (@() 1) : (@() 2), 1))
The function colon is a built-in MATLAB function. Please use a different function name.
3 Fail
assert(isequal((1 > 2) : (@() 1) : (@() 2), 2))
The function colon is a built-in MATLAB function. Please use a different function name.
4 Fail
fib = @(f, n) (n > 2) : (@() f(f, n - 1) + f(f, n - 2)) : (@() 1); assert(fib(fib, 20) == 6765)
The function colon is a built-in MATLAB function. Please use a different function name.
5 Fail
x = magic(3); [m,I] = (x(1) > 0) : (@() max(x)) : (@() min(x)) assert(isequal(m, [8 9 7]) && isequal(I, [1 3 2]))
The function colon is a built-in MATLAB function. Please use a different function name.
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 369 | 1,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-50 | latest | en | 0.592749 |
https://www.dsprelated.com/showthread/comp.dsp/22749-1.php | 1,713,129,958,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00870.warc.gz | 705,914,415 | 18,578 | # Hilbert transform & analytic signals
Started by July 14, 2003
```Hi Guys,
I've been modeling (with MATLAB) the Hilbert transform
for use in generating the analytic signal (a complex
signal) corresponding to a real signal x(n). That is,
I'm computing a complex signal whose real part is x(n)
and whose imaginary part is the Hilbert transform of x(n).
I can use MATLAB's remez() command specifying 'Hilbert'
to get an FIR Hilbert filter that will generate the imaginary
part of my desired complex signal.
OK, no big deal.
Another idea for generating an analytic signal, that's
bounced around this newsgroup for quite a while, is
designing a lowpass filter and translating its center
frequency to Fs/4 by multiplying the lowpass coefficients
by exp(j*2*pi*Fs/4*Time), resulting in a complex filter
centered at Fs/4. That complex filter would then be
implemented as a dual-path filter where the I-path is a real-
coefficient filter having coefficients equal to the real
part of exp(j*2*pi*Fs/4*Time), and the Q-path is a real-
coefficient filter having coefficients equal to the imaginary
part of exp(j*2*pi*Fs/4*Time).
This all fine.
Here's what I've seen: The passband magnitude response ripple
using a remez-designed FIR Hilbert filter (floating point
numbers) having around 30 taps is roughly 0.1 dB peak-peak.
To get that same peak-peak passband ripple using the
complex filter scheme seems to require 3-4 times as many taps!!
If my modeling is correct, the complex-filter scheme for
generating an analytic signal requires *many* more multiplies
than the traditional FIR Hilbert method.
Have any of you guys modeled these two methods and seen the
same effects that I've seen?
Have I screwed something up here?
Thanks much,
[-Rick-]
```
```Rick,
Have you tried other methods to do the filter development, other than
the Remez, I have noticed with certain characteristics for inputs the
Remez has a difficulty getting get results, however, if I go straight
to the actual design equations for the Remez or Park's Mcclellon
(however it is spelled) I get a crisp a nice crisp answer. You might
play around with this fact, if I get time I may see if I can generate
some code that I will post later that may show this point, but work is
pretty busy right now, so it may not be anytime soon.
good luck,
Craig
ricklyon@REMOVE.onemain.com (Rick Lyons) wrote in message news:<3f135667.60818218@news.earthlink.net>...
> Hi Guys,
>
> I've been modeling (with MATLAB) the Hilbert transform
> for use in generating the analytic signal (a complex
> signal) corresponding to a real signal x(n). That is,
> I'm computing a complex signal whose real part is x(n)
> and whose imaginary part is the Hilbert transform of x(n).
> I can use MATLAB's remez() command specifying 'Hilbert'
> to get an FIR Hilbert filter that will generate the imaginary
> part of my desired complex signal.
> OK, no big deal.
>
> Another idea for generating an analytic signal, that's
> bounced around this newsgroup for quite a while, is
> designing a lowpass filter and translating its center
> frequency to Fs/4 by multiplying the lowpass coefficients
> by exp(j*2*pi*Fs/4*Time), resulting in a complex filter
> centered at Fs/4. That complex filter would then be
> implemented as a dual-path filter where the I-path is a real-
> coefficient filter having coefficients equal to the real
> part of exp(j*2*pi*Fs/4*Time), and the Q-path is a real-
> coefficient filter having coefficients equal to the imaginary
> part of exp(j*2*pi*Fs/4*Time).
> This all fine.
>
> Here's what I've seen: The passband magnitude response ripple
> using a remez-designed FIR Hilbert filter (floating point
> numbers) having around 30 taps is roughly 0.1 dB peak-peak.
> To get that same peak-peak passband ripple using the
> complex filter scheme seems to require 3-4 times as many taps!!
>
> If my modeling is correct, the complex-filter scheme for
> generating an analytic signal requires *many* more multiplies
> than the traditional FIR Hilbert method.
>
> Have any of you guys modeled these two methods and seen the
> same effects that I've seen?
> Have I screwed something up here?
>
> Thanks much,
> [-Rick-]
```
```"Rick Lyons" <ricklyon@REMOVE.onemain.com> wrote in message
> Here's what I've seen: The passband magnitude response ripple
> using a remez-designed FIR Hilbert filter (floating point
> numbers) having around 30 taps is roughly 0.1 dB peak-peak.
> To get that same peak-peak passband ripple using the
> complex filter scheme seems to require 3-4 times as many taps!!
To compare the two approaches, make the complex filter corresponding to the
Hilbert transform method -- multiply the hilbert filter by j, add 1 (real)
to the zeroth coefficient, and phasor-modulate the passband down to 0Hz.
You can then compare the features of this "hilbert halfband" with the
halfband prototype you designed directly. I think you'll notice the
passband isn't the same shape as the stopband. You'll also notice that the
filter is complex and not quite linear-phase. Each of these flaws increases
the number of degrees of freedom available to the remez design process,
which is why imposing the complementary restrictions leads to a requirement
of 4 times as many taps.
If you really don't care about those things, though, you can still use your
directly designed halfband prototype -- just dispense with the real-part
filtering like you do with the Hilbert method, and notice that half the
imaginary part coefficients are zero, leading to a 4x speedup in the
implementation, on par with the Hilbert method.
```
```On Tue, 15 Jul 2003 01:18:27 GMT, ricklyon@REMOVE.onemain.com (Rick
Lyons) wrote:
Hi Guys,
Shoot! I didn't explain myself properly.
Here's what I wrote:
>
>Here's what I've seen: The passband magnitude response ripple
>using a remez-designed FIR Hilbert filter (floating point
>numbers) having around 30 taps is roughly 0.1 dB peak-peak.
>To get that same peak-peak passband ripple using the
>complex filter scheme seems to require 3-4 times as many taps!!
That last sentence should have said:
To get that same peak-peak passband ripple *difference*
between the two real filters using the
complex filter scheme seems to require 3-4 times as many taps
in each filter!!
If my modeling is correct, the complex-filter scheme for
generating an analytic signal requires *many* more multiplies
than the traditional single-FIR-filter Hilbert method.
Thanks,
[-Rick-]
>
>
```
```On 15 Jul 2003 05:37:57 -0700, crrea2@umkc.edu (Craig) wrote:
>Rick,
>Have you tried other methods to do the filter development, other than
>the Remez, I have noticed with certain characteristics for inputs the
>Remez has a difficulty getting get results, however, if I go straight
>to the actual design equations for the Remez or Park's Mcclellon
>(however it is spelled) I get a crisp a nice crisp answer. You might
>play around with this fact, if I get time I may see if I can generate
>some code that I will post later that may show this point, but work is
>pretty busy right now, so it may not be anytime soon.
>
>good luck,
>
>Craig
Thanks Craig.
[-Rick-]
```
```In article 82396605.0307150437.7703c267@posting.google.com, Craig at
crrea2@umkc.edu wrote on 07/15/2003 08:37:
> Rick,
> Have you tried other methods to do the filter development, other than
> the Remez,
yeah, check out the firls() function in MATLAB, Rick.
r b-j
```
```Rick,
Can you post your MATLAB filter design script?
When you say 3-4 times more taps are you counting the half of the real
and half of the imaginary taps that should be 0 after the mix?
Dirk
Dirk Bell
ricklyon@REMOVE.onemain.com (Rick Lyons) wrote in message news:<3f13fc8e.103352734@news.earthlink.net>...
> On Tue, 15 Jul 2003 01:18:27 GMT, ricklyon@REMOVE.onemain.com (Rick
> Lyons) wrote:
>
> Hi Guys,
>
> Shoot! I didn't explain myself properly.
>
> Here's what I wrote:
> >
> >Here's what I've seen: The passband magnitude response ripple
> >using a remez-designed FIR Hilbert filter (floating point
> >numbers) having around 30 taps is roughly 0.1 dB peak-peak.
> >To get that same peak-peak passband ripple using the
> >complex filter scheme seems to require 3-4 times as many taps!!
>
>
> That last sentence should have said:
>
> To get that same peak-peak passband ripple *difference*
> between the two real filters using the
> complex filter scheme seems to require 3-4 times as many taps
> in each filter!!
>
> If my modeling is correct, the complex-filter scheme for
> generating an analytic signal requires *many* more multiplies
> than the traditional single-FIR-filter Hilbert method.
>
> Thanks,
> [-Rick-]
> >
> >
```
```Rick:
[snip]
"Rick Lyons" <ricklyon@REMOVE.onemain.com> wrote in message
> Hi Guys,
:
:
> Another idea for generating an analytic signal, that's
> bounced around this newsgroup for quite a while, is
> designing a lowpass filter and translating its center
:
:
> If my modeling is correct, the complex-filter scheme for
> generating an analytic signal requires *many* more multiplies
> than the traditional FIR Hilbert method.
>
> Have any of you guys modeled these two methods and seen the
> same effects that I've seen? Have I screwed something up here?
[snip]
I use this technique often, but I use IIR filters rather than FIR!
It works really well with an IIR filter translated/rotated to place its'
transition band at the origin and the IIR filter consumes a lot fewer
multiply-adds and memory than either of the FIR [Hilbert
or Filter] techniques.
A relatively simple low order IIR complex filter does a very efficient
job of separating positive frequencies from negative frequencies
and can be readily used to generate analytic signals to any prescribed
degree of "purity" [measured in dB of negative frequency suppression]
desired.
--
Peter
Consultant
Indialantic By-the-Sea, FL.
```
```On Wed, 16 Jul 2003 00:47:40 -0400, "Peter Brackett"
<ab4bc@ix.netcom.com> wrote:
>Rick:
>
>[snip]
>"Rick Lyons" <ricklyon@REMOVE.onemain.com> wrote in message
>> Hi Guys,
>:
>:
>> Another idea for generating an analytic signal, that's
>> bounced around this newsgroup for quite a while, is
>> designing a lowpass filter and translating its center
>:
>:
>> If my modeling is correct, the complex-filter scheme for
>> generating an analytic signal requires *many* more multiplies
>> than the traditional FIR Hilbert method.
>>
>> Have any of you guys modeled these two methods and seen the
>> same effects that I've seen? Have I screwed something up here?
>[snip]
>
>I use this technique often, but I use IIR filters rather than FIR!
>
>It works really well with an IIR filter translated/rotated to place its'
>transition band at the origin and the IIR filter consumes a lot fewer
>multiply-adds and memory than either of the FIR [Hilbert
>or Filter] techniques.
>
>A relatively simple low order IIR complex filter does a very efficient
>job of separating positive frequencies from negative frequencies
>and can be readily used to generate analytic signals to any prescribed
>degree of "purity" [measured in dB of negative frequency suppression]
>desired.
>
>--
>Peter
>Consultant
>Indialantic By-the-Sea, FL.
Hi Peter,
Interesting. I've never used an IIR filter to implement a
Hilbert transform. However, I've read that IIR Hilbert
transformers don't yield the desired exact 90-degree phase
shift.
By the way, how do you go about translating a lowpass
IIR filter's passband center frequency up to, say, Fs/4?
Thanks,
[-Rick-]
```
```On 15 Jul 2003 11:53:58 -0700, dirkman@erols.com (Dirk Bell) wrote:
>Rick,
>
>Can you post your MATLAB filter design script?
>
>When you say 3-4 times more taps are you counting the half of the real
>and half of the imaginary taps that should be 0 after the mix?
>
>Dirk
>
Hi Dirk,
yep the numbers I quoted include the zero-valued
coefficients.
What I've found is that a Hilbert FIR filter of 27 coefficients
will have a magnitude peak-peak ripple of roughly 0.1 dB.
The two real filters (making up the single complex
filter used to generate an analytic signal) had to have
over 100 coefficients each to keep the difference in their
magnitude responses within 0.1 dB.
If my modeling is correct, that 'complex filter' method
for generating an analytic signal, from a real signal, requires
about eight times as many multiplies as an equivalent single Hilbert
FIR filter!!
Here's my MATLAB code:
clear
%%% Design a Hilbert transformer (FIR)
Ntaps = 27; % Number of taps in Hilbert FIR
% Zero -out the coeffs that should be zero
MagB = 20*log10(MagB/max(MagB));
MagB(MagB<-1) = -1; % Threshold
figure(1), plot(MagB,'-b'), grid on, zoom on
xlabel('Freq'), ylabel('dB')
title('Mag of Hilb-FIR filter')
% Now design the complex FIR filter
% **** Design the lowpass FIR filter first ******
F_pass = 0.4; % Filter's cutoff freq
F_stop = 0.45; % Beginning of stopband
F = [0,F_pass,F_stop ,1 ]; % Freq vector for remez
M = [1,1,0,0 ]; % Mag vector for remez
N_taps = 107; % Number of FIR filter taps
N = N_taps-1;
B_LP = remez(N,F,M); % Calc the lowpass filter's coeffs
%B_LP = B_LP.*hanning(N_taps)'; % Window if you want.
%B_LP = 2*B_LP; % Gain compensation, if you want
% **** Translate LP filter to a BP filter (complex) ***
Mix_Seq = exp(j*(pi/2)*(0:N_taps-1)); % complex expon. at Fs/4.
B_BP = B_LP.*Mix_Seq(1:N_taps); % Freq translation.
B_Inphase = real(B_BP); % BP filter coeffs real parts
B_Quad = imag(B_BP); % BP filter coeffs imag parts
MagReal = abs(fft(B_Inphase,512));
MagReal = 20*log10(MagReal/max(MagReal)); % Log ref at 0 dB.
MagImag = 20*log10(MagImag/max(MagImag));
MagReal(MagReal<-1) = -1; % Threshold
MagImag(MagImag<-1) = -1;
figure(2), plot(MagReal,'-b'),hold on, plot(MagImag,'-r')
plot(MagReal-MagImag,'-k'), hold off, grid on, zoom on
title('Magnitudes: I filter-blue, Q filter-red, difference-black,')
xlabel('Freq'), ylabel('dB')
Thanks Dirk,
[-Rick-]
``` | 3,700 | 13,930 | {"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} | 2.859375 | 3 | CC-MAIN-2024-18 | latest | en | 0.924062 |
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http://forum.enjoysudoku.com/the-hardest-sudokus-new-thread-t6539-225.html | 1,531,792,727,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589537.21/warc/CC-MAIN-20180717012034-20180717032034-00183.warc.gz | 134,922,583 | 13,800 | ## The hardest sudokus (new thread)
Everything about Sudoku that doesn't fit in one of the other sections
### Re: The hardest sudokus (new thread)
At the start, Cigarette has a variant of the sk-loop that I don't recall seeing before.
____
In particular, note that the clues in r8c9 and r5c5 differ. Ditto for the clues in r9c6 and r6c9.
ronk
2012 Supporter
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA
### Re: The hardest sudokus (new thread)
For my understanding of the SK loop this is just a basic one, the pairs loop
r4c56=13 -> r4c89=89 -> r56c7=24 -> r89c7=67 -> r7c89=15 -> r7c56=58 -> r89c4=15 -> r56c4=79 -> r4c56=13
is all you need for the eliminations.
