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http://amsterjob.com/zip-code-miami-map/
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Zip Code Miami Map
16 views
5 / 5 ( 1votes )
At the time you think of Zip Code Miami Map, exactly what you thinking of? In essence a map is a representation of a topology or function. Pertaining to example a formula such as X=2Y maps a worth of Y to each value of Back button. Of course you know that mathematicians are weird and sometimes hard to understand but they have you ever seen a schematic map of a subway (underground railway) system? Maybe you have ever seen the same network of rails descriptive on a more "normal" Zip Code Miami Map of the location in which it is located? Different Zip Code Miami Map of the extremely same thing can look quite different.
At the time you make a Zip Code Miami Map of your flat area - a "plan" or "elevation" - things are quite simple, but when you make an effort to map a larger area, like the surface of an whole planet, things can get quite complicated if you would like your map to be toned. It truly is all very well to make a world, but try turning the of that globe into a set Zip Code Miami Map! Yikes!
However you start it, you ending plan edge-effects. As I write this post I am actually included in programming map-generating programs meant to generate maps of fictional landscapes. I happen to be examining the map-generators that are included in the free, open-source (GNU GPL licensed) strategy game, FreeCiv. Edge results are incredibly apparent in such maps. The Zip Code Miami Map are basically rectangular, but you can choose to obtain them act like cylinders by "wrapping" left to right or top to lower part, or you may even have "wrap" in both guidelines. Most often people determine on "wrap" only remaining to right, and obstruct the very best and bottom with "polar regions". Such easy "wrapping" makes for quite extreme distortion though if you give it a try with a real Zip Code Miami Map worldwide!
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https://example-a.com/answer/77516549
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excelvbaexcel-2007
# Count average value of every 6th column in one row via VBA (Excel)
Right now I'm writing macro on VBA , I almost finished it, but right now i'm stuck on one issue. I can't understand how should I count average value of range that consists of every 6th column of my table. In excel this formula looks like "=AVG(I3, O3, U3...HE3)"
I tried to use macro recorder but it was just hard-coding average value by repeating the whole formula and every cell address. But my table updates everyday and everyday I add 6 new columns at the end of the table, left from last column and last column is column that stores average value of every 6th column of a row. I tried to do it through loop and I think that I'm pretty close to solving this issue, but I don't understand how to add looped values to a particular cell with formula.
Here is my code:
``````Dim i As Integer
i = 8
For i = 8 To rng.Columns.Count Step 6
rng.Cells(3, rng.Columns.Count) = Application.WorksheetFunction.Average(rng.Cells(3,i))
Next i
``````
rng - is Range variable that stores my table.
With above code I managed to loop through all cells that I need to count avg value, but I don't understand how to get values of these cells to a particular cell that counts avg value. Could you please help me?
Solution
• You can just sum the values then divide by the count of (columns-8)/6.
``````Dim i As Integer
Dim sum as long
i = 8
sum = 0
For i = 8 To rng.Columns.Count Step 6
sum = sum + rng.Cells(3,i).value
Next i
rng.Cells(3, rng.Columns.Count).offset(0,1).value = sum/((rng.columns.count-8)/6) ' I assumed you want the value in rng.columns.count +1, since but change it if you want.
``````
EDIT, can't use rng.columns.count+1 I think since the range ends at the count. Changed it to offset.
Also note that we have no error checking in this code.
I would suggest something like `If Isnumber(rng.Cells(3,i).value) then` before adding to the sum since a string value will break the code
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CC-MAIN-2024-10
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https://www.stata.com/statalist/archive/2011-04/msg00214.html
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|
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| 943,813,707
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Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: egen rowmean, loops and if
From Nick Cox <[email protected]> To [email protected] Subject Re: st: egen rowmean, loops and if Date Wed, 6 Apr 2011 00:37:43 +0100
```This would be a lot easier if you -reshape-d, even temporarily.
Otherwise, with this data structure: -egen, rowmean()- is a
non-starter here and I think you need to work at a lower level,
building up sums and counts and deriving means.
A side-detail is that -foreach- is not needed here: use -forval- instead.
Nick
On Tue, Apr 5, 2011 at 10:09 PM, Thomas Speidel <[email protected]> wrote:
> I have a wide dataset:
> +-------------------------------------+
> | id occ_~1_1 occ_~2_1 occ_~3_1 |
> |-------------------------------------|
> | 1 4 7 . |
> | 2 1.5 . . |
> | 3 2.3 3.3 . |
> | 4 3.3 2.3 3.5 |
> | 5 1.5 . . |
> |-------------------------------------|
> | 6 1.5 . . |
> | 7 2.3 . . |
> | 8 1.5 . . |
> | 9 1.5 2.3 3.3 |
> | 10 1.5 2.3 3.3 |
> +-------------------------------------+
>
> where occ_~1_1 = occ_met1_1
> occ_~2_1 = occ_met2_1
> occ_~3_1 = occ_met3_1
>
> Of course, the data is much wider (and taller), with
> occ_metj_i j = 1 to <=3
> i = 1 to <=8
>
> I need to create summary measures for each i that takes the mean of the
> three j's. Two means are to be created: one that only evaluates values
>>1.5, and one that only evaluates values <=1.5:
>
> +-----------------------------------------------------+
> | id occ_~1_1 occ_~2_1 occ_~3_1 mean1_1 mean2_1 |
> |-----------------------------------------------------|
> | 1 4 7 . 5.50 . |
> | 2 1.5 . . . 1.50 |
> | 3 2.3 3.3 . 2.80 . |
> | 4 3.3 2.3 3.5 3.03 . |
> | 5 1.5 . . . 1.50 |
> |-----------------------------------------------------|
> | 6 1.5 . . . 1.50 |
> | 7 2.3 . . 2.30 . |
> | 8 1.5 . . . 1.50 |
> | 9 1.5 2.3 3.3 2.80 1.50 |
> | 10 1.5 2.3 3.3 2.80 1.50 |
> +-----------------------------------------------------+
>
> The main issue is that the two commands I thought of using, egen and cond,
> do not allow replace and egen, respectively.
> This loop is clearly wrong, but it was an attempt at producing what I need:
> foreach j of num 1/3 {
> foreach i of num 1/8 {
> egen mean1_`i' = rowmean(occ_met1_`i' occ_met2_`i'
> occ_met3_`i') if (occ_met`j'_`i'>1.5)
> egen mean2_`i' = rowmean(occ_met1_`i' occ_met2_`i'
> occ_met3_`i') if (occ_met`j'_`i'<=1.5)
> }
> }
>
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
```
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https://www.teachstarter.com/teks/math-k-3b/
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Teks
# Math K.3(B)
solve word problems using objects and drawings to find sums up to 10 and differences within 10; and
19 teaching resources for those 'aha' moments
• #### teaching resource
A set of 16 task cards that include word problems practicing addition and subtraction with 10.
6 pages Grades: K - 1
• #### teaching resource
Practice addition and subtraction within 10 with this set of 18 word problem task cards.
9 pages Grades: K - 1
### Bump! Subtraction to 10 Game
A board game to practice subtraction facts to 10.
5 pages Grades: K - 1
### Bump! Addition to 10 Game
A board game to practice addition facts to 10.
5 pages Grades: K - 1
### Addition and Subtraction Mat - Apple Tree
A hands-on activity to use with your students when learning to solve addition and subtraction problems up to 10.
2 pages Grades: PK - 1
### My Tens Frames - Template
A template to use when working on counting, addition and subtraction, odd and even numbers, or place value.
1 page Grades: K - 1
### Number Chart – 1-50
A bright number chart representing numbers 1-50.
1 page Grades: K - 2
### Number Chart – 1-20
A bright number chart representing numbers 1-20.
1 page Grades: PK - 1
• #### teaching resource
1 page Grades: PK - K
### Goal Labels - Number and Place Value (Lower Elementary)
Forty-five Number and Place Value Goal Labels for lower elementary.
47 pages Grades: K - 1
### Drawing the Difference Cards
Use these cards when working with students on finding and calculating the difference between two amounts.
8 pages Grades: K - 1
### 'What is the Difference?' Cards
Cards for students to visually identify the differences.
5 pages Grades: K - 1
### Number Line - 10's to 100 - Space
A number line going up by 10's to 100.
1 page Grades: K - 3
### 0 - 50 Space Number Line - Labeled
A number line going up by 5s to 50.
1 page Grades: K - 3
### Number Line 0-30 - Zebras
A 0-30 number line with zebras on it.
1 page Grades: PK - 1
### Number Line 0-30 - Frogs
A 0-30 number line with frogs on it.
1 page Grades: PK - 1
### Number Line 0-30 - Fish
A 0-30 number line with fish on it.
1 page Grades: PK - 1
### Number Line 0-30 - Bees
A 0-30 number line with bees on it.
1 page Grades: PK - 1
| 610
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|
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http://easy-ciphers.com/glacification
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|
Easy Ciphers Tools:
Caesar cipher
Caesar cipher, is one of the simplest and most widely known encryption techniques. The transformation can be represented by aligning two alphabets, the cipher alphabet is the plain alphabet rotated left or right by some number of positions.
When encrypting, a person looks up each letter of the message in the 'plain' line and writes down the corresponding letter in the 'cipher' line. Deciphering is done in reverse.
The encryption can also be represented using modular arithmetic by first transforming the letters into numbers, according to the scheme, A = 0, B = 1,..., Z = 25. Encryption of a letter x by a shift n can be described mathematically as
Plaintext: glacification
cipher variations: hmbdjgjdbujpo incekhkecvkqp jodflilfdwlrq kpegmjmgexmsr lqfhnknhfynts mrgioloigzout nshjpmpjhapvu otikqnqkibqwv pujlrorljcrxw qvkmspsmkdsyx rwlntqtnletzy sxmouruomfuaz tynpvsvpngvba uzoqwtwqohwcb vaprxuxrpixdc wbqsyvysqjyed xcrtzwztrkzfe ydsuaxauslagf zetvbybvtmbhg afuwczcwuncih bgvxdadxvodji chwyebeywpekj dixzfcfzxqflk ejyagdgayrgml fkzbhehbzshnm
Decryption is performed similarly,
(There are different definitions for the modulo operation. In the above, the result is in the range 0...25. I.e., if x+n or x-n are not in the range 0...25, we have to subtract or add 26.)
Atbash Cipher
Atbash is an ancient encryption system created in the Middle East. It was originally used in the Hebrew language.
The Atbash cipher is a simple substitution cipher that relies on transposing all the letters in the alphabet such that the resulting alphabet is backwards.
The first letter is replaced with the last letter, the second with the second-last, and so on.
An example plaintext to ciphertext using Atbash:
Plain: glacification Cipher: tozxrurxzgrlm
Baconian Cipher
To encode a message, each letter of the plaintext is replaced by a group of five of the letters 'A' or 'B'. This replacement is done according to the alphabet of the Baconian cipher, shown below.
```a AAAAA g AABBA m ABABB s BAAAB y BABBA
b AAAAB h AABBB n ABBAA t BAABA z BABBB
c AAABA i ABAAA o ABBAB u BAABB
d AAABB j BBBAA p ABBBA v BBBAB
e AABAA k ABAAB q ABBBB w BABAA
f AABAB l ABABA r BAAAA x BABAB
```
Plain: glacification Cipher: AABBA ABABA AAAAA AAABA ABAAA AABAB ABAAA AAABA AAAAA BAABA ABAAA ABBAB ABBAA
Affine Cipher
In the affine cipher the letters of an alphabet of size m are first mapped to the integers in the range 0..m - 1. It then uses modular arithmetic to transform the integer that each plaintext letter corresponds to into another integer that correspond to a ciphertext letter. The encryption function for a single letter is
where modulus m is the size of the alphabet and a and b are the key of the cipher. The value a must be chosen such that a and m are coprime.
Considering the specific case of encrypting messages in English (i.e. m = 26), there are a total of 286 non-trivial affine ciphers, not counting the 26 trivial Caesar ciphers. This number comes from the fact there are 12 numbers that are coprime with 26 that are less than 26 (these are the possible values of a). Each value of a can have 26 different addition shifts (the b value) ; therefore, there are 12*26 or 312 possible keys.
Plaintext: glacification
cipher variations:
hmbdjgjdbujpo
tibhzqzhbgzro
feblpaplbspto
rabpfkfpbefvo
dwbtvuvtbqvxo
psbxlelxbclzo
nkbfryrfbardo
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hafxzyzxfuzbs
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cbyimxmiypmql
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atyqsrsqynsul
mpyuibiuyziwl
khycovocyxoal
wdygefegyjecl
izykupukyvuel
uvyokzkoyhkgl
grysajasytail
snywqtqwyfqkl
fkzbhehbzshnm
rgzfxoxfzexpm
dczjnynjzqnrm
pyzndidnzcdtm
buzrtstrzotvm
nqzvjcjvzajxm
lizdpwpdzypbm
xezhfgfhzkfdm
jazlvqvlzwvfm
vwzplalpzilhm
hsztbkbtzubjm
tozxrurxzgrlm
glacification
shagypygafyqn
edakozokarosn
cvasutusapuwn
orawkdkwabkyn
mjaeqxqeazqcn
yfaighgialgen
kbamwrwmaxwgn
wxaqmbmqajmin
itauclcuavckn
upaysvsyahsmn
The decryption function is
where a - 1 is the modular multiplicative inverse of a modulo m. I.e., it satisfies the equation
The multiplicative inverse of a only exists if a and m are coprime. Hence without the restriction on a decryption might not be possible. It can be shown as follows that decryption function is the inverse of the encryption function,
ROT13 Cipher
Applying ROT13 to a piece of text merely requires examining its alphabetic characters and replacing each one by the letter 13 places further along in the alphabet, wrapping back to the beginning if necessary. A becomes N, B becomes O, and so on up to M, which becomes Z, then the sequence continues at the beginning of the alphabet: N becomes A, O becomes B, and so on to Z, which becomes M. Only those letters which occur in the English alphabet are affected; numbers, symbols, whitespace, and all other characters are left unchanged. Because there are 26 letters in the English alphabet and 26 = 2 * 13, the ROT13 function is its own inverse:
ROT13(ROT13(x)) = x for any basic Latin-alphabet text x
An example plaintext to ciphertext using ROT13:
Plain: glacification Cipher: tynpvsvpngvba
Polybius Square
A Polybius Square is a table that allows someone to translate letters into numbers. To give a small level of encryption, this table can be randomized and shared with the recipient. In order to fit the 26 letters of the alphabet into the 25 spots created by the table, the letters i and j are usually combined.
1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z
Basic Form:
Plain: glacification Cipher: 22131131421242311144424333
Extended Methods:
Method #1
Plaintext: glacification
method variations: mqfholohfyots rvlntqtnldtyx waqsyvysqiydc bfvxdadxvodih
Method #2
Bifid cipher
The message is converted to its coordinates in the usual manner, but they are written vertically beneath:
```g l a c i f i c a t i o n
2 1 1 3 4 1 4 3 1 4 4 4 3
2 3 1 1 2 2 2 1 1 4 2 3 3 ```
They are then read out in rows:
21134143144432311222114233
Then divided up into pairs again, and the pairs turned back into letters using the square:
Plain: glacification Cipher: bldoqthcfgain
Method #3
Plaintext: glacification
method variations: bclqbrmaqtrnh clqbrmaqtrnhb lqbrmaqtrnhbc qbrmaqtrnhbcl brmaqtrnhbclq rmaqtrnhbclqb maqtrnhbclqbr aqtrnhbclqbrm qtrnhbclqbrma trnhbclqbrmaq rnhbclqbrmaqt nhbclqbrmaqtr hbclqbrmaqtrn
Permutation Cipher
In classical cryptography, a permutation cipher is a transposition cipher in which the key is a permutation. To apply a cipher, a random permutation of size E is generated (the larger the value of E the more secure the cipher). The plaintext is then broken into segments of size E and the letters within that segment are permuted according to this key.
In theory, any transposition cipher can be viewed as a permutation cipher where E is equal to the length of the plaintext; this is too cumbersome a generalisation to use in actual practice, however.
The idea behind a permutation cipher is to keep the plaintext characters unchanged, butalter their positions by rearrangement using a permutation
This cipher is defined as:
Let m be a positive integer, and K consist of all permutations of {1,...,m}
For a key (permutation) , define:
The encryption function
The decryption function
A small example, assuming m = 6, and the key is the permutation :
The first row is the value of i, and the second row is the corresponding value of (i)
The inverse permutation, is constructed by interchanging the two rows, andrearranging the columns so that the first row is in increasing order, Therefore, is:
Total variation formula:
e = 2,718281828 , n - plaintext length
Plaintext: glacification
first 5040 cipher variations(6187239489 total)
glacification
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# Trapezoid
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In trapezoid $PQRS,$ $\overline{PQ} \parallel \overline{RS}$. Let $X$ be the intersection of diagonals $\overline{PR}$ and $\overline{QS}$. The area of triangle $PQX$ is $2,$ and the area of triangle $RSX$ is $2.$ Find the area of trapezoid $PQRS$.
Feb 23, 2024
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## Systematic Treatment of Molar Solubility
How do we systematically calculate the molar solubility of a substance with regards to its Ksp? In other words, where does the value of "molar solubility" fit in to the equilibrium calculations?
This should build up so that I can understand how to systematically treat the common ion effect, among other equilibria, if necessary. For example, if I want to know what the molar solubility is of my salt when, if it dissolves, it will undergo hydrolysis of the anion, or complex formation, or both (so we would need to consider acid/base equilibria and Kf at the same time as Ksp), how do I set up the equilibrium expressions to solve to get my molar solubility? This would come from an understanding of how exactly to calculate "molar solubility" in relation to equilibria in general.
I suspect that a major step is to note that Ksp = the equilibrium concentrations of the two ions of the salt, to the power of their coefficients in the molecular formula respectively, multiplied together, ONLY when the molar solubility is actually reached - until then, their equilibrium concentrations will multiply to a lower value than the Ksp only. That still leaves me unclear as to exactly what "molar solubility" is and how we would go about calculating it in situations with many equilibria going around.
PhysOrg.com chemistry news on PhysOrg.com >> Scientists make breast cancer advance that turns previous thinking on its head>> Cradle turns smartphone into handheld biosensor>> New filtration material could make petroleum refining cheaper, more efficient
Admin This is not different from the general equilibrium calculations that we discussed in the past. Ksp is just one of the equilibrium equations, that has to be used together wit all other equilibria, mass balances and charge balance. Molar solubility is something that you can easily calculate AFTER you know all equilibrium concentrations. If, for any reason, solution is not saturated then yes, Q < Ksp.
Quote by Borek This is not different from the general equilibrium calculations that we discussed in the past. Ksp is just one of the equilibrium equations, that has to be used together wit all other equilibria, mass balances and charge balance. Molar solubility is something that you can easily calculate AFTER you know all equilibrium concentrations. If, for any reason, solution is not saturated then yes, Q < Ksp.
1.) OK, so if the solution is not saturated then we cannot use the Ksp expression in the equilibrium calculations. However, this would not affect our attempt to find molar solubility anyway as the solution will be saturated and only then do we work out molar solubility (and then we can also use the Ksp expression in our equilibrium calculations).
2.) Will the molar solubility simply be the analytical concentration of the cation (in the same way as defined for the mass balance) divided by the coefficient of the cation in the molecular formula of the salt (and also the same thing, for the anion, both of these methods giving the same value)?
3.) Let's say I've added Ag2SO4 to solution and this has established an equilibrium not according to Ksp (the Ksp value is known but the solution is not saturated). Now I want to find out how much AgCl (with a known Ksp value) I can add, i.e. molar solubility of AgCl. Is there any way/how would we do this calculation?
## Systematic Treatment of Molar Solubility
Quote by Big-Daddy 2.) Will the molar solubility simply be the analytical concentration of the cation (in the same way as defined for the mass balance) divided by the coefficient of the cation in the molecular formula of the salt (and also the same thing, for the anion, both of these methods giving the same value)?
Can be, but doesn't have to. Your cation can be complexed and then to calculate the molar solubility you will need to sum concentrations of all forms of the cation (followed by division by the stoichiometric coefficient).
3.) Let's say I've added Ag2SO4 to solution and this has established an equilibrium not according to Ksp (the Ksp value is known but the solution is not saturated). Now I want to find out how much AgCl (with a known Ksp value) I can add, i.e. molar solubility of AgCl. Is there any way/how would we do this calculation?
If the solution is not saturated, the only way to calculate concentration of Ag+ is from the mass balance - you should know how much Ag2SO4 were put into the mix. If you don't know it, you can't solve the problem
Quote by Borek Can be, but doesn't have to. Your cation can be complexed and then to calculate the molar solubility you will need to sum concentrations of all forms of the cation (followed by division by the stoichiometric coefficient).
OK so let's redefine as the molar solubility is the initial concentration of the cation added to the solution (which is equal to the cation's mass balance, which would include all species containing the cation multiplied by the stoichiometric coefficient of the cation in those species, and this total sum of concentrations is the initial concentration of the cation), which can be divided by the stoichiometric coefficient of the cation in the original salt's molecular formula to reach the molar solubility. Or the same thing for the anion.
Quote by Borek If the solution is not saturated, the only way to calculate concentration of Ag+ is from the mass balance - you should know how much Ag2SO4 were put into the mix. If you don't know it, you can't solve the problem
You can have how many moles of Ag2SO4 were originally put in the solution, and the Ksp of Ag2SO4, but Ag2SO4 is not being added until saturation. Can you work it out then? Since Ksp ≠ [Ag+]2*[SO42-] unless the solution is saturated, which it here is not ...
Help?
If im correctly understanding [Ag+] before saturation as part of your question : Then can you get at this with electrode potentials ? Where E∅ saturated = [Ag+(aq)] ,E value,[?]/Ag electrode + .059 lg[Ag+aq] Where [Ag(+)aq] on right side is silver concentration in saturated solution. From the Nernst equation. E = E∅ + (RT/zF) ln[Mz+] E=E∅+(.059)/z) lg[Agz+] E∅ at electrode potential at 1mole/liter Each 10x lowering of [Ag+] lowers potential by (.059/z) volts
Quote by Big-Daddy You can have how many moles of Ag2SO4 were originally put in the solution, and the Ksp of Ag2SO4, but Ag2SO4 is not being added until saturation. Can you work it out then? Since Ksp ≠ [Ag+]2*[SO42-] unless the solution is saturated, which it here is not ...
If you know initial moles of Ag2SO4 you put them into the mass balance for silver and if the solution is not saturated, you simply drop Ksp from the set of equations.
Quote by Borek If you know initial moles of Ag2SO4 you put them into the mass balance for silver and if the solution is not saturated, you simply drop Ksp from the set of equations.
Thank you.
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# MECHANICAL DESIGN ENGINEERING – Geometrical Dimensioning and Tolerancing_What is the CIRCULAR RUNOUT tolerance?
Runout is one of the oldest and simplest concepts used in GD & T. Maybe as a child you stood your bicycle upside down on the ground and spun a wheel. If you fixed your stare on the shiny rim where it passed a certain part of the frame, you could see the rim wobble from side to side and undulate inward and outward. Instead of the rim running in a perfect circle, it, well—run-out. Runout, then, is the variation in the surface elements of a round feature relative to an axis. There are 2 types of Run-Out tolerances which are used in GD &T. According to the measurement method we have:
1. THE CIRCULAR RUN-OUT TOLERANCE = checked at each individual point and distributed arbitrarily over the tolerated element.
2. THE TOTAL RUN-OUT TOLERANCE = During the measurement, the measuring device is guided over the entire tolerated element.
According to measuring direction we have:
• Circular Run-Out/ Axial Run-Out = The measuring direction is perpendicular or parallel to the reference axis (both for a circular run and for a total run)
• Run-out in any or specified direction = The measurement direction goes perpendicularly to the direction of the tolerated surface respectiely to the drawing view where it is specified (standardized only for circular run-out)
THE RUN-OUT TOLERANCES (circular and total) come from mechanical measurement technics and are clearly conceivable. Such mechanical tolerances restrict several functionally important types of tolerances togehter, therefore run-out tolerances have a wide area of applications. In the text below, I discuss THE CIRCULAR RUN-OUT TOLERANCE characteristics. Yet most of what I expose about Circular Run-Out is applicable for Total Run-Out as well.
Description
Definition: The Circular Runout is how much one a given reference feature or features vary with respect to another datum when the part is rotated 360° around the datum axis. It is essentially a control of a circular feature, and how much variation it has with the rotational axis.
Runout can be called out on any feature that is rotated about an axis. It is essentially how much “wobble” occurs in the one part feature when referenced to another.
Symbol: The Symbol for Circular Run-Out tolerance is a slanting arrow pointing upwards to the right, representing the pointer of a dial gauge.
The symbol does not differentiate between round, axial or any other type of runout; the measuring direction results solely from the direction of the tolerance arrow.
Why Do We Use It? – In precision assemblies, runout causes misalignment and/or balance problems. In Fig. 2, runout of the ring groove diameters relative to the piston’s diameter might cause the ring to squeeze unevenly around the piston or force the piston off center in its bore. A motor shaft that runs out relative to its bearing journals will cause the motor to run out-of-balance, shortening its working life. A designer can prevent such wobble and lopsidedness by specifying a runout tolerance.
The Tolerance call out: A circular runout tolerance is specified using a feature control frame displaying the characteristic symbol. As illustrated in Fig. 3, the arrowhead may be drawn filled or unfilled. The feature control frame includes the runout tolerance value followed by 1 or 2 (but never 3) datum references.
How Does It Work? – As shown in Fig. 2, for as long as piston ring grooves and motor shafts have been made, manufacturers have been finding ways to spin a part about its functional axis while probing its surface with a dial indicator. As the indicator’s tip surfs up and down over the undulating surface, its dial swings gently back and forth, visually displaying the magnitude of runout. Thus, measuring runout can be very simple as long as we agree on 3 things:
• What surface(s) establish the functional axis for spinning—datums?
• Where the indicator is to probe?
• How much swing of the indicator’s dial is acceptable?
The whole concept of “indicator swing” is somewhat dated. Draftsmen used to annotate it on drawings as TIR for “Total Indicator Reading.” The Y14.5 standard briefly called it FIR for “Full Indicator Reading.” Then, in 1973, Y14.5 adopted the international term, FIM for “Full Indicator Movement.”
Full Indicator Movement (FIM) is the difference (in millimeters or inches) between an indicator’s most positive and most negative excursions. Thus, if the lowest reading is -0.01mm and the highest is +0.02mm, the FIM (or TIR or FIR) is 0.03mm. Just because runout tolerance is defined and discussed in terms of FIM doesn’t mean runout tolerance can only be applied to parts that spin in assembly. Neither does it require the part to be rotated, nor use of an antique 20th century, jewel-movement, dial indicator to verify conformance. The “indicator swing” standard is an ideal meant to describe the requirements for the surface. Conformance can be verified using a CMM, optical comparator, laser scanning with computer modeling, process qualification by SPC, or any other method that approximates the ideal.
Considering the purpose for runout tolerance and the way it works, there’s no interaction between a feature’s size and its runout tolerance that makes any sense. In our piston ring groove diameter example from fig.2, an MMC modifier would be counterproductive, allowing the groove diameter’s eccentricity to increase as it gets smaller. That would only aggravate the squeeze and centering problems we’re trying to correct. Thus, material condition modifier symbols, MMC and LMC, are prohibited for both circular and total runout tolerances and their datum references. If you find yourself wishing you could apply a runout tolerance at MMC, you’re not looking at a genuine runout tolerance application; you probably want positional tolerance instead.
Tolerated geometry elements:
Datums for Runout Control = A runout tolerance controls surface elements of a round feature relative to a datum axis. GD&T modernized runout tolerancing by applying the rigors and flexibility of the DRF (Datum Reference Frame). Every runout tolerance shall reference a datum axis. Fig. 4 shows 3 different methods for doing this.
Since a designer wishes to control the runout of a surface as directly as possible, it’s important to select a functional feature(s) to establish the datum axis. During inspection of a part such as that shown in Fig. 4 (a), the datum feature might be placed in a V-block or fixtured in a precision spindle so that the part can be spun about the axis of the datum feature’s TGC (True Geometric Counterpart). This requires that the datum feature be long enough and that its form be well controlled (perhaps by its own size limits or form tolerance). In addition, the datum feature must be easily accessible for such fixturing or probing.
Where a single datum feature or co-datum feature pair establishes the axis, further datum references are meaningless and confusing. However, there are applications where a shoulder or end face exerts more leadership over the part’s orientation in assembly while the diametral datum feature merely establishes the center of revolution. In Fig. 4(c), for example, the face is identified as primary datum feature A and the bore is labeled secondary datum feature B. In inspection, the part will be spun about datum axis B which, remember, is restrained perpendicular to datum plane A. A run-out deviation which can make the testing stylus to move, can have 2 different causes (see Figure 11 below, except for the axial runout b):
The measured cross-section is not rounded.
The tolerated form feature is not coaxial.
The circular runout deviation results from the overlaping of circularity and concentricity deviation, therefore is not simply as addition. In this case the dependecies are quite complex.
Circular Runout and Circularity: The perimeter of the tolerated circular cross-section in Figure 11a must lie entirely within the tolerance zone, which – like the circularity tolerance zone – forms a circular ring with a width of tCRo = 0.1 mm. From here the following rule applies:
RULE #1 = Runout tolerance and circularity: In the runout tolerance, the roundness (circularity/circular shape) of all cross-sections is included, meaning that the circularity deviation cannot be greater than the circular run-out tolerance tCRo.
Figure 5 shows a workpiece similar to example (a) from fig.6 (shown below). In limiting case in fig. 5a, the concentricity (coaxiality) deviation between the reference and the tolerated element should be dCO= 0. A circularity tolerance of tC = 0.1 mm is specified for the journal, which use it fully. Such deviations can occur approximately if the workpiece is machined on a worn lathe machine (with circularity deviation) in one setting (without concentricity deviation). The dial gauge obviously deflects by the amount of circular runout tolerance dCRO = tC = 0.1 mm. This gives the next rule:
RULE #2 = Runout and Circularity deviations: When there is no concentricity deviation, the measured runout deviation is equal to the circularity deviation.
Circularity measuring devices respectively form measuring devices work according to this principle. They compensate for coaxiality deviations mathematically or mechanically by aligning the reference axis.
Circular Runout and Concentricity tolerance: = In the other limiting case, Figure 5b, the tolerated cross-section should not have a circularity deviation dC, but a coaxiality deviation (axis offset) dCo; this is equal to the limit deviation tCo/2= 0.05 mm, meaning it makes full use of the specified concentricity tolerance tCo= 0.1 mm. Such deviations can be encountered when the workpiece is re-clamped in a worn lathe machine (with an offset) from a very good lathe (without circularity deviation). The dial gauge again shows a runout deviation of dCRo = 0.1 mm.
RULE # 3 = Runout tolerance and concentricity: If the tolerated cross-section is accurately circular, then the radial runout deviation dCRo is twice as large as the axis offset, just like the concentricity deviation dCo. So if circularity deviations are negligible, the runout tolerance tCRo includes the concentricity with the same tolerance value.In other words: the circular runout tolerance corresponds to a concentricity tolerance with the same tolerance value as long as there are no deviations from the circular shape.
In practice, there will always be a concentricity deviation in addition to the circularity deviation. This usually increases the measured runout deviation,(at least does not decreases it). In Figure 5 the case c is actualy the case b additionally accompanied by a circularity deviation dC = 0.05 mm. Nevertheless, in this case the displayed runout deviation remains, as the greatest difference in the distances to the reference axis A, at dCRo = 0.1 mm (as in case b).
However, it is also possible that the additional circularity deviation to make the measured runout deviation smaller. With an ideally circular cross-section, the concentricity deviation is equal to half the runout deviation (see rule #3). Practical measurements and my own investigations have shown that under real conditions the actual concentricity deviation can be greater than half the measured circular runout deviation by a deviation factor U ~ 1.1 to 1.2; in special cases, U can theoretically even approach the value 2. Figure 6 explains the relationships based on two idealized cross-section profiles.
Nominal constant thickness profiles often resemble the profile in Figure 6a. It has 3 lines of symmetry and consists of 6 circular arcs with 2 different radii r and R, which emanate from the corner points of an internal equilateral triangle (or ideal 3-arc equidistance). Calculations show that the largest excess factor with U = 1.1547 occurs when the reference axis A is in a corner point; in this case, dCo = 0.058 mm (instead of 0.05 mm as in Figure 5(b)). This does not depend on the ratio of the concentricity deviation dCo to the uniform thickness s (where s = r + R), meaning that this exceeding can also occur with small values for axis offset and circularity deviation.
In the case of profiles with 2 lines of symmetry (e.g. ellipses, see Figure 5a), there is no exceeding (i.e. U = 1), but this can happen in the case of profiles with only 1 line of symmetry (mirror symmetry), for example as in Figure 6b . In this case, the excess factor U can theoretically reach a value of 2, while under more practical conditions it is around 1.1 to 1.2. In contrast to the 3-sided profiles, the results here also depend on the calculation of the profile center point M. For this there is still a need for further measurement results and standardization. In summary, we must proceed with the following 2 rules:
RULE #4 = Included tolerances for circular runout: In the circular runout tolerance tCRo the circularity tolerance tC is included with the same tolerance value. The concentricity tolerance tCo, however, is only with an exceeding factor U. Currently, U ~ 1.2 appears to be a usable value (but not U = 1, as is often assumed).
RULE # 5 = Runout instead of coaxiality test: If instead of a specified concentricity tolerance tCo, the circular runout is checked, then the permissible runout deviation is:
dCRo ≤ tCo/U (with U ≈ 1.2)
Note: If a circularity tolerance tC and a concentricity tolerance tCo are specified, the runout deviation dCRo, cannot be greater than their sum: dCRo ≤ tC+tCo
Runout in any direction, circularity and concentricity: With a measuring angle ɑ (acc. to Figure 11 c and d; where d has ɑ = 60°), each individual tolerance zone forms a cone with the angle of inclination ɑ. However, since circularity and concentricity are always checked perpendicular to the reference axis, the included runout tolerances must be slightly modified for any runout direction according to the above rules #1 to #5. If a runout tolerance tCRo is specified, then the following formula applies (according to rules #1 to #5):
Circularity deviation: dC ≤ tCRo/sin ɑ
Concentricity deviation: dCo≤ U*tCRo/2sin ɑ
GD&T Tolerance Zone
The 2-Dimensional circular tolerance zone is an area that is defined by a datum axis where all points on the called surface must fall into. The zone is a direct reference to the datum feature. Circular Runout is the total variation that the reference surface can have when the part is rotated around the datum’s true axis. (as shown in figure 7)
For deviations the following general rule applies:
RULE #6 = Runout deviations: The nominal deviation coresponds to the largest one that occur as display difference during the measurement (apart from the inaccuracy of the measuring device). The limit deviation is equal to the runout tolerance.
When the Circular RunOut tolerance is used?
Circular Runout tolerance is the lesser level of runout control. Its tolerance applies to the FIM (Full Indicator Reading) while the indicator probes over a single circle on the part surface. That means the indicator’s body is to remain stationary both axially and radially relative to the datum axis as the part is spun at least 360° about its datum axis. The tolerance applies at every possible circle on the feature’s surface, but each circle may be evaluated separately from the others. Therefore we have 2 types of such cases as shown in Figure 8 below:
• a) Circular Runout on radius
• b) Circular Runout on axis
Circular runout can be applied to any feature that is nominally cylindrical, spherical, toroidal, conical, or any revolute having round cross sections (perpendicular to the datum axis). When evaluating noncylindrical features, the indicator shall be continually realigned so that its travel is always normal to the surface at the subject circle. See Fig. 9. Circular runout can also be applied to a face or face groove that is perpendicular to the datum axis. Here, the surface elements are circles of various diameters, each concentric to the datum axis and each evaluated separately from the others.
Runout examples of use:
Example 1: Two coaxial features establishing a datum axis for runout control
Let’s evaluate the 0,5mm circular runout tolerance of Fig.10. We place an indicator near the left end of the controlled diameter and spin the part 360°. We see that the farthest counterclockwise excursion of the indicator dial reaches -0.1mm and the farthest clockwise excursion reaches +0.2mm. The circular runout deviation at that circle is 0.3mm. We move the indicator to the right and probe another circle. Here, the indicator swings between -0.3mm and +0.1mm. The difference, 0.4mm, is calculated without regard for the readings we got from the first circle. The FIM for each circle is compared with the 0.5mm tolerance separately. Obviously, we can’t spend all day trying to measure infinitely many circles, but after probing at both ends of the feature and various places between, we become confident that no circle along the feature would yield an FIM greater than, perhaps, 0.4. Then, we can conclude the feature conforms to the 0.5mm circular runout tolerance.
There are many cases where the part itself is a spindle or rotating shaft that, when assembled, will be restrained in 2 separate places by 2 bearings or 2 bushings. See Fig. 10. If the 2 bearing journals have ample axial separation, it’s unrealistic to try to fixture on just one while ignoring the other. We could better stabilize the part by identifying each journal as a datum feature and referencing both as equal co-datum features. In the feature control frame, the datum reference letters are placed in a single box, separated by a hyphen. The hyphenated co-datum features work as a team. Neither co-datum feature has precedence over the other. We can’t assume the 2 journals will be made perfectly coaxial. To get a decent datum axis from them, we should add a runout tolerance for each journal, referencing the common datum axis they establish, meaning to apply a Total Runout Tolerance which I discuss more in details in a dedicated article.
Example 2: Circular Runuout – Radial, Axial, In any direction and In a specified direction
Figure 11a)Circular runout – radial: The tolerance arrow is perpendicular to the reference A. The tolerance zone is initially the gap of 0,1mm by which the tip of the probe can move radially with the tolerance tCRo = 0,1mm. This applies to all points of the circular cross-section; meaning that the rotation turns it into a circular ring with the width of the runout tolerance tCRo .The shape of the tolerance zone corresponds to that of circularity, except that its center is now defined by the reference axis. Since each circular ring is checked individually, the convex or conical form of the lateral surface thus the cylindrical shape is not recorded. (This workpiece has a “long” reference axis.)
Figure 11b)Circular runout – axial: The tolerance arrow is parallel to reference A-B. Individual circular rings are checked concentrically to the axis. The tolerance zones form individual circular cylinder rings of the axial length of the run-out tolerance tCRo. This will only limit the wobble of the frontal surface but not its flatness; it could be for instance either convex or conical. For this reason its perpendicularity deviation can be bigger than its planar run-out tolerance. (The reference axis is in this case the common axis of the 2 bearing seats)
Figure 11c)Runout in any direction: The runout is checked in the normal direction on the tolerated surface, meaning on local perpendicular direction. For a mechanical measurement this is the simplest way to do it. This means that any surface of revolution can be tolerated (for instance cone, sphere, fillet). Each individual tolerance zone form a cone; this distinguishes from the roundness (circularity) tolerance which is always circular to the ring perpendicular to the axis. (This is a “short” axis which is not suitable as one-time reference. With the locating surface it forms a reference system that allows both checking and function.)
Figure 11d)- Runout in a specified direction: This is used for instance when a copy template must be tolerated in the direction of associated probe. The measuring direction is entered as angular dimension without tolerance (as nominal angle), which applies as theoretical dimension. It would make sense to enter this theoretical dimension in a square. (In this case the common axis is defined by the 2 forehead centereing surfaces.)
Example 3: Rotating Shaft
As shown in Figure 12, a shaft that is rotated at very high speeds is prone to oscillations if the right edge of the shaft is too far offset from the left side. To control how much wobble this part will have runout is used to ensure that the smaller diameter surface is relatively controlled to datum surface A.
To control this without GD&T would be nearly impossible. The small amount of variation in the shaft, straightness of the shaft, and roundness of the individual surfaces would be unrealistic to control.