Different to your last one, which needed more than a pure pairs loop.
For the record:
In the other thread Coloin has posted extremely high suexrat ratings for many puzzles. A part of these puzzles (including the hardest) had already been published by me in this or that thread without suexrat ratings, so i claim them for me. To honour Coloins findings i want to give the hardest his name:
Code: Select all
` +-------+-------+-------+ | 1 2 . | 3 . . | 4 . . | | 5 . . | 1 . . | 3 . . | | . . . | . 6 . | . 2 . | +-------+-------+-------+ | 3 . . | 4 . . | 5 . . | | . 7 . | . . . | . . . | | . . 8 | . . . | . 3 . | +-------+-------+-------+ | 9 . . | . . . | . . . | | . . . | 9 . . | . 5 3 | | . . . | . 4 1 | 9 . . | +-------+-------+-------+ Coloin`
It seems to be very asymmetric with its 6 minrow/colums with 2 clues, maybe a reason, why it has not been found before.
btw what are the options for gsf's program for finding a "best" symmetry ?
So these are the puzzles with suexrat9 > 6800 and/or suexratt > 4000 (both limits higher than the current top puzzles) with ED,q1,q2:
eleven
Code: Select all
`10378 5710 ..3......4...8..36..8...1...4..6..73...9..........2..5..4.7..686........7..6..5.. 11.0 11.0 2.6 95078 98434 Coloin 9547 5174 ..3......4...8..36..8...1...4..6..73...9..........2..5..4.7..686....4...7.....5.. 11.0 11.0 2.6 95226 98414 9220 5804 ..3......4...8..36..83..1...4..6..73...9..........2..5..4.7..686........7.....5.. 10.9 10.9 2.6 95183 98587 8917 5333 ...4......5..8.2.6.....7...2...4....3......1...5.3.8.25...6.3.8..6....95..8...... 10.4 10.4 2.6 95273 67100 8568 4143 1....6.8...71....66.....15..3.9.....7....184.....2........9.41.5....4..8...8..5.. 10.3 10.3 10.0 95209 97527 8508 4880 .2....7..4....9.3.6..2.3.4.....1......89.....9....4.6..94....5.5.....6.3.....5... 10.9 10.9 2.6 95178 96787 8507 4354 ..3......4...8..36..8...1...4..6..73...9...1......2.....4.7..686........7..6..5.. 11.0 11.0 2.6 98810 98383 8491 4603 1...5.7.9..7.......6.......2...........5.1..2....2.39.3.4.9...15...1...3...8...4. 10.3 10.3 2.6 95219 72396 8446 4487 ..3......4...8..36..83..1...4..6..73...9...1......2.....4.7..686........7.....5.. 10.9 10.9 2.6 96128 98323 8042 4534 ..34.....4...8..36..8...1...4..6..73...9..........2..5..4.7..686........7.....5.. 11.0 11.0 2.6 95316 98797 7927 3770 ...4......5...9...6...2..1.2...7.9.1..5....7......8.3...6....9.7...3.1......9.327 10.3 10.3 2.6 59168 25378 7799 4027 ...4......5..8.2.6.....71..2...4....3......1...5.3.8.25...6.3.8..6....9...8...... 10.4 10.4 2.6 95223 74804 7515 5454 1...5.7.9..71......6.......2...........5.1..2....2.39.3...9...15...1...3...8...4. 10.4 10.4 2.6 95099 49425 7270 3085 1...56....5718...66...7.....9.3............9.8....5..35...17..8..2....1.......4.. 10.2 10.2 2.6 58616 22275 7140 4410 1....6.8...7.....66.....15..3.9.........2....7....854.....9.81.8...41..5.1....4.. 10.3 10.3 10.0 95140 21349 7052 3603 ..3......4...8..36..8...1...4..6..73...9..........2..5....7..686....4...7.....54. 11.0 1.2 1.2 95498 98798 6993 4979 ...4......5..8.2.6.....7...2...4....3......1...5.3.8.25...6.3.8..6..8.9...8...... 10.4 10.4 2.6 95111 58050 6952 4241 ...4......5..8.2.6.....7...2........3.....51...5.3.8.25...6.3.8..6....9...8.7.... 10.4 10.4 2.6 95327 95086 6903 3436 12..5...9.57...2...9..2..1....8..9.47...6.1.......4............57..9.6....6..3... 10.3 10.3 2.6 45827 77805 6753 4018 .2.4...8...7.....3.8.237.1.2.1....9..9....8.4...9......1.8...4.5............6...8 10.9 1.2 1.2 46997 97421 6752 4026 .2.4...8...7.....3...237.1.2.1....9..9....8.48..9......1.8...4.5.8..........6.... 10.9 1.2 1.2 46997 97421 6747 4052 .2.4...8...7.....3.8.237.1.2.1....9..9....8.4...9......1.8...4.5.8..........6.... 10.9 1.2 1.2 46997 97421 6678 4348 ..3..67.........2.79.2......3....6..5....4..76.7..3.453.5..74...............1...8 10.6 10.6 9.5 70048 91994 6645 4192 ..3.5.....5.1..2.66...2..4....8...9..8..1.6.5...6.....7.......4..........6...18.2 10.9 1.2 1.2 69047 97688 6628 4182 ..3.56....5.1..2.6....2..4...68...9..8..1.6.5...6.....7.......4..........6...18.2 10.9 1.2 1.2 69047 97688 6602 4877 ..3.....945...92....9..3.54....6....6..9..8....5..8.2..1.7................4..5.92 10.2 10.2 9.5 38405 1758 6374 4140 .....6..9..67...3.79...3....1...7.5...752......5.....2..167..2......14..8........ 10.2 10.2 2.6 63468 4946 6344 4397 ..34.........8..36..8...1.4.4..6..73...9..........2..5..4.7..686........7.....5.. 11.0 1.2 1.2 95358 98816 6338 4433 ...4......5..8.2.6.....71..2........3.....51...5.3.8.25...6.3.8..6..8.9...8...... 10.7 10.7 2.6 97677 98790 6069 4284 .2.4...8......9..2..9.3............5..8..7....4.5..82...46..21.6.21..4...1......8 10.4 10.4 9.4 95596 97645 6067 4552 ........9.5.7...2.7.9..2....1.67..5.......4..8....5....7.31....6....7.3..3..6...1 10.2 1.2 1.2 9037 26212 5995 4032 ....5.7..4.7..9.3..8....1..2.8..7.....4..8....7..9...3......6...4...2.97..2.....1 10.4 1.2 1.2 77612 42046 5973 4360 .2.4.6..9..7......6...2..5......15...1.64...28......1..3.26.....6..1...39....3... 10.2 1.2 1.2 95257 41099 5812 4278 1....67.9.5.........9.....4....9..3.....1....9..6..8.1..27.....7..8...4.8...6.1.7 10.4 10.4 2.6 21741 95184 5778 4081 .2..5...9...7.....7.....5..2..........4..8....6..2..91.3.2..9..6...9..13..1.6..2. 10.4 1.2 1.2 51889 95234 5708 4095 .......8...67......8...36.5.4..3..5......4..66....83.4..1.9..2..3...25..9........ 10.2 1.2 1.2 64720 23051 `
Coloin
Code: Select all
` 9456 4909 ..3.9....4...8..36..8...1...4..6..73...9..........2.....4.7..686........7.....5.4 10.4 10.4 2.6 95298 50600 8900 4556 ..3.9....4...8..36..8...1...4..6..73...9..........2.....4.7..686........7.....54. 10.4 10.4 2.6 95274 69328 8649 3845 ..39.....4...8..36..8...1...4..6..738......1......2.....4.7..686........7.....5.. 11.0 11.0 2.6 46938 95442 8538 4092 ..3......4...8..36..8...1...4..6..73...9..........2.....4.7..686........7..6..59. 10.8 10.8 2.6 36455 95608 8399 4167 ..39.....4...8..36..8...1...4..6..73...3...1......2.....4.7..686........7.....5.. 10.9 10.9 2.6 55568 97517 8342 3752 ..3.9....4...8..36..8...1...4..6..73...9....1.....2.....4.7..686........7.....5.. 10.3 10.3 2.6 95217 74239 8168 4488 ..3......4...8..36..83..1...4..6..73...9..........2.....4.7..686........7.....59. 10.8 10.8 2.6 49382 95472 8065 3924 ..3......4...8..36..8...1...4..6..73...9..........7..5..4.7..686....2...7.....5.. 10.9 10.9 2.6 95174 96815 7849 3471 ..3.2....4...8..36..8...1...4..6..73...9....1.....2.....4.7..686........7.....5.. 10.4 10.4 2.6 95224 74804 7607 3356 ..3......4...8..36..8...1...4..6..73...9..........2.....4.7..686...2....7..6..5.. 10.4 10.4 2.6 95122 37130 7522 3782 ..3......4...8..36..8...1...4..6..73...9..........4..5..4.7..686....2...7.....5.. 11.1 11.1 2.6 96569 98832 7506 2900 ..3..2...4...8..36..8...1...4..6..73...9.......6........4.7..686........7.....59. 10.9 1.2 1.2 95184 78304 7443 4799 ..3.9....4...8..36..8...1...4..6..73...9..........2.....4.7..686........7..6..5.. 10.4 10.4 2.6 95098 50898 7392 3990 ..3......4...8..36..8...1...4..6..73...9...1......2.....4.7..686....4...7.....5.. 10.9 10.9 2.6 97628 98508 7273 3943 ..34.....4...8..36..8...1...4..6..73...9..........2.....4.7..686........7.....59. 10.8 10.8 2.6 56053 96586 7256 4302 ..3..2...4...8..36..8...1...4..6..73...9..........4.....4.7..686........7.....59. 10.9 10.9 2.6 54251 85324 7227 3243 ..3.9....1...4..36..8..21...4..8..73...9................4.7..686........7.....5.4 9.7 9.7 2.6 13330 706 7069 2829 ..3......4...8..36..8...1...4..6..73...9.........42..5..4.7..686........7.....5.. 10.9 10.9 2.6 46041 97341 7034 3623 ..3.9....4...8..36..8...1...4..6..73...9..........2.....4.7..686..4.....7.....5.. 10.4 10.4 2.6 95264 94486 6997 2761 ..3......4...8..36..8...1...4..6..73...9........7.2..5..4.7..686........7.....5.. 10.8 10.8 2.6 14550 42333 6957 4279 ..3.9....4...8..36..8...1...4..6..73...9..........2.....4.7..686........7....65.. 10.4 10.4 2.6 95112 60280 6853 4096 ..3..61..4...8..36..89......4..6..73..............2..1..4.7..686........7.....5.. 10.9 1.2 1.2 69049 97687 `
eleven
Posts: 1799
Joined: 10 February 2008
### Re: The hardest sudokus (new thread)
eleven wrote:So these are the puzzles with suexrat9 > 6800 and/or suexratt > 4000 (both limits higher than the current top puzzles) with ED,q1,q2:
eleven
Code: Select all
`10378 5710 ..3......4...8..36..8...1...4..6..73...9..........2..5..4.7..686........7..6..5.. 11.0 11.0 2.6 95078 98434 Coloin`
Code: Select all
` *-----------------------------------------------------------------------------* | 1259 125679 3 | 12457 12459 145679 | 24789 24589 2479 | | 4 12579 12579 | 1257 8 1579 | 279 3 6 | | 259 25679 8 | 23457 23459 345679 | 1 2459 2479 | |-------------------------+-------------------------+-------------------------| | 12589 4 1259 | 158 6 158 | 289 7 3 | | 12358 12357 12567 | 9 1345 134578 | 2468 1248 124 | | 1389 1379 1679 | 13478 134 2 | 4689 1489 5 | |-------------------------+-------------------------+-------------------------| | 12359 12359 4 | 1235 7 1359 | 239 6 8 | | 6 123589 1259 | 123458 123459 134589 | 23479 1249 12479 | | 7 12389 129 | 6 12349 13489 | 5 1249 1249 | *-----------------------------------------------------------------------------*`
Just checked for the first one: double BB parttern (1259)r13c1/r89c3 => r2c3=7 & r7c1=3
ttt
ttt
Posts: 185
Joined: 20 October 2006
Location: vietnam
### Re: The hardest sudokus (new thread)
eleven
Posts: 1799
Joined: 10 February 2008
### Re: The hardest sudokus (new thread)
Code: Select all
` *-----------------------------------------------------------------------------* | 1259 125679 3 | 12457 12459 145679 | 24789 24589 2479 | | 4 12579 12579 | 1257 8 1579 | 279 3 6 | | 259 25679 8 | 23457 23459 345679 | 1 2459 2479 | |-------------------------+-------------------------+-------------------------| | 12589 4 1259 | 158 6 158 | 289 7 3 | | 12358 12357 12567 | 9 1345 134578 | 2468 1248 124 | | 1389 1379 1679 | 13478 134 2 | 4689 1489 5 | |-------------------------+-------------------------+-------------------------| | 12359 12359 4 | 1235 7 1359 | 239 6 8 | | 6 123589 1259 | 123458 123459 134589 | 23479 1249 12479 | | 7 12389 129 | 6 12349 13489 | 5 1249 1249 | *-----------------------------------------------------------------------------*`
Code: Select all
`1 1 . 1 1 1 . . . . 1B 1 1 . 1 . . . . . . . . . X . . 1A . 1 1 . 1 . . . 1 1 1 . 1 1 . 1 1 1 1 1 1 1 . . 1 . 1 1 . 1 . 1 . . . . 1 1T 1 1 1 . 1 1 . 1 1T . 1 1 . 1 1 `
A short explanation about BB pattern (1259)r89c3 for above puzzle:
- Look at candidates 1’s as above: if T=1 AND A&B<>1 => no 1’s on R7 or B9 => if T=1 then A or B must contain candidate 1’s. The same for candidates 2’s, 5’s, 9’s => Any extra candidates at A and B can be eliminated => r2c2<>7, r4c1<>8
Now, back to the puzzle then see that T must be one of (1,2), (1,5), (1,9), (2,5), (2,9), (5,9):
- If T=(1,2) => (A=1 AND B=2) OR (A=2 AND B=1) => r13c1=59 => r2c3 & r7c1<>1259, r2c3=7, r7c1=3
The same result for other cases T=(1,5), (1,9)…
Conclusion: r2c3=7, r7c1=3
I’m sorry by my poor English...,
Hope you can understand what I explained (that took me very long time than solving the puzzle ).
ttt
ttt
Posts: 185
Joined: 20 October 2006
Location: vietnam
### Re: The hardest sudokus (new thread)
Thanks for the explanation. I start to understand this pattern without ever having read the article
This is my alternative illustration:
For each of the 4 numbers we have a potential impossible pattern, XY for 15, XZ for 2, YZ for 9. So one of the starred cells (guardians) must hold each number to prevent that. But this implies, that none of the 4 numbers can be both in r13c1 and in r89c3, i.e. the #-cells must have all of them - and you can eliminate them in r123c3 and r789c1.
Code: Select all
` *-----------------------------------------------------------------------------* |#1259 1259 | 125 1259 159 | 29 259 29 | | *1259 *1259 |X125 Y159 |Z29 | |#259 259 | 25 259 59 | 1 259 29 | |-------------------------+-------------------------+-------------------------| |*1259 *1259 |X15 Y15 |Z29 | | 125 125 125 | 9 15 15 | 2 12 12 | | 19 19 19 | 1 1 2 | 9 19 5 | |-------------------------+-------------------------+-------------------------| |*1259 *1259 |X125 Y159 |Z29 | | 1259 #1259 | 125 1259 159 | 29 129 129 | | 129 #129 | 129 19 | 5 129 129 | *-----------------------------------------------------------------------------*`
[Added:]The first 3 puzzles in my "eleven" list and many in the "coloin" list can be solved this way.
eleven
Posts: 1799
Joined: 10 February 2008
### Re: The hardest sudokus (new thread)
Thanks eleven - there is no problem with the attribution of all your and subsequently trivially produced puzzles.
but thanks for he name ! I think I found a higher suexrat9 value - will update it !
Indeed the clue distribution may well be the key to suexrats apparent over rating. The 3 clues of the same value in a shute is a prominent feature.
I just wonder if we can tweak suexrat9 furthur.
dukuso helped by giving us the xxx 2 function in suexratt - this gave us the "node" count for the easiest 2 % of tries. The higher the relative ratio suexratt/suexrat9 - tends to indicate harder puzzles.
The way that suexrat9 locates the paired loci is unclear to me but if we can understand the apparent over- rating - we might be able to improve on suexratt. The discrepancy between isomorphic puzzles was never addressed - and perhaps didnt need to be.
What might be interesting is the lowest node count out of 1000 tries say.
I hope i'm not being over simplistic !
Does this indicate the easiest way for the program to guess/solve the puzzle ?
C
coloin
Posts: 1709
Joined: 05 May 2005
### Re: The hardest sudokus (new thread)
For me the basic problem is, that suexrat is a pure machine rating, having not implemented any sudoku solution techniques but singles. So there seems to be a quite good correlation to singles chain ratings, but the more other techniques can crack a puzzle, the higher is the chance, that suexrat overrates.
Now many of these puzzles in the lists above can "easily" be solved over BB patterns. I just checked the next one. Same solution:
Code: Select all
`+----------------------------+----------------------------+----------------------------+| 16789 1236789 12379 | 4 1259 123569 | 1579 3578 1379 || 1479 5 13479 | 139 8 139 | 2 347 6 || 14689 1234689 12349 | 123569 1259 7 | 1459 3458 1349 |+----------------------------+----------------------------+----------------------------+| 2 16789 #179 | 156789 4 15689 | 5679 3567 379 || 3 46789 #479 | 256789 2579 25689 | 45679 1 479 || 14679 14679 5 | 1679 3 169 | 8 467 2 |+----------------------------+----------------------------+----------------------------+| 5 12479 12479 | 1279 6 1249 | 3 247 8 ||#147 12347 6 | 12378 127 12348 | 147 9 5 ||#1479 123479 8 | 123579 12579 123459 | 1467 2467 147 |+----------------------------+----------------------------+----------------------------+`
You can use the impossible patterns in r267c46 for 1, r26c8 for 4, r267c48 for 7, r267c46 for 9. [corrected typos, but nobody noticed it anyway]
So nr 5 in my list is the first one, where a solution cannot be found quickly this way (the BB pattern 2379 in r12c7/r79c9 is useless for me - ok it places 7 in r1c7, but thats harder to see and does not solve it).
eleven
Posts: 1799
Joined: 10 February 2008
### Re: The hardest sudokus (new thread)
q1 99495, q2 99587, ED 10.8/10.8/3.4
Code: Select all
` +-------+-------+-------+ | 1 2 . | 3 . . | . . . | | 3 4 . | . . . | 1 . . | | . . 5 | . . . | . . . | +-------+-------+-------+ | 6 . 2 | 4 . . | 5 . . | | . . . | . 6 . | . 7 . | | . . . | . . 8 | . . 6 | +-------+-------+-------+ | . . 4 | 2 . . | 3 . . | | . . . | . 7 . | . . 9 | | . . . | . . 9 | . 8 . | +-------+-------+-------+ Catboat`
A grid with only 1 strong link (ER 10.1)
Code: Select all
` +-------+-------+-------+ | 1 2 . | 3 . . | 4 . . | | 5 . . | . 1 . | . . 3 | | . . 3 | . . 6 | . . . | +-------+-------+-------+ | 7 . . | . . 8 | 9 . . | | . . 2 | 5 . . | . . 4 | | . . . | . . . | . . . | +-------+-------+-------+ | . . 1 | 6 . . | . 4 . | | . . . | . 9 . | 8 . . | | . 9 . | . . 4 | . 7 . | +-------+-------+-------+ *------------------------------------------------------------------------* | 1 2 6789 | 3 578 579 | 4 5689 5678 | | 5 4678 46789 | 24789 1 279 | 267 2689 3 | | 489 478 3 | 24789 24578 6 | 1257 12589 12578 | |-----------------------+----------------------+-------------------------| | 7 13456 456 | 124 2346 8 | 9 12356 1256 | | 3689 1368 2 | 5 367 1379 | 1367 1368 4 | | 34689 134568 45689 | 12479 23467 12379 | 123567 123568 125678 | |-----------------------+----------------------+-------------------------| | 238 3578 1 | 6 23578 2357 | 235 4 9 | | 2346 34567 4567 | 127 9 12357 | 8 12356 1256 | | 2368 9 568 | 128 2358 4 | 12356 7 1256 | *------------------------------------------------------------------------*`
eleven
Posts: 1799
Joined: 10 February 2008
### Re: The hardest sudokus (new thread)
q1 99529, q2 99578, ED 10.4/10.4/10.0
Code: Select all
` +-------+-------+-------+ | 1 2 . | 4 . . | 3 . . | | 3 . . | . 1 . | . 5 . | | . . 6 | . . . | 1 . . | +-------+-------+-------+ | 7 . . | . 9 . | . . . | | . 4 . | 6 . 3 | . . . | | . . 3 | . . 2 | . . . | +-------+-------+-------+ | 5 . . | . 8 . | 7 . . | | . . 7 | . . . | . . 5 | | . . . | . . . | . 9 8 | +-------+-------+-------+ Discrepancy`
[Added:] Compare: ER 10.4, q1 17445, q2 147 (!)
Code: Select all
` +-------+-------+-------+ | 1 2 . | . . . | . . . | | 3 . . | 4 . . | . . . | | . . 5 | . 3 6 | 7 . . | +-------+-------+-------+ | . . 2 | . . . | . 6 . | | . . 8 | . 6 5 | . . 7 | | . . . | 7 . . | . 8 . | +-------+-------+-------+ | . . 3 | . 5 8 | . 7 . | | . . . | 9 . . | . . 3 | | . . . | . . . | . . 5 | +-------+-------+-------+`
eleven
Posts: 1799
Joined: 10 February 2008
### Re: The hardest sudokus (new thread)
My list of currently 4151 puzzles with ER 11+
(including a few new 11.8's).
[Edit:]
I deleted the list, because it contained duplicates.
Last edited by eleven on Sat Mar 26, 2011 4:11 pm, edited 1 time in total.
eleven
Posts: 1799
Joined: 10 February 2008
### Re: The hardest sudokus (new thread)
eleven wrote:My list of currently 4151puzzles with ER 11+ can be downloaded here (including a few new 11.8's).
According to my program, there are only 4093 unique sudokus in that list, for instance, puzzles #3853, #3854 and #3855 are the same.
Mauricio
Posts: 1174
Joined: 22 March 2006
### Re: The hardest sudokus (new thread)
Yes, sorry for that. I was sloppy, when i put the list together.
On the other hand some of my early puzzles (also with high rating) are missing.
I will post a corrected list in the next weeks, at the latest when champagne comes back.
Thanks for being so careful to check the list for duplicates.
I did not know, if there was any interest for it.
eleven
Posts: 1799
Joined: 10 February 2008
### Re: The hardest sudokus (new thread)
Now i made a new list with 4506 ER 11+ puzzles, hopefully all minimal and without duplicates. It can be downloaded here.
To my knowledge together with the puzzles in champagne's list here these are all ER 11.x puzzles, which have been published.
A few new puzzles for the suexrat rating, which i had not posted yet:
Code: Select all
`1...5......7..9.3...9..754...4..3.7..6........9.8........79..2......24.3..2...... 9924 6480 .23.....94.....1...9..3..4.2..81...4.....78..9...4...23...9...1.6..........5..... 8433 5517 1..4.6..........2..8..3.5.6.6...48.5............5..2.......3.9...7..8....4.6..3.8 8096 4859 ..3.5.....567....27..2...4......18..3...2...66...7...453..4...7.......9.......4.. 7712 4114 ...4.67....7.....6....7..512......9...5.6.1...91.426..3...........8.......4.1.5.. 7442 4377 1..4....9.56..9.......1..6..6....8..5....4.9.9....5.1..7....2..6....1.5....3..... 7141 5356 1.....7....6.8.1...89.....6.....4.5.6..59...19...1...8..2.........3.....8...6...7 7067 5315 1.....78...6.8.1....9.....6.....4.5.6..59...19...1...8..2.........3.....8...6...7 7046 5297 `
In sum i have checked more than 10 mio "step 1" puzzles (puzzles, which are not solvable with cell forcing chains using basics+jellyfish) now, all of them could be solved with T&E(NS+HS,2) (level 2 nested singles contradiction chains).
eleven
Posts: 1799
Joined: 10 February 2008
### Re: The hardest sudokus (new thread)
A couple of 11.3's
post63875.html#p63875
post64025.html#p64025
From patterns game 126, 3 puzzles
post204859.html#p204859
From patterns game 124, 3 puzzles
post204214.html#p204214
Mauricio
Posts: 1174
Joined: 22 March 2006
PreviousNext | 9,891 | 23,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-30 | latest | en | 0.803918 |
https://calculat.io/en/length/inches-to-mm/1771.65 | 1,701,624,718,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00532.warc.gz | 180,803,441 | 21,673 | # Convert 1771.65 inches to mm
## How many mm is 1771.65 inches?
1771.65 Inches is equal to 44999.91 Millimeters
## Explanation of 1771.65 Inches to Millimeters Conversion
Inches to Millimeters Conversion Formula: mm = in × 25.4
According to 'inches to mm' conversion formula if you want to convert 1771.65 (one thousand seven hundred seventy-one point six five) Inches to Millimeters you have to multiply 1771.65 by 25.4.
Here is the complete solution:
1771.65″ × 25.4
=
44999.91 mm
(forty-four thousand nine hundred ninety-nine point nine one millimeters)
## About "Inches to Millimeters" Calculator
This converter will help you to convert Inches to Millimeters (in to mm). For example, it can help you find out how many mm is 1771.65 inches? (The answer is: 44999.91). Enter the number of inches (e.g. '1771.65') and hit the 'Convert' button.
## Inches to Millimeters Conversion Table
InchesMillimeters
1771.5 Inches44996.1 mm
1771.51 Inches44996.354 mm
1771.52 Inches44996.608 mm
1771.53 Inches44996.862 mm
1771.54 Inches44997.116 mm
1771.55 Inches44997.37 mm
1771.56 Inches44997.624 mm
1771.57 Inches44997.878 mm
1771.58 Inches44998.132 mm
1771.59 Inches44998.386 mm
1771.6 Inches44998.64 mm
1771.61 Inches44998.894 mm
1771.62 Inches44999.148 mm
1771.63 Inches44999.402 mm
1771.64 Inches44999.656 mm
44999.91 mm
1771.66 Inches45000.164 mm
1771.67 Inches45000.418 mm
1771.68 Inches45000.672 mm
1771.69 Inches45000.926 mm
1771.7 Inches45001.18 mm
1771.71 Inches45001.434 mm
1771.72 Inches45001.688 mm
1771.73 Inches45001.942 mm
1771.74 Inches45002.196 mm
1771.75 Inches45002.45 mm
1771.76 Inches45002.704 mm
1771.77 Inches45002.958 mm
1771.78 Inches45003.212 mm
1771.79 Inches45003.466 mm
## FAQ
### How many mm is 1771.65 inches?
1771.65″ = 44999.91 mm | 612 | 1,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-50 | latest | en | 0.614278 |
https://philoid.com/question/8192-a-cuboid-is-of-dimensions-60-cm-xd7-54-cm-xd7-30-cm-how-many-small-cubes-with-side6-cm-can-be-placed-in-the-given-cuboid | 1,719,236,617,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00684.warc.gz | 394,113,267 | 7,869 | ##### A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side6 cm can be placed in the given cuboid?
Number of cubes =
= 450 cubes
15 | 47 | 160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-26 | latest | en | 0.868268 |
http://tilde.club/~matei/20211029144038-relative_formula_mass_and_moles.html | 1,669,781,165,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710719.4/warc/CC-MAIN-20221130024541-20221130054541-00471.warc.gz | 46,722,587 | 3,376 | # Relative formula mass and moles
• Relative formula mass (Mr) is the sum of the relative atomic masses of all atoms in a compound
• For example, the Ar of a hydrogen atom is 1, and the Ar of an oxygen atom is 16
• Therefore, a H2O compound, which has 2 hydrogen atoms, and 1 oxygen atoms, has a Mr value of 18 (2 x 1 + 1 x 16 = 18)
• A mole is the measurement of an amount of substance
• There are 6.023x1023 particles in a mole
• This is called the Avogadro constant
• To calculate the mass of a mole, you have to calculate the Ar or Mr of the element/compound, and add units (grams)
• Example, to calculate a mole of CO2:
• 1 x 12 + 2 x 16 = 44g
• To calculate the number of moles present in a sample, you have to divide the ass by Mr
• Example, to calculate the number of moles in 11g of CO2:
• 11g / 44g = 0.25
Created: 2021-11-13 Sat 19:33
Validate | 264 | 857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-49 | latest | en | 0.899603 |
http://nasawavelength.org/resource-search?facetSort=1&topicsSubjects=Physical+sciences&smdForumPrimary=Astrophysics&resourceType%5B%5D=Instructional+materials%3AData&resourceType%5B%5D=Instructional+materials%3ALesson+or+lesson+plan&materialsCost=1+cent+-+%241 | 1,547,998,163,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583722261.60/warc/CC-MAIN-20190120143527-20190120165527-00422.warc.gz | 164,687,342 | 14,980 | NASA Wavelength is transitioning to a new location on Jan. 30, 2019, read notice »
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# What's Hot in Yellowstone National Park
In this activity, learners discover new perspectives on geothermal features, such as geysers, mudpots, hot springs, and hot spring terraces by exploring infrared images. Learners will gain an understanding of infrared light and infrared imaging, as... (View More)
# The Crawl of the Crab
In this activity, students compare two images of the Crab Nebula taken more than 40 years apart. By measuring the motion of some of the knots of glowing gas in the neubla, students will be able to determine the date of the supernova explosion that... (View More)
# Newton's Law of Gravitation
The two activities included in this guide introduce students to the principles behind Newton's Law of Gravitation. The guide provides teachers with background information, preparation suggestions for each activity, as well as material lists, and... (View More)
# Newton's Second Law of Motion: Force, Acceleration and Velocity
This educational wallsheet provides several simple illustrations of Newton's Second Law. In the activity included, students study the motion shown in the drawings to decide how it relates to the object's velocity, whether or not the velocity is... (View More)
# Modeling Mass in the Solar System and a Galaxy
In this activity, students will discover how mass is distributed in the solar system and a galaxy by using kitty litter. Students compare the distribution of mass in a solar system to the distribution of mass in a galaxy. This is Activity 6a in the... (View More)
# Sorting out the Cosmic Zoo
In this hands-on activity, students analyze the data on Mystery Object Cards, observe that astronomical objects have many observable properties, and discover that these properties allow scientists to categorize astronomical objects into different... (View More)
# The Electromagnetic Spectrum: Resonating Atmosphere
Using a paper and tape device, students experience how atoms and molecules of gas in Earth’s atmosphere absorb electromagnetic energy through resonance. This activity is part of Unit 2 in the Space Based Astronomy guide that contains background... (View More)
1 | 735 | 3,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.87048 |
https://github.polettix.it/ETOOBUSY/2021/07/30/orthogonal-vectors/ | 1,726,516,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00790.warc.gz | 256,134,446 | 3,712 | TL;DR
Additional notes on the orthogonality of two vectors.
In previous post PWC123 - Square Points there is this statement, which might appear as having been drawn out of thin air:
Checking for orthogonality can be done calculating their regular scalar (or dot) product:
$v \cdot w = v_x w_x + v_y w_y$
This is 0 if and only if the two vectors are orthogonal, so it’s exactly the condition we are after.