With runout, you have your final rotational condition that you want controlled without needing to specify unnecessary tight control on the entire part. By constraining with runout as shown on the drawing you are ensuring that when the shaft rotating, with datum A fixed in housing, the reference surface will not go outside of a perfect central rotation by more than 30 microns. This will ensure that only a limited vibration is made and that both parts will wear evenly. To ensure this condition is met, you must measure the parts with a gauge.
B is now controlled in relation to A, ensuring a smooth, near-perfect rotational system.
Note: this runout must be controlled on any cross-section along the reference surface. You must gauge each cross-section separately though (Gauging the entire cylinder at once would be Total Runout).
Gauging / Measurement
Runout is measured using a simple height gauge on the reference surface. The datum axis is controlled by fixing all datum points and rotating the central datum axis. The part is usually constrained with V-blocks, or a spindle, on each datum that is required to be controlled. The part is then rotated around this axis and the variation is measured using the height gauge held perpendicular to the part surface. As long as the gauge does not vary by more than the runout tolerance, the part is in spec.
Geometry elements and checking: Runout tolerances are only applicable to rotationally symmetrical parts.
All runout tolerances have an axis (axis of rotation) as a reference. During the measurement, the workpiece rotates around this axis while a measuring device (dial gauge, precision gauge) is placed on the surface to be tested. Nominal features with a circular cross-section (cylinders, cones, etc.) and plane surfaces (frontal surfaces of revolution) are therefore always tolerated. Of course, the measuring device can also rotate around the workpiece. The reference axes have the runout tolerances in common with the coaxiality tolerance. The tolerated area, e.g. the cone or the frontal surface of a shaft must comply with the running tolerance at all points:
RULE # 7 = Testing of the circular runout: The runout is to be checked at a sufficient number of individual positions. Each individual measurement result is compared with the limit deviation (i.e. with the runout tolerance tCRo).
As it is said in practice, between the measurements “the clock is again reset to zero” . Because each individual measurement checks a circular ring of the toleranced element, hence the designation “circular runout” in ISO 1101.
Relation to Other GD&T Symbols
RunOut tolerances have circular tolerance zones of various shapes; therefore they are related to Circularity and Concentricity tolerances.
A great way to relate this symbol to others is through this equation:
Circular Runout = Total Circularity (out of round) + Concentricity (axis offset)
Runout captures both of these in a single measurement when you are comparing the surface to another datum. Runout can also be constrained using a face as well as another circular surface. If this is the case the perpendicularity of the datum face to the reference surface can add into the runout of the surface as well, since if the part is tilted at an angle, the part would runout higher due to the tilting of the part.
Circular Runout is the 2D version of Total Runout. While it is measured in individual cross-sections, total runout takes the measurement around and across the surface of the entire part in a 3D tolerance zone.
Final Notes
Circular Name: Runout as a GD&T symbol is often referred to as circular runout to differentiate it from total runout.
Two similar versions: Runout is a relation of surface to datum surface or surface to datum axis. When the datum is a surface, any out of round on the datum surface can impact the runout of the part, depending on if the high and low spots on the datum correspond to the high and low spots on the reference feature. (Remember relating the axis to a datum axis is Concentricity)
Regardless of Feature Size: Circular Runout is always RFS (Regardless of Feature Size) meaning that the boundary formed by the dimensions is the entire part envelope that the part can exist regardless of how large the tolerance is. No MMC or bonus tolerance is ever used with it.
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# Therefore f y y d dy f y y f x y f x y 1 2 π e y 2 1
• Notes
• 5
This preview shows page 3 - 5 out of 5 pages.
Therefore,FY(y) =ddyFY(y) =fX(y) +fX(-y) =12πe-y2+12πe-y2=r2πe-y2.Thus,EY=Z0yr2πe-y2dy=r2πZ0e-udu=r2πh-e-uflflfl0i=r2π,whereu=y22.EY2=Z0y2r2πe-y2dy=r2π"-ye-y2flflfl0+Z0e-y2dy#=r2πrπ2= 1.This was done using integration by part withu=yanddv=ye-y2dy. ThenV ar(Y) = 1-2π.2.12 We have tanx=y/d, thereforetan-1(y/d) =xandddytan-1(y/d) =11+(y/d)21ddy=dx. Thus,fY(y) =2πd11 + (y/d)2,0< y <.This is the Cauchy distribution restricted to (0,), and the mean is infinite.2.14Z0(1-FX(x))dx=Z0P(X > x)dx=Z0ZxfX(y)dydx=Z0Zy0dxfX(y)dy=Z0yfX(y)dy=EX,where the last equality follows from changing the order of integration.3
2.16 From Exercise 2.14,ET=Z0[ae-λt+ (1-a)e-μt]dt=-ae-λtλ-(1-a)e-μtμflflflflfl0=aλ+1-aμ.2.17 a.Rm03x2dx=m3set=12m= (12)1/3=.794b. The function is symmetric about zero, thereforem= 0 as long as the integral is finite.1πZ-∞11 +x2dx=1πtan-1(x)flflflflfl-∞=1π(π2+π2) = 1.This is the Cauchy pdf.2.23 a. Use Theorem 2.1.8 withA0= 0,A1= (-1,0) andA2= (0,1). Theng1(x) =x2onA1andg2(x) =x2onA2. ThenfY(y) =12y-1/2,0< y <1.b.EY=R10yfY(y)dy=13EY2=R10y2fY(y)dy=15V arY=15-(13)2=445.2.25 a.Y=-Xandg-1(y) =-y. ThusfY(y) =fX(g-1(y))flflflddyg-1(y)flflfl=fX(-y)| -1|=fX(y) foreveryy.b. To show thatMX(t) is symmetric about 0 we must show thatMX(0 +²) =MX(0-²) for all² >0.MX(0 +²)=Z-∞e(0+²)xfX(x)dx=Z0-∞e²xfX(x)dx+Z0e²xfX(x)dx=Z0e²(-x)fX(-x)dx+Z0-∞e²(-x)fX(-x)dx=Z-∞e-²xfX(x)dx=Z-∞e(0-²)xfX(x)dx=MX(0-²).
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Find Concavity and Inflection Points Using Second Derivatives — Practice Questions - dummies
# Find Concavity and Inflection Points Using Second Derivatives — Practice Questions
One purpose of the second derivative is to analyze concavity and points of inflection on a graph. The following figure shows a graph with concavity and two points of inflection.
Concavity and points of inflection.
Using this figure, here are some points to remember about concavity and inflection points:
• The section of curve between A and B is concave down — like an upside-down spoon or a frown; the sections on the outsides of A and B are concave up — like a right-side up spoon or a smile; and A and B are inflection points.
• A positive second derivative means that section is concave up, while a negative second derivative means concave down. And where the concavity switches from up to down or down to up (like at A and B), you have an inflection point, and the second derivative there will (usually) be zero.
## Practice questions
1. Find the intervals of concavity and the inflection points of f(x) = –2x3 + 6x2 – 10x + 5.
2. Find the intervals of concavity and the inflection points of g(x) = x4 – 12x2.
1. For f(x) = –2x3 + 6x2 – 10x + 5, f is concave up from negative infinity to the inflection point at (1, –1), then concave down from there to infinity.
To solve this problem, start by finding the second derivative.
Now set it equal to 0 and solve.
Check for x values where the second derivative is undefined. (There are none.)
Now test your two regions, to the left and to the right of x = 1.
Make a sign graph, as shown here.
Because the concavity switches at x = 1 and because
equals zero there, there’s an inflection point at x = 1.
Find the height of the inflection point.
Thus f is concave up from negative infinity to the inflection point at (1, –1), and then concave down from there to infinity. As always, you should check your result on your graphing calculator. Hint: To get a good feel for the look of this function, you need a fairly odd graphing window — try something like xmin = –2, xmax = 4, ymin = –20, ymax = 20.
2. For g(x) = x4 – 12x2, g is concave up from negative infinity to the inflection point at
then concave down to an inflection point at
then concave up again to infinity.
To solve this problem, start by finding the second derivative.
Now set it to 0 and solve.
Is the second derivative undefined anywhere? No.
Now it’s time to test the three regions.
Make a sign graph, as shown here.
Because the concavity switched signs at the two zeros of
there are inflection points at these two x values.
Find the heights of the inflection points.
g is concave up from negative infinity to the inflection point at
concave down from there to another inflection point at
and then concave up again from there to infinity.
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https://www.teachoo.com/8096/2709/Converting-paisa-and-rupees/category/Decimal-Numbers/
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We know that
100 paise = ₹ 1
1 paise = ₹ 1/100
### Convert 86 paise into rupees
86 paise = 86 × 1 paise
= 86 × 1/100 rupees
= (86 )/100 rupees
= 086/100
= 0.86 rupees
### Convert 123 paise into rupees
123 paise = 123 × 1 paise
= 123 × 1/100 rupees
= 123/100 rupees
= 1.23 rupees
Subscribe to our Youtube Channel - https://you.tube/teachoo
1. Chapter 2 Class 7 Fractions and Decimals
2. Concept wise
3. Decimal Numbers
Transcript
We know that 100 paise = 1 1 paise = 1/100 Convert 86 paise into rupees 86 paise = 86 1 paise = 86 1/100 rupees = (86 )/100 rupees = 086/100 = 0.86 rupees Convert 123 paise into rupees 123 paise = 123 1 paise = 123 1/100 rupees = 123/100 rupees = 1.23 rupees
Decimal Numbers
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https://community.appinventor.mit.edu/t/calculating-reaction-time/80954
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Calculating Reaction Time
Actually, I am making a game where random texts are appearing with random colors and the person actually has to select the answer which can be either red, yellow, blue or green colored buttons. If the button color matches with the color of the random text and not the text itself, he gets a SCORE. For each answer he gives, I need the reaction time from the onset of that random colored random text till the person presses the answer button. This has to be done for each and every random text separately and in the end , mean of all reaction times has to be displayed. Any help is this regard would be highly appreciated.
The text random appears 40 times so 40 reaction times need to be calculated and towards the end, the mean of those 40 reaction times need to be calculated. Thanks
welcome...
this seems to be a great task for some homework...
as a start it would help to learn some basics
concerning the reaction time take a look into the Clock component
http://ai2.appinventor.mit.edu/reference/components/sensors.html#Clock
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### 3.1132 $$\int \frac{(a+b x+c x^2)^2}{(b d+2 c d x)^7} \, dx$$
Optimal. Leaf size=37 $\frac{\left (a+b x+c x^2\right )^3}{3 d^7 \left (b^2-4 a c\right ) (b+2 c x)^6}$
[Out]
(a + b*x + c*x^2)^3/(3*(b^2 - 4*a*c)*d^7*(b + 2*c*x)^6)
________________________________________________________________________________________
Rubi [A] time = 0.0143671, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.042, Rules used = {682} $\frac{\left (a+b x+c x^2\right )^3}{3 d^7 \left (b^2-4 a c\right ) (b+2 c x)^6}$
Antiderivative was successfully verified.
[In]
Int[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^7,x]
[Out]
(a + b*x + c*x^2)^3/(3*(b^2 - 4*a*c)*d^7*(b + 2*c*x)^6)
Rule 682
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*
a*c, 0] && EqQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0] && NeQ[p, -1]
Rubi steps
\begin{align*} \int \frac{\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^7} \, dx &=\frac{\left (a+b x+c x^2\right )^3}{3 \left (b^2-4 a c\right ) d^7 (b+2 c x)^6}\\ \end{align*}
Mathematica [A] time = 0.0290149, size = 65, normalized size = 1.76 $-\frac{16 a^2 c^2-3 \left (b^2-4 a c\right ) (b+2 c x)^2-8 a b^2 c+b^4+3 (b+2 c x)^4}{192 c^3 d^7 (b+2 c x)^6}$
Antiderivative was successfully verified.
[In]
Integrate[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^7,x]
[Out]
-(b^4 - 8*a*b^2*c + 16*a^2*c^2 - 3*(b^2 - 4*a*c)*(b + 2*c*x)^2 + 3*(b + 2*c*x)^4)/(192*c^3*d^7*(b + 2*c*x)^6)
________________________________________________________________________________________
Maple [B] time = 0.044, size = 74, normalized size = 2. \begin{align*}{\frac{1}{{d}^{7}} \left ( -{\frac{16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4}}{192\,{c}^{3} \left ( 2\,cx+b \right ) ^{6}}}-{\frac{1}{64\,{c}^{3} \left ( 2\,cx+b \right ) ^{2}}}-{\frac{4\,ac-{b}^{2}}{64\,{c}^{3} \left ( 2\,cx+b \right ) ^{4}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^7,x)
[Out]
1/d^7*(-1/192*(16*a^2*c^2-8*a*b^2*c+b^4)/c^3/(2*c*x+b)^6-1/64/c^3/(2*c*x+b)^2-1/64*(4*a*c-b^2)/c^3/(2*c*x+b)^4
)
________________________________________________________________________________________
Maxima [B] time = 1.18392, size = 221, normalized size = 5.97 \begin{align*} -\frac{48 \, c^{4} x^{4} + 96 \, b c^{3} x^{3} + b^{4} + 4 \, a b^{2} c + 16 \, a^{2} c^{2} + 12 \,{\left (5 \, b^{2} c^{2} + 4 \, a c^{3}\right )} x^{2} + 12 \,{\left (b^{3} c + 4 \, a b c^{2}\right )} x}{192 \,{\left (64 \, c^{9} d^{7} x^{6} + 192 \, b c^{8} d^{7} x^{5} + 240 \, b^{2} c^{7} d^{7} x^{4} + 160 \, b^{3} c^{6} d^{7} x^{3} + 60 \, b^{4} c^{5} d^{7} x^{2} + 12 \, b^{5} c^{4} d^{7} x + b^{6} c^{3} d^{7}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^7,x, algorithm="maxima")
[Out]
-1/192*(48*c^4*x^4 + 96*b*c^3*x^3 + b^4 + 4*a*b^2*c + 16*a^2*c^2 + 12*(5*b^2*c^2 + 4*a*c^3)*x^2 + 12*(b^3*c +
4*a*b*c^2)*x)/(64*c^9*d^7*x^6 + 192*b*c^8*d^7*x^5 + 240*b^2*c^7*d^7*x^4 + 160*b^3*c^6*d^7*x^3 + 60*b^4*c^5*d^7
*x^2 + 12*b^5*c^4*d^7*x + b^6*c^3*d^7)
________________________________________________________________________________________
Fricas [B] time = 1.88119, size = 351, normalized size = 9.49 \begin{align*} -\frac{48 \, c^{4} x^{4} + 96 \, b c^{3} x^{3} + b^{4} + 4 \, a b^{2} c + 16 \, a^{2} c^{2} + 12 \,{\left (5 \, b^{2} c^{2} + 4 \, a c^{3}\right )} x^{2} + 12 \,{\left (b^{3} c + 4 \, a b c^{2}\right )} x}{192 \,{\left (64 \, c^{9} d^{7} x^{6} + 192 \, b c^{8} d^{7} x^{5} + 240 \, b^{2} c^{7} d^{7} x^{4} + 160 \, b^{3} c^{6} d^{7} x^{3} + 60 \, b^{4} c^{5} d^{7} x^{2} + 12 \, b^{5} c^{4} d^{7} x + b^{6} c^{3} d^{7}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^7,x, algorithm="fricas")
[Out]
-1/192*(48*c^4*x^4 + 96*b*c^3*x^3 + b^4 + 4*a*b^2*c + 16*a^2*c^2 + 12*(5*b^2*c^2 + 4*a*c^3)*x^2 + 12*(b^3*c +
4*a*b*c^2)*x)/(64*c^9*d^7*x^6 + 192*b*c^8*d^7*x^5 + 240*b^2*c^7*d^7*x^4 + 160*b^3*c^6*d^7*x^3 + 60*b^4*c^5*d^7
*x^2 + 12*b^5*c^4*d^7*x + b^6*c^3*d^7)
________________________________________________________________________________________
Sympy [B] time = 3.89161, size = 172, normalized size = 4.65 \begin{align*} - \frac{16 a^{2} c^{2} + 4 a b^{2} c + b^{4} + 96 b c^{3} x^{3} + 48 c^{4} x^{4} + x^{2} \left (48 a c^{3} + 60 b^{2} c^{2}\right ) + x \left (48 a b c^{2} + 12 b^{3} c\right )}{192 b^{6} c^{3} d^{7} + 2304 b^{5} c^{4} d^{7} x + 11520 b^{4} c^{5} d^{7} x^{2} + 30720 b^{3} c^{6} d^{7} x^{3} + 46080 b^{2} c^{7} d^{7} x^{4} + 36864 b c^{8} d^{7} x^{5} + 12288 c^{9} d^{7} x^{6}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x**2+b*x+a)**2/(2*c*d*x+b*d)**7,x)
[Out]
-(16*a**2*c**2 + 4*a*b**2*c + b**4 + 96*b*c**3*x**3 + 48*c**4*x**4 + x**2*(48*a*c**3 + 60*b**2*c**2) + x*(48*a
*b*c**2 + 12*b**3*c))/(192*b**6*c**3*d**7 + 2304*b**5*c**4*d**7*x + 11520*b**4*c**5*d**7*x**2 + 30720*b**3*c**
6*d**7*x**3 + 46080*b**2*c**7*d**7*x**4 + 36864*b*c**8*d**7*x**5 + 12288*c**9*d**7*x**6)
________________________________________________________________________________________
Giac [B] time = 1.24655, size = 117, normalized size = 3.16 \begin{align*} -\frac{48 \, c^{4} x^{4} + 96 \, b c^{3} x^{3} + 60 \, b^{2} c^{2} x^{2} + 48 \, a c^{3} x^{2} + 12 \, b^{3} c x + 48 \, a b c^{2} x + b^{4} + 4 \, a b^{2} c + 16 \, a^{2} c^{2}}{192 \,{\left (2 \, c x + b\right )}^{6} c^{3} d^{7}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^7,x, algorithm="giac")
[Out]
-1/192*(48*c^4*x^4 + 96*b*c^3*x^3 + 60*b^2*c^2*x^2 + 48*a*c^3*x^2 + 12*b^3*c*x + 48*a*b*c^2*x + b^4 + 4*a*b^2*
c + 16*a^2*c^2)/((2*c*x + b)^6*c^3*d^7)
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https://www.riddledude.com/brain-riddle/old-james/
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# Old James
James said that he was born on February 29, 1900. What birthday did he celebrate in the year 2000?
### 5 guesses to Old James
• Boesner
my older brother is born on a leap year, so he is only a few years old, but James mentioned above, is 25 years old. Leap year is every 4 years
• anant
I don’t think James could have been born on February 29, 1900 as February 1900 did not have 29 days
• Ryver
Actually, 1900 is a leap year (it’s divisible by 4). My guess is he didn’t celebrate his birthday because he probably died.
• Dude
You are correct, Anant.
James was lying, 1900 was not a leap year.
You are today’s winner.
• Dude
And for the record:
There is a leap year every year whose number is perfectly divisible by four – except for years which are both divisible by 100 and not divisible by 400.
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https://feet-to-inches.appspot.com/266-feet-to-inches.html
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Feet To Inches
# 266 ft to in266 Feet to Inches
ft
=
in
## How to convert 266 feet to inches?
266 ft * 12.0 in = 3192.0 in 1 ft
A common question is How many foot in 266 inch? And the answer is 22.1666666667 ft in 266 in. Likewise the question how many inch in 266 foot has the answer of 3192.0 in in 266 ft.
## How much are 266 feet in inches?
266 feet equal 3192.0 inches (266ft = 3192.0in). Converting 266 ft to in is easy. Simply use our calculator above, or apply the formula to change the length 266 ft to in.
## Convert 266 ft to common lengths
UnitUnit of length
Nanometer81076800000.0 nm
Micrometer81076800.0 µm
Millimeter81076.8 mm
Centimeter8107.68 cm
Inch3192.0 in
Foot266.0 ft
Yard88.6666666667 yd
Meter81.0768 m
Kilometer0.0810768 km
Mile0.0503787879 mi
Nautical mile0.0437779698 nmi
## What is 266 feet in in?
To convert 266 ft to in multiply the length in feet by 12.0. The 266 ft in in formula is [in] = 266 * 12.0. Thus, for 266 feet in inch we get 3192.0 in.
## Alternative spelling
266 ft to Inch, 266 ft in Inch, 266 Feet to Inch, 266 Feet in Inch, 266 Foot to in, 266 Foot in in, 266 Foot to Inches, 266 Foot in Inches, 266 Feet to in, 266 Feet in in, 266 Foot to Inch, 266 Foot in Inch, 266 ft to in, 266 ft in in
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Re: Unconstrained Arrays problem Rick Duley <[log in to unmask]> Fri, 5 Feb 1999 12:17:29 +0800 text/plain (157 lines) ``` I received suggestions from Ben Brosgol, Steven Deller, David Hoos Sr, Sam Mize and Richard Riehle. Thanks folks for your rapid and helpful responses to my query. 1. On the matter of declaration of Unconstrained Arrays as 'limited private' there are two schools of thought: Ben Brosgol reported: You can declare a private type with an "unknown discriminant part". The following example compiles and runs (I tried it under Aonix ObjectAda 7.1.2): --- package P is type T(<>) is private; function F(N : Integer) return T; private type T is array(Positive range <>) of Integer; end P; package body P is function F(N : Integer) return T is begin return T'(1..N => 0); end F; end P; with P; use P; procedure Test_Unconstrained is X : T := F(10); begin null; end Test_Unconstrained; and yet Steven Deller's response was: For item 1, my compiler tells me that for the following: package Trial3 is type Case_Type is (Open, Shut); type Uncon is private; procedure Dumb (X : in out Uncon); private type Uncon is array (Positive range <>) of Case_Type; end Trial3; " >>> Line 6: type Uncon is array (Positive range <>) of Case_Type" "*** array (Positive range <>) of Case_Type must be constrained [RM_95 7.3(12)]" -- oops! missed that! :( When I read 7.3(12) it states that "..., then the full_type_declaration shall define a definite subtype" (and a definite subtype requires array bounds). That makes sense to me, since if code that used this package declared an object of type Uncon, there would be no way to determine the array bounds for that object. It is possible to use access types for Uncon, such that only the package body does any actual allocations of array entities, which may satisfy the original need you had: package Trial2 is type Case_Type is (Open, Shut); type Uncon is private; private type Uncon_array is array (Positive range <>) of Case_Type; type Uncon is access Uncon_array ; end Trial2; It is also possible to defer the definition of the Uncon_array into the body of the package: package Trial2 is type Case_Type is (Open, Shut); type Uncon is private; procedure Dumb (X : in out Uncon); private type Uncon_Array; type Uncon is access Uncon_Array; end Trial2; package body Trial2 is type Uncon_Array is array (Positive range <>) of Case_Type; procedure Dumb (X : in out Uncon) is begin if X = null then X := new Uncon_Array'(1 => Open, 2 => Shut); end if; end Dumb; end Trial2; 2. Declaration of a constant: There were two main variations on solutions. Always_Guilty : constant Court_Type := (2 .. 1 => Shut); as David Hoos commented "You have to specify a value for the array elements that aren't going to be there!" Alternatively, as Sam Mize suggested: "...you can just declare it as a variable. After all, it's effectively constant -- you can't change any of the (non-existent) values! Never_Guilty: Court_Type (2..1); Richard Riehle made the point: "..., good object-oriented programming style would suggest that we abandon the use of deferred constants in our package designs. The usual form is something such as, package P is type T is private; Empty_T : constant T; -- more public stuff private full definitions of private stuff end P; "The more object-oriented way to approach this is to export a function for Empty_T in place of the deferred constant. function Empty_T return T; "This allows a variety of implementations on Empty_T, eliminates problems of dependencies when we change the values of the constants later, and improves maintainability. It also eliminates the need for the designer to fool around with syntax better left to the package implementor." It seems there _are_ ways around the matter. Once again, thanks to all! ----------------------------------------------------------------------- ______ Rick Duley / \ Edith Cowan University (____/\ ) Perth, Western Australia |___ U?(____ _\L. | \ ___ ECU: +61 (08) 9370 6619 / /"""\ /.-' | |\ | mob: 0416 365 619 ( / _/u | \___|_)_| \| \\ / / \_(___ __) | \\ / / | | | | ) _/ / ) | | _\__/.-' /___( | | I think, _/ __________/ \ | | Therefore I am... // / ( ) | | ( \__|___\ \______ /__|____| I Think! \ (___\ |______)_/ \ |\ \ \ / \ | \__ ) )___/ \ \ )/ /__( ___ | /_//___| \_________ _/ ( / OUuuu \ `----'(____________) ```
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# Search by Topic
#### Resources tagged with Combinations similar to Spell by Numbers:
Filter by: Content type:
Stage:
Challenge level:
### There are 105 results
Broad Topics > Decision Mathematics and Combinatorics > Combinations
### Six Is the Sum
##### Stage: 2 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Five Coins
##### Stage: 2 Challenge Level:
Ben has five coins in his pocket. How much money might he have?
### Two Egg Timers
##### Stage: 2 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
### Train Carriages
##### Stage: 1 and 2 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
##### Stage: 2 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### Finding Fifteen
##### Stage: 2 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Super Value Shapes
##### Stage: 2 Challenge Level:
If each of these three shapes has a value, can you find the totals of the combinations? Perhaps you can use the shapes to make the given totals?
### Polo Square
##### Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Sealed Solution
##### Stage: 2 Challenge Level:
Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?
### Magic Triangle
##### Stage: 2 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### On Target
##### Stage: 2 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Hubble, Bubble
##### Stage: 2 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
### Plants
##### Stage: 1 and 2 Challenge Level:
Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
### More and More Buckets
##### Stage: 2 Challenge Level:
In this challenge, buckets come in five different sizes. If you choose some buckets, can you investigate the different ways in which they can be filled?
### Button-up Some More
##### Stage: 2 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### Combining Cuisenaire
##### Stage: 2 Challenge Level:
Can you find all the different ways of lining up these Cuisenaire rods?
### New House
##### Stage: 2 Challenge Level:
In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
### Rod Measures
##### Stage: 2 Challenge Level:
Using 3 rods of integer lengths, none longer than 10 units and not using any rod more than once, you can measure all the lengths in whole units from 1 to 10 units. How many ways can you do this?
### Three by Three
##### Stage: 1 and 2 Challenge Level:
Arrange 3 red, 3 blue and 3 yellow counters into a three-by-three square grid, so that there is only one of each colour in every row and every column
### Calcunos
##### Stage: 2 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
### Sweets in a Box
##### Stage: 2 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### Making Cuboids
##### Stage: 2 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
### Penta Place
##### Stage: 2 Challenge Level:
Penta people, the Pentominoes, always build their houses from five square rooms. I wonder how many different Penta homes you can create?
### Halloween Investigation
##### Stage: 2 Challenge Level:
Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
### Calendar Cubes
##### Stage: 2 Challenge Level:
Make a pair of cubes that can be moved to show all the days of the month from the 1st to the 31st.
### Delia's Routes
##### Stage: 2 Challenge Level:
A little mouse called Delia lives in a hole in the bottom of a tree.....How many days will it be before Delia has to take the same route again?
### Penta Primes
##### Stage: 2 Challenge Level:
Using all ten cards from 0 to 9, rearrange them to make five prime numbers. Can you find any other ways of doing it?
### Cereal Packets
##### Stage: 2 Challenge Level:
How can you put five cereal packets together to make different shapes if you must put them face-to-face?
### Cat Food
##### Stage: 2 Challenge Level:
Sam sets up displays of cat food in his shop in triangular stacks. If Felix buys some, then how can Sam arrange the remaining cans in triangular stacks?
### Chocoholics
##### Stage: 2 Challenge Level:
George and Jim want to buy a chocolate bar. George needs 2p more and Jim need 50p more to buy it. How much is the chocolate bar?
### Penta Post
##### Stage: 2 Challenge Level:
Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g?
### Wag Worms
##### Stage: 2 Challenge Level:
When intergalactic Wag Worms are born they look just like a cube. Each year they grow another cube in any direction. Find all the shapes that five-year-old Wag Worms can be.
### 3 Rings
##### Stage: 2 Challenge Level:
If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
### Three Way Mix Up
##### Stage: 1 and 2 Challenge Level:
Jack has nine tiles. He put them together to make a square so that two tiles of the same colour were not beside each other. Can you find another way to do it?
### Map Folding
##### Stage: 2 Challenge Level:
Take a rectangle of paper and fold it in half, and half again, to make four smaller rectangles. How many different ways can you fold it up?
### A Rod and a Pole
##### Stage: 2 Challenge Level:
A lady has a steel rod and a wooden pole and she knows the length of each. How can she measure out an 8 unit piece of pole?
### Tiles on a Patio
##### Stage: 2 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
### Chocs, Mints, Jellies
##### Stage: 2 Challenge Level:
In a bowl there are 4 Chocolates, 3 Jellies and 5 Mints. Find a way to share the sweets between the three children so they each get the kind they like. Is there more than one way to do it?
### Waiting for Blast Off
##### Stage: 2 Challenge Level:
10 space travellers are waiting to board their spaceships. There are two rows of seats in the waiting room. Using the rules, where are they all sitting? Can you find all the possible ways?
### Ice Cream
##### Stage: 2 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### Throw a 100
##### Stage: 2 Challenge Level:
Can you score 100 by throwing rings on this board? Is there more than way to do it?
### Team Scream
##### Stage: 2 Challenge Level:
Seven friends went to a fun fair with lots of scary rides. They decided to pair up for rides until each friend had ridden once with each of the others. What was the total number rides?
### Octa Space
##### Stage: 2 Challenge Level:
In the planet system of Octa the planets are arranged in the shape of an octahedron. How many different routes could be taken to get from Planet A to Planet Zargon?
### Newspapers
##### Stage: 2 Challenge Level:
When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
### Street Party
##### Stage: 2 Challenge Level:
The challenge here is to find as many routes as you can for a fence to go so that this town is divided up into two halves, each with 8 blocks.
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## Monday, August 19, 2013
### VECTORS - What is a UNIT VECTOR?
By vector definition, a UNIT VECTOR is one in which the Magnitude equals 1. (To review how to calculate Magnitude, refer to thenormalgenius.blogspot.com article "Vectors: What is Magnitude?," published on 8/19/13.)
In working with vectors, it is sometimes simpler to use, or the equation calls for, a unit vector in standard position, i.e., originating from (0, 0). Transforming a given vector into a friendly unit vector is easily envisioned by relating it to common algebra and geometry.
Think of a vector as the hypotenuse of a right triangle. Draw sides parallel to the x axis and y axis.
If you start with a right triangle (5,12,13 for example) you would need to multiply the hypotenuse by its reciprocal (the multiplicative inverse - one of the properties of real numbers) in order to get it back to 1 (the multiplicative identity). Dividing each of the legs by 13 (the hypotenuse) would give you the coordinates that would create a special right triangle with hypotenuse of 1 (5/13, 12/13, 1).
Try it with any right triangle:
3,4,5 transformed to a right triangle with hypotenuse of
length 1 would be 3/5, 4/5, 1.
x, x√3, 2x gives you sides of length (x / 2x) or 1/2
and (x√3 / 2x) or √3 / 2.
These numbers sound a little familiar, don’t they? I’m thinking Trigonometry -- Sine and Cosine of 30° and 60° angles. And they coincide with the 30°- 60° - 90° right triangle that has sides equal to our example and the hypotenuse terminating on the circumference of a UNIT circle.
So when the study of vectors comes up and you are asked to determine whether a vector is a UNIT VECTOR, if the magnitude is 1, you know it is. For scalar or vector components, you will want to change a vector INTO a UNIT VECTOR: divide x and y by the magnitude. In the not-so-distant future, as part of the study of vectors in Precalculus, you'll discover equations which contain terms that look like this...
and you'll recognize the unit vectors. A discussion of components will highlight these terms in an upcoming blog article.
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For faster navigation, this Iframe is preloading the Wikiwand page for 弓形.
# 弓形
## 维基百科，自由的百科全书
半径 圆心角 ${\displaystyle r\,}$ ${\displaystyle \alpha \,}$ 高度 ${\displaystyle h=r-(r\cdot \cos \left({\frac {\alpha }{2))\right)),}$ 弦长 ${\displaystyle s=2r\cdot \sin \left({\frac {\alpha }{2))\right),}$ 弧长 ${\displaystyle b=2\pi r\cdot {\frac {\alpha }{360^{\circ ))},}$ 弓形面积 ${\displaystyle A=\pi r^{2}\cdot {\frac {\alpha }{360^{\circ ))}-{\frac {1}{2))r^{2}\sin \alpha }$
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The SAT loves to ask a particular kind of question about averages that can pretty confusing without a nice, easy way to organize your information. Enter The Average Table. KNEEL WHEN IT ENTERS THE ROOM, KNAVE! Seriously, this thing kicks ass.
To build it, just remember what you have known for a long time about averages: how to calculate them. If I gave you a set of 5 test scores and asked you to average them, what would you do? You’d add them up, and then divide the total by 5. That’s because…
$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;Average\:&space;of\:&space;Values&space;=&space;\frac{Sum\:&space;of\:&space;Values}{Number\:&space;of\:&space;Values}$
That’s just how averages work. But what if we multiplied both sides of that equation by [Number of Values]?
$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
You can use this to set up a very handy little table, which will help you solve even the hairiest looking average questions. I’m going to use colors to help you see how! Aww yiss.
Let’s illustrate with a problem that looks like it sucks:
1. A delivery truck is loaded with seven packages weighing an average of 30 pounds. At his first stop, the delivery man drops off three packages weighing a total of 60 pounds. He also picks up one package weighing 15 pounds. He makes one more stop to deliver two more packages, which weigh 42 and 48 pounds. What is the average weight, in pounds, of the packages that remain on the truck?
(A) 15
(B) 17
(C) 19
(D) 25
(E) 30
OK. So let’s set up the average table, using the colors in the problem above to show what came from where (if you’re colorblind I’m so sorry):
$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
7 packages 30 pounds -3 packages -60 pounds +1 package +15 pounds -2 packages -90 pounds
See where everything’s coming from? When we have an average (30 pounds for the initial 7 packages) we put it in the average column. When we have a sum, we put it in the sum column. We keep track of whether the packages are being delivered or picked up with + and – signs. Now let’s fill in the rest of the table just to see how everything works together (calculated values are in bold type…make sure you understand where they come from):
$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
7 packages 30 pounds 210 pounds -3 packages 20 pounds -60 pounds +1 package 15 pounds +15 pounds -2 packages 45 pounds -90 pounds
We filled everything in just for practice, but for the next step we’re only going to need the values in the outer columns. So we know we started with a total weight in the truck of 210 pounds. We dropped off 3 packages weighing 60 pounds, picked up 1 package weighing 15 pounds, and dropped off 2 more weighing 90 pounds. Using our table, we can easily see that the number of packages left on the truck is 7-3+1-2 = 3, and the total weight on the truck is 210-60+15-90 = 75. So the average weight of the 3 packages left on the truck is 25 pounds! That’s choice (D)!
$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
7 packages 30 pounds 210 pounds -3 packages 20 pounds -60 pounds +1 package 15 pounds +15 pounds -2 packages 45 pounds -90 pounds = 3 packages 25 pounds = 75 pounds
Some last notes about the average table before I give you a few more practice problems:
• You can only add or subtract up and down the outer columns. Try adding and subtracting averages and you’ll get all screwed up. You can only use the middle column for
• calculating the sum by multiplying the number by the average, or
• calculating the average by dividing the sum by the number.
• This will work with questions that have variables instead of numbers, as long as you follow the rules (but it’s a good idea to substitute real numbers to make your life easier whenever possible).
##### Try these two for practice:
You need to be registered and logged in to take this quiz. Log in or Register
## Comments (13)
The question says the SUM of the 3 numbers is g, so you don’t need to multiply g by anything.
Also, I would be remiss if I didn’t take this opportunity to say that plug-in can really clarify a question like this. Say f = 10 and g = 20. If the average of 5 numbers is 10, then the sum of them is 50. If the sum of 3 of them is 20, then the sum of the other 2 must be 30. So the average of those 2 would be 15. Which answer choice gives you 15 when f = 10 and g = 20?
Sure, although it’s maybe even easier to understand in table form, so I’ve included a screenshot of the solution in my book below. The basic idea, as in most average questions, is that you need to work with the sums. If 12 students had an average score of 74, then the sum of their scores is 12×74=888. The 4 kids who averaged 96 contributed 4×96=384 to that total.
If you want to know the average of the 8 other kids, you first find the sum of their scores. 888-384=504. Now, to get the average, you divide that by the number of kids. 504/8=63.
Jonnylagenteestamuyloca says:
WOW. The table helps tremendously. Thanks dude. I’ll be eternally thankful for it.
To find the average of the remaining two numbers, find their sum! You know that all five numbers have an average of f, so their sum must be 5f. Then you know three of the numbers have a sum of g. So the sum of the remaining two numbers must be 5f – g. Since there are two numbers, you divide that by 2 to get their average.
You might find it helpful to plug in here, as well. I talked through a solution that uses that technique in this comment.
Mary says:
Can all average problems be solved with this table? What about this one:
A baseball team averages 5 runs scored per game over a span of 6 games. The team scored one run in the first game, two runs in the second game, three runs in the third game and four runs in the fourth game. What is the average of the runs they scored in games 5 and 6?
Sure, that can be done. I’ve attached an image below. Look at it, and follow along. Each game is one game, so you keep putting 1 in the number of numbers column for each of the first four games, where the team scored 1, 2, 3, and 4 games, respectively. Then you put a 2 for number of games, and an x for their sum total of runs.
You know the number of numbers times the average equals the sum, so if the team averages 5 runs a game for 6 games, you know they score a total of 30 runs.
1 + 2 + 3 + 4 + x = 30
x = 20
So the team scored 20 runs in its last two games—that’s an average of 10 runs per game for games 5 and 6.
Mary says:
Can you explain the question on Averages in your Diagnostic Drill #1? I think it’s question 6.
In a class of 6 students, the average height is 5 feet and 8 inches. If a student joins the class and causes the average height of the class to increase by 1 inch, what is the height of the new student? (1 foot = 12 inches)
Can it be solved with the Average Table?
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Cm To M
# 265 cm to m265 Centimeters to Meters
cm
=
m
## How to convert 265 centimeters to meters?
265 cm * 0.01 m = 2.65 m 1 cm
A common question is How many centimeter in 265 meter? And the answer is 26500.0 cm in 265 m. Likewise the question how many meter in 265 centimeter has the answer of 2.65 m in 265 cm.
## How much are 265 centimeters in meters?
265 centimeters equal 2.65 meters (265cm = 2.65m). Converting 265 cm to m is easy. Simply use our calculator above, or apply the formula to change the length 265 cm to m.
## Convert 265 cm to common lengths
UnitLength
Nanometer2650000000.0 nm
Micrometer2650000.0 µm
Millimeter2650.0 mm
Centimeter265.0 cm
Inch104.330708661 in
Foot8.6942257218 ft
Yard2.8980752406 yd
Meter2.65 m
Kilometer0.00265 km
Mile0.0016466337 mi
Nautical mile0.0014308855 nmi
## What is 265 centimeters in m?
To convert 265 cm to m multiply the length in centimeters by 0.01. The 265 cm in m formula is [m] = 265 * 0.01. Thus, for 265 centimeters in meter we get 2.65 m.
## Alternative spelling
265 Centimeters to m, 265 Centimeters in m, 265 Centimeter to m, 265 Centimeter in m, 265 Centimeters to Meters, 265 Centimeters in Meters, 265 Centimeters to Meter, 265 Centimeters in Meter, 265 Centimeter to Meters, 265 Centimeter in Meters, 265 cm to m, 265 cm in m, 265 cm to Meters, 265 cm in Meters
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# Some 70 percent of consumers polled at an upscale
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Some 70 percent of consumers polled at an upscale [#permalink]
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Some 70 percent of consumers polled at an upscale supermarket said that they did not care for the taste of Apex toothpaste. Nonetheless, sales of Apex accounted for a full 75 percent of toothpaste sold at that supermarket. The conclusion that at least 75 percent of shoppers buying toothpaste did in fact enjoy the taste of Apex toothpaste relies on the assumption that ___________________.