So we have two vectors $\vec{v} = (v_1, …, v_n)$ and $\vec{w} = (w_1, …, w_n)$ (where $n = 2$ in the case of the previous post, but we’re aiming for the big, general case here) and we want to understand whether they’re orthogonal or not, based on their scalar product. Let’s go!
If they are, then $\vec{v}$ is also orthogonal to $-\vec{w}$, because $\vec{w}$ and $-\vec{w}$ are 180° apart from each other.
Let’s now consider a triangle $\overset{\triangle}{ABC}$ where:
$A = O + \vec{w} \\ B = O - \vec{w} \\ C = O + \vec{v}$
Sub-triangles $\overset{\triangle}{AOC}$ and $\overset{\triangle}{BOC}$ are congruent, which practically speaking means that they’re the same triangle with some translation and/or rotation and/or flipping (but no scaling). They satisfy the so-called Side-Angle-Side (SAS) condition for congruence, because:
• $\overline{OA}$ and $\overline{OB}$ have the same length:
$L_{\overline{OA}} = |\vec{w}| = |-\vec{w}| = L_{\overline{OB}}$
• $\overline{OC}$ is in common;
• angles in between $\widehat{AOC}$ and $\widehat{BOC}$ are both 90°.
This implies that segments $\overline{AC}$ and $\overline{BC}$ have the same length so, by extension, the square of their lengths:
$L_{\overline{AC}}^2 = |\vec{v} - \vec{w}|^2 = \sum_{i=1}^{n}(v_i - w_i)^2 = \sum_{i=1}^{n}(v_i^2 - 2 v_i w_i + w_i^2) \\ L_{\overline{BC}}^2 = |\vec{v} - (-\vec{w})|^2 = \sum_{i=1}^{n}(v_i + w_i)^2 = \sum_{i=1}^{n}(v_i^2 + 2 v_i w_i + w_i^2)$
are the same, i.e. their difference is $0$:
$L_{\overline{BC}}^2 - L_{\overline{AC}}^2 = 0 \\ \sum_{i=1}^{n}(v_i^2 + 2v_i w_i + w_i^2 - v_i^2 + 2v_i w_i - w_i^2) = 0 \\ 4 \sum_{i=1}^{n}v_i w_i = 0 \\ \sum_{i=1}^{n}v_i w_i = 0 \iff \vec{v} \cdot \vec{w} = 0$
i.e. the scalar product (a.k.a. dot product) between $\vec{v}$ and $\vec{w}$ is $0$.
On the other hand, it’s easy to go in the opposite direction: if the scalar product is $0$, then either one of the two vectors is the null vector (which is assumed to be orthogonal to any other vector, by definition), or they must form a triangle that satisfy the relation above where the length of $\overline{AC}$ is the same as the length of $\overline{BC}$, which in turn implies that segment $\overline{OC}$ is the height of the isosceles triangle and that angle $\widehat{AOC}$ is 90°.
Summing up, then:
$\vec{v} \perp \vec{w} \iff \vec{v} \cdot \vec{w} = 0$
which is the property we used in the previous post. What a ride, whew!
I find this fact awesome: by just doing some very simple arithmetics over the coordinates of the vectors we can easily establish if they’re perpendicular as geometric objects.
Enough for today… stay safe folks!
Comments? Octodon, , GitHub, Reddit, or drop me a line! | 1,044 | 3,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.814199 |
https://stats.stackexchange.com/questions/160222/a-closed-form-formula-for-the-normalizing-constant-in-standard-normal-auto-regre | 1,611,026,647,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517559.41/warc/CC-MAIN-20210119011203-20210119041203-00722.warc.gz | 591,429,343 | 35,703 | # A closed form formula for the normalizing constant in standard normal auto-regressive series?
Let $Z_t = c_1Z_{t-1} + c_2Z_{t-2} + ... + c_nZ_{t-n} + c\epsilon_t$
where $Z_t, \epsilon_t \sim \mathtt{N}(0,1)$ are iid variables and $Z_s \sim \mathtt{N}(0,1)$ for all $s$.
Given the values of $c_i$ for $i = 1 ...n$ is there a closed form formula for $c$?
We can derive that $c = \sqrt{1 - c_1^2}$ for the case $n = 1$ and that $c = \sqrt{1 - c_1^2 - c_2^2 -\frac{2c_1^2c_2}{1-c_2}}$ for case $n=2$.
So it seems like there should be a nice closed form formula for $c$. However I was unable to work through the algebra for $n = 3$.
But I think it might be that I don't know enough in this topic and perhaps this is a standard piece of work and someone has already worked it all out.
• The part $Z_s \sim N(0,1)$ is misleading to me. I don't see why the variance of $Z_s$ is fixed to one. In a first order autoregressive model we have: $\hbox{Var}(Z_t) = c_1^2 \hbox{Var}(Z_{t-1}) + c^2 \hbox{Var}(\epsilon_t)$. Denoting the variance of $Z_t$ $\hbox{Var}(Z_t) = \sigma^2_Z$ and assuming a stationary process where the variance of $Z_t$ is the same as the variance of $Z_{t-1}$, we have $(1 - c_1^2)\sigma^2_Z = c^2\cdot 1$ and therefore $\sigma^2_Z = c^2/(1 - c_1^2)$, which is not equal to $1$ for any value of $c_1$ and $c$. – javlacalle Jul 11 '15 at 9:02
• Are the values of the series $Z_t$, $t=1,2,...,n$ known? If so, can these values be part of the closed form solution you are looking for? – javlacalle Jul 11 '15 at 9:06
• Denote $\gamma(h) = C[Z_t, Z_t+h]$. Then, because of the linearity of the covariance, $$\gamma(0) = C[Z_t, Z_t] = C[c_1 Z_{t-1} + \cdots, c_1 Z_{t-1} + \cdots] = c^2_1\gamma(1) + c^2_2\gamma(2) + \cdots c^2_p\gamma(p) + c^2.$$ That is, $$c^2 = \gamma(0) - \sum_{k=1}^p c^2_k\gamma(k).$$ So what you need is an expression for $\gamma$. There are a few different ways to compute it, see for example econweb.ucsd.edu/muendler/teach/00s/ps1-prt1.pdf – Hunaphu Jul 15 '15 at 9:38
It's not strange you didn't calculate the AR(3) case. It's rather complicated! And no, there is no closed form for the AR(n)-case. For the AR(3) we start with the Yule-Walker equationsAR-model wikipedia(where $\gamma_j=\gamma_{-j}$):
• $\gamma_1=c_1\gamma_0+c_2\gamma_{-1}+c_3\gamma_{-2}=c_1\gamma_0+c_2\gamma_{1}+c_3\gamma_{2}\\ \gamma_2=c_1\gamma_1+c_2\gamma_0+c_3\gamma_{-1}=c_1\gamma_1+c_2\gamma_0+c_3\gamma_{1}\\ \gamma_3=c_1\gamma_2+c_2\gamma_1+c_3\gamma_{0}$
• Then we multiply the AR(3) expression by $Z_t$ and take the expectation: $E[Z_t^2]=c_1E[Z_tZ_{t-1}]+c_2E[Z_tZ_{t-2}]+c_3E[Z_tZ_{t-3}]+c^2\Rightarrow \\ \gamma_0=c_1\gamma_1+c_2\gamma_2+c_3\gamma_3+c^2$
• From Yule-Walker do we get: $\gamma_1=\frac{c_1+c_2c_3}{1-c_2-c_1c_3-c_3^2}\gamma_0\\ \gamma_2=\left(\frac{c_1(c_1+c_2c_3)+c_3(c_1+c_2c_3)}{1-c_2-c_1c_3-c_3^2}+c_2\right)\gamma_0\\ \gamma_3=\left(\frac{(c_1^ 2+c_3c_1+c_2)(c_1+c_2c_3)}{1-c_2-c_1c_3-c_3^2}+c_2+c_3\right)\gamma_0$
• Plugging these into the equation above (the second point) give what you want by setting $Var[Z_s]=\gamma_0=1$.
• In Hamiltons book on time series (p59) he writes that the solutions for $\gamma_j$ takes the form: $\gamma_j=g_1\lambda_1^j+g_2\lambda_2^j+\cdots+g_p\lambda_p^j$
where the eigenvalues are the solutions of $\lambda^p-c_1\lambda^{p-1}-c_2\lambda^{p-2}-\cdots-c_p=0$ (This is exactly the method outlined in the paper user Hunaphu has added). So my guess is that these equations will be very complicated for $p=5$ or above since no formula exists for the solution of an equation of degree five or higher. It has to be solved by e.g. elliptic functions or theta functions. So to find a general solution for the AR(n)-case you must find a general solution for the n'th-degree algebraic equation which does not exist.
• (+1) This is the idea that I had in mind when I asked the OP whether the data of $Z_t$ can be part of the closed form formula. This formula would then be given by your second point: $c = \sqrt{\gamma_0 + c_1\gamma_1 + c_2\gamma_2 + \cdots + c_n\gamma_n}$, where the terms $\gamma_i$ are the expressions of the sample autocovariances of order $i$ (and therefore depend on $Z_t$). – javlacalle Jul 14 '15 at 16:47
• I don't see any reason to restrict the sample variance of $Z_t$ to 1, $\gamma_0=1$. In an AR process of order $n$ it will in general be different from zero. The OP did not clarify this issue. – javlacalle Jul 14 '15 at 16:48 | 1,664 | 4,437 | {"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} | 3.46875 | 3 | CC-MAIN-2021-04 | latest | en | 0.874329 |
https://la.mathworks.com/matlabcentral/profile/authors/13882981?s_tid=cody_local_to_profile | 1,606,162,906,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141164142.1/warc/CC-MAIN-20201123182720-20201123212720-00591.warc.gz | 375,609,756 | 19,139 | Community Profile
# Michael Neiman
##### Last seen: alrededor de 2 meses ago
11 total contributions since 2020
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What is the distance from point P(x,y) to the line Ax + By + C = 0?
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alrededor de 2 meses ago
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Angle between Two Vectors
The dot product relationship, a dot b = | a | | b | cos(theta), can be used to determine the acute angle between vector a and ve...
alrededor de 2 meses ago
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Is the Point in a Circle?
Check whether a point or multiple points is/are in a circle centered at point (x0, y0) with radius r. Points = [x, y]; c...
alrededor de 2 meses ago
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Find third Side of a right triangle given hypotenuse and a side. No * - or other functions allowed
Find the remaining side of a triangle given the hypotenuse and a side. However, the normal functions and symbols are not allowe...
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Volume of a Parallelepiped
Calculate the volume of a Parallelepiped given the vectors for three edges that meet at one vertex. A cube is a special case ...
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Are all the three given point in the same line?
In this problem the input is the coordinate of the three points in a XY plane? P1(X1,Y1) P2(X2,Y2) P3(X3,Y3) how can...
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Calculate the area of a triangle between three points
Calculate the area of a triangle between three points: P1(X1,Y1) P2(X2,Y2) P3(X3,Y3) these three points are the vert...
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Television Screen Dimensions
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Right Triangle Side Lengths (Inspired by Project Euler Problem 39)
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Compute the dilation of a binary image
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Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
2 meses ago | 629 | 2,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-50 | latest | en | 0.781611 |
http://6choice.eu/read-four-views-on-free-will-pdf/ | 1,521,345,834,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645513.14/warc/CC-MAIN-20180318032649-20180318052649-00319.warc.gz | 5,687,396 | 5,634 | In this publication, as well as some of Hinton’s later work, the word was occasionally spelled “tessaract”. The tesseract can be constructed in a number of ways. Each pair of non-parallel hyperplanes intersects to form 24 square read four views on free will pdf in a tesseract. Three cubes and three squares intersect at each edge.
There are four cubes, six squares, and four edges meeting at every vertex. All in all, it consists of 8 cubes, 24 squares, 32 edges, and 16 vertices. Two points A and B can be connected to a line, giving a new line segment AB. Two parallel line segments AB and CD can be connected to become a square, with the corners marked as ABCD. Two parallel squares ABCD and EFGH can be connected to become a cube, with the corners marked as ABCDEFGH. Two parallel cubes ABCDEFGH and IJKLMNOP can be connected to become a hypercube, with the corners marked as ABCDEFGHIJKLMNOP. It is possible to project tesseracts into three- and two-dimensional spaces, similarly to projecting a cube into two-dimensional space.
Projections on the 2D-plane become more instructive by rearranging the positions of the projected vertices. A tesseract is in principle obtained by combining two cubes. The scheme is similar to the construction of a cube from two squares: juxtapose two copies of the lower-dimensional cube and connect the corresponding vertices. Each edge of a tesseract is of the same length. 4 and there are many different paths to allow weight balancing. The nearest and farthest cells are projected onto the cube, and the remaining 6 cells are projected onto the 6 square faces of the cube. Two pairs of cells project to the upper and lower halves of this envelope, and the 4 remaining cells project to the side faces.
Six cells project onto rhombic prisms, which are laid out in the hexagonal prism in a way analogous to how the faces of the 3D cube project onto 6 rhombs in a hexagonal envelope under vertex-first projection. The two remaining cells project onto the prism bases. Two vertices of the tesseract are projected to the origin. This projection is also the one with maximal volume. The nondiagonal elements represent the number of row elements are incident to the column element. The configurations for dual polytopes can be seen by rotating the matrix elements by 180 degrees. | 504 | 2,315 | {"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} | 3.390625 | 3 | CC-MAIN-2018-13 | longest | en | 0.938325 |
https://gamedev.stackexchange.com/questions/31546/find-nearest-tile-of-type-x | 1,717,107,876,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971684053.99/warc/CC-MAIN-20240530210614-20240531000614-00035.warc.gz | 229,974,240 | 41,544 | # Find nearest tile of type x
I currently have an array of tiles which makes up the tilemap (stored as an int array) and I want an NPC to be able to move itself to the nearest tile of type X (ie find the nearest tree and chop it down). What's the best algorithm to use to find the nearest tile of a certain type given a location (x,y)?
Problems like this can easily be solved by using Dijkstra's algorithm. See this answer for an explanation of how it works.
The others suggested a flood-fill, but that won't work (or at least won't be optimal) if there are obstacles (not trees) somewhere along the path. If there are only a few items to find, you could also use A* with an appropriate heuristic (eg. min. straight-line distance to items).
Update: If you don't want to implement Djikstra's algorithm yourself but have an A* implementation in your path-finding library, you could also set the A* heuristic to a constant value (eg. that it always returns 0). That way, A* will behave the same as Djikstra's.
Also note that in the examples I linked to, there's usually one goal-node to find. But it works just as well when searching for a "type" of node (just end the search whenever you find a tile of your desired type).
• Dijkstra's is the right way to go for simple pathfinding. It works even better if the tiles have different travel speeds, like moving over a paved path is faster than moving over grassland. If, on the other hand, you have a huge tileset and Dijkstra's is taking too long, then use A*. Jul 3, 2012 at 13:21
• @stevehb It's not really about the size of the tileset (amount of nodes) or "simple" vs. "complex" pathfinding. A* is using a heuristic to predict the best direction to go, thus potentially speeding up the search drastically. If you're searching for any tile of type X, then applying a good heuristic might be less efficient or even impossible compared to an uninformed search such as Djikstra's. Jul 3, 2012 at 13:36
If you know absolutely nothing about where such a tile could be I would suggest a sort of breadth first style floodfill:
PathNode start;
PathNodeType query; // the type we are looking for
Queue<PathNode> queue; // gives us a breadth first style search
HashSet<PathNode> visited; // tiles we already visited
queue.Push(start)
while(queue.Count > 0){
PathNode current = queue.Pop();
foreach(PathNode neighbour in current.Neighbours)
{
if(!visited.Contains(neighbour))
{
if(neighbour.Type == query)
{
return neighbour;
}
else if(neighbour.IsWalkable)
{
queue.Push(neighbour);
}
}
}
}
This algorithm first searches the tiles closest to start and then starts circling wider and wider over the walkable tiles until a tile of the given tile is found. By first checking if a neighbour tile is of the given type and only then checking if it is walkable we can also found tiles that are next to walkable tiles but arent walkable itself.