Question
Which of the following best completes the passage?
A)most of the shoppers polled did not make any toothpaste purchases that day.
B)most shoppers who bought Apex did so because of its proven efficacy in fighting tooth decay.
C)nearly every shopper who bought Apex did so because he or she did not like its taste.
D)a large number of shoppers polled were dental professionals.
E)shoppers will not buy a certain brand of toothpaste if they do not care for its taste.
[Reveal] Spoiler: OA
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19 Jan 2011, 07:09
Is it E ?
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19 Jan 2011, 07:18
prac wrote:
Is it E ?
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### Show Tags
19 Jan 2011, 14:42
A)most of the shoppers polled did not make any toothpaste purchases that day.
most of shoppers----------we don't talk about
B)most shoppers who bought Apex did so because of its proven efficacy in fighting tooth decay.
idem
C)nearly every shopper who bought Apex did so because he or she did not like its taste.
who care nearly
D)a large number of shoppers polled were dental professionals.
we don't know
E)shoppers will not buy a certain brand of toothpaste if they do not care for its taste.
correct
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19 Jan 2011, 15:07
Between A and E, I'm leaning toward E.
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19 Jan 2011, 15:17
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Woot woot! Nearly 6 weeks after I took my gmat I am still able to get the correct answer: E
ABD are out of context, they could be true or false it won't change the outcome.
E is a double negation, that addresses the issue of the taste.
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20 Jan 2011, 16:46
E it is.
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21 Jan 2011, 07:05
E is correct.
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21 Jan 2011, 11:46
E is right b/c E is saying that if someone doesn't care for the taste of toothpaste then he wouldn't buy it (or if someone cares for the taste of the toothpaste he would buy it) - sort rearranged statement in the passage that says that 75% of the customers bought Apex for the taste (ie b/c they cared about the taste)
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22 Jan 2011, 10:35
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this is one the weirdest question i have ever come across !!
and going by the methods people have used they have taken this question for an assumption question which is not the case . This is a complete the passage question !!
ans is E coz tht's the only thing that logically follows
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09 Sep 2011, 17:55
Some 70 percent of consumers polled at an upscale supermarket said that they did not care for the taste of Apex toothpaste. Nonetheless, sales of Apex accounted for a full 75 percent of toothpaste sold at that supermarket. The conclusion that at least 75 percent of shoppers buying toothpaste did in fact enjoy the taste of Apex toothpaste relies on the assumption that ___________________.
Question
Which of the following best completes the passage?
A)most of the shoppers polled did not make any toothpaste purchases that day.
B)most shoppers who bought Apex did so because of its proven efficacy in fighting tooth decay.
C)nearly every shopper who bought Apex did so because he or she did not like its taste.
D)a large number of shoppers polled were dental professionals.
E)shoppers will not buy a certain brand of toothpaste if they do not care for its taste.
E is the only choice that at least related to the last part of the passage.
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10 Sep 2011, 08:04
E
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10 Sep 2011, 10:55
I agree with E, when looking at the question independent of the first premise given regarding poll... but i don't understand how this premise is connected to answer E. can we say this premise is no way realted and can be ignored?
if 70% of the shoppers did not care for Apex taste (means 70% don't enjoy the Apex taste), how would 75% of the total toothpaste buyers enjoyed the Apex taste. Only possible situation is that Poll wasn't conducted with all kinds of people, selected evenly for pool, who buy the toothpaste. So, A seems to be a good contender. If most of the folks who don't enjoy the taste did not buy any toothpaste that day, only then it is possible to sell Apex based on its taste.
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10 Sep 2011, 11:21
yea.. E it is... A follows closely but E wins the head-to-head i guess
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10 Sep 2011, 19:55
E it is ...... C is the opposite answer.
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10 Sep 2011, 21:42
+1 for E.
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12 Sep 2011, 08:39
How is it possible that 75% bought Apex when 70% of those polled didn't like the taste?
That means those who were polled was not an accurate representation of the cross section of ppl who bought APEX.
And those who bought Apex bought it because they liked the taste.
I guess A explains the discrepancy whereas E is an assumption that supports the conclusion. Anyway, this is a confusing question, and thus annoying!
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14 Sep 2011, 01:08
Thankfully E!
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02 Nov 2011, 11:39
+1 E
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02 Nov 2011, 12:11
E is correct.
Re: Upscale Supermarket [#permalink] 02 Nov 2011, 12:11
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So if the wind blows on a tilted turbine blade, what's the linear force perpendicular to the wind? So basically let's say we have a wind going 5m/s and a blad tilted 60 degrees from flat, what would be the force making the turbine spin.
The problem i'm having is the force increases until it hits 45 degrees then begins to decrease again. My first thought was it was just Fsin(theta), but that would continue to increase until 90 degrees. Could it just be Fsin(2theta)?
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# A car is moving on a straight road due north with a uniform speed of 50km/h when it turns left through 90* .If the speed remains unchanged after turning than the change of the velocity of car is
1
by Ark2406
2015-05-17T14:47:38+05:30
### This Is a Certified Answer
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Let the North be represented by Y axis. The when the car turns left, it moves along the -ve x axis, towards West.
Let
Change in velocity is =
Magnitude of change in velocity=
Direction of the change in velocity = South west , 45 deg. from South towards West.
please click on thanks blue button above
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## Solver for Rinaldi's "Laura and Petrarch" Model
Warren Weckesser
This web page allows you to solve the system of differential equations created by Sergio Rinaldi to model the dynamics of unrequited love between the poet Petrarch and his muse Laura ("Laura and Petrarch: An Intriguing Case of Cyclical Love Dynamics", SIAM J. Appl. Math, Vol. 58, No. 4, pp. 1205-1221, August 1998). The equations solved here are given in equations (6), (7) and (8) on page 1210 of the paper. The default parameter values are given on page 1212 of the paper. With these parameters, the solution plotted here recreates the upper plot of Figure 3 in the paper (page 1214).
Enter the following data, select the Output Options, then click on Show Solution below.
α1,2,3 = β1,2,3 = γ = δ = AL = AP =
Initial conditions:
L(0) = P(0) = Z(0) =
Duration of solution: (Maximum duration is 1000.)
Output Options
Phase Space Plots
Plot Horizontal Axis Vertical Axis L P Z L P Z L P Z L P Z L P Z L P Z
Time Series Plots
Plot Vertical Axis L P Z L P Z L P Z
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# Mechanical and Electrical Systems
## Presentation on theme: "Mechanical and Electrical Systems"— Presentation transcript:
Mechanical and Electrical Systems
SAB 2032/SAM 3012 (Sistem Mekanikal dan Elektrikal)
Mechanical and Electrical System
SAB 2032/SAM 3012 Lecturer: Dr. Khurram Kamal Dept. of Electrical Engineering (Faculty of Electrical Engineering/FKE) P07-405, Phone:
Objective: To give basic information about electrical principle, electrical machinery, distribution system, wiring and protection Mendedahkan Pelajar kepada Bekalan Elektrik, Mesin Elektrik, Sistem Pengagihan, Pendawaian dan Perlindungan Lecture hours: 14 hrs (electrical part)
Syllabus and Lecture Plan
Power Supply (AC and DC) 4 hrs 1.1 Current, Voltage, Power and their relationships 1.2 Single and Three Phase System (star and delta) 1.3 Source of Supply, Transmission and Distribution 2. Electrical Machinery (Transformer and Three Phase Induction Motor) 6 hrs 2.1 TransformerPrinciple of operation and application, Rating, Losses and Efficiency 2.2 Induction Motor Principle of operation and application,Synchronous speed, Rotor speed and sleep, Rating and starting circuits.
3. Electrical Distribution and Wiring 4 hrs
3.1 Wiring system, Types and size of cables 3.2 Protections and Grounding 3.3 Electrical Load (Estimation) 3.4 Substation, Switchboard and Distribution Board 3.5 Symbols and Single line diagram
Marking Total 50% 1. Task : 10% 2. Test : 15% 3. Final Exam : 25%
Familiarization with electricity
Electric shock
Electrical Engineer design systems objective:
To gather, store, process, transport, and present information To distribute, store, and convert energy between various form Manipulation of energy interdependent Manipulation of information
Basic Electrical System
Electricity is a form of energy Examples of energy source – hydro, coal, wind, nuclear and solar Electrical systems permits us easily to transmit energy from a source of supply to a point of application Electrical engineering is the profession concerned with systems that produce, transmit and measure electrical signals Examples of electrical systems – power system, communication system, computer system, control system and signal processing system
…Basic Electrical System
1. The source - to provide energy for the electrical system, eg. Battery, generator, socket outlet 2. The load - to absorb the electrical energy supplied by the source, eg. Lamps, air-cond 3. The transmission system - conducts energy from the source to the load, eg. Insulated wire 4. The control apparatus - permits energy to flow or interrupts the flow, eg. switch
…Basic Electrical System
Example of Electrical System
…Basic Electrical System
Example of Electrical System
1. Power Supply (AC and DC)
Electricity is the movement of free electrons in a material toward an area of positive (+) charges. The conduction of those electrons is determined by the type of material. Some conduct well, while other materials prevent the movement of electrons. Electricity can take the form of static electricity, direct current (DC) electricity, or alternating current (AC) electricity.
What are free electrons?
What determines the conduction of electricity? What are the different types of electricity?
Free electrons Most electrons are bound in orbit around atoms. But in many substances, there are electrons that are not connected to any atom and are roaming freely throughout the material. These electrons may have been knocked free in the creation of ions or may be the result of a collision of a high energy particle, such as from radioactive materials or cosmic rays. Electrons have a negative (-) electrical charge and protons have a positive (+) charge. Atoms with an excess of electrons are called negative ions and those that are missing electrons in the shells or orbits are called positive ions. An electric force field causes particles with opposite charges to attract each other. A buildup of opposite charges creates an electric potential. Release of the potential energy results in the movement of free electrons, which is called electricity.
Valence electrons are the electrons contained in the outermost, or valence, electron shell of an atom. Valence electrons are important in determining how an element reacts chemically with other elements: The fewer valence electrons an atom holds, the less stable it becomes and the more likely it is to react. Proton charge= x Coulomb Elec. charge= x Coulomb This helium (He) model displays two valence electrons located in its outermost energy level. Helium is a member of the noble gases and contains two protons, neutrons, and electrons.
Solid metals are good conductors of electricity, because electrons are allowed to move freely throughout the material. Copper and gold are some of the best conductors of electricity. Although iron is a good conductor, iron oxide (rust) is not. In the solid state, the atoms of metals are held in place and only vibrate. This allows free electrons to roam about the material.
Free electrons among metal atoms
In semiconductors—such as materials used in computer chips—the electrons have limitations to their movement, such as only being allow to move in one direction or in one plane. Nonconductors inhibit the movement of electrons within the material. But they often do allow electrons and ions to collect on their surfaces. Examples of nonconductors or electrical insulators are: Plastic, Rubber, Glass, Most metal oxides (like rust), Air, Oil, Pure, de-ionized water Gases are not good conductors of electricity because of the distances between atoms. Electrons have difficulty moving through gases, unless the gas is ionized or heated to higher temperatures.
Conductivity (Ohm. m) Silver 1.59×10−8 Copper 1.68×10−8 Gold 2.44×10−8 Aluminium 2.82×10−8 Tungsten 5.60×10−8 Nickel 6.99×10−8 Brass 0.8×10−7 Iron 1.0×10−7 Tin 1.09×10−7 Platinum 1.1×10−7 Lead 2.2×10−7 Manganin 4.82×10−7 Constantan 4.9×10−7 Mercury 9.8×10−7
Nichrome 1.10×10−6 Carbon 3.5×10−5 Germanium 4.6×10−1 Silicon 6.40×102 Glass 1010 to 1014 Hard rubber approx. 1013 Sulfur 1015 Paraffin 1017 Quartz (fused) 7.5×1017 PET 1020 Teflon 1022 to 1024
Insulator Materials Material Dielectric constant Air (vacuum) Teflon Paper Oil Mica Glass Ceramic
Types of electricity Common types of electricity are static electricity, direct current (DC) electricity, and alternating current (AC) electricity. Static electricity Static electricity is the collection of free electrons on the surface of a material, giving it a negative (-) charge. Atoms on the surface of another material that have lost one more of their electrons are called positive (+) ions. Often the electrons are pulled from the atoms on one surface and allowed to collect on the surface of another material. Static electricity is caused by rubbing the two different materials together. Since opposite charges attract, there is a tendency for the electrons to attract toward the positive ions, resulting in static electricity.
DC and AC electricity In a metal or other conducting material, electrons will flow from an area of an excess negative (-) charges to an area of positive (+) charges. This flow of electrons through the conductor is electricity. If the opposite charges are constant, such as with the terminals in a battery, the current is called direct current or DC electricity, because it is going one direction. If the terminals constantly switch their polarity from (+) to (-) and back again, the direction of the electrons alternates and is call alternating current or AC electricity. This is the type of electricity that comes from the outlets in most homes. DC can be created by a battery or DC generator. AC requires an AC generator for its creation. DC is used in many devices that do not require high voltages for their operation, such that batteries are used for power. AC can be used in higher voltages. It has the advantage of being able to have its voltage easily changed to a higher or lower level. AC is required for many electronic devices.
Conclusion Electricity is the movement of free electrons in a material. It moves the best though metals. Static electricity is the collection of electrons and positive ions on the surface of a material--usually a non-conductor. Direct current electricity moves in one direction and usually is created in batteries. Alternating current electricity is most commonly used in homes and can have its voltage changed to suit the need.
Electric Current Electric current is the rate of charge flow past a given point in an electric circuit, measured in coulombs/second which is named amperes. In most DC electric circuits, it can be assumed that the resistance to current flow is a constant so that the current in the circuit is related to voltage and resistance by Ohm's law.
Voltage Voltage is the electrical potential energy and is measured in volts. A good analogy is to think of a water hose. There is water pressure or potential energy on the other side of the faucet or outlet valve. Once you open the faucet, the pressure causes the water to rush through the hose. The unit symbol for volts is V, as in 110V. Current Current indicates the amount of electrons passing through the wire and is measured in amperes or amps for short. For some reason, they use I to indicate current instead of a different letter. The unit symbol for amps is A, as in 2.0A. Electrical current is similar to the rate of water flowing through a hose.
Resistance Electrical resistance can be thought of as the "friction" on the movement of electrons in a wire. Resistance is measured in ohms, and the unit symbol for it is the Greek letter omega, Ω. Thus 3 ohms is often written as 3 Ω. Most devices in an electrical circuit can be considered resistors, including light bulbs and electric motors. Even the wire itself provides some resistance. Just as you get some heat from friction, electrical resistance also results in heat. That is why the light bulb filament gets hot and glows. Following the water hose analogy, resistance is similar to the friction inside the hose. But also, the resistance increases with a narrower hose, just like a thin copper wire has more electrical resistance than a thick wire.
Electrical Quantities & Units
Electric Charge, Q Energy exists at proton and electron Unit - Coulomb (C) 1 C - electrical quantity when 1 Ampere current flows for 1 second in a conductor Current, I Rate of charge flows I = Q / t Ampere (A) 1 A = transfer of 1 C charge in 1 s DC and AC
… Electrical Quantities & Units
Energy, W Capacity for doing work Unit - Joule (J) Voltage, E Potential between 2 points in a circuit Energy needed to transfer 1 unit charge V = W/Q Volt (V) 1V = Energy needed to transfer 1 C charge through an element
… Electrical Quantities & Units
Power, P Rate for doing work P = W / t Watt (W) 1 W – Power used when 1 A current flowing through a potential of 1 V P = VI = (W/Q)(Q/t) Resistance, R All conductors have their own resistance To limit the flow of current in a circuit Unit – Ohm () – Element with resistance of 1 will allow 1 A to pass through if 1 V voltage is applied across the element R = 0 Ω – short circuit (large current flow) R = - open circuit (no current flow)
… Electrical Quantities & Units
No Quantity Symbol Unit Formula 1 Charge Q Coulomb, C I x t 2 Current I Ampere, A Q/t 3 Voltage V Volt, V W/Q 4 Energy W Joule, J P x t 5 Power P Watt, W W/t 6 Resistance R Ohm, Ω V/I 7 Capacitance C Farad, F Q/V 8 Inductance L Henry, H Φ/I 9 Frequency F Hertz, Hz 1/t 10 Impedance Z 11 Admittance X
Ohm's Law for Electrical Circuits
Ohm's Law states that in a simple electrical circuit, the voltage equals the electrical current times the resistance. (The current flowing in a circuit is directly proportional to the voltage applied and inversely proportional to the resistance at a constant temperature). where: V is the voltage in volts I is the current in amperes or amps R is the resistance in ohms IR is I times R V = IR
…Basic Electrical Laws – Ohm’s Law
V=IR V I R
Example: How many amperes of current are in the circuit below? R 100 V
Vs 20 Ω I = Vs R = Using Ohm’s law: 100 V 20 Ω 5 A
…Basic Electrical Laws – Ohm’s Law
Examples: 1. An electric bulb uses 0.5 A of current with voltage generated being 120 V. Determine the value of resistance. 2. If a current of 0.5 A flows through resistor of 15 Ω, calculate the voltage drop across the resistor. 1. Ans; R = V/I = 120/0.5 = 240 Ω 2. Ans; V = IR = 0.5 x 15 = 7.5 V
…Basic Electrical Laws – Ohm’s Law
More Examples: 3. (i) For the circuit shown, determine current flowing and power absorbed by the resistor if the resistance is 1 kΩ and voltage across it is 10 V (ii) If the current flowing through the circuit is 3A and power absorbed is 72 W, determine the resistor value and voltage across it.
Voltage, Current and Resistance
Direct current or DC electricity is the continuous movement of electrons from negative to positive through a conducting material such as a metal wire. A DC circuit is necessary to allow the current or steam of electrons to flow. In a circuit, the direction of the current is opposite the flow of electrons. DC electricity in a circuit consists of voltage, current and resistance. The flow of DC electricity is similar to the flow of water through a hose. Batteries and DC generators are the sources to create DC electricity.
Power (watts) = Voltage (volts) x Current (amperes)
DC Power The electric power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. For a resistor in a D C Circuit the power is given by the product of applied voltage and the electric current : P = VI Power (watts) = Voltage (volts) x Current (amperes) Power consumed = kilowatt hours (kWh) x charge (dollars, RM, etc)
Calculating your electric bill
Computer : 500 watts (12 hours) TV : 400 watts (5 hours) Lighting : 3 x 15 watt = 45 watts (6 hours) Others : 100 watts (2 hours) P = VI Power consumed : (500 x 12) + (400 x 5) + (45 x 6) + (100 x 2) = 8470 wh = 8.47kWh Electric bill per month : 8.47 kWh x RM 0.218/kWh x 30 = RM 55.50
DC Circuit Water Analogy
Voltage-Pressure Analogy
A battery is analogous to a pump in a water circuit. A pump takes in water at low pressure and does work on it, ejecting it at high pressure. A battery takes in charge at low voltage, does work on it and ejects it at high voltage.
Current-Flow rate Analogy
Connecting a battery to an appliance through a wire is like using a large pipe for water flow. Very little voltage drop occurs along the wire because of its small resistance. You can operate most appliances at the end of an extension cord without noticeable effects on performance.
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Econ 57 Spring 1997 Final Exam
1. Explain why the following visual display of household income data in Sierra Madre, California, is misleading and then redraw the figure correctly.
2. There are 12 astrological signs (Aquarius, Scorpio, and so on.) If a person's birthday is equally likely to be in each of these 12 signs, what is the probability that in a group of 4 persons, at least 2 will have the same sign?
3. Each semester, Professor Smith gives students a copy of the final exam from the semester before to help them study for their final. Once, as he passed out the previous semester's exam, he warned, "There is inevitably variation in the difficulty of tests. This exam was one standard deviation below the mean." Assuming normality, how would you interpret this remark?
Say 10 tickets are numbered 1 through 10 in a drawing. Half the numbers are even and half are odd. The first ticket is drawn, and it's No. 3 which is odd. That leaves five even numbers and four odd ones. Doesn't this mean that the next ticket to be drawn is more likely to be even? If I buy a ticket at this point, wouldn't I have a better chance of winning the next draw by choosing an even number?
5. Use regression toward the mean to explain why a movie sequel is usually not as good as the original.
6. Suppose that the damage award favored by individual potential jurors in a particular personal injury case can be described by a normal distribution with a mean of \$5,000,000 and a standard deviation of \$2,000,000. (This probability distribution is across randomly selected jurors.) What percentage of potential jurors favor an award of more than \$5,500,000? Of less than \$4,500,000?
7. Continuing with the preceding exercise, assume that a jury is a random sample from this distribution and that the jury's damage award is the average of the awards favored by those who serve on the jury. Carefully explain the differences in rewards that can be anticipated with a 6-person jury system versus a 12-person system.
8. Without doing any calculations, explain why you believe the two-sided P value for a test of H0: m1 = m2 to be either larger than 0.05 or smaller than 0.05 for these data:
Sample 1 Sample 2 4.0 4.8 3.4 3.5 4.1 7.2 6.7 5.8 6.5 6.7 4.6 4.0 4.4 3.9 4.3 6.7 5.9 5.8 5.8 6.0 4.8 3.4 4.6 4.0 4.4 5.8 6.0 6.7 5.8 6.2 4.8 3.4 4.0 3.5 4.1 6.5 6.7 7.2 6.7 5.8 3.9 4.3 4.6 4.0 4.4 6.0 5.8 6.7 5.9 5.8 3.4 4.6 4.8 4.0 4.4 5.8 6.0 6.2 5.8 6.7 3.4 4.0 4.8 3.5 4.1 7.2 6.5 6.7 6.7 5.8 4.0 4.4 3.9 4.3 4.6
9. Use these data on the number of households (in millions) of different sizes to draw two histograms, one for 1890 and one for 1990. (Calculate the relative frequency for the 7-or-more category, but do not include this category in your histogram.) What important differences do you detect in these two histograms?
Size 1 2 3 4 5 6 7 or more 1890 0.457 1.675 2.119 2.132 1.916 1.472 2.919 1990 23 30.2 16.1 14.6 6.2 2.2 1.5
10. The preceding exercise shows the number of households of different sizes in 1990.
a. Looking at the 93.8 million households, ordered by size, what is the size of the median household?
b. Looking at all 264 million individuals, ordered by the size of the household they live in, what is the size of the household that the median individual lives in?
c. Pomona College reports its average class size as 14 students per class. However, a survey that asked Pomona College students to give the size of the classes they were enrolled in found that the average class size was 38 students and that 70 percent of the respondents' classes had more than 14 students. Use the insights gained in the first two parts of this exercise to explain this disparity.
11. In the United States, 36 of 50 states have a death penalty. To examine the relationship between the death penalty and murder rates, the following equation was estimated by least squares using 1987 data:
where y = murder rate (murders per 100,000 persons); x = 1 if state has a death penalty, 0 if it doesn't; and the standard errors are in parentheses.
a. What does R-squared mean and how is it measured?
b. Is the relationship statistically significant at the 5 percent level?
c. What is the average value of the murder rate?
d. The researcher added the District of Columbia to his data and obtained these results:
The murder rate in D.C. is 36.2; the state with the highest murder rate is Michigan, with 12.2 murders per 100,000 persons. Explain how a comparison of the two estimated equations persuades you that D.C. either does or does not have a death penalty.
e. Explain the error in this interpretation of the results including D.C.: “Those of us who oppose the death penalty can breathe a sigh of relief, now armed with statistically significant evidence that the death penalty is only cruel and not effective.”
12. In a 1982 racial-discrimination lawsuit, the court accepted the defendant's argument that racial differences in hiring and promotion should be separated into eight job categories. In hiring, it turned out that blacks were underrepresented by statistically significant amounts (at the 5 percent level) in four of the eight job categories. In the other four categories, whites were underrepresented in two cases and blacks were underrepresented in two cases, though the differences were not statistically significant at the 5 percent level. The court concluded that four of eight categories was not sufficient to establish a prima facie case of racial discrimination. Assume that the data from these eight job categories are independent random samples.
a. What is the null hypothesis?
b. If the null hypothesis is true and the court requires statistical significance at the 5 percent level in all eight job categories, what is the probability that the null hypothesis will be rejected?
c. If the null hypothesis is true and the court requires statistical significance at the 5 percent level in at least one of the eight job categories, what is the probability that the null hypothesis will be rejected?
13. Continuing with the preceding exercise:
a. Explain why data that are divided into eight job categories might not show statistical significance in any of these job categories, even though there is a statistically significant relationship when the data are not disaggregated.
b. Explain why data that are divided into eight job categories might show statistical significance in each of the eight categories, even though there is not a statistically significant relationship when the data are not disaggregated.
14. A researcher used data on these three variables for 60 universities randomly selected from the 229 national universities in U.S. News & World Report's 1995 rankings of U.S. colleges and universities: y = graduation rate, mean 59.65; x1 = student body's median SAT, mean 1,030.5; and x2 = percent of student body with GPAs among the top 10 percent at their high school, mean = 46.6. Here are the results:
Interpret these results. Be sure to explain why the estimated coefficients of x1 and x2 are lower in the third equation than in the first two.
15. Explain the error in this critique of a statistics paper in which the z-value turned out to be 2.8: "Even though the z-value was larger than 2, the sample size was only 24. The results could be invalidated by sampling error due to the small sample."
16. If six standard six-sided dice are rolled simultaneously, what is the probability that all six numbers will appear, not necessarily in order?
17. Here are the predicted and actual daily high and low temperatures at the Los Angeles Civic Center for every fourth day during the period from November 7, 1996 to February 15, 1997:
Sample 1 Sample 2 Daily Low Daily High Predicted Actual Error Predicted Actual Error 11/7/96 50 51 -1 81 73 8 11/11/96 60 63 -3 87 91 -4 11/15/96 58 56 2 73 62 11 11/19/96 56 54 2 71 71 0 11/23/96 54 59 -5 67 67 0 11/27/96 58 54 4 83 80 3 12/1/96 46 46 0 67 66 1 12/5/96 51 50 1 70 59 11 12/9/96 55 53 2 67 57 10 12/13/96 52 56 -4 72 68 4 12/18/96 48 50 -2 71 77 -6 12/21/96 47 50 -3 68 59 9 12/25/96 52 50 2 70 75 -5 12/29/96 52 53 -1 62 66 -4 1/2/97 56 60 -4 67 64 3 1/6/97 47 52 -5 60 62 -2 1/10/97 48 47 1 70 62 8 1/14/97 48 47 1 56 57 -1 1/18/97 47 51 -4 74 79 -5 1/22/97 49 51 -2 62 56 6 1/26/97 55 54 1 60 58 2 1/30/97 58 56 2 83 78 5 2/3/97 49 55 -6 67 66 1 2/7/97 51 49 2 66 70 -4 2/11/97 51 50 1 64 62 2 2/15/97 50 49 1 75 77 -2
Test the null hypothesis that the probability that the daily low temperature will be at least 50 degrees Fahrenheit is the same in each of these months
18. Use two box plots to display the data in Exercise 17 on the actual low and high temperatures.
19. Use the data in Exercise 17 to determine a 95 percent confidence interval for the population mean error predicting the daily low temperature and to test at the 5 percent level the null hypothesis that the population mean error predicting the daily low temperature is zero.
20. In March of 1992, The Wall Street Journal reported that, "Foreign stocks and foreign-stock mutual funds have been miserable performers since early 1989, which suggests a rebound is long overdue." The article went on to quote the chief investment officer of the G.T. Global mutual funds group: "The U.S. stock market has outperformed the EAFE index [of European, Australian, and Far East stocks] for three consecutive years. If the U.S. were to outperform the EAFE index in 1992, it would be the first [such] four-year stretch in the 20 years that the EAFE index has been compiled. If history is any guide, it's about time that foreign markets started doing better." Explain the error in their statistical reasoning.
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Law of Conservation of Energy
Law of conservation of energy states that energy can neither be created nor destroyed, but can be transferred from one form to another. The total energy before and after the transformation remains constant.
For example: consider a ball falling freely from a height.
Potential energy + Kinetic energy = Constant (Mechanical energy)
A body of mass ‘m’ is raised to height ‘h’ at A its potential energy is maximum and kinetic energy is 0 as it is stationary.
When body falls at B, h is decreasing hence potential energy decreases and V is increasing hence kinetic energy is increasing.
When the body is about to reach the ground level, h = 0, v will be maximum hence kinetic energy –> potential energy
Decrease in potential energy = Increase in kinetic energy
This shows the continual transformation of gravitational potential energy into kinetic energy.
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Layout: Blue and Brown (Default) Author's Creation
Home > the bills and math..
the bills and math..
January 23rd, 2008 at 05:54 pm
I was cold yesterday, so I thought I would walk out to the mailbox..to remind myself it could be worse, I could be without a house!
In the mail was a gas bill, perfect timing (gas heat) I figured I could look at the bill suffer some sticker shock at the price and be thankful for a warm sweater.
Instead I found a bill of 67.66 Now no it isn't cheap, but compared to previous years of 80 dollar gas bills in January, it was nice. I peeked at the graph to see how we compared 'thermwise' to last year. 55 last year and only 42 this year! Wow I thought pretty nifty. The average temp (also info on the bill) was 50 degrees last year and 47 this year, so life was colder and we still spent less!
Then I caught the 'days' entry, last January was a bill for 39 days, this year only 31 (I have no idea why they bill such odd cycles)
Phooey, now I have to do the math to see if we are really spending less. Errr let me go get my math brain son....
Last year... 55therms/39 days = 1.41th/dy
This year... 42 therms/31 days = 1.35th/dy
So a smidgen less this year than last. Nothing to write home about really.
7 Responses to “the bills and math..”
1. littlemama Says:
1201120243
At least it hasn't gone up.
2. M E Says:
1201138754
I know you live in a *warmer* climate, but IMO, \$67.66 for a gas bill in the dead of winter is C H E A P! My bill is \$250. Wanna trade???
I didn't think so! :-)
3. Joan.of.the.Arch Says:
1201145515
My bill- \$161.
The price of natural gas is lower this year than last, but we had 34% more degree heating days this year than last. Nonetheless our consumption rose by only 6.2%. We conserved better than last year.
4. nanamom Says:
1201182301
my bill was 197, and I was excited about that! I remember 397 a couple winters ago. This is the coldest winter yet.
5. princessperky Says:
1201191851
Ok yeah, I do live in NC so no I don't want to trade
6. HelpMeFriend Says:
1201227658
Don't they offer budgets or assistance with problems in the winter with not being able to pay? Here, they won't turn you off if you have elderly or children in the home. They try to help with assistance, but usually they just build it up and after the frost, shut you off. My bill began very high, but after one year here, the company offered a budget ammount of only \$84.00. Half what I was paying before was a much better deal.
7. monkeymama Says:
1201372100
Reminds me my bill always shows how many therms per day compared to last month and last year. I guess we are lucky, we don't have to do all that math!
Our bill is still averaging \$30/month (I saw you don't get the same budget billing plan - bummer). We may have a \$100 bill or 2 in the winter but the rest of the year it is \$10 or something. I am enjoying the \$30 plan. Our weather is milder though. So don't kill me. {A lot of it is the energy efficiency of our house. Plenty of people out here getting \$300-\$500 bills in the winter, regardless. They just charge more per therm here since it is milder}.
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http://asvabpracticetest.org/good-asvab-composite-scores-asvab-marines-asvabpracticetestorg.html
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25. Use the following passage to answer questions 25 and 26. Sarah is about to fly from New York to Los Angeles. She packs the night before and checks her list three times. Anxiety keeps her from sleeping well. The next morning, she accidentally puts salt in her coffee instead of sugar. She forgets her purse and has to go back to the house to get it. Finally, she arrives at the airport just in time to catch her flight. The flight is uneventful, and soon Sarah is landing at LAX. Her sister is waiting for her. What was the main idea of this passage?
These tests, again the high quality ones, are formatted like the real thing so you can get used to the question and answer formats and the time limits so nothing will be a surprise on test day. You’ll know what to expect and you’ll be used to going from different concept to different concept as is often required on the ASVAB. For example, on the math section you may have a problem using one popular math principle followed by another problem that relies on a completely different principle. This is common on a broad test like the ASVAB and preparing your mind to make these leaps can allow you to answer more questions in less time and boost your score.
The Electronics Information section of the practice test gauges your knowledge of electrical equipment and parts, including circuits, currents, batteries, and resistors. An example may be, “Because solid state diodes have no filament, they: don’t work, are less efficient than tubes, require less operating power, or require more operating power?” The CAT-ASVAB has 16 questions in 8 minutes; the paper-and-pencil version has 20 questions in 9 minutes.
#### The Mechanical Comprehension section of the ASVAB practice test measures your understanding of basic mechanical principles and mechanisms. You may be asked why an intake valve on a pump opens when the piston goes down, or what direction friction is going when shown a diagram of a skier. The CAT-ASVAB has 16 questions in 20 minutes; the paper-and-pencil version has 25 questions in 19 minutes.
##### As of 2017, the U.S. spends about US\$610 billion annually to fund its military forces and Overseas Contingency Operations.[6] Put together, the U.S. constitutes roughly 40 percent of the world's military expenditures. The U.S. Armed Forces has significant capabilities in both defense and power projection due to its large budget, resulting in advanced and powerful technologies which enables a widespread deployment of the force around the world, including around 800 military bases outside the United States.[12] The U.S. Air Force is the world's largest air force, the U.S. Navy is the world's largest navy by tonnage, and the U.S. Navy and the U.S. Marine Corps combined are the world's second largest air arm. In terms of size, the U.S. Coast Guard is the world's 12th largest naval force.[13] [14][15]
The Armed Services Vocational Aptitude Battery (ASVAB) is a comprehensive test that helps determine students’ eligibility and suitability for careers in the military. Students who score in the 31st percentile are eligible for a wide variety of careers (combined with standards such as physical condition and personal conduct). Students and schools are also encouraged to take advantage of a free career exploration program that links personal interests with demonstrated aptitudes from the 9 ASVAB subtests, half of which relate to vocational skills, in addition to math and verbal skills.
Sergeant Leigh Ann Hester became the first woman to receive the Silver Star, the third-highest U.S. decoration for valor, for direct participation in combat. In Afghanistan, Monica Lin Brown was presented the Silver Star for shielding wounded soldiers with her body.[45] In March 2012, the U.S. military had two women, Ann E. Dunwoody and Janet C. Wolfenbarger, with the rank of four-star general.[46][47] In 2016, Air Force General Lori Robinson became the first female officer to command a major Unified Combatant Command (USNORTHCOM) in the history of the United States Armed Forces.[48]
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https://oeis.org/A122601
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A122601 a(n)=(n-th prime +1) modulo 7. 1
3, 4, 6, 1, 5, 0, 4, 6, 3, 2, 4, 3, 0, 2, 6, 5, 4, 6, 5, 2, 4, 3, 0, 6, 0, 4, 6, 3, 5, 2, 2, 6, 5, 0, 3, 5, 4, 3, 0, 6, 5, 0, 3, 5, 2, 4, 2, 0, 4, 6, 3, 2, 4, 0, 6, 5, 4, 6, 5, 2, 4, 0, 0, 4, 6, 3, 3, 2, 5, 0, 4, 3, 4, 3, 2, 6, 5, 6, 3, 4, 0, 2, 5, 0, 6, 3, 2, 3, 0, 2, 6, 4, 5, 2, 3, 0, 6, 4, 6, 3, 2, 5, 4, 3, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(n)=1 only for n=4; frequences f(m) of other values are almost equal, e.g., for n=1..1000, f(m=0..6): 166,1,165,166,168,164,170. LINKS FORMULA a(n)= (A039705(n)+1) mod 7 MATHEMATICA Table[Mod[(Prime[n]+1), 7], {n, 1000}] CROSSREFS Cf. A039705. Sequence in context: A021293 A275979 A246859 * A125776 A066977 A007551 Adjacent sequences: A122598 A122599 A122600 * A122602 A122603 A122604 KEYWORD nonn AUTHOR Zak Seidov, Sep 24 2006 STATUS approved
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Last modified May 5 18:16 EDT 2021. Contains 343572 sequences. (Running on oeis4.)
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https://discusstest.codechef.com/t/wa-maximize-it/21565
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# WA :Maximize It
My Solution : https://www.codechef.com/viewsolution/21692278
It does not work for the following simple test case:
``````1
5 2
1 1 3 3 8
``````
Your code generates 3, while it should be 4.
The code has multitude of problems, starting with
`````` int n,k;
char numbers[n+1];
int sum=0,sum1=0,prod=1,fprod=1,res=0,p;
cin>>n>>k;
for(int i=0;i<n;i++){
int num=1;
cin>>p;
for(int j=0;j<p;j++) {
num=num*k;
}
numbers[i]=num;
}
``````
1. You define array numbers[n+1] at the moment when n is not properly initialized yet.
2. You store values of powers of p in char which is limited by 127.
What’s the most interesting is that your code is still able to solve some of the test cases.
@oleg_b Thank You ! Even ‘long long int’ won’t work for storing the values of the (integer)^(power) , Please correct me if I am wrong ?
Yes, ‘long long int’ does not have sufficient capacity. The approach should be quite different - you need to represent the numbers in the base of k. For example, if k is 10, then 10^100 is a vector with 101 elements, all of them are zeros, except the last one which is 1.
//
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https://gmatclub.com/forum/a-challenge-151404-40.html?sort_by_oldest=true
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# A challenge. Edit. : Completed. Debrief 760 (Q50 V 44)
Author Message
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Re: A challenge. Edit. : Completed, almost. 760 (Q50 V 44) [#permalink]
### Show Tags
31 Jul 2013, 09:37
Great work dude...congrats
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Re: A challenge. Edit. : Completed, almost. 760 (Q50 V 44) [#permalink]
### Show Tags
01 Aug 2013, 23:16
Great work dude...congrats
Sure, thanks
doesnt let me submit this reply for some reason.
Re: A challenge. Edit. : Completed, almost. 760 (Q50 V 44) [#permalink] 01 Aug 2013, 23:16
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# Khan Academy Addition Andction Word Problems With Zeroscting Integers Using Counters Fractions Practice
By Lea Burger at November 09 2018 16:12:33
Who can help me reach this goal? This is a very important question, and your answer is also very important. An unachieved goal usually means we lack the self discipline to get there alone. So we need to lean on the discipline and accountability of another person. In some cases they might be partners who are moving toward a similar goal; in other cases they are mentors who are leading us and coaching us to go where they have already gone. Either way, this person is often the difference between success and failure in goal setting. What are all the steps I need to take to reach this goal? I like to simply write these things out as they come to mind, with no real regard for order or priority. Just get every logical step down so you can see exactly what is required. This is another reality check stage, but it can also be quite encouraging since your large goal has been reduced to bite-sized chunks!
At the grassroots level, teachers in schools are given a packed curriculum for the year. Schools try to teach the students a number of procedures without delving much into its finer details. Hence, the student is left in a confounding position as to when a particular procedure must be used. The key ingredient to understanding math is constant practice and math assignment help. Unfortunately, this is not a common scenario among the popular math classes.Connect The Letter To The Correct Sound/Word: These are activities where you draw a line between a letter and the picture items that start with that letter. For instance, you'd draw a line from the letter A to the word "Apple" and the letter L to the word "Lemon". This activity is good, but takes a lot of monitoring to make sure that students are correctly connecting the letters. It's best as a homework activity, where parents can help to make sure their children are correctly connecting the letters to the words.
## Gallery of Khan Academy Subtraction
In math bingo, each student is given a bingo card (also known as a "bingo worksheet" or "bingo board") printed with numbers. These aren't necessarily the standard bingo numbers, but rather are the answers to a number of different math problems. The game is then played exactly like a normal game of bingo, with the teacher playing the part of the bingo caller, but instead of the teacher calling out the numbers printed on the cards, the teacher instead calls out math problems (the teacher may also write the problem on the blackboard). The students' task is to solve each problem, and then look for the number on their bingo card. If you are looking for an article that describes the basics of Excel and introduces the interface and concepts for beginners, you have come to the right place. Microsoft Excel is a powerful business application that is organized into a structural hierarchy of Workbooks, Worksheets, and Cells.
The present generation seems to be blessed immensely with intellect and the benefits of mastering math are something worth considering. It is a well-known fact that math is not a subject that one learns by simply reading the problems and its solutions. There are a lot of websites offering free phonics worksheets, but when selecting good kindergarten phonics worksheet it's important to keep in mind the fact that the group your teaching is extremely young. What that means is, don't try and throw anything too tricky into the worksheet. I'll look at some of the different kinds of worksheets that are available so that you can get some ideas for designing your own worksheets and using worksheets that are actually relevant to kindergarten students.
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html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
ArcSech
http://functions.wolfram.com/01.30.06.0058.01
Input Form
ArcSech[z] \[Proportional] Log[2/z] - (z^2/4) (1 + (3 z^2)/8 + (5 z^4)/24 + O[z^6])
Standard Form
Cell[BoxData[RowBox[List[RowBox[List["ArcSech", "[", "z", "]"]], "\[Proportional]", RowBox[List[RowBox[List["Log", "[", FractionBox["2", "z"], "]"]], "-", RowBox[List[FractionBox[SuperscriptBox["z", "2"], "4"], RowBox[List["(", RowBox[List["1", "+", FractionBox[RowBox[List["3", SuperscriptBox["z", "2"]]], "8"], "+", FractionBox[RowBox[List["5", " ", SuperscriptBox["z", "4"]]], "24"], "+", RowBox[List["O", "[", SuperscriptBox["z", "6"], "]"]]]], ")"]]]]]]]]]]
MathML Form
sech - 1 ( z ) log ( 2 z ) - z 2 4 ( 1 + 3 z 2 8 + 5 z 4 24 + O ( z 6 ) ) Proportional z 2 z -1 -1 z 2 4 -1 1 3 z 2 8 -1 5 z 4 24 -1 O z 6 [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["ArcSech", "[", "z_", "]"]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List["Log", "[", FractionBox["2", "z"], "]"]], "-", RowBox[List[FractionBox["1", "4"], " ", SuperscriptBox["z", "2"], " ", RowBox[List["(", RowBox[List["1", "+", FractionBox[RowBox[List["3", " ", SuperscriptBox["z", "2"]]], "8"], "+", FractionBox[RowBox[List["5", " ", SuperscriptBox["z", "4"]]], "24"], "+", SuperscriptBox[RowBox[List["O", "[", "z", "]"]], "6"]]], ")"]]]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2007-05-02
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https://www.coursehero.com/file/ptf79g/We-may-overestimate-our-contributions-because-our-own-contributions-are-more/
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psychexam4.docx
# We may overestimate our contributions because our own
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combined, the answer ended up being more than 100%. We may overestimate our contributions, because our own contributions are more readily available to us You can be very unaware of the chores done by the other o Problem 4 – Class evaluation study:
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Representative heuristics: Judging something by the extent to which it resembles a typical case. (Ex. Being hired not because of your resume, but because the boss glanced at you and thought you resembled other coworkers. The boss relied on resemblance and not probability of how well you’d do the work if hired.) o Problem 1 – Birth order problem: 72 families of six were surveyed, where the exact order was: G B G B B G Participants were asked what’s the estimate of the number of families surveyed where the order was: B G B B B B Statistically, both birth orders are equally likely but since the ratio is different, the second order is judged to be less likely The ratio does not represent the population o Problem 2 – Gambler’s fallacy & hot hand fallacy: The gambler’s fallacy is the idea that during a losing streak, it’s likely that the gamblers luck will turn around and they’ll start winning The hot hand fallacy is the idea that when on a winning streak, the streak is more likely to continue (based on the idea that having already won a number improves the probability of winning the next) o Problem 3 – Generalization from a single case to the population/prison guard study: This is an error that will make you believe the entire category will have the same properties as one individual Participants were shown where a prison guard discussed his job; in
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A136665 Triangle of coefficients of Hermite-like analog of A053120 Chebyshev's T(n, x) polynomials (powers of x in increasing order): p(x,n)=2*x*p(x,n-1)-n*p(x,n-2). 0
1, 0, 1, -2, 0, 2, 0, -7, 0, 4, 8, 0, -22, 0, 8, 0, 51, 0, -64, 0, 16, -48, 0, 234, 0, -176, 0, 32, 0, -453, 0, 916, 0, -464, 0, 64, 384, 0, -2778, 0, 3240, 0, -1184, 0, 128, 0, 4845, 0, -13800, 0, 10656, 0, -2944, 0, 256, -3840, 0, 37470, 0, -60000, 0, 33152, 0, -7168, 0, 512 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,4 COMMENTS Row sums: {1, 1, 0, -3, -6, 3, 42, 63, -210, -987, 126} LINKS FORMULA p(x,n)=2*x*p(x,n-1)-n*p(x,n-2). EXAMPLE {1}, {0, 1}, {-2, 0, 2}, {0, -7, 0, 4}, {8, 0, -22, 0, 8}, {0, 51, 0, -64, 0, 16}, {-48, 0, 234, 0, -176, 0, 32}, {0, -453, 0, 916, 0, -464,0, 64}, {384, 0, -2778, 0, 3240, 0, -1184, 0, 128}, {0, 4845, 0, -13800, 0, 10656, 0, -2944, 0,256}, {-3840, 0, 37470, 0, -60000, 0, 33152, 0, -7168, 0, 512} MATHEMATICA P[x, 0] = 1; P[x, 1] = x; P[x_, n_] := P[x, n] = 2*x*P[x, n - 1] - n*P[x, n - 2]; Table[ExpandAll[P[x, n]], {n, 0, 10}]; a = Table[CoefficientList[P[x, n], x], {n, 0, 10}]; Flatten[a] CROSSREFS Cf. A053120. Sequence in context: A175950 A066285 A327873 * A047765 A068463 A099554 Adjacent sequences: A136662 A136663 A136664 * A136666 A136667 A136668 KEYWORD uned,tabl,sign AUTHOR Roger L. Bagula, Apr 02 2008 STATUS approved
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A178231 Decimal expansion of sqrt(181). 1
1, 3, 4, 5, 3, 6, 2, 4, 0, 4, 7, 0, 7, 3, 7, 1, 0, 3, 1, 7, 1, 6, 3, 0, 8, 5, 4, 6, 2, 1, 7, 0, 4, 0, 4, 1, 9, 3, 8, 7, 0, 9, 6, 2, 3, 8, 5, 5, 2, 0, 1, 2, 0, 2, 7, 5, 2, 0, 0, 4, 8, 7, 2, 8, 3, 3, 1, 8, 0, 3, 8, 7, 1, 8, 4, 2, 6, 3, 8, 8, 4, 1, 8, 2, 9, 7, 3, 5, 1, 5, 5, 4, 8, 2, 9, 1, 5, 8, 8, 2, 0, 9, 6, 4, 4 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 2,2 COMMENTS Continued fraction expansion of sqrt(181) is A010224, has period length 21. LINKS EXAMPLE sqrt(181) = 13.45362404707371031716... CROSSREFS Cf. A178230 (decimal expansion of sqrt(1086)), A010224. Sequence in context: A165565 A033706 A121890 * A298734 A137926 A218342 Adjacent sequences: A178228 A178229 A178230 * A178232 A178233 A178234 KEYWORD cons,nonn AUTHOR Klaus Brockhaus, May 23 2010 STATUS approved
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Last modified August 20 04:46 EDT 2019. Contains 326139 sequences. (Running on oeis4.)
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https://www.gatecseit.in/gcd-and-lcm-using-recursion-multiple-choice-questions-and-answers-mcqs/
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# GCD and LCM using Recursion Multiple choice Questions and Answers (MCQs)
## GCD and LCM using Recursion Multiple choice Questions and Answers (MCQs)
Question 1 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
Which of the following is not an alias for GCD?
A LCM B GCM C GCF D HCF
Question 1 Explanation:
LCM (Least Common Multiple) is not an alias for GCD while GCM (Greatest Common Measure), GCF (Greatest Common Factor), HCF (Highest Common Factor).
Question 2 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
What is the GCD of 8 and 12?
A 8 B 12 C 2 D 4
Question 2 Explanation:
GCD is largest positive integer that divides each of the integer. So the GCD of 8 and 12 is 4.
Question 3 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
If GCD of two number is 8 and LCM is 144, then what is the second number if first number is 72?
A 24 B 2 C 3 D 16
Question 3 Explanation:
As A * B = GCD (A, B) * LCM (A, B). So B= (144 * 8)/72.
Question 4 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
Which of the following is also known as GCD?
A Highest Common Divisor B Highest Common Multiple C Highest Common Measure D Lowest Common Multiple
Question 4 Explanation:
GCM (Greatest Common Measure), GCF (Greatest Common Factor), HCF (Highest Common Factor) and HCF (Highest Common Divisor) are also known as GCD.
Question 5 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
Which of the following is coprime number?
A 54 and 24 B 4 and 8 C 6 and 12 D 9 and 28
Question 5 Explanation:
Coprime numbers have GCD 1. So 9 and 28 are coprime numbers. While 54 and 24 have GCD 6, 4 and 8 have GCD 4, 6 and 12 have GCD 6.
There are 5 questions to complete.
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# Introduction to Enterprise Valuations
### Sarah Martin
30 years: Corporate Valuations
In this video in the series on corporate valuations, Sarah introduces enterprise value multiples and first covers the background to multiple valuations and then walks us through the process of carrying out a multiple valuation.
In this video in the series on corporate valuations, Sarah introduces enterprise value multiples and first covers the background to multiple valuations and then walks us through the process of carrying out a multiple valuation.
Speak to an expert
Speak to an expert today to access this and all of the content on our platform.
### Introduction to Enterprise Valuations
16 mins 27 secs
Overview
A multiple valuation simply involves multiplying a company’s financial variable, such as EBITDA or net profit by a corresponding multiple or ratio observed in a peer group of similar firms. The main benefit of a multiple valuation is that it allows us to estimate valuations without having to go through the rigour of a DCF calculation. It is important to understand the the background to it along with the execution.
Key learning objectives:
• Understand what are multiple valuations and the background to them
• Understand the mechanics of multiple valuations
• Comprehend why multiples vary between firms and sectors
## Speak to an expert
Speak to an expert today to access this and all of the content on our platform.
Summary
#### What are enterprise valuation multiples and where are they used?
Taking a metric such as a firm’s forecast EBITDA and multiplying this variable by the average or median EBITDA multiple at which a peer group of comparable firms trades gives us an estimate of the EV of a firm. This estimate is produced by using only two variables and hence is straightforward.
Multiples can be used to value a private company or to test whether an existing value is high or low relative to the chosen peer group. It is a relative value method – valuing one firm, based on the current valuation of comparable firms.
#### How do multiple valuations work?
1. Select a peer group which has current valuations. These valuations are usually a share price but an M&A valuation could also work. Ideally, peer group members have the same key characteristics as the firm you are valuing. It is usually better to have a small peer group of reasonably comparable companies, rather than a large peer group where the constituents might technically be in the same sector but in practice have very different business profiles and return outlooks.
2. Calculate the enterprise value of the peer group companies. For listed firms, we normally take the market capitalisation and add on net debt and equivalents to reach enterprise value. It is important to note that valuation experts often work out net debt differently from each other.
3. Calculate EBITDA for the firm being valued and all the peer group members. As EBITDA is undefined by IFRS and US GAAP, there are differing definitions. Underlying EBITDA should be used, not reported EBITDA. Normally, it is preferred to use forecast numbers and if the peer group constituents have different fiscal year ends, it is better to use next 12-month forecasts.
4. Calculate the EV/EBITDA multiple for each firm. Usually use 3 multiples for each firm - a historic multiple, this year’s multiple and a forecast multiple. EV comes from the market capitalisation, which changes daily with the share price. Hence, the multiples will change every day as the share prices change every day.
5. Calculate the average or median EBITDA multiple for your peer group and apply this to the EBITDAs of the firm you are valuing. Using the 2023 forecast multiple for the peer group and multiplying this with the forecast 2023 EBITDA for the firm you are valuing would provide a valuation today for the firm. This is because the peer group average multiple was based on today’s share prices of the peer group.
#### Why do multiples differ between firms?
Financial factors:
• Different growth outlooks – high-growth firms usually have higher multiples
• Risk outlook – lower-risk firms, with lower WACC, usually have higher multiples.
• Outlook for the return after tax on capital employed, also known as the return on invested capital
Non-financial factors – Some of the non-financial factors include Poor corporate governance, ineffective management, adverse regulatory changes, a poor track record of M&A, lack of value creation, worsening event risks and geo-political risks, technological threats, a poor innovation record and high customer concentration. A firm may also be impacted by a low free float (only a small percentage of the shares are listed, its trading multiple can be affected)
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### Sarah Martin
Sarah Martin has a degree in economics from the London School of Economics and stock exchange and regulatory qualifications from London and New York. She has worked in investment banking for 17 years, as well as private equity transactions and as an expert witness in financial trials. She became a financial trainer 15 years ago and specialises in credit, distressed debt, and valuation. Recent assignments have included the European Central Bank, the European Investment Bank, the EBRD, Gibbs Business School in Johannesburg, the Bahrain Institute of Business Finance, the Bank of China, BBVA, the African Development Bank, Siemens, Carnegie Bank, Rand Merchant Bank, the Hamburg Central Bank, and Mizuho Bank.
There are no available videos from "Sarah Martin"
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# MAT 171 Pre-Calculus
Exit Skills:
The following exit skills were developed with the assumption that the student has a good grasp of all intermediate algebra skills and that word problems and applications are an integral part of any algebra course.
1. Solve quadratic equations, equations of quadratic form and miscellaneous equations.
2. Solve quadratic, rational, compound and absolute value inequalities writing solutions in interval notation.
3. Graph families of functions using transformations and state domain and range in interval notation.
4. Solve higher-degree polynomial equations using the Remainder, Factor, Rational Root, Conjugate Root Theorems, Bounds Theorem, and the Intermediate Value Theorem.
5. Simplify difference quotients including polynomial, rational and radical functions.
6. Graph rational functions identifying vertical and horizontal asymptotes, including use of limits.
7. Find the rule for the inverse of a function and graph the inverse.
8. Graph exponential and logarithmic functions.
9. Solve exponential and logarithmic equations.
10. Solve systems of non-linear equations and inequalities.
11. Solve 3 x 3 systems of linear equations using Gaussian elimination, inverse of a matrix and Cramer’s Rule.
12. Perform matrix algebra operations including addition, subtraction, products and the evaluation of determinants by the minor-cofactor method.
13. Graph conic sections including circles, parabolas, ellipses and hyperbolas.
14. Find terms of a sequence using recursion and explicit formulas, and find the nth term formula for arithmetic and geometric sequences.
15. Use summation notation, find the sum of a finite arithmetic series and find the sum of finite and infinite geometric series.
16. Use the binomial theorem to expand a binomial.
17. Define the trigonometric functions using the circular and triangular definitions.
18. Convert between degree and radian measure.
19. Locate and find trigonometric functions of special angles and angles corresponding by reference.
20. Use amplitude, period, phase shift and vertical shift to graph trigonometric equations.
21. Identify the range restrictions for the inverse trigonometric functions and find inverses.
22. Prove and use basic, Pythagorean, angle sum and difference, double angle and half angle identities to simplify expressions, verify identities and solve conditional equations.
23. Graph, add and subtract vectors and solve associated applications.
24. Prove and use the Law of Sines and the Law of Cosines, and solve associated applications.
25. Convert between rectangular and trigonometric form of complex numbers.
26. Multiply and divide complex numbers in trigonometric form, and use DeMoivre’s theorem and the Nth-Root theorem to find integer powers and roots of complex numbers.
27. Convert between rectangular and polar coordinates and equations and graph common polar equations.
#### Office Hours
8 a.m.-4:30 p.m. Monday-Friday
#### Office Location
Administration Building, Room 1242
#### Office Phone Number
636-922-8496
Phyllis Marchand
636-922-8496
phyllis_marchand@stchas.edu
Joseph (Joe) W. Howe
636-922-8318
jhowe@stchas.edu
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https://goprep.co/ex-11.a-q2d-the-9th-term-of-the-ap-3-4-5-4-7-4-9-4-l-find-i-1njyg0
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Q. 2 D
# Find:the 9th term of the AP
Here, First term = a = 3/4
Common difference = d = 5/4 - 3/4 = 2/4
To find = 9th term, n = 9
Using the formula for finding nth term of an A.P.,
an = a + (n - 1) × d
an = (3/4) + (9 - 1) × (2/4)
an = 3/4 + 8 × (2/4) = 3/4 + 16/4 = 19/4
9th term of the given AP is 19/4.
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https://stats.stackexchange.com/questions/331040/conditional-probability-that-one-of-three-random-variables-is-the-largest/331043
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# Conditional probability that one of three random variables is the largest
Suppose I have three random variables $X_1$, $X_2$, and $X_3$ that are i.i.d $N(0,1)$ distributed. The chance that $X_3$ is larger than $X_2$, given that $X_3$ is also larger than $X_1$ should equal $\frac{2}{3}$ in theory. However, when trying to calculate this conditional probability, I always end up with an probability equal to 1, which obviously is not correct. Perhaps you could point out where my mistake is?
$\Pr[X_3 > X_2 \mid X_3 > X_1] = \Pr[X_2 < X_1]\Pr[X_3 > X_2 \mid X_3 > X_1, X_2 < X_1]+\Pr[X_2 \geq X_1]\Pr[X_3 > X_2 \mid X_3 > X_1, X_2 \geq X_1]$.
I then rewrite the last term as
$\Pr[X_2 \geq X_1]\Pr[X_3 > X_2 \mid X_3 > X_1, X_2 \geq X_1] = \Pr[X_2 \geq X_1]\frac{\Pr[X_3 > X_2]}{\Pr[X_3 > X_1]}$.
This gives me
$\Pr[X_3 > X_2 \mid X_3 > X_1] = 0.5 * 1 + 0.5\frac{0.5}{0.5} = 1 \neq \frac{2}{3}$.
The easiest thing is to look at it combinatorically. If you rank the numbers $X_i$ by 1,2 and 3 according to their magnitude (3 is the largest), then you have 6 ranked combinations of ($X_1,X_2,X_3$) in total: (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2) and (3,2,1).
The ones with $X_3>X_1$: (1,2,3),(1,3,2) and (2,1,3).
Out of these two combinations also have $X_3>X_2$: (1,2,3) and (2,1,3).
Hence your probability $P(X_3>X_2\mid X_3>X_1)$: 2/3
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```Question 357958
<pre><font size = 3 color = "indigo"><b>
Hi,
Multiply. (x^2+x+5)(x-5)
Take it a step at a time: for ex: b(x-5) = x*b - 5*b
first Multiplying thru with x and then multiplying thru with -5
x^3 + x^2 + 5x - 5x^2 - 5x - 25
combining like terms
x^3 - 4x^2 - 25```
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files/ Torres9 -- Blogmeister
Bmx is SIK
by Torres9 teacher: Professor McGonagall
Assignments
05/21 05/07 04/22 04/13 02/03 01/05 12/17 12/08 12/01 11/18 11/05 11/03 10/30 10/29 10/27 10/21 10/13 10/08 10/08 10/08 10/07 09/18 09/12 09/11 09/10 08/24
Blog Entries
5/8 Fave Sculpture 4/13 Welcome Back Spring Break 3/16 News Article 3/5 Non-fiction story 3/5 Non fiction story 1/22 choice writing 12/17 LET IT SNOW!!! 12/10 Word study 6 12/9 dialogue 12/3 Word study5 12/1 Thanksgiving food 11/20 word study 4 11/17 Cordinate grids 11/5 Word study 2 11/5 Word study 10/30 mountain biking 10/30 Number puzzles 10/30 Choice assignment 10/30 Make our class better 10/27 My weekend 10/23 hockey 10/22 water water and more water!!! 10/15 Reading response 10/8 Paintball 10/8 Reading response 10/8 Number puzzle List 25, 50, all
Number puzzles
Hey world today I am going to give you 2 number puzzles to solve. I hope you can.
My number is a multiple of 2.
My number is higher than 45.
My number is lower than 90.
Andt he sum of my number is 8.
I hope you can figure the number out.
-Torres9
Article posted October 30, 2008 at 02:35 PM • comment • Reads 1419 • see all articles
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# Simulating the Spread of a Disease Using a Computer
The rapid spread of COVID-19 around the globe has caused a great deal of worry, and for good reason. At-risk segments of the population, including the elderly and those with preexisting health issues, face life-threatening dangers as the disease spreads. Although younger and healthier people tend to fare far better with COVID-19, the fact that they can carry the disease so easily to others who are much more impacted by it have led universities, entertainment venues, and even an entire country to abandon business as usual in an abundance of caution. Why do diseases such as COVID-19 spread, and why do they do so with such virulence? Computer simulations can give us some valuable insights.
The SIR (Susceptible-Infected-Recovered) model provides a common and effective approach to studying the spread of a disease. SIR divides a population into at least three categories: those susceptible to the disease, those infected by the disease, and those who have recovered from it. Some percentage of those who have “recovered” may have done so by passing on to the next life, a rather macabre twist on the notion of recovery, but one that gives rise to a fourth population, the deceased. Those who are susceptible to the disease can transition to the infected camp by coming in contact with an infected person, but not all such interactions will pass the sickness to the susceptible person. Rather, some fraction – call it alpha – of those who encounter an infected person will contract the disease. With this, alpha represents the rate of infection and is an important measure of how easily a disease spreads. Once infected, we will assume that person will either become part of the recovered group – those permanently immune to the disease – or to the deceased group. Suppose the rate of recovery is defined as beta, and that the death rate of those who have “recovered” from the disease is denoted as gamma.
We’ve seen a lot of definitions so far, including a few Greek letters thrown in just to complicate things. But we’re getting close to being able to use all these things. Suppose we are dealing with a population of 50,000 people, 10 of whom are already part of the “I” group (infected), and 49,990 of which are currently part of the “S” group (susceptible). No one is part of the the R (recovered) or D (deceased) groups yet, so their initial populations are 0. A key step in modeling the various segments of this population is to identify how and why the sizes of the S, I, R, and D groups will change over time. In other words, we need to identify and quantify those forces that will make S, I, R, and D increase, as well as those forces that will make them decrease. Let’s consider each of these, in turn:
S: The number of susceptible people can only decrease, because some of them will become infected. How many of them become infected by an interaction with an infected person is controlled by the rate alpha. Once infected and recovered, we assume a person is immune and cannot become susceptible again, so there is nothing that will make S increase.
I: The number of infected people will increase as a result of susceptible people coming into contact with a currently infected person, although not all such interactions will result in a new infection. The number of infected people will decrease as a result of an infected person recovering, which they will do at the beta rate. Although some of these recovered people will die (the fraction gamma), such unfortunate cases still result in a decrease in the number of infected people, as, to put it bluntly and obviously, someone who has died is no longer infected.
R: An infected person can recover and will do so at the rate beta, and this results in an increase in the number of people who have recovered. Unfortunately, some of these people – the fraction gamma – who have recovered will pass, and these people result in a decrease in the size of the R population.
D: The number of deceased can only grow over time as a percentage of people who have “recovered” from the infection. Since a person cannot come back to life, no factors can result in an increase in the D population.
Let’s get back to some math. The rate of change of a quantity that changes over time can always be expressed as
rate_of_change_of_quantity =
changes_that_result_in_an_increase – changes_that_result_in_a_decrease
We have four quantities – the populations S, I, R, and D – and a lot of words that describe their rates of change in terms of the things that make them increase and the factors that make them decrease. Let’s reduce these words to the following differential equations. As you read these equations, compare them with the verbal explanations above and see if they make sense. For example, the rate of change of S is due entirely to a decrease: the number of susceptible people will decrease because they will become infected. How quickly? It depends on the number of people who are susceptible and the number of people who are already infected, because they will mingle with each other, and some fraction alpha of those interactions will actually pass the disease. That is what the first of these equations represent. See if you can interpret the other equations in an identical manner.
rate_of_change_of_S = – alpha * S * I
rate_of_change_of_I = alpha * S * I – beta * I
rate_of_change_of_R = beta * I – beta * gamma * I
rate_of_change_of_D = beta*gamma*I
So now we have equations – differential equations – that describe how slowly or quickly the sizes of the various populations change. What do we do with them? Differential equations like these are useful because they help us learn the values of the various quantities – in this case, population sizes – as time moves on. To solve a differential equation means to consider these kinds of rates of change, together with the initial values of the quantities before they start changing, to calculate the actual values of those quantities as time advances. Undergraduate math, physics, and engineering majors typically take a course in differential equations, but they learn how to solve only certain categories of rather specific and relatively simple differential equations. More complicated differential equations require numerical techniques to solve them. That’s where computers come in especially handy.
Computers solve differential equations using a technique called numerical integration. There are many ways to perform numerical integration. The simplest way is to use a technique called Forward Euler. The Forward Euler approach calculates the next value of a quantity – in other words, its value at the next moment in time – from its current value and its rate of change over the time that elapsed. In other words:
the_next_value = the_current_value + the_rate_of_change * time_gone_by
For example, since we know the rate of change of S, the susceptible population, we can compute how S changes over time using Forward Euler like so:
the_next_value_of_S = the_current_value_of_S + the_rate_of_change_of_S *
time_gone_by
But we know that “the_rate_of_change_of_S” is – alpha * the_current_value_of_S * the_current_value_of_I. So, plug this into the equation above, and don’t forget that leading minus sign.
the_next_value_of_S =
the_current_value_of_S – alpha * the_current_value_of_S *
the_current_value_of_I
We can apply exactly the same approach to I, R, and D, the other three populations.
the_next_value_of_I =
the_current_value_of_I + alpha * the current_value_of_S *
the_current_value_of_I – beta * the current_value_of_I
the_next_value_of_R =
the_current_value_of_R + beta * I – beta * gamma * I
the next_value_of_D =
the_current_value_of_D + beta * gamma * I
We then have a system of four differential equations to solve. Assuming we knew the initial values of S, I, R, and D (and we do – S = 49,990, I = 10, R = 0, and D = 0 at the start), we could use these four equations to predict how S, I, R, and D vary over time. We could then determine if a community would be able to weather the spread of the disease as it worked its way through the population.
The good news is that this approach is rather easy to computerize. Although Forward Euler is very susceptible to the size of the time step you use and can go unstable and grossly inaccurate if you use too large a time step, it is particularly easy to program. Here, for example, is Python code that computes the four populations over time and then plots them.
``````# Ray Klump
import matplotlib.pyplot as plt
alpha = 0.000005
beta = 0.025
chi = 0.1
t = [0]
s0 = 49990
i0 = 10
r0 = 0
d0 = 0
sus = [s0]
inf = [i0]
rec = [r0]
for i in range(1,100):
t.append(i)
sus.append(sus[i-1] - alpha * sus[i-1]*inf[i-1])
inf.append(inf[i-1] + alpha*sus[i-1]*inf[i-1] - beta*(1-chi)*inf[i-1])
rec.append(rec[i-1] + beta*(1-chi)*inf[i-1])
plt.plot(t,sus,"r")
plt.plot(t,inf,"k")
plt.plot(t,rec,"g")
plt.xlabel("Day")
plt.ylabel("Number of people")
plt.title("Populations")
plt.show()
``````
Running this code will produce the following plot of the four populations over time.
You can simulate various strategies, such as quarantining, deploying masks, or just doing business-as-usual, by adjusting the parameters alpha, beta, and gamma. For example, if the rate of infection (alpha) doubles, the spread occurs more frequently, with the disease firmly taking hold in the population by day 20, and more than 42,000 contract the disease, up from about 35,000 with the original infection rate. Plots like these could demonstrate to a skeptical public the benefits to be had by canceling public gatherings and other events that could enhance the disease’s ability to spread.
Of course, there is guesswork involved in adjusting the parameters in ways that accurately capture the effects of these strategies. Still, simulations like these can help you understand how a disease like COVID-19 spreads, why it poses such significant risks, and which mitigating measures might be most effective in stopping it.
## About Ray Klump
Associate Dean, College of Aviation, Science, and Technology at Lewis University Director, Master of Science in Information Security Lewis University http://online.lewisu.edu/ms-information-security.asp, http://online.lewisu.edu/resource/engineering-technology/articles.asp, http://cs.lewisu.edu. You can find him on Google+.
### One thought on “Simulating the Spread of a Disease Using a Computer”
1. April 6, 2020 at 9:39 am
thank you
But
It must be said that the simulation is painful for these days
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which of the following most logically completes the argument : GMAT Critical Reasoning (CR)
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# which of the following most logically completes the argument
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which of the following most logically completes the argument [#permalink]
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which of the following most logically completes the argument below?
According to promotional material published by the city of Springfield, more tourists stay in hotels in Springfield than stay in the neighboring city of Harristown. A brochure from the largest hotel in Harristown claims that more tourists stay in that hotel than stay in the Royal Arms Hotel in Springfield. If both of these sources are accurate, however, the “Report on Tourism” for the region must be in error in stating that ___________
A. the average length of stay is longer at the largest hotel in Harristown than it is at the Royal Arms Hotel.
B. There is only one hotel in Harristown that is larger than the Royal Arms Hotel.
C. More tourists stay in hotels in Harristown than stay in the Royal Arms Hotel.
D. The Royal Arms hotel is the largest hotel in Springfield
E. The royal arms hotel is the only hotel in Springfield.
Last edited by singh_amit19 on 10 Oct 2007, 13:35, edited 1 time in total.
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10 Oct 2007, 13:23
How u guys rate this problem???? Easy/Medium/Tough???
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Re: CR - SET 24 Q16 [#permalink]
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10 Oct 2007, 22:32
singh_amit19 wrote:
which of the following most logically completes the argument below?
According to promotional material published by the city of Springfield, more tourists stay in hotels in Springfield than stay in the neighboring city of Harristown. A brochure from the largest hotel in Harristown claims that more tourists stay in that hotel than stay in the Royal Arms Hotel in Springfield. If both of these sources are accurate, however, the “Report on Tourism” for the region must be in error in stating that ___________
A. the average length of stay is longer at the largest hotel in Harristown than it is at the Royal Arms Hotel.
B. There is only one hotel in Harristown that is larger than the Royal Arms Hotel.
C. More tourists stay in hotels in Harristown than stay in the Royal Arms Hotel.
D. The Royal Arms hotel is the largest hotel in Springfield
E. The royal arms hotel is the only hotel in Springfield.
I'd rate it a very easy logic question.
We are given two facts:
1. More people say in Springfield hotels than Harristown hotels
2. The largest hotel in Harristown has more guests than Royal Arms
From this we can conclude that Royal Arms is not the only hotel in Springfield.
E.
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10 Oct 2007, 22:34
I forgot to add: keyword in this question is which of the following must be in error.
The other answers are either unknown or definitely not in error.
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10 Oct 2007, 22:38
JingChan wrote:
I forgot to add: keyword in this question is which of the following must be in error.
The other answers are either unknown or definitely not in error.
BINGO.....great man! OA is E
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Re: CR - SET 24 Q16 [#permalink]
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10 Oct 2007, 22:42
JingChan wrote:
singh_amit19 wrote:
which of the following most logically completes the argument below?
According to promotional material published by the city of Springfield, more tourists stay in hotels in Springfield than stay in the neighboring city of Harristown. A brochure from the largest hotel in Harristown claims that more tourists stay in that hotel than stay in the Royal Arms Hotel in Springfield. If both of these sources are accurate, however, the “Report on Tourism” for the region must be in error in stating that ___________
A. the average length of stay is longer at the largest hotel in Harristown than it is at the Royal Arms Hotel.
B. There is only one hotel in Harristown that is larger than the Royal Arms Hotel.
C. More tourists stay in hotels in Harristown than stay in the Royal Arms Hotel.
D. The Royal Arms hotel is the largest hotel in Springfield
E. The royal arms hotel is the only hotel in Springfield.
I'd rate it a very easy logic question.
We are given two facts:
1. More people say in Springfield hotels than Harristown hotels
2. The largest hotel in Harristown has more guests than Royal Arms
From this we can conclude that Royal Arms is not the only hotel in Springfield.
E.
Agree w/ E.
The only reason this question may be a bit more difficult than normal CR's is because the question type is somewhat unfamiliar.
Took me a lil longer for this one b/c I didn't quite get what the question wanted.
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Re: CR - SET 24 Q16 [#permalink]
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11 Oct 2007, 07:38
This was easy compared to many CRs. I played with numbers and obvious answer is E.
More ppl stay in S hotels say 500 then H hotels say 400
Lh (largest hotel in H) claims it has more tourists(100) than in Hotel RS=50 (in S)
So the obvious error is E. RA cant be the only hotel in S to add up to 500. There have to be other hotels.
Re: CR - SET 24 Q16 [#permalink] 11 Oct 2007, 07:38
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# which of the following most logically completes the argument
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# confidence intervals
1. ### Confidence intervals: normal vs t-distribution (central limit theorem)
Hi all! We did a serological survey for a disease taking 300 samples in a population of 50 000. If our sample mean gives a 2% prevalence. How do we work out the confidence intervals? I have all the formula's and there are great CI calculators on the web BUT seeing that is only one survey...
2. ### help!! I have a test and none of this makes sense.
Hello! I have an exam on inferential statistics, etc and need serious help.
3. ### Reaching a 95% Confidence level within an estimated population.
How can I reach a 95% Confidence level after estimating the population size? I have a population of 367 products of which I have a sample of 50. Products can have a damaged package or can be damaged itself. We know for a fact that products can only be damaged if the package is damaged. From the...
4. ### Non-Parametric Analyses. Report medians and 95% CI in graphs?
Okay so, I'm currently writing my thesis, and I have to do non-parametric tests. I know that I have to report the medians in text. However, I'm doing graphs. Just wanted to ask whether anyone knows if I report the 95% CI and the medians in the graphs? When i do this it looks super weird and...
5. ### Probability that Bernoulli p-parameter is greater than some value, given N samples?
Hello everybody, I have the following question. Assume I observe N outcomes from a Bernoulli random variable with unknown success parameter p. Based on these outcomes (or samples), how likely is it that the real parameter p is larger (or smaller) than a given value eps (e.g., 1%)? Any...
6. ### Clarification on Error Propagation
I thought I understood the basic operations for error propagation (addition, multiplication, etc), but I have a few questions: 1. Is variance the only measure you can use? I only have 95% CI available to me. 2. Does multiplying/dividing a mean +/- CI by a constant preserve the error (ie CI...
7. ### Confidence intervals for eta squared?
Hello, is it possible to calculate confidence intervals for eta-squared manually? I found a paper which lists the formula for this: http://www.jstatsoft.org/v20/i08/paper (pages 12 and 13), but the paper doesn't say how to calculate ^L and ^D, which are a part of the formula. Can anyone please...
8. ### Help with confidence interval and interpreting data!?
Someone please please help me make sense of these stats! I really need help with interpreting outcome of studies! How do i calculate CIs (no need to know actual calculation as can use calculator) and make sense of outcome... Any help would make me forever grateful! STUDY 1. Pre- and...
9. ### Determining an indirect confidence interval for justifying a non-inferiority margin
I need to establish the efficacy of a test treatment (T) by means of an indirect comparison to placebo (P). Therefore a study of T vs. reference (R) will be performed, and there is 'historical' data for the comparison between R and P. The outcome measure of interest is a responder rate (i.e...
10. ### Does Sqrt of confidence intervals require a correction factor?
I have the following formula for confidence intervals on samples from a Rayleigh process: \frac{2(n-1)\overline{r^2}}{\chi_2^2} \leq \widehat{\sigma^2} \leq \frac{2(n-1)\overline{r^2}}{\chi_1^2} I want to give confidence intervals in terms of \sigma, not \sigma^2. If [x, y] are the 95%...
11. ### Need help: Can i genaralize to the population, or do i have to weight my data?
I have a sample of 194 households from 3 villages. The number of hh from each village is 61, 67 and 66. I used a 90% confidence level and 10% margin of error to determine sample sizes. (total household population for all three villages is 4194 households). The total sample size for all three...
12. ### 95% Confidence Interval for Regression Coefficients Question
Hello all, I have been trying to figure out how to answer these homework questions for far too long and happened upon this forum with hopes that I could get a nudge in the right direction…Here is the problem as an attachment. Also included is my work thus far…I think my issue is...
13. ### Confidence Intervals for r: Why must we standardize?
Hello! I know we are supposed to convert our r-values to z-values (and standard error) in order to calculate confidence intervals. I am a bit confused why we must do this. The sampling distribution for Pearson's r is a t-distribution and we use a t-crit in confidence intervals when our N...
14. ### [SaTScan] relative risk confidence interval calculation
Thank you for reading this. My question is regards the methodology used by SatScan for purely temporal cluster analysis using a discrete Poisson model satScan's output gives me the following: How do I calculate confidence intervals around the relative risk? Thank you!
15. ### Accounting for confidence intervals in determining a mean
I have received some elemental analysis of a metal which I want to use to determine the effects of radiation. I want to report my results for the mean and 90% CL. The analysis only presents 8 data points for each element so I was planning on using the t distrobution as follows...
16. ### Statictics Assistance needed
Hi Im really stuck with a particular question regarding calculating confidence intervals. I dont expect anyone to solve it for me but just a push in the right direction would help heaps :) Recent research published in the Proceedings of the National Academy of Sciences has examined whether...
17. ### Is there a test to see if a sample (one result, many parameters) is an outlier?
I am posting here because I am not sure really how to look for my answer. I am used to comparing means of two samples, but what happens when you have only one sample but 20 parameters? How do you determine if this is significantly outside the data cloud for a data set that has about 100 samples...
18. ### Using the Right Test to Confirm Separation of Confidence Intervals
Hello, I'm working on a research project (not homework) and I could benefit from a quick consult since some of my stats are a bit rusty. I have a series of experiments that qualify as Bernoulli trials, so I am estimating 95% confidence intervals on the success rates assuming the binomial...
19. ### Confidence limits for logistic regression equation
I have generated a logistic model for a disease status Y (unaffected-affected) using several independent variables on a very large (600K+) dataset. I then want to apply that formula to a much smaller subset of that data set, having a particular characteristic not included in the model. I then...
20. ### Trying to evaluate/measure how much 1 variable affects another.
Hello. So I've been helping look at some data for a fundraising event my school ran. We used an app and the school wants to evaluate if it is worth it. From the event I have the data of all donations (how much it was and who it went to). So what was done is for all fundraisers the sum of all...
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# Bivectors/Trivectors: How to imagine?
A bivector is a wedge product of two independent vectors which can be imagined as a parallelogram and a trivector is a wedge product of three independent vectors which can be imagined as a paralleopiped. So am I correct in thinking that this parallelogram and paralleopiped is the bivector and the trivector itself, respectively? (With its magnitude as the area and the volume, respectively?).
If so how can you imagine the orientation of a bivector? (its even more harder to imagine the orientation of a trivector!).
robphy
Homework Helper
Gold Member
A bivector is a wedge product of two independent vectors which can be imagined as a parallelogram and a trivector is a wedge product of three independent vectors which can be imagined as a paralleopiped. So am I correct in thinking that this parallelogram and paralleopiped is the bivector and the trivector itself, respectively? (With its magnitude as the area and the volume, respectively?).
If so how can you imagine the orientation of a bivector? (its even more harder to imagine the orientation of a trivector!).