• My words exactly. I'll just add a nice way of visualizing the algorithm: it behaves as if you started pouring water on your starting tile, which then flows into adjacent (accessible) tiles in all directions. (Hence, 'floodfill'.) The first tile of the sought type to be flooded is your nearest match.
– ver
Jul 3, 2012 at 6:32
• You should keep in mind, though, the special case here - when there are multiple equidistant tiles of the target type. (I.e. they are the same distance from your starting tile.) The algorithm provided by @RoyT just returns whichever one is first chosen by the PathNode.Neighbours implementation.
– ver
Jul 3, 2012 at 6:36
• You need some sort of "visited" flag, otherwise your code will loop endlessly. Also this approach will fall short (won't find the optimal path) if there are obstacles to consider. Jul 3, 2012 at 6:42
• @bummzack yes, it may not find the optimal path, but that's irrelevant. The asker wanted an algorithm for finding the nearest tile that specifies the given condition; not a pathfinding one.
– ver
Jul 3, 2012 at 7:08
• In a tile-based environment, this flood-fill is basically an implementation of Djikstra's anyway. It just needs to keep track of costs to also get the path to the found tile (to satisfy the requirement of the OP: "I want an NPC to be able to move itself to the nearest tile of type X"). Jul 3, 2012 at 7:35
One approach is a simple breadth first search. If you've heard of A*, this is sort of the basis for that algorithm. Except you don't have a known goal location, so you can't direct the search in any one direction So you search all directions at once! There's some implementation details and nice little examples for tile based games here.
Could you add some more details like what language you are in? I mean I guess there could be a general algorithm.
If you have these different tiles in classes per se, you could add a move function in your NPC class or method. It would have to "see" the tile, so you could give it a tile radius and say if anything within X amount of squares of the NPC is this ID type of tile, then move to its x and y location.
An example in an OO language like Java:
public Tile Tree;
Tree.cutTree(NPC.getX, NPC,getY, Tree.getX, Tree.getY)
public void cutTree(int NPCX, int NPCY, int treeX, int treeY)
{
if(treeX >= NPCX >= 5)
{
if(treeX >= NPCX >= 5)
{
NPC.move(treeX - 1, treeY - 1);
}
}
}
Again I don't know EXACTLY what numbers to add and whether you have a move() function, but basically I would use the x and y position of the two tiles (if you can access each tree tile in the array).
The if statements are where I would use like the edge of the frame or screen but you have to remember if the map of tiles is big enough you could be accessing a tree tile across the map. | 1,376 | 5,687 | {"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-2024-22 | latest | en | 0.937881 |
http://www.shogun-toolbox.org/static/notebook/current/multiclass_reduction.html | 1,556,223,987,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578733077.68/warc/CC-MAIN-20190425193912-20190425215912-00200.warc.gz | 296,391,935 | 239,168 | # Multiclass Reductions¶
#### by Chiyuan Zhang and Sören Sonnenburg
This notebook demonstrates the reduction of a multiclass problem into binary ones using Shogun. Here, we will describe the built-in One-vs-Rest, One-vs-One and Error Correcting Output Codes strategies.
In SHOGUN, the strategies of reducing a multiclass problem to binary classification problems are described by an instance of CMulticlassStrategy. A multiclass strategy describes
• How to train the multiclass machine as a number of binary machines?
• How many binary machines are needed?
• For each binary machine, what subset of the training samples are used, and how are they colored? In multiclass problems, we use coloring to refer partitioning the classes into two groups: $+1$ and $-1$, or black and white, or any other meaningful names.
• How to combine the prediction results of binary machines into the final multiclass prediction?
The user can derive from the virtual class CMulticlassStrategy to implement a customized multiclass strategy. But usually the built-in strategies are enough for general problems. We will describe the built-in One-vs-Rest, One-vs-One and Error-Correcting Output Codes strategies in this tutorial.
The basic routine to use a multiclass machine with reduction to binary problems in shogun is to create a generic multiclass machine and then assign a particular multiclass strategy and a base binary machine.
## One-vs-Rest and One-vs-One¶
The One-vs-Rest strategy is implemented in CMulticlassOneVsRestStrategy. As indicated by the name, this strategy reduce a $K$-class problem to $K$ binary sub-problems. For the $k$-th problem, where $k\in\{1,\ldots,K\}$, the samples from class $k$ are colored as $+1$, and the samples from other classes are colored as $-1$. The multiclass prediction is given as
$$f(x) = \operatorname*{argmax}_{k\in\{1,\ldots,K\}}\; f_k(x)$$
where $f_k(x)$ is the prediction of the $k$-th binary machines.
The One-vs-Rest strategy is easy to implement yet produces excellent performance in many cases. One interesting paper, Rifkin, R. M. and Klautau, A. (2004). In defense of one-vs-all classification. Journal of Machine Learning Research, 5:101–141, it was shown that the One-vs-Rest strategy can be
as accurate as any other approach, assuming that the underlying binary classifiers are well-tuned regularized classifiers such as support vector machines.
Implemented in CMulticlassOneVsOneStrategy, the One-vs-One strategy is another simple and intuitive strategy: it basically produces one binary problem for each pair of classes. So there will be $\binom{K}{2}$ binary problems. At prediction time, the output of every binary classifiers are collected to do voting for the $K$ classes. The class with the highest vote becomes the final prediction.
Compared with the One-vs-Rest strategy, the One-vs-One strategy is usually more costly to train and evaluate because more binary machines are used.
In the following, we demonstrate how to use SHOGUN's One-vs-Rest and One-vs-One multiclass learning strategy on the USPS dataset. For demonstration, we randomly 200 samples from each class for training and 200 samples from each class for testing.
The CLibLinear is used as the base binary classifier in a CLinearMulticlassMachine, with One-vs-Rest and One-vs-One strategies. The running time and performance (on my machine) is reported below:
------------------------------------------------- Strategy Training Time Test Time Accuracy ------------- ------------- --------- -------- One-vs-Rest 12.68 0.27 92.00% One-vs-One 11.54 1.50 93.90% ------------------------------------------------- Table: Comparison of One-vs-Rest and One-vs-One multiclass reduction strategy on the USPS dataset.
First we load the data and initialize random splitting:
In [1]:
%pylab inline
%matplotlib inline
import numpy as np
from numpy import random
Xall = mat['data']
#normalize examples to have norm one
Xall = Xall / np.sqrt(sum(Xall**2,0))
Yall = mat['label'].squeeze()
# map from 1..10 to 0..9, since shogun
# requires multiclass labels to be
# 0, 1, ..., K-1
Yall = Yall - 1
N_train_per_class = 200
N_test_per_class = 200
N_class = 10
# to make the results reproducable
random.seed(0)
# index for subsampling
index = np.zeros((N_train_per_class+N_test_per_class, N_class), 'i')
for k in range(N_class):
Ik = (Yall == k).nonzero()[0] # index for samples of class k
I_subsample = random.permutation(len(Ik))[:N_train_per_class+N_test_per_class]
index[:, k] = Ik[I_subsample]
idx_train = index[:N_train_per_class, :].reshape(N_train_per_class*N_class)
idx_test = index[N_train_per_class:, :].reshape(N_test_per_class*N_class)
random.shuffle(idx_train)
random.shuffle(idx_test)
Populating the interactive namespace from numpy and matplotlib
import SHOGUN components and convert features into SHOGUN format:
In [2]:
from modshogun import RealFeatures, MulticlassLabels
from modshogun import LibLinear, L2R_L2LOSS_SVC, LinearMulticlassMachine
from modshogun import MulticlassOneVsOneStrategy, MulticlassOneVsRestStrategy
from modshogun import MulticlassAccuracy
import time
feats_train = RealFeatures(Xall[:, idx_train])
feats_test = RealFeatures(Xall[:, idx_test])
lab_train = MulticlassLabels(Yall[idx_train].astype('d'))
lab_test = MulticlassLabels(Yall[idx_test].astype('d'))
define a helper function to train and evaluate multiclass machine given a strategy:
In [3]:
def evaluate(strategy, C):
bin_machine = LibLinear(L2R_L2LOSS_SVC)
bin_machine.set_bias_enabled(True)
bin_machine.set_C(C, C)
mc_machine = LinearMulticlassMachine(strategy, feats_train, bin_machine, lab_train)
t_begin = time.clock()
mc_machine.train()
t_train = time.clock() - t_begin
t_begin = time.clock()
pred_test = mc_machine.apply_multiclass(feats_test)
t_test = time.clock() - t_begin
evaluator = MulticlassAccuracy()
acc = evaluator.evaluate(pred_test, lab_test)
print "training time: %.4f" % t_train
print "testing time: %.4f" % t_test
print "accuracy: %.4f" % acc
Test on One-vs-Rest and One-vs-One strategies.
In [4]:
print "\nOne-vs-Rest"
print "="*60
evaluate(MulticlassOneVsRestStrategy(), 5.0)
print "\nOne-vs-One"
print "="*60
evaluate(MulticlassOneVsOneStrategy(), 2.0)
One-vs-Rest
============================================================
training time: 0.3293
testing time: 0.0060
accuracy: 0.9265
One-vs-One
============================================================
training time: 0.3598
testing time: 0.0402
accuracy: 0.9465
/usr/local/lib/python2.7/dist-packages/ipykernel/__main__.py:7: RuntimeWarning: [WARN] In file /home/buildslave/nightly_default/build/src/shogun/multiclass/MulticlassOneVsOneStrategy.cpp line 34: MulticlassOneVsOneStrategy::CMulticlassOneVsOneStrategy(): register parameters!
LibLinear also has a true multiclass SVM implemenentation - so it is worthwhile to compare training time and accuracy with the above reduction schemes:
In [5]:
from modshogun import MulticlassLibLinear
mcsvm = MulticlassLibLinear(5.0, feats_train, lab_train)
mcsvm.set_use_bias(True)
t_begin = time.clock()
mcsvm.train(feats_train)
t_train = time.clock() - t_begin
t_begin = time.clock()
pred_test = mcsvm.apply_multiclass(feats_test)
t_test = time.clock() - t_begin
evaluator = MulticlassAccuracy()
acc = evaluator.evaluate(pred_test, lab_test)
print "training time: %.4f" % t_train
print "testing time: %.4f" % t_test
print "accuracy: %.4f" % acc
training time: 1.0422
testing time: 0.0061
accuracy: 0.9305
As you can see performance of all the three is very much the same though the multiclass svm is a bit faster in training. Usually training time of the true multiclass SVM is much slower than one-vs-rest approach. It should be noted that classification performance of one-vs-one is known to be slightly superior to one-vs-rest since the machines do not have to be properly scaled like in the one-vs-rest approach. However, with larger number of classes one-vs-one quickly becomes prohibitive and so one-vs-rest is the only suitable approach - or other schemes presented below.
## Error-Correcting Output Codes¶
Error-Correcting Output Codes (ECOC) is a generalization of the One-vs-Rest and One-vs-One strategies. For example, we can represent the One-vs-Rest strategy with the following $K\times K$ coding matrix, or a codebook:
$$\begin{bmatrix} +1 & -1 & -1 & \ldots & -1 & -1 \\\\ -1 & +1 & -1 & \ldots & -1 & -1\\\\ -1 & -1 & +1 & \ldots & -1 & -1\\\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\\\ -1 & -1 & -1 & \ldots & +1 & -1 \\\\ -1 & -1 & -1 & \ldots & -1 & +1 \end{bmatrix}$$
Denote the codebook by $B$, there is one column of the codebook associated with each of the $K$ classes. For example, the code for class $1$ is $[+1,-1,-1,\ldots,-1]$. Each row of the codebook corresponds to a binary coloring of all the $K$ classes. For example, in the first row, the class $1$ is colored as $+1$, while the rest of the classes are all colored as $-1$. Associated with each row, there is a binary classifier trained according to the coloring. For example, the binary classifier associated with the first row is trained by treating all the examples of class $1$ as positive examples, and all the examples of the rest of the classes as negative examples.
In this special case, there are $K$ rows in the codebook. The number of rows in the codebook is usually called the code length. As we can see, this codebook exactly describes how the One-vs-Rest strategy trains the binary sub-machines.
In [6]:
OvR=-np.ones((10,10))
fill_diagonal(OvR, +1)
_=gray()
_=imshow(OvR, interpolation='nearest')
_=gca().set_xticks([])
_=gca().set_yticks([])
A further generalization is to allow $0$-values in the codebook. A $0$ for a class $k$ in a row means we ignore (the examples of) class $k$ when training the binary classifiers associated with this row. With this generalization, we can also easily describes the One-vs-One strategy with a $\binom{K}{2}\times K$ codebook:
$$\begin{bmatrix} +1 & -1 & 0 & \ldots & 0 & 0 \\\\ +1 & 0 & -1 & \ldots & 0 & 0 \\\\ \vdots & \vdots & \vdots & \ddots & \vdots & 0 \\\\ +1 & 0 & 0 & \ldots & -1 & 0 \\\\ 0 & +1 & -1 & \ldots & 0 & 0 \\\\ \vdots & \vdots & \vdots & & \vdots & \vdots \\\\ 0 & 0 & 0 & \ldots & +1 & -1 \end{bmatrix}$$
Here each of the $\binom{K}{2}$ rows describes a binary classifier trained with a pair of classes. The resultant binary classifiers will be identical as those described by a One-vs-One strategy.
Since $0$ is allowed in the codebook to ignore some classes, this kind of codebooks are usually called sparse codebook, while the codebooks with only $+1$ and $-1$ are usually called dense codebook.
In general case, we can specify any code length and fill the codebook arbitrarily. However, some rules should be followed:
• Each row must describe a valid binary coloring. In other words, both $+1$ and $-1$ should appear at least once in each row. Or else a binary classifier cannot be obtained for this row.
• It is good to avoid duplicated rows. There is generally no harm to have duplicated rows, but the resultant binary classifiers are completely identical provided the training algorithm for the binary classifiers are deterministic. So this can be a waste of computational resource.
• Negative rows are also duplicated. Simply inversing the sign of a code row does not produce a "new" code row. Because the resultant binary classifier will simply be the negative classifier associated with the original row.
Though you can certainly generate your own codebook, it is usually easier to use the SHOGUN built-in procedures to generate codebook automatically. There are various codebook generators (called encoders) in SHOGUN. However, before describing those encoders in details, let us notice that a codebook only describes how the sub-machines are trained. But we still need a way to specify how the binary classification results of the sub-machines can be combined to get a multiclass classification result.
Review the codebook again: corresponding to each class, there is a column. We call the codebook column the (binary) code for that class. For a new sample $x$, by applying the binary classifiers associated with each row successively, we get a prediction vector of the same length as the code. Deciding the multiclass label from the prediction vector (called decoding) can be done by minimizing the distance between the codes and the prediction vector. Different decoders define different choices of distance functions. For this reason, it is usually good to make the mutual distance between codes of different classes large. In this way, even though several binary classifiers make wrong predictions, the distance of the resultant prediction vector to the code of the true class is likely to be still smaller than the distance to other classes. So correct results can still be obtained even when some of the binary classifiers make mistakes. This is the reason for the name Error-Correcting Output Codes.
In SHOGUN, encoding schemes are described by subclasses of CECOCEncoder, while decoding schemes are described by subclasses of CECOCDecoder. Theoretically, any combinations of encoder-decoder pairs can be used. Here we will introduce several common encoder/decoders in shogun.
• CECOCRandomDenseEncoder: This encoder generate random dense ($+1$/$-1$) codebooks and choose the one with the largest minimum mutual distance among the classes. The recommended code length for this encoder is $10\log K$.
• CECOCRandomSparseEncoder: This is similar to the random dense encoder, except that sparse ($+1$/$-1$/$0$) codebooks are generated. The recommended code length for this encoder is $15\log K$.