Those figures are accurate geometrical representations of those multivectors.
Orientation is specified by indicating an ordering of the vectors, which you could draw in as a directed-arc from the first vector to the other vector... or as a directed-helix. Look at Burke's "Applied Differential Geometry" or Schouten's "Tensor Analysis for Physicists".
I'm working on a paper and computer program to visualize tensors [metrics, differential forms, and multivectors]. A while back I visualized differential forms in electromagnetism using VRML http://www.phy.syr.edu/courses/vrml/electromagnetism/ [Broken] ... but since no one uses VRML any more, I need to revise it. Check out the references.
You might be interested in this thread.
Last edited by a moderator:
I was reading this article recently and it says that bivectors need not be paralellograms; in fact they can be of any shape! Here's the exact quote:
We should remark that the shape of the plane segment representing a bivector is not relevant. Thus the representation of a bivector as the outer product of two vectors (one could say ‘factorisation’) is highly non-unique. The plane segment need not even be a parallelogram. It may be any flat figure with an orientation.
Here's the original article:
Last edited by a moderator:
robphy
Homework Helper
Gold Member
Yes, that's correct. They need not be parallelograms... just any similarly oriented directed area. ..and, I believe, this is restricted to simple bivectors (those that can be written as the wedge of two vectors). For visualization purposes, it might be useful to choose a shape that conveys some of the symmetries of the problem.
Jancewicz is one of the authors in my references.
BTW, thanks for those programs robphy. I'll check em out this weekend.
Chris Hillman
Visualizing multivectors
Hi, Swapnil,
Just thought I'd chime in with a more elaborate answer.
First, note that k-multivector is to vector as k-multivector field is to vector field. That is, you are asking about exterior algebra (due to Hermann Grassmann), not exterior calculus (due to Elie Cartan).
Swapnil said:
So am I correct in thinking that this parallelogram and paralleopiped is the bivector and the trivector itself, respectively? (With its magnitude as the area and the volume, respectively?).
The answer I'd give to your question is that a simple k-multivector in the vector space R^n represents a k-dimensional subspace of R^n together with an oriented k-volume element. As Cartan realized, this is just what we need to capture the notion of "a surface element of a k-dimensional submanifold". In exterior calculus, when we integrate a k-form over a k-dimensional surface, we automatically account for the Jacobian determinant factor familiar from the change of variables formula for a multidimensional integral as treated in vector calculus.
As you know, the key algebraic rule for exterior algebra is that the wedge product is antisymmetric: $\vec{e}_1 \wedge \vec{e}_2 = -\vec{e}_2 \wedge \vec{e}_1$ and so on. Thus for the vector space R^2 with basis $\vec{e}_1, \; \vec{e}_2$, the wedge product of two vectors (a simple bivector) is
$$\left( a \, \vec{e}_1 + b \, \vec{e}_2 \right) \wedge \left( p \, \vec{e}_1 + q \, \vec{e}_2 \right) = \left( a q - b p \right) \; \vec{e}_1 \wedge \vec{e}_2$$
where I expanded out the product using the antisymmetry and simplified.
As a bit of thought shows, in R^2 the exterior algebra, considered as a four dimensional real algebra, is the direct sum of a one dimensional space (the zero-vectors or real constants), a two dimensional space (the vectors), and a one dimensional space (the bivectors), so our bivector is completely determined by the value of the determinant. In this sense, we have infinitely many representations of a given simple bivector in terms of parallelograms.
Similarly, you can decompose the exterior algebra on R^n into subspaces whose dimension is given by binomial coefficients; the first few are the space of constants, the space of vectors, the space of bivectors, the space of trivectors, and so on. The series is symmetric and the last subspace is the space of n-multivectors $\omega = a \, \vec{e}_1 \wedge \vec{e}_2 \wedge \dots \vec{e}_n$, which is one dimensional. Any n-fold wedge product of n vectors in R^n is thus completely characterized by an n by n determinant. Thus, in a sense exterior algebra naturally gives rise to determinants and offers a geometric interpretation of determinants in terms of volume factors. Indeed, in exterior calculus, n-multivectors give rise to n-forms on n-dimensional manifolds, or "volume forms".
Simple k-multivectors are k-fold wedge products of vectors. For example, a typical simple bivector in R^3 is
$$\left( a \, \vec{e}_1 + b \, \vec{e}_2 + c \, \vec{e}_3 \right) \wedge \left( p \, \vec{e}_1 + q \, \vec{e}_2 + r \, \vec{e}_3 \right) = (aq-bp) \, \vec{e}_1 \wedge \vec{e}_2 + (br - cq) \, \vec{e}_2 \wedge \vec{e}_3 + (cp-ar) \, \vec{e}_3 \wedge \vec{e}_1$$
The subspace of bivectors in R^3 is three dimensional, since $\vec{e}_1 \wedge \vec{e}_2, \; \vec{e}_2 \wedge \vec{e}_3, \; \vec{e}_3 \wedge \vec{e}_1$ forms a basis for this subspace. Note that 1+3+3+1=8, the dimension of the exterior algebra over R^3.
It probably will not be clear why not every k-multivector need be simple (expressible as a k-fold wedge product of vectors). To understand this, you need to know a bit about Grassmann-Plucker sygygies. The simplest case arises when we consider the six dimensional subspace of bivectors in R^4. Then if we write a generic simple bivector
$$\left( x^1 \, \vec{e}_1 + x^2 \, \vec{e}_2 + x^3 \, \vec{e}_3 + x^4 \, \vec{e}_4 \right) \wedge \left( y^1 \, \vec{e}_1 + y^2 \, \vec{e}_2 + y^3 \, \vec{e}_3 + y^4 \, \vec{e}_4 \right)$$
as a linear combination of the six basis bivectors,
$$u_{12} \, \vec{e}_1 \wedge \vec{e}_2 + u_{13} \, \vec{e}_1 \wedge \vec{e}_3 + \dots u_{34} \, \vec{e}_3 \wedge \vec{e}_4$$
where $u_{12} = x^1 \, y^2 - x^2 \, y^1$ and so on, then the Grassmann-Plucker syzygy is the (rather sophisticated!) "high school algebra" identity
$$u_{12} \, u_{34} - u_{13} \, u_{24} + u_{23} \, u_{14} = 0$$
This tells us that the six components of a simple bivector in R^4 are not rationally independent: any one can be determined from the other five using the syzygy. Thus, not every bivector in R^4 is simple; the set of simple bivectors forms a kind of hypersurface (an algebraic variety, not a linear subspace!) in the six dimensional subspace of bivectors. Sometimes every k-multivector is simple, for example bivectors in R^3 or n-multivectors in R^n, but these cases are exceptional.
For another example, try trivectors in R^5. (If you are very good at high school algebra, you should find three independent syzygies, not all obviously guessed. Note that the syzygies give algebraic relations among the components, but when there are several syzygies, these may themselves be related by an algebraic identity. This suggests the notion of a kind of sequence which keeps track of relations among coefficients, relations among these relations, relations among the previous relations, and so on.)
The textbook by Harris, Algebraic Geometry: A First Course, provides a wealth of information on some very interesting topics closely related to topics discussed above, including Grassmann manifolds, Stiefel manifolds, and Plucker embeddings of Grassmannians in higher dimensional projective spaces. In addition, if you have followed the This Week postings of John Baez, you may sense (correctly!) that this stuff is related to the Schubert calculus, which concerns incidence relations among k-flats in (complex) projective n-space. I note too that Grassmann and Stiefel manifolds are classic examples of homogeneous spaces (coset spaces for a closed subgroup of a Lie group). The textbook An Introduction To Differentiable Manifolds And Riemannian Geometry by Boothby offers a good introduction to homogeneous spaces.
The textbook by Spivak, Calculus on Manifolds offers a very good introduction to exterior algebra and exterior calculus. The more elementary approach in the classic textbook by Birkhoff and Mac Lane, Modern Algebra, and in the classic monograph by Flanders, Differential Forms with Applications to the Physical Sciences, are well worth studying, in addition to the books already mentioned.
Last edited:
Hurkyl
Staff Emeritus
Gold Member
A bivector is a wedge product of two independent vectors which can be imagined as a parallelogram and a trivector is a wedge product of three independent vectors which can be imagined as a paralleopiped. So am I correct in thinking that this parallelogram and paralleopiped is the bivector and the trivector itself, respectively? (With its magnitude as the area and the volume, respectively?).
If so how can you imagine the orientation of a bivector? (its even more harder to imagine the orientation of a trivector!).
You can always imagine them as algebraic objects!
mathwonk
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Tooltip
These resources have been reviewed and selected by STEM Learning’s team of education specialists for factual accuracy and relevance to teaching STEM subjects in UK schools.
# Probability
Designed to improve confidence in mathematics, these resources from the Centre for Innovation in Mathematics Teaching, focus on probability and were developed particularly for primary teachers and those non-specialists who teach mathematics in the lower secondary years.
Each section offers an overview of a particular topic and the key issues that will be covered, as well as highlighting some common misconceptions. Worked examples are provided to support learning and these are followed by focused exercises to practice skills. Activities to reinforce learning and offer extension opportunities are also supplied, as are topic related tests.
Probability part A - Probabilities ▪ Simple Probability ▪ Outcome of Two Events ▪ Finding Probabilities Using Relative Frequency ▪ Determining Probabilities ▪ Probability of Two Events ▪ Use of Tree Diagrams ▪ Multiplication for Independent Events ▪ Mutually Exclusive Events.
Probability part B - Misconceptions ▪ Evens and Odds ▪ Experimental Probability ▪ Tossing Three Coins ▪ Throwing Two Dice ▪ A Russian Fable ▪ Open and Shut Case ▪ Fruit Machines ▪ Birthdays.
#### Show health and safety information
Please be aware that resources have been published on the website in the form that they were originally supplied. This means that procedures reflect general practice and standards applicable at the time resources were produced and cannot be assumed to be acceptable today. Website users are fully responsible for ensuring that any activity, including practical work, which they carry out is in accordance with current regulations related to health and safety and that an appropriate risk assessment has been carried out.
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# How do I calculate friction?
The coefficient of friction (fr) is a number that is the ratio of the resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together. It is represented by the equation: fr = Fr/N.
## How do you calculate static friction?
The formula to calculate the static friction is given as: Static Friction = Normal Force x Static Friction coefficient. Static friction = 60 N.
## How do you find energy lost due to friction?
The energy “lost” due to friction will be W = Fd = 500 N x 1.0 m = 500 J. This is only half the kinetic energy he has, so he makes it to home plate.
## How do you find acceleration without time?
If the acceleration is constant, it is possible to find acceleration without time if we have the initial and final velocity of the object as well as the amount of displacement. The formula v2=u2+2as where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement is used.
## What are the types friction?
There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction.
## What is limiting friction?
Limiting friction occurs when the moving force and the force opposing motion are equal; any addition to the moving force will cause slipping. The limiting frictional force is proportional to the normal reaction between the contacting surfaces and is independent of the area of contact.
## What is an example of mechanical energy to thermal energy?
Rubbing your hands together converts mechanical energy to thermal energy. Turning on a light switch converts mechanical energy to electrical and radiant energy.
## How do you find initial velocity and final velocity?
Work out which of the displacement (S), initial velocity (U), acceleration (A) and time (T) you have to solve for final velocity (V). If you have U, A and T, use V = U + AT. If you have S, U and T, use V = 2(S/T) – U. If you have S, A and T, use V = (S/T) + (AT/2).
## How do I calculate kinetic energy?
If you know the mass and velocity of an object, use the kinetic energy calculator to find it’s energy in movement.
To calculate kinetic energy:
1. Find the square of the velocity of the object.
2. Multiply this square by the mass of the object.
3. The product is the kinetic energy of the object.
## How do you change friction?
Another way to reduce friction is with a lubricant like grease or oil. Machines and engines use grease and oil to reduce friction and wear so they can last longer. Another way to reduce friction is to change the types of materials in contact with one another.
## What is sliding friction for Class 8?
In Class 8 Science, sliding friction is a frictional force that appears when two bodies come into sliding contact. This friction can also be referred to as kinetic friction and is relatively weaker than static friction. For instance, it makes it simpler to slide the furniture over the floor after starting to move it.
## How many types of friction are there?
Friction is the force that opposes the motion of a solid object over another. There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction.
See also How do you fix a RC car that won't move?
## What are the types of kinetic friction?
Friction occurs essentially in two different modes: sliding and rolling friction. The relationships between the frictional force and the load or weight of the sliding object differ for dry, or unlubricated, surfaces and lubricated surfaces.
## How do you calculate thermal energy from friction?
For the force of gravity use F=ma where a is the acceleration of gravity. For the frictional force, multiply this F by the coefficient to get Ff=μF=0.105F. For the thermal energy use W=Ffd, where W is the work and d is the distance.
## What are heat transfer methods?
Heat can be transferred in three ways: by conduction, by convection, and by radiation.
• Conduction is the transfer of energy from one molecule to another by direct contact. …
• Convection is the movement of heat by a fluid such as water or air. …
• Radiation is the transfer of heat by electromagnetic waves.
## How do you find change in kinetic energy after a collision?
[Once v’1 is chosen, v’2 is determined by conservation of momentum.] because the final momentum is constrained to be p’ = m1v’1 + m2v’2 = kg m/s . Final kinetic energy KE = 1/2 m1v’12 + 1/2 m2v’22 = joules. For ordinary objects, the final kinetic energy will be less than the initial value.
## How do you find velocity without time?
Examine the problem to find the displacement of the object and its initial velocity. Plug the acceleration, displacement and initial velocity into this equation: (Final Velocity)^2 = (Initial Velocity) ^2 + 2_(Acceleration)_(Displacement). Solve the problem using pen, paper and calculator.
## What form of energy is present in a rolling ball?
Kinetic energy depends on an object’s mass and its speed. Ignoring frictional losses, the total amount of energy is conserved. For a rolling object, kinetic energy is split into two types: translational (motion in a straight line) and rotational (spinning).
## What is the unit for work?
Scientists use the joule to measure work. One joule is equal to the work done by a force of one newton to move an object one meter in the direction of the force.
## How long does a friction disc last?
It snows 1-7 times per year, 2″-7″ with occasional 12″-25″ snows. A friction disk will last 30+ years.
## What is a friction plate?
A friction clutch plate is used in vehicles to allow the transmission input shaft and engine to run at the same speed when rotating. The friction that is created between the engine and the transmission is what provides the force required to move the vehicle.
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## Measuring Up: Prototypes for Mathematics Assessment (1993) Mathematical Sciences Education Board (MSEB)
### Citation Manager
. "Mystery Graphs." Measuring Up: Prototypes for Mathematics Assessment. Washington, DC: The National Academies Press, 1993.
Page 24
The following HTML text is provided to enhance online readability. Many aspects of typography translate only awkwardly to HTML. Please use the page image as the authoritative form to ensure accuracy.
Measuring Up: Prototypes for Mathematics Assessment
Name _________________________________________ Date ______________
Look at the five graphs on the next pages. Each graph shows something about a classroom of fourth graders.
1. Which of the five graphs do you think shows:
1. The number of cavities that the fourth graders have?
2. The ages of the fourth graders' mothers?
3. The heights of the fourth graders, in inches?
4. The number of people in the fourth graders' families?
1. Explain why you think the graph you picked for c is the one that shows the heights of fourth graders.
2. Why do you think the other graphs don't show the fourth graders' heights?
Page 24
Front Matter (R1-R10) Introduction (1-3) The Challenge (4-4) The Criteria (5-6) The Caveats (7-7) The Audience (8-8) The Prototypes (9-11) The Tryouts (12-12) The Format (13-13) The Protorubrics (14-15) The Standards (16-18) The Future (19-20) The Prototypes (21-22) Mystery Graphs (23-30) The Checkers Tournament (31-42) Bridges (43-52) Hexarights (53-64) Bowl-A-Fact (65-74) Point of View (75-84) The Quilt Designer (85-94) How Many Buttons? (95-100) The Taxman (101-114) Lightning Strikes Again (115-124) Comparing Grizzly Bears and Black Bears (125-132) The Towers Problem (133-140) The Hog Game (141-156) Resources (157-160) Mathematical Sciences Education Board (161-164) Credits (165-166)
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## What is a multiphase sampling method?
Definition: Multistage sampling is defined as a sampling method that divides the population into groups (or clusters) for conducting research. During this sampling method, significant clusters of the selected people are split into sub-groups at various stages to make it simpler for primary data collection.
## What is the difference between multistage sampling and multiphase sampling?
Multiphase sampling must be distinguished from multistage sampling since, in multiphase sampling, the different phases of observation relate to sample units of the same type, while in multistage sampling, the sample units are of different types at different stages.
What are the sampling procedures?
Sample: a portion of the entire group (called a population) • Sampling procedure: choosing part of a population to use to test hypotheses about the entire population. Used to choose the number of participants, interviews, or work samples to use in the assessment process. used, e.g. random or stratified sampling.
Which is the first phase of multiphase sampling?
The first phase of the multiphase sampling technique is to randomly choose the large sample. As we did before, we can choose our six schools by flipping a coin or drawing names out of a hat. The second phase of the multiphase sampling technique is to choose the small sample from the large sample.
### Which is an advantage of multistage sampling method?
The advantage of multistage sampling is that it allows us to choose more clusters. Instead of four schools, we drew students from eight schools. We wouldn’t be measuring every student at each of those schools, so we are free to choose more schools.
### When to use two phase or double sampling?
With only one sub-sample, the design is called two-phase or double sampling. Double sampling is used in some situations where making use of an auxiliary variable is desirable.
How is the sample chosen in probability sampling?
The sample of a study is the group of participants in the study, and it is chosen in a process known as sampling. One of the main types of sampling techniques is probability sampling, which involves choosing your sample randomly. Let’s look closer at three different types of probability sampling: cluster, multistage, and multiphase sampling.
What is a multiphase sampling method? Definition: Multistage sampling is defined as a sampling method that divides the population into groups (or clusters) for conducting research. During this sampling method, significant clusters of the selected people are split into sub-groups at various stages to make it simpler for primary data collection. What is the difference…
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Valuing FRAs
I find this topic extremely confusing!
Post some questions. There are many here that are good at explaining them.
its basically the first step of Annualizing and de-annualizing that i don’t get. Here’s an example from SchweserPro: 30 days ago, J. Klein took a short position in a \$10 million 90-day forward rate agreement (FRA) based on the 90-day London Interbank Offered Rate (LIBOR) and priced at 5%. The current LIBOR curve is: 30-day = 4.8% 60-day = 5.0% 90-day = 5.1% 120-day = 5.2% 150-day = 5.4% The current value of the FRA, to the short, is closest to: A) - \$15,495. B) - \$15,154. C) - \$15,280. D) - \$15,331
http://www.analystforum.com/phorums/read.php?12,708128 And that has a link to an earlier thread on this same question. Hope that helps
thanks.
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# Penrose Tilings and the Golden Ratio
## Introduction
In mathematics, the tiling of a plane is a collection of tiles that cover the plane without gaps or overlaps. There are different types of tilings. You may use regular or irregular polygons to tile a plane by using a repeating pattern.
These repeating patterns are called periodic tiling.
A Penrose tiling is a pattern of tiles, discovered by Roger Penrose and Robert Ammann, which can completely cover a plane, but only in a pattern that is non-repeating. This type of tiling is called aperiodic tiling.
## Kites and Darts
The two main tiles of Penrose tiling are the red Kite and the blue Dart. Explore these shapes below – you can measure the side lengths and internal angles using the ruler and protractor tools:
All the shorter sides are 1 cm long, while the longer sides have a measure of approximately 1.6 - 1.7. This number is called Phi. It is an irrational number with the value of:
$(1+√5)/2$ = 1.6180339887 ...
This number is referred to as the Golden Ratios and might help us understand the beauty of the designs created by the Dart and the Kite. Let's create our own designs of aperiodic tiling. The tiles are put together with one rule:
No two tiles can be touching so as to form a single parallelogram.
Given this rule, there are many ways to tile an infinite plane with no gaps, overlaps, or holes, but the tiling is always guaranteed to be aperiodic. It also will not have translational symmetry, which means the pattern cannot be shifted to match itself over the entire plane.
Cover the entire Polypad canvas as much as possible using these tiles. Does your initial pattern continuously repeat itself?
Now try answering these questions:
• How can you prove that they tile the plane?
• What are the other patterns do you realize?
Here are some examples:
## Further Reasing
• Search for "quasicrystals" to learn about where Penrose tilings occur in nature
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DETERMINING EQUATION TO A DRAWN CURVE????
Experimenter
Hi everyone,
I am seeking help in determining method(s) to determining the equation (e.g. polynomial equation) to a curve that's been drawn.
For example, with the curve drawn in the attached example figure, how can I determine the equation to define that curve?
Thank you in advance for any insight!
Re: DETERMINING EQUATION TO A DRAWN CURVE????
Honored Contributor
This is a good question for a math teacher. I know it can be done but I can't tell you the steps.
My concerns would be best described as analog vs digital.
Do you want to take a sample point every 1% of the line and use straight lines between? This would be the digital version.
Analog method:
The other method that comes to mind is creating an equation and graph in excel, then revision the equation until it matches. In this case the equation would be an up-side down perabola y=f(X^2). I would start with coordinate system in the bottom center, and the two tend points. Create a perabola equation that matches the ends. then play with the variables until the curve matches.
This would be really cool if you could say, display the excel graph on top of the cad view while revising the formula.
And since SE effectively has a spreadsheet, and you can link variables to dimensions. It would be a lot of work, but could be done by programming equation into SE.
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## Saturday, May 21, 2016
### 39 47 111 119 666 | 12-year old girl ties paying tug of war at school, May 19, 2016 (May 23 predictive programming?)
So a 12-year old died playing 'tug of war' in 'Pell City' at her school's 'field day'?
Of course 39 has numerology of '12' like her age.
39 = 3+9 = 12
What is this bullshit story really about, eh?
Notice the dates on the sign, May 19 and May 23. In case you forgot already, May 19 was the day the plane disappeared, also typically the 139th day of the year.
Personally, I'm looking forward to 'Honors Day'.
Let's examine our numbers; 39, 47, 119. Do I need to elaborate?
Jews go by the 39-Books in the Old Testament....
http://www.cnn.com/2016/05/20/health/girl-death-tug-of-war-pell-city-alabama-trnd/index.html
Tug of War also connects to '111' and '666' in addition to '39'. Think about how New York connects to all of these letter.
New York = 5+5+5+7+6+9+2 = 39 (Tug of War) (Field Day)
1. Tug of war---111
Every election year is a tug of war, as a tug of war(non-gaming event) is when two equally matched factions are struggling to obtain control over the same objective.
New York, of course, is also 111 and where Hillary is coming from. Could this signify her political defeat?
Of course, with 171 days until the election, could just be a Coincidence.
2. Now the girl is 13.
3. Upon further review, I think this is a finacial collapse piece. 111 for the Tug of War and New York, Madison Avenue is in NY, Wentworth makes me think of financial giant JG Wentworth.
The say she collapsed suddenly with no known health issues, tug of war wasn't even really part of it. Of course there is a GoFraudMe page to pay the family.
4. 12 year old - 1+2+7+5+1+9+6+3+4 = 38
12 year old girl - 1+2+7+5+1+9+6+3+4+7+9+9+3 = 66
5. WILLIAMS INTERMEDIATE SCHOOL" in the English Reduction system equals 122
GOLDEN GATE= 122
POPE FRANCIS= 122
6. ELOUISE HAROLD WILLIAMS" in the English Reduction system equals 98
"ELOUISE AND HAROLD WILLIAMS" in the English Reduction system equals 108
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Nash–Moser theorem
The Nash–Moser theorem, attributed to mathematicians John Forbes Nash and Jürgen Moser, is a generalization of the inverse function theorem on Banach spaces to a class of "tame" Fréchet spaces.
Introduction
In contrast to the Banach space case, in which the invertibility of the derivative at a point is sufficient for a map to be locally invertible, the Nash–Moser theorem requires the derivative to be invertible in a neighborhood. The theorem is widely used to prove local uniqueness for non-linear partial differential equations in spaces of smooth functions.
History
While Nash (1956) originated the theorem as a step in his proof of the Nash embedding theorem, Moser (1966a, 1966b) showed that Nash's methods could be successfully applied to solve problems on periodic orbits in celestial mechanics.
Formal statement
The formal statement of the theorem is as follows:[1]
Let $F$ and $G$ be tame Frechet spaces and let $P:U\subseteq F\rightarrow G$ be a smooth tame map. Suppose that the equation for the derivative $DP(f)h = k$ has a unique solution $h=VP(f)k$ for all $f \in U$ and all $k$, and that the family of inverses $VP: U \times G\rightarrow F$ is a smooth tame map. Then P is locally invertible, and each local inverse $P^{-1}$ is a smooth tame map.
References
1. ^ Hamilton, Richard S. (1982). "The inverse function theorem of Nash and Moser" (PDF-12MB). Bulletin of the American Mathematical Society 7 (1): 65–222. doi:10.1090/S0273-0979-1982-15004-2. MR 0656198.. (A detailed exposition of the Nash–Moser theorem and its mathematical background.)
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# Variation of parameters
mathematics
Variation of parameters, general method for finding a particular solution of a differential equation by replacing the constants in the solution of a related (homogeneous) equation by functions and determining these functions so that the original differential equation will be satisfied.
To illustrate the method, suppose it is desired to find a particular solution of the equation y″ + p(x)y′ + q(x)y = g(x).To use this method, it is necessary first to know the general solution of the corresponding homogeneous equation—i.e., the related equation in which the right-hand side is zero. If y1(x) and y2(x) are two distinct solutions of the equation, then any combination ay1(x) + by2(x)will also be a solution, called the general solution, for any constants a and b.
The variation of parameters consists of replacing the constants a and b by functions u1(x) and u2(x) and determining what these functions must be to satisfy the original nonhomogeneous equation. After some manipulations, it can be shown that if the functions u1(x) and u2(x) satisfy the equations u1y1 + u2y2 = 0and u1y1′ + u2y2′ = g,then u1y1 + u2y2will satisfy the original differential equation. These last two equations can be solved to give u1′ = −y2g/(y1y2′ − y1y2)and u2′ = y1g/(y1y2′ − y1y2). These last equations either will determine u1 and u2 or else will serve as a starting point for finding an approximate solution.
Computing device for solving differential equations. Its principal components perform the mathematical operation of integration (see also integrator). The American electrical engineer...
Mathematical statement containing one or more derivatives —that is, terms representing the rates of change of continuously varying quantities. Differential equations are very common...
Way of graphically representing the solutions of a first-order differential equation without actually solving the equation. The equation y ′ = f (x, y) gives a direction, y ′,...
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# String making introduction mega-thread. New to this craft? Start here!
This is a megathread with information and resources to get anyone started on this craft. I will list some resources, recipes and available materials at the bottom of this post.
There are many different kinds of strings, many different types of thread, many different types of rigs, and many different types of recipes. I will share some basic recipes and rig setup ideas to get you started, and from there you can experiment as you please until you land on a recipe that suits your preferences. If you are like me, you will develop a basic recipe that you occasionally make tweaks to because the process is fun, and there are always new ideas to tweak or improve recipes and rig setups.
A summary of the process:
So you have two anchor points that are 12 feet apart, and a center anchor that is a little bit less than halfway from your starting anchor point. This is your most basic rig.
You tie off your thread to your starting anchor point and wrap the thread around the farthest anchor point and back 6 to 8 times. From your starting anchor, to the farthest anchor and back is 1 full wrap. So usually anywhere from 6 to 8 of those.
Wraps – the number of wraps is one aspect of developing a recipe that is easy to play around with. The fewer wraps, the thinner the final string will be. The more wraps, the thicker the final string will be. You can even do half wraps. You’d be surprised at how much difference a half wrap can make.
Let’s talk reduction. A typical string reduction is anywhere from 10-15%. For the purpose of this summary we will just go with about a 12% reduction. This is where math comes in. So 12 feet is 144 inches. 12% of 144 is 17.28. For Simplicity sake, we can round 17.28 down to 17.25, which is 17 1/4 inches. Simply use your tape measure to make a mark 17 1/4 inches. away from your second anchor point.
Now we reduce. Make sure your drill has a hook loaded in the tip, and is set to spin clockwise, and spin spin spin! You will notice that the string begins to reduce in length, hence “reduction”. You want to keep the string taught but not too tight while it’s reducing, just enough tension to keep the string from binding up on itself. And stop the drill when you reach your reduction mark.
Reduction – reduction is another aspect of the recipe that can be played with and adjusted. some people like a really high amount of reduction and others prefer less reduction. More reduction will give you a tighter wound completed string, and less reduction will give you a looser wound completed string.
Remember the third Anchor Point that was placed slightly off-center between the two main anchors? Here’s more math for you (yay!). The location of that center anchor is going to depend on how much you choose to reduce your string by, because it is going to be half of your reduced string length. In this case, we reduce the string by 17.25 inches. Subtract 17.25 from 144 and we now have a string 126.75 (3/4) inches long. Half of 126.75 is 63.375, which is 63 3/8 inches. So, measure from your starting anchor 63 3/8 inches, and that’s the center point for your reduced string. Still with me? Hahaha
You are going to take the end of the string attached to your drill and double it around your center anchor and secure it to the starting anchor point. Keep the string taut during this phase, this is very important. If you let any slack in the string during this phase, you will end up with a rat’s nest of kinked and twisted string that probably won’t be salvageable. Set your drill to reverse and hook the string from the center anchor, lift it off the anchor and begin to relieve the torsion with your drill. You will feel when the string torsion has been completely released because while you are keeping the string taut, you will feel it stretch out and then begin to reduce again. By that point you have adequately relieved the torsion.
Doubling – this is what gives you that closed loop on the end of the string that slips around the bearing or axle.
Now pinch the string close to your starting Anchor Point and cut it with your scissors. If you don’t hold on to the string, it will just unravel quickly so you need to hold on to the string while you cut it. Tie a knot in it and vualla! That’s basically how a yo-yo string is made. The final string will most likely be several inches longer than you actually need, so put it on your yoyo and tie it off at its proper length with a loop.
If you want extra credit, figure out how long you typically like your strings to be, and make a mark that distance from your starting anchor so that you can just cut it at the length you like it to be right off the rig.
1. So, where to start…
Here’s a list of things you’ll need to make a yoyo string and/or a rig
• A tape measure.
• Something to mark with, like a marker, pen or pencil.
• At least 16 ft of space (you will need 12 feet of space for the string itself, so any extra wiggle room would be nice).
• A few hooks, or something to anchor the threads to. A nail, screw, or even a doorknob would work.
• A drill of somekind. I used both a battery drill and a Dremel. Just be aware that a Dremel will spin much much faster than a drill.
• Some hooks. I use the tiny little screw in hooks and small S hooks
• Some thread. You can use any thread, but there are specific threads that are tried-and-true in this craft. More on that below.
• Something to cut the thread with, like some scissors or a razor knife.
• Some people make their rig using a long 2Ă—4, I made mine on an actual countertop in my garage. I just screwed all my anchor points right into the countertop
• And one last thing, it is not mandatory but I would suggest keeping a journal or a log to keep up with you’re recipes. There’s nothing more frustrating than spinning up a string that performs beautifully, but you cannot replicate it because you can’t remember exactly what you did.
2. How to set up a basic rig
You will need to start by establishing two anchor points. these two anchor points need to be exactly 12 feet apart. So if you need to, have someone help you hold the end of your tape measure right on your first anchor, then make a mark at 12 ft with your marking utensil and then place your second anchor right on that spot. I use screws as anchor points (3 inch green deck screws have a nice section of threadless shaft just below the screwhead).
Once you have your end anchors in place, figure out how much you want to reduce your strings by, and use the formula I laid out above to put a mark at your reduction sweet spot. The formula is basically this- What is [x %] of [length of rig in inches]. For clarification, the example I shared above would look like- What is 12% of 144 inches? The answer is 17.28 inches.
• Pro tip- If you suck at math like I do, you can always ask Google, lol.
Next, find your halfway point of your fully reduced string and put your third (center) anchor in place.
And that’s basically it as far as setting up the rig goes.
So, for unresponsive play there are pretty much two main types of thread used- polyester and nylon. There are also several different types of each, all with unique performance properties. I’ll share a few of each, and this little list isn’t exhaustive by any means:
There are several different brands of spun poly, and the spun poly is all more or less the same afaik. Now, trilobal on the other hand should be in a category all of its own. Spun poly is your basic bulk string material. I’ll list some pros and cons of each below.
The nylons listed are all what is known as “textured”, and are more or less unique brand specific spools. These nylons are all technically the same, but they all act and feel noticeably different during play. A lot of people say they don’t like bulky nylon, but that’s all I had access to from my local store for a while and I learn to like it in a blend.
### Spun poly
PROS
• Cheap
• Classic feel
• Largest selection of colors, especially neons
• Produces a soft string
• Slip knot stays pretty secure during play
CONS
• Wears out pretty fast
• Okay for whips and slacks
### Trilobal
PROS
• Lasts. For. Ever. Seriously, this stuff does not wear out!
• Decent selection of neon colors
• Holds neutral torsion very well during play
• Whips and slacks beautifully
CONS
• Very course on the fingers, not pleasant
• Slip knot doesn’t like to stay tight
• Kind of pricey
### Textured nylon
PROS
• Very soft on the fingers
• Has this lively springiness to it
• Holds neutral torsion pretty well during play
CONS
• Wears out kind of fast (not as fast as spun poly, though)
• Somewhat weak whips and slacks
A few more thoughts on thread before we move on. Threads come in different thicknesses, so if you make a recipe that is 8 wraps of trilobal poly, the same recipe using spun poly might actually give you a thinner completed string. You will have to play around with your recipes and find your preferred number of wraps for each different material.
Okay, so now that I have given a basic run through of typical thread used for spinning, let’s explore blending these threads.
• just a quick note on blending before we go any further- if you are new to this or if you’ve never spun strings before, I would suggest you go for a pure string first. That is, 100% polyester or 100% nylon. Blending materials can get pretty complicated pretty quick. But hey, if you feel led to start with blending, by all means go for it. Don’t let me stop you.
In short, the best reason for blending different threads is because it gives you the advantages of each of these threads in one string. For instance, my personal favorite recipe is a 50\50 blend of trilobal polyester and textured nylon. This gives the advantage of the longevity and whippiness of the trilobal, and the silky soft springy liveliness of textured nylon. You can explore with different ratios of each thread and also different thread types too.
Alright, this is where it gets kind of complicated and time-consuming. Pre twisting your blended thread actually makes a pretty noticeable difference. It allows the two materials to blend more evenly before you start reducing it. This seems like a good place for me to share my personal recipe, feel free to use it and tweak it as you please. I will explain the pre twisting idea below.
To start, I use a doubled rig set up at 12 feet, with a 17 1/4 inch (about 12%) reduction. Basically, two runs right next to each other, with only one center anchor. This rig will make 1 string at a time, I’ll explain below.
My recipe
My recipe does not yield the softest strings on the planet, but my working hands don’t mind them. These are high performance, long lasting strings. They keep their torsion well during play and also keep their shape nicely during slacks and whips. Don’t believe me? Try them.
• 50\50 trilobal and textured nylon (I’m currently using maxi lock, SO to Spool Thread for the recommendation)
• 7 wraps
• Pre twisted in halves (I’ll explain this below)
• 17 1/4 inch reduction
So, remember the “two runs right next to each other” thing I mentioned above? To start I do 3 wraps on one set, and 4 wraps on the other (3+4=7, get it? 7 total wraps). I run both, the nylon and polly, at the same time. So remember that each wrap is going to count as 2 since you’re using twice as much string. So, to clarify, tie off both threads (trilobal and nylon) to a starting anchor, and do your 3 wraps (from your starting anchor, to the second anchor, back to homebase, and back to the second anchor counts as 3 full wraps). Then go to your next set and tie off both threads to a starting anchor, and do 4 wraps (from your starting anchor, to your second anchor 2 times counts as 4 wraps).
wraps – you don’t want your thread sagging, but you also don’t want your threads too tight. Find a nice happy medium where they are snug
Now the pre twist. Take your drill with the hook in it and set it for clockwise. Hook the first cluster of threads with the drill and spin it at full speed for about 10 seconds, keeping the threads taut but not pulled too tight. There should be no visible reduction at this phase, only twisting. Repeat this step for the other run.
Next, consolidate you runs together by carefully transferring one end at a time so that you now have 1 string.
Spin spin spin! Reduce it by 17 1/4 inches, double it, and relieve the torsion. this is where you can use the S hook- you can put the S hook on the end of the string before you reduce it, and just link it to the hook in the end of your drill. this makes it easier to attach to your starting anchor after you double the string in half.
Boom, cut and tie it off and there you have it! A pre twisted, 50\50 blended, high performance, long lasting string.
5. Fixie strings
So, fixed axle yo-yos need string made out of 100% cotton. The reason is because polyester and nylon are actually both forms of plastic, and cotton is a natural fiber. The friction of the fix axle on the string will actually cause nylon and polyester to melt, eventually sending your yoyo flying. Cotton does not melt under friction. Now, I’ve heard of people blending cotton and polyester though, these are known as slick strings. I have only spun 100% cotton strings and cannot personally vouch for the process of blending cotton and poly. I’m assuming that you could blend two separate spools, one of each, or you could just use a cotton poly thread. I’m pretty sure they sell blended spools of thread.
anyway, the process for making fixie strings is pretty much the same, but remember that the thread will probably be a different thickness than what you’re used to, so you may have to do a couple of trial runs before you get a string that is satisfactory. Fixie strings need to be right in that sweet spot- too thin and the yoyo won’t respond, too thick and you won’t be able to do tricks very easily because it’s too responsive. So play around with it and see what you can create!
6. Some final thoughts and resources
I was introduced to spinning after watching Dylan Kowalski’s video on how to make the best yo-yo string. It’s a great example of resourcefulness.
The owner of the late great (but not forgotten) boutique string company Spool Thread was awesome enough to share his recipe after he closed down operations. These strings were highly coveted for a season of time, and this is the most in depth recipe I’ve ever seen. He also has a very in-depth list of materials and even has a section that talks about dying strings if you want to go that route. Please respect his Attribution Creative Commons Liscense if you decide to glean anything from his recipe.
Airetic strings actually has part of their website devoted to string school with several really great videos showing how to make strings with a rig and without a rig.
And nd here is a ridiculously ingenuitive rig that you may be able to glean some ideas from.
Head over to r/StringMakerz on Reddit if you don’t already know about them, it is a sub dedicated to string making kind of similar to this forum. It’s just another means of interacting with this community of string makers. Go post some pictures of your stuff, go ask questions, go share new ideas, go get inspired!
If you want to go the advanced string route, I suggest staying away from blending materials. Try to use the same material thread in different colors. But hey, if you feel led to blend different materials for any of these advanced processes, go for it!
shokata made a really epic walkthrough of making fade strings and shared it to Reddit. SunsetRiderRadi made a video explaining shokata’s walkthrough of fades. Here’s an example of a fade string. It fades in one way or another from end-to-end
### Twists
Twists are pretty cool looking too. They look like a candy cane stripe.
8. Final thoughts
Pro tip – if you make a string that has a slippy Slipknot that will not stay secure to your finger during play, take the top two inches of string (including the knot) and rub it over a piece of beeswax a couple of times. That will make it nice and tacky and snug during play.
I am not a professional string maker, but I do make all my own strings. the processes that I’ve shared may be kind of biased to my own methods and preferences, but they work. The purpose of this was to give you a thorough introduction to string making. There are methods and materials that are tried-and-true, but don’t let that stop you from exploring new possibilities! If you can get the basic process down, you can use that as a foundation to try new things. String making is a craft, an art form, a form of expression. One wouldn’t think that a string can really make a significant difference in performance, but once you start exploring different materials and different recipes, it’s actually pretty surprising how different strings can be.
If you are new to this craft, welcome! I hope you found this inspiring! If you are a seasoned string maker, I also hope that this was inspiring to you!
33 Likes
You don’t need to a rig as long as you’re making normal length string and have a relatively high anchor point (I use a cabinet handle). After twisting, take a stringless yoyo, unhook the string from the drill, and carefully hold the yoyo low while you bring the string ends together. Then unhook the other end from the anchor and just hold it high and let it twist. Measuring becomes a bit tricky, but if you blend completely different color threads you can use those to measure how twisted it is (e.g. spin until it’s 8 twists along the face of a quarter).
I won’t claim this is a superior method, but it is easier to get started with to start trying out some blends. Also, when experimenting, don’t stick to a single reduction% (or measurement). A blend that’s terrible at one twist level can play beautifully when twisted a bit less.
-Trilobal all purpose poly (different from embroidery poly) is a good base - it’s heavy, binds well, whips well.
-Embroidery poly is nice, colorful, but really slippery. A little bit is nice to add color and keep your string a little slippery, but too much is bad.
-Bulky Nylon is good for getting some nylon into a blend, which adds durability, good tension, and is a bit soft/slippery.
-Rayon is good to add just a bit of to a blend (2 threads), it has basically no stretch at all, it adds whippiness and will make a slippery blend bind better. Try to make the rayon thread section slightly longer than the other threads (e.g. stretch the other threads and use that much rayon).
2 Likes
I mean, I basically covered this in the OP. I just offered a base example that I have tested that works, so as not to confuse people with an endless amount of variables. I also encourage people to experiment with different variables.
I also covered pros and cons of all of those thread types except for rayon. I’ve never used rayon so I don’t know what it’s like.
I’m just not sure what the purpose of your comment was
3 Likes
I was just sharing my rigless method which doesn’t rely on keeping track of any lengths or percentages. I didn’t see it being covered, I guess maybe in one of the related videos?
As far as the threads, I was just listing the ones I’ve used based on my experiences (also All purpose poly thread and embroidery poly thread are significantly different, which wasn’t mentioned above). And if they overlap or if they don’t it doesn’t matter because I’m just sharing my personal experiences with different threads, there’s no right or wrong.
1 Like
Question! Question from the back of the class!
I have a suspicion that trilobal polyester might wear a response pad faster than say wooly nylon because it feels a bit more rough.
Secondly, trilobal polyester thread by itself feels SIGNIFICANTLY different as a yoyo string to me. Extremely “whippy”. When I do any kind of slack trick like a hook or anything that has a whipping motion I can hear the string zooming through the air in a way that, for example, polyester Kitty string just doesn’t do. Is this how it should feel? It’s even to the point that sometimes if I do a slack bind it’ll slap against my hand painfully (like a whip!).
I suspect that the 50/50 blend of trilobal polyester and nylon mentioned above might reduce that effect slightly.
Love the thread though. Extremely useful information all around.
1 Like
I’m sure that courser threads wear the response faster, but I had never actually thought about it. You should do an experiment! Make a string out of pure nylon and a string out of pure trilobal string them up on different yo-yos both with fresh new response pads and see which one last longer. Science!
And as far as using trilobal, I don’t find it very pleasant at all by itself. It is extremely whippy, but it is really rough. It is so dense that it offers the perfect amount of backbone to a nylon when used in a blend. Because nylon is lightweight and silky soft, the two play off of each other very well, magnifying one another’s strengths. The trilobal also tames the nylon a little bit, making it less springy.
Another reason I like to use these two in a blend is because they both have similar life spans. It feels weird to me using a regular poly in a blend with nylon because the regular poly wears out so quickly and it just feels like a waste. It also doesn’t give nylon very much extra weight, it mainly just reduces the springiness of the nylon.
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The different thread types effect the whip as well as tension or how much you reduce it. Also the thread count and if your mixing thread types the ratios. Everything effects how it whips- fast or slower. The fun part is finding what works best for you and your yoyos.
Example I make a thin version that works better and is less snaggy for my older clyw and generals.
You can transform a good yoyo into an amazing one with the right string.
3 Likes
So…Mr. Smileypants.
I’ve done a bit of testing with a few different strings, to determine if there’s a difference.
Based on my initial testing there seems to be a difference…but frankly, it’s probably not enough to really matter. What I think I found might surprise you though…
100% trilobal poly seems to be the most grabby kind of string. It’s very coarse, but snappy on the returns. 100% nylon (maxi-lock brand) seems to be a LITTLE bit more slippy (slippy? that’s a word right?) on the binds. 100% woolly nylon is even more so.
As such I set up four yoyos with new pads and four strings (4 skyvas, since it’s the only one I had multiple of…not the heaviest yoyo, but enough) 1 trilobal poly, 1 nylon, 1 woolly nylon, and 1 kitty fat.
I threw them for about 3 weeks straight just about two hours a day so .5 hours on each at least.
It seems to me that the nylon strings actually wear the pad quickest. I suspect it’s because the binds actually cause the string to drag on the pads more than with the poly…the poly just catches and snaps back…I’m guessing though.
4 Likes
Cool man! I glad you’re experimenting! I never considered there being a noticeable difference of pad wear, that was interesting.
2 Likes
Great guide! good job!
Some Corrections:
Embroidery polyester is slicker, smoother, and has more sheen than both wooly and textured nylons, that is the design of trilobal fibers and the reason it’s used for embroidery. (And yes, almost all embroidery polyester is trilobal, and it’s also available in almost ever color, including neons) if you’re finding that it’s grabbing too much/course then you’re using too much tension.
All brands are different, with different wt. plys, so you might think your blending 50/50 but you’re actually not, it goes by the wt. and string should be balanced. Also pay attention to the ply on your thread.
The pre-blending theory is interesting, but the finish actually depends more on the initial tension on the string. Say one string was less taught when you first tied it on the hook, that color/fiber will be more abundant in the final product. Also things like keeping the loop continuous as to crossing threads can change the final product.
General info (not related to fiber material)
The heavier the string the faster the whip.
Less tension-slick and floppy, less wear on pad
More tension-rough and stiff, more wear on pad
If you’re 5’8” or taller, you can cut out that middle “anchor” by using outstretched arms, just hand over the string.
String making is so fun and easy, anyone who throws should make their own here’s some of my strings, and lots of seconds if anyone wants some free ones
8 Likes
Excellent info man! Thanks for sharing!
2 Likes
First time posting my strings. These are a few from a lazy Sunday. All are 33% wooly nylon 66% spun poly. Thanks for lookin!
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Is that a nickel plated OD rebirth?
2 Likes
Markmont Next
3 Likes
Where do you normally order your supplies?
2 Likes
Right here. They even have a six pack of neon/uv reactive.
6 Likes
Wow. Not a bad deal. You could string a lot of yoyos for \$20.
2 Likes
It’s yield was a great number- something like 60+ from each spool(I don’t remember exactly how many).
One thing to note is it has 3 shades of pink.
1 Like
im gonna have to try this thread out, I use maxi lock, mainly because I have hard time finding bulk thread with neon colors. If yall have any suggestions to were I could find a wide range pls let me know, f your willing o share your secrets, thx
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I loved making my own string. It’s draining though and gets boring fast when you can only make one at a time
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# kbit/Min to Tibit/Day - 512 kbit/Min to Tibit/Day Conversion
expand_more
Input Kilobits per Minute (kbit/Min) - and press Enter.
kbit/Min
Sec
Min
Hr
Day
Sec
Min
Hr
Day
S = Second, M = Minute, H = Hour, D = Day
label_important RESULT sentiment_satisfied_alt
512 kbit/Min =0.00067055225372314453125 Tibit/Day
( Equal to 6.7055225372314453125E-4 Tibit/Day )
content_copy
Calculated as → 512 x 1000 ÷ 10244 x 60 x 24 smart_display Show Stepsexpand_more
Below chart table shows the amount of data that can be transferred at a constant speed of 512 kbit/Min in various time frames.
Transfer RateAmount of Data can be transferred
@ 512 kbit/Minin 1 Second0.0000000077610214551289876302083333333333 Tebibits
in 1 Minute0.0000004656612873077392578125 Tebibits
in 1 Hour0.00002793967723846435546875 Tebibits
in 1 Day0.00067055225372314453125 Tebibits
## Kilobits per Minute (kbit/Min) to Tebibits per Day (Tibit/Day) Conversion - Formula & Steps
The kbit/Min to Tibit/Day Calculator Tool provides a convenient solution for effortlessly converting data rates from Kilobits per Minute (kbit/Min) to Tebibits per Day (Tibit/Day). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Kilobit) and target (Tebibit) data units.
Source Data Unit Target Data Unit
Equal to 1000 bits
(Decimal Unit)
Equal to 1024^4 bits
(Binary Unit)
The conversion from Data per Minute to Day can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
The formula for converting the Kilobits per Minute (kbit/Min) to Tebibits per Day (Tibit/Day) can be expressed as follows:
diamond CONVERSION FORMULA Tibit/Day = kbit/Min x 1000 ÷ 10244 x 60 x 24
Now, let's apply the aforementioned formula and explore the manual conversion process from Kilobits per Minute (kbit/Min) to Tebibits per Day (Tibit/Day). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Tebibits per Day = Kilobits per Minute x 1000 ÷ 10244 x 60 x 24
STEP 1
Tebibits per Day = Kilobits per Minute x 1000 ÷ (1024x1024x1024x1024) x 60 x 24
STEP 2
Tebibits per Day = Kilobits per Minute x 1000 ÷ 1099511627776 x 60 x 24
STEP 3
Tebibits per Day = Kilobits per Minute x 0.0000000009094947017729282379150390625 x 60 x 24
STEP 4
Tebibits per Day = Kilobits per Minute x 0.0000000009094947017729282379150390625 x 1440
STEP 5
Tebibits per Day = Kilobits per Minute x 0.00000130967237055301666259765625
By applying the previously mentioned formula and steps, the conversion from 512 Kilobits per Minute (kbit/Min) to Tebibits per Day (Tibit/Day) can be processed as outlined below.
1. = 512 x 1000 ÷ 10244 x 60 x 24
2. = 512 x 1000 ÷ (1024x1024x1024x1024) x 60 x 24
3. = 512 x 1000 ÷ 1099511627776 x 60 x 24
4. = 512 x 0.0000000009094947017729282379150390625 x 60 x 24
5. = 512 x 0.0000000009094947017729282379150390625 x 1440
6. = 512 x 0.00000130967237055301666259765625
7. = 0.00067055225372314453125
8. i.e. 512 kbit/Min is equal to 0.00067055225372314453125 Tibit/Day.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Kilobits per Minute to Tebibits per Day using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Kilobit ?
A Kilobit (kb or kbit) is a decimal unit of digital information that is equal to 1000 bits. It is commonly used to express data transfer speeds, such as the speed of an internet connection and to measure the size of a file. In the context of data storage and memory, the binary-based unit of Kibibit (Kibit) is used instead.
arrow_downward
#### What is Tebibit ?
A Tebibit (Tib or Tibit) is a binary unit of digital information that is equal to 1,099,511,627,776 bits and is defined by the International Electro technical Commission(IEC). The prefix 'tebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'terabit' (Tb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Kilobits per Minute (kbit/Min) to Tebibits per Day (Tibit/Day)
Apply the formula as shown below to convert from 512 Kilobits per Minute (kbit/Min) to Tebibits per Day (Tibit/Day).
A B C
1 Kilobits per Minute (kbit/Min) Tebibits per Day (Tibit/Day)
2 512 =A2 * 0.0000000009094947017729282379150390625 * 60 * 24
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Kilobits per Minute (kbit/Min) to Tebibits per Day (Tibit/Day) Conversion
You can use below code to convert any value in Kilobits per Minute (kbit/Min) to Kilobits per Minute (kbit/Min) in Python.
kilobitsperMinute = int(input("Enter Kilobits per Minute: "))
tebibitsperDay = kilobitsperMinute * 1000 / (1024*1024*1024*1024) * 60 * 24
print("{} Kilobits per Minute = {} Tebibits per Day".format(kilobitsperMinute,tebibitsperDay))
The first line of code will prompt the user to enter the Kilobits per Minute (kbit/Min) as an input. The value of Tebibits per Day (Tibit/Day) is calculated on the next line, and the code in third line will display the result.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation.
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# Daily Puzzle Questions (29-06-2020)
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Here you will get 2 to 5 Puzzle Questions Daily Monday To Sunday. Daily Visit PuzzlesHuB to solve these puzzle questions on a daily basis to improve your score in the reasoning section. All the Puzzles provided here are good quality puzzles and based on the real exam level. If you like these puzzles & want us to continue to provide you quality puzzles, please share our website with all your Aspirants friends.
## Daily Puzzle Questions In English Practice Sets :
### Pre Level Puzzle :
Eight friends, B, C, D, E, F, G, H and I, are sitting around a square table in such a way that four of them sit on four corners of the square while four sit in the middle of each of the four sides. Each of them likes different colours, viz Red, Blue, Green, White, Pink, Black, Brown and Orange. The ones who sit on the four corners face the centre while those who sit in the middle of the sides face outside, but not necessarily in the same order.
B, who likes Blue, faces the centre and sits third to the right of G. F, who faces the centre, likes Orange and, is not an immediate neighbour of G. Only one person sits between G and H and they like Pink and Green respectively. E, who does not like Black, sits second to the right of C, who likes neither Red nor Black. E faces the centre. D is not an immediate neighbour of B. I likes Brown.
Solution :
Click Here To Watch Solution
### Mains Level Puzzle :
Eight persons i.e. B, J, P, E, T, V, R and M are sitting around a square table. After completion of their graduation, they are going to join the different company in different months of the same year. These companies are GT, HCL, CTS, HP, MS, CP, Infy and TCS but not in the same order. Now different joining months are March, June, August, December, January, April, February and November but not in the same order. Four persons are seated at the corner of this table and face inside and Rest four persons are seated at the middle of each side of this table and faces outside.
T is not going to join in June. The one who is going to join TCS does not sit near to V. The one who sits at one of the corners of this table is going to join in August and sits 3rd left of the one who is going to join MS. There are two persons sit between P and the one who is going to join in August. B sits 2nd right of P who is going to join in April. V is not the immediate neighbour of the one who is going to join in August. The one who is going to join in March sits 2nd left of the person who is going to join in that month which has fewer days compared to all other months. P sits opposite to the one who is going to join in February. E who is going to join in December sits immediate right of P. The one who is going to join CTS sits opposite to the one who is going to join HP. There are two members sit between E and V who is going to join in January. M is going to join TCS but sits diagonally opposite to the one who is going to join CP. The one who is going to join in November sits immediate right of R. The one who is going to join HP sits near to E. The one who is going to join in March does not want to join Infy and GT. The one who is going to join GT sits just near to R.
Solution :
Click Here To Watch Solution
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# Area of a Polar Graph
This is an extra credit problem for a take home test, so i will understand if no one feels comfortable helping me out, but any advice is greatly appreciated
## Homework Statement
Compute the area enclosed by one loop of the graph given by r = sqrt(sin(3{theta}))
see above
## The Attempt at a Solution
The graph makes 3 loops, so i tried finding the area from (0, 2{pi}) but all come up with is 0
Thanks again!
Related Calculus and Beyond Homework Help News on Phys.org
Mark44
Mentor
You get one loop as theta ranges from 0 to pi/3.
Thank you Mark. Is finding the finding the area of the entire graph, then dividing that answer by the number of loops a viable method for this type of problem?
Mark44
Mentor
You could do it that way, I suppose, but it makes more sense to me to get the area within one loop and multiply it by the number of loops. Keep in mind that the integrand is undefined for theta in [pi/3, 2pi/3], because of the square root.
The formula for the area [ A = (1/2)*r2 ] effectively eliminates the square root, so i dont quite understand how its undefined from [pi/3, 2pi/3] because of the square root.
However after proving to myself that r = 0 at 0 and pi/3, I have successfully found the area of one loop. Thanks again for all the help Mark
Mark44
Mentor
I was just looking at r in your first post, and wasn't thinking about the integral.
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If x is positive, which of the following could be the
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If x is positive, which of the following could be the [#permalink]
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If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2
I. x ^2 < 2x < 1/x
ii. x ^2 < 1/x < 2x
iii. 2x < x ^2 < 1/x
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09 Apr 2006, 17:33
bewakoof wrote:
If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2
I. x^2 < 2x < 1/x
ii. x^2 < 1/x < 2x
iii. 2x < x^2 < 1/x
i. x^2 < 2x < 1/x is a correct order if x <0.5
ii. x^2 < 1/x < 2x is not a correct order no matter what the value of x is.
iii. 2x < x^2 < 1/x is a correctone if 2>x >1.
So I and III.
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09 Apr 2006, 19:01
sm176811 wrote:
I and II
OA?
what set of numbers satisfied II?
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09 Apr 2006, 19:15
bewakoof wrote:
prof, check again
ok, now i go with I, II, and III.
i. x^2 < 2x < 1/x is a correct order if x <0.5
ii. x^2 < 1/x < 2x is also a correct order if 0.70 < x < 1.00.
iii. 2x < x^2 < 1/x is a correctone if 2>x >1.
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09 Apr 2006, 19:20
Professor wrote:
bewakoof wrote:
prof, check again
ok, now i go with I, II, and III.
i. x^2 < 2x < 1/x is a correct order if x <0.5
ii. x^2 < 1/x < 2x is also a correct order if 0.70 < x < 1.00.
iii. 2x < x^2 < 1/x is a correctone if 2>x >1.[/quote]
thats not right
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09 Apr 2006, 19:42
Whats OA... I can provide the nos for II.. will post back l8r
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09 Apr 2006, 21:01
bewakoof wrote:
Professor wrote:
bewakoof wrote:
prof, check again
ok, now i go with I, II, and III.
i. x^2 < 2x < 1/x is a correct order if x <0.5
ii. x^2 < 1/x < 2x is also a correct order if 0.70 < x < 1.00.
iii. 2x < x^2 < 1/x is a correctone if 2>x >1.[/quote]
thats not right
oh, sorry buddy. i overlooked as to the reverse order.
i and ii are correct.....
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09 Apr 2006, 21:09
We could solve the set in the following way:
Case I: Solve for x^^2 < 2x and 2x < 1/x.
====
You would get x < 2 and x < 1/sqrt(2).
So, x < 1/sqrt(2) satisfies the set.
Case II: Solve for x^^2 < 1/x and 1/x < 2x.
=====
You would get x^^3 < 1 and 2x^^2 > 1
So, for all real x, x <1 and x > 1/sqrt(2).
So, 1/sqrt(2) < x < 1 satsifies the set.
Case III: Solve for 2x < x^^2 and x^^2 <1/x.
=====
You would get x >2 and x^^3 < 1.
For all real x, you must have x < 1 (from the second equation).
Now x < 1 and x >2 are incompatible.
So III does not hold good for all real x.
So I & II are correct.
Caveat:
=====
Be cautious not to select values such as x =1, x = 2.
For these values, some of the functions yield the same value e.g. for x =1, x^^2 = 1/x = 1, for x = 2, 2x = x^^2 = 4.
So you would not get the desired relationships.
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Thanks,
Zooroopa
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10 Apr 2006, 02:07
I. x ^2 < 2x < 1/x
passes If x = 1/2
but fails if x = 2
ii. x ^2 < 1/x < 2x
agree with Zooroopa on this
iii. 2x < x ^2 < 1/x
This is also not possible if x = 3/2
3<>2.5<0.66
if x = 5
10<25<>0.2
so I will go with I and II
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10 Apr 2006, 18:17
Whats the trick here to save time?
Should we try different values like 1/2, 1, 3/2, 4 for x and then figure out what satisfies.
I guess these kind of questions are included so that we spend maximum time on them and end up in a mess.
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10 Apr 2006, 18:48
If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2
I. x ^2 < 2x < 1/x
ii. x ^2 < 1/x < 2x
iii. 2x < x ^2 < 1/x
II is possible
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10 Apr 2006, 19:27
Choosing x = 1 and x = 2 are going to violate both I and II.
For x = 1, we would get
I: 1 < 2 < 1 (this is meaningless)
II: 1 < 1 < 2 (this is meaningless as well)
Similar pitfalls exist if you choose x = 2.
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11 Apr 2006, 07:02
choose 3 numbers:
a)one whole number > 2 (aka 3!)
b)fraction < 1 (3/4)
c)fraction > 1 (3/2)
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11 Apr 2006, 07:02
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#### Provide solution RD Sharma maths class 12 chapter 10 differentiation exercise 10.6 question 1 maths textbook solution
Answer: $\inline \frac{d y}{d x}=\frac{1}{2 y-1}$
Hint: The value of y is given as infinite series. If a term is deleted from an infinite series, it remains the same in this case.
Given: $\inline y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \ldots+\infty}}}$
Solution:
Here it is given that,
$\inline y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \ldots+\infty}}}$
This can be written as:
$\inline y=\sqrt{x+y}$
Squaring on both sides, we get:
$\inline y^{2}=x+y$ …(1)
Differentiating (1) w.r.t x,
\inline \begin{aligned} &2 y \frac{d y}{d x}=1+\frac{d y}{d x} \\ &\frac{d y}{d x}(2 y-1)=1 \\ &\therefore \frac{d y}{d x}=\frac{1}{2 y-1} \end{aligned}
Hence, it is proved.
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Showing that a map on a tubular neighbourhood is a diffeomorphism
Let $$S$$ be an embedded submanifold of a smooth manifold $$M$$. Then there exists a tubular neighbourhood $$U$$ of $$S$$ in $$M$$, that is, $$U$$ is the diffeomorphic image of $$\exp$$ restricted to the normal bundle $$NS$$ of $$S$$ of a subset $$V \subseteq \mathcal{E} \cap NS$$, where $$\mathcal{E} \subseteq TM$$ is the domain of the exponential map, of the form $$V = \{(x,v) \in NS : |v| < \delta(x)\}$$ for some positive continuous function $$\delta : S \to \mathbb{R}$$.
Let $$0 \leq t \leq 1$$. Then we can define $$\psi_t : U \to U$$ by $$\psi_t(\exp(x,v)) := \exp(x,tv).$$
Now the book Introduction to Symplectic Topology by McDuff and Salamon claims that this map is a diffeomorphism for $$t > 0$$. I do not quite see why. I mean, this map is injective and smooth, but I fail to see surjectivity or the explicit form of an inverse. Moreover, I guess that this map is a local diffeomorphism by considering its derivative. Can anyone help?
After a quick correspondence with the second author, the proof goes as follows: $$\psi_t$$ is a diffeomorphism onto its image, because $$\psi_t$$ is injective since $$\exp_S$$ is. Moreover, an explicit inverse is given by $$\exp_S(x, tv) \mapsto \exp_S(x, v)$$.
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+0
# inverse functions
+1
128
3
+500
What is the inverse of $f(x)=4-5x$?
May 6, 2021
#1
+964
+1
$$f(x) = y = -5x + 4$$
$$y - 4 = -5x$$
$$\frac{4 - y}{5} = x$$
---> $$\boxed{f^{-1}(x) = \frac{4 - x}{5}}$$
May 6, 2021
#3
+2209
0
Hello, I haven't seen you on in a long time. :))
=^._.^=
catmg May 6, 2021
#2
+2209
+1
replace f(x) with x
x = 4 - 5f(x)
4-x = 5f(x)
f(x) = (4-x)/5
=^._.^=
May 6, 2021
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# asterixdb-notifications mailing list archives
##### Site index · List index
Message view
Top
From "Taewoo Kim (JIRA)" <j...@apache.org>
Subject [jira] [Commented] (ASTERIXDB-1778) Calculating the edit-distance in SimilarityMetricEditDistance class can be improved.
Date Wed, 01 Feb 2017 05:36:52 GMT
```
[ https://issues.apache.org/jira/browse/ASTERIXDB-1778?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=15848035#comment-15848035
]
Taewoo Kim commented on ASTERIXDB-1778:
---------------------------------------
I thought more and one or more conditions need to be applied. Let me get back to you once
I get the condition(s).
> Calculating the edit-distance in SimilarityMetricEditDistance class can be improved.
> ------------------------------------------------------------------------------------
>
> Key: ASTERIXDB-1778
> URL: https://issues.apache.org/jira/browse/ASTERIXDB-1778
> Project: Apache AsterixDB
> Issue Type: Improvement
> Reporter: Taewoo Kim
> Priority: Minor
>
> Local optimization of the current edit-distance handling code in AsterixDB: Can we terminate
the edit distance calculation based on a given edit-distance threshold T? I think the answer
is yes.
> Basic Algorithm:
> For two strings X and Y,
> Init:
> Construct D(len(X)+1=M, len(Y)+1=N).
> D(i,0) = i
> D(j,0) = j
> Iteration:
> For each i = 1 to M
> For each j = 1 to N
> D(i,j) = min( D(i-1,j) + 1 // deletion case
> D(i,j-1) + 1 // insertion case
> D(i-1,j-1) + cost (cost is 2 if X(i) != Y(j),
> cost is 0 if X(i) = Y(j))
> )
> Result:
> D(M,N) is the edit distance value.
> Example:
> Early Termination condition:
> So, for the given threshold T, we can stop the computation early if the values of three
cases are greater than T. That is,
> min (D(i-1,j), D(i,j-1), D(i-1,j-1)) > T
> This holds since if the cost of all possible cases (insertion, deletion, and substitution)
is greater than T, all future operations will be greater than T in any cases.
--
This message was sent by Atlassian JIRA
(v6.3.15#6346)
```
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# Removing Zeros (between non-zero values) and maintaining the Tensor dimensions
Hi Guys,
I would like to remove zero values of a tensor and “join” the non-zero values in each row of a tensor in format [B, C, H, W]. A naive way would to do `out_x = x[x!=0]`, this approach is bad because would destruct the Tensor dimensions. For resume, I would like to transform an input tensor like this:
``````in_x = torch.tensor([[[[0., 0., 2., 20., 250., 0., 0., 0., 0., 0., 250.,
20., 2., 0., 0., 0.],
[0., 0., 2., 20., 0., 0., 20., 250., 0., 0., 250.,
20., 20., 2., 0., 0.],
[0., 0., 0., 0., 0., 0., 2., 20., 250., 0., 250.,
20., 20., 2., 0., 0.],
[0., 2., 20., 0., 20., 250., 0., 250., 20., 0., 20.,
2., 0., 0., 0., 0.]]]])
``````
In an output tensor like this:
``````out_x = torch.tensor([[[[ 2., 20., 250., 250., 20., 2., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0.],
[ 2., 20., 20., 250., 250., 20., 20., 2., 0., 0., 0.,
0., 0., 0., 0., 0.],
[ 2., 20., 250., 250., 20., 20., 2., 0., 0., 0., 0.,
0., 0., 0., 0., 0.],
[ 2., 20., 20., 250., 250., 20., 20., 2., 0., 0., 0.,
0., 0., 0., 0., 0.]]]])
``````
Note that both Tensors have the same shape: (1,1,4,16)
The solution must not contain For Loops or anything else that degrades performance.
Thanks.
Hi Luiz!
At the cost of `n log (n)` time complexity, you can use `argsort()`* and
then use `gather()` to index back into your input tensor:
``````>>> import torch
>>> torch.__version__
'1.9.0'
>>> in_x = torch.tensor([[[[0., 0., 2., 20., 250., 0., 0., 0., 0., 0., 250.,
... 20., 2., 0., 0., 0.],
... [0., 0., 2., 20., 0., 0., 20., 250., 0., 0., 250.,
... 20., 20., 2., 0., 0.],
... [0., 0., 0., 0., 0., 0., 2., 20., 250., 0., 250.,
... 20., 20., 2., 0., 0.],
... [0., 2., 20., 0., 20., 250., 0., 250., 20., 0., 20.,
... 2., 0., 0., 0., 0.]]]])
>>>
>>> in_x.gather (3, (in_x == 0.0).sort (dim = 3, stable = True)[1])
tensor([[[[ 2., 20., 250., 250., 20., 2., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0.],
[ 2., 20., 20., 250., 250., 20., 20., 2., 0., 0., 0.,
0., 0., 0., 0., 0.],
[ 2., 20., 250., 250., 20., 20., 2., 0., 0., 0., 0.,
0., 0., 0., 0., 0.],
[ 2., 20., 20., 250., 250., 20., 20., 2., 0., 0., 0.,
0., 0., 0., 0., 0.]]]])
``````
*) Except you have to use `sort()` because `argsort()` doesn’t support
`stable = True`).
Best.
K. Frank
1 Like
Thank you very much K. Frank, this works fine. Have a Merry Christmas and a Happy New Year!
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Thursday
May 5, 2016
# Homework Help: math
Posted by Anonymous on Wednesday, May 2, 2012 at 5:20pm.
What would the unit rates be for a store that had 30 customers in 12 minutes?
• math - Ms. Sue, Wednesday, May 2, 2012 at 6:12pm
30/12 = 2.5 customer per minute
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# trig
posted by .
As a hot-air balloon rises vertically, its angle of elevation from a point P on level ground d = 140 kilometers from the point Q directly underneath the balloon changes from 15°10' to 29°30' (see the figure). Approximately how far does the balloon rise during this period? (Give the answer to one decimal place.)
• trig -
For height at lower angle:
tan 15°10' = height1/140
height1 = 140tan15°10'
in the same way:
height2 = 140tan29°30'
rise of the balloon in that change of angle
= 140tan29°30' - 140tan15°10'
= 41.256
I am sure your units are not correct.
How can they even see the balloon from 140 km away ?
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Technical Articles and Newsletters
# Designing a Torque Controller for a PMSM through Simulation on a Virtual Dynamometer
By Dakai Hu, MathWorks
Controlling the torque of a permanent magnet synchronous machine (PMSM) to achieve high levels of accuracy and efficiency is one of the most important targets in high-performance motor drives design. In this article you will learn how simulation performed on a finite element analysis (FEA)-based high-fidelity PMSM machine can help you design motor control algorithms that achieve high torque control accuracy while maximizing the machine’s efficiency.
Motor control engineers commonly use simulation only for algorithm design proof-of-concept studies. These algorithms typically include lookup tables (LUTs) that must be calibrated to achieve the desired performance. Most controller LUTs are obtained by running tests on actual hardware using a dynamometer (dyno). Typically, these include machine validation, characterization, and efficiency tests. While testing on a dyno is the ultimate goal, it can be impractical at times. Several factors need to be considered when it comes to utilizing dyno time―for example, the dyno operation time, operation costs, safety issues, and failures in the machine, the inverter, or the load bank.
Minimizing dyno time is inevitably a high priority for motor control engineers. Model-Based Design helps engineers perform more testing by simulating a “virtual dyno” in Simulink® to reduce hardware testing and overall development time. At a high level, the purpose of performing simulation using the virtual dyno approach is to characterize the PMSM and obtain the machine’s nonlinear flux linkage and torque data, which can then be used to design and implement flux-weakening torque control LUTs.
This article introduces a workflow that uses a virtual dyno to design and test a torque controller for an FEA-based high-fidelity PMSM machine. We’ll consider the following questions:
• What is a virtual dynamometer?
• Why use an FEA-based high-fidelity PMSM machine model?
• How to characterize the high-fidelity PMSM machine?
• How to design a torque controller using the characterization data?
The initial FEA data that we’ll be using for the high-fidelity PMSM machine model was generated from ANSYS® Maxwell® and JMAG®, and provided courtesy of ANSYS and JMAG.
## What is a Virtual Dynamometer?
A virtual dynamometer is a model that brings the concept of the motor dynamometer to desktop simulation. Dynamometers are used to test torque or power of combustion engines or electric machines. Typically, the dyno can operate in all four quadrants on the torque-speed plane, enabling steady-state or transient motoring and generating tests of the coupled machine. Figure 1 shows a schematic of a dyno setup. The machine under test is an interior PMSM (IPM), while the dyno could be a PMSM, an induction machine, or any other machine capable of four-quadrant operation.
Figure 1. Schematic of an actual dyno setup.
On the virtual dyno, a virtual speed or torque source replaces the dyno machine. The machine under test is driven to operate in torque mode by the speed source or in speed mode by the torque source, just as it would be on an actual dyno. In this way, all machine characterization and testing can be performed through simulation.
## Why Use an FEA-Based PMSM Model?
Traditionally, the FEA-based motor design workflow and the motor control development workflow have proceeded separately because motor control engineers do not use FEA data for closed-loop control system simulation. Today, however, FEA simulation data can be imported into Simulink and Simscape Electrical™ for high-fidelity PMSM modeling. The high-fidelity PMSM model contains nonlinear characteristics due to saturation, as well as rotor-position-dependent spatial harmonic components in the back EMF, flux linkage, and torque.
Unlike a conventional linear lumped-parameter PMSM model, the FEA-based PMSM model behaves like an actual motor. This is because instead of having constant parameters for inductances and permanent magnet flux linkage, it has a nonlinear mapping between rotor position, flux linkage, current, and torque.
The FEA-based PMSM model enables control engineers to build a realistic closed-loop simulation and obtain the nonlinear operating characteristics of the machine even before it is manufactured, aligning control engineers with the motor design engineers during the early stages of development. In addition, the model gives motor control engineers the freedom to explore extreme operating conditions without out-of-bounds worries, since all tests are done using simulation in Simulink. The simulation results can guide actual dyno testing once the machine is manufactured. The Design of Experiments (DoE) setup requires an understanding of the machine’s characteristics, and the simulation helps engineers determine the minimum number of points to test.
## How to Characterize the FEA-Based PMSM Model?
The purpose of characterizing the FEA-based PMSM model is to obtain the nonlinear flux linkage information under different operation points. In our case, the operation points are specified by steady-state currents on the synchronous reference frame—that is, by steady-state id and iq operation points.
With the virtual dyno, the speed of the PMSM model can be kept constant, and always below base speed (the speed at which the terminal voltage of the machine reaches its rated modulation index). In the example shown in Figure 2, under 500 V dc bus voltage the base speed is around 1800 rpm.
During the DoE setup, current id and iq commands are executed by a current controller (Figure 2). In the plant model, a speed source acting as the virtual dyno controls the speed of the PMSM model.
Figure 2. DoE setup under the virtual dyno.
For every [id, iq] combination that is commanded, we run the simulation, allow the current response to reach steady state, and then log the following dataset: [id, iq, flux_d, flux_q, torque]. Because there are harmonics and ripples in the FEA-based PMSM machine, it is a good idea to take the average value of a certain duration under steady state before logging the dataset.
For example, in order to characterize the PMSM machine in the motoring region, all [id, iq] combinations specified in Figure 3 are swept. The red curve in Figure 3 indicates the current operation limit, or current-limit circle, for this particular PMSM machine. Although the machine itself will never operate beyond the current-limit circle during normal operation, under the virtual dyno we can push beyond this limit and sweep all marked operation points shown in Figure 3 without worrying about thermal issues in the actual machine.
Figure 3. FEA-based high-fidelity PMSM machine sweeping points.
We could complete characterization by scripting in MATLAB®. Alternatively, we can use Model-Based Calibration Toolbox™ to set up the DoE, automate the sweeping process, and collect data.
## How to Design a Torque Controller Using the Characterization Data?
Now that we have characterization data for the high-fidelity PMSM machine, we can begin designing the torque controller. This involves three steps:
1. Find the optimal operation boundary.
2. Choose lookup table points.
3. Test the torque controller performance.
## Finding the Optimal Operation Boundary
The optimal operation boundary is defined such that under a specific torque command and speed feedback, it encompasses the machine’s optimal operating points. For a linear lumped-parameter PMSM model, the optimal operation boundary can be calculated mathematically using the PMSM machine’s parameters. This calculation will not be accurate for the real machine, however, since the real machine’s parameters change according to the operation points.
There are two ways to calculate a more accurate optimal operation boundary for the high-fidelity PMSM machine model. It can be calculated using the characterized dataset [id, iq, flux_d, flux_q, torque] and MATLAB scripts, or it can be derived using Model-Based Calibration Toolbox. With Model-Based Calibration Toolbox we can design experiments, set up objectives, and log data that meets those objectives. For example, one section of the optimal operation boundary is known as the maximum torque per ampere (MTPA) curve. To calculate this curve, we can use Model-Based Calibration Toolbox to set up a DoE that lets us sweep the current operation points along a current circle and monitor the torque until the maximum torque point is reached. Similar approaches can be used for calculating the maximum current and maximum torque per volt (MTPV) boundaries.
Figure 4 shows the calculated optimal operation boundary. We also plot contours of torque and speed because they serve as either objectives or constraints in the calculation process. We use Curve Fitting Toolbox™ to smooth out the optimal operation boundary and remove outliers resulting from either the nonlinearity of the machine or harmonics in the sweeping data.
Figure 4. Calculation of the optimal operation boundary.
## Choosing Lookup Table Points
The second step for the torque controller design is to locate each operation point inside the optimal operation boundary according to each torque command and speed feedback. The goal is to locate operation points that not only meet different torque commands and voltage constraints but also minimize the stator winding copper losses. In Model-Based Calibration Toolbox we can set maximum torque per ampere (MTPA) as the objective, set the maximum phase current Is_max and voltage Vs_max as constraints, and then run the optimization.
Figure 5 shows a cluster of optimized operation points that meet these objectives and constraints. These optimized operation points will serve as the lookup table data points in the proposed torque controller shown in Figure 6.
Figure 5. Optimized operation points inside the optimal operation boundary.
Figure 6. Schematic of the open-loop torque controller with LUTs.
## Testing the Torque Controller Performance
To test the controller, we run simulations with our virtual dyno. During the test, we initially keep the speed of the machine at 1500 rpm, which is below the base speed of around 1800 rpm. After 1 second, we increase the speed to the point where the machine enters the flux-weakening region. We give independent torque step commands, which are executed by the open-loop torque controller. Figure 7a shows the simulation results.
We can see from the performance waveform in Figure 7a that the torque is controlled to follow the torque step commands very well both below and above base speed.
Figure 7a. Torque controller performance.
Figure 7b. Torque controller performance (with torque ripple zoomed-in).
Figure 7b gives a zoomed-in view of the torque ripple waveform that resulted from utilizing the high-fidelity PMSM model as the plant. (Note that the torque ripple is usually dampened by the mechanical system connecting to the PMSM, and does not pose any concern.)
We’re satisfied by these simulation results. The optimized torque control lookup tables, which are the final results of the proposed workflow, can now be tested on an actual dyno once the machine is manufactured.
By adopting this model-based virtual dynamometer approach, we can start the motor control development work almost concurrently with motor design, and provide useful insights for DoE and initial control lookup tables. The closed-loop simulation platform in this article can also be used to quickly verify motor drives performance without running an actual dyno.
Published 2017 - 93100V00
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# What Is an Example of a Biconditional Statement?
One example of a biconditional statement is "a triangle is isosceles if and only if it has two equal sides." A biconditional statement is true when both facts are exactly the same, either both true or both false. Biconditional statements are created to form mathematical definitions.
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# confused
• Jul 22nd 2012, 02:22 PM
confused
Bob deposited $25k in a savings account @ 10% interest compounded semiannually. At the beginning of year 4 he deposited another$40k @10% compounded semiannually. At the end of 6yrs, what is the balance in his account? I have worked it out several ways, but now I am unsure of which way to go....(Worried)
3yrsx2=6 periods
10% /2=5%
25000x1.7716 (per table in my book)=44290-25000=$19290 interest for 4yrs off$25000
2yrsx2=4 periods
40000x1.4641=58564-40000=18564 interest for 2 yrs off $40000 25000+40000+19290+18564=102,854 I also worked it out this way....... 4yrsx2=8 periods; 5% 25000x2.1436(book table)=53590-25000=$28590 interest for 4 yrs off $25000 40000x1.4641=5864-40000=18564 interest for 2yrs off$40000
25000+40000+28590+18564=$112154 • Jul 22nd 2012, 03:25 PM skeeter Re: confused really? ... does$19290 interest on a 3 year investment of $25000 at 10% make sense? if so, let me know where that institution is ... I'll deposit all I have ASAP. general equation for compounded interest is$\displaystyle A = P\left(1 + \frac{r}{n}\right)^{nt}$where A = account balance after t years paid at an APR of r, compounded n times a year. after 3 years ...$\displaystyle A = 25000(1.05)^6 = 33502.39$add 40000 at the end of year 3 ...$\displaystyle A = 73502.39(1.05)^6 = 98500.23$at the end of 6 years • Jul 22nd 2012, 03:45 PM HallsofIvy Re: confused You could also do this by calculating the balance on the first amount for the full 6 years:$\displaystyle 25000(1+ .05)^{12}= (25000)(1.7958)= 44896.41$the the balance on the second amount for three years:$\displaystyle 40000(1.05)^{6}= (40000)(1.3401)= 53603.82$and then adding:$98500.23.