• CECOCOVREncoder, CECOCOVOEncoder: These two encoders mimic the One-vs-Rest and One-vs-One strategies respectively. They are implemented mainly for demonstrative purpose. When suitable decoders are used, the results will be equivalent to the corresponding strategies, respectively.
Using ECOC Strategy in SHOGUN is similar to ordinary one-vs-rest or one-vs-one. You need to choose an encoder and a decoder, and then construct a ECOCStrategy, as demonstrated below:
In [7]:
from modshogun import ECOCStrategy, ECOCRandomDenseEncoder, ECOCLLBDecoder
print "\nRandom Dense Encoder + Margin Loss based Decoder"
print "="*60
evaluate(ECOCStrategy(ECOCRandomDenseEncoder(), ECOCLLBDecoder()), 2.0)
Random Dense Encoder + Margin Loss based Decoder
============================================================
training time: 2.8417
testing time: 0.0141
accuracy: 0.8865
### Using a kernel multiclass machine¶
Expanding on the idea of creating a generic multiclass machine and then assigning a particular multiclass strategy and a base binary machine, one can also use the KernelMulticlassMachine with a kernel of choice.
Here we will use a GaussianKernel with LibSVM as the classifer. All we have to do is define a new helper evaluate function with the features defined as in the above examples.
In [8]:
def evaluate_multiclass_kernel(strategy):
from modshogun import KernelMulticlassMachine, LibSVM, GaussianKernel
width=2.1
epsilon=1e-5
kernel=GaussianKernel(feats_train, feats_train, width)
classifier = LibSVM()
classifier.set_epsilon(epsilon)
mc_machine = KernelMulticlassMachine(strategy, kernel, classifier, lab_train)
t_begin = time.clock()
mc_machine.train()
t_train = time.clock() - t_begin
t_begin = time.clock()
pred_test = mc_machine.apply_multiclass(feats_test)
t_test = time.clock() - t_begin
evaluator = MulticlassAccuracy()
acc = evaluator.evaluate(pred_test, lab_test)
print "training time: %.4f" % t_train
print "testing time: %.4f" % t_test
print "accuracy: %.4f" % acc
print "\nOne-vs-Rest"
print "="*60
evaluate_multiclass_kernel(MulticlassOneVsRestStrategy())
One-vs-Rest
============================================================
training time: 2.8974
testing time: 2.0434
accuracy: 0.9270
So we have seen that we can classify multiclass samples using a base binary machine. If we dwell on this a bit more, we can easily spot the intuition behind this.
The MulticlassOneVsRestStrategy classifies one class against the rest of the classes. This is done for each and every class by training a separate classifier for it.So we will have total $k$ classifiers where $k$ is the number of classes.
Just to see this in action lets create some data using the gaussian mixture model class (GMM) from which we sample the data points.Four different classes are created and plotted.
In [9]:
from modshogun import *
from numpy import *
num=1000;
dist=1.0;
gmm=GMM(4)
gmm.set_nth_mean(array([-dist*4,-dist]),0)
gmm.set_nth_mean(array([-dist*4,dist*4]),1)
gmm.set_nth_mean(array([dist*4,dist*4]),2)
gmm.set_nth_mean(array([dist*4,-dist]),3)
gmm.set_nth_cov(array([[1.0,0.0],[0.0,1.0]]),0)
gmm.set_nth_cov(array([[1.0,0.0],[0.0,1.0]]),1)
gmm.set_nth_cov(array([[1.0,0.0],[0.0,1.0]]),2)
gmm.set_nth_cov(array([[1.0,0.0],[0.0,1.0]]),3)
gmm.set_coef(array([1.0,0.0,0.0,0.0]))
x0=array([gmm.sample() for i in xrange(num)]).T
x0t=array([gmm.sample() for i in xrange(num)]).T
gmm.set_coef(array([0.0,1.0,0.0,0.0]))
x1=array([gmm.sample() for i in xrange(num)]).T
x1t=array([gmm.sample() for i in xrange(num)]).T
gmm.set_coef(array([0.0,0.0,1.0,0.0]))
x2=array([gmm.sample() for i in xrange(num)]).T
x2t=array([gmm.sample() for i in xrange(num)]).T
gmm.set_coef(array([0.0,0.0,0.0,1.0]))
x3=array([gmm.sample() for i in xrange(num)]).T
x3t=array([gmm.sample() for i in xrange(num)]).T
traindata=concatenate((x0,x1,x2,x3), axis=1)
testdata=concatenate((x0t,x1t,x2t,x3t), axis=1)
l0 = array([0.0 for i in xrange(num)])
l1 = array([1.0 for i in xrange(num)])
l2 = array([2.0 for i in xrange(num)])
l3 = array([3.0 for i in xrange(num)])
trainlab=concatenate((l0,l1,l2,l3))
testlab=concatenate((l0,l1,l2,l3))
In [10]:
_=jet()
_=scatter(traindata[0,:], traindata[1,:], c=trainlab, s=100)
Now that we have the data ready , lets convert it to shogun format features.
In [11]:
feats_tr=RealFeatures(traindata)
labels=MulticlassLabels(trainlab)
The KernelMulticlassMachine is used with LibSVM as the classifer just as in the above example.
Now we have four different classes, so as explained above we will have four classifiers which in shogun terms are submachines.
We can see the outputs of two of the four individual submachines (specified by the index) and of the main machine. The plots clearly show how the submachine classify each class as if it is a binary classification problem and this provides the base for the whole multiclass classification.
In [12]:
from modshogun import KernelMulticlassMachine, LibSVM, GaussianKernel
width=2.1
epsilon=1e-5
kernel=GaussianKernel(feats_tr, feats_tr, width)
classifier=LibSVM()
classifier.set_epsilon(epsilon)
mc_machine=KernelMulticlassMachine(MulticlassOneVsRestStrategy(), kernel, classifier, labels)
mc_machine.train()
size=100
x1=linspace(-10, 10, size)
x2=linspace(-10, 10, size)
x, y=meshgrid(x1, x2)
grid=RealFeatures(array((ravel(x), ravel(y)))) #test features
out=mc_machine.apply_multiclass(grid) #main output
z=out.get_labels().reshape((size, size))
sub_out0=mc_machine.get_submachine_outputs(0) #first submachine
sub_out1=mc_machine.get_submachine_outputs(1) #second submachine
z0=sub_out0.get_labels().reshape((size, size))
z1=sub_out1.get_labels().reshape((size, size))
figure(figsize=(20,5))
subplot(131, title="Submachine 1")
c0=pcolor(x, y, z0)
_=contour(x, y, z0, linewidths=1, colors='black', hold=True)
_=colorbar(c0)
subplot(132, title="Submachine 2")
c1=pcolor(x, y, z1)
_=contour(x, y, z1, linewidths=1, colors='black', hold=True)
_=colorbar(c1)
subplot(133, title="Multiclass output")
c2=pcolor(x, y, z)
_=contour(x, y, z, linewidths=1, colors='black', hold=True)
_=colorbar(c2)
The MulticlassOneVsOneStrategy is a bit different with more number of machines. Since it trains a classifer for each pair of classes, we will have a total of $\frac{k(k-1)}{2}$ submachines for $k$ classes. Binary classification then takes place on each pair. Let's visualize this in a plot.
In [13]:
C=2.0
bin_machine = LibLinear(L2R_L2LOSS_SVC)
bin_machine.set_bias_enabled(True)
bin_machine.set_C(C, C)
mc_machine1 = LinearMulticlassMachine(MulticlassOneVsOneStrategy(), feats_tr, bin_machine, labels)
mc_machine1.train()
out1=mc_machine1.apply_multiclass(grid) #main output
z1=out1.get_labels().reshape((size, size))
sub_out10=mc_machine1.get_submachine_outputs(0) #first submachine
sub_out11=mc_machine1.get_submachine_outputs(1) #second submachine
z10=sub_out10.get_labels().reshape((size, size))
z11=sub_out11.get_labels().reshape((size, size))
no_color=array([5.0 for i in xrange(num)])
figure(figsize=(20,5))
subplot(131, title="Submachine 1") #plot submachine and traindata
c10=pcolor(x, y, z10)
_=contour(x, y, z10, linewidths=1, colors='black', hold=True)
lab1=concatenate((l0,l1,no_color,no_color))
_=scatter(traindata[0,:], traindata[1,:], c=lab1, cmap='gray', s=100)
_=colorbar(c10)
subplot(132, title="Submachine 2")
c11=pcolor(x, y, z11)
_=contour(x, y, z11, linewidths=1, colors='black', hold=True)
lab2=concatenate((l0, no_color, l2, no_color))
_=scatter(traindata[0,:], traindata[1,:], c=lab2, cmap="gray", s=100)
_=colorbar(c11)
subplot(133, title="Multiclass output")
c12=pcolor(x, y, z1)
_=contour(x, y, z1, linewidths=1, colors='black', hold=True)
_=colorbar(c12)
/usr/local/lib/python2.7/dist-packages/ipykernel/__main__.py:7: RuntimeWarning: [WARN] In file /home/buildslave/nightly_default/build/src/shogun/multiclass/MulticlassOneVsOneStrategy.cpp line 34: MulticlassOneVsOneStrategy::CMulticlassOneVsOneStrategy(): register parameters!
The first two plots help us visualize how the submachines do binary classification for each pair. The class with maximum votes is chosen for test samples, leading to a refined multiclass output as in the last plot. | 5,822 | 21,934 | {"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": 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} | 3.3125 | 3 | CC-MAIN-2019-18 | latest | en | 0.847879 |
https://www.physicsforums.com/threads/the-solid-angle-of-coffee.115386/ | 1,606,394,806,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00703.warc.gz | 829,638,861 | 15,456 | # The solid angle of coffee
This is the hardest brain teaser I could think of:
----------------
The solid angle of coffee is quite complicated. It involves some kind of resistance to change, and the density parameter for the brightness of stars is suprisingly inadequate. Unlike most people think, standard angular velocity is not the answer. The issue has to do with asymptotic behaviour of functions rather that the first transfinite ordinal number. Even a simple arithmetic function could tell you that - it is not like it involves baryons and a meson. This is not the end..but it is.
----------------
It is not random things, but actually highly structural! The solution is in the text. You may ask for hints if you wish.
Last edited:
## Answers and Replies
Related General Discussion News on Phys.org
The answer is Omega.
CORRECT! And I though this would be a hard brainteaser...What did you solve it on?
Mk
So what is the solid angle of coffee?
Mk said:
So what is the solid angle of coffee?
The coffee is probably irrelevant. In physics books, upper case Omega is a commonly used symbol for solid angle just as it is for electrical resistance. But in math books lower case is commonly used for the first transfinite ordinal and in physics books for angular velocity. The phrase "This is not the end" is not the end of the puzzle, there is a little bit of text after it. However, omega is the end of the greek alphabet. I'm not completely sure about the other clues, but I felt that I had enough to answer the problem and I got lucky.
Last edited: | 338 | 1,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | latest | en | 0.944425 |
http://www.traditionaloven.com/tutorials/distance/convert-china-cun-unit-to-fingers-fingerbreadth-measure.html | 1,529,413,054,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267862929.10/warc/CC-MAIN-20180619115101-20180619135101-00101.warc.gz | 510,387,463 | 11,712 | Convert 市寸 to finger | Chinese cùn to fingerbreadths
# length conversion
## Amount: 1 Chinese cùn (市寸) of length Equals: 0.98 fingerbreadths (finger) in length
Converting Chinese cùn to fingerbreadths value in the length units scale.
TOGGLE : from fingerbreadths into Chinese cùn in the other way around.
## length from Chinese cùn to fingerbreadth Conversion Results:
### Enter a New Chinese cùn Amount of length to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other length measuring units - complete list.
Conversion calculator for webmasters.
## Length, Distance, Height & Depth units
Distance in the metric sense from any two A to Z points (interchangeable with Z and A), also applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units.
Convert length measuring units between Chinese cùn (市寸) and fingerbreadths (finger) but in the other reverse direction from fingerbreadths into Chinese cùn.
conversion result for length: From Symbol Equals Result To Symbol 1 Chinese cùn 市寸 = 0.98 fingerbreadths finger
# Converter type: length units
This online length from 市寸 into finger converter is a handy tool not just for certified or experienced professionals.
First unit: Chinese cùn (市寸) is used for measuring length.
Second: fingerbreadth (finger) is unit of length.
## 0.98 finger is converted to 1 of what?
The fingerbreadths unit number 0.98 finger converts to 1 市寸, one Chinese cùn. It is the EQUAL length value of 1 Chinese cùn but in the fingerbreadths length unit alternative.
How to convert 2 Chinese cùn (市寸) into fingerbreadths (finger)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
0.984251968504 * 2 (or divide it by / 0.5)
QUESTION:
1 市寸 = ? finger
1 市寸 = 0.98 finger
## Other applications for this length calculator ...
With the above mentioned two-units calculating service it provides, this length converter proved to be useful also as a teaching tool:
1. in practicing Chinese cùn and fingerbreadths ( 市寸 vs. finger ) values exchange.
2. for conversion factors training exercises between unit pairs.
3. work with length's values and properties.
International unit symbols for these two length measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for Chinese cùn is:
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for fingerbreadth is:
finger
### One Chinese cùn of length converted to fingerbreadth equals to 0.98 finger
How many fingerbreadths of length are in 1 Chinese cùn? The answer is: The change of 1 市寸 ( Chinese cùn ) unit of length measure equals = to 0.98 finger ( fingerbreadth ) as the equivalent measure for the same length type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in 市寸 - Chinese cùn for length amount, the rule is that the Chinese cùn number gets converted into finger - fingerbreadths or any other length unit absolutely exactly.
Conversion for how many fingerbreadths ( finger ) of length are contained in a Chinese cùn ( 1 市寸 ). Or, how much in fingerbreadths of length is in 1 Chinese cùn? To link to this length Chinese cùn to fingerbreadths online converter simply cut and paste the following.
The link to this tool will appear as: length from Chinese cùn (市寸) to fingerbreadths (finger) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 944 | 3,932 | {"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} | 2.953125 | 3 | CC-MAIN-2018-26 | latest | en | 0.783442 |
https://www.physicsforums.com/threads/linearly-polarized-electric-field.743370/ | 1,721,237,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.19/warc/CC-MAIN-20240717151625-20240717181625-00083.warc.gz | 830,891,979 | 19,319 | # Linearly polarized electric field
• rogeralms
In summary: The equation for a harmonic wave is: ##E(\vec r,t) =\cdots##. When dealing with the z-direction, you need to use the equation for a vector: ##\vec k = k_x\hat x + k_y\hat y +...##. This will give you the direction of the electric field.
rogeralms
## Homework Statement
Consider a harmonic, electromagnetic plane wave traveling along a line from the origin to the point (3,5,6). It is linearly polarized and its electric field lies in a plane perpendicular to the direction of travel of the wave. The wavelength of the wave is 2.0mm and it has a frequency of 100 GHz.
a) Write an expression for the magnitude (i.e. don't worry about the orientation) of the electric field of this plane wave if the amplitude of the electric field is 10V/m.
b) What is the index of refraction of the medium in which the wave is traveling?
c) Compute the irradiance of the wave.
d) Determine a unit vector for a possible direction that would define the direction of the electric field.
## Homework Equations
k = 3i + 5j + 6k
|k| = (32 + 52 + 62)(1/2)
unit vector in k direction = k / (70)(1/2)
n= v/c
E (dot) k = 0
I = (c * epsilonzero /2) * (E0)2
## The Attempt at a Solution
a) I am really confused as to how to find the magnitude of the electric field
As you see, I have the magnitude of the direction vector k.