The powers of "12" and "6" are, of course, the 12 "half years" in 6 years and 6 "half years" in 3 years. And, of course, the "1.05" is the 10% per year divided by 2 since we are dealing with half years, together with the "1" for the original amount.
• Jul 23rd 2012, 09:42 AM
Wilmer
Re: confused
Quote:
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# How do we convert days to years?
## How many years makes 1000 days?
Convert 1,000 Days to Years
d y
1,000 2.7379
1,010 2.7653
1,020 2.7927
1,030 2.8200
## How do you convert 146 days into years?
Answer: 146 days = 2/5 years.
## How long is 1k hour?
1000 Hours is 41 Days and 16 Hours.
## How old is 10000days?
29.5 years
10,000 Days Is Roughly The Orbital Period Of The Planet Saturn. Well, technically speaking, it’s 10,759 days – which equates to nearly 29.5 years. And it’s this philosophical notion of the Saturn Return that Maynard James Keenan once revealed as the meaning behind the title.
## Who many minutes are in a day?
Obviously if we don’t rotate at all, all 24⋅60=1440 times are valid, so the total is 1440+576+288+288=2592 valid rotations out of 4⋅24⋅60=5760, slightly less than half.
## How do you convert days into weeks and years?
Approach :
1. Number of years will be the quotient when number of days will be divided by 365 i.e days / 365 = years.
2. Number of weeks will be the result of (Number_of_days % 365) / 7.
3. Number of days will be the result of (Number_of_days % 365) % 7.
## How many days have you lived if your 14?
At 14 years old, you’ll be 5110 days old! Enter your birthdate to get your exact age in days!
## How many hours are in the year?
8,760 hours
Common year: 8,760 hours x 60 minutes per hour = approximately 525,600 minutes. Leap year: 8,784 hours x 60 minutes per hour = approximately 527,040 minutes.
That is 7/12.
## What fraction of 1 year is 4 months?
1 3
The required fraction is 1 3 Thus 4 months is 1 3 of a year .
People also asking: Can I feed my dog white rice everyday?
## How many days are in a year in fractions?
Thus the average year has 365 + 1/4 = 365.25 days.
## What is 0.4 in a fraction?
2/5
Answer: 0.4 can be written in a fraction as 4/10 or 2/5.
## What is the simplest form of 146 365?
What is 146/365 Simplified? – 2/5 is the simplified fraction for 146/365.
365/146=73:73
## How do you convert days into months and years in Excel?
2. In the Formulas Helper dialog, please select Date from the Formula Type drop-down list, click to highlight Convert days to year month day in the Choose a formula list box, then specify the number cell in the Number box, and finally click the Ok button. Now the calculation result is output in the selected cell.
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# How do you solve (4N+5)/7= -1?
Feb 6, 2016
$\frac{4 N + 5}{7} = - 1$
Multiply both sides of the equation by 7:
$4 N + 5 = - 7$
Subtract 5 to both sides of the equation:
$4 N = - 12$
Divide by 4 to both sides of the equation:
$N = - 3$
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| 4.1875
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CC-MAIN-2019-51
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longest
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http://www.reproducibility.org/RSF/book/tccs/sgk/paper_html/node5.html
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Fast dictionary learning by sequential generalized K-means
Although K-SVD can obtain very successful performance in a number of sparse representation based approaches, since there involves many SVD operations in the K-SVD algorithm, it is very computationally expensive. Especially when utilized in multidimensional seismic data processing (e.g. 3D or 5D processing), the computational cost is not tolerable. The sequential generalized K-means algorithm (SGK) was proposed to increase the computational efficiency (Sahoo and Makur, 2013). SGK tries to solve slightly different iterative optimization problem in sparse coding as equation 3:
(8)
indicates that has all 0s except 1 in the th position. The dictionary updating in SGK algorithm is also different. In SGK, equation 6 also holds. Instead of using SVD to minimize the objective function, which is computationally expensive, SGK turns to use least-squares method to solve the minimization problem. Taking the derivative of with respect to and setting the result to 0 gives the following equation:
(9)
solving equation 9 leads to
(10)
It can be derived further that
(11)
Here, has the same meaning as shown in equation 5 except for a smaller size due to the selection set that selects the entries in that are non-zero.
Since , as constrained in equation 8 then
(12)
Since is a smaller version of row vector and all its entries are all equal to 1, is simply a summation over all the column vectors in . Considering that ,
(13)
Following equation 13, equation 11 becomes
(14)
It is simple to derive that , where denotes the number of elements in the set , or the number of training signals associated with the atom . The th atom in is
(15)
Thus, in SGK, one can avoid the use of SVD. Instead the trained dictionary can be simply expressed as an average of several training signals. In this way, SGK can obtain significantly higher efficiency than K-SVD. In the next section, I will use several examples to show that the overall denoising performance does not degrade when one can obtain a much faster implementation.
2020-04-03
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CC-MAIN-2020-50
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https://www.scribd.com/document/1068557/description-tags-GA200403
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# United States Department of Education Office of Federal Student Aid National Student Loan Data System
Loan Processing and Issuance Fee Rate Change Technical Update GA–2004–03 January 29, 2004 This information is intended for the person in your organization who is responsible for working with NSLDS. If that person is not you, please forward this update to the appropriate person.
Changes to LPIF Rate
As announced in the Dear Financial Partner letter dated July 24, 2003, the Loan Processing and Issuance Fee (LPIF) rate for loans originated during fiscal years beginning on or after October 1, 2003, will change to 0.40%, as required under Section 428(f)(1) of the Higher Education Act (HEA). Beginning with the Fiscal Year (FY) 2004 Quarter 1 (Q1) calculation, NSLDS will use the date of guaranty to determine whether disbursement activity for the loan should be paid at the 0.40% or 0.65% fee rate. The fee rates are as follows: • • Loans guaranteed between October 1, 1999, and September 30, 2003, will continue to be paid at the 0.65% fee rate. Loans guaranteed on or after October 1, 2003, will be paid at the 0.40% fee rate.
The calculation also considers changes to the date of guaranty, which can impact the LPIF rate, and makes adjustments to ensure that the appropriate fee is paid.
1
Examples of LPIF Calculations
To better understand LPIF calculations, the following examples illustrate how NSLDS determines LPIF payments for the FY 2004 Q1 reporting period. Example 1 A new loan was reported to NSLDS during the current reporting period with a guaranty date of September 1, 2003, and total amount disbursed of \$1000. Because the loan was guaranteed between October 1, 1999, and September 30, 2003, the fee rate is 0.65%. The fee amount for this loan is calculated as follows: • \$1000 x 0.0065 = \$6.50.
Example 2 A new loan was reported to NSLDS during the current reporting period with a guaranty date of November 1, 2003, and total amount disbursed of \$1000. Because the loan was guaranteed on or after October 1, 2003, the fee rate is 0.40%. The fee amount for this loan is calculated as follows: • \$1000 x 0.004 = \$4.
Example 3 During the current reporting period, GA ABC reported to NSLDS that a loan’s guaranty date changed from September 30, 2003, to October 1, 2003. The total of \$1000 in disbursement remained the same. Because the guaranty date changed and impacted the fee rate, the following calculations are necessary to correct the fee amount: • • The fee amount for the prior period needs to be “backed-out” for the loan: (\$1000) x 0.0065 = (\$6.50). The fee amount needs to be calculated at the new rate even though the total amount disbursed is unchanged: \$1000 x 0.004 = \$4.
2
Example 4 During the current reporting period, GA XYZ reported to NSLDS changes to a loan’s guaranty date and disbursement amount. The guaranty date for the loan changed from October 1, 2003, to September 1, 2003. And the disbursement amount changed from \$1000 to \$2000. Because the guaranty date changed, which impacted the fee rate, and the disbursement amount increased, the following calculations are necessary to correct the fee amount: • • The fee amount for the period needs to be “back-out” for the loan: (\$1000) x 0.004 = (4). The fee amount needs to be calculated at the new rate and for the increased disbursement: \$2000 x 0.0065 = \$13.
In addition to the above examples, NSLDS will continue, as it has in the past, to look for changes in the total amount disbursed in the current quarter as compared to the previous quarter. If the total amount disbursed decreased, the difference will be calculated as follows: • Prior Period Total Disbursement Amount \$2000 - Current Period Total Disbursement Amount \$1000 = Difference Total Disbursement Amount (\$1000), which is then multiplied by the fee rate derived by the date of guaranty.
Changes to LPIF Files
The LPIF file format changes are detailed in Attachment A, Quarterly Difference Activity LPIF, and Attachment B, Cumulative LPIF. The Quarterly Difference Activity LPIF file, Attachment A, includes two new fields: • • Fee Rate (positions 274 through 278)—The Fee Rate, as determined by the current value of the date of guaranty, is displayed as either 65000 or 40000. Change in Date of Guaranty Flag (position 279)—The Change in Date of Guaranty Flag is displayed as a ‘Y’ when there is a change in the date of guaranty that impacts the fee rate.
In cases where there is a change in the date of guaranty that impacts the fee rate, NSLDS provides both a prior period and a current period record with the Change in Date of Guaranty Flag set to ‘Y’ for the same loan. The prior period record includes current amounts for guaranty, cancellation, and disbursement fields set to “zeroes” and a prior period amount equal to the total value of each amount from the previous quarter.
3
As a result, the different amounts are negative so the fee paid in prior periods can be fully “backed-out” at the old fee rate. A current period record provides the total amount for guaranty, cancellation, and disbursement in the current period fields with prior period values all set to “zeroes” so the fee paid in the current period can be fully assessed at the current period fee rate. In addition, the Quarterly Difference Activity LPIF file will continue to provide detail records where there is only a change in either the guaranty or cancellation amounts. The Quarterly Difference file no longer reports consolidation loans. The trailer record now provides a total guaranty, cancellation, net guaranty, and disbursement amounts for both fee rates. The Total Disbursement Amount for each fee rate provides the summary for the fee payment by fee rate. This fee payment can then be compared to invoicing information provided by ED’s Financial Management System. The Cumulative LPIF file, Attachment B, includes one new field: • Fee Rate (positions 274 through 278)—The Fee Rate, as determined by the current value of the date of guaranty, is displayed as either 65000 or 40000.
The Cumulative LPIF file provides the most current record for each loan with a guaranty date on or after October 1, 1999. The file also provides the fee rate associated with each loan. In addition, the Cumulative LPIF file will continue to provide detail records where there is only a change in either the guaranty or cancellation amounts. The Cumulative LPIF file no longer reports consolidation loans. The trailer record now provides a Total Guaranty, Cancellation, Net Guaranty, and Disbursement amounts for both fee rates. The Total Disbursement Amount for each fee rate provides the summary for the fee payment by fee rate.
If you have any questions, please contact the NSLDS Customer Service Center at (800) 999-8219 or e-mail nsldscoe@raytheon.com.
4
Attachment A Quarterly Difference Activity LPIF
Quarterly Difference Activity LPIF Header Record Layout
Pos FR 1 Pos TO 9 Attribute Record Sequence Number Description Positional sort field for sorting the header record to the top of the file. Set the value to all zeroes; ‘000000000’. Identification code for guaranty agency. Start date of fiscal year or fiscal quarter; CCYYMMDD format. End of date of fiscal year or fiscal quarter; CCYYMMDD format. Spaces Date on which the file is created; CCYYMMDD format. Time when the file is created; HH:MM:SS format. ‘Quarterly Difference Activity LPIF Detail Header’ Spaces
Length = 300
Field Format Num. Lth 9
10 13 21 29 30 38 46 76
12 20 28 29 37 45 75 300
Code for GA Report Begin Date Report End Date Filler Create Date Create Time File Description Filler
Char. Date Date Char. Date Char. Char. Char.
3 8 8 1 8 8 30 225
Quarterly Difference Activity LPIF Detail Record Layout
Pos FR 1 10 18 30 65 Pos TO 9 17 29 64 66 Attribute Student’s Social Security Number Date of Student’s Birth Student’s First Name Student’s Last Name Type of Loan Description Title IV aid recipient or beneficiary’s SSN. Date when a Title IV aid recipient or beneficiary was born; CCYYMMDD. Title IV aid recipient or beneficiary’s first name. Title IV aid recipient or beneficiary’s last name. Code indicating type of aid received or guaranteed: SF–Stafford SU–Unsubsidized Stafford PL–PLUS
Length = 300
Field Format Char. Date Char. Char. Char. Lth 9 8 12 35 2
5
Pos FR 67
Pos TO 74
Attribute Date of Guaranty
Description Date the FFEL loan was originally guaranteed in CCYYMMDD format. Only loans with date of guaranty on or after 10/1/1999 are extracted for LPIF Process. An indicator used to differentiate among multiple loans of the same type, which have the same guaranty date for same student attending the same school. OPE code for school in which student was enrolled or accepted for enrollment at the time the loan was made. Unique ID that has been assigned to a loan by the GA (optional). Applicable to PLUS (PL) Loans only. It will be set to spaces otherwise. Applicable to PLUS (PL) Loans only. It will be set to spaces otherwise. Applicable to PLUS (PL) Loans only. It will be set to spaces otherwise. Applicable to PLUS (PL) Loans only. It will be set to spaces otherwise. The amount of loan guaranty as reported by the GA prior to the current reporting period end date. The amount of loan guaranty reported by GA prior to the previous reporting period end date. Difference between amount of guaranty for the current period and amount of guaranty from the previous period. Date of last cancellation reported by GA prior to the end of the current reporting period. CCYYMMDD format. The cumulative amount of loan cancellation last reported by GA prior to the end of the current reporting period. The cumulative amount of loan cancellation last reported by GA prior to the end of the previous reporting period. Difference between amount of cancellation for the current period and amount of cancellation from the previous period.
Field Format Date
Lth 8
75
75
Indicator of Separate Loan
Char.
1
76
83
Code for Original School
Char.
8
84 105 114 122 134 169
104 113 121 133 168 174
GA Unique Loan ID PLUS Borrower’s Social Security Number PLUS Borrowers’ Date of Birth PLUS Borrower’s First Name PLUS Borrower’s Last Name Amount of Guaranty Current Period Amount of Guaranty Prior Period
Char. Char. Date Char. Char. Num.
21 9 8 12 35 6
175
180
Num.
6
181
186
Difference Guaranty Amount
Num.
6
187
194
Date of Cancellation
Date
8
195
200
Amount of Cancellation Current Period Amount of Cancellation Prior Period Difference Cancellation Amount
Num.
6
201
206
Num.
6
207
212
Num.
6
6
Pos FR 213
Pos TO 220
Attribute Date of Disbursement
Description Date of last disbursement reported by GA prior to the end of the current reporting period. CCYYMMDD format. The cumulative amount of loan disbursement last reported by GA prior to the end of the current reporting period. The cumulative amount of loan disbursement last reported by GA prior to the end of the previous reporting period. Difference between amount of disbursement for the current period and amount of cancellation from the previous period. Responsible Begin Date for Loan Guarantor. Responsible End Date for Loan Guarantor. Start Date of Quarter CCYYMMDD. End Date of Quarter CCYYMMDD. Identification code for GA. LPIF rate (e.g. 65000 or 40000). Populated with a ‘Y’ for prior period and current period records included in the difference activity file as a result of a change in date of guaranty that impacts change in fee rate. Records without a fee rate impact or no change in date of guaranty are filled with a space. Spaces
Field Format Date
Lth 8
221
226
Amount of Disbursement Current Period Amount of Disbursement Prior Period Difference Disbursement Amount
Num.
6
227
232
Num.
6
233
238
Num.
6
239 247 255 263 271 274 279
246 254 262 270 273 278 279
Responsibility Begin Date Responsibility End Date Report Begin Date Report End Date Code for GA Fee Rate Change in Date of Guaranty Flag
Date Date Date Date Char. Num. Char.
8 8 8 8 3 5 1
280
300
Filler
Char.
21
Quarterly Difference Activity LPIF Trailer Record Layout
Pos FR 1 Pos TO 9 Attribute Record Sequence Number Description Positional sort field for sorting the trailer record to the bottom of the file. Set the value to all nines; ‘999999999’. Identification Code for GA. ‘Quarterly Difference Activity LPIF Detail Trailer’. This number will reflect the number of detail records contained in the current file.
Length = 300
Field Format Num. Lth 9
10 13 43
12 42 51
Code for GA File Description Total Number of Detail Records
Char. Char. Num.
3 30 9
7
Pos FR 52
Pos TO 66
Attribute
Description
Field Format Num.
Lth 15
Total Difference Guaranty Amount The sum of all difference guaranty amounts at 0.65 percent fee rate in this file for loans with a fee rate of 0.65 percent. Total Difference Cancellation Amount at 0.65 percent fee rate Total Net Guaranty Amount at 0.65 percent fee rate The sum of all difference cancellation amounts in this file with a fee rate of 0.65 percent. The difference between the Total Difference Guaranty Amount and the Total Difference Cancellation Amount with a fee rate of 0.65 percent. The sum of all difference disbursement amounts in this file with fee rate of 0.65 percent. The sum of all difference guaranty amounts in this file with fee rate of 0.40 percent.
67
81
Num.
15
82
96
Num.
15
97
111
Total Difference Disbursement Amount at 0.65 percent fee rate Total Guaranty Amount at 0.40 percent fee rate
Num.
15
112 127
126 141
Num. Num.
15 15
Total Cancellation Amount at 0.40 The sum of all difference cancellation percent fee rate amounts in this file with a fee rate of 0.40 percent. Total Net Guaranty Amount at 0.40 percent fee rate The difference between the Total Difference Guaranty Amount and the Total Difference Cancellation Amount with a fee rate of 0.40 percent. The sum of all difference disbursement amounts in this file with a fee rate of 0.40 percent. Spaces
142
156
Num.
15
157
171
Total Disbursement Amount at 0.40 percent fee rate Filler
Num.
15
172
300
Char.
129
8
Attachment B Cumulative LPIF
Pos FR 1 Pos TO 9 Attribute Record Sequence Number Description Positional sort field for sorting the header record to the top of the file. Set the value to all zeroes; ‘000000000’. Identification code for guaranty agency. Start date of fiscal quarter; CCYYMMDD format. End date of fiscal quarter; CCYYMMDD format. Spaces Date on which the file is created; CCYYMMDD format. Time when the file is created; HH:MM:SS format. ‘Cumulative LPIF Detail Header’ Spaces Identification code for guaranty agency. Spaces
Length = 300
Field Format Num. Lth 9
10 13 21 29 30 38 46 96 271 274
12 20 28 29 37 45 95 270 273 300
Code for GA Report Begin Date Report End Date Filler Create Date Create Time File Description Filler Code for GA Filler
Char. Date Date Char. Date Char. Char. Char. Char. Char.
3 8 8 1 8 8 50 175 3 27
Cumulative LPIF Detail Record Layout
Pos FR 1 10 18 30 65 Pos TO 9 17 29 64 66 Attribute Student’s Social Security Number Date of Student’s Birth Student’s First Name Student’s Last Name Type of Loan Description Title IV aid recipient or beneficiary’s SSN. Date when a Title IV aid recipient or beneficiary was born; CCYYMMDD. Title IV aid recipient or beneficiary’s first name. Title IV aid recipient or beneficiary’s last name. Code indicating type of aid received or guaranteed: SF–Stafford SU–Unsubsidized Stafford PL–PLUS
Length = 300
Field Format Char. Date Char. Char. Char. Lth 9 8 12 35 2
9
Pos FR 67
Pos TO 74
Attribute Date of Guaranty
Description Date the FFEL loan was originally guaranteed in CCYYMMDD format. Only loans with date of guaranty on or after 10/1/1999 are extracted for LPIF Process. An indicator used to differentiate among multiple loans of the same type, which have the same guaranty date for same student attending the same school. OPE code for school in which student was enrolled or accepted for enrollment at the time the loan was made. Unique ID that has been assigned to a loan by the GA (optional). Applicable to PLUS (PL) Loans only. It will be set to spaces otherwise. Applicable to PLUS (PL) Loans only. It will be set to spaces otherwise. Applicable to PLUS (PL) Loans only. It will be set to spaces otherwise. Applicable to PLUS (PL) Loans only. It will be set to spaces otherwise. The amount of loan guaranty as reported by the GA prior to the current reporting period end date. Zeroes Zeroes Date of last cancellation reported by GA prior to the end of the current reporting period. CCYYMMDD format. The cumulative amount of loan cancellation last reported by GA prior to the end of the current reporting period. Zeroes Zeroes Date of last disbursement reported by GA prior to the end of the current reporting period. CCYYMMDD format. The cumulative amount of loan disbursement last reported by GA prior to the end of the current reporting period. Zeroes Zeroes
Field Format Date
Lth 8
75
75
Indicator of Separate Loan
Char.
1
76
83
Code for Original School
Char.
8
84 105 114 122 134 169
104 113 121 133 168 174
GA Unique Loan ID PLUS Borrower’s Social Security Number PLUS Borrowers’ Date of Birth PLUS Borrower’s First Name PLUS Borrower’s Last Name Amount of Guaranty Current Period Filler Filler Date of Cancellation
Char. Char. Date Char. Char. Num.
21 9 8 12 35 6
175 181 187
180 186 194
Num. Num. Date
6 6 8
195
200
Amount of Cancellation Current Period Filler Filler Date of Disbursement
Num.
6
201 207 213
206 212 220
Num. Num. Date
6 6 8
221
226
Amount of Disbursement Current Period Filler Filler
Num.
6
227 233
232 238
Num. Num.
6 6
10
Pos FR 239 247 255 263 271 274 279 280
Pos TO 246 254 262 270 273 278 279 300
Attribute Responsibility Begin Date Responsibility End Date Report Begin Date Report End Date Code for GA Fee Rate Filler Filler
Description Responsible Begin Date for Loan Guarantor. Responsible End Date for Loan Guarantor. Start Date of Quarter: CCYYMMDD. End of Quarter Date: CCYYMMDD. Identification Code for guaranty agency. LPIF rate (e.g. 65000 or 40000). Space Spaces
Field Format Date Date Date Date Char. Num. Char. Char.
Lth 8 8 8 8 3 5 1 21
Cumulative LPIF Trailer Record Layout
Pos FR 1 Pos TO 9 Attribute Record Sequence Number Description Positional sort field for sorting the trailer record to the bottom of the file. Set the value to all nines; ‘999999999’. Identification Code for guaranty agency. ‘Cumulative LPIF Detail Trailer’. This number will reflect the number of detail records contained in the current file excluding consolidation loans.
Length = 300
Field Format Num. Lth 9
10 13 63
12 62 71
Code for GA File Description Total Number of Detail Records
Char. Char. Num.
3 50 9
72 87 102
86 101 116
Total Difference Guaranty Amount The sum of all guaranty amounts in this file at 0.65 percent fee rate for loans with a fee rate of 0.65 percent. Total Difference Cancellation Amount at 0.65 percent fee rate Total Net Guaranty Amount at 0.65 percent fee rate Total Difference Disbursement Amount at 0.65 percent fee rate Total Guaranty Amount at 0.40 percent fee rate The sum of all cancellation amounts in this file with a fee rate of 0.65 percent. The difference between the Total Guaranty Amount and the Total Cancellation Amount with a fee rate of 0.65 percent. The sum of all disbursement amounts in this file with fee rate of 0.65 percent. The sum of all guaranty amounts in this file with fee rate of 0.40 percent.
Num. Num. Num.
15 15 15
117 132 147 162
131 146 161 176
Num. Num. Num. Num.
15 15 15 15
Total Cancellation Amount at 0.40 The sum of all cancellation amounts in this percent fee rate file with a fee rate of 0.40 percent. Total Net Guaranty Amount at 0.40 percent fee rate The difference between the Total Guaranty Amount and the Total Cancellation Amount with a fee rate of 0.40 percent.
11
Pos FR 177 192 270 273
Pos TO 191 269 272 300
Attribute Total Difference Disbursement Amount at 0.40 percent fee rate. Filler Code for GA Filler
Description The sum of all disbursement amounts in this file with a fee rate of 0.40 percent. Spaces Identification code for guaranty agency Spaces
Field Format Num. Char. Char. Char.
Lth 15 78 3 28
12
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CC-MAIN-2018-39
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latest
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https://help.scilab.org/docs/6.0.1/en_US/feedback.html
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Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
Change language to: Français - Português - 日本語 - Русский
See the recommended documentation of this function
# feedback
feedback operation
### Syntax
`Sl=Sl1/.Sl2`
### Arguments
Sl1,Sl2
linear systems (`syslin` list) in state-space or transfer form, or ordinary gain matrices.
Sl
linear system (`syslin` list) in state-space or transfer form
### Description
The feedback operation is denoted by `/.` (slashdot). This command returns `Sl=Sl1*(I+Sl2*Sl1)^-1`, i.e the (negative) feedback of `Sl1` and `Sl2`. `Sl` is the transfer `v -> y` for `y = Sl1 u`, `u = v - Sl2 y`.
The result is the same as `Sl=LFT([0,I;I,-Sl2],Sl1)`.
Caution: do not use with decimal point (e.g. `1/.1` is ambiguous!)
### Examples
```S1=ssrand(2,2,3);S2=ssrand(2,2,2);
W=S1/.S2;
ss2tf(S1/.S2)
//Same operation by LFT:
ss2tf(lft([zeros(2,2),eye(2,2);eye(2,2),-S2],S1))
//Other approach: with constant feedback
BigS=sysdiag(S1,S2); F=[zeros(2,2),eye(2,2);-eye(2,2),zeros(2,2)];
Bigclosed=BigS/.F;
W1=Bigclosed(1:2,1:2); //W1=W (in state-space).
ss2tf(W1)
//Inverting
ss2tf(S1*inv(eye()+S2*S1))```
• lft — linear fractional transformation
• sysdiag — Create a block diagonal matrix from provided inputs or block diagonal system connection
• augment — augmented plant
• obscont — observer based controller
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| 2.640625
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CC-MAIN-2020-40
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longest
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https://www.bankersadda.com/reasoning-quiz_18/
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# Reasoning : Quiz
1. Ramnik walked 55 metres towards South, took a left turn and walked 45 metres. He then took a right turn and walked 40 metres. He again took a right turn and walked 45 metres. How far is he from the starting point?
(1) 95 metre
(2) 15 metre
(3) 185 metre
(4) Can’t say
(5) None of these
Directions: (2 – 6) In the following questions, the symbols @, \$, *, # and used with the following meaning as illustrated below:
‘P \$ Q’ means ‘P is not smaller than Q’
‘P@Q’ means ‘P is neither smaller than nor equal to Q’
‘P # Q’ means ‘P is neither greater than nor equal to Q’
‘P Q means ‘P is neither greater than nor smaller than Q’
‘P *Q’ means ‘P is not greater than Q’
Now, in each of the following questions assuming the given statements to be true, find which of the four conclusions I, II, III and IV given below them is/are definitely true and give your answer accordingly.
2. Statements:
N € B
B \$ W
W # H
H * M
Conclusions:
I. M @ W
II. H @ N
III. W € N
IV. W # N
(1) Only I is true
(2) Only III is true
(3) Only IV is true
(4) Only either III or IV is true
(5) Only either III or IV and I are true
3. Statements:
R * D
D \$ J
J # M
M @ K
Conclusions:
I. K # J
II. D @ M
III. R # M
IV. D @ K
(1) None is true
(2) Only I is true
(3) Only II is true
(4) Only III is true
(5) Only IV is true
4. Statements:
H @ T
T # F
F € E
E * V
Conclusions:
I. V \$ F
II. E @ T
III. H @ V
IV. T # V
(1) Only I, II and III are true
(2) Only I, II and IV are true
(3) Only II, III and IV are true
(4) Only I, III and IV are true
(5) All I, II, III and IV are true
5. Statements:
D # R
R * K
K @ F
F \$ J
Conclusions:
I. J # R
II. J # K
III. R # F
IV. K @ D
(1) Only II and III are true
(2) Only III and IV are true
(3) Only I, III and IV are true
(4) All I, II, III and IV are true
(5) Only II and IV are true.
6. Statements:
M \$ K
K @ N
N * R
R # W
Conclusions:
I. W @ K
II. M \$ R
III. K @ W
IV. M @ N
(1) Only I and II are true
(2) Only I is true
(3) Only III and IV are true
(4) Only III is true
(5) Only IV is true
Directions (7-10) : Study the following information to answer the given questions:
In a certain code, ‘za la ka ga’ is code for ‘must obey traffic rules’ ,“ za fa sa na ‘ is code for ‘we obey the elders’ , ‘ na la da sa’ is a code for ‘ we must be elders’ , and ‘wa sa za da’ is code for ‘be elders obey younger’.
7. Which of the following is the code for ‘must’?
(l) sa
(2) da
(3) la
(4) na
(5) None of these
8. What does the code ‘za’ stand for ?
(1) traffic
(2) must
(3) be
(4) younger
(5) obey
9. Which of the following is the code for ‘elders’?
(1) sa
(2) wa
(3) za
(4) la
(5) fa
10. How does ’rules of younger’ coded in the code language?
(l) ka da fa
(2) wa ka sa
(3) ja da wa
(4) ka wa ya
(5) Cannot be determined
1. (1)
Ramnik is 95 m away from the starting point towards south.
Solution: (30 – 34)
\$ – ≥, @ – >, # – <, – =, ¬¬¬¬* – ≤
2. (5) N = B ≥ W < H ≤ M
I. M > W (True) II. H > N (False)
III. W = N (False) IV. W < N (False)
So : N ≥ W (Only either III or IV and I are true)
3. (1) R ≤ D ≥ J < M > K
I. K < J (False) II. D > M (False)
III. R < M (False) IV. D > K (False)
So none is true.
4. (2) H > T < F = E ≤ V
I. V ≥ F (True) II. E > T (True)
III. H > V (False) IV. T < V (True)
So only I, II and IV are true.
5. (5) D < R ≤ k > F ≥ J
I. J < R (False) II. J < K (True)
III. R < F (False) IV. K > D (True)
So only II and IV are true.
6. (5) M ≥ K > N ≤ R < W
I. W > K (False) II. M ≥ R (False)
III. K > W (False) IV. M > N (True)
So only IV is true.
Solutions (7-10):
Word obey Must elders we be The younger traffic rules
Code za La sa na da fa wa ka/ga ga/ka
7. (3)
8. (5)
9. (1)
10. (5)
×
Thank You, Your details have been submitted we will get back to you.
×
OR
×
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The Content Of The Article:
The benchmark index of mortgages is crucial for the calculation of the interest rate on variable and variable rate loans and mixed rate loans. A fixed rate, never changing during the term of the loan. What are these clues, how are they fixed, how can we see them clearly?
## The main indices currently under way
The most commonly used benchmark since the changeover to the euro has been Euribor (Euro Interbank Offered Rate)
It's the short-term money rate practiced between banks in the euro area. There are two main variants: the 3-month Euribor and the 12-month Euribor. Swap 3 years / 5 years / 7 years / 10 years are also in force.
These indices are used to calculate the interest rate of a variable rate, revisable or blended rate loan. Always add the margin earned by the bank. This one is fixed and depends on your profile. It will be even smaller than your bank considers it good. This interest rate represents a portion of the overall loan cost or TEG: the overall effective rate.
If you decide to move from a floating rate to a fixed rate, provided that this clause is stipulated in the loan agreement, then the indexes used are the long-term private sector rate 1st end of month balance or the TMO
To know: The Euribor is also called Tibeur in its French translation (Interbank Rate Offered in Euros) and replaced the Pibor (Paris Interbank Offered Rate) or Tiop (Interbank Offered Rate in Paris).
## Fixed rate, revisable rate, variable rate, some reminders
The fixed rate of a home loan depends on the long-term market and can be modified during the term of the loan. This is the choice of security since you are immune to a rise in interest rates.
The floating rate tracks the change in a financial index at each maturity. While the revisable rate varies only after a reference period (generally one year) depending on the overall evolution of the benchmark.
According to the formula provided for in the loan agreement, in the event of a rise in the revisable or variable ratethis can first be passed on over the term of the loan, which can be extended by up to 5 years, with constant monthly payments. If the adjustment of the duration is not enough, the amount of the monthly payments is increased. In the event of a rate cut, the duration of the loan generally decreases.
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### One step hybrid block method for solving nonlinear second order Dirichlet value problems of ordinary differential equations directly
Document Type : Research Paper
Author
Department of Mathematics, College of Art and Sciences-Tabarjal, Jouf University, Saudi Arabia
10.22075/ijnaa.2023.24666.2794
Abstract
The aim of this article is to approximate the solution of nonlinear second-order Dirichlet boundary value problems of ordinary differential equations directly using the hybrid block method. To derive this method, we first transform the boundary value problem to its corresponding second-order initial value problem via the nonlinear shooting method. Then, a direct one-step hybrid block with three off-step points is derived using a collocation and interpolation approach. The numerical results clearly show that the developed method is able to generate good results when it is compared with the existing method in terms of error.
Keywords
[1] R. Abdelrahim, and Z. Omar Direct solution of second-order ordinary differential equation using a single-step hybrid block method of order five, Math. Comput. Appl. 21 (2016), no. 2, 12.
[2] R. Abdelrahim and Z.Omar Solving third order ordinary differential equations using hybrid block method of order five, Int. J. Appl. Engin. Res. 10 (2016), no. 24, 44307–44310.
[3] L.R. Burden Numerical Analysis, Brooks/Cole Cengage Learning, 2011.
[4] U. Erdogan and T.A. Ozis, Smart nonstandard finite difference scheme for second order nonlinear boundary value problems, J. Comput. Phys. 230 (2011), no. 17, 6464–6474.
[5] S. N. Ha, A nonlinear shooting method for two point boundary value problems, Comput. Math. Appl. 42 (2001), 1411–1420.
[6] M.D. Jafri, M. Suleiman, Z.A. Majid, and Z.B. Ibrahim Solving directly two point boundary value problems using direct multistep method, Sains Malay. 38 (2009), no. 5, 723–728.
[7] S.P. Phang, A.Z. Majid, K.I. Othman, F. Ismail, and M. Suleman, New algorithm of two-point block method for solving boundary value problem with Dirichlet and Neumann boundary conditions, Math. Prob. Engin. 2013 (2013), 1–10.
[8] S.P. Phang, A.Z. Majid, and M. Suleman, Solving nonlinear two point boundary value problem using two step direct method, J. Qual. Measur. Anal. 7 (2011), no. 1, 129–140.
[9] S.M. Roberts, J.S. Shipman, and M. Suleman On the closed form solution of Troesch’s problem, J. Comput. Phys. 21 (1976), 291–304..
[10] J. Yun, A note on three-step iterative method for nonlinear equations, Appl. Math. Comput. 202 (2008), 401–405.
###### Articles in Press, Corrected Proof Available Online from 29 January 2024
• Receive Date: 26 September 2021
• Revise Date: 30 September 2021
• Accept Date: 07 December 2023
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Conversion formula
The conversion factor from kilometers per hour to knots is 0.53995680345662, which means that 1 kilometer per hour is equal to 0.53995680345662 knots:
1 km/h = 0.53995680345662 kt
To convert 281 kilometers per hour into knots we have to multiply 281 by the conversion factor in order to get the velocity amount from kilometers per hour to knots. We can also form a simple proportion to calculate the result:
1 km/h → 0.53995680345662 kt
281 km/h → V(kt)
Solve the above proportion to obtain the velocity V in knots:
V(kt) = 281 km/h × 0.53995680345662 kt
V(kt) = 151.72786177131 kt
The final result is:
281 km/h → 151.72786177131 kt
We conclude that 281 kilometers per hour is equivalent to 151.72786177131 knots:
281 kilometers per hour = 151.72786177131 knots
Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 knot is equal to 0.0065907473309499 × 281 kilometers per hour.
Another way is saying that 281 kilometers per hour is equal to 1 ÷ 0.0065907473309499 knots.
Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred eighty-one kilometers per hour is approximately one hundred fifty-one point seven two eight knots:
281 km/h ≅ 151.728 kt
An alternative is also that one knot is approximately zero point zero zero seven times two hundred eighty-one kilometers per hour.
Conversion table
kilometers per hour to knots chart
For quick reference purposes, below is the conversion table you can use to convert from kilometers per hour to knots
kilometers per hour (km/h) knots (kt)
282 kilometers per hour 152.268 knots
283 kilometers per hour 152.808 knots
284 kilometers per hour 153.348 knots
285 kilometers per hour 153.888 knots
286 kilometers per hour 154.428 knots
287 kilometers per hour 154.968 knots
288 kilometers per hour 155.508 knots
289 kilometers per hour 156.048 knots
290 kilometers per hour 156.587 knots
291 kilometers per hour 157.127 knots
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Created: 2023-03-29
Updated: 2023-12-15
The birthday paradox is a surprising statistical phenomenon that shows how even in a small group, it’s very likely that two people share the same birthday.
In this blog post, we’ll explore the math behind the paradox and its practical applications in computer science and cryptography.
As a flagship example, it states that in a group of just 23 people, there is a 50% change that two of them will share the same birthday.
This probability, which seems quite high, may come as a surprise to many, however it can be easily explained by the principles of probability.
The paradox is a classical example of how probability can lead to seemingly counterintuitive conclusions and is a great illustration of how probability can be used in everyday life.
### Probability
The probability of the Birthday Paradox is computed by considering the number of possible pairs of people in a group and the probability that any of these people will have different birthdays.
To calculate the probability that any two people in a group of n will have different birthdays, we can use the formula:
p(different birthdays) = 365/365 ⋅ 364/365 ⋅ ... ⋅ (365-n+1)/365
= 365! / [(365-n)!⋅365ⁿ]
To find the probability that at least two people in the group will have the same birthday, we can take 1 minus the probability that they all have different birthdays
p(same birthday) = 1 - p(different birthdays)
Some results:
n p(n)
10 0.117
20 0.411
23 0.507
40 0.891
70 0.999
## Generalized Problem
Given a year of d days we want to determine the smaller number n such that the probability of a birthday coincidence is at least 50%.
In other words n is the minimal integer such that:
1 - (1-1/d)⋅(1-2/d)..(1-(n-1)/d) ≥ 1/2
Which is equivalent to:
(1-1/d)⋅(1-2/d)..(1-(n-1)/d) ≤ 1/2
As an approximation, ∏ (1 - i/d) ≈ 1/2 for:
n ≈ ⌈√(2·d·ln(2))⌉
This approximation is derived using the Taylor expansion of the exponential function and is valid for large d.
Proof
∏ (1 - i/d) ≈ exp[ ln( ∏ (1 - i/d) ) ] = exp[ ∑ ln(1 - i/d) ]
Taylor expansion for ln(1 - x) around x = 0 which is ln(1-x) ≈ -x.