I know that the divergence of the E and k vectors is zero. But I don't know how the amplitude is used in the magnitude. Please I do not want the final answer, just a reference as to how to find the magnitude.
b) v = lambda * nu = 2*10-3 * 100*109 = 2*108 m/s
n = c/v = 3*108 / 2*108 = 1.5
c) I = (c*ep0 / 2)* E0 2 =
150 ep0 *108 V/m
d) I'm not sure on this one
It has something to do with divergence of E k = 0
Would it be something like
(3-x)Ex + (5-y)Ey + (6-z)Ez = 0?
Thanks for any help. I just need some hints.
rogeralms said:
a) I am really confused as to how to find the magnitude of the electric field
You are not asked to find the magnitude - you are asked to write an expression for it.
You are told that the wave is "harmonic" and "polarized".
Do you understand what these terms mean?
You are given the amplitude, the frequency, and the wavelength ... you know the wave speed.
Do you know the equation for a harmonic wave?
I'll start you off: ##E(\vec r,t) =\cdots##
Note: you have used k to mean two different things.
Please makes sure that each symbol has only one meaning.
Last edited:
Simon,
Thank you so much for answering and sorry for the double use of k. That was the convention used in my textbook and in my professor's lectures.
E(r,t) = E0 cos($\omega$ (t - r/c) + $\epsilon$)
where k = ω / c
Since cos varies between -1 and 1 and the amplitude is given as 10 V/m, then would the magnitude just be the expression
10 * cos (ω *t - k ° r) ?
Then would r = (x-3)i + (y-5)j + (z-6)k ?
I could not find a definition for magnitude of an electric field in the book, the posted lecture notes, or the internet. Again sorry for the double use of k as the direction of motion vector and the unit vector in the z direction, but that is the way my book and the professor use it.
rogeralms said:
Simon,
Thank you so much for answering and sorry for the double use of k. That was the convention used in my textbook and in my professor's lectures.
Such sloppiness is quite common - but you have to be careful about it or you'll get confused.
If you have to use the wave vector as ##\vec k## then you will need an alternate form for the z-direction unit vector. ##\hat z## or ##\hat e_z## are common, with similar ones for x and y unit vectors.
E(r,t) = E0 cos($\omega$ (t - r/c) + $\epsilon$)
where k = ω / c
Can a vector be equal to a scalar?
Be careful here - it is usual to use c as the speed of light in a vacuum, the problem statement implies that this wave is traveling through a medium with a different speed.
(Mention of "refractive index"?)
Since cos varies between -1 and 1 and the amplitude is given as 10 V/m, then would the magnitude just be the expression
10 * cos (ω *t - k ° r) ?
I see you are struggling with the symbol chart as well as the mathematical description. Suggest you try LaTeX. If you use the "quote" button (below this post) you will get to see how I get the symbols to come out nicely.
i.e. ##\vec E(\vec r , t)=\hat n E_0\cos(\omega t - \vec k \cdot\vec r) = ## ... where ##\hat n## is the polarization direction.
Then would r = (x-3)i + (y-5)j + (z-6)k ?
##\vec r## usually just points to some location in space.
This may help... p7-8
... how does ##\vec k = k_x\hat x + k_y\hat y + k_z\hat z## come into this?
I could not find a definition for magnitude of an electric field in the book, the posted lecture notes, or the internet.
The magnitude of the electric field is defined the same way as the magnitude of any vector.
Basically the exercise is trying to get you to learn how to describe waves in 3D.
Thanks
I finally understand part a)
k=3i+5j+6z so unit k vector = k/(70)^(1/2)
E(t) = E0 cos(k(dot)r - ωt)
E(t) = 10 V/m * cos ( (3x + 5y + 6z)/ 70(1/2) - ωt)
part d)
k(dot)<x,y,z> = 0 then 3x + 5y + 6z = 0. So choose y = 0 gives 3x + 6z = 0.
This makes x = -2z, so choose z = 1, then x = -2
A perpendicular vector to <3,5,6> is <-2,0,1>. To get the unit vector just divide by 5(1/2)
Thanks so much for your patience and gentle nudging!
... well done - I think you are on your way.
## 1. What is a linearly polarized electric field?
A linearly polarized electric field is a type of electromagnetic wave that oscillates in a single plane. This means that the electric field vector, which describes the direction and magnitude of the electric field, only changes in one direction as the wave propagates through space.
## 2. How is a linearly polarized electric field created?
A linearly polarized electric field can be created by passing an electromagnetic wave through a polarizing filter, which only allows waves with a certain orientation of the electric field vector to pass through. It can also be created by using specialized antennas or by reflecting light off of certain surfaces.
## 3. What is the significance of a linearly polarized electric field?
The linear polarization of an electric field is important because it allows for the manipulation and control of electromagnetic waves. This is crucial in many modern technologies, such as telecommunications, radar, and satellite communication.
## 4. Can a linearly polarized electric field be converted into a circularly polarized one?
Yes, a linearly polarized electric field can be converted into a circularly polarized one by using a device called a waveplate. This device changes the phase difference between the two components of the electric field, resulting in circular polarization.
## 5. How does the orientation of a linearly polarized electric field affect its interaction with matter?
The orientation of a linearly polarized electric field can determine how it interacts with matter. For example, certain materials may absorb or reflect the wave more effectively depending on the orientation of the electric field vector. This is the principle behind polarized sunglasses, which are designed to block horizontally polarized light from entering the eyes.
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https://www.coursehero.com/file/6816198/5-2-Clicker-Second-Order-Linear-ODEs/ | 1,498,241,935,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320077.32/warc/CC-MAIN-20170623170148-20170623190148-00312.warc.gz | 862,672,583 | 124,902 | 5_2-Clicker-Second Order Linear ODEs
# 5_2-Clicker-Second Order Linear ODEs - where ω is...
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1 How many initial conditions are required to fully determine the particular solution to a 2nd order differential equation?
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2 How many initial conditions are required to fully determine the general solution to a 2nd order linear differential equation?
How many initial conditions are required to fully determine the particular solution to a 2nd order differential equation? A) 1 B) 2 C) 3 D) 4 E) Cannot tell
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4 Since cos(ωt) and cos(-ωt) are both solutions of can we express the general solution as y(t)=C1cos(ωt) + C2cos(-ωt) ? A) yes B) no C) ???/depends Ý Ý x ( t ) =-ϖ 2 x ( t )
5 Ý Ý x ( t ) =-ϖ 2 x ( t ) What is the general solution to the ODE
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Unformatted text preview: where ω is some (known) constant? A) x ( t ) = A cos ϖ t + B sin t B) x ( t ) = Ce i t + De-i t C) x ( t ) = A cos t + B sin t + Ce i t + De-i t D) None of these is fully general! E) More than one of these is fine Challenge question: Is there any OTHER general form for this solution? 6 Are sin(x) and cos(x) linearly independent? A) yes B) no C) ??? 7 Are eiωx and e-iωx linearly independent? A) yes B) no C) It depends on omega D) ??? 8 Are e-t and e-4t linearly independent? A) yes B) no C) ???...
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Title: Lecture 10 Finite State Machine Design
1
Lecture 10Finite State Machine Design
• Hai Zhou
• ECE 303
• Spring 2002
2
Outline
• Review of sequential machine design
• Moore/Mealy Machines
• FSM Word Problems
• Finite string recognizer
• Traffic light controller
• READING Katz 8.1, 8.2, 8.4, 8.5, Dewey 9.1, 9.2
3
Example Odd Parity Checker
Assert output whenever input bit stream has odd
of 1's
Reset
Present State
Input
Next State
Output
Even
0
Even
0
Even
1
Odd
0
0
Even
Odd
0
Odd
1
0
Odd
1
Even
1
1
1
Symbolic State Transition Table
Odd
Output
Next State
Input
Present State
1
0
0
0
0
0
0
1
1
0
1
1
0
1
State Diagram
1
0
1
1
Encoded State Transition Table
4
Odd Parity Checker Design
Next State/Output Functions
NS PS xor PI OUT PS
Input
Output
T
Q
NS
Input
CLK
D
Q
PS/Output
Q
CLK
R
Q
R
\Reset
\Reset
T FF Implementation
D FF Implementation
Input
1
0
0
1
1
0
1
0
1
1
1
0
Clk
1
1
0
1
0
0
1
1
0
1
1
1
Output
Timing Behavior Input 1 0 0 1 1 0 1 0 1 1 1 0
5
Basic Design Approach
1. Understand the statement of the
Specification 2. Obtain an abstract
specification of the FSM 3. Perform a state
mininimization 4. Perform state
assignment 5. Choose FF types to implement FSM
state register 6. Implement the FSM
1, 2 covered now 3, 4, 5 covered later 4, 5
generalized from the counter design procedure
6
Example Vending Machine FSM
General Machine Concept
deliver package of gum after 15 cents
deposited single coin slot for dimes,
nickels no change
Step 1. Understand the problem
Draw a picture!
N
Block Diagram
Coin
Vending
Gum
Open
D
Sensor
Machine
Release
FSM
Mechanism
Reset
Clk
7
Vending Machine Example
Step 2. Map into more suitable abstract
representation
Reset
S0
Tabulate typical input sequences
N
D
three nickels nickel, dime dime, nickel two
dimes two nickels, dime
S1
S2
D
N
D
N
Draw state diagram
S4
S6
S3
S5
Inputs N, D, reset Output open
open
open
open
N
D
S8
S7
open
open
8
Vending Machine Example
Step 3 State Minimization
Inputs
Reset
0
N
5
N
10
N, D
15
open
reuse states whenever possible
Symbolic State Table
9
Vending Machine Example
Step 4 State Encoding
Inputs
10
Vending Machine Example
Step 5. Choose FFs for implementation
D FF easiest to use
Q1
Q1
Q1
Q1 Q0
Q1 Q0
Q1 Q0
D N
D N
D N
0 1 1 0
0 0 1 1
0 0 1 0
1 0 1 1
0 0 1 0
0 1 1 1
N
N
N
X X X X
X X X X
X X X X
D
D
D
1 1 1 1
0 1 1 1
0 0 1 0
Q0
Q0
Q0
K-map for D0
K-map for D1
K-map for Open
D
D
Q
D1 Q1 D Q0 N D0 N Q0 Q0 N Q1 N
Q1 D OPEN Q1 Q0
CLK
Q
R
N
\reset
N
OPEN
D
Q
CLK
Q
8 Gates
N
R
\reset
D
11
Moore and Mealy Machine Design Procedure
Moore Machine Outputs are function solely of the
current state Outputs change synchronously
with state changes
State
Register
Comb.
X
Combinational
i
Logic for
Inputs
Logic for
Outputs
Next State
Z
(Flip-flop
k
Outputs
Inputs)
Clock
state
feedback
Mealy Machine Outputs depend on state AND
inputs Input change causes an immediate
output change Asynchronous signals
Z
X
k
i
Combinational
Outputs
Inputs
Logic for
Outputs and
Next State
State
Feedback
State Register
Clock
12
Equivalence of Moore and Mealy Machines
Moore Machine
N D Reset
Mealy Machine
(N D Reset)/0
Reset/0
Reset
0
0
0
Reset
Reset/0
N
N/0
5
5
D/0
N D/0
0
N
N/0
10
10
D/1
0
N D/0
ND
ND/1
15
15
1
Reset
Reset/1
Outputs are associated with State
Outputs are associated with Transitions
13
States vs Transitions
Mealy Machine typically has fewer states than
Moore Machine for same output sequence
0
0/0
0
0
0
Same I/O behavior Different of states
1/0
0/0
0
1
0
1
1
1/1
0
1
2
1
1
14
Analyze Behavior of Moore Machines
Reverse engineer the following
J
X
A
Q
Input X Output Z State A, B Z
C
X
\A
K
Q
R
\B
FFa
\Reset
Clk
J
X
Z
Q
C
X
\B
K
Q
R
\A
FFb
\Reset
Two Techniques for Reverse Engineering Ad
Hoc Try input combinations to derive transition
table Formal Derive transition by
analyzing the circuit
15
Behavior in response to input sequence 1 0 1 0 1
0
100
X
Clk
A
Z
\Reset
Partially Derived State Transition Table
16
Formal Reverse Engineering
Derive transition table from next state and
output combinational functions presented to
the flipflops!
Z B
Ka X B Kb X xor A
Ja X Jb X
FF excitation equations for J-K flipflop
A Ja A Ka A X A (X B)
A B Jb B Kb B X B (X A
X A) B
Next State K-Maps
A
State 00, Input 0 -gt State 00 State 01, Input 1
-gt State 01
B
17
Behavior of Mealy Machines
Clk
X
A
B
D
Q
J
Q
DA
C
C
Q
K
Q
R
R
\Reset
\Reset
A
DA
X
B
B
Z
X
A
Input X, Output Z, State A, B
State register consists of D FF and J-K FF
18
Signal Trace of Input Sequence 101011
100
Note glitches in Z! Outputs valid at following
falling clock edge
X
Clk
A
B
Z
\Reset
Partially completed state transition table based
on the signal trace
19
Formal Reverse Engineering
A B (A X) A B B X B Jb
B Kb B (A xor X) B X B
A B X A B X B X Z A X
B X
Missing Transitions and Outputs
State 01, Input 0 -gt State 01, Output 1 State 10,
Input 0 -gt State 00, Output 0 State 11, Input 1
-gt State 11, Output 1
A
B
Z
20
Finite State Machine Word Problems
Mapping English Language Description to Formal
Specifications
Case Studies Finite String Pattern
Recognizer Traffic Light Controller
We will use state diagrams and ASM Charts
21
Finite String Pattern Recognizer
A finite string recognizer has one input (X) and
one output (Z). The output is asserted whenever
the input sequence 010 has been observed, as
long as the sequence 100 has never
been seen. Step 1. Understanding the problem
statement Sample input/output
behavior
X 00101010010 Z 00010101000 X
11011010010 Z 00000001000
22
Finite String Recognizer
Step 2. Draw State Diagrams/ASM Charts for the
strings that must be recognized.
I.e., 010 and 100.
Reset
S0
0
Moore State Diagram Reset signal places FSM in
S0
S4
S1
0
0
S2
S5
0
0
S3
S6
Loops in State
Outputs 1
1
0
23
Finite String Recognizer
Exit conditions from state S3 have recognized
010 if next input is 0 then have 0100!
if next input is 1 then have 0101 01
(state S2)
Reset
S0
0
S4
S1
0
0
S2
S5
0
0
S3
S6
1
0
24
Finite String Recognizer
Exit conditions from S1 recognizes strings of
form 0 (no 1 seen) loop back to S1 if
input is 0 Exit conditions from S4 recognizes
strings of form 1 (no 0 seen) loop back to
S4 if input is 1
Reset
S0
0
S4
S1
0
0
S2
S5
0
0
S3
S6
1
0
25
Finite String Recognizer
S2, S5 with incomplete transitions S2 01 If
next input is 1, then string could be prefix of
(01)1(00) S4 handles just this
case! S5 10 If next input is 1, then string
could be prefix of (10)1(0) S2
handles just this case!
Final State Diagram
26
Review of Design Process
Write down sample inputs and outputs to
understand specification Write down
sequences of states and transitions for the
missing transitions reuse states as much as
possible Verify I/O behavior of your state
diagram to insure it functions like the
specification
27
Traffic Light Controller
A busy highway is intersected by a little used
farmroad. Detectors C sense the presence of cars
waiting on the farmroad. With no car on
farmroad, light remain green in highway
direction. If vehicle on farmroad, highway
lights go from Green to Yellow to Red, allowing
the farmroad lights to become green. These stay
green only as long as a farmroad car is detected
but never longer than a set interval. When
these are met, farm lights transition from Green
green. Even if farmroad vehicles are waiting,
highway gets at least a set interval as
green. Assume you have an interval timer that
generates a short time pulse (TS) and a long time
pulse (TL) in response to a set (ST) signal.
TS is to be used for timing yellow lights and TL
for green lights.