∑ ln(1- i/d) = -∑ i/d = -n·(n-1)/(2·d)
→ ∏ (1 - i/d) ≈ exp[ -n·(n-1)/(2·d) ]
Equate to 1/2 which is the target value
exp[ -n·(n-1)/(2·d) ] ≈ 1/2
→ -n·(n-1)/(2·d) ≈ ln(1/2) = -ln(2)
→ n²-n ≈ 2·d·ln(2)
For large enough n we can omit the subtraction
→ n² ≈ 2·d·ln(2)
→ n ≈ √(2·d·ln(2))
Note that the classical birthday problem is an instance of this problem with d = 365 and gives as a result n = 23.
### Collision Probability
Given n items drawn from a set of d elements, we look for the probability p(n) that at least two numbers are equal.
The generic formula is derived using the same argument given in the previous section:
p(n) = 1 - (1-1/d)⋅(1-2/d)..(1-(n-1)/d)
Conversely, n(p) denotes the number of items drawn from a set of d elements to obtain a probability p that at least two numbers are the same, then:
n(p) ≈ √(2·d·ln(1/(1-p)))
The proof of this formula is similar to the one given in the previous section for p = 1/2.
When applied to hash functions this is the expected number of N-bit hashes that can be generated before getting a collision with probability p.
Surprisingly, for p = 1/2 this is not O(2ᴺ), but rather only O(√(2ᴺ)).
There are a number of attacks to crypto systems which leverages this fact.
## Attack
The birthday paradox can be particularly insidious if the hash is used as a primitive building block for a more complex scheme which heavily relies on the collision resistance property of some hash function.
For example, if an attacker discovers two messages m₁ and m₂ such that H(m₁) = H(m₂), he can submit m₁ to the victim in order to have it signed, thus obtaining a signature for H(m₁), but this is a valid signature for H(m₂) as well.
If the number of possible outputs of H is 2ᴺ, a technique to find a collision is:
• Construct k = 2^(N/2) variants of a legit message m₁.
• Use the same technique to construct k variants of a malicious message m₂.
• Construct two sets of digests: A = {H(m₁ᵢ)}, B = {H(m₂ᵢ)}.
• Search for any item in A ∩ B.
The variants can be constructed with a technique which generates messages with changes which are not detectable from a typical renderer. For example insert the same number of spaces and backspaces after a word.
By inserting 100 spaces around the message the attacker can construct 2^100 different variants.
The probability of a collision obviously depends on k.
Can be proven that
Pr(A ∩ B ≠ ∅) ≥ 1/2 if k ≥ 2^(N/2)
The number of the elements in each set corresponds to the expected number of elements to withdraw from a set of 2ᴺ before observing a collision with probability 1/2.
As |A ∪ B| ≥ 2^(N/2) then the probability of finding a collision is still ≈ 1/2 (it doesn’t double). Thus, there is a good chance (≥ 1/2) to find a collision for two elements which belong to different sets.
### Mitigations
To reduce birthday attack risk we have to choose a value for N that is sufficiently large.
Hashes like MD5 (N=128) or sha1 (N=160) are not considered secure anymore.
Another way to counter this attack is to use a MAC (a keyed hash) which bounds the output also to a secret key k.
In this way an attacker cannot pre-compute offline a table of collisions, as they do not know the secret key used in the particular context.
As a final note, always keep in mind that for pigeon-hole principle if the hash possible outputs are k, then after k+1 extractions we’re going to have a collision with probability 1.
## References
• Simple PoC in Rust here
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## Introduction to Basel Capital Accord
What is the Basel Committee for Banking Supervision? The Basel Committee on Banking Supervision is a committee of banking supervisory authorities, which was established by the central bank governors of the Group of Ten (G10) countries in 1975. It consists of senior...
## CFA Exam Results for June 2010
CFA Exam Results for June 2010 are out. For Level 1, Pass rate is 42%. That's up from 34% in Dec. '09 and down from 46% in June '09. For Level 2, Pass rate is 39% We would like to congratulate everyone who has passed the exam. For others, best of luck for the December...
## eBook – Risk Management Essentials
We are happy to announce the launch of our eBook, Risk Management Essentials. And yes, this eBook is completely free. About This eBook The management of risk has emerged as the key challenge for every participant in the financial markets. Risk management is at the...
## CFA Careers: Options for CFA Charterholders
According to CFA Institute, the CFA charter provides you with a strong foundation for a variety of career choices in the investment profession. It opens up opportunities in various areas such as portfolio management, investment research, advisory services, and investment banking careers.
## CFA Calculator: Financial Calculator for CFA Exam
In order to work with quantitative sections of the study material, it is important that you be able to work with a financial calculator. The exam questions are construction with the assumption that the candidates have the ability to work with a financial calculator. CFA Institute allows two types of calculators in the exam.
## Calculating VaR using Monte Carlo Simulation
Computing VaR with Monte Carlo Simulations very similar to Historical Simulations. The main difference lies in the first step of the algorithm – instead of using the historical data for the price (or returns) of the asset and assuming that this return (or price) can re-occur in the next time interval, we generate a random number that will be used to estimate the return (or price) of the asset at the end of the analysis horizon.
## Monte Carlo Simulation – Example
In the previous post, we learned the algorithm to compute VaR using Monte Carlo Simulation. Let us compute VaR for one share to illustrate the algorithm.
We apply the algorithm to compute the monthly VaR for one stock. We will only consider the share price and thus work with the assumption we have only one share in our portfolio. Therefore the value of the portfolio corresponds to the value of one share.
## The Energy Risk Professional Exam from GARP
Recently, the Global Association of Risk Professionals (GARP) started a new certification program for energy risk professionals. This article provides an overview of the ERP certification, its benefits and the uniqueness of the program.
An energy risk professional is someone who deals with the diverse range of energy commodities and is responsible (directly or indirectly) for managing the risks inherent in dealing with these commodities.
## Calculating VaR Using Historical Simulation
The fundamental assumption of the Historical Simulations methodology is that you base your results on the past performance of your portfolio and make the assumption that the past is a good indicator of the near-future.
The below algorithm illustrates the straightforwardness of this methodology. It is called Full Valuation because we will re-price the asset or the portfolio after every run. This differs from a Local Valuation method in which we only use the information about the initial price and the exposure at the origin to deduce VaR.
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# Lecture07 - CSE 421 Algorithms Richard Anderson Lecture 7...
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CSE 421 Algorithms Richard Anderson Lecture 7 Greedy Algorithms: Homework Scheduling and Optimal Caching
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Greedy Algorithms Solve problems with the simplest possible algorithm The hard part: showing that something simple actually works Today’s problems (Sections 4.2, 4.3) Homework Scheduling Optimal Caching
Homework Scheduling Tasks to perform Deadlines on the tasks Freedom to schedule tasks in any order
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Scheduling tasks Each task has a length t i and a deadline d i All tasks are available at the start One task may be worked on at a time All tasks must be completed Goal: minimize maximum lateness – Lateness = f i – d i if f i >= d i
Example 2 3 2 4 Deadline Time 2 3 2 3 Lateness 1 Lateness 3
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Determine the minimum lateness 2 3 4 5 6 4 5 12 Show the schedule 2, 3, 4, 5 first and compute lateness Deadline Time
Greedy Algorithm Earliest deadline first Order jobs by deadline This algorithm is optimal This result may be surprising, since it ignores the job lengths
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Lecture07 - CSE 421 Algorithms Richard Anderson Lecture 7...
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Session_15_Powerpoint - pass over a grid of equally spaced...
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PHYS 1100 – SI Session 15 Leader – Mel Triay
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Group Activity A Many of the solutions to problems in this class are hinted by certain ‘buzz’ words or phrases. In a group (3 or more), come up with a list of these ‘buzz’ words and phrases.
Group Activity B Which types of problems are giving you the most trouble in this class so far?
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Kinematics Vectors vs. scalars Displacement vs. distance travelled Velocity vs. speed Average vs. Instantaneous
Average quantities - Vector
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Example 1
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Practice Set 1
Practice Set 1 b. When the legal speed limit for the New York Thruway was increased from 55 mi/h to 65 mi/h, how much time was saved by a motorist who drove the 700 km between the Buffalo entrance and the New York City exit at the legal speed limit?
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Conceptual The figure below shows four paths along which objects move from a starting point to a final point, all in the same time interval. The paths
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Unformatted text preview: pass over a grid of equally spaced straight lines. Rank the paths according to (a) the average velocity of the objects and (b) the average speed of the objects, greatest first. Instantaneous Kinematics • Calculus – Did you really think you’d never have to use it? Position Velocity Acceleration Position Velocity Acceleration Key Phrases and Concepts • Speeding up / Slowing down • Max. / min. velocity and displacement • ‘Turnaround’ points Example 2 A particle's position is given by x = 4 - 12 t + 3 t 2 (where t is in seconds and x is in meters). a. What is its velocity at t = 1 s? b. Is it moving in the positive or negative direction of x just then? c. What is its speed just then? d. Is it speeding up or slowing down just Answer to Example 2 Practice Set 2 Answers to Practice Set 2...
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Sum Of The Digits (Posted on 2004-01-27)
Find sum of digits of:(1999)^1999.
[The final answer should be a single digit number, for example, (2)^16 = 65536 and the sum of its digits will be given by (6 + 5 + 5 + 3 + 6 = 25, which again will be reduced to 2 + 5 = 7].
See The Solution Submitted by Ravi Raja Rating: 2.1667 (6 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): super easy | Comment 6 of 13 |
(In reply to re: super easy by SilverKnight)
http://mathforum.org/library/drmath/sets/select/dm_mod.html
Posted by Richard on 2004-01-27 12:03:26
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# Area Lessons
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### Area Formulas
• Learn about formulas for the area of shapes. Understand why those formulas work as well as how t...
### Area of a Right Triangle
• Find the area of a right triangle
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### Areas part 2
• Area and circumference of circle and areas of composite shapes involving parts of circles
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### Basic Areas
• Basic math of area, units, rectangle, parallelogram, triangle
00:09:23
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### Finding Area
• How to find area of figures.
### Finding Area of a Circle given the Diameter
• Find the area of a circle given the diameter
### Finding Area of an Oblique Triangle
• Find the area of an oblique triangle
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### Finding the Area of a Trapezoid
• Learn how to find the area of a trapezoid.
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### Finding the Area of a Trapezoid (Alternate Approach)
• Finding the Area of a Trapezoid by summing the areas of a rectangle and a right triangle.
### Finding the Surface Area of a 3D Pyramid
• Finding the surface area of a 3D pyramid
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What is the purpose of the project Hapax?
“hapax” means “only once” in ancient Greek. You may have seen it in the form hapax legomenon, quote: ” a word that occurs only once within a context, either in the written record of an entire language, in the works of an author, or in a single text”.
After a bit of research I found the correct, I hope, form that I use for this project:
It reads “hapax cheon” and it means, again in ancient Greek, “poured only once”.
Why this? Because, according to this wiki link, “the Greek word χυμεία khumeia originally meant “pouring together””.
The motivation of the project hapax comes from the realization that we only explored a tiny drop in the sea of possibilities. As an example, just look at lambda calculus, one of the two pillars of computation. Historically there are many reasons to consider lambda calculus something made in heaven, or a platonic ideal.
But there are 14400 = 5! X 5! alternatives to the iconic beta rewrite only. Is the original beta special or not?
By analogy with the world of CA, about a third of cellular automata are Turing universal. My gues is that a significant fraction of the alternatives to the beta rewrite are as useful as the original beta.
When we look at lambda calculus from this point of view, we discover that all the possible alternatives, not only of beta, but of the whore graph rewriting formalism, say in the form of chemlambda, all these alternative count a huge number, liek 10^30 in the most conservative estimates.
Same for interaction combinators. Same for knot theory. Same for differential calculus (here I use em).
I started to collect small graph rewrite systems which can be described with the same formalism.
The formalism is based on a formulation which uses exclusively permutations (for the “chemistry” and Hamiltonian mechanics side) and a principle of dissipation which accounts for the probabilistic side.
The goal of the project hapax is to build custom worlds (physics and chemistry)
“poured only once”
which can be used to do universal computation in a truly private way. Because once the rules of computation are private, this leads to the fact that the who;le process of computation becomes incomprehensible.
Or is it so? Maybe yes, maybe not. How can we know, without trying?
That is why I starded to make the hapax stuff.
For the moment is not much, only demos like this one, but the rest will pass from paper to programs, then we’ll play.
Stick-and-ring graphs (I)
Until now the thread on small graph rewrite systems (last post here) was about rewrites on a family of graphs which I call “unoriented stick-and-ring graphs”. The page on small graph rewrite systems contains several formalisms, among them IC2, SH2 and system X are on unoriented stick-and-ring graphs and chemlambda strings is with oriented edges. Emergent algebras and Interaction Combinators are with oriented nodes. Pseudoknots are stick-and-ring graphs with oriented nodes and edges.
In this post I want to make clear what unoriented stick-and-ring graphs are, with the help of some drawings.
Practically an unoriented stick-and-ring graph is a graph with colored nodes, of valence 1, 2 or 3, which admit edges with the ends on the same node. We imagine that the nodes have 1, 2, or 3 ports and any edge between two nodes joins a port of one with a port of another one. Supplementary, we accept loops with no nodes and moreover any 3-valent node has a marked port.
If we split each 3-valent node into two half-nodes, one of them with the one marked port, the other with the remaing two ports, then we are left with a collection of disjoint connected graphs made of 1-valent or 2-valent nodes.
These graphs can be either sticks, i.e. they have 2 ends which are 1-valent nodes, or they can be rings, i.e. they are made entirely of 2-valent nodes.
It follows that we can recover our initial graph by gluing along the sticks ends on other sticks or rings. We use dotted lines for gluing in the next figure.
A drawing of an unoriented stick-and-ring graph is an embedding of the graph in the plane. Only the combinatorial information matters. Here is another depiction of the same graph.
__________________________________________
Small graph rewrite systems (5)
Here are some more tentative descriptions of system X and a play with the trefoil knot. This post comes after the intermezzo and continues the series on small graph rewrite systems.
Recall that system X is a proposal to decompose a crossing into two trivalent nodes, which transforms a knot diagram into an unoriented stick-and-ring graph.
The rewrites are the following, written both with the conventions from the stick-and-ring graphs and also with the more usual conventions which resemble the slide equivalence or spin braids mentioned at the intermezzo.
The first rewrite is GL (glue), which is a Reidemeister 1 rewrite in only one direction.
The second rewrite is RD2, which is a Reidemeister 2 rewrite in one direction.
There is a DIST rewrite, the kind you encounter in interaction combinators or in chemlambda.
And finally there are two SH rewrites, the patterns as in chemlambda or appearing in the semantics of interaction combinators.
One Reidemeister 3 rewrite appears from these ones, as explaned in the following figure (taken from the system X page).
Let’s play with the trefoil knot now. The conversion to stick-and rings
is practically the Gauss code. But when we apply some sequences of rewrites
we obtain more complex graphs, where
• either we can reverse some pairs of half-crossings into crossings, thus we obtain knotted Gauss codes (?!)
• or we remark that we get fast out of the Gauss codes graphs…
thus we get sort of a recursive Gauss codes.
Finally, remark that any knot diagram has a ring into it. Recall that lambda terms translated to chemlambda don’t have rings.
Small graph rewrite systems (4)
This post follows Problems with slide equivalence. A solution is to replace slide equivalence with System X.
This supposes to change the decomposition of a crossing like this:
I let you discover system X (or will update later) but here I want to show you that the Reidemeister 3 rewrite looks like that:
There is now a page dedicated to small graph rewrite systems and stick-and-rings graphs.
Problems with slide equivalence
UPDATE: System X is a solution.
_________
After the Intermezzo, in this post I’ll concentrate on the slide equivalence for unoriented (virtual) links, as defined in L.H. Kauffman, Knots and Physics, World Scientific 1991, p. 336.
Later on we shall propose a small graph rewrite system which is different from this, but we first need to understand that there are some problems with slide equivalence.
Kauffman rule I’ is half a definition, half a rewrite rule. He gives two decompositions of a crossing into two 3-valent nodes. The rewrite is that we can pass from one decomposition to the other.
Problem 1. How many types of 3-valent nodes are used? My guess is just one.
Problem 2. Is the rule II’ needed at all? Why not use instead the rule III’, with the price of a loop:
Problem 3. Is the rule I’ too strong? Maybe, look at the following configuration made of two crossings.
Neighboring crossings dissappear.
We don’t even need two neighboring crossings. In the next figure I took the left pattern from the rule IV’, first part. It is also a pattern where the rules I’, then III’ apply.
The result is very different from the application of IV’.
The same happens for the right pattern of the rule IV’, first part.
We can use again I’ and III’ to obtain a very different configuration than expected.
Conclusion. The slide equivalence rewrites with a “dumb” algorithm of rewrites application behaves otherwise than expected. By “dumb” I mean my favorite algorithms, like greedy deterministic or random.
Used with intelligence, the slide equivalence rewrites have interesting computational aspects, but what about the “intelligent” algorithm? Kauffman brains are rare.
Intermezzo (small graph rewrite systems)
Between the last post on small graph rewrite systems and a new one to follow, here are some other, real world examples of such systems.
Where is this from? Answer: M. Khovanov, New geometrical constructions in low-dimensional topology, preprint 1990, fig. 20
Where is this from? Answer: L.H. Kauffman, Knots and Physics, World Scientific 1991, p. 336.
How can we put this in order?
Small graph rewrite systems (3)
Previous posts on the same subject are (1) (2). Related is this post.
In this post I update the small rewrite system SH-2.1 to SH-2.2. If you look at SH-2.1, it has 3 rewrites: SH, GL and RM.
None of these rewrites allow two “sticks” to merge or one stick to transform into a ring.
Compare with the interaction combinators inspired IC-2.1, with the rewrites DIST, RW1 and RW2. Is true, that system is too reactive, but it has one rewrite, namely RW2, which allows two sticks to merge.
A rewrite which has this property (sticks merge) is essential for computational purposes. The most famous of such rewrites is the BETA rewrite in lambda calculus, or the $\gamma \gamma$ and the $\delta \delta$ rewrites from interaction combinators:
(figure from Lafont article).
In the oriented sticks and rings version of chemlambda we have the rewrites BETA (or A-L) and FI-FOE, with the same property.
We shall modify therefore one of the rewrites from SH-2.1.
The SH-2.2 system
We keep the rewrites SH and GL from the SH-2.1 system:
and we replace the rewrite RM with the new rewrite R2:
The new rewrite R2 needs a ring!
Let’s show that SH-2.2 is better than SH-2.1. All we need is to be able to do the rewrite RM from SH-2.1 in SH-2.2. Here is it.
Mind that the ring from the upper right graph is not the same as the ring from the bottom graph. Indeed, in the rewrite R2 the ring from the bottom is consumed and a new ring appears from the merging of the ends of the stick with two blue nodes which sits on the top of the other stick with two yellow ends from the bottom graph.
Compared with the original RM rewrite
we have an extra ring at the left and at the right of the rewrite RM, as it appears in SH-2.2. Otherwise said the ring plays the role of an enzyme.
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x
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# Millennium Chess - The Bishop
Jul 8, 2015, 9:33 PM 4
In this third installment of my blog series on Millennium Chess, I will explain the 3D motion of the bishop. In my last post, I dealt with the simplest piece: the knight. The bishop is a little more complicated, but please bear with me.
The bishops are always quite powerful pieces, in that their reach is hampered only by the edges of the board, and of course, any pieces that stand in their way. When playing in a 3D field, this could become unwieldy, so certain limitations must be placed on their travel.
When moving between levels, the bishop continues to move diagonally, and still remains on the same color square throughout the game. The bishop travels one square along its normal diagonal path for each level it moves. This is a little difficult to grasp in writing, so I will once again utilize diagrams to illustrate my point. (Boards 1, 2, and 3. The white bishops represent your pieces, and the black bishops represent the squares to which they can move.)
That concludes my explanation of the motion of the bishop in Millennium Chess. I hope I demonstrated the concept clearly. As always, if you have any thoughts or questions, please post them in the comments section! And be sure to look out for my next post on the rook.
Online Now
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# Unit Rates with Complex Fractions Quick Color
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This ready to use product is a quick, fun way to have your students practice calculating unit rates with complex fractions. Students will be given 15 different problems to solve. Instead of writing their answers, students will color them! You can either choose the colors for them to make it unified across the classroom or let them choose their own. This makes a great visual for your visual learners. This is an extremely easy assignment to grade! It is perfect for guided practice, homework, or re-teaching.
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### Standards
to see state-specific standards (only available in the US).
Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units. For example, if a person walks 1/2 mile in each 1/4 hour, compute the unit rate as the complex fraction ½/¼ miles per hour, equivalently 2 miles per hour.
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Roman Numerals
# Roman Numerals 701 - 800
## The numbers 701 to 800 in roman numerals
The right column shows how each roman numeral adds up to the total.
701 = DCCI = 500 + 100 + 100 + 1 702 = DCCII = 500 + 100 + 100 + 1 + 1 703 = DCCIII = 500 + 100 + 100 + 1 + 1 + 1 704 = DCCIV = 500 + 100 + 100 + 5 − 1 705 = DCCV = 500 + 100 + 100 + 5 706 = DCCVI = 500 + 100 + 100 + 5 + 1 707 = DCCVII = 500 + 100 + 100 + 5 + 1 + 1 708 = DCCVIII = 500 + 100 + 100 + 5 + 1 + 1 + 1 709 = DCCIX = 500 + 100 + 100 + 10 − 1 710 = DCCX = 500 + 100 + 100 + 10 711 = DCCXI = 500 + 100 + 100 + 10 + 1 712 = DCCXII = 500 + 100 + 100 + 10 + 1 + 1 713 = DCCXIII = 500 + 100 + 100 + 10 + 1 + 1 + 1 714 = DCCXIV = 500 + 100 + 100 + 10 + 5 − 1 715 = DCCXV = 500 + 100 + 100 + 10 + 5 716 = DCCXVI = 500 + 100 + 100 + 10 + 5 + 1 717 = DCCXVII = 500 + 100 + 100 + 10 + 5 + 1 + 1 718 = DCCXVIII = 500 + 100 + 100 + 10 + 5 + 1 + 1 + 1 719 = DCCXIX = 500 + 100 + 100 + 10 + 10 − 1 720 = DCCXX = 500 + 100 + 100 + 10 + 10 721 = DCCXXI = 500 + 100 + 100 + 10 + 10 + 1 722 = DCCXXII = 500 + 100 + 100 + 10 + 10 + 1 + 1 723 = DCCXXIII = 500 + 100 + 100 + 10 + 10 + 1 + 1 + 1 724 = DCCXXIV = 500 + 100 + 100 + 10 + 10 + 5 − 1 725 = DCCXXV = 500 + 100 + 100 + 10 + 10 + 5 726 = DCCXXVI = 500 + 100 + 100 + 10 + 10 + 5 + 1 727 = DCCXXVII = 500 + 100 + 100 + 10 + 10 + 5 + 1 + 1 728 = DCCXXVIII = 500 + 100 + 100 + 10 + 10 + 5 + 1 + 1 + 1 729 = DCCXXIX = 500 + 100 + 100 + 10 + 10 + 10 − 1 730 = DCCXXX = 500 + 100 + 100 + 10 + 10 + 10 731 = DCCXXXI = 500 + 100 + 100 + 10 + 10 + 10 + 1 732 = DCCXXXII = 500 + 100 + 100 + 10 + 10 + 10 + 1 + 1 733 = DCCXXXIII = 500 + 100 + 100 + 10 + 10 + 10 + 1 + 1 + 1 734 = DCCXXXIV = 500 + 100 + 100 + 10 + 10 + 10 + 5 − 1 735 = DCCXXXV = 500 + 100 + 100 + 10 + 10 + 10 + 5 736 = DCCXXXVI = 500 + 100 + 100 + 10 + 10 + 10 + 5 + 1 737 = DCCXXXVII = 500 + 100 + 100 + 10 + 10 + 10 + 5 + 1 + 1 738 = DCCXXXVIII = 500 + 100 + 100 + 10 + 10 + 10 + 5 + 1 + 1 + 1 739 = DCCXXXIX = 500 + 100 + 100 + 10 + 10 + 10 + 10 − 1 740 = DCCXL = 500 + 100 + 100 + 50 − 10 741 = DCCXLI = 500 + 100 + 100 + 50 − 10 + 1 742 = DCCXLII = 500 + 100 + 100 + 50 − 10 + 1 + 1 743 = DCCXLIII = 500 + 100 + 100 + 50 − 10 + 1 + 1 + 1 744 = DCCXLIV = 500 + 100 + 100 + 50 − 10 + 5 − 1 745 = DCCXLV = 500 + 100 + 100 + 50 − 10 + 5 746 = DCCXLVI = 500 + 100 + 100 + 50 − 10 + 5 + 1 747 = DCCXLVII = 500 + 100 + 100 + 50 − 10 + 5 + 1 + 1 748 = DCCXLVIII = 500 + 100 + 100 + 50 − 10 + 5 + 1 + 1 + 1 749 = DCCXLIX = 500 + 100 + 100 + 50 − 10 + 10 − 1 750 = DCCL = 500 + 100 + 100 + 50 751 = DCCLI = 500 + 100 + 100 + 50 + 1 752 = DCCLII = 500 + 100 + 100 + 50 + 1 + 1 753 = DCCLIII = 500 + 100 + 100 + 50 + 1 + 1 + 1 754 = DCCLIV = 500 + 100 + 100 + 50 + 5 − 1 755 = DCCLV = 500 + 100 + 100 + 50 + 5 756 = DCCLVI = 500 + 100 + 100 + 50 + 5 + 1 757 = DCCLVII = 500 + 100 + 100 + 50 + 5 + 1 + 1 758 = DCCLVIII = 500 + 100 + 100 + 50 + 5 + 1 + 1 + 1 759 = DCCLIX = 500 + 100 + 100 + 50 + 10 − 1 760 = DCCLX = 500 + 100 + 100 + 50 + 10 761 = DCCLXI = 500 + 100 + 100 + 50 + 10 + 1 762 = DCCLXII = 500 + 100 + 100 + 50 + 10 + 1 + 1 763 = DCCLXIII = 500 + 100 + 100 + 50 + 10 + 1 + 1 + 1 764 = DCCLXIV = 500 + 100 + 100 + 50 + 10 + 5 − 1 765 = DCCLXV = 500 + 100 + 100 + 50 + 10 + 5 766 = DCCLXVI = 500 + 100 + 100 + 50 + 10 + 5 + 1 767 = DCCLXVII = 500 + 100 + 100 + 50 + 10 + 5 + 1 + 1 768 = DCCLXVIII = 500 + 100 + 100 + 50 + 10 + 5 + 1 + 1 + 1 769 = DCCLXIX = 500 + 100 + 100 + 50 + 10 + 10 − 1 770 = DCCLXX = 500 + 100 + 100 + 50 + 10 + 10 771 = DCCLXXI = 500 + 100 + 100 + 50 + 10 + 10 + 1 772 = DCCLXXII = 500 + 100 + 100 + 50 + 10 + 10 + 1 + 1 773 = DCCLXXIII = 500 + 100 + 100 + 50 + 10 + 10 + 1 + 1 + 1 774 = DCCLXXIV = 500 + 100 + 100 + 50 + 10 + 10 + 5 − 1 775 = DCCLXXV = 500 + 100 + 100 + 50 + 10 + 10 + 5 776 = DCCLXXVI = 500 + 100 + 100 + 50 + 10 + 10 + 5 + 1 777 = DCCLXXVII = 500 + 100 + 100 + 50 + 10 + 10 + 5 + 1 + 1 778 = DCCLXXVIII = 500 + 100 + 100 + 50 + 10 + 10 + 5 + 1 + 1 + 1 779 = DCCLXXIX = 500 + 100 + 100 + 50 + 10 + 10 + 10 − 1 780 = DCCLXXX = 500 + 100 + 100 + 50 + 10 + 10 + 10 781 = DCCLXXXI = 500 + 100 + 100 + 50 + 10 + 10 + 10 + 1 782 = DCCLXXXII = 500 + 100 + 100 + 50 + 10 + 10 + 10 + 1 + 1 783 = DCCLXXXIII = 500 + 100 + 100 + 50 + 10 + 10 + 10 + 1 + 1 + 1 784 = DCCLXXXIV = 500 + 100 + 100 + 50 + 10 + 10 + 10 + 5 − 1 785 = DCCLXXXV = 500 + 100 + 100 + 50 + 10 + 10 + 10 + 5 786 = DCCLXXXVI = 500 + 100 + 100 + 50 + 10 + 10 + 10 + 5 + 1 787 = DCCLXXXVII = 500 + 100 + 100 + 50 + 10 + 10 + 10 + 5 + 1 + 1 788 = DCCLXXXVIII = 500 + 100 + 100 + 50 + 10 + 10 + 10 + 5 + 1 + 1 + 1 789 = DCCLXXXIX = 500 + 100 + 100 + 50 + 10 + 10 + 10 + 10 − 1 790 = DCCXC = 500 + 100 + 100 + 100 − 10 791 = DCCXCI = 500 + 100 + 100 + 100 − 10 + 1 792 = DCCXCII = 500 + 100 + 100 + 100 − 10 + 1 + 1 793 = DCCXCIII = 500 + 100 + 100 + 100 − 10 + 1 + 1 + 1 794 = DCCXCIV = 500 + 100 + 100 + 100 − 10 + 5 − 1 795 = DCCXCV = 500 + 100 + 100 + 100 − 10 + 5 796 = DCCXCVI = 500 + 100 + 100 + 100 − 10 + 5 + 1 797 = DCCXCVII = 500 + 100 + 100 + 100 − 10 + 5 + 1 + 1 798 = DCCXCVIII = 500 + 100 + 100 + 100 − 10 + 5 + 1 + 1 + 1 799 = DCCXCIX = 500 + 100 + 100 + 100 − 10 + 10 − 1 800 = DCCC = 500 + 100 + 100 + 100
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CC-MAIN-2024-30
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http://codeforces.com/problemset/problem/848/D
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D. Shake It!
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
A never-ending, fast-changing and dream-like world unfolds, as the secret door opens.
A world is an unordered graph G, in whose vertex set V(G) there are two special vertices s(G) and t(G). An initial world has vertex set {s(G), t(G)} and an edge between them.
A total of n changes took place in an initial world. In each change, a new vertex w is added into V(G), an existing edge (u, v) is chosen, and two edges (u, w) and (v, w) are added into E(G). Note that it's possible that some edges are chosen in more than one change.
It's known that the capacity of the minimum s-t cut of the resulting graph is m, that is, at least m edges need to be removed in order to make s(G) and t(G) disconnected.
Count the number of non-similar worlds that can be built under the constraints, modulo 109 + 7. We define two worlds similar, if they are isomorphic and there is isomorphism in which the s and t vertices are not relabelled. Formally, two worlds G and H are considered similar, if there is a bijection between their vertex sets , such that:
• f(s(G)) = s(H);
• f(t(G)) = t(H);
• Two vertices u and v of G are adjacent in G if and only if f(u) and f(v) are adjacent in H.
Input
The first and only line of input contains two space-separated integers n, m (1 ≤ n, m ≤ 50) — the number of operations performed and the minimum cut, respectively.
Output
Output one integer — the number of non-similar worlds that can be built, modulo 109 + 7.
Examples
Input
3 2
Output
6
Input
4 4
Output
3
Input
7 3
Output
1196
Input
31 8
Output
64921457
Note
In the first example, the following 6 worlds are pairwise non-similar and satisfy the constraints, with s(G) marked in green, t(G) marked in blue, and one of their minimum cuts in light blue.
In the second example, the following 3 worlds satisfy the constraints.
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https://writinguniverse.com/risk-management/
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risk management
Wait no more. Let us write you an essay from scratch
On the Loss Funding Option Slot. Smith’s family should buy the homeowner’s insurance at 50% of the property worth (Coverage A). The personal property cover (coverage c) would be 50% of that number. Annual fee of \$900 and deductible of \$250. And will be extended once through the coverage (A, B, C, and D). They were supposed to buy a 50 per cent homeowner’s policy instead of a full one, so they would have had to spend an additional \$600 as a premium, and still the house is worth \$1667 per year. Probability= Loss value (based on replacement costs)/25% value of dwelling
Loss value= 25%Xdwelling replacement costs loss value=25%X175, 000=43750.
Dwelling cost = \$25,000+\$150,000=175,000
dwelling cost=25%X175,000=43750.
Probability=43750/43750=1
Stage 2
Loss financing decisions should purchase \$25, 00 liability for section II of homeowner’s insurance policy with an annual premium of \$500.No need to buy any of the others in section 2, 3. To manage liabilities, the dog should be enrolled in professional dog obedience classes.
Probability of potential losses option 1(Smiths sued because of their dog bites a visitor home
Would be less value/defense costs
=\$ 50,000/\$ 240,000=0.208
Stage 3
No purchase of any supplemental LTD insurance, the family should consider HMO health insurance was covering the entire family. Loss control option John should enroll in skiing lessons and snow sports helmet protect his head. Cost \$ 300.
Probability of distribution of potential losses
Medical expenses/lost wages
=\$13,000/\$30,000
=0.43
Stage 4
The risk of premature death, an additional Purchase 100,000 life insurance on John’s life and \$50, 00 on Karen’s life. Sine they are planning to get another kid.
Probability of potential loss distribution if
John dies: loss value/\$400,000
\$72,000/ \$400,000=0.18
If Karen dies probability= loss value/100,000
72,000/100,000=0.72
In conclusion, risk management should be focused on critical and most probable risks that could happen in life and order of their severity. House, car, health are but some of the things we should focus on insuring. (“Risk Managment Tools,” 2006). On financing decision, the probability is one, house owner insurance 0.208, health insurance 0.43 and life insurance 0.18 and 0.72 for John and Karen respectively.
References
Risk Managment Tools. (2006). Guidelines for Mechanical Integrity Systems, 203-230. doi:10.1002/0470048085.ch11
This sample could have been used by your fellow student... Get your own unique essay on any topic and submit it by the deadline.
Let a professional writer get your back and save some time!
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https://zivid.atlassian.net/wiki/spaces/ZividKB/pages/95944793/8.+Hand-eye+calibration+residuals
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# 8. Hand-eye calibration residuals
Zivid Knowledge Base has been moved! This site is getting deprecated and will no longer be updated. Please head to our new website to find the latest and greatest in Zivid documentation.
In order to evaluate the performance of the hand-eye calibration, we need a method to check the residuals. Here we explain what hand-eye calibration residuals represent and how they are calculated.
For each checkerboard point cloud in the dataset, Zivid software extracts a certain number of feature points. We shall refer to this collection of feature points as a feature point set.
Using the result of hand-eye calibration, it is possible to compute coordinate transformation that can convert the feature point sets from the camera coordinate frame to the robot base frame. If each element of the robotic system, i.e. camera, robot, and hand-eye calibration algorithm were perfect, feature points from one transformed set would have the same coordinates as their counterparts from the other sets in the dataset. Visually, this means that all the feature point sets in 3D space would overlap. This is never the case in reality and there are always some residuals. Visually, this means that the same feature points from different sets don't fully overlap, as can be seen from the image below, which is the visualization of one of the in-house hand-eye calibration experiments.
We will now explain how Zivid software calculates the residuals.
With all feature point sets from the dataset, a set of reference feature points is found that represents the arithmetic mean of all other feature point sets. This means that each feature point of the reference set has such coordinates so that the sum of Euclidean distances from its counterpart feature points from the other sets is minimized (the black enlarged sphere in the zoomed in view). Zivid software then estimates the pose of each feature point set, including the reference set. Finally, translational and rotational residuals are calculated as the relative position and orientation between the reference feature point set and all other feature point sets. The translational residual is given as the Euclidean distance between the reference frames that represent the two feature point sets. The rotational residual is given as the angle of the angle-axis representation between the two reference frames.
Continue reading on how to use the result of the hand-eye calibration.
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http://www.algebra-class.com/algebra-equations.html
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Home
#### Algebra and Pre-Algebra Lessons
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» » Two-Step Equations
# Solving Two-Step Algebra Equations
##### Help is Here! Easy to understand explanations on solving two-step algebra equations.
Now that you know all the rules for solving one-step equations, solving two-step algebra equations will be a piece of cake! You will need prior knowledge of solving one-step equations in order to understand this lesson.
There are 5 examples to this lesson, so make sure that you keep scrolling down to learn how to solve all types of two step equations.
Each equation is going to be solved in two separate steps. Let's take a look:
2x + 3 = 43
Notice that in order to get x on the left hand side by itself, (x = ) we need to remove the 3 and the 2. Therefore, this equation will involve two separate steps.
Remember that you must use the opposite mathematical operation in order to remove a number from one side of an equation.
There's one rule to remember when solving two-step equations:
• Always remove the constant first using addition or subtraction.
• Your second step will be to remove any coefficients or divisors using multiplication or division.
Let's solve this example together. Notice how two separate steps are involved in solving this equation!
## Example 1
We'll look at another example similar to this one.
## Example 2
For more clarification on example 2, take a look at the following video...
If you want to see this example on video, click here to visit my You Tube channel.
## Are You Still Having Trouble Solving Equations?
The Algebra Class E-course goes through each step of solving two-step equations within video lessons. Plus you get detailed, step-by-step solutions to all practice problems, quizzes, and tests!
The next two examples will demonstrate how to solve two-step equations when the coefficient is a fraction or when you have a divisor. Let's take a look!
## Example 4
The video below will help explain the two-step process. If you are still having trouble, check it out.
If you want to see this example on video, click here to visit my You Tube channel.
Ok... one more example! This example involves an additional step before starting the two-step process.
If you have any terms in the equation that are like terms, then you will want to combine those like terms first before solving. Let's take a look....
## Example 5
Wow! You did it! You just applied two different rules for solving equations in order to solve two-step equations. I know you are feeling better now!
## Having Trouble with Your Homework?
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Just type in your problem and click "solve".
## Other Lessons You Might Like on Solving Equations
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https://betterlesson.com/lesson/resource/2485313/exit-slip-piecewise-functions-pdf
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Exit slip- piecewise functions.pdf - Section 4: Exit Slip
Exit slip- piecewise functions.pdf
Writing, Graphing, and Describing Piecewise Linear Functions
Unit 3: Linear Functions
Lesson 14 of 20
Big Idea: In this lesson we apply function notation to piecewise defined functions and assess students' ability evaluate functions in different forms.
Print Lesson
1 teacher likes this lesson
Standards:
Subject(s):
Algebra, Math, Graphing (Algebra), piecewise-defined function, Math, function, building functions
55 minutes
Rhonda Leichliter
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Question of Exercise 2(Subjective)
# Question Water flows out through a circular pipe whose internal radius is 1 cm, at the rate of 80 cm/second into an empty cylindrical tank, the radius of whose base is 40 cm. By how much will the level of water rise in the tank in half an hour ?
The horizontal distance between two towers is 140m
The angle of elevation of the
top of the first tower, when seen from the top of the second tower is 30º If the height of the second tower is 60m , find the height of the first tower.
Solution:
Explanation:
Let draw the figure according to teh given question
The height of the tower is found to be 140.83m.
Prove that in a right triangle the square on the
hypotenuse is equal to the sum of the squares on the other two sides.
Solution:
Explanation:
Given: Right angle triangle ABC, right angled at B
Hence, In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides (AC2=AB2+BC2).
Find the roots of the following quadratic equations by factorisation
Solution:
Explanation:
We have to find the roots of the following quadratic equations by factorisation:
If two tangents inclined at an angle of 60º
are drawn to a circle of radius 3 cm, then length of each tangent (in cm) is equal to :
Solution:
Hence, the length of each tangent is3√3cm
The length of each tangent is 3√3cm
A polynomial of degree ___ is called linear polynomial
A: 0
B: 2
C: 1
D: None of these
Solution:
Explanation:
The general form of linear polynomial is P(x)=ax+b where x is a variable and b is a constant.
The first term of this polynomial has power 1 and therefore, the degree of the polynomial is 1.
A polynomial of degree 1 is called linear polynomial.
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CC-MAIN-2023-06
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latest
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en
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