28
Traffic Light Controller
C
HL
FL
Highway
Highway
FL
HL
C
29
Traffic Light Controller
Tabulation of Inputs and Outputs
Input Signal reset C TS TL Output Signal HG, HY,
HR FG, FY, FR ST
Description place FSM in initial state detect
vehicle on farmroad short time interval
expired long time interval expired Description as
sert green/yellow/red highway lights assert
green/yellow/red farmroad lights start timing a
short or long interval
Tabulation of Unique States Some light
configuration imply others
Description Highway green (farmroad red) Highway
State S0 S1 S2 S3
30
Traffic Light Controller
Compare with state diagram
TL C
Reset
S0 HG S1 HY S2 FG S3 FY
S0
TLC/ST
TS/ST
TS
S1
S3
TS
TS/ST
TL C/ST
S2
TL C
on paths and conditions for exiting a state
Exit conditions built up incrementally, later
combined into single Boolean
condition for exit Easier to understand
the design as an algorithm
31
Summary
• Review of sequential machine design
• Moore/Mealy Machines
• FSM Word Problems
• Finite string recognizer
• Traffic light controller
• NEXT LECTURE Finite State Machine Optimization
• READING Katz 9.1, 2.2.1, 9.2.2, Dewey 9.3 | 3,078 | 9,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.677841 |
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# Two's Complement Calculator
Enter any logical number (binary, decimal, or hexadecimal) and select the order of binary digits and hit the calculate button to find the two's complement with detailed information.
Decimal
Binary
Hex
No. of Binary Digits:
Number of Bits
Decimal Number Range
-128 to 127
Binary Number Range
Hex Number Range
0-9 and A-F (16-Digits)
Two's complement Binary 1:
Two's complement Binary 2:
Table of Content
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This free two’s complement calculator allows you to calculate 2’s complement of the given decimal, binary or hexadecimal number.
No doubt, number conversion is complicated to express but the 2s complement calculator changes the entered number into one’s complement, two’s complement, signed binary to decimal, and hexadecimal.
## What is Two’s Complement?
There is a simple mathematical operation on different binary numbers and used in computing as the method of signed number representation. To get 2’s complement of a binary system, just transpose the certain number and add one to the LSB (Least Significant Bit) of given results.
For instance, to convert decimal to 2’s complement, we have a number (20)10 which is equal to (0001 0100)2. Now convert 1 to 0 and 0 to 1, so number is 1110 1011 and add 1 in one’s complement to get two’s 1110 1011 +1 = 1110 1100.
How Two’s Complement Calculator Works?
The 2’s complement calculator works as follows to find the 2s complement to decimal of entered values, add, or subtract them.
Input:
• First of all, you need to select the option either you want to find twos complement to decimal with decimal, binary or hexadecimal number form.
• Once selected, you have to add the values into the given fields.
• Select any number of binary digits from the drop-down menu according to your need.
• Once done, hit the calculate button.
Output:
• Two complements of a given number.
• Also shows the results in binary, decimal, hexadecimal, and 1’s complement.
Input:
• Write down the first two’s complement binary number
• Repeat the same for second number
• Select the operation from the middle
• Tap the calculate button
Output:
• Addition of Two’s complement binary numbers
• Subtraction of Two’s complement binary numbers
## Reference:
From the source of Wikipedia: Converting from two’s complement representation, From the ones’ complement, Working from LSB towards MSB. | 628 | 2,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2023-50 | longest | en | 0.844151 |
https://www.numbersaplenty.com/15624827 | 1,719,233,119,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00021.warc.gz | 805,296,678 | 3,188 | Search a number
15624827 = 36434289
BaseRepresentation
bin111011100110…
…101001111011
31002101211020022
4323212221323
512444443302
61314521055
7246544301
oct73465173
932354208
1015624827
118902189
12529618b
133310b7a
14210a271
1515898a2
hexee6a7b
15624827 has 4 divisors (see below), whose sum is σ = 15632760. Its totient is φ = 15616896.
The previous prime is 15624809. The next prime is 15624863. The reversal of 15624827 is 72842651.
It is a semiprime because it is the product of two primes, and also a brilliant number, because the two primes have the same length, and also an emirpimes, since its reverse is a distinct semiprime: 72842651 = 710406093.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-15624827 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (15624227) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1499 + ... + 5787.
It is an arithmetic number, because the mean of its divisors is an integer number (3908190).
Almost surely, 215624827 is an apocalyptic number.
15624827 is a deficient number, since it is larger than the sum of its proper divisors (7933).
15624827 is an equidigital number, since it uses as much as digits as its factorization.
15624827 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 7932.
The product of its digits is 26880, while the sum is 35.
The square root of 15624827 is about 3952.8251921885. The cubic root of 15624827 is about 249.9990773299. Note that the first 3 decimals are identical.
The spelling of 15624827 in words is "fifteen million, six hundred twenty-four thousand, eight hundred twenty-seven".
Divisors: 1 3643 4289 15624827 | 548 | 1,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.880682 |
https://www.zometric.com/products/statistical-software/u-chart/ | 1,718,320,144,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00440.warc.gz | 982,064,038 | 15,879 | # U Chart
## What is U Chart?
U Chart is a statistical process control chart that is used to monitor and control the variation in the number of defects per unit during the production process. It is used to track defects or non-conformances in a sample of items, where the sample size may vary from period to period.
The U Chart plots the number of defects per unit over time, and determines if the process is stable or not. It is commonly used in manufacturing and quality control to identify and correct issues in the production line.
## When to use U Chart?
A U chart is typically used to analyze data that reflects the count of defects in a set of items. It is particularly useful for assessing the presence of special causes of variation in data. A U chart is used when the sample size is constant and the defects are measured in discrete units. Therefore, it is commonly used in manufacturing processes, quality control, and similar settings where the occurrence of defects needs to be monitored and tracked over time.
## Guidelines for correct usage of U Chart
• Count defects on each item or unit, or use P Chart Diagnostic for binary data
• Collect data in time order
• Collect data at appropriate time interval
• Collect data in subgroups that are representative of the process output and subject to the same process conditions
• Subgroups must be large enough to accurately estimate control limits
• Collect enough subgroups to obtain precise control limits
## Alternatives: When not to use U chart
If you are only able to classify items as defective or non defective, then you should use the P Chart Diagnostic.
## Example of U Chart?
The director of quality for a group of hospitals wants to assess the medication error rate. Examples of errors include delivering medication at the wrong time, delivering the wrong dose, and delivering the wrong medication. The director records the number of patients and the number of medication errors each week for 32 weeks. The average subgroup size is more than 7500. Because of the large number of patients, the director uses a U chart diagnostic test to test for over dispersion. She has performed this in following steps:
1. She worked all day and gathered the data.
1. After gathering the data, she uses mathematical formula for finding the u bar, n bar, Upper Control Limit (UCL) and Lower Control Limit(LCL).
1. Now, after calculating u bar, UCL and LCL, she analyzes the data with the help of https://qtools.zometric.com/
2. After using the above mentioned tool, she fetches the useful graph as follows:
## How to generate U Chart?
The guide is as follows:
1. Login in to QTools account with the help of https://qtools.zometric.com/
2. On the home page, one will see U Chart under control charts.
3. Click on U Chart and will reach the dashboard.
4. Next, update the data manually or can completely copy (Ctrl+C) the data from excel sheet and paste (Ctrl+V) it here.
5. Next, you need to select the desired Check Rules.
6. Finally, click on calculate at the bottom of the page and you will get desired results.
On the dashboard of U Chart, the window is separated into two parts.
On the left part, Data Pane is present. Data can be fed manually or the one can completely copy (Ctrl+C) the data from excel sheet and paste (Ctrl+V) it here.
On the right part, there are many options present as follows:
• Process mean: If process mean is provided, this value is considered to be the centerline. If not, Zometric Q-Tools calculates the centerline from the data provided.
• Check Rule 1: 1 point > K Stdev from center line: If a data point is K standard deviations from the center line, it means that it is K times the standard deviation away from the mean. This is important in statistical process control because it indicates whether a data point is within acceptable limits or whether there may be a problem with the process that needs to be addressed. Typically, data points that are more than three standard deviations from the center line are considered outliers and may require further investigation.
• Check Rule 2: K points in a row on same side of center line: If there are K points in a row on the same side of the center line in a dataset, it suggests that there may be a bias or trend in the data that is causing the values to cluster together. This could be due to a variety of factors, such as measurement error, sampling bias, or a true underlying pattern in the data.
• Check Rule 3: K points in a row, all increasing or all decreasing: If there are K points in a row, it is certain that at least one of two things must be true:
• The points are all increasing (i.e. each point has a greater y-coordinate than the one before it)
• The points are all decreasing (i.e. each point has a smaller y-coordinate than the one before it)
• Check Rule 4: K points in a row, alternating up and down:
• If the trend is upwards, it indicates that the process is becoming less consistent and more variable over time. This can be caused by factors such as equipment deterioration, operator error or changes in raw material quality.
• If the trend is downwards, it indicates that the process is becoming more consistent and less variable over time. This could be due to process improvements or tighter control measures being implemented. | 1,131 | 5,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-26 | longest | en | 0.917928 |
https://www.coursehero.com/file/84711/151-Quiz-3A-Sp-08/ | 1,516,640,189,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891485.97/warc/CC-MAIN-20180122153557-20180122173557-00581.warc.gz | 895,519,238 | 31,330 | 151 Quiz 3A Sp 08
# 151 Quiz 3A Sp 08 - MATH151 Mrs.Bonny Tighe QUIZ 3A 1 1-1 3...
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MATH 151 Mrs. Bonny Tighe QUIZ 3A 1.1- 1.3 25 points NAME Section 2/18/08 1. Estimate the length ofthe curve /(:r) =x2 +2 onthe interval -1<.r < 1 andn=4 line sesments. .-.---. ..-.- I \L .' l/t\t, /91 | -- .llJ t L<r .l- t-t4 I-# -'t -'l- o 'L \ (",u) (l,rqh) (o, uy (f ,rtt"1 (, ,2) 3. Evaluate the hdicated a) lim (r'? -e'z.1= ( - .9- o 2 . For the frrnction g whose graph is given, state the value of the given quantity, if it exists. a) e(l)=--al 1 b) lims(x) 5 -l c) lrmg(.r) = | 'r+0"- v d) g(-2): O'b,, 'T' lL e) lime(x) ' x+l--- . ,tI fl lim s(x)= aL-.f - /t- ll e) lim e(x) = h) limg(x) 0 ,ts. ,.* x2 +2x-3 (-'tl ) Jli ,.; --\=--z- *J "(i)' (u-,{vt
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h;+ r D 1- (1r") G;-(mr'^)' h, ?1# h'zo 1 @a*J-je lvtH -t ,/\,k&+1 Pj \ {96{n1 - t( \xr rr F t^l tx-zt / . I e) lim' '= t -l I '-+2 x-2 -t-----t--7- \-----'- c) lim h-+0 (7 + h)-t -7 | ') x' +-l -l =() ^ . (t 1+t\ I) nn{ -- , I= r+o\r t. ) +' - ( --t' I r-t,tf ,<-r e) f(x)=1x" -1, if -r<x<1 lJ'-1, if x>7 r,o) ( t, "-; t\- L-Q,-. }. fuo& .19 /(') = \rg.f(x)= i; hn1(x csc x)
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151 Quiz 3A Sp 08 - MATH151 Mrs.Bonny Tighe QUIZ 3A 1 1-1 3...
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Ask a homework question - tutors are online | 737 | 1,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-05 | latest | en | 0.442506 |
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# Perfect solution to all problems
## Tips, Tricks, General Knowledge, Current Affairs, Latest Sample,
Previous Year, Practice Papers with solutions.
## CBSE 12th Mathematics 2012 Unsolved Paper
Outside Delhi
Note
This pdf file is downloaded from www.4ono.com. Editing the content or publicizing this on any blog or
website without the written permission of Rewire Media is punishable, the suffering will be decided under
DMC
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## CBSE 12th Mathematics 2012 Unsolved Paper
Outside Delhi
TIME - 3HR. | QUESTIONS - 29
## THE MARKS ARE MENTIONED ON EACH QUESTION
__________________________________________________________________________
SECTION A
Question numbers 1 to 10 carry 1 mark each.
Q.1. The binary operation *: R x R R, is defined as a*b = 2a + b. find (2*3)*4. 1 mark
Q.2. Find the principle value of 1 mark
().
Q.3. find the value of x + y from the following equation: 1 mark
[ ]+[ ]=[ ]
Q.4. if [ ] = [ ] . . 1 mark
Q.5. Let A be a square matrix of order 3 x 3. Write the value of ||, where ||=4. 1 mark
Q.6. Evaluate:
.
Q.7. Given ( + ) = () +
Write f(x) satisfying above 1 mark
+ . . 1 mark
Q.8. write the value of ( ).
Q.9. Fine the scalar components of the vector with initial A (2, 1) and terminal point B
(-5, 7). 1 mark
Q.10. Find the of the plane 3x 4y + 12z = 3 from the origin. 1 mark
SECTION B
Question numbers 11 to 22 carry 4 marks each.
Q.11. prove the following: 4 marks
( + ) =
4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com
## Q.12. Using properties of determinates, show that 4 marks
+
| + | =
+
Q.13. Show that : , given by, 4 marks
+ ,
() = { .
,
OR
## Consider the binary operations* : R x R R defined as a * b = | | and a o b
= a for all a, b R. show that * is commutative but not associative o is
associative but not commutative.
## Q.14. If, 4 marks
= , =
, = .
OR
Differentiate
+
[ ] .
Q.15. If x = a ( cos t + t sin t ) and 4 marks
= ( ), < < ,
, .
Q.16. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along
the ground, away from the wall, at the rate of 2 cm/s. how fast is its height on the
wall decreasing when the fool of the ladder is 4 m away from the wall? 4 marks
Q.17. Evaluate: 4 marks
| |
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OR
Evaluate:
+
Q.18. From the differential equation of the family of circles in the second quadrant and
touching the coordinate axes. 4 marks
OR
Find the particular solution of the differential equation
( )
= ; = = .
Q.19. solve the following differential equation: 4 marks
( + ) + = ;
= + + ,
Q. 20. = +
= + .
## Find a vector and and
which is perpendicular to both .
= . 4 marks
Q. 21. Find the coordinates of the point where the line through the point A (3, 4, 1) and B
(5, 1, 6) crosses the XY-plane. 4 marks
Q. 22. Two cards are drawn simultaneously (without replacement) from a well- shuffled
pack of 52 cards. Find the mean and variance of the number of red cards. 4 marks
SECTION-C
Question numbers 23 to 29 carry 6 marks each.
Q. 23. Using matrices, solve the following system of equation: 6 marks
+ + = , + = ,
=
Q. 24. Prove that the radius of the right circular cylinder of greatest curved surface area
which can be inscribed in a given cone is half of that of the cone. 6 marks
OR
An open box with a square base is to be made out of a given quantity of cardboard
of area square units. Show that maximum volume of the box is
.
4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com 4ono.com
## Q. 25. Evaluate: 6 marks
OR
Evaluate:
+
.
( ) ( + )
Q. 26. Find the area of the region 6 marks
{(, ): + , + }.
Q. 27.
= = = = ,
Find the value of k and hence find the equation of plane containing these
lines. 6 marks
Q. 28. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin 3 times and notes
the number of heads. If she gets 1,2,3 or 4 she tosses a coin once and notes whether a
head or tail is obtained. If she obtained exactly one head, what is the probability that
she threw 1,2,3 or 4 with the die? 6 marks
Q. 29. A dietician wishes to mix two types of Foods in such a way that the vitamin contents
of the mixture contains at least 8 units of vitamin A and 10 units of vitamin C. Food I
contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C while Food II contains 1
unit/kg of vitamin A and 2 unit/kg of vitamin C. It costs Rs5 per kg to purchase food
I and Rs7 per kg to purchase Food II. Determine the minimum cost of such a
mixture. Formulate the above as a LPP and solve it graphically. 6 marks | 1,511 | 4,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-24 | latest | en | 0.726682 |
